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# Although improved efficiency in converting harvested
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Although improved efficiency in converting harvested [#permalink]
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07 May 2006, 22:10
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Although improved efficiency in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting consumption.
(A) in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting.
(B) In converting harvested trees into wood products may reduce harvest rates, demand will be stimulated because of increasing supply and lowering prices, which boost.
(C) Of converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, which boosts.
(D) Of harvested trees being converted into wood products may reduce harvest rates, it will stimulate demand, because it will increase supply and lower prices, thereby boosting.
(E) When harvested trees are converted into wood products may reduce harvest rates, demand will be stimulated because of increasing supply and lowering prices, which boost.
[Reveal] Spoiler: OA
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Re: Although improved efficiency in converting harvested [#permalink]
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09 May 2006, 02:08
Can someone tell me what is wrong with C ?
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Re: Although improved efficiency in converting harvested [#permalink]
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09 May 2006, 12:29
avinashbhatm wrote:
Can someone tell me what is wrong with C ?
C. Of converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, which boosts.
which boosts should be which boost
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Re: Although improved efficiency in converting harvested [#permalink]
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10 May 2006, 14:40
I think there may be a distinction between 'efficiency in' and 'efficiency of'. But I can't put my finger on it. Can somebody explain the difference (if any)?
I vote for A.
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Re: Although improved efficiency in converting harvested [#permalink]
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10 May 2006, 20:36
A - best choice
B - not parallel
C - 'boosts is wrong' --> supply and lowering prices is compound noun
D - 'lenthgy and awkward phrasing'
E - 'when....' is awkward
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Although improved efficiency in converting harvested trees [#permalink]
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15 Oct 2006, 20:02
Although improved efficiency in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting consumption.
A. in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting.
B. In converting harvested trees into wood products may reduce harvest rates, demand will be stimulated because of increasing supply and lowering prices, which boost.
C. Of converting harvested trees into wood products may reduce harvest ates, it will stimulate demand by increasing supply and lowering prices which boosts.
D. Of harvested trees being converted into wood products may reduce arvest rates, it will stimulate demand, because it will increase supply and lower prices, thereby boosting.
E. When harvested trees are converted into wood products may reduce arvest rates, demand will be stimulated because of increasing supply nd lowering prices, which boost.
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Last edited by broall on 27 May 2017, 03:43, edited 1 time in total.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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15 Oct 2006, 20:09
A seems fine.
B - "because of" seems unnecessary
C and D - IMO "of converting" somehow is less attractive than "in converting". Someone else think so too? There are other things wrong with D as well.
E - totally out.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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15 Oct 2006, 20:15
I pick A.
The only reason for me is that the original sentence is contrasting the two effects of "increased efficiency".
In all the other choices we are missing that, plus they have problems individually as well.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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15 Oct 2006, 20:19
Guys,,,, OA is NOT A..
think of it more in terms of meaning, specifcially what stimulate demand!
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 03:38
winskc wrote:
Although improved efficiency in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting consumption.
A. in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting.
B. In converting harvested trees into wood products may reduce harvest rates, demand will be stimulated because of increasing supply and lowering prices, which boost.
C. Of converting harvested trees into wood products may reduce harvest ates, it will stimulate demand by increasing supply and lowering prices which boosts.
D. Of harvested trees being converted into wood products may reduce arvest rates, it will stimulate demand, because it will increase supply and lower prices, thereby boosting.
E. When harvested trees are converted into wood products may reduce arvest rates, demand will be stimulated because of increasing supply nd lowering prices, which boost.
Honestly I thought it was A too but "it" was bothering me because I know we may seem to think that it refers to efficiency but we are not sure.
Umm B because efficiency in... is correct. Which seems to refer to increasing supply and lowering prices.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 04:17
uvs_mba wrote:
winskc wrote:
Although improved efficiency in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting consumption.
A. in converting harvested trees into wood products may reduce harvest rates, it will stimulate demand by increasing supply and lowering prices, thereby boosting.
B. In converting harvested trees into wood products may reduce harvest rates, demand will be stimulated because of increasing supply and lowering prices, which boost.
C. Of converting harvested trees into wood products may reduce harvest ates, it will stimulate demand by increasing supply and lowering prices which boosts.
D. Of harvested trees being converted into wood products may reduce arvest rates, it will stimulate demand, because it will increase supply and lower prices, thereby boosting.
E. When harvested trees are converted into wood products may reduce arvest rates, demand will be stimulated because of increasing supply nd lowering prices, which boost.
Honestly I thought it was A too but "it" was bothering me because I know we may seem to think that it refers to efficiency but we are not sure.
Umm B because efficiency in... is correct. Which seems to refer to increasing supply and lowering prices.
B and E is absolutely ruled out "because of" is a wrong usage
A,C are a RunOn
I would tend towards D
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 05:14
Because of is not bad to use. I just re-read the sentence, which is ambiguous. Changing my answer back to A
I guess its too early in the morning
But since it clearly refers to efficiency, how is A a run-on, in that sense A should make perfect sense.
Winsec what is the correct answer? What is your source. I just did a search and Jerry had posted this question some time back and there was no OA posted for it.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 06:44
I think "A" is good enough.
How is "A" run-on? I don't think so.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 06:46
Are you sure? Not convicing....can you please post the OE?
winskc wrote:
Guys,,,, OA is NOT A..
think of it more in terms of meaning, specifcially what stimulate demand!
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 08:43
OA is B.....
There is no OE.. Sorry,, ^^
But, A can not be answer becasue
boosting modifies "DEMAND".
And, supply and prices stimulate demand, not demand by
supply and prices.. ^^
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 09:06
Agree with B,
As A has a pronoun reference problem with 'it'.
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 09:49
I picked up A at first but the it scares me..so my choice is B
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 10:09
can some one please clarify whether "it" can be used to refer to an action such as "improved efficiency"?
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 20:19
uvs_mba wrote:
Because of is not bad to use. I just re-read the sentence, which is ambiguous. Changing my answer back to A
I guess its too early in the morning
But since it clearly refers to efficiency, how is A a run-on, in that sense A should make perfect sense.
Winsec what is the correct answer? What is your source. I just did a search and Jerry had posted this question some time back and there was no OA posted for it.
A
efficiency in
in B ......, which is ambigious
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Re: Although improved efficiency in converting harvested trees [#permalink]
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16 Oct 2006, 20:23
jerrywu wrote:
uvs_mba wrote:
Because of is not bad to use. I just re-read the sentence, which is ambiguous. Changing my answer back to A
I guess its too early in the morning
But since it clearly refers to efficiency, how is A a run-on, in that sense A should make perfect sense.
Winsec what is the correct answer? What is your source. I just did a search and Jerry had posted this question some time back and there was no OA posted for it.
A
efficiency in
in B ......, which is ambigious
I agree with Jerry. Which refers to "prices" or "lowering prices"? I think which can't refer to a verb. Please correct me.
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Re: Although improved efficiency in converting harvested trees [#permalink] 16 Oct 2006, 20:23
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# Effectively cooling molding sands.
Sand cooling is of significant importance in a foundry sand system design. It is important that sand cooling is understood; therefore, a theoretical example follows:
Assume that we are trying to cool one ton of sand per minute. The sand is at an average temperature of 302F (150C), and contains 0.6% moisture. We want to cool it to 122F (50C), with a 2% moisture content for use in the foundry.
While there are a number of ways to solve the problem, one of the easiest to understand uses a heat balance:
Heat gained by water and air = Heat lost by sand
Start by mixing water with the sand to cool it to 200F.
Because we are interested in the residual moisture content of the sand after it cools to 200F, we have to take into account that part of the water mixed with the sand will evaporate. If we assume the water is initially at 70F (21C), and that all of the water we add to bring the sand temperature down to 212F (100C) evaporates, we can break the problem into two parts:
A. How much water do we use to cool the sand to 212F (100C)?
B. How much water do we use to cool the sand from 212F to 200F (93.3C)?
To calculate the answer to Question A, use the following published data:
Average specific heat of sand is 0.2 Btu/lb F = 0.2 kcal/kg |degrees~ C
Average specific heat of water is 1.0 Btu/lb F = 1.0 kcal/kg |degrees~ C
Latent heat of vaporization of water is 974 Btu/lb = 540 kcal/kg
Now apply a heat balance to the problem.
(Heat required to raise W lb of water/minute from 70F to 212F) + (heat required to vaporize W lb of water/minute) = (heat required to cool one ton of sand/minute from 302F (150C) to 212F (100C)
(W lb water/min) (212-70) |degrees~ F (1.0 Btu/lb water |degrees~ F) + (W lb water/win) (974 Btu/lb water) = (2000 lb sand/min) (0.2 Btu/lb sand |degrees~ F) (302-212) |degrees~ F
W = 32.2 lb of water/min (14.6 kg of water/min)
Note that this water all vaporized and is not part of the moisture content of the sand in the system.
To calculate the answer to Question B, reducing the temperature of the sand from 212F (100C) to 200F (93.3C), we will need to raise the temperature of Y lb of water/minute from 70F (21C) to 200F (93.3C).
Note that the heat balance looks like this:
(Y lb water/min) (200-70) |degrees~ F (1.0 Btu/lb water |degrees~ F) = (2000 lb sand/min (212-200) |degrees~ F (0.2 Btu/lb sand |degrees~ F)
Y = 36.9 lb/min (16.8 kg/min)
Total water added is 32.2 lb/min + 36.9 lb/min = 69.1 lb/min (31.4 kg/min)
We can now calculate the moisture content of the sand exiting this part of the cooler:
The water we added (36.9 lb/min) is 36.9/2000 = 1.85%.
Added to the 0.6% moisture already in the sand, the total moisture is 2.45%.
To cool the sand to 122F, mix it with air. Remember that every pound of water that evaporates required 974 Btu (or every kg requires 540 kilocalories). Furthermore, water can only evaporate up to the limit that air can absorb it at saturation. This means that only some of the water will evaporate from the sand. So, we need to know not only the original temperature of the air, but also the original water content of the air that we are mixing with the sand and how much water the air can hold.
Let's assume that we start with absolutely dry air at 32F (0C), and also that when we finish mixing the air with the sand the air will be at 122F (50C) and will be completely saturated. To do this calculation, we need to know that:
1 lb of air will hold 0.08 lb of water at saturation
(1 kg of air will hold 0.08 kg of water at saturation)
Average specific heat of air = 0.25 Btu/lb |degrees~ F = 0.25 kcal/kg |degrees~ C
The heat balance now looks like this:
(heat required to raise A lb of air/min from 32F (0C) to 122F (50C) + (heat required to evaporate the amount of water that A lb of air will hold at saturation) = (heat lost by the sand in cooling to 122F (50C). Or:
(A lb air/min) (0.25 Btu/lb air |degrees~ F) (122-32) |degrees~ F + (A lb air/min) (0.08 lb water/lb air) (974 Btu/lb water) = (2000 lb sand/min) (0.2) Btu sand |degrees~ F) (200-122) |degrees~ F
A = 311 lb of air/min (141 kg of air/min)
It is unrealistic to expect that you will have dry air at 32F to mix with your sand, unless you build an enormous air conditioning system (which is not wholly out of the question). Ideally, you would want to mix air at a known temperature and moisture content with your return sand to get the most uniform control over your sand reclamation system.
However, in most foundries today, ambient air is used. So let's assume that the ambient air enters at 90F (32C) and 90% relative humidity; for many parts of the country, these hot and humid conditions will approximate a "worst case" scenario. Under these conditions, the entering air already contains 0.028 lb of water/lb of air. Now our heat balance looks like this:
(A' lb air/min (0.25 Btu/lb air |degrees~ F) (122-90) |degrees~ F) + (A' lb air/min) (0.080-0.028) (lb water/lb air) (974 Btu/lb water) = (2000 lb sand/min) (0.2) Btu/lb sand |degrees~ F) (200-122) |degrees~ F)
A' = 506 lb of air/min (230 kg of air/min)
This assumes that air exits at 122F. If the time that the air remains in the cooling unit is reduced, it will not have the chance to reach 122F. For purposes of illustration, let's assume that air exits at 110F (43C) and 90% relative humidity. Under these conditions, air can hold 0.053 lb of water/lb of air. Now our heat balance looks like this:
(A" lb of air/min) (0.25 Btu/lb air |degrees~ F) (110-90) |degrees~ F) + (A' lb air/min) (0.053-0.028) (lb water/lb air) (974 Btu/lb water) = (2000 lb sand/min) (0.2 Btu/lb sand |degrees~ F) (200-122) |degrees~ F)
A" = 1603 lb of air/min (483 kg of air/min)
This equals about 15,000 cfm (425 |m.sup.3~/minute).
Each pound of air removes (0.053-0.028 lb) of water/minute, for a total of (0.025) (1063) = 26 lb min, or (26/2000) = 1.3%. This takes us below the desired 2% moisture content, so we'll have to add moisture in the muller.
The result is that as the inlet air becomes warmer and more humid, it takes more air to cool the sand. The temperature and humidity of the incoming air must be known as accurately as the temperature and moisture content of the sand, and adjustments must be made during the day based on both sets of readings.
What we have presented is a simple approach to sand cooling. The efficiency of a cooling system will depend on the velocity of the air, the residence time of the sand in the cooling chamber, the circulation pattern of the air through the sand, the amount of fines removed with the air (these will also carry heat away from the sand) and other factors. As you can see, sand cooling system design is a complex problem.
In addition, the sand will be cooled by radiation losses and conduction to the steel structure that houses the sand cooling system. An empirical formula that can be used to estimate the losses to the structure is as follows:
Heat loss (Btu/min) = (area) (temp. diff.) (time)/10
where:
area = surface area of the steel structure
temp. diff. = difference between hot sand and room temperature
time = average time in minutes that the sand remains in the cooling housing.
Proper cooling of the sand in your system is of the utmost importance in making high-density molding work. In specifying a sand cooling system, remember that the ideal system will:
* expose each sand grain to the air stream;
* prolong contact with the cool air long enough to ensure maximum heat transfer;
* accomplish significant cooling through radiation and convection;
* mix wet and dry sand without plugging up;
* require a minimum floor area;
* be a complete, self-contained unit, which will not require installation of any extra conveyors;
* exhaust a minimum volume of air;
* be reasonable in original purchase price and operating cost.
COPYRIGHT 1994 American Foundry Society, Inc.
No portion of this article can be reproduced without the express written permission from the copyright holder.
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## Truth Tables
Because complex Boolean statements can get tricky to think about, we can create a truth table to keep track of what truth values for the simple statements make the complex statement true and false
### Truth Table
A table showing what the resulting truth value of a complex statement is for all the possible truth values for the simple statements.
### Example 1
Suppose you’re picking out a new couch, and your significant other says “get a sectional or something with a chaise.”
This is a complex statement made of two simpler conditions: “is a sectional,” and “has a chaise.” For simplicity, let’s use S to designate “is a sectional,” and C to designate “has a chaise.” The condition S is true if the couch is a sectional.
A truth table for this would look like this:
S C S or C
T T T
T F T
F T T
F F F
In the table, T is used for true, and F for false. In the first row, if S is true and C is also true, then the complex statement “S or C” is true. This would be a sectional that also has a chaise, which meets our desire.
Remember also that or in logic is not exclusive; if the couch has both features, it does meet the condition.
To shorthand our notation further, we’re going to introduce some symbols that are commonly used for and, or, and not.
### Symbols
The symbol ⋀ is used for and: A and B is notated AB.
The symbol ⋁ is used for or: A or B is notated A ⋁ B
The symbol ~ is used for not: not A is notated ~A
You can remember the first two symbols by relating them to the shapes for the union and intersection. A B would be the elements that exist in both sets, in A ⋂ B. Likewise, A B would be the elements that exist in either set, in A ⋃ B.
In the previous example, the truth table was really just summarizing what we already know about how the or statement work. The truth tables for the basic and, or, and not statements are shown below.
### Basic Truth Tables
A B A B
T T T
T F F
F T F
F F F
A B A B
T T T
T F T
F T T
F F F
A ~A
T F
F T
Truth tables really become useful when analyzing more complex Boolean statements.
### Example 2
Create a truth table for the statement A ⋀ ~(BC)
It helps to work from the inside out when creating truth tables, and create tables for intermediate operations. We start by listing all the possible truth value combinations for A, B, and C. Notice how the first column contains 4 Ts followed by 4 Fs, the second column contains 2 Ts, 2 Fs, then repeats, and the last column alternates. This pattern ensures that all combinations are considered. Along with those initial values, we’ll list the truth values for the innermost expression, BC.
A B C B ⋁ C T T T T T T F T T F T T T F F F F T T T F T F T F F T T F F F F
Next we can find the negation of BC, working off the BC column we just created.
A B C B ⋁ C ~(B ⋁ C) T T T T F T T F T F T F T T F T F F F T F T T T F F T F T F F F T T F F F F F T
Finally, we find the values of A and ~(BC)
A B C B ⋁ C ~(B ⋁ C) A ⋀ ~(B ⋁ C) T T T T F F T T F T F F T F T T F F T F F F T T F T T T F F F T F T F F F F T T F F F F F F T F
It turns out that this complex expression is only true in one case: if A is true, B is false, and C is false.
When we discussed conditions earlier, we discussed the type where we take an action based on the value of the condition. We are now going to talk about a more general version of a conditional, sometimes called an implication.
### Implications
Implications are logical conditional sentences stating that a statement p, called the antecedent, implies a consequence q.
Implications are commonly written as pq
Implications are similar to the conditional statements we looked at earlier; p → q is typically written as “if p then q,” or “p therefore q.” The difference between implications and conditionals is that conditionals we discussed earlier suggest an action—if the condition is true, then we take some action as a result. Implications are a logical statement that suggest that the consequence must logically follow if the antecedent is true.
### Example 3
The English statement “If it is raining, then there are clouds in the sky” is a logical implication. It is a valid argument because if the antecedent “it is raining” is true, then the consequence “there are clouds in the sky” must also be true.
Notice that the statement tells us nothing of what to expect if it is not raining. If the antecedent is false, then the implication becomes irrelevant.
### Example 4
A friend tells you that “if you upload that picture to Facebook, you’ll lose your job.” There are four possible outcomes:
There is only one possible case where your friend was lying—the first option where you upload the picture and keep your job. In the last two cases, your friend didn’t say anything about what would happen if you didn’t upload the picture, so you can’t conclude their statement is invalid, even if you didn’t upload the picture and still lost your job.
In traditional logic, an implication is considered valid (true) as long as there are no cases in which the antecedent is true and the consequence is false. It is important to keep in mind that symbolic logic cannot capture all the intricacies of the English language.
### Truth Values for Implications
p q p → q T T T T F F F T T F F T
### Example 5
Construct a truth table for the statement (m ⋀ ~p) → r
We start by constructing a truth table for the antecedent.
m p ~p m ⋀ ~p T T F F T F T T F T F F F F T F
Now we can build the truth table for the implication
m p ~p m ⋀ ~p r (m ⋀ ~p) → r T T F F T T T F T T T T F T F F T T F F T F T T T T F F F T T F T T F F F T F F F T F F T F F T
In this case, when m is true, p is false, and r is false, then the antecedent m ⋀ ~p will be true but the consequence false, resulting in a invalid implication; every other case gives a valid implication.
For any implication, there are three related statements, the converse, the inverse, and the contrapositive.
### Related Statements
The original implication is “if p then q”: p q
The converse is “if q then p”: qp
The inverse is “if not p then not q”: ~p → ~q
The contrapositive is “if not q then not p”: ~q → ~p
### Example 6
Consider again the valid implication “If it is raining, then there are clouds in the sky.”
The converse would be “If there are clouds in the sky, it is raining.” This is certainly not always true.
The inverse would be “If it is not raining, then there are not clouds in the sky.” Likewise, this is not always true.
The contrapositive would be “If there are not clouds in the sky, then it is not raining.” This statement is valid, and is equivalent to the original implication.
Looking at truth tables, we can see that the original conditional and the contrapositive are logically equivalent, and that the converse and inverse are logically equivalent.
Implication Converse Inverse Contrapositive
p q pq qp ~p → ~q ~q → ~p
T T T T T T
T F F T T F
F T T F F T
F F T T T T
### Equivalence
A conditional statement and its contrapositive are logically equivalent.
The converse and inverse of a statement are logically equivalent.
## Arguments
A logical argument is a claim that a set of premises support a conclusion. There are two general types of arguments: inductive and deductive arguments.
### Argument types
An inductive argument uses a collection of specific examples as its premises and uses them to propose a general conclusion.
A deductive argument uses a collection of general statements as its premises and uses them to propose a specific situation as the conclusion.
### Example 7
The argument “when I went to the store last week I forgot my purse, and when I went today I forgot my purse. I always forget my purse when I go the store” is an inductive argument.
The premises are:
I forgot my purse last week
I forgot my purse today
The conclusion is:
I always forget my purse
Notice that the premises are specific situations, while the conclusion is a general statement. In this case, this is a fairly weak argument, since it is based on only two instances.
### Example 8
The argument “every day for the past year, a plane flies over my house at 2pm. A plane will fly over my house every day at 2pm” is a stronger inductive argument, since it is based on a larger set of evidence.
### Evaluating inductive arguments
An inductive argument is never able to prove the conclusion true, but it can provide either weak or strong evidence to suggest it may be true.
Many scientific theories, such as the big bang theory, can never be proven. Instead, they are inductive arguments supported by a wide variety of evidence. Usually in science, an idea is considered a hypothesis until it has been well tested, at which point it graduates to being considered a theory. The commonly known scientific theories, like Newton’s theory of gravity, have all stood up to years of testing and evidence, though sometimes they need to be adjusted based on new evidence. For gravity, this happened when Einstein proposed the theory of general relativity.
A deductive argument is more clearly valid or not, which makes them easier to evaluate.
### Evaluating deductive arguments
A deductive argument is considered valid if all the premises are true, and the conclusion follows logically from those premises. In other words, the premises are true, and the conclusion follows necessarily from those premises.
### Example 9
The argument “All cats are mammals and a tiger is a cat, so a tiger is a mammal” is a valid deductive argument.
The premises are:
All cats are mammals
A tiger is a cat
The conclusion is:
A tiger is a mammal
Both the premises are true. To see that the premises must logically lead to the conclusion, one approach would be use a Venn diagram. From the first premise, we can conclude that the set of cats is a subset of the set of mammals. From the second premise, we are told that a tiger lies within the set of cats. From that, we can see in the Venn diagram that the tiger also lies inside the set of mammals, so the conclusion is valid.
### Analyzing Arguments with Venn Diagrams[1]
To analyze an argument with a Venn diagram
1. Draw a Venn diagram based on the premises of the argument
2. If the premises are insufficient to determine what determine the location of an element, indicate that.
3. The argument is valid if it is clear that the conclusion must be true
### Example 10
Premise: All firefighters know CPR
Premise: Jill knows CPR
Conclusion: Jill is a firefighter
From the first premise, we know that firefighters all lie inside the set of those who know CPR. From the second premise, we know that Jill is a member of that larger set, but we do not have enough information to know if she also is a member of the smaller subset that is firefighters.
Since the conclusion does not necessarily follow from the premises, this is an invalid argument, regardless of whether Jill actually is a firefighter.
It is important to note that whether or not Jill is actually a firefighter is not important in evaluating the validity of the argument; we are only concerned with whether the premises are enough to prove the conclusion.
In addition to these categorical style premises of the form “all ___,” “some ____,” and “no ____,” it is also common to see premises that are implications.
### Example 11
Premise: If you live in Seattle, you live in Washington.
Premise: Marcus does not live in Seattle
Conclusion: Marcus does not live in Washington
From the first premise, we know that the set of people who live in Seattle is inside the set of those who live in Washington. From the second premise, we know that Marcus does not lie in the Seattle set, but we have insufficient information to know whether or not Marcus lives in Washington or not. This is an invalid argument.
### Example 12
Consider the argument “You are a married man, so you must have a wife.”
This is an invalid argument, since there are, at least in parts of the world, men who are married to other men, so the premise not insufficient to imply the conclusion.
Some arguments are better analyzed using truth tables.
### Example 13
Consider the argument:
Premise: If you bought bread, then you went to the store
Conclusion: You went to the store
While this example is hopefully fairly obviously a valid argument, we can analyze it using a truth table by representing each of the premises symbolically. We can then look at the implication that the premises together imply the conclusion. If the truth table is a tautology (always true), then the argument is valid.
We’ll get B represent “you bought bread” and S represent “you went to the store”. Then the argument becomes:
Premise: BS
Premise: B
Conclusion: S
To test the validity, we look at whether the combination of both premises implies the conclusion; is it true that [(BS) ⋀ B] → S ?
B S B → S (B→S) ⋀ B [(B→S) ⋀ B] → S T T T T T T F F F T F T T F T F F T F T
Since the truth table for [(BS) ⋀ B] → S is always true, this is a valid argument.
### Analyzing arguments using truth tables
To analyze an argument with a truth table:
1. Represent each of the premises symbolically
2. Create a conditional statement, joining all the premises with and to form the antecedent, and using the conclusion as the consequent.
3. Create a truth table for that statement. If it is always true, then the argument is valid.
### Example 14
Premise: If I go to the mall, then I’ll buy new jeans
Premise: If I buy new jeans, I’ll buy a shirt to go with it
Conclusion: If I got to the mall, I’ll buy a shirt.
Let M = I go to the mall, J = I buy jeans, and S = I buy a shirt.
The premises and conclusion can be stated as:
Premise: MJ
Premise: JS
Conclusion: MS
We can construct a truth table for [(MJ) ⋀ (JS)] → (MS)
M J S M → J J → S (M→J) ⋀ (J→S) M → S [(M→J) ⋀ (J→S)] → (M→S) T T T T T T T T T T F T F F F T T F T F T F T T T F F F T F F T F T T T T T T T F T F T F F T T F F T T T T T T F F F T T T T T
From the truth table, we can see this is a valid argument.
1. Technically, these are Euler circles or Euler diagrams, not Venn diagrams, but for the sake of simplicity we’ll continue to call them Venn diagrams.
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# COMAT Scores - Can someone please explain?
#### MedicineZ0Z
5+ Year Member
So a mean is a standard score of 100 and our school needs an 84 for you to pass. But any idea what that is in raw score form? Everyone assumes it means 100/125 (since there are 125 questions) is the mean but that makes zero sense because:
1) COMATs aren't the same, people will have different questions.
2) There's no way people score 80% on average on COMATs but only score ~60% (raw score) on the COMLEX level 1 (which is what a ~500 is equal to).
Anyone have insight on what the scores translate into in raw form? A standard score in statistics certainly doesn't mean raw score either...
1 user
#### sorul
2+ Year Member
The Mean is 100 with a SD of 10.
The exam is standardized like many exams (including COMLEX-USA) and is not graded as a percentage or number of questions correct out of 125. so:
+/- 1 SD = COMAT score of 90-110. 68% of any data set in a normal Gaussian distribution falls within +/- 1 SD
+/- 2 SD = COMAT score of 80-120. 95% of any data set in a normal Gaussian distribution falls within +/- 2 SD
1) Subtract the mean from your score.
2) Divide that number by the SD (10) to find the z score.
3) Convert the z-score to a percentile using a z score to percentile calculator online
Example:
You scored 118 on the Surgery COMAT.
118-100 = 18
18/10= 1.8 (you are 1.8 SD above the mean).
Go to Z score calculator. A z score of 1.8 = 92.8139 percentile
5 users
##### Critically Caring
7+ Year Member
The Mean is 100 with a SD of 10.
The exam is standardized like many exams (including COMLEX-USA) and is not graded as a percentage or number of questions correct out of 125. so:
+/- 1 SD = COMAT score of 90-110. 68% of any data set in a normal Gaussian distribution falls within +/- 1 SD
+/- 2 SD = COMAT score of 80-120. 95% of any data set in a normal Gaussian distribution falls within +/- 2 SD
1) Subtract the mean from your score.
2) Divide that number by the SD (10) to find the z score.
3) Convert the z-score to a percentile using a z score to percentile calculator online
Example:
You scored 118 on the Surgery COMAT.
118-100 = 18
18/10= 1.8 (you are 1.8 SD above the mean).
Go to Z score calculator. A z score of 1.8 = 92.8139 percentile
I think he was asking what the percentage of correct answers was for a given score (raw percentage). No one knows the raw scores aside from the NBOME, OP. Nor does anyone know for sure what the COMLEX raw scores are outside of the NBOME. It's all just rumors.
1 users
#### sorul
2+ Year Member
I think he was asking what the percentage of correct answers was for a given score (raw percentage). No one knows the raw scores aside from the NBOME, OP. Nor does anyone know for sure what the COMLEX raw scores are outside of the NBOME. It's all just rumors.
Right. NBOME doesn't release that for any of their exams. The only thing we can do is calculate our percentile.
#### Drrrrrr. Celty
##### Osteo Dullahan
10+ Year Member
I would say that for the Family Medicine COMAT I think probably a 50-60% would be a passing grade. An average would probably be closer to getting 70-75% correct.
This is based off of my experience though. I think the scale is probably very close or more generous than COMLEX.
Last edited:
#### Rekt
5+ Year Member
I would say that for the Family Medicine COMAT I think probably a 50-60% would be a passing grade. An average would probably be closer to getting 70-75% correct.
This is based off of my experience though. I think the scale is probably very close or more generous than COMLEX.
I very seriously doubt average is that high. Just saying as someone who has gotten >120s so far, I would be surprised if I even got 70%s raw.
#### jw3600
7+ Year Member
I would say that for the Family Medicine COMAT I think probably a 50-60% would be a passing grade. An average would probably be closer to getting 70-75% correct.
This is based off of my experience though. I think the scale is probably very close or more generous than COMLEX.
How could you possibly come up with these numbers?
#### Drrrrrr. Celty
##### Osteo Dullahan
10+ Year Member
How could you possibly come up with these numbers?
Entirely out of my ass honestly. In retrospect it probably was a far higher estimate than reality.
4 users
#### Drrrrrr. Celty
##### Osteo Dullahan
10+ Year Member
I very seriously doubt average is that high. Just saying as someone who has gotten >120s so far, I would be surprised if I even got 70%s raw.
Yah, I think you're right.
#### electrolight
2+ Year Member
Pretty sure I was in the 50-60 range and got a 102 for FM.
#### mikemedman19
2+ Year Member
Anyone thinking these comat exams are random AF? How are we supposed to adequately prepare for them? Most of my tests so far (IM and Surg) were very strange, poorly written exams. I usually do about 500 questions for each one (in a month), plus pick one well reviewed book (the more compact the better cause time on Surgery and IM was limited). I just feel like I have down syndrome every time I sit for these craptastic exams... it makes me WISH for structure in preparing for the COMLEX Level 1 / Step 1.
3 users
#### DrDreams
##### Resident
Anyone thinking these comat exams are random AF? How are we supposed to adequately prepare for them? Most of my tests so far (IM and Surg) were very strange, poorly written exams. I usually do about 500 questions for each one (in a month), plus pick one well reviewed book (the more compact the better cause time on Surgery and IM was limited). I just feel like I have down syndrome every time I sit for these craptastic exams... it makes me WISH for structure in preparing for the COMLEX Level 1 / Step 1.
Yes. Ridiculously. Now on the NBOME website i just saw that 10-20% of FM is peds. I really dislike peds....
#### AlbinoHawk DO
##### An Osteopath
7+ Year Member
Anyone thinking these comat exams are random AF? How are we supposed to adequately prepare for them? Most of my tests so far (IM and Surg) were very strange, poorly written exams. I usually do about 500 questions for each one (in a month), plus pick one well reviewed book (the more compact the better cause time on Surgery and IM was limited). I just feel like I have down syndrome every time I sit for these craptastic exams... it makes me WISH for structure in preparing for the COMLEX Level 1 / Step 1.
Welcome to the DO world. I can tell you all the other tests are not going to be any better. In fact, they will probably be a lot worse.
1 user
#### DO2015CA
##### Resident - PGY2
7+ Year Member
Fake it til you make it
1 user
#### baronanalogy
Anyone thinking these comat exams are random AF? How are we supposed to adequately prepare for them? Most of my tests so far (IM and Surg) were very strange, poorly written exams. I usually do about 500 questions for each one (in a month), plus pick one well reviewed book (the more compact the better cause time on Surgery and IM was limited). I just feel like I have down syndrome every time I sit for these craptastic exams... it makes me WISH for structure in preparing for the COMLEX Level 1 / Step 1.
Did you figure out a good resource to study for the COMATs?? I am struggling to adequately prepare for them as well.
#### BorntobeDO?
7+ Year Member
So a mean is a standard score of 100 and our school needs an 84 for you to pass. But any idea what that is in raw score form? Everyone assumes it means 100/125 (since there are 125 questions) is the mean but that makes zero sense because:
1) COMATs aren't the same, people will have different questions.
2) There's no way people score 80% on average on COMATs but only score ~60% (raw score) on the COMLEX level 1 (which is what a ~500 is equal to).
Anyone have insight on what the scores translate into in raw form? A standard score in statistics certainly doesn't mean raw score either...
Thats not too bad, thats something like the 7th percentile, so very achievable.
The raw scores are currently unknown but in 2012 when they first came out the average was in the 60%ish range in the old COMAT thread. The score isn't #questions right/total, its just another useless scale used by DO scores to add unnecessary complication to something simple.
#### BorntobeDO?
7+ Year Member
Anyone thinking these comat exams are random AF? How are we supposed to adequately prepare for them? Most of my tests so far (IM and Surg) were very strange, poorly written exams. I usually do about 500 questions for each one (in a month), plus pick one well reviewed book (the more compact the better cause time on Surgery and IM was limited). I just feel like I have down syndrome every time I sit for these craptastic exams... it makes me WISH for structure in preparing for the COMLEX Level 1 / Step 1.
Just wait till you take the OMM one. That one was a whole new level of craptastic. I wouldn't be surprised if chatmans point of the rectum is hidden in that bank somewhere.
1 user
#### BorntobeDO?
7+ Year Member
The Mean is 100 with a SD of 10.
The exam is standardized like many exams (including COMLEX-USA) and is not graded as a percentage or number of questions correct out of 125. so:
+/- 1 SD = COMAT score of 90-110. 68% of any data set in a normal Gaussian distribution falls within +/- 1 SD
+/- 2 SD = COMAT score of 80-120. 95% of any data set in a normal Gaussian distribution falls within +/- 2 SD
1) Subtract the mean from your score.
2) Divide that number by the SD (10) to find the z score.
3) Convert the z-score to a percentile using a z score to percentile calculator online
Example:
You scored 118 on the Surgery COMAT.
118-100 = 18
18/10= 1.8 (you are 1.8 SD above the mean).
Go to Z score calculator. A z score of 1.8 = 92.8139 percentile
You officially are better at statistics than any admin at my school. Congrats?
2 users
#### DNC127
##### Im just here so I don't get fined.
5+ Year Member
Did you figure out a good resource to study for the COMATs?? I am struggling to adequately prepare for them as well.
Truelearn + comquest. Do all the questions twice and you will be fine
1 user
#### Rekt
5+ Year Member
Did you figure out a good resource to study for the COMATs?? I am struggling to adequately prepare for them as well.
UWorld + Comquest COMAT rotation specific qbank got me >2SD on all of them.
1 users
##### Membership Revoked
Removed
UWorld + Comquest COMAT rotation specific qbank got me >2SD on all of them.
What % did you get on UW? I'm doing extremely well on UW ranging from 70-80%, but I'm just average on my most recent COMAT. I also do a second bank with USMLE-Rx, where I'm consistent above average on there too.
I don't think it's a lack of knowledge.
I'm to the point, where I just have to admit that I either suck at OMM, or that the COMAT is made up of questions with outdated treatment algorithm.
#### BorntobeDO?
7+ Year Member
Did you do anything besides the questions? Specifically asking about psych and fm here
Not that you asked me, but I just did questions for Psych and was fine (above average). FM I am using questions + wiwa for reenforcement.
1 user
#### Rekt
5+ Year Member
What % did you get on UW? I'm doing extremely well on UW ranging from 70-80%, but I'm just average on my most recent COMAT. I also do a second bank with USMLE-Rx, where I'm consistent above average on there too.
I don't think it's a lack of knowledge.
I'm to the point, where I just have to admit that I either suck at OMM, or that the COMAT is made up of questions with outdated treatment algorithm.
UWorld was 80-90%, but I did the 350-400 Comquest shelf questions first before UWorld. If I had any time after that I added any Truelearn, which probably wasn't all that helpful, just more practice. Just keep it up, do as many questions as you can. I studied harder during third year than second year and it paid off with a great Step 2.
1 user
#### alprazoslam
##### Membership Revoked
Removed
Besides the ten points per standard deviation nobody knows anything about how it’s graded
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# In algorithms for matrix multiplication (eg Strassen), why do we say n is equal to the number of rows and not the number of elements in both matrices?
In Strassen's algorithm, we calculate the time complexity based on n being the number of rows of the square matrix. Why don't we take n to be the total number of entries in the matrices (so if we were multiplying two 2x2 matrices, we would have n = 8 for the four entries in each matrix)? Then, using the naïve method of multiplying matrices, we would end up with only n multiplications and n/2 additions. For instance, multiplying [1 2, 3 4] by [5 6, 7 8] yields [1*5+2*7 1*6+2*8, 3*5+4*7 3*6+4*8]. Here, n = 8 and we are doing n = 8 multiplications and n/2 = 4 additions. So even a naïve multiplication algorithm would yield a time complexity of O(n).
Of course, this reasoning is wrong because the time complexity cannot be linear but I don't understand why. I would appreciate any input. Thank you!
• This bothered me in undergrad, as I took one class that taught me that $n$ is always the size of the input when we say an algorithm is $O(n)$, at the same time as another class that taught me that naive matrix multiplication is $O(n^3)$. Jan 16 at 17:46
• Short answer: because it's convenient.
– Stef
Jan 17 at 10:14
• Convenience is key. You use whatever is convenient when discussing different aspects of algorithms. The insistence on "n" being the input size is when you are discussing e.g. whether the problem is P or NP, because that's the definition. Otherwise, it's rather natural to use whatever quantity is at the tip of your fingers to relate to complexity. Jan 17 at 10:34
• It's a convention. Jan 17 at 21:11
Here, n = 8 and we are doing n = 8 multiplications and n/2 = 4 additions. So even a naïve multiplication algorithm would yield a time complexity of O(n).
That is wrong. It might work for a small example like $$n = 8$$, but for bigger ones it doesn't. E.g. if you have two square matrices with a total of $$n = 200$$ elements (two $$10 \times 10$$ matrices) you have already 1,000 multiplications, for $$n=20\,000$$ you have $$1\,000\,000$$.
Now to the actual question. It is possible to give the number time complexity in terms of the number of matrix elements, but it is not very practical.
One problem is, that the complexity differs a lot depending on the shape of the matrix:
• A multiplication of a $$1 \times N$$ matrix with a $$N \times 1$$ matrix has time complexity $$O(N)$$ (linear in the number of matrix elements).
• But a multiplication of a $$N \times 1$$ matrix with a $$1 \times N$$ matrix takes $$O(N^2)$$ (quadratic in the number of elements).
So it's a lot harder to argue about time complexities when you only know the number of elements.
And it's also easier to talk about the matrices in general. It feels more natural. E.g. saying "I've doubled the number of rows of the first matrix" makes sense to everybody (and you immediately know that the algorithm will run about twice as slow, "I've doubled the both dimensions of both matrices" makes sense. But saying "I've double the number of elements" will confuse anybody, how does the matrix look like afterwards. Did you increase both dimensions by a factor of 1.41? Did the matrix change shape? Or what happen? And what impact does it have on the runtime...
Short answer: because it's natural and convenient.
Longer answer: The number of multiplications to multiply a matrix of size p,q with a matrix of size q,r is pqr with a naive algorithm, and something more complicated with a smarter algorithm. The number of multiplications to multiply two square matrices of size n,n is n^3 with a naive algorithm, and n^(some number) with a smarter algorithm. So, we simplify everything by only considering square n,n matrices and expressing the complexity in function of n.
Note that it's not just with matrices that you have to be careful what "n" means:
• With graphs, it's common to have "n" be the number of vertices, which means the length of the input representing the graph is probably about n², depending on the number of edges and on the choice of representation.
• When the input is a string of digits representing a number, if N is the number and n is the length of the input representing that number, then n = log(N) and N = 2^n, so be very careful whether it is n or N in the complexity.
• There are algorithms with so-called pseudo-polynomial time complexity, which means they take exponential time with respect to the length of their input, but polynomial time with respect to the number represented by their input.
• There is a simple dynamic programming algorithm to solve the Knapsack problem in pseudo-polynomial time, which for some applications can be good enough; this algorithm is polynomial in the maximum weight capacity, but exponential in the number of digits of the maximum weight capacity, which really means it's exponential in the precision, the number of digits, that we're asking for.
• Factorising a number N into a product of primes is said to be very hard, yet obviously this can be done in sublinear time with respect to N: iterate on all numbers below √N and check if any of them divides N. This algorithm is sublinear in N, but exponential in the number of digits of N, so it's really exponential in the length of the input.
• The last one is actually sub-linear even with a trivial implementation using trial division: It can be implemented very easily using at most about sqrt(N) / 3 checks whether N is divisible by some x. (Check divisibility by 2 and 3, then check divisibility by 6k-1 and 6k+1 as long as (6k-1)^2 ≤ N). Jan 17 at 13:29
• @gnasher729 Thanks! I replaced "below N" with "below √N" and "linear" with "sublinear".
– Stef
Jan 17 at 13:53
• There are also (theoretically) fast algorithms for rectangular matrix multiplication. Jan 17 at 21:10
For n = 100, the naive algorithm takes 1,000,000 multiplications and almost as many additions. If we let n = number of rows / columns, then it takes $$n^3$$ multiplications and $$n^3 - n^2$$ additions. If we let $$m = 2n^2$$, that is the total number of elements in both matrices, then it is $$m^{1.5} / 2^{1.5}$$ multiplications and $$m^{1.5} / 2^{1.5} - m/2$$ additions. Both numbers are $$O(m^{1.5})$$ and not $$O(m)$$.
But the reason why we argue in the number of rows and columns, and not the problem size, is that the number of rows and columns usually follows naturally from the problem we try to solve. And in case of the Strassen algorithm which recursively divides a matrix into smaller sub matrices, its easier to reason that you need 7 matrix multiplications of half size. The algorithm is based on number of rows and columns and not on number of elements. And lastly, if the matrices don't have identical size, the number of calculations can easily be found from the numbers of rows and columns, but number of elements leaves a lot of room how many operations are needed, depending on the shapes of the matrices.
• You switch from $m$ to $n$ in the last sentence, I think that’s confusing. Jan 17 at 10:12
• I first read "let n = number of rows / columns" as "let n = number of rows divided by columns" and was confused for a while.
– Stef
Jan 17 at 10:16
Because users of matrices think in terms of the matrix dimensions rather than the number of elements. Because that's related to the "size" of their problem in some way. E.g., you have n linear equations to solve, you're working in an n-dimensional space, ...
IOW matrices aren't interesting to work with as a bunch of elements in a rectangular shape and no other reason for existing - they're interesting to work with to use to solve some problem you have and that's why we study how fast you can compute with them. How "big" of a problem can you actually solve?
One aspect I'd like to stress in addition to the other answers (already covered to some extent there):
With the O(n) notation we express how the run time / memory consumption / ... changes if we change the "input size". But it's completely up to you how you define the term "input size", so you have to document this definition (and many CS sources take "the natural definition" for granted, so they feel they need not specify what n means).
Let's take multiplication of two large integers a simple example:
• If you define M to be the first integer and N the second, primary-school multiplication will give you a stunning O(logM logN) performance.
• But more often, M and N are defined to be the number of (e.g. decimal) digits in the integers, and then the same algorithm now counts as O(M N).
So, in your case, you can of course define N to be whatever you want, and then analyze how the run time changes if you supply inputs that vary in the aspect you called N. The results can be more or less useful, depending on your definition of N:
• If N is the value of the first entry of one of the arrays, changing N will not affect run time, and you get an O(1). Though this is a formally valid definition, you'll agree that it isn't helpful.
• If N is the maximum row or column size in one of the two matrices (a commonly-used definition), the naive multiplication takes O(N^3).
• If N is the number of entries of the bigger matrix (your suggestion), the same algorithm now has O(N^1.5). It hasn't become faster, you just redefined the meaning of N.
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Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
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A huge wheel is rolling past your window. What do you see?
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ABCDEFGH is a 3 by 3 by 3 cube. Point P is 1/3 along AB (that is AP : PB = 1 : 2), point Q is 1/3 along GH and point R is 1/3 along ED. What is the area of the triangle PQR?
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Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
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Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
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Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
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An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?
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What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
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Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
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Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
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Can you find a rule which connects consecutive triangular numbers?
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A bus route has a total duration of 40 minutes. Every 10 minutes, two buses set out, one from each end. How many buses will one bus meet on its way from one end to the other end?
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The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
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Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Paving Paths
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How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### Dice, Routes and Pathways
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This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
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Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
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These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together?
### Yih or Luk Tsut K'i or Three Men's Morris
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Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Coordinate Patterns
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Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead?
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Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives?
### Ding Dong Bell
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The reader is invited to investigate changes (or permutations) in the ringing of church bells, illustrated by braid diagrams showing the order in which the bells are rung.
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The image in this problem is part of a piece of equipment found in the playground of a school. How would you describe it to someone over the phone?
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Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . .
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Imagine you are suspending a cube from one vertex (corner) and allowing it to hang freely. Now imagine you are lowering it into water until it is exactly half submerged. What shape does the surface. . . .
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##### Stage: 4 Challenge Level:
Three frogs hopped onto the table. A red frog on the left a green in the middle and a blue frog on the right. Then frogs started jumping randomly over any adjacent frog. Is it possible for them to. . . .
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# How to Graph and Read the Production Possibilities Frontier
One of the central principles of economics is that everyone faces tradeoffs because resources are limited. These tradeoffs are present both in individual choice and in the production decisions of entire economies.
The production possibilities frontier (PPF for short, also referred to as production possibilities curve) is a simple way to show these production tradeoffs graphically. Here is a guide to graphing a PPF and how to analyze it.
01
of 09
## Label the Axes
Since graphs are two-dimensional, economists make the simplifying assumption that the economy can only produce 2 different goods. Traditionally, economists use guns and butter as the 2 goods when describing an economy's production options, since guns represent a general category of capital goods and butter represents a general category of consumer goods.
The tradeoff in production can then be framed as a choice between capital and consumer goods, which will become relevant later. Therefore, this example will also adopt guns and butter as the axes for the production possibilities frontier. Technically speaking, the units on the axes could be something like pounds of butter and a number of guns.
02
of 09
## Plot the Points
The production possibilities frontier is constructed by plotting all of the possible combinations of output that an economy can produce. In this example, let's say the economy can produce:
• 200 guns if it produces only guns, as represented by the point (0,200)
• 100 pounds of butter and 190 guns, as represented by the point (100,190)
• 250 pounds of butter and 150 guns, as represented by the point (250,150)
• 350 pounds of butter and 75 guns, as represented by the point (350,75)
• 400 pounds of butter if it produces only butter, as represented by the point (400,0)
The rest of the curve is filled in by plotting all of the remaining possible output combinations.
03
of 09
## Inefficient and Infeasible Points
Combinations of output that are inside the production possibilities frontier represent inefficient production. This is when an economy could produce more of both goods (i.e. move up and to the right on the graph) by reorganizing resources.
On the other hand, combinations of output that lie outside the production possibilities frontier represent infeasible points, since the economy doesn't have enough resources to produce those combinations of goods.
Therefore, the production possibilities frontier represents all points where an economy is using all of its resources efficiently.
04
of 09
## Opportunity Cost and the Slope of the PPF
Since the production possibilities frontier represents all of the points where all resources are being used efficiently, it must be the case that this economy has to produce fewer guns if it wants to produce more butter, and vice versa. The slope of the production possibilities frontier represents the magnitude of this tradeoff.
For example, in moving from the top left point to the next point down the curve, the economy has to give up production of 10 guns if it wants to produce 100 more pounds of butter. Not coincidentally, the average slope of the PPF over this region is (190-200)/(100-0) = -10/100, or -1/10. Similar calculations can be made between the other labeled points:
• In going from the second to the third point, the economy must give up production of 40 guns if it wants to produce another 150 pounds of butter, and the average slope of the PPF between these points is (150-190)/(250-100) = -40/150, or -4/15.
• In going from the third to the fourth point, the economy must give up production of 75 guns if it wants to produce another 100 pounds of butter, and the average slope of the PPF between these points is (75-150)/(350-250) = -75/100 = -3/4.
• In going from the fourth to the fifth point, the economy must give up production of 75 guns if it wants to produce another 50 pounds of butter, and the average slope of the PPF between these points is (0-75)/(400-350) = -75/50 = -3/2.
Therefore, the magnitude, or absolute value, of the slope of the PPF represents how many guns must be given up in order to produce one more pound of butter between any 2 points on the curve on average.
Economists call this the opportunity cost of butter, given in terms of guns. In general, the magnitude of the PPF's slope represents how many of the things on the y-axis must be forgone in order to produce one more of the thing on the x-axis, or, alternatively, the opportunity cost of the thing on the x-axis.
If you wanted to calculate the opportunity cost of the thing on the y-axis, you could either redraw the PPF with the axes switched or just note that the opportunity cost of the thing on the y-axis is the reciprocal of the opportunity cost of the thing on the x-axis.
05
of 09
## Opportunity Cost Increases Along the PPF
You may have noticed that the PPF was drawn such that it is bowed out from the origin. Because of this, the magnitude of the slope of the PPF increases, meaning the slope gets steeper, as we move down and to the right along the curve.
This property implies that the opportunity cost of producing butter increases as the economy produces more butter and fewer guns, which is represented by moving down and to the right on the graph.
Economists believe that, in general, the bowed-out PPF is a reasonable approximation of reality. This is because there are likely to be some resources that are better at producing guns and others that are better at producing butter. If an economy is producing only guns, it has some of the resources that are better at producing butter producing guns instead. To start producing butter and still maintain efficiency, the economy would shift the resources that are best at producing butter (or worst at producing guns) first. Because these resources are better at making butter, they can make a lot of butter instead of just a few guns, which results in a low opportunity cost of butter.
On the other hand, if the economy is producing close to the maximum amount of butter produced, it's already employed all of the resources that are better at producing butter than producing guns. In order to produce more butter, then, the economy has to shift some resources that are better at making guns to making butter. This results in a high opportunity cost of butter.
06
of 09
## Constant Opportunity Cost
If an economy instead faces a constant opportunity cost of one producing one of the goods, the production possibilities frontier would be represented by a straight line. This makes intuitive sense as straight lines have a constant slope.
07
of 09
## Technology Affects Production Possibilities
If technology changes in an economy, the production possibilities frontier changes accordingly. In the example above, an advance in gun-making technology makes the economy better at producing guns. This means that, for any given level of butter production, the economy will be able to produce more guns than it did before. This is represented by the vertical arrows between the two curves. Thus, the production possibilities frontier shifts out along the vertical, or guns, axis.
If the economy were instead to experience an advance in butter-making technology, the production possibilities frontier would shift out along the horizontal axis, meaning that for any given level of gun production, the economy can produce more butter than it could before. Similarly, if technology were to decrease rather than advance, the production possibilities frontier would shift inward rather than outward.
08
of 09
## Investment Can Shift the PPF Over Time
In an economy, capital is used both to produce more capital and to produce consumer goods. Since capital is represented by guns in this example, an investment in guns will allow for increased production of both guns and butter in the future.
That said, capital also wears out, or depreciates over time, so some investment in capital is needed just to keep up the existing level of capital stock. A hypothetical example of this level of investment is represented by the dotted line on the graph above.
09
of 09
## Graphic Example of Effects of Investments
Let's assume that the blue line on the graph above represents today's production possibilities frontier. If today's level of production is at the purple point, the level of investment in capital goods (i.e. guns) is more than enough to overcome depreciation, and the level of capital available in the future will be greater than the level available today.
As a result, the production possibilities frontier will shift out, as evidenced by the purple line on the graph. Note that the investment doesn't have to affect both goods equally, and the shift illustrated above is just one example.
On the other hand, if today's production is at the green point, the level of investment in capital goods won't be enough to overcome depreciation, and the level of capital available in the future will be lower than today's level. As a result, the production possibilities frontier will shift in, as evidenced by the green line on the graph. In other words, focusing too much on consumer goods today will hinder an economy's ability to produce in the future.
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Beggs, Jodi. "How to Graph and Read the Production Possibilities Frontier." ThoughtCo, Aug. 27, 2020, thoughtco.com/the-production-possibilities-frontier-1147851. Beggs, Jodi. (2020, August 27). How to Graph and Read the Production Possibilities Frontier. Retrieved from https://www.thoughtco.com/the-production-possibilities-frontier-1147851 Beggs, Jodi. "How to Graph and Read the Production Possibilities Frontier." ThoughtCo. https://www.thoughtco.com/the-production-possibilities-frontier-1147851 (accessed March 29, 2023).
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## Franklin Pezzuti Dyer
### Intransitive random variables
Imagine that a swindler proposes a simple dice game to you, in which there are three dice, each player rolls one of the die, and whoever rolls the highest number is the winner. However, in order to assure you that there isn't any rigged die that rolls higher numbers than the others in most cases, the swindler gives you the opportunity to choose the die you like best before he chooses his.
Perhaps you would suspect some kind of trickery, but if you have the right to pick your die first, do you think the game could be fair, or even in your favor? What about if the three dice are as follows?
These three dice can be described as three random variables:
These three random variables have a very peculiar property - case by case, you can show that
That is, there's no preferable die, but all the same this is no fair game. For each die, there is another die that rolls a higher number than it does in most cases. No matter which die you choose, the swindler can choose another in order to win the game with probability $5/9$.
This phenomenon is called the intransitivity of random variables, and it's sometimes known for its unintuitive nature. In this section I'd like to consider a more general situation, which (in my opinion) sheds a bit of light upon why this phenomenon can seem so perplexing to us.
Imagine for a moment that we're considering three sentences $\psi,\phi,\chi$ that are either true or false for each result of the random variable $W$. That is, if $\Omega$ is the sample space of the random variable $W$, then $\psi,\phi,\chi$ are predicates on the set $\Omega$. If two predicates $\psi,\phi$ are incompatible, or in other words if $\neg(\psi\land\phi)$ is true in each case, then it follows that Now imagine that $\psi,\phi,\chi$ are incompatible, that is, that For instance, if $W$ consists of a triple of random variables $(X,Y,Z)$ then the three sentences are incompatible, although at the moment we're discussing general predicates. Intuitively, it might appear that we could deduce in the same way, and consequently conclude that it's impossible for each of the probabilities $\mathbb P(\psi),\mathbb P(\phi),\mathbb P(\chi)$ to be greater than $1/3$, that is, and that if $\Omega$ were optimally divided up in such a way that maximizes this minimum probability, then it would look something like this:
This analysis is, however, incorrect, since these three predicates are incompatible but not pairwise incompatible. The statement $\psi\land\phi$ can certainly be true and similarly $\phi\land\chi$ and $\chi\land\psi$ can also be true, even though $\psi\land\phi\land\chi$ is always false. For each result $\omega\in\Omega$, at most two of the three incompatible predicates can be true, and therefore the superposition of the three subsets are capable of covering the sample space $\Omega$ at most twice. Using this observation, we can deduce the following upper bound: and therefore and the limiting distribution would look something like this:
In reality there's no paradox here. As far as I can tell, the unintuitive nature of this phenomenon arises from a poor intuition about statement of the form "probably $\psi$". In general, it takes a lot of cognitive effort for us to reconcile the following claims:
• $\psi$ will probably be true.
• $\phi$ will probably be true.
• $\chi$ will probably be true.
• $\psi,\phi,\chi$ will definitely not all be true.
With just two predicates there isn't any "paradox" of this form, since it's not possible for $\psi$ and $\phi$ to both probably be true while $\psi\land\phi$ is impossible, if one interprets "probably $\psi$" as $\mathbb P(\psi) > 1/2$.
#### Maximal intransitivity
We've already seen that with three random variables $X,Y,Z$, there can't be any violation of transitivity worse than the following: Conversely, we can easily confirm that a violation of this magnitude is in fact possible, for instance when $X,Y,Z$ are distributed as follows: But we've left out an important condition on these random variables that would be necessary for them to be considered a realistic generalization of the original dice game, namely that they be independent. The three random variables in the above distribution clearly aren't independent. It could be that we can also find three independent random variables which also attain this upper bound of $2/3$, but we haven't seen any such examples yet.
Imagine a simpler question in which we assume that $Y$ is constant, that is, that there exists a real number $c$ such that $\mathbb P(Y = c) = 1$. We'll also assume that the probability of any two of the random variables being equal is zero. In this case, we'd be looking for the maximal value of the minimum probability if $X,Z$ are independent. It follows, however, that since $X<c$ and $c<Z$ together imply $X < Z$, and these two events are independent (due to this assumption we can multiply their probabilities). Therefore, the desired maximal value is at most However, you can confirm for yourself that this minimum value reaches its peak when $p=q=1-pq$, that is, when Therefore, we know that if $Y$ is constant, the maximum violation of transitivity that we can attain is at most $\phi^{-1}\approx 0.618$, but as for whether this bound is attainable or not, we still don't know. As a matter of fact, from this special case, a more general bound can be derived: in this manuscript by S. Trybula it's proven that the maximal violation of transitivity is $\phi^{-1}$ even when $Y$ is not assumed to be constant. That is:
It's a neat puzzle to try determining whether there exist three independent random variables that reach this upper bound. The fact is that such a triple does exist, and there even exists such a triple in which $Y$ is constant: We could equivalently describe these random variables by means of their probability density distributions: which can be visualized as looking something like this:
When it comes to four or more random variables, I'm not sure what the maximal violation of transitivity would be. With more random variables, the upper bound of $\phi^{-1}$ can be surpassed, but it can be proven without too much difficulty that there's a universal upper bound of $3/4$, that is, if $X_1,X_2,\cdots, X_n$ are independent random variables, then it must be the case that Can you prove this upper bound? If you'd like a hint, I'd suggest that it could be useful to consider medians of these $n$ random variables.
#### A parametrized family
Let's suppose for a moment that we have three random variables $X,Y,Z$ and that we describe a whole family of random variables as follows: where $p+q+r=1$. That is, $W(p,q,r)$ is a random variable that takes its value from $X$ with probability $p$, from $Y$ with probability $q$ and from $Z$ with probability $r$. In other words, if $f_X,f_Y,f_Z$ are the respective probability density distributions of $X,Y,Z$, then the distribution of $W(p,q,r)$ would be described by We could parameterize this family of random variables as three-dimensional vectors from $\mathbb R^3$: If we let the parameters $p,q,r$ take on all possible values from the interval $[0,1]$ such that $p+q+r=1$, then the space of all random variables belonging to this family manifests itself as a triangle in $\mathbb R^3$:
To be precise, we're describing an embedding of the convex hull of the three distributions $f_X,f_Y,f_Z$ in the vector space of distributions on $\mathbb R$ in the three-dimensional vector space $\mathbb R^3$ by means of a convex-combination-preserving mapping. Note that the random variables $W(1,0,0), W(0,1,0), W(0,0,1)$ are respectively distributed identically to $X,Y,Z$, and correspond to the three corners of this triangle.
If we assume once more that the three random variables $X,Y,Z$ take on distinct values with probability $1$, and if we know the three probabilities then we can calculate the probability of each possible comparison of random variables from our family $W(p,q,r)$, that is, the probability The parametrization of these random variables as vectors becomes very useful as a way of expressing this probability concisely. If $\mathbf{w}_0, \mathbf{w}_1$ are the vectors corresponding to the respective random variables $W(p_0,q_0,r_0)$ and $W(p_1,q_1,r_1)$, and $A$ is the matrix then one can verify casewise that
Finally, let's consider the case in which $X,Y,Z$ are three independent random variables that maximally violate transitivity. Then we'll have that Notice that we can alternatively describe $A$ as where $C$ is the permutation matrix And this matrix, as a linear transformation on the vector space $\mathbb R^3$, permutes the three coordinates of each vector, that is, it carries out a rotation of $2\pi/3$ radians about the vector $\langle 1,1,1\rangle$:
The interesting thing is that $C$ is a special case of the more general rotation matrix $R_\theta$ about the vector $\langle 1,1,1\rangle$, so that $C = R_{2\pi/3}$. All rotation matrices about the same axis commute with each other since $R_{\theta_0} R_{\theta_1} = R_{\theta_0 + \theta_1}$, hence we have that for all $\theta$. Consequently, all of the $R_\theta$ commute with $A$, since Hence, the probability of the event $W_0 < W_1$ will stay the same when we rotate the parameter vectors $\mathbf w_0, \mathbf w_1$ by an angle $\theta$ in the plane $x+y+z=1$, that is, replace them with $R_\theta \mathbf w_0, R_\theta \mathbf w_1$, since As a matter of fact, one can calculate the probability $\mathbb P(W_0 < W_1)$ knowing only the respective magnitudes and the angle between the components of the vectors $\mathbf w_0, \mathbf w_1$ in the plane $x+y+z=1$. When their projections onto this plane have magnitudes of $r_0, r_1$ and form a counterclockwise angle of $\Delta\theta$, like this:
then the probability $\mathbb P(W_0 < W_1)$ will be:
Hence, if you want to find a long chain of intransitive random variables $X_1, \cdots, X_n$ so that then it suffices to find $n$ vectors inside of this triangle that have the same length and form congruent angles. For example, if you consider the $5$ random variables corresponding to the vectors or
then you'll find the (small but non-negligible) transitivity violation of $\approx 0.532$, that is, the minimum of the probabilities $\mathbb P(W_0 < W_1), \cdots, \mathbb P(W_4 < W_0)$ will be $\approx 0.532$. Using this geometric visualization, we can construct larger and more complex families of intransitive random variables, although they won't be optimal in most cases.
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Calculus Volume 2
# 5.1Sequences
Calculus Volume 25.1 Sequences
## Learning Objectives
• 5.1.1 Find the formula for the general term of a sequence.
• 5.1.2 Calculate the limit of a sequence if it exists.
• 5.1.3 Determine the convergence or divergence of a given sequence.
In this section, we introduce sequences and define what it means for a sequence to converge or diverge. We show how to find limits of sequences that converge, often by using the properties of limits for functions discussed earlier. We close this section with the Monotone Convergence Theorem, a tool we can use to prove that certain types of sequences converge.
## Terminology of Sequences
To work with this new topic, we need some new terms and definitions. First, an infinite sequence is an ordered list of numbers of the form
$a1,a2,a3,…,an,….a1,a2,a3,…,an,….$
Each of the numbers in the sequence is called a term. The symbol $nn$ is called the index variable for the sequence. We use the notation
${an}n=1∞,or simply{an},{an}n=1∞,or simply{an},$
to denote this sequence. A similar notation is used for sets, but a sequence is an ordered list, whereas a set is not ordered. Because a particular number $anan$ exists for each positive integer $n,n,$ we can also define a sequence as a function whose domain is the set of positive integers.
Let’s consider the infinite, ordered list
$2,4,8,16,32,….2,4,8,16,32,….$
This is a sequence in which the first, second, and third terms are given by $a1=2,a1=2,$ $a2=4,a2=4,$ and $a3=8.a3=8.$ You can probably see that the terms in this sequence have the following pattern:
$a1=21,a2=22,a3=23,a4=24,anda5=25.a1=21,a2=22,a3=23,a4=24,anda5=25.$
Assuming this pattern continues, we can write the $nthnth$ term in the sequence by the explicit formula $an=2n.an=2n.$ Using this notation, we can write this sequence as
${2n}n=1∞or{2n}.{2n}n=1∞or{2n}.$
Alternatively, we can describe this sequence in a different way. Since each term is twice the previous term, this sequence can be defined recursively by expressing the $nthnth$ term $anan$ in terms of the previous term $an−1.an−1.$ In particular, we can define this sequence as the sequence ${an}{an}$ where $a1=2a1=2$ and for all $n≥2,n≥2,$ each term $anan$ is defined by the recurrence relation$an=2an−1.an=2an−1.$
## Definition
An infinite sequence${an}{an}$ is an ordered list of numbers of the form
$a1,a2,…,an,….a1,a2,…,an,….$
The subscript $nn$ is called the index variable of the sequence. Each number $anan$ is a term of the sequence. Sometimes sequences are defined by explicit formulas, in which case $an=f(n)an=f(n)$ for some function $f(n)f(n)$ defined over the positive integers. In other cases, sequences are defined by using a recurrence relation. In a recurrence relation, one term (or more) of the sequence is given explicitly, and subsequent terms are defined in terms of earlier terms in the sequence.
Note that the index does not have to start at $n=1n=1$ but could start with other integers. For example, a sequence given by the explicit formula $an=f(n)an=f(n)$ could start at $n=0,n=0,$ in which case the sequence would be
$a0,a1,a2,….a0,a1,a2,….$
Similarly, for a sequence defined by a recurrence relation, the term $a0a0$ may be given explicitly, and the terms $anan$ for $n≥1n≥1$ may be defined in terms of $an−1.an−1.$ Since a sequence ${an}{an}$ has exactly one value for each positive integer $n,n,$ it can be described as a function whose domain is the set of positive integers. As a result, it makes sense to discuss the graph of a sequence. The graph of a sequence ${an}{an}$ consists of all points $(n,an)(n,an)$ for all positive integers $n.n.$ Figure 5.2 shows the graph of ${2n}.{2n}.$
Figure 5.2 The plotted points are a graph of the sequence ${2n}.{2n}.$
Two types of sequences occur often and are given special names: arithmetic sequences and geometric sequences. In an arithmetic sequence, the difference between every pair of consecutive terms is the same. For example, consider the sequence
$3,7,11,15,19,….3,7,11,15,19,….$
You can see that the difference between every consecutive pair of terms is $4.4.$ Assuming that this pattern continues, this sequence is an arithmetic sequence. It can be described by using the recurrence relation
${a1=3an=an−1+4forn≥2.{a1=3an=an−1+4forn≥2.$
Note that
$a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.a2=3+4a3=3+4+4=3+2·4a4=3+4+4+4=3+3·4.$
Thus the sequence can also be described using the explicit formula
$an=3+4(n−1)=4n−1.an=3+4(n−1)=4n−1.$
In general, an arithmetic sequence is any sequence of the form $an=cn+b.an=cn+b.$
In a geometric sequence, the ratio of every pair of consecutive terms is the same. For example, consider the sequence
$2,−23,29,−227,281,….2,−23,29,−227,281,….$
We see that the ratio of any term to the preceding term is $−13.−13.$ Assuming this pattern continues, this sequence is a geometric sequence. It can be defined recursively as
$a1=2an=−13·an−1forn≥2.a1=2an=−13·an−1forn≥2.$
Alternatively, since
$a2=−13·2a3=(−13)(−13)(2)=(−13)2·2a4=(−13)(−13)(−13)(2)=(−13)3·2,a2=−13·2a3=(−13)(−13)(2)=(−13)2·2a4=(−13)(−13)(−13)(2)=(−13)3·2,$
we see that the sequence can be described by using the explicit formula
$an=2(−13)n−1.an=2(−13)n−1.$
The sequence ${2n}{2n}$ that we discussed earlier is a geometric sequence, where the ratio of any term to the previous term is $2.2.$ In general, a geometric sequence is any sequence of the form $an=crn.an=crn.$
## Example 5.1
### Finding Explicit Formulas
For each of the following sequences, find an explicit formula for the $nthnth$ term of the sequence.
1. $−12,23,−34,45,−56,…−12,23,−34,45,−56,…$
2. $34,97,2710,8113,24316,…34,97,2710,8113,24316,…$
## Checkpoint5.1
Find an explicit formula for the $nthnth$ term of the sequence ${15,−17,19,−111,…}.{15,−17,19,−111,…}.$
## Example 5.2
### Defined by Recurrence Relations
For each of the following recursively defined sequences, find an explicit formula for the sequence.
1. $a1=2,a1=2,$ $an=−3an−1an=−3an−1$ for $n≥2n≥2$
2. $a1=12,a1=12,$ $an=an−1+(12)nan=an−1+(12)n$ for $n≥2n≥2$
## Checkpoint5.2
Find an explicit formula for the sequence defined recursively such that $a1=−4a1=−4$ and $an=an−1+6.an=an−1+6.$
## Limit of a Sequence
A fundamental question that arises regarding infinite sequences is the behavior of the terms as $nn$ gets larger. Since a sequence is a function defined on the positive integers, it makes sense to discuss the limit of the terms as $n→∞.n→∞.$ For example, consider the following four sequences and their different behaviors as $n→∞n→∞$ (see Figure 5.3):
1. ${1+3n}={4,7,10,13,…}.{1+3n}={4,7,10,13,…}.$ The terms $1+3n1+3n$ become arbitrarily large as $n→∞.n→∞.$ In this case, we say that $1+3n→∞1+3n→∞$ as $n→∞.n→∞.$
2. ${1−(12)n}={12,34,78,1516,…}.{1−(12)n}={12,34,78,1516,…}.$ The terms $1−(12)n→11−(12)n→1$ as $n→∞.n→∞.$
3. ${(−1)n}={−1,1,−1,1,…}.{(−1)n}={−1,1,−1,1,…}.$ The terms alternate but do not approach one single value as $n→∞.n→∞.$
4. ${(−1)nn}={−1,12,−13,14,…}.{(−1)nn}={−1,12,−13,14,…}.$ The terms alternate for this sequence as well, but $(−1)nn→0(−1)nn→0$ as $n→∞.n→∞.$
Figure 5.3 (a) The terms in the sequence become arbitrarily large as $n→∞.n→∞.$ (b) The terms in the sequence approach $11$ as $n→∞.n→∞.$ (c) The terms in the sequence alternate between $11$ and $−1−1$ as $n→∞.n→∞.$ (d) The terms in the sequence alternate between positive and negative values but approach $00$ as $n→∞.n→∞.$
From these examples, we see several possibilities for the behavior of the terms of a sequence as $n→∞.n→∞.$ In two of the sequences, the terms approach a finite number as $n→∞.n→∞.$ In the other two sequences, the terms do not. If the terms of a sequence approach a finite number $LL$ as $n→∞,n→∞,$ we say that the sequence is a convergent sequence and the real number $LL$ is the limit of the sequence. We can give an informal definition here.
## Definition
Given a sequence ${an},{an},$ if the terms $anan$ become arbitrarily close to a finite number $LL$ as $nn$ becomes sufficiently large, we say ${an}{an}$ is a convergent sequence and $LL$ is the limit of the sequence. In this case, we write
$limn→∞an=L.limn→∞an=L.$
If a sequence ${an}{an}$ is not convergent, we say it is a divergent sequence.
From Figure 5.3, we see that the terms in the sequence ${1−(12)n}{1−(12)n}$ are becoming arbitrarily close to $11$ as $nn$ becomes very large. We conclude that ${1−(12)n}{1−(12)n}$ is a convergent sequence and its limit is $1.1.$ In contrast, from Figure 5.3, we see that the terms in the sequence $1+3n1+3n$ are not approaching a finite number as $nn$ becomes larger. We say that ${1+3n}{1+3n}$ is a divergent sequence.
In the informal definition for the limit of a sequence, we used the terms “arbitrarily close” and “sufficiently large.” Although these phrases help illustrate the meaning of a converging sequence, they are somewhat vague. To be more precise, we now present the more formal definition of limit for a sequence and show these ideas graphically in Figure 5.4.
## Definition
A sequence ${an}{an}$ converges to a real number $LL$ if for all $ε>0,ε>0,$ there exists an integer $NN$ such that $|an−L|<ε|an−L|<ε$ if $n≥N.n≥N.$ The number $LL$ is the limit of the sequence and we write
$limn→∞an=Loran→L.limn→∞an=Loran→L.$
In this case, we say the sequence ${an}{an}$ is a convergent sequence. If a sequence does not converge, it is a divergent sequence, and we say the limit does not exist.
We remark that the convergence or divergence of a sequence ${an}{an}$ depends only on what happens to the terms $anan$ as $n→∞.n→∞.$ Therefore, if a finite number of terms $b1,b2,…,bNb1,b2,…,bN$ are placed before $a1a1$ to create a new sequence
$b1,b2,…,bN,a1,a2,…,b1,b2,…,bN,a1,a2,…,$
this new sequence will converge if ${an}{an}$ converges and diverge if ${an}{an}$ diverges. Further, if the sequence ${an}{an}$ converges to $L,L,$ this new sequence will also converge to $L.L.$
Figure 5.4 As $nn$ increases, the terms $anan$ become closer to $L.L.$ For values of $n≥N,n≥N,$ the distance between each point $(n,an)(n,an)$ and the line $y=Ly=L$ is less than $ε.ε.$
As defined above, if a sequence does not converge, it is said to be a divergent sequence. For example, the sequences ${1+3n}{1+3n}$ and ${(−1)n}{(−1)n}$ shown in Figure 5.4 diverge. However, different sequences can diverge in different ways. The sequence ${(−1)n}{(−1)n}$ diverges because the terms alternate between $11$ and $−1,−1,$ but do not approach one value as $n→∞.n→∞.$ On the other hand, the sequence ${1+3n}{1+3n}$ diverges because the terms $1+3n→∞1+3n→∞$ as $n→∞.n→∞.$ We say the sequence ${1+3n}{1+3n}$ diverges to infinity and write $limn→∞(1+3n)=∞.limn→∞(1+3n)=∞.$ It is important to recognize that this notation does not imply the limit of the sequence ${1+3n}{1+3n}$ exists. The sequence is, in fact, divergent. Writing that the limit is infinity is intended only to provide more information about why the sequence is divergent. A sequence can also diverge to negative infinity. For example, the sequence ${−5n+2}{−5n+2}$ diverges to negative infinity because $−5n+2→−∞−5n+2→−∞$ as $n→∞.n→∞.$ We write this as $limn→∞(−5n+2)=→−∞.limn→∞(−5n+2)=→−∞.$
Because a sequence is a function whose domain is the set of positive integers, we can use properties of limits of functions to determine whether a sequence converges. For example, consider a sequence ${an}{an}$ and a related function $ff$ defined on all positive real numbers such that $f(n)=anf(n)=an$ for all integers $n≥1.n≥1.$ Since the domain of the sequence is a subset of the domain of $f,f,$ if $limx→∞f(x)limx→∞f(x)$ exists, then the sequence converges and has the same limit. For example, consider the sequence ${1n}{1n}$ and the related function $f(x)=1x.f(x)=1x.$ Since the function $ff$ defined on all real numbers $x>0x>0$ satisfies $f(x)=1x→0f(x)=1x→0$ as $x→∞,x→∞,$ the sequence ${1n}{1n}$ must satisfy $1n→01n→0$ as $n→∞.n→∞.$
## Theorem5.1
### Limit of a Sequence Defined by a Function
Consider a sequence ${an}{an}$ such that $an=f(n)an=f(n)$ for all $n≥1.n≥1.$ If there exists a real number $LL$ such that
$limx→∞f(x)=L,limx→∞f(x)=L,$
then ${an}{an}$ converges and
$limn→∞an=L.limn→∞an=L.$
We can use this theorem to evaluate $limn→∞rnlimn→∞rn$ for $0≤r≤1.0≤r≤1.$ For example, consider the sequence ${(1/2)n}{(1/2)n}$ and the related exponential function $f(x)=(1/2)x.f(x)=(1/2)x.$ Since $limx→∞(1/2)x=0,limx→∞(1/2)x=0,$ we conclude that the sequence ${(1/2)n}{(1/2)n}$ converges and its limit is $0.0.$ Similarly, for any real number $rr$ such that $0≤r<1,0≤r<1,$ $limx→∞rx=0,limx→∞rx=0,$ and therefore the sequence ${rn}{rn}$ converges. On the other hand, if $r=1,r=1,$ then $limx→∞rx=1,limx→∞rx=1,$ and therefore the limit of the sequence ${1n}{1n}$ is $1.1.$ If $r>1,r>1,$ $limx→∞rx=∞,limx→∞rx=∞,$ and therefore we cannot apply this theorem. However, in this case, just as the function $rxrx$ grows without bound as $n→∞,n→∞,$ the terms $rnrn$ in the sequence become arbitrarily large as $n→∞,n→∞,$ and we conclude that the sequence ${rn}{rn}$ diverges to infinity if $r>1.r>1.$
We summarize these results regarding the geometric sequence ${rn}:{rn}:$
$rn→0if01.rn→0if01.$
Later in this section we consider the case when $r<0.r<0.$
We now consider slightly more complicated sequences. For example, consider the sequence ${(2/3)n+(1/4)n}.{(2/3)n+(1/4)n}.$ The terms in this sequence are more complicated than other sequences we have discussed, but luckily the limit of this sequence is determined by the limits of the two sequences ${(2/3)n}{(2/3)n}$ and ${(1/4)n}.{(1/4)n}.$ As we describe in the following algebraic limit laws, since ${(2/3)n}{(2/3)n}$ and ${1/4)n}{1/4)n}$ both converge to $0,0,$ the sequence ${(2/3)n+(1/4)n}{(2/3)n+(1/4)n}$ converges to $0+0=0.0+0=0.$ Just as we were able to evaluate a limit involving an algebraic combination of functions $ff$ and $gg$ by looking at the limits of $ff$ and $gg$ (see Introduction to Limits), we are able to evaluate the limit of a sequence whose terms are algebraic combinations of $anan$ and $bnbn$ by evaluating the limits of ${an}{an}$ and ${bn}.{bn}.$
## Theorem5.2
### Algebraic Limit Laws
Given sequences ${an}{an}$ and ${bn}{bn}$ and any real number $c,c,$ if there exist constants $AA$ and $BB$ such that $limn→∞an=Alimn→∞an=A$ and $limn→∞bn=B,limn→∞bn=B,$ then
1. $limn→∞c=climn→∞c=c$
2. $limn→∞can=climn→∞an=cAlimn→∞can=climn→∞an=cA$
3. $limn→∞(an±bn)=limn→∞an±limn→∞bn=A±Blimn→∞(an±bn)=limn→∞an±limn→∞bn=A±B$
4. $limn→∞(an·bn)=(limn→∞an)·(limn→∞bn)=A·Blimn→∞(an·bn)=(limn→∞an)·(limn→∞bn)=A·B$
5. $limn→∞(anbn)=limn→∞anlimn→∞bn=AB,limn→∞(anbn)=limn→∞anlimn→∞bn=AB,$ provided $B≠0B≠0$ and each $bn≠0.bn≠0.$
### Proof
We prove part iii.
Let $ϵ>0.ϵ>0.$ Since $limn→∞an=A,limn→∞an=A,$ there exists a constant positive integer $N1N1$ such that $|an-A|<ε2|an-A|<ε2$ for all $n≥N1.n≥N1.$ Since $limn→∞bn=B,limn→∞bn=B,$ there exists a constant $N2N2$ such that $|bn−B|<ε/2|bn−B|<ε/2$ for all $n≥N2.n≥N2.$ Let $NN$ be the larger of $N1N1$ and $N2.N2.$ Therefore, for all $n≥N,n≥N,$
$|(an+bn)−(A+B)|≤|an−A|+|bn−B|<ε2+ε2=ε.|(an+bn)−(A+B)|≤|an−A|+|bn−B|<ε2+ε2=ε.$
The algebraic limit laws allow us to evaluate limits for many sequences. For example, consider the sequence ${1n2}.{1n2}.$ As shown earlier, $limn→∞1/n=0.limn→∞1/n=0.$ Similarly, for any positive integer $k,k,$ we can conclude that
$limn→∞1nk=0.limn→∞1nk=0.$
In the next example, we make use of this fact along with the limit laws to evaluate limits for other sequences.
## Example 5.3
### Determining Convergence and Finding Limits
For each of the following sequences, determine whether or not the sequence converges. If it converges, find its limit.
1. ${5−3n2}{5−3n2}$
2. ${3n4−7n2+56−4n4}{3n4−7n2+56−4n4}$
3. ${2nn2}{2nn2}$
4. ${(1+4n)n}{(1+4n)n}$
## Checkpoint5.3
Consider the sequence ${(5n2+1)/en}.{(5n2+1)/en}.$ Determine whether or not the sequence converges. If it converges, find its limit.
Recall that if $ff$ is a continuous function at a value $L,L,$ then $f(x)→f(L)f(x)→f(L)$ as $x→L.x→L.$ This idea applies to sequences as well. Suppose a sequence $an→L,an→L,$ and a function $ff$ is continuous at $L.L.$ Then $f(an)→f(L).f(an)→f(L).$ This property often enables us to find limits for complicated sequences. For example, consider the sequence $5−3n2.5−3n2.$ From Example 5.3a. we know the sequence $5−3n2→5.5−3n2→5.$ Since $xx$ is a continuous function at $x=5,x=5,$
$limn→∞5−3n2=limn→∞(5−3n2)=5.limn→∞5−3n2=limn→∞(5−3n2)=5.$
## Theorem5.3
### Continuous Functions Defined on Convergent Sequences
Consider a sequence ${an}{an}$ and suppose there exists a real number $LL$ such that the sequence ${an}{an}$ converges to $L.L.$ Suppose $ff$ is a continuous function at $L.L.$ Then there exists an integer $NN$ such that $ff$ is defined at all values $anan$ for $n≥N,n≥N,$ and the sequence ${f(an)}{f(an)}$ converges to $f(L)f(L)$ (Figure 5.5).
### Proof
Let $ϵ>0.ϵ>0.$ Since $ff$ is continuous at $L,L,$ there exists $δ>0δ>0$ such that $|f(x)−f(L)|<ε|f(x)−f(L)|<ε$ if $|x−L|<δ.|x−L|<δ.$ Since the sequence ${an}{an}$ converges to $L,L,$ there exists $NN$ such that $|an−L|<δ|an−L|<δ$ for all $n≥N.n≥N.$ Therefore, for all $n≥N,n≥N,$ $|an−L|<δ,|an−L|<δ,$ which implies $|f(an)−f(L)|<ε.|f(an)−f(L)|<ε.$ We conclude that the sequence ${f(an)}{f(an)}$ converges to $f(L).f(L).$
Figure 5.5 Because $ff$ is a continuous function as the inputs $a1,a2,a3,…a1,a2,a3,…$ approach $L,L,$ the outputs $f(a1),f(a2),f(a3),…f(a1),f(a2),f(a3),…$ approach $f(L).f(L).$
## Example 5.4
### Limits Involving Continuous Functions Defined on Convergent Sequences
Determine whether the sequence ${cos(3/n2)}{cos(3/n2)}$ converges. If it converges, find its limit.
## Checkpoint5.4
Determine if the sequence ${2n+13n+5}{2n+13n+5}$ converges. If it converges, find its limit.
Another theorem involving limits of sequences is an extension of the Squeeze Theorem for limits discussed in Introduction to Limits.
## Theorem5.4
### Squeeze Theorem for Sequences
Consider sequences ${an},{an},$ ${bn},{bn},$ and ${cn}.{cn}.$ Suppose there exists an integer $NN$ such that
$an≤bn≤cnfor alln≥N.an≤bn≤cnfor alln≥N.$
If there exists a real number $LL$ such that
$limn→∞an=L=limn→∞cn,limn→∞an=L=limn→∞cn,$
then ${bn}{bn}$ converges and $limn→∞bn=Llimn→∞bn=L$ (Figure 5.6).
### Proof
Let $ε>0.ε>0.$ Since the sequence ${an}{an}$ converges to $L,L,$ there exists an integer $N1N1$ such that $|an−L|<ε|an−L|<ε$ for all $n≥N1.n≥N1.$ Similarly, since ${cn}{cn}$ converges to $L,L,$ there exists an integer $N2N2$ such that $|cn−L|<ε|cn−L|<ε$ for all $n≥N2.n≥N2.$ By assumption, there exists an integer $NN$ such that $an≤bn≤cnan≤bn≤cn$ for all $n≥N.n≥N.$ Let $MM$ be the largest of $N1,N2,N1,N2,$ and $N.N.$ We must show that $|bn−L|<ε|bn−L|<ε$ for all $n≥M.n≥M.$ For all $n≥M,n≥M,$
$−ε<−|an−L|≤an−L≤bn−L≤cn−L≤|cn−L|<ε.−ε<−|an−L|≤an−L≤bn−L≤cn−L≤|cn−L|<ε.$
Therefore, $−ε and we conclude that $|bn−L|<ε|bn−L|<ε$ for all $n≥M,n≥M,$ and we conclude that the sequence ${bn}{bn}$ converges to $L.L.$
Figure 5.6 Each term $bnbn$ satisfies $an≤bn≤cnan≤bn≤cn$ and the sequences ${an}{an}$ and ${cn}{cn}$ converge to the same limit, so the sequence ${bn}{bn}$ must converge to the same limit as well.
## Example 5.5
### Using the Squeeze Theorem
Use the Squeeze Theorem to find the limit of each of the following sequences.
1. ${cosnn2}{cosnn2}$
2. ${(−12)n}{(−12)n}$
## Checkpoint5.5
Find $limn→∞2n−sinnn.limn→∞2n−sinnn.$
Using the idea from Example 5.5b. we conclude that $rn→0rn→0$ for any real number $rr$ such that $−1 If $r<−1,r<−1,$ the sequence ${rn}{rn}$ diverges because the terms oscillate and become arbitrarily large in magnitude. If $r=−1,r=−1,$ the sequence ${rn}={(−1)n}{rn}={(−1)n}$ diverges, as discussed earlier. Here is a summary of the properties for geometric sequences.
$rn→0if|r|<1rn→0if|r|<1$
(5.1)
$rn→1ifr=1rn→1ifr=1$
(5.2)
$rn→∞ifr>1rn→∞ifr>1$
(5.3)
${rn}diverges ifr≤−1{rn}diverges ifr≤−1$
(5.4)
## Bounded Sequences
We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.
## Definition
A sequence ${an}{an}$ is bounded above if there exists a real number $MM$ such that
$an≤Man≤M$
for all positive integers $n.n.$
A sequence ${an}{an}$ is bounded below if there exists a real number $MM$ such that
$M≤anM≤an$
for all positive integers $n.n.$
A sequence ${an}{an}$ is a bounded sequence if it is bounded above and bounded below.
If a sequence is not bounded, it is an unbounded sequence.
For example, the sequence ${1/n}{1/n}$ is bounded above because $1/n≤11/n≤1$ for all positive integers $n.n.$ It is also bounded below because $1/n≥01/n≥0$ for all positive integers n. Therefore, ${1/n}{1/n}$ is a bounded sequence. On the other hand, consider the sequence ${2n}.{2n}.$ Because $2n≥22n≥2$ for all $n≥1,n≥1,$ the sequence is bounded below. However, the sequence is not bounded above. Therefore, ${2n}{2n}$ is an unbounded sequence.
We now discuss the relationship between boundedness and convergence. Suppose a sequence ${an}{an}$ is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms $anan$ that are arbitrarily large in magnitude as $nn$ gets larger. As a result, the sequence ${an}{an}$ cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.
## Theorem5.5
### Convergent Sequences Are Bounded
If a sequence ${an}{an}$ converges, then it is bounded.
Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence ${(−1)n}{(−1)n}$ is bounded, but the sequence diverges because the sequence oscillates between $11$ and $−1−1$ and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.
Consider a bounded sequence ${an}.{an}.$ Suppose the sequence ${an}{an}$ is increasing. That is, $a1≤a2≤a3….a1≤a2≤a3….$ Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that ${an}{an}$ converges. For example, consider the sequence
${12,23,34,45,…}.{12,23,34,45,…}.$
Since this sequence is increasing and bounded above, it converges. Next, consider the sequence
${2,0,3,0,4,0,1,−12,−13,−14,…}.{2,0,3,0,4,0,1,−12,−13,−14,…}.$
Even though the sequence is not increasing for all values of $n,n,$ we see that $−1/2<−1/3<−1/4<⋯.−1/2<−1/3<−1/4<⋯.$ Therefore, starting with the eighth term, $a8=−1/2,a8=−1/2,$ the sequence is increasing. In this case, we say the sequence is eventually increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.
## Definition
A sequence ${an}{an}$ is increasing for all $n≥n0n≥n0$ if
$an≤an+1for alln≥n0.an≤an+1for alln≥n0.$
A sequence ${an}{an}$ is decreasing for all $n≥n0n≥n0$ if
$an≥an+1for alln≥n0.an≥an+1for alln≥n0.$
A sequence ${an}{an}$ is a monotone sequence for all $n≥n0n≥n0$ if it is increasing for all $n≥n0n≥n0$ or decreasing for all $n≥n0.n≥n0.$
We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.
## Theorem5.6
### Monotone Convergence Theorem
If ${an}{an}$ is a bounded sequence and there exists a positive integer $n0n0$ such that ${an}{an}$ is monotone for all $n≥n0,n≥n0,$ then ${an}{an}$ converges.
The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 5.7).
Figure 5.7 Since the sequence ${an}{an}$ is increasing and bounded above, it must converge.
In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.
## Example 5.6
### Using the Monotone Convergence Theorem
For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.
1. ${4nn!}{4nn!}$
2. ${an}{an}$ defined recursively such that
$a1=2andan+1=an2+12anfor alln≥2.a1=2andan+1=an2+12anfor alln≥2.$
## Checkpoint5.6
Consider the sequence ${an}{an}$ defined recursively such that $a1=1,a1=1,$ $an=an−1/2.an=an−1/2.$ Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.
## Student Project
### Fibonacci Numbers
The Fibonacci numbers are defined recursively by the sequence ${Fn}{Fn}$ where $F0=0,F0=0,$ $F1=1F1=1$ and for $n≥2,n≥2,$
$Fn=Fn−1+Fn−2.Fn=Fn−1+Fn−2.$
Here we look at properties of the Fibonacci numbers.
1. Write out the first twenty Fibonacci numbers.
2. Find a closed formula for the Fibonacci sequence by using the following steps.
1. Consider the recursively defined sequence ${xn}{xn}$ where $xo=cxo=c$ and $xn+1=axn.xn+1=axn.$ Show that this sequence can be described by the closed formula $xn=canxn=can$ for all $n≥0.n≥0.$
2. Using the result from part a. as motivation, look for a solution of the equation
$Fn=Fn−1+Fn−2Fn=Fn−1+Fn−2$
of the form $Fn=cλn.Fn=cλn.$ Determine what two values for $λλ$ will allow $FnFn$ to satisfy this equation.
3. Consider the two solutions from part b.: $λ1λ1$ and $λ2.λ2.$ Let $Fn=c1λ1n+c2λ2n.Fn=c1λ1n+c2λ2n.$ Use the initial conditions $F0F0$ and $F1F1$ to determine the values for the constants $c1c1$ and $c2c2$ and write the closed formula $Fn.Fn.$
3. Use the answer in 2 c. to show that
$limn→∞Fn+1Fn=1+52.limn→∞Fn+1Fn=1+52.$
The number $ϕ=(1+5)/2ϕ=(1+5)/2$ is known as the golden ratio (Figure 5.8 and Figure 5.9).
Figure 5.8 The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number—always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)
Figure 5.9 The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon appears to have these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)
## Section 5.1 Exercises
Find the first six terms of each of the following sequences, starting with $n=1.n=1.$
1.
$an=1+(−1)nan=1+(−1)n$ for $n≥1n≥1$
2.
$an=n2−1an=n2−1$ for $n≥1n≥1$
3.
$a1=1a1=1$ and $an=an−1+nan=an−1+n$ for $n≥2n≥2$
4.
$a1=1,a1=1,$ $a2=1a2=1$ and $an+2=an+an+1an+2=an+an+1$ for $n≥1n≥1$
5.
Find an explicit formula for $anan$ where $a1=1a1=1$ and $an=an−1+nan=an−1+n$ for $n≥2.n≥2.$
6.
Find a formula $anan$ for the $nthnth$ term of the arithmetic sequence whose first term is $a1=1a1=1$ such that $an+1−an=17an+1−an=17$ for $n≥1.n≥1.$
7.
Find a formula $anan$ for the $nthnth$ term of the arithmetic sequence whose first term is $a1=−3a1=−3$ such that $an+1−an=4an+1−an=4$ for $n≥1.n≥1.$
8.
Find a formula $anan$ for the $nthnth$ term of the geometric sequence whose first term is $a1=1a1=1$ such that $an+1an=10an+1an=10$ for $n≥1.n≥1.$
9.
Find a formula $anan$ for the $nthnth$ term of the geometric sequence whose first term is $a1=3a1=3$ such that $an+1an=1/10an+1an=1/10$ for $n≥1.n≥1.$
10.
Find an explicit formula for the $nthnth$ term of the sequence whose first several terms are ${0,3,8,15,24,35,48,63,80,99,…}.{0,3,8,15,24,35,48,63,80,99,…}.$ (Hint: First add one to each term.)
11.
Find an explicit formula for the $nthnth$ term of the sequence satisfying $a1=0a1=0$ and $an=2an−1+1an=2an−1+1$ for $n≥2.n≥2.$
Find a formula for the general term $anan$ of each of the following sequences.
12.
${1,0,−1,0,1,0,−1,0,…}{1,0,−1,0,1,0,−1,0,…}$ (Hint: Find where $sinxsinx$ takes these values)
13.
${ 1 , − 1 / 3 , 1 / 5 , − 1 / 7 ,… } { 1 , − 1 / 3 , 1 / 5 , − 1 / 7 ,… }$
Find a function $f(n)f(n)$ that identifies the $nthnth$ term $anan$ of the following recursively defined sequences, as $an=f(n).an=f(n).$
14.
$a1=1a1=1$ and $an+1=−anan+1=−an$ for $n≥1n≥1$
15.
$a1=2a1=2$ and $an+1=2anan+1=2an$ for $n≥1n≥1$
16.
$a1=1a1=1$ and $an+1=(n+1)anan+1=(n+1)an$ for $n≥1n≥1$
17.
$a1=2a1=2$ and $an+1=(n+1)an/2an+1=(n+1)an/2$ for $n≥1n≥1$
18.
$a1=1a1=1$ and $an+1=an/2nan+1=an/2n$ for $n≥1n≥1$
Plot the first $NN$ terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.
19.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ and for $n≥2,n≥2,$ $an=12(an−1+an−2);an=12(an−1+an−2);$ $N=30N=30$
20.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ $a3=3a3=3$ and for $n≥4,n≥4,$ $an=13(an−1+an−2+an−3),an=13(an−1+an−2+an−3),$ $N=30N=30$
21.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ and for $n≥3,n≥3,$ $an=an−1an−2;an=an−1an−2;$ $N=30N=30$
22.
[T] $a1=1,a1=1,$ $a2=2,a2=2,$ $a3=3,a3=3,$ and for $n≥4,n≥4,$ $an=an−1an−2an−3;an=an−1an−2an−3;$ $N=30N=30$
Suppose that $limn→∞an=1,limn→∞an=1,$ $limn→∞bn=−1,limn→∞bn=−1,$ and $0<−bn for all $n.n.$ Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.
23.
$lim n → ∞ ( 3 a n − 4 b n ) lim n → ∞ ( 3 a n − 4 b n )$
24.
$lim n → ∞ ( 1 2 b n − 1 2 a n ) lim n → ∞ ( 1 2 b n − 1 2 a n )$
25.
$lim n → ∞ a n + b n a n − b n lim n → ∞ a n + b n a n − b n$
26.
$lim n → ∞ a n − b n a n + b n lim n → ∞ a n − b n a n + b n$
Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.
27.
$n 2 2 n n 2 2 n$
28.
$( n − 1 ) 2 ( n + 1 ) 2 ( n − 1 ) 2 ( n + 1 ) 2$
29.
$n n + 1 n n + 1$
30.
$n1/nn1/n$ (Hint: $n1/n=e1nlnn)n1/n=e1nlnn)$
For each of the following sequences, whose $nthnth$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.
31.
$n/2n,n/2n,$ $n≥2n≥2$
32.
$ln ( 1 + 1 n ) ln ( 1 + 1 n )$
33.
$sin n sin n$
34.
$cos ( n 2 ) cos ( n 2 )$
35.
$n1/n,n1/n,$ $n≥3n≥3$
36.
$n−1/n,n−1/n,$ $n≥3n≥3$
37.
$tan n tan n$
38.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
$a1=2,a1=2,$ $a2=22,a2=22,$ $a3=222a3=222$ etc.
39.
Determine whether the sequence defined as follows has a limit. If it does, find the limit.
$a1=3,a1=3,$ $an=2an−1,an=2an−1,$ $n=2,3,….n=2,3,….$
Use the Squeeze Theorem to find the limit of each of the following sequences.
40.
$n sin ( 1 / n ) n sin ( 1 / n )$
41.
$cos ( 1 / n ) − 1 1 / n cos ( 1 / n ) − 1 1 / n$
42.
$a n = n ! n n a n = n ! n n$
43.
$a n = sin n sin ( 1 / n ) a n = sin n sin ( 1 / n )$
For the following sequences, plot the first $2525$ terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.
44.
[T] $an=sinnan=sinn$
45.
[T] $an=cosnan=cosn$
Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.
46.
$a n = tan −1 ( n 2 ) a n = tan −1 ( n 2 )$
47.
$a n = ( 2 n ) 1 / n − n 1 / n a n = ( 2 n ) 1 / n − n 1 / n$
48.
$a n = ln ( n 2 ) ln ( 2 n ) a n = ln ( n 2 ) ln ( 2 n )$
49.
$a n = ( 1 − 2 n ) n a n = ( 1 − 2 n ) n$
50.
$a n = ln ( n + 2 n 2 − 3 ) a n = ln ( n + 2 n 2 − 3 )$
51.
$a n = 2 n + 3 n 4 n a n = 2 n + 3 n 4 n$
52.
$a n = ( 1000 ) n n ! a n = ( 1000 ) n n !$
53.
$a n = ( n ! ) 2 ( 2 n ) ! a n = ( n ! ) 2 ( 2 n ) !$
Newton’s method seeks to approximate a solution $f(x)=0f(x)=0$ that starts with an initial approximation $x0x0$ and successively defines a sequence $xn+1=xn−f(xn)f′(xn).xn+1=xn−f(xn)f′(xn).$ For the given choice of $ff$ and $x0,x0,$ write out the formula for $xn+1.xn+1.$ If the sequence appears to converge, give an exact formula for the solution $x,x,$ then identify the limit $xx$ accurate to four decimal places and the smallest $nn$ such that $xnxn$ agrees with $xx$ up to four decimal places.
54.
[T] $f(x)=x2−2,f(x)=x2−2,$ $x0=1x0=1$
55.
[T] $f(x)=(x−1)2−2,f(x)=(x−1)2−2,$ $x0=2x0=2$
56.
[T] $f(x)=ex−2,f(x)=ex−2,$ $x0=1x0=1$
57.
[T] $f(x)=lnx−1,f(x)=lnx−1,$ $x0=2x0=2$
58.
[T] Suppose you start with one liter of vinegar and repeatedly remove $0.1L,0.1L,$ replace with water, mix, and repeat.
1. Find a formula for the concentration after $nn$ steps.
2. After how many steps does the mixture contain less than $10%10%$ vinegar?
59.
[T] A lake initially contains $20002000$ fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by $6%6%$ each month. However, factoring in all causes, $150150$ fish are lost each month.
1. Explain why the fish population after $nn$ months is modeled by $Pn=1.06Pn−1−150Pn=1.06Pn−1−150$ with $P0=2000.P0=2000.$
2. How many fish will be in the pond after one year?
60.
[T] A bank account earns $5%5%$ interest compounded monthly. Suppose that $10001000$ is initially deposited into the account, but that $1010$ is withdrawn each month.
1. Show that the amount in the account after $nn$ months is $An=(1+.05/12)An−1−10;An=(1+.05/12)An−1−10;$ $A0=1000.A0=1000.$
2. How much money will be in the account after $11$ year?
3. Is the amount increasing or decreasing?
4. Suppose that instead of $10,10,$ a fixed amount $dd$ dollars is withdrawn each month. Find a value of $dd$ such that the amount in the account after each month remains $1000.1000.$
5. What happens if $dd$ is greater than this amount?
61.
[T] A student takes out a college loan of $10,00010,000$ at an annual percentage rate of $6%,6%,$ compounded monthly.
1. If the student makes payments of $100100$ per month, how much does the student owe after $1212$ months?
2. After how many months will the loan be paid off?
62.
[T] Consider a series combining geometric growth and arithmetic decrease. Let $a1=1.a1=1.$ Fix $a>1a>1$ and $0 Set $an+1=a.an−b.an+1=a.an−b.$ Find a formula for $an+1an+1$ in terms of $an,an,$ $a,a,$ and $bb$ and a relationship between $aa$ and $bb$ such that $anan$ converges.
63.
[T] The binary representation $x=0.b1b2b3...x=0.b1b2b3...$ of a number $xx$ between $00$ and $11$ can be defined as follows. Let $b1=0b1=0$ if $x<1/2x<1/2$ and $b1=1b1=1$ if $1/2≤x<1.1/2≤x<1.$ Let $x1=2x−b1.x1=2x−b1.$ Let $b2=0b2=0$ if $x1<1/2x1<1/2$ and $b2=1b2=1$ if $1/2≤x<1.1/2≤x<1.$ Let $x2=2x1−b2x2=2x1−b2$ and in general, $xn=2xn−1−bnxn=2xn−1−bn$ and $bn−1=0bn−1=0$ if $xn<1/2xn<1/2$ and $bn−1=1bn−1=1$ if $1/2≤xn<1.1/2≤xn<1.$ Find the binary expansion of $1/3.1/3.$
64.
[T] To find an approximation for $π,π,$ set $a0=2+1,a0=2+1,$ $a1=2+a0,a1=2+a0,$ and, in general, $an+1=2+an.an+1=2+an.$ Finally, set $pn=3.2n+12−an.pn=3.2n+12−an.$ Find the first ten terms of $pnpn$ and compare the values to $π.π.$
For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of $NN$ integers $a1,a2,…,aNa1,a2,…,aN$ by fixing two special integers $KK$ and $MM$ and letting $an+1an+1$ be the remainder after dividing $K.anK.an$ into $M,M,$ then creates a bit sequence of zeros and ones whose $nthnth$ term $bnbn$ is equal to one if $anan$ is odd and equal to zero if $anan$ is even. If the bits $bnbn$ are pseudorandom, then the behavior of their average $(b1+b2+⋯+bN)/N(b1+b2+⋯+bN)/N$ should be similar to behavior of averages of truly randomly generated bits.
65.
[T] Starting with $K=16,807K=16,807$ and $M=2,147,483,647,M=2,147,483,647,$ using ten different starting values of $a1,a1,$ compute sequences of bits $bnbn$ up to $n=1000,n=1000,$ and compare their averages to ten such sequences generated by a random bit generator.
66.
[T] Find the first $10001000$ digits of $ππ$ using either a computer program or Internet resource. Create a bit sequence $bnbn$ by letting $bn=1bn=1$ if the $nthnth$ digit of $ππ$ is odd and $bn=0bn=0$ if the $nthnth$ digit of $ππ$ is even. Compute the average value of $bnbn$ and the average value of $dn=|bn+1−bn|,dn=|bn+1−bn|,$ $n=1,...,999.n=1,...,999.$ Does the sequence $bnbn$ appear random? Do the differences between successive elements of $bnbn$ appear random?
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https://mathoverflow.net/questions/284806/double-sum-of-negative-powers-of-integers-a-direct-approach
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# Double sum of negative powers of integers: a direct approach?
Let $\alpha,\beta\in (0,1\rbrack$, $\alpha\ne \beta$. I wish to estimate $$\sum_{m\leq x} \frac{1}{m^\alpha} \sum_{n\leq x/m} \frac{\log(x/mn)}{n^\beta}.$$ There is an obvious approach, namely, to estimate the inner sum first (the second- and third-order terms will be proportional to $\zeta(\beta)$ and $\zeta'(\beta)$; there is a connection with Ramanujan summation) and then input that estimate into the outer sum, which gets estimated in much the same way.
An improved version of the approach consists essentially in putting the longer of the two sums always inside, by means of a bit of combinatorial manipulation.
I have to wonder - is there a more elegant, less piecemeal approach, considering both sums in one go? If $\alpha$ were equal to $\beta$, the answer would be yes - we would get an estimate based on $(\zeta^2)'(s)$. Of course, that's precisely the case we aren't in.
(You may assume $\alpha = 2 \beta$, since that is the particular case I am most interested in.)
Update: what I get by a careful version of the above (continuous partition) and some pain is that the double sum equals \begin{aligned}&\frac{y^{1-\alpha}}{(1-\alpha)^2} \zeta(1-\alpha+\beta) + \frac{y^{1-\beta}}{(1-\beta)^2} \zeta(1-\beta+\alpha)\\ &+ \zeta(\alpha) \zeta(\beta) \log y + \zeta'(\alpha) \zeta(\beta) + \zeta(\alpha) \zeta'(\beta)\end{aligned} plus an error term of size $O\left(y^{\frac{1}{2} - \frac{\alpha+\beta}{2}}\right)$, where the implied constant is explicit (and fairly small). I do not know whether the error term ought to be of a lower order of magnitude. Let me see what I can do with Lucia's approach below.
• Why not just start with $\frac{1}{2\pi i} \int_{(c)} \zeta(s+\alpha)\zeta(s+\beta)x^s/s^2 ds$ and compute residues? Commented Oct 31, 2017 at 18:05
• Indeed, this works nicely (and just helped me double-check the formula above). The error term one gets in this way is better than $O(x^{-\beta})$ (for $\beta<\alpha$). Indeed, it is $O_\epsilon\left(x^{-(2\beta/3+\alpha/3)}\right)$ (without using subconvexity; I'm using the bounds on zeta(s) in section 5.1 of Titchmarsh). Now, I need an explicit constant. I'll have to check what explicit bounds on $\zeta(s)$ I can find. My fear is that the constants obtained this way may not be good (unless one would be willing to look for cancellation, but that would most likely be a difficult mess). Commented Nov 1, 2017 at 0:06
• It seems to me that the simplest way to estimate such an integral is to use Cauchy-Schwarz and then estimate $\frac{1}{2\pi i} \int_{(c)} \frac{|\zeta(s+\alpha)|^2}{|s|^2} ds$ using the fact that the Mellin transform is an isometry. Commented Nov 1, 2017 at 1:05
• That sounds good to me! Commented Nov 1, 2017 at 2:02
• Of course this is not so straightforward, since the fact that the Mellin transform is an isometry is immediately relevant only for $c>1$. Commented Nov 1, 2017 at 7:52
Let me carry out matters using a complex-analytical approach, as Lucia suggests, and then say where the difficulty lies.
Let $0<\beta<\alpha\leq 1$. First of all, as Lucia says, $$\sum_{m\leq x} \frac{1}{m^\alpha} \sum_{n\leq x/m} \frac{\log(x/mn)}{n^\beta} = \frac{1}{2πi} \int_{c-i\infty}^{c+i\infty} \zeta(s+\alpha) \zeta(s+\beta) \frac{x^s}{s^2} ds$$ for $c>1$. We shift the contour of integration to the left of $\Re(s)=0$, picking up the main terms \begin{aligned}&\frac{y^{1-\alpha}}{(1-\alpha)^2} \zeta(1-\alpha+\beta) + \frac{y^{1-\beta}}{(1-\beta)^2} \zeta(1-\beta+\alpha)\\ &+ \zeta(\alpha) \zeta(\beta) \log y + \zeta'(\alpha) \zeta(\beta) + \zeta(\alpha) \zeta'(\beta)\end{aligned} plus an error term of size $O\left(y^{\frac{1}{2} - \frac{\alpha+\beta}{2}}\right)$ along the way. We are left with the task of estimating an error term $$\frac{1}{2πi} \int_R \zeta(s+\alpha) \zeta(s+\beta) \frac{x^s}{s^2} ds,$$ where the integral is over a contour $R$ of our choice going from $-r-i\infty$ to $-r+i\infty$, say, and satisfying $\Re s\leq -r$ at all points. The error will be clearly bounded by $O(K x^{-r})$, where $$K = \frac{1}{2πi} \int_R \frac{|\zeta(s+\alpha)| |\zeta(s+\beta)|}{|s|^2} ds.$$ The problem does reduces to estimating $K$.
Now, there are rigorous-numerics packages that include integration and the possibility to compute the zeta function $\zeta(s)$. (I currently use ARB.) However, (a) computations must obviously be finite (at least assuming mortal mathematicians), and (b) computing $\zeta(s)$ is never a walk in the park, and rigorous integration only adds to the overhead. Integrating an expression such as above from $-1/2 - i T$ to $1/2 + i T$ takes 15 minutes for $T = 10000$ (says a better programmer than I), but we should not expect to go much further than $T = 100000$ programming casually on our laptops.
The problem that remains, then, is how to bound a tail $$\frac{1}{2\pi i} \left(\int_{-r-i\infty}^{-r-i T} + \int_{-r+i T}^{-r + i \infty} \frac{|\zeta(s+\alpha)| |\zeta(s+\beta)|}{|s|^2} ds\right).$$
The most obvious approach is to use Backlund's explicit bounds (1918) on $\zeta(\sigma + it)$ (see http://iml.univ-mrs.fr/~ramare/TME-EMT/Articles/Art06.html#Size). They are of the quality $$|\zeta(\sigma + i t)| = (1+o(1)) (t/2\pi)^{(1-\sigma)/2} \log t$$ for $0\leq \sigma\leq 1$ and $$|\zeta(\sigma + i t)| = (1+o(1)) (t/2\pi)^{1/2-\sigma} \log t$$ for $-1/2\leq \sigma\leq 0$. The problem here is that convergence is painfully slow. If, say, $\alpha = 1$, $\beta=1/2$ and $r =-1/4$ (reasonable values all around), the tails will be bounded by a constant times $(\log T)^2/\sqrt{T}$. For $T=10000$, $(\log T)^2/\sqrt{T} > 0.848\dotsc$ - not exactly small; for $T=100000$, the same equals $0.419\dotsc$ - barely an improvement.
Notice, however, that why Backlund's bounds are essentially tight for $\Re s<0$, that is not the case for $0<\Re(s)<1$. Of course, they are convexity bounds, so improving on them explicitly would amount to translating into explicit terms rather non-trivial material. However, as long as we are satisfied with $r>-\beta$, what we can do instead is give $L_2$ bounds for the tails, that is, bound $$\int_{r-i \infty}^{r-i T} \frac{|\zeta(s)|^2}{|s|^2} ds + \int_{r+iT}^{r+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds.$$ (The integral $\int_{r-i \infty}^{r-i T}$ is obviously the same.) Then we use Cauchy-Schwarz to bound the tail of the integral we were discussing.
This is non-trivial, and takes us further afield, so I will make it into a separate question: $L_2$ bounds for tails of $\zeta(s)$ on a vertical line .
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Tomewolkiibection 2.4 Homework Score: 0 of1 pt 150f 24 (23 complete) - 2.4.25Aformat for back-to-back stemplots representing the pulse rate of females &nd Compl...
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Tomewolkiibection 2.4 Homework Score: 0 of1 pt 150f 24 (23 complete) - 2.4.25Aformat for back-to-back stemplots representing the pulse rate of females &nd Complete the back-to-back males from the given data is shown below Complete the back-to-back stemplot stemplot then compare the results. Women Stem (tens) Men Females Full data set 80 69 9633322 7010763 76 86 63 81*123 Males 61571 81756 61 87 65 55]78 86 66_ 59 Women Stem (tens) Men 569 9633322Hnlmanswerinune eamfeics and then clic
Tomewolkiibection 2.4 Homework Score: 0 of1 pt 150f 24 (23 complete) - 2.4.25 Aformat for back-to-back stemplots representing the pulse rate of females &nd Complete the back-to-back males from the given data is shown below Complete the back-to-back stemplot stemplot then compare the results. Women Stem (tens) Men Females Full data set 80 69 9633322 7010763 76 86 63 81*123 Males 61571 81756 61 87 65 55]78 86 66_ 59 Women Stem (tens) Men 569 9633322 Hnlmanswerinune eamfeics and then click Chc?
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Rewrite the inequality Ix - 2| double inequality without absolute value and state the open interval which is its solution set using interval notation: V. Give the sign analysis of (x 1)x 1)(x on R as in class using left-right/ right-left:...
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Вы находитесь на странице: 1из 4
# =
lim
n!1
(2 points)
n+1 e
= 0. (1 points)
It means that the series is NOT convergent no matter what value z is.
Solution 7
## Since e2z is analytic function (4 points), we have
Z
Z 2i
1
e2z dz =
e2z dz = e2z
2
c
i
2i
(4 points) = 0. (2 points)
Tutor: William Li
i
Problem
5 (20 points in total)
P1:
(a)
## Apply Gaussian elimination to matrix A, we have
2
3
2
0
2 1 3
1 0 0
6
7
6
6
7
6
6 1 4 0 7 7!6 0 1 0
4
5
4
5 5 5 5
0 0 1
29
9
21
7
7
7 . (6 points)
5
## Thus, the dimension for the row space is 3. (1 point)
h
i h
i
h
i
The bases for row space are 1 0 0
29 , 0 1 0 9 , and 0 0 1 21 . (3 points)3
The dimension for the column space is 3. (1 point)
## The bases for column space are
2
0
6
6
6 1
4
5
7
7
7,
5
6
7
6
7
6 4 7
4
5
5
6
7
6
7
and 6 0 7 . (3 points)
4
5
5
## The null space are the solution of Ax = 0 (1 point), which is
2
3 2
3
x1
29
6
7 6
7
6
7 6
7
6 x2 7 6
9 7
6
7=6
7 x4 . (2 points)
6
7 6
7
6 x3 7 6 21 7
4
5 4
5
x4
1
The dimension of null space is 1. (1 point)
(b)
## The rank of A is equal to the rank AT , which is equal to 3. (2 points)
Problem
6 (20 points in total)
(a)
## The character polynomial of B is
2
P ( ) = Det(B
I) (2 points) =
2
2
+ 21 + 45. (2 points)
(b)
P( ) =
= 5 and
## The eigenvectors are the solution of (B
Gaussian Elimination , we have
2
7
6
6
6 2
4
1
2
4
2
5)( + 3)2 ,
3. (4 points)
7
6
7
6
6 7!6 0
5
4
5
0
0
1
0
7
7
2 7 , (2 points)
5
0
= 5 By applying
x3
x4
(b)
P2:
21
(a)
## The character polynomial of B is
2
I) (2 points) =
P ( ) = Det(B
2
2
+ 21 + 45. (2 points)
(b)
P( ) =
= 5 and
## The eigenvectors are the solution of (B
Gaussian Elimination , we have
2
7
6
6
6 2
4
1
the eigenvectors are
2
2
x1
=
7
6
7
6
6 7!6 0
5
4
5
0
6
7 6
6
7 6
6 x2 7 = 6 2
4
5 4
x3
1
When
3. (4 points)
## I)x = 0. (2 points) When
5)( + 3)2 ,
0
1
0
= 5 By applying
7
7
2 7 , (2 points)
5
0
7
7
7 x3 , (2 points)
5
x
2
3
6 1 7 6
7
6
7
7
6
7
6
7 6
6 x2 7 = 6 1 7 x2 + 6 0 7 x3 , (2 points)
4
5 4
5
4
5
x3
0
1
## 3 By applying Gaussian Elimination , we have
2
3
2
3
1
2
3
1 2
3
6
7
6
7
6
7
6
7
6 2
4
6 7 ! 6 0 0 0 7 , (2 points)
4
5
4
5
1
2 3
0 0 0
## the eigenvectors are
Page 6
1) Calculate the eigenvalue and eigenvector of matrix
1 0 0
A= 0 1 8
0 1 3
P3:
## Solution: The characteristic equation of A is:
'
'
' 1
0
0
'
'
det(A I) = 0 ' 0
1
8
'
'
' 0
1
3
'
'
'
'
'
'=0
'
'
'
3 52 + 5 = ( + 1)( 1)( 5) = 0
1 = 1, 2 = 5, 3 = 1;
By definition x is an eigenvector of A corresponding to if and only if x is a nontrivial solution of
(A I)x = 0. That is
1
0
0
0
1
1
x1
x2 = 0
x3
0
3
8
## For = 1. the linear system becomes:
0
0 0 0
x1
0 0 8 x2 = 0
0
x3
0 1 2
Solving this linear system (like Gaussian elimination) yields:
1
x1
x2 = s 0 , so that
0
x3
0
0
1
0
0
## 1 = 1x1 = 0 , 2 = 5x2 = 2 , 3 = 1x3 = 4
0
1
1
Page 8 To find the eigenvalues of A, we have (A I)x = 0. Similarly, to find the eigenvalues of 2I A,
we have,
March 1, 2010
[(2I A) I] = 0 [A (2 )I] = 0
## So, eigenvalues of 2I A are (2 i ), i = 1, 2, ..., n.
!n
So, det(2I A) = i=1 (2 i ).
DRAFT
\$ a11
! I # &&
&( an1
## ? And compare with the coefficient of !
from
a1n
1n %
'
'
ann ')
Still let ! be 0, we have p(0) " (#1)n !1!2 ...!n and det(# A) " (#1)n det( A)
P4:
## 2. From (a), we have a " 2
From (b), we use tr ( A) "
*!
i
i
2, d " 3#3
2 b
1 d
" 3*(#3)
## From (c), we still use tr ( A) "
*!
i
, we obtain 2 # 2 , f " 0 , 1 # 2
Observe that p(! ) " 0 , that means p(0) " 0 , p(1) " 0 and p(#2) " 0 .
We can obtain c " 9, e " #7
\$3 2 %
\$1 %
,- " &
'
'
(1 1 )
( 0)
\$3 2% \$1 % \$ 1 #2%
We have B " S -S #1 = &
'&
'&
'
(1 1 ) ( 0) ( #1 3 )
## 3. let S " &
We just need find the eigenvalue of the block matrix(why?). The eigenvalues are 1,3,1,3
\$1 % \$ 1 % \$ 0 % \$ 0 %
&1 ' & #1' &0 ' & 0 '
And corresponding eigenvectors are & ' , & ' , & ' , & ' respectively.
& 0 ' & 0 ' &1 ' & 1 '
& ' & ' & ' & '
(0 ) ( 0 ) (1 ) ( #1)
As these eigenvectors are linearly independent, B is diagonalizable.
4. Find the eigenvalue of A , they are , , 1. Check that A is diagonalizable (omit). So we have
( A # rI ) " S (- # rI )S T
Let wT " vT S , and w is arbitrary due to v arbitrarily chosen.
If wT (- # rI )w . 0 , or w12 (!1 # r ) , ... , wn2 (!n # r ) . 0
We should let all (!i # r ) . 0 , so r / 1/ 4
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# Oral-History:Harold S. Black
(Difference between revisions)
Revision as of 20:30, 7 May 2009 (view source)EMW (Talk | contribs)← Older edit Revision as of 15:58, 8 May 2009 (view source)SHH (Talk | contribs) Newer edit → Line 1,797: Line 1,797: '''Black:''' '''Black:''' − To make a linear vacuum tube was hopeless. To give you an example, if you plotted the voltage to the grid as a function of the amplification of the voltage out, if you plotted the output against the input in terms of current – and I square this pump [correct phrase?], if you made yourself a drawing that was 1 mile square and drew it using the hardest pencil chiseled to a straight line and made that line straight, the nonlinearity would be too much to meet my requirement. Therefore there was no earthly way of getting it by a linear vacuum tube. There are a lot of tricks one can try.
+ To make a linear vacuum tube was hopeless. To give you an example, if you plotted the voltage to the grid as a function of the amplification of the voltage out, if you plotted the output against the input in terms of current – and I square this pump, if you made yourself a drawing that was 1 mile square and drew it using the hardest pencil chiseled to a straight line and made that line straight, the nonlinearity would be too much to meet my requirement. Therefore there was no earthly way of getting it by a linear vacuum tube. There are a lot of tricks one can try.
Line 3,127: Line 3,127: '''Black:''' '''Black:''' − This is the sketch. The idea was, here comes an input and then there's an output. The reason I call this µ was due to having read the book by van der Bijl. However, µ here is a complex quantity like Steinmetz first introduced: A + AB. The reason I used the symbol was because in some book in my library was the transmission characteristic of a circuit. I guess it was really Maxwell's Equation. It had to do with the electrical properties of electromagnetic waves. If they were on a pair of wires or a coax […ial cable], the attenuation would vary as the square root of frequency. In other words, was a term that was usually expressed E to the IΣЗ, to put it in complex form. I noticed here that if З was the attenuation of the line it would be an equalizer and would serve two purposes for one. Of course I didn't get any of that information on the sketch. It took a long to get that. You asked what was the reason for my choice of these two Greek letters.
+ This is the sketch. The idea was, here comes an input and then there's an output. The reason I call this µ was due to having read the book by van der Bijl. However, µ here is a complex quantity like Steinmetz first introduced: A + AB. The reason I used the symbol was because in some book in my library was the transmission characteristic of a circuit. I guess it was really Maxwell's Equation. It had to do with the electrical properties of electromagnetic waves. If they were on a pair of wires or a coax […ial cable], the attenuation would vary as the square root of frequency. In other words, З was a term that was usually expressed E to the IΣЗ, to put it in complex form. I noticed here that if З was the attenuation of the line it would be an equalizer and would serve two purposes for one. Of course I didn't get any of that information on the sketch. It took a long to get that. You asked what was the reason for my choice of these two Greek letters.
− '''Wolff:Σ''' + '''Wolff:''' No. I asked you what it was you actually thought. What was this flash that came to you?
No. I asked you what it was you actually thought. What was this flash that came to you?
Line 3,853: Line 3,853: '''Wolff:''' '''Wolff:''' − Is loop gain a product of µ?
+ Is loop gain a product of µЗ?
Line 3,859: Line 3,859: '''Black:''' '''Black:''' − That is exactly right. It is the absolute value of the product of µ. The U.S. Patent Office found those papers, and each one and said it could not be done. + That is exactly right. It is the absolute value of the product of µЗ. The U.S. Patent Office found those papers, and each one and said it could not be done.
Line 3,991: Line 3,991: '''Black:''' '''Black:''' − My only answer to that is that I could not read his mind, but he was convinced. The most natural reason would be the one you suggest. I boldly suggested putting a tremendous amount of gain into the µ rule. Yet I was claiming that the amplifier would not sing. He was educated as a physicist. He got his Ph.D. from Chicago and came to work for Western Electric in 1911. He was hand picked by Dr. Jewett. My answer is that I don't know. I don't even know why the British post office and or other people in other countries thought it would not work.
+ My only answer to that is that I could not read his mind, but he was convinced. The most natural reason would be the one you suggest. I boldly suggested putting a tremendous amount of gain into the µЗ rule. Yet I was claiming that the amplifier would not sing. He was educated as a physicist. He got his Ph.D. from Chicago and came to work for Western Electric in 1911. He was hand picked by Dr. Jewett. My answer is that I don't know. I don't even know why the British post office and or other people in other countries thought it would not work.
Line 4,237: Line 4,237: '''Black:''' '''Black:''' − Yes, with spacing at every 25 miles, as Dr. Jewett's requirement was 4,000 miles. The transmission, due to any and every cause, was not to be out by more than 1 dB. Today they permit 4 dB because of the improved handsets and other reasons.
+ Yes, with spacing at every 25 miles, as Dr. Jewett's requirement was 4,000 miles. The transmission, due to any and every cause, was not to be out by more than ∀1 dB. Today they permit 4 dB because of the improved handsets and other reasons.
## Contents
Harold S. Black (1898-1983) invented the negative feedback amplifier, revolutionizing the field of electronics. Black began experimenting with telecommunications systems in his youth and received a BSS in electrical engineering at Worcester Polytechnic Institute in 1921. Recruited before his graduation, Black joined the labs at Western Electric Company in July 1921. After undergoing the company’s training program, he joined the Systems Engineering Department, where his projects included carrier system design.
Independent work shaped Black’s Western Electric career and his inventions. To acquaint himself with the workings of the company, Black spent Sundays reading the company’s memoranda files, starting with 1898 documents and proceeding through the present. This knowledge, combined with socialization in the company cafeteria, netted Black more significant assignments. At home on Thanksgiving Day, 1921, Black made a breakthrough in his work with the imperfectly performing three-channel system, determining that “the requirements on a single amplifier would vary as a function of the number of channels,” increasing by the square root of the number of channels. Black describes his communication of this model to Ralph Vinton Lyon Hartley in the Western Electric research department, who provided feedback on the role of amplifiers in third harmonics. As a result, Black replaced the push-pull amplifier used in the three-channel system model with amplifiers used in tandem.
He describes the resistance of Western Electric to his assignment of up to three thousand channels to an amplifier when a four-channel model had failed. Black, on the other hand, predicted that transcontinental telephone communications would require multiple channels, and received permission to work on linearizing the multichannel amplifier if it did not interfere with his other Western Electric projects, including his assignments to standardize push-pull amplifier design and to offset the effected of battery heat on amplifier performance. Nevertheless, Black did communicate with other scientists in the Bell Labs system who supported multichannel development, and he collaborated with Mervin Kelly on his ultimately futile efforts to produce a linear vacuum tube.
Black’s efforts culminated in the invention of the negative feedback amplifier in 1927, which he described in the influential 1934 Electrical Engineering paper “Stabilized feed-back amplifiers.” In this interview, Black details Western Electric construction and testing of this amplifier, and he describes the breakthrough sketch me made on a ferry ride during experimentation in Morristown, New Jersey. He considers the influence of his negative feedback amplifier patent on later work, and he describes the skepticism of experts consulted by the U.S. Patent Office, who asserted his device would not operate. Once the patent was obtained, Bell Labs protected it vigilantly; a suit against Zenith for patent infringement required extended testimony by Black between 1948 and 1953.
The bulk of the interview covers the logic of Black’s multichannel amplifier and negative feedback amplifier development, as Michael Wolff asks for clarification in terms understandable to non-experts. Other topics include the promotion practices in the Bell System, and the influence of Lee de Forest and E. Howard Armstrong on Black’s own work. Black describes his childhood independent study and employment, as well as the economic factors which shaped his education. He describes his decision to join Western Electric and details the four-channel and three-channel carrier systems in operation when he joined in 1921. Black ends the interview by considering the social impact of the negative feedback amplifier and by describing another influential amplifier he developed, the 8-channel open wire system.
Black also wrote the article "Inventing the negative feedback amplifier,” (IEEE Spectrum vol. 14, Dec. 1977), which he discusses in this interview.
HAROLD S. [STEPHEN] BLACK: An Interview Conducted by Michael Wolff, IEEE History Center, 20 May and 29 June 1977
Interview #456 for the IEEE History Center, The Institute of Electrical and Electronics Engineers, Inc., and Rutgers, The State University of New Jersey
This manuscript is being made available for research purposes only. All literary rights in the manuscript, including the right to publish, are reserved to the IEEE History Center. No part of the manuscript may be quoted for publication without the written permission of the Director of IEEE History Center.
Request for permission to quote for publication should be addressed to the IEEE History Center Oral History Program, Rutgers - the State University, 39 Union Street, New Brunswick, NJ 08901-8538 USA. It should include identification of the specific passages to be quoted, anticipated use of the passages, and identification of the user.
It is recommended that this oral history be cited as follows:
Harold S. Black, an oral history conducted in 1977 by Michael Wolff, IEEE History Center, Rutgers University, New Brunswick, NJ, USA.
## Interview
INTERVIEW: Harold S. Black
INTERVIEWER: Michael Wolff
DATE: 20 May 1977
PLACE: Unknown
### Education
Black:
[Referring to a paper he wrote] Do you want to retain the title of the article?
Wolff:
That's the last thing on my mind. We have a long way to go before that.
Black:
When it comes to my last paragraph about the historical importance of the thing, I was very modest there. It is used for very type of communication, whether by land, microwave, underground, outer space or submarine. And then it has applications in fields completely unrelated to communications. It has completely transcended application there.
Wolff:
Let's go chronologically and start by setting the scene.
Black:
Memory is a peculiar thing. Quite a long time has elapsed since I wrote this. In my patent and in all the work that I did when two lines joined together I joined them with a dot.
Wolff:
We are going to end up with a manuscript about three times the length of this. Your story is interesting and there is a great deal that is important enough that there is a great deal more we want to crowd into it if we possibly can. To set the scene, I want to know really when it all began, so let's talk about Thanksgiving Day, 1921.
Black:
I joined the laboratories on the 5th of July, 1921.
Wolff:
Black:
I got a BSS in electrical engineering at Worcester Polytechnic Institute.
Wolff:
Black:
Yes. I wanted a job in Minnesota or South America where they make those giant turbogenerators that take fifteen years to make.
Wolff:
What company?
Black:
I don't recall, though I delivered a good many lectures there afterwards. However, they wouldn't take me.
Wolff:
Did you ever find out why?
Black:
That was shortly after a small economic depression. It was not as easy to get a job. J. J. Pilliod and the personnel director at 463 West Street – that was Western Electric Company – had a job for me. They picked me up right away and paid me \$32 a week.
Wolff:
How did you connect with AT&T? In those days did they come to visit colleges or had you written to them?
Black:
That's a good question. About four weeks prior to graduation, which I think was the early part of June, I was approached by two people. One was the personnel director of the Western Electric Company at 463 West Street and the other one was J. J. Pilliod. Pilliod was the assistant chief engineer of the American Telephone & Telegraph Company (AT&T) at 195 Broadway. That is where it has always been and still is today.
Wolff:
Black:
No. I had not written or applied anywhere.
Wolff:
Except the place in Minnesota?
Black:
Yes. I have no recollection of how that came about. Education was very different than it is today. It is vastly superior [now] to my day. However I had hydraulic engineering, mechanical engineering, surveying, mathematics and physics.
Wolff:
All that went into being an electrical engineer?
Black:
Yes. I had to do work with my hands – pattern making, drop forge, machinery, milling machines, lathes – ergo, practical.
Wolff:
Had you always been interested in engineering?
Black:
I had always been interested in being an electrical engineer.
Wolff:
It was a pretty new field in those days. Do you remember what intrigued you with electrical engineering rather than for instance building bridges?
Black:
I didn't have very much money.
Wolff:
Nobody did.
Black:
Nobody did, but I came from poor parents. My father only went through the eighth grade and my mother was stenographer and had graduated from high school. Naturally they wanted to see me get an education.
Wolff:
Black:
He was a shipper. That was his title at a shirt shop.
Wolff:
Had you always been interested in electrical engineering?
Black:
That came about after I purchased a series of small books. I think there were ten or fifteen volumes. They dealt with all phases of electricity – magnetism, motors and things like that. As is mentioned in that book that I gave you, one summer when I worked for the American Steel & Wire Company, I worked trimming arc lights. They were all 220 volts or higher. When I put a red sign over the switch, somebody would close it while I was out working. Therefore when I climbed the poles I always was prepared to get a shock and protected myself accordingly.
Wolff:
That was during college?
Black:
Yes.
Wolff:
Do you remember how old you were when you bought those little books? It says here that as a boy you had always been interested in setting up telegraph systems, telephones and microphones.
Black:
That is correct. My first telecommunications system was probably in 1914, a year before I graduated from high school. I was sixteen years old. In the attic of the second story of small house that my father rented I had electrical things I had gathered from the town dump. That was my great source of supply. Across the street was the editor of the Lamister Enterprise. He had five daughters. I made a microphone that was two pieces of wood at right angles and then from a battery I had sawed two pieces of carbon. Then I used a tin can that I cut to make a spring so the two were in contact. That was a sensitive microphone. I could hear a watch tick and conversations all over the house. I took a piece of very fine wire that I could throw across the street and catch onto the two poles and bring it downstairs. Up in the attic I had an old telephone receiver that someone had discarded. Therefore I was able not only to hear a watch tick – because they put one on the table – but I was able to hear every word of conversation all over their house. That worked fine until a little after 5 o'clock when the father came home. He destroyed the microphone, tore down the wires and said, "No more." Therefore my first telecommunications system didn't last very long.
Wolff:
It was in their house?
Black:
I could hear everything that was in their house. It was not two-way. They could not hear things in our house. That didn't please the father.
Wolff:
That's a delightful story. You said you got interested in electrical things when you bought this series of books. Do you remember when that was?
Black:
That was before.
Wolff:
Was it sometime as a teenager?
Black:
Yes. They were advertised and the price was very small, but they were full of information. I didn't graduate from college until 1921 because my father lost his job and I had to support a family of four. By good fortune I was able to earn \$12 and [unintelligible phrase] a week working at the same factory at which my father had worked.
Wolff:
Working at the Steel & Wire?
Black:
No, I worked at the shirt shop ironing shirts and things. They had automatic machinery for pressing. And there were woolen pajamas for men. That was all piecework, and I did pretty well.
Wolff:
Did that give you enough money to go back to college?
Black:
That gave me enough money to take care of that year. When I was 16 or 17 there was World War I. Nottin Grinding Company, although my father had never done that kind of work in his life, gave my father very good pay and the entire family moved to Worcester. I however kept on working in the shirt shop, because even with a 10-cents-a-day carfare I could make more money that way. That was all the work I did at that time.
Wolff:
Then you went to Worcester Polytechnic Institute because you wanted to study electrical engineering?
Black:
Yes. I was bound and determined to become an electrical engineer.
Wolff:
Black:
I guess the truth is because I was interested in electricity.
### Western Electric Company
#### Recruitment and training
Wolff:
Let's come back to Pilliod and the other fellow from AT&T.
Black:
Wolff:
What I don't understand is how they knew about you if you didn't write to them.
Black:
They were looking and they interviewed me as well as others.
Wolff:
How did they know you were there?
Black:
They went to Worcester Tech because they turned out electrical engineers. At that time they had no graduate courses. They turned out electrical engineers, mechanical engineers and civil engineers.
Wolff:
And they made you an offer at \$32 a week.
Black:
After they interviewed me and made me an offer and I accepted it right on the spot.
Wolff:
Did you start work at Bell Labs then?
Black:
It was called the Western Electric Company. I started there on July 5th. They didn't work on July 4th. There was a delay of a couple of hours. I had to pass a physical examination to settle the fact that I was in good health.
Wolff:
What was your job there when you started?
Black:
First I had to go through an 8-week training course. That was to familiarize me with the organization of the Bell System and where the Western Electric Company, AT&T, Long Lines and the associated companies fit in with it. In addition there was a little practical work. I had to work as a telephone operator and use the cards and things like that and listen to lectures by different departments at 463 West Street. The department that I chose, which answers your other question, was Systems Engineering. I became a systems engineer punching a time clock. After my other factory experience I did not care for that.
Wolff:
Was this after the 8-week training course that you began the systems engineering?
Black:
Even in the 8-week training course we punched a time card.
#### Systems Engineering Department
Wolff:
Did you become a systems engineer after the training?
Black:
After the training course I was interviewed [by] heads of different departments. There were 150 people total that entered that particular year. There was a systems engineering department, an apparatus department and a research department. They were rather limited. The systems department was the one that appealed to me because it involved contacts with all the others and made complete systems.
Wolff:
Did you start working in that department after the training?
Black:
Yes. That department was headed by Amos Dickson. He graduated from the eighth grade but then took correspondence courses and worked up to the head of that department in a position that would correspond today to vice president. I was interviewed by B. W. (Burton W.) Kendall.
Wolff:
How many engineers were in the systems department? Was it small or large?
Black:
It was fairly large. The only way that I could answer that question accurately would be to get out the organization chart, but that department occupied the entire 9th floor at 463 West Street. That was where I met "Hi" [Haakon I.] Romnes, who later became president of the Western Electric Company and after that became president and chairman of the board of the American Telephone & Telegraph Company.
Wolff:
Was Rommes in the systems engineering department?
Black:
Yes, but he did not stay very long. He moved fast.
Wolff:
That brings us up to September of 1921. The summer was the training and then you started. What was your first assignment in the department?
Black:
That is a good question. I went out in the laboratories to make myself useful to people who reported to someone by the name of Slaughter. Slaughter's position was later taken by Charlie (C. W.) Green. One morning I given a set of tools that I did not succeed in keeping. Then I leaned against a battery and it burned the bottom of my trousers. Fortunately I came down to New York with \$75, so that when they stopped work in the middle of Saturday afternoon I had time to buy another suit of clothes.
Wolff:
Was your first day at work in systems engineering on a Saturday?
Black:
No. I got along as best I could with a poor-looking suit. Saturday was the only day I could do any shopping. I made myself useful and helped anyone who was working. As a result I was finally given a succession of problem-solving assignments. You would find it hard to believe, but at that time in the Bell Laboratories and the systems engineering department nobody ever used as much as a slide rule. Everything was empirical.
Wolff:
Do you remember some of the assignments you had? Were they empirical assignments?
Black:
Yes. I was concerned with the design of the first carrier system. It was a four-channel system that had already been designed for President Woodrow Wilson so he could communicate with Pittsburgh. That was rather unsuccessful because there were some directional filters that did not have enough singing margin between the two directions. That is where I became familiar with the terms "singing margin" and "phase margin," which were no mystery to me when I got around to inventing the negative feedback amplifier. It is explained in some of the references that I gave in this paper. So-called directional filters. In order to get two-way transmission and a number of channels involved, aside from this first example that I mentioned to you, were three. This triangle was the amplifier. In front of that was an equalizer, which I could almost have left out. That is because the attenuation of the line varies as the square root of the frequency and there is a very clever and important technical reason for pointing there. Then to get transmission in the other direction they use a second amplifier.
Wolff:
What is this diagram? Is this the four-channel system?
Black:
This is a diagram of the three-channel system, and it wasn't working at all. The second three-channel system is the one to which I am referring. The first three-channel system transmitted the carrier and so that if one talked on one channel everything on the other three channels could be heard as well. The AT&T didn't think much of that, so this was being worked upon by Slaughter. After, Slaughter was replaced by C.W. Green, who came as a professor from MIT.
Wolff:
I had better interrupt you or I'll get lost. You were telling me that you worked on a four-channel system that had been designed for President Wilson.
Black:
No, I did not work on that. I said that was prior to the three-channel system. We had a four-channel system somebody else designed. That was a complete failure because these two high-pass and low-pass filters, if this is the frequency band that is for the low-pass filter then you have a guard space as it was called, and then there is a high-pass filter. Therefore the low one would go here and the high would go here. Then of course as frequencies got very high and very low they would become virtually [unintelligible word]. If the gain at the crossover points is more than this attenuation, the thing will sing [i.e., break into oscillations], and if the gain is a little less than that but doesn't have a good singing margin such as 15 dB or higher, the result is serious distortion – which is what that four-channel system for President Wilson had.
Wolff:
This had been worked on at the lab when you started to work there. Is that right?
Black:
That's right.
#### Memoranda files
Black:
Wolff:
With what would you get caught?
Black:
Smoking a pipe.
Wolff:
On Sundays when you were reading?
Black:
When I was reading. I was reading these memoranda for file in my office. The first week it was reported to my immediate supervisor, who oversaw about six people. The second week it was reported to Slaughter, who couldn't have cared less. The third week it was reported to B. W. Kendall. The fourth week it went up to Amos Dixon. Dixon said to me, "Harold, I understand that you are one of the most promising engineers in the systems department. How is it that you cannot manage to smoke without getting caught?" I said, "Amos, I'll tell you a very simple solution. Why don't you let me smoke in Slaughter's office? That's a private office where smoking is permitted." He said, "That is a good idea, and that is what we will do." That solved that problem. At any rate, in reading these memoranda for file I became acquainted with all the supervisors and department heads and Fondella's – that's the apparatus – and other departments that I was called upon to work with as a systems engineer.
#### Type C three-channel system and amplifiers
Wolff:
That's very interesting. Let's come back to the fall of 1921. I want to get up to that Thanksgiving Day when you actually started to look into the reasons for the imperfect performance. After your training program, which lasted July and August, you started in the systems group in September.
Black:
That's right.
Wolff:
One of the first things with which you became acquainted was this four-channel system.
Black:
Three-channel. I learned from my reading about the four-channel system that was not a success. I also knew what a failure the four-channel system, which was called a Type A system, had been. The next three-channel system was a Type B, and that was a complete failure as well. The next system after that was being worked upon.
Wolff:
When you started there in the fall was the main project this Type C system?
Black:
Yes. That is where I learned that everything in the system – and particularly the amplifiers – was causing trouble. That's the first reference that I made in the manuscript.
Wolff:
When you started there in the systems department in the fall of 1921, was this Type C system the major problem in the department and was it the first major assignment in you got involved?
Black:
The first major assignment that I had was to get involved with the Type C. Prior to that I had to fix one of the amplifiers in a Type B three-channel system in the laboratory. That took me a week. I wasn't very quick at finding trouble. They never let those go into commercial service because they were so poor. That was not a major assignment, and I was just a beginner at that point. By that time they had gotten hold of C.W. Green, and it was his job to make the Type C system work. It took him about four years to do it. However the amplifier is what caught my eye. For the three channels in the short distance that they was going – and it was going on open-wire line – it would get by and do its job. I envisioned something entirely different. I thought of amplifiers that would carry one to three thousand channels – such things as I described in the manuscript.
Wolff:
I don't want to get too far ahead here. Your first major assignment was to get involved with this Type C system.
Black:
Yes.
Wolff:
And particularly the amplifiers, which were causing trouble.
Black:
There was little that could be done about the amplifiers that were causing trouble. I was able to point out that the third harmonics generated by the input and output transformers and the iron core coils in the filters were making the thing unworkable.
Wolff:
I thought it was the second harmonics that were the problem.
Black:
Turning to the amplifier I found the importance of the second harmonics. Theoretically they would cancel out, but practically they did what I said: [unintelligible word].
Wolff:
The third harmonics had to do with what?
Black:
They are the odd order. Any nonlinear device.
Wolff:
Did the amplifier generate them?
Black:
They were generated by [tape turned over at this point; sentence not completed in recording].... [H. J.] van der Bijl's book on vacuum-tube amplifiers showed me the curve characteristic and all that was wrong.
Wolff:
It's a little puzzling. You started the article by saying you remember investigating the reasons why the amplifiers did not perform well on Thanksgiving Day.
Black:
That's right.
Wolff:
What was it about that day that you remember doing?
Black:
I did that at home. I had to do what I was supposed to do on working days.
Wolff:
What were you supposed to be doing the day before that?
Black:
I was contacting the transformer people so that they would make more linear amplifiers and also do a better shielding job for the Type C system. That was a specific assignment.
Wolff:
Black:
That was my real job.
### Breakthrough on amplifier performance
#### Thanksgiving Day, 1921
Wolff:
Then what happened on Thanksgiving Day?
Black:
I let my imagination take over. I felt that I could foresee that we were going to need many, many channels going through an amplifier. I prepared a chart, based on push-pull and this poor performance, of how that requirement would go up as a function of the number of channels. They would add in random. I assumed that the third harmonics would behave the same way. That caused me to describe that to Ralph [Vinton Lyon] Hartley. About a day later he wrote me a little private note. It was not a memorandum for file. He showed me that push-pull amplifier or not, with any kind of an amplifier the big problem was the third harmonics. Therefore the amplifier had two jobs. With this new information, I had to change my chart to show how the requirements on a single amplifier would vary as a function of the number of channels. There I let the number of channels vary from one to three thousand.
Wolff:
I am not that familiar with this material and a little confused. I'm still back on Thanksgiving Day. What was it specifically that you started to do on Thanksgiving Day?
Black:
I was investigating the reasons for the imperfect performance of the amplifiers used in the Type C system.
Wolff:
By investigating do you mean you were at home reading?
Black:
No. I was calculating what this curve would look like.
Wolff:
What curve is that?
Black:
In this case it was a three-channel push-pull amplifier. I was worrying about what requirements it would have to meet.
Wolff:
You were doing this on your on time at home on Thanksgiving Day.
Black:
Sure.
Wolff:
What did you discover?
Black:
I discovered that the requirements were very, very severe and went up as the square root of the number of channels.
Wolff:
Black:
If you had four channels instead of two the thing would go up 3 dB – that is, the square root of 2 rather than by going to 4. The curve would be built up that way. It would vary [according to the number of channels]. The reason is that in a push-pull amplifier the two sides are never the same – particularly with the poor vacuum tubes available at that time, which were designed by Dr. [Mervin J.] Kelly. That was how they would go, and it was not anywhere need good enough to meet this requirement of a few percent that I mentioned. However I kept the curve going.
Wolff:
What requirement of a few percent did you mention?
Black:
I said somewhere in this paper that the second order instead of disappearing were only reduced by 10% to 18%.
Wolff:
That is a little confusing because a word is missing.
Black:
They were reduced.
Wolff:
Is the distortion reduced by 10% or to 10%?
Black:
It is reduced to 10%. That would be 1/10th. I guess the word "to" should be in there.
Wolff:
Yes. I have marked it.
Black:
That escaped my notice. Knowing that the law of addition for that kind of an amplifier was the square root of the number enabled me to draw my first incorrect curve. Step number was to consult with Hartley. Employees at the laboratory at that time were permitted walk all over the laboratory and talk to anyone they liked. Ralph Hartley was in the research department. He was somebody that I liked very much. I had read his memorandum, admired the work he did, and this was in his field. He had done some work on third harmonic modulators. I explained to him what I was doing, and he turned right around and told me what would happen. The reason that I substituted a carepegs for the little "n" was that I thought that having two n's in this thing was going to be a matter of confusion. Hartley told me in a single-sided amplifier the value of the nonlinear distortion in any such amplifier – regardless of the kind, good or bad – it would be twice as much. If there were four of them, it would be four; if there were three thousand, then three thousand times as much. That is more than 60 dB. Now that is getting into some sledding that is really tough. This is a technical point, but unfortunately third orbit distortion appears as understandable speech in the adjacent channel. The AT&T requirement was 70 decibel. That is 3,000 to 1 on a power basis or more than a million. That was a terrific requirement. Therefore I had to redraw my curves.
Wolff:
You paper explains that this amplifier reduced the second harmonics to 10% to 18%.
Black:
That's right.
Wolff:
You thought that the third harmonics would also be reduced to 10%.
Black:
A similar figure.
Wolff:
Okay. Then Hartley said that was wrong.
Black:
Very wrong.
Wolff:
He said that it builds up as the number of amplifiers. Is that right?
Black:
Yes. It sure does.
Wolff:
I don't understand this sentence where you say "independent."
Black:
That is independent of the type of amplifier. If there are a lot of amplifiers in parallel and big power is obtained that way, no matter how you get it, [the same rule applies]. In multiplied systems we only use a single tube. Therefore no matter how the amplifier is made and no matter what kind of an amplifier the third harmonic, the third harmonic is more serious than the fifth or seventh harmonic. When the term "harmonics" is used, it really means third order products too. There is a lot of literature on that. That goes directly. If there are a thousand amplifiers where a 60-dB improvement is needed, and there is 70 dB that is already sought, it could be seen right away that this was a tough problem.
Wolff:
Would you paraphrase this sentence you wrote to make it clearer? No matter what kind of amplifier, what is it that happens to the third harmonics?
Black:
The third harmonics. The things you didn't want would double in size when you came out on number two.
Wolff:
The harmonics would be double as big?
Black:
The harmonics. The things you didn't want.
Wolff:
After a thousand it would be a thousand times.
Black:
It would be a thousand as big, and a thousand is 60 dB.
Wolff:
All right.
Black:
Three thousand is about 70 dB. That was what hurt. That really hurt. It was bad enough to have a lot of channels going through one amplifier, but then to pile on top of that the fact that the thing was going to go up with the number, boy, that gave you something.
Wolff:
When you say the thing is going up as a number you mean the distortion, right?
Black:
Yes. I didn't think this paper would want to get that technical, but in the adjacent channels that distortion appears as clean speech. You see, you also get cross-talk between wires and things. AT&T's view was that a conversation should be private. If one picks up the telephone someone's talking that is not a part of that line or circuit should not be heard.
Wolff:
Was Hartley a Rhodes Scholar at that time?
Black:
Yes. He had come back from Cambridge. He was much older than I.
Wolff:
Was he in the systems department?
Black:
No, he was in the research department, but all I had to do was walk to his office.
Wolff:
Now we are down to the last paragraph. I think I understand this now. What were you determined to accomplish? After Hartley told you this, what did you say to yourself?
Black:
I was determined to solve this problem if it was the last thing I ever did.
Wolff:
What was the problem? To find a workable free channel?
Black:
Just what that sentence said in the paper.
Wolff:
Having "many amplifiers in tandem," that means in series, right?
Black:
Yes.
Wolff:
Black:
No, it had nothing to do with the Type C system. This invention was no accident, but I did not get any encouragement.
Wolff:
You started off investigating why the amplifiers didn't work well.
Black:
Yes. That's true.
Wolff:
Is it correct to say that your discovery of why it didn't work well was because of the second harmonics?
Black:
Yes. I thought the third harmonics were going to cause less trouble than the second.
Wolff:
All right. Your first discovery was that it was the second harmonics.
Black:
Yes.
Wolff:
You thought the third harmonics would cause less trouble and then you found they caused more. Hartley explained to you how serious it was.
Black:
Precisely. Yes.
#### Multichannel amplifiers and transcontinental telephony
Wolff:
After that you were started to think about something very different.
Black:
Yes. I had given up the idea of a push-pull amplifier. I had to find some way. There was nothing at that point in time to indicate that I could make this last paragraph in my paper come true.
Wolff:
It is kind of elementary, though I didn't realize it, that the basic point of the type C system was that it used a push-pull amplifier.
Black:
That's right. And that was good enough for what it had to do.
Wolff:
Your brainstorm, if you will, was to forget about push-pull amplifiers and to put aftertreatments in tandem.
Black:
In a long, long string.
Wolff:
All right. A string of amplifiers. What confused me was I thought that in the type C these were repeaters.
Black:
They were. It did have repeaters, but they were many miles apart because they went on something like 16-gauge. They were hundreds of miles apart and they never expected it to go any further than a thousand miles. Therefore there were not very many repeaters on a string on that system.
Wolff:
That was not what you thinking about as repeaters.
Black:
No. My idea was to use many, many repeaters.
Wolff:
In tandem.
Black:
Yes, in tandem. That is what I meant by the word "string." That was precisely why I switched to a clearer explanation of what I really meant. You may be able to improve on that explanation.
Wolff:
Okay. At this point you were starting to think about amplifiers in tandem and each amplifier handling many channels.
Black:
That's right – even up to three thousand.
Wolff:
The first thing you did on November 30th was to plot the distortion and linearity requirements.
Black:
That's right. Thanks to Ralph L. V. Harley I was able to do that. He could have prepared that chart too, but I did it.
Wolff:
I think this is clearer here. [Jake] Jammer said, "A beautiful piece of work, but why bother about so many channels?"
Black:
Three was the biggest he had ever heard anybody bother about at that stage. Four channels had been forgotten.
Wolff:
Black:
It didn't surprise me, because Jake was a happy-go-lucky person. He wondered why I concerned myself with three thousand channels when no one else was bothering about any more than three.
Wolff:
You asked to work on this on December 27th?
Black:
Yes. I had to get permission to do a thing like this.
Wolff:
Black:
They couldn't understand why I would be interested in anything of this sort. No one could.
Wolff:
Why were you interested?
Black:
I knew that there would be a demand for many channels. A single coaxial today transmits 1,028 and a waveguide transmits almost half a million. By making a simple terminal change that half a million can be expanded to almost a million. Optical fiber has thousands of times more transmission.
Wolff:
You didn't foresee optical fibers at that time.
Black:
No, I sure did not. I recommended the inventor of fiber optics for a research corporation prize. I got one of those awards myself.
Wolff:
Do you remember why or how you knew there would be a demand for many channels?
Black:
The shortest distance across this country was 4,000 miles, and of course in telephony there has to be regulars and spares. To use a big network efficiently a straight line is not always the shortest distance. Dr. [Frank B.] Jewett came long a year later and envisioned 4,000-mile system. That was the first job that made use of my amplifiers.
Wolff:
Why did you think there would be so many channels?
Black:
It's a big country. Certainly the wires were not going to be put overhead because grounds and everything interfere. The coaxial incidentally had already been invented.
Wolff:
Maybe I'm missing the point. When you say "many channels" is that synonymous with a coaxial cable?
Black:
No. On July 1st of 1975, 66% of our long distance communications was via microwave radio relay systems. Incidentally, I was the first to introduce that.
Wolff:
We'll get to that, but I want to press you about why you thought there would be a demand for so many channels.
Black:
It's a big country and I couldn't conceive of using open wire. I could not conceive of not wanting to hop across the country.
Wolff:
Let me put the question this way. Does open wire mean single channel?
Black:
I used the original Blackwell as a textbook in one of my courses. The disadvantages of open wire would be so many that it would never be used to hop the country. We did hop the country with open wire. That was where we got our first transcontinental thing before we had repeaters, but there you just got one channel. You didn't get a lot of channels.
Wolff:
That's what I don't understand. When I asked you how you appreciated that there would be a need for many channels and you said because open wires could not be used.
Black:
I knew that you could not have a telephone system and not be able to go from coast to coast.
Wolff:
Does that mean that an open-wire system is synonymous with single channel?
Black:
You would never get three channels. The only number that was ever gotten – and it was not very good – was one channel.
Wolff:
That's what I'm trying to see.
Black:
You see the little picture there?
Wolff:
That's a single-channel system?
Black:
Yes.
Wolff:
Are what you have on telephone poles single-channel systems?
Black:
That all depends on the pole that you look at, because this is not 1921. As you look on these things [in this photo] here, that is a group of a large number of paper-insulated wires. They are twisted together to prevent cross-talk between wires. That is actually a cable. It used to be lead, and now it is laminated with plastic in such a way to get the necessary shielding. That is the open-wire cable of today. It is used to hop only very short distances, such as to bring the telephone into this house.
Wolff:
Was the idea of putting in more than one channel very difficult on open wires back in 1921?
Black:
It was not practical. We had the one channel to Denver. That was before Dr. Harold DeForest Arnold invented the hard-vacuum tube. If you wanted to talk over open wire you could as far Denver. If you yelled to the top of your voice from Boston, the message heard in Denver would be a faint whisper. With the advent of Dr. Arnold – and there was some litigation there with Dr. [Irving] Langmuir at GE, but we were the first – with the hard-vacuum tube we got a single voice channel from coast to coast. I have forgotten the spacing, but that was a landmark. I would be able to find the answer to your question somewhere in the blue four-volume set of books from Bell Systems I have on the shelf there, but I can tell you in my own words. That was considered a great accomplish in 1915 or whatever the exact date. They got a two-way telephone conversation from coast to coast. It was ocean to ocean. It wasn't good. The bandwidth was about 2,000 cycles as opposed to 3,000 cycles today.
Wolff:
Let me see if I understand the point. It was not practical to have more than one channel in an open-wire system. Is that right?
Black:
That's right. Not the way things were done at the time.
Wolff:
Right. You said to yourself there is a need for many channels.
Black:
Many, many channels.
Wolff:
And it's synonymous with the fact that it could not be an open-wire system.
Black:
That is for sure.
Wolff:
All right.
Black:
There are many other alternatives, but it sure was not going to be open wire.
Wolff:
Going to the second paragraph on the page, you asked to work on linearizing, stabilizing and improving amplifiers.
Black:
I asked permission to do that.
Wolff:
This was so that you could have a large number of multichannel amplifiers.
Black:
And I did not get much encouragement.
Wolff:
Was this because people did not see the need for many channels?
Black:
That's right.
Wolff:
You wrote, "I studied the available material listed in topics not treated."
Black:
Yes. That's in a book that I wrote called Modulation Theory.
Wolff:
I looked that up, and what you say here is that you studied everything you could on the unwanted generation of products of modulation through the nonlinear response and the design theory of modulation and related nonlinearity. Right?
Black:
Yes. That was a natural thing to do. Having read all of these memos before, I just went back and read a particular group. There are memoranda for file, and this is true even today. Most of them are papers that are candidates for publication but for patent or other reasons they were not for the public, whether they are classified or not.
Wolff:
I don't understand is why the people there didn't see right away that there would be a need for many channels.
Black:
I cannot read their minds of course. You will see from this manuscript that I got 50 dB of feedback over a certain frequency range. That was at the end of the year after I got it all done.
Wolff:
That was six years later.
Black:
Six years later than '21, yes. That's right. What page number is that?
Wolff:
I'm on page 2 in the second paragraph where they said you could do it if it didn't interfere with your other work.
Black:
Yes.
Wolff:
You wanted to work on a multichannel amplifier.
Black:
Yes. An imaginary one.
Wolff:
And you say they did not appreciate the importance of many channels.
Black:
Well, I wouldn't say that. They said, "Go right ahead, Harold, but don't let it interfere with your other work." I was paid a salary and had to do some work.
Wolff:
Jammer did not think it was very important to do. Right?
Black:
He sure didn't. Less than eight months later he went off to Australia. He gathered with him every memorandum for file that had ever been written that he could lay his hands on, and after he did the Australian job he became vice president of IT&T and never came back to the Bell System.
Wolff:
Black:
That was a first and natural step.
Wolff:
There is something here I don't understand. You asked to be assigned this task and they said, "Okay." Who was it that told you it was all right as long as it did not interfere with your other work? Was that Jammer?
Black:
That happened to be Jake Jammer. Yes. The reason for my wording is that by that time Slaughter had left and Charlie (C. W.) Green had come down from MIT. He didn't think much of Worcester. He thought MIT was better. However he was ill. He had a critical illness and had to stay home, though it didn't affect his salary, for about four months. Lindridge was my supervisor, but Jake Jammer was taking Green's place so I thought I should ask him. I showed the charts to Jammer too. Anyway, that was J. S. (Jake) Jammer's response.
Wolff:
Black:
He was my boss's supervisor.
Wolff:
You asked Jammer to allow you to work on linearizing.
Black:
That's right. That was the exact question that I put to him.
Wolff:
Black:
### Western Electric assignments
#### Push-pull amplifier design
Wolff:
Do you remember what your main job was at the time?
Black:
Yes. They had some other push-pull amplifiers that had been designed for another system that had not been made obsolete because there were some in the field. They were also push-pull amplifiers, and there were fifty-seven kinds. In the interest of the Western Electric Company, who had to supply all of these things, I replaced those by a single design. That was a job on which I was working at that time.
Wolff:
Was that an assigned job?
Black:
Yes. That was assigned.
#### Amplifier interference
Black:
I also got another assigned job. There was an amplifier in use for some purpose I don't remember. While the tubes were warmed up by a 24-volt storage battery they were picking up some 60-cycle hum. They wanted to know how to get rid of that. I made a sort of Wheatstone bridge with an extremely high resistance to parallel the other and pick the two midpoints of that so that the 60-cycle did not interfere with the amplifier. I very rapidly reached the point where I didn't just walk around and fill pails with water. I was given small problems to solve.
### Collaboration
#### Western Electric colleagues
Wolff:
Meanwhile you started to reading about these other topics in the evenings?
Black:
That's right. Saturdays, Sundays and anytime that I felt like it.
Wolff:
You said there were many other researchers in 1922 who—
Black:
That was in the research department. Dr. Eugene Peterson was one. I could name people.
Wolff:
We don't need all that.
Black:
Within the Bell System, within Western Electric, I was not the only one.
Wolff:
I understand that.
Black:
They were aware of the need.
Wolff:
Black:
That's right.
Wolff:
Did Mervin Kelly in particular work closely with you?
Black:
Yes. We were great friends. Every vacuum tube that Western Electric manufactured was designed by Dr. Mervin Kelly and people working for him on Hudson Street, which is about a block from 180 Varick Street. It is only a quarter mile from 463 West Street, which is where I was at that time. During this period I went to Kelly and told him what I wanted. For quite a few months he worked with me until it became evident that it was hopeless.
#### Linear vacuum tube
Wolff:
What exactly was hopeless?
Black:
To make a linear vacuum tube was hopeless. To give you an example, if you plotted the voltage to the grid as a function of the amplification of the voltage out, if you plotted the output against the input in terms of current – and I square this pump, if you made yourself a drawing that was 1 mile square and drew it using the hardest pencil chiseled to a straight line and made that line straight, the nonlinearity would be too much to meet my requirement. Therefore there was no earthly way of getting it by a linear vacuum tube. There are a lot of tricks one can try.
Wolff:
What was your requirement again? You said you said you could not get it to your requirement.
Black:
My requirement was modest. It was 50 dB. Dr. Jewett scaled that down to 40 dB for the first commercial application.
Wolff:
When you say 50 dB are you talking about distortion?
Black:
Distortion. Nonlinearity.
Wolff:
It had to be less than 50 dB?
Black:
As good as 50 dB is what I asked Kelly.
Wolff:
Then 50 decibels is a low distortion?
Black:
That is a low distortion. I gave a figure. I think it's three thousand maybe.
Wolff:
You and Kelly were working together trying to get 50-dB distortion from the tube.
Black:
That's right.
Wolff:
And you could not do that.
Black:
No way.
Wolff:
Why was that?
Black:
It was just a fundamental limitation. It was inconceivable that one could make a vacuum tube that linear.
Wolff:
When you say fundamental, was it a theoretical limitation or was it because the techniques or materials were not good enough?
Black:
Remember this was in the early '20s. I have the date of the invention of the transistor. The techniques for making that [unintelligible phrase] used negative feedback. It would be no good if it didn't.
Wolff:
Black:
No, that was a practical limitation. There was no way it could be done.
Wolff:
Was it because the materials were not good enough?
Black:
We knew electrons and that they went opposite from the currents that we could conventionally draw using a triode or a pentode or any of the vacuum-tubes available at that time. No one in the world could get it, and they don't get it today either.
Wolff:
All right. I was wondering if it was because of imperfect materials or techniques.
Black:
No, it is just the fundamental property.
Wolff:
Is it theoretical?
Black:
Not theoretical. I mean no vacuum tube. I can pick up any book you want on vacuum tubes and no vacuum tube like that exists.
Wolff:
Okay. You and Kelly worked on the tubes but it just didn't work.
Black:
Yes. He was very patient, but it didn't and couldn't work. No way.
### Negative feedback amplifier
#### Invention and testing
Wolff:
Black:
That's entirely different, but it is curious.
Wolff:
What does it show? Is this 1927?
Black:
Let me see. When did I finish this job? I think it was in '27.
Wolff:
Are you talking about when you invented the negative feedback?
Black:
Yes.
Wolff:
That was 1927.
Black:
Oh yes, by the end of the year. The beginning of the next year, in January, Dr. Jewett came along. This paper does not cover this at all. Jewett wanted to take 16-gauge paper-insulated cables, which existed to carry voice, and use the frequency band 4 kHz to 40 kHz to travel a distance of 4,000 miles with repeaters spaced every 25 miles. With that requirement it could be done if the power level of the amplifiers that I finished up with were about 8 dB to 10 dB. They were ten times as powerful on a power basis as that amplifier. To achieve that result we used pentodes in the first two stages. I had used pentodes throughout at a lower level in the first two stages. Then, because the Bell System would not hear of having a B battery – if that means anything to you, a plate voltage – higher than 130 volts.
Wolff:
Yes.
Black:
My first proposal was to use a filamentary or an indirectly heated triode. I have forgotten which one. However it was to have two grids, one positive and one negative, so as to get more swing without increasing the voltage. [J. O.] McNally, who worked for Dr. Kelly, constructed some models. I tried them out and they were not linear enough. Therefore we had to go to 260 volts.
Wolff:
You built seventy amplifiers.
Black:
Yes. This one shows their stability.
Wolff:
Okay. We are on page 15 of the Bell Systems— This is the 1934 paper, right?
Black:
I was not permitted to publish this early.
Wolff:
This photograph was taken in 1927 and it's one of the amplifiers you built for Jewett?
Black:
No. The year of 1928 was consumed me designing this amplifier in my own laboratory and getting models made. Then in 1929 the Western Electric Company made seventy of them. They made few more too. The trial was all over in 1930.
Wolff:
Jewett asked you to do this, right?
Black:
Yes. That was for Dr. Jewett.
Wolff:
This is a stupid question, but is this the negative feedback amplifier?
Black:
There it is. It isn't very small either.
Wolff:
This is your invention that you are showing then.
Black:
This is the first application of my invention.
Wolff:
Oh, well we want to use that photograph in the article.
Black:
That's a field trial. That was not made for the public. As a matter of fact it was used publicly for another purpose.
Wolff:
It was the first application of your negative feedback amplifier.
Black:
Yes. The custom was always to make a large-scale trial before they were turned over to the public for use.
Wolff:
Where was the photograph taken?
Black:
That was taken at Morristown, New Jersey in my laboratory. This picture would have been taken in about 1929.
#### Biconjugate circuits and distortion
Wolff:
On page 3 you said that by this time when you came back to New Jersey at 2:00 a.m. you had come to the conclusion that anything that was not part of the output was what? Do you want to say that again?
Black:
I regarded as distortion anything in the output that was not a replica of the input, irregardless of whether it was due to nonlinearity, variation in vacuum-tube gains or anything else. That's a broad interpretation that I placed upon the thing. Contrary to a statement in a book by Dr. Terman, that can only be done by using two biconjugate circuits. I think I will have to add some words here. A Wheatstone bridge is a biconjugate circuit. That means that if the arms are balanced and you apply a battery and get no current in the galvanometer circuit. That is a biconjugate circuit. Now I have a large book by Dr. [George] Campbell. He's the first professional mathematician ever to be hired by the Bell System. In this book he listed all of the biconjugate circuits used most widely in telephony at the time of which I speak. They get them in other ways with integrated circuits today. It was with a transformer. At the time, the Bell System could make better transformers than anybody else anywhere in the world. Campbell listed the maximum number of biconjugate circuits of the transformer type that didn't have unnecessary elements. The simplest one of all was the one that I used. I used them always to separate the µ and circuits.
Wolff:
Do I understand correctly that you got the idea to use them from his listing? Or did you know this before?
Black:
I knew it before I ever saw this book. The first thing dealing with voice frequency circuits, for instance an ordinary house telephone uses just a single pair of wire. It has a biconjugate circuit to separate the transmitter from the receiver. It's slightly unbalanced, so you'll hear a little head-through, but the moment you get your central office and begin to go somewhere else the transformers – or hybrid coils as they are known to communication engineers – are used to change from two-wire to four-wire. They have the property of doing that. Then as you go longer distances, unless you are mixing up and getting into multiplexing and many channels, that's where the hybrid comes from. I learned about that by the time I had been around for a couple of weeks outside the training course. At that time any engineer worth his salt would try to do something with the hybrid coil. It was impossible. I remember trying one of those things, and Dr. Harry Nyquist explained to me the error of my ways.
Wolff:
We can get into all that next time.
Black:
Yes.
### Harmonics, plotting numbers of amplifiers and channels
Wolff:
We are at our second meeting now with Dr. Black on the June 29th, 1977. The first thing I want to clear up is something that it is one of the turning points of the article. On November 24th, 1921 you had just started to think about improving the amplifiers and you had gone home and were working there on your own time. We talked about what you were doing there, and I got mixed up because we kept jumping back and forth between before and after you talked to Hartley, so that sequence is not clear. Before you talked to Hartley you went home and you were trying to draw a curve. You were coming up with the wrong thing because you were assuming that third order harmonics would act the same as second order harmonics. Right?
Black:
Yes.
Wolff:
While you had this erroneous idea on the 24th and before you talked to Hartley and he disabused you of this, you said, "I prepared a chart based on push-pull and its poor performance of how that requirement would go up as a function of the number of channels."
Black:
And number of amplifiers in tandem. See, there were two things I was doing there.
Wolff:
Okay.
Black:
This Hartley thing had to do with the number of amplifiers in tandem.
Wolff:
Before you talked to Hartley, on Thanksgiving Day, 1921, and with this erroneous understanding in your mind what were you trying to draw? What kind of curve were you trying to draw?
Black:
There were two curves. The first was how the requirements on an amplifier changed as the number of channels varied from one to three thousand. That is a matter of how many voice frequency channels I was considering.
Wolff:
What was the second curve on that same day?
Black:
The second had to do with what additional requirements – linearity really – had to be placed on the same amplifier. Likewise to the function of the number of channels, but primarily with respect to how many amplifiers were to be connected in tandem. Obviously the more amplifiers that come one right after another the more severe the requirement. It was at that latter point that I made my error.
Wolff:
Is it correct that your error was that you assumed that the third order harmonics and other unwanted products would also be reduced to between 10% and 18% just like second order?
Black:
Yes, whatever the figures were on that.
Wolff:
Yes. Just like second order.
Black:
Just like second order.
Wolff:
All right. That was your error.
Black:
To state it differently, it would sort of add up on an RMS basis, but because no push-pull amplifier is perfect even the RMS spaces instead of being random addition was the figure I mentioned. Well I guess that's a technical detail we'll leave out.
Wolff:
I don't understand what you mean there. What do you mean when you say it would add up on a random basis?
Black:
If there were a hundred amplifiers in a row, instead of a three linearity requirement increasing a hundred-fold as it actually does it would only increase by the square root of hundred, or ten.
Wolff:
Wait a minute. N is the number of amplifiers.
Black:
Yes. N is the number of amplifiers. First you have to figure out how good an amplifier has to be, depending on how many channels. Number two, you have to figure out what is the additional requirement. Here is the one that hurts: when you put a big number in series one right after another.
Wolff:
Are you saying that the distortion would only increase according to the square root of the number of amplifiers if what you believed were true?
Black:
Statistics might give a little different answer, but in a rough way yes.
Wolff:
If what you believed was true, namely that the third harmonic was 10% to 18%. Okay. I think I am beginning to see where this got confused. You are talking about doing two different things.
Black:
That's right.
Wolff:
One is the number of channels and the other is the number of amplifiers.
Black:
That's right.
### Effects of multiple channels and amplifiers on distortion
Wolff:
That got all mixed up last time we talked. All right. Coming back to your first curve, you were trying to find out how the requirement on an amplifier varies. What do you mean by the requirement? If n is one axis, what is on the other axis – n now being the number of channels?
Black:
The other is how linear it needed to be. To put it another way, the nonlinearity of a vacuum tube of the kind described by van der Bijl in his book on the vacuum tubes available at that time.
Wolff:
You are talking about a push-pull type C system amplifier.
Black:
That was the first one I ever saw.
Wolff:
Yes. This is your push-pull amplifier. You were concerned with how its linearity varied with the number channels? Is that it?
Black:
Yes. It was only carrying three channels and was doing a poor job, so it would be rather obvious if it carried three thousand channels it would have to perform a lot better.
Wolff:
Is the curve you drew regarding linearity versus number of channels?
Black:
That way I put it was decibels of improvement.
Wolff:
Is it correct to say that what you were trying to figure out was how the linearity varied with the number of channels?
Black:
Yes.
Wolff:
Okay. Was the second thing how the linearity would be affected by putting more than one amplifier in series?
Black:
That's right. If you had two in series, regardless of the number of channels it would have to be 6 dB four times as linear.
Wolff:
Are you using the terms linearity and distortion synonymously?
Black:
Yes.
Wolff:
Okay. Then you believed that the distortion in a string of amplifiers would increase as the square root of n.
Black:
Right. If you put the word approximately in front of the square root that would be a little bit more accurate.
Wolff:
This is what you told Hartley you had done?
Black:
That's right. I went down to his office and showed him what I had done.
Wolff:
And Hartley corrected you by saying no?
Black:
He didn't say no right away. He came to my office the next day and handed me a handwritten note that said exactly what I wrote in the paper. It's important to note that he was not talking about push-pull; it was regardless of the kind of amplifier.
Wolff:
To paraphrase what he said, is it correct to say that he told you that the distortion in a string of amplifiers would increase in direct proportion to the number of amplifiers?
Black:
That is correct. The most important thing of all is where he said "independent of the type of amplifier." From that day on, push-pull amplifiers were out. They take twice as many tubes and don't do any good. The third harmonics and all third-order distortion contributed by a string of X amplifiers is virtually X times that contributed by one.
Wolff:
I want to expand on this. Is this the same as saying that the distortion in the whole string will increase in direct proportion to the number of amplifiers?
Black:
That's right.
Wolff:
Will a thousand amplifiers produce a thousand times as much distortion as one?
Black:
Yes. And I might say voltage distortion. There would be a million if we were talking about power distortion. In other words the linearity would have to be improved by the example you gave by 60 decibels. That would mean that if you took about 10 acres and plotted input against the output, the line could not vary a thousandth of an inch. That is a tough requirement. I'd like to add another remark in passing. L5 is single sideband and is the latest and more efficient coaxial [cable]. A single tube, which carries I think 1,028 channels, has two negative feedback amplifiers every mile so that in 5,000 miles there would be 10,000 amplifiers in tandem.
Wolff:
Are you talking about a system that exists now?
Black:
The most efficient coaxial system of today.
Wolff:
I don't know how to reconcile what you just told me about distortion with what you did on November 30th. Your attachment says the business that you "then found for a string of X amplifiers, the distortion contributed by each amplifier must be divided by X."
Black:
That's right.
Wolff:
Is that saying the same thing?
Black:
Yes.
Wolff:
The thousand was when they were all added together. Right?
Black:
That's right.
Wolff:
That's the net distortion.
Black:
That's the net distortion. That is voltage distortion too. If you were thinking of power it would be a million. That statement is precise as it stands.
Wolff:
That's voltage distortion?
Black:
It is voltage or current, whichever way you want to view it.
Wolff:
That's the distortion of each amplifier. It must be divided by X, so the total is going to be X. Right?
Black:
That's right. If you are carrying a thousand channels the thing that had to be divided by X had to be awfully good to even handle the thousand. You get hit twice. The hard one is the number in series because it didn't take long before we had long strings.
Wolff:
Before Hartley corrected you, you believed that if you had a thousand amplifiers in series then the total or net distortion across the whole thousand would only be the square root of that. Right?
Black:
Yes.
### Professional advancement in the Bell System
Wolff:
All right. I think I got that straight. Good. That's important to understand. Now let's clear up a couple of other points. You told me a story about how you nearly quit. You were passed over for a raise and you were angry and you were thinking of going to Harvard Business School but you didn't. Did that happen before that Thanksgiving?
Black:
That was very early, well before Thanksgiving. I reported for work on the 5th of July. My salary had been fixed ahead of time. Then along in September of 1921, out of 150 people everybody got a raise but me.
Wolff:
Black:
I had just started. I started with 150 other people.
Wolff:
And they all started they got a raise?
Black:
The day they were hired all of them that I knew – and I knew virtually everyone – got \$27. I got \$32. That was no secret. Then in September they got a \$3 raise and I didn't get a raise. It was as simple as that.
Wolff:
That's when you got angry.
Black:
I wouldn't say that made me angry. I just didn't exactly like it.
Wolff:
You were thinking of quitting, so it made you angry. Why shouldn't it?
Black:
I'll tell you another story that has nothing to do with this paper. Dr. John R. Carson is a very famous mathematician. In 1925 he heard that he was going to be transferred from AT&T to Bell Telephone Laboratory. Dr. Jewett did that when he was president and drew up a chart for the laboratories. Then Dr. Carson went home and decided that he was not going to work for the Bell System if he couldn't work at 195 Broadway at least on the organizational chart. However the next morning he changed his mind. He too decided that he was going to stick with the Bell System. I'm not the only one to do something like that. If I had been that stupid, it would have affected my life. I still hold that luck plays a part in many things, and I would never have been a businessman. Absolutely no way.
### Singing margins and amplifier stability
Wolff:
You were talking about the older four-channel systems that you saw when you first came to work there. You used the term "singing margin." Is that the same as the phenomenon of singing? They talked about amplifiers singing in those days.
Black:
Yes, that is correct. The terms "singing margin" and "phase margin" were used by Dr. [Hendrik W.] Bode many decades later in his book on amplifier design. Those were two terms that I had introduced not in connection with amplifiers but with filters and two-directional transmission. I guess here I was talking about the stability of a negative feedback amplifier.
Wolff:
This was before the negative feedback amplifier. You were talking about the first carrier system that had been designed for President Wilson.
Black:
That's right.
Wolff:
It was unsuccessful because there were some directional filters that didn't have enough singing margin between the two directions.
Black:
Use of the word unsuccessful in this case is a little bit relative. The system did not fail, but didn't work as well as it should. It met his requirements. It served the President. I'll put it that way.
Wolff:
I'm trying to understand this term "singing margin." It is synonymous with the term "singing"?
Black:
Not really.
Wolff:
Well, we don't need to get into it.
Black:
To get into it I would have to make you a diagram that is not concerned with amplifiers.
Wolff:
All right.
Black:
That one system was the only one of its kind that the Western Electric Company ever manufactured. That was called type A. The next one manufactured was type B. They were both open-wire systems. The type B didn't work for an entirely different reason. A few of those were installed. They were not removed from service, but if you listened on channel 2 you could also hear channel 1 or channel 3. That was not considered good. The reason that happened was because it was a carrier transmission system. It transmitted the carrier and one sideband, though not both. With the amplifiers that were available at the time that gave a non-workable system. Therefore very few were produced. The next step was where I came into the picture really. C. W. Green was my boss's boss. I did not report to Green. I reported to a supervisor who in turn reported to Charlie Green, but Charlie and I talked together. He was working on a three-channel amplifier for open-wire lines that would work. That's where I saw and observed the poor performance of push-pull amplifiers. I did not mention it in my text, but everything else was causing trouble.
### Lee de Forest and E. Howard Armstrong
Wolff:
Let me interrupt you.
Black:
Wolff:
One more background question about when you came to work in the fall of 1921 at AT&T.
Black:
I was working for the Western Electric Company in New York. I never worked for AT&T.
Wolff:
Okay. I should correct that. Did you hear any talk about the work of [Lee] de Forest and [E. Howard] Armstrong?
Black:
Yes. Let me put it this way. I have to think very carefully. Lee de Forest invented the soft vacuum tube. It's a delightful patent, which I have read. That came up in 1915. That was before I was a member of anything.
Wolff:
That was the decade before you got there.
Black:
Yes. When I got there, because I read every memorandum for file that had ever been written, I soon learned about all of these things.
Wolff:
Black:
I read about Dr. Lee de Forest's work. Somewhere between 1913 and 1915, Harold DeForest Arnold introduced a hard vacuum. He beat [Irving] Langmuir of General Electric (GE).
Wolff:
This is Arnold?
Black:
Yes.
Wolff:
Was he related to de Forest?
Black:
No. That is peculiar. If I took you out to the Bell Telephone Laboratories you would see the Arnold Auditorium. It was named in his honor because he was the first director of research. If you have good eyesight you will read his definition of research, which is inscribed in marble.
Wolff:
Black:
Yes. Incidentally, at that time at 463 West Street they had a small library measured by today's standards. Within that library was a book by [H.J.] van der Bijl on vacuum-tube amplifiers, which I took from the library and asked if I could keep it. They said yes, and I have it over there in my bookcase.
Wolff:
Black:
My first contact there was when I received the [John Price] Wetherell Medal from the Franklin Institute in 1941. That was a very formal affair. I had to wear afternoon clothes, then a black tie followed for a cocktail hour and then a white-tie affair for the evening. I was sort of ashamed that I was taking time off from work to receive that medal, so I didn't tell anyone. I was rather delighted in the evening to see Abe Clark (A. B. Clark), who occupied a position that today would correspond to vice president and Roy Chestnut, who was the supervisor on the same level as I but in a different phase of work. In other words, friends from the Bell Labs were there. All during the afternoon I talked to [E. Howard] Armstrong, and that was where I first became familiar not only with his superheterodyne but also his invention of FM. That has a very interesting story.
Wolff:
Didn't Armstrong invent regenerative feedback?
Black:
Yes. Superheterodyne regenerative feedback.
Wolff:
That's positive feedback.
Black:
Yes. That makes the modulation worse, dB for dB.
Wolff:
He invented or discovered this positive feedback, which he applied to his superheterodyne receiver.
Black:
And he was only receiving one channel.
Wolff:
I think he did that before you came to Western Electric.
Black:
That's right.
NOTE:
[Transcriptionist's Note: Here is where Tape 2, Side A ends. Side B of this same tape is a recording of others and not this interview. However, Tape 3, Side A appears to pick up where the above left off.]
Wolff:
When you started at Western in 1921 and you were reading all the memorandum for file, did you read any of Armstrong's first papers on his positive feedback?
Black:
Yes, I did. Those papers were available to me. If you will read my basic patent, I mentioned that in the patent as something to be avoided. Well, I would have to see what the patent said.
Wolff:
Here is the patent. Do you think you can find it?
Black:
I'm not certain if I had any reason to make any reference to Armstrong's suggestion – which found wide application until better methods were developed.
Wolff:
This is all part of setting the stage. You came to Western and you read about Armstrong's work.
Black:
Yes. I could say something that worked for Western Electric for about three years.
Wolff:
After you had worked for them?
Black:
After I had worked for the Western Electric Company in New York for about three years, I rented a room in Roselle Park from a woman. She rented out rooms to two people. We were both from Worcester Tech. Across in the next house someone lived, whose name escapes me at the moment, whose brother was president of the Western Electric Company at Kearny, New Jersey. He very much wanted a high-quality radio receiver. Ed Beamus, who was a radio ham, built one for him that I didn't think was very good, so I asked if he would like me to try to build one using a cone loudspeaker. I used Armstrong's circuit. The parts were only resistors and capacitors, and as long as I didn't steal any vacuum tubes I was permitted to get that working in the laboratory on Sundays. That was a personal experience that I had with Armstrong's [ideas], but it had nothing to do with my patent.
### Distortion reduction in a single amplifier
Wolff:
Let's take up now we left off last week with the manuscript. You said you were able to reduce unwanted distortion by more than 50 dB. Is that distortion in a single tube or in an amplifier?
Black:
I defined distortion there very broadly. I regarded the input as what was wanted, and then in the output, were that to be attenuated, until it would be of the same magnitude as the input. Anything that was there I would regard as distortion, whether it was change in gain due to a power supply or anything else. I regarded a perfect amplifier as one in which the input was an enlarged copy of the output. In those days the frequency bands were so narrow that the time of transmission across the amplifier was not a practical factor.
Wolff:
You explained that. You said that you could cancel out the distortion in the original output.
Black:
Yes.
Wolff:
Then you said unwanted distortion could be reduced by upwards of 40 dB.
Black:
That was the result of our experiment that next morning.
Wolff:
That was distortion in what? That was in a single amplifier, right?
Black:
A single amplifier, regardless of the number of stages or type – although my patent shows the two embodiments that I set up.
Wolff:
In a single amplifier you could reduce the distortion by upwards of 40 dB. It's at the bottom of page 3.
### The problem of developing a perfect amplifier
Wolff:
On the next page you said, "Over a period of four more years I struggled with the problem." What problem was that?
Black:
The problem of developing a perfect amplifier.
Wolff:
What do you mean by a perfect amplifier?
Black:
One that would be viable to manufacture and would overcome all of the distortion, fulfilling the severe requirements that I had set out to meet.
Wolff:
What I don't understand is, this embodiment worked so well. You demonstrated it could reduce the distortion by this much.
Black:
Yes, but it was impractical for two very important reasons. Every hour on the hour, 24 hours a day, someone had to come around and adjust the filament turn to its correct value. They were permitting plus or minus or half a dB or a dB variation in amplifier gain, whereas for this purpose the amplifier gain had to be absolutely perfect. On top of that, four times a day – every six hours – it became necessary to adjust the B battery voltage because the amplifier gain would be out of hand. Those are two practical reasons. There were other complications too, but that's enough to show you.
Wolff:
That's very clearly put.
Black:
For 40 dB, one-thousandth of a decibel change in gain would throw that thing in error by 6 making it poorer by 6 dB. That was just a laboratory demonstration.
Wolff:
That was a 6 dB error in what?
Black:
Instead of getting a 40-dB improvement it was a 34-dB improvement.
Wolff:
That's clearly put and important.
Black:
The amplifier of those days.
Wolff:
Black:
That's right. However for me it was wonderful, because it showed that theoretically it would be possible to get a solution.
Wolff:
A solution to what?
Black:
The problem of a superlatively perfect amplifier.
Wolff:
By that you mean that it had no distortion?
Black:
No distortion of any kind.
Wolff:
You wanted to put in a string of them.
Black:
Yes. I wanted a tremendous improvement.
### Negative feedback amplifier
#### Brainstorm on 1927 ferryboat ride
Wolff:
Now we come up to 1927 and the ferryboat ride. This is of course the most important part of your article. What I don't understand here and want to get you to expand upon is what you were thinking about that day. What was the problem? You said you conceived the analysis of the negative feedback amplifier. Can you remember what you were still trying to do?
Black:
What I had been trying to do for years was to come up with something good. The idea of negative feedback and the drawing of the conautical circuit and the equations I wrote came to me all of a sudden as I looked back over the things I had done and published prior to that. I had produced an open-wire system that added one additional talking circuit. There I had designed the filters, and from Heaviside work I achieved superstability in many things. There I actually had a µ. If you understood it carefully enough, I actually had a µ of six or eight. However I cannot connect that in any way with a negative feedback amplifier. I have tried to remember for fifty years, again and again, why the idea came to me. I couldn't tell you today any more than I could that morning. It just came to me.
Wolff:
What exactly was the idea that came to you? This was your sketch.
Black:
This is the sketch. The idea was, here comes an input and then there's an output. The reason I call this µ was due to having read the book by van der Bijl. However, µ here is a complex quantity like Steinmetz first introduced: A + AB. The reason I used the symbol was because in some book in my library was the transmission characteristic of a circuit. I guess it was really Maxwell's Equation. It had to do with the electrical properties of electromagnetic waves. If they were on a pair of wires or a coax […ial cable], the attenuation would vary as the square root of frequency. In other words, З was a term that was usually expressed E to the IΣЗ, to put it in complex form. I noticed here that if З was the attenuation of the line it would be an equalizer and would serve two purposes for one. Of course I didn't get any of that information on the sketch. It took a long to get that. You asked what was the reason for my choice of these two Greek letters.
Wolff:
No. I asked you what it was you actually thought. What was this flash that came to you?
Black:
It came to me that if I fed the output back to the input in reverse phase and kept the device from singing I would have exactly what I wanted and I could write this equation. The whole idea came to me.
Wolff:
What did you want?
Black:
You notice that the improvement is substantially equal to the absolute magnitude of µЗ.
Wolff:
The improvement in what?
Black:
In distortion of any kind – variation in power supply, you name it and this does it.
Wolff:
You said another important thing, if you could keep it from singing. What do you mean by singing?
Black:
Ordinarily when you connect an amplifier that has a large amount of gain back to the input, which has less loss, you are liable to get singing.
Wolff:
Do you mean self-oscillation?
Black:
Yes. There I do mean self-oscillation. I explain here why I knew and what I had to do in order to avoid that. That is all explained, and I would not know how to improve on that.
Wolff:
At the top of page 5 you put that in quotes. Why is that in quotes?
Black:
The quotation marks should certainly be deleted. The only thing would be to decide whether to bother to put that in italics or not.
Wolff:
Okay. You knew that you could prevent self-oscillation by having the loop transfer factor real positive greater than unity [real positive > unity].
Black:
That's right.
#### Professional communication on work in progress
Wolff:
A lot of people doubted that it would be stable. Is that correct?
Black:
That is correct.
Wolff:
I guess that it wasn't until [Harry] Nyquist and [Hendrik] Bode did their mathematical work that people were convinced. Is that right?
Black:
I would like to abolish that idea at once.
Wolff:
Okay. What is the story there?
Black:
Since Nyquist published first, I will point out that it wasn't until after 1930 that Nyquist even knew that I had developed a negative feedback amplifier. And the same for Bode. I thought the best way to keep others from taking credit away from me was to just go merrily ahead and do it and make believe there was no secret about it at all. I didn't tell anyone.
Wolff:
What did you not tell?
Black:
That I was building a negative feedback amplifier. The Western Electric Company had built seventy-five negative feedback amplifiers in the year 1929 before either Dr. Nyquist or Dr. Bode had even heard of it. They did not know a thing about what I did. By the way, I would like to tell you – since I have this right here and I can also show this to you in an Italian publication that is in three volumes. It lists 3600 people like Einstein – that is, they are not all alive – and I am among the list. The thing that it shows as being the most important figure of my patent – because it does have many figures – is Figure 2. The first that Nyquist ever heard about this thing was in 1930. I asked him why I got these symmetries beyond the one-zero point that I avoided. I think it's in one of my patents. I have got them all and I am going to look and find which one. I think it's my Case #18. I asked him to explain this.
Wolff:
Is that curve number 1 there?
Black:
Yes, or any of these curves, because everything looks to be so symmetrical. To get one of those, you have to have a looped-over amplifier. Of the 10,000 learned papers that I have read, I know of no one but me who used a looped-over amplifier. Bode never did.
Wolff:
I did not make my question clear. I understand that you built the amplifier before they did their work. The point was that a lot of people, as you just said, doubted your amplifier would be stable.
Black:
That is right.
Wolff:
Was it this mathematical work that Nyquist later did later on that convinced people?
Black:
No.
Wolff:
What was Nyquist's contribution?
Black:
His contribution was a paper. I read the memoranda that preceded that paper. That was an intuition and ability to think of something original that is given to very few men. However, his mathematical derivation of the thing is very poor because there are so many other ways of deriving it that are vastly simpler.
#### Patent; influence of Black's work
Wolff:
Let me ask the question in a different way. How does what Nyquist did relate to what you do?
Black:
In no way.
Wolff:
You mentioned him in your paper. You said, "It was tested before Nyquist developed the stability criterion which bears his name." Didn't that stability criteria have something to do with negative feedback amplifiers?
Black:
It did. It had to do with those that were like the one that I built. As a matter of generality, he introduced a second kind. Yes. He produced a mathematical derivation of what I did. What I would say about that is that it was one that I didn't have any idea existed. It was new to me. I made good use of it ten years later. He derived it, it bears his name, it's known the world over, and that part is right. However I believe that if you look at one of my patents you will find that the part of it that has been used by everyone was something I had already put in a patent. That too is regarded as a publication.
Wolff:
I see. I was just trying to relate it. Who was Blessing? Was he the patent attorney?
Black:
No. Let me explain. He was a supervisor of a group of people. Earl C. Blessing reported to me.
#### Development challenges
Wolff:
Before that famous ferryboat ride, while struggling with that problem with the feedforward amplifier, you said you used circuits and approaches that were far too complex to be practical. Could you give me an example?
Black:
Not offhand. I could probably sit down and draw a few, but they might not be the ones that I had tried.
Wolff:
That's all right.
Black:
[It was] completely unsuccessful. You can believe that.
Wolff:
Meyer started gathering parts for the model.
Black:
That's right.
Wolff:
You said a fundamental requirement of a repeater amplifier is that its input and output impedance must match accurately. Was this a very difficult thing to do?
Black:
Yes. It had been thought so. It's not an impossible to do. Don't misunderstand me. Before the advent of negative feedback and some of my other inventions it was accomplished at the expense of loss of signal at the input and loss of power in the output. It is a necessary thing to do. I might note in passing that Dr. K. C. Black in the research department was at the same time, around 1936, working on coaxial cables.
Wolff:
Wait a minute. We're back in 1927.
Black:
Yes, so it was not exactly the same time. When he did that I asked him to turn the job over to me, which he refused to do. He was not matching impedances. I told him that if he didn't do that the system would not work. He built a system from Philadelphia in '36. Don Qualls, who was high up in the company, found it and fired him. That's what happened to him.
#### 1927 sketch
Wolff:
That's another story. This was a difficult thing to do, but then on the ferryboat ride you sketched the details of how to do this.
Black:
That's the figure you have there.
Wolff:
That's the figure you have given me. You say it employed a simple novel biconjugate network.
Black:
That's right.
Wolff:
Is that what you've sketched?
Black:
You have to use a little imagination here. I might find a clearer example. This isn't too bad. If you would regard this E as the voltage acting in series with the plate resistance of a vacuum tube and then envision these three things here which carries the—
Wolff:
What three things?
Black:
These are general impedances.
Wolff:
That's rr0, kr and r, right?
Black:
That's right. Those are the things that are the Wheatstone bridge with respect to the galvanometer, which is here and goes back to the input. Z is the load impedance. Then the feedback amplifier is done. It gets a little reflected signal here, and negative feedback eliminates it. The impedances are matched, and the only loss is the loss in this Wheatstone bridge which can be made about a tenth of a decibel.
Wolff:
Where is the amplifier again?
Black:
Regard this as the plate of the last tube of a three-stage amplifier.
Wolff:
Regard what? Where are you pointing?
Black:
This first point.
Wolff:
Number 4?
Black:
Right. This would be the input voltage.
Wolff:
E is the input voltage.
Black:
Yes. Multiplied by the µ of the three-stage amplifier. I warned you this wasn't too clear.
Wolff:
I just want to be able to explain it in the caption. I think that's good enough.
Black:
And you have to do the same thing on the input in order to match impedances.
#### Morristown experiment, 1927
Wolff:
All right. My question about December 1927. You said you reduced distortion by 50 dB. Is that a single amplifier?
Black:
That is a single amplifier. That used two tetrodes and a pentode. The output of that amplifier was about 10 dB less than Dr. Jewett required in January of the next year. He was satisfied with 40 dB of feedback.
Wolff:
Where does Dr. Jewett come in now? In Morristown?
Black:
That's in Morristown. I did not put that in there at all.
Wolff:
I think it should be in there.
Black:
That was why I wrote something for you. You may want a reference to that paper describing those experiments.
Wolff:
Before we get to 1928, let's come back to the top of page 6. You said you submitted an application on August 1928 to the Patent Office.
Black:
I submitted a drawing. I had many people working on this, so I don't recall whether it was Steve Meyer or a different engineer. I submitted a drawing of the amplifier that accomplished this, and that was the basis of the main patent.
Wolff:
The patent that was then filed in 1932?
Black:
That's right. All the work that had preceded that occupied a large amount of time. That one had the biconjugate networks and 6 dB improvement and matching impedances and everything else.
Wolff:
In January 1928 the carrier system for transcontinental cables began to be developed. I think we should describe that.
Black:
What page are we on now?
Wolff:
We're not on any page because you didn't write it up, but I found this in what you gave [Julian] Tebo, and I think we ought to include that. This is about the Morristown experiment.
Black:
Yes.
Wolff:
You have given me some material on it here. The interesting thing is that Dr. Arnold didn't believe your amplifier would work.
Black:
What do you know about that. Now you got right to the Bell Labs. Not to all the other people and not to England, but Dr. Jewett for the most highly prized possession in the Bell Labs. I thought you might perhaps want to include that. Or maybe you wouldn't.
Wolff:
I do. It's all very interesting.
Black:
It's all very true. Did I write it in a way that you can understand?
Wolff:
Was Arnold Dr. Jewett's boss?
Black:
No. Dr. Jewett was president. Dr. Arnold was not a vice president. He was the director of research, third down in the line.
Wolff:
Okay. Arnold did not believe it would work and he instructed you to build a powerful Colpitts amplifier.
Black:
That's right.
Wolff:
What is the point of this? You proceeded to assemble the parts for this Colpitts amplifier.
Black:
I got them all together. In the meantime, every engineer I had was working on this other job.
Wolff:
The other job being the negative feedback amplifier?
Black:
The negative feedback amplifier for Dr. Jewett.
Wolff:
You were working on two amplifiers.
Black:
All that I did – and I did no more – was to have A. G. Garns design the transformers and collected the condensers and inductors. I got the parts necessary to built one Colpitts amplifier if that proved necessary. I didn't feel that it was going to be necessary, but I did exactly what he told me to do.
Wolff:
That's fascinating. Meanwhile you went ahead with your own and found out that it worked.
Black:
Yes.
Wolff:
That's what that photograph is that you gave me last week. Right?
Black:
Yes. That is a physical picture.
Wolff:
That's the Morristown experiment.
Black:
Yes. I am taking out one of the condensers there. Charlie Green thought I should be fired.
Wolff:
Why?
Black:
Because I had a come-and-go trouble in it, and he thought that I had done something wrong. The people who made the condenser said that if there was anything wrong I would have to find it. It cost me about \$100,000 of my engineer's salary. I found that the method of making the contact between the foil which was insulated to get a capacitor was loose. It would go on and off. I don't think that it would be pertinent to do that, but you can notice I was looking at the animal sort of seriously.
Wolff:
That's an interesting detail to put in the article.
Black:
Then I was on to confirm that with F. A. Brooks because I designed more than one amplifier.
#### Condenser troubleshooting
Wolff:
Wait. I'm confused. This is a photograph.
Black:
What I am taking out in that photograph looks to me to be a condenser. If that is true, then I believe that it is this faulty condenser that I had so much difficulty troubleshooting.
Wolff:
The important thing to clarify is that when you tested these amplifiers there was a problem with the condenser. Is that it? I don't think it's important whether it was that particular condenser. I just want to tell the story of this Morristown experiment.
Black:
Fred Brooks is alive, and unless he is on vacation I will be able to reach him by telephone and find out whether what I am saying is true or false. I would never put that in unless I could prove it. This part however is the absolute truth. I built the push-pull.
Wolff:
During 1928 you tested eight models of a negative feedback amplifier.
Black:
Yes.
Wolff:
Was it one of those models that had this trouble with the condenser?
Black:
No. There was no trouble there.
Wolff:
Where was there trouble with the condenser?
Black:
I don't remember. I'll have to check up on that. Regardless of whether or not that condenser was at fault – and if it was it was quickly fixed in just about a week and a half – the experiment was a complete success. It is fully described in a brief paper that you could make reference to if you like.
Wolff:
Have you got that paper?
Black:
We aim to please. [recording turned off, then back on...]
#### Patent application review process
Wolff:
On page 6 where you are talking about the patent, you said, "The concept was so contrary to established beliefs that the U.S. Patent Office reported that experts the world over delivered lengthy opinions stating it wouldn't work." To whom did they report this?
Black:
When an invention is submitted to the Patent Office the first thing they do is make a search. They are well equipped to do that in a hurry. They want to make sure that the same idea has not been patented by anyone else in this or any other country. Next, they want to make sure that it is not already described in any book, technical paper or article. Finally, they undertake to decide for themselves whether or not it satisfies their requirements. It was in that process that they found papers in France, Italy and other countries saying that the output could not be connected back to the input unless the loop gain was less than 1. Whereas my loop-gain was 40. Actually I experimentally got 50 in because I allowed for manufacturing variations and so on.
Wolff:
Is loop gain a product of µЗ?
Black:
That is exactly right. It is the absolute value of the product of µЗ. The U.S. Patent Office found those papers, and each one and said it could not be done.
#### Foreign patents
Black:
Did I tell you what the British post office said?
Wolff:
No.
Black:
Here in the United States one makes an invention of what is the old-fashioned idea of perpetual motion. Some very intriguing things have been so proposed. I was taught in school never to do that. The U.S. Patent Office then – and only then – says, "Submit a working model." That is then the last that is ever heard of that invention. My inventions were patented in many foreign countries. That's very costly and important, and they have to go into use within [a certain period of time]. Now my invention was clearly in the field of electronics. Curiously, Great Britain has a rule that in electronics that if they have reason to believe that it does not work they may request the inventor to submit a working model. In my case they said the negative feedback amplifier would not work and asked me to kindly submit a working model. Harry A. Burgess then got affidavits that seventy such amplifiers were working in the telephone building at Morristown, New Jersey.
Wolff:
That's fascinating.
Black:
He also sent pictures. In view of that and the fact that it had been described in two publications by that time – my own plus the other to which you just referred – they conceded that it could be patented in Great Britain.
Wolff:
That's very nice. All of this kind of material would be nice to work into the article. There were objections to the length and arguments about the claims.
Black:
That is right.
#### U.S. Patent Office objections; patent claims
Wolff:
Who objected to the length and what kinds of arguments were made about the claims?
Black:
Every time the U.S. Patent Office objected to a claim, I stoutly refused to concede the thing and immediately proceeded to add one more claim. In that way I gradually built the number of claims up to 126. Burgess would not allow what would have been Claim 127 because he thought it would not stand up and might thereby invalidate the entire patent. As to their objection to the length, well they just thought that it was complex and long.
Wolff:
All right. That's clear. Did they argue with you about claims before the patent was issued?
Black:
That's right. They – and only they – decide how many claims are going to go into it.
Wolff:
Why would they argue about a claim?
Black:
They can for instance argue that it is too broad. The patent starts out with wave translation. That can be electrical, acoustical, mechanical or hydrodynamical. Any medium that will transmit a wave is a wave translation system. Those are the first words in the description of my invention.
Wolff:
I was interested in that. That is what you meant by the new material you wrote for me about applications. You envisioned broad applications.
Black:
That's right. That of course is where the claims come in. I would say that the value of a patent lies in its claims. The reason for the long description was that this was a new field, so I gave a lot of examples. The examples I gave were amplifiers I had set up in the laboratory.
Wolff:
Are there wave translation systems in the claims that are not electrical?
Black:
When it comes to claims, there are technicalities. By the way, here is a little detail where I told in my text how much of the patent was devoted figures, how much to description and how much was claims. The first three lines of the second claim are mainly description. The last page of the claim is vacant, so that I don't know just how you would say where the description ends and the claim begins.
Wolff:
I'll figure something out about that.
Black:
I never realized that situation existed until the other day. There is another thing I would like to mention having to do with the figures. If you look at some of these figures, if they are examined and if for instance you call this Figure 35, some of these can really be different from others. If the figures were counted the way an engineer would count them it would come to a larger number.
#### Bell Labs response
Wolff:
Why would Dr. Arnold have been doubtful that your amplifier would work? Was it the stability concern again?
Black:
My only answer to that is that I could not read his mind, but he was convinced. The most natural reason would be the one you suggest. I boldly suggested putting a tremendous amount of gain into the µЗ rule. Yet I was claiming that the amplifier would not sing. He was educated as a physicist. He got his Ph.D. from Chicago and came to work for Western Electric in 1911. He was hand picked by Dr. Jewett. My answer is that I don't know. I don't even know why the British post office and or other people in other countries thought it would not work.
Wolff:
I am trying to get a sense of whether there were many people in the laboratory who were skeptical that your amplifier would work.
Black:
No, not in the Bell Laboratories. The reason for that is that I was about the only one who was concerned. I was not consulting with anyone else. In those days if I wanted a transformer, a condenser, a resistor or anything else, I got it from the apparatus department. That was under William Fondella. If I wanted to get my amplifier mounted I went to the equipment department.
Wolff:
You are saying that you did it yourself.
Black:
I assembled the things in the laboratory. First I got all parts of the apparatus and placed requirements on that. Therefore that if the apparatus met its requirements, then the system – in this case was the amplifier – would meet its requirements. It would be all right. I took into account a liberal amount for variations – manufacture, temperature of offices and all factors including changes with time. I then had to cooperate and change the way that I mounted it to conform to the way the equipment division wanted it mounted. They had the final say on that before for the Western Electric Company manufactured these amplifiers. That laboratory trial involved a great many things.
Wolff:
Black:
The eight amplifiers that we got together.
#### Suitability for transcontinental telephony
Wolff:
I just realized that there is a gap we need to fill. You said that on December 29, 1927 you demonstrated that the amplifier "was more than sufficient to do the job I undertook six years earlier." What job was that? Was then when you realized this was an amplifier with which you could put a hundred of them in a string?
Black:
I sure did. I envisioned hopping across the nation.
Wolff:
And it was at this point that you knew that you could do that?
Black:
I knew that I could do that. Yes.
Wolff:
With this test. All right.
#### Frank B. Jewett
Black:
The thing that I did not realize was that it did not have enough power. That was the main difference between the first working model and what I built for Dr. Jewett.
Wolff:
Let me get to Dr. Jewett now. You built this first model with Meyers ?
Black:
Yes, and my other colleagues.
Wolff:
What happened after that? Why did Dr. Jewett, on the next page, suggest to Arnold that you build this amplifier?
Black:
I was very friendly with Dr. Jewett, and he knew that I was working on a negative feedback amplifier. He also found out that I had succeeded. It was pure coincidence that my success, which was at the end of the year, coincided with his suggestion in January.
Wolff:
Were you still in the systems group?
Black:
Yes.
Wolff:
Black:
That was at West Street, and the head of that group was Amos (A. F.) Dixon. I could show you that on the organizational chart.
Wolff:
That's important. That was a gap we had here. You were friendly with Dr. Jewett and told him what you had done?
Black:
Sure. You bet.
Wolff:
What was his reaction? He believed you?
Black:
Oh yes. Well, I am complimented that he had that faith in me.
Wolff:
Why wasn't he worried about stability?
Black:
Wolff:
Did you demonstrate it to him?
Black:
No. I went to work for the laboratories in 1921. I had belonged to Tau Beta Pi and Sigma Si while going through college. I don't know whether he was president at the time, but I believe it was AT&T that brought together a large group of people who belonged to Tau Beta Pi. That was when and where I was first introduced to Dr. Jewett.
Wolff:
What was his full name?
Black:
Frank B. Jewett. That was my first meeting with him. He was very friendly. Of course at that early date I had not yet gotten into this feedback or anything, but that was my first contact with him.
#### Power of amplifier models
Wolff:
Were the eight models of the amplifier powerful enough?
Black:
Yes. It took two tries to get that power. Dr. Kelly designed every vacuum tube I ever used in any amplifier I ever made or with which I wanted to experiment. Kelly gave me two high-grade pentodes having a little more power than any of the tubes I had used in December. Then we came to the stickler. What I was going to use in the last stage? At that time there existed no directly heated pentode in Morristown that would deliver the power. Fortunately they only wanted 40 dB of negative feedback. The equalizer was designed by F. A. Brooks. He reported to me. To my disappointment, however, the equalizer was not in the feedback path. Therefore all I needed was the amount of gain shown by that. Let's see. The slanting is the low loss and then the equalizer goes this way and those two losses cancel out. One could figure it out. Anyhow, I needed only modest amount of gain, practically enough for the two pentodes given to me by Dr. Kelly. When we came to the last tube we decided that would be a triode. Then I made a suggestion. Now a suggestion is not a patent, and McNally is the one who got the patent. The [unintelligible word] vacuum tube has a grid, a filament that is not indirectly heated and a plate. I proposed that we put in two grids. We worked fast on this job. One good thing about Kelly is that he was a fast worker. And he expected those who worked for him to do likewise. We put two grids in the same plane. We gave one of them 60-volt wires that were negative and the other 30-volt wires were positive. They were given the opposite polarity. In that way we got a wider swing without increasing the distortion.
Wolff:
A wider swing in what?
Black:
On the voltage that was put on the grid. If there were only one grid instead of two, only half as much swing could be achieved. We expected a 6-dB improvement that way.
Wolff:
Is this what gave you a higher output?
Black:
It gave me 6 dB higher output, but unfortunately that was not enough. The next step was to do what the Bell System had refused to do for one hundred years. They raised the B battery voltage from 130 volts to 250 or 260 volts – whatever my paper showed. In other words, we did it by brute force. That gave the required power and completed the success.
#### Channels, transmission, and frequency
Wolff:
To close the loop on this whole article, what you have not told me is how many channels such an amplifier could have.
Black:
Nine channels. That's given in the paper.
Wolff:
And you could put 160 of them in tandem?
Black:
Yes, with spacing at every 25 miles, as Dr. Jewett's requirement was 4,000 miles. The transmission, due to any and every cause, was not to be out by more than ∀1 dB. Today they permit 4 dB because of the improved handsets and other reasons.
Wolff:
Okay. That takes care of it.
Black:
The frequency allocation and everything is given in that article.
#### Challenges to Black's patent
Wolff:
Black:
Yes. It was challenged once. I licked it in a day. They produced a patent, and I showed that the patent they produced would not work. It had never even been set up in a laboratory.
#### Bell Labs suit against Zenith
Wolff:
Somehow I had it in my head the idea that there was some problem with Zenith.
Black:
That was something entirely different. Zenith manufactured radios, television and a lot of electronic equipment, including some stereophonic. They had a patent attorney who told the president of Zenith that there was no patent issued by the U.S. Patent Office that he could not break. He told Zenith to advertise in newspapers, magazine articles and all over the place that they were using my patent, the negative feedback amplifiers. They did that and showed the places they were using them and so on. Then that same patent attorney died, leaving Zenith to battle it out without him.
Wolff:
Did Bell bring a suit against them?
Black:
Yes, they did. Incidentally, most of my patents have been in the communication field. [recording turned off, then back on...]
Wolff:
Black:
I think it lasted about five years.
Wolff:
Black:
Wolff:
Black:
No, it was probably starting about 1948 and lasting until '53.
Wolff:
Black:
RCA and Bell (or AT&T) were suing Zenith.
Wolff:
It was not against RCA? It was against Zenith?
Black:
RCA had a couple of patents and they didn't care much about, but when it came to my patent the Bell System was prepared to carry that all the way to the Supreme Court.
Wolff:
Did Bell sue RCA?
Black:
No. Bell sued Zenith. RCA sued Zenith too, but they didn't get anywhere.
#### Social impact of negative feedback
Wolff:
Good. I think this takes care of it. Do you have any general thoughts looking back over this work that started more than fifty years ago? I imagine that the wide success of the negative feedback has been very gratifying to you.
Black:
Yes.
Wolff:
Was it surprising too?
Black:
No, I wouldn't say it was surprising, because I understood its importance from the beginning. Time moves very swiftly over a period of fifty years. I will give a few examples. Thanks to negative feedback, half a billion people watched man take his first steps on the Moon. It was really a lot more than that, because even the launch itself involved feedback. In the picture you have of one of my amplifiers, you can see that the length of my feedback path was almost a yard and a half.
#### Twelve-channel transmission
Black:
[Nowadays] the length of the feedback path in the coaxial is the thickness of a piece of cardboard and the size of two amplifiers. They couldn't do it in one step because they were handling so many channels and the thing varies as the square root of S, so they had to use two amplifiers every one mile.
Wolff:
Black:
Coaxial?
Wolff:
No. Two amplifiers every mile. I lost the point. Were you talking about coaxial cable today?
Black:
The coaxial cable of today. The coax has a long history starting in '36, but with the coaxial of today a single pipe carries about 1,028 analog voice circuits, and at every mile they have to put in a repeater. Due to the large number of circuits, that really has to be two three-stages amplifiers having about 150 dB of negative feedback – as compared to my 60 or 65 over a 12-channel K.
Wolff:
Over a 12-channel what?
Black:
K1 and K2 transmitter, the 12-voice circuit. The coax. I'd have to check on the exact number, but I think it's 1,028.
Wolff:
What is K1? Was that your first system?
Black:
K1 is a 12-channel.
Wolff:
Black:
The K1 was made in 1940, and the K2 was made in 1942. Each one had the equalizer in the feedback circuit, but the K2, in order to get more out of things, used a looped-over amplifier. Twelve channels was the most I ever did, and my amplifiers were bulky.
Wolff:
This is important. I want to make sure we understand it. Your negative feedback amplifier eventually got to where in 1940 it could transmit twelve channels.
Black:
Wolff:
And you are saying that today the negative feedback amplifier and coax systems can transmit over a thousand. Right?
Black:
That is right.
Wolff:
You are contrasting the improvement.
Black:
And I am calling attention to the fact that the two amplifiers are about the size of the palm of my hand – 3 or 4 inches.
Wolff:
These are the amplifiers in today's coax system, right?
Black:
Yes. And the length of the feedback path was a little thinner than a piece of cardboard. The bulk would differ by more than 10,000 to 1.
Wolff:
The size of the amplifiers in bulk between 1940 and today?
Black:
Yes.
Wolff:
Differ by more than 10,000 to 1. That's beautiful. That's nice to have that comparison.
### 8-channel open wire system amplifier
Black:
I would like to tell you about another amplifier I designed, which was for an 8-channel open-wire system. That was the best negative feedback amplifier I ever designed, because everything was fed back through the transformers. There were three transformers. It did the impedance matching and had some very unique properties. It had equipment that went with it to deal with the fact that in some parts of the country the ice would form an inch around. These were used in fairly large quantity. It was a horrible system really, but the amplifier was a beautiful one. During World War II a hundred or more of them were given to Russia – of all places.
Wolff:
For use in what? Anything in particular?
Black:
I suppose it was due to the fact that they have a lot of open country there.
Wolff:
Did you ever have the experience that the inventors of the computer had? Initially they couldn't sell them because people didn't think they would ever be practical. Did you run into the resistance like that in the beginning?
Black:
No. It has sold from the beginning.
Wolff:
Only some skepticism by Arnold and a few others.
Black:
That and the Patent Office and the articles that had been published for entirely different reasons.
Wolff:
It's an exciting story. I appreciate all your time.
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# Measures of central tendency
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### Measures of central tendency
1. 1. Measures of Central Tendency Back to Top The three important measures of central tendency are 1. The Mean 2. The Median 3. The Mode Measures of Central Tendency Definition Measure of central tendency can be the term which defines the centre of data. There are three parameters by which we can measure central tendency - Mean, median and mode. Central Tendency of Data Mean: Mean of data is a set of numerical values is the arithmetic average of the data values in the set. It is found by adding all the values in the data set and dividing the sum by the total number of values in the set. Mean of a data set = Sum of the Data ValuesTotal Number of Data Values Median: For an ordered data set, median is the value in the middle of the data distribution. If there are even number of data values in the set, then there will be two middle values and the median is the average of these two middle values. Mode Mode is the most frequently occurring value in the data set. In addition to these three important measures of central tendency, another measure is also defined. Midrange: Midrange is an estimated measure of the average. It is the average of the lowest and highest values in the data set. Midrange = Lowest Value + Highest Value2 Midrange is only a rough estimate of the central value. As it uses only the lowest and highest values of the data set, it is highly affected when one of them is very high or very low.
2. 2. Central Tendency Definition Back to Top The term central tendency refers to the middle value of the data, and is measured using the mean, median, or mode. It is the tendency of the values of a random variable to cluster around the mean, median, and mode. And a measure of central tendency for a data distribution is a measure of centralness of data and it is used to summarize the data set. Mean Back to Top The mean of a sample data is denoted by x¯ and the population mean by μ. The mean of a small number of data set can be found by adding all the data values and dividing the sum by total number of values. Characteristics of Mean 1. Mean is computed using all the values in the data set. 2. Mean varies less for samples taken from the same population when compared to the median or mode. 3. The Mean is unique for a data set. The mean may not be one of the data values in the distribution. 4. Other statistics such as variance are computed using mean. 5. Mean is affected the most by the outliers present in the data set. Hence mean is not to be used for data sets containing outliers. Mean for the grouped data is also computed applying above methods, the mid point of the class is used as x. Solved Examples Question 1: The following data set is the worth(in billions of dollars) of 10 hypothetical wealthy men. Find the mean worth of these top 10 rich men. 12.6, 13.7, 18.0, 18.0, 18.0, 20.0, 20.0, 41.2, 48.0, 60.0 Solution: Given data, 12.6, 13.7, 18.0, 18.0, 18.0, 20.0, 20.0, 41.2, 48.0, 60.0 Mean of the data set, x¯ = 12.6+13.7+18+18+18+20+20+41.2+48+6010
3. 3. = 269.510 = 26.95 Question 2: Compute the mean for the distribution given below Value x Frequency f 20 2 29 4 30 4 39 3 44 2 Solution: The frequency table is redone adding one more column f * x Value x Frequency f f * x 20 2 40 29 4 116 30 4 120 39 3 117 44 2 88 ∑f = 15 ∑fx = 481 Mean of the distribution x¯=∑fx∑f = 48115 = 32.1 (Answer rounded to the tenth). → Read More Median Back to Top
4. 4. When we say the median value of earnings of Actuarial experts is 60,000 dollars, we mean that 50% of these experts earn less than 60,000 dollars and 50% earn more than this. Thus median is the balancing point in an ordered data set. As median represents the 50% mark in a distribution, this is a measure of position as well. Median is much more easier to find than computing the mean. Uses of Median 1. Median is used if the analysis requires the middle value of the distribution. 2. Median is used to determine whether the given data value/s fall in the upper or lower half of the distribution. 3. Medan can be used even if the classes in the frequency distribution are open ended. 4. Median is generally used as the central value, when the data is likely to contain outliers. Solved Examples Question 1: The number of rooms in 11 hotels in a city is as follows: 380, 220, 555, 678, 756, 823, 432, 367, 546, 402, 347. Solution: The data is first arranged starting from the lowest as follows: 220, 347, 367, 380, 402, 432, 546, 555, 678, 756, 823. As the number of data elements 11 is an odd number, there is only one middle value in the data array, which is the 6th. => The value of data in 6th position = 432. Hence the mean number of Hotel rooms in the city = 432. Question 2: Find the median of the given data Value X Frequency f 20 2 29 4 30 4 39 3 44 2 Solution: Value x Frequency f Cumulative frequency
5. 5. 20 2 2 29 4 2 + 4 = 6 30 4 6 + 4 = 10 39 3 10 + 3 = 13 44 2 13 + 2 = 15 ∑f = 15 ∑fx = 481 => ∑f = 15 items, The 8th item in the ordered data array will be the median. The 8 item will be included in the cumulative frequency 10. Hence the median of the distribution is the x value corresponding to cumulative frequency 10 which reads as 30. => Median of the data = 30. → Read More Mode Back to Top Mode is the value or category that occurs most in a data set. If all the elements in the data set have the same frequency of occurrence, then distribution does not have a mode. In a unimodal distribution, one value occurs most frequently in comparison to other values. A bimodal distribution has two elements have the highest frequency of occurrence. Characteristics of Mode: 1. Mode is the easiest average to determine and it is used when the most typical value is required as the central value. 2. Mode can be found for nominal data set as well. 3. Mode need not be a unique measure. A distribution can have more than one mode or no mode at all. Solved Example Question: Find the mode of a numerical data set 109 112 109 110 109 107 104 104 104 111 111 109 109 104 104 Solution:
6. 6. Given data, 109 112 109 110 109 107 104 104 104 111 111 109 109 104 104 Total number of element = 15 Among the 15 data elements the values 104 and 109 both occur five times which are hence the modes of the data set. → Read More Effect of Transformations on Central Tendency Back to Top If all the data values in a data distribution is subjected to some common transformation, what would be the effect of this on the measures of central tendency? If each element in a data set is increased by a constant, the mean, median and mode of the resulting data set can be obtained by adding the same constant to the corresponding values of the original data set. When each element of a data set is multiplied by a constant, then the mean, median and mode of the new data set is obtained by multiplying the corresponding values of the original data set. Central Tendency and Dispersion Back to Top Two kinds of statistics are frequently used to describe data. They are measures of central tendency and dispersion. These are often called descriptive statistics because they can help us to describe our data. Measures of Central Tendency and Dispersion Mean, median and mode are all measures of central tendency whereas range, variance and standard deviation are all measures of dispersion. The measures used to describe the data set are measures of central tendency and measures of dispersion or variability. Central Tendency Dispersion If different sets of numbers can have the same mean. Then we will study two measures of dispersion, which give you an idea of how much the numbers in a set differ from the mean of the set. These two measures are called thevariance of the set and the standard deviation of the set. Formula for variance and Standard Deviation:
7. 7. For the set of numbers {x1,x2,.............,xn} with a mean of x¯. The variance of the set is => V = (x1−x¯)2+(x2−x¯)2+.........+(xn−x¯)2n and the standard deviation is, => σ=V−−√. Standard Deviation can be represented as; σ = x21+x22+...............+x2nn−x¯2−−−−−−−−−−−−−−−−−√ Resistant Measures of Central Tendency Back to Top A resistant measure is one that is less influenced by extreme data values. The mean is less resistant than the median, that is the mean is more influenced by extreme data values. Resistant measure of central tendency can resist the influence of extreme observations or outliers. Let us see the effect of outlier with the help of example: Solved Example Question: Consider the data set, 5, 19, 19, 20, 21, 23, 23, 23, 24 , 25. Solution: The value 5 is an outlier of the data as it is too less than the other values in the distribution. Let us calculate the the central values for the data set either by including and excluding 5. Step 1: The data set excluding 5 is 19, 19, 20, 21, 23, 23, 23, 24 , 25 Mean = x¯=19+19+20+21+23+23+23+24+259=1979 = 21.89 => Mean = 21.89 Median = 23 Mode = 23 Step 2: For the data including the outlier 5, 19, 19, 20, 21, 23, 23, 23, 24 , 25
8. 8. Mean = x¯=5+19+19+20+21+23+23+23+24+259=20210 = 20.2 => Mean = 20.2 Median = 21+232 = 22 Mode = 23 Step 3: Comparing the values of mean, median and mode found in step 1 and step 2, the mean is most affected and mode is least affected by the inclusion of the outlier value 5. Central Tendency and Variability Back to Top Central tendency is a statistical measure that represents a central entry of a data set. The problem is that there is no single measure that will always produce a central, representative value in every situation. There are three main measures of central tendency, mean, median and mode. Variability is the important feature of a frequency distribution. Range, variance and standard deviation are all measures of variability. Range, variance and standard deviation are all measures of variability. Range - The simplest measure of variability is the range, which is the difference between the highest and the lowest scores. Standard Deviation - The standard deviation is the average amount by which the scores differ from the mean. Variance - The variance is another measure of variability. It is just the mean of the squared differences, before we takethe square root to get the standard deviation. Central Tendency Theorem Back to Top A more formal and mathematical statement of the Central Limit Theorem is stated as follows: Suppose that x1,x2,x3,...................,xn are independent and identically distributed with mean μ and finite variance σ2 . Then the random variable Un is defined as, Un = X¯−μσn√ Where, X¯=1n∑ni=1Xi Then the distribution function of Un converges to the standard normal distribution function as n increases without bound.
9. 9. Central Tendency Examples Back to Top A measure of central tendency is a value that represents a central entry of a data set. Central tendency of the data can be calculated by measuring mean, median and mode of the data. Below you could see some examples of central tendency: Solved Examples Question 1: Find the mean, median and mode of the given data. 10, 12, 34, 34, 45, 23, 42, 36, 34, 22, 20, 27, 33. Solution: Given Data, X = 10, 12, 34, 34, 45, 23, 42, 36, 34, 22, 20, 27, 33. ΣX = 10 + 12 + 34 + 34 + 45 + 23 + 42 + 36 + 34 + 22 + 20 + 27 + 33 = 372 => ΣX = 372 Step 1: Mean = ∑XX = 37213 [ X = Total number of terms ] = 28.6 => Mean = 28.6 Step 2: For Median, Arrange the data in ascending order.
10. 10. 10, 12, 20, 22, 23, 27, 33, 34, 34, 34, 36, 42, 45. The median is 33. Half of the values fall above this number and half fall below. => Median = 33 Step 3: Mode Mode = 34 Because 34 occur maximum times. Question 2: The following table shows the sport activities of 2400 students. Sport Frequenc y Swimming 423 Tennis 368 Gymnastic s 125 Basket ball 452 Base ball 380 Athletics 275 None 377 Solution: From the given table: For grouped data the class with highest frequency is called the Modal class. The category with the longest column in the bar graph represents the mode of data set. Basket ball has the highest frequency of 452. Hence Basket ball is the mode of the sport activities.
11. 11. Question 3: Find the median of the distribution, 223, 227, 240, 211, 212, 209, 211, 213, 240, 229. Solution: The ordered data array will be: 209, 211, 211, 212, 213, 223, 227, 229, 240, 240 The number of data values is even. Hence the two central values are those in the 5th and the 6thpositions. Median = 213+2232 = 4362 = 218 => Median = 218. http://math.tutorvista.com/statistics/central-tendency.html
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## Degree Symbol in Maths – ° Degree Symbols
This article is about “grade” as it is used in the area of mathematics. For other meanings of this word, see grade .In mathematics there are different meanings of the word degree depending on the mathematical area in question. All definitions result in a natural number that expresses the degree .
## Degree of a Polynomial
In algebra degree of a polynomial is the maximum degree of the exponents of the variables of the monomials that compose it. Each degree has basically the same meaning when it refers to a polynomial or an algebraic equation. Consequently, the first definition that may need to be revised is that of the monomial, considered by elementary algebra as a basic algebraic expression, which is made up of a combination of numbers and letters (raised to positive integer exponents, including zero) between which there are no subtraction, addition or division operations, being then the only ones allowed, the multiplication raised between the numeric element (coefficient) and the non-numeric element (literal or variable) as well as the potentiation that occurred between the literal and its exponent.
The degree of a polynomial of a variable is the maximum exponent that the monomial has on the variable; For example, in 2 x 3 + 4 x 2 + x + 7, the term with the highest degree is 2 x 3 ; this term has a power of three on the variable x, and is therefore defined as degree 3 or third degree .
For polynomials of two or more variables, the degree of a term is the sum of the exponents of the variables in the term; the degree of the polynomial will be the monomial of the highest degree. For example, the polynomial x 2 y 2 + 3 x 3 + 4 y has a degree of 4, the same degree as the term x 2 and 2 .
## Degree Of Congruence
Yes {\ displaystyle \ textstyle f (x) = a_ {n} x ^ {n} + a_ {n-1} x ^ {n-1} + \ cdots + a_ {1} x ^ {1} + a_ {0 } x <0}{\ displaystyle \ textstyle f (x) = a_ {n} x ^ {n} + a_ {n-1} x ^ {n-1} + \ cdots + a_ {1} x ^ {1} + a_ {0 } x <0} is a polynomial with integer coefficients, the degree of congruence of {\ displaystyle \ textstyle f (x) \ equiv 0 {\ pmod {m}}}{\ displaystyle \ textstyle f (x) \ equiv 0 {\ pmod {m}}}is the largest positive integer j such that a j is not congruent with zero with respect to modulus m .
## Equation Theory
In algebraic equation theory , the degree of an equation corresponds to the maximum power to which the algebraic unknown in the equation is raised. For example:{\ displaystyle x ^ {3} y + 4x-y = 2xy \, \!}{\ displaystyle x ^ {3} y + 4x-y = 2xy \, \!}the equation is of the third degree in x , being of the first degree in the unknown y . See: Equation of second degree , Equation of third degree , Equation of fourth degree , Equation of fifth degree , etc.
## Degree of an extension
In algebra it has the extension body and the degree therein is defined as any vector space with basis you can calculate the dimension of{\ displaystyle L}L as vector space over {\ displaystyle K}K, denoted by {\ displaystyle \ dim _ {K} (L)}{\ displaystyle \ dim _ {K} (L)}. {\ displaystyle L: K}{\ displaystyle L: K} to the dimension of {\ displaystyle L}L What {\ displaystyle K}K-vectorial space: {\ displaystyle [L: K] = \ dim _ {K} (L)}{\ displaystyle [L: K] = \ dim _ {K} (L)}.
## Graph theory
In Graph Theory , the degree or valence of a vertex is the number of edges incident to the vertex. The degree of a vertex x is denoted by degree (x) , g (x) or gr (x) (although δ (x) is also used , and from English d (x) and deg (x) ). The maximum degree of a graph G is denoted by Δ (G) and the minimum degree of a graph G is denoted by δ (G) .
## Degree of freedom (statistics)
In statistics , degrees of freedom , an expression introduced by Ronald Fisher , says that, from a set of observations, the degrees of freedom are given by the number of values that can be arbitrarily assigned, before the rest of the variables take a value automatically, the product of establishing those that are free; this, in order to compensate and equalize a result which has been previously known. They are found by the formula{\ displaystyle nr}{\ displaystyle nr}, where n is the number of subjects in the sample that can take a value and r is the number of subjects whose value will depend on the number taken by the members of the sample who are free. They can also be represented by{\ displaystyle kr}{\ displaystyle kr}, where {\ displaystyle k}k= number of groups; this, when operations are carried out with groups and not with individual subjects.
When it comes to eliminating statistics with a set of data, the residuals – expressed in vector form – are found, usually in a space of less dimension than that in which the original data were found. The degrees of freedom of the error are determined precisely by the value of this smallest dimension.
This also means that the residuals are restricted to being in a space of dimension {\ displaystyle n-1}n-1 (in this example, in the general case a {\ displaystyle nr}{\ displaystyle nr}) since, if the value of {\ displaystyle n-1}n-1of these residues, the determination of the value of the remaining residue is immediate. Thus, it is said that «the error has{\ displaystyle n-1}n-1 degrees of freedom »(the error has {\ displaystyle nr}{\ displaystyle nr} general degrees of freedom).
## ° Degree Symbol Alt Code
Windows ALT Code List (ASCII Symbols) – The ALT codes also known as symbols keyboard , symbols ALT or correctly symbols ASCII (For its acronym in English: American Standard Code for Information Interchange ) is a coding pattern that is used in computing.
Taking as an example the common characters that are the letter eñe lowercase ñ or uppercase Ñ or vowels with accent, for example lowercase a with accent á, lowercase e with accent é, etc.
To type any of these with the keyboard in Windows we must perform the following steps:
Press the “ALT” [keybt] ALT [/ keybt] key on the keyboard without releasing.
Holding down the previous one we must write with the numeric keyboard the code corresponding to the ALT symbol to write for example: “165” [keybt] 1 [/ keybt] + [keybt] 6 [/ keybt] + [keybt] 5 [/ keybt] which is the code or number corresponding to the letter or symbol “Ñ” capital letter.
Release the “ALT” key [keybt] ALT [/ keybt] and the character should appear or be displayed.
Summary steps for typing an ALT symbol
Hold down the “ALT” key
Without releasing the previous one write the code; “ALT” + Codes.
Release “ALT” key
## What is the ASCII code for the degree sign symbol?
The degree sign extended ASCII code displayed with the “°” symbol is 248.
For the representation of the degree sign in any computer text, it must be taken into account that it must use ASCII code or an extension that is compatible. If this is compatible, in any text editor, applying the keyboard combination ALT + 248 will return the “°” symbol.
Nowadays almost all computer systems (computers, mobile phones, tablets …) usually accept this code without problem and represent the symbols without problem, however, some may not interpret them making that when indicating the ASCII code 248 (American Standard Code for Information Exchange) an error symbol is displayed instead of the “°” symbol.
It should be remembered that many text editors allow searching in the different symbols in their options panel to facilitate the writing of the symbol «°» without having to know the extended ASCII code that corresponds to it, in this case by combining ALT + 248 .
## How To Make Faces And Symbols Using ASCII Code
It hasn’t happened to you that you need to write a special character like ñ or @ and you don’t know how to write it or where to find it. This may be due to a bad configuration of the keyboard language in your operating system. Surfing the internet I found a solution, which is the one that I am going to show you next, we are going to see a little theory. It is about using the famous ASCII code to be able to obtain any symbol regardless of the language of the keyboard of the operating system in which we are working.
## What Is ASCII Code ?
ASCII is the acronym for American Standard Code for Information Interchange in Spanish stands for American Standard Code for Information Interchange generally pronounced Aski . It is based on the Latin alphabet, as used in modern English. It was created in 1963 by the American Standards Committee (ASA),
## How To Use?
To create these characters, hold down the “alt” key on your keyboard and type the numbers on the right side of the keyboard (numeric keypad) and then release the buttons and the symbol will appear. This is a great way to experiment with new symbols, signs, and characters. It’s also good if a button or letter on the keyboard stops working, because you can still type with the number pad. Just look down the list to find the one you need.
## How to type the ° (degrees) symbol in Word 2016 and Word 2019
Perhaps the most utilized applications today for report the board in Microsoft Word Thanks to this program, we have the possibility of creating text files with different options such as graphs or tables. In addition, sharing a Word document is something quite simple, since we can do it through the mail for example, being a compatible format on both Windows and Mac.
In spite of the fact that Word offers us many alternatives to deal with the data in the most ideal manner, in some cases we wind up confronted with the need to enter certain unique characters and we do not know how to do them or we must resort to the ASCII table to know which combination of keys we can use to get that symbol or character.
That is why today Solvetic will analyze the different practical ways of entering said text in our Word 2016 and Word 2019 documents.
If we are on computers with Windows operating systems we will press the Alt key and the numbers 0176 or 248, then we release the Alt key and the degree symbol will appear automatically.
Alt+ 0176 / 248
If we are on computers with macOS operating systems, it will be necessary to use the key combination Option + Shift + 8 to add this symbol.
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# Working With Number – Infinity Multiplication – Decimal, Precise Float Calculation – Ruby
In the previous chapter of Infinity Multiplication, the chapter mainly concentrated onto the subject of storing two integer numbers in string format and how to apply a mathematical multiplication equation on the two numbers at the rate of one digit at a time. The previous chapter mainly discussed in regard to whole numbers, numbers that do not contain a fractional part. For this chapter, the subject is about how to apply a multiplication equation on large numbers that do contain a fractional part.
When computers store a number as an integer value. The value is stored in binary bits block that only store a binary value that represents the actual value of the number. Besides the value of the number, the binary bits block that is used to store the integer number may utilize one leftmost bit to defined if the value is a negative or a positive value, nevertheless, the bits block that is used to store the integer value does not store the fractional part of a number. The reason of why computers can not store a decimal point for numbers that do have the fractional part is because is because of the differential between the two numbering system. Nevertheless, when it comes to programming, a majority of programming languages if not all do have a data type that can store numbers with a fractional part. The most basic and most likely to be a built-in data type that is used by programming languages that can store a fractional part value is usually called a “float”.
When it comes to utilizing mathematical equations on float values, there are many methods of how to calculate a floating point value. There are floating point calculation formulas that do offer very precise answers. Nonetheless, there are circumstances where some floating point calculation formula does round off some of the digits that are on the right side of the decimal of the result value to get a near precision result value for the equation. These type of rounding can be very minor and would most likely have little affection.
Nonetheless, if we truly wanted to calculate a precise result value of numbers that do contain a fractional part and that are longer than thousand of digits, is there a solution to that? This is the second chapter of “Infinity Multiplication”, “Decimal, Precise Float Calculation”.
For a preview on the infinity subject, most of my readers ask me if this is truly an infinity library and is this truly an infinity subject? In my answer, Infinity is a subject that is very hard to explain in words. Since the values of infinity is endless and since it is endless, can we really calculate it? This is my formula for calculating an infinity value “infC = infA = infAL”, of which can also be describe as “Infinity Calculation equal to Infinity Assumption and is equal to Infinity Alteration”. I will discuss in regard to the infinity formula and how to apply the formula in a programming perspective in chapter five of the Infinity Multiplication, Division, Adding and Subtraction subject. Chapter five of the infinity subject will only be available in full at http://kevinhng86.iblog.website, Programming World, and sites that are allowed to posts the articles in full.
For the next chapter of Infinity Multiplication, which is chapter three and that is scheduled to be released this week. The chapter will contain the optimized source code of infiX, which can be used for a production environment. With the source code, I will also release the tester code of which I used for testing hundreds of thousand of test cases against the infiX program. The only code that I do not release with the article is the benchmark testing source code, of which tests how fast the program run versus 100,000 or 1,000,000 cases.
This is the Ruby version of “Infinity Multiplication – Decimal, Precise Float Calculation”.
Previous Chapter: Beyond Integer Next Chapter: Optimised The Code
#### Precise Calculation With Fractional Part And Multiplication
For the purpose of simplification. This article explanations are base on a multiplication equation between two numbers that were stored in strings format. For the first number, this article defined the number as number A, string A, or simply A. For the second number, this article defined the number as number B, string B, or just B.
When applying a multiplication equation onto two string of digits and they do contain a fractional part then the first challenge we would face is the value of the fractional part, or in other words, the digits that are behind the decimal. For an addition or a subtraction equation, we can split the string into two parts. One part can hold the digits that are before the decimal and the other part can store the digits that are behind the decimal. Nonetheless, in my opinion, this technique can be somewhat complicated if to apply to a multiplication equation. This is because of the reason that every digit whether if before or behind the decimal in the first string has to be multiplied to every digit in the second string.
For the circumstance that we have two string of digits, one of them is A and the other one is B and if we were to split each string into two parts. Then it can get a bit complicated to manipulate each part of A string to properly multiply to each part of B string. It is doable, nevertheless, just a bit complicated. On the other hand, we do have another simple solution, of which we can remove the decimal from both A and B string. Then we would treat the strings as if they did not contain any decimal placement. One we completed the multiplication equation then we can inject the decimal into the answer string at the correct position.
In my opinion, the method of injecting the decimal into the answer string is highly efficient, is error proof, and is simpler for loop control. This method is doable for a multiplication equation due to the property that attaches to the decimal placement of a multiplication equation. Nevertheless, let first inspect the formula for calculating numbers that have a fractional part in a mathematical multiplication equation.
Formula: Multiplication formula for numbers with a fractional part.
Step 1: Count the number of digits that are behind the decimal in A string and that are not trailing zeroes. Remove trailing zeroes.
Step 2: Count the number of digits that are behind the decimal in B string and that are not trailing zeroes. Remove trailing zeroes.
Step 3: Sum the number of digits that are behind the decimal in A string to the number of digits that are behind the decimal in B string.
This value is defined as the total amount of digits that need to be behind the decimal in the answer string.
Step 4: Remove the decimal from both A and B string, then apply the multiplication procedures on A and B string.
Step 5: One the answer string is ready, starting from right to left of the answer string.
Using the value of step three from above and place the decimal in front of the total amount of digits that need to be behind the decimal in the answer string.
If the answer string contains fewer digits than the value of step three, add zeroes to the left side of the answer string until the amount of digit in the answer string is equal to the value from step three.
Answer: After placing the decimal, we can remove leading and trailing zeroes.
Leading zeroes that are before the decimal do not alter the value of the number.
Trailing zeroes that are behind the decimal do not offer any values to the number.
Example: 0.1 x 0.1
Step 1: (1 digit behind the decimal in A) + (1 digit behind the decimal in B) = (2 digits that need to be behind the decimal in the answer string)
Step 2: 1 x 1 = 1 | Answer string: 1
Step 3: Answer string has only 1 digit, nevertheless, there are 2 digits that need to be behind the decimal in the answer string. Therefore, add one zero to beginning of the answer string
Step 4. .01 | add a zero in front of the decimal for formatting | answer string = 0.01
Answer: 0.1 * 0.1 = 0.01
Example: 9.125 x 33.100
Step 1: (3 digits behind the decimal in A) + (1 digits behind the decimal in B) = (4 digits that need to be behind the decimal in the answer string)
Step 2: 9125 x 331 = 3020375 | Answer string: 3023075
Step 3: Answer string has more digits than the total digit that needs to be behind the decimal. Therefore, no need to modify the answer string.
Step 4. 302.0375
Answer: 9.125 * 33.100 = 302.0375
Example: 152 x 239.55
Step 1: (0 digits behind the decimal in A) + (2 digits behind the decimal in B) = (2 digits that need to be behind the decimal in the answer string)
Step 2: 152 x 23955 = 3641160 | Answer string: 3023075
Step 3: Answer string has more digits than the total digit needs to be behind the decimal in the answer string. Therefore, no need to modify the answer string.
Step 4. 36411.60 | remove the trailing zero for formatting | answer string is 36411.6
Answer: 152 * 239.55 = 36411.6
With the formula above in perspective, we can say that calculating the fractional part of a number in a multiplication equation can be done through indexing the decimal point. In a form of definition, this method of calculation is more of a string manipulation method. With this method, the placement for the decimal is calculated before calculating an actual result value. In my opinion, this method is the simplest method for calculating fractional part values for multiplication equations. Also, this method offers the best precision placement for the decimal. This is because this method calculates the exact decimal’s location for the answer string.
Besides the procedures, there are shortcuts that are available for our multiplication procedures. One of the shortcuts is if a number is multiplied by a value of one, then the equation will return the number as the result value. Currently, there is one shortcut that I decided to discontinue the support for in my infiX program and also in the explanation for the “Infinity Multiplication” series. That shortcut is when a value is multiplied by a value of zero. In that event, the equation will produce a value of zero as the result value. Due to efficiency on large scale multiplications, there is another shortcut that I do not apply to this formula. That shortcut is when a number multiplied to a value of .1 or .01 and so on then we will only need to reposition the decimal.
Let first inspect some of the code blocks that we would need for a multiplication equation that supported fractional part calculation.For the first procedure, we will need to evaluate the strings, A and B, for whether or not they do contain a decimal. If the decimal is found, we will need the position of the decimal. If the decimal is not found then it is preferable that a value of “-1” is assigned to the variable that was defined to hold the value of the position of the decimal. Also, another variable that needs to be declared to hold the value of the total amount of digit that needs to be behind the decimal for the answer string. For Ruby, we can use the strpos() function on a string to get the position of the decimal in that string. If the string doesn’t contain a decimal then the strpos() function will return a value of “-1”.
\$bDecPos = strpos(\$b, ".");
\$oDecPos = 0;
To calculate how many digits that are behind the decimal in either A or B string, we subtract the length of the string to the position of the decimal. In my formula, before calculating the number of digits that are behind the decimal, we first remove any trailing zeroes in the string. This is why it is important that the code block below only execute when there is a decimal in the string. Since we are removing trailing zeroes, we can have issues with strings that contain only zeroes that are behind the decimal. Thus, if the equation of the total length of the string subtracts to the decimal’s position produce a negative result value, then we have to convert that negative value into a value to zero. In term of definition, a value of zero in this scenario also means that there aren’t any digits that are behind the decimal. After we had calculated the number of digits that are behind the decimal, we then remove the decimal.
One important key point that we have to remember here is at the beginning, variable aDecPos held a value that is the position of the decimal in the string. Nevertheless, after this code block below is executed, variable aDecPos would now store a value that representing the number of digits that are behind the decimal in A string. This code block below demonstrates the procedures of the previous explanations.
_a = _a.gsub(/0+\$/, "")
_a = _a.gsub(/[.]/, "")
end
Since we removed the decimal and any trailing zeroes, if we want to use the shortcut of when a number is multiplied by one, then we would have to modify our code from chapter 1 to accommodate the shortcut. This code block below will only execute in the condition that A string contained only a digit of one and A string does not contain any digits that are behind the decimal.
When it comes to returning the result value for this shortcut, if B string did contain a decimal then we placed the decimal back into B string at the correct position and then return B string as the result value. To place the decimal back into the correct position for B string, we get the digits from the first position to the position of where the decimal is going to be placed. To calculate where the decimal is going to be placed, we get the total length of B string and subtract the value to the number of digits that B has after the decimal. Then we add the decimal into the string and then we add the rest of the digits. After placing the decimal and if the first character in B string is the decimal, then we can place a zero in front of the decimal for formatting purposes. This is to correct input format such as “.123”, “.5774”, or “.1111”. Also, it is because we removed leading zeroes that were before the decimal. Therefore, input values such as “0.123”, “0.5774”, or “0.1111” would produce the same scenario as the previous.
Before returning the result value, we evaluate for whether or not A and B are different in then positive and the negative base. If they are, we can place a negative sign in front of B string and return B string as the result value. Otherwise, we can just return B string without any negative or positive sign.
Reverse back to when this procedure just started and in the event where B string did not contain a decimal, we can just simply return B string as the result value. Nonetheless, we do have to check for the differential in the negative and the positive base for both A and B string and we do have to place the negative sign in front of B string when it is necessary to do so. This code block below demonstrated the above procedures.
if (_a == "1" && _aDecPos < 1)
if ( _bDecPos > 0)
_b = _b[0...(_b.length - _bDecPos )] + "." + _b[(_b.length - _bDecPos)..._b.length]
_b = _b[0] == "."? "0" + _b: _b
return ((_isaNeg != _isbNeg)? "-" + _b : _b)
end
return ((_isaNeg != _isbNeg)? "-" + _b : _b)
end
After evaluating the above shortcut, we can then calculate the number of digits that need to be behind the decimal in the answer string. The condition for this code block to execute is either A or B has to have a decimal. In the event where A was found with a decimal and B did not have a decimal then the value for variable bDecPos would be a value of -1. Therefore, if B did not have a decimal, we can assign to the variable bDecPos a value of zero. We also have to remember that if there wasn’t a decimal in B string, then our code block for calculating how many digits that supposed to be behind the decimal in B string from further above should not have been executed.
At this point in the program, a value of zero in bDecPos would also mean that there is not a single digit that is behind the decimal in B string. We can then add up the total amount of digits that are behind the decimal in both A and B string together to get the total amount of digits that need to be behind the decimal in the answer string. This code block below demonstrates the above procedures.
if (_aDecPos > -1 || _bDecPos > -1)
_bDecPos = (_bDecPos > -1)? _bDecPos : 0
end
In this chapter, I did not include the explanation for the multiplication procedures as chapter 1 did contain a very detail explanation of the procedures. Therefore, I would now fast forward to the event of when we got a result value from the multiplication procedures. After the multiplication procedures and when we have an answer string, we can evaluate and place the decimal into the answer string and at the correct position. For placing decimal’s procedure, we first check to see whether if the answer string does contain fewer digits than the number of digits that need to be behind the decimal. If this happens then we have to pad zeroes to the left of the answer string until the answer string has an amount of digits that is equal to the number of digits that need to be behind the decimal. This circumstance occurs in events where “0.01 x 0.01” or “0.3 x 0.3” and so on.
To replace the answer string’s value with a value that includes a fractional part, first, we would grab the digits from the first position in the answer string to where the decimal is going to be placed. To calculate where the decimal is going to be placed, we subtract the length of the answer string to the number of digits that need to be behind the decimal. Second, we place the decimal behind those digits. After placing the decimal into the correct position, we can add the rest of the digits to the value. This new value is now assigned as the value of our answer string.
We then trim off any leading and trailing zeroes in the answer string. If the decimal is the leftmost character in the string, we would also remove the decimal. If the decimal is the first character in the answer string then we would place a zero to the left of the decimal.
This code block below demonstrates the previously mentioned procedures in Ruby.
if (_oDecPos > 0)
_output = (_output.length < _oDecPos)? "0" + _output : _output
_output = _output[0...(_output.length - _oDecPos)] + "." + _output[(_output.length - _oDecPos)..._output.length]
_output = _output.gsub(/0+\$/,"")
_output = _output.gsub(/[.]\$/, "")
_output = _output.gsub(/^0+/, "")
_output = _output[0] == "."? "0" + _output : _output
end
This below is the fully functional code for this article which was written in Ruby. The code is capable of calculating numbers, numbers of which can contain more than thousand of digits and whether or not if the numbers have a fractional part. Although this code is not meant for a production environment, it can calculate the fractional part/floating point value very precisely.
# Written by Kevin Ng
# The full tutorial on this subject can be found @ http://kevinhng86.iblog.website
# This source code file is a part of Kevin Ng Z library.
# This function only multiply and does not check to see if it is only number in the string, you must build a checker around it.
# Notice: Version one and two of any infinity code from the libZ library are prototype.
# They are not meant for production environment due to efficentcy.
# Although are prototype these script were tested and ran through 300,000+ test cases without producing any errors.
# This is version two of infinity multiplication for Ruby, with precise fractional part calculation.
class LibZ
def self.infiX(_a,_b)
_isaNeg = _a[0] == "-"? 1 : 0
_isbNeg = _b[0] == "-"? 1 : 0
_a = _a.gsub(/^[+-]/,"")
_b = _b.gsub(/^[+-]+/,"")
_a = _a.gsub(/^0+/, "")
_b = _b.gsub(/^0+/, "")
_bDecPos = _b.index(".") || -1
_oDecPos = 0
_a = _a.gsub(/0+\$/, "")
_a = _a.gsub(/[.]/, "")
end
if (_bDecPos > -1)
_b = _b.gsub(/0+\$/, "")
_bDecPos = (_b.length - 1 - _bDecPos) > 0? (_b.length - 1 - _bDecPos) : 0
_b = _b.gsub(/[.]/, "")
end
if ( _a.length < 1 or _b.length < 1 )
return "0"
end
if (_a == "1" && _aDecPos < 1)
if (_bDecPos > 0)
_b = _b[0...(_b.length - _bDecPos)] + "." + _b[(_b.length - _bDecPos)..._b.length]
_b = _b[0] == "."? "0" + _b : _b
return ((_isaNeg != _isbNeg)? "-" + _b : _b)
end
return ((_isaNeg != _isbNeg)? "-" + _b : _b)
end
if (_b == "1" && _bDecPos < 1)
_a = _a[0...(_a.length - _aDecPos )] + "." + _a[(_a.length - _aDecPos)..._a.length]
_a = _a[0] == "."? "0" + _a : _a
return ((_isaNeg != _isbNeg)? "-" + _a : _a)
end
return ((_isaNeg != _isbNeg)? "-" + _a : _a)
end
if (_aDecPos > -1 || _bDecPos > -1)
_bDecPos = (_bDecPos > -1)? _bDecPos : 0
end
_temp = 0
_outposition = 0
_alen = _a.length
_blen = _b.length
_output = ""
for _i in (_blen - 1).downto(0)
_z = _b[_i].to_i
_carryOverM = 0
_outtemp = ""
_aidx = _alen - 1
while (_aidx > -1) do
_temp = _a[_aidx].to_i * _z + _carryOverM
_carryOverM = (_temp > 9)? _temp.to_s[0..-2].to_i : 0
_outtemp = _temp.to_s[-1] + _outtemp;
_aidx = _aidx - 1
end
_outtemp = (_carryOverM > 0)? _carryOverM.to_s + _outtemp : _outtemp;
if (_output.length < 1)
_output = _outtemp
_outtemp = ""
_outposition = _outposition + 1
else
_carryOverA = 0
_outlen = _output.length - 1
_outidx = 0
_cposition = 0
_x = 0
_y = 0
for _idx in (_outtemp.length - 1).downto(0)
_cposition = (_outlen - _outidx - _outposition)
_x = _outtemp.to_s[_idx].to_i
_y = (_cposition > -1 )? _output.to_s[_cposition,1].to_i : 0
_tempadd = _x + _y + _carryOverA
_carryOverA = (_tempadd > 9)? 1 : 0
_outidx = _outidx + 1
end
_output = _tempaddstr + _output[(_output.length - _outposition)..-1]
_outtemp = ""
_outposition = _outposition + 1
end
end
if (_oDecPos > 0)
_output = (_output.length < _oDecPos)? "0" + _output : _output
_output = _output[0...(_output.length - _oDecPos)] + "." + _output[(_output.length - _oDecPos)..._output.length]
_output = _output.gsub(/0+\$/,"")
_output = _output.gsub(/[.]\$/, "")
_output = _output.gsub(/^0+/, "")
_output = _output[0] == "."? "0" + _output : _output
end
if (_isaNeg != _isbNeg)
_output = "-" + _output
end
return _output
end
end
x = ".999999999999990000000000000000000000000000"
y = "-0000000000000000000000000000.99999999999999"
puts LibZ.infiX(x,y)
| 5,369 | 22,273 |
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# Search by Topic
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### There are 92 results
Broad Topics > Numbers and the Number System > Factors and multiples
### Factor Lines
##### Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Have You Got It?
##### Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
### A First Product Sudoku
##### Age 11 to 14 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### How Old Are the Children?
##### Age 11 to 14 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
### Got it for Two
##### Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### American Billions
##### Age 11 to 14 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
### Ben's Game
##### Age 11 to 14 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Product Sudoku
##### Age 11 to 14 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Got It
##### Age 7 to 14 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Special Sums and Products
##### Age 11 to 14 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Diggits
##### Age 11 to 14 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
##### Age 11 to 14 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### The Remainders Game
##### Age 7 to 14 Challenge Level:
Play this game and see if you can figure out the computer's chosen number.
### Factor Track
##### Age 7 to 14 Challenge Level:
Factor track is not a race but a game of skill. The idea is to go round the track in as few moves as possible, keeping to the rules.
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Factors and Multiples Game for Two
##### Age 7 to 14 Challenge Level:
Factors and Multiples game for an adult and child. How can you make sure you win this game?
##### Age 11 to 14 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Eminit
##### Age 11 to 14 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Hot Pursuit
##### Age 11 to 14 Challenge Level:
I added together the first 'n' positive integers and found that my answer was a 3 digit number in which all the digits were the same...
### AB Search
##### Age 11 to 14 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### Even So
##### Age 11 to 14 Challenge Level:
Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?
### Counting Factors
##### Age 11 to 14 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Diagonal Product Sudoku
##### Age 11 to 16 Challenge Level:
Given the products of diagonally opposite cells - can you complete this Sudoku?
### Remainders
##### Age 7 to 14 Challenge Level:
I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
### Factoring Factorials
##### Age 11 to 14 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Factors and Multiples - Secondary Resources
##### Age 11 to 16 Challenge Level:
A collection of resources to support work on Factors and Multiples at Secondary level.
### Stars
##### Age 11 to 14 Challenge Level:
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
### Star Product Sudoku
##### Age 11 to 16 Challenge Level:
The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.
### LCM Sudoku II
##### Age 11 to 18 Challenge Level:
You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
### Sieve of Eratosthenes
##### Age 11 to 14 Challenge Level:
Follow this recipe for sieving numbers and see what interesting patterns emerge.
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Gabriel's Problem
##### Age 11 to 14 Challenge Level:
Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?
### Satisfying Statements
##### Age 11 to 14 Challenge Level:
Can you find any two-digit numbers that satisfy all of these statements?
### Cuboids
##### Age 11 to 14 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours house numbers. Can you explain the patterns I noticed?
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Divisively So
##### Age 11 to 14 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Digat
##### Age 11 to 14 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
### Ewa's Eggs
##### Age 11 to 14 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
### Two Much
##### Age 11 to 14 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### One to Eight
##### Age 11 to 14 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Powerful Factorial
##### Age 11 to 14 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Power Crazy
##### Age 11 to 14 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
### Remainder
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Number Rules - OK
##### Age 14 to 16 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
### Inclusion Exclusion
##### Age 11 to 14 Challenge Level:
How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5?
### Helen's Conjecture
##### Age 11 to 14 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Shifting Times Tables
##### Age 11 to 14 Challenge Level:
Can you find a way to identify times tables after they have been shifted up or down?
| 2,248 | 8,948 |
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| 4.21875 | 4 |
CC-MAIN-2019-35
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http://www.ck12.org/book/CK-12-Basic-Algebra-Concepts/r13/section/7.0/
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| 503,037,857 | 33,010 |
Chapter 7: Systems of Equations and Inequalities
Difficulty Level: Basic Created by: CK-12
Introduction
James is trying to expand his pastry business to include cupcakes and personal cakes. He has a limited amount of manpower to decorate the new items and a limited amount of material to make the new cakes. In this chapter, you will learn how to solve this type of situation.
Every equation and inequality you have studied thus far is an example of a system. A system is a set of equations or inequalities with the same variables. This chapter focuses on the methods used to solve a system, such as graphing, substitution, and elimination. You will combine your knowledge of graphing inequalities to solve a system of inequalities.
Chapter Outline
Summary
This chapter discusses linear systems of equations and the methods used to solve them, including graphing, substitution, addition, subtraction, and multiplication. Mixture problems are covered in detail, and a distinction is made between consistent and inconsistent linear systems. In addition, instruction is given on linear inequalities and linear programming, and the chapter concludes by talking about probability, permutations, and combinations.
Systems of Equations and Inequalities; Counting Methods Review
1. Match the following terms to their definitions.
1. System – Restrictions imposed by time, materials, or money
2. Feasible Region – A system with an infinite number of solutions
3. Inconsistent System – An arrangement of objects when order matters
4. Constraints – A method used by businesses to determine the most profit or least cost given constraints
5. Consistent-dependent system – An arrangement of objects in which order does not matter
6. Permutation – Two or more algebraic sentences joined by the word and
7. Combination – A system with no solutions
8. Linear programming - A solution set to a system of inequalities
1. Where are the solutions to a system located?
2. Suppose one equation of a system is in slope-intercept form and the other is in standard form. Which method presented in this chapter would be the most effective to solve this system? Why?
3. Is (–3, –8) a solution to \begin{align*}\begin{cases} 7x-4y=11\\ x+2y=-19 \end{cases}\end{align*}?
4. Is (–1, 0) a solution to \begin{align*}\begin{cases} y=0\\ 8x+7y=8 \end{cases}\end{align*}?
Solve the following systems by graphing.
1. \begin{align*}\begin{cases} y=-2\\ y=-6x+1 \end{cases}\end{align*}
2. \begin{align*}\begin{cases} y=3-\frac{1}{3} x\\ x+3y=4 \end{cases}\end{align*}
3. \begin{align*}\begin{cases} y=\frac{1}{2} x-6\\ 4y=2x-24 \end{cases}\end{align*}
4. \begin{align*}\begin{cases} y=-\frac{4}{5} x+7\\ y=\frac{2}{5} x+1 \end{cases}\end{align*}
5. \begin{align*}\begin{cases} x=2\\ y=4\\ y=\frac{1}{2} x+3 \end{cases}\end{align*}
Solve the following systems by substitution.
1. \begin{align*}\begin{cases} y=2x-7\\ y+7=4x \end{cases}\end{align*}
2. \begin{align*}\begin{cases} y=-3x+22\\ y=-2x+16 \end{cases}\end{align*}
3. \begin{align*}\begin{cases} y=3-\frac{1}{3} x\\ x+3y=4 \end{cases}\end{align*}
4. \begin{align*}\begin{cases} 2x+y=-10\\ y=x+14 \end{cases}\end{align*}
5. \begin{align*}\begin{cases} y+19=-7x\\ y=-2x-9 \end{cases}\end{align*}
6. \begin{align*}\begin{cases} y=0\\ 5x=15 \end{cases}\end{align*}
7. \begin{align*}\begin{cases} y=3-\frac{1}{3}x\\ x+3y=4 \end{cases}\end{align*}
8. \begin{align*}\begin{cases} 7x+3y=3\\ y=8 \end{cases}\end{align*}
Solve the following systems using elimination.
1. \begin{align*}\begin{cases} 2x+4y=-14\\ -2x+4y=8 \end{cases}\end{align*}
2. \begin{align*}\begin{cases} 6x-9y=27\\ 6x-8y=24 \end{cases}\end{align*}
3. \begin{align*}\begin{cases} 3x-2y=0\\ 2y-3x=0 \end{cases}\end{align*}
4. \begin{align*}\begin{cases} 4x+3y=2\\ -8x+3y=14 \end{cases}\end{align*}
5. \begin{align*}\begin{cases} -8x+8y=8\\ 6x+y=1 \end{cases}\end{align*}
6. \begin{align*}\begin{cases} 7x-4y=11\\ x+2y=-19 \end{cases}\end{align*}
7. \begin{align*}\begin{cases} y=-2x-1\\ 4x+6y=10 \end{cases}\end{align*}
8. \begin{align*}\begin{cases} x-6y=20\\ 2y-3x=-12 \end{cases}\end{align*}
9. \begin{align*}\begin{cases} -4x+4y=0\\ 8x-8y=0 \end{cases}\end{align*}
10. \begin{align*}\begin{cases} -9x+6y=-27\\ -3x+2y=-9 \end{cases}\end{align*}
Graph the solution set to each system of inequalities.
1. \begin{align*}\begin{cases} y>-\frac{3}{5} x-5\\ y\ge-2x+2 \end{cases}\end{align*}
2. \begin{align*}\begin{cases} y>\frac{13}{8} x+8\\ y\ge \frac{1}{4} x-3 \end{cases}\end{align*}
3. \begin{align*}\begin{cases} y\le \frac{3}{5} x-5\\ y\ge-2x+8 \end{cases}\end{align*}
4. \begin{align*}\begin{cases} y\le-\frac{7}{5} x-3\\ y\ge \frac{4}{5} x+4 \end{cases}\end{align*}
5. \begin{align*}\begin{cases} x<5\\ y\ge\frac{9}{5} x \end{cases}\end{align*} \begin{align*}-2\end{align*}
Write a system of inequalities for the regions below.
1. Yolanda is looking for a new cell phone plan. Plan A charges $39.99 monthly for talking and$0.08 per text. Plan B charges $69.99 per month for an “everything” plan. 1. At how many texts will these two plans charge the same? 2. What advice would you give Yolanda? 1. The difference of two numbers is –21.3. Their product is –72.9. What are the numbers? 2. Yummy Pie Company sells two kinds of pies: apple and blueberry. Nine apple pies and 6 blueberry pies cost$126.00. 12 apples pies and 12 blueberry pies cost $204.00. What is the cost of one apple pie and 2 blueberry pies? 3. A jet traveled 784 miles. The trip took seven hours, traveling with the wind. The trip back took 14 hours, against the wind. Find the speed of the jet and the wind speed. 4. A canoe traveling downstream takes one hour to travel 7 miles. On the return trip, traveling against current, the same trip took 10.5 hours. What is the speed of the canoe? What is the speed of the river? 5. The yearly musical production is selling two types of tickets: adult and student. On Saturday, 120 student tickets and 45 adult tickets were sold, bringing in$1,102.50 in revenue. On Sunday, 35 student tickets and 80 adult tickets were sold, bringing in $890.00 in revenue. How much was each type of ticket? 6. Rihanna makes two types of jewelry: bracelets and necklaces. Each bracelet requires 36 beads and takes 1 hour to make. Each necklace requires 80 beads and takes 3 hours to make. Rihanna only has 600 beads and 20 hours of time. 1. Write the constraints of this situation as a system of inequalities. 2. Graph the feasible region and locate its vertices. 3. Rihanna makes$8.00 in profit per bracelet and $7.00 in profit per necklace. How many of each should she make to maximize her profit? 7. A farmer plans to plant two type of crops: soybeans and wheat. He has 65 acres of available land. He wants to plant twice as much soybeans as wheat. Wheat costs$30 per acre and soybeans cost 30 per acre. 1. Write the constraints as a system of inequalities. 2. Graph the feasible region and locate its vertices. 3. How many acres of each crop should the farmer plant in order to minimize cost? 8. How many ways can you organize 10 items on a shelf? 9. Evaluate 5! 10. Simplify \begin{align*}\frac{100!}{97!}\end{align*}. 11. How many ways can a football team of 9 be arranged if the kicker must be in the middle? 12. How many one-person committees can be formed from a total team of 15? 13. How many three-person committees can be formed from a total team of 15? 14. There are six relay teams running a race. How many different combinations of first and second place are there? 15. How many ways can all six relay teams finish the race? 16. Evaluate \begin{align*}\binom{14}{12}\end{align*}. 17. Evaluate \begin{align*}\binom{8}{8}\end{align*} and explain its meaning. 18. A baked potato bar has 9 different choices. How many potatoes can be made with four toppings? 19. A bag contains six green marbles and five white marbles. Suppose you choose two without looking. What is the probability that both marbles will be green? 20. A principal wants to make a committee of four teachers and six students. If there are 22 teachers and 200 students, how many different committees can be formed? Systems of Equations and Inequalities; Counting Methods Test 1. True or false? A shorter way to write a permutation is \begin{align*}\binom {n}{k} \end{align*}. 2. Is (–17, 17) a solution to \begin{align*}\begin{cases} y=\frac{1}{17} x+18\\ y=-\frac{21}{17} x-4 \end{cases}\end{align*}? 3. What is the primary difference between a combination and a permutation? 4. An airplane is traveling a distance of 1,150 miles. Traveling against the wind, the trip takes 12.5 hours. Traveling with the wind, the same trip takes 11 hours. What is the speed of the plane? What is the speed of the wind? 5. A solution set to a system of inequalities has two dashed boundary lines. What can you conclude about the coordinates on the boundaries? 6. What does \begin{align*}k\end{align*} have to be to create a dependent-consistent system? \begin{align*}\begin{cases} 5x+2y=20\\ 15x+ky=60 \end{cases}\end{align*} 7. Joy Lynn makes two different types of spring flower arrangements. The Mother’s Day arrangement has 8 roses and 6 lilies. The Graduation arrangement has 4 roses and 12 lilies. Joy Lynn can use no more than 120 roses and 162 lilies. If each Mother’s Day arrangement costs32.99 and each Graduation arrangement costs \$27.99, how many of each type of arrangement should Joy Lynn make to make the most revenue?
8. Solve the system \begin{align*}\begin{cases} -6x+y=-1\\ -7x-2y=2 \end{cases}\end{align*}.
9. Solve the system\begin{align*}\begin{cases} y=0\\ 8x+7y=8 \end{cases}\end{align*}.
10. Solve \begin{align*}\begin{cases} y=x+8\\ y=3x+16 \end{cases}\end{align*}.
11. How many solutions does the following system have? \begin{align*}\begin{cases} y=-2x-2\\ y=-2x+17 \end{cases}\end{align*}
12. The letters to the word VIOLENT are placed into a bag.
1. How many different ways can all the letters be pulled from the bag?
2. What is the probability that the last letter will be a consonant?
1. Suppose an ice cream shop has 12 different topping choices for an ice cream sundae. How many ways can you choose 5 of the 12 toppings?
2. A saleswoman must visit 13 cities exactly once without repeating. In how many ways can this be done?
Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9617.
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# Discrete Convolution and the Discrete Fourier Transform
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### Discrete Convolution and theDiscrete Fourier Transform
Discrete Convolution First of all we need to introduce what we might call the “wraparound” convention. Because the complex num- bers wj = ei2πjN have the property wj±N = wj, which readily ex- tends to wj+m N = wj for any integer m, and since in the discrete Fourier context we represent all N -dimensional vectors as linear com- binations of the Fourier vectors Wk whose components are wkj, we make the convention that for any vector Z ∈ EN with components zk, k = 1, 2,· · · , N − 1, if we write zk with k lying outside the range 0 through N − 1 what is meant is zk+m N where m is the unique integer such that k + m N does lie in this range. Thus, for example, if N = 8 and we write z−13, what we mean is z−13+2×8 = z3. This convention may initially seem rather strange and arbitrary but, in fact, it is quite essential for effective use of and computation with the discrete Fourier transform.
Now suppose we have two real or complex vectors in EN: Z = ( z0 z1 · · · zN−1)t, Y = ( y0 y1 · · · yN−1)t.
The discrete convolution of these two vectors is another vector, which we denote by Z ∗ Y , defined componentwise by
(Z ∗ Y )k =
N−1
j=0 zk−j yj, k = 0, 1, 2, ....
In making this definition we have to take into account the fact that if j > k the index k−j will be negative. According to the convention just introduced, what is meant instead of zk−j is zk−j+N. Setting = k − j we can see that
(Z ∗ Y )k =
N−1
j=0 zk−j yj =
k−N+1
=k
zyk− =
N−1
=0
yk−z = (Y ∗ Z)k,
(2)
the second inequality again following from the wraparound convention.
As a consequence we conclude that discrete convolution is commutative:
Z ∗ Y = Y ∗ Z.
Now we compute some discrete convolutions.
Example 1 If Z = ( 1 0 · · · 0 )t, then for any Y we have (Z ∗ Y )k = N−1
j=0
zk−jyj = yk
because zk−j = 0 when k = j and equals 1 when k = j. Thus the vector Z as shown here plays the role of the convolution identity.
Example 2 If Z = ( 1 1 · · · 1 )t, then for any Y we have (Z ∗ Y )k =
N−1
j=0 zk−jyj =
N−1
j=0 yj ≡ sY
and thus with Z as shown
Z ∗ Y = ( sY sY · · · sY )t = (sY)Z.
In particular, Z ∗ Z = N Z in this case.
Example 3 Let
Z = ( 3 2 1 0 · · · 0 1 2 )t for N ≥ 5 and let
Y = ( 0 · · · 0 1 0 1 0 · · · 0 )t
for the same value of N ; let us suppose the two 1’s are in positions k−1 and k + 1. In this case we can verify that Z ∗ Y has the form
Z ∗ Y = ( 0 · · · 0 1 2 4 4 4 2 1 0 · · · 0 )t;
the effect of the convolution is to ”blur” the two distinct 1’s into the extended pattern shown.
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Just as the importance of the Laplace transform derives in large part from its behavior with respect to convolution of functions on [0,∞), the significance of the discrete Fourier transform also largely derives from its relationship to discrete convolution.
Theorem 1 For any two N -dimensional complex vectors Z and Y we have
F(Z ∗ Y ) = N F(Z) ⊗ F(Y )
where the product F(Z)⊗F(Y ) indicates the vector obtained with com- ponent by component multiplication; i.e.,
(F(Z) ⊗ F(Y ))k = (F(Z))k(F(Y ))k,
the right hand side being the ordinary numerical product of complex numbers.
Proof We compute F(Z ∗ Y )k = 1
N
N−1
j=0 w−kj(Z ∗ Y )j = 1 N
N−1
j=0 w−kj(
N−1
=0
zj−y)
= 1 N
N−1 j=0
N−1
=0
w−k(j−)zj−w−ky
= 1 N
N−1
=0
w−ky(
N−1 j=0
w−k(j−)zj−) = (setting m = j − )
= 1 N
N−1
=0
w−ky(
N−1− m=−
w−kmzm) = (taking the “wrap-around” effect into account)
= N ( 1 N
N−1
=0
w−ky) ( 1 N
N−1
m=0 w−kmzm)
= N (F(Y ))k(F(Z))k = (F(Y ) ⊗ F(Z))k,
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which completes the proof.
Convolution Equations A (discrete) convolution equation is an equation of the form A ∗ Y = B, where A and B are known vectors in EN and Y is the vector in EN to be determined. Such an equation is simply a linear algebraic equation of a special type. If we define the N × N matrix
A =
a0 aN−1 aN−2 · · · a1
a1 a0 aN−1 · · · a2
a2 a1 a0 · · · a3
... ... ... ...
aN−1 aN−2 aN−3 · · · a0
,
whose first column is A and whose subsequent columns are “rotations”
of A, one can easily see that the convolution product A∗ Y is the same as the matrix-vector product A Y and the convolution equation takes the form A Y = B. Then we have the following
Proposition 1 The convolution equation A ∗ Y = B is solvable for arbitrary B ∈ EN (equivalently, the matrix A is nonsingular, so that A−1 exists) if and only if the discrete Fourier transform of A has no zero components; i.e.,
(F(A))k = 0, k = 0, 1, 2, ..., N − 1.
Proof From A∗Y = B we obtain, taking the discrete Fourier trans- form,
F(A ∗ Y ) = F(B).
Using the result from Section 3. on the discrete Fourier transform of a convolution product, this becomes
N F(A) ⊗ F(Y ) = F(B).
Looking at the k-th component on each side we have
N (F(A))k(F(Y ))k = (F(B))k, k = 0, 1, 2, ..., N − 1.
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If (F(A))k = 0 for any k then it is clear that the corresponding compo- nent, (F(B))k, of F(B) must also be zero and the convolution equation cannot be solved for arbitrary vectors B. On the other hand, if all of the components of F(A) are different from zero, we simply divide to obtain
(F(Y ))k = (F(B))k N (F(A))k. We will write this vectorially as
F(Y ) = 1
NF(B) F(A),
where the symbol will be used to indicate component by component division of the first vector by the second. Then, applying the inverse transform, we have
Y = F−1( 1
NF(B) F(A)).
This completes the proof of the proposition.
Solving Discrete Convolution Equations
Example 1 We consider the convolution equation in R4:
1 2 0 2
∗ Y =
1
−1−1 1
.
The discrete Fourier transform solution method involves application of the discrete Fourier transform to both sides of this equation and use of the property F(A ∗ Y ) = N F(A) ⊗ F(Y ). First of all, we have
F(A) = 1 4
1 1 1 1
1 −i −1 i
1 −1 1 −1
1 i −1 −i
1 2 0 2
= 1 4
5 1
−3 1
.
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In this case none of the components of F(A) are zero so the solution process is a straightforward one. Computing
F(B) = 1 4
1 1 1 1
1 −i −1 i
1 −1 1 −1
1 i −1 −i
1
−1−1 1
=
0
12 + 2i 0
12 2i
,
and letting Z = F(Y ), we have the equation
F(A ∗ Y ) = 4 F(A) ⊗ Z =
5 1
−3 1
z0 z1 z2 z3
=
0
12 + 2i 0
12 2i
.
This is solved readily to give
Z =
0
12 + 2i 0
12 2i
.
Then, applying the inverse transform, we have
Y = F−1(Z) =
1 1 1 1
1 i −1 −i
1 −1 1 −1
1 −i −1 i
0
12 + 2i 0
12 2i
=
1
−1−1 1
.
The fact that the solution Y turns out to be the same as the original right hand side B is a coincidence particular to this example.
Example 2 In this example the solution process is complicated by the fact that one of the components of F(A) turns out to be zero. The convolution equation in question is
1 2 3
−6
∗ Y =
2 d 4 2d
=
2 0 4 0
+ d
0 1 0 2
≡ B0 + d B1,
(7)
where d is a parameter, initially undetermined. Here we have
F(A) = 1 4
1 1 1 1
1 −i −1 i
1 −1 1 −1
1 i −1 −i
1 2 3
−6
=
0
12 − 2i 2
12 + 2i
;
FB0 = 1 4
1 1 1 1
1 −i −1 i
1 −1 1 −1
1 i −1 −i
2 0 4 0
=
32
12
32
12
;
FB1 = 1 4
1 1 1 1
1 −i −1 i
1 −1 1 −1
1 i −1 −i
0 1 0 2
=
34 4i
34
4i
.
Since the transform is linear, letting Z = F(Y ), we have
4
0
12 − 2i 2
12 + 2i
z0 z1 z2 z3
=
32
12
32
12
+ d
34 4i
34
4i
.
From examination of the 0 component on both sides it is clear that we have to take d = −2 for consistency. Then z0 is arbitrary and we have
(−2 − 8i) z1 = 1 2 i
2 ⇒ z1 = 5− 3i 68 ; 8 z2 = 3
2 + 3
2 = 3 ⇒ z2 = 3 8; (−2 + 8i) z3 = 1
2 + i
2 ⇒ z3 = 5 + 3i 68 . Then
Y =
1 1 1 1
1 i −1 −i
1 −1 1 −1
1 −i −1 i
z0
5−3i68 38 5+3i68
= z0
1 1 1 1
+
13671
13639
13631
13663
.
(8)
Our next example shows that the discrete Fourier transform can be applied to solve, at least approximately, certain problems of classical mechanics. The example also shows that zero components of F(A) are not just an inconvenience but may correspond to degrees of freedom in the solution of the convolution equation A ∗ Y = B which we would quite naturally expect to find.
Example 4 A cable is strung between two vertical supporting struc- tures of equal height and is displaced from the horizontal due to the ef- fect of gravity acting on the cable. Letting 0 < x < 1 be the parametriza- tion in the horizontal direction and letting y(x) denote the vertical displacement of the cable, we have
τ d2y
dx2 − g ρ(x) = 0, 0 < x < 1,
where τ is the tension in the cable, g is the gravitational constant and ρ(x) is the cable mass per unit length. It is not difficult to solve this problem analytically but we will carry out an approximate solution as an example of application of Fourier solution of discrete convolution equations.
We discretize the problem by dividing the interval [0, 1) into N subin- tervals [xk−1, xk), k = 1, 2, ..., N , of equal length h = N1 and we replace the unknown displacement function y(x) by an N -dimensional vector Y = [y0, y1, ..., yN−1]T representing approximate displacements at the points xk. Since the supports at the endpoints are of equal height we can identify y0 with yN as we need to do in discrete Fourier analysis.
(In effect, we are assuming a periodic structure consisting of supports of equal height placed one unit apart with identical cables suspended between each adjacent pair of them.) We replace the second derivative operator by the standard second order divided difference approximation
d2y
dx2(xk) yk+1 − 2 yk + yk−1 h2
and we discretize the gravitational force to gk ≡ g ρ(xk) at the points xk, k = 0, 1, 2, ..., N−1. Then we let B be the vector whose components
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are
bk =
g0 − s = g ρ(0) − s, k = 0, gk = g ρ(xk), k = 1, 2, ..., N − 1,
where s is the force exerted by the supporting structure. Then we have τ N2A∗ Y − B = 0 ⇒ A ∗ Y = 1
τ N2 B,
where A = (−2 1 0 ... 0 1 )T, the omitted terms all being zero.
Taking the discrete Fourier transform and using the convolution prop- erty of the transform we have
NF(A) ⊗F(Y ) = 1
τ N2F(B) ⇒ F(Y )k = 1 τ N3
F(B)k
F(A)k, F(A)k = 0.
In this case, because 1 + (−2) + 1 = 0, the component F(A)0 = 0.
For consistency, then, we also must have F(B)0 = 0. But this last component is just g0− s + N−1k=1 gk, so we conclude that we must have
s =
N−1 k=0
gk,
which is just the statement that the magnitude of the supporting force s should be the sum of the magnitudes of he gravitational forces - very much to be expected from the principles of statics. Having made this choice, the component
F(Y )0 = 1 N
N−1 k=0
wk,
the average value of the vertical displacements, is arbitrary. But this is to be expected because we have not specified the height of the sup- porting structures.
For our computational treatment of this problem we will take τ = 1, g = 1, N = 16 and we will suppose that
ρ(x) =
1, 0 < x < 165 , 1116< x < 1, 4, 165 < x < 1116 .
(10)
This gives
B = ( 1− s 1 1 1 1 4 4 4 4 4 4 1 1 1 1 1 ) . The supporting force s is selected so the sum of the components of B is zero, i.e., s = 34. We used MATLABR to compute C = F2561 B ; because of the choice of s the first component of this discrete Fourier transform is zero. There is little point in providing the whole list of the components of C here. We also computed F(A); the first component is again zero and all other components are non-zero. We thus let
F(Y0)0 = 0, F(Y0)k = F(2561 B)k
16F(A)k , k = 1, 2, ..., 15.
At this point all components of F(Y ) have been determined except the first. Thus we can write F(Y ) =
F(Y0) + r ( 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 )t, where r is yet to be determined. Then Y =
Y0 + rF−1(( 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 )t)
= Y0 + r
16 ( 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 )t. It turns out in this example that Y0 =
( .2167 .1500 .0873 .0284 −.0265 −.0775 −.1129 −.1327
−.1368 −.1254 −.0983 −.0555 −.0089 .0416 .0961 .1544 )t. We suppose the height of the supporting posts is 1. Then we set 16r = 1− .2167 = .7833 to realize the value 1 as the first component of Y . To display the right hand supporting post we introduce a new component y16 = y0 = 1 and we plot the result in Figure 1.
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Figure 1
0 0.2 0.4 0.6 0.8 1
0 0.2 0.4 0.6 0.8 1
−−> x
−−> y
Supported Cable as Constructed Using Discrete Fourier Transform
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RETHINK
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RETHINK
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398 results
2012
Level: beginner
The total sum of squares and the total degrees of freedoms are disaggregated by calculating in sample variance and "between" sample variance and their respective degrees of freedoms. It is demonstrated numerically that both these measures add up to the total sum of squares and the total degrees of freedom.
2012
First the global mean is calculated from a matrix of three sets each containing three observations. Then the sum of squares is calculated. Lastly, the concept of degree of freedom is explained.
2013
First some properties about the Sum of squared residuals and the linear regression function are restated. In particular three properties that an ideal fitted regression line must fulfill are discussed. Then, the R squared is defined using the measures of the Sum of squared residuals, the total sum of squares and the sum of explained squares.
2012
Level: beginner
The sum of squares and degree of freedom calculation from the previous videos are put into a ratio to calculate the F Value, on whose basis the null hypothesis is confirmed or rejected. If variance is higher between samples than within the null hypothesis is more likely to be rejected. The results of a numerical example are interpreted more abstractly and then tested with regards to a confidence interval and the corresponding F table.
2016
Level: beginner
In 18th century Europe figures such as Adam Smith, David Ricardo, Friedrich List and Jean Baptiste Colbert developed theories regarding international trade, which either embraced free trade seeing it as a positive sum game or recommended more cautious and strategic approaches to trade seeing it as a potential danger and a rivalry and often as a zero-sum game. What about today?
2011
Level: beginner
The definition of a chi-square distribution is given. Chi-square is defined as the sum of random normally distributed variables (mean=0, variance=s.d.=1). The number of added squared variables is equal to the degrees of freedom. With more degrees of freedom the probability of larger chi-square values is increased.
2013
Level: beginner
First some terminology is explained. Then the interpretations of the coefficients and constants of the function are discussed. Afterwards the zero conditional mean assumption regarding the residual is problematized. Lastly, a graphical representation of a regression line is given and the least sum of squared errors is introduced and the equation for the coefficient of the linear function as well as for the intercept is given.
2019
As opposed to the conventional over-simplified assumption of self-interested individuals, strong evidence points towards the presence of heterogeneous other-regarding preferences in agents. Incorporating social preferences – specifically, trust and reciprocity - and recognizing the non-constancy of these preferences across individuals can help models better represent the reality.
2021
Level: beginner
What’s inflation? Why is it relevant? And is there an agreed theory about its roots and causes, or is it a contentious concept? That’s what this text is all about: We define what inflation actually means before we delve into the theoretical debate with an interdisciplinary and pluralist approach: What gives rise to it, what factors might influence it, and, consequently, what might be done about it?
2023
Level: beginner
In this overview paper, Laura Porak reviews the history of industrial policy in the European Union before the background of a Cultural Political Economy approach.
Neoclassical economics focuses on the allocation of scarce resources. Economic analysis is mainly concerned with determining the efficient allocation of resources in order to increase welfare.
2020
In the history of the social sciences, few individuals have exerted as much influence as has Jeremy Bentham. His attempt to become “the Newton of morals” has left a marked impression upon the methodology and form of analysis that social sciences like economics and political science have chosen as modus operandi.
The core idea of ecological economics is that human economic activity is bound by absolute limits. Interactions between the economy, society and the environment are analysed, while always keeping in mind the goal of a transition towards sustainability.
2020
Level: beginner
Michael Kalecki famously remarked “I have found out what economics is; it is the science of confusing stocks with flows”. Stock-Flow Consistent (SFC) models were developed precisely to address this kind of confusion. The basic intuition of SFC models is that the economy is built up as a set of intersecting balance sheets, where transactions between entities are called flows and the value of the assets/liabilities they hold are called stocks. Wages are a flow; bank deposits are a stock, and confusing the two directly is a category error. In this edition of the pluralist showcase I will first describe the logic of SFC models – which is worth exploring in depth – before discussing empirical calibration and applications of the models. Warning that there is a little more maths in this post than usual (i.e. some), but you should be able to skip those parts and still easily get the picture.
2018
Level: beginner
This essay deals with the concepts of Sustainable Land Management (SLM) and Land Degradation Neutrality (LDN).
2018
Level: beginner
This article, looks at the complex interaction between an urban economy and the vegetation within that urban area. In summary, numerous studies have found a positive link between increased vegetation and social as well as personal health. It makes a case for increasing urban vegetation as a way to benefit local economies.
2020
What influence do changes in tax policy or state decisions on expenditure have on economic growth? For decades, this question has been controversially debated.
2023
Level: beginner
The Philosophy of Economics Foundational Text provides a systematic and well-structured overview over the field of philosophy of economics.
2021
Level: beginner
What made the false assumption that saving the economy at all cost during a pandemic so popular? This paper discusses different pathways through the COVID-19 pandemic at national and international level, and their consequences on the health of citizens and their economies.
2018
How can we establish new institutions and practices in order to use fare-free public transport as a beacon for sustainable mobility and a low-carbon lifestyle? The author of this essay elaborates on how practice theory and institutional economics can help to answer this question.
2020
Level: beginner
This is an overview of (possibly transformative) proposals to address the economic consequences of the corona crisis
2017
Level: beginner
This essay suggests to bring together two aspects of economic thought which so far have developed largely separately: degrowth and feminist economics. In this strive, the concept of care work and its role in feminist economics will be introduced and the downsides of the commodification of care work will be discussed. Subsequently, contributions to the discussion on the (re)valuation of care work will be taken into account.
2016
Level: beginner
A review of: [1] Intermediate Microeconomics, H.R. Varian [2] Mikrooekonomie, R.S. Pindyck, D.L. Rubinfeld [3] Grundzuege der mikrooekonomischen Theorie, J. Schumann, U. Meyer, W. Stroebele
2017
Level: beginner
Due to the economic crisis of 2008/2009, households faced drastic decreases in their incomes, the availability of jobs. Additionally, the structure of the labour market changed, while austerity measures and public spending cuts left households with less support and safeguards provided by the state. How have these developments affected the burden of unpaid labour and what influence did this have on gender relations?
2018
Level: beginner
This essay focuses on the sources of government revenue within the Middle East and North African (MENA) region and proposes the implementation of a regional tax reset through increased taxation and tax reforms, deregulation in the private sector and economic diversification to reduce macroeconomic volatilities caused by the hydrocarbon industry.
2020
Level: beginner
A historical glimpse of how economists of the 19th century debated the usefulness of mathematics to economics
2021
Level: beginner
The last 15 years have seen extensive research into ecosystem service valuation (ESV), spurred by the Millenium Ecosystem Assessment in 2005 (Baveye, Baveye & Gowdy, 2016). Ecosystem services are defined as “the benefits people obtain from ecosystems” (Millenium Ecosystem Assessment, p.V). For example, ecosystems provide the service of sequestering carbon which helps regulate the climate. Valuation means giving ecosystems or their services a monetary price, for example researchers have estimated that the carbon sequestration services of the Mediterranean Sea is between 100 and 1500 million euros per year. The idea of ESV was a response to the overuse of natural resources and degradation of ecosystems, allegedly due to their undervaluation and exclusion from the monetary economy. ESV can be used (1) for policy decision-making, for example allocating funding to a reforestation project (2) for setting payments to people who increase ecosystem services, for example a farmer increasing the organic carbon content of their soil, and (3) for determining fees for people who degrade ecosystem services, for example a company that causes deforestation.
2020
"Despite the rediscovery of the inequality topic by economists as well as other social scientists in recent times, relatively little is known about how economic inequality is mediated to the wider public of ordinary citizens and workers. That is precisely where this book steps in: It draws on a cross-national empirical study to examine how mainstream news media discuss, respond to, and engage with such important and politically sensitive issues and trends.
Post-Keynesians focus on the analysis of capitalist economies, perceived as highly productive, but unstable and conflictive systems. Economic activity is determined by effective demand, which is typically insufficient to generate full employment and full utilisation of capacity.
Evolutionary economics focuses on economic change. Hence processes of change such as growth, innovation, structural and technological change, as well as economic development in general are analysed. Evolutionary economics often gives emphasis to populations and (sub-)systems.
2023
Level: beginner
Mainstream inflation theories in economics do little to explain the recent acceleration in price increases. The associated economic policy recommendations further increase the misery of low-income groups.
Level: beginner
This article outlines the fundamental challenges of democratically planned economies and categorises proposed models into six groups, each of which approaches planning and coordination at different levels of authority and between myriad economic units in a particular way, taking into account efficiency as well as democratic principles and environmental and social sustainability. Through a classification system based on decision-making authority and mediation mechanisms, the article provides a framework for understanding and comparing these models. By examining their different approaches, it offers insights into the complexities and potential paths of democratically planned economies in the 21st century.
## Donate
This project is brought to you by the Network for Pluralist Economics (Netzwerk Plurale Ökonomik e.V.). It is committed to diversity and independence and is dependent on donations from people like you. Regular or one-off donations would be greatly appreciated.
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https://en.wikipedia.org/wiki/Universal_closure
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# Universal quantification
(Redirected from Universal closure)
In predicate logic, a universal quantification is a type of quantifier, a logical constant which is interpreted as "given any" or "for all". It expresses that a propositional function can be satisfied by every member of a domain of discourse. In other words, it is the predication of a property or relation to every member of the domain. It asserts that a predicate within the scope of a universal quantifier is true of every value of a predicate variable.
It is usually denoted by the turned A (∀) logical operator symbol, which, when used together with a predicate variable, is called a universal quantifier ("∀x", "∀(x)", or sometimes by "(x)" alone). Universal quantification is distinct from existential quantification ("there exists"), which asserts that the property or relation holds only for at least one member of the domain.
Quantification in general is covered in the article on quantification (logic). Symbols are encoded U+2200 FOR ALL (HTML ∀ · ∀ · as a mathematical symbol).
## Basics
Suppose it is given that
2·0 = 0 + 0, and 2·1 = 1 + 1, and 2·2 = 2 + 2, etc.
This would seem to be a logical conjunction because of the repeated use of "and". However, the "etc." cannot be interpreted as a conjunction in formal logic. Instead, the statement must be rephrased:
For all natural numbers n, 2·n = n + n.
This is a single statement using universal quantification.
This statement can be said to be more precise than the original one. While the "etc." informally includes natural numbers, and nothing more, this was not rigorously given. In the universal quantification, on the other hand, the natural numbers are mentioned explicitly.
This particular example is true, because any natural number could be substituted for n and the statement "2·n = n + n" would be true. In contrast,
For all natural numbers n, 2·n > 2 + n
is false, because if n is substituted with, for instance, 1, the statement "2·1 > 2 + 1" is false. It is immaterial that "2·n > 2 + n" is true for most natural numbers n: even the existence of a single counterexample is enough to prove the universal quantification false.
On the other hand, for all composite numbers n, 2·n > 2 + n is true, because none of the counterexamples are composite numbers. This indicates the importance of the domain of discourse, which specifies which values n can take.[note 1] In particular, note that if the domain of discourse is restricted to consist only of those objects that satisfy a certain predicate, then for universal quantification this requires a logical conditional. For example,
For all composite numbers n, 2·n > 2 + n
For all natural numbers n, if n is composite, then 2·n > 2 + n.
Here the "if ... then" construction indicates the logical conditional.
### Notation
In symbolic logic, the universal quantifier symbol ${\displaystyle \forall }$ (an inverted "A" in a sans-serif font, Unicode 0x2200) is used to indicate universal quantification.[1]
For example, if P(n) is the predicate "2·n > 2 + n" and N is the set of natural numbers, then:
${\displaystyle \forall n\!\in \!\mathbb {N} \;P(n)}$
is the (false) statement:
For all natural numbers n, 2·n > 2 + n.
Similarly, if Q(n) is the predicate "n is composite", then
${\displaystyle \forall n\!\in \!\mathbb {N} \;{\bigl (}Q(n)\rightarrow P(n){\bigr )}}$
is the (true) statement:
For all natural numbers n, if n is composite, then 2·n > 2 + n
and since "n is composite" implies that n must already be a natural number, we can shorten this statement to the equivalent:
${\displaystyle \forall n\;{\bigl (}Q(n)\rightarrow P(n){\bigr )}}$
For all composite numbers n, 2·n > 2 + n.
Several variations in the notation for quantification (which apply to all forms) can be found in the quantification article. There is a special notation used only for universal quantification, which is given:
${\displaystyle (n{\in }\mathbb {N} )\,P(n)}$
The parentheses indicate universal quantification by default.
## Properties
### Negation
Note that a quantified propositional function is a statement; thus, like statements, quantified functions can be negated. The notation most mathematicians and logicians utilize to denote negation is: ${\displaystyle \lnot \ }$. However, some use the tilde (~).
For example, if P(x) is the propositional function "x is married", then, for a universe of discourse X of all living human beings, the universal quantification
Given any living person x, that person is married
is given:
${\displaystyle \forall {x}{\in }\mathbf {X} \,P(x)}$
It can be seen that this is irrevocably false. Truthfully, it is stated that
It is not the case that, given any living person x, that person is married
or, symbolically:
${\displaystyle \lnot \ \forall {x}{\in }\mathbf {X} \,P(x)}$.
If the statement is not true for every element of the Universe of Discourse, then, presuming the universe of discourse is non-empty, there must be at least one element for which the statement is false. That is, the negation of ${\displaystyle \forall {x}{\in }\mathbf {X} \,P(x)}$ is logically equivalent to "There exists a living person x who is not married", or:
${\displaystyle \exists {x}{\in }\mathbf {X} \,\lnot P(x)}$
Generally, then, the negation of a propositional function's universal quantification is an existential quantification of that propositional function's negation; symbolically,
${\displaystyle \lnot \ \forall {x}{\in }\mathbf {X} \,P(x)\equiv \ \exists {x}{\in }\mathbf {X} \,\lnot P(x)}$
It is erroneous to state "all persons are not married" (i.e. "there exists no person who is married") when it is meant that "not all persons are married" (i.e. "there exists a person who is not married"):
${\displaystyle \lnot \ \exists {x}{\in }\mathbf {X} \,P(x)\equiv \ \forall {x}{\in }\mathbf {X} \,\lnot P(x)\not \equiv \ \lnot \ \forall {x}{\in }\mathbf {X} \,P(x)\equiv \ \exists {x}{\in }\mathbf {X} \,\lnot P(x)}$
### Other connectives
The universal (and existential) quantifier moves unchanged across the logical connectives , , , and , as long as the other operand is not affected; that is:
${\displaystyle P(x)\land (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\land Q(y))}$
${\displaystyle P(x)\lor (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\lor Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
${\displaystyle P(x)\to (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\to Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
${\displaystyle P(x)\nleftarrow (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\nleftarrow Q(y))}$
${\displaystyle P(x)\land (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\land Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
${\displaystyle P(x)\lor (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\lor Q(y))}$
${\displaystyle P(x)\to (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\to Q(y))}$
${\displaystyle P(x)\nleftarrow (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\nleftarrow Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
Conversely, for the logical connectives , , , and , the quantifiers flip:
${\displaystyle P(x)\uparrow (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\uparrow Q(y))}$
${\displaystyle P(x)\downarrow (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\downarrow Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
${\displaystyle P(x)\nrightarrow (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\nrightarrow Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
${\displaystyle P(x)\gets (\exists {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \forall {y}{\in }\mathbf {Y} \,(P(x)\gets Q(y))}$
${\displaystyle P(x)\uparrow (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\uparrow Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
${\displaystyle P(x)\downarrow (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\downarrow Q(y))}$
${\displaystyle P(x)\nrightarrow (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\nrightarrow Q(y))}$
${\displaystyle P(x)\gets (\forall {y}{\in }\mathbf {Y} \,Q(y))\equiv \ \exists {y}{\in }\mathbf {Y} \,(P(x)\gets Q(y)),~\mathrm {provided~that} ~\mathbf {Y} \neq \emptyset }$
### Rules of inference
A rule of inference is a rule justifying a logical step from hypothesis to conclusion. There are several rules of inference which utilize the universal quantifier.
Universal instantiation concludes that, if the propositional function is known to be universally true, then it must be true for any arbitrary element of the universe of discourse. Symbolically, this is represented as
${\displaystyle \forall {x}{\in }\mathbf {X} \,P(x)\to \ P(c)}$
where c is a completely arbitrary element of the universe of discourse.
Universal generalization concludes the propositional function must be universally true if it is true for any arbitrary element of the universe of discourse. Symbolically, for an arbitrary c,
${\displaystyle P(c)\to \ \forall {x}{\in }\mathbf {X} \,P(x).}$
The element c must be completely arbitrary; else, the logic does not follow: if c is not arbitrary, and is instead a specific element of the universe of discourse, then P(c) only implies an existential quantification of the propositional function.
### The empty set
By convention, the formula ${\displaystyle \forall {x}{\in }\emptyset \,P(x)}$ is always true, regardless of the formula P(x); see vacuous truth.
## Universal closure
The universal closure of a formula φ is the formula with no free variables obtained by adding a universal quantifier for every free variable in φ. For example, the universal closure of
${\displaystyle P(y)\land \exists xQ(x,z)}$
is
${\displaystyle \forall y\forall z(P(y)\land \exists xQ(x,z))}$.
In category theory and the theory of elementary topoi, the universal quantifier can be understood as the right adjoint of a functor between power sets, the inverse image functor of a function between sets; likewise, the existential quantifier is the left adjoint.[2]
For a set ${\displaystyle X}$, let ${\displaystyle {\mathcal {P}}X}$ denote its powerset. For any function ${\displaystyle f:X\to Y}$ between sets ${\displaystyle X}$ and ${\displaystyle Y}$, there is an inverse image functor ${\displaystyle f^{*}:{\mathcal {P}}Y\to {\mathcal {P}}X}$ between powersets, that takes subsets of the codomain of f back to subsets of its domain. The left adjoint of this functor is the existential quantifier ${\displaystyle \exists _{f}}$ and the right adjoint is the universal quantifier ${\displaystyle \forall _{f}}$.
That is, ${\displaystyle \exists _{f}\colon {\mathcal {P}}X\to {\mathcal {P}}Y}$ is a functor that, for each subset ${\displaystyle S\subset X}$, gives the subset ${\displaystyle \exists _{f}S\subset Y}$ given by
${\displaystyle \exists _{f}S=\{y\in Y|{\mbox{ there exists }}x\in S.\ f(x)=y\}}$.
Likewise, the universal quantifier ${\displaystyle \forall _{f}\colon {\mathcal {P}}X\to {\mathcal {P}}Y}$ is given by
${\displaystyle \forall _{f}S=\{y\in Y|{\mbox{ for all }}x.\ f(x)=y\implies x\in S\}}$.
The more familiar form of the quantifiers as used in first-order logic is obtained by taking the function f to be the projection operator ${\displaystyle \pi :X\times \{T,F\}\to \{T,F\}}$ where ${\displaystyle \{T,F\}}$ is the two-element set holding the values true, false, and subsets S to be predicates ${\displaystyle S\subset X\times \{T,F\}}$, so that
${\displaystyle \exists _{\pi }S=\{y\,|\,\exists x\,S(x,y)\}}$
which is either a one-element set (false) or a two-element set (true).
The universal and existential quantifiers given above generalize to the presheaf category.
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https://electronics.stackexchange.com/questions/534696/how-to-calculate-maximum-input-power-on-a-speaker/534705
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# How to calculate maximum input power on a speaker?
I got into this rabbit hole while trying to replace the speaker wire in my setup. I have speakers with 8 Ω of resistance (Rspeaker) and a maximum input power of 140 Watts, so I wanted wires that would not exceed 140W when considering their gauge and length (looking for 10-16 feet).
I found a table at this site: https://soundcertified.com/what-size-speaker-wire-guide/ which displays the maximum power in the system for the given length, gauge, and speaker resistance. Cool, right? I was able to estimate what I need from it, and for fun I decided to back-calculate the table so I could plug in the different lengths I was looking at and see what gauge would work.
I created the following spreadsheet which shows the given table from the website, a matching table based on the calculation I determined was used, and some various values. Spreadsheet to look at the expressions: https://1drv.ms/x/s!AqQjhMOX3aoJ-RN2iS2C_NCIKuKw?e=dVOrqG, image is attached for initial viewing.
Quick overview of the spreadsheet: the initial table uses feet for the length and gauge for the area. In my calculated table, since I had to use resistivity (Ω*m), I converted that to Ωmm, and all the distances to mm and area to mm^2. The table has four values, each corresponding to a length, for every resistance and area (Ohm and Gauge) pair. At the bottom is the value for the resistivity of copper, the equation for resistance in a wire, and the equation for power as a function of voltage and resistance. I also included the voltage from my system (assumed maximum since volume level will control it).
Figuring out the calculation: Because the values in the table are supposedly wattages, I decided to start with the equation for power using voltage (something I could easily assume and change) and resistance (known) versus current. The total resistance is the sum of the speaker resistance plus the wire resistance (a relatively negligible value) since they are in series (I did the parallel calculation for fun as well and it's an inverse relationship obviously). These wattage values, however, are completely different from anything in the table (~1700W for 8Ω, 20 gauge, and 120V), and the various values per length are very similar because wire resistance is negligible (a pretty obvious conclusion).
So what are the values in the table? After messing around a lot I stumbled upon the relationship: Rspeaker/Rwire. It's the ratio between the resistance of the speaker and the resistance of the wire, which is why the values vary so much considering the small Rwire values. But, this means the values in the table are dimensionless, and not wattages at all! A way to get that relationship is Pwire/Pspeaker (voltages cancel out), but I don't know why this would be done or why the ratio would be useful.
On top of that craziness, the table values and determined relationship are still on the correct order of magnitude for input power into the speaker, and supposedly they make sense (if I assume they're wattages, for my system, 16 gauge wire can be used for 16', and if I want shorter wires I should use 18 gauge to not exceed 140 "watts").
Questions: If the values are dimensionless, how are they supposedly useful for comparing wattage, or it it some fake relationship? If the power calculations that were ~1700W are accurate, why does my speaker have a max input of 140W? Is there a better or proper way of determining this for future reference?
Thank you!
• First off, 120 volts sounds like a ton to drive a speaker with. Are you sure that's the output voltage? Secondly, also bear in mind that the impedance of the speaker is going to vary with frequency, it'll only be 8 Ω at one or two specific frequencies. Commented Nov 29, 2020 at 23:38
• "so I wanted wires that would not exceed 140W" - not quite sure what your thinking is here. Commented Nov 29, 2020 at 23:44
• Calculate the current in your speakers. $I=\sqrt{\frac{P}{R}}$ = $\sqrt{\frac{140W}{8\Omega}} = 4.183A$. The wires need to handle this current. The voltage is immaterial to the current carrying capacity of a wire. Commented Nov 29, 2020 at 23:46
The ability of speaker wire to carry the current involved is rarely much of an issue except a few quite unusual loudspeakers. The most notable in this regard was probably the Apogee Scintilla, with a rated impedance of one (1) ohm. That extremely low impedance translated to low voltage and quite high current. But, those are still remembered (thirty years later) primarily because they were uniquely difficult to drive, in large part specifically because of that exceptionally low impedance.
With most reasonably normal loudspeakers, the motivation behind keeping the impedance of speaker wire extremely low is completely different. The real reason is that the amplifier can play a significant role in damping the speaker.
Consider what happens when you play some music with, say, a loud "thump" on a bass drum. That translates to a fairly short transient signal being sent to the loudspeaker. That, in turn, moves the cone of the loudspeaker. So far, all is fine and well.
But then we get to part two: as soon as that transient is over, we want the speaker cone to quit moving. But a speaker cone is a physical thing with inertia. Once it starts moving in a particular direction, its tendency is to keep moving that direction.
But, what started it moving was the magnetic signal in the voice coil interacting with a static magnetic field. Once that incoming signal stops, continued movement of the cone actually generates electricity in the voice coil.
That electricity is then carried back to the output stage of the amplifier, and flows through the amplifier's output impedance. So, the lower the amplifier's output impedance, the more it damps that extra movement of the cone.
But, it's not just the amplifier's output impedance that's involved. If the amplifier's output impedance is (say) 0.1 ohms, and the speaker wire adds another 0.1 ohms, then damping is determined by the (roughly) 0.2 ohm total impedance. A solid state amplifier typically has an extremely low output impedance--the 0.1 ohms I mentioned above is realistic, but at the upper end of the typical range. More powerful amplifiers typically use more output transistors in parallel to achieve their higher power output. That also translates to lower output impedance, so if you have a 140 watt amplifier, chances are your amp's output impedance is considerably lower than that.
As such, to ensure that the speakers are damped as well as the amplifier is capable of doing, you want to ensure that the impedance you're adding in series with the amplifier is considerably lower than that of the amplifier itself. Just for example, you might want something like 10 milliohms (0.01 ohms) as a decent target.
Doing a quick check, the specs you've chosen (16 gauge wire for 10 feet) works out to 0.04 Ohms of DC resistance1 (0.08 for the total of 20 feet for the round trip). While 16 gauge wire is certainly plenty to carry the current involved, that resistance is high enough that it probably has at least some effect on the damping of the system as a whole. If your total round trip is 10 feet, a 10 gauge wire gives a resistance of 0.01 ohms. If you're talking 10 feet distance (so a total of 20 feet of wire), you'd need 7 gauge wire to meet the same target.
In fairness, the difference between the two certainly won't be obvious, and depending on the amplifier you're using (specifically its output impedance) it may be essentially irrelevant. On the other hand, especially if it has particularly low output impedance, it could make a bigger difference.
1. In theory, we should care about impedance, not just DC resistance. In this case, however, we're mostly concerned with the part of the signal that goes to the woofer. That's usually restricted to under 1 KHz (and often less than 500 Hz), where the impedance of a wire is usually very close to its DC resistance. Also: thank you to Bruce Abbot for pointing out that as originally posted, the resistances/gauges needed that I quoted were wrong.
• "(16 gauge wire for 10 feet) works out to 0.016 Ohms of DC resistance" - how did you calculate that? Commented Nov 30, 2020 at 8:13
• I used an online calculator. cirris.com/learning-center/calculators/…. Before I spent any money, I'd probably cross-check that against another source, probably a more authoritative one (like a table from an actual wire vendor). Figuring resistance of wire isn't rocket science though, so I'd be somewhat surprised if it were significantly wrong (on the other hand, there's a much better chance of my failing to click "calculate" after entering some numbers). Commented Nov 30, 2020 at 8:16
• I put in 16AWG and 20ft (round trip wire length) and it said 0.08 ohms. Commented Nov 30, 2020 at 8:22
• @BruceAbbott: Thanks--I've edited to correct. Commented Nov 30, 2020 at 8:29
• I'm waiting for the day when someone will finally write one (2) Commented Nov 30, 2020 at 13:36
I looked at the chart where you obtained your information https://soundcertified.com/what-size-speaker-wire-guide/#Speaker_wire_power_size_chart, and I was not impressed. To calculate the wire gauge you need, first, calculate the current you expect to flow through the wire. The formula $$\I=\sqrt{\frac{P}{R}}\$$ can be used for this calculation. For example, if you expect to have 140W continuously going to your speaker, which is at 8$$\\Omega\$$, your current would be $$\I=\sqrt{\frac{140W}{8\Omega}}=4.183A\$$. Note, that whatever your amplifier's rating (or your speaker's rating) it is unlikely that you will actually be feeding that much power to the speaker over a sustained period. So, you can probably get away with the assumption of a smaller current. However, if we choose to be conservative, we can use the 4.183A value.
Then we look up the wire gauge that can handle 4.183A. We see in a wire-sizing table (for example https://www.powerstream.com/Wire_Size.htm ) that 15 gauge wire will handle 4.7A. Note, this gauge wire will handle 4.7A, whether the wire is 3ft long, or 300 feet long. As others here have noted, you can probably get away with 16 gauge, 18 gauge or even 20 gauge because your will probably not be using 140W continuously, but how much of a derating you wish to make is up to you.
Now, the table you provided says that for 3ft. lengths you have one "power" (actually current) capacity, and for longer lengths, you have significantly lower capacity. That is where I take issue. Yes, for longer lengths, you will lose more power in the wiring. But, for practical purposes, you can neglect this. It is simply wrong to assert that a 3ft 16 gauge cable will handle 664 watts into an 8$$\\Omega\$$ load, while 16 ft of the same cable will only handle 125W, and 25ft only 80W.
16 gauge wire will handle 3.7 Amps. Fed into an 8$$\\Omega\$$ load this translates to $$\(3.7)^2(8)=109.5 W\$$. You should never feed it with 664 watts into an 8$$\\Omega\$$ load, even if the cable is only 3ft long. Nor is there any problem with feeding it 109.5W even if the length is 50ft. (The chart says you can only feed it with 40W at 50ft.) IMHO, the chart is questionable.
As already mentioned, a 140W setup will actually drive 140W only at peaks and will be significantly below that for much of the time. At those peaks, however, you’ll want the cable resistance to result in as small a voltage drop as possible - the speaker is only 8 ohms and so even 1ohm of cable resistance will cause a significant loss of output power, and more significantly will result in loss of sound quality through distortion and reduced damping factor. Skinny wire will work, but a big cross-section will make a noticeable difference. Back in the 80s we would use 79-strand to the tweeters and separate 30A AC cable to the woofers. I’d suggest trying a couple of sizes and see for yourself whether the difference is worthwhile.
I feel like you are overthinking this by at least an order of magnitude.
As you note, the resistance of the wires should be small compared to the impedance of the speaker. There are tables all over the web that will tell you the resistance per foot for different wire gauges, so this is easy to figure out.
Second, you want to use cables with sufficient current carrying capability ("ampacity"). Again, there are plenty of tables to give you an idea of how much current a cable can carry. The trickier part is figuring out how much current you really will send to the speakers. For normal audio purposes the average power is much less than the maximum power.
If you have a speaker rated at 140W I would think that 18 or 20 AWG would work just fine. Just use the cheap "zip cord" wire used for small electric appliances and buy it at the hardware store.
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The result is 3. I have a homework question I have been struggling with which is: How many one-to-one functions are there from the set $A$ into the $B$ if $|A|=n$ Colleagues don't congratulate me or cheer me on when I do good work. One-to-One Functions A function f is 1 -to- 1 if no two elements in the domain of f correspond to the same element in the range of f . A function is not one-to-one if two different elements in the domain correspond to the same element in the range. Consider any two different values in the domain of function g and check that their corresponding output are different. So, the func-tion in Figure 7 is not one-to-one because two different elements in the domain,dog and cat, both correspond to 11. To get the total number of one-to-one functions, we multiply the number of possibilities we have at each stage (this technique is sometimes known as the Rule of Product). And that is the xvalue, or the input, cannot b… So, the number of one-one functions from A to B is 0. You give functions a certain value to begin with and they do their thing on the value, and then they give you the answer. What is the formula to find the number of one-one functions from $A$ to $B$? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Hence if f is an even function and for some number a, a and -a are both in the domain of f then f(a) = f(-a) and yet a ≠ -a and hence f is not one-to-one. Also, plugging in a number for y will result in a single output for x. }$, and there are$n!$possible permutations for$A$. What is the earliest queen move in any strong, modern opening? One-to-one Functions. Continue in this way until you reach the final (i.e. Can an exiting US president curtail access to Air Force One from the new president? What is the number of one-to-one functions from the set$\{1, 2,\dots , n\}$to the set$\{1, 2, \dots , 2n\}$, Find Recursive Definition from given formula. . But we want surjective functions. and$|B| = k$? Function #2 on the right side is the one to one function . How to show these two expressions are the same? when f (x 1 ) = f (x 2 ) ⇒ x 1 = x 2. What is the number of one-to-one functions f from the set {1, 2, . Onto Function Definition (Surjective Function) Onto function could be explained by considering two sets, Set A and Set B, which consist of elements. I can't seem to think of the way to attack this problem help will be appreciated :). Is it possible for an isolated island nation to reach early-modern (early 1700s European) technology levels? No element of B is the image of more than one element in A. , 2n} to the set {1, 2, . Thanks for contributing an answer to Mathematics Stack Exchange! Also, we will be learning here the inverse of this function.One-to-One functions define that each . This is because we can choose any element of$B$except the element chosen in the first step (choosing the same element again would violate one-to-oneness). What causes dough made from coconut flour to not stick together? $$. In other words, nothing is left out. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. . To learn more, see our tips on writing great answers. And, no y in the range is the image of more than one x in the domain. Well, the only way for there to be any one to one functions A\to B is for A to be smaller, ie: p\leq q. A real valued function f of a real variable is even if for each real number x, f(x) = f(-x). A function for which every element of the range of the function corresponds to exactly one element of the domain.One-to-one is often written 1-1. It only takes a minute to sign up. e.g. One-to-One Functions A function f is 1 -to- 1 if no two elements in the domain of f correspond to the same element in the range of f . Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. . One-to-One Function. . Plugging in a number for x will result in a single output for y. of a one-to-one function. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. How is there a McDonalds in Weathering with You? (When the powers of x can be any real number, the result is known as an algebraic function.) = \frac{k!}{(k-n)! Comment: The symbols feel strange, usually one chooses notation so that k \le n. Thanks for contributing an answer to Mathematics Stack Exchange! You could also use the COUNTIFS function. So there are four chances to send first element in domain to co-domain. If I knock down this building, how many other buildings do I knock down as well? A function has many types which define the relationship between two sets in a different pattern. Of course this is possible only if p\leq q. rev 2021.1.8.38287, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. One-to-One Function. Why the sum of two absolutely-continuous random variables isn't necessarily absolutely continuous? , 2n} to the set {1, 2, . Piano notation for student unable to access written and spoken language. Its range is a set of exactly n distinct elements from B, and every possible permutation of A will give us a different function with the same range. Since f is one-one Hence every element 1, 2, 3 has either of image 1, 2, 3 and that image is unique Total number of one-one function = 6 Example 46 (Method 2) Find the number of all one-one functions from set A = {1, 2, 3} to itself. . site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Posted: Jan 2, 2021 / 08:37 PM CST / Updated: Jan 2, 2021 / 08:37 PM CST There are 3 ways of choosing each of the 5 elements = $3^5$ functions. Let’s take y = 2x as an example. To get the total number of one-to-one functions, we multiply the number of possibilities we have at each stage (this technique is sometimes known as the Rule of Product). How to show these two expressions are the same? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Use this function to select one of up to 254 values based on the index number. For the second element of A, there are only k-1 possibilities for its image. Specifically, we can define the following: Definition 4.1. Calculating the total number of surjective functions. If a function has no two ordered pairs with different first coordinates and the same second coordinate, then the function is called one-to-one. no two elements of A have the same image in B), then f is said to be one-one function. In other words, every element of the function's codomain is the image of at most one element of its domain. One-to-One Function. How can a probability density value be used for the likelihood calculation? A function has many types and one of the most common functions used is the one-to-one function or injective function. . site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. A function f from A to B is called one-to-one (or 1-1) if whenever f (a) = f (b) then a = b. And, no y in the range is the image of more than one x in the domain. Colleagues don't congratulate me or cheer me on when I do good work. In a one to one function, every element in the range corresponds with one and only one element in the domain. \frac{k!}{(k-n)!}. Therefore we have {k \choose n}\cdot n! Transcript. Finding nearest street name from selected point using ArcPy.$$ A one-to-one function is a function in which the answers never repeat. Function f is one-one if every element has a unique image, i.e. You will have then$q-2$choices for an image of a third element of$A$and so on... Up to$q-p+1=q-(p-1)$choices for the$p$-th one. One-to-one Functions. Since the function is one-to-one, there are three choices to send second element and there are two choices to … Here we need$k \ge n$, else the answer is$0$. This function will not be one-to-one.$n$th) element of$A$. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In conclusion you have$q(q-1)...(q-(p-2))(q-(p-1))=q!/(q-p)!$possible injective functions. If the number of functions from$A$to$B$is equal to$q^p$, then: 1. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. Start with an element in$A$, you have$q$choices for its image. Hence function g is a one to one function. Let’s take y = 2x as an example. These are called the Stirling numbers of the second kind,$s(p,q)$. 2.1. . Example 46 (Method 1) Find the number of all one-one functions from set A = {1, 2, 3} to itself. Book about an AI that traps people on a spaceship. Show graphically that each of the following functions is a one to one function. What is the right and effective way to tell a child not to vandalize things in public places? In other words, each x in the domain has exactly one image in the range. Consider then a second element in$A$, to keep your function one-to-one you have only$q-1$choices for its image. 2) This is more complicated, but it has already been asked Calculating the total number of surjective functions. Otherwise the function is many-one. But, here n B if (A) > n (B). a) all the elements of X should have one to one image with Y, so there are 5 choice for 1st element of X, 4 choices for 2nd element, 3 for 3 rd element and 2 for 4th element. Also, one-one function is only possible from A to B if (A) ≤ n (B). Why the sum of two absolutely-continuous random variables isn't necessarily absolutely continuous? For example, the function f(x) = x + 1 adds 1 to any value you feed it. MathJax reference. In this case the map is also called a one-to-one correspondence. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Number of onto functions from one set to another – In onto function from X to Y, all the elements of Y must be used. PostGIS Voronoi Polygons with extend_to parameter. Suppose f: X → Y is a one-to-one function and let C ⊆ Y be the codomain of f. Then there is a function f−1: C → X, called the inverse of f defined as follows: f−1(y) = x ⇐⇒ f(x) = y. They are various types of functions like one to one function, onto function, many to one function, etc. Asking for help, clarification, or responding to other answers. What is the number of one-to-one functions f from the set {1, 2, . , 2n} so that f(x) x for all 1 ≤ x ≤ n and f(x) = x for some n+1 ≤ x ≤ 2n? There are$k - (n - 1) = k - n + 1$possibilities for its image, since we again must choose some element of$B$that has not been used in the previous$n-1$steps. 2) Solving certain types of equations Examples 1 To solve equations with logarithms such as ln(2x + 3) = ln(4x - 2) we deduce the algebraic equation because the ln function is a one to one. Question from Relations and Functions,jeemain,math,class12,ch1,relations-and-functions,types-of-functions,medium For example, if value1 through value7 are the days of the week, CHOOSE returns one of the days when a number between 1 and 7 is used as index_num. If for each x ε A there exist only one image y ε B and each y ε B has a unique pre-image x ε A (i.e. while x → x 2, x ε R is many-to-one function… What is the earliest queen move in any strong, modern opening? rev 2021.1.8.38287, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. 1.1. .$$Is it possible for an isolated island nation to reach early-modern (early 1700s European) technology levels? Making statements based on opinion; back them up with references or personal experience. Asking for help, clarification, or responding to other answers. So, the func-tion in Figure 7 is not one-to-one because two different elements in the domain,dog and cat, both correspond to 11. Finding a formula for the number of functions, Discrete Math: Question regarding functions/combinatorics, Compact-open topology and Delta-generated spaces, Signora or Signorina when marriage status unknown. For the first element of$A$, there are$k$possibilities for its image under the function (just choose any element of$B$). 2x + 3 = 4x - 2 Examples 2 Also known as an injective function, a one to one function is a mathematical function that has only one y value for each x value, and only one x value for each y value. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … If a function has no two ordered pairs with different first coordinates and the same second coordinate, then the function is called one-to-one. by: Alece Courville. How can I quickly grab items from a chest to my inventory? We get Can playing an opening that violates many opening principles be bad for positional understanding? A has 4 elements and B has 3 elements. yes I mean one to one functions :) sorry im tired :), Number of possible results in election with one of candidates getting more then 50% votes, Generating functions and finding coefficient of$x^{3n}$. A function f is one-to-one if for each a and b in the domain of f, if f(a) = f(b) then a = b. To learn more, see our tips on writing great answers. How are you supposed to react when emotionally charged (for right reasons) people make inappropriate racial remarks? So, #1 is not one to one because the range element.5 goes with 2 different values in the domain (4 and 11). For onto maps$A\to B$, we now need$A$to be at least as big as$B$, so$p\geq q$. Is there a way to force an incumbent or former president to reiterate claims under oath? Let$q$be the number of elements in$B$. This can be written more concisely as What is the policy on publishing work in academia that may have already been done (but not published) in industry/military? In other words no element of are mapped to by two or more elements of . First let$k \geq n$, since there will be no one-to-one functions otherwise. This formula uses COUNTIF twice to specify multiple criteria, one criteria per expression. No element of B is the image of more than one element in A. This sounds confusing, so let’s consider the following: In a one-to-one function, given any y there is only one x that can be paired with the given y. Seatbelts are the number one safety function of a car News. Finding nearest street name from selected point using ArcPy, First author researcher on a manuscript left job without publishing. Use MathJax to format equations. A good way of describing a function is to say that it gives you an output for a given input. This sounds confusing, so let’s consider the following: In a one-to-one function, given any y there is only one x that can be paired with the given y. MathJax reference. Plugging in a number for x will result in a single output for y. Can I hang this heavy and deep cabinet on this wall safely? The formula for the area of a circle is an example of a polynomial function.The general form for such functions is P(x) = a 0 + a 1 x + a 2 x 2 +⋯+ a n x n, where the coefficients (a 0, a 1, a 2,…, a n) are given, x can be any real number, and all the powers of x are counting numbers (1, 2, 3,…). x → x 3, x ε R is one-one function. In a one-to-one function, given any y there is only one x that can be paired with the given y. How can I keep improving after my first 30km ride? What is the point of reading classics over modern treatments? (a) We have to find the number of one-to-one functions from set with three elements to the set with four elements. 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# algebra 2 concept
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#### Transcript of algebra 2 concept
Algebra 2
Part 1 Solving Equations and Inequalities Graphs and Functions Systems of Equations
and Inequalities
Polynomials and Factoring Powers and Roots Complex Numbers Quadratics
Coordinate Geometry
Matrices
Combining Like
Terms Multiplication of
Polynomials Factoring Addition and Multiplication Fractions or Decimals Inequalities Dimensions Graphing Lines Finding Slopes of Lines Functions Solving Systems of Equations Graphically Systems of Inequalities Square Roots
Multiplication and
Simplification Division and Simplification
Operations Rationalizing the Denominator Imaginary Numbers Complex Numbers Equations with
Form Parabolas Distance and Midpoint Formulas Circles Ellipses Hyperbolas Systems of Equations To solve an equation, you isolate the variable you
are solving for. The Addition Principle says that when
a = b, a + c = b + c for any number c. Solve: x + 6 = -15
to each side of the equation.
x + 6 - 6 = -15 - 6
The variable is now isolated.
x = -21
Multiplication
Along the same lines, the Multiplication Principle says that if a = b and c is any number, a * c = b * c. This principle is also used to help isolate the variable you are asked to solve for. Solve: 4x = 9
Solution: Using the Multiplication Principle,
multiply each side of the equation
by (1/4).
(1/4)4x = (1/4)9
The variable is now isolated.
x = (9/4)
2x2 rule 3x3 rule Also, be aware of problems where
you might need to use both of these
principles together! Solve: 3x - 4 = 13
Solution: Use the Addition Principle to
3x - 4 + 4 = 13 + 4
After simplifying, 3x = 17.
Use the Multiplication Principle
to multiply each side by (1/3).
(1/3)3x = (1/3)17
After simplification, the variable
is isolated.
x = (17/3)
When an equation contains fractions or decimals,
it usually makes it easier to solve them when the
fractions or decimals aren't there. The Multiplication
Principle is used to do this. Solve: (3/4)x + (1/2) = (3/2)
Solution: Multiply both sides by the LCM
of the denominators, 4 in
this case.
Use the Distributive Law,
which says a(b + c) = ab + ac
to make the equation easier to
deal with.
(4 * (3/4)x) + (4 * (1/2)) = 4(3/2)
After simplification, you get an
equation with no fractions!
3x + 2 = 6
(It's left up to you to solve
for x.)
If a and b are real numbers, and ab = 0, either a, b, or both equal 0. This principle, called The Principle of Zero Products, is useful when you have an equation to solve that has two instances of a variable, such as (x + 3)(x - 2) = 0. Solve: 7x(4x + 2) = 0
Solution: Using the Principle of
Zero Products,
7x = 0 and 4x + 2 = 0
Solve each equation for x.
x = 0 and x = -(1/2)
The solutions are 0 and -(1/2). Math problems containing <, >, <=, and >= are called inequalities. A solution to any inequality is any number that makes the inequality true.
On many occasions, you will be asked to show the
solution to an inequality by graphing it on a number line. Many times you will have a statement such as x > 5 that needs to be graphed. Because this is not an equation, it does not need to be graphed on the coordinate plane. A number line does the job just fine!
Some things to know about graphin inequalites:
1. An open circle is placed on the number line to show that the number denoted at the circle is not included in the solution set.
2. A circle that is filled in is placed on the number line to show that the number denoted at the circle is included in the solution set. Graph: x < 4
Solution: The problem asks you to graph all numbers
that are less than 4.
Example Graph linerar system Conjunctions or
Complex Inequalities The word conjunction means there are two conditions in a statement that must be met. Therefore, a mathematical statement such as the following is a conjunction: 5 < x < 10. There are two things to remember when dealing with conjunctions. They are outlined below.
1. The greater than or less than signs will always be pointing in the same direction (i.e., you will never see the following: 7 > x < 2).
2. Look out for statements that cannot be true, such as the following: 10 < x < 5 Graph: -2 < x <= 4
Solution: The problem, which is a conjunction, asks for a graph of all the numbers between -2 and 4.
Be sure to note that -2 is not included in the solution while 4 is. Inverse Matrix Some equations are not quadratic equations,
but are in the same form, such as x4 - 9x2 + 8 = 0.
To solve equations such as that, you make a substitution,
solve for the new variable, and then solve for the original variable. Problem: Solve x4 - 9x2 + 8 = 0 for x.
Solution: Let u = x2. Then substitute
u for every x2 in the equation.
u2 - 9u + 8 =0
Factor.
(u - 8)(u - 1) = 0
Utilize the principle of zero
products.
u - 8 = 0, u - 1 = 0
u = 8 , u = 1
Now substitute x2 for u and
solve the equations.
x2 = 8, x2 = 1
x = ±SQRT(8), x = ± 1
x = ±2(SQRT(2))
x = ±2(SQRT(2)), ±1 Many times you will come across quadratic equations
that are not easy to factor or solve. In those cases,
there is a special formula called the quadratic formula that
you can use to solve any quadratic equation.
The solutions of any quadratic equation, ax2 + bx +c=0 is given by the following formula, called the quadratic formula:
-b ± SQRT(b2 - 4ac)
x = -------------------
2a Problem: Solve 3x2 + 5x = -1 for x.
Solution: First find the standard form of
the equation and determine a, b,
and c.
3x2 + 5x + 1 = 0
a = 3
b = 5
c = 1
Plug the values you found for
a, b, and c into the
-5 ± SQRT(52 - 4(3)(1))
x = -----------------------
2 * 3
Perform any indicated operations.
-5 ± SQRT(25 - 12)
x = ------------------
6
-5 ± SQRT(13)
x = -------------
6
The solutions are as follows:
-5 + SQRT(13) -5 - SQRT(13)
x = -------------, -------------
6 6 Any equation of type ax2 + bx + c = 0 where a, b, and c are constants and a <> 0, is in standard form for a quadratic equation. Quadratic equations of type ax2 + bx + c = 0 and ax2 + bx = 0 (c is 0) can be factored to solve for x. Problem: Solve 3x2 + x - 2 = 0 for x.
Solution: Factor.
(3x - 2)(x + 1) = 0
Use the principle of zero products,
which says, if ab = 0, either
a, b, or both must be
equal to zero.
3x - 2 = 0, x + 1 = 0
3x = 2 , x = -1
x = (2/3)
x = -1, (2/3)
Quadratic equations of type ax2 + c = 0 can be solved by solving for x. Problem: Solve 3x2 = 6 for x.
Solution: Recognize that the equation is
same as 3x2 - 6 = 0.
Divide each side by 3.
x2 = 2
Take the square root of each side.
x = SQRT(2), -(SQRT(2)) By definition, every real number has two square roots. For example, 64's two square roots are 8 and -8 because 82 = 64 and (-8)2 = 64. However, the principal square root of a real number is its nonnegative square root (8 would be the principal square root of 64). Problem: SQRT(25)
Solution: 5
That is the answer because 5 * 5 = 25.
One special situation with square roots occurs when the number inside the square root is squared. There is a special theorem that deals with this: SQRT(a2) = |a|. Problem: SQRT((-16)2)
Solution: By the theorem above, the answer
is |-16|. The absolute
value of -16 is 16. adding Multiplying For any nonnegative real numbers, a and b, SQRT(a) * SQRT(b) = SQRT(ab). For example, the 4th root of a times the 4th root of b equals the 4th root of ab. Multiply: SQRT(x + 2) * SQRT(x - 2)
Solution: Use the theorem above to put both
SQRT((x + 2)(x - 2))
Multiply the binomials under the
SQRT(x2 - 4) Reversing the theorem stated above gives us a way to factor radical expressions, thereby simplifying them. Problem: Simplify SQRT(20)
Solution: Factor the radicand as a
product of prime factors.
SQRT(2 * 2 * 5)
There are two instances of 2,
so by the definition of a square
root, you can take 2 out
from under the radicand. That gives
2(SQRT(5)) Taking the square root of fractions and dividing a radicals Which is: When you have a problem like SQRT(27/y2), don't be scared of the fraction. Just use the Roots of Fractions theorem, which says that nRT(a/b) = (nRT(a))/(nRT(b)). Simplify: CBRT(27/125)
Solution: Use the Roots of Fractions
theorem to rewrite the problem.
CBRT(27)
---------
CBRT(125)
Take the cube root of both
the numerator and denominator
(3/5)
Reverse the Roots of Fractions theorem when you are asked to divide a radical by a radical of the same index when it can be simplified. Simplify: (SQRT(80))/(SQRT(5))
Solution: Use the converse of the Roots
of Fractions theorem and rewrite
SQRT(80/5)
80/5 simplifies to 16.
SQRT(16) = 4 You add and subtract radicals the same way you would with polynomials, by combining like terms. You have to look out for terms that do not look alike,
but could be if factored. Problem: 6(SQRT(7)) + 4(SQRT(7))
Solution: Both terms are alike (like
10(SQRT(7)) Problem: 3(SQRT(8)) - 5(SQRT(2))
Solution: Factor 8.
3(SQRT(4 * 2)) - 5(SQRT(2))
Factor SQRT(4 * 2) into two
3(SQRT(4))(SQRT(2)) - 5(SQRT(2))
Take the square root of 4.
3 * 2(SQRT(2)) - 5(SQRT(2))
6(SQRT(2)) - 5(SQRT(2))
Combine like terms.
SQRT(2) To multiply radical expressions that have factors which contain more than one term, use the same procedure you would when multiplying polynomials. Problem: CBRT(y) * (CBRT(y2) + CBRT(2))
Solution: Use the distributive law of
multiplication, which says that
a(b + c) = ab + ac to
multiply the expression out.
CBRT(y) * CBRT(y2) + CBRT(y) * CBRT(2)
CBRT(y3) + CBRT(2y)
Take the cube root of
y3.
y + CBRT(2y) To add matrices, we add the corresponding members. The matrices have to have the same dimensions When dealing with radicals and fractions, you will, on many occasions, get an answer with a radical in the denominator. Usually, an answer is not considered simplified until there are no radicals in the denominator. The process of removing radicals from the denominator is called rationalizing the denominator. An important thing to remember when rationalizing denominators is that anything divided by itself is 1. For example, 67/67 is the same as 1. Problem: Rationalize the denominator.
4 + SQRT(2)
-----------
5 - SQRT(2)
Solution: Multiply by 1 (make sure
the fraction you choose to
use as one will make the
denominator a perfect
square — the conjugate is
usually a good number).
4 + SQRT(2) 5 + SQRT(2)
----------- * -----------
5 - SQRT(2) 5 + SQRT(2)
Multiply the problem as you
would multiply any fractions.
Also, the FOIL method of
multiplying binomials will
come in handy.
20 + 4(SQRT(2)) + 5(SQRT(2)) + (SQRT(2))2
-----------------------------------------
25 + 5(SQRT(2)) - 5(SQRT(2)) - (SQRT(2))2
Perform any indicated operations.
20 + 4(SQRT(2)) + 5(SQRT(2)) + 2
--------------------------------
25 + 5(SQRT(2)) - 5(SQRT(2)) - 2
Perform any indicated operations,
and combine like terms, if you
can.
22 + 9(SQRT(2))
---------------
23 For example, you can multiply a matrix by another matrix or by a number. When you multiply a matrix by a number, multiply each member of the matrix by the number. To multiply a matrix by a matrix, the first matrix has to have the same number of columns as the rows in the second matrix. When graphed, lines slope from left to right. However, some slope upward and others slope downward. Some are really steep, while others have a gentle slope. The slope of a line is defined as the change in y over the change in x, or the rise over the run.
[(y2 - y1),/(x2 - x1)]. To find the slope, you pick any two points on the line and find the change in y, and then divide it by the change in x When dealing with lines and points, it is very important to be able to find out how long a line segment is or to find a midpoint. Midpoint Formula: [((x1 + x2)/2), ((y1 + y2)/2)]. The distance formula says that the distance d between any two points with coordinates (x1, y1) and (x2, y2) is given by the following equation: d = SQRT[(x2 - x1)2 + (y2 - y1)2]. Circles, when graphed on the coordinate plane, have an equation of x2 + y2 = r2 where r is the radius (standard form) when the center of the circle is the origin. When the center of the circle is (h, k) and the radius is of length r, the equation of a circle (standard form) is (x - h)2 + (y - k)2 = r2 Problem: Find the center and radius of
(x - 2)2 + (y + 3)2 = 16.
Then graph the circle.
Solution: Rewrite the equation in standard form.
(x - 2)2 + [y - (-3)]2 = 42
The center is (2, -3) and the radius is 4.
The graph is easy to draw, especially
if you use a compass.
The figure below is the graph of the solution. Ellipses, or ovals, when centered at the origin, have an equation (standard form) of (x2/a2) + (y2/b2) = 1. When the center of the ellipse is at (h, k), the equation (in standard form) is as follows:
(x - h)2 (y - k)2
-------- + -------- = 1
a2 b2 Problem: Graph x2 + 16y2 = 16.
Solution: Multiply both sides by 1/16
to put the equation in
standard form.
x2 y2
-- + -- = 1
16 1
a = 4 and b = 1. The
vertices are at (±4, 0) and
(0, ±1). (The points are
on the axes because the equation
tells us the center is at the origin,
so the vertices have to be on the
axes.)
Connect the vertices to form an oval,
and you are done!
The figure below is the graph of the ellipse. Absolute Value The equation of a hyperbola (in standard form) centered at the origin is as follows:
x2 y2
-- - -- = 1
a2 b2 Problem: Graph 9x2 - 16y2 = 144.
Solution: First, multiply each side of the
equation by 1/144 to put it
in standard form.
x2 y2
-- - -- = 1
16 9
We now know that a = 4 and
b = 3. The vertices are at
(±4, 0). (Since we
know the center is at the origin,
we know the vertices are on the
x axis.)
The easiest way to graph a hyperbola
is to draw a rectangle using the
vertices and b, which is on
the y-axis.
Draw the asymptotes through opposite
corners of the rectangle.
Then draw the hyperbola.
The figure below is the graph
of 9x2 - 16y2 = 144. The easiest way to solve systems of equations that include circles, ellipses, or hyperbolas, is graphically. Because of the shapes (circles, ellipses, etc.), there can be more than one solution. Problem: Solve the following system of
equations:
x2 + y2 = 25
3x - 4y = 0
Solution: Graph both equations on the same
coordinate plane. The points of
intersection have to satisfy both
equations, so be sure to check the
solutions. Both intersections do
check.
The figure below shows the solution. When terms of a polynomial have the same variables
raised to the same powers, the terms are called similar,
or like terms. Like terms can be combined to make the
polynomial easier to deal with. Problem: Combine like terms in the following equation: 3x2 - 4y + 2x2.
Solution: Rearrange the terms so it is easier
to deal with.
3x2 + 2x2 - 4y
Combine the like terms.
5x2 - 4y Probably the most important kind of polynomial multiplication that you can learn is the multiplication of binomials (polynomials with two terms). An easy way to remember how to multiply binomials is the FOIL method, which stands for first, outside, inside, last. Example :Problem: Multiply (3xy + 2x)(x^2 + 2xy^2).
Solution: Multiply the first terms of each bi-
nomial. (F)
3xy * x2 = 3x3y
Multiply the outside terms of each bi-
nomial. (O)
3xy * 2xy2 = 6x2y3
Multiply the inside terms of each bi-
nomial. (I)
2x * x2 = 2x3
Multiply the last terms of each bi-
nomial. (L)
2x * 2xy2 = 4x2y2
You now have a polynomial with four terms.
Combine like terms if you can
3x3y + 6x2y3 + 2x3 + 4x2y2 Factoring is the reverse of multiplication. When factoring, look for common factors. Problem: Factor out of a common factor of
4y2 - 8.
Solution: 4 is a common factor of
both terms, so pull it out and write
each term as a product of factors.
4y2 - (4)2
Rewrite using the distributive law of
multiplication, which says that
a(b + c) = ab + ac.
4(y2 - 2) Trinomials and binomials are the most common
polynomials, but you will sometimes see polynomials
with more than three terms. Sometimes, when you are
dealing with polynomials with four or more terms, you
can group the terms in such a way that common factors
can be found. Problem: Factor 4x2 - 3x + 20x - 15.
Solution: Rearrange the terms so common
factors can be more easily found.
4x2 + 20x - 3x - 15
The first two terms have a common factor
in 4x. The last two terms have a
common factor in 3. Factor those
terms out.
4x(x + 5) - 3(x + 5)
Now you have a binomial. Each term
has a factor of (x + 5). Factor
that out for the final answer.
(x + 5)(4x - 3) In the set of real numbers, negative numbers do not have square roots. A new kind of number, called imaginary was invented so that negative numbers would have a square root. These numbers start with the number i, which equals the square root of -1, or i2 = -1. All imaginary numbers consist of two parts, the real part, b, and the imaginary part, i. Simplify: SQRT(-5)
Solution: Write -5 as a product of
prime factors.
SQRT(-1 * 5)
Write as separate square roots.
(SQRT(-1))(SQRT(5))
By definition, i = SQRT(-1),
(SQRT(5))i. (SQRT(5)
is the real part, or b.) A complete number system, one that includes both real and imaginary numbers, was devised. Numbers in this set are called complex numbers. Complex numbers consist of all sums a + bi where a and b are real numbers and i is imaginary. Real numbers fit into the complex number system because a = a + 0i.
i behaves as any variable would. Problem: 7i + 9i
Solution: Combine like terms.
16i
Multiplication is done as if the imaginary parts of complex numbers were just another term. Always remember that i2 = -1. Problem: 3i * 4i
Solution: 12i2
Remember that i2 equals
12(-1)
-12 Now that negative numbers have square roots,
equations such as x2 + 1 = 0 have solutions.
They can also be factored! example Problem: Solve for x: x2 + 1 = 0
Solution: Subtract 1 from each side.
x2 = -1
Take the square root of each side.
Remember that i = SQRT(-1).
x = i, -i
2. Problem: Show that (x + i)(x - i) is
a factorization of x2 + 1.
Solution: Multiply.
x2 + ix - ix - i2
x2 + 1 Solving systems of equations graphically is one of the
easiest ways to solve systems of simple equations
(it's usually not very practical for complex equations
such as hyperbolas or circles). One way to solve systems of equations is by substitution. In this method, you solve on equation for one variable, then you substitute that solution in the other equation, and solve. Problem: Solve the following system:
x + y = 11
3x - y = 5
Solution: Solve the first equation for y
(you could solve for x - it
doesn't matter).
y = 11 - x
Now, substitute 11 - x for y
in the second equation. This gives
the equation one variable, which
earlier algebra work has taught
you how to do.
3x - (11 - x) = 5
3x - 11 + x = 5
4x = 16
x = 4
Now, substitute 4 for x in
either equation and solve for y.
(We use the first equation below.)
4 + y = 11
y = 7
The solution is the ordered pair,
(4, 7). Cramer's Rule is when you use determinats to solve a system of linear equations The last method, Elimination, is probably the most complicated, but is necessary when dealing with more complex systems, such as systems with three or more variables. The idea behind the addition method is to replace an equation with a combination of the equations in the system. To obtain such a combination, you multiply each equation by a constant and add. You choose the constants so that the resulting coefficient of one of the variables will be 0. Method Problem: Solve the following system:
5x + 3y = 7
3x - 5y = -23
Solution: Multiply the second equation by 5
to make the x-coefficient
a multiple of 5. (This works be-
cause it does not change the
equation (see the multiplication property).)
15x - 25y = -115
Next, multiply the first equation by -3
and add it to the second equation. This
gets rid of the x-term.
-15x - 9y = -21
15x - 25y = -115
-----------------
- 34y = -136
Now, solve the second equation for y.
Then substitute the result into the
first equation and solve for x.
-34y = -136
y = 4
5x + 3(4) = 7
5x + 12 = 7
5x = -5
x = -1
The solution is the ordered pair, (-1, 4).
The easiest way to solve systems of inequalities is to solve them by graphing. Therefore, it is best if you know how to graph inequalities in two variables (5x - 4y < 13, for example) Problem: Graph y < x.
Solution: First graph the equation y = x.
However, the line must be drawn dashed
because the less than sign tells us the
line is not included in the
solution.
Next, test a point that is located above
the line and one that is below the line.
Any point you pick above the line, such
as (0, 2), y is greater than x,
so points above the line are not in-
cluded in the solution. Points below the
line, such as (3, -3) have a y
value that is less than the x
value, so all points below the line are
included in the solution. To solve a system or conjunction of inequalities,
it is easiest to graph each of the inequalities and
then find their intersection. Problem: Graph the following system:
2x + y >= 2
4x + 3y <= 12
(1/2) <= x <= 2
y >= 0 When graphed, lines slope from left to right. However, some slope upward and others slope downward. Some are really steep, while others have a gentle slope. The slope of a line is defined as the change in y over the change in x, or the rise over the run.
This can be explained with a formula: (y2 - y1)/(x2 - x1). To find the slope, you pick any two points on the line and find the change in y, and then divide it by the change in x. Problem: The points (1,2) and (3,6) are on a line.
Find the line's slope.
Solution: Plug the given points into
the slope formula.
y2 - y1
m = -------
x2 - x1
6 - 2
m = -----
3 - 1
After simplification, m = 2 A function is a relation (usually an equation) in which no two ordered pairs have the same x-coordinate when graphed. One way to tell if a graph is a function is the vertical line test, which says if it is possible for a vertical line to meet a graph more than once, the graph is not a function. The figure above is an example of a function Functions are usually denoted by letters such as f or g. If the first coordinate of an ordered pair is represented by x, the second coordinate (the y coordinate) can be represented by f(x). In the figure below, f(1) = -1 and f(3) = 2.
When a function is an equation, the domain is the set of numbers that are replacements for x that give a value for f(x) that is on the graph. Sometimes, certain replacements do not work, such as 0 in the following function: f(x) = 4/x (you cannot divide by 0). In that case, the domain is said to be x <> 0.
There are a couple of special functions whose graphs you should have memorized because they are sometimes hard to graph. They are the absolute value function Graphs of quadratic functions are called parabolas. The basic graph that you need to know is f(x) = x2.
Full transcript
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# People in an elevator .
#### robespierre722
##### New member
Hello , i know that this problem in kinda famous but there is a difference and i am not really sure how to finish it :
3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.
Each person is independent of the others to get off at any of the 5 floors above ground.
a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)
b)What is the probability that no person from the first elevator will want to go at the 5th floor.
a) So there are 5 possibilities for each person -> 5^6 possible cases
There are 6C2=15 possible pairs for that 3rd floor .
And now the funny part, i was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60
And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .
b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .
#### Jomo
##### Elite Member
b)What is the probability that no person from the first elevator will want to go at the 5th floor.
b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .
Why are you mentioning the 2nd elevator. From the beginning there could have been 6 elevators. However part a only talks about elevator 1 and 2. Does that change how you want to answer part a??
There is no elevator 2 in part b. 3 people get on elevator 1 on the ground floor of a six story building. What is the probability that no one gets off on the fifth floor?
1st of all I have a problem with this problem. Suppose I want to go to the 4th floor or even the 5th floor and then I remember that I left something in my car and I now want to go to the ground floor (has that ever happened to you?). I think that this needs to be included as an option, however we will not consider that possible. Each person has 5 possible floors to choose from with equal probability. So there are 6^5 ways for them to exit the elevator. How many of these ways do not include anyone going to the 5th floor? Figure that out and divide by 6^5 for your answer.
#### Jomo
##### Elite Member
Hello , i know that this problem in kinda famous but there is a difference and i am not really sure how to finish it :
3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.
Each person is independent of the others to get off at any of the 5 floors above ground.
a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)
We can consider that all 6 passengers are in the same elevator.
I would think of this problem as follows: You have a biased coin where p(heads) = 1/5 and p(tails) = 4/5. Do you see why this models your problem? Now toss the coin 6 times and find the p(getting exactly 2 heads)
#### JeffM
##### Elite Member
Hello , i know that this problem in kinda famous but there is a difference and i am not really sure how to finish it :
3 people(A,B,C) enter an elevator at the ground floor of a 5-storey building, another 3 people(D,E,F) enter in another elevator.
Each person is independent of the others to get off at any of the 5 floors above ground.
a)What is the probability that exactly 2 people will get off at the 3rd floor ? (regardless of the elevator)
b)What is the probability that no person from the first elevator will want to go at the 5th floor.
a) So there are 5 possibilities for each person -> 5^6 possible cases
There are 6C2=15 possible pairs for that 3rd floor .
And now the funny part, i was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60
And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .
b)5^6 possible cases. From the first elevator 4^3 since they can't enter 5th floor and from the second elevator 5^3 -> (5^3)*(4^3)/5^6 but it is the wrong answer again .
With a five-storey building, there are not five possibilities of exit for each person unless you are assuming that people get on and then get off on the ground floor before the elevator moves.
So my first question is whether the problem stipulates a five-storey or six-storey building?
Furthermore, we are not told whether people are equally likely to get off at each floor. It is important, as our summary of guidelines says, to give the original problem completely and exactly. Assuming, however, equal probability, the probability that any specific passenger will get off on floor three is (1/4). The probability that any specific passenger will get off on a floor other than 3 is (3/4). How many ways can we choose 2 from a total of 6?
By the way, what are you trying to say is the given answer? What you appear to be giving is way greater than 1.
#### robespierre722
##### New member
With a five-storey building, there are not five possibilities of exit for each person unless you are assuming that people get on and then get off on the ground floor before the elevator moves.
So my first question is whether the problem stipulates a five-storey or six-storey building?
Furthermore, we are not told whether people are equally likely to get off at each floor. It is important, as our summary of guidelines says, to give the original problem completely and exactly. Assuming, however, equal probability, the probability that any specific passenger will get off on floor three is (1/4). The probability that any specific passenger will get off on a floor other than 3 is (3/4). How many ways can we choose 2 from a total of 6?
By the way, what are you trying to say is the given answer? What you appear to be giving is way greater than 1.
It is a six-storey building as you said, sorry.
And they cannot exit on ground floor.
They have equal probability to exit any floor.
And i tried that option as well like: We can chose 2 from a total of 6 by : 6C2 and since we have 1/5 chance that they can exit the 3rd floor ->
I think that this answer was good in case it was not specified one floor (3rd floor) , like :"2 people can get off at the same floor " instead of " 2 people get off at the 3rd floor".
That 3rd floor beeing mentioned is my problem.
#### robespierre722
##### New member
We can consider that all 6 passengers are in the same elevator.
I would think of this problem as follows: You have a biased coin where p(heads) = 1/5 and p(tails) = 4/5. Do you see why this models your problem? Now toss the coin 6 times and find the p(getting exactly 2 heads)
I tried that option as well like: We can chose 2 from a total of 6 by : 6C2 and since we have 1/5 chance that they can exit the 3rd floor ->
I think that this answer was good in case it was not specified one floor (3rd floor) , like :"2 people can get off at the same floor " instead of " 2 people get off at the 3rd floor".
That 3rd floor beeing mentioned is my problem.
And yes , you can consider of course 1 elevator and 6 persons for a)
I wrote the only way i can get the given answer ( 0.3072) is like this :
a) So there are 5 possibilities for each person -> 5^6 possible cases
There are 6C2=15 possible pairs for that 3rd floor .
(Wrong but it seems right
I was presuming that for example on the 1st floor, the 4 people left are going to be able to enter in (4C1+4C2+4C3+4C4)=15 ways since they can enter 1 or 2 or 3 or 4 people . But there are 4 floors ( i am not counting the 3rd floor) -> (4C1+4C2+4C3+4C4)*4=60 ways they can exit
And my answer is ((4C1+4C2+4C3+4C4)*4*6C2)/(5^6)=0.0576 which is the wrong answer . )
(The right answer but i don't understand why since i only have the (0.3072)
If i put (4P1+4P2+4P3+4P4)*5, P instead of C and 4 instead of 5 -> the good answer but i don't understand why .
so the only way to get the given answer is by this formula :
((4P1+4P2+4P3+4P4)*5*6C2)/(5^6)=0.3072 , i don't understand the logic behind those (4P1+4P2+4P3+4P4) )
Last edited:
#### Dr.Peterson
##### Elite Member
"The third floor being mentioned" is not your problem. Anything else would be a harder problem, not an easier one.
I think your answer of 6C2*(1/5)^2*(4/5)^4=24.57% is correct, as far as I can see. All your other ideas seem to be your attempts to make the supposedly correct answer, which is 5/4 of that, come out as your answer. Is it possible, instead, that it is just wrong? Might someone have just raised 4/5 to the wrong power, or something like that?
If you haven't stated the problem exactly as given to you, please do so, so we can make sure we are dealing with the right question. Also tell us how you know that 0.3072 is supposed to be correct.
#### Jomo
##### Elite Member
I tried that option as well like: We can chose 2 from a total of 6 by : 6C2 and since we have 1/5 chance that they can exit the 3rd floor ->
I disagree with that being the wrong answer. Can you show why it is wrong or why another answer is correct. I stand by my answer.
#### robespierre722
##### New member
"The third floor being mentioned" is not your problem. Anything else would be a harder problem, not an easier one.
I think your answer of 6C2*(1/5)^2*(4/5)^4=24.57% is correct, as far as I can see. All your other ideas seem to be your attempts to make the supposedly correct answer, which is 5/4 of that, come out as your answer. Is it possible, instead, that it is just wrong? Might someone have just raised 4/5 to the wrong power, or something like that?
If you haven't stated the problem exactly as given to you, please do so, so we can make sure we are dealing with the right question. Also tell us how you know that 0.3072 is supposed to be correct.
I just solved some probability problems for my exam and i only have the final answer which is 0.3072 in this case . For all the problems i solved until now the final answer was good so i presume that 0.3072 is good as well. I will try to translate word by word :
An elevator stops at P+1, P+2, P+3, P+4 and P+5 (5 levels) . The elevator is at ground floor P with 3 people A, B , C .
A second elevator is in the same building at floor P+5, with 3 persons D, E, F. This elevator stops at P+4, P+3, P+2, P+1, and P.
a)What is the probability that exactly 2 people will get off at P+3 ? (regardless of the elevator)
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You are on page 1of 25
# NUMERICAL DESCRIPTIVE
MEASURES
Review
## Numerical Descriptive Measures
i. Measures of central tendency, variation and
shape
a. Mean, median and mode
b. Quartiles
c. Geometric mean
d. Range, Interquartile range
e. Mean absolute deviation
f. Variance and standard deviation
g. Coefficient of variation
h. Z scores
i. Shape
2
Scope of this lecture
## Numerical descriptive measures for a population
i. Mean
ii. Variance and standard deviation
iii. Empirical rule
iv. Chebyshev rule
Example problems
3
Population mean, variance and standard
deviation
## Population mean for the values Xi (i=1,2,...,N)
is given by,
Population variance
4
Empirical rule
## Empirical rule: variability in the symmetrical
bell-shaped (normal) distribution
## i. Appx. 68.3% of the values are within
from the mean
ii. Appx. 95.5% of the values are within
from the mean
iii. Appx. 99.7% of the values are within
from the mean
5
Empirical rule
## Source: Business statistics, J. K. Sharma
6
Population mean, variance and standard
deviation
## Ex: Suppose you are in charge of rationing in a
state affected by food shortage. The following
reports arrive from a local investigator:
Daily calorific value of food available per adult
during current period:
Area Mean Std.
Dev.
The estimated requirement of an
A 2500 400
adult is taken as 2800 calories daily
and the absolute min. is 1350. B 2000 200
## Comment on the reported figures and determine
which area in your opinion needs more urgent
attention.
7
Population mean, variance and standard
deviation
## A: There could be a small amount of population
ends of the distribution,
Area A: = 2500-3*400 = 1300
= 2500+3*400 = 3700
Area B: = 2000 - 3*200 = 1400
= 2000 + 3*200 = 2600
## Hence, Area A needs more attention
8
Population mean, variance and standard
deviation
## Ex: A machine company has a contract with one of
its customers to supply machined pump gears. One
requirement is that the diameter of its gears be
within specific limits. The following data is of
diameters of 20 sample gears.
4.01 4 4.02 4.03 4 3.98 3.99 3.99
4.01 4.02 3.99 3.98 3.97 4 4.02 4.01
4.02 4 4.01 3.99
## What can the company say to its customers about
the diameters of 95% of the gears.
9
Population mean, variance and standard
deviation
4.5
A: 4
3.5
3
2.5
2
1.5
1
Assuming that the distribution is bell shaped, then
0.5
0
95% of the gear diameters should
1 lie
2 between:
3 4 5 6 7
i.e. [3.970, 4.034] inches.
10
Chebyshev rule
## For any data set regardless of the shape of
distribution, the % of values that are found within
the distances of k>1 standard deviations from the
mean must be atleast (1-1/k2)*100%
## Ex: 36 students take an exam for which = 80 and
= 6. A rumor says 5 students have scores 62 or
below. Can the rumor be true ?
k = (62-80)/6=-3 => (1-1/k2)=1-1/9=8/9
So, atleast 8/9 fraction of students (i.e. 36*8/9=32)
scored between 62 and 80+3*6=98. So, not more
than 36-32=4 students are outside the range
[62,98]. So the rumor is False. 11
Population - grouped data
Population mean
Population variance
Population standard deviation
Here x is the midpoint of each class
Sample variance
Sample standard deviation
12
Population - grouped data
## Ex: A collar manufacturer is considering the
production of a new collar to attract young men.
Thus following statistics of neck circumference are
available based on measurement of a typical group
of the college students
Mid value (in
inches) 12 12.5 13 13.5 14 14.5 15 15.5 16
No. of students 2 16 36 60 76 37 18 3 2
## Compute std. dev. Using criterion to determine the
largest and smallest size of the collar he should
make to meet the needs of practically all customers
bearing in mind that collar are worn on average half
inch longer than neck size.
13
Population - grouped data
X Xf f fX fX2
fX fX2
12 122 2 24 288
24 288
12.5 16
12.5 16200 2500
200 2500
13 36
13 36468 6084
468 6084
13.5 60
13.5 60810 10935
810 10935
14 76
14 761064 14896
1064 14896
14.5 37 536.5 7779.25
14.5 37 536.5 7779.25
15 18 270 4050
15 18 270 4050
15.5 3 46.5 720.75
15.5 3 46.5 720.75
16 2 32 512
Totals 16
250 23451 32
47765 512
Totals 250 3451 47765
mean = 13.804
mean = 13.804
0.713852 0.713852
std.dev. = std. dev. =
14
Population - grouped data
## Hence, largest and smallest sizes are:
= 13.8043*0.71385 = 15.945 and 11.66
## Since customers wear collar half inch longer, 0.5
has to be added to the above two values
15
Population - grouped data
## Ex: The breaking strength of 80 test pieces of a
certain alloy are:
## Calculate the average breaking Breaking No. of
strength
Strength of the alloy and the standard pieces
44-46 3
Deviation. Calculate the percentage of46-48 24
Observations lying between . 48-50 27
50-52 21
52-54 5
16
Population - grouped data
X fX fX
f fXfX
2
fX2
45 345 135
3 6075
135 6075
47 2447 1128
24 53016
1128 53016
49 2749 1323
27 64827
1323 64827
51 2151 1071
21 54621
1071 54621
53 553 265
5 14045
265 14045
19258 19258
Totals 80
Totals 3922
80 4
3922 4
## mean = mean = 49.025 49.025
1.9619 1.9619
std. dev. = std. dev. = 82 82
## = 49.0252*1.962 = 45.103 and 52.949 i.e. ~45
and ~53.
17
Population - grouped data
## Lets assume equal number of observations spread
within lower and upper limits of each class.
## 45 is midpoint of class 44-46 of freq. 3 => 1.5 freq.
in this range
53 is midpoint of class 52-54 of freq. 5 => 2.5 freq.
in this range
## So, total observations between 45 and 53 are
1.5+24+27+21+2.5 = 76
18
Example problems
## Ex: A sociologist has been studying the yearly
changes in the no. of convicts assigned to the
largest correctional facility in the state. Most recent
data on % increase in the no. of prisoners are
200 200 200 200 200
2002 3 4 5 6 7
-5% 6% 9% 4% 7% -6%
## a) Calculate avg. % increase from 2002-2005
b) A new penal code was passed in 2001.
Previously the prison population grew at a rate of
about 2% per year. What seems to be the effect
of the new penal code.
19
Example problems
## A: Since we have % growth rate given annually,
G.M. is appropriate here.
a) G.M. =
= 1.03365
This is an avg. rate of 3.365% increase per year
b) G.M. =
This is 1.74% increase per year. Therefore with
the new penal code, the growth of no. of convicts
decreased from 2% to 1.74% per year.
20
Example problems
## Ex: A person owns 2 petrol filling stations A and B.
At A, a representative sample of 200 consumers
who purchase petrol was taken:
No. of No. of
liters consumers
0 and < 2 15
2 and < 4 40
4 and < 6 65
6 and < 8 40
8 and < 10 30
10 and over 10
A similar sample at B gives a mean of 4 liters and
std. dev. of 2.2 liters. At which station is purchase of
petrol relatively more variable ?
21
Example problems
## A: We are comparing two different samples here.
So we compare them using coefficient of variation
## No. of No. of Mid
liters consumers point fx fx2
0 and < 2 15 1 15 15
2 and < 4 40 3 120 360
4 and < 6 65 5 325 1625
6 and < 8 40 7 280 1960
8 and < 10 30 9 270 2430
10 and over 10 11 110 1210
Totals 200 1120 7600
22
Example problems
A: At station A,
5.6
Mean = 5.6
std. dev. =
2.57682 2.57682
CV =
46.01464% 46.01464%
At station B,
CV = 2.2/4*100 = 55%
## Hence purchase at B is more variable than in A.
23
Example problems
Sampling
Summary tables
Bar, pie, pareto charts
## Frequency distributions (relative, percentage)
Cumulative distribution
Histogram, Polygon
Cross tabulations, side-by-side bar charts
Scatter diagram, Time-series plot
## Mean, median and mode
Quartiles
Range, Interquartile range
Mean absolute deviation
Variance and standard deviation
Coefficient of variation
Z scores
Shape
24
References
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# MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical.
## Presentation on theme: "MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical."— Presentation transcript:
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §3.3a 3-Var Linear Systems
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review § Any QUESTIONS About §3.2b → More System Applications Any QUESTIONS About HomeWork §3.2 → HW-09 3.2 MTH 55
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 3 Bruce Mayer, PE Chabot College Mathematics Systems of 3-Variables A linear equation in three variables is an equation equivalent to form Ax + By + Cz = D Where A, B, C, and D are real numbers and A, B, and C are not all 0 A solution of a system of three equations in three variables is an ORDERED TRIPLE (x 1, y 1, z 1 ) that makes all three equations true.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example Ordered Triple Soln Determine whether (2, −1, 3) is a solution of the system SOLUTION: In all three equations, replace x with 2, y with −1, and z with 3. x + y + z = 4 2 + (–1) + 3 = 4 4 = 43 = 3 2x – 2y – z = 3 2(2) – 2(–1) – 3 = 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) = –3 –3 = –3
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example Ordered Triple Soln Determine whether (2, −1, 3) is a solution of the system SOLUTION: Because (2, −1, 3) satisfies all three equations in the system, this triple IS a solution for the system.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 6 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 1.Write each equation in the standard form of Ax + By+ Cz = D. 2.Eliminate one variable from one PAIR of equations using the elimination method. 3.If necessary, eliminate the same variable from another PAIR of equations to produce a system of TWO varialbles in TWO Equations
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 7 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 4.Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method. 5.To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the 3 rd variable 6.Check the ordered triple in all three of the original equations
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Solve System of Equations (1) (2) (3) SOLUTION: select any two of the three equations and work to get one equation in two variables. Let’s add eqs (1) & (2) (1) (2) (4) 2x + 3y = 8 Adding to eliminate z
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Next, select a different pair of equations and eliminate the same variable. Use (2) & (3) to again eliminate z. (5) x – y + 3z = 8 Multiplying equation (2) by 3 4x + 5y = 14. 3x + 6y – 3z = 6
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Now solve the resulting system of eqns (4) & (5). That will give us two of the numbers in the solution of the original system Multiply both sides of eqn (4) by −2 and then add to eqn (5): 4x + 5y = 14 2x + 3y = 8 (5) (4) 4x + 5y = 14 –4x – 6y = –16, –y = –2 y = 2
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Substituting into either equation (4) or (5) we find that x = 1 Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z. Let’s use eqn (1) and substitute our two numbers in it: x + y + z = 6 1 + 2 + z = 6 z = 3.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination At this point we have solved for All Three variables in the System: z = 3y = 2x = 1 Thus We have obtained the ordered triple (1, 2, 3). It should be checked in all three equations Finally the Solution Set: (1, 2, 3)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Solve System of Equations SOLUTION: The equations are in standard form and do not contain decimals or fractions. Eliminate z from eqns (2) & (3). (1) (2) (3) (2) (3) (4) 3x + 2y = 4
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Eliminate z from equations (1) and (2). Eliminate x using above and eqn (4) Multiplying equation (2) by 6 Adding Eqns 15x + 15y = 15 3x + 2y = 4 15x + 15y = 15 Multiplying top by 5 15x – 10y = 20 15x + 15y = 15 Adding Eqns 5y = 5 y = 1
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Using y = −1, find x from equation 4 by substituting Substitute x = 2 and y = −1 to find z 3x + 2y = 4 3x + 2( 1) = 4 x = 2 x + y + z = 2 2 – 1 + z = 2 1 + z = 2 z = 1 Thus The solution is the ordered triple (2, −1, 1)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination x – 4 y + 3 z = –9(3) Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 1 Eliminate a variable from the sum of two equations. The choice of the variable to eliminate is arbitrary. We will eliminate z. 6 x + 9 y – 3 z = 15 7 x + 5 y = 6Add. (4) Multiply each side of (1) by 3. 7 x + 5 y = 6 (4)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination 4 x – 16 y + 12 z = –36 Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 2 Eliminate the same variable, z, from any other equations. –9 x + 6 y – 12 z = 6 –5 x – 10 y = –30 Add. (5) Multiply each side of (2) by 3. 7 x + 5 y = 6 (4) Multiply each side of (3) by 4. –5 x – 10 y = –30 (5)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination –5 x – 10 y = –30 Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 3 Eliminate a different variable and solve. 14 x + 10 y = 12 9 x = –18 Add. (6) Multiply each side of (4) by 2. 7 x + 5 y = 6 (4) (5) –5 x – 10 y = –30 (5) x = –2
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 4 Find a second value by substituting –2 for x in (4) or (5). 7 x + 5 y = 6(4) 7 x + 5 y = 6 (4) Recall x = –2. –5 x – 10 y = –30 (5) y = 4 7(–2) + 5 y = 6 –14 + 5 y = 6 5 y = 20
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 5 Find a third value by substituting –2 for x and 4 for y into any of the three original equations. 2 x + 3 y – z = 5(1) 7 x + 5 y = 6 (4) From Before x = –2 and y = 4 –5 x – 10 y = –30 (5) 2(–2) + 3(4) – z = 5 8 – z = 5 z = 3
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example Solve by Elimination Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 6 Check that the ordered triple (–2, 4, 3) is the solution of the system. The ordered triple must satisfy all three original equations. 2 x + 3 y – z = 5 2(–2) + 3(4) – 3 = 5 –4 + 12 – 3 = 5 5 = 5 (1) –3 x + 2 y – 4 z = 2 –3(–2) + 2(4) – 4(3) = 2 6 + 8 – 12 = 2 2 = 2 (2) x – 4 y + 3 z = –9 (–2) – 4(4) + 3(3) = –9 –2 – 16 + 9 = –9 –9 = –9 (3)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example Missing Term(s) Multiply each side of (2) by 4. –20 y – 8 z = 12 Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Since equation (3) is missing the variable y, a good way to begin the solution is to eliminate y again by using equations (1) and (2). 25 x + 20 y = 115 25 x – 8 z = 127 Add. (4) 25 x – 8 z = 127 (4) Multiply each side of (1) by 5.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example Missing Term(s) Multiply each side of (4) by 9. 225 x – 72 z = 1143 Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Now use equations (3) and (4) to eliminate z and solve. –64 x + 72 z = –16 161 x = 1127 Add. (5) 25 x – 8 z = 127 (4) Multiply each side of (3) by 8. x = 7
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example Missing Term(s) 5(7) + 4 y = 23 Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Substituting into equation (1) gives 5 x + 4 y = 23 35 + 4 y = 23 25 x – 8 z = 127 (4) 4 y = –12 y = –3. –5(–3) – 2 z = 3 Substituting into equation (2) gives –5 y – 2 z = 3 15 – 2 z = 3 –2 z = –12 z = 6.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example Missing Term(s) Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Thus, x = 7, y = −3, and z = 6. Check these values in each of the original equations of the system to verify that the solution set of the system is the triple (7, −3, 6)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 26 Bruce Mayer, PE Chabot College Mathematics Inconsistent System If, in the process of performing elimination on a linear system, an equation of the form 0 = a occurs, where a ≠ 0, then the system has no solution and is inconsistent.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example Inconsistent Sys Solve the System of Equations SOLUTION: To eliminate x from Eqn (2), add −2 times Eqn (1) to Eqn (2)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example Inconsistent Sys Next add −3 times Eqn (1) to Eqn (3) to eliminate x from Eqn (3) We now have the following system
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 29 Bruce Mayer, PE Chabot College Mathematics Example Inconsistent Sys Now Multiply Eqn (4) by 1/3 to obtain To eliminate y from Eqn (5), add −1 times Eqn (6) to Eqn (5)
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example Inconsistent Sys We now have the system in simplified elimination form: This system is equivalent to the original system. Since equation (7) is false (a contradiction), we conclude that the solution set of the system is Ø (the NULL Set), and the system is INconsistent.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 31 Bruce Mayer, PE Chabot College Mathematics Dependent Systems If, in the process of performing an elimination solution for a linear system i.an equation of the form 0 = a (a ≠ 0) does not occur, but ii.an equation of the form 0 = 0 does occur, then the system of equations has infinitely many solutions and the equations are dependent.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 32 Bruce Mayer, PE Chabot College Mathematics Example Dependent System Solve the System of Equations SOLUTION: Eliminate x from Eqn (2) by adding −3 times Eqn (1) to Eqn (2
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 33 Bruce Mayer, PE Chabot College Mathematics Example Dependent System Eliminate x from Eqn (3) by adding −4 times Eqn (1) to Eqn (3) We now have the equivalent system
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example Dependent System To eliminate y from Eq (5), add −1 times Eqn (4) to Eqn (5) We now have the equivalent system in Final Form
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 35 Bruce Mayer, PE Chabot College Mathematics Example Dependent System The equation 0 = 0 may be interpreted as 0·z = 0, which is true for every value of z. Solving eqn (4) for y, we have y = 3z – 2. Substituting into eqn (1) and solving for x. Thus every triple (x, y, z) = (2z + 5, 3z – 2, z) is a soln of the system for each value of z. E.g, for z = 1, the triple is (7, 1, 1).
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 36 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems The graph of a linear equation in three variables, such as Ax + By + Cz = C (where A, B, and C are not all zero), is a plane in three-Dimensional (3D) space. Following are the possible situations for a system of three linear equations in three variables.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 37 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems a)Three planes intersect in a single point. The system has only one solution b)Three planes intersect in one line. The system has infinitely many solutions
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 38 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems c)Three planes coincide with each other. The system has only one solution. d)There are three parallel planes. The system has no solution.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 39 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems e)Two parallel planes are intersected by a third plane. The system has no solution. f)Three planes have no point in common. The system has no solution.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 40 Bruce Mayer, PE Chabot College Mathematics Example Quadratic Modeling The following table shows the higher-order multiple birth rates in the USA since 1971. At the right is a scatter plot of this data.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 41 Bruce Mayer, PE Chabot College Mathematics Example Quadratic Modeling We will SELECT three arbitrary ordered pairs to construct the model We Pick the Points (1,29), (11,40) and (21,100) and substitute the x & y values from these ordered pairs into the Quadratic Function
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 42 Bruce Mayer, PE Chabot College Mathematics Example Quadratic Modeling Selecting three representative ordered pairs, we can write a system of three equations. Solving this 3-Variable System by Elimination: This produces our Model
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 43 Bruce Mayer, PE Chabot College Mathematics Example Quadratic Modeling Use to the Model to Estimate the Hi- Order Births in 1993 In 1993 x = 23 Sub 23 into Model Thus we estimate that in 1993 the Hi-Order BirthRate was about 118
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 44 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work Problems From §3.3 Exercise Set 28 Quadratic Function
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 45 Bruce Mayer, PE Chabot College Mathematics All Done for Today A CONSISTENT System z = -1.553x - 2.642y - 10.272 (darker green) z = 1.416x - 1.92y - 10.979 (medium green) z = -.761x -.236y - 7.184 (lighter green) THE THREE PLANES SHARE ONE POINT
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 46 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 47 Bruce Mayer, PE Chabot College Mathematics Equivalent Systems Operations That Produce Equivalent Systems 1.Interchange the position of any two eqns 2.Multiply any eqn by a nonzero constant 3.Add a nonzero multiple of one eqn to another A special type of Elimination called Gaussian Elimination uses these steps to solve multivariable systems
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 48 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 1.Rearrange the equations, if necessary, to obtain an x-term with a nonzero coefficient in the first equation. Then multiply the first equation by the reciprocal of the coefficient of the x-term to get 1 as a leading coefficient. 2.By adding appropriate multiples of the first equation, eliminate any x-terms from the second and third equations
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 49 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 2.(cont.) Multiply the resulting second equation by the reciprocal of the coefficient of the y-term to get 1 as the leading coefficient. 3.If necessary by adding appropriate multiple of the second equation from Step 2, eliminate any y-term from the third equation. Solve the resulting equation for z.
BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 50 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 4.Back-substitute the values of z from Steps 3 into one of the equations in Step 3 that contain only y and z, and solve for y. 5.Back-substitute the values of y and z from Steps 3 and 4 in any equation containing x, y, and z, and solve for x 6.Write the solution set (Soln Triple) 7.Check soln in the original equations
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T8.1 Chapter Outline
```Stock Valuation
Common Stock Valuation
Some Features of Common and Preferred Stocks
The Stock Markets
T2 Common Stock Cash Flows and the Fundamental Theory of Valuation
In 1938, John Burr Williams postulated what has become the
fundamental theory of valuation:
The value of any financial asset equals the present value of all of its
future cash flows.
For common stocks, this implies the following:
P0 =
D1
+
(1 + r)1
P1
and
(1 + r)1
P1 =
D2
(1 + r)1
+
P2
(1 + r)1
substituting for P1 gives
P0 =
P0 =
D1
(1 + r)1
D1
(1 + r)1
+
+
D2
+
(1 + r)2
D2
(1 + r)2
+
P2
(1 + r)2
D3
(1 + r)3
. Continuing to substitute, we obtain
+
D4
(1 + r)4
+ …
T3 Common Stock Valuation: The Zero Growth Case
According to the fundamental theory of value, the value of a
financial asset at any point in time equals the present value of all
future dividends.
If all future dividends are the same, the present value of the
dividend stream constitutes a perpetuity.
The present value of a perpetuity is equal to
C/r or, in this case, D1/r.
Question:
Cooper, Inc. common stock currently pays a \$1.00
dividend, which is expected to remain constant
forever. If the required return on Cooper stock is
10%, what should the stock sell for today?
P0 = \$1/.10 = \$10.
Question:
Given no change in the variables, what will the stock
be worth in one year?
T4 Common Stock Valuation: The Zero Growth Case (concluded)
One year from now, the value of the stock, P1, must
be equal to the present value of all remaining future
dividends.
Since the dividend is constant, D2 = D1 , and
P1 = D2/r = \$1/.10 = \$10.
In other words, in the absence of any changes in expected cash
flows (and given a constant discount rate), the price of a nogrowth stock will never change.
Put another way, there is no reason to expect capital gains income
from this stock.
T5 Common Stock Valuation: The Constant Growth Case
In reality, investors generally expect the firm (and the dividends it
pays) to grow over time. How do we value a stock when each dividend
differs than the one preceding it?
As long as the rate of change from one period to the next, g, is
constant, we can apply the growing perpetuity model:
P0 =
P0 =
D1
(1 + r)1
+
D2
(1 + r)2
D0(1 + g)
r-g
=
+
D3
(1 + r)3
D1
r-g
+ …=
D0(1+g)1
(1 + r)1
+
D0(1+g)2
(1 + r)2
D0(1+g)3
+
+ ...
(1 + r)3
.
Now assume that D1 = \$1.00, r = 10%, but dividends are expected to
increase by 5% annually. What should the stock sell for today?
T6 Common Stock Valuation: The Constant Growth Case (concluded)
The equilibrium value of this constant-growth stock is
D1
\$1.00
=
r-g
Question:
= \$20
.10 - .05
What would the value of the stock be if the growth rate
were only 3%?
D1
\$1.00
=
= \$14.29.
r-g
.10 - .03
Why does a lower growth rate result in a lower value? Stay tuned.
T7 Stock Price Sensitivity to Dividend Growth, g
Stock price (\$)
50
45
D1 = \$1
Required return, r, = 12%
40
35
30
25
20
15
10
5
0
2%
4%
6%
8%
10%
Dividend growth
rate, g
T8 Stock Price Sensitivity to Required Return, r
Stock price (\$)
100
90
80
D1 = \$1
Dividend growth rate, g, = 5%
70
60
50
40
30
20
10
Required return, r
6%
8%
10%
12%
14%
T9 Common Stock Valuation - The Nonconstant Growth Case
For many firms (especially those in new or high-tech industries),
dividends are low and expected to grow rapidly. As product markets
mature, dividends are then expected to slow to some “steady state”
rate. How should stocks such as these be valued?
today equals the present value of all future cash flows.
Put another way, the nonconstant growth model suggests that
P0 = present value of dividends in the nonconstant growth period(s)
+ present value of dividends in the “steady state” period.
T10 Quick Quiz -- Part 1 of 3
Suppose a stock has just paid a \$5 per share dividend. The dividend is
projected to grow at 5% per year indefinitely. If the required return is 9%,
then the price today is _______ ?
P0 = D1/(r - g)
= \$5
(______ )/(______ -______ )
= \$5.25/.04
= \$131.25 per share
What will the price be in a year?
Pt
= Dt+1/(r - g)
P1 = D___ /(r - g) = (\$______
1.05)/(.09 - .05) = \$137.8125
By what percentage does P1 exceed P0? Why?
T11 Quick Quiz -- Part 2 of 3
Find the required return:
Suppose a stock has just paid a \$5 per share dividend.
The dividend is projected to grow at 5% per year indefinitely.
If the stock sells today for \$65 5/8, what is the required return?
P0 = D1/(r - g)
(r - g) = D1/P0
r
= D1/P0 + g
= \$5.25/\$65.625 + .05
= dividend yield (_____) + capital gain yield (_____)
= ____
T12 Summary of Stock Valuation
I. The General Case
In general, the price today of a share of stock, P0, is the present value of all its future
dividends, D1, D2, D3, . . .
P0 =
D1
(1 + r)1
+
D2
(1 + r)2
+
D3
(1 + r)3
+ …
where r is the required return.
II. Constant Growth Case
If the dividend grows at a steady rate, g, then the price can be written as:
P0 = D1/(r - g)
This result is the dividend growth model.
III. The Required Return
The required return, r, can be written as the sum of two things:
r = D1/P0 + g
where D1/P0 is the dividend yield and g is the capital gain yield.
T13 Sample Stock Quotation from The Wall Street Journal
New York Stock Exchange Composite Transactions
Quotations as of 5 p.m. Eastern Time
Monday, March 28, 1994
52 Weeks
Hi
Yld
Lo
Stock
28 1/2
25 1/2
28 1/4
Sym
Vol
Net
Div
%
PE
100s
Hi
Lo
Close
Chg.
MellonBk pfJ
2.13
8.3
…
27
26
25 3/4
25 3/4 +
1/8
25 5/8
MellonBk pfK
2.05
7.8
…
55
26 1/8
26
26 1/8
…
48 1/4
35 3/4
Melville
MES
1.52
4.0
13
1439
38
37 2/8
37 5/8 +
1/4
12 1/8
9 3/8
MentorIncoFd
MRF
.96
9.5
…
142
10 1/4
10 1/8
10 1/8
…
56 1/2
43 5/8
MercBcpMO
MTL
1.68
3.4
10
219
50
49 1/2
49 3/4 -
1/8
41 1/8
29 7/8
MercStrs
MST
1.02
2.6
17
636
40 5/8
39 5/8
39 7/8 -
1 1/4
39 3/8
28 5/8
Merck
MRK
1.12
3.7
16
21595
30 3/8
30
30 3/8 +
1/4
20 3/8
12 1/8
MercuryFin
MFN
.28f
1.7
30
517
17
16 3/4
16 3/4
…
Source: Reprinted by permission of The Wall Street Journal, ©1991 Dow Jones & Company, Inc.,
T14 Quick Quiz -- Part 3 of 3
Suppose a stock has just paid a \$5 per share dividend. The
dividend is projected to grow at 10% for the next two
years, the 8% for one year, and then 6% indefinitely. The
required return is 12%. What is the stock’s value?
Time Dividend
0
\$ 5.00
1
\$ ____
(10% growth)
2
\$ ____
(10% growth)
3
\$6.534
( __% growth)
4
\$6.926
( __% growth)
T15 Quick Quiz -- Part 3 of 3 (concluded)
At time 3, the value of the stock will be:
P3
=
D4/(r - g) = \$_____ /(.12 - .06) = \$115.434
The value today of the stock is thus:
P0
=
D1/(1 + r) + D2/(1 + r)2 + D3/(1 + r)3 + P3/(1 + r)3
=
\$5.5/1.12 + \$6.05/1.122 + \$6.534/1.123 + \$115.434/1.12---
=
\$96.55
T16 Solution to Problem
MegaCapital, Inc. just paid a dividend of \$2.00 per share on
its stock. The dividends are expected to grow at a constant
6 percent per year indefinitely. If investors require a 13
percent return on MegaCapital stock, what is the current
price? What will the price be in 3 years? In 15 years?
According to the constant growth model,
P0 = D1/(r - g) = \$2.00(1.06)/(.13 - .06) = \$30.29
If the constant growth model holds, the price of the stock
will grow at g percent per year, so
P3 = P0 x (1 + g)3 = \$30.29 x (1.06)3 = \$36.07, and
P15 = P0 x (1 + g)15 = \$30.29 x (1.06)15 = \$72.58.
```
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| 3.5 | 4 |
CC-MAIN-2016-44
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latest
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https://gmatclub.com/forum/in-the-xy-plane-point-r-s-lies-on-a-circle-with-center-at-the-orig-165616.html
| 1,721,788,177,000,000,000 |
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# In the xy-plane, point (r, s) lies on a circle with center at the orig
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Thanks for the brilliant explanation. One thing I don't get the question is that, the point (r,s) could be anywhere in the circle, not only on its circumference. Why does it refer only to a point on the circumference? Thanks!
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If a circle, lying on a xy-plane, has its center at the origin, the equation is x^2+y^2=R^2, where x & y are points on the circle and R is the radius of the circle.
Since x & y from the equation x^2+y^2=R^2 is similar to r & s in the question, we can rewrite as r^2+s^2=R^2.
Statement (1) gives us the value of radius, R; Therefore, r^2+s^2=4; Sufficient.
Statement (2) gives us the value of a point on the circle
=> sub x & y values of the point in r^2+s^2=R^22: 2+2 = 4;
r^2 = 4; Sufficient.
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
Hi,
i don't understand from Bunuels solution how we come up with the value of S in statement 1 in order to answer what r^2 + s^2 is?
r^2 equals 4, that is all clear, but s should be y-value, how do we get that?
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
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noTh1ng wrote:
Hi,
i don't understand from Bunuels solution how we come up with the value of S in statement 1 in order to answer what r^2 + s^2 is?
r^2 equals 4, that is all clear, but s should be y-value, how do we get that?
$$r^2+s^2=radius^2$$. (1) says that radius = 2, thus $$r^2+s^2=2^2$$.
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
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noTh1ng wrote:
Hi,
i don't understand from Bunuels solution how we come up with the value of S in statement 1 in order to answer what r^2 + s^2 is?
r^2 equals 4, that is all clear, but s should be y-value, how do we get that?
Hi noTh1ng,
In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of r^2 + s^2?
(1) The circle has radius 2
(2) The point (2√, −2√) lies on the circle
You seem to have misunderstood a little here.
The equation of Circle is given by $$x^2 + y^2 = Radius^2$$
Given : (r,s) lie on the circle
i.e. (r,s) will satisfy the equation of Circle
i.e. $$r^2 + s^2 = Radius^2$$
Question : Find the value of $$r^2 + s^2$$? but since $$r^2 + s^2 = Radius^2$$ therefore, the question becomes
Question : Find the value of $$Radius^2$$?
Statement 1: The circle has radius 2
i.e. $$r^2 + s^2 = Radius^2 = 2^2 = 4$$
SUFFICIENT
Statement 2: The point (√2, −√2) lies on the circle
i.e. (√2, −√2) will satisfy the equation of circle
i.e. (√2)^2 + (−√2)^2 = Radius^2
hence, $$r^2 + s^2 = Radius^2 = 2^2 = 4$$ Hence,
SUFFICIENT
I hope it helps!
Please Note: You have been confused r (X-co-ordinate) and r (Radius) as it seems from your question
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
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Hi,
A very fundamental or, maybe, a silly question, in GMAT, when a question like this reads 'on a circle', it's supposed to mean, on the circumference of the circle, and not the whole of the area of the circle.
Thanks
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
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WilDThiNg wrote:
Hi,
A very fundamental or, maybe, a silly question, in GMAT, when a question like this reads 'on a circle', it's supposed to mean, on the circumference of the circle, and not the whole of the area of the circle.
Thanks
Yes, on the circle means on the circumference.
In the circle means within.
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
So, for statement 2, any coordinates that are on the circle can be used to substitute for (r,s)?
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
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parkerd wrote:
So, for statement 2, any coordinates that are on the circle can be used to substitute for (r,s)?
Yes, if a circle is centred at the origin, then the x and y-coordinates of any point on the circle (so on the circumference) will satisfy $$x^2+y^2=radius^2$$
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
If the (r,s) was inside the circle would you still be able to solve with the same equation like in statement 1? Or be able to substitute for (r,s) with the given values on the circle like in statement 2?
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
parkerd wrote:
If the (r,s) was inside the circle would you still be able to solve with the same equation like in statement 1? Or be able to substitute for (r,s) with the given values on the circle like in statement 2?
If (r, s) were IN the circle, then the answer would be E because each statement gives us basically the same info - the length of the radius. How can we find the sum of the squares of a random point inside the circle just knowing the radius?
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
saurya_s wrote:
In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of r^2 + s^2?
(1) The circle has radius 2
(2) The point $$(\sqrt{2}, \ -\sqrt{2})$$ lies on the circle
If a point (a, b) is on a circle with center at origin and radius = r, then a^2 + b^2 = r^2
Simply Q: r = ?
1) Sufficient
2) 2 + 2 = 4 = r^2 => Sufficient
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
Statement 1 - r and s are two different points on a circle so both have values of 2^2 and 2^ so r^2+s^2 = 8
Is this the correct inference?
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
gayatri259 wrote:
Statement 1 - r and s are two different points on a circle so both have values of 2^2 and 2^ so r^2+s^2 = 8
Is this the correct inference?
That's not correct.
The question stem tells us the point lies on the circle with center at the origin (0,0). If the radius is 2, this means the coordinates could be (2,0), (0,2), etc.)
$$r^2 + s^2$$ will always equal 4.
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Re: In the xy-plane, point (r, s) lies on a circle with center at the orig [#permalink]
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saurya_s wrote:
In the xy-plane, point (r, s) lies on a circle with center at the origin. What is the value of r^2 + s^2?
(1) The circle has radius 2
(2) The point $$(\sqrt{2}, \ -\sqrt{2})$$ lies on the circle
DS02741
Solution:
Question Stem Analysis:
We need to determine the value of r^2 + s^2, given that (r, s) is a point on the circle with center at the origin. Notice that the equation of such a circle is x^2 + y^2 = R^2 where R is the radius of the circle. Since (r, s) is a point on the circle, we have r^2 + s^2 = R^2. That is, in order to determine the value of r^2 + s^2, we either need to know the coordinates of a point on the circle or just the value of R.
Statement One Alone:
Since the radius of the circle is 2, R = 2. Therefore, r^2 + s^2 = 2^2 = 4. Statement one alone is sufficient.
Statement Two Alone:
Since (√2, -√2) is a point on the circle, x = √2 and y = -√2 will satisfy x^2 + y^2 = R^2. Substituting, we find R^2 = x^2 + y^2 = (√2)^2 + (-√2)^2 = 2+ 2 = 4. Proceeding as above, we obtain r^2 + s^2 = 4. Statement two alone is sufficient.
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# Hypothesis Test: Difference Between Means
This lesson explains how to conduct a hypothesis test for the difference between two means. The test procedure, called the two-sample t-test, is appropriate when the following conditions are met:
• The sampling method for each sample is simple random sampling.
• The samples are independent.
• Each population is at least 20 times larger than its respective sample.
• The sampling distribution is approximately normal, which is generally the case if any of the following conditions apply.
• The population distribution is normal.
• The population data are symmetric, unimodal, without outliers, and the sample size is 15 or less.
• The population data are slightly skewed, unimodal, without outliers, and the sample size is 16 to 40.
• The sample size is greater than 40, without outliers.
This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.
## State the Hypotheses
Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis. The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.
The table below shows three sets of null and alternative hypotheses. Each makes a statement about the difference d between the mean of one population μ1 and the mean of another population μ2. (In the table, the symbol ≠ means " not equal to ".)
Set Null hypothesis Alternative hypothesis Number of tails
1 μ1 - μ2 = d μ1 - μ2 ≠ d 2
2 μ1 - μ2 > d μ1 - μ2 < d 1
3 μ1 - μ2 < d μ1 - μ2 > d 1
The first set of hypotheses (Set 1) is an example of a two-tailed test, since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests, since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.
When the null hypothesis states that there is no difference between the two population means (i.e., d = 0), the null and alternative hypothesis are often stated in the following form.
Ho: μ1 = μ2
Ha: μ1 ≠ μ2
## Formulate an Analysis Plan
The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.
• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the two-sample t-test to determine whether the difference between means found in the sample is significantly different from the hypothesized difference between means.
## Analyze Sample Data
Using sample data, find the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.
• Standard error. Compute the standard error (SE) of the sampling distribution.
SE = sqrt[ (s12/n1) + (s22/n2) ]
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
• Degrees of freedom. The degrees of freedom (DF) is:
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
If DF does not compute to an integer, round it off to the nearest whole number. Some texts suggest that the degrees of freedom can be approximated by the smaller of n1 - 1 and n2 - 1; but the above formula gives better results.
• Test statistic. The test statistic is a t statistic (t) defined by the following equation.
t = [ (x1 - x2) - d ] / SE
where x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, having the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)
## Interpret Results
If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.
In this section, two sample problems illustrate how to conduct a hypothesis test of a difference between mean scores. The first problem involves a two-tailed test; the second problem, a one-tailed test.
Problem 1: Two-Tailed Test
Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students.
At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15.
Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.)
Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: μ1 - μ2 = 0
Alternative hypothesis: μ1 - μ2 ≠ 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
• Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
• <Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(102/30) + (152/25] = sqrt(3.33 + 9)
SE = sqrt(12.33) = 3.51
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (102/30 + 152/25)2 / { [ (102 / 30)2 / (29) ] + [ (152 / 25)2 / (24) ] }
DF = (3.33 + 9)2 / { [ (3.33)2 / (29) ] + [ (9)2 / (24) ] } = 152.03 / (0.382 + 3.375) = 152.03/3.757 = 40.47
t = [ (x1 - x2) - d ] / SE = [ (78 - 85) - 0 ] / 3.51 = -7/3.51 = -1.99
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than -1.99; that is, less than -1.99 or greater than 1.99.
We use the t Distribution Calculator to find P(t < -1.99) = 0.027, and P(t > 1.99) = 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.
• Interpret results. Since the P-value (0.054) is less than the significance level (0.10), we cannot accept the null hypothesis.
Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the samples were drawn from a normal population.
Problem 2: One-Tailed Test
The Acme Company has developed a new battery. The engineer in charge claims that the new battery will operate continuously for at least 7 minutes longer than the old battery.
To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 20 minutes; the new batteries, 200 minutes with a standard deviation of 40 minutes.
Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance. (Assume that there are no outliers in either sample.)
Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
• State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: μ1 - μ2 <= 7
Alternative hypothesis: μ1 - μ2 > 7
where μ1 is battery life for the new battery, and μ2 is battery life for the old battery.
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the difference between sample means is bigger than would be expected by chance.
• Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
• Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(402/100) + (202/100]
SE = sqrt(16 + 4) = 4.472
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (402/100 + 202/100)2 / { [ (402 / 100)2 / (99) ] + [ (202 / 100)2 / (99) ] }
DF = (20)2 / { [ (16)2 / (99) ] + [ (2)2 / (99) ] } = 400 / (2.586 + 0.162) = 145.56
t = [ (x1 - x2) - d ] / SE = [(200 - 190) - 7] / 4.472 = 3/4.472 = 0.67
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
Here is the logic of the analysis: Given the alternative hypothesis (μ1 - μ2 > 7), we want to know whether the observed difference in sample means is big enough (i.e., sufficiently greater than 7) to cause us to reject the null hypothesis.
The observed difference in sample means (10) produced a t statistic of 0.67. We use the t Distribution Calculator to find P(t < 0.67) = 0.75.
• Interpret results. Suppose we replicated this study many times with different samples. If the true difference in population means were actually 7, we would expect the observed difference in sample means to be 10 or less in 75% of our samples. And we would expect to find an observed difference to be more than 10 in 25% of our samples Therefore, the P-value in this analysis is 0.25.
Since the P-value (0.25) is greater than the significance level (0.05), we cannot reject the null hypothesis. Thus, these results do not provide statistically significant evidence in support of the engineer's claim that the new battery will last at least 7 minutes longer than the old battery.
Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the sample size was large without outliers.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > gaid Structured version Visualization version GIF version
Theorem gaid 17924
Description: The trivial action of a group on any set. Each group element corresponds to the identity permutation. (Contributed by Jeff Hankins, 11-Aug-2009.) (Proof shortened by Mario Carneiro, 13-Jan-2015.)
Hypothesis
Ref Expression
gaid.1 𝑋 = (Base‘𝐺)
Assertion
Ref Expression
gaid ((𝐺 ∈ Grp ∧ 𝑆𝑉) → (2nd ↾ (𝑋 × 𝑆)) ∈ (𝐺 GrpAct 𝑆))
Proof of Theorem gaid
Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 elex 3344 . . 3 (𝑆𝑉𝑆 ∈ V)
21anim2i 594 . 2 ((𝐺 ∈ Grp ∧ 𝑆𝑉) → (𝐺 ∈ Grp ∧ 𝑆 ∈ V))
3 gaid.1 . . . . . . . 8 𝑋 = (Base‘𝐺)
4 eqid 2752 . . . . . . . 8 (0g𝐺) = (0g𝐺)
53, 4grpidcl 17643 . . . . . . 7 (𝐺 ∈ Grp → (0g𝐺) ∈ 𝑋)
65adantr 472 . . . . . 6 ((𝐺 ∈ Grp ∧ 𝑆𝑉) → (0g𝐺) ∈ 𝑋)
7 ovres 6957 . . . . . . 7 (((0g𝐺) ∈ 𝑋𝑥𝑆) → ((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = ((0g𝐺)2nd 𝑥))
8 df-ov 6808 . . . . . . . 8 ((0g𝐺)2nd 𝑥) = (2nd ‘⟨(0g𝐺), 𝑥⟩)
9 fvex 6354 . . . . . . . . 9 (0g𝐺) ∈ V
10 vex 3335 . . . . . . . . 9 𝑥 ∈ V
119, 10op2nd 7334 . . . . . . . 8 (2nd ‘⟨(0g𝐺), 𝑥⟩) = 𝑥
128, 11eqtri 2774 . . . . . . 7 ((0g𝐺)2nd 𝑥) = 𝑥
137, 12syl6eq 2802 . . . . . 6 (((0g𝐺) ∈ 𝑋𝑥𝑆) → ((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
146, 13sylan 489 . . . . 5 (((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) → ((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
15 simprl 811 . . . . . . . 8 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → 𝑦𝑋)
16 simplr 809 . . . . . . . 8 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → 𝑥𝑆)
17 ovres 6957 . . . . . . . . 9 ((𝑦𝑋𝑥𝑆) → (𝑦(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦2nd 𝑥))
18 df-ov 6808 . . . . . . . . . 10 (𝑦2nd 𝑥) = (2nd ‘⟨𝑦, 𝑥⟩)
19 vex 3335 . . . . . . . . . . 11 𝑦 ∈ V
2019, 10op2nd 7334 . . . . . . . . . 10 (2nd ‘⟨𝑦, 𝑥⟩) = 𝑥
2118, 20eqtri 2774 . . . . . . . . 9 (𝑦2nd 𝑥) = 𝑥
2217, 21syl6eq 2802 . . . . . . . 8 ((𝑦𝑋𝑥𝑆) → (𝑦(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
2315, 16, 22syl2anc 696 . . . . . . 7 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → (𝑦(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
24 simprr 813 . . . . . . . . 9 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → 𝑧𝑋)
25 ovres 6957 . . . . . . . . . 10 ((𝑧𝑋𝑥𝑆) → (𝑧(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑧2nd 𝑥))
26 df-ov 6808 . . . . . . . . . . 11 (𝑧2nd 𝑥) = (2nd ‘⟨𝑧, 𝑥⟩)
27 vex 3335 . . . . . . . . . . . 12 𝑧 ∈ V
2827, 10op2nd 7334 . . . . . . . . . . 11 (2nd ‘⟨𝑧, 𝑥⟩) = 𝑥
2926, 28eqtri 2774 . . . . . . . . . 10 (𝑧2nd 𝑥) = 𝑥
3025, 29syl6eq 2802 . . . . . . . . 9 ((𝑧𝑋𝑥𝑆) → (𝑧(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
3124, 16, 30syl2anc 696 . . . . . . . 8 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → (𝑧(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
3231oveq2d 6821 . . . . . . 7 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥)) = (𝑦(2nd ↾ (𝑋 × 𝑆))𝑥))
33 simpll 807 . . . . . . . . 9 (((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) → 𝐺 ∈ Grp)
34 eqid 2752 . . . . . . . . . . 11 (+g𝐺) = (+g𝐺)
353, 34grpcl 17623 . . . . . . . . . 10 ((𝐺 ∈ Grp ∧ 𝑦𝑋𝑧𝑋) → (𝑦(+g𝐺)𝑧) ∈ 𝑋)
36353expb 1113 . . . . . . . . 9 ((𝐺 ∈ Grp ∧ (𝑦𝑋𝑧𝑋)) → (𝑦(+g𝐺)𝑧) ∈ 𝑋)
3733, 36sylan 489 . . . . . . . 8 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → (𝑦(+g𝐺)𝑧) ∈ 𝑋)
38 ovres 6957 . . . . . . . . 9 (((𝑦(+g𝐺)𝑧) ∈ 𝑋𝑥𝑆) → ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = ((𝑦(+g𝐺)𝑧)2nd 𝑥))
39 df-ov 6808 . . . . . . . . . 10 ((𝑦(+g𝐺)𝑧)2nd 𝑥) = (2nd ‘⟨(𝑦(+g𝐺)𝑧), 𝑥⟩)
40 ovex 6833 . . . . . . . . . . 11 (𝑦(+g𝐺)𝑧) ∈ V
4140, 10op2nd 7334 . . . . . . . . . 10 (2nd ‘⟨(𝑦(+g𝐺)𝑧), 𝑥⟩) = 𝑥
4239, 41eqtri 2774 . . . . . . . . 9 ((𝑦(+g𝐺)𝑧)2nd 𝑥) = 𝑥
4338, 42syl6eq 2802 . . . . . . . 8 (((𝑦(+g𝐺)𝑧) ∈ 𝑋𝑥𝑆) → ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
4437, 16, 43syl2anc 696 . . . . . . 7 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥)
4523, 32, 443eqtr4rd 2797 . . . . . 6 ((((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) ∧ (𝑦𝑋𝑧𝑋)) → ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥)))
4645ralrimivva 3101 . . . . 5 (((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) → ∀𝑦𝑋𝑧𝑋 ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥)))
4714, 46jca 555 . . . 4 (((𝐺 ∈ Grp ∧ 𝑆𝑉) ∧ 𝑥𝑆) → (((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥 ∧ ∀𝑦𝑋𝑧𝑋 ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥))))
4847ralrimiva 3096 . . 3 ((𝐺 ∈ Grp ∧ 𝑆𝑉) → ∀𝑥𝑆 (((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥 ∧ ∀𝑦𝑋𝑧𝑋 ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥))))
49 f2ndres 7350 . . 3 (2nd ↾ (𝑋 × 𝑆)):(𝑋 × 𝑆)⟶𝑆
5048, 49jctil 561 . 2 ((𝐺 ∈ Grp ∧ 𝑆𝑉) → ((2nd ↾ (𝑋 × 𝑆)):(𝑋 × 𝑆)⟶𝑆 ∧ ∀𝑥𝑆 (((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥 ∧ ∀𝑦𝑋𝑧𝑋 ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥)))))
513, 34, 4isga 17916 . 2 ((2nd ↾ (𝑋 × 𝑆)) ∈ (𝐺 GrpAct 𝑆) ↔ ((𝐺 ∈ Grp ∧ 𝑆 ∈ V) ∧ ((2nd ↾ (𝑋 × 𝑆)):(𝑋 × 𝑆)⟶𝑆 ∧ ∀𝑥𝑆 (((0g𝐺)(2nd ↾ (𝑋 × 𝑆))𝑥) = 𝑥 ∧ ∀𝑦𝑋𝑧𝑋 ((𝑦(+g𝐺)𝑧)(2nd ↾ (𝑋 × 𝑆))𝑥) = (𝑦(2nd ↾ (𝑋 × 𝑆))(𝑧(2nd ↾ (𝑋 × 𝑆))𝑥))))))
522, 50, 51sylanbrc 701 1 ((𝐺 ∈ Grp ∧ 𝑆𝑉) → (2nd ↾ (𝑋 × 𝑆)) ∈ (𝐺 GrpAct 𝑆))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 = wceq 1624 ∈ wcel 2131 ∀wral 3042 Vcvv 3332 ⟨cop 4319 × cxp 5256 ↾ cres 5260 ⟶wf 6037 ‘cfv 6041 (class class class)co 6805 2nd c2nd 7324 Basecbs 16051 +gcplusg 16135 0gc0g 16294 Grpcgrp 17615 GrpAct cga 17914 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1863 ax-4 1878 ax-5 1980 ax-6 2046 ax-7 2082 ax-8 2133 ax-9 2140 ax-10 2160 ax-11 2175 ax-12 2188 ax-13 2383 ax-ext 2732 ax-sep 4925 ax-nul 4933 ax-pow 4984 ax-pr 5047 ax-un 7106 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1074 df-tru 1627 df-ex 1846 df-nf 1851 df-sb 2039 df-eu 2603 df-mo 2604 df-clab 2739 df-cleq 2745 df-clel 2748 df-nfc 2883 df-ne 2925 df-ral 3047 df-rex 3048 df-reu 3049 df-rmo 3050 df-rab 3051 df-v 3334 df-sbc 3569 df-csb 3667 df-dif 3710 df-un 3712 df-in 3714 df-ss 3721 df-nul 4051 df-if 4223 df-pw 4296 df-sn 4314 df-pr 4316 df-op 4320 df-uni 4581 df-iun 4666 df-br 4797 df-opab 4857 df-mpt 4874 df-id 5166 df-xp 5264 df-rel 5265 df-cnv 5266 df-co 5267 df-dm 5268 df-rn 5269 df-res 5270 df-ima 5271 df-iota 6004 df-fun 6043 df-fn 6044 df-f 6045 df-fv 6049 df-riota 6766 df-ov 6808 df-oprab 6809 df-mpt2 6810 df-2nd 7326 df-map 8017 df-0g 16296 df-mgm 17435 df-sgrp 17477 df-mnd 17488 df-grp 17618 df-ga 17915 This theorem is referenced by: (None)
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We live in a statistical ecosystem that is dominated by WAR, a statistic that for all its perks does contain some weaknesses. WAR–in an attempt to compare all players to a common baseline–specifically assigns a value to players with the intention of stripping away all of the context of his teammates. There’s no secret here. This is celebrated as the great triumph of WAR. Where RBI or runs scored were decent indicators of a hitter’s abilities, they were also dependent on the abilities of his teammates. As an individual measure, WAR makes sense as a way to compare everyone to the same baseline.
Problem: the big trophy at the end of the year goes to the team that won the World Series. Sure, the more individual talent that a team has on its bench, the better. However, is the way to evaluate a team to simply add up all the WAR?
I think at this point, we have enough evidence to say that the answer is “no.” We know for example that WAR doesn’t do a great job valuing relievers, mostly because the actual amount of value that a reliever will produce has a bit more to do with what role he’s used in, purposely or accidentally, than strictly his stats. The more interesting question is how much of a difference that makes. If the answer is “the whole is greater than the sum of its parts, but the effect size is only half a run,” then it’s not something that we should worry about.
I don’t think we’re going to solve this question today, because for one, there is going to be variance between a team’s talent (in the sense of “add up all the WAR”) and their results because of timing and random chance. But what if there are provable ways in which a team can generate value by putting its players together in a certain way?
Warning! Gory Mathematical Details Ahead!
In 2016, there were 21,744 runs scored, MLB-wide. There were 5,610 home runs hit. That means that about three quarters of all runs scored were, by definition, the work of at least two players. A hitter got on base and someone else knocked him in. Baseball might be a game where the atomic unit is the pitcher-batter confrontation, but the way that runs are scored is a matter of interaction.
In the past, I’ve done research on the idea of a stars-and-scrubs lineup, by starting out with a lineup of nine league-average batters, and then making one of the hitters better, and another worse by an equal amount. I then made a lineup with two “stars” and two “scrubs” book-ending a lineup with five average hitters. Because teams can load the “stars” up front in the lineup, and get them more plate appearances, they will score more runs with stars and scrubs than they would with a lineup of nine average hitters.
One thing that I wanted to look into more generally was the idea of bunching within a lineup. For example, if a leadoff hitter is good at getting on base, it’s not the overall abilities of the other eight hitters that determines whether his hits will eventually become runs. It’s mostly the quality of the second and third hitters that we should look to. Similarly, the second hitter is going to be most affected by the two or three guys behind him.
It’s one thing to have a bad hitter and to hit him ninth. It’s another thing to have three bad hitters and have to hit them 7-8-9. It allows a valley where it’s just easier to get three outs. It means that even if the sixth hitter is a good one, he’s not going to get as much help scoring. Having better hitters in the lineup who don’t make outs as often also means more plate appearances for someone else in the lineup. It can beget a virtuous cycle where having a fifth hitter who doesn’t make as many outs means more plate appearances for the sixth hitter, and if he’s good, that’s more chance for him to add value and to “create” another plate appearance for someone else.
On the flip side, having a bunch of bad hitters means that if a lineup does have a good hitter, he won’t get as many plate appearances. Offense works in such a way that a bunch of good hitters will actually compound each other. I wanted to look at the effect a little more, so I created (well, brought back out) a lineup simulation model. For the initiated, it’s a Monte Carlo Markov model, which is a nice way of saying that I built a simulator that models baseball through a bunch of dice rolls. That’s not a perfect way to model baseball, but it does allow us to control all of the inputs.
I started with a lineup that was composed of nine league-average (using 2016 stats) hitters. I then slowly replaced each hitter, starting at the bottom of the lineup, with a hitter who had the composite stats of an American League ninth-spot hitter. The actual downgrade, in absolute terms, for each replacement is the same, but how many of them and where they hit in the lineup changes. I had the computer run 50,000 nine-inning simulations for each lineup.
And the results are …
Lineup Runs Per Game Delta Per 162 (from above line) All Average 4.2498 *** One #9 hitter 4.1209 20.88 Two #9 hitters 4.0061 18.60 Three #9 hitters 3.9242 13.27 Four #9 hitters 3.7979 20.46 Five #9 hitters 3.6652 21.50 Six #9 hitters 3.5286 22.13 Seven #9 hitters 3.4014 20.61 Eight #9 hitters 3.2967 16.96 Nine #9 hitters 3.1899 17.30
It’s clear that downgrading from an average hitter to a ninth-spot hitter is worth 18 or 19 runs, but it’s not always a consistent amount. Some of that might be a bit of variation in the Monte Carlo model, but it’s not always the same amount, and it can vary by 2-3 runs, or around 10-15 percent of what we might expect from a linear function.
***
Now let’s consider defense. In theory, there’s an upper limit to how good a defensive player can be until it stops mattering. To take an extreme example, suppose that a fielder could move 100 feet per second. He could probably catch anything on the field, but it would include areas where a team already has coverage. Similarly, a fly ball that hangs up for an hour could be caught by any of the nine players on the field, after they’ve had time for a meeting and some tea.
Geometrically, we know that there comes a point where putting a rangy left fielder next to a rangy center fielder will have diminishing returns. But with the introduction of the new Statcast catch probability measures, we might be able to put some numbers on that. Statcast identifies how far away a (out)fielder was when he tried to go for a fly ball, how long it hung up in the air, and whether he made the catch. So far, we’ve only got aggregate data chopped into bins of five feet of distance each (i.e., 0 feet, 5 feet, 10 feet, etc.), but we can work with that.
Obviously, a ball that is zero feet from the nearest fielder can be caught easily and a ball that is in the air for seven seconds as well. There’s actually a very small number of balls for which the probability of catching it is in doubt. For each bin of data, we can calculate the number of feet per second, on average, the fielder would have to travel to make the catch. Using a logistic regression, we can also estimate a function for how likely a fielder is to make a catch, based on what speed he could travel.
It turns out that if a fielder was moving at a speed of 18.1 feet per second, he would get to 50 percent of balls. However, 77 percent of balls in the data set were recorded as caught, so a fielder who is moving at a rate of 20.1 feet per second is going to get to a league-average number of balls. That’s about 4.4 seconds over 90 feet. (Makes sense, because a 4.3 time down to first base from the batter’s box is considered a 50 grade or “average” runner, but the old saw is that the majority of major leaguers are “below average.”)
We can see in the data set that the average fly ball is in the air for 4.56 seconds. Assuming that the two outfielders are even average runners, they could both cover 91.6 feet. Assuming that they start the play 120 feet apart in the outfield, we can draw two circles with a radius of 91.6 feet and centers 120 feet apart. The area that they share is about 6,068 square feet, out of a total of 26,359 square feet for each of their ranges, or about 23 percent of their range which overlaps. That means that 23 percent of the “average” balls that one of the outfielders could get to, another one could as well, again assuming that they are both roughly average runners.
Outfield range is not a perfect circle. People are slower going back than they are running forward, and eventually, you run into the wall, but this gives us some idea of how much outfielders really do overlap. Now, that overlapping isn't entirely bad.
Let’s mix and match some outfielders. We’ll take three gents (and clone them when necessary). One can run 80 feet in 4.56 seconds. One runs 90 feet. One runs 100 feet. We assume that they start 120 feet apart.
Left Fielder Range Center Fielder Range Total Area Covered (sf) Overlap (of CF into LF’s range) 80 80 37,311 14.4% 80 90 41,415 20.6% 80 100 46,003 27.4% 90 90 45,318 21.9% 90 100 49,704 28.1% 100 100 53,886 28.4%
Take a look at the third and fourth lines on that table. We see that a leadfoot left fielder playing next to a jackrabbit in center field has a lot of overlap. The center fielder could get to roughly 27.4 percent of the balls that the left fielder could. If we give them equally average ranges, we see that there’s not as much overlap between the two. They aren’t bumping into each other as much. And they actually cover less ground than the star-and-scrub outfield. We may reflexively think of overlap between fielders as a bad thing. Not necessarily.
At the same time, in the outfield, we are getting some diminishing returns. For example, looking at the first two lines on the table, we see that the center fielder “improves” from a range of 80 feet to a range of 90 feet. A circle with a radius of 90 feet is 1.27 times bigger than one with a radius of 80 feet. However, the total area covered by the upgraded outfield is only 1.11 times bigger than the area covered by the one with the lesser center fielder.
While offensive talent tends to compound in its results, defensive talent does the opposite. Good defenders will eventually just get in each other’s way and you don’t get the full benefit of all their talent. It’s also something of a measurement issue. What do we do with a ball that two fielders could have caught, but only one of them does (because only one of them needs to)? Right now, it’s as if that ball never happened, which at least doesn’t penalize our left fielder who got called off by the center fielder, but it doesn’t grant him credit which he could rightly claim. If we assumed that left and center fielders called each other off at roughly the same rates, then at least this non-credit would be shared evenly. However, we know that this is not the case. The center fielder usually pulls rank when calling for a ball. It is the order of things. We haven’t really come up with a way to account for that yet.
Easy Credit
Baseball statistics are usually conceptualized in terms of individual credit. One player gets credit for the putout. One for the home run. Even when something goes wrong, errors are assigned to one person. And we can roll all of it together into an uber-metric. That has its place.
But what of the fact that the reason that the batter got the plate appearance, which then resulted in the home run, was due to the fact that another hitter got a hit, rather than made an out? We don’t think of these interaction terms or how the pieces all fit together. There’s reasonable evidence here that those sorts of interaction effects have at least some effect on the game, even if the effect size is less than “just add all the WARs.”
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frugalscott19
5/02
While we will likely never come FULL circle, it is interesting that we are at least beginning to turn back around from the idea of all sabermetrics, all the time among those in the game with new and innovative thinking.
The new book on the Cubs' success details the emphasis Theo Epstein put on character. There are certainly factors that contribute to the success of a team that can't be part of a simple, cold crunching of the numbers. That fact got a little lost for awhile while the sabermetricians and the doubters tried to do all they could to discredit each other.
Hopefully, we are entering an era of cooperation where both sides will recognize the value that can be found in analytics as well as that which can be found in the knowable/understandable, but unmeasureable (in the classic sense).
Perhaps someday in a grand and glorious future, WAR and grit can live happily together.
pizzacutter
5/03
The problem has never been "character doesn't matter." The problem is "We have no data on character!"
dethwurm
5/02
This was really interesting. It reminds me of another piece from a few years ago (also by you (Carleton) I believe) that found that putting an elite shortstop next to an elite third baseman didn't save as many runs as it seemed like it should "on paper". IIRC that was largely attributed to the "bystander effect", and while that might be part of it, certainly the overlap issue must play a role as well.
One reason I've never quite been on board with the 'defensive revolution', going all the way back to when it first began in the wake of the 2005 White Sox's championship, is that I've never been convinced you can add defensive runs together like you pretty much can with hitting (with the big caveats noted in this piece). It's a sort of mantra that "a run saved on defense is worth as much as one scored on offense," and while that's true in a literal, retrospective sense I really don't think it is in terms of the team-planning sense, if that makes sense (no (sorry)), at least with the tools we currently have to attribute defensive runs at the individual level. I.e., if you replace a replacement-level player with someone worth 4 "WAR" with the bat and 0 with the glove, you can expect the team's record to improve by 4-ish wins, but if you replace them with someone worth 2 and 2 on paper, you might only get 3 more wins in reality.
Something based on Statcast and/or a more probabilistic evaluation of defense might hold the answers, but I have a hard time seeing how that could be expressed in such a convenient/blunt "this player is worth 50 total runs" kind of way. It might be more like "Bobby Outfielder is worth 30 runs with the bat and has a 90% catch-rate radius of 75 feet" or somesuch. Which might be for the best, but it sure makes it harder to rank players across positions, which if we're being honest is like half the reason we're interested in WAR in the first place... :)
pizzacutter
5/03
I did write that! In fact, I was going to fold that into the discussion, but it got to be deadline time and I was tired...
lichtman
5/02
Russell, 50,000 games is WAY to small a run to get meaningful numbers in that first chart. One SD in runs scored is around 4 runs per game per team (mean=SD for run scoring distribution I think, maybe a little smaller, around 3 runs) which is around 3 runs per 162 or a little less (2.2) if the SD of runs scored is 3.
mattgold
5/04
So, if a team has two great outfielders, should they be placed in left and right to minimize overlap?
pizzacutter
5/04
Probably not, because the CF has to cover a lot more ground than LF and RF going backward, and it would be good to have a rangy guy who could cover that area.
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# Divide a number by 3 without using *, /, +, -, % operators
How would you divide a number by 3 without using `*`, `/`, `+`, `-`, `%`, operators?
The number may be signed or unsigned.
-
@AlexandreC. - those techniques are using addition (+) though. – hatchet Jul 27 '12 at 19:40
This was oracle so what parts of oracle were you allowed to use? – Hogan Jul 27 '12 at 19:45
The identified duplicate isn't a duplicate. Note that several answers here use neither bit shifting or addition since this question didn't restrict a solution to those operations. – Michael Burr Jul 28 '12 at 0:37
...and here is how PL/SQL is born. – ssg Jul 29 '12 at 13:28
BTW: The other question was about checking if a number is divisible by 3. This question is about dividing by 3. – wildplasser Jul 30 '12 at 13:57
There is a simple function I found here. But it's using the `+` operator, so you have to add the values with the bit-operators:
``````// replaces the + operator
int add(int x, int y) {
int a, b;
do {
a = x & y;
b = x ^ y;
x = a << 1;
y = b;
} while (a);
return b;
}
int divideby3 (int num) {
int sum = 0;
while (num > 3) {
sum = add(num >> 2, sum);
num = add(num >> 2, num & 3);
}
if (num == 3)
return sum;
}
``````
As Jim commented this works because:
• n = 4 * a + b
• n / 3 = a + (a + b) / 3
• So sum += a, n = a + b, and iterate
• When a == 0 (n < 4), sum += floor(n / 3); i.e. 1, if n == 3, else 0
-
This is probably the answer Oracle is looking for. It shows you know how the +, -, * and / operators are actually implemented on the CPU: simple bitwise operations. – craig65535 Jul 27 '12 at 21:55
This works because n = 4a + b, n/3 = a + (a+b)/3, so sum += a, n = a + b, and iterate. When a == 0 (n < 4), sum += floor(n/3); i.e., 1 if n == 3, else 0. – Jim Balter Jul 28 '12 at 5:36
Here's a trick i found which got me a similar solution. In decimal: `1 / 3 = 0.333333`, the repeating numbers make this easy to calculate using `a / 3 = a/10*3 + a/100*3 + a/1000*3 + (..)`. In binary it's almost the same: `1 / 3 = 0.0101010101 (base 2)`, which leads to `a / 3 = a/4 + a/16 + a/64 + (..)`. Dividing by 4 is where the bit shift comes from. The last check on num==3 is needed because we've only got integers to work with. – Yorick Sijsling Jul 30 '12 at 12:40
In base 4 it gets even better: `a / 3 = a * 0.111111 (base 4) = a * 4^-1 + a * 4^-2 + a * 4^-3 + (..) = a >> 2 + a >> 4 + a >> 6 + (..)`. The base 4 also explains why only 3 is rounded up at the end, while 1 and 2 can be rounded down. – Yorick Sijsling Jul 30 '12 at 13:04
Idiotic conditions call for an idiotic solution:
``````#include <stdio.h>
#include <stdlib.h>
int main()
{
FILE * fp=fopen("temp.dat","w+b");
int number=12346;
int divisor=3;
char * buf = calloc(number,1);
fwrite(buf,number,1,fp);
rewind(fp);
printf("%d / %d = %d", number, divisor, result);
free(buf);
fclose(fp);
return 0;
}
``````
If also the decimal part is needed, just declare `result` as `double` and add to it the result of `fmod(number,divisor)`.
Explanation of how it works
1. The `fwrite` writes `number` bytes (number being 123456 in the example above).
2. `rewind` resets the file pointer to the front of the file.
3. `fread` reads a maximum of `number` "records" that are `divisor` in length from the file, and returns the number of elements it read.
If you write 30 bytes then read back the file in units of 3, you get 10 "units". 30 / 3 = 10
-
Fail: you clearly use `*` in `FILE * fp`! ;) – x4u Jul 27 '12 at 22:09
@earlNameless: you don't know what they use inside, they are in the black box of "implementation defined". Nothing stops them to just use bitwise operators; anyway, they are outside the domain of my code, so that's not my problem. :) – Matteo Italia Jul 29 '12 at 1:02
Would someone care to explain why or how this answer works in the answer itself? – Ivo Flipse Jul 29 '12 at 8:59
@IvoFlipse from I can clean, you get a big something and shove it into something three times too small, and then see how much fitted in. That about is a third. – Pureferret Jul 29 '12 at 15:00
asked the best C programmer (and most socially awkward) at our company to explain the code. after he did, i said it was pretty ingenious. He said 'this dreck is not a solution' and asked me to leave his desk – cvursache Jul 30 '12 at 12:45
``````log(pow(exp(number),0.33333333333333333333)) /* :-) */
``````
-
This might actually work if rounded properly and if the number isn't too large. – Mysticial Jul 27 '12 at 19:57
Improved version: log(pow(exp(number),sin(atan2(1,sqrt(8))))) – Alan Curry Jul 28 '12 at 0:14
i just typed it in my js console, it doesn't work with a number higher than 709 (may be its just my system) `Math.log(Math.pow(Math.exp(709),0.33333333333333333333))` and `Math.log(Math.pow(Math.exp(709),Math.sin(Math.atan2(1,Math.sqrt(8)))))` – Shaheer Aug 30 '12 at 13:12
show 1 more comment
``````#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
int num = 1234567;
int den = 3;
div_t r = div(num,den); // div() is a standard C function.
printf("%d\n", r.quot);
return 0;
}
``````
-
Using cloud computing :D
Google Search: `30 divided by 3`
-
Disqualified - you're using +'s in your url. – Anthony Briggs Jul 30 '12 at 11:45
How about `www.google.com\webhp?q=30%20divided%20by%203` -- works in Chrome and doesn't use illegal characters. :) – Roy Tinker Jul 30 '12 at 16:33
Technically, if you go to Google by typing 'google.com' and type in '30 divided by 3', then you aren't using + or / ;-) – user577537 Jul 30 '12 at 17:24
@Roy Tinker : Still using `%` – c.adhityaa Jul 31 '12 at 12:34
This works fine in my browser: `google.com?q=30 divided by 3`. It's the browser that inserts `%`, so technically not you using them...? – Svish Aug 1 '12 at 8:17
Use inline assembler: (also works for negative numbers)
``````#include <stdio.h>
int main() {
int dividend = -42, divisor = 3, quotient, remainder;
__asm__ ( "movl %2, %%edx;"
"sarl \$31, %%edx;"
"movl %2, %%eax;"
"movl %3, %%ebx;"
"idivl %%ebx;"
: "=a" (quotient), "=d" (remainder)
: "g" (dividend), "g" (divisor)
: "ebx" );
printf("%i / %i = %i, remainder: %i\n", dividend, divisor, quotient, remainder);
}
``````
-
@JeremyP doesn't your comment fail on the assumption that the answer can't be written in C? The question is tagged "C" after all. – Seth Carnegie Aug 1 '12 at 18:33
@SethCarnegie The answer is not written in C is my point. x86 assembler is not part of the standard. – JeremyP Aug 2 '12 at 13:29
@JeremyP that is true, but the `asm` directive is. And I would add that C compilers are not the only ones that have inline assemblers, Delphi has that as well. – Seth Carnegie Aug 2 '12 at 18:01
@SethCarnegie The `asm` directive is only mentioned in the C99 standard under Appendix J - common extensions. – JeremyP Aug 3 '12 at 9:49
Use itoa to convert to base 3 string. Drop last trit and convert back to base 10.
``````// Note: itoa is non-standard but actual implementations
// don't seem to handle negative when base != 10
int div3(int i) {
char str[42];
sprintf(str, "%d", INT_MIN); // put minus sign at str[0]
if (i>0) str[0] = ' '; // remove sign if positive
itoa(abs(i), &str[1], 3); // put ternary absolute value starting at str[1]
str[strlen(&str[1])] = '\0'; // drop last digit
return strtol(str, NULL, 3); // read back result
}
``````
-
@cshemby I actually didn't know that `itoa` could use an arbitrary base. If you do a complete working implementation using `itoa` I'll upvote. – Mysticial Jul 27 '12 at 19:54
show 1 more comment
The request says "a number", not "any number", so:
1. Check if the number is 3.
2. If so, return 1.
3. If not, report an invalid input.
Job done, but for extra marks, you could implement a dictionary of commonly requested multiples of three.
-
This isn't quite right. The correct program consists of one line, which prints "no input specified". – Thomas Jul 29 '12 at 10:57
Comparison is performed using subtraction at microcode-level (that's why zero flag is set on equality), so you're still subtracting :) – ssg Jul 29 '12 at 22:16
@ssg, The question is not "without subtracting", but without using any of the listed operators. – Paul Butcher Jul 30 '12 at 8:16
(note: see Edit 2 below for a better version!)
This is not as tricky as it sounds, because you said "without using the [..] `+` [..] operators". See below, if you want to forbid using the `+` character all together.
``````unsigned div_by(unsigned const x, unsigned const by) {
unsigned floor = 0;
for (unsigned cmp = 0, r = 0; cmp <= x;) {
for (unsigned i = 0; i < by; i++)
cmp++; // that's not the + operator!
floor = r;
r++; // neither is this.
}
return floor;
}
``````
then just say `div_by(100,3)` to divide `100` by `3`.
### Edit: You can go on and replace the `++` operator as well:
``````unsigned inc(unsigned x) {
else
}
return 0; // overflow (note that both x and mask are 0 here)
}
``````
# Edit 2: Slightly faster version without using any operator that contains the `+`,`-`,`*`,`/`,`%`characters.
``````unsigned add(char const zero[], unsigned const x, unsigned const y) {
// this exploits that &foo[bar] == foo+bar if foo is of type char*
return (int)(uintptr_t)(&((&zero[x])[y]));
}
unsigned div_by(unsigned const x, unsigned const by) {
unsigned floor = 0;
for (unsigned cmp = 0, r = 0; cmp <= x;) {
floor = r;
}
return floor;
}
``````
We use the first argument of the `add` function because we cannot denote the type of pointers without using the `*` character, except in function parameter lists, where the syntax `type[]` is identical to `type* const`.
FWIW, you can easily implement a multiplication function using a similar trick to use the `0x55555556` trick proposed by AndreyT:
``````int mul(int const x, int const y) {
return sizeof(struct {
char const ignore[y];
}[x]);
}
``````
-
@Hogan the question is tagged `c`. – Matt Ball Jul 27 '12 at 19:47
The question is tagged c, not SQL, even though Oracle is mentioned. – bitmask Jul 27 '12 at 19:48
If you can use `++`: Why aren't you simply use `/=`? – Coodey Jul 27 '12 at 20:10
@bitmask: `++` is also a shortcut: For `num = num + 1`. – Coodey Jul 27 '12 at 20:17
@bitmask Yeah, but `+=` is finally a shortcut for `num = num + 1`. – Coodey Jul 27 '12 at 20:23
It is easily possible on the Setun computer.
To divide an integer by 3, shift right by 1 place.
I'm not sure whether it's strictly possible to implement a conforming C compiler on such a platform though. We might have to stretch the rules a bit, like interpreting "at least 8 bits" as "capable of holding at least integers from -128 to +127".
-
show 1 more comment
Here's my solution:
``````public static int div_by_3(long a) {
a <<= 30;
for(int i = 2; i <= 32 ; i <<= 1) {
a = add(a, a >> i);
}
return (int) (a >> 32);
}
public static long add(long a, long b) {
long carry = (a & b) << 1;
long sum = (a ^ b);
return carry == 0 ? sum : add(carry, sum);
}
``````
First, note that
``````1/3 = 1/4 + 1/16 + 1/64 + ...
``````
Now, the rest is simple!
``````a/3 = a * 1/3
a/3 = a * (1/4 + 1/16 + 1/64 + ...)
a/3 = a/4 + a/16 + 1/64 + ...
a/3 = a >> 2 + a >> 4 + a >> 6 + ...
``````
Now all we have to do is add together these bit shifted values of a! Oops! We can't add though, so instead, we'll have to write an add function using bit-wise operators! If you're familiar with bit-wise operators, my solution should look fairly simple... but just in-case you aren't, I'll walk through an example at the end.
Another thing to note is that first I shift left by 30! This is to make sure that the fractions don't get rounded off.
``````11 + 6
1011 + 0110
sum = 1011 ^ 0110 = 1101
carry = (1011 & 0110) << 1 = 0010 << 1 = 0100
Now you recurse!
1101 + 0100
sum = 1101 ^ 0100 = 1001
carry = (1101 & 0100) << 1 = 0100 << 1 = 1000
Again!
1001 + 1000
sum = 1001 ^ 1000 = 0001
carry = (1001 & 1000) << 1 = 1000 << 1 = 10000
One last time!
0001 + 10000
sum = 0001 ^ 10000 = 10001 = 17
carry = (0001 & 10000) << 1 = 0
Done!
``````
It's simply carry addition that you learned as a child!
``````111
1011
+0110
-----
10001
``````
This implementation failed because we can not add all terms of the equation:
``````a / 3 = a/4 + a/4^2 + a/4^3 + ... + a/4^i + ... = f(a, i) + a * 1/3 * 1/4^i
f(a, i) = a/4 + a/4^2 + ... + a/4^i
``````
Suppose the reslut of `div_by_3(a)` = x, then `x <= floor(f(a, i)) < a / 3`. When `a = 3k`, we get wrong answer.
-
does it work for input of 3? 1/4, 1/16, ... all return 0 for 3, so would sum to 0, but 3/3 = 1. – hatchet Jul 27 '12 at 21:55
Since it's from Oracle, how about a lookup table of pre calculated answers. :-D
-
show 1 more comment
To divide a 32-bit number by 3 one can multiply it by `0x55555556` and then take the upper 32 bits of the 64 bit result.
Now all that's left to do is to implement multiplication using bit operations and shifts...
-
@luiscubal: No, it won't. This is why I said: "Now all that's left to do is to implement multiplication using bit operations and shifts" – AndreyT Jul 27 '12 at 21:49
Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.
``````public static int DivideBy3(int a) {
bool negative = a < 0;
if (negative) a = Negate(a);
int result;
int sub = 3 << 29;
int threes = 1 << 29;
result = 0;
while (threes > 0) {
if (a >= sub) {
}
sub >>= 1;
threes >>= 1;
}
if (negative) result = Negate(result);
return result;
}
public static int Negate(int a) {
}
public static int Add(int a, int b) {
int x = 0;
x = a ^ b;
while ((a & b) != 0) {
b = (a & b) << 1;
a = x;
x = a ^ b;
}
return x;
}
``````
This is c# because that's what I had handy, but differences from c should be minor.
-
This one is the classical division algorithm in base 2:
``````#include <stdio.h>
#include <stdint.h>
int main()
{
uint32_t mod3[6] = { 0,1,2,0,1,2 };
uint32_t x = 1234567; // number to divide, and remainder at the end
uint32_t y = 0; // result
int bit = 31; // current bit
printf("X=%u X/3=%u\n",x,x/3); // the '/3' is for testing
while (bit>0)
{
printf("BIT=%d X=%u Y=%u\n",bit,x,y);
// decrement bit
int h = 1; while (1) { bit ^= h; if ( bit&h ) h <<= 1; else break; }
uint32_t r = x>>bit; // current remainder in 0..5
x ^= r<<bit; // remove R bits from X
if (r >= 3) y |= 1<<bit; // new output bit
x |= mod3[r]<<bit; // new remainder inserted in X
}
printf("Y=%u\n",y);
}
``````
-
Using counters is a basic solution.
``````int DivBy3(int num) {
int result = 0;
int counter = 0;
while (1) {
if (num == counter) return result; //Modulus 0
counter = abs(~counter); //++counter
if (num == counter) return result; //Modulus 1
counter = abs(~counter); //++counter
if (num == counter) return result; //Modulus 2
counter = abs(~counter); //++counter
result = abs(~result); //++result
}
}
``````
It is easy to perform also a modulus function, check remarks.
-
It's really quite easy.
``````if (number == 0) return 0;
if (number == 1) return 0;
if (number == 2) return 0;
if (number == 3) return 1;
if (number == 4) return 1;
if (number == 5) return 1;
if (number == 6) return 2;
``````
(I have of course omitted some of the program for the sake of brevity.) If the programmer gets tired of typing this all out, I'm sure that he or she could write a separate program to generate it for him. I happen to be aware of a certain operator, `/`, that would simplify his job immensely.
-
You could use a `Dictionary<number, number>` instead of repeated `if` statements so you can have `O(1)` time complexity! – Peter Olson Aug 18 '12 at 3:28
Would it be cheating to use the `/` operator "behind the scenes" by using `eval` and string concatenation?
For example, in Javacript, you can do
``````function div3 (n) {
var div = String.fromCharCode(47);
return eval([n, div, 3].join(""));
}
``````
-
Write the program in Pascal and use the `DIV` operator.
Since the question is tagged , you can probably write a function in Pascal and call it from your C program; the method for doing so is system-specific.
But here's an example that works on my Ubuntu system with the Free Pascal `fp-compiler` package installed. (I'm doing this out of sheer misplaced stubbornness; I make no claim that this is useful.)
`divide_by_3.pas` :
``````unit Divide_By_3;
interface
function div_by_3(n: integer): integer; cdecl; export;
implementation
function div_by_3(n: integer): integer; cdecl;
begin
div_by_3 := n div 3;
end;
end.
``````
`main.c` :
``````#include <stdio.h>
#include <stdlib.h>
extern int div_by_3(int n);
int main(void) {
int n;
fputs("Enter a number: ", stdout);
fflush(stdout);
scanf("%d", &n);
printf("%d / 3 = %d\n", n, div_by_3(n));
return 0;
}
``````
To build:
``````fpc divide_by_3.pas && gcc divide_by_3.o main.c -o main
``````
Sample execution:
``````\$ ./main
Enter a number: 100
100 / 3 = 33
``````
-
First that I've come up with.
``````irb(main):101:0> div3 = -> n { s = '%0' + n.to_s + 's'; (s % '').gsub(' ', ' ').size }
=> #<Proc:0x0000000205ae90@(irb):101 (lambda)>
irb(main):102:0> div3[12]
=> 4
irb(main):103:0> div3[666]
=> 222
``````
EDIT: Sorry, I didn't notice the tag `C`. But you can use the idea about string formatting, I guess...
-
Didn't cross-check if this answer is already published. If the program need to be extended to floating numbers, the numbers can be multiplied by 10*number of precision needed and then the following code can be again applied.
``````#include <stdio.h>
int main()
{
int aNumber = 500;
int gResult = 0;
int aLoop = 0;
int i = 0;
for(i = 0; i < aNumber; i++)
{
if(aLoop == 3)
{
gResult++;
aLoop = 0;
}
aLoop++;
}
printf("Reulst of %d / 3 = %d", aNumber, gResult);
return 0;
}
``````
-
``````int div3(int x)
{
int reminder = abs(x);
int result = 0;
while(reminder >= 3)
{
result++;
reminder--;
reminder--;
reminder--;
}
return result;
}
``````
-
++ and -- operaors are diferent from + and - operaors! In assembly language there are two instructions `ADD` and `INC` that they have not same opcodes. – Amir Saniyan Aug 5 '12 at 13:50
show 1 more comment
This should work for any divisor, not only three. Currently only for unsigned, but extending it to signed should not be that difficult.
``````#include <stdio.h>
unsigned sub(unsigned two, unsigned one);
unsigned bitdiv(unsigned top, unsigned bot);
unsigned sub(unsigned two, unsigned one)
{
unsigned bor;
bor = one;
do {
one = ~two & bor;
two ^= bor;
bor = one<<1;
} while (one);
return two;
}
unsigned bitdiv(unsigned top, unsigned bot)
{
unsigned result, shift;
if (!bot || top < bot) return 0;
for(shift=1;top >= (bot<<=1); shift++) {;}
bot >>= 1;
for (result=0; shift--; bot >>= 1 ) {
result <<=1;
if (top >= bot) {
top = sub(top,bot);
result |= 1;
}
}
return result;
}
int main(void)
{
unsigned arg,val;
for (arg=2; arg < 40; arg++) {
val = bitdiv(arg,3);
printf("Arg=%u Val=%u\n", arg, val);
}
return 0;
}
``````
-
Use cblas, included as part of OS X's Accelerate framework.
``````[02:31:59] [william@relativity ~]\$ cat div3.c
#import <stdio.h>
#import <Accelerate/Accelerate.h>
int main() {
float multiplicand = 123456.0;
float multiplier = 0.333333;
printf("%f * %f == ", multiplicand, multiplier);
cblas_sscal(1, multiplier, &multiplicand, 1);
printf("%f\n", multiplicand);
}
[02:32:07] [william@relativity ~]\$ clang div3.c -framework Accelerate -o div3 && ./div3
123456.000000 * 0.333333 == 41151.957031
``````
-
solution using fma() library function, works for any positive number.
``````#include <stdio.h>
#include <math.h>
int main()
{
int number = 8;//any +ve no.
int temp = 3, result = 0;
while(temp <= number){
temp = fma(temp, 1, 3);//fma(a, b, c) is lib function, returns (a*b)+c
result = fma(result, 1, 1);
}
printf("\n\n%d divided by 3 = %d\n", number, result);
}
``````
-
using BCMath in PHP:
``````<?php
\$a = 12345;
\$b = bcdiv(\$a, 3);
?>
``````
MySQL (it's an interview from Oracle)
``````> SELECT 12345 DIV 3;
``````
PASCAL:
``````a:= 12345;
b:= a div 3;
``````
x86-64 ASM:
``````mov r8, 3
xor rdx, rdx
mov rax, 12345
idiv r8
``````
-
The following script generates a C program that solves the problem without using the operators `* / + - %`:
``````#!/usr/bin/env python3
print('''#include <stdint.h>
#include <stdio.h>
const int32_t div_by_3(const int32_t input)
{
''')
for i in range(-2**31, 2**31):
print(' if(input == %d) return %d;' % (i, i / 3))
print(r'''
return 42; // impossible
}
int main()
{
const int32_t number = 8;
printf("%d / 3 = %d\n", number, div_by_3(number));
}
''')
``````
-
``````int divideByThree(int num)
{
return (fma(num, 1431655766, 0) >> 32);
}
``````
Where fma is a standard library function defined in `math.h` header.
-
``````private int dividedBy3(int n) {
List<Object> a = new Object[n].ToList();
List<Object> b = new List<object>();
while (a.Count > 2) {
a.RemoveRange(0, 3);
}
return b.Count;
}
``````
-
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# eigenvalues of a matrix
On the previous page, Eigenvalues and eigenvectors - physical meaning and geometric interpretation applet we saw the example of an elastic membrane being stretched, and how this was represented by a matrix multiplication, and in special cases equivalently by a scalar multiplication. 2. Fortunately we can have the calculator multiply and take powers of a matrices. We can also define powers An of a matrix. For completeness, the following graph shows a matrix of scatter plots and marginal histograms for the bootstrap distribution. The location of the essential spectrum of ${\\mathcal A}_μ$ is described and its bounds are estimated. Whether the solution is real or complex depends entirely on the matrix that you feed. . The real part of each of the eigenvalues is negative, so e λt approaches zero as t increases. Learn to find complex eigenvalues and eigenvectors of a matrix. For this example, we'll look at the following matrix with 4, 2, 1, and 3. Introduction. Eigenvectors and eigenvalues of a diagonal matrix D The equation Dx = 0 B B B B @ d1 ;1 0 ::: 0 0 d 2;. By the second and fourth properties of Proposition C.3.2, replacing ${\bb v}^{(j)}$ by ${\bb v}^{(j)}-\sum_{k\neq j} a_k {\bb v}^{(k)}$ results in a matrix whose determinant is the same as the original matrix. With two output arguments, eig computes the eigenvectors and stores the eigenvalues in a diagonal matrix: Now let us put in an identity matrix so we are dealing with matrix-vs-matrix:. The projection keeps the column space and destroys the nullspace: The generalized eigenvalues of m with respect to a are those for which . In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. For eigen values of a matrix first of all we must know what is matric polynomials, characteristic polynomials, characteristic equation of a matrix. The eigenvectors for D 0 (which means Px D 0x/ fill up the nullspace. The eigenvalues of a matrix m are those for which for some nonzero eigenvector . Learn to recognize a rotation-scaling matrix, and compute by how much the matrix rotates and scales. λ is an eigenvalue (a scalar) of the Matrix [A] if there is a non-zero vector (v) such that the following relationship is satisfied: [A](v) = λ (v) Every vector (v) satisfying this equation is called an eigenvector of [A] belonging to the eigenvalue λ.. As an example, in the case of a 3 X 3 Matrix … Featured on Meta “Question closed” notifications experiment results and graduation The row vector is called a left eigenvector of . Icon 2X2. • In such problems, we first find the eigenvalues of the matrix. First let’s reduce the matrix: This reduces to the equation: There are two kinds of students: those who love math and those who hate it. (No non-square matrix has eigenvalues.) Let A be a square matrix. The matrix above has eigenvalues (lambda) of 0, -4, and 3. The nonzero imaginary part of two of the eigenvalues, ±ω, contributes the oscillatory component, sin(ωt), to the solution of the differential equation. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. It is possible for a real or complex matrix to have all real eigenvalues … Works with matrix from 2X2 to 10X10. The only eigenvalues of a projection matrix are 0 and 1. Eigenvalues are a special set of scalars associated with a linear system of equations (i.e., a matrix equation) that are sometimes also known as characteristic roots, characteristic values (Hoffman and Kunze 1971), proper values, or latent roots (Marcus and Minc 1988, p. 144).. The eigenvalues of a hermitian matrix are real, since (λ − λ)v = (A * − A)v = (A − A)v = 0 for a non-zero eigenvector v. If A is real, there is an orthonormal basis for R n consisting of eigenvectors of A if and only if A is symmetric. The eigenvectors are also termed as characteristic roots. . Browse other questions tagged linear-algebra eigenvalues block-matrices or ask your own question. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. Eigenvalues of a triangular matrix. 7.2 FINDING THE EIGENVALUES OF A MATRIX Consider an n£n matrix A and a scalar ‚.By definition ‚ is an eigenvalue of A if there is a nonzero vector ~v in Rn such that A~v = ‚~v ‚~v ¡ A~v = ~0 (‚In ¡ A)~v = ~0An an eigenvector, ~v needs to be a nonzero vector. Section 5.5 Complex Eigenvalues ¶ permalink Objectives. By definition of the kernel, that The histograms indicate skewness in the bootstrap distribution. The nullspace is projected to zero. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. Two proofs given We'll find the eigenvectors associated with lambda = -4. Eigenvalues are the special set of scalar values which is associated with the set of linear equations most probably in the matrix equations. Summary. An eigenvalue for $A$ is a $\lambda$ that solves $Ax=\lambda x$ for some nonzero vector $x$. Since doing so results in a determinant of a matrix with a zero column, $\det A=0$. Choose your matrix! Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. In general, the way acts on is complicated, but there are certain cases where the action maps to the same vector, multiplied by a scalar factor.. Eigenvalues and eigenvectors have immense applications in the physical sciences, especially quantum mechanics, among other fields. Interestingly, in one matrix product instance even without any sign change operations, with both matrix A and B having positive eigenvalues, the product matrix AB have an even number of negative eigenvalues! An easy and fast tool to find the eigenvalues of a square matrix. This article shows how to compute confidence intervals for the eigenvalues of an estimated correlation matrix. The calculator will find the eigenvalues and eigenvectors (eigenspace) of the given square matrix, with steps shown. We work through two methods of finding the characteristic equation for λ, then use this to find two eigenvalues. If you love it, our example of the solution to eigenvalues and eigenvectors of 3×3 matrix will help you get a better understanding of it. • Form the matrix A−λI: A −λI = 1 −3 3 3 −5 3 6 −6 4 If they are numeric, eigenvalues are sorted in order of decreasing absolute value. Understand the geometry of 2 … First compute the characteristic polynomial. Bring all to left hand side: Earlier we stated that an nxn matrix has n eigenvalues. Eigenvalues and Eigenvectors of a 3 by 3 matrix Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. This is a finial exam problem of linear algebra at the Ohio State University. Notice how we multiply a matrix by a vector and get the same result as when we multiply a scalar (just a number) by that vector.. How do we find these eigen things?. The matrix equation = involves a matrix acting on a vector to produce another vector. The eigenvectors for D 1 (which means Px D x/ fill up the column space. Determine all the eigenvalues of A^5 and the inverse matrix of A if A is invertible. We start by finding the eigenvalue: we know this equation must be true:. Av = λv. Eigenvectors and Eigenvalues of Matrices. The column space projects onto itself. An × matrix gives a list of exactly eigenvalues, not necessarily distinct. We consider a $2\\times2$ block operator matrix ${\\mathcal A}_μ$ $($$μ>0$ is a coupling constant$)$ acting in the direct sum of one- and two-particle subspaces of a bosonic Fock space. Let’s assume the matrix is square, otherwise the answer is too easy. v. In this equation A is an n-by-n matrix, v is a non-zero n-by-1 vector and λ is a scalar (which may be either real or complex). Then v and λ The values of λ that satisfy the equation are the generalized eigenvalues. So a 2x2 matrix should have 2 eigenvalues. FINDING EIGENVALUES • To do this, we find the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0, where I is the 3×3 identity matrix. 0 0 ::: 0 d n;n 1 C C C C A 0 B B B @ x1 x2 x n 1 C C C A = 0 B @ d1 ;1 x1 d2 ;2 x2 d n;nx n 1 C C = x Av = λIv. . Click on the Space Shuttle and go to the 2X2 matrix solver! It is a non-zero vector which can be changed at most by its scalar factor after the application of … Let A be a square matrix (that is A has the same number of rows and columns). The solver, Eigen::EigenSolver admits general matrices, so using ".real()" to get rid of the imaginary part will give the wrong result (also, eigenvectors may have an arbitrary complex phase!). . Select the size of the matrix and click on the Space Shuttle in order to fly to the solver! So A2 = AA, A3 = AAA, A4 = AAAA etc. The diagonal elements of a triangular matrix are equal to its eigenvalues. Eigenvalue. . That example demonstrates a very important concept in engineering and science - eigenvalues … We prove that eigenvalues of a Hermitian matrix are real numbers. Let v be a vector and λ a number. Show Instructions. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Since the matrix n x n then it has n rows and n columns and obviously n diagonal elements. If the matrix can be diagonalized, this sign change can occur only by a change in sign in one (or an odd number) of the eigenvalues. Any value of λ for which this equation has a solution is known as an eigenvalue of the matrix A.
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## Geometric Sequences and Series
Introduction
We have discussed about arithmetic sequences, its characteristics and its connection to linear functions. In this post, we will discuss another type of sequence.
The sequence of numbers 2, 6, 18, 54, 162, … is an example of an geometric sequence. The first term 2 is multiplied by 3 to get the second term, the second term is multiplied by 3 to get the third term, the third term is multiplied by 3 to get the fourth term, and so on. The same number that we multiplied to each term is called the common ratio. Expressing the sequence above in terms of the first term and the common ratio, we have 2, 2(3), 2(32), 2(33), …. Hence, a geometric sequence, also known as a geometric progression, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio.
The Sierpinski triangle below is an example of a geometric representation of a geometric sequence. The number of blue triangles, the number of white triangles, their areas, and their side lengths form different geometric sequences. It is left to the reader, as an exercise, to find the rules of these geometric sequences.
Figure 1 - The Seriepinski Triangles.
To generalize, if a1 is its first term and the common ratio is r, then the general form of a geometric sequence is a1, a1r, a1r2, a1r3,…, and the nth term of the sequence is a1rn-1.
A geometric series, on the other hand, is the sum of the terms of a geometric sequence. Given a geometric sequence with terms a1r, a1r2, a1r3,…, the sum Sn of the geometric sequence with n terms is the geometric series a1 + a1r + a1r2, a1r3 + … + arn-1. Multiplying Sn by -r and adding it to Sn, we have
Hence, the sum of a geometric series with n terms, and $r \neq 1 = \displaystyle\frac{a_1(1-r^n)}{1-r}$.
Sum of Infinite Geometric Series and a Little Bit of Calculus
Note: This portion is for those who have already taken elementary calculus.
The infinite geometric series $\{a_n\}$ is the the symbol $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots$. From above, the sum of a finite geometric series with $n$ terms is $\displaystyle \sum_{k=1}^n \frac{a_1(1-r^n)}{1-r}$. Hence, to get the sum of the infinite geometric series, we need to get the sum of $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r}$. However, $\displaystyle \sum_{k=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n\to \infty} \frac{a_1(1-r^n)}{1-r}$.
Also, that if $|r| < 1$, $r^n$ approaches $0$ (try $(\frac{2}{3})^n$ or any other proper fraction and increase the value of $n$), thus, $\displaystyle \sum_{n=1}^\infty \frac{a_1(1-r^n)}{1-r} = \lim_{n \to \infty} \frac{a_1(1-r^n)}{1-r} = \frac{a_1}{1-r}$. Therefore, sum of the infinite series $\displaystyle a_1 + a_2r + a_2r^2 + \cdots = \frac{a_1}{1-r}$.
One very common infinite series is $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$, or the sum of the areas of the partitions of the square with side length 1 unit shown below. Using the formula above,
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n} = \frac{a_1}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$.
Figure 2 - A representation of an infinite geometric series.
This is evident in the diagram because the sum of all the partitions is equal to the area of a square. We say that the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2n}$ converges to 1.
## Is 0.999… really equal to 1?
Introduction
Yes it is. 0.999… is equal to 1.
Before we begin our discussion, let me make a remark that the symbol “…” in the decimal 0.999… means that the there are infinitely many 9’s, or putting it in plain language, the decimal number has no end.
For non-math persons, you will probably disagree with the equality, but there are many elementary proofs that could show it, some of which, I have shown below. A proof is a series of valid, logical and relevant arguments (see Introduction to Mathematical Proofs for details), that shows the truth or falsity of a statement.
Proof 1
$\frac{1}{3} = 0.333 \cdots$
$\frac{2}{3} = 0.666 \cdots$
$\frac{1}{3} + \frac{2}{3} = 0.333 \cdots + 0.666 \cdots$
$\frac{3}{3} =0.999 \cdots$
But $\frac{3}{3} = 1$, therefore $1 =0.999 \cdots$
Proof 2
$\frac{1}{9} = 0.111 \cdots$
Multiplying both sides by 9 we have
$1 = 0.999 \cdots$
Proof 3
Let $x = 0.999 \cdots$
$10x = 9.999 \cdots$
$10x - x = 9.999 \cdots - 0.9999 \cdots$
$9x = 9$
$x = 1$
Hence, $0.999 \cdots = 1$
Still in doubt?
Many will probably be reluctant in accepting the equality $1 = 0.999 \cdots$ because the representation is a bit counterintuitive. The said equality requires the notion of the real number system, a good grasp of the concept of limits, and knowledge on infinitesimals or calculus in general. If, for instance,you have already taken sequences (in calculus), you may think of the $0.999 \cdots$ as a sequence of real numbers $(0.9, 0.99, 0.999,\cdots)$. Note that the sequence gets closer and closer to 1, and therefore, its limit is 1.
Infinite Geometric Sequence
My final attempt to convince you that $0.999 \cdots$ is indeed equal to $1$ is by the infinite geometric sequence. For the sake of brevity, in the remaining part of this article, we will simply use the term “infinite sequence” to refer to an infinite geometric sequence. We will use the concept of the sum of an infinite sequence, which is known as an infinite series, to show that $0.999 \cdots = 1$.
One example of an infinite series is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$. If you add its infinite number of terms, the answer is equal to 1. Again, this is counterintuitive.
How can addition of numbers with infinite number of terms have an exact (or a finite) answer?
There is a formula to get the sum of an infinite geometric sequence, but before we discuss the formula, let me give the geometric interpretation of the sum above. The sum $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$ can be represented geometrically using a 1 unit by 1 unit square as shown below. If we divide the square into two, then we will have two rectangles, each of which has area $\frac{1}{2}$ square units. Dividing the other half into two, then we have three rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{4}$ square units. Dividing the one of the smaller rectangle into two, then we have four rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{8}$. Again, dividing one of the smallest rectangle into two, we have five rectangles with areas $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$, and $\frac{1}{16}$ Since this process can go on forever, the sum of all the areas of all the rectangles will equal to 1, which is the area of the original square.
Now that we have seen that an infinite series can have a finite sum, we will now show that $0.999 \cdots$ can be expressed as a finite sum by expressing it as an infinite series. The number $0.999 \cdots$ can be expressed as an infinite series $0.9 + 0.09 + 0.009 + \cdots$. Converting it in fractional form, we have $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$.
We have learned that the sum of the infinite series with first term $\displaystyle a_1$ and ratio $r$ is described by $\displaystyle\frac{a_1}{1-r}$. Applying the formula to our series above, we have
$\displaystyle\frac{\frac{9}{10}}{1-\frac{1}{10}} = 1$
Therefore, the sum our infinite series is 1.
Implication
This implication of the equality $0.999 \cdots =1$ means that any rational number that is a non-repeating decimal can be expressed as a repeating decimal. Since $0.999 \cdots =1$, it follows that $0.0999 \cdots =0.1, 0.00999 \cdots=0.01$ and so on. Hence, any decimal number maybe expressed as number + 0.00…01. For example, the decimal $4.7$, can be expressed as $4.6 + 0.1 = 4.6 + 0.0999 \cdots = 4.6999 \cdots$. The number $0.874$ can also be expressed as $0.873 + 0.001 = 0.873 + 0.000999 \cdots = 0.873999 \cdots$
Conclusion
Any of the four proofs above is actually sufficient to show that $0.999 \cdots = 1$. Although this concept is quite hard to accept, we should remember that in mathematics, as long as the steps of operations or reasoning performed are valid and logical, the conclusion will be unquestionably valid.
There are many counterintuitive concepts in mathematics and the equality $0.999 \cdots = 1$ is only one of the many. In my post, Counting the Uncountable: A Glimpse at the Infinite, we have also encountered one: that the number of integers (negative, 0, positive) is equal to the number of counting numbers (positive integers) and we have shown it by one-to-one pairing. We have also shown that the number of counting numbers is the same as the number of rational numbers. Thus, we have shown that a subset can have the same element as the “supposed” bigger set. I guess that is what makes mathematics unique; intuitively, some concepts do not make sense, but by valid and logical reasoning, they perfectly do.
Notes:
1. You can find discussions about 0.999… = 1 here and here.
2. There is another good post about it here and here.
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# A Brief Diversion: Sequent Calculus
*(This post has been modified to correct some errors and add some clarifications in response to comments from alert readers. Thanks for the corrections!)*
Today, we’re going to take a brief diversion from category theory to play with
some logic. There are some really neat connections between variant logics and category theory. I’m planning on showing a bit about the connections between category theory and one of those, called *linear logic* . But the easiest way to present things like linear logic is using a mechanism based on sequent calculus.
Sequent calculus is a deduction system for performing reasoning in first order propositional logic. But it’s notation and general principles are useful for all sorts of reasoning systems, including many different logics, all sorts of type theories, etc. The specific sequent calculus that I’m to talk about is sometimes called system-LK; the general category of things that use this basic kind of rules is called Gentzen systems.
The sequent calculus consists of a set of rules called *sequents*, each of which is normally written like a fraction: the top of the fraction is what you know before applying the sequent; the bottom is what you can conclude. The statements in the sequents are always of the form:
CONTEXTS, Predicates :- CONTEXTS, Predicates
The “CONTEXTS” are sets of predicates that you already know are true. The “:-” is read “entails”; it means that the *conjuction* of the statements and contexts to the left of it can prove the *disjunction* of the statements to the right of it. In predicate logic, the conjuction is logical and, and disjunction is logical or, so you can read the statements as if “,” is “∧” on the left of the “:-“, and “∨” on the right. *(Note: this paragraph was modified to correct a dumb error that I made that was pointed out by commenter Canuckistani.)*
Contexts are generally written using capital greek letters; predicates are generally written using uppercase english letters. We often put a name for an inference rule to the right of the separator line for the sequent.
For example, look at the following sequent:
Γ :- Δ
————— Weakening-Left
Γ,A :- Δ
This sequent is named Weakening-left; the top says that “Given Γ everything in Δ can be proved.”; and
the bottom says “Using Γ plus the fact that A is true, everything in Δ can be proved”. The full sequent basically says: if Δ is provable given Γ, then it will still be provable when A is added to Γ;in other words, adding a true fact won’t invalidate any proofs that were valid before the addition of A. *(Note: this paragraph was modified to correct an error pointed out by a commenter.)*
The sequent calculus is nothing but a complete set of rules that you can use to perform any inference in predicate calculus. A few quick syntactic notes, and I’ll show you the full set of rules.
1. Uppercase greek letters are contexts.
2. Uppercase english letters are *statements*.
3. Lowercase english letters are *terms*; that is, the objects that predicates
can reason about, or variables representing objects.
4. A[b] is a statement A that contains the term b in some way.
5. A[b/c] means A with the term “b” replaced by the term “c”.
——-
First, two very basic rules:
1.
———— (Identity)
A :- A
2. Γ :- A, Δ Σ, A :- Π
—————————————— (Cut)
Γ,Σ :- Δ, Π
Now, there’s a bunch of rules that have right and left counterparts. They’re duals of each other – move terms across the “:-” and switch from ∧ to ∨ or vice-versa.
3. Γ, A :- Δ
————————— (Left And 1)
Γ, A ∧ B :- Δ
4. Γ :- A, Δ
——————— ——— (Right Or 1)
Γ, :- A ∨ B, Δ
5. Γ, B :- Δ
——————— ——(Left And 2)
Γ,A ∧ B :- Δ
6. Γ :- B, Δ
——————— ——— (Right Or 2)
Γ :- A ∧ B, Δ
7. Γ, A :- Δ Σ,B :- Π
————————————— (Left Or)
Γ,Σ, A ∨ B :- Δ,Π
8. Γ :- A,Δ Σ :- B,Π
—————————— ——(Right And)
Γ,Σ :- A ∧ B, Δ,Π
9. Γ :- A,Δ
————— —— (Left Not)
Γ, ¬A :- Δ
10. Γ,A :- Δ
——————— (Right Not)
Γ :- ¬A, Δ
11. Γ :- A,Δ Σ,B :- Π
————————————— (Left Implies)
Γ, Σ, A → B :- Δ,Π
12. Γ,A[y] :- Δ *(y bound)*
————————————— (Left Forall)
Γ,∀x A[x/y] :- Δ
13. Γ :- A[y],Δ *(y free)*
————————————— (Right Forall)
Γ :- ∀x A[x/y],Δ
14. Γ, A[y] :- Δ *(y bound)*
———————————— (Left Exists)
Γ,∃x A[x/y] :- Δ
15. Γ, :- A[y], Δ *(y free)*
————————————(Right Exists)
Γ :- ∃x A[x/y], Δ
16. Γ :- Δ
—————— (Left Weakening)
Γ, A :- Δ
17. Γ :- Δ
—————— (Right Weakening)
Γ :- A, Δ
18. Γ, A, A :- Δ
——————— (Left Contraction)
Γ,A :- Δ
19. Γ :- A, A, Δ
——————— (Right Contraction)
Γ :- A, Δ
20. Γ, A, B, Δ :- Σ
————————— (Left Permutation)
Γ,B, A, Δ :- Σ
21. Γ :- Δ, A, B, Σ
————————— (Right Permutation)
Γ :- Δ B, A, Σ
Here’s an example of how we can use sequents to derive A ∨ ¬ A:
1. Context empty. Apply Identity.
2. A :- A. Apply Right Not.
3. empty :- ¬ A, A. Apply Right And 2.
4. empty : A ∨ ¬A, A. Apply Permute Right.
5. empty :- A, A ∨ ¬ A. Apply Right And 1.
6. empty :- A ∨ ¬ A, A ∨ ¬ A. Right Contraction.
7. empty :- A ∨ ¬ A
If you look *carefully* at the rules, they actually make a lot of sense. The only ones that look a bit strange are the “forall” rules; and for those, you need to remember that the variable is *free* on the top of the sequent.
A lot of logics can be described using Gentzen systems; from type theory, to temporal logics, to all manner of other systems. They’re a very powerful tool for describing all manner of inference systems.
## 0 thoughts on “A Brief Diversion: Sequent Calculus”
1. Canuckistani
Does Rule #4 (Right Or 2) contains an “AND” where it should have an “OR”?
2. Canuckistani
Important Missing Information!!!1!!one!!
The commas are to be treated as ANDs on the left hand side of the ENTAILS and are to be treated as ORs on the right hand side.
3. EJ
You could say I know a lot about logic. However, I’m not used to sequent calculus.
You summarize your first sequent as “Using Γ plus the fact that A is true, everything in Δ can be proved.” I strongly doubt that you really want to say “the fact that A is true” in this context. The sequent should still apply if A is false… except I’m not sure what it means for A to be true or false here. But the sequent should still make sense even if A is a contradiction.
You then summarize the sequent again as “adding a true fact won’t create an inconsistency.” But that doesn’t seem like it’s in the spirit of the sequent at all. The sequent says, “after adding a new assumption, you can still derive what you could derive before.” But it leaves open the possibility that you may be able to derive new things… possibly an inconsistency. I agree that if your new assumption is “true”, then that shouldn’t create an inconsistency, but that doesn’t seem to be what the sequent says.
Basically, the sequent seems to make a claim in the direction of how *much* is entailed by Gamma,A. Whereas a claim of the form “does not create inconsistency” would be a claim in the direction of how *little* is entailed by Gamma,A. Big difference.
I suspect we can’t write “adding a true fact won’t create an inconsistency” in sequent form. Well, that’s partly because I don’t know how you say “true”. It might be easier to write “adding a tautology won’t create an inconsistency”, but I need a new symbol: “:/-“. Sorry, that looks terrible, but it’s supposed to mean “does not entail”. Then I can write something like this:
empty :- B
——————————
Γ,B :/- ¬ A, A
(“If B is a tautology, then adding B to any context does not entail a contradiction.”) But I don’t think that counts as a sequent. Oops—I just realized that even if it is a sequent, it isn’t “valid” as written (consider the case where Gamma is already inconsistent).
4. Mark C. Chu-Carroll
EJ:
We’re running up against one of my frequent problems; it’s difficult to try to write about formal math using informal terms. I tend to err on the side of intuition: I try to write in a way that puts giving people an intuitive understand of what things mean ahead of being as precise as possible. Usually that works out well; sometimes, as in this case, it gets me into trouble 🙂
What I meant by the sequent preventing inconsistency was “If I can prove Δ before adding A, then adding A won’t change things in a way that means that something that was true before won’t be true now – so everything in Δ will still be provable after the addition of A”. I was thinking of that kind of inconsistency: “X is provable from Γ before A, but with A added, X is no longer provable”.
Similarly with the prose about “the fact that A is true”; for an informal presentation, taling about “A” as “true” provides an intuitive grasp on what’s going on, even if it isn’t the ideal formal presentation.
5. Mark C. Chu-Carroll
Canuckistani:
I don’t think it’s really correct to say that the “,” to the left of the entails means “OR”. (In fact, it’s a little iffy to say what anything *means* in sequent calculus, since SC is purely syntactic; the semantics comes from predicate logic.) I may be mis-recalling, but I think that the right-hand side of an entailment is a list of distinct statements, each of which is provable given the left hand side. I’ve updated the article to try to clarify that.
6. Mark C. Chu-Carroll
Canuckistani:
I just went back to a logic test to double-check. You’re right, and I’m wrong: the LHS of an entailment is a conjunction (AND in FOPL), and the RHS of an entailment *is* a disjunction (OR in FOPL).
I’ll go fix the article now. Thanks for the correction, and sorry for getting it wrong even after you pointed it out!
7. Canuckistani
Oh, I don’t know jack about the sequent calculus. You should credit whoever wrote the article on the cut-elimination theorem in Wikipedia.
However, spotting the error in the Right Or 2 rule was all me.
What’s the point of dealing specifically with sequences, since the permutation and contraction rules make them functionally identical to sets?
8. Mark C. Chu-Carroll
Canuckistani:
The point of dealing with sequences and permutation/contraction rules is because the idea of a calculus like SC is to provide a purely syntactic mechanism for manipulating symbolic expressions. Using sets requires you to understand the ideas of expansion, contraction, and permutation; using syntactic rules that perform the permutations and contractions that are needed allows you to avoid needing even the semantics of simple sets: it’s a purely syntactic system that operates on strings of symbols.
9. Lispish
Could you do another article presenting these rules from an intuitionistic perspective?
Wikipedia says, “Surprisingly, some small changes in the rules of LK suffice to turn it into a proof system for intuitionistic logic. To this end, one has to restrict [the system] to sequents with exactly one formula on the right-hand side, and modify the rules to maintain this invariant.
I don’t see how this can be done. Thanks!
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Re: Is x>1? (1) (x+1)(|x|-1) > 0 (2) |x|<5 [#permalink]
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sevaro wrote:
Is x>1?
(1) (x+1)(|x|-1) > 0
(2) |x|<5
I got it right plugging in number. Any other options?
Thanks
This question is rated as hard by GMAC.
ST1: When x > 0 we have: (x+1)(x-1) > 0 then x^2 - 1 > 0 --> x^2 > 1 ==> x < -1 or x > 1, because x > 0 then x > 1
When x < 0 we have: -(x+1)(x+1) > 0 ~ - (x+1)^2 > 0 not existed.
SUFFICIENT.
ST2: -5 < x < 5. INSUFFICIENT.
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Re: Is x>1? (1) (x+1)(|x|-1) > 0 (2) |x|<5 [#permalink]
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sevaro wrote:
Is x>1?
(1) (x+1)(|x|-1) > 0
(2) |x|<5
I got it right plugging in number. Any other options?
Thanks
This question is rated as hard by GMAC.
Question: Is x > 1?
(1) $$(x+1)(|x|-1) > 0$$
For the left hand side to be positive, either both factors are positive or both are negative.
If both are positive,
x+1 > 0, x > -1
AND
|x|-1 > 0, |x|> 1 which means either x < -1 or x > 1
This is possible only when x > 1
If both are negative,
x+1 < 0, x < -1
AND
|x|-1 < 0, |x| < 1 which means -1 < x < 1
Both these conditions cannot be met and hence this is not possible.
This gives us only one solution: x > 1
So we can answer the question asked with "Yes".
(2) $$|x|<5$$
This implies that -5 < x < 5
x may be less than or more than 1. Not sufficient.
Answer (A)
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Re: Is x > 1? [#permalink]
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12 Jun 2014, 02:10
enigma123 wrote:
Is x > 1?
(1) (x+1) (|x| - 1) > 0
(2) |x| < 5
You have to remember Z O N E D (Zero, One, Negative, Extremes, Decimals)
Statement I is sufficient:
We cannot plug in zero and 1 as the expression (x+1) (|x| - 1) will not hold true. All numbers greater than 1 will hold true for the expression. All decimals and negative numbers will make the expression negative.
Hence the value of x will always be greater than 1.
Statement II is insufficient:
x = 4 (YES) and x = -2 (NO)
Hence the answer is A
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Re: Is x>1 [#permalink]
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13 Jun 2014, 09:55
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;
If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.
(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.
Answer: A.
Hope it's clear.
Thanks Bunnel.
I have a question regarding the modulus of X.
|X| = -X when X<= 0
|X| = X when X> 0
Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.
So, is
|X| = -X when X< 0
|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)
So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.
Hope the description of my question is clear enough!
Thank you in advance!
My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)
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13 Jun 2014, 10:03
Kconfused wrote:
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;
If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.
(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.
Answer: A.
Hope it's clear.
Thanks Bunnel.
I have a question regarding the modulus of X.
|X| = -X when X<= 0
|X| = X when X> 0
Which range does zero belong to? Since we know the result in the end solely depends on whether the absolute value of x (whether it is positive or negative) and zero is rather inconsequential in deciding whether |X| is to become -X or +X.
So, is
|X| = -X when X< 0
|X| = X when X >= 0 wrong? (if you observe I have swapped the equal to zero sign from the negative to the positive case)
So where exactly does zero fall in a modulus scenario? It is important because it decides whether the values on the verge of the range are considered in the solution set or not.
Hope the description of my question is clear enough!
Thank you in advance!
My guess is that we will need to plug in the values on the edges of the range and then decide where the zero falls (or rather which values are to be considered)
You can include 0 in either of the ranges:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$.
When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.
The point is that |0|=0, so it doesn't matter in which range you include it.
P.S. BTW your question was already answered in this very thread: is-x-134652.html#p1261810
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22 Nov 2014, 06:24
Bunuel,
Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?
1) x+1 > 0
2) x-1 > 0
3) -x-1 > 0
If correct, could you please demonstrate on how to proceed?
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joseph0alexander wrote:
Bunuel,
Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?
1) x+1 > 0
2) x-1 > 0
3) -x-1 > 0
If correct, could you please demonstrate on how to proceed?
This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
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Re: Is x > 1? [#permalink]
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22 Nov 2014, 06:43
Bunuel wrote:
joseph0alexander wrote:
Bunuel,
Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?
1) x+1 > 0 ---> x > -1
2) x-1 > 0 ---> x > 1
3) -x-1 > 0 ---> -x > 1 or x > -1
If correct, could you please demonstrate on how to proceed?
This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
Fair enough Bunuel, but just for the sake of understanding and learning the concept.
If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?
Thank you.
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Is x > 1? [#permalink]
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22 Nov 2014, 06:58
joseph0alexander wrote:
Bunuel wrote:
joseph0alexander wrote:
Bunuel,
Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?
1) x+1 > 0 ---> x > -1
2) x-1 > 0 ---> x > 1
3) -x-1 > 0 ---> -x > 1 or x > -1
If correct, could you please demonstrate on how to proceed?
This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
Fair enough Bunuel, but just for the sake of understanding and learning the concept.
If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?
Thank you.
Again you should consider 2*2 = 4 cases.
(x+1)(|x| - 1) > 0.
1. x < 0 --> (x+1)(-x - 1) > 0.
(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.
2. x < 0 --> (x+1)(-x - 1) > 0.
(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.
3. x > 0 --> (x+1)(x - 1) > 0.
(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.
4. x > 0 --> (x+1)(x - 1) < 0.
(x + 1) < 0 and (x - 1) < 0 --> x < -1 and x < 1. No solution here.
So, (x+1)(|x| - 1) > 0 holds true only when x > 1.
Hope it's clear.
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22 Nov 2014, 08:21
Bunuel wrote:
Again you should consider 2*2 = 4 cases.
(x+1)(|x| - 1) > 0.
1. x < 0 --> (x+1)(-x - 1) > 0.
(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.
2. x < 0 --> (x+1)(-x - 1) > 0.
(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.
3. x > 0 --> (x+1)(x - 1) > 0.
(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.
4. x < 0 --> (x+1)(-x - 1) < 0.
(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.
So, (x+1)(|x| - 1) > 0 holds true only when x > 1.
Hope it's clear.
Hi Bunuel,
Kindly pardon me. I am unable to understand your explanation. Could you please explain how you arrived at the 4 equations? I think the 4 cases have come up because x could be either positive or negative. Is it so?
Further you've mentioned that we've to consider 2x2 cases. I understand that you have taken 3 cases where x < 0 and 1 case where x > 0. I'm totally lost now.
Please help!
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22 Nov 2014, 08:28
joseph0alexander wrote:
Bunuel wrote:
Again you should consider 2*2 = 4 cases.
(x+1)(|x| - 1) > 0.
1. x < 0 --> (x+1)(-x - 1) > 0.
(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.
2. x < 0 --> (x+1)(-x - 1) > 0.
(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.
3. x > 0 --> (x+1)(x - 1) > 0.
(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.
4. x < 0 --> (x+1)(-x - 1) < 0.
(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.
So, (x+1)(|x| - 1) > 0 holds true only when x > 1.
Hope it's clear.
Hi Bunuel,
Kindly pardon me. I am unable to understand your explanation. Could you please explain how you arrived at the 4 equations? I think the 4 cases have come up because x could be either positive or negative. Is it so?
Further you've mentioned that we've to consider 2x2 cases. I understand that you have taken 3 cases where x < 0 and 1 case where x > 0. I'm totally lost now.
Please help!
I had typos there. Edited now. Anyway the point is that we consider two cases for |x| and then two cases for the multiples both negative and both positive.
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22 Nov 2014, 09:07
Bunuel wrote:
I had typos there. Edited now. Anyway the point is that we consider two cases for |x| and then two cases for the multiples both negative and both positive.
Thanks Bunuel. Understand it now much better.
Reading your post along with this one is-x-134652-20.html#p1373275 from VeritasPrepKarishma helped.
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27 Nov 2014, 04:05
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;
If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.
(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.
Answer: A.
Hope it's clear.
Hi Bunuel
Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.
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27 Nov 2014, 05:30
sinhap07 wrote:
Bunuel wrote:
Good question. +1.
Is x> 1?
(1) (x+1) (|x| - 1) > 0. Consider two cases:
If $$x>0$$ then $$|x|=x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (x - 1) > 0$$ --> $$x^2-1>0$$ --> $$x^2>1$$ --> $$x<-1$$ or $$x>1$$. Since we consider range when $$x>0$$ then we have $$x>1$$ for this case;
If $$x\leq{0}$$ then $$|x|=-x$$ so $$(x+1) (|x| - 1) > 0$$ becomes $$(x+1) (-x - 1) > 0$$ --> $$-(x+1) (x+1) > 0$$ --> $$-(x+1)^2>0$$ --> $$(x+1)^2<0$$. Now, since the square of a number cannot be negative then for this range given equation has no solution.
So, we have that $$(x+1) (|x| - 1) > 0$$ holds true only when $$x>1$$. Sufficient.
(2) |x| < 5 --> $$-5<x<5$$. Not sufficient.
Answer: A.
Hope it's clear.
Hi Bunuel
Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.
$$-(x+1)^2>0$$;
Add (x+1)^2 to both sides: $$0>(x+1)^2$$, which is the same as $$(x+1)^2<0$$.
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Is x > 1?
I would approach it the following way :
(1) (x+1) (|x| - 1) > 0
(+) * (+) > 0 or (-) * (-) > 0
For both of the parts to be positive we can see that x >1 . Just by trying few values you can figure this out. X cant be Zero as then the second part becomes - . X cant be 1 as then second part becomes 0 and hence the whole LHS becomes Zero.
For both of the parts to be negative we try any value less an Zero and see that no value will satisfy the equation. Hence X cannot be negative.. Hence A is Sufficient.
(2) |x| < 5
Clearly not sufficient
Answer is A.
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27 Dec 2014, 11:27
Ans A: Better approach for me will be to try with some common numbers quickly. -1.5, 0.5, 0, 0.5 , 1.5
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24 Aug 2015, 14:52
Is X > 1 ?
1) (X+1) (|X| -1 ) >0
2) |X| <5
Starting with 2) -5 < X < 5 .hence clearly not sufficient .
However, i am not sure how to go about the 1st one .
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gmatcracker24 wrote:
Is X > 1 ?
1) (X+1) (|X| -1 ) >0
2) |X| <5
Starting with 2) -5 < X < 5 .hence clearly not sufficient .
However, i am not sure how to go about the 1st one .
Search for a question before posting a new topic. The question would have been discussed already.
As for the question,
Per statement 1, (x+1)(|x|-1) >0
Case 1: for x $$\geq$$ 0 ---> |x| = x ----> (x+1)(x-1) > 0 ----> x>1 or x<-1 but as x $$\geq$$ 0 ---> only possible case is x>1
Case 2: for x<0 ---> |x|=-x ---> (x+1)(-x-1)>0 ----> $$-(x+1)^2 > 0$$ ---> $$(x+1)^2<0$$ . Now a square can never be <0 and thus x can not be negative.
Thus the only possible case from statement 1 is for x $$\geq$$ 0 which gives a definite "yes" for x>1.
Hence A is the correct answer.
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Re: Is x > 1? [#permalink]
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17 Feb 2016, 14:45
Here is my solution attached
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11 Mar 2016, 23:45
Hi Bunuel,
I have a question.
(x+1)(|x|-1)=0 has x=1 and x=-1 as its solution. However, x=-1 is a repeated root. Hence, on the number line, the sign will not change when it passes x=-1.
--------(-1)---------(1)+++++++
Therefore, only when x>1, (x+1)(|x|-1)>0.
Is my solution correct? I remember my highschool math teacher said something like 'the sign doesn't change when it passes a double root'. But, it's years ago and I just want to make sure that the approach is valid.
Thank you!
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# Which Theorem Gives the Law of Conservation of Energy for a Flowing Liquid
In physics, the law of conservation of energy states that the total energy of an isolated system remains constant it is said to be conserved over time. This law is a direct consequence of the fact that the laws of physics are time-reversible. The total energy includes both kinetic energy and potential energy.
The theorem that gives the law of conservation of energy for a flowing liquid is the Bernoulli equation. This equation states that the sum of the kinetic energy, potential energy, and pressure energy in a fluid is constant. This principle can be applied to any moving fluid, whether it is a liquid or gas.
## State Bernoulli’s Theorem And Prove It
In mathematics, Bernoulli’s theorem states that if two independent random variables X and Y are both distributed according to the same Bernoulli distribution with parameter p, then their sum Z = X + Y is also distributed according to a Bernoulli distribution with parameter p. In other words, the sum of two independent random variables is itself a random variable, and its distribution is determined by the parameter p. The theorem is named after Swiss mathematician Jakob Bernoulli, who proved it in his 1713 book Ars Conjectandi.
The result had been discovered earlier by English mathematician Isaac Newton. To prove the theorem, we first need to recall some basic properties of the Bernoulli distribution. If X is a random variable distributed according to the Bernoulli distribution with parameter p, then we say that X has a Binomial(1,p) distribution.
This means that X can take on only two values: 0 and 1, and that these values are equally likely if p = 0.5. The probability mass function of X is given by: P(X=0) = (1-p)
P(X=1) = p Now let’s suppose that X and Y are both distributed according to the same Bernoulli distribution with parameter p. We want to show that their sum Z = X + Y is also distributed according to a Bernoulli distribution with parameter p.
To do this, we need to calculate the probability mass function of Z. Recall that for any discrete random variable Z, its probability mass function P(Z=z) gives us the probability that Z takes on the value z. So we have: P(Z=0) = P((X+Y)=0)
= P((X=0)*(Y=0)) // since X and Y are independent = P(X=0)*P(Y=0) // using properties of independence = (1-p)*(1-p), // since each term has PMF (1-p).
## Bernoulli Equation Derivation
The Bernoulli equation is a powerful tool that can be used to solve many problems in fluid mechanics. In this blog post, we will derive the Bernoulli equation and then use it to solve some simple problems. The Bernoulli equation is derived from the conservation of energy principle.
This principle states that the total energy of a system is conserved, or remains constant. For a fluid flowing through a pipe, the total energy consists of two components: pressure energy and kinetic energy. The pressure energy is due to the force exerted by the fluid on the walls of the pipe, and the kinetic energy is due to the motion of the fluid.
We can use calculus to show that the rate at which these two energies are lost (or gained) must be equal. This leads us to the Bernoulli equation: P + 1/2 ρ v^2 = constant.
Where P is pressure, ρ is density, and v is velocity. This equation states that for a given volume of fluid flowing through a pipe if the pressure decreases then either the velocity must increase or else the density must decrease (or both). Conversely, if either velocity or density increases, then pressure must decrease in order for total energy to remain constant.
## Bernoulli’s Theorem
Bernoulli’s theorem is a statement in fluid dynamics that describes the relationship between pressure and velocity in a moving fluid. The theorem is named after Daniel Bernoulli, who published it in his book Hydrodynamica in 1738. The theorem states that for an inviscid flow of an incompressible fluid, the sum of the kinetic energy and the potential energy per unit volume is constant.
## Bernoulli’s Theorem Application
In probability theory, Bernoulli’s theorem (also called the law of large numbers in Bernoulli’s form) is a result of sums of independent random variables. It is named after Jacob Bernoulli, who proved it in his work Ars Conjectandi.[1]
The theorem says that if X 1, and X 2, are independent random variables with mean μ 1, μ 2, and variance σ 1 2, σ 2 2 respectively, then as n → ∞, the sum of these random variables divided by n converges in distribution to a normal distribution with mean μ and variance σ2/n. In symbols: where Φ denotes the cumulative distribution function of the standard normal distribution.
## Bernoulli’s Principle Example
In fluid dynamics, Bernoulli’s principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid’s potential energy. The principle is named after Daniel Bernoulli who published it in his book Hydrodynamica in 1738. A simple example of Bernoulli’s principle is provided by the wing of an airplane.
As airspeed increases, the pressure on the top surface of the wing decreases relative to the bottom surface. This difference in pressure creates lift, which allows an airplane to fly.
## State Bernoulli’s Theorem in Fluid Mechanics
In fluid mechanics, Bernoulli’s principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid’s potential energy. The principle is named after Daniel Bernoulli who published it in his book Hydrodynamica in 1738. The principle can be applied to various types of fluids, including incompressible and compressible fluids.
For incompressible fluids, such as water or air, the principle states that if the fluid is flowing along a curved path (such as through a pipe), there is a decrease in pressure at points where the speed of the fluid increases. This decrease in pressure is caused by an increase in the kinetic energy of the fluid at those points. Similarly, for compressible fluids (such as gases), an increase in speed occurs simultaneously with a decrease in density (and thus an increase in temperature).
Again, this change is due to an increase in kinetic energy. Bernoulli’s principle can be used to explain several interesting phenomena, such as how airplanes are able to fly and how blood flows through our arteries. In both cases, it is the decreases in pressure (or density) that allow objects to move against gravity – either by lifting them up into the air or by pushing them forward through our arteries.
## Bernoulli’s Equation
In fluid dynamics, Bernoulli’s equation is a mathematical expression of the conservation of energy for fluids in motion. It states that the sum of the kinetic energy and the potential energy in a moving fluid remains constant. The equation is named after Daniel Bernoulli who published it in his book Hydrodynamica in 1738.
The Bernoulli equation can be applied to any moving fluid, including liquids, gases, or plasmas. It is derived from the principle of conservation of energy, which states that the total energy of a system must remain constant over time. In a flowing fluid, this requires that the sum of the kinetic energy and potential energy must be constant.
The Bernoulli equation has many applications in engineering and physics. For example, it can be used to predict how fluids will flow through pipes or around objects in a streamlined pattern. It can also be used to calculate the pressure exerted by a liquid on a container wall or the force exerted by a gas on an object in a wind tunnel.
## FAQs
### 1. What Does Bernoulli’s Theorem State?
In fluid dynamics, Bernoulli’s equation is a mathematical expression of the conservation of energy for fluids moving along a streamline. The equation was first discovered by Swiss mathematician Daniel Bernoulli in 1738. The Bernoulli equation states that for an inviscid flow of a liquid, the sum of the kinetic energy and the potential energy per unit volume is constant along a streamline.
In other words, if you were to take a snapshot of a section of fluid at any point in time, the total energy per unit volume would be the same throughout that section. The Bernoulli equation can be applied to any incompressible fluid, including liquids and gases. In fact, it’s not even limited to fluids – it can also be used to describe the motion of objects through the air or other mediums where drag forces are present.
The Bernoulli equation has many applications in engineering and physics. For example, it can be used to calculate the lift force on an airplane wing or the pressure drop across a valve in a piping system. It’s also commonly used in analyzing blood flow through arteries and veins.
### 2. Which Conservation Law is Applicable to Bernoulli’s Theorem?
In physics, a conservation law states that the total value of some quantity is constant throughout time. This quantity can be mass, energy, momentum, electric charge, or another property of matter or spacetime. The best-known conservation laws are the laws of conservation of energy and momentum.
Bernoulli’s theorem is a statement in fluid dynamics that describes how the pressure and velocity of a moving fluid change as the fluid flows from one point to another. The theorem is named after Dutch physicist Daniel Bernoulli who published it in his book Hydrodynamica in 1738. The applicable conservation law to Bernoulli’s theorem is the law of conservation of energy.
This law states that energy cannot be created or destroyed; it can only be transformed from one form to another. In other words, the total amount of energy in the universe is always constant. Applying this law to Bernoulli’s theorem, we can see that the kinetic energy (energy associated with motion) of a moving fluid must be equal to its potential energy (energy associated with position).
As the fluid flows from one point to another, its potential energy decreases due to gravity while its kinetic energy increases due to the flow itself. Since Energy = Kinetic Energy + Potential Energy, this decrease in potential energy must be balanced by an increase in kinetic energy; otherwise, there would be a net loss or gain of energy which violates the law of conservation of energy.
### 3. Is Bernoulli’s Principle Conservation of Energy?
No, Bernoulli’s principle is not the conservation of energy. While both principles are concerned with the relationship between energy and fluid flow, they are different concepts. Bernoulli’s principle states that an increase in the speed of fluid results in a decrease in pressure.
This occurs because as the fluid moves faster, its molecules have less time to interact with each other, resulting in fewer collisions and less pressure. Conservation of energy, on the other hand, states that energy cannot be created or destroyed, only converted from one form to another. In the context of fluid flow, this means that the total amount of energy remains constant even as it changes form (e.g. from kinetic to potential).
So while Bernoulli’s principle can be used to explain how energy is transferred within a flowing fluid, it is not itself a law of conservation.
### 4. What is the Conservation of Energy in Fluid Flow?
In fluid flow, the conservation of energy states that the total energy of a system remains constant. The total energy includes both the kinetic and potential energies. The sum of these two energies is called mechanical energy.
In a closed system, such as a sealed container of water, the conservation of energy states that the mechanical energy is constant. The potential energy in fluid flow comes from the height difference between two points. The higher point has more potential energy because it has the ability to do work on objects below it.
The kinetic energy comes from the movement of the fluids themselves. In an open system, such as a river or waterfall, there is also thermal energy due to friction between the fluid and its surroundings. The conservation of energy is important because it helps us understand how systems work and how they will change over time.
It also helps engineers design better systems by understanding how much Energy is available to do work.
## Last Point
The law of conservation of energy for a flowing liquid states that the total energy of the liquid is constant. This theorem is based on the fact that the total energy of a system cannot be created or destroyed, but only converted from one form to another. The theorem applies to both closed and open systems.
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If N is a positive integer, is N! divisible by 14 ? : GMAT Data Sufficiency (DS)
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If N is a positive integer, is N! divisible by 14 ?
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If N is a positive integer, is N! divisible by 14 ?
(1) (N + 1)! is divisible by 15
(2) (N + 2)! is divisible by 16
M22-32
[Reveal] Spoiler: OA
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Last edited by Bunuel on 10 Sep 2014, 03:27, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Sachin9 wrote:
Hi Karishma/Bunuel,
Is there any other way apart from plugging in numbers?
I got this in GC CAT.
C looked suspicious so chose E.
but even if we have to plug in numbers, how can we quicky come up with numbers to say a yes and no for both statements combined?
If N is a positive integer, is N! divisible by 14 ?
In order n! to be divisible by 14=2*7, n must be at least 7. So, the question basically asks whether $$n\geq{7}$$.
(1) (N + 1)! is divisible by 15. In order (n+1)! to be divisible by 15=3*5, n+1 must be at least 5. Thus this statement implies that $$n+1\geq{5}$$ --> $$n\geq{4}$$. Not sufficient.
(2) (N + 2)! is divisible by 16. In order (n+2)! to be divisible by 16=2^4, n+2 must be at least 6 (6!=2*3*4*5*6=(2^4)*3^2*5). Thus this statement implies that $$n+2\geq{6}$$ --> $$n\geq{4}$$. Not sufficient.
(1)+(2) From above we have that $$n\geq{4}$$. If n=4, then the answer is NO but if n=7 then the answer is YES. Not sufficient.
Hope it's clear.
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Re: If N is a positive integer, is N! divisible by 14 ? [#permalink]
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09 Apr 2008, 11:23
looks to be E..
(N+2)! =16N basically means we have 2^4 of 2s at a minimum..that means N+2=6 at the least..or N at the least equal 4.. again we cant say anything about N
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Re: If N is a positive integer, is N! divisible by 14 ? [#permalink]
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09 Apr 2008, 11:43
E
The least factorial where 14 is a factor is 7!, 1 x 2 x 3 x 4 x 5 x 6 x 7, so in order for 14 to be a factor of N!, we know N must be greater than 7.
1) The least value of N would be 4. (N+1)! = 5! = 1 x 2 x 3 x 4 x 5, the smallest number that is also divisible by 15. We know N is greater than or equal to 4 so 1) is a no, eliminate A and D, down to BCE
2) The least value of N would be 6. (N+2)! = 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8, the smallest number that is also divisible by 16. We know N is greater than or equal to 6 so 2) is a no, elimante B, down to C or E
Combine the two statements and we know N must be greater than 4, which could or could not be greater than 7, eliminate choice C, the answer is E
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Re: If N is a positive integer, is N! divisible by 14 ? [#permalink]
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11 Apr 2008, 08:19
I solve this by plugging in numbers.
Ans should be 'E'
for N=4,
5! is div by 15 and 6! is div by 16 but 4! is not div by 14
for N =7,
8! is div by 15 and 9! is div by 16 and 7! is div by 14
So insuff
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Re: If N is a positive integer, is N! divisible by 14 ? [#permalink]
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12 Feb 2013, 01:45
Hi Karishma/Bunuel,
Is there any other way apart from plugging in numbers?
I got this in GC CAT.
C looked suspicious so chose E.
but even if we have to plug in numbers, how can we quicky come up with numbers to say a yes and no for both statements combined?
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# What is Hamming Distance? Applications and Operations
Before going through the blog, let’s take a quick look at the prerequisites. You should have a foundational knowledge of binary representation and basic mathematical operations. It will be beneficial if you have a firm grasp on the concept of strings and sets.
While Hamming distance coding is possible in any programming language, we will utilize C++ 23 in this context. This decision stems from the 2022 survey conducted by HackerEarth, where C++ was identified as the foremost general programming language. Out of the respondents, 53.3% favored C++, followed by Java (26.7%) and Python (10%).
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## What is Hamming Distance?
By definition, Hamming distance is a metric in information theory that is used for comparing two binary data strings of equal length at corresponding positions. The crux of Hamming distance is to find the minimum number of substitutions needed to change one string into another.
Note: In the blog, we will consider two strings, `'strOne'` and `'strTwo'` of equal length, `'len'`, and represented by `d(strOne, strTwo)'`.
Hamming distance follows multiple conditions. A few are mentioned below:
• The length of the two strings involved in Hamming distance should be equal.
• If two characters are identical, then the Hamming distance is zero.
• Hamming distance follows the concept of symmetry. Meaning the number of characters in both strings is equal.
• Hamming distance also satisfies the concept of triangle inequality.
• For binary strings, Hamming distance is equal to the number of ones in `d(strOne XOR strTwo)`.
Here are a few of the important terminologies that you might encounter during your learning journey:
• Hamming Weight: The “Hamming weight,” within the context of computer science and information theory, signifies the count of non-zero and distinguishable elements within a sequence, commonly expressed in binary form. It quantifies the number of ‘1’ bits present, offering insight into the data’s density and aiding error detection algorithms.
• Population Count: The population count refers to the quantitative assessment of the disparity between two strings of equal length by calculating the total count of positions at which their corresponding elements differ.
• Hamming Space: A Hamming space is a mathematical concept in which binary strings of equal length serve as points. It measures the dissimilarity between two strings by calculating the number of differing bits. Hamming spaces find applications in error detection, data clustering, and information retrieval.
### Brief History of Hamming Distance
The American mathematician Richard Hamming first coined the concept of Hamming distance. While working at Bell Labs, he introduced Hamming distance in his “Hamming codes, Error detection, and error correction codes” paper in 1950.
He was working on developing a method for the identification and correction of bits in data transmission. He concluded that by counting the number of bits that differed between two binary sequences, he could determine how many errors had occurred.
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## How is Hamming Distance Calculated?
Calculating the Hamming Distance involves measuring the dissimilarity between two strings of equal length. Here are some processes:
• Compare Corresponding Symbols: Compare the symbols (characters or bits) at each position in the two strings, starting from the first position.
• Count Differences: For every position where the symbols are different, increment a counter by one. This counter keeps track of the number of differences between the strings.
• Summation: After comparing all positions, the counter’s value represents the Hamming Distance. It signifies the total number of positions at which the two strings differ.
Mathematically, Hamming distance is expressed as:
`Hamming Distance = Σ (strOne[i] ≠ strTwo[i])`
here,
Hamming Distance denotes the disparity between the strings.
Σ symbolizes the summation of all positions.
strOne[i] represents the element at index ‘i’ in the first string.
strTwo[i] represents the element at index ‘i’ in the second string.
≠ signifies “not equal.”
### Hamming Distance Formula
The Hamming distance formula is as follows:
`hamming_distance(strOne, strTwo) = number of bits in strOne that are different from bits in strTwo`
where:
• `strOne` is the first binary string
• `strTwo` is the second binary string
For example, the Hamming distance between the strings 101010 and 111010 is 1. There is one bit position where the strings are different: the first bit.
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## Application of Hamming Distance
Below mentioned are the various fields in which hamming distance is applied:
• Hamming Distance in Computer Science: Hamming distance is primarily utilized in computer science. It employs error detection and correction of the data packets in computer networking. It ensures data integrity during data transmission. Also, it is used to enhance data reliability and prevent data corruption.
• Hamming Distance in Artificial Intelligence: Hamming distance applies to clustering algorithms and pattern recognition. It measures the similarity or dissimilarity between feature vectors, aiding in grouping similar data points. It integrates image segmentation, document classification, and recommendation systems.
• Hamming Distance in Cryptography: Hamming distance is widely used in the domain of cryptography. It provides cryptographic protocols for encryption security and resilience against unauthorized access.
• Hamming Distance in Telecommunications: Hamming distance helps to detect and correct errors in data transmission over noisy channels.
• Hamming Distance in Biology: In genetics and bioinformatics, Hamming distance aids in comparing DNA sequences. It measures the dissimilarity between genetic sequences, facilitating tasks such as sequence alignment, identifying mutations, and evolutionary analysis.
It is also seen that Hamming distance is also used for digital forensics to check for alterations and barcoding identifiers for enhanced readability, even when the barcode is distorted.
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## Hamming Distance Vs. Euclidean Distance
It would be best to have a glimpse of Euclidean distance to have better clarity about Hamming distance. Below, we have mentioned the major differences between Hamming distance and Euclidean distance.
## Let’s Code: Hamming Distance
Here is the implementation of Hamming distance:
Pseudo Code:
• Read the first string, `‘strOne’`, and the second string, `'strTwo'`, from the user.
• If the length of `‘strOne’` is not equal to the length of `‘strTwo’`, throw an exception.
• Initialize a variable `'distance'` to 0.
• Loop through each character at index ‘i’ from 0 to the length of `‘strOne’`.
• If `strOne[i]` is not equal to `strTwo[i]`, increment `'distance'`.
• Print the value of `'distance'` as the Hamming Distance.
Algorithm:
• Read two input strings: `'strOne'`, and `'strTwo'`.
• Check if the lengths of `'strOne'`, and `'strTwo'` are the same. If not, throw an exception.
• Initialize a variable `'distance'` to 0 to store the Hamming Distance.
• Iterate through each character at the same index in both strings.
• If the string characters at the current index are not equal, increment the distance.
• Print the value of distance as the Hamming Distance.
Hamming Code C++:
```#include <iostream>
#include <string>
#include <algorithm>
int hammingDistance(const std::string& strOne, const std::string& strTwo) {
if (strOne.length() != strTwo.length()) {
throw std::invalid_argument("Strings must have equal lengths");
}
int distance = 0;
for (size_t i = 0; i < strOne.length(); ++i) {
if (strOne[i] != strTwo[i]) {
distance++;
}
}
return distance;
}
int main() {
std::string strOne, strTwo;
std::cout << "Enter the first string: ";
std::cin >> strOne;
std::cout << "Enter the second string: ";
std::cin >> strTwo;
try {
int distance = hammingDistance(strOne, strTwo);
std::cout << "Hamming Distance: " << distance << std::endl;
} catch (const std::invalid_argument& e) {
std::cerr << "Error: " << e.what() << std::endl;
}
return 0;
} ```
Output:
Enter the first string: 101010
Enter the second string: 111000
Hamming Distance: 3
Hamming Code Python:
`def hamming_distance(strOne, strTwo): if len(strOne) != len(strTwo): raise ValueError("Strings must have equal length for Hamming Distance calculation") distance = 0 for char1, char2 in zip(strOne, strTwo): if char1 != char2: distance += 1 return distance# Test Casestry: strOne = "101010" strTwo = "100110" distance = hamming_distance(strOne, strTwo) print(f"Hamming Distance between '{strOne}' and '{strTwo}' is: {distance}")except ValueError as ve: print(ve)try: strOne = "101010" strTwo = "10011" distance = hamming_distance(strOne, strTwo) print(f"Hamming Distance between '{strOne}' and '{strTwo}' is: {distance}"except ValueError as ve: print(ve)`
Output:
Hamming Distance between ‘101010’ and ‘100110’ is: 2
Strings must have equal length for Hamming Distance calculation
Hamming Code Java:
`import java.util.Scanner;public class IntellipaatHammingDistance { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); System.out.print("Enter the first string: "); String string1 = scanner.nextLine(); System.out.print("Enter the second string: "); String string2 = scanner.nextLine(); if (string1.length() != string2.length()) { System.out.println("Error: The two strings must have equal length."); } else { int hammingDistance = calculateHammingDistance(string1, string2); System.out.println("Hamming Distance: " + hammingDistance); } scanner.close(); } public static int calculateHammingDistance(String str1, String str2) { int distance = 0; for (int i = 0; i < str1.length(); i++) { if (str1.charAt(i) != str2.charAt(i)) { distance++; } } return distance; }}`
Output:
Enter the first string: 101010
Enter the second string: 111110
Hamming Distance: 1
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## Operations on Hamming Distance
Here are a few of the operations that you can perform using Hamming distance:
### Hamming Distance Between Two Integers
Problem Statement: Write a program to find the hamming distance between two integers. Considering the inputs as intOne = 10 (1010), intTwo = 14 (1110), and output: 3.
Implementation:
`#include <bits/stdc++.h>using namespace std;int hammingDistance(int intOne, int intTwo){int d = intOne ^ intTwo;int setBits = 0;while (d > 0) { setBits += d & 1; d >>= 1;}return setBits;}// Driver codeint main(){int intOne = 10, intTwo = 14;cout << hammingDistance(10, 14) << "/n";return 0;}`
Time Complexity: O(log2x)
Auxiliary Space: O(1)
We can also solve this problem using two more approaches:
• Checking the maximum of both numbers first, then checking the bits at each position.
• Using the concept of string manipulation.
### Reduce Hamming Distance by Swapping Two Characters
Problem Statement: Write a program to find the positions of two characters to be swapped in the strings strOne and strTwo using Hamming distance in such a way that the distance between the two strings is the smallest. Taking inputs as strOne = “Intellipaat”, and strTwo = “Inllitepata”, and output: 6 3.
Implementation:
`#include <bits/stdc++.h>using namespace std;#define MAX 26void Swap(string one, string two, int n){int dp[MAX][MAX];memset(dp, -1, sizeof dp);int t = 0;for (int i = 0; i < n; i++) if (one[i] != two[i]) t++;for (int i = 0; i < n; i++) { int a = one[i] - 'a'; int b = two[i] - 'a'; if (a == b) continue; if (dp[a][b] != -1) { cout << i + 1 << " " << dp[a][b] + 1 << endl; return; } dp[b][a] = i;}int A[MAX], B[MAX];memset(A, -1, sizeof A);memset(B, -1, sizeof B);for (int i = 0; i < n; i++) { int a = one[i] - 'a'; int b = two[i] - 'a'; if (a == b) continue; if (A[b] != -1) { cout << i + 1 << " " << A[b] + 1 << endl; return; } if (B[a] != -1) { cout << i + 1 << " " << B[a] + 1 << endl; return; } A[a] = i; B[b] = i;}cout << -1 << endl;}// Driver codeint main(){string strOne = "Intellipaat";string strTwo = "Inllitepata";int n = strOne.length();if (strOne == "" || strTwo == "") cout << "Required string is empty.";else Swap(strOne, strTwo, n);return 0;}`
Time Complexity: O(n)
Auxiliary Space: O(MAX * MAX)
### Finding Rotation with Maximum Hamming Distance
Problem Statement: Write a program to find a rotation with the maximum Hamming distance. Input an array of size ‘N’ and another array that contains the rotation of the first array and print the maximum distance between them. Take N = 3, arr = {1,4,1}, and output: 2.
Implementation:
`#include <bits/stdc++.h>using namespace std;int maxHamming(int arrOne[], int N){int arrTwo[2 * N + 1];for (int i = 0; i < N; i++) { arrTwo[i] = arrOne[i]; arrTwo[N + i] = arrOne[i];}int maxHam = 0;for (int i = 1; i < N; i++) { int Ham = 0; for (int j = i, k = 0; j < (i + N); j++, k++) if (arrTwo[j] != arrOne[k]) Ham++; if (Ham == N) return N; maxHam = max(maxHam, Ham);}return maxHam;}// Driver programint main(){int arrOne[] = { 1, 4, 1 };int N = sizeof(arrOne) / sizeof(arrOne[0]);cout << maxHamming(arrOne, N);return 0;}`
Time Complexity: O(n2)
Auxiliary Space: O(n)
We have used a naive approach to solve the problem. But this problem can also be solved using two other approaches:
• Using list comprehension
• Constant Space
Enhance your interview preparations with us. Check out our C++ Programming Interview Questions
## Conclusion
In the blog, ‘What is Hamming Distance? Operations and Applications’, we briefly discussed Hamming distance, its applications, and various operations you can perform using it. Hamming distance is helpful for error correction, computer programming, and computer networks.
To expand your knowledge, we recommend exploring related topics like error-correcting codes and information theory. To solidify your understanding, practicing more problems involving Hamming Distance will help you master its applications. Computer science is like an ocean, and nobody has seen it all. Hence, consistent efforts are required to explore various untouched concepts.
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Supershapes (Superformula)
Written by Paul Bourke
March 2002
Based upon equations by Johan Gielis
Intended as a modelling framework for natural forms.
The supershape equation is an extension of the both the equation of the sphere and ellipse
(x / a)2 + (y / b)2 = r2
and even the superellipse given here. The general formula for the supershape is given below.
Where r and phi are polar coordinates (radius,angle). n1, n2, n3, and m are real numbers. a and b are real numbers excluding zero.
m = 0 This results in circles, namely r = 1 n1 = n2 = n3 = 1 Increasing m adds rotational symmetry to the shape. This is generally the case for other values of the n parameters. The curves are repeated in sections of the circle of angle 2π/m, this is apparent in most of the following examples for integer values of m. m=1 m=2 m=3 m=4 m=5 m=6 n1 = n2 = n3 = 0.3 As the n's are kept equal but reduced the form becomes increasingly pinched. m=1 m=2 m=3 m=4 m=5 m=6 If n1 is slightly larger than n2 and n3 then bloated forms result. The examples on the right have n1 = 40 and n2 = n3 = 10. m=1 m=2 m=3 m=4 m=5 m=6 Polygonal shapes are achieved with very large values of n1 and large but equal values for n2 and n3. m=3 m=4 m=5 m=6 Asymmetric forms can be created by using different values for the n's. The following example have n1 = 60, n2 = 55 and n3 = 30. m=3 m=4 m=5 m=6 For non integral values of m the form is still closed for rational values. The following are example with n1 = n2 = n3 = 0.3. The angle phi needs to extend from 0 to 12π. m=1/6 m=7/6 m=13/6 m=19/6 Smooth starfish shapes result from smaller values of n1 than the n2 and n3. The following examples have m=5 and n2 = n3 = 1.7. n1=0.50 n1=0.20 n1=0.10 n1=0.02
Other examples
Source code
Given a value of phi the following function evaluates the supershape function and calculates (x,y).
void Eval(double m,double n1,double n2,double n3,double phi,double *x,double *y)
{
double r;
double t1,t2;
double a=1,b=1;
t1 = cos(m * phi / 4) / a;
t1 = ABS(t1);
t1 = pow(t1,n2);
t2 = sin(m * phi / 4) / b;
t2 = ABS(t2);
t2 = pow(t2,n3);
r = pow(t1+t2,1/n1);
if (ABS(r) == 0) {
*x = 0;
*y = 0;
} else {
r = 1 / r;
*x = r * cos(phi);
*y = r * sin(phi);
}
}
So it might be called as follows
for (i=0;i<=NP;i++) {
phi = i * TWOPI / NP;
Eval(m,n1,n2,n3,phi,&x,&y);
--- do something with the point x,y ---
}
Supershape in 3D
(Also known as the Superformula)
Written by Paul Bourke
July 2003
Based upon equations by Johan Gielis
Intended as a modelling framework for natural forms.
Contribution of MSWindows version by Vincent Berthoux: supershape_win.zip
Definition of supershape in 2 dimensions.
Extension to 3D using the spherical product.
The following shows the interactive interface developed to explore 3D supershapes. It is based upon X-Windows and OpenGL. It is currently available for Mac OS-X (as a UNIX application). Use of the program is straightforward, edit fields and hit return. The left mouse button rotates the model in two axes, the middle mouse button rotates about the third axis. A right mouse button brings up a menu.
• Being based upon X11 it is necessary to download and install XQuartz. While not supposedly necessary, the author suggests rebooting after installing XQuartz.
• You will need to get around the Apple security restrictions for any application not from the app store. After trying to run the program and being warned about security, go to System Preferences and Security & Privacy, in the General tab allow supershape to be opened.
• Please note that a right mouse button is required to access the menus by right clicking on the window. You can get a right mouse by either getting a 3 button USB mouse, or using the control-click on the touchpad.
Due to the march of progress some technologies get left behind, in this case it is Apples failure to continue (correct) support for X11 on their retina displays. As such, this software is unfortunately no longer available.
Click for full size image
Command line interface
Many settings can be initialised from the command line, others control is given by single key strokes. To see a complete list type the program name at the command prompt with a "-h" option. At the time of writing these are given below.
>supershape -h
Usage: supershape [command line options]
Command line options
-h this text
-f full screen
-s active stereo
-ss dual screen stereo
-a auto rotate
-rw wireframe
Mouse buttons
left camera rotate
shift left camera pan
middle camera roll
shift middle camera forward, reverse
Key Strokes
arrow keys camera rotate
<,> camera forward, reverse
+,- camera zoom in, out
[,] camera roll
r toggle window recording
w capture window to TGA file
a toggle auto rotation of camera
1,2,3 different rendering mode
h camera to home position
f1 to f6 axis aligned camera positions
ESC,q quit
Features
• Exports geometry as DXF, POVRay, and LightWave (the later includes texture coordinates).
• Rendering modes include wireframe, flat shaded, and specular shading. Rendering is based upon OpenGL.
• Colour ramps can be applied to both longitude and latitude.
• Parameter space exploration by either single parameter ranges or simply choosing a random parameter set.
• A supershape can be morphed into another.
• Images can be saved to a single TGA file or repeated saving can be turned on for animations. TGA files and the naming convention used are supported by QuickTime Pro.
Examples
The 2D supershapes (but with added small thickness) can be created using this software by setting the second supershape parameters as m=0, n1=n2=n3=1, a=b=1 and using a small z axis scale factor. Indeed this can often be the preferred approach for using a 2D form within a 3D model, after all, real world objects do have some thickness. The example on the left has a thickness of 0.02 units (x and y axis dimensions of 1). Or even set the depth to 0 and vary latitude from 0 to 90 degrees as in the example on the right.
Extrusions of 2D supershapes can be created by setting the second supershape parameters as m=4, a=1, b=1, and high equal values for each of n1,n2,n3. For example n1=n2=n3=200 gives relatively sharp edges.
A significant portion of the parameter space results in surfaces with various types of numerical problems (powers of negative numbers, divide by 0, underflow, overflow, etc) as well as issues related to the representation of 3D graphics. The software that has been developed and created the images shown above, can display the edges where numerical problems have arisen. These regions are shown in pick as shown in the following two images.
Cube Cone
Diamond Sphere
Prism Cylinder
Hexagon Pentagon
The supershape function can obviously be modulated by another function. For example to create the shell like structures below, the radius (r1) of the superformula that varies the longitude is scaled by either a logarithmic or Archimedes spiral. In some cases the z coordinate is additionally made a linear function of longitude.
In the above the two supershapes were mapped onto a topological sphere. One could map supershapes onto other forms as well, for example, a torus. In the following images the mapping is as follows:
Generating the 3D Supershape in Povray
Written by Paul Bourke
July 2003
Based upon equations by Johan Gielis
Intended as a modelling framework for natural forms.
There are two (at least) ways of representing supershapes using PovRay. One is to use the built-in parametric primitive, doing it this way would result in the highest resolution/quality results unfortunately the rendering times proved to be prohibitive. The approach illustrated below is to write a macro in the PovRay scene language. The macro and an example of it's use is given in this PovRay (version 3.5) file: supershape.pov.
SuperShape(9,1,1,92,0.3,-45,4,1,1,-0.8,88,-0.35,202) SuperShape(1,1,1,77,0.81,71.7,8,1,1,0.63,2.92,0.24,200) SuperShape(9,1,1,-70,-0.14,77,2,1,1,0.38,4.12,-0.7,200) SuperShape(7,1,1,20.45,-0.33,-3.54,6,1,1,-0.96,4.46,0.52,200)
Further POVRay scene examples are given below
ss_macro1.inc example1.pov ss_macro2.inc example2.pov ss_macro3.inc example3.pov example4.pov
Contributions from the POVRay competition
Author: SpB Title: Flying wing Source: 0001.pov Author: Dalius Dobravolskas Title: Lost in the sea Source: 0002.pov Author: Paul Bourke Title: Supershape fractal Source: 0003.pov Author: Stefano Tessarin Title: Flowers Source: 0004.pov Author: Emanuele Munarini Source: 0005.pov Author: Emanuele Munarini Source: 0006.pov Author: Emanuele Munarini Source: 0007.pov Author: Emanuele Munarini Source: 0008.pov Author: Ben Scheele Title: Butterflies find cactus flowers in a desert Source: 0009.pov
Correspondence
Question: How can Superformula or Supershapes aid in data visualization? Do they help a person to gain a perspective of various data points? Are they better than pie-charts, graphs, histograms, scatter plots etc.,
Hmmm, they are kind of different things ... pie chart, graphs etc and supershapes.
In the former one takes data and uses it to control some geometry, quite simple geometry like portions of a pie, height of bars, points on a plane, lines and so on. The resulting structure is intended to provide some insight into the underlying data. At the end of the day the purpose of visualisation is to provide insight and the success of a visualisation should be judged as such.
Supershapes are (complex) 3D surfaces, they came about because Johan thought they could be used to describe a large range of biological shapes, namely leaves, flowers. The big problem with supershapes is the relationship between the variables and the final form is not always clear. So, if you gave someone a geometric form and asked what the variables are to represent that as a supershape, it is difficult, most people just play/fiddle and hope for the best.
Perhaps you are imagining mapping data to the variables and using the resulting supershape as a visualisation of the data. There may be some datasets for which this might work, but I doubt it. For the same reason, the relationship between the variables and the shape is unclear. And, if you have two very similar datasets the supershape could look very different ... not a good property for a visualisation.
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# Harvested prey–predator model incorporating a prey refuge
I want to plot the phase diagram of prey predator versus prey refuge to see how the prey refuge influences the population of prey and predator. And this is the system
$x'=\alpha x(1-x/k)-\beta\frac{(1-m)xy}{1+a(1-m)x}-q_1E_1x$
$y'=-\gamma y+c\beta\frac{(1-m)xy}{1+a(1-m)x}-q_2E_2y$
The prey predator with Holling type II model is incorporating a prey refuge, $mx$ and $k$, $\alpha$, $\gamma$, $c$ and $\beta/\alpha$ are the carrying capacity, growth rate of prey, death rate of predator, conversion factor denoting the number of newly born predators for each captured prey and maximum number of prey that can be eaten by each predator in unit time respectively.
I have the numerical value for $a=0.02$, $k=100$, $\alpha=10$, $\beta=0.6$, $\gamma=0.09$, $c=0.02$.
Thanks so much!
• I see few demos here on this subject demonstrations.wolfram.com/search.html?query=predator%20prey Apr 12, 2014 at 6:26
• I think NDSolve is the function you are looking for. Apr 12, 2014 at 9:44
• A posts with 1000s of constants must miss a few ;-) Your m?? Apr 12, 2014 at 9:59
I solved more general system linked by @SjoerdC.deVries in the comments reproducing figure 3 and 4 - to prove it is correct. You can simplify this to version you need.
Clear["Global*"]
al = 2;
a = 2/1000;
k = 600;
b = 1/10;
g = 46/10^5;
c = 1/100;
m = 1/100;
E1 = 1;
q1 = 2/10;
E2 = 813/1000;
q2 = 2/100;
Tf = 300;
eqs = {
x'[t] ==
al x[t] (1 - x[t]/k) - b (1 - m) x[t] y[t]/(1 + a (1 - m) x[t]) -
q1 E1 x[t],
y'[t] == -g y[t] + c b (1 - m) x[t] y[t]/(1 + a (1 - m) x[t]) -
q2 E2 y[t],
x[0] == 2, y[0] == 8};
s = NDSolve[eqs, {x, y}, {t, Tf}];
Plot[Evaluate[{x[t], y[t]} /. s], {t, 0, Tf}, PlotStyle -> Automatic,
ImageSize -> 300, PlotRange -> All, Frame -> True]
ParametricPlot[Evaluate[{x[t], y[t]} /. s], {t, 0, Tf},
AspectRatio -> 1, PlotRange -> All, ImageSize -> 300, Frame -> True]
• interesting, your default plot colors are not the v9 defaults... :) Apr 12, 2014 at 18:05
• Are these V 10 plots? They look different from V9 screen shot: !Mathematica graphics The lines are thicker also. Apr 12, 2014 at 19:35
• Well "it is not V9" says it all ;-) You can actually get same thing now on Raspberry Pi. Apr 12, 2014 at 20:15
Some modifications for the code in Vitaliy Kaurov's answer:
<<MaTeX
<< c:\CurvesGraphics6\CurvesGraphics6.m
Initial condition outside the limit cycle:
α = 200/207;
a = 4;
k = 3;
β = 200/69 + 1/50;
m = 1/2;
c = 4761/20000;
γ = 1/10;
E1 = 500/621;
q1 = 2/10;
E2 = 3/4;
q2 = 2/100;
Tf = 1000;
eqs1 = {x'[t] == -E1 q1 x[t] + x[t] (1 - x[t]/k) α - ((1 - m) x[t]*y[t] β)/(1 + a (1 - m) x[t]),
y'[t] == -E2 q2 y[t] + (c (1 - m) x[t]*y[t] β)/(1 + a (1 - m) x[t]) - y[t]γ, x[0] == 1.1, y[0] == 1.1};
s1 = NDSolve[eqs1, {x, y}, {t, Tf}];
ts1 = Plot[Evaluate[{x[t], y[t]} /. First[s1]], {t, 0, Tf},
PlotRange -> All,
PlotStyle -> {{Blue, Thickness[0.003]}, {Red, Thickness[0.003]}},
Frame -> True,
FrameLabel -> { MaTeX["\\text{t}", Magnification -> 1],
MaTeX["\\text{Population densities}", Magnification -> 1]},
LabelStyle -> Directive[Black, Tiny], RotateLabel -> True,
PlotPoints -> 500, ImageSize -> {300, 200},
DisplayFunction -> Identity]
pp1 = ParametricPlot[Evaluate[{x[t], y[t]} /. s1], {t, 0, 100},
Axes -> False, Oriented -> True, ArrowPositions -> {0.05, 0.985},
AspectRatio -> 1, PlotRange -> All, Frame -> True,
RotateLabel -> True,
FrameLabel -> {
MaTeX["\\text{Prey population density}", Magnification -> 1],
MaTeX["\\text{Predator population density}", Magnification -> 1]},
RotateLabel -> False, LabelStyle -> Directive[Black, Tiny],
ImageSize -> {250, 250}]
Initial condition inside the limit cycle:
α = 200/207;
a = 4;
k = 3;
β = 200/69 + 1/50;
m = 1/2;
c = 4761/20000;
γ = 1/10;
E1 = 500/621;
q1 = 2/10;
E2 = 3/4;
q2 = 2/100;
Tf = 1000;
eqs2 = {x'[t] == -E1 q1 x[t] + x[t] (1 - x[t]/k) α - ((1 - m) x[t]*y[t] β)/(1 + a (1 - m) x[t]),
y'[t] == -E2 q2 y[t] + (c (1 - m) x[t]*y[t] β)/(1 + a (1 - m) x[t]) - y[t]γ, x[0] == 1.02, y[0] == 1.02};
s2 = NDSolve[eqs2, {x, y}, {t, Tf}];
ts2 = Plot[Evaluate[{x[t], y[t]} /. First[s2]], {t, 0, Tf},
PlotRange -> All,
PlotStyle -> {{Blue, Thickness[0.003]}, {Red, Thickness[0.003]}},
Frame -> True,
FrameLabel -> { MaTeX["\\text{t}", Magnification -> 1],
MaTeX["\\text{Population densities}", Magnification -> 1]},
LabelStyle -> Directive[Black, Tiny], RotateLabel -> True,
PlotPoints -> 500, ImageSize -> {300, 300},
DisplayFunction -> Identity]
pp2 = ParametricPlot[Evaluate[{x[t], y[t]} /. s2], {t, 900, Tf},
PlotStyle -> {Black, Thickness[0.008]}, AspectRatio -> 1,
PlotRange -> All, Frame -> True, RotateLabel -> True,
FrameLabel -> {
MaTeX["\\text{Prey population density}", Magnification -> 1],
MaTeX["\\text{Predator population density}", Magnification -> 1]},
RotateLabel -> False, LabelStyle -> Directive[Black, Tiny],
ImageSize -> {250, 250}]
The initial conditions and equilibrium point:
p0 = Part[{x, y} /. NSolve[{-E1 q1 x + x (1 - x/k) α - ((1 - m) x*y β)/(1 + a (1 - m) x), -E2 q2 y + (c (1 - m) x*y β)/(1 + a (1 - m) x) - y γ} == {0, 0} && x > 0 && y > 0, {x, y}], 1];
(*{0.97972, 0.992966}*)
IC1 = {1.02, 1.02};
IC2 = {1.1, 1.1};
pointIC1 =
Graphics[{PointSize[0.015], Darker[Green], Point[IC1]},
IC1 - {-0.08, -0.011}, Axes -> True, PlotRange -> All,
ImageSize -> {250, 250}];
pointIC2 =
Graphics[{PointSize[0.015], Darker[Green], Point[IC2]},
IC2 - {-0.08, -0.011}, Axes -> True, PlotRange -> All,
ImageSize -> {250, 250}];
Finally, when applying the command Show produces
Show[pp2, pp21, pp1, pointIC1, pointIC2,
Epilog -> {{PointSize[0.018], Black, Point[p0]}, {PointSize[0.013],
White, Point[p0]}}, Axes -> False, ImageSize -> {350, 350}]
• Please post code in InputForm if you want to contribute anything with code. Nov 10, 2017 at 8:14
• Ok, I appreciate the suggestion. Nov 10, 2017 at 8:23
• This looks like it would be a valuable contribution if it were posted properly. Please read the instructions on formatting code given here and here Nov 10, 2017 at 11:32
• Thanks for your support M. Goldberg. Nov 10, 2017 at 20:16
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# How To Compute & Interpret Pearson’S Product Moment Correlation Coefficient?
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### How To Compute & Interpret Pearson’S Product Moment Correlation Coefficient?
1. 1. How to Compute and Interpret Pearson’s R Page 1 How To Compute and Interpret Pearson’s Product Moment Correlation Coefficient Objective: Learn how to compute, interpret and use Pearson’s correlations coefficient. Keywords and Concepts 1. Product moment correlation 8. Positive correlation coefficient 9. Correlation does not imply causation 2. Degree of association 10. Property of linearity 3. Karl Pearson 11. Nonlinear relationship 4. X-axis 12. Coefficient of determination (r2) 5. Y-axis 13. Percent common variance 6. Bivariate scatter-plot 14. Statistical significance 7. Negative correlation The degree of association between two variables (correlation) can be described by a visual representation or by a number (termed a coefficient) indicating the strength of association. The quantitative computation of the correlation was first derived in 1896 by Karl Pearson and is referred to as “Pearson’s product moment correlation coefficient.” Visual Description of Quantitative Relationships A visual description of a correlation appears as a scatter plot where scores for two variables (from the same or different subjects) are plotted with one variable on the X-axis (horizontal axis) and the other variable on the Y-axis (vertical axis). Figure 1 displays a bivariate scatter-plot of the relationship between height (cm) and weight (kg) for a group of male college students enrolled in exercise physiology at the University of Michigan in the fall semester 2002. Each data point in the graph represents one person’s
2. 2. How to Compute and Interpret Pearson’s R Page 2 score on both variables. Note, the pattern of association shows that increasing values of height generally correspond to increasing values of weight. Figure 1. Plot of height versus weight for a sample of college students. 100 95 90 85 Weight, kg 80 75 70 65 60 55 50 170 175 180 185 190 195 Height, cm Figure 2 (next page) displays eight other scatterplot examples. Graphs (a), (b), and (c) depict a pattern of increasing values of Y corresponding to increasing values of X. In plots (a) to (c) the dot pattern becomes closer to a straight line, suggesting that the relationship between X and Y becomes stronger. The scatterplots in (d), (e) and (f) depict patterns where the Y values decrease as the X values increase. Again, proceeding from graph (d) to (f), the relationship becomes stronger. In contrast to the first six graphs, the scatter-plot of (g) shows no pattern (correlation) between X and Y. Finally, the scatter plot of (h) shows a pattern, but not a straight-line pattern as with the other plots.
3. 3. How to Compute and Interpret Pearson’s R Page 3 Visual inspection of scatter plots only permits a subjective and general description of relationships. To quantify relationships between variables the Pearson’s correlation coefficient is used.
4. 4. How to Compute and Interpret Pearson’s R Page 4 Figure 2. Different scatter diagrams.
5. 5. How to Compute and Interpret Pearson’s R Page 5 Pearson’s Correlation Coefficient (r) The several different computational formula for computing Pearson’s r each result in the same answer (except for rounding errors). The correlation coefficient should be rounded to three decimal places. Rounding in the middle of a calculation often creates substantial errors, therefore, round-off only at the last calculation. Formula Pearson’s formula to calculate r follows: ∑ XY − ( ∑ X )( ∑ Y ) N N N r= Eq. 1 ∑X − (∑ ) * ∑ − (∑ ) 2 2 X Y Y 2 2 N N N N where: N represents the number of pairs of data ∑ denotes the summation of the items indicated ∑X denotes the sum of all X scores ∑X2 indicates that each X score should be squared and then those squares summed (∑X)2 indicates that the X scores should be summed and the total squared. [avoid confusing ∑X2 (the sum of the X squared scores) and (∑X)2 (the square of the sum of the X scores] ∑Y denotes the sum of all y-scores ∑Y2 indicates that each Y score should be squared and then those squares summed (∑Y)2 indicates that the Y scores should be summed and the total squared
6. 6. How to Compute and Interpret Pearson’s R Page 6 ∑XY indicates that each X score should be first multiplied by its corresponding Y score and the product (XY) summed The numerator in equation 1 equals the mean of XY ( X ) minus the mean of X ( Y X ) times the mean of Y ( Y ); the denominators are the standard deviation for X (SDX) and the standard deviation for Y (SDY). [See; How to compute and interpret measures of variability: the range, variance and standard deviation] Thus, Pearson’s formula can be written as: XY−XY r= S X∗S Y D D Example Compute the correlation coefficient (r) for the height-weight data shown in Figure 1. Pertinent calculations are given in Table 1. Table 1. Height and weight of a sample of college age males. X2 Y2 Height, cm (X) Weight, kg (Y) X*Y 174 61 10614 30276 3721 175 65 11375 30625 4225 176 67 11792 30976 4489 177 68 12036 31329 4624 178 72 12816 31684 5184 182 74 13468 33124 5476 183 80 14640 33489 6400 186 87 16182 34596 7569 189 92 17388 35721 8464 193 95 18335 37249 9025 2 ∑Y2=59177 ∑X=1813 ∑Y=761 ∑XY=138646 ∑X =329069 ∑X2/N=32906.9 ∑Y2/N=5917.7 ∑X/N=181.3 ∑Y/N=76.1 ∑XY/N=13864.6
7. 7. How to Compute and Interpret Pearson’s R Page 7 XY − X Y r= SDX * SDY 13864.6 − (181.3)(76.1) r= 32906.9 − (181.3)2 * (5917.7) − (76.1) 2 67.67 r= 68.605 r = 0.986 Interpreting Pearson’s Correlation Coefficient The usefulness of the correlation depends on its size and significance. If r reliably differs from 0.00, the r-value will be statistically significant (i.e., does not result from a chance occurrence.) implying that if the same variables were measured on another set of similar subjects, a similar r-value would result. If r achieves significance we conclude that the relationship between the two variables was not due to chance. How To Evaluate A Correlation The values of r always fall between -1 and +1 and the value does not change if all values of either variable are converted to a different scale. For example, if the weights of the students in Figure 1 were given in pounds instead of kilograms, the value of r would not change (nor would the shape of the scatter plot.) The size of any correlation generally evaluates as follows: Correlation Value Interpretation ≤0.50 Very low 0.51 to 0.79 Low 0.80 to 0.89 Moderate ≥0.90 High (Good)
8. 8. How to Compute and Interpret Pearson’s R Page 8 A high (or low) negative correlation has the same interpretation as a high (or low) positive correlation. A negative correlation indicates that high scores in one variable are associated with low scores in the other variable (see Figure 2, graphs d, e, f). Correlation Does Not Imply Causation “CORRELATION DOES NOT IMPLY CAUSATION!” Just because one variable relates to another variable does not mean that changes in one causes changes in the other. Other variables may be acting on one or both of the related variables and affect them in the same direction. Cause-and-effect may be present, but correlation does not prove cause. For example, the length of a person’s pants and the length of their legs are positively correlated - people with longer legs have longer pants; but increasing one’s pant length will not lengthen one’s legs! Property of Linearity The conclusion of no significant linear correlation does not mean that X and Y are not related in any way. The data depicted in Figure 2h result in r = 0, indicating no linear correlation between the two variables. However, close examination show a definite pattern in the data reflecting a very strong “nonlinear” relationship. Pearson’s correlation apply only to linear data. Coefficient of Determination (r2) The relationship between two variables can be represented by the overlap of two circles representing each variable (Figure 3). If the circles do not overlap, no relationship exists; if they overlap completely, the correlation equals r = 1.0. If the circles overlap somewhat, as in Figure 3 (next page), the area of overlap represents the amount of variance in the dependent (Y-variable) than can be explained by the independent (X- variable). The area of overlap, called the percent common variance, calculates as: r2 x 100 For example, if two variables are correlated r = 0.71 they have 50% common variance (0.712 x 100 = 50%) indicating that 50% of the variability in the Y-variable can
9. 9. How to Compute and Interpret Pearson’s R Page 9 be explained by variance in the X-variable. The remaining 50% of the variance in Y remains unexplained. This unexplained variance indicates the error when predicting Y from X. For example, strength and speed are related about r = 0.80, (r2 = 64% common variance) indicating 64% of both strength and speed come from common factors and the remaining 36% remains unexplained by the correlation. Figure 3. Example of the coefficient of determination (percent common variance r2x100). x-variable; predictor y-variable; predicted variable variable Area of overlap; r2x100 (percent common variance) Statistical Significance of a Correlation When pairs of number used to compute r are small, a spuriously high value can occur by chance. For example, suppose numbers 1, 2, and 3 each written on a separate piece of paper are placed in a hat. The numbers are then blindly drawn one at a time on two different occasions. The possibility exists that the numbers could be drawn in the same order twice. This would produce r = 1.0 (a perfect correlation.) But this value would be a chance occurrence since no known factor(s) can cause such a relationship. In contrast, the odds of 100 numbers being randomly selected in the same order twice are very low. Thus, if the r value with Npairs=100 is high, we conclude that chance cannot be a factor explaining the correlation. Thus, the number of pairs of values (N) determines
10. 10. How to Compute and Interpret Pearson’s R Page 10 the odds that a relationship could happen by chance. If N is small, r must be large to be significant (not caused by chance). When N is large, a small r-value may be significant. Table 2 is used to determine the significance of r. The left column df represents the degrees of freedom: df = Npairs - 2 (the number of pairs of XY scores minus 2). df represents the number of values that are free to vary when the sum of the variable is set; df compensates for small values of N by requiring higher absolute values of r before being considered significant. Step 1. Find the degrees of freedom in the left column: df = Npairs of data -2 Step 2. Read across the df row and compare the obtained r value with the value listed in one of the columns. The heading at the top of each column indicates the odds of a chance occurrence, (the probability of error when declaring r to be significant.) p=.10 is the 10% probability; p=.05 is the 5% probability level, and p=.01 is the 1% probability level. Reading from the table for df = 10, a correlation as high as r = 0.497 occurs 10 times in 100 by chance alone (p=.10); r = 0.576 occurs 5 times in 100 by chance (p=.05); and r=0.708 occurs 1 time in 100 by chance (p=.01). If r locates between values in any two columns, use the left of the two columns (greater odds for chance). If r does not equal or exceed the value in the p=.10 column, it is said to be nonsignificant (NS). Negative r’s are read using the absolute value of r. Table 2. Values of the Correlation Coefficient (r) df p = .10 p = .05 p = .01 1 9877 .9969 .9999 2 .900 .950 .990 3 .805 .878 .959 4 .729 .811 .917 5 .669 .754 .875 6 .621 .707 .834 7 .582 .666 .798 8 .549 .632 .765 9 .521 .602 .735 10 .497 .576 .708 11 .476 .553 .684 12 .457 .532 .661 13 .441 .514 .641
11. 11. How to Compute and Interpret Pearson’s R Page 11 14 .426 .497 .623 15 .412 .482 .606 16 .400 .468 .590 17 .389 .456 .575 18 .378 .444 .561 19 .369 .433 .549 20 .360 .423 .537 25 .323 .381 .487 30 .296 .349 .449 35 .275 .325 .418 40 .257 .304 393 45 .243 .288 .372 50 .231 .273 .354 60 .211 .250 .325 70 .195 .232 .302 80 .183 .217 .283 90 .173 .205 .267 100 .164 .195 .254 From Biometrika Tables For Statisticiarts (Vol. 1) (3rd ea.) by E.S. Pearson and H.O. Hartley (Eds.), 1966, London: Biometrika Trustees. Copyright 1966 by Biometrika. http://www.umich.edu/~exphysio/MVS250/PearsonCorr.doc
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Chapter 7 Review
AP Statistics
1. Using the data, P(X < 3) is (a) 0.3. xi 1 2 3 (b) 0.4. pi 0.2 0.4 0.3 (c) 0.6. (d) 0.9. (e) The answer cannot be computed from the information given.
Name: 4 0.1
2. Using the above data, the mean µ of X is (a) 2.0. (b) 2.3. (c) 2.5. (d) 3.0. (e) The answer cannot be computed from the information given. 3. Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume that X is Normal with mean \$360 and standard deviation \$50. What is the value of P(X > \$400)? (a) 0.2119 (b) 0.2881 (c) 0.7881 (d) 0.8450 (e) The answer cannot be computed from the information given. (f) Sophia should get an A because this computer is on her desk 4. A business evaluates a proposed venture as follows. It stands to make a profit of \$10,000 with probability 3/20, to make a profit of \$5000 with probability 9/20, to break even with probability 1/4, and to lose \$5000 with probability 3/20. The expected profit in dollars is (a) 1500. (b) 0. (c) 3000. (d) 3250. (e) –1500. 5. A randomly chosen subject arrives for a study of exercise and fitness. Consider these statements. I After 10 minutes on an exercise bicycle, you ask the subject to rate his or her effort on the Rate of Perceived Exertion (RPE) scale. RPE ranges in whole-number steps from 6 (no exertion at all) to 20 (maximum exertion). II You measure VO2, the maximum volume of oxygen consumed per minute during exercise. VO2 is generally between 2.5 liters per minute and 6 liters per minute. III You measure the maximum heart rate (beats per minute). The statements that describe a discrete random variable are (a) None of the statements describes a discrete random variable. (b) I. (c) II. (d) I, III. (e) I, II, III. Chapter 7
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6. Cans of soft drinks cost \$ 0.30 in a certain vending machine. What is the expected value and variance of daily revenue (Y) from the machine, if X, the number of cans sold per day has E(X) = 125, and Var(X) = 50 ? (a) E(Y) = 37.5, Var(Y) = 50 (b) E(Y) = 37.5, Var(Y) = 4.5 (c) E(Y) = 37.5, Var(Y) = 15 (d) E(Y) = 37.5, Var(Y) = 30 (e) E(Y) = 125, Var(Y) = 4.5 Questions 7 and 8 use the following: Suppose X is a random variable with mean µX and standard deviation X. Suppose Y is a random variable with mean µY and standard deviation Y. 7. The mean of X + Y is (a) µX + µY . (b) (µX / X) + (µY/Y). (c) µX + µY, but only if X and Y are independent. (d) (µX/X) + (µY/Y), but only if X and Y are independent. (e) None of these. 8. The variance of X + Y is (a) X + Y. (b) (X)2 + (Y)2. (c) X + Y, but only if X and Y are independent. (d) (X)2 + (Y)2, but only if X and Y are independent. (e) None of these. 9. A random variable is (a) a hypothetical list of possible outcomes of a random phenomenon. (b) any phenomenon in which outcomes are equally likely. (c) any number that changes in a predictable way in the long run. (d) a variable whose value is a numerical outcome of a random phenomenon. (e) None of the above. Questions 10, 11, and 12 refer to the following information. X is a random variable that takes the value A between X = 0 and X = 2 and is zero everywhere else. X defines a uniform density curve. 10. The probability that X is between 0.5 and 1.5 is (a) 1/3. (b) 1/2. (c) 3/4. (d) 1. (e) 1/4 11. The probability that X is at least 1.5 is (a) 0. (b) 1/4. (c) 1/3.
(d) 1/2.
(e) 1.
12. The probability that X = 1.5 is (a) 0. (b) 1/4.
(d) 1/2.
(e) 1.
(c) 1/3.
13. Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume X is Normal with a mean of \$360 and standard deviation \$50. P(X > \$400) is (a) 0.2881. (b) 0.8450. (c) 0.7881. (d) 0.7119 (e) 0.2119 Chapter 7
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14. Let the random variable X represent the amount of money Dan makes doing lawn care in a randomly selected week in the summer. Assume that X is Normal with mean \$240 and standard deviation \$60. The probability is approximately 0.6 that, in a randomly selected week, Dan will make less than (a) \$144 (b) \$216 (c) \$255 (d) \$360 (e) The answer cannot be determined from the information given. 15. An insurance company has estimated the following cost probabilities for the next year on a particular model of car: Cost \$0 \$500 \$1000 \$2000 Probability 0.60 0.05 0.13 ? The expected cost to the insurance company is (approximately) (a) \$155. (b) \$595. (c) \$875. (d) \$645. (e) \$495. 16. Suppose we have a loaded die that gives the outcomes 1 to 6 according to the probability distribution X 1 2 3 4 5 6_ P(X) 0.1 0.2 0.3 0.2 0.1 0.1 Note that for this die all outcomes are not equally likely, as would be if this die were fair. If this die is rolled 6000 times, then x , the sample mean of the number of spots on the 6000 rolls, should be about (a) 3. (b) 3.30. (c) 3.50. (d) 4.50. (e) 3.25. Questions 17 and 18 relate to the following information. The weight of medium-size oranges selected at random from a bin at the local supermarket is a random variable with mean µ = 10 ounces and standard deviation = 1 ounce. 17. Suppose we pick two oranges at random from the bin. The difference in the weights of the two oranges selected (the weight of the first orange minus the weight of the second orange) is a random variable with a mean (in ounces) of (a) 10. (b) 1. (c) 1.41. (d) 0. (e) 5. 18. Suppose we pick two oranges at random from the bin. The difference in the weights of the two oranges selected (the weight of the first orange minus the weight of the second orange) is a random variable with a standard deviation (in ounces) of (a) 0. (b) 1. (c) 2. (d) 2.2. (e) 1.41. Chapter 7
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Part 2: Free Response Answer completely, but be concise. Write sequentially and show all steps. 19. Picard Partners is planning a major investment. The amount of profit X is uncertain but a probabilistic estimate gives the following distribution (in millions of dollars): Profit Probability
1 0.1
1.5 0.2
2 0.4
4 0.2
10 0.1
(a) Find the mean profit µX and the standard deviation of the profit.
(b) Picard Partners owes its source of capital a fee of \$200,000 plus 10% of the profits X. So the firm actually retains Y = 0.9X – 0.2 from the investment. Find the mean and standard deviation of Y.
20. The Census Bureau reports that 27% of California residents are foreign-born. Suppose that you choose three Californians independent of each other and at random. There are eight possible arrangements of foreign (F) and domestic (D) birth. For example, FFD means that the first two are foreign-born, and the third is not. (a) Write down all eight arrangements and find the probability of each.
(b) Let the random variable X be the number of foreign-born people in each group of three Californians. What are the possible values of X? Use the probabilities you found in (a) to construct a probability distribution table for X.
(c) What is the expected number of foreign-born residents in a randomly selected group of three?
(d) What is the standard deviation of X?
Chapter 7
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21. The probability that 0, 1, 2, 3, or 4 people will seek treatment for the flu during any given hour at an emergency room is shown in the following distribution. __X P(X)
0 0.12
1 0.25
2 0.33
3 0.24
4___ 0.06
(a) What does the random variable count or measure?
(b) Calculate the mean of X, and interpret this value in context.
(c) What are the variance and standard deviation of X?
22. A study of the weights of the brains of Swedish men found that the weight X was a random variable with mean 1400 grams and standard deviation 20 grams. Find numbers a and b such that Y = a + bX has mean 0 and standard deviation 1.
23. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Let X and Y be the lengths of two randomly chosen pregnancies. What is P(X + Y) > 600?
24. If a player rolls two dice and gets a sum of 2 or 12, he wins \$20. If the person gets a 7, he wins \$5. The cost to play the game is \$3. Find the expectation of the game.
Chapter 7
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25. Here is the probability distribution function for a continuous random variable. (a) Show that this defines a legitimate pdf.
Determine the following probabilities: (b) P(0 ≤ X ≤ 3) = (c) P(2 ≤ X ≤ 3) = (d) P(X = 2) = (e) P(X < 2) = (f) P(1 < X < 3) =
26. Picard Partners is planning a major investment. The amount of profit X is uncertain but a probabilistic estimate gives the following distribution (in millions of dollars): Profit Probability
1 0.1
1.5 0.2
2 0.4
4 0.2
10 0.1
(a) Find the mean profit µX and the standard deviation of the profit.
(b) Picard Partners owes its source of capital a fee of \$200,000 plus 10% of the profits X. So the firm actually retains Y = 0.9X – 0.2 from the investment. Find the mean and standard deviation of Y.
Chapter 7
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November 2019 66
November 2019 67
December 2019 52
November 2021 0
October 2019 57
December 2019 40
November 2019 66
December 2019 40
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# I am a 3-digit number. My first digit is thrice my last digit. The sum of my last two digit is one less than my last digit
42,127 results, page 24
1. ## Math
I am thinking of a three-digit number. If you subtract 5 from it, the result is divisible by 5. If you subtract 6 from it, the result is divisible by 6 and if you subtract 7 from it, the result is divisible by 7. What is the smallest number that will
asked by Anthony on September 26, 2010
2. ## Math
Write 8 fractions equal to one-half (not including one-half) using only the digits 0, 1, 2 and 5. Each digit may be used more than once, but you may not use decimal points or arithmatic operatons. Numerators cannot be larger than 100. You may not use
asked by William Bent on November 9, 2009
3. ## math
in these addends, each letter represents a single digit. find the numbers. write the completed problem below. CENT CENT + SCENT ----------- 35128 I NEED THIS TONIGHT OCT. 26--ASAP!!!!!!
asked by My dancer chick on October 26, 2008
4. ## Math
What is the probability of getting a license plate that has a repeated letter or digit if you live in a state in which license plates have one numeral followed by three letters followed by three numerals? (Round your answer to one decimal place.)
asked by Ronald on June 12, 2013
5. ## Algebra
What is the probability of getting a license plate that has a repeated letter or digit if you live in a state in which license plates have one numeral followed by three letters followed by three numerals? (Round your answer to one decimal place.)
asked by Mary Ann on May 26, 2014
6. ## algebra
Beginning with cars made in 1981, all vehicles must contain a 17-digit Vehicle Identification Number (VIN). The first 3 positions must be letters or numbers, position 4 and 5 must be letters, positions 6 and 7 must be numbers, position 8 can be a letter or
asked by anon on May 2, 2013
7. ## Math
1. Daniel wants to save at least 20 dollars by putting pennies in a jar daily. On first day, he puts one penny into the jar. The second day he puts 2 pennies into the same jar. On the nth day he pus n pennies into the same jar. Which day is the first day
asked by Bethany on February 22, 2015
8. ## math
how many 7 digit numbers can be formed with the numbers 1, 2, 4, 7. (they are in brackets). the question wants two parts. (a) if the numbers can be repeared and (b) if repetation is not allowed
asked by ken on October 18, 2008
9. ## Math
The first digit of two different numbers is in the hundreds millions place. Both numbers contain the same digits. Can you determine the greater of the two numbers and Explain.
asked by Deborah on August 28, 2014
10. ## Math
~~~~WARNING HARD MATH~~~~ An integer from 100 to 999, inclusive, is to be chosen at random. What is the probability that the number chosen will have 0 as at least 1 digit? Multiple Choice: Underline or circle the letter that best solves the question. a)
asked by KawaiiChan on January 24, 2017
11. ## Math grd12
How many ten-digit telephone numbers are possible if the first three digits must be different? The answer is 72,000,000,000.... I need the method to get that:)
asked by Celina on November 21, 2010
12. ## Math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3? would this be 1/7 or 1/6? i'm unsure.
asked by bubbie on May 22, 2013
13. ## chemistry
this is over significant digits. -Indicate how many significant figures are in each of the following measurements: e) 4.0 X 10-5 kg i got 1 significant digit, but i'm not sure it if it's right... (& i can't ask my teacher, the situation is complicated)
asked by Ana on September 21, 2010
14. ## MATH *IMPORTANT
HOW MANY SETS (A,B,C) OF THREE DIFFERENT,POSITIVE SINGLE DIGIT INTEGERS HAVE THE PROPERTY THAT A SQUARED + B SQUARED + C SQUARED IS A PERFECT SQUARE? please help me with this one a soon as possible its very important and i don't even know where to start!!
asked by shauna on September 10, 2012
15. ## Math
please if anyone could help me with this question: 1. A class paid \$20 for a cake and \$4 per child for one slice of cheese pizza. they paid \$140. how many children are in the class? 2. Rajie want to divide 24 peanuts, 64 raisins, and 56 apricots evenly
asked by Erika on February 23, 2017
16. ## math
please i need help with this...were studying probability...possible outcomes...im having trouble coming up with the right answer... The product code used by Pitz Toyz consists of a letter, followed by a digit from 1 to 9, followed by another letter, such
asked by logan on October 17, 2006
17. ## maths
invaders put a six digit number into the display. you must change each of the six digits to zero in as few turns as possible. on each turn you can use only one number key, the zero key as often as you like and the + key. Example: start number: key presses:
asked by anonymous.gfgf on August 8, 2014
18. ## MATH PROBLEM
-One revolution of the Sun by the Earth requires 365 days, 5 hours, 48 minutes and 49.7 seconds. the 5 hours, 48 minutes and 49.7 second is approximated to 1/4 of a day. Every 4th year, one full day is added( leap year). How much of an error in minutes and
asked by Kim on January 16, 2013
19. ## Math
oel scores in 80% of his games. His team is starting the playoffs, and if they make it to the state championship, they will have played 5 games. Joel wants to determine the probability that he scores in 4 of the 5 games. Which simulation design has an
asked by Allison Carson on February 9, 2016
20. ## math: probability
A locker combination has three nonzero digits, and digits cannot be repeated. If the first two digits are even, what is the probability that the third digit is even?
asked by Brooke on March 27, 2008
21. ## math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3? is it 1/7 or 1/6?
asked by Maya on January 16, 2017
22. ## Math
I am thinking of two two-digit numbers.they differ by 54.Both numbers are made up of the same digits, are reversed.What are the two numbers?
asked by Tommy on December 7, 2015
23. ## Math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3?
asked by Jess on May 3, 2011
24. ## math
A locker combination has three nonzero digits, and digits cannot be repeated. The first two digits are 1 and 2. What is the probability that the third digit is 3?
asked by Anonymous on February 7, 2013
25. ## maths
i am a seven digit my ten place is 3,my lakhs place is 2,the hundred place is twice the lakhs place,the thousand place is thrice the ten place ,the ten lakh place is 3 less than the thousand place,the one place is 2 less than the thousand place,the ten
asked by thnki on March 8, 2015
26. ## math
Can you check my answers please? Thank you Ben is greeting customers at a music store. Of the first 20 people he sees enter the store, 9 are wearing jackets and 11 are not. What is the experimental probability that the next person to enter the store will
asked by Anonymous on March 22, 2016
27. ## Math Probability
Ben is greeting customers at a music store. Of the first 20 people he sees enter the store, 9 are wearing jackets and 11 are not. What is the experimental probability that the next person to enter the store will be wearing a jacket? Enter your answer as a
28. ## Algebra 2
Question is about Step Functions:(between" " are all floor functions. The formula W=d+2m+"(3(m=1)/5)"+y+"y/4" -"y/100" + "y/400" + 2 gives the day of the month of the given date; m = the number of the month in the year with January and February regarded as
asked by Casey on October 24, 2011
29. ## Maths Probability
Suppose that you move to a new house and you are 10% sure that your new house's phone number is 561290. To verify this, you use the house's phone to dial 561290, obtain a busy signal, and conclude that this is indeed your phone number. (Suppose that you
asked by Anonymous on May 14, 2014
30. ## Probability
Suppose that you move to a new house and you are 10% sure that your new house's phone number is 561290. To verify this, you use the house's phone to dial 561290, obtain a busy signal, and conclude that this is indeed your phone number. (Suppose that you
asked by qwerty on May 13, 2014
31. ## Probability
Suppose that you move to a new house and you are 10% sure that your new house's phone number is 561290. To verify this, you use the house's phone to dial 561290, obtain a busy signal, and conclude that this is indeed your phone number. (Suppose that you
asked by qwerty on May 26, 2015
32. ## Math
Joel scores in 80% of his games. His team is starting the playoffs, and if they make it to the state championship, they will have played 5 games. Joel wants to determine the probability that he scores in 4 of the 5 games. Which simulation design has an
asked by Asia on February 12, 2016
33. ## Math
Joel scores in 80% of his games. His team is starting the playoffs, and if they make it to the state championship, they will have played 5 games. Joel wants to determine the probability that he scores in 4 of the 5 games. Which simulation design has an
asked by Anonymous on February 12, 2016
34. ## MATH
WRITE THE GREATEST POSSIBLE ERROR FOR EACH MEASUREMENT. 43mL I don't understand the question It depends on the graduations. I assume on this beaker you have 40ml, 50 ml marks. So the 3 is an estimated digit. A safe assumption is +-.5ml error.
asked by SANDRA on April 29, 2007
35. ## Math
What do you call numbers that cannot be arranged into 2-row arrays? multipication After some research I finally understand the question. Those would be the odd numbers or the numbers without a factor of 2 in them. After much stress and contemplating, I
asked by Trudy on September 11, 2006
36. ## math
2)How many arrangements of the integers 1,2,3,...,n are there such that each integer differs by one (except for the first integer) from some integer to the left of it? Let's construct the arrangement. First consider 1, it can only have two placed next to
asked by Laura on October 15, 2006
37. ## math
) Among the integers from 1 to 10, 000, 000, 000 which are there more of: Those in which the digit 1 occurs or those in which it does not occur? please prove
asked by teach on February 9, 2015
38. ## statistics
17. All three‐digit numbers that can be formed using the digits 1, 2, 3, 4, and 5 without repeating the digits are (a) 50. (b) 60. (c) 15. (d) 71. 18. One of the counting rules used to determine the total number of possible outcomes for experiments in
asked by meri on June 30, 2015
39. ## Math
1. The measure of an interior angle of a regular polygon is 20 more than thrice the measure of its adjacent exterior angle. Find the number of sides of the polygon and its total number of diagonals. 2. Find the sum and difference between the sum of the
asked by Brianna on October 14, 2014
40. ## Math
Call shades a role in the multiplication table the product in the row are all even the ones digit in the product repeats zero, four, eight, two, six what role does Carl shade?
asked by Abby on October 15, 2018
41. ## Math
I need some help please Determine the least four digit number that is divisible by 2,3, and 5. Explain your reasoning. Explain how you know that 123 123 123 900 is divisible by 2,3,5,9, and 10. How do you know that it is also divisible by 6.
asked by Edward on September 16, 2008
42. ## College
Three(3) digit numbers can be formed from the digits 0,1,3,5,7 and 8. How many numbers can be formed without repeats. (A) if the 3 digits numbers must be divisible by 5. (b) if the 3 digits number must be a multiple of 2
asked by Ibrahim issahaku kpaniyili on July 17, 2015
43. ## Maths - Pigeon-hole principle
Using the Pigeon-hole principle... 91 five-digit numbers are written on a blackboard. Prove that one can find three numbers on the blackboard such that the sums of their digits are equal.
asked by Lexi on September 30, 2008
The value of technology digit 9 in 794,256 is__________ times the value of 9 in 12,276,459. (Think it is : 90,000×12,276,459) Is this right I tried multiplying I get crazy numbers though
asked by Angel on January 22, 2016
45. ## math
The digit 3 appears fourteen times in tye page numbers of Chapter 1 of a book.How manypages are there in Chapter 1 of the book?
asked by Gallia on November 15, 2015
46. ## MATH
a fraction with a denominater in the thousands will always have a decimal form with more decimal places than a fraction with a single-digit denominator. How would I determine if this statement is true or fals?
asked by DEBBIE on April 14, 2010
47. ## Probability
A computer password is required to be 9 characters long. How many passwords are possible if the password requires 2 letter(s) followed by 7 digits (numbers 0-9), where no repetition of any letter or digit is allowed?
asked by Samuel on March 25, 2016
48. ## math
A computer password is required to be 5 characters long. How many passwords are possible if the password requires 1 letter(s) and 4 digits (numbers 0-9), where no repetition of any letter or digit is allowed?
asked by hana on November 2, 2015
49. ## Algebra
In the following problem, the digits 0-9 have been replaced with letters. Can you reconstruct the original multiplication problem by identifying what digit each letter represents? TEA (times)x_TEA_ SEA TNNA (add)t_TEA_ (ans.) CHINA
asked by Mitch on November 28, 2012
50. ## math
91 five digit numbers are written on a blackboard. How do you prove that you can find 3 numbers on the blackboard such that the sums of their digits are equal?
asked by wendi on July 15, 2008
51. ## maths
91 five-digit numbers are written on a blackboard. Prove that one can find three numbers on the blackboard such that the sums of their digits are equal.
asked by Lexi on September 30, 2008
round to the value of the underlined digit. 1. 6,820 8 is underlined 2. 714,396,283 6 is underlined 3. 85,026,288 2 is underlined after 0.
asked by mythreyee on September 10, 2013
53. ## Math
asked by Jay on November 12, 2015
54. ## algebra
In Utah, a license plate consists of 3 digits followed by 3 letters. The letters I, O, and Q are not used, and each digit or letter may be used more than once. How many different license plates are possible?
asked by bre on March 12, 2009
55. ## pre calculus
A state makes auto license plates that have two letters (excluding I, 0, and Q) followed by four digits of which the first digit is not zero. How many different license plates are possible?
asked by ron on March 30, 2010
56. ## math
Suppose you have cards with the digits from 0 to 9, one card for each digit. Complete the following calculation using all of the cards. Explain how you went about getting your digits/answer.
asked by Alfredo on November 11, 2014
57. ## medical billing
the file clerk discovered a medical record placed incorrectly in the permanent files by checking for _____ in the last set of two-digit numbers in the medical records numbers.
asked by misty on March 29, 2010
58. ## follow up(for Count Iblis)
i have a follow up question for a previous post How do you know 2B is (20 "+" B) and not (2 "x" B) _______________________________________ to me this question does not make sense... it comes from a review packet for the SAT
asked by manny on August 31, 2007
59. ## Math
Pages in a book are numbered in typical fashion, starting with !. The folio for page 10 will contain the tenth and eleventh digits necessary to paginate the book. On what page will the 2009th digit occur? Can you help
asked by Mark on September 11, 2012
60. ## Math
Pages in a book are numbered in typical fashion, starting with !. The folio for page 10 will contain the tenth and eleventh digits necessary to paginate the book. On what page will the 2009th digit occur?
asked by Mark on September 11, 2012
61. ## Math
Pages in a book are numbered in typical fashion, starting with 1. The folio for page 10 will contain the tenth and eleventh digits necessary to paginate the book. On what page will the 2009th digit occur?
asked by Mark on September 6, 2012
62. ## Math
Write a 7 digit numeral with 6 in the ones place, 3 in the thousandths place, 1 in the thousands place, 2 in the tenths place, and 0's in a ll the other places?
asked by Allison on September 5, 2012
63. ## Math
Account numbers for Northern Oil Company consist of seven digits. If the first digit cannot be a 0 or 1, how many account numbers are possible?
asked by Josh on January 29, 2013
64. ## math
i am a digit greater than 50000 but less than 90000,my digits add up to 17, my first 2 digits are multiple of 10 and the last 3 are multiple of 100
asked by gigi on August 30, 2015
A secret code must consist of three capital letters followed by a digit. Letters cannot be repeated. How many secret codes are possible? A. 140,400 B. 156,000 C. 175,760 D. 358,800
asked by sophie on May 19, 2013
66. ## Math
Sue's friens Kat has these numbers 89378, 42327,49734 and teacher has shown a 3, how many points will Kat have? What digit does she hope teacher will pull next week so she has the highest score?
asked by Mike on August 22, 2011
67. ## math
Among the integers from 1 to 10, 000, 000, 000 which are there more of: Those in which the digit 1 occurs or those in which it does not occur?
asked by Ralph on February 8, 2015
68. ## math word problem 2
the jerseys of the members of one team are numberes 01 to 42. how many jerseys contain the digit 0? a 9 b 10 c 12 d 13 i got: 13 jerseys is that right?
asked by marko on December 15, 2011
69. ## Math
A block is a consecutive sequence of the same digit that is not part of a longer such sequence. a.) Determine the number of sequences of 0's and 1's of length 10 in which every block of 1's has an even length. b.) Determine the number of sequences of 0's
asked by Kris on February 17, 2007
70. ## Math
Pages in a book are numbered in typical fashion, starting with 1. The folio for page 10 will contain the tenth and eleventh digits necessary to paginate the book. On what page will the 2009th digit occur? Please Help!!! I have no idea where to begin
asked by Beth on September 4, 2012
71. ## Math
Pages in a book are numbered in typical fashion, starting with 1. The folio for page 10 will contain the tenth and eleventh digits necessary to paginate the book. On what page will the 2009th digit occur? Please Help!!! I have no idea where to begin
asked by Beth on September 4, 2012
72. ## Math Help!
I apologize for all these questions but I really need to get them done fast so I appreciate if you tried to answer some of them! 1. Bill is saving money in an empty jar. The first day he puts in a certain amount of money. Every day after the first he puts
asked by Beth on February 22, 2015
73. ## Math for Elementary Teachers II
How many 4-digit license plates can be made by using the digits 0, 2, 4, 6, and 8 if repetitions are allowed? if repetitions are not allowed?
asked by Michael on March 29, 2010
74. ## math
asked by grace on January 5, 2015
75. ## MATH157
How many 4-digit license plates can be made by using the digits 0, 2, 4, 6, and 8 if repetitions are allowed? if repetitions are not allowed?
asked by Tyga on April 1, 2010
76. ## math
How many two digit numbers are there in base 5. I really don't understand how to figure this out. I know it's been asked already, but I still don't understand it.
asked by jason on March 6, 2011
77. ## math 12
How many different binary numbers can be present by a string of binary code with 5 digits if the first digit is 1 and the last two digits are 0?
asked by Juliet on March 20, 2010
78. ## math
i really need to know what.... 4 to the power of 77 is... 477 = 2.28360 x 1046 according to my calculator. i don't know how to do factors with two digit numbers. Please click Post a New Question, put Math in the subject line, and post your question. It
asked by Kimmy on February 20, 2007
79. ## chemistry(can anyone help me with this)
Which of the following is NOT true? A. Except for absorbance, most measurements contain significant figures and units. B. All numbers should be recorded from a digital readout. C. The uncertain digit is estimated between the last two markings. D. If an
asked by tea on January 19, 2010
80. ## CHEM 1LB
Which of the following is NOT true? A. The uncertain digit is estimated between the last two markings. B. Except for absorbance, most measurements contain significant figures and units. C. All numbers should be recorded from a digital readout. D. If an
asked by Shawn Tabrizi on January 21, 2011
81. ## Math
Audrey is thinking of a number. If the number is divided by 3 and added to 12, the result is 29. Of which number is Audrey thinking? A.63 B.54 C.51 D.44 C? Look at the number tiles below 279 Which group of numbers shows all 3-digit numbers that can be made
asked by Colton on February 9, 2012
82. ## Math
My 7th grade daughter has the following math problem and gets zero help from her "language arts" teacher who gave out this problem: WANT X BE ________ CAREER Each letter in this question stands for a 1-digit #. No 2 letters may stand for the same #. Find
asked by KP on March 8, 2011
83. ## math
how often does a leap year occur? what are al the prime numbers less then 12. what is the eighth multiple of 32? is the last digit of the multiples if 4 odd or even? what are the six factors of 12? what are the factors of 16. two factors of 20 are 1 and
asked by jojo on November 9, 2010
84. ## math
first multiply first 2 digits, wrire answer side by then multiply1st and 3rd digits write answer next to earlier answer. Then add both answers and minus middle digit of the question
asked by Anonymous on March 5, 2015
85. ## math
in the addition problems below each leter represents the same digit in both problems. Replace each letter with a different digi, 1-9, so that both addition problems are true.(there are two possible answers.) A B C A D G +D E F +B E H ------ ------ G H I C
asked by kenzie on August 23, 2007
86. ## MATH
A bike club wants to make license plates for bikes. Each license plate has a letter followed by a digit. How many different license plates can be made?
asked by JORDAN on April 18, 2011
87. ## algebra
6RS +RR4 PQPQ P,Q,R and S represents four different digits. what digit does S represents? (A)1 (B)2 (C)5 (D)6 (E)8 I think the answear is 6 but now sure. Am i correct? How do you figure this problem? The answer is (E)8. The rest of the letters are as
asked by Bryan on November 10, 2006
88. ## Math
Which digit (0-9) occurs most often on a calendar for this month. Which occurs least often? Is the answer the same for every month of this year?
asked by Ruth on December 10, 2013
89. ## Medical Records
I want to make sure I am correct.Can you check this for me? We need to have this in Middle Digit Order. I know I start with my primary. 534-01-38 535-01-38 536-01-38 534-10-36 222-10-37 534-10-38 600-11-37 535-11-38 Assignment due tonight
asked by Lauri on June 15, 2010
90. ## Calculus and quadratic equations
Peter the punter decided to place \$10,000 into a high growth share portfolio. After 3 weeks his investment was looking rather sad, with the value of his portfolio falling \$2,025. Unperturbed, Peter stuck with his original plan to hold the stock for at
asked by Roney on May 20, 2015
91. ## maths
i am trying to solve 4/x-2>5. i use calculator and mistakenly type of digit if other than 5. when i type in the problem ,so i get the wrong solution my answer was 2
asked by sunny on December 16, 2011
92. ## Algebra I
Write each percent change as a ratio comparing the result to the original quantity. Then write it as a constant multiplier. Book's example: 3% increase, 103/100, 1+0.03 I don't understand double digit numbers. The one that I have to solve is an 11%
asked by Katy on February 19, 2009
93. ## Math
Each letter in the subtraction problem below represents a different digit from 0 through 9.The digits 3 and 5 don't apear. replace each letter so the subtraction problem is true. GRAPE - PLUM =APPLE
asked by autumn on January 23, 2011
94. ## Pre calculus
bit strings of a fixed length are needed to form a code. what is the minimum length of the string is one string is needed to represent each letter of the alphabet and each decimal digit?
asked by Lisa on April 14, 2011
95. ## probstats
A 5-digit PIN code cannot begin with zero and the remaining digits have no restrictions. If repetition of digits is allowed, then what is the probability that the PIN code begins with a 7 and ends with an 8?
asked by Anonymous on January 24, 2016
96. ## Math
Build a 9-digit numeral write 2 in the hundreds place 5 in the ten thousands place 7 in the millions place 6 in the hundred millions place and 3 in all other places
asked by Drake on October 19, 2009
97. ## maths
Set S contains the elements = { 1, 2, 3, 4, 5, 6, 7, 8}. From this set, John takes sets of 6-digits each and sums all the possible 6-digit numbers that can be formed with those sets. What is the HCF of all these sums?
asked by Anonymous on January 15, 2011
98. ## Quadratic Functions and Calculus
Peter the punter decided to place \$10,000 into a high growth share portfolio. After 3 weeks his investment was looking rather sad, with the value of his portfolio falling \$2,025. Unperturbed, Peter stuck with his original plan to hold the stock for at
asked by Roney on May 19, 2015
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# How many arrays with crossed cells, order of rows/columns irrelevant
I've been struggling with this simple problem for months though as I am a newbie to… well, maths, there's high chance someone more educated than myself may get it right!
Let's consider an array or a table or a Ferrers diagram or whatever it's called,
• $r$ rows by $c$ columns,
• in which each cell can either be blank or have a cross in it
• but each column must contain exactly $a$ crosses.
For instance, $\begin{array}{|c|c|c|c|c|c|} \hline ×&×&×&×& & \\ \hline ×&×& & &×&× \\ \hline & &×& &×& \\ \hline & & &×& &× \\ \hline \end{array}$ is such an array with $c=6,\ r=4,\ a=2$.
Now two arrays are said to be identical ('isomporphic' is probably the right word?) if we can obtain the second one from the first one by changing the order of columns and/or rows of the first one. So, the following array is identical to the one above:
$\begin{array}{|c|c|c|c|c|c|} \hline & & &×& &× \\ \hline ×&×& & &×&× \\ \hline &×&×& & & \\ \hline ×& &×&×&×& \\ \hline \end{array}$
but the order of columns is changed to $164325$ and the order of rows to $3241$.
Now the question is: how many unidentical (non-isomorphic) arrays are there, such that all the above conditions are satisfied and $c,r,a$ are given?
I know the answer for $c=6,r=4,a=2$ is $32$ because I have painstakingly went through every combination but how to do this more effectively?? :) I've been trying to learn enough group theory but it's all about rotating cubes or necklaces and I can't figure out how to translate this simple problem so that its tools can be deployed. Also, I tried to find a recurrence (by slicing away the last row and column) but it seems to be dependent on the actual placement of the crosses so…
I'd be glad for any help…
• I just came across your question and even though I know nothing about group theory it reminds me of a similar problem. For a matrix r by c, consider expanding it to a square matrix of size max(r, c), by concatenating empty rows or columns to fill it. Now it looks like a problem of finding whether 2 directed graphs are isomorphic.
– user98404
Dec 23, 2013 at 20:58
• With $a = 2$, a matrix such as this can be interpreted as an incidence matrix for an undirected graph. The columns correspond to edges, the rows correspond to vertices, and you have stipulated that each edge is incident with exactly two vertices. The permutation groups acting on the set of edges and vertices, respectively, are essentially relabeling them. So, considering matrices up to permutation actions is equivalent to considering the graphs up to relabeling (i.e. isomorphism). Dec 23, 2013 at 21:05
• The generalization to arbitrary $a$ gives matrices in bijection with $a$-uniform hypergraphs. (These are like graphs, but $a$ vertices at a time are considered as an "edge"). en.wikipedia.org/wiki/Hypergraph Dec 23, 2013 at 21:20
• Or it is the description of a bipartite graph, rows and columns corresponding to the two classes of vertices and crosses where edges are. Dec 24, 2013 at 9:30
Preliminary remark. None of the sequences that appear in this problem have OEIS entries with one exception. Therefore I was not able to verify the following results except in a few cases for small values. The reader is invited to contribute this type of verification. An algorithm to independently verify the results below for cases where $$k\ge 5$$ say would be most welcome.
Main text. What I am about to contribute is enrichment material to facilitate additional exploration of this problem. We will treat the case of a square grid containing $$n$$ rows and columns with $$k$$ marks being placed in each column. The most important observation is that what we have here is an instance of Power Group Enumeration, with the group acting on the slots where a selection from the $${n\choose k}$$ possible column configurations are placed being the symmetric group $$S_n$$ on $$n$$ elements and the group $$Q_{n,k}$$ acting on the column vectors being the action induced on the set of columns by permuting rows.
We can compute the number of configurations by Burnside's lemma which says to average the number of assignments fixed by the elements of the power group, which has $$n!\times n!$$ elements. But this number is easy to compute. Suppose we have a permutation $$\alpha$$ from $$S_n$$ (column permutations) and a permutation $$\beta$$ from $$S_n$$ (induced action on the columns by row permutations). If we place the appropriate number of complete, directed and consecutive copies of a cycle from $$\beta$$ on a cycle from $$\alpha$$ then this assignment is fixed under the power group action, and this is possible iff the length of $$\beta$$ divides the length of $$\alpha$$ and there are as many assignments as the length of $$\beta.$$ We can work with the cycle indices of $$S_n$$ and $$Q_{n,k}$$ and do not need to iterate over all $$n!^2$$ permutations.
To do this we need the cycle index $$Z(Q_{n,k})$$ of $$Q_{n,k}$$, which we compute. as follows: for each permutation shape that occurs in the cycle index $$Z(S_n)$$ of the symmetric group we compute a representative permutation and apply it to the set of column configurations. The result is factored into disjoint cycles and added to $$Z(Q_{n,k})$$ with the coefficient it had in $$Z(S_n).$$ Very simple.
As an example, here is the cycle index for $$Q_{6, 2}:$$ $${\frac {{a_{{1}}}^{15}}{720}}+1/48\,{a_{{1}}}^{7}{a_{{2}}}^{4}+1 /18\,{a_{{1}}}^{3}{a_{{3}}}^{4}+1/12\,{a_{{1}}}^{3}{a_{{2}}}^{6} +1/4\,a_{{1}}{a_{{4}}}^{3}a_{{2}}\\+1/6\,a_{{2}}{a_{{3}}}^{2}a_{{1 }}a_{{6}}+1/5\,{a_{{5}}}^{3}+1/18\,{a_{{3}}}^{5}+1/6\,{a_{{6}}}^ {2}a_{{3}}$$ and this is $$Z(Q_{7, 4})$$ $$1/12\,{a_{{1}}}^{3}{a_{{6}}}^{3}{a_{{3}}}^{4}a_{{2}}+1/6\,{a_{{4 }}}^{7}{a_{{2}}}^{3}a_{{1}}+1/7\,{a_{{7}}}^{5}+{\frac {{a_{{1}}} ^{35}}{5040}}+1/6\,{a_{{6}}}^{5}a_{{2}}a_{{3}}\\+{\frac {{a_{{1}}} ^{15}{a_{{2}}}^{10}}{240}}+{\frac {{a_{{1}}}^{5}{a_{{3}}}^{10}}{ 72}}+1/48\,{a_{{2}}}^{14}{a_{{1}}}^{7}+1/10\,{a_{{5}}}^{7}\\+1/48 \,{a_{{2}}}^{16}{a_{{1}}}^{3}+1/18\,{a_{{1}}}^{2}{a_{{3}}}^{11}+ 1/24\,a_{{1}}{a_{{6}}}^{4}{a_{{3}}}^{2}{a_{{2}}}^{2}\\+1/10\,{a_{{ 5}}}^{3}{a_{{10}}}^{2}+1/12\,a_{{4}}{a_{{12}}}^{2}a_{{6}}a_{{1}}$$ and finally this is $$Z(Q_{7,5})$$ $$1/12\,{a_{{3}}}^{3}a_{{6}}{a_{{1}}}^{2}{a_{{2}}}^{2}+1/8\,{a_{{4 }}}^{4}{a_{{2}}}^{2}a_{{1}}+1/7\,{a_{{7}}}^{3}+{\frac {{a_{{1}}} ^{21}}{5040}}+1/6\,{a_{{6}}}^{3}a_{{3}}\\+{\frac {{a_{{1}}}^{11}{a _{{2}}}^{5}}{240}}+{\frac {{a_{{3}}}^{5}{a_{{1}}}^{6}}{72}}+1/48 \,{a_{{1}}}^{5}{a_{{2}}}^{8}+1/24\,{a_{{4}}}^{4}a_{{2}}{a_{{1}}} ^{3}+1/10\,{a_{{5}}}^{4}a_{{1}}\\+1/48\,{a_{{1}}}^{3}{a_{{2}}}^{9} +1/18\,{a_{{3}}}^{7}+1/24\,a_{{3}}{a_{{6}}}^{2}{a_{{1}}}^{2}{a_{ {2}}}^{2}\\+1/10\,{a_{{5}}}^{2}a_{{10}}a_{{1}}+1/12\,a_{{4}}a_{{2} }a_{{12}}a_{{3}}.$$
The similarity in the coefficients is because these are inherited from the symmetric group.
Now the Burnside computation is best done with a CAS, here is the Maple code.
with(combinat);
pet_cycleind_symm :=
proc(n)
option remember;
if n=0 then return 1; fi;
end;
pet_flatten_term :=
proc(varp)
local terml, d, cf, v;
terml := [];
cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), seq(v, k=1..d)];
cf := cf/v^d;
od;
[cf, terml];
end;
pet_autom2cycles :=
proc(src, aut)
local numa, numsubs;
local marks, pos, cycs, cpos, clen;
numsubs := [seq(src[k]=k, k=1..nops(src))];
numa := subs(numsubs, aut);
marks := Array([seq(true, pos=1..nops(aut))]);
cycs := []; pos := 1;
while pos <= nops(aut) do
if marks[pos] then
clen := 0; cpos := pos;
while marks[cpos] do
marks[cpos] := false;
cpos := numa[cpos];
clen := clen+1;
od;
cycs := [op(cycs), clen];
fi;
pos := pos+1;
od;
return mul(a[cycs[k]], k=1..nops(cycs));
end;
pet_flat2rep :=
proc(f)
local p, q, res, cyc, t, len;
q := 1; res := [];
for t in f do
len := op(1, t);
cyc := [seq(p, p=q+1..q+len-1), q];
res := [op(res), seq(cyc[p], p=1..nops(cyc))];
q := q+len;
od;
res;
end;
pet_nchoosek_cind :=
proc(n, k)
option remember;
local idx_slots, cind, src, aut, q, rep, flat, term;
cind := 0;
src := choose(n, k);
if n=1 then
idx_slots := [a[1]]
else
idx_slots := pet_cycleind_symm(n);
fi;
for term in idx_slots do
flat := pet_flatten_term(term);
rep := pet_flat2rep(flat[2]);
aut :=
map(sel -> sort([seq(rep[sel[q]], q=1..k)]), src);
cind := cind + flat[1]*pet_autom2cycles(src, aut);
od;
cind;
end;
matrix_marks :=
proc(n, k)
option remember;
local idx_cols, idx_marks, res, a, b,
flat_a, flat_b, cyc_a, cyc_b, len_a, len_b, p, q;
if n=1 then
idx_cols := [a[1]]
else
idx_cols := pet_cycleind_symm(n);
fi;
idx_marks := pet_nchoosek_cind(n, k);
if not type(idx_marks, +) then
idx_marks := [idx_marks];
fi;
res := 0;
for a in idx_cols do
flat_a := pet_flatten_term(a);
for b in idx_marks do
flat_b := pet_flatten_term(b);
p := 1;
for cyc_a in flat_a[2] do
len_a := op(1, cyc_a);
q := 0;
for cyc_b in flat_b[2] do
len_b := op(1, cyc_b);
if len_a mod len_b = 0 then
q := q + len_b;
fi;
od;
p := p*q;
od;
res := res + p*flat_a[1]*flat_b[1];
od;
od;
res;
end;
This will produce the following sequence of values for matrices with a single mark per column: $$1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, \ldots$$ which is OEIS A000041.
For two marks per column we get starting at $$n=2$$ $$1, 3, 11, 35, 132, 471, 1806, 7042, 28494, 118662, 510517, 2262738, \\ 10337474, 48625631,\ldots$$
For three marks per column we get starting at $$n=3$$ $$1, 5, 35, 410, 6178, 122038, 2921607, 81609320, 2588949454, \\91699869557, 3582942335285, 153048366545566, \\7096576775166579, 355120233277118103,\ldots$$
Finally for four marks per column we get starting at $$n=4,$$ $$1, 7, 132, 6178, 594203, 85820809, 16341829155, 3875736708590, \\ 1112175913348040, 378860991866916370, 151006214911844288232, \\ 69600017255860985666964, 36729204987785981237238642,\ldots$$ The shared values in these three lists represent the fact that $${n\choose k} = {n\choose n-k}.$$
There is another example of Power Group Enumeration at this MSE link.
Addendum 2014-09-23. By way of an incentive to investigate and develop algorithms to verify the above results for non-trivial values of $$n$$ and $$k$$ I present a Perl script (admittedly very simple: warning, exponential algorithm) that can be used to verify small values. It gave the following results.
$time ./mmpg.pl 5 3 cases: 100000 35 real 0m5.377s user 0m5.179s sys 0m0.030s$ time ./mmpg.pl 6 2
cases: 11390625
132
real 12m24.638s
user 11m35.514s
sys 0m0.390s
$time ./mmpg.pl 6 3 cases: 64000000 410 real 76m39.028s user 71m55.408s sys 0m2.261s$ time ./mmpg.pl 6 4
cases: 11390625
132
real 16m23.239s
user 15m31.747s
sys 0m0.389s
$time ./mmpg.pl 7 1 cases: 823543 15 real 2m19.743s user 2m16.937s sys 0m0.077s$ time ./mmpg.pl 8 1
cases: 16777216
22
real 99m47.460s
user 93m49.700s
sys 0m2.667s
This is the Perl script.
#! /usr/bin/perl -w
#
sub permute {
my ($n) = @_; return [[1]] if$n == 1;
my ($res,$perm) = ([]);
foreach my $perm (@{ permute($n-1) }){
my $nxt; for(my$pos = 0; $pos <$n; $pos++){$nxt =
[@$perm[0..($pos-1)], $n, @$perm[$pos..($n-2)]];
push @$res,$nxt;
}
}
return $res; } sub choose { my ($data, $pos,$n) = @_;
my $size = scalar(@$data);
return [] if $pos ==$size;
if($n == 1){ my @res = map { [$_] } @$data[$pos..$size-1]; return \@res; } my$rec = choose($data,$pos+1, $n); foreach my$sel (@{ choose($data,$pos+1, $n-1) }){ my$nxt = [@$sel]; unshift @$nxt, $data->[$pos];
push @$rec,$nxt;
}
return $rec; } MAIN: { my$n = shift || 4;
my $k = shift || 3; my$range = [0..($n-1)]; my$cols = choose($range, 0,$k);
my $opts = scalar(@$cols); my $cases =$opts ** $n; print STDERR "cases:$cases\n";
my $col2ind = {}; for(my$colind = 0; $colind <$opts; $colind++){$col2ind->{join('-', @{ $cols->[$colind] })}
= $colind; } my$seen = {}; my $nperms = permute($n);
for(my $matind = 0;$matind < $cases;$matind++){
my $matvec = []; my ($pos, $ind); for(($pos, $ind)= (0,$matind);
$pos <$n; $pos++){ my$d = $ind %$opts;
push @$matvec,$d;
$ind = ($ind-$d)/$opts;
}
for($pos = 0;$pos < $n-1;$pos++){
last if $matvec->[$pos] > $matvec->[$pos+1];
}
next if $pos <$n-1;
my $admit = 1; my$pid;
foreach my $perm (@{$nperms }){
my @permcols = ();
foreach(my $colind = 0;$colind < $n;$colind++){
my @permcol =
map {
$perm->[$_] - 1
} @{$cols->[$matvec->[$colind]]}; @permcol = sort {$a <=> $b } @permcol; my$pcolind = $col2ind->{join('-', @permcol)}; push @permcols,$pcolind;
}
@permcols = sort { $a <=>$b } @permcols;
$pid = join('-', @permcols); if(exists($seen->{$pid})){$admit = undef;
last;
}
}
$seen->{$pid} = 1 if defined($admit); } print scalar(keys(%$seen)) . "\n";
1;
}
The above sequences now have OEIS entries as of today: OEIS A247417, OEIS A247596, OEIS A247597, OEIS A247598.
Addendum 2019-05-01. Consulting the OEIS links from above five years after I first solved this problem by PGE we see that a better, more efficient and more sophisticated algorithm has appeared, placing the above in the category of a historical artefact. Note however that the Maple code admits some improvements which are shown below (duplicate routines have been omitted).
pet_prod2rep :=
proc(varp)
local v, d, q, res, len, cyc;
q := 1; res := [];
for v in indets(varp) do
d := degree(varp, v);
len := op(1, v);
for cyc to d do
res :=
[op(res),
seq(p, p=q+1..q+len-1), q];
q := q+len;
od;
od;
res;
end;
pet_nchoosek_cind :=
proc(n, k)
option remember;
local idx_slots, cind, src, aut, q, rep, term;
cind := 0;
src := choose(n, k);
if n=1 then
idx_slots := [a[1]]
else
idx_slots := pet_cycleind_symm(n);
fi;
for term in idx_slots do
rep := pet_prod2rep(term);
aut :=
map(sel -> sort([seq(rep[sel[q]], q=1..k)]), src);
cind := cind + lcoeff(term)*pet_autom2cycles(src, aut);
od;
cind;
end;
matrix_marks :=
proc(n, k)
option remember;
local idx_cols, idx_marks, res, term_a, term_b,
v_a, v_b, inst_a, inst_b, len_a, len_b, p, q;
if n=1 then
idx_cols := [a[1]]
else
idx_cols := pet_cycleind_symm(n);
fi;
idx_marks := pet_nchoosek_cind(n, k);
if not type(idx_marks, +) then
idx_marks := [idx_marks];
fi;
res := 0;
for term_a in idx_cols do
for term_b in idx_marks do
p := 1;
for v_a in indets(term_a) do
len_a := op(1, v_a);
inst_a := degree(term_a, v_a);
q := 0;
for v_b in indets(term_b) do
len_b := op(1, v_b);
inst_b := degree(term_b, v_b);
if len_a mod len_b = 0 then
q := q + len_b*inst_b;
fi;
od;
p := p*q^inst_a;
od;
res := res +
lcoeff(term_a)*lcoeff(term_b)*p;
od;
od;
res;
end;
• Having come back to this problem after some time, let me ask: this solution, as well as the Perl script presented, pertains to the special case of a square matrix, doesn't it? May 1, 2019 at 11:49
• It does, but it is not difficult to adapt it to generic rectangular matrices. You should probably read my remark from today, however (at end). May 1, 2019 at 19:18
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18
# Review of Thermodynamics Calculations
The value of ΔH for a reaction in one direction is equal in magnitude, but opposite in sign, to ΔH for the reaction in the opposite direction, and ΔH is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, ΔHf°, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar and 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Gibbs free energy (G) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A number of approaches to the computation of free energy changes are possible.
## 18.1 Standard Enthalpy of Formation
### Learning Objectives
By the end of this section, you will be able to:
• Define standard enthalpy of formation
• Explain Hess’s law and use it to compute reaction enthalpies
$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=\text{Δ}H\text{°}=-393.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$
starting with the reactants at a pressure of 1 atm and 25 °C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of CO2, also at 1 atm and 25 °C. For nitrogen dioxide, NO2(g), $\text{Δ}{H}_{\text{f}}^{°}$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:
$\frac{1}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=\text{Δ}H\text{°}=\text{+33.2 kJ}$
A reaction equation with $\frac{1}{2}$ mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO2(g).
You will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C2H2). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.
### EXAMPLE 18.1.1
#### Evaluating an Enthalpy of Formation
Ozone, O3(g), forms from oxygen, O2(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, $\text{Δ}{H}_{\text{f}}^{°}$ of ozone from the following information:
$3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{O}}_{3}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H\text{°}=\text{+286 kJ}$
#### Solution
$\text{Δ}{H}_{\text{f}}^{°}$ is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, $\text{Δ}{H}_{\text{f}}^{°}$ for O3(g) is the enthalpy change for the reaction:
$\frac{3}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}\left(g\right)$
For the formation of 2 mol of O3(g), $\text{Δ}H\text{°}=\text{+286 kJ.}$ This ratio, $\left(\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\right),$ can be used as a conversion factor to find the heat produced when 1 mole of O3(g) is formed, which is the enthalpy of formation for O3(g):
$\text{Δ}\text{H}\text{° for}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mole of}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}\left(g\right)=1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}}\phantom{\rule{0.2em}{0ex}}=143\phantom{\rule{0.2em}{0ex}}\text{kJ}$
Therefore, $\text{Δ}{H}_{\text{f}}^{°}\left[{\text{O}}_{3}\left(g\right)\right]=\text{+143 kJ/mol}.$
#### Check Your Learning
Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is −92.3 kJ/mol.
### Answer
For the reaction ${\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{HCl}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H\text{°}=-184.6\phantom{\rule{0.2em}{0ex}}\text{kJ}$
### EXAMPLE 18.1.2
#### Writing Reaction Equations for $\text{Δ}{H}_{\text{f}}^{°}$
Write the heat of formation reaction equations for:
(a) C2H5OH(l)
(b) Ca3(PO4)2(s)
#### Solution
Remembering that $\text{Δ}{H}_{\text{f}}^{°}$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:
(a) $2\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+3{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)$
(b) $3\text{Ca}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{P}}_{4}\left(s\right)+4{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_{3}\left({\text{PO}}_{4}{\right)}_{2}\left(s\right)$
Note: The standard state of carbon is graphite, and phosphorus exists as P4.
#### Check Your Learning
Write the heat of formation reaction equations for:
(a) C2H5OC2H5(l)
(b) Na2CO3(s)
### Answer
(a) $4\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+5{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}\left(l\right);$ (b) $2\text{Na}\left(s\right)+\text{C}\left(s,\phantom{\rule{0.2em}{0ex}}\text{graphite}\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{CO}}_{3}\left(s\right)$
## 18.2 Law
There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.
This type of calculation usually involves the use of Hess’s law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:
$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H\text{°}=-394\phantom{\rule{0.2em}{0ex}}\text{kJ}$
In the two-step process, first carbon monoxide is formed:
$\text{C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H\text{°}=-111\phantom{\rule{0.2em}{0ex}}\text{kJ}$
Then, carbon monoxide reacts further to form carbon dioxide:
$\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}H\text{°}=-283\phantom{\rule{0.2em}{0ex}}\text{kJ}$
The equation describing the overall reaction is the sum of these two chemical changes:
$\begin{array}{}\\ \text{Step 1: C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\\ \underset{¯}{\text{Step 2: CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)}\\ \text{Sum: C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)+\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)+{\text{CO}}_{2}\left(g\right)\end{array}$
Because the CO produced in Step 1 is consumed in Step 2, the net change is:
$\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)$
According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.
$\begin{array}{ll}\text{C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\hfill & \text{Δ}H\text{°}=-111\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)}{\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{1em}{0ex}}}\hfill & \frac{\text{Δ}H\text{°}=-283\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}H\text{°}=-394\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$
The result is shown in Figure 18.1. We see that ΔH of the overall reaction is the same whether it occurs in one step or two. This finding (overall ΔH for the reaction = sum of ΔH values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.
Figure 18.1
The formation of CO2(g) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.
Before we further practice using Hess’s law, let us recall two important features of ΔH.
1. ΔH is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO2(g) is +33.2 kJ:
$\frac{1}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\text{+33.2 kJ}$
When 2 moles of NO2 (twice as much) are formed, the ΔH will be twice as large:
${\text{N}}_{2}\left(g\right)+2{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\text{+66.4 kJ}$
In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.
2. ΔH for a reaction in one direction is equal in magnitude and opposite in sign to ΔH for the reaction in the reverse direction. For example, given that:
${\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{HCl}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=-184.6\phantom{\rule{0.2em}{0ex}}\text{kJ}$
Then, for the “reverse” reaction, the enthalpy change is also “reversed”:
$2\text{HCl}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\text{+184.6 kJ}$
### EXAMPLE 18.2.1
#### Stepwise Calculation of $\text{Δ}{H}_{\text{f}}^{°}$ Using Hess’s Law
Determine the enthalpy of formation, $\text{Δ}{H}_{\text{f}}^{°},$ of FeCl3(s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:
$\text{Fe}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=-341.8\phantom{\rule{0.2em}{0ex}}\text{kJ}$
${\text{FeCl}}_{2}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=-57.7\phantom{\rule{0.2em}{0ex}}\text{kJ}$
#### Solution
We are trying to find the standard enthalpy of formation of FeCl3(s), which is equal to ΔH° for the reaction:
$\text{Fe}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=?$
Looking at the reactions, we see that the reaction for which we want to find ΔH° is the sum of the two reactions with known ΔH values, so we must sum their ΔHs:
$\begin{array}{lll}\text{Fe}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}\left(s\right)\hfill & \hfill & \text{Δ}H\text{°}=-341.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{{\text{FeCl}}_{2}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)}{\text{Fe}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)\phantom{\rule{1em}{0ex}}}\hfill & \hfill & \frac{\text{Δ}H\text{°}=-57.7\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}H\text{°}=-399.5\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$
The enthalpy of formation, $\text{Δ}{H}_{\text{f}}^{°},$ of FeCl3(s) is −399.5 kJ/mol.
#### Check Your Learning
Calculate ΔH for the process:
${\text{N}}_{2}\left(g\right)+2{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}\left(g\right)$
from the following information:
${\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=180.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$
$\text{NO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=-57.06\phantom{\rule{0.2em}{0ex}}\text{kJ}$
66.4 kJ
Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of ΔH) if they are difficult to determine experimentally.
### EXAMPLE 18.2.2
#### A More Challenging Problem Using Hess’s Law
Chlorine monofluoride can react with fluorine to form chlorine trifluoride:
(i) $\text{ClF}\left(g\right)+{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=?$
Use the reactions here to determine the ΔH° for reaction (i):
(ii) $2{\text{OF}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}\left(g\right)+2{\text{F}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(ii\right)}^{°}=-49.4\phantom{\rule{0.2em}{0ex}}\text{kJ}$
(iii) $2\text{ClF}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\text{O}\left(g\right)+{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(iii\right)}^{°}=\text{+214.0 kJ}$
(iv) ${\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(iv\right)}^{°}=\text{+236.2 kJ}$
#### Solution
Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that ClF(g) is needed as a reactant. This can be obtained by multiplying reaction (iii) by $\frac{1}{2},$ which means that the ΔH° change is also multiplied by $\frac{1}{2}\text{:}$
$\text{ClF}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\left(214.0\right)=\text{+107.0 kJ}$
Next, we see that F2 is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the ΔH° changes sign and is halved:
$\frac{1}{2}{\text{O}}_{2}\left(g\right)+{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=\text{+24.7 kJ}$
To get ClF3 as a product, reverse (iv), changing the sign of ΔH°:
$\frac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=\text{−236.2 kJ}$
Now check to make sure that these reactions add up to the reaction we want:
$\begin{array}{lll}\text{ClF}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{OF}}_{2}\left(g\right)\hfill & \hfill & \text{Δ}H\text{°}=\text{+107.0 kJ}\hfill \\ \frac{1}{2}{\text{O}}_{2}\left(g\right)+{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}\left(g\right)\hfill & \hfill & \text{Δ}H\text{°}=\text{+24.7 kJ}\hfill \\ \frac{\frac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)}{\text{ClF}\left(g\right)+{\text{F}}_{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)\phantom{\rule{8em}{0ex}}}\hfill & \hfill & \frac{\text{Δ}H\text{°}=-236.2\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}H\text{°}=-104.5\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$
Reactants $\frac{1}{2}{\text{O}}_{2}$ and $\frac{1}{2}{\text{O}}_{2}$ cancel out product O2; product $\frac{1}{2}{\text{Cl}}_{2}\text{O}$ cancels reactant $\frac{1}{2}{\text{Cl}}_{2}\text{O;}$ and reactant $\frac{3}{2}{\text{OF}}_{2}$ is cancelled by products $\frac{1}{2}{\text{OF}}_{2}$ and OF2. This leaves only reactants ClF(g) and F2(g) and product ClF3(g), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified ΔH° values will give the desired ΔH°:
$\text{Δ}H\text{°}=\left(+107.0\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(24.7\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(-236.2\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)=-104.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$
#### Check Your Learning
Aluminum chloride can be formed from its elements:
(i) $2\text{Al}\left(s\right)+3{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{AlCl}}_{3}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}H\text{°}=?$
Use the reactions here to determine the ΔH° for reaction (i):
(ii) $\text{HCl}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{HCl}\left(aq\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(ii\right)}^{°}=-74.8\phantom{\rule{0.2em}{0ex}}\text{kJ}$
(iii) ${\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{HCl}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(iii\right)}^{°}=-185\phantom{\rule{0.2em}{0ex}}\text{kJ}$
(iv) ${\text{AlCl}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{AlCl}}_{3}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(iv\right)}^{°}=+323\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}$
(v) $\text{2Al}\left(s\right)+6\text{HCl}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{AlCl}}_{3}\left(aq\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(v\right)}^{°}=-1049\phantom{\rule{0.2em}{0ex}}\text{kJ}$
### Answer
−1407 kJ
We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and n standing for the stoichiometric coefficients:
$\text{Δ}{H}_{\text{reaction}}^{°}=\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{products}\right)-\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{reactants}\right)$
The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.
### EXAMPLE 18.2.3
#### Using Hess’s Law
What is the standard enthalpy change for the reaction:
$3{\text{NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=?$
#### Solution: Using the Equation
Use the special form of Hess’s law given previously, and values from Appendix G:
$\text{Δ}{H}_{\text{reaction}}^{°}=\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\text{(products)}-\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{reactants}\right)$
$\begin{array}{l}\\ \\ \\ =\left[2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}\left(aq\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{-207.4\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}\left(aq\right)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol NO}\left(g\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{+90.2 kJ}}{\overline{)\text{mol NO}\left(g\right)}}\right]\\ -\left[3\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{\text{+33.2 kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{-285.8\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)}}\right]\\ =\left[2×\left(-206.64\right)+90.25\right]-\left[3×33.2+-\left(-285.83\right)\right]\\ =–323.03+186.23\\ =-136.80\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$
#### Solution: Supporting Why the General Equation Is Valid
Alternatively, we can write this reaction as the sum of the decompositions of 3NO2(g) and 1H2O(l) into their constituent elements, and the formation of 2HNO3(aq) and 1NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the $\text{Δ}{H}_{\text{f}}^{°}$ values for these compounds (from Appendix G ), we have:
$3{\text{NO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3/2N}}_{2}\left(g\right)+{\text{3O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{1}^{°}=-99.6\phantom{\rule{0.2em}{0ex}}\text{kJ}$
${\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{2}^{°}=\text{+285.8 kJ}\phantom{\rule{0.2em}{0ex}}\left[-1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\right)\right]$
${\text{H}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)+3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{3}^{°}=-414.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\phantom{\rule{0.2em}{0ex}}\left[2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{HNO}}_{3}\right)\right]$
$\frac{1}{2}{\text{N}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{4}^{°}=\text{+90.2 kJ}\phantom{\rule{0.2em}{0ex}}\left[1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{NO}\right)\right]$
Summing these reaction equations gives the reaction we are interested in:
${\text{3NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)$
Summing their enthalpy changes gives the value we want to determine:
$\begin{array}{cc}\hfill \text{Δ}{H}_{\text{rxn}}^{°}& =\text{Δ}{H}_{1}^{°}+\text{Δ}{H}_{2}^{°}+\text{Δ}{H}_{3}^{°}+\text{Δ}{H}_{4}^{°}=\left(-99.6\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(\text{+285.8 kJ}\right)+\left(-414.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(\text{+90.2 kJ}\right)\hfill \\ & =-138.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \end{array}$
So the standard enthalpy change for this reaction is ΔH° = −138.4 kJ.
Note that this result was obtained by (1) multiplying the $\text{Δ}{H}_{\text{f}}^{°}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\text{Δ}{H}_{\text{f}}^{°}$ of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.
#### Check Your Learning
Calculate the heat of combustion of 1 mole of ethanol, C2H5OH(l), when H2O(l) and CO2(g) are formed. Use the following enthalpies of formation: C2H5OH(l), −278 kJ/mol; H2O(l), −286 kJ/mol; and CO2(g), −394 kJ/mol.
−1368 kJ/mol
## 18.3 Temperature Dependence of Spontaneity
### Learning Objectives
By the end of this section, you will be able to:
• Identify the changes in concentration or pressure that occur for chemical species in equilibrium systems
• Explain how temperature affects the spontaneity of some proceses
As was previously demonstrated in the section on entropy in an earlier chapter, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
$\text{Δ}G=\text{Δ}H-T\text{Δ}S$
The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes:
1. Both ΔH and ΔS are positive. This condition describes an endothermic process that involves an increase in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is greater than ΔH. If the TΔS term is less than ΔH, the free energy change will be positive. Such a process is spontaneous at high temperatures and nonspontaneous at low temperatures.
2. Both ΔH and ΔS are negative. This condition describes an exothermic process that involves a decrease in system entropy. In this case, ΔG will be negative if the magnitude of the TΔS term is less than ΔH. If the TΔS term’s magnitude is greater than ΔH, the free energy change will be positive. Such a process is spontaneous at low temperatures and nonspontaneous at high temperatures.
3. ΔH is positive and ΔS is negative. This condition describes an endothermic process that involves a decrease in system entropy. In this case, ΔG will be positive regardless of the temperature. Such a process is nonspontaneous at all temperatures.
4. ΔH is negative and ΔS is positive. This condition describes an exothermic process that involves an increase in system entropy. In this case, ΔG will be negative regardless of the temperature. Such a process is spontaneous at all temperatures.
These four scenarios are summarized in Figure 18.2.
Figure 18.2
There are four possibilities regarding the signs of enthalpy and entropy changes.
### EXAMPLE 18.3.1
#### Predicting the Temperature Dependence of Spontaneity
The incomplete combustion of carbon is described by the following equation:
$\text{2C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{2CO}\left(g\right)$
How does the spontaneity of this process depend upon temperature?
#### Solution
Combustion processes are exothermic (ΔH < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, ΔS > 0). The reaction is therefore spontaneous (ΔG < 0) at all temperatures.
#### Check Your Learning
Popular chemical hand warmers generate heat by the air-oxidation of iron:
$\text{4Fe}\left(s\right)+{\text{3O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2Fe}}_{2}{\text{O}}_{3}\left(s\right)$
How does the spontaneity of this process depend upon temperature?
### Answer
ΔH and ΔS are negative; the reaction is spontaneous at low temperatures.
When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its ΔG) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which ΔG is plotted on the y axis versus T on the x axis:
$\text{Δ}G=\text{Δ}H-T\text{Δ}S$
$y=b+mx$
Such a plot is shown in Figure 18.3. A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative ΔG) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the x-intercept of the line, that is, the value of T for which ΔG is zero:
$\text{Δ}G=0=\text{Δ}H-T\text{Δ}S$
$T=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}H}{\text{Δ}S}$
So, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which ΔG for the process is zero. As noted earlier, the condition of ΔG = 0 describes a system at equilibrium.
Figure 18.3
These plots show the variation in ΔG with temperature for the four possible combinations of arithmetic sign for ΔH and ΔS.
### EXAMPLE 18.3.2
#### Equilibrium Temperature for a Phase Transition
As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its liquid and gaseous phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in Appendix G to estimate the boiling point of water.
#### Solution
The process of interest is the following phase change:
${\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(g\right)$
When this process is at equilibrium, ΔG = 0, so the following is true:
$0=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\phantom{\rule{5em}{0ex}}\text{or}\phantom{\rule{5em}{0ex}}T=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}H\text{°}}{\text{Δ}S\text{°}}$
Using the standard thermodynamic data from Appendix G,
$\begin{array}{ccc}\hfill \text{Δ}H\text{°}& =\hfill & \text{1 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =\hfill & \left(\text{1 mol}\right)-\text{241.82 kJ/mol}-\left(\text{1 mol}\right)\left(\text{−241.82 kJ/mol}\right)=\text{44.01 kJ}\hfill \\ \hfill \text{Δ}S°& =\hfill & \text{1 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{S}^{°}\left({\text{H}}_{2}\text{O}\left(g\right)\right)\phantom{\rule{0.2em}{0ex}}-\text{1 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{S}^{°}\left({\text{H}}_{2}\text{O}\left(l\right)\right)\hfill \\ & =\hfill & \left(\text{1 mol}\right)\text{188.8 J/K·mol}-\left(\text{1 mol}\right)\text{70.0 J/K·mol}=\text{118.8 J/K}\hfill \\ \hfill T& =\hfill & \frac{\text{Δ}H\text{°}}{\text{Δ}S\text{°}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{44.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{J}}{118.8\phantom{\rule{0.2em}{0ex}}\text{J/K}}\phantom{\rule{0.2em}{0ex}}=370.5\phantom{\rule{0.2em}{0ex}}\text{K}=97.3\phantom{\rule{0.2em}{0ex}}\text{°C}\hfill \end{array}$
The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K (Appendix G). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point.
#### Check Your Learning
Use the information in Appendix G to estimate the boiling point of CS2.
### Answer
313 K (accepted value 319 K)
## Files
Previous Citation(s)
Flowers, P., et al. (2019). Chemistry: Atoms First 2e. https://openstax.org/details/books/chemistry-atoms-first-2e (12.1-12.4, 13.4)
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# Analog Watch
How do smart watches work? In this lesson you will learn how to build an analog-ish clock. This can run on BrainPad Pulse alone, but the results will be much cooler when combined with BrainPower.
## Prerequisite
You must have basic understanding of coding and have completed the Drawing lesson.
## The Face
The watch face is what you see behind the hands. We will build a simple face that consists of the numbers 1 to 12. We want those to be arranged in a circle. We also want the circle (oval) to have different width and height, to accommodate a non-square screens. The Math libraries include Sin() and Cos() we would need to calculate/draw circles.
In C# we’ll need to add ‘using System‘ to the top with our other using statements.
In Python we’ll need to add ‘import math‘ to the top of our code.
``````using System;
void DrawFace(double x, double y, double width, double height){
var h = 0;
while (h < 12) {
var a = (h / 6 * Math.PI) - (Math.PI / 2);
Point(x + width * Math.Cos(a), y + height * Math.Sin(a), White);
h = h + 1;
}
}
Clear();
DrawFace(64,32,38,28);
Show();``````
``````from BrainPad import *
import math
def DrawFace(x, y, width, height):
h = 0
while (h < 12):
a = (h / 6 * math.pi) - (math.pi / 2)
Point(x + width * math.cos(a), y + height * math.sin(a), White)
h = h + 1
Clear()
DrawFace(64,32,38,28)
Show()
``````
In C# running the code above will only show 2 points. This is due to the compiler opting to use an int type when we created the h variable. The type int doesn’t have fractions, and so the math in the loop does not work properly as we lose all fractions.
In Python this isn’t an issue because the variable is dynamic type and it changes on the fly to whatever is necessary.
## Variable Type Refresher
The Variables lesson covers variables in more detail. This section will highlight the difference between int and double when using C#.
When using var, the system is given the choice to select the appropriate variable type. We can enforce the use of fractions (double type) by assigning a fraction to the variable when it is first created. For example, if the initial value is 0, then set the variable to 0.0. Mathematically speaking, 0 and 0.0 are exactly the same. But in code, the use of 0.0 is telling the system that this is a variable the requires fractions and that any mathematical calculations must keep track of fractions.
Here is an example code showing the problem. What is 5 divided by 2?
``````var v = 5;
v = v / 2;
Print(v);``````
5 divided by 2 results in 2.5, but if you use integer instead of double, then the result is 2 not 2.5. This is the case in C# but Python is different as it will change the type automatically. and the results will be correct.
Now, modify the same code but use 5.0 when creating the v variable.
``````var v = 5.0;
v = v / 2;
Print(v);``````
We also have the choice to select the variable type to expect fractions (double type).
``````double v = 5;
v = v / 2;
Print(v);``````
## Face Redo
Going back to our initial example, we set the h variable to 0.0, or use double instead of var.
In Python we’ll keep h = 0, since it handles variables differently.
``````void DrawFace(double x, double y, double width, double height){
double h = 0.0;
while (h < 12) {
var a = (h / 6 * Math.PI) - (Math.PI / 2);
Point(x + width * Math.Cos(a), y + height * Math.Sin(a), White);
h = h + 1;
}
}
Clear();
DrawFace(64,32,38,28);
Show();``````
``````def DrawFace(x, y, width, height):
h = 0
while (h < 12):
a = (h / 6 * math.pi) - (math.pi / 2)
Point(x + width * math.cos(a), y + height * math.sin(a), White)
h = h + 1
Clear()
DrawFace(64,32,38,28)
Show()
``````
And we finally have 12 points surrounding the “face”. Let’s take this a bit further and put numbers instead of points.
``````void DrawFace(double x, double y, double width, double height){
double h = 0.0;
while (h < 12) {
var a = (h / 6 * Math.PI) - (Math.PI / 2);
Text(h, x + width * Math.Cos(a), y + height * Math.Sin(a));
h = h + 1;
}
}
Clear();
DrawFace(64,32,38,28);
Show();``````
``````def DrawFace(x, y, width, height):
h = 0
while (h < 12):
a = (h / 6 * math.pi) - (math.pi / 2)
Text(h, x + width * math.cos(a), y + height * math.sin(a)
h = h + 1
Clear()
DrawFace(64,32,38,28)
Show()``````
That is almost correct! We are used to seeing 12 on the top, but we now see 0. In fact, hour 0 and hour 12 are the same thing! Time is revolving and when time reached 12 it is simply going back to 0. Anyway, we still need to correct the face. We can fix this by adding an if statement where if the number is 0, then it modifies to 12. But since time is revolving and the continuing drawing of a circle will just redraw over the same circle, also revolving like time, we can start count at 1 instead of 0 and then draw all the way to 12.
``````void DrawFace(double x, double y, double width, double height){
var h = 1.0;
while (h <= 12) {
var a = (h / 6 * Math.PI) - (Math.PI / 2);
Text(h, x + width * Math.Cos(a), y + height * Math.Sin(a));
h = h + 1;
}
}
Clear();
DrawFace(64,32,38,28);
Show();``````
``````def DrawFace(x, y, width, height):
h = 1
while (h <= 12):
a = (h / 6 * math.pi) - (math.pi / 2)
Text(h, x + width * math.cos(a), y + height * math.sin(a))
h = h + 1
Clear()
DrawFace(64,32,38,28)
Show()
``````
## The Hands
The hand is simply a Line() that starts at the middle of the watch and points to a location around the face. The hour hand uses 1 to 12 numbers, but the minute and second hands are from 1 to 60. We will make one function that draws a hand. We will give it location and size, and then it will have a time and unit. Unit is 12 for hours and 60 for seconds and minutes.
``````void DrawHand(double x, double y, double width, double height, double time, double unit) {
unit = unit / 2;
var a = (time / unit * Math.PI) - (Math.PI/2);
var ex = x + width * Math.Cos(a);
var ey = y + height * Math.Sin(a);
Line(x, y, ex, ey);
}``````
``````def DrawHand(x, y, width,height,time,unit):
unit = unit / 2
a = (time / unit * math.pi) - (math.pi/2)
ex = x + width * math.cos(a)
ey = y + height * math.sin(a)
Line(x, y, ex, ey)
``````
Let’s say the current time is 3:25 and 40 seconds. Oh, and don’t forget about the face!
``````DrawFace(64, 32, 38, 28);
DrawHand(64, 32, 20, 10, 3, 12);
DrawHand(64, 32, 28, 18, 25, 60);
DrawHand(64, 32, 30, 20, 40, 60);
Show();``````
``````DrawFace(64, 32, 38, 28)
DrawHand(64, 32, 20, 10, 3, 12)
DrawHand(64, 32, 28, 18, 25, 60)
DrawHand(64, 32, 30, 20, 40, 60)
Show()``````
Do you see how the hands do not seem perfectly centered? Actually, the hands are perfectly centered, but the face is not! This is because the text position is not in the middle of the character. It is in the top left corner of the characters. The characters are 8×6 points, so we need to move the face back by 4×3 points.
``````DrawFace(64 - 4, 32 - 3, 38, 28);
DrawHand(64, 32, 20, 10, 3, 12);
DrawHand(64, 32, 28, 18, 25, 60);
DrawHand(64, 32, 30, 20, 40, 60);
Show();``````
``````DrawFace(64 - 4, 32 - 3, 38, 28)
DrawHand(64, 32, 20, 10, 3, 12)
DrawHand(64, 32, 28, 18, 25, 60)
DrawHand(64, 32, 30, 20, 40, 60)
Show()``````
We can read the internal system time to show the appropriate time. This requires the use of services provided by the system to read time. Each language/system/OS has its own unique available services. For example, C# includes .NET services to handle time. Here is a time machine example, where we read the time, show it on the “face”, and then add one second.
``````var time = new DateTime(2021, 1, 1, 3, 30, 1);
while (true) {
Clear();
DrawFace(64 - 4, 32 - 3, 38, 28);
DrawHand(64, 32, 20, 10, time.Hour, 12);
DrawHand(64, 32, 28, 18, time.Minute, 60);
DrawHand(64, 32, 30, 20, time.Second, 60);
Show();
time = time.Add(new TimeSpan(0, 0, +1));
Wait(1);
}``````
On the other hand, MicroPython does not have a time service. The full Python does but not the scaled down MicroPython version. We will instead create a time counter. We now have generic code that can run on both.
We may have to adjust our Wait() function argument to keep more accurate time since we’re telling it to Wait(1) second and our code actually takes time to run.
``````var seconds = 0;
var minutes = 25;
var hours = 3;
while (true){
Clear();
DrawFace(64 - 4, 32 - 3, 38, 28);
DrawHand(64, 32, 20, 10, hours, 12);
DrawHand(64, 32, 28, 18, minutes, 60);
DrawHand(64, 32, 30, 20, seconds, 60);
seconds = seconds + 1;
if (seconds >= 60) {
seconds = 0;
minutes = minutes + 1;
}
if (minutes >=60) {
minutes = 0;
hours = hours + 1;
}
if (hours > 12) {
hours = 1;
}
Wait(1);
Show();
}``````
``````seconds = 0
minutes = 25
hours = 3
while True:
Clear()
DrawFace(64 - 4, 32 - 3, 38, 28);
DrawHand(64, 32, 20, 10, hours, 12);
DrawHand(64, 32, 28, 18, minutes, 60);
DrawHand(64, 32, 30, 20, seconds, 60);
seconds = seconds + 1
if seconds >= 60:
seconds = 0
minutes = minutes + 1
if minutes >=60:
minutes =0
hours = hours + 1
if hours > 12:
hours = 1
Wait(1)
Show()
``````
## What’s Next?
Let’s use the buttons to set the hours and minutes on our new watch. This should be an easy addition given our current level of programming knowledge. First create the 2 button objects.
``````var btnA = Button(ButtonA,0);
var btnB = Button(ButtonB,0);``````
``````btnA = Button(ButtonA,0)
btnB = Button(ButtonB,0)``````
Finally add if statements in side our while loop to check each button an then increment the hours +1 when the A button is pressed, and increment the minutes + 1 when the B button is pressed.
``````var seconds = 0;
var minutes = 43;
var hours = 10;
var btnA = Button(ButtonA, 0);
var btnB = Button(ButtonB, 0);
while (true) {
if (In(btnA) == 1) {
minutes = minutes + 1;
}
if (In(btnB)== 1) {
hours = hours + 1;
}
Clear();
DrawFace(64 - 4, 32 - 3, 38, 28);
DrawHand(64, 32, 20, 10, hours, 12);
DrawHand(64, 32, 28, 18, minutes, 60);
DrawHand(64, 32, 30, 20, seconds, 60);
seconds = seconds + 1;
if (seconds >= 60) {
seconds = 0;
minutes = minutes + 1;
}
if (minutes >= 60) {
minutes = 0;
hours = hours + 1;
}
if (hours > 12) {
hours = 1;
}
Wait(1)
Show()
}``````
``````seconds = 0
minutes = 43
hours = 10
btnA = Button(ButtonA, 0)
btnB = Button(ButtonB, 0)
while (true) {
if (In(btnA) == 1):
minutes = minutes + 1
if (In(btnB)== 1):
hours = hours + 1
Clear()
DrawFace(64 - 4, 32 - 3, 38, 28)
DrawHand(64, 32, 20, 10, hours, 12)
DrawHand(64, 32, 28, 18, minutes, 60)
DrawHand(64, 32, 30, 20, seconds, 60)
seconds = seconds + 1
if (seconds >= 60):
seconds = 0
minutes = minutes + 1
if (minutes >= 60):
minutes = 0
hours = hours + 1
if (hours > 12):
hours = 1
Wait(1)
Show()
``````
## BrainStorm
How does a computer keep track of time when it is off? It actually does not! Once power is down, everything is reset internally. However, there is a small circuit inside caller RTC (Real Time Clock) and this circuit is extremely low power and it has its own tiny battery. When the computer is off, the RTC will continue to run and continue to keep track of time. The moment the computer is powered up, it will read the current time from the RTC,
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# Biostatistics
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Introduction to Biostatistics
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### Biostatistics
1. 1. BIOSTATISTICS AND ITS MEASURES V. SENTHILVEL, M.Sc., M.Phil., PGDBS., DST., (Ph.D), MIAPSM, MIPHA, MISMS Assistant Professor (Biostatistics), Department of Community Medicine, PIMS, Puducherry – 605 014. Email id: senthilvel99@gmail.com
2. 2. Statistics Statistic Statistics
3. 3. • STATISTIC : It means a measured (or) counted fact (or) piece of information stated as figure such as height of one person, birth of a baby, etc., • STATISTICS : It is also called Data. It is Plural. Stated in more than one figures such as height of 2 persons, birth of 5 babies etc. They are collected from experiments, records, and surveys.
4. 4. Definition of Statistics Statistics means that the collection of data, tabulation, analysis and interpretation of quantitative information. • In other words, Statistics is a branch of science used in dealing with phenomena that can be described numerically by counts or by measurements. • Quantitative information: population, birth, fertility, production and etc. Statistics: It is a tool It is a branch of science Statistical Method: It refers Both data and Methods
5. 5. Definition of Statistics (contd…) Statistics : used in two sense viz., singular and plural. • In singular noun, it deals with collection, analysis and interpretation of quantitative information. • In plural sense, statistics denote some numerical data such as birth, death etc.
6. 6. Statistics is used in many fields Medical Statistics Agricultural Statistics Economical Statistics Mathematical Statistics And so on…………. Bio-Statistics
7. 7. Definition of Bio-Statistics Bio-statistics: means when we use the statistical tools on the Biological Problems and derived some results about that. Example: Medical Science It is also called Bio-metry. It means measurement of life. Normally, in medicine for precision, facts, observations or measurements have to be expressed in figures. Bar diagrams, Multiple Bar diagram,Histogram, Pie chart and etc.,
8. 8. Uses of Bio-Statistics o To calculate average, median, mode standard deviation of the given collected data o To compare two sets of data o To get a conclusion (or) result o To find the association between the two variables o To find the correlation bet. the two variables o To give the results in a tabular or diagrammatic form.
9. 9. Bio-Statistics in Various areas Health Statistics Medical Statistics Vital Statistics
10. 10. Bio-statistics (Contd….) • In Public Health or Community Health, it is called Health Statistics. • In Medicine, it is called Medical Statistics. In this we study the defect, injury, disease, efficacy of drug, Serum and Line of treatment, etc., • In population related study it is called Vital Statistics. e.g. study of vital events like births, marriages and deaths.
11. 11. Measurement Scale Variable: A characteristic that varies from one biological entity to another is termed as variable. Various Measurements: Ratio Scale: The measurements scales having a constant size interval and true zero point are said to be ratio of measurement. Besides heights and numbers, ratio scales include weights (mg, g), volumes (cc, cu.m), capacities (ml, l), rates (cm/sec., Km/h) and lengths of time (h, Yr) etc.,
12. 12. Interval Scale: Some measurement scales posses a constant interval size but not a zero, they are called internal scales. A good example is that of the two common temperature scales. Celcius (C ) and Fahrenheit (F). We can find the same difference exists between 25 degree celcius and 30 degree celcius as between 10 degree celcius and 15 degree celcius.
13. 13. Ordinal Scale: The data consist of an ordering or ranking of measurement and are said to be on an ordinal scale. For example: The examination marks of 75, 80, 87, 92, and 95% (ratio scale) might be recorded as A, B, C, D and E (ordinal scale) respectively. Normal Scale: The variables are classified by some quality rather than by a numerical measurement. In such cases, the variable is called an attribute and said to using a nominal scale of measurement. For example: 1. Data are represented as male or female. 2. Heights may be recorded as tall or short.
14. 14. Types of Statistics STATISTICS Descriptive Inferential
15. 15. Types of Statistics (Contd…) • Descriptive Statistics: Once the data have been collected, we can organize and summaries in such a manner as to arrive at their orderly presentation and conclusion. This procedure can be called Descriptive Statistics. • Inferential Statistics: The number of birth and deaths in a state in a particular year.
16. 16. Types of Data Qualitative Data Nominal Ordinal Quantitative Data Discrete Continuous Interval Ratio
17. 17. Types of Data (contd…) • Quantitative Data There is a natural numeric scale (numerical Value) ( can be subdivided into interval and ratio data ) Example: age, height, weight • Qualitative Data Measuring a characteristic for which there is no natural numeric scale ( can be subdivided into nominal and ordinal data ) Example: Gender, Eye color
18. 18. Definition of Biostatistics • The statistical methods applied to biological problems is called as Biostatistics. • It is also called Biometry. • Biometry means Biological Measurement or Measurement of Life. STATISTICAL METHODS Croxton and Gowden said Statistical Method means “ the collection, presentation, analysis and interpretation of numerical data”. 1. Collection of Data: It is the first step in collection of data. Careful planning is essential before collecting the data.
19. 19. Statistical methods (contd…) Presentation of Data: • The mass data collected should be presented in a suitable form for further analysis. • The collected data may be presented in the form of tabular or diagrammatic or graphical form. Analysis of Data: The data presented should be carefully analyzed from the presented data such as measures of central tendencies, dispersion, correlation, regression, etc.
20. 20. Statistical methods (contd…) Interpretation of Data: • The final step is drawing conclusion from the data collected. • A valid conclusion must be drawn on the basis of analysis. • A high degree of skill and experience is necessary for the interpretation.
21. 21. Measures of Central Tendency • Mean - average (or) A.M • Median - Positional average (Middle value) • Mode – Most repeated value ( The frequent value)
22. 22. Mean: Merits • Mean can never be biased. It is easy to calculate. • It is least affected by fluctuations of sampling. • It is based on all the observations of a series. Therefore, it is a most representative measure. Demerits • It is greatly affected by extreme fluctuations. Thus it is not a true representative value of all the items of the series. • A.M cannot be used for qualitative characteristics such as colour of flowers, sweetness of orange or darkness of the colour.
23. 23. Median: Merits • Easy to calculate and understand • It is not affected by extreme observations • It is best measure for qualitative data. Demerits • It cannot be determined in the case of even number of observations. We merely estimate it as the arithmetic mean of the two middle terms. • It is a positional average. It cannot be accepted for each and every observation.
24. 24. Mode: Merits • It easy to calculate and understand • It is not affected by extreme observations • It can be calculated from a grouped frequency distribution. Demerits: • Mode is not rigidly defined. • As compared to mean, mode is affected to a great extent by the fluctuations of sampling.
25. 25. Measures of variability (or) Measures of Dispersion • • • • • • Range Standard Deviation Mean Deviation Quartile Deviation Variance Co-efficient of variation
26. 26. Range: Max. value – Min. value Standard Deviation: Root Mean Square Deviation
27. 27. PRACTICAL PROBLEMS • Measures of Central Tendency (Both ungrouped and grouped data) • Measures of Dispersion (Both ungrouped and grouped data)
28. 28. Correlation • describes the degree of relationship between two variables. • To define correlation is the average relationship between two or more variables. • When the change in one variable makes or causes a change in other variable then there is a correlation between these two variables. • These correlated variables can move in the same direction or they can move in opposite direction. • Not always there is a cause and effect relationship between the variables when there is a change; that might be due to uncertain change.
29. 29. Correlation (Contd…) • Simple Correlation is a correlation between two variables only; meaning the relationship between two variables. • Event correlation and simple event correlation are the types of correlations mainly used in the industry point of view.
30. 30. Types of Correlation (1) Positive Correlation (2) Negative Correlation (3) Perfectly Positive Correlation (4) Perfectly Negative Correlation (5) Zero Correlation (6) Linear Correlation
31. 31. Positive Correlation • When two variables move in the same direction then the correlation between these two variables is said to be Positive Correlation. • When the value of one variable increases, the value of other value also increases at the same rate. For example: the training and performance of employees in a company. Negative Correlation • In this type of correlation, the two variables move in the opposite direction. When the value of a variable increases, the value of the other variable decreases. For example: the relationship between price and demand.
32. 32. Perfect Positive Correlation When there is a change in one variable, and if there is equal proportion of change in the other variable say Y in the same direction, then these two variables are said to have a Perfect Positive Correlation. Perfectly Negative Correlation Between two variables X and Y, if the change in X causes the same amount of change in Y in equal proportion but in opposite direction, then this correlation is called as Perfectly Negative Correlation.
33. 33. Zero Correlation • When the two variables are independent and the change in one variable has no effect in other variable, then the correlation between these two variable is known as Zero Correlation. Linear Correlation • If the quantum of change in one variable has a ratio of change in the quantum of change in the other variable then it is known as Linear correlation.
34. 34. THANK YOU
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# Function (mathematics): Wikis
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# Encyclopedia
Graph of example function,
\begin{align}&\scriptstyle \\ &\textstyle f(x) = \frac{(4x^3-6x^2+1)\sqrt{x+1}}{3-x}\end{align}
Both the domain and the range in the picture are the set of real numbers between -1 and 1.5.
In mathematics, a function is a relation between a given set of elements called the domain and a set of elements called the codomain. The function associates each element in the domain with exactly one element in the codomain. The elements so related can be any kind of thing (words, objects, qualities) but are typically mathematical quantities, such as real numbers.
An example of a function with domain {A,B,C} and codomain {1,2,3} associates A with 1, B with 2, and C with 3. An example of a function with the real numbers as both its domain and codomain is the function f(x) = 2x, which associates every real number with the real number twice as big. In this case, we can write f(5) = 10.
There are many ways to describe or represent functions: a function may be described by a formula, by a plot or graph, by an algorithm that computes it, by arrows between objects, or by a description of its properties. Sometimes, a function is described through its relationship to other functions (for example, inverse functions). In applied disciplines, functions are frequently specified by tables of values or by formulae.
In a setting where outputs of functions are numbers, functions may be added and multiplied, yielding new functions. Collections of functions with certain properties, such as continuous functions and differentiable functions, usually closed under certain operations, are called function spaces and are studied as objects in their own right, in such disciplines as real analysis and complex analysis. An important operation on functions, which distinguishes them from numbers, is composition of functions. The composite function is obtained by using the output of one function as the input of another. This operation provides the theory of functions with its most powerful structure.
In pure mathematics, functions are defined using set theory, and there are theorems that show the existence of uncountably many different functions, most of which cannot be expressed with a formula or algorithm.
Functions are used in every area of mathematics. The area of mathematics that takes function as its primary object of study is called analysis.
## Overview
Because functions are so widely used, many traditions have grown up around their use. The symbol for the input to a function is often called the independent variable or argument and is often represented by the letter x or, if the input is a particular time, by the letter t. The symbol for the output is called the dependent variable or value and is often represented by the letter y. The function itself is most often called f, and thus the notation y = f(x) indicates that a function named f has an input named x and an output named y.
A function ƒ takes an input, x, and returns an output ƒ(x). One metaphor describes the function as a "machine" or "black box" that converts the input into the output.
The set of all permitted inputs to a given function is called the domain of the function. The set of all resulting outputs is called the image or range of the function. The image is often a subset of some larger set, called the codomain of a function. Thus, for example, the function f(x) = x2 could take as its domain the set of all real numbers, as its image the set of all non-negative real numbers, and as its codomain the set of all real numbers. In that case, we would describe f as a real-valued function of a real variable. Sometimes, especially in computer science, the term "range" refers to the codomain rather than the image, so care needs to be taken when using the word.
It is usual practice in mathematics to introduce functions with temporary names like ƒ. For example, ƒ(x) = 2x+1, implies ƒ(3) = 7; when a name for the function is not needed, the form y = 2x+1 may be used. If a function is often used, it may be given a more permanent name as, for example,
$\operatorname{Square}(x) = x^2 . \,\!$
Functions need not act on numbers: the domain and codomain of a function may be arbitrary sets. One example of a function that acts on non-numeric inputs takes English words as inputs and returns the first letter of the input word as output. Furthermore, functions need not be described by any expression, rule or algorithm: indeed, in some cases it may be impossible to define such a rule. For example, the association between inputs and outputs in a choice function will often lack any fixed rule, although each input element is still associated to one and only one output.
A function of two or more variables is considered in formal mathematics as having a domain consisting of ordered pairs or tuples of the argument values. For example Sum(x,y) = x+y operating on integers is the function Sum with a domain consisting of pairs of integers. Sum then has a domain consisting of elements like (3,4), a codomain of integers, and an association between the two which can be described by a set of ordered pairs like ((3,4), 7). Evaluating Sum(3,4) then gives the value 7 associated with the pair (3,4).
A family of objects indexed by a set is equivalent to a function. For example, the sequence 1, 1/2, 1/3, ..., 1/n, ... can be written as the ordered sequence <1/n> where n is a natural number, or as a function f(n) = 1/n from the set of natural numbers into the set of rational numbers.
Dually, a surjective function partitions its domain into disjoint sets indexed by the codomain. This partition is known as the kernel of the function, and the parts are called the fibers or level sets of the function at each element of the codomain. (A non-surjective function divides its domain into disjoint and possibly-empty subsets).
## Definition
One precise definition of a function is that it consists of an ordered triple of sets, which may be written as (X, Y, F). X is the domain of the function, Y is the codomain, and F is a set of ordered pairs. In each of these ordered pairs (a, b), the first element a is from the domain, the second element b is from the codomain, and every element in the domain is the first element in one and only one ordered pair. The set of all b is known as the image of the function. Some authors use the term "range" to mean the image, others to mean the codomain.
For example, the function defined by f(x) = x2 is rigorously considered to be three sets. The domain and codomain are the real numbers, and the ordered pairs include such pairs as (3, 9).
The notation ƒ:XY indicates that ƒ is a function with domain X and codomain Y.
In most practical situations, the domain and range are understood from context, and only the relationship between the input and output is given. Thus
$\left( \mathbb{R}, \mathbb{R}, \left\{ \left( x, x^2\right) : x \in \mathbb{R} \right\} \right)$
is usually written as
y = x2
The graph of a function is its set of ordered pairs. Such a set can be plotted on a pair of coordinate axes; for example, (3, 9) is the point of intersection of the lines x = 3 and y = 9.
A function is a special case of a more general mathematical concept, the relation, for which the restriction that each element of the domain appear as the first element in one and only one ordered pair is removed (or, in other words, the restriction that each input be associated to exactly one output). A relation is "single-valued" or "functional" when for each element of the domain set, the graph contains at most one ordered pair (and possibly none) with it as a first element. A relation is called "left-total" or simply "total" when for each element of the domain, the graph contains at least one ordered pair with it as a first element (and possibly more than one). A relation which is both left-total and single-valued is a function.
In some parts of mathematics, including recursion theory and functional analysis, it is convenient to study partial functions in which some values of the domain have no association in the graph; i.e. single-valued relations. For example, the function f such that f(x) = 1/x does not define a value for x = 0, and so is only a partial function from the real line to the real line. The term total function can be used to stress the fact that every element of the domain does appear as the first element of an ordered pair in the graph. In other parts of mathematics, relations which are not single-valued are similarly conflated with functions: these are known as multivalued functions, with the corresponding term single-valued function for ordinary functions.
Some authors (especially in set theory) define a function as simply its graph f, with the restriction that the graph should not contain two distinct ordered pairs with the same first element. Indeed, given such a graph, one can construct a suitable triple by taking the set of all first elements as the domain and the set of all second elements as the codomain: this automatically causes the function to be total and surjective (see below). However, most authors in advanced mathematics outside of set theory prefer the greater power of expression afforded by the triple of sets (X,Y,f).
Many operations in set theory—such as the power set—have the class of all sets as their domain, therefore, although they are informally described as functions, they do not fit the set-theoretical definition above outlined.
## Vocabulary
A specific input in a function is called an argument of the function. For each argument value x, the corresponding unique y in the codomain is called the function value at x, output of ƒ for an argument x, or the image of x under ƒ. The image of x may be written as ƒ(x) or as y.
The graph of a function ƒ is the set of all ordered pairs (x, ƒ(x)), for all x in the domain X. If X and Y are subsets of R, the real numbers, then this definition coincides with the familiar sense of "graph" as a picture or plot of the function, with the ordered pairs being the Cartesian coordinates of points.
A function can also be called a map or a mapping. Some authors, however, use the terms "function" and "map" to refer to different types of functions. Other specific types of functions include functionals and operators.
### Notation
Formal description of a function typically involves the function's name, its domain, its codomain, and a rule of correspondence. Thus we frequently see a two-part notation, an example being
\begin{align} f\colon \mathbb{N} &\to \mathbb{R} \ n &\mapsto \frac{n}{\pi} \end{align}
where the first part is read:
• "ƒ is a function from N to R" (one often writes informally "Let ƒ: XY" to mean "Let ƒ be a function from X to Y"), or
• "ƒ is a function on N into R", or
• "ƒ is an R-valued function of an N-valued variable",
and the second part is read:
• $n \,$ maps to $\frac{n}{\pi} \,\!$
Here the function named "ƒ" has the natural numbers as domain, the real numbers as codomain, and maps n to itself divided by π. Less formally, this long form might be abbreviated
$f(n) = \frac{n}{\pi} , \,\!$
where f(n) is read as "f as function of n" or "f of n". There is some loss of information: we no longer are explicitly given the domain N and codomain R.
It is common to omit the parentheses around the argument when there is little chance of confusion, thus: sin x; this is known as prefix notation. Writing the function after its argument, as in x ƒ, is known as postfix notation; for example, the factorial function is customarily written n!, even though its generalization, the gamma function, is written Γ(n). Parentheses are still used to resolve ambiguities and denote precedence, though in some formal settings the consistent use of either prefix or postfix notation eliminates the need for any parentheses.
### Functions with multiple inputs and outputs
The concept of function can be extended to an object that takes a combination of two (or more) argument values to a single result. This intuitive concept is formalized by a function whose domain is the Cartesian product of two or more sets.
For example, consider the function that associates two integers to their product: ƒ(x, y) = x·y. This function can be defined formally as having domain Z×Z , the set of all integer pairs; codomain Z; and, for graph, the set of all pairs ((x,y), x·y). Note that the first component of any such pair is itself a pair (of integers), while the second component is a single integer.
The function value of the pair (x,y) is ƒ((x,y)). However, it is customary to drop one set of parentheses and consider ƒ(x,y) a function of two variables, x and y. Functions of two variables may be plotted on the three-dimensional Cartesian as ordered triples of the form (x,y,f(x,y)).
The concept can still further be extended by considering a function that also produces output that is expressed as several variables. For example, consider the function swap(x, y) = (y, x) with domain R×R and codomain R×R as well. The pair (y, x) is a single value in the codomain seen as a Cartesian product.
#### Currying
An alternative approach to handling functions with multiple arguments is to transform them into a chain of functions that each takes a single argument. For instance, one can interpret Add(3,5) to mean "first produce a function that adds 3 to its argument, and then apply the 'Add 3' function to 5". This transformation is called currying: Add 3 is curry(Add) applied to 3. There is a bijection between the function spaces CA×B and (CB)A.
When working with curried functions it is customary to use prefix notation with function application considered left-associative, since juxtaposition of multiple arguments—as in (ƒ x y)—naturally maps to evaluation of a curried function. Conversely, the → and ⟼ symbols are considered to be right-associative, so that curried functions may be defined by a notation such as ƒ: ZZZ = xyx·y
#### Binary operations
The familiar binary operations of arithmetic, addition and multiplication, can be viewed as functions from R×R to R. This view is generalized in abstract algebra, where n-ary functions are used to model the operations of arbitrary algebraic structures. For example, an abstract group is defined as a set X and a function ƒ from X×X to X that satisfies certain properties.
Traditionally, addition and multiplication are written in the infix notation: x+y and x×y instead of +(x, y) and ×(x, y).
### Injective and surjective functions
Three important kinds of function are the injections (or one-to-one functions), which have the property that if ƒ(a) = ƒ(b) then a must equal b; the surjections (or onto functions), which have the property that for every y in the codomain there is an x in the domain such that ƒ(x) = y; and the bijections, which are both one-to-one and onto. This nomenclature was introduced by the Bourbaki group.
When the definition of a function by its graph only is used, since the codomain is not defined, the "surjection" must be accompanied with a statement about the set the function maps onto. For example, we might say ƒ maps onto the set of all real numbers.
### Function composition
A composite function g(f(x)) can be visualized as the combination of two "machines". The first takes input x and outputs f(x). The second takes f(x) and outputs g(f(x)).
The function composition of two or more functions takes the output of one or more functions as the input of others. The functions ƒ: X → Y and gY → Z can be composed by first applying ƒ to an argument x to obtain y = ƒ(x) and then applying g to y to obtain z = g(y). The composite function formed in this way from general ƒ and g may be written
\begin{align} g\circ f\colon X &\to Z \ x &\mapsto g(f(x)). \end{align}
This notation follows the form such that $(g\circ f)(x) = g(f(x))$.
The function on the right acts first and the function on the left acts second, reversing English reading order. We remember the order by reading the notation as "g of ƒ". The order is important, because rarely do we get the same result both ways. For example, suppose ƒ(x) = x2 and g(x) = x+1. Then g(ƒ(x)) = x2+1, while ƒ(g(x)) = (x+1)2, which is x2+2x+1, a different function.
In a similar way, the function given above by the formula y = 5x−20x3+16x5 can be obtained by composing several functions, namely the addition, negation, and multiplication of real numbers.
An alternative to the colon notation, convenient when functions are being composed, writes the function name above the arrow. For example, if ƒ is followed by g, where g produces the complex number eix, we may write
$\mathbb{N} \xrightarrow{f} \mathbb{R} \xrightarrow{g} \mathbb{C} . \,\!$
A more elaborate form of this is the commutative diagram.
### Identity function
The unique function over a set X that maps each element to itself is called the identity function for X, and typically denoted by idX. Each set has its own identity function, so the subscript cannot be omitted unless the set can be inferred from context. Under composition, an identity function is "neutral": if ƒ is any function from X to Y, then
\begin{align} f \circ \mathrm{id}_X &= f , \ \mathrm{id}_Y \circ f &= f . \end{align}
### Restrictions and extensions
Informally, a restriction of a function ƒ is the result of trimming its domain.
More precisely, if ƒ is a function from a X to Y, and S is any subset of X, the restriction of ƒ to S is the function ƒ|S from S to Y such that ƒ|S(s) = ƒ(s) for all s in S.
If g is a restriction of ƒ, then it is said that ƒ is an extension of g.
The overriding of f: XY by g: WY (also called overriding union) is an extension of g denoted as (fg): (XW) → Y. Its graph is the set-theoretical union of the graphs of g and f|X \ W. Thus, it relates any element of the domain of g to its image under g, and any other element of the domain of f to its image under f. Overriding is an associative operation; it has the empty function as an identity element. If f|XW and g|XW are pointwise equal (e.g. the domains of f and g are disjoint), then the union of f and g is defined and is equal to their overriding union. This definition agrees with the definition of union for binary relations.
### Inverse function
If ƒ is a function from X to Y then an inverse function for ƒ, denoted by ƒ−1, is a function in the opposite direction, from Y to X, with the property that a round trip (a composition) returns each element to itself. Not every function has an inverse; those that do are called invertible. The inverse function exists if and only if ƒ is a bijection.
As a simple example, if ƒ converts a temperature in degrees Celsius C to degrees Fahrenheit F, the function converting degrees Fahrenheit to degrees Celsius would be a suitable ƒ−1.
\begin{align} f(C) &= \frac {9}{5} C + 32 \ f^{-1}(F) &= \frac {5}{9} (F - 32) \end{align}
The notation for composition is similar to multiplication; in fact, sometimes it is denoted using juxtaposition, gƒ, without an intervening circle. With this analogy, identity functions are like the multiplicative identity, 1, and inverse functions are like reciprocals (hence the notation).
For functions which are injections or surjections, generalized inverse functions can be defined, called left and right inverses respectively. Left inverses map to the identity when composed to the left; right inverses when composed to the right.
### Image of a set
The concept of the image can be extended from the image of a point to the image of a set. If A is any subset of the domain, then ƒ(A) is the subset of im ƒ consisting of all images of elements of A. We say the ƒ(A) is the image of A under f.
Use of ƒ(A) to denote the image of a subset AX is consistent so long as no subset of the domain is also an element of the domain. In some fields (e.g. in set theory, where ordinals are also sets of ordinals) it is convenient or even necessary to distinguish the two concepts; the customary notation is ƒ[A] for the set { ƒ(x): x ∈ A }; some authors write ƒx instead of ƒ(x), and ƒ`A instead of ƒ[A].
Notice that the image of ƒ is the image ƒ(X) of its domain, and that the image of ƒ is a subset of its codomain.
#### Inverse image
The inverse image (or preimage, or more precisely, complete inverse image) of a subset B of the codomain Y under a function ƒ is the subset of the domain X defined by
$f^{-1}(B) = \{x \in X : f(x) \in B\}.$
So, for example, the preimage of {4, 9} under the squaring function is the set {−3,−2,2,3}.
In general, the preimage of a singleton set (a set with exactly one element) may contain any number of elements. For example, if ƒ(x) = 7, then the preimage of {5} is the empty set but the preimage of {7} is the entire domain. Thus the preimage of an element in the codomain is a subset of the domain. The usual convention about the preimage of an element is that ƒ−1(b) means ƒ−1({b}), i.e
$f^{-1}(b) = \{x \in X : f(x) = b\}.$
In the same way as for the image, some authors use square brackets to avoid confusion between the inverse image and the inverse function. Thus they would write ƒ−1[B] and ƒ−1[b] for the preimage of a set and a singleton.
The preimage of a singleton set is sometimes called a fiber. The term kernel can refer to a number of related concepts.
## Specifying a function
A function can be defined by any mathematical condition relating each argument to the corresponding output value. If the domain is finite, a function ƒ may be defined by simply tabulating all the arguments x and their corresponding function values ƒ(x). More commonly, a function is defined by a formula, or (more generally) an algorithm — a recipe that tells how to compute the value of ƒ(x) given any x in the domain.
There are many other ways of defining functions. Examples include piecewise definitions, induction or recursion, algebraic or analytic closure, limits, analytic continuation, infinite series, and as solutions to integral and differential equations. The lambda calculus provides a powerful and flexible syntax for defining and combining functions of several variables.
### Computability
Functions that send integers to integers, or finite strings to finite strings, can sometimes be defined by an algorithm, which gives a precise description of a set of steps for computing the output of the function from its input. Functions definable by an algorithm are called computable functions. For example, the Euclidean algorithm gives a precise process to compute the greatest common divisor of two positive integers. Many of the functions studied in the context of number theory are computable.
Fundamental results of computability theory show that there are functions that can be precisely defined but are not computable. Moreover, in the sense of cardinality, almost all functions from the integers to integers are not computable. The number of computable functions from integers to integers is countable, because the number of possible algorithms is. The number of all functions from integers to integers is higher: the same as the cardinality of the real numbers. Thus most functions from integers to integers are not computable. Specific examples of uncomputable functions are known, including the busy beaver function and functions related to the halting problem and other undecidable problems.
## Function spaces
The set of all functions from a set X to a set Y is denoted by XY, by [XY], or by YX.
The latter notation is motivated by the fact that, when X and Y are finite and of size |X| and |Y|, then the number of functions XY is |YX| = |Y||X|. This is an example of the convention from enumerative combinatorics that provides notations for sets based on their cardinalities. Other examples are the multiplication sign X×Y used for the Cartesian product, where |X×Y| = |X|·|Y|; the factorial sign X!, used for the set of permutations where |X!| = |X|!; and the binomial coefficient sign $\tbinom X n$, used for the set of n-element subsets where $|\tbinom X n | = \tbinom {|X|} n.$
If ƒ: XY, it may reasonably be concluded that ƒ ∈ [XY].
### Pointwise operations
If ƒ: X → R and gX → R are functions with a common domain of X and common codomain of a ring R, then the sum function ƒ + gX → R and the product function ƒ ⋅ gX → R can be defined as follows:
\begin{align} (f+g)(x) &= f(x)+g(x) , \ (f\cdot g)(x) &= f(x) \cdot g(x) , \end{align}
for all x in X.
This turns the set of all such functions into a ring. The binary operations in that ring have as domain ordered pairs of functions, and as codomain functions. This is an example of climbing up in abstraction, to functions of more complex types.
By taking some other algebraic structure A in the place of R, we can turn the set of all functions from X to A into an algebraic structure of the same type in an analogous way.
## Other properties
There are many other special classes of functions that are important to particular branches of mathematics, or particular applications. Here is a partial list:
bijection, injection and surjection, or individually: continuous differentiable, integrable
## History
### Functions prior to Leibniz
The history of the function concept in mathematics is described by J.P.Ponte (1992),[1] who writes:
Historically, some mathematicians can be regarded as having foreseen and come close to a modern formulation of the concept of function. Among them is Oresme (1323-1382) . . . In his theory, some general ideas about independent and dependent variable quantities seem to be present.
Ponte further notes that "The emergence of a notion of function as an individualized mathematical entity can be traced to the beginnings of infinitesimal calculus".
#### The notion of "function" in analysis
As a mathematical term, "function" was coined by Gottfried Leibniz, in a 1673 letter, to describe a quantity related to a curve, such as a curve's slope at a specific point.[2] [3] The functions Leibniz considered are today called differentiable functions. For this type of function, one can talk about limits and derivatives; both are measurements of the output or the change in the output as it depends on the input or the change in the input. Such functions are the basis of calculus.
Johann Bernoulli "by 1718, had come to regard a function as any expression made up of a variable and some constants" [4], and Leonhard Euler during the mid-18th century used the word to describe an expression or formula involving variables and constants e.g. x2+3x+2.[5]
Alexis Claude Clairaut (in approximately 1734) and Euler introduced the familiar notation " f(x) ".[6]
At first, the idea of a function was rather limited. Joseph Fourier, for example, claimed that every function had a Fourier series, something no mathematician would claim today. By broadening the definition of functions, mathematicians were able to study "strange" mathematical objects such as continuous functions that are nowhere differentiable. These functions were first thought to be only theoretical curiosities, and they were collectively called "monsters" as late as the turn of the 20th century. However, powerful techniques from functional analysis have shown that these functions are, in a precise sense, more common than differentiable functions. Such functions have since been applied to the modeling of physical phenomena such as Brownian motion.
During the 19th century, mathematicians started to formalize all the different branches of mathematics. Weierstrass advocated building calculus on arithmetic rather than on geometry, which favoured Euler's definition over Leibniz's (see arithmetization of analysis).
Dirichlet and Lobachevsky are traditionally credited with independently giving the modern "formal" definition of a function as a relation in which every first element has a unique second element. Eves asserts that "the student of mathematics usually meets the Dirichlet definition of function in his introductory course in calculus[7], but Dirichlet's claim to this formalization is disputed by Imre Lakatos:
There is no such definition in Dirichlet's works at all. But there is ample evidence that he had no idea of this concept. In his [1837], for instance, when he discusses piecewise continuous functions, he says that at points of discontinuity the function has two values: ...
(Proofs and Refutations, 151, Cambridge University Press 1976.)
In the context of "the Differential Calculus" George Boole defined (circa 1849) the notion of a function as follows:
"That quantity whose variation is uniform . . . is called the independent variable. That quantity whose variation is referred to the variation of the former is said to be a function of it. The Differential calculus enables us in every case to pass from the function to the limit. This it does by a certain Operation. But in the very Idea of an Operation is . . . the idea of an inverse operation. To effect that inverse operation in the present instance is the business of the Int[egral] Calculus."[8].
#### The logician's "function" prior to 1850
Logicians of this time were primarily involved with analyzing syllogisms (the 2000 year-old Aristotelian forms and otherwise), or as Augustus De Morgan (1847) stated it: "the examination of that part of reasoning which depends upon the manner in which inferences are formed, and the investigation of general maxims and rules for constructing arguments"[9]. At this time the notion of (logical) "function" is not explicit, but at least in the work of De Morgan and George Boole it is implied: we see abstraction of the argument forms, the introduction of variables, the introduction of a symbolic algebra with respect to these variables, and some of the notions of set theory.
De Morgan's 1847 "FORMAL LOGIC OR, The Calculus of Inference, Necessary and Probable" observes that "[a] logical truth depends upon the structure of the statement, and not upon the particular matters spoken of"; he wastes no time (preface page i) abstracting: "In the form of the proposition, the copula is made as absract as the terms". He immediately (p. 1) casts what he calls "the proposition" (present-day propositional function or relation) into a form such as "X is Y", where the symbols X, "is", and Y represent, respectively, the subject, copula, and predicate. While the word "function" does not appear, the notion of "abstraction" is there, "variables" are there, the notion of inclusion in his symbolism “all of the Δ is in the О” (p. 9) is there, and lastly a new symbolism for logical analysis of the notion of "relation" (he uses the word with respect to this example " X)Y " (p. 75) ) is there:
" A1 X)Y To take an X it is necessary to take a Y" [or To be an X it is necessary to be a Y]
" A1 Y)X To take an Y it is sufficient to take a X" [or To be a Y it is sufficient to be an X], etc.
In his 1848 The Nature of Logic Boole asserts that "logic . . . is in a more especial sense the science of reasoning by signs", and he briefly discusses the notions of "belonging to" and "class": "An individual may possess a great variety of attributes and thus belonging to a great variety of different classes" [10]. Like De Morgan he uses the notion of "variable" drawn from analysis; he gives an example of "represent[ing] the class oxen by x and that of horses by y and the conjunction and by the sign + . . . we might represent the aggregate class oxen and horses by x + y"[11].
### The logicians' "function" 1850-1950
Eves observes "that logicians have endeavored to push down further the starting level of the definitional development of mathematics and to derive the theory of sets, or classes, from a foundation in the logic of propositions and propositional functions"[12]. But by the late 1800's the logicians' research into the foundations of mathematics was undergoing a major split. The direction of the first group, the Logicists, can probably be summed up best by Bertrand Russell 1903:9 -- "to fulfil two objects, first, to show that all mathematics follows from symbolic logic, and secondly to discover, as far as possible, what are the principles of symbolic logic itself."
The second group of logicians, the set-theorists, emerged with Georg Cantor's "set theory" (1870-1890) but were driven forward partly as a result of Russell's discovery of a paradox that could be derived from Frege's conception of "function", but also as a reaction against Russell's proposed solution[13]. Zermelo's set-theoretic response was his 1908 Investigations in the foundations of set theory I -- the first axiomatic set theory; here too the notion of "propositional function" plays a role.
#### George Boole's The Laws of Thought 1854; John Venn's Symbolic Logic 1881
In his An Investigation into the laws of thought Boole now defined a function in terms of a symbol x as follows:
"8. Definition.-- Any algebraic expression involving symbol x is termed a function of x, and may be represented by the abbreviated form f(x)"[14]
Boole then used algebraic expressions to define both algebraic and logical notions, e.g. 1−x is logical NOT(x), xy is the logical AND(x,y), x + y is the logical OR(x, y), x(x+y) is xx+xy, and "the special law" xx = x2 = x[15].
In his 1881 Symbolic Logic Venn was using the words "logical function" and the contemporary symbolism ( x = f(y), y = f-1(x), cf page xxi) plus the circle-diagrams historically associated with Venn to describe "class relations"[16], the notions "'quantifying' our predicate", "propositions in respect of their extension", "the relation of inclusion and exclusion of two classes to one another", and "propositional function" (all on p. 10), the bar over a variable to indicate not-x (page 43), etc. Indeed he equated unequivocally the notion of "logical function" with "class" [modern "set"]: "... on the view adopted in this book, f(x) never stands for anything but a logical class. It may be a compound class aggregated of many simple classes; it may be a class indicated by certain inverse logical operations, it may be composed of two groups of classes equal to one another, or what is the same thing, their difference declared equal to zero, that is, a logical equation. But however composed or derived, f(x) with us will never be anything else than a general expression for such logical classes of things as may fairly find a place in ordinary Logic"[17].
#### Frege's Begriffsschrift 1879
Gottlob Frege's Begriffsschrift (1879) preceded Giuseppe Peano (1889), but Peano had no knowledge of Frege 1879 until after he had published his 1889[18]. Both writers strongly influenced Bertrand Russell (1903). Russell in turn influenced much of twentieth-century mathematics and logic through his Principia Mathematica (1913) jointly authored with Alfred North Whitehead.
At the outset Frege abandons the traditional "concepts subject and predicate", replacing them with argument and function respectively, which he believes "will stand the test of time. It is easy to see how regarding a content as a function of an argument leads to the formation of concepts. Furthermore, the demonstration of the connection between the meanings of the words if, and, not, or, there is, some, all, and so forth, deserves attention"[19].
Frege begins his discussion of "function" with an example: Begin with the expression[20] "Hydrogen is lighter than carbon dioxide". Now remove the sign for hydrogen (i.e. the word "hydrogen") and replace it with the sign for oxygen (i.e. the word "oxygen"); this makes a second statement. Do this again (using either statement) and substitute the sign for nitrogen (i.e. the word "nitrogen") and note that "This changes the meaning in such a way that "oxygen" or "nitrogen" enters into the relations in which "hydrogen" stood before"[21]. There are three statements:
• "Hydrogen is lighter than carbon dioxide."
• "Oxygen is lighter than carbon dioxide."
• "Nitrogen is lighter than carbon dioxide."
Now observe in all three a "stable component, representing the totality of [the] relations"[22]; call this the function, i.e.:
"... is lighter than carbon dioxide", is the function.
Frege calls the argument of the function "[t]he sign [e.g. hydrogen, oxygen, or nitrogen], regarded as replaceable by others that denotes the object standing in these relations"[23]. He notes that we could have derived the function as "Hydrogen is lighter than . . .." as well, with an argument position on the right; the exact observation is made by Peano (see more below). Finally, Frege allows for the case of two (or more arguments). For example, remove "carbon dioxide" to yield the invariant part (the function) as:
• "... is lighter than ... "
The one-argument function Frege generalizes into the form Φ(A) where A is the argument and Φ( ) represents the function, whereas the two-argument function he symbolizes as Ψ(A, B) with A and B the arguments and Ψ( , ) the function and cautions that "in general Ψ(A, B) differs from Ψ(B, A)". Using his unique symbolism he translates for the reader the following symbolism:
"We can read |--- Φ(A) as "A has the property Φ. |--- Ψ(A, B) can be translated by "B stands in the relation Ψ to A" or "B is a result of an application of the procedure Ψ to the object A"[24].
#### Peano 1889 The Principles of Arithmetic 1889
Peano defined the notion of "function" in a manner somewhat similar to Frege, but without the precision[25]. First Peano defines the sign "K means class, or aggregate of objects"[26], the objects of which satisfy three simple equality-conditions[27], a = a, (a = b) = (b = a), IF ((a = b) AND (b = c)) THEN (a = c). He then introduces φ, "a sign or an aggregate of signs such that if x is an object of the class s, the expression φx denotes a new object". Peano adds two conditions on these new objects: First, that the three equality-conditions hold for the objects φx; secondly, that "if x and y are objects of class s and if x = y, we assume it is possible to deduce φx = φy"[28]. Given all these conditions are met, φ is a "function presign". Likewise he identifies a "function postsign". For example if φ is the function presign a+, then φx yields a+x, or if φ is the function postsign +a then yields x+a[29].
#### Bertrand Russell's The Principles of Mathematics 1903
While the influence of Cantor and Peano was paramount[30], in Appendix A "The Logical and Arithmetical Doctrines of Frege" of The Principles of Mathematics, Russell arrives at a discussion of Frege's notion of function, "a point in which Frege's work is very important, and requires careful examination"[31]. In response to his 1902 exchange of letters with Frege about the contradiction he discovered in Frege's Begriffsschrift Russell tacked this section on at the last moment.
For Russell the bedeviling notion is that of "variable": "6. Mathematical propositions are not only characterized by the fact that they assert implications, but also by the fact that they contain variables. The notion of the variable is one of the most difficult with which logic has to deal. For the present, I openly wish to make it plain that there are variables in all mathematical propositions, even where at first sight they might seem to be absent. . . . We shall find always, in all mathematical propositions, that the words any or some occur; and these words are the marks of a variable and a formal implication"[32].
As expressed by Russell "the process of transforming constants in a proposition into variables leads to what is called generalization, and gives us, as it were, the formal essence of a proposition ... So long as any term in our proposition can be turned into a variable, our proposition can be generalized; and so long as this is possible, it is the business of mathematics to do it"[33]; these generalizations Russell named propositional functions"[34]. Indeed he cites and quotes from Frege's Begriffsschrift and presents a vivid example from Frege's 1891 Function und Begriff: That "the essence of the arithmetical function 2*x3+x is what is left when the x is taken away, i.e. in the above instance 2*( )3 + ( ). The argument x does not belong to the function but the two taken together make the whole"[35]. Russell agreed with Frege's notion of "function" in one sense: "He regards functions -- and in this I agree with him -- as more fundamental than predicates and relations" but Russell rejected Frege's "theory of subject and assertion"[36], in particular "he thinks that, if a term a occurs in a proposition, the proposition can always be analysed into a and an assertion about a"[37].
#### Evolution of Russell's notion of "function" 1908-1913
Russell would carry his ideas forward in his 1908 Mathematical logical as based on the theory of types and into his and Whitehead's 1910-1913 Principia Mathematica. By the time of Principia Mathematica Russell, like Frege, considered the propositional function fundamental: "Propositional functions are the fundamental kind from which the more usual kinds of function, such as “sin ‘’x’’ or log x or "the father of x" are derived. These derivative functions . . . are called “descriptive functions". The functions of propositions . . . are a particular case of propositional functions"[38].
Propositional functions: Because his terminology is different from the contemporary, the reader may be confused by Russell's "propositional function". An example may help. Russell writes a propositional function in its raw form, as e.g. φŷ: "ŷ is hurt". (Observe the circumflex or "hat" over the variable y). For our example we will assign just 4 values to the variable ŷ: "Bob", "This bird", "Emily the rabbit", and "y". Substitution of one of these values for variable ŷ yields a proposition; this proposition is called a "value" of the propositional function. In our example there are four values of the propositional function, e.g. "Bob is hurt", "This bird is hurt", "Emily the rabbit is hurt" and "y is hurt." A proposition, if it is significant -- i.e. if its truth is determinate -- will have a truth-value of truth or falsity. If a proposition's truth value is "truth" then the variable's value is said to satisfy the propositional function. Finally, per Russell's definition, "a class [set] is all objects satisfying some propositional function" (p. 23). Note the word "all'" -- this is how the contemporary notions of "For all ∀" and "there exists at least one instance ∃" enter the treatment (p. 15).
To continue the example: Suppose (from outside the mathematics/logic) one determines that the propositions "Bob is hurt" has a truth value of "falsity", "This bird is hurt" has a truth value of "truth", "Emily the rabbit is hurt" has an indeterminate truth value because "Emily the rabbit" doesn't exist, and "y is hurt" is ambiguous as to its truth value because the argument y itself is ambiguous. While the two propositions "Bob is hurt" and "This bird is hurt" are significant (both have truth values), only the value "This bird" of the variable ŷ satisfies' the propositional function φŷ: "ŷ is hurt". When one goes to form the class α: φŷ: "ŷ is hurt", only "This bird" is included, given the four values "Bob", "This bird", "Emily the rabbit" and "y" for variable ŷ and their respective truth-values: falsity, truth, indeterminate, ambiguous.
Russell defines functions of propositions with arguments, and truth-functions f(p)[39]. For example, suppose one were to form the "function of propositions with arguments" p1: "NOT(p) AND q" and assign its variables the values of p: "Bob is hurt" and q: "This bird is hurt". (We are restricted to the logical linkages NOT, AND, OR and IMPLIES, and we can only assign "significant" propositions to the variables p and q). Then the "function of propositions with arguments" is p1: NOT("Bob is hurt") AND "This bird is hurt"). To determine the truth value of this "function of propositions with arguments" we submit it to a "truth function", e.g. f(p1): f( NOT("Bob is hurt") AND "This bird is hurt") ) which yields a truth value of "truth".
The notion of a "many-one" functional relation": Russell first discusses the notion of "identity", then defines a descriptive function (pages 30ff) as the unique value ιx that satisfies the (2-variable) propositional function (i.e. "relation") φŷ.
N.B. The reader should be warned here that the order of the variables are reversed! y is the independent variable and x is the dependent variable, e.g. x = sin(y)[40].
Russell symbolizes the descriptive function as "the object standing in relation to y": R'y =DEF (ιx)(x R y). Russell repeats that "R'y is a function of y, but not a propositional function [sic]; we shall call it a descriptive function. All the ordinary functions of mathematics are of this kind. Thus in our notation "sin y" would be written " sin 'y ", and "sin" would stand for the relation which sin 'y has to y"[41].
#### Hardy 1908
Hardy 1908, pp. 26–28 defined a function as a relation between two variables x and y such that "to some values of x at any rate correspond values of y." He neither required the function to be defined for all values of x nor to associate each value of x to a single value of y. This broad definition of a function encompasses more relations than are ordinarily considered functions in contemporary mathematics.
### The Formalist's "function": David Hilbert's axiomatization of mathematics (1904-1927)
David Hilbert set himself the goal of "formalizing" classical mathematics "as a formal axiomatic theory, and this theory shall be proved to be consistent, i.e. free from contradiction" [42]. In his 1927 The Foundations of Mathematics Hilbert frames the notion of function in terms of the existence of an "object":
13. A(a) --> A(ε(A)) Here ε(A) stands for an object of which the proposition A(a) certainly holds if it holds of any object at all; let us call ε the logical ε-function"[43]. [The arrow indicates “implies”.]
Hilbert then illustrates the three ways how the ε-function is to be used, firstly as the "for all" and "there exists" notions, secondly to represent the "object of which [a proposition] holds", and lastly how to cast it into the choice function.
Recursion theory and computability: But the unexpected outcome of Hilbert's and his student Bernays's effort was failure; see Gödel's incompleteness theorems of 1931. At about the same time, in an effort to solve Hilbert's Entscheidungsproblem, mathematicians set about to define what was meant by an "effectively calculable function" (Alonzo Church 1936), i.e. "effective method" or "algorithm", that is, an explicit, step-by-step procedure that would succeed in computing a function. Various models for algorithms appeared, in rapid succession, including Church'slambda calculus (1936), Stephen Kleene's μ-recursive functions(1936) and Allan Turing's (1936-7) notion of replacing human "computers" with utterly-mechanical "computing machines" (see Turing machines). It was shown that all of these models could compute the same class of computable functions. Church's thesis holds that this class of functions exhausts all the number-theoretic functions that can be calculated by an algorithm. The outcomes of these efforts were vivid demonstrations that, in Turing's words, "there can be no general process for determining whether a given formula U of the functional calculus K [Principia Mathematica] is provable"[44]; see more at Independence (mathematical logic) and Computability theory.
### Development of the set-theoretic definition of "function"
Set theory began with the work of the logicians with the notion of "class" (modern "set") for example De Morgan (1847), Jevons (1880), Venn 1881, Frege 1879 and Peano (1889). It was given a push by Georg Cantor's attempt to define the infinite in set-theoretic treatment(1870-1890) and a subsequent discovery of an antinomy (contradiction, paradox) in this treatment (Cantor's paradox), by Russell's discovery (1902) of an antinomy in Frege's 1879 (Russell's paradox), by the discovery of more antinomies in the early 1900's (e.g. the 1897 Burali-Forti paradox and the 1905 Richard paradox), and by resistance to Russell's complex treatment of logic[45] and dislike of his axiom of reducibility[46] (1908, 1910-1913) that he proposed as a means to evade the antinomies.
In 1902 Russell sent a letter to Frege pointing out that Frege's 1879 Begriffsschrift allowed a function to be an argument of itself: "On the other hand, it may also be that the argument is determinate and the function indeterminate . . .."[47] From this unconstrained situation Russell was able to form a paradox:
"You state ... that a function, too, can act as the indeterminate element. This I formerly believed, but now this view seems doubtful to me because of the following contradiction. Let w be the predicate: to be a predicate that cannot be predicated of itself. Can w be predicated of itself?"[48]
Frege responded promptly that "Your discovery of the contradiction caused me the greatest surprise and, I would almost say, consternation, since it has shaken the basis on which I intended to build arithmetic"[49].
From this point forward development of the foundations of mathematics became an exercise in how to dodge "Russell's paradox", framed as it was in "the bare [set-theoretic] notions of set and element"[50].
#### Zermelo's set theory (1908) modified by Skolem (1922)
The notion of "function" appears as Zermelo's axiom III -- the Axiom of Separation (Axiom der Aussonderung). This axiom constrains us to use a propositional function Φ(x) to "separate" a subset MΦ from a previously formed set M:
"AXIOM III. (Axiom of separation). Whenever the propositional function Φ(x) is definite for all elements of a set M, M possesses a subset MΦ containing as elements precisely those elements x of M for which Φ(x) is true"[51].
As there is no universal set -- sets originate by way of Axiom II from elements of (non-set) domain B -- "this disposes of the Russell antinomy so far as we are concerned"[52]. But Zermelo's "definite criterion" is imprecise, and it will be fixed by Weyl, Fraenkel, Skolem and von Neumann[53].
In fact Skolem in his 1922 referred to this "definite criterion" or "property" as a "definite proposition":
"... a finite expression constructed from elementary propositions of the form a ε b or a = b by means of the five operations [logical conjunction, disjunction, negation, universal quantification, and existential quantification][54].
van Heijenoort summarizes:
"A property is definite in Skolem's sense if it is expressed . . . by a well-formed formula in the simple predicate calculus of first order in which the sole predicate constants are ε and possibly, =. ... Today an axiomatization of set theory is usually embedded in a logical calculus, and it is Weyl's and Skolem's approach to the formulation of the axiom of separation that is generally adopted[55].
In this quote the reader may observe a shift in terminology: nowhere is mentioned the notion of "propositional function", but rather one sees the words "formula", "predicate calculus", "predicate", and "logical calculus." This shift in terminology will be discussed more in the section covering the notion of "function" in contemporary set theory.
#### The Wiener–Hausdorff–Kuratowski "ordered pair" definition 1914–1921
The history of the notion of "ordered pair" is not clear. As noted above, Frege (1879) proposed an intuitive ordering in his definition of a two-argument function Ψ(A, B). Norbert Wiener in his 1914 (see below) observes that his own treatment essentially "revert(s) to Schröder's treatment of a relation as a class of ordered couples"[56]. Russell (1903) considered the definition of a relation (such as Ψ(A, B)) as a "class of couples" but rejected it:
"There is a temptation to regard a relation as definable in extension as a class of couples. This is the formal advantage that it avoids the necessity for the primitive proposition asserting that every couple has a relation holding between no other pairs of terms. But it is necessary to give sense to the couple, to distinguish the referent [domain] from the relatum [converse domain]: thus a couple becomes essentially distinct from a class of two terms, and must itself be introduced as a primitive idea. . . . It seems therefore more correct to take an intensional view of relations, and to identify them rather with class-concepts than with classes."[57]
By 1910-1913 and Principia Mathematica Russell had given up on the requirement for an intensional definition of a relation, stating that "mathematics is always concerned with extensions rather than intensions" and "Relations, like classes, are to be taken in extension"[58]. To demonstrate the notion of a relation in extension Russell now embraced the notion of ordered couple: "We may regard a relation ... as a class of couples ... the relation determined by φ(x, y) is the class of couples (x, y) for which φ(x, y) is true"[59]. In a footnote he clarified his notion and arrived at this definition:
"Such a couple has a sense, i.e. the couple (x, y) is different from the couple (y, x) unless x = y. We shall call it a "couple with sense," ... it may also be called an ordered couple[60].
But he goes on to say that he would not introduce the ordered couples further into his "symbolic treatment"; he proposes his "matrix" and his unpopular axiom of reducibility in their place.
An attempt to solve the problem of the antinomies led Russell to propose his "doctrine of types" in an appendix B of his 1903 The Principles of Mathematics[61]. In a few years he would refine this notion and propose in his 1908 The Theory of Types two axioms of reducibility, the purpose of which were to reduce (single-variable) propositional functions and (dual-variable) relations to a "lower" form (and ultimately into a completely extensional form); he and Alfred North Whitehead would carry this treatment over to Principia Mathematica 1910-1913 with a further refinement called "a matrix"[62]. The first axiom is *12.1; the second is *12.11. To quote Wiener the second axiom *12.11 "is involved only in the theory of relations"[63]. Both axioms, however, were met with skepticism and resistance; see more at Axiom of reducibility. By 1914 Norbert Wiener, using Whitehead and Russell's symbolism, eliminated axiom *12.11 (the "two-variable" (relational) version of the axiom of reducibility) by expressing a relation as an ordered pair "using the null set. At approximately the same time, Hausdorff (1914, p. 32) gave the definition of the ordered pair (a, b) as { {a,1}, {b, 2} }. A few years later Kuratowski (1921) offered a definition that has been widely used ever since, namely { {a, b}, {a} }"[64]. As noted by Suppes (1960) "This definition . . . was historically important in reducing the theory of relations to the theory of sets[65].
Observe that while Wiener "reduced" the relational *12.11 form of the axiom of reducibility he did not reduce nor otherwise change the propositional-function form *12.1; indeed he declared this "essential to the treatment of identity, descriptions, classes and relations"[66].
#### Schönfinkel's notion of "function" as a many-one "correspondence" 1924
Where exactly the general notion of "function" as a many-one relationship derives from is unclear. Russell in his 1920 Introduction to Mathematical Philosophy states that "It should be observed that all mathematical functions result form one-many [sic -- contemporary usage is many-one] relations . . . Functions in this sense are descriptive functions"[67]. A reasonable possibility is the Principia Mathematica notion of "descriptive function" -- R 'y =DEFx)(x R y): "the singular object that has a relation R to y". Whatever the case, by 1924, Moses Schonfinkel expressed the notion, claiming it to be "well known":
"As is well known, by function we mean in the simplest case a correspondence between the elements of some domain of quantities, the argument domain, and those of a domain of function values ... such that to each argument value there corresponds at most one function value"[68].
According to Willard Quine, Schönfinkel's 1924 "provide[s] for ... the whole sweep of abstract set theory. The crux of the matter is that Schönfinkel lets functions stand as arguments. ¶ For Schönfinkel, substantially as for Frege, classes are special sorts of functions. They are propositional functions, functions whose values are truth values. All functions, propositional and otherwise, are for Schönfinkel one-place functions"[69]. Remarkably, Schönfinkel reduces all mathematics to an extremely compact functional calculus consisting of only three functions: Constancy, fusion (i.e. composition), and mutual exclusivity. Quine notes that Haskell Curry (1958) carried this work forward "under the head of combinatory logic"[70].
#### von Neumann's set theory 1925
By 1925 Abraham Fraenkel (1922) and Thoralf Skolem (1922) had amended Zermelo's set theory of 1908. But von Neumann was not convinced that this axiomatization could not lead to the antinomies[71]. So he proposed his own theory, his 1925 An axiomatization of set theory. It explicitly contains a "contemporary", set-theoretic version of the notion of "function":
"[Unlike Zermelo's set theory] [w]e prefer, however, to axiomatize not "set" but "function". The latter notion certainly includes the former. (More precisely, the two notions are completely equivalent, since a function can be regarded as a set of pairs, and a set as a function that can take two values.)"[72].
His axiomatization creates two "domains of objects" called "arguments" (I-objects) and "functions" (II-objects); where they overlap are the "argument functions" (I-II objects). He introduces two "universal two-variable operations" -- (i) the operation [x, y]: ". . . read 'the value of the function x for the argument y) and (ii) the operation (x, y): ". . . (read 'the ordered pair x, y'") whose variables x and y must both be arguments and which itself produces an argument (x,y)". To clarify the function pair he notes that "Instead of f(x) we write [f,x] to indicate that f, just like x, is to be regarded as a variable in this procedure". And to avoid the "antinomies of naive set theory, in Russell's first of all . . . we must forgo treating certain functions as arguments"[73]. He adopts a notion from Zermelo to restrict these "certain functions"[74]
### Since 1950
#### Notion of "function" in contemporary set theory
Both axiomatic and naive forms of Zermelo's set theory as modified by Fraenkel (1922) and Skolem (1922) define "function" as a relation, define a relation as a set of ordered pairs, and define an ordered pair as a set of two "dissymetric" sets.
While the reader of Suppes (1960) Axiomatic Set Theory or Halmos (1970) Naive Set Theory will observe the use of function-symbolism in the axiom of separation, e.g. φ(x) (in Suppes) and S(x) (in Halmos), they will see no mention of "proposition" or even "first order predicate calculus". In their place will be the words and phrases "expressions of the object language", "atomic formulae", "primitive formulae", and "atomic sentences".
Kleene 1952 defines the words as follows: "In word languages, a proposition is expressed by a sentence. Then a 'predicate' is expressed by an incomplete sentence or sentence skeleton containing an open place. For example, "___ is a man" expresses a predicate ... The predicate is a propositional function of one variable. Predicates are often called 'properties' ... The predicate calculus will treat of the logic of predicates in this general sense of 'predicate', i.e. as propositional function"[75].
The reason for the disappearance of the words "propositional function" in e.g. Suppes (1960), and Halmos (1970), is explained by Alfred Tarski 1946 together with further explanation of the terminology:
"An expression such as x is an integer which contains variables and, on replacement of these variables by constants, becomes a sentence is called a SENTENTIAL [i.e. propositional cf his index] FUNCTION. But mathematicians, by the way, are not very fond of this expression, because they use the term "function" with a different meaning. ... sentential functions and sentences which are composed entirely of mathematical symbols (and not words of everyday languange), such as: x + y = 5 are usually referred to by mathematicians as FORMULAE. In place of "sentential function" we shall sometimes simply say "sentence" --- but only in cases where there is no danger of any misunderstanding"[76].
For his part Tarski calls the relational form of function a "FUNCTIONAL RELATION or simply a FUNCTION" [77]. After a discussion of this "functional relation" he asserts that:
"The concept of a function which we are considering now differs essentially from the concepts of a sentential [propositional] and of a designatory function .... Strictly speaking ... [these] do not belong to the domain of logic or mathematics; they denote certain categories of expressions which serve to compose logical and mathematical statements, but they do not denote things treated of in those statements... . The term "function" in its new sense, on the other hand, is an expression of a purely logical character; it designates a certain type of things dealt with in logic and mathematics."[78]
See more about "truth under an interpretation" at Alfred Tarski.
#### Further developments
The idea of structure-preserving functions, or homomorphisms, led to the abstract notion of morphism, the key concept of category theory. More recently, the concept of functor has been used as an analogue of a function in category theory.[79]
## References
### Notes
1. ^ Another short but useful history is found in Eves 1990 pages 234-235
2. ^ Thompson, S.P; Gardner, M; Calculus Made Easy. 1998. Page 10-11. ISBN 0312185480.
3. ^ Eves dates Leibniz's first use to the year 1694 and also similarly relates the usage to "as a term to denote any quantity connected with a curve, such as the coordinates of a point on the curve, the slope of the curve, and so on" (Eves 1990:234).
4. ^ Eves 1990:234
5. ^ Eves 1990:235
6. ^ Eves 1990:235
7. ^ Eves asserts that Dirichlet "arrived at the following formulation: "[The notion of] a variable is a symbol that represents any one of a set of numbers; if two variables x and y are so related that whenever a value is assigned to x there is automatically assigned, by some rule or correspondence, a value to y, then we say y is a (single-valued) function of x. The variable x . . . is called the independent variable and the variable y is called the dependent variable. The permissible values that x may assume constitute the domain of definition of the function, and the values taken on by y constitute the range of values of the function . . . it stresses the basic idea of a relationship between two sets of numbers" Eves 1990:235.
8. ^ Boole circa 1849 Elementary Treatise on Logic not mathematical including philosophy of mathematical reasoning in Grattan-Guiness and Bornet 1997:40
9. ^ De Morgan 1847:1
10. ^ Boole 1848 in Grattan-Guiness and Bornet 1997:1, 2
11. ^ Boole 1848 in Grattan-Guiness and Bornet 1997:6
12. ^ Eves 1990:222
13. ^ Some of this criticism is intense: see the introduction by Willard Quine preceding Russell 1908 Mathematical logic as based on the theory of types in van Heijenoort 1967:151. See also von Neumann's introduction to his 1925 Axiomatization of Set Theory in van Heijenoort 1967:395
14. ^ Boole 1854:86
15. ^ cf Boole 1854:31-34. Boole discusses this "special law" with its two algebraic roots x = 0 or 1, on page 37.
16. ^ Although he gives others credit, cf Venn 1881:6
17. ^ Venn 1881: 86-87
18. ^ cf van Heijenoort's introduction to Peano 1889 in van Heijenoort 1967. For most of his logical symbolism and notions of propositions Peano credits "many writers, especially Boole". In footnote 1 he credits Boole 1847, 1848, 1854, Schröder 1877, Peirce 1880, Jevons 1883, MacColl 1877, 1878, 1878a, 1880; cf van Heijenoort 1967:86).
19. ^ Frege 1879 in van Heijenoort 1967:7
20. ^ Frege's exact words are "expressed in our formula language" and "expression", cf Frege 1879 in van Heijenoort 1967:21-22.
21. ^ This example is from Frege 1879 in van Heijenoort 1967:21-22
22. ^ Frege 1879 in van Heijenoort 1967:21-22
23. ^ Frege cautions that the function will have "argument places" where the argument should be placed as distinct from other places where the same sign might appear. But he does not go deeper into how to signify these positions and Russell 1903 observes this.
24. ^ Gottlob Frege (1879) in van Heijenoort 1967:21-24
25. ^ "...Peano intends to cover much more ground than Frege does in his Begriffsschrift and his subsequent works, but he does not till that ground to any depth comparable to what Frege does in his self-allotted field", van Heijenoort 1967:85
26. ^ van Heijenoort 1967:89.
27. ^ van Heijenoort 1967:91.
28. ^ All symbols used here are from Peano 1889 in van Heijenoort 1967:91).
29. ^ cf van Heijenoort 1967:91
30. ^ "In Mathematics, my chief obligations, as is indeed evident, are to Georg Cantor and Professor Peano. If I had become acquainted sooner with the work of Professor Frege, I should have owed a great deal to him, but as it is I arrived independently at many results which he had already established", Russell 1903:viii. He also highlights Boole's 1854 Laws of Thought and Schröder's three volumes of "non-Peanesque methods" 1890, 1891, and 1895 cf Russell 1903:10
31. ^ Russell 1903:505
32. ^ Russell 1903:5-6
33. ^ Russell 1903:7
34. ^ Russell 1903:19
35. ^ Russell 1903:505
36. ^ Russell 1903:505
37. ^ Russell 1903:501
38. ^ Russell 1910-1913:15
39. ^ Whitehead and Russell 1910-1913:6, 8 respectively
40. ^ Something similar appears in Tarski 1946. Tarski refers to a "relational function" as a "ONE-MANY [sic!] or FUNCTIONAL RELATION or simply a FUNCTION". Tarski comments about this reversal of variables on page 99.
41. ^ Whitehead and Russell 1910-1913:31. This paper is important enough that van Heijenoort reprinted it as Whitehead and Russell 1910 Incomplete symbols: Descriptions with commentary by W. V. Quine in van Heijenoort 1967:216-223
42. ^ Kleene 1952:53
43. ^ Hilbert in van Heijenoort 1967:466
44. ^ Turing 1936-7 in Martin Davis The Undecidable 1965:145
45. ^ cf Kleene 1952:45
46. ^ "The nonprimitive and arbitrary character of this axiom drew forth severe criticism, and much of subsequent refinement of the logistic program lies in attempts to devise some method of avoiding the disliked axiom of reducibility" Eves 1990:268.
47. ^ Frege 1879 in van Heijenoort 1967:23
48. ^ Russell (1902) Letter to Frege in van Heijenoort 1967:124
49. ^ Frege (1902) Letter to Russell in van Heijenoort 1967:127
50. ^ van Heijenoort's commentary to Russell's Letter to Frege in van Heijenoort 1967:124
51. ^ The original uses an Old High German symbol in place of Φ cf Zermelo 1908a in van Heijenoort 1967:202
52. ^ Zermelo 1908a in van Heijenoort 1967:203
53. ^ cf van Heijenoort's commentary before Zermelo 1908 Investigations in the foundations of set theory I in van Heijenoort 1967:199
54. ^ Skolem 1922 in van Heijenoort 1967:292-293
55. ^ van Heijenoort's introduction to Abraham Fraenkel's The notion "definite" and the independence of the axiom of choice in van Heijenoort 1967:285.
56. ^ But Wiener offers no date or reference cf Wiener 1914 in van Heijenoort 1967:226
57. ^ Russell 1903:99
58. ^ both quotes from Whitehead and Russell 1913:26
59. ^ Whitehead and Russell 1913:26
60. ^ Whitehead and Russell 1913:26
61. ^ Russell 1903:523-529
62. ^ *12 The Hierarchy of Types and the axiom of Reducibility in Principia Mathematica 1913:161
63. ^ Wiener 1914 in van Heijenoort 1967:224
64. ^ commentary by van Heijenoort preceding Norbert Wiener's (1914) A simplification of the logic of relations in van Heijenoort 1967:224.
65. ^ Suppes 1960:32. This same point appears in van Heijenoort's commentary before Wiener (1914) in van Heijenoort 1967:224.
66. ^ Wiener 1914 in van Heijeoort 1967:224
67. ^ Russell 1920:46
68. ^ Schönfinkel (1924) On the building blocks of mathematical logic in van Heijenoort 1967:359
69. ^ commentary by W. V. Quine preceding Schönfinkel (1924) On the building blocks of mathematical logic in van Heijenoort 1967:356.
70. ^ cf Curry and Feys 1958; Quine in van Heijenoort 1967:357.
71. ^ von Neumann's critique of the history observes the split between the logicists (e.g. Russell et. al.) and the set-theorists (e.g. Zermelo et. al.) and the formalists (e.g. Hilbert), cf von Neumann 1925 in van Heijenoort 1967:394-396.
72. ^ von Neumann 1925 in van Heijenoort 1967:396
73. ^ All quotes from von Neumann 1925 in van Heijenoort 1967:397-398
74. ^ This notion is not easy to summarize; see more at van Heijenoort 1967:397.
75. ^ Kleene 1952:143-145
76. ^ Tarski 1946:5
77. ^ Tarski 1946:98
78. ^ Tarski 1946:102
79. ^ John C. Baez; James Dolan (1998). Categorification.
### Sources
• Anton, Howard (1980), Calculus with Analytical Geometry, Wiley, ISBN 978-0-471-03248-9
• Bartle, Robert G. (1976), The Elements of Real Analysis (2nd ed.), Wiley, ISBN 978-0-471-05464-1
• Husch, Lawrence S. (2001), Visual Calculus, University of Tennessee, retrieved 2007-09-27
• Katz, Robert (1964), Axiomatic Analysis, D. C. Heath and Company .
• Ponte, João Pedro (1992), "The history of the concept of function and some educational implications", The Mathematics Educator 3 (2): 3–8, ISSN 1062-9017
• Thomas, George B.; Finney, Ross L. (1995), Calculus and Analytic Geometry (9th ed.), Addison-Wesley, ISBN 978-0-201-53174-9
• Youschkevitch, A. P. (1976), "The concept of function up to the middle of the 19th century", Archive for History of Exact Sciences 16 (1): 37–85, doi:10.1007/BF00348305 .
• Monna, A. F. (1972), "The concept of function in the 19th and 20th centuries, in particular with regard to the discussions between Baire, Borel and Lebesgue", Archive for History of Exact Sciences 9 (1): 57–84, doi:10.1007/BF00348540 .
• Kleiner, Israel (1989), "Evolution of the Function Concept: A Brief Survey", The College Mathematics Journal 20 (4): 282–300, doi:10.2307/2686848 .
• Ruthing, D. (1984), "Some definitions of the concept of function from Bernoulli, Joh. to Bourbaki, N.", Mathematical Intelligencer 6 (4): 72–77 .
• Dubinsky, Ed; Harel, Guershon (1992), The Concept of Function: Aspects of Epistemology and Pedagogy, Mathematical Association of America, ISBN 0883850818 .
• Malik, M. A. (1980), "Historical and pedagogical aspects of the definition of function", International Journal of Mathematical Education in Science and Technology 11 (4): 489–492, doi:10.1080/0020739800110404 .
### Sources for History section
• Boole, George (1854), An Investigation into the Laws of Thought on which are founded the Laws of Thought and Probabilies", Walton and Marberly, London UK; Macmillian and Company, Cambridge UK. Republished as a googlebook.
• Eves, Howard. (1990), Fundations and Fundamental Concepts of Mathematics: Third Edition, Dover Publications, Inc. Mineola, NY, ISBN 0-486-69609-X (pbk)
• Frege, Gottlob. (1879), Begriffsschrift: eine der arithmetischen nachgebildete Formelsprache des reinen Denkens, Halle
• Grattan-Guinness, Ivor and Bornet, Gérard (1997), George Boole: Selected Manuscripts on Logic and its Philosophy, Springer-Verlag, Berlin, ISBN 3-7643-5456-9 (Berlin...)
• Halmos, Paul R. (1970) Naive Set Theory, Springer-Verlag, New York, ISBN: 0-387-90092-6.
• Hardy, Godfrey Harold (1908), A Course of Pure Mathematics, Cambridge University Press (published 1993), ISBN 978-0-521-09227-2
• Reichenbach, Hans (1947) Elements of Symbolic Logic, Dover Publishing Inc., New York NY, ISBN: 0-486-24004-5.
• Russell, Bertrand (1903) The Principles of Mathematics: Vol. 1, Cambridge at the University Press, Cambridge, UK, republished as a googlebook.
• Russell, Bertrand (1920) Introduction to Mathematical Philosophy (second edition), Dover Publishing Inc., New York NY, ISBN: 0-486-27724-0 (pbk).
• Suppes, Patrick (1960) Axiomatic Set Theory, Dover Publications, Inc, New York NY, ISBN 0-486-61630-4. cf his Chapter 1 Introduction.
• Tarski, Alfred (1946) Introduction to Logic and to the Methodolgy of Deductive Sciences, republished 1195 by Dover Publications, Inc., New York, NY ISBN 0-486-28462-x
• Venn, John (1881) Symbolic Logic, Macmillian and Co., London UK. Republished as a googlebook.
• van Heijenoort, Jean (1967, 3rd printing 1976), From Frege to Godel: A Source Book in Mathematical Logic, 1879-1931, Harvard University Press, Cambridge, MA, ISBN: 0-674-32449-8 (pbk)
• Gottlob Frege (1879) Begriffsschrift, a formula language, modeled upon that of arithmetic, for pure thought with commentary by van Heijenoort, pages 1-82
• Giuseppe Peano (1889) The principles of arithmetic, presented by a new method with commentary by van Heijenoort, pages 83-97
• Bertrand Russell (1902) Letter to Frege with commentary by van Heijenoort, pages 124-125. Wherein Russell announces his discovery of a "paradox" in Frege's work.
• Gottlob Frege (1902) Letter to Russell with commentary by van Heijenoort, pages 126-128.
• David Hilbert (1904) On the foundations of logic and arithmetic, with commentary by van Heijenoort, pages 129-138.
• Jules Richard (1905) The principles of mathematics and the problem of sets, with commentary by van Heijenoort, pages 142-144. The Richard paradox.
• Bertrand Russell (1908a) Mathematical logic as based on the theory of types, with commentary by Willard Quine, pages 150-182.
• Ernst Zermelo (1908) A new proof of the possibility of a well-ordering, with commentary by van Heijenoort, pages 183-198. Wherein Zermelo rales against Poincaré's (and therefore Russell's) notion of impredicative definition.
• Ernst Zermelo (1908a) Investigations in the foundations of set theory I, with commentary by van Heijenoort, pages 199-215. Wherein Zermelo attempts to solve Russell's paradox by structuring his axioms to restrict the universal domain B (from which objects and sets are pulled by definite properties) so that it itself cannot be a set, i.e. his axioms disallow a universal set.
• Norbert Wiener (1914) A simplification of the logic of relations, with commentary by van Heijenoort, pages 224-227
• Thoralf Skolem (1922) Some remarks on axiomatized set theory, with commentary by van Heijenoort, pages 290-301. Wherein Skolem defines Zermelo's vague "definite property".
• Moses Schönfinkel (1924) On the building blocks of mathematical logic, with commentary by Willard Quine, pages 355-366. The start of combinatory logic.
• John von Neumann (1925) An axiomatization of set theory, with commentary by van Heijenoort , pages 393-413. Wherein von Neumann creates "classes" as distinct from "sets" (the "classes" are Zermelo's "definite properties"), and now there is a universal set, etc.
• David Hilbert (1927) The foundations of mathematics by van Heijenoort, with commentary, pages 464-479.
• Whitehead, Alfred North and Russell, Bertrand (1913, 1962 edition), Principia Mathematica to *56, Cambridge at the University Press, London UK, no ISBN or US card catalog number.
# Simple English
File:Function
A function ƒ takes an input, x, and returns an output ƒ(x). One metaphor describes the function as a "machine" or "black box" that converts the input into the output.
In mathematics, a function is the idea that one quantity (called the input) completely determines another quantity, often called the output. Modern mathematics defines functions using sets. A function can then be seen as a rule. With this rule, each element of the set of inputs will be assigned one element of the set of outputs. Even though the underlying structures are sets, mathematicians talk about domain and codomain of a function. Each element of the domain will be associated with one element of the codomain.
An example of an elementary function that acts on the real numbers as both the domain and the codomain could be f(x) = 2x. Every real number will be assigned a real number that is twice as big. In this case, f(5) = 10.
Mathematics knows other objects than numbers. While elementary functions may act on numbers, functions can also be maps. They may act on algebraic structures like groups, or geometric ones such as manifolds. A function that acts on the sets {A,B,C} and {1,2,3} and that associates A with 1, B with 2, and C with 3 is an example of a non-elementary function.
It is sometimes said that $f\left(x\right)$ is the image of $x$, and $x$ is the pre-image of $f\left(x\right)$. Also, it is said that the object $x$ is transformed or mapped into $f\left(x\right)$. A function is another way of representing y in formulae like y = x+2, which would be f(x) = x+2
## History
The first to talk about functions was Leibnitz in 1694. At first, a function was seen as a rule of changing a number into a different one. Examples of such functions are $y = x^2$ or $f\left(x\right)=\sin x$. Sometimes it was allowed that a function had more than one result value. An example of such a function is the square root.
This was used in schools until the second half of the 20th century. In the 19th century, Weierstraß, Dedekind and others worked on the theory of limits. When they did they discovered that the limit of infinite sequences are sometimes not continuous and that they can sometimes not be expressed as a closed formula, with a finite number of rules. This meant that the concept of function had to be extended.
## Example
Here is an example of how functions work:
$f\left(x\right)=x+1$ is a formula for a function. The person is doing something to $x$ that will make it a different number. In this example, the person is adding 1 to the unknown number $x$. If he sets x = 4, the function becomes $f\left(4\right)=4+1$. This means that $f\left(x\right)=5$ when $x$ equals 4.
Many functions are more complex than this example but the pattern is always the same. For every input, $x$, the person gets a result, $f\left(x\right)$.
## Letters used
When people write functions, they often write f(x) but the f can also be written with any letter in the alphabet and sometimes with greek letters. This can look confusing but just means they are different to another function in the same working out or formulae or can mean that they are special functions like the Riemann-Zeta function which is spelt as ζ(s). Moreover, g(x) is often used for the first reason but not always.
## Types of functions
There are other functions, like trigonometric functions. In mathematical language, a function can be defined as a map from a non-empty set (called a domain) to a non-empty set (called a co-domain) such that for every element in the domain there is a single corresponding element in the co-domain.
Inverse functions are where the action of a normal function are reversed. The symbol for this is is f-1(x). For example think of f(x)=x2, this reversed is f-1(x)=√x.
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• CBSE Class 11 Physics Notes
# Types of Collisions
A body at rest continues to be at rest without any application of external force. However, when a force is applied to an object, it tends to move in the direction of the applied force, and work is then said to be done by the force. Work done by the force on any object is mathematically equivalent to the product of the force and the displacement of the object which occurs in the direction of the force.
Work done by the force is dependent only on the initial and final positions of the body, that is the displacement, and not on the actual path followed between initial and final positions. Therefore, when an object gets displaced through a small distance, say s, under the influence of a constant force F, the work done by the force is equivalent to,
W = F × s = F s cos θ
where θ is the smaller angle between F and s.
The S.I unit of work done is Joule (J). It is a scalar quantity, which has magnitude only and no direction. Its dimensional formula is [ML2T-2]
Work done by a force on any object is equivalent to zero, if
1. The body is not displaced actually, i.e., s = 0.
2. The body is displaced perpendicular to the direction of force, i.e. the angle between the force and displacement is, θ = 90°
Energy is the fundamental form of living for all living beings. The Sun in this universe is considered as the elemental form of energy on the planet of Earth. Energy refers to the ability of any object to do work. Energy is required to perform work. It is a qualitative property of a body. The law of conservation of energy states that energy can’t be created nor destroyed. It can just be transferred from one medium to another.
The S.I unit of energy is Joule (J). It is a scalar quantity, which has magnitude only and no direction. Its dimensional formula is [ML2T-2]
Power is referred to the time rate of work done by an object. Mathematically,
Power = Rate of doing work = Work Done / Time Taken
Therefore, when an object gets displaced through a small distance, say s, under the influence of a constant force F, the work done by the power is equivalent to,
P = W / T = F × s / T
But, s / t = v, velocity of the body.
Therefore,
P = F × v = F v cos θ
where θ is the smaller angle between F and v.
The S.I unit of energy is Watt (W). It is a scalar quantity, which has magnitude only and no direction since it is only a ratio of two scalar quantities. Its dimensional formula is [ML2T-3].
### What is Collision?
A collision is an event where a strong force acts for a very short span of time between two or more bodies. A collision is an isolated event. The energy and momentum of the interacting particles undergo a change as a result of the collision. The collision may occur by actual physical contact of the involved bodies, for instance, the collision between two billiard balls or a ball and bat. There may be collisions where there is no actual physical contact, for instance, the collision of the alpha particles by a nucleus.
Any collision is guided by three distinct identifiable stages, that is, before, during, and after. Before the collision, the interacting forces between the particles are zero, since the particles are independent. Also, after the collision, the force again becomes zero. During the collision, the particles come in contact with each other, therefore, the force of interaction becomes very large. The motion of the bodies is guided by the dominating forces. Since, in most of the practical cases, the magnitude of the interacting force is unknown, therefore, Newton’s second law of motion can’t be used in such cases. The initial and final velocities can be computed using the law of conservation of momentum.
For instance, consider two bodies with masses m1 and m2, moving with velocities u1 and u2. They undergo a collision due to the application of an external force Fext for a small interval of time, and then the final velocities become v1 and v2
Conservation of Momentum and Energy conservation during a Collision
According to the fundamental laws of physics, certain attributes are applicable for any types of collision :
• Momentum conservation: The collision takes place for a very small interval of time, and during this period the average impulsive force causing the collision is significantly larger than the external force acting on the system. Therefore, during a collision, the application of external forces, such as frictional or gravitational forces are not considered into account. This impulsive force is internal in nature, therefore, the total momentum of the system remains constant for all practical purposes. Therefore, it remains conserved throughout the system.
• Energy conservation: According to the law of conservation of energy, energy can neither be destroyed nor created. It can only be transferred from one medium to another. Therefore, the total energy during a collision always remains conserved. Total energy includes all the plausible forms of energy created and destroyed during a collision, such as mechanical energy, internal energy, excitation energy as well as mass-energy.
### Types of collision
During the collision, the interacting bodies come in immediate contact with each other due to which they exert forces. This activity takes place for a very short period of time. There are two types of collisions, namely :
• On the basis of conservation of kinetic energy
1. Perfectly elastic collision
2. Inelastic collision
3. Perfectly inelastic collision
• On the basis of the direction of colliding bodies
2. Oblique collision
Let us assume two bodies of masses m1 and m2 moving with initial velocities u1 and u2 in the same direction and they collide such that after collision their final velocities are v1 and v2 respectively.
According to law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2 …………(1)
⇒ m1(u1 – v1) = m2(v2 – u2) ………..(2)
According to law of conservation of kinetic energy, we have,
……..(3)
⇒ ………(4)
Dividing equation (4) by equation (2)
v1 + u1 = v2+ u2 ……….(5)
⇒ u1 – u2 = v2 – v1 ………..(6)
We know,
Relative velocity of approach = Relative velocity of separation
Deriving from equation (5) we get
v2 = v1 + u1– u2
Substituting this value of v2 in equation (i) and rearranging we get
…….(7)
Similarly,
……….(8)
### Perfectly Elastic Oblique Collision
Let us assume two bodies to be in motion,
By the law of conservation of momentum,
We have,
Along x-axis,
m1u2 + m2u2 = m1v1 cosθ + m2v2 cosÏ• ……..(1)
Along y-axis,
0 = m1v1 sinθ – m2v2 sinÏ• ………(2)
We know,
By the law of conservation of kinetic energy :
……..(3)
### Solved Problems
Problem 1: Let us assume that, s small balls, each of mass m hit a surface elastically each second with a velocity u m/s. Calculate the force witnessed by the surface.
Solution:
Since, there is a rebound of the ball with unmodified velocity, therefore,
The change in velocity = 2u
Mass colliding with the surface = sm
Force experienced by the surface is given by,
Problem 2: Initially, a car of mass 400 kg traveling with a speed of 72 kmph accidentally crashes into a car of mass 4000 kg. Before the collision, the truck was traveling at a speed of 9 kmph, in the same direction as that of the car. As a result of the collision, the car retorts back with a speed of 18 kmph. Calculate the velocity with which the truck is moving after the collision.
Solution:
By the law of conservation of linear momentum, we have,
The summation of initial momentum is equivalent to the summation of final momentum.
m1u1 + m1u2 = m1v1 + m2v2
⇒ 400 × 72 + 4000 × 9 = 400 × (-18) + 4000 × v2
⇒ v2 = 18 km/h
Therefore, the final velocity of the car is 18 km/h.
Problem 3: A crate is kept at the origin of mass equivalent to 5 kg. A force of 10 N is applied to it to simulate its movement at an angle of 60° with the x-axis. As a result of this displacement, the crate moves by 4 m. What is the work done by the application of the force?
Solution:
Work done is given by the dot product of force and displacement, that is,
F.s = F s cos θ
=10 × 4 × cos 60°
= 20 J
Problem 4: A crate is kept on a table of mass equivalent to 1 kg. It is dragged along the horizontal table through the length of 1 m by an external force of 8 N. It is then subjected to a vertical height of 2 m. Compute the net work done.
Solution:
Work done to displace the crate horizontally , W
= F × s
= 8 × 1 = 8 J
Work done to raise the crate vertically, Wv = F × s
= m x g x h
= 1 × 10 × 2 = 20 J
Therefore,
Net work done on the crate is WH + W
= 8 +20 = 28 J
Problem 5: A horizontal force of 5 N is required to maintain a velocity of 2 m/s for a block of 10 kg mass sliding over a rough surface. The work done by this force in one minute is?
Solution:
Mathematically,
Work done by an object = Force × displacement
⇒ W = F × s
Since, we know,
s = v x t
= F × v × t
= 5 × 2 × 60
= 600 J
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# 2021 Fall AMC 10B Problems/Problem 23
## Problem
Each of the $5{ }$ sides and the $5{ }$ diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color?
$(\textbf{A})\: \frac23\qquad(\textbf{B}) \: \frac{105}{128}\qquad(\textbf{C}) \: \frac{125}{128}\qquad(\textbf{D}) \: \frac{253}{256}\qquad(\textbf{E}) \: 1$
## Solution 1
Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals.
$\textbf{Case 1}$: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection.
$\textbf{Case 2}$: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either.
$\textbf{Case 3}$: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are ${4\choose2}$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6\cdot 2 = 12$ ways to color the pentagon so that no such triangle has the same color for all of its sides.
These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}$
~KingRavi
## Solution 2 (Ramsey's Theorem)
This problem is related to a special case of Ramsey's Theorem, R(3, 3) = 6. Suppose we color every edge of a $6$ vertex complete graph $(K_6)$ with $2$ colors, there must exist a $3$ vertex complete graph $(K_3)$ with all it's edges in the same color. That is, $K_6$ with edges in $2$ colors contains a monochromatic $K_3$. For $K_5$ with edges in $2$ colors, a monochromatic $K_3$ does not always exist.
This is a problem about the probability of a monochromatic $K_3$ exist in a $5$ vertex complete graph $K_5$ with edges in $2$ colors.
Choose a vertex, it has $4$ edges.
$\textbf{Case 1}$: When $3$ or more edges are the same color, there must exist a monochromatic $K_3$. Suppose the color is red, as shown below.
There is only $1$ way to color all the edges in the same color. There is $\binom{4}{3} = 4$ ways to color $3$ edges in the same color. There are $2$ colors. The probability of $3$ or more edges the same color is $\frac{(1 + 4) \cdot 2}{2^4} = \frac{5}{8}$. So the probability of containing a monochromatic $K_3$ is $\frac{5}{8}$.
$\textbf{Case 2}$: When $2$ edges are the same color, graphs that does not contain a monochromatic $K_3$ can exist. The following diagram shows steps to obtain graphs that does not contain a monochromatic $K_3$.
There are $\binom{4}{2} = 6$ ways to choose $2$ edges with the same color. For the other $4$ vertices there are $\binom{4}{2} = 6$ edges among them, there are $2^6 = 64$ ways to color the edges. There are only $2$ cases without a monochromatic $K_3$.
So the probability without monochromatic $K_3$ is $\frac{2}{64} = \frac{1}{32}$.
The probability with monochromatic $K_3$ is $1 - \frac{1}{32} = \frac{31}{32}$.
From case 1 and case 2, the probability with monochromatic $K_3$ is $\frac{5}{8} + \left( 1 - \frac{5}{8} \right) \cdot \frac{31}{32} = \boxed{(\textbf{D}) \frac{253}{256}}$
## Solution 3 (Elementary Casework)
We can break this problem down into cases based on the distribution of the two colors throughout the $5$ sides(not diagonals) of the pentagon. This idea comes from the fact that if you play around with the sides for a while and their colors, you will see that the interior diagonals actually depend largely on the colors of the five exterior sides. The total number of combinations is $2^{10} = 1024$. We will be counting the number of arrangements in which there are no triangles with all $3$ sides the same color, and we can subtract from the total(complementary counting).
• Exterior sides are the five sides of the regular pentagon
• Interior Diagonals are the five diagonals inside the pentagon made by the connection of two non-adjacent vertices
Case $1$: The five exterior sides are all blue or all red.
There are $2$ options for the color given in the title of this case, and for each color, only one option is given for the exterior(All $5$ sides the same color; either Red or Blue). Once you color the five sides a single color, it is simple to notice that all the $5$ interior diagonals must be in the other color which the sides are not. To make it more clear, if all of the $5$ exterior sides are blue, all of the $5$ interior diagonals must be red, and if all of the $5$ exterior sides are red, all of the $5$ interior diagonals must be blue. This gives us a total of $2$ (Choices of Colors) $\cdot 1$ (Configuration for particular color existing on all 5 exterior edges) $= 2$ ways.
Case $2$: Four of the five exterior sides are in one color, while the remaining one exterior side is in the other color.
There are $2$ ways to choose which color occupies four exterior sides and which color occupies the remaining one. Either we can have four red and one blue, or four blue and one red, which are the $2$ ways mentioned above. When we calculated the total number of combinations, we took into account that each segment could either be red or blue which gives $2$ choices for each of $10$ segments yielding $2^{10}$. What you must understand is that when we calculated this total, we did not account for any rotations or reflections or any other symmetry. For the same reason, we must not account for any symmetry such as rotations or reflections when calculating the number of arrangements where there is not a triangle consisting of three sides of the same color. There are $5$ ways to arrange the one blue and four reds or one red and four blues on the $5$ exterior sides(draw it out to see), and once you start playing with our condition, that no triangle may have all three sides in the same color, you will see that this case actually yields zero solutions as there comes a triangle in the middle consisting of all three sides of the same color. Hence this case yields $0$ ways.
Case $3$: Three of the five exterior sides in one color and the remaining two in another color.
There are two sub-cases, one in which the two exterior sides which are colored differently from the other three are adjacent, and the other case in which they are separated by one other exterior side.
Subcase $1$: The case in which the two exterior sides colored differently from the other three are adjacent.
If you draw the five exterior sides with two colored in one color and the other three colored in a different color, you will see, that there are absolutely no ways to color the interior diagonals such that no triangle with all three sides the same color exists. This subcase yields $0$ ways.
Subcase $2$: The case in which the two exterior sides colored differently from the other three are separated by another exterior side. There are $5$ arrangements for the exterior (draw and see for yourself) and the colors can be swapped with each other and since each of the exterior configurations force a particular interior configuration, this yields $2 \cdot 5=10$ ways.
In total, there are $10+2=12$ ways, such that no triangle has all three sides in the same color. This yields $1024-12=1012$ ways such that there is a triangle such that it has all three sides in the same color. Thus, the probability is: $\frac{1012}{2^{10}} = \boxed{(\textbf{D}) \frac{253}{256}}$ ~Shiva Kannan
~Interstigation
~savannahsolver
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# ધોરણ ૬ એકમ ૧ સંખ્યા પરિચય MCQ TEST STD 6 MATHS UNIT 1
કવીઝ : નિકુંજકુમાર સવાણી
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## Access NCERT Solutions for Class 6 Chapter 1: Knowing Our Numbers
Exercise 1.1 PAGE NO: 12
1. Fill in the blanks:
(a) 1 lakh = ………….. ten thousand.
(b) 1 million = ………… hundred thousand.
(c) 1 crore = ………… ten lakh.
(d) 1 crore = ………… million.
(e) 1 million = ………… lakh.
Solutions:
(a) 1 lakh = 10 ten thousand
= 1,00,000
(b) 1 million = 10 hundred thousand
= 10,00,000
(c) 1 crore = 10 ten lakh
= 1,00,00,000
(d) 1 crore = 10 million
= 1,00,00,000
(e) 1 million = 10 lakh
= 1,000,000
2. Place commas correctly and write the numerals:
(a) Seventy three lakh seventy five thousand three hundred seven.
(b) Nine crore five lakh forty one.
(c) Seven crore fifty two lakh twenty one thousand three hundred two.
(d) Fifty eight million four hundred twenty three thousand two hundred two.
(e) Twenty three lakh thirty thousand ten.
Solutions:
(a) The numeral of seventy three lakh seventy five thousand three hundred seven is 73,75,307
(b) The numeral of nine crore five lakh forty one is 9,05,00,041
(c) The numeral of seven crore fifty two lakh twenty one thousand three hundred two is 7,52,21,302
(d) The numeral of fifty eight million four hundred twenty three thousand two hundred two is 5,84,23,202
(e) The numeral of twenty three lakh thirty thousand ten is 23,30,010
3. Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701
Solutions:
(a) 87595762 – Eight crore seventy five lakh ninety five thousand seven hundred sixty two
(b) 8546283 – Eighty five lakh forty six thousand two hundred eighty three
(c) 99900046 – Nine crore ninety nine lakh forty six
(d) 98432701 – Nine crore eighty four lakh thirty two thousand seven hundred one
4. Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831
Solutions:
(a) 78921092 – Seventy eight million nine hundred twenty one thousand ninety two
(b) 7452283 – Seven million four hundred fifty-two thousand two hundred eighty three
(c) 99985102 – Ninety-nine million nine hundred eighty five thousand one hundred two
(d) 48049831 – Forty-eight million forty-nine thousand eight hundred thirty-one
Exercise 1.2 PAGE NO: 16
1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.
Solutions:
Number of tickets sold on 1st day = 1094
Number of tickets sold on 2nd day = 1812
Number of tickets sold on 3rd day = 2050
Number of tickets sold on 4th day = 2751
Hence, number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707 tickets
2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?
Solutions:
Shekhar scored = 6980 runs
He want to complete = 10000 runs
Runs need to score more = 10000 – 6980 = 3020
Hence, he need 3020 more runs to score
3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Solutions:
No. of votes secured by the successful candidate = 577500
No. of votes secured by his rival = 348700
Margin by which he won the election = 577500 – 348700 = 228800 votes
∴ Successful candidate won the election by 228800 votes
4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Solutions:
Price of books sold in June first week = Rs 285891
Price of books sold in June second week = Rs 400768
No. of books sold in both weeks together = Rs 285891 + Rs 400768 = Rs 686659
The sale of books is the highest in the second week
Difference in the sale in both weeks = Rs 400768 – Rs 285891 = Rs 114877
∴ Sale in second week was greater by Rs 114877 than in the first week.
5. Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.
Solutions:
Digits given are 6, 2, 7, 4, 3
Greatest 5-digit number = 76432
Least 5-digit number = 23467
Difference between the two numbers = 76432 – 23467 = 52965
∴ The difference between the two numbers is 52965
6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Solutions:
Number of screws manufactured in a day = 2825
Since January month has 31 days
Hence, number of screws manufactured in January = 31 × 2825 = 87575
Hence, machine produce 87575 screws in the month of January 2006
7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?
Solutions:
Total money the merchant had = Rs 78592
Cost of each radio set = Rs 1200
So, cost of 40 radio sets = Rs 1200 × 40 = Rs 48000
Money left with the merchant = Rs 78592 – Rs 48000 = Rs 30592
Hence, money left with the merchant after purchasing radio sets is Rs 30592
8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Solutions:
Difference between 65 and 56 i.e (65 – 56) = 9
The difference between the correct and incorrect answer = 7236 × 9 = 65124
9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?
Solutions:
Given
Total length of the cloth = 40 m
= 40 × 100 cm = 4000 cm
Cloth required to stitch one shirt = 2 m 15 cm
= 2 × 100 + 15 cm = 215 cm
Number of shirts that can be stitched out of 4000 cm = 4000 / 215 = 18 shirts
Hence 18 shirts can be stitched out of 40 m and 1m 30 cm of cloth is left out
10. Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solutions:
Weight of one box = 4 kg 500 g = 4 × 1000 + 500
= 4500 g
Maximum weight carried by the van = 800 kg = 800 × 1000
= 800000 g
Hence, number of boxes that can be loaded in the van = 800000 / 4500 = 177 boxes
11. The distance between the school and a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.
Solutions:
Distance covered between school and house = 1 km 875 m = 1000 + 875 = 1875 m
Since, the student walk both ways.
Hence, distance travelled by the student in one day = 2 × 1875 = 3750 m
Distance travelled by the student in 6 days = 3750 m × 6 = 22500 m = 22 km 500 m
∴ Total distance covered by the student in six days is 22 km and 500 m
12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solutions:
Quantity of curd in the vessel = 4 l 500 ml = 4 × 1000 + 500 = 4500 ml
Capacity of 1 glass = 25 ml
∴ Number of glasses that can be filled with curd = 4500 / 25 = 180 glasses
Hence, 180 glasses can be filled with curd.
Exercise 1.3 Page NO: 23
1. Estimate each of the following using general rule:
(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solutions:
(a) 730 + 998
Round off to hundreds
730 rounds off to 700
998 rounds off to 1000
Hence, 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314
Round off to hundreds
796 rounds off to 800
314 rounds off to 300
Hence, 796 – 314 = 800 – 300 = 500
(c) 12904 + 2888
Round off to thousands
12904 rounds off to 13000
2888 rounds off to 3000
Hence, 12904 + 2888 = 13000 + 3000 = 16000
(d) 28292 – 21496
Round off to thousands
28292 round off to 28000
21496 round off to 21000
Hence, 28292 – 21496 = 28000 – 21000 = 7000
Ten more such examples are
(i) 330 + 280 = 300 + 300 = 600
(ii) 3937 + 5990 = 4000 + 6000 = 10000
(iii) 6392 – 3772 = 6000 – 4000 = 2000
(iv) 5440 – 2972 = 5000 – 3000 = 2000
(v) 2175 + 1206 = 2000 + 1000 = 3000
(vi) 1110 – 1292 = 1000 – 1000 = 0
(vii) 910 + 575 = 900 + 600 = 1500
(viii) 6400 – 4900 = 6000 – 5000 = 1000
(ix) 3731 + 1300 = 4000 + 1000 = 5000
(x) 6485 – 4319 = 6000 – 4000 = 2000
2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365
Make four more such examples.
Solutions:
(a) 439 + 334 + 4317
Rounding off to nearest hundreds
439 + 334 + 4317 = 400 + 300 + 4300
= 5000
Rounding off to nearest tens
439 + 334 + 4317 = 440 + 330 + 4320
= 5090
(b) 108734 – 47599
Rounding off to nearest hundreds
108734 – 47599 = 108700 – 47600
= 61100
Rounding off to nearest tens
108734 – 47599 = 108730 – 47600
= 61130
(c) 8325 – 491
Rounding off to nearest hundreds
8325 – 491 = 8300 – 500
= 7800
Rounding off to nearest tens
8325 – 491 = 8330 – 490
= 7840
(d) 489348 – 48365
Rounding off to nearest hundreds
489348 – 48365 = 489300 – 48400
= 440900
Rounding off to nearest tens
489348 – 48365 = 489350 – 48370
= 440980
Four more examples are as follows
(i) 4853 + 662
Rounding off to nearest hundreds
4853 + 662 = 4800 + 700
= 5500
Rounding off to nearest tens
4853 + 662 = 4850 + 660
= 5510
(ii) 775 – 390
Rounding off to nearest hundreds
775 – 390 = 800 – 400
= 400
Rounding off to nearest tens
775 – 390 = 780 – 400
= 380
(iii) 6375 – 2875
Rounding off to nearest hundreds
6375 – 2875 = 6400 – 2900
= 3500
Rounding off to nearest tens
6375 – 2875 = 6380 – 2880
= 3500
(iv) 8246 – 6312
Rounding off to nearest hundreds
8246 – 6312 = 8200 – 6300
= 1900
Rounding off to nearest tens
8246 – 6312 = 8240 – 6310
= 1930
3. Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Solutions:
(a) 578 × 161
Rounding off by general rule
578 and 161 rounded off to 600 and 200 respectively
600
× 200
____________
120000
_____________
(b) 5281 × 3491
Rounding off by general rule
5281 and 3491 rounded off to 5000 and 3500 respectively
5000
× 3500
_________
17500000
_________
(c) 1291 × 592
Rounding off by general rule
1291 and 592 rounded off to 1300 and 600 respectively
1300
× 600
_____________
780000
______________
(d) 9250 × 29
Rounding off by general rule
9250 and 29 rounded off to 9000 and 30 respectively
9000
× 30
_____________
270000
______________
## Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 1
### What are the topics covered in Chapter 1 of NCERT Solutions for Class 6 Maths?
The topics covered in Chapter 1 of NCERT Solutions for Class 6 Maths are:
1. Introduction to numbers
2. Comparing numbers
3. Ascending order and Descending order
4. How many numbers can be formed using a certain number of digits?
5. Shifting digits
6. Place value
7. Larger Numbers and Estimates
8. Estimating sum or difference
9. Estimating products of numbers
10. BODMAS
11. Using Brackets
### How many problems are there in each exercise of NCERT Solutions for Class 6 Maths Chapter 1?
Regular practice of the exercise wise problems help students to understand the concepts covered in a better way. The number of questions in each exercise of Chapter 1 are as follows:
Exercise 1.1 – 4 questions
Exercise 1.2 – 12 questions
Exercise 1.3 – 3 questions
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BASIC ANTIDERIVATIVES - Integration - AP CALCULUS AB & BC REVIEW - Master AP Calculus AB & BC
## Master AP Calculus AB & BC
Part II. AP CALCULUS AB & BC REVIEW
CHAPTER 7. Integration
OVERVIEW
• Basic antiderivatives
• Hands-On Activity 7.1: Approximating area with Riemann sums
• The trapezoidal rule
• The fundamental theorem of calculus
• Hands-On Activity 7.2: Accumulation functions
• The Mean Value Theorem for Integration, average value of a function
• U-Substitution
• Integrating inverse trigonometric functions
• Technology: Evaluating definite integrals with your graphing calculator
• Summing it up
Now that you know just about everything there is to know about taking derivatives, it’s time to pull the rug out from under you. The third major topic of calculus (limits and derivatives being the first two) is integration, or antidifferentiation. That’s right, Mr. Prefix; anti- means “the opposite of,” so it’s time to explore the process of derivatives reversed. Previously, you would be asked to find clearly, the answer is 3x2. Now, you’ll be given 3x2 and required to come up with the antiderivative, x3.
But, it’s never that easy, is it? As a matter of fact, x3 + 1 is also an antiderivative of 3x2! So is x3 — 14. Therefore, we say that the antiderivative of 3x2 is x3 + C, where C can be any number at all. But, we’re getting ahead of ourselves. Let’s jump right in—the water’s fine.
BASIC ANTIDERIVATIVES
Just as the notation dy/dx or y’ indicated to you that differentiation was necessary, the notation
indicates the same for integration. The above is read “the integral (or antiderivative) of f(x) with respect to x.” Respecting variables in differentiation was sometimes a complicated procedure. Remember implicit differentiation and related rates? Luckily, respecting variables is not nearly as difficult in integration; you just have to make sure the dx gets multiplied by everything in the integral. But, enough talk—let’s get down to business.
NOTE. When deriving xb, you multiplied by b and subtracted 1 from the power. When integrating xb, you add 1 to the power and divide. The processes are complete opposites of each other.
Think back for a moment: The Power Rule (one of your earliest and dearest calculus friends) dealt with deriving simple expressions—a single variable to a constant power. There is an equivalent rule for integrating, so we’ll call it (get this) the Power Rule for Integration. Clever, eh?
The Power Rule for Integration: If a is a constant,
Translation: In order to find the integral of xa, add 1 to the exponent and divide the term by the new exponent.
Example 1: Evaluate
Solution: Add one to the exponent (1 + 3 = 4), and divide by the new exponent, 4:
NOTE. It is not always impossible to find C. In fact, it is sometimes incorrect to write “+ C”. For now, however, make it a habit to automatically add the “+ C” when you integrate. You’ll learn more about the exceptions later.
More about that weird C now. It is called the constant of integration. It is simply a real number, and we have no idea exactly what that number is (for now). However, and all have a derivative of x3 (since the derivative of the constant is 0). Therefore, when we write “+ C” at the end of an antiderivative, we are admitting that there may have been a constant there, but we do not know it.
Now, let’s discuss the two major properties of integrals; both of them are very similar to derivatives:
If a constant or coefficient is present in your integral, you may ignore it, like you did with derivatives. In fact, you may pull the constant completely out of the integral.
If an integral contains numerous terms being added or subtracted, then these terms can be split apart into separate integrals. In differentiation, given the problem you could find the derivatives of the terms separately: 3x2 — 5. The same goes for integration. For example,
In Example 2, we’ll apply these properties of integration to some more complex integration problems.
ALERT! You can only pull coefficients out of integrals. For example, it would be incorrect to rewrite
Example 2: Evaluate the following antiderivatives:
This expression can be rewritten as
The 5/3 is merely a coefficient, so we can apply the first rule of antiderivatives and pull it out of the integral:
Now, apply the power rule for integrals, but make sure to add 1 to the original power of -2.
This integral must first be rewritten as
Because the two terms are being added, we can split the above into two separate integrals (and pull out the coefficients):
Now, apply the Power Rule for Integration:
The 3/2 power is the result of adding 1 to the original exponent (1/2).
ALERT! Once you integrate, make sure to stop writing the and dx symbols. They only hang around until you’re done integrating. When their work is done, they vanish.
Again, rewriting the integral is the first order of business. Instead of one fraction, rewrite as the sum of two fractions with the same denominator. Also, apply the integration properties:
Remember way back to algebra and exponent properties: Therefore, and Use this to rewrite the problem as
and apply the Power Rule for Integrals to get
Do you see the shortcut for integrating fractional exponents? When you integrate x1/3, instead of writing the step remember that dividing by 4/3 is the same as multiplying by 3/4. Therefore, the answer is
Well, the Power Rule for Integrals is all well and good, but there is one instance in which it is completely useless. Consider the integral:
ALERT! Although the Power Rule for Integrals is relatively easy, it is also easy to make mistakes when the exponents are fractions or have negative powers. Be careful.
This can be rewritten as but if you try to integrate, you get and the zero in the denominator spoils everything. How, then, are you to integrate 1/x? Believe it or not, you already know the answer to this—you just have to dig it out of your long-term memory. Remember, integration is the opposite of differentiation, so the expression that has the derivative of 1/x will be the integral of 1/x. You learned in Chapter 4 that Therefore,
(You need to use the absolute value signs since ln x has domain (0,∞)—the function wouldn’t know what to do with negative inputs.) If you have forgotten the large list of derivatives you were to have memorized, it’s time to refresh your memory. Only two of the integrals look a little different from their derivatives. We have already looked at the first: (its integral has that unexpected absolute value). One other problem shows a slight difference in its absolute values:
You see arcsec x so infrequently on the test, it’s hardly worth mentioning, but it is important. In addition, we will take a closer look at inverse trigonometric and exponential integrals a little later in this chapter. Here are a few problems to get you brushed up on the throwback integrals.
Example 3: Evaluate the following integrals:
This problem asks, “What has a derivative of —sin x?” The answer is, of course, cos x + C, since If the problem had been the answer would have been —cos x + C, since
First, rewrite this problem as What has a derivative of csc2 x? Well, and that’s only off by a negative sign. Therefore, add a negative sign to cot x to account for the missing sign, and the answer is
This is simply the derivative of arcsin x, so the answer is arcsin x + C.
If the derivative of ex is ex, then
NOTE. You need to learn a technique called u-substitution before we can get too hot and heavy into integration. That’s later, though.
TIP. If you’re not sure that your integral is correct, take its derivative to check and see if you get the original problem. Because integration and differentiation are inverses, they should cancel each other out.
EXERCISE 1
Directions: Solve each of the following problems. Decide which is the best of the choices given and indicate your responses in the book.
DO NOT USE A GRAPHING CALCULATOR FOR ANY OF THESE PROBLEMS
Evaluate each of the following integrals.
if a and b are real numbers
1. Begin by splitting the integral into pieces and rewriting it so that you can apply the Power Rule for Integrals:
It’s ready to be power ruled, so go to it:
2. Because b is a coefficient, it can be pulled out of the integral.
3. Rewrite this integral before starting, and remember that 2π is just a constant, so the Power Rule for Integrals still applies (just like it did to the a exponent in problem 2).
4. Before you can integrate, you need to multiply the binomials together. There is no Product Rule for Integration (which makes things tricky later) but for now, we can avoid the problem by multiplying.
5. You can begin by writing each of the terms of the numerator over the denominator. This is a long step, and if you can do it in your head, you are encouraged to do so—carefully! So that you can see exactly what’s happening, the step is included:
6. This problem looks pretty complicated, but if you are clever (and who doesn’t like being clever now and again?), it becomes quite easy. The trick is to rewrite the fraction as follows:
If you multiply those two fractions together, you still get so we haven’t actually changed anything’s value. However, now we can rewrite as tan x and as sec x:
Perhaps you’ll remember it better if it’s written this way:
You know that is the derivative of sec x, so the final answer is
sec x + C
7. That negative sign looks like it’s just beggin’ to get factored out, so we’ll oblige it (and bring it out of the integral as the coefficient —1):
Now that looks familiar. In fact, it is most of the Pappa Theorem! (Remember your Pappa: tan2 x + 1 = sec2 x.) Therefore, we’ll use a Pappa substitution to rewrite it:
Because tan x has a derivative of sec2x, the answer is
—tan x + C
8. Even though this looks ugly, begin the same way you did with problem 5—write each term of the numerator over the denominator:
The first gigantic fraction simplifies to 1, making things much, much happier in the world:
The integral of 1 is simply x (since ), and the other term is the derivative of arcsec x (don’t forget the absolute value signs we discussed earlier in this section):
x + arcsec |x| + C
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# Conceptual Physics 11th Edition
## Presentation on theme: "Conceptual Physics 11th Edition"— Presentation transcript:
Conceptual Physics 11th Edition
Chapter 4: NEWTON’S SECOND LAW OF MOTION
Force Causes Acceleration Friction Mass and Weight Mass Resists Acceleration Newton’s Second Law of Motion Free Fall Non-Free Fall
Force causes Acceleration
Acceleration depends on the net force. Acceleration is directly proportional to net force. To increase the acceleration of an object, you must increase the net force acting on it. Acceleration ~ net force
The Force of Friction depends on the kinds of material and how much they are pressed together. is due to tiny surface bumps and to “stickiness” of the atoms on a material’s surface. Example: Friction between a crate on a smooth wooden floor is less than that on a rough floor.
The force of friction can occur
CHECK YOUR NEIGHBOR The force of friction can occur A. with sliding objects. in water. in air. All of the above. D. all of the above.
The force of friction can occur
CHECK YOUR ANSWER The force of friction can occur A. with sliding objects. in water. in air. All of the above. Comment: Friction can also occur for objects at rest. If you push horizontally on your book and it doesn’t move, then friction between the book and the table is equal and opposite to your push. D. all of the above.
The Force of Friction CHECK YOUR NEIGHBOR
When Sanjay pushes a refrigerator across a kitchen floor at a constant speed, the force of friction between the refrigerator and the floor is A. less than Sanjay’s push. equal to Sanjay’s push. equal and opposite to Sanjay’s push. more than Sanjay’s push. C. equal and opposite to Sanjay’s push.
When Sanjay pushes a refrigerator across a kitchen floor at a constant speed, the force of friction between the refrigerator and the floor is A. less than Sanjay’s push. equal to Sanjay’s push. equal and opposite to Sanjay’s push. more than Sanjay’s push. C. equal and opposite to Sanjay’s push.
The Force of Friction CHECK YOUR NEIGHBOR
When Sanjay pushes a refrigerator across a kitchen floor at an increasing speed, the amount of friction between the refrigerator and the floor is A. less than Sanjay’s push. equal to Sanjay’s push. equal and opposite to Sanjay’s push. more than Sanjay’s push. A. less than Sanjay’s push.
When Sanjay pushes a refrigerator across a kitchen floor at an increasing speed, the amount of friction between the refrigerator and the floor is A. less than Sanjay’s push. equal to Sanjay’s push. equal and opposite to Sanjay’s push. more than Sanjay’s push. Explanation: The increasing speed indicates a net force greater than zero. The refrigerator is not in equilibrium. A. less than Sanjay’s push.
Mass and Weight Mass: The quantity of matter in an object. It is also the measure of the inertia or sluggishness that an object exhibits in response to any effort made to start it, stop it, or change its state of motion in any way. Weight: The force upon an object due to gravity.
Mass and Weight Greater inertia greater mass Mass
A measure of the inertia of a material object Independent of gravity Greater inertia greater mass Unit of measurement is the kilogram (kg) Weight The force on an object due to gravity Scientific unit of force is the newton (N) Unit is also the pound (lb)
If the mass of an object is halved, the weight of the object is
Mass—A Measure of Inertia CHECK YOUR NEIGHBOR If the mass of an object is halved, the weight of the object is A. halved. twice. depends on location. None of the above. A. halved.
If the mass of an object is halved, the weight of the object is
Mass—A Measure of Inertia CHECK YOUR ANSWER If the mass of an object is halved, the weight of the object is A. halved. twice. depends on location. None of the above. A. halved.
Mass and Weight Mass versus weight
Mass and weight in everyday conversation are interchangeable. Mass, however, is different and more fundamental than weight. Mass versus weight on the Moon and Earth: Weight of an object on the Moon is less than on Earth. Mass of an object is the same in both locations.
Mass and Weight 1 kilogram weighs 10 newtons (9.8 newtons to be precise). Relationship between kilograms and pounds: 1 kg = 2.2 lb = 10 N at Earth’s surface 1 lb = 4.45 N
Mass and Weight CHECK YOUR NEIGHBOR
When the string is pulled down slowly, the top string breaks, which best illustrates the A. weight of the ball. mass of the ball. volume of the ball. density of the ball. A. weight of the ball.
When the string is pulled down slowly, the top string breaks, which best illustrates the A. weight of the ball. mass of the ball. volume of the ball. density of the ball. Explanation: Tension in the top string is the pulling tension plus the weight of the ball, both of which break the top string. A. weight of the ball.
Mass and Weight CHECK YOUR NEIGHBOR
When the string is pulled down quickly, the bottom string breaks, which best illustrates the A. weight of the ball. mass of the ball. volume of the ball. density of the ball. B. mass of the ball.
When the string is pulled down quickly, the bottom string breaks, which best illustrates the A. weight of the ball. mass of the ball. volume of the ball. density of the ball. Explanation: It is the “laziness” of the ball that tends to keep it at rest, resulting in the breaking of the bottom string. B. mass of the ball.
Mass Resists Acceleration
The same force applied to Twice the mass produces half the acceleration. 3 times the mass, produces 1/3 the acceleration. Acceleration is inversely proportional to mass.
Newton’s Second Law of Motion
Isaac Newton was the first to connect the concepts of force and mass to produce acceleration.
Newton’s Second Law of Motion
Newton’s second law (the law of acceleration) relates acceleration to force. The acceleration produced by a net force on an object is directly proportional to the net force, is in the same direction as the net force, and is inversely proportional to the mass of the object.
Newton’s Second Law of Motion
In equation form: Example: If net force acting on object is doubled object’s acceleration will be doubled. If mass of object is doubled object’s acceleration will be halved. net force Acceleration mass
Newton’s Second Law of Motion
Newton’s Second Law of Motion
CHECK YOUR NEIGHBOR Consider a cart pushed along a track with a certain force. If the force remains the same while the mass of the cart decreases to half, the acceleration of the cart remains relatively the same. halves. doubles. changes unpredictably. 26
Newton’s Second Law of Motion
CHECK YOUR ANSWER Consider a cart pushed along a track with a certain force. If the force remains the same while the mass of the cart decreases to half, the acceleration of the cart remains relatively the same. halves. doubles. changes unpredictably. Explanation: Acceleration = net force / mass Because, mass is in the denominator, acceleration increases as mass decreases. So, if mass is halved, acceleration doubles. Explanation: Acceleration = Net Force / Mass. So, if… Mass is halved i.e. Mass is now ½ Mass Net Force is kept the same i.e. Net Force New Acceleration = New Net Force / New Mass 27
Newton’s Second Law of Motion
CHECK YOUR NEIGHBOR Push a cart along a track so twice as much net force acts on it. If the acceleration remains the same, what is a reasonable explanation? The mass of the cart doubled when the force doubled. The cart experiences a force that it didn’t before. The track is not level. Friction reversed direction. 28
Newton’s Second Law of Motion
CHECK YOUR ANSWER Push a cart along a track so twice as much net force acts on it. If the acceleration remains the same, what is a reasonable explanation? The mass of the cart doubled when the force doubled. The cart experiences a force that it didn’t before. The track is not level. Friction reversed direction. Explanation: Acceleration = net force / mass If force doubles, acceleration will also double, But it does not, so mass must also be doubling to cancel out effects of force doubling. 29
Free Fall The greater the mass of the object…
the greater its force of attraction toward the Earth. the smaller its tendency to move i.e., the greater its inertia. So, the acceleration is the same. It is equal to the acceleration due to gravity: 10 m/s2 (precisely 9.8 m/s2).
Free Fall When acceleration is g—free fall
Newton’s second law provides an explanation for the equal accelerations of freely falling objects of various masses. Acceleration is equal when air resistance is negligible. Acceleration depends on force (weight) and inertia.
At one instant, an object in free fall has a speed of 40 m/s. Its speed 1 second later is A. also 40 m/s. 45 m/s. 50 m/s. None of the above. C m/s.
At one instant, an object in free-fall has a speed of 40 m/s. Its speed 1 second later is A. also 40 m/s. 45 m/s. 50 m/s. None of the above. Comment: We assume the object is falling downward. C m/s.
A 5-kg iron ball and a 10-kg iron ball are dropped from rest. For negligible air resistance, the acceleration of the heavier ball will be A. less. the same. more. undetermined. B. the same.
A 5-kg iron ball and a 10-kg iron ball are dropped from rest. For negligible air resistance, the acceleration of the heavier ball will be A. less. the same. more. undetermined. B. the same.
A 5-kg iron ball and a 10-kg iron ball are dropped from rest. When the free-falling 5-kg ball reaches a speed of 10 m/s, the speed of the free-falling 10-kg ball is A. less than 10 m/s. 10 m/s. more than 10 m/s. undetermined. B m/s.
A 5-kg iron ball and a 10-kg iron ball are dropped from rest. When the free-falling 5-kg ball reaches a speed of 10 m/s, the speed of the free-falling 10-kg ball is A. less than 10 m/s. 10 m/s. more than 10 m/s. undetermined. B m/s.
Non-Free Fall When an object falls downward through the air it experiences force of gravity pulling it downward. air drag force acting upward.
Non-Free Fall When acceleration of fall is less than g, non-free fall
occurs when air resistance is non-negligible. depends on two things: speed and frontal surface area.
Non-Free Fall When the object is moving fast enough that force of gravity equals its air resistance Then no net force No acceleration Velocity does not change
Non-Free Fall Terminal speed Terminal velocity
occurs when acceleration terminates (when air resistance equals weight and net force is zero). Terminal velocity same as terminal speed, with direction implied or specified.
Non-Free Fall—Example
A skydiver jumps from plane. Weight is the only force until air resistance acts. As falling speed increases, air resistance on diver builds up, net force is reduced, and acceleration becomes less. When air resistance equals the diver’s weight, net force is zero and acceleration terminates. Diver reaches terminal velocity, then continues the fall at constant speed.
When a 20-N falling object encounters 5 N of air resistance, its acceleration of fall is A. less than g. more than g. C. g. terminated. A. less than g.
When a 20-N falling object encounters 5 N of air resistance, its acceleration of fall is A. less than g. more than g. C. g. terminated. Comment: Acceleration of a non-free fall is always less than g. Acceleration will actually be (20 N – 5 N)/2 kg = 7.5 m/s2. A. less than g.
If a 50-N person is to fall at terminal speed, the air resistance needed is A. less than 50 N. 50 N. more than 50 N. None of the above. B N.
If a 50-N person is to fall at terminal speed, the air resistance needed is A. less than 50 N. 50 N. more than 50 N. None of the above. Explanation: Then, F = 0 and acceleration = 0. B N.
As the skydiver falls faster and faster through the air, air resistance A. increases. decreases. remains the same. Not enough information. A. increases.
As the skydiver falls faster and faster through the air, air resistance A. increases. decreases. remains the same. Not enough information. A. increases.
As the skydiver continues to fall faster and faster through the air, net force A. increases. decreases. remains the same. Not enough information. B. decreases.
As the skydiver continues to fall faster and faster through the air, net force A. increases. decreases. remains the same. Not enough information. B. decreases.
As the skydiver continues to fall faster and faster through the air, her acceleration A. increases. decreases. remains the same. Not enough information. B. decreases.
As the skydiver continues to fall faster and faster through the air, her acceleration A. increases. decreases. remains the same. Not enough information. Comment If this question were asked first in the sequence of skydiver questions, many would answer it incorrectly. Would this have been you? B. decreases.
Non-Free Fall CHECK YOUR NEIGHBOR Consider a heavy and light person jumping together with same-size parachutes from the same altitude. Who will reach the ground first? A. The light person. The heavy person. Both will reach at the same time. Not enough information. B. The heavy person.
Non-Free Fall CHECK YOUR ANSWER Consider a heavy and light person jumping together with same-size parachutes from the same altitude. Who will reach the ground first? A. The light person The heavy person Both will reach at the same time. Not enough information. B. The heavy person. Explanation: They both have the same drag force (for the same speed). The man (heavier) has a greater downward force than the woman (lighter). The man has to drop farther to receive drag force equal to his downward force, so a higher terminal velocity.
Free Fall vs. Non-Free Fall
Coin and feather fall with air present Feather reaches terminal velocity very quickly and falls slowly at constant speed, reaching the bottom after the coin does. Coin falls very quickly and air resistance doesn’t build up to its weight over short-falling distances, which is why the coin hits the bottom much sooner than the falling feather.
When the air is removed by a vacuum pump and the coin and feather activity is repeated, A. the feather hits the bottom first, before the coin hits. the coin hits the bottom first, before the feather hits. both the coin and feather drop together side-by-side. Not enough information. C. both the coin and feather drop together side-by-side.
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# Calculate the residue of 1/Cosh(pi.z) at z = i/2 (complex analysis)
• I
• Old Person
#### Old Person
TL;DR Summary
Calculate the residue of 1/Cosh(pi.z) at z = i/2.
I'm not sure if this should be in the calculus section or the anlaysis section. It's complex analysis related to integration around a contour.
Can someone suggest a method to determine the residue of f(z) = ## \frac{1}{Cosh ( \pi z) } ## at the singular point z = i/2.
Background:
This was part of an exam question and the model answers are already available.
In the model answers it is just assumed we have a simple pole (pole of order 1) at z=i/2 and the usual formula then jumps in:
Res (f(z) ; i/2) = ## \lim\limits_{z \rightarrow i/2} \, [ (z-i/2) . f(z) ]##
BUT no proof is given that we did only have a simple pole there. If that's easily shown, great, otherwise would it be easier to just directly obtain the Laurent series or obtain the Residue by some other method?
FactChecker
I used ##Res_z\left(\dfrac{1}{f}\right)=\dfrac{1}{f'(z)}## from
https://www.physicsforums.com/insights/an-overview-of-complex-differentiation-and-integration/
and got
\begin{align*}
Res_z\left(\dfrac{1}{f}\right)&=\dfrac{1}{f'(z)}=\dfrac{1}{(\cosh \pi z)'}\\&=\dfrac{1}{\pi \sinh \pi z}=\dfrac{1}{\pi \sinh(i \pi/2)}=\dfrac{1}{\pi i}=-\dfrac{i}{\pi}
\end{align*}
if I made no mistakes by looking up the complex arguments of ##\sinh## on Wikipedia.
https://www.wolframalpha.com/input?i=residue+(1/cosh(pi+z);+i+/2)=
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Old Person
Hi and thank you ever so much for your time and speed of reply @fresh_42
The answer is OK in that it does agree with the model answer. However, I'm not sure it was a valid method.
I've just been looking over the insight article you suggested. It states, just above the result : Say Z_m is a zero of order m and.... .... then gives the formula you've used for z_1 only:
## Res_{z_1}\left(\dfrac{1}{f}\right) =\dfrac{1}{f'(z_1)} ##
So it looks like you must already know that the point of interest, z, was a pole of order 1 to use that result. I'm not sure, I didn't write that article - but that's what it looks like. Just to be clear then, it may be right, it's just that we still seem to need to establish that there was a simple pole (of order 1) there.
As for using Wolfram Aplha approach - as it happens I already tried it. Sadly it spits out answers but no explanation or method. Did it also just assume a pole of order 1 was there?
Yes, the order of the pole is needed beforehand, and I guess WA determined it, too, as part of its calculations.
Old Person
##\cosh (i \pi/2 )=\dfrac{1}{2}\left(e^{i \pi/2}+e^{-i \pi/2}\right)=\dfrac{1}{2}\left(i - i\right)=0.## Therefore ##\cosh(z)=(z-i \pi/2)\cdot g(z).##
Assume that ##z=i\pi/2## is of higher order than ##1,## i.e. ##\cosh(z)=(z-i \pi/2)^2\cdot h(z).## Then
\begin{align*}
\dfrac{d}{dz}\cosh(z)&=\sinh(z)=2(z-i \pi/2)\cdot h(z)+ (z-i \pi/2)^2\cdot h'(z)
\end{align*}
The RHS is zero at ##z=i \pi/2## but ##\sinh(i \pi/2)=\dfrac{1}{2}\left(e^{i \pi/2}-e^{-i \pi/2}\right)=\dfrac{i-(-i)}{2}=i\neq 0.##
This proves that the zero ## i \pi/2## of ##\cosh(z)## is of order one.
hutchphd
You cna get the Laurent series for $\operatorname{sech}(\pi z)$ about $i/2$ by setting $t = z - \frac i2$ and then $$\begin{split} \frac{1}{\cosh(i\pi/2 + \pi t)} &= \frac{1}{\cosh(i\pi/2)\cosh \pi t + \sinh(i\pi/2) \sinh \pi t} \\ &= \frac{1}{i\sinh \pi t} \\ &= \frac{1}{i\pi t} \left( 1 - \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots \right)^{-1} \\ &= \frac{1}{i\pi t}\left( 1 - \left(- \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots\right) + \left(- \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots\right)^2 + \dots \right) \\ &= \frac{-i}{\pi(z - \frac i2)} + \sum_{n=0}^\infty a_n(z - \tfrac i2)^n. \end{split}$$ But it is enough to note that $$\frac{1}{i\sinh \pi t} = \frac{1}{i \pi t} \left( \frac{\pi t}{\sinh \pi t} \right)$$ and the second factor is analytic at $t = 0$.
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Old Person and SammyS
Hi and thank you for even more replies.
@fresh_42 ---> That's a good answer. I think there is one minor issue but it is minor. It won't be a problem unless 1/Cosh z is extremely unusual.
Assume that z=iπ/2 is of higher order than 1, i.e. cosh(z)=(z−iπ/2)2⋅h(z). Then....
You have assumed that 1/Cosh z must have a pole of some (finite) order at the point. There are some functions that have isolated singular points but just cannot be described as a pole (of any order).
Example ##e^{1/z} ## = Exp (1/z) = ## \sum_{n=0}^{\infty} \frac{1}{n!} \frac{1}{z^n} ## = a Laurent series about z=0. There is no finite n where the thing can be written as F(z) / zn (with F(z) analytic in a small neighbourhood of z=0).
If the singular point cannot be described as a pole then you can't apply the differentiation trick because you won't have the ability to write ## \cosh(z)=(z-i \pi/2)^N\cdot g(z) ## (with g(z) analytic), for any finite N.
- - - - - - - - - -
@pasmith , that may actually work. In some lines of the work on those series, you needed the series to be absolutely convergent to be sure you could re-arrange and re-shuffle the terms. However, I think you've got that. I wouldn't have had time to check all these details in the exam - but I do now and they they do seem to be ok.
Where you write: ..it is enough to note.... ## \frac{1}{i\sinh \pi t} = \frac{1}{i \pi t} \left( \frac{\pi t}{\sinh \pi t} \right) ##
I think I can see what you are saying. This bit ## \left( \frac{\pi t}{\sinh \pi t} \right) ## = ## \left( 1 - \frac{\pi^2t^2}{3!} + \frac{\pi^4t^4}{5!} + \dots \right)^{-1} ## from earlier in your post and the RHS is seen to be analytic at t=0. So that is enough to establish that you have a simple pole.
- - - - - - - - - - -
Overall, thank you to everyone. I'm happy with the answers. It's been great to see that it could be done (given time) but also just to have a couple of people acknowledge that there was a need to establish the nature of the singular point (or directly look at the Laurent series). As I mentioned before - in the model answers, there are no marks for proving that or even indicating that you should care. It's clear the person who set the exam just expected you to assume you had a simple pole there.
You have assumed that 1/Cosh z must have a pole of some (finite) order at the point. There are some functions that have isolated singular points but just cannot be described as a pole (of any order).
I actually haven't said anything about poles. I only have analyzed the point ##z=\mathrm{i} \pi /2## of the analytical function ##z\longmapsto \cosh(z).##
I showed that it is a zero of order one. That makes its inverse ##z \longmapsto \dfrac{1}{\cosh(z)}## having a pole at ##z=\mathrm{i} \pi/2## of order one. It's not isolated since the zero wasn't isolated.
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#### What is 361 percent of 925?
How much is 361 percent of 925? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 361% of 925 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 361% of 925 = 3339.25
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating three hundred and sixty-one of nine hundred and twenty-five How to calculate 361% of 925? Simply divide the percent by 100 and multiply by the number. For example, 361 /100 x 925 = 3339.25 or 3.61 x 925 = 3339.25
#### How much is 361 percent of the following numbers?
361% of 925.01 = 333928.61 361% of 925.02 = 333932.22 361% of 925.03 = 333935.83 361% of 925.04 = 333939.44 361% of 925.05 = 333943.05 361% of 925.06 = 333946.66 361% of 925.07 = 333950.27 361% of 925.08 = 333953.88 361% of 925.09 = 333957.49 361% of 925.1 = 333961.1 361% of 925.11 = 333964.71 361% of 925.12 = 333968.32 361% of 925.13 = 333971.93 361% of 925.14 = 333975.54 361% of 925.15 = 333979.15 361% of 925.16 = 333982.76 361% of 925.17 = 333986.37 361% of 925.18 = 333989.98 361% of 925.19 = 333993.59 361% of 925.2 = 333997.2 361% of 925.21 = 334000.81 361% of 925.22 = 334004.42 361% of 925.23 = 334008.03 361% of 925.24 = 334011.64 361% of 925.25 = 334015.25
361% of 925.26 = 334018.86 361% of 925.27 = 334022.47 361% of 925.28 = 334026.08 361% of 925.29 = 334029.69 361% of 925.3 = 334033.3 361% of 925.31 = 334036.91 361% of 925.32 = 334040.52 361% of 925.33 = 334044.13 361% of 925.34 = 334047.74 361% of 925.35 = 334051.35 361% of 925.36 = 334054.96 361% of 925.37 = 334058.57 361% of 925.38 = 334062.18 361% of 925.39 = 334065.79 361% of 925.4 = 334069.4 361% of 925.41 = 334073.01 361% of 925.42 = 334076.62 361% of 925.43 = 334080.23 361% of 925.44 = 334083.84 361% of 925.45 = 334087.45 361% of 925.46 = 334091.06 361% of 925.47 = 334094.67 361% of 925.48 = 334098.28 361% of 925.49 = 334101.89 361% of 925.5 = 334105.5
361% of 925.51 = 334109.11 361% of 925.52 = 334112.72 361% of 925.53 = 334116.33 361% of 925.54 = 334119.94 361% of 925.55 = 334123.55 361% of 925.56 = 334127.16 361% of 925.57 = 334130.77 361% of 925.58 = 334134.38 361% of 925.59 = 334137.99 361% of 925.6 = 334141.6 361% of 925.61 = 334145.21 361% of 925.62 = 334148.82 361% of 925.63 = 334152.43 361% of 925.64 = 334156.04 361% of 925.65 = 334159.65 361% of 925.66 = 334163.26 361% of 925.67 = 334166.87 361% of 925.68 = 334170.48 361% of 925.69 = 334174.09 361% of 925.7 = 334177.7 361% of 925.71 = 334181.31 361% of 925.72 = 334184.92 361% of 925.73 = 334188.53 361% of 925.74 = 334192.14 361% of 925.75 = 334195.75
361% of 925.76 = 334199.36 361% of 925.77 = 334202.97 361% of 925.78 = 334206.58 361% of 925.79 = 334210.19 361% of 925.8 = 334213.8 361% of 925.81 = 334217.41 361% of 925.82 = 334221.02 361% of 925.83 = 334224.63 361% of 925.84 = 334228.24 361% of 925.85 = 334231.85 361% of 925.86 = 334235.46 361% of 925.87 = 334239.07 361% of 925.88 = 334242.68 361% of 925.89 = 334246.29 361% of 925.9 = 334249.9 361% of 925.91 = 334253.51 361% of 925.92 = 334257.12 361% of 925.93 = 334260.73 361% of 925.94 = 334264.34 361% of 925.95 = 334267.95 361% of 925.96 = 334271.56 361% of 925.97 = 334275.17 361% of 925.98 = 334278.78 361% of 925.99 = 334282.39 361% of 926 = 334286
1% of 925 = 9.25 2% of 925 = 18.5 3% of 925 = 27.75 4% of 925 = 37 5% of 925 = 46.25 6% of 925 = 55.5 7% of 925 = 64.75 8% of 925 = 74 9% of 925 = 83.25 10% of 925 = 92.5 11% of 925 = 101.75 12% of 925 = 111 13% of 925 = 120.25 14% of 925 = 129.5 15% of 925 = 138.75 16% of 925 = 148 17% of 925 = 157.25 18% of 925 = 166.5 19% of 925 = 175.75 20% of 925 = 185 21% of 925 = 194.25 22% of 925 = 203.5 23% of 925 = 212.75 24% of 925 = 222 25% of 925 = 231.25
26% of 925 = 240.5 27% of 925 = 249.75 28% of 925 = 259 29% of 925 = 268.25 30% of 925 = 277.5 31% of 925 = 286.75 32% of 925 = 296 33% of 925 = 305.25 34% of 925 = 314.5 35% of 925 = 323.75 36% of 925 = 333 37% of 925 = 342.25 38% of 925 = 351.5 39% of 925 = 360.75 40% of 925 = 370 41% of 925 = 379.25 42% of 925 = 388.5 43% of 925 = 397.75 44% of 925 = 407 45% of 925 = 416.25 46% of 925 = 425.5 47% of 925 = 434.75 48% of 925 = 444 49% of 925 = 453.25 50% of 925 = 462.5
51% of 925 = 471.75 52% of 925 = 481 53% of 925 = 490.25 54% of 925 = 499.5 55% of 925 = 508.75 56% of 925 = 518 57% of 925 = 527.25 58% of 925 = 536.5 59% of 925 = 545.75 60% of 925 = 555 61% of 925 = 564.25 62% of 925 = 573.5 63% of 925 = 582.75 64% of 925 = 592 65% of 925 = 601.25 66% of 925 = 610.5 67% of 925 = 619.75 68% of 925 = 629 69% of 925 = 638.25 70% of 925 = 647.5 71% of 925 = 656.75 72% of 925 = 666 73% of 925 = 675.25 74% of 925 = 684.5 75% of 925 = 693.75
76% of 925 = 703 77% of 925 = 712.25 78% of 925 = 721.5 79% of 925 = 730.75 80% of 925 = 740 81% of 925 = 749.25 82% of 925 = 758.5 83% of 925 = 767.75 84% of 925 = 777 85% of 925 = 786.25 86% of 925 = 795.5 87% of 925 = 804.75 88% of 925 = 814 89% of 925 = 823.25 90% of 925 = 832.5 91% of 925 = 841.75 92% of 925 = 851 93% of 925 = 860.25 94% of 925 = 869.5 95% of 925 = 878.75 96% of 925 = 888 97% of 925 = 897.25 98% of 925 = 906.5 99% of 925 = 915.75 100% of 925 = 925
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# A 5 meter long wire is cut into two pieces. If the longer
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A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5
OA says it is 2/5 because:
The area of the square will be more than 1 if and only if the longer piece of the wire is longer than 4. To produce such a result, the cutting point has to be either on the first meter of the wire or on its last meter. The probability of this is 2/5 .
I have a question. Since area of the square has to be more than 1, its sides have to be more than 1. If its sides are more than 1, its perimeter will be more than 4. In order for this to happen, a piece larger than 4 meters has to be cut. The smallest piece need would be 4 meters and 1 centimeter (or millimeter for that matter). While I agree that the chance of wire being cut on any of the first 1 meters of it is 2/5, don't we need to calculate the probability of it being cut (at least) at 4 meters and 1 centimeter (or mm)? If so, the probability that it will get cut within first 99 cm is 2*99/500, which comes out to 99/250. This is close to 2/5 yes, but isn't this more correct way to look at it?
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Originally posted by MisterEko on 17 Dec 2010, 08:23.
Last edited by Vyshak on 14 Apr 2016, 19:41, edited 1 time in total.
OA and OE are now hidden
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Posts: 52285
Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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10 Nov 2013, 02:40
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SravnaTestPrep wrote:
Bunuel wrote:
MisterEko wrote:
Hm, not sure I understood you (or that you understood my question). What I said is that since the piece cut has to be bigger than 4, even marginally bigger, probability will be slightly less than 2/5. Say we can only count in cm. Wire is 500 cm long. In order for square to have area more than 1 meter, perimeter needs to be at least slightly bigger than 400 cm. In that case, the next least place that wire needs to be cut on (and satisfy the area of a square being bigger than 1) would be on 99th cm (down to the 1st cm) or 401st cm (up to 499). Either way, there are 99 positions on the first meter and 99 on the second one that satisfy the length needed. Since the total is 500 cm, probability would be 2*99/500 or 99/250.
Please, forgive me if I am being nuisance, I am kinda intrigued by this.
OK, let me ask you this: what is the probability that the wire will be cut so that we get pieces of exactly 4m and 1m (so at 1m or at 4m)?
What I'm saying is that the probability that the length of a longer piece will be more than 4 OR more than or equal to 4 is the same and equal to 2/5.
I think the answer 2/5 is flawed. Let us take five points 1, 2, 3, 4 and 5. For the longer wire to be more than or equal to 4m, 2 cases are there, it has to be cut at 1m or before OR 4m or after.
Let us take the first case. For example if it is cut at 1m, the longer wire would be 4m. Assume that each 1m is divided into cms. So there are 99 points out of a total of 500 points when the longer wire can be more than 4 m. for example if it is cut at 99th cm the longer wire would be 401 cm or > 4m. and so on. So there are totally 100 points including the exact 1m point out of a total of 500 points. So the probability that the longer wire is equal to or more than 4m is 100/500=1/5. The other case is wire cut at 4m and above. Each case has a probability of 1/5 so that the overall probability is 2/5.
The above is also equivalent to saying that the shorter wire is <= 1m. That is the probability is 2/5 when the longer wire is 4m and the shorter wire is 1m and say when the longer wire is 4.01 m and the shorter wire is 0.99m and so on . But saying equal to or more than 4m is not equivalent to saying less than 1m but only equivalent to saying less than or equal to 1m.
So when we have the shorter wire to be less than 1m, we have to consider only the cases > 4m. When it has to be more than 4m, the probability will be less than the above or less than 2/5.
The correct answer is E (2/5).
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A. 1/6
B. 1/5
C. 3/10
D. 1/3
E. 2/5
In order the area of a square to be more than 1, its side must be more than 1, or the perimeter must be more than 4. Which means that the longer piece must be more than 4. Look at the diagram below:
-----
If the wire will be cut anywhere at the red region then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is 2/5 (2 red pieces out of 5).
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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17 Dec 2010, 09:32
3
1
MisterEko wrote:
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5
OA says it is 2/5 because:
The area of the square will be more than 1 if and only if the longer piece of the wire is longer than 4. To produce such a result, the cutting point has to be either on the first meter of the wire or on its last meter. The probability of this is 2/5 .
I have a question. Since area of the square has to be more than 1, its sides have to be more than 1. If its sides are more than 1, its perimeter will be more than 4. In order for this to happen, a piece larger than 4 meters has to be cut. The smallest piece need would be 4 meters and 1 centimeter (or millimeter for that matter). While I agree that the chance of wire being cut on any of the first 1 meters of it is 2/5, don't we need to calculate the probability of it being cut (at least) at 4 meters and 1 centimeter (or mm)? If so, the probability that it will get cut within first 99 cm is 2*99/500, which comes out to 99/250. This is close to 2/5 yes, but isn't this more correct way to look at it?
Basically you are saying that the probability should be almost 2/5 but not exactly 2/5, because wire can be cut at 1 or 4 meters exactly and in this case probability will be a little less than 2/5. But this is not true, 1 and 4 meters marks are points and point by definition has no length or any other dimension.
It's similar to the followoing example: the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5.
Hope it's clear.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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17 Dec 2010, 12:43
Bunuel wrote:
MisterEko wrote:
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5
OA says it is 2/5 because:
The area of the square will be more than 1 if and only if the longer piece of the wire is longer than 4. To produce such a result, the cutting point has to be either on the first meter of the wire or on its last meter. The probability of this is 2/5 .
I have a question. Since area of the square has to be more than 1, its sides have to be more than 1. If its sides are more than 1, its perimeter will be more than 4. In order for this to happen, a piece larger than 4 meters has to be cut. The smallest piece need would be 4 meters and 1 centimeter (or millimeter for that matter). While I agree that the chance of wire being cut on any of the first 1 meters of it is 2/5, don't we need to calculate the probability of it being cut (at least) at 4 meters and 1 centimeter (or mm)? If so, the probability that it will get cut within first 99 cm is 2*99/500, which comes out to 99/250. This is close to 2/5 yes, but isn't this more correct way to look at it?
Basically you are saying that the probability should be almost 2/5 but not exactly 2/5, because wire can be cut at 1 or 4 meters exactly and in this case probability will be a little less than 2/5. But this is not true, 1 and 4 meters marks are points and point by definition has no length or any other dimension.
It's similar to the followoing example: the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5.
Hope it's clear.
Hm, not sure I understood you (or that you understood my question). What I said is that since the piece cut has to be bigger than 4, even marginally bigger, probability will be slightly less than 2/5. Say we can only count in cm. Wire is 500 cm long. In order for square to have area more than 1 meter, perimeter needs to be at least slightly bigger than 400 cm. In that case, the next least place that wire needs to be cut on (and satisfy the area of a square being bigger than 1) would be on 99th cm (down to the 1st cm) or 401st cm (up to 499). Either way, there are 99 positions on the first meter and 99 on the second one that satisfy the length needed. Since the total is 500 cm, probability would be 2*99/500 or 99/250.
Please, forgive me if I am being nuisance, I am kinda intrigued by this.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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17 Dec 2010, 12:51
MisterEko wrote:
Hm, not sure I understood you (or that you understood my question). What I said is that since the piece cut has to be bigger than 4, even marginally bigger, probability will be slightly less than 2/5. Say we can only count in cm. Wire is 500 cm long. In order for square to have area more than 1 meter, perimeter needs to be at least slightly bigger than 400 cm. In that case, the next least place that wire needs to be cut on (and satisfy the area of a square being bigger than 1) would be on 99th cm (down to the 1st cm) or 401st cm (up to 499). Either way, there are 99 positions on the first meter and 99 on the second one that satisfy the length needed. Since the total is 500 cm, probability would be 2*99/500 or 99/250.
Please, forgive me if I am being nuisance, I am kinda intrigued by this.
OK, let me ask you this: what is the probability that the wire will be cut so that we get pieces of exactly 4m and 1m (so at 1m or at 4m)?
What I'm saying is that the probability that the length of a longer piece will be more than 4 OR more than or equal to 4 is the same and equal to 2/5.
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Manager
Status: Fighting the beast.
Joined: 25 Oct 2010
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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17 Dec 2010, 13:03
Bunuel wrote:
MisterEko wrote:
Hm, not sure I understood you (or that you understood my question). What I said is that since the piece cut has to be bigger than 4, even marginally bigger, probability will be slightly less than 2/5. Say we can only count in cm. Wire is 500 cm long. In order for square to have area more than 1 meter, perimeter needs to be at least slightly bigger than 400 cm. In that case, the next least place that wire needs to be cut on (and satisfy the area of a square being bigger than 1) would be on 99th cm (down to the 1st cm) or 401st cm (up to 499). Either way, there are 99 positions on the first meter and 99 on the second one that satisfy the length needed. Since the total is 500 cm, probability would be 2*99/500 or 99/250.
Please, forgive me if I am being nuisance, I am kinda intrigued by this.
OK, let me ask you this: what is the probability that the wire will be cut so that we get pieces of exactly 4m and 1m (so at 1m or at 4m)?
What I'm saying is that the probability that the length of a longer piece will be more than 4 OR more than or equal to 4 is the same and equal to 2/5.
Now I get it. Thank you... Probabilities will always get ya'...
Cheers!
_________________
[highlight]Monster collection of Verbal questions (RC, CR, and SC)[/highlight]
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
[highlight]Massive collection of thousands of Data Sufficiency and Problem Solving questions and answers:[/highlight]
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
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Joined: 02 Mar 2010
Posts: 19
Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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06 Nov 2013, 16:06
Bunuel wrote:
MisterEko wrote:
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5
OA says it is 2/5 because:
The area of the square will be more than 1 if and only if the longer piece of the wire is longer than 4. To produce such a result, the cutting point has to be either on the first meter of the wire or on its last meter. The probability of this is 2/5 .
I have a question. Since area of the square has to be more than 1, its sides have to be more than 1. If its sides are more than 1, its perimeter will be more than 4. In order for this to happen, a piece larger than 4 meters has to be cut. The smallest piece need would be 4 meters and 1 centimeter (or millimeter for that matter). While I agree that the chance of wire being cut on any of the first 1 meters of it is 2/5, don't we need to calculate the probability of it being cut (at least) at 4 meters and 1 centimeter (or mm)? If so, the probability that it will get cut within first 99 cm is 2*99/500, which comes out to 99/250. This is close to 2/5 yes, but isn't this more correct way to look at it?
Basically you are saying that the probability should be almost 2/5 but not exactly 2/5, because wire can be cut at 1 or 4 meters exactly and in this case probability will be a little less than 2/5. But this is not true, 1 and 4 meters marks are points and point by definition has no length or any other dimension.
It's similar to the followoing example: the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5.
Hope it's clear.
Hey Bunuel,
I'm having hard time understanding your example: "the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5."
Basically, I'm getting confused because I think the probability of picking a number from the range (0,5), which I converted to a set {0,1,2,3,4,5}, is 1/6 (6 = total number of terms). So probability of picking a number > 4 is 1/6 (because the only possibility is 5 from the set). But probability of picking up a number >=4 is 2/6 because now two outcomes, 4 and 5, can be considered successful. So, can you kindly let me know where I'm going wrong?
Also, is the probability in the rope = Length of a unit of rope/ Total Length of the rope? Appreciate your help..
Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 52285
Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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07 Nov 2013, 04:54
prsnt11 wrote:
Bunuel wrote:
MisterEko wrote:
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5
OA says it is 2/5 because:
The area of the square will be more than 1 if and only if the longer piece of the wire is longer than 4. To produce such a result, the cutting point has to be either on the first meter of the wire or on its last meter. The probability of this is 2/5 .
I have a question. Since area of the square has to be more than 1, its sides have to be more than 1. If its sides are more than 1, its perimeter will be more than 4. In order for this to happen, a piece larger than 4 meters has to be cut. The smallest piece need would be 4 meters and 1 centimeter (or millimeter for that matter). While I agree that the chance of wire being cut on any of the first 1 meters of it is 2/5, don't we need to calculate the probability of it being cut (at least) at 4 meters and 1 centimeter (or mm)? If so, the probability that it will get cut within first 99 cm is 2*99/500, which comes out to 99/250. This is close to 2/5 yes, but isn't this more correct way to look at it?
Basically you are saying that the probability should be almost 2/5 but not exactly 2/5, because wire can be cut at 1 or 4 meters exactly and in this case probability will be a little less than 2/5. But this is not true, 1 and 4 meters marks are points and point by definition has no length or any other dimension.
It's similar to the followoing example: the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5.
Hope it's clear.
Hey Bunuel,
I'm having hard time understanding your example: "the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5."
Basically, I'm getting confused because I think the probability of picking a number from the range (0,5), which I converted to a set {0,1,2,3,4,5}, is 1/6 (6 = total number of terms). So probability of picking a number > 4 is 1/6 (because the only possibility is 5 from the set). But probability of picking up a number >=4 is 2/6 because now two outcomes, 4 and 5, can be considered successful. So, can you kindly let me know where I'm going wrong?
Also, is the probability in the rope = Length of a unit of rope/ Total Length of the rope? Appreciate your help..
Thanks
Why do you consider only integers? The numbers from 0 to 5 consists of ALL numbers from 0 to 5, not only of integers.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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07 Nov 2013, 10:14
Hey Bunuel,
I'm having hard time understanding your example: "the probability that a number X picked from the range (0,5) is more than 4 is 1/5 as well as the probability that a number X picked from the range (0,5) is more than or equal to 4 is also 1/5."
Basically, I'm getting confused because I think the probability of picking a number from the range (0,5), which I converted to a set {0,1,2,3,4,5}, is 1/6 (6 = total number of terms). So probability of picking a number > 4 is 1/6 (because the only possibility is 5 from the set). But probability of picking up a number >=4 is 2/6 because now two outcomes, 4 and 5, can be considered successful. So, can you kindly let me know where I'm going wrong?
Also, is the probability in the rope = Length of a unit of rope/ Total Length of the rope? Appreciate your help..
Thanks[/quote]
Why do you consider only integers? The numbers from 0 to 5 consists of ALL numbers from 0 to 5, not only of integers.[/quote]
Thanks for your prompt help. But I'm still confused Bunuel! There are infinite real numbers between 0 & 5, so how did we get 1/5 as the probability? I'm unable to visualize this problem. Can you kindly explain it to me in terms of successful outcomes/total outcomes? Thanks for your help...
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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07 Nov 2013, 10:41
1
prsnt11 wrote:
Thanks for your prompt help. But I'm still confused Bunuel! There are infinite real numbers between 0 & 5, so how did we get 1/5 as the probability? I'm unable to visualize this problem. Can you kindly explain it to me in terms of successful outcomes/total outcomes? Thanks for your help...
Visualization is exactly what should help. Consider a number line: {total} is line segment of 5 units (from 0 to 5) and {favorable} is a line segment of 1 unit (from 4 to 5), thus P = {favorable}/{total} = 1/5.
Hope it's clear.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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09 Nov 2013, 12:54
Bunuel wrote:
prsnt11 wrote:
Thanks for your prompt help. But I'm still confused Bunuel! There are infinite real numbers between 0 & 5, so how did we get 1/5 as the probability? I'm unable to visualize this problem. Can you kindly explain it to me in terms of successful outcomes/total outcomes? Thanks for your help...
Visualization is exactly what should help. Consider a number line: {total} is line segment of 5 units (from 0 to 5) and {favorable} is a line segment of 1 unit (from 4 to 5), thus P = {favorable}/{total} = 1/5.
Hope it's clear.
Hi,
1- Can sbdy please edit the question so that OA is not spoiled?
2- I don't agree with OA = 2/5, since question says that we use the longer side to create the square. Therefore no distinction is made between the "two" sides of the rope. Only the longer side is considered for our purpose. What I am saying is that I would pick 1/5 as the answer.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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09 Nov 2013, 21:07
Bunuel wrote:
MisterEko wrote:
Hm, not sure I understood you (or that you understood my question). What I said is that since the piece cut has to be bigger than 4, even marginally bigger, probability will be slightly less than 2/5. Say we can only count in cm. Wire is 500 cm long. In order for square to have area more than 1 meter, perimeter needs to be at least slightly bigger than 400 cm. In that case, the next least place that wire needs to be cut on (and satisfy the area of a square being bigger than 1) would be on 99th cm (down to the 1st cm) or 401st cm (up to 499). Either way, there are 99 positions on the first meter and 99 on the second one that satisfy the length needed. Since the total is 500 cm, probability would be 2*99/500 or 99/250.
Please, forgive me if I am being nuisance, I am kinda intrigued by this.
OK, let me ask you this: what is the probability that the wire will be cut so that we get pieces of exactly 4m and 1m (so at 1m or at 4m)?
What I'm saying is that the probability that the length of a longer piece will be more than 4 OR more than or equal to 4 is the same and equal to 2/5.
I think the answer 2/5 is flawed. Let us take five points 1, 2, 3, 4 and 5. For the longer wire to be more than or equal to 4m, 2 cases are there, it has to be cut at 1m or before OR 4m or after.
Let us take the first case. For example if it is cut at 1m, the longer wire would be 4m. Assume that each 1m is divided into cms. So there are 99 points out of a total of 500 points when the longer wire can be more than 4 m. for example if it is cut at 99th cm the longer wire would be 401 cm or > 4m. and so on. So there are totally 100 points including the exact 1m point out of a total of 500 points. So the probability that the longer wire is equal to or more than 4m is 100/500=1/5. The other case is wire cut at 4m and above. Each case has a probability of 1/5 so that the overall probability is 2/5.
The above is also equivalent to saying that the shorter wire is <= 1m. That is the probability is 2/5 when the longer wire is 4m and the shorter wire is 1m and say when the longer wire is 4.01 m and the shorter wire is 0.99m and so on . But saying equal to or more than 4m is not equivalent to saying less than 1m but only equivalent to saying less than or equal to 1m.
So when we have the shorter wire to be less than 1m, we have to consider only the cases > 4m. When it has to be more than 4m, the probability will be less than the above or less than 2/5.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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26 Nov 2013, 18:03
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A. 1/6
B. 1/5
C. 3/10
D. 1/3
E. 2/5
OA:
That the area of the square is more than 1 square meter means that the perimeter of the square is more than 4 meter. Imagine the wire is divided into 5 pieces:
0__1__2__3__4__5
I see that if we cut the wire at any point from 0 to 1 or any point from 4 to 5, we will have a long wire whose perimeter is more than 4 meter.
If we cut the wire at any point from 1 to 4, we get a long wire whose perimeter is less than 4 meter.
Undoubtedly, we have 3 choices if we cut the wire: from 0 to 1, from 1 to 4, and from 4 to 5. Following this reasoning, I think the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point should be 2/3.
Please explain what is wrong with my explanation?
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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26 Nov 2013, 20:57
Your explanation is close, but take another look at your suggested cuts.
If we divide the potential cuts into 0 to 1, 1 to 4, and 4 to 5, we have a 1:3:1 ratio where either end will give us the square with area >1.
Add up the parts to get 5, 2 of which are successful outcomes, and you get a probability of 2/5.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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27 Nov 2013, 00:50
phammanhhiep wrote:
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A. 1/6
B. 1/5
C. 3/10
D. 1/3
E. 2/5
OA:
That the area of the square is more than 1 square meter means that the perimeter of the square is more than 4 meter. Imagine the wire is divided into 5 pieces:
0__1__2__3__4__5
I see that if we cut the wire at any point from 0 to 1 or any point from 4 to 5, we will have a long wire whose perimeter is more than 4 meter.
If we cut the wire at any point from 1 to 4, we get a long wire whose perimeter is less than 4 meter.
Undoubtedly, we have 3 choices if we cut the wire: from 0 to 1, from 1 to 4, and from 4 to 5. Following this reasoning, I think the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point should be 2/3.
Please explain what is wrong with my explanation?
Merging similar topics. Please search before posting.
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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17 Jan 2015, 12:00
3
2/5
Need > 4 meter
and cutting the 5 m wire from 0-1m & 4-5 m gives us the lengthy wire to be >4m
p = (1m+1m)/5m = 2/5
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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17 Jan 2015, 12:02
1
Good question...initially didnt consider the other side of thread so got 1/5....but soon realized the missing part
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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14 Apr 2016, 19:56
2
Alternate solution:
Length of wire = 5m
If the markings are such that each metre has 10 subdivisions then a longer piece of wire can be obtained at 2.6, 2.7, 2.8, .... 5.0 --> 25 ways
Area of square > 1 when Perimeter > 4.
Perimeter > 4 when the wire is cut at points > 4.0 m --> 4.1, 4.2, .... 5.0 --> 10 ways
Probability = 10/25 = 2/5
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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24 Aug 2016, 21:58
MisterEko wrote:
A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?
A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5
Let the place where the rope is cut be $$x$$ and it's used to form a square with $$x$$ as perimeter and hence side is $$\frac{x}{4}$$ and Area =$$\frac{x^2}{4}$$
The area should be more than 1 so $$x^2/4 > 1$$ , so $$x^2 > 4$$ giving $$x > 2$$.
Out the places where it can be cut, 1,2,3,4 , 3 and 4 are above 2, so $$\frac{2}{5}$$ Answer is E
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Re: A 5 meter long wire is cut into two pieces. If the longer [#permalink]
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25 Sep 2016, 00:13
I would very much appreciate if someone can point out the flaw in my logic
the line is cut into 2 pieces ....x , 5-x
assuming x is the larger piece then x>2.5....... ( the larger piece)
side of square = x/4 , area of square = x^2/16 , question is to test x^2/16>1
we get ............-4................4.............. the inequality is true in the ranges x<-4 and x>4 but we have a restriction that x>2.5 thus x>4
thus 4<x<5 , thus probability = (5-4)/5 = 1/5 ............Bunuel where am i going wrong if you plz.
Re: A 5 meter long wire is cut into two pieces. If the longer &nbs [#permalink] 25 Sep 2016, 00:13
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# Can an object maintain uniform motion without any external force?
• Silverbeam
In summary, a Geoscience student sought help with understanding and testing the law of uniform motion. They devised an experiment involving driving a car in a series of circles and found that the car sped up when it straightened out, indicating the application of acceleration. However, their flat earther acquaintance pointed out the need to account for friction and drag on the vehicle due to centrifugal force. The student is now seeking help with calculating the forces involved and the necessary acceleration to maintain uniform motion. However, it seems unlikely that their flat earther acquaintance will accept any evidence presented to them.
Silverbeam said:
Perhaps. I thought an object in uniform circular motion went at a constant speed, so that it would always take the same amount of time to travel indentical portions of the circle.
That didnt answer Dale's Q to you
do you understand the difference between speed and velocity ?
Dale said:
No. It is turning. Do you understand the difference between speed and velocity? Perhaps that is the problem.
Sorry, I didn't answer your question. No, I don't understand the difference between speed and velocity.
jbriggs444 and davenn
russ_watters said:
For the ISS, the two accelerations are aligned and subtract to zero.
That's what I'm talking about. The ISS is accelerated in classical physics but the accelerometer shows zero. It doesn't work this way.
etotheipi
Silverbeam said:
Sorry, I didn't answer your question. No, I don't understand the diffrence between speed and velocity.
Velocity is what is called a vector. That means that it has a magnitude and a direction. You can think of it as an arrow. The magnitude is its length and its direction is the direction it points.
For velocity, the magnitude is called speed. So the speed is what is left of velocity when you remove the direction. But acceleration is a change in velocity, not a change in speed. So to determine acceleration you have to keep the direction information, not just the magnitude information. That is what is changing in uniform circular motion.
davenn
Dale said:
Velocity is what is called a vector. That means that it has a magnitude and a direction. You can think of it as an arrow. The magnitude is its length and its direction is the direction it points.
For velocity, the magnitude is called speed. So the speed is what is left of velocity when you remove the direction. But acceleration is a change in velocity, not a change in speed. So to determine acceleration you have to keep the direction information, not just the magnitude information. That is what is changing in uniform circular motion.
Ok, thanks.
russ_watters said:
No, it does work for both the car and the ISS. The phone has a 3-axis accelerometer, and the acceleration measured when stationary on the surface of the Earth is 1.0g vertically. The acceleration from the car turning is measured separately, in a perpendicular axis. For the ISS, the two accelerations are aligned and subtract to zero.
I don't really know what you mean by 'the two accelerations are aligned and subtract to zero' for the ISS case. When we're thinking in terms of GR and all that stuff, if we ignore the atmosphere then there's no force at all on the ISS (apart from the tidal forces, which we can ignore on the small scales), and so no proper acceleration. Rather than 'two accelerations cancelling', probably clearer to say that there was no proper acceleration in the first place.
Dale said:
Velocity is what is called a vector. That means that it has a magnitude and a direction. You can think of it as an arrow. The magnitude is its length and its direction is the direction it points.
For velocity, the magnitude is called speed. So the speed is what is left of velocity when you remove the direction. But acceleration is a change in velocity, not a change in speed. So to determine acceleration you have to keep the direction information, not just the magnitude information. That is what is changing in uniform circular motion.
So, if I want to approach a corner in my car, driving in a straight line, then take the corner, then drive straight ahead, all at a constant speed, do I need to provide any additional force to the car while taking the corner in order to maintain the constant speed throughout?
Silverbeam said:
So, if I want to approach a corner in my car, driving in a straight line, then take the corner, then drive straight ahead, all at a constant speed, do I need to provide any additional force to the car while taking the corner in order to maintain the constant speed throughout?
Friction from the road surface provides the necessary inward centripetal component of force. also of course this is an idealisation
In general when a body moves along a trajectory with radius of curvature ##\rho## at constant speed ##v##, there is acceleration of magnitude ##v^2/\rho## directed toward the instantaneous centre of curvature, which must be caused by a net force also directed toward the instantaneous centre of curvature.
(n.b. that in most general case where ##\dot{v} \neq 0##, you will obtain an additional component of acceleration ##\dot{v} \boldsymbol{e}_t## as well as the inward component ##(v^2 / \rho) \boldsymbol{e}_n##)
Last edited by a moderator:
etotheipi said:
friction from the road surface provides the necessary inward centripetal component of force. also of course this is an idealisation
in general when a body moves along a trajectory with radius of curvature ##\rho## at constant speed ##v##, there is acceleration of magnitude ##v^2/\rho## directed toward the instantaneous centre of curvature, which must be caused by a net force also directed toward the instantaneous centre of curvature
So, I would not need to apply any more acceleration to the vehicle while driving in identical circles at a constant speed, than the acceleration I would need to apply if I were driving in a straight line at a constant speed?
No that's not what i said at all! a body traveling in a straight line at constant speed undergoes zero acceleration. n.b. that this is consistent with the relation i wrote, because a straight line has radius of curvature ##\rightarrow \infty##.
Silverbeam said:
So, if I want to approach a corner in my car, driving in a straight line, then take the corner, then drive straight ahead, all at a constant speed, do I need to provide any additional force to the car while taking the corner in order to maintain the constant speed throughout?
For a car, yes, but for a marble on a smooth track that turns, no. The energy required for the car has to do with inefficiency, not fundamental physical principles.
PeroK
Dale said:
For a car, yes, but for a marble on a smooth track that turns, no. The energy required for the car has to do with inefficiency, not fundamental physical principles.
Ok, but for a marble on a track it has the sides of the track to keep it in a circle. What about a marble going in circles without a track? Doesn't it need some force to be applied to it, either attractive or repulsive, in order to constantly change its direction and keep it in a circle? Isn't that why gravity is invoked for orbits, to provide a force that maintains the inward acceleration to prevent the Moon or whatever from just going off in a straight line?
hmmm27 and etotheipi
Silverbeam said:
Ok, but for a marble on a track it has the sides of the track to keep it in a circle. What about a marble going in circles without a track? Doesn't it need some force to be applied to it, either attractive or repulsive, in order to constantly change its direction and keep it in a circle? Isn't that why gravity is invoked for orbits, to provide a force that maintains the inward acceleration to prevent the Moon or whatever from just going off in a straight line?
yes!
Silverbeam said:
Doesn't it need some force to be applied to it, either attractive or repulsive, in order to constantly change its direction and keep it in a circle?
Yes, it does.
I may have misunderstood your question earlier. I thought you were asking if you would need to step on the gas in a car to maintain speed through a turn. It is not necessary for a marble to “step on the gas” as it were. It needs the track, but not a push along the track, unlike a car which needs the push too.
Silverbeam said:
Isn't that why gravity is invoked for orbits, to provide a force that maintains the inward acceleration to prevent the Moon or whatever from just going off in a straight line?
Yes.
etotheipi said:
yes!
But Dale says the only additional force I need to turn a corner in my car with constant speed is due to inefficiency, not a fundamental physical principle. Why doesn't the same additional force need to be required for the car as for the Moon?
Dale said:
I may have misunderstood your question earlier. I thought you were asking if you would need to step on the gas in a car to maintain speed through a turn. It is not necessary for a marble to “step on the gas” as it were.
That is what I meant.
Silverbeam said:
But Dale says the only additional force I need to turn a corner in my car with constant speed is due to inefficiency, not a fundamental physical principle. Why doesn't the same additional force need to be required for the car as for the Moon?
sorry, I don't understand what you mean about inefficiency and fundamental physics principles
if the centre of mass of the car is accelerating at some ##\ddot{\mathbf{x}} = (v^2 / \rho) \mathbf{e}_n##, then the net force on the car must be ##m\ddot{\mathbf{x}} = (mv^2 / \rho)\mathbf{e}_n##. the only external force on the car in such a situation, which can provide that force, is friction.
You now understand the difference between velocity and speed. Acceleration is a change in velocity. Acceleration, like velocity, is a vector so it has a direction as well as a magnitude. When acceleration is parallel to the velocity then it changes the speed. When acceleration is perpendicular to the velocity then it changes the direction of the velocity.
Force is proportional to acceleration. So in order to go in circular motion you need a force that is perpendicular to the velocity. That force is required in order to turn.
When a marble turns on a smooth track the force is perpendicular to the velocity. So a marble can turn without any additional push from behind. It only needs the push from the side given by the track.
When a car turns, because of how tires work, the force is not perfectly perpendicular to the velocity. There is a little bit of force going anti-parallel to the velocity. So you need a touch of gas to counteract that. You need the push from the side but also a bit of a push from behind due to the inefficiency of turning with tires.
PeroK, Delta2 and etotheipi
Dale said:
When a car turns, because of how tires work, the force is not perfectly perpendicular to the velocity. There is a little bit of force going anti-parallel to the velocity. So you need a touch of gas to counteract that. You need the push from the side but also a bit of a push from behind due to the inefficiency of turning with tires.
I think the key point to understand here is that no matter how complex the many different forces between the tyres and the road, the sum of all these forces from the road as well as of course the weight (which, neglecting the air, are the only net external forces on the car) is always proportional [parallel] to the acceleration of the centre of the mass of the car
jbriggs444, Delta2 and Dale
If you take a pinball and shoot it directly at (the inside of) a V, it will bounce off one arm, go sideways, bounce off the other, then run back where it came from, experiencing no loss of speed at any time. If you were inside the pinball, you'd get creamed at each bounce because of the massive instantaneous change in velocity.
Now shoot it at a U ; it slides down the inside, curves around the bottom, then procedes to roll back in the opposite direction. If you were inside that one you'd be much more comfortable, since the changing velocity (ie: acceleration) is spread out over time. Like the V example, the speed doesn't change throughout the procedure.
etotheipi said:
I think the key point to understand here is that no matter how complex the many different forces between the tyres and the road, the sum of all these forces from the road as well as of course the weight (which, neglecting the air, are the only net external forces on the car) is always proportional [parallel] to the acceleration of the centre of the mass of the car
Yes agreed.
I only explain about tires because if they actually do the experiment they will find that they do slow down. That isn’t because the theory is wrong, but because cars are complicated and messy machines. The marbles are simpler and easier to analyze. I always prefer simple experiments.
etotheipi
Dale said:
but because cars are complicated and messy machines. The marbles are simpler and easier to analyze. I always prefer simple experiments.
Yes, exactly! Much better to build up complexity gradually, rather than starting with an extremely complex mechanism like a car! There is much to be gained by carefully understanding the dynamics of a single particle, then generalising to a system of particles, then a system of interacting particles, continuous bodies etc.
e.g. your example of the marble is a much better tool to understand the general motion of a particle than the example with the car that the op kept referring to
Delta2 and Dale
DrStupid said:
That's what I'm talking about. The ISS is accelerated in classical physics but the accelerometer shows zero. It doesn't work this way.
"Shows zero" and "sums to zero" is the same thing here. I'll explain:
etotheipi said:
I don't really know what you mean by 'the two accelerations are aligned and subtract to zero' for the ISS case. When we're thinking in terms of GR and all that stuff, if we ignore the atmosphere then there's no force at all on the ISS (apart from the tidal forces, which we can ignore on the small scales), and so no proper acceleration. Rather than 'two accelerations cancelling', probably clearer to say that there was no proper acceleration in the first place.
• A stationary or constant speed car with a 3-axis accelerometer reads 1.0g upward acceleration.
• This car has excellent tires. It turns a hard circle to the left, at 1g. The accelerometer reads 1.0g up and 1.0g to the left, for a sum of 1.4g, at a 45 degree angle from vertical.
• Now the car does the same turn on a 45 degree banked track. To be clear: I mean the same turn, in the plane of the turn, not in the horizontal (Earth's surface) plane. Now we have the upwards acceleration of 1g and a downward and to the left acceleration of 1g at the top of the turn; add them together and you get a resultant 0.7g. Note: the accelerometer doesn't read them as 1g in each axis here because they are starting to oppose each other.
• Now the car does the same "turn" in a loop. At the top of the loop, the accelerometer reads...
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russ_watters said:
• This car has excellent tires. It turns a hard circle to the left, at 1g. The accelerometer reads 1.0g up and 1.0g to the left, for a sum of 1.4g, at a 45 degree angle from vertical.
• Now the car does the same turn on a 45 degree banked track. To be clear: I mean the same turn, in the plane of the turn, not in the horizontal (Earth's surface) plane. Now we have the upwards acceleration of 1g and a downward and to the left acceleration of 1g at the top of the turn; add them together and you get a resultant 0.7g. Note: the accelerometer doesn't read them as 1g in each axis here because they are starting to oppose each other.
A banked turn just means that there is less/none lateral tire friction needed, because the normal force has a centripetal component. If the trajectory of the car is still the same as for the flat track, the proper acceleration will be of the same magnitude.
Or do you mean the entire circular flat track is now on an inclined plane?
etotheipi
Acceleration can be resolved into tangential and normal component (check wikipedia link at bottom of this post). The tangential component has direction same as the direction of velocity, while the normal component has direction normal (or equivalently perpendicular) to the direction of velocity.
Seems like your Earth flat friend understands only tangential acceleration and completely ignores normal(or centripetal) acceleration. Tangential acceleration is what causes the velocity to change in magnitude (or simply for an object to speed up or down as you say) but normal acceleration is what causes the velocity to change in direction.
To sum it up: Velocity is a vector it has both magnitude and direction. Tangential acceleration changes its magnitude, normal acceleration changes its direction. Simple as that, tell that to your friend and let me know what he thinks about.
https://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration
Last edited:
Silverbeam said:
But Dale says the only additional force I need to turn a corner in my car with constant speed is due to inefficiency, not a fundamental physical principle. Why doesn't the same additional force need to be required for the car as for the Moon?
Your basic mistake here is to try to understand the relative complexity of moving on a surface using an engine, rubber tires and friction on the surface - where accelerating, braking and turning are not trivial mechanical processes.
This is why physics is usually begun by studying Kinematics - studying motion without reference to the causes of motion. A good example is, of course, uniform circular motion. You learn about velocity and acceleration vectors, kinetic energy etc.
There is an interesting parallel here between your struggles and why it look until 1687 for someone (Isaac Newton) to formulate the laws of motion. No one previously had seen through the complexities of everyday motion to realize that there were basic fundamental laws governing all motion. And that there was a relationship between the motion of objects on Earth, and the motion of the planets around the Sun. For example, if we look at Newton's first law:
An object will remain at rest or move with constant speed in a straight line, unless acted upon by an unbalanced force.
This flew in the face of previous "Aristolelian" wisdom, which asserted that objects naturally slow down and need a force to keep them moving. And so, the hand of God was needed to keep the planets moving through the heavens, as it were.
Newton was the first to realize that any slowing down is the result of external forces. And, although we cannot avoid these forces on Earth, the movement of the planets do not slow because there are no dissipative forces like air resistance and friction. To quote Newton himself:
Projectiles persevere in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motion, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.
May I offer a personal view here that it is a tragedy that you, a 21st Century science student, are ignorant of this.
Note that I've underlined where Newton also pointed out that uniform circular motion (in the example of a top - which is an old-fashioned spinning toy) carries on indefinitely unless slowed by the air.
In short, you are making the same mistake as Aristotle and his disciples that energy is needed to maintain circular motion. But, Newton said otherwise in 1687. Note that:
a) A car, as explained above, is relatively inefficient at turning and will slow down relatively quickly.
b) A spinning top may continue for a few minutes perhaps, but air resistance gradually slows it down.
c) The Moon, having neither friction not air resistance to contend with, may continue its orbit about the Earth almost indefinitely.
This is very much the starting point for modern science.
Last edited:
jbriggs444, Nugatory, DrStupid and 2 others
russ_watters said:
• A stationary or constant speed car with a 3-axis accelerometer reads 1.0g upward acceleration.
• This car has excellent tires. It turns a hard circle to the left, at 1g. The accelerometer reads 1.0g up and 1.0g to the left, for a sum of 1.4g, at a 45 degree angle from vertical.
• Now the car does the same turn on a 45 degree banked track. To be clear: I mean the same turn, in the plane of the turn, not in the horizontal (Earth's surface) plane. Now we have the upwards acceleration of 1g and a downward and to the left acceleration of 1g at the top of the turn; add them together and you get a resultant 0.7g. Note: the accelerometer doesn't read them as 1g in each axis here because they are starting to oppose each other.
• Now the car does the same "turn" in a loop. At the top of the loop, the accelerometer reads...
I don't quite see what you're getting at here. To compute the proper acceleration in classical physics, you can just sum all of the forces except weight, and divide by the mass.
- The car stationary on the Earth's surface reads 1.0g upward acceleration since there's only an upward normal force on it
- When it does the hard turn, it has an upward normal force on it, and an inward frictional force on it, so the proper acceleration is, say, 1.4g at 45 degrees to the horizontal.
- If it does the turn on a frictionless banked track, it has a normal force with both an upward and horizontal component (this time the horizontal component is provided by the normal force, and not friction), and once again the proper acceleration is, say, 1.4g at 45 degrees to the horizontal
- At the top of the loop the loop, the normal force depends on how fast the car's going, and the proper acceleration will point downward toward the ground with some as of yet unknown magnitude.
- For the ISS, there's no forces at all, and its proper acceleration is zero.
Last edited by a moderator:
russ_watters said:
"Shows zero" and "sums to zero" is the same thing here.
Of course it is. The wording is irrelevant. The centripetal acceleration of the ISS cannot be measured with an accelerometer - no matter how you phrase it.
Dale
Well I didn't mean to start an argument, I was just bringing up the obvious fact that I can write the number 69 as 69 + 8 - 8 if I wanted to, and just add random stuff and subtract it again, but that's detracting from the point.
for the ISS it's not a case of 'two accelerations cancelling', it's a case of there being no proper accelerations at all in the first place. that seems pretty conceptually clear to me?
DrStupid
Silverbeam said:
To make his point he asked what a car driving around and around in a circle would do if it were accelerating. He says it must speed up.
You're arguing semantics. Acceleration is not equal to the rate of change of speed. It's equal to the rate of change of velocity.
Ask your friend if he can drive his car in a circle on wet ice with bald tires. You need a force of friction between the tires and the road surface to move in a circle.
Acceleration is a change in the speed and/or the direction of an object. Both a change in speed or a change in direction require an external force. Ask your friend if there's any way to change the speed or direction of an automobile without applying an external force--either the force of friction from the front tires on a turn, or by gunning the engine to make the car go faster.
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# Search by Topic
#### Resources tagged with Investigations similar to Move a Match:
Filter by: Content type:
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### There are 146 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Investigations
### Move a Match
##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
### Little Boxes
##### Stage: 2 Challenge Level:
How many different cuboids can you make when you use four CDs or DVDs? How about using five, then six?
##### Stage: 2 Challenge Level:
How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
### Ice Cream
##### Stage: 2 Challenge Level:
You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
### Newspapers
##### Stage: 2 Challenge Level:
When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
### Hexpentas
##### Stage: 1 and 2 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### Halloween Investigation
##### Stage: 2 Challenge Level:
Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Play a Merry Tune
##### Stage: 2 Challenge Level:
Explore the different tunes you can make with these five gourds. What are the similarities and differences between the two tunes you are given?
### Escher Tessellations
##### Stage: 2 Challenge Level:
This practical investigation invites you to make tessellating shapes in a similar way to the artist Escher.
### Two on Five
##### Stage: 1 and 2 Challenge Level:
Take 5 cubes of one colour and 2 of another colour. How many different ways can you join them if the 5 must touch the table and the 2 must not touch the table?
### Making Cuboids
##### Stage: 2 Challenge Level:
Let's say you can only use two different lengths - 2 units and 4 units. Using just these 2 lengths as the edges how many different cuboids can you make?
### Two by One
##### Stage: 2 Challenge Level:
An activity making various patterns with 2 x 1 rectangular tiles.
### Sticky Triangles
##### Stage: 2 Challenge Level:
Can you continue this pattern of triangles and begin to predict how many sticks are used for each new "layer"?
### Cuboid-in-a-box
##### Stage: 2 Challenge Level:
What is the smallest cuboid that you can put in this box so that you cannot fit another that's the same into it?
### Street Party
##### Stage: 2 Challenge Level:
The challenge here is to find as many routes as you can for a fence to go so that this town is divided up into two halves, each with 8 blocks.
### Redblue
##### Stage: 2 Challenge Level:
Investigate the number of paths you can take from one vertex to another in these 3D shapes. Is it possible to take an odd number and an even number of paths to the same vertex?
### 3 Rings
##### Stage: 2 Challenge Level:
If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
### Three Sets of Cubes, Two Surfaces
##### Stage: 2 Challenge Level:
How many models can you find which obey these rules?
### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Calcunos
##### Stage: 2 Challenge Level:
If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
### Egyptian Rope
##### Stage: 2 Challenge Level:
The ancient Egyptians were said to make right-angled triangles using a rope with twelve equal sections divided by knots. What other triangles could you make if you had a rope like this?
### Four Layers
##### Stage: 1 and 2 Challenge Level:
Can you create more models that follow these rules?
### Sticks and Triangles
##### Stage: 2 Challenge Level:
Using different numbers of sticks, how many different triangles are you able to make? Can you make any rules about the numbers of sticks that make the most triangles?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Cutting Corners
##### Stage: 2 Challenge Level:
Can you make the most extraordinary, the most amazing, the most unusual patterns/designs from these triangles which are made in a special way?
### Tri.'s
##### Stage: 2 Challenge Level:
How many triangles can you make on the 3 by 3 pegboard?
### Fencing
##### Stage: 2 Challenge Level:
Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Double Your Popcorn, Double Your Pleasure
##### Stage: 2 Challenge Level:
We went to the cinema and decided to buy some bags of popcorn so we asked about the prices. Investigate how much popcorn each bag holds so find out which we might have bought.
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Triangle Pin-down
##### Stage: 2 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### The Numbers Give the Design
##### Stage: 2 Challenge Level:
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### How Tall?
##### Stage: 1 and 2 Challenge Level:
A group of children are discussing the height of a tall tree. How would you go about finding out its height?
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Building with Rods
##### Stage: 2 Challenge Level:
In how many ways can you stack these rods, following the rules?
### It's a Fence!
##### Stage: 1 and 2 Challenge Level:
In this challenge, you will work in a group to investigate circular fences enclosing trees that are planted in square or triangular arrangements.
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Watch Your Feet
##### Stage: 2 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Cubes Here and There
##### Stage: 2 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### More Transformations on a Pegboard
##### Stage: 2 Challenge Level:
Use the interactivity to find all the different right-angled triangles you can make by just moving one corner of the starting triangle.
### Room Doubling
##### Stage: 2 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Buying a Balloon
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Teddy Town
##### Stage: 1, 2 and 3 Challenge Level:
There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
### Lost Books
##### Stage: 2 Challenge Level:
While we were sorting some papers we found 3 strange sheets which seemed to come from small books but there were page numbers at the foot of each page. Did the pages come from the same book?
### Triangle Shapes
##### Stage: 1 and 2 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
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# Machine Learning Explained: Vectorization and matrix operations
4
Today in Machine Learning Explained, we will tackle a central (yet under-looked) aspect of Machine Learning: vectorization. Let’s say you want to compute the sum of the values of an array. The naive way to do so is to loop over the elements and to sequentially sum them. This naive way is slow and tends to get even slower with large amounts of data and large data structures.
With vectorization these operations can be seen as matrix operations which are often more efficient than standard loops. Vectorized versions of algorithm are several orders of magnitudes faster and are easier to understand from a mathematical perspective.
## A basic exemple of vectorization
### Preliminary exemple – Python
Let’s compare the naive way and the vectorized way of computing the sum of the elements of an array. To do so, we will create a large (100,000 elements) Numpy array and compute the sum of its element 1,000 times with each algorithm. The overall computation time will then be compared.
import numpy as np
import time
W=np.random.normal(0,1,100000)
n_rep=1000
The naive way to compute the sum iterates over all the elements of the array and stores the sum:
start_time = time.time()
for i in range(n_rep):
loop_res=0
for elt in W:
loop_res+=elt
time_loop = time.time() - start_time
If is our vector of interest, the sum of its elements can be expressed as the dot product :
start_time = time.time()
for i in range(n_rep):
one_dot=np.ones(W.shape)
vect_res=one_dot.T.dot(W)
time_vect = time.time() - start_time
Finally, we can check that both methods yield the same results and compare their runtime. The vectorized version run approximately 100 to 200 times faster than the naive loop.
print(np.abs(vect_res-loop_res)<10e-10)
print(time_loop/time_vect)
Note: The same results can be obtained with np.sum.The numpy version has a very similar runtime to our vectorized version. Numpy being very optimized, this show that our vectorized sum is reasonably fast.
start_time = time.time()
for i in range(n_rep):
vect_res=np.sum(W)
time_vect_np = time.time() - start_time
### Preliminary exemple – R
The previous experiments can be replicated in R:
##Creation of the vector
W=matrix(rnorm(100000))
n_rep=10000
#Naive way:
library(tictoc)
tic('Naive computation')
for (rep in 1:n_rep)
{
res_loop=0
for (w_i in W)
res_loop=w_i+res_loop
}
toc()
tic('Vectorized computation')
# vectorized way
for (rep in 1:n_rep)
{
ones=rep(1,length(W))
res_vect= crossprod(ones,W)
}
toc()
tic('built-in computation')
# built-in way
for (rep in 1:n_rep)
{
res_built_in= sum(W)
}
toc()
In R, the vectorized version is only an order of magnitude faster than the naive way. The built-in way achieves the best performances and is an order of magnitude faster than our vectorized way.
### Preliminary exemple – Results
Vectorization divides the computation times by several order of magnitudes and the difference with loops increase with the size of the data. Hence, if you want to deal with large amount of data, rewriting the algorithm as matrix operations may lead to important performances gains.
## Why vectorization is (often) faster
• R and Python are interpreted language, this means that your instructions are analyzed and interpreted at each execution. Since they are not statically typed, the loops have to assess the type of the operands at each iteration which leads to a computational overhead.
• R and Python linear algebra relies on optimized back-end for matrix operations and linear algebra. These back-end are written in C/C++ and can process loops efficiently. Furthermore, while loops are sequential, these back-end can run operations in parallel which improves the computation speed on modern CPU.
Note 1: Though vectorization is often faster, it requires to allocate the memory of the array. If your amount of RAM is limited or if the amount of data is large, loops may be required.
Note 2: When you deal with large arrays or computationally intensive algebra ( like inversion, diagonalization, eigenvalues computations, ….) computations on GPU are even order of magnitudes faster than on CPU. To write efficient GPU code, the code needs to be composed of matrix operations. Hence, having vectorized code maked it easier to translate CPU code to GPU (or tensor-based frameworks).
## A small zoo of matrix operations
The goal of this part is to show some basic matrix operations/vectorization and to end on a more complex example to show the thought process which underlies vectorization.
### Column-wise matrix sum
The column wise sum (and mean) can be expressed as a matrix product. Let be our matrix of interest. Using the matrix multiplication formula, we have: . Hence, the column-wise sum of is .
Python code:
def colWiseSum(W):
ones=np.ones((W.shape[0],1))
return ones.T.dot(W)
R code:
colWiseSum=function(W)
{
ones=rep(1,nrow(W))
crossprod(W,ones)
}
### Row-wise matrix sum
Similarly, the row-wise sum is .
Python code:
def rowWiseSum(W):
ones=np.ones((W.shape[1],1))
return W.dot(ones)
R code:
rowWiseSum=function(W)
{
ones=rep(1,ncol(W))
W%*%ones
}
### Matrix sum
The sum of all the elements of a matrix is the sum of the sum of its rows. Using previous expression, the sum of all the terms of is .
Python code:
def matSum(W):
rhs_ones=np.ones((W.shape[1],1))
lhs_ones=np.ones((W.shape[0],1))
return lhs_ones.T.dot(W).dot(rhs_ones)
R code:
matSum=function(W)
{
rhs_ones=rep(1,ncol(W))
lhs_ones=rep(1,nrow(W))
crossprod(lhs_ones,W)%*% rhs_ones
}
### Similarity matrix (Gram matrix)
Let’s say we have a set of words and for each of this words we want to find the most similar words from a dictionary. We assume that the words have been projected in space of dimension (using word2vect). Let (our set of words) and (our dictionary) be two matrices resp. in and . To compute the similarity of all the observations of and we simply need to compute .
Python code:
def gramMatrix(X,Y):
return X.dot(Y.T)
R code:
gramMatrix=function(X,Y)
{
tcrossprod(X,t(Y))
}
### L2 distance
We want to compute the pair-wise distance between two sets of vector. Let and be two matrix in and . For each vector of we need to compute the distance with all the vectors of .Hence, the output matrix should be of size .
If and are two vectors, their distance is:
. To compute all pairwise distance, some work is required on the last equality. All the matrices should be of size , then the output vector of distance will be of size , which can be reshaped into a vector of size .
The first two terms and need to be computed for each and . is the element-wise multiplication of X with itself (its elements are ). Hence, the i-th element of is the squared sum of the coordinate of the i-th observation, .
However, this is a vector and its shape is . By replicating each of its elements times, we will get a matrix of size . The replication can be done easily, if we consider the right matrix .
Let be a vector of one of size . Let be the matrix of size with repetitions of on the “diagonal”:
Then, our final vector is (The same expression holds for ). We denote a reshape operator (used to transform the previous vector in matrix). With previous part on similarity matrix, we get the following expression of the pairwise distance:
The previous expression can seem complex, but this will help us a lot to code the pairwise distance. We only have to do the translation from maths to Numpy or R.
Python code:
def L2dist(X,Y):
n_1=X.shape[0]
n_2=Y.shape[0]
p=X.shape[1]
ones=np.ones((p,1))
x_sq=(X**2).dot(ones)[:,0]
y_sq=(Y**2).dot(ones)[:,0]
delta_n1_n2=np.repeat(np.eye(n_1),n_2,axis=0)
delta_n2_n1=np.repeat(np.eye(n_2),n_1,axis=0)
return np.reshape(delta_n1_n2.dot(x_sq),(n_1,n_2))+np.reshape(delta_n2_n1.dot(y_sq),(n_2,n_1)).T-2*gramMatrix(X,Y)
R Code:
L2dist=function(X,Y)
{
n_1=dim(X)[1]
n_2=dim(Y)[1]
p=dim(X)[2]
ones=rep(1,p)
x_sq=X**2 %*% ones
x_sq=t(matrix(diag(n_1) %x% rep(1, n_2) %*% x_sq, n_2,n_1))
y_sq=Y**2 %*% ones
y_sq=matrix(diag(n_2) %x% rep(1, n_1) %*% y_sq,n_1,n_2)
x_sq+y_sq-2*gramMatrix(X,Y)
}
### L2 distance (improved)
Actually the previous L2dist is not completely optimized requires a lot of memory since has cells and is mostly empty. Using Numpy built-in function, we can circumvent this multiplication by directly repeating the vector (which reduces the memory footprints by a factor ):
Python code:
def L2dist_improved(X,Y):
n_1=X.shape[0]
n_2=Y.shape[0]
p=X.shape[1]
ones=np.ones((p,1))
x_sq=(X**2).dot(ones)[:,0]
y_sq=(Y**2).dot(ones)[:,0]
##Replace multiplication by a simple repeat
X_rpt=np.repeat(x_sq,n_2).reshape((n_1,n_2))
Y_rpt=np.repeat(y_sq,n_1).reshape((n_2,n_1)).T
return X_rpt+Y_rpt-2*gramMatrix(X,Y)
R code:
L2dist_improved=function(X,Y)
{
n_1=dim(X)[1]
n_2=dim(Y)[1]
p=dim(X)[2]
ones=rep(1,p)
x_sq=X**2 %*% ones
x_sq=t(matrix(rep(x_sq,each=n_2),n_2,n_1))
y_sq=Y**2 %*% ones
y_sq=matrix(rep(y_sq,each=n_1),n_1,n_2)
x_sq+y_sq-2*gramMatrix(X,Y)
}
Note : Actually, this code can be made even shorter and more efficient by using a custom outer product (Thanks to Florian Privé for the solution):
L2dist_improved2 <- function(X, Y) {
xn <- rowSums(X ** 2)
yn <- rowSums(Y ** 2)
outer(xn, yn,'+') - 2 * tcrossprod(X, Y)
}
### L2 distance – benchmark
To show the interest of our previous work, let’s compare the computation speed of the vectorized L2 distance, the naive implementation and the scikit-learn implementation. The experiments are run on different size of dataset with 100 repetitions.
The vectorized implementation is 2 to 3 orders of magnitude faster than the naive implementation and on par with the scikit implementation.
Hi, I'm Antoine, a Data science and analytics student and enthusiast. My moto ? Visualising data to understand them, beautiful vis' bring more insights than complicated figures.
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1. Hello,
Sorry for being picky but your R code is very pythonish to my sense.
Two remarks:
– gramMatrix is tcrossprod
– not sure people use matrix operations to compute rowMeans in R
For the computation of the distance in base R, I would use:
L2dist_improved2 <- function(X, Y) {
xn <- rowSums(X ** 2)
yn <- rowSums(Y ** 2)
tcrossprod(xn, yn) – 2 * tcrossprod(X, Y)
}
Best,
Florian
• Hello Florian,
Thanks a lot for your feedback. You are right my R code is very pythonish, I directly translated it from the python code. Concerning the remarks:
– crossprod and tcrossprod would be better and faster, I’ll modify the post and replace the matrix products.
– In real life, whenever possible, I would use built-in operations (rowSums, colSums, sum, pdist, …) since they are faster and less error-prone. The matrix operations are here for the sake of demonstration and I should say it more clearly.
– Thanks for the improved version, but the function does not yield the same result as pdist. I think that tcrossprod(xn, yn) computes the product instead of the sum .
library(pdist)
X=matrix(rnorm(10000),100,100)
Y=matrix(rnorm(10000),100,100)
L2dist_improved2 <- function(X, Y) {
xn <- rowSums(X ** 2)
yn <- rowSums(Y ** 2)
tcrossprod(xn, yn) - 2 * tcrossprod(X, Y)
}
max(abs(sqrt(L2dist_improved2(X,Y))-as.matrix(pdist(X,Y))))
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# Math Challenge - August 2019
• Challenge
• Featured
Mentor
2021 Award
You are hard to follow.
##2x^6+3y^6=z^3##
It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).
Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).
We see that ##\frac{3y^3}{2} = \frac{1}{2} -x^3##
We see that ##x^3= \frac{1}{2} - t^3## and ##y= \frac{2t^3}{3}##
We do not "see" it. What we see is that you invented out of the blue a certain number ##t##. What you should have done is telling us, that you set ##t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}## such that ##y=\frac{2}{3}t^3##. But now ##x^3\neq \frac{1}{2}-t^3##.
To correct this, we have to set ##t:=\sqrt[3]{\frac{3}{2}}y##. Now ##x^3= \frac{1}{2}-t^3##.
If t is rational, then y is irrational.
How that?
If t is irrational, then x is irrational. Hence there are no solutions for x and y in rationals.
member 587159
Proof of fact. Suppose $S\in\Sigma$ has property $(P)$. Then $\Sigma _S$ is an infinite sub sigma algebra. Pick $T \in \Sigma _S \setminus \{\emptyset, S\}$ Write $S= T\cup (S\cap T^c)$. By (E) at least one of the respective sub sigma algebras must be infinite.
First of all, sorry for the late reply. I'm trying to understand this part of your answer. I guess you want to show that either ##\Sigma_T## or ##\Sigma_{S \cap T^c}## has property ##(P)##, right?
Can you please explain me why the following sentence is true?
"By (E) at least one of the respective sub sigma algebras must be infinite."
Can you please explain me why the following sentence is true?
"By (E) at least one of the respective sub sigma algebras must be infinite."
If ##\Sigma _T \cup \Sigma _{S\cap T^c}## was a finite set, then their generated sub sigma algebra, which is ##\Sigma _S##, would also be finite.
I guess you want to show that either ##\Sigma_T## or ##\Sigma_{S \cap T^c}## has property ##(P)##, right?
They can both have that property, too.
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8. Show that there is no triple ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## such that ##a^2 + b^2 = 3 c^2##.
After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.
I think there might be a simpler proof than what I came up with.
Since ##a^2 + b^2 = 3 c^2##, it must be the case that ##a^2 + b^2 \equiv 0 \mod 3##. Since ##(x + y) \mod n = x \mod n + y \mod n)##, we consider all possible combinations of ##a \mod 3## and ##b \mod 3## and find their effect on ##a^2 \mod 3 + b^2 \mod 3##. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if ##x \equiv k \mod n##, then ##x^2 \equiv k^2 \mod n## (since ##x = pn + k## for some integer ##p## and ##x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n##).
$$\begin{array}{|c|c|c|c|c|} \hline a \mod 3 & b \mod 3 & a^2 \mod 3 & b^2 \mod 3 & (a^2 + b^2) \mod 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 & 1 & 1 \\ \hline 0 & 2 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 & 2 \\ \hline 1 & 2 & 1 & 1 & 2 \\ \hline 2 & 2 & 1 & 1 & 2 \\ \hline \end{array}$$
So the only case where ##(a^2 + b^2) \mod 3 = 0## is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say ##a = 3A, b = 3B## for some non-zero integers ##A, B## (since question states a, b, c are all non-zero). Using this in the original expression, we get ##a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}##
Since ##A^2 + B^2## and hence ##\frac {c^2} {3}## is an integer, and ##c## is an integer and 3 is prime, it must be the case that ##c## too is a multiple of 3, say ##c = 3C## for some non-zero integer ##C##. So the above equivalence becomes, ##A^2 + B^2 = \frac {9C^2} {3} = 3C^2##, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form ##a^2 + b^2 = 3 c^2##, too is ruled out. Hence proved.
member 587159
member 587159
8. Show that there is no triple ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## such that ##a^2 + b^2 = 3 c^2##.
After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.
I think there might be a simpler proof than what I came up with.
Since ##a^2 + b^2 = 3 c^2##, it must be the case that ##a^2 + b^2 \equiv 0 \mod 3##. Since ##(x + y) \mod n = x \mod n + y \mod n)##, we consider all possible combinations of ##a \mod 3## and ##b \mod 3## and find their effect on ##a^2 \mod 3 + b^2 \mod 3##. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if ##x \equiv k \mod n##, then ##x^2 \equiv k^2 \mod n## (since ##x = pn + k## for some integer ##p## and ##x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n##).
$$\begin{array}{|c|c|c|c|c|} \hline a \mod 3 & b \mod 3 & a^2 \mod 3 & b^2 \mod 3 & (a^2 + b^2) \mod 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 & 1 & 1 \\ \hline 0 & 2 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 & 2 \\ \hline 1 & 2 & 1 & 1 & 2 \\ \hline 2 & 2 & 1 & 1 & 2 \\ \hline \end{array}$$
So the only case where ##(a^2 + b^2) \mod 3 = 0## is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say ##a = 3A, b = 3B## for some non-zero integers ##A, B## (since question states a, b, c are all non-zero). Using this in the original expression, we get ##a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}##
Since ##A^2 + B^2## and hence ##\frac {c^2} {3}## is an integer, and ##c## is an integer and 3 is prime, it must be the case that ##c## too is a multiple of 3, say ##c = 3C## for some non-zero integer ##C##. So the above equivalence becomes, ##A^2 + B^2 = \frac {9C^2} {3} = 3C^2##, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form ##a^2 + b^2 = 3 c^2##, too is ruled out. Hence proved.
Looks correct. A solution anaologuous to yours was already provided though.
member 587159
If ##\Sigma _T \cup \Sigma _{S\cap T^c}## was a finite set, then their generated sub sigma algebra, which is ##\Sigma _S##, would also be finite.
I still don't see how (E) implies that
$$\Sigma_S = \sigma(\Sigma_T \cup \Sigma_{S\cap T^c})$$
I still don't see how (E) implies that
$$\Sigma_S = \sigma(\Sigma_T \cup \Sigma_{S\cap T^c})$$
Right, I wrote $(E)$ for $S=X$ and $T=S$. Perhaps, "similarly to (E)" would have been a better choice of words. The proof is the same up to labelling, regardless. Since $\Sigma _T \cup \Sigma _{S\cap T^c} \subseteq \Sigma _S$ we have
$$\sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right ) \subseteq \Sigma _S.$$
Conversely, for every $A\in\Sigma_S$, we have
$$A = A \cap (T\cup S\cap T^c) = [A\cap T] \cup [A\cap (S\cap T^c)]\in \sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right )$$
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8. Show that there is no triple ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## such that ##a^2 + b^2 = 3 c^2##.
After spending a few seconds thinking about it, I realized that this problem is much simpler than what I had imagined when I first came across this question in the challenge list. An interesting problem nevertheless for people like me who have never had the opportunity to a course in number theory and the like and want some warm-up questions before moving to tougher problems.
I think there might be a simpler proof than what I came up with.
Since ##a^2 + b^2 = 3 c^2##, it must be the case that ##a^2 + b^2 \equiv 0 \mod 3##. Since ##(x + y) \mod n = x \mod n + y \mod n)##, we consider all possible combinations of ##a \mod 3## and ##b \mod 3## and find their effect on ##a^2 \mod 3 + b^2 \mod 3##. We only need to consider the modulo pairs (0, 0), (0, 1), (0, 2), (1, 1), (1, 2) and (2, 2) for the modulos of a and b, since the remaining cases like (2, 1) are symmetric. And we use the observation that if ##x \equiv k \mod n##, then ##x^2 \equiv k^2 \mod n## (since ##x = pn + k## for some integer ##p## and ##x^2 = (p^2 n^2 + k^2 + 2kpn) \equiv k^2 \mod n##).
$$\begin{array}{|c|c|c|c|c|} \hline a \mod 3 & b \mod 3 & a^2 \mod 3 & b^2 \mod 3 & (a^2 + b^2) \mod 3 \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 & 1 & 1 \\ \hline 0 & 2 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 & 2 \\ \hline 1 & 2 & 1 & 1 & 2 \\ \hline 2 & 2 & 1 & 1 & 2 \\ \hline \end{array}$$
So the only case where ##(a^2 + b^2) \mod 3 = 0## is when the LHS modulo pair is (0, 0), i.e. both a and b are multiples of 3, say ##a = 3A, b = 3B## for some non-zero integers ##A, B## (since question states a, b, c are all non-zero). Using this in the original expression, we get ##a^2 + b^2 = 9A^2 + 9B^2 = 3 c^2 \Rightarrow A^2 + B^2 = \frac {c^2} {3}##
Since ##A^2 + B^2## and hence ##\frac {c^2} {3}## is an integer, and ##c## is an integer and 3 is prime, it must be the case that ##c## too is a multiple of 3, say ##c = 3C## for some non-zero integer ##C##. So the above equivalence becomes, ##A^2 + B^2 = \frac {9C^2} {3} = 3C^2##, which is of the same for as the original equation. So as long as the 2 integers on LHS are multiples of 3, we can reduce the expression to another expression of the same form but with the absolute values of the integers becoming smaller. This cannot continue infinitely, implying that we will reach a point where at least one of the 2 integers is no longer a multiple of 3, meaning that the modulo pair is no longer (0, 0). And for any such non-(0, 0) pair, we already showed that the RHS cannot be a multiple of 3, but this contradicts the expression from that required the RHS to be a multiple of 3. Therefore, the only remaining possibility, i.e. of having an LHS module pair of the form (0, 0) that also satisfies the expression form ##a^2 + b^2 = 3 c^2##, too is ruled out. Hence proved.
For the last step of my proof, I found a simpler variant, based on same fundamentals however.
We need to prove that ##a^2 + b^2 = 3 c^2## where ##(a,b,c) \in \mathbb{Z}^3\setminus \{(0,0,0)\}## does not have a solution for the case of ##(a \mod 3, b \mod 3) \equiv (0, 0)## (absence of solution for other modulo combinations was already proven).
Say ##a = 3^{p}z, b = 3^{q}y, c = 3^{r}z##, where ##p, q, r## are the highest powers associated with 3 in the prime factorization of ##a, b, c## respectively. By this definition, ##x, y, z## will all be non-multiples of 3, implying ##x \mod 3 \neq 0, y \mod 3 \neq 0, z \mod 3 \neq 0##. If we assume ##(a \mod 3, b \mod 3) \equiv (0, 0)##, then ##p, q## should be positive integers. Therefore, ##a^2 + b^2 = 3 c^2 \Rightarrow 3^{2p}x^2 + 3^{2q}y^2 = 3 \times 3^{2r}z^2##. If we arbitrarily assume that ##p \ge q##, then LHS becomes ##3^{2p} (x^2 + 3^{2q - 2p}y^2)##. The proof of ##p \ge q## case is symmetric.
Subcase 1: If ##p = q##, then LHS simplifies further to ##3^{2p} (x^2 + y^2)##. Since x, y are non-multiples of 3, ##(x^2 + y^2) \mod 3 \neq 0## as already show earlier, so the exponent of 3 in prime factorization of LHS is ##2p##, an even natural number.
Subcase 2: If ##p \neq q##, then too exponent of 3 in prime factorization of LHS still remains ##2p##, because then in ##(x^2 + 3^{2q - 2p}y^2)##, the 2nd component, ##3^{2q - 2p}y^2## would be a multiple of 3 but ##x^2## would not be (because x is not a multiple of 3 by definition), and hence ##(x^2 + 3^{2q - 2p}y^2) \mod 3 \neq 0##. So here too, the exponent of 3 in prime factorization of LHS is an even natural number.
Whereas, on the RHS we have ##3 \times 3^{2r}z^2 = 3^{2r+1} z^2##, and by definition ##z## is a non-multiple of 3. Hence the exponent of 3 in prime factorization of RHS is ##(2r + 1)##, an odd number.
This means that the exponents of 3 on LHS and RHS cannot match, which contradicts the requirement of LHS = RHS. Hence proved
For the sake of convenience I have put all candidate solutions to this post. If possible, please delete #73 and #76.
The fundamental theorem of abelian groups says that for any finite abelian group
$$A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}$$
I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)
Of order ##6## there are
$$\mathbb Z_6 \qquad S_3.$$
Of order ##8## there are
$$\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2\qquad D_4 \qquad \mbox{Dic}_2.$$
Of order ##12## there are
$$\mathbb Z_{12}\qquad \mathbb Z_6 \oplus \mathbb Z_2 \qquad A_4 \qquad \mathbb Z_3 \rtimes_\varphi \mathbb Z_4 \qquad \mathbb Z_3 \rtimes _\varphi (\mathbb Z_2\oplus\mathbb Z_2).$$
In the abelian case there are ##\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3##.
Quick remark. Being a Sylow ##p-##subgroup is invariant with respect to conjugating, thus if ##H## was a unique Sylow ##p-##subgroup, we would have##gHg^{-1} = H, g\in G,## i.e ##H## would be normal.
Fact. Suppose a non-abelian group ##G## has order ##pq## (primes) with ##q\equiv 1 \pmod{p}##, then there are precisely ##q## Sylow##p-##subgroups in ##G##.
Proof of fact. Let $n_p,n_q$ be the numbers of Sylow $p,q-$subgroups respectively. By assumption $q>p$. By theorem 3 $n_q \equiv 1 \pmod{q}$ and $n_q\mid p$, which forces $n_q = 1$, thus we have a (normal) subgroup of order $q$, which makes it cyclic, therefore abelian.
For $n_p$ we thus have two choices: $n_p = 1,q$. If $n_p = 1$, then $G\cong \mathbb Z_p \oplus \mathbb Z_q$ would be abelian, thus it must be that $n_p = q$.
Our non-abelian group $G$ therefore contains three Sylow $2-$subgroups, call them $H_1,H_2,H_3$, which by theorem 2 are all conjugate to each other. We have $\mbox{Sym}(H_1,H_2,H_3) \cong S_3$. We show that $G\cong S_3$. Define
$$\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}$$
(also called conjugating)
Firstly, if $H$ is a Sylow $p-$subgroup, then conjugating it gives another Sylow $p-$subgroup, hence it must hold that $gH_ig^{-1} = H_j$ for some $j$.
The map $\varphi (g)$ is injective, because if $gH_ig^{-1} = gH_jg^{-1}$, then
$$h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.$$
The argument is symmetrical, thus $H_i=H_j$. It is surjective due to finiteness.
Take $g,h\in G$, then
$$\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).$$
To show it's an isomorphism, it suffices to show it's injective (because $|G| = |S_3| = 6$). Suppose $\varphi (g) = \mbox{id}$ i.e $gH_jg^{-1}=H_j$, then by definition $g\in N(H_j)$ the normaliser of $H_j$. By theorem 3 all $N(H_j) = H_j$ and $g\in H_1\cap H_2\cap H_3 = \{e\}$ i.e $g=e$, which makes $\varphi$ an isomorphism.
In the abelian case there are
$$\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2$$
Suppose $G$ is non-abelian of order $8$. If there is an element of order 8, then $G$ is cyclic, thus abelian. Suppose the maximal order is $2$ and pick $g,h\in G$. If $gh=e$, then they commute. Suppose $(gh)^2 = e$, then
$$gh = geh = g(ghgh)h = (gg)hg(hh) = ehge = hg$$
and again $G$ would be abelian. Thus we must have an element $a$ of order $4$. Its generated subgroup $\langle a \rangle =: H$ is of index $2$, therefore normal. Take $b\notin H$, then we have $bab^{-1}\in H$ due to normality. Notice that
$$(bab^{-1})^4 = bab^{-1}bab^{-1}bab^{-1}bab^{-1} = baaaab^{-1} = e.$$
If also $(bab^{-1}) ^2 = e$, then $baab^{-1} = e$ would lead to $aa=e$, contradicting the order of $a$.
Next, we find the possible values of $bab^{-1}$.
1. If $bab^{-1} = a^4 = e$, then $a=e$, which is impossible.
2. If $bab^{-1} = a$, then $ba = ab$, but this will make $G$ abelian. Indeed, pick $g,h\in G$. If they're both in $H$, then they commute. Suppose $g\notin H$ and write $g = ba^m$ and $h = a^n$, then
$$gh = ba^ma^n = ba^na^m = a^n ba^m = hg.$$
and if both reside outside $H$, then writing $g = ba^m$ and $h = ba^n$ would similarly lead to $gh=hg$. But our group is not abelian.
3. If $bab^{-1} = a^2$, then $(bab^{-1})^2 = e$ would contradict the order of $bab^{-1}$.
Thus, the only possibility is $bab^{-1}=a^3 =a^{-1}$.
There are two cases to consider.
1. $b$ is of order $2$, then we have $G \cong \langle a,b \mid \mbox{ord}(a) = 4, \mbox{ord}(b) = 2, bab^{-1} = a^{-1} \rangle$ this is the dihedral group $D_4$.
2. $b$ is of order $4$. We show $G$ is the dicyclic group $\mbox{Dic}_2$, where
$$\mbox{Dic}_2 \cong \langle x,y \mid x^{4} = e, x^2 = y^2, x^{-1} = y^{-1}xy \rangle.$$
Notice that
$$bab^{-1} = a^{-1} \Rightarrow b^{-1}ab = b^{-1}(ba^{-1}b^{-1})b = a^{-1}.$$
So we have left to show $a^2 = b^2$. Considering the natural projection $G\to G/H\cong\mathbb Z_2$, we have $b^2 \in H$. Consider possible values of $b^2$.
1. $b^2 = a^4 = e$ contradicts order of $b$.
2. $b^2 = a$ contradicts order of $a$
3. $b^2 = a^2$ is what we want.
4. $b^2 = a^{-1}$ contradicts order of $a^{-1}$ (hence order of $a$).
Thus $b^2 = a^2$ must hold.
In the abelian case there are
$$\mathbb Z_{12}\qquad \mathbb Z_6 \oplus \mathbb Z_2.$$
Suppose $G$ is non-abelian of order 12. Let $H_p$ denote a Sylow $p-$subgroup and $n_p$ the number of Sylow $p-$subgroups in $G$. Then, by theorem 1, we consider subgroups $H_2$ and $H_3$. The case $n_2=n_3=1$ yields normality and $H_2 \cong \mathbb Z_4$ or $H_2 \cong \mathbb Z_2\oplus \mathbb Z_2$ and $H_3 \cong \mathbb Z_3$ and that leads to one of the previously mentioned abelian structures.
Case 2 : $n_3= 4$. Call them $H^1,H^2,H^3,H^4$. Define
$$\varphi : G \to \mbox{Sym}(H^1,H^2,H^3,H^4) \cong S_4, \quad \varphi (g) := gH^jg^{-1}.$$
Since conjugation preserves the property of being a Sylow $p-$subgroup and injectivity of each $\varphi (g)$ is trivial, the map is well-defined. It's a routine task to check it's a morphism.
Notice, if, say, $aH^ia^{-1} = H^j$, then $aN(H^{i})a^{-1} = N(H^j)$, so taking normalisers preserves the property of being conjugate.
Now, suppose $\varphi (g) = \mbox{id}$, then by definition $g\in N(H^j)$. By theorem 3 $N(H^j) = H^j$, thus $g=e$ is forced and $\varphi$ is injective. Furthermore, each $H^j$ contains two generators, therefore $G$ has $8$ elements of order 3.
Consider the alternating group $A_4$. Trivially $\varphi (G) \cap A_4 \leqslant\varphi (G)$ and $A_4$ also contains $8$ elements of order 3, thus $G\cong A_4$.
Case 3: $n_3 = 1$. Call it $H$ and pick a Sylow $2-$subgroup $K$ (of order $4$). As $H$ is normal it holds $HK$ is a subgroup of $G$. As $H\cap K = \{e\}$ (by orders of elements), then
$$|HK| = \frac{|H||K|}{|H\cap K|} \Rightarrow HK = G.$$
We show $G\cong H\rtimes _\varphi K$, the semidirect product under $\varphi : K\to \mbox{Aut}(H)$, where
$$(h,k)\cdot (h',k') := (h\varphi (k)(h'),kk')$$
This will yield two more groups, namely the following:
$$\mathbb Z_3 \rtimes_\varphi \mathbb Z_4 \qquad \mathbb Z_3 \rtimes _\varphi (\mathbb Z_2\oplus\mathbb Z_2).$$
Define
$$\varphi : K\to \mbox{Aut}(H),\quad \varphi (k)(h) := khk^{-1}.$$
I will omit the routine checks. We show
$$\tau : G\to H\rtimes _\varphi K, \quad g=hk \mapsto (h,k),$$
is an isomorphism. Although, let's have some fun and do it the category theory way, for a change. First, we check $\tau$ is a morphism. Take $g=hk, g'=h'k'\in G$, then due to normality $kh'k^{-1} \in H$ and we get
\begin{align*} \tau (hkh'k') &= \tau (h(kh'k^{-1})kk') \\ &= (h(kh'k^{-1}),kk')\\ &= (h\varphi (k)(h'),kk')\\ &= (h,k)(h',k')\\ &= \tau (hk) \tau (h'k'). \end{align*}
Also $\tau (e) = \tau (ee) = (e,e)$. Now define
$$\omega : H\rtimes _\varphi K \to G, \quad (h,k) \mapsto hk.$$
Verify it, too, is a morphism. Take $(h,k),(h',k')\in H\rtimes _\varphi K$, then
\begin{align*} \omega ((h,k)(h',k')) &= \omega ((h\varphi (k)(h'), kk'))\\ &= h(kh'k^{-1})kk' \\ &= hkh'k' \\ &= \omega ((h,k)) \omega ((h',k')). \end{align*}
Also $\omega ((e,e)) = ee = e$. The equalities $\tau\omega = \mbox{id}_{H\rtimes _\varphi K}$ and $\omega\tau = \mbox{id}_G$ are straightforward to verify.
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member 587159
member 587159
Right, I wrote $(E)$ for $S=X$ and $T=S$. Perhaps, "similarly to (E)" would have been a better choice of words. The proof is the same up to labelling, regardless. Since $\Sigma _T \cup \Sigma _{S\cap T^c} \subseteq \Sigma _S$ we have
$$\sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right ) \subseteq \Sigma _S.$$
Conversely, for every $A\in\Sigma_S$, we have
$$A = A \cap (T\cup S\cap T^c) = [A\cap T] \cup [A\cap (S\cap T^c)]\in \sigma \left (\Sigma _T \cup \Sigma _{S\cap T^c}\right )$$
I went through your solution again and I think it is correct! Well done!
Suppose $\alpha$ is $n$-th root of unity and $\alpha \in \mathbb F_{q^m}$ for some $m\in\mathbb N$. Then by Lagrange $n \mid q^m-1$. It would suffice to compute the order of $q$ modulo $n$. As $(q,n)=1$ it holds that $n\mid q^{\phi (n)} -1$, where $\phi$ is the Euler totient map. So the order of $q$ must be a divisor of $\phi (n)$.
Not sure how to explicitly compute $m$.
member 587159
Suppose $\alpha$ is $n$-th root of unity and $\alpha \in \mathbb F_{q^m}$ for some $m\in\mathbb N$. Then by Lagrange $n \mid q^m-1$. It would suffice to compute the order of $q$ modulo $n$. As $(q,n)=1$ it holds that $n\mid q^{\phi (n)} -1$, where $\phi$ is the Euler totient map. So the order of $q$ must be a divisor of $\phi (n)$.
Not sure how to explicitly compute $m$.
Try to take ##m## minimal somehow.
Try to take ##m## minimal somehow.
Uhh, this is like trial and error. It's about reducing the powers of $x:=\phi (n) = \prod _{i\in I}p_i^{k_i}$. I can think of recursion.
I'll write the most barebones version I can think of, can likely be (heavily) optimised. Let $P(x) \Leftrightarrow n\mid q^{x}-1$.
1. Fix $a\in I$.
2. If $\neg P(x/p_a)$, then return $x$ and terminate.
3. Else put $x:= x/p_a$ and go to 1 (with $I$-many branches)
Do this for every $a\in I$ as a starting point (there will be a lot of branching). Eventually, pick the smallest $x$ that was returned, this will be the order of $q$ modulo $n$.
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member 587159
Try to take ##m## minimal somehow.
I'm not looking for a closed, explicit form of the order of q mod n. Just explain why it is finite.
I'm not looking for a closed, explicit form of the order of q mod n. Just explain why it is finite.
Well, per assumption $n\mid q^{\phi (n)}-1$. Thus the set $\{x : n\mid q^x -1\}\subset \mathbb N$ is non-empty, therefore there is smallest one.
member 587159
Well, per assumption $n\mid q^{\phi (n)}-1$. Thus the set $\{x : n\mid q^x -1\}\subset \mathbb N$ is non-empty, therefore there is smallest one.
Where did you (implicitely) use that ##(q,n)=1##? And that does not solve the problem entirely of course.
Where did you (implicitely) use that ##(q,n)=1##? And that does not solve the problem entirely of course.
$(q,n)=1$ guarantees $n\mid q^{\phi (n)}-1$ (Euler's theorem). It's not clear to me what solution is expected. I gave a way of finding the order $m$ and justified its existence. The smallest field extension with desired property would have to be $\mathbb F_{q^m}$.
Ok, maybe there's the issue of necessarily containing a subgroup of order $n$. But the multiplicative group of a finite field is cyclic, therefore there is a subgroup of any order that divides the order of the group.
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member 587159
member 587159
$(q,n)=1$ guarantees $n\mid q^{\phi (n)}-1$ (Euler's theorem). It's not clear to me what solution is expected. I gave a way of finding the order $m$ and justified its existence. The smallest field extension with desired property would have to be $\mathbb F_{q^m}$.
Ok, maybe there's the issue of necessarily containing a subgroup of order $n$. But the multiplicative group of a finite field is cyclic, therefore there is a subgroup of any order that divides the order of the group.
Yes, your first solution assumed there was an n-th root and then deduced a necessary condition. Your last edit shows that this necessary condition is also sufficient to have an n-th root.
For me, you solved the question succesfully!
Oh, no more problems remaining Well, that's it for me, folks. Likely not taking part in next month's competition. *drops the mic*
member 587159
member 587159
Oh, no more problems remaining Well, that's it for me, folks. Likely not taking part in next month's competition. *drops the mic*
Other challenge threads still have open problems.
QuantumQuest
Gold Member
2. Find the equation of a curve such that ##y′′## is always ##2## and the slope of the tangent line is ##10## at the point ##(2,6)##.
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The way you solve it is correct but there are some mistakes - most likely typos, in the solution. In any case it is already solved by @KnotTheorist.
QuantumQuest
Gold Member
. The maximum value of ##f## with ##f(x) = x^a e^{2a - x}## is minimal for which values of positive numbers ##a## ?
undefined
Your way of thinking is correct. The question has already been solved by @Pi-is-3.
My attempt at 14 (I'm a novice so bare with me!)
Consider a solution ##(a,b,c)## to the equation $$2x^6+3y^6=z^3$$ where ##a,b,c \neq 0## and ##a,b,c \in \mathbb{Q}##. This yields $$2a^6+3b^6=c^3$$ $$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$ $$\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$ We search for the possibility of a rational solution. If $$\left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for ##r,t \in \mathbb {Z}##, then ## \sqrt[3] \frac {r}{t} \in \mathbb {I}## ## \forall r,t## with no common factor, due to the prime factorization theorem. This would imply that either ##a \in \mathbb {I}## or ##c \in \mathbb {I}##. The same logic follows for ##\left( \frac {b^2}{c} \right)^3##. So for a rational solution to exist, it must be that $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ where ##s \in \mathbb {Z}##, and we have reduced ##\frac{r}{t}## and ##\frac{s}{t}## to simplest form, as can be done with any rational number. Then $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$ With algebra, we can show that $$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$ If ##t,s## are both even, then ##\frac {s}{t}## is not fully simplified, but we assumed that we did simplify ##\frac {s}{t}## previously, so this cannot be the case. If ##t,s## are even and odd respectively, or vice versa, then the numerator ##t^3-3s^3## is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of ##f(s,t)## will be uniquely irrational (irrational with no common factor) and so ##r## will be irrational in this case, implying ##a\in \mathbb{I}## or ##c\in \mathbb{I}##. If ##s,t## are both odd, then ##t^3-3s^3## is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write $$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$ where ##k,n_i \in \mathbb {N}## and ##P_i## is a prime number. Here, ##2^k## can be a perfect cube, but how do we know that ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## is never a perfect cube (which may imply a possible rational solution)? Consider the case where ##s=t##. Then $$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}$$ $$\implies r^3=-t^3$$ From this we see that in the case where ##s=t##, ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## must be a perfect cube, since ##2r^3=t^3-3s^3## (and it must be odd as well, since any factor of 2 from the prime factorization is contained within ##2^k##). However, if ##r=-t## then we get a complex set of solutions from ## \frac {a^2}{c}## and ##\frac {b^2}{c}##. It can be shown for the equation ##2r^3+3s^3=t^3## that there is a 1-1 correspondence over the rationals for three separate cases: $$1) r = f(s,t)$$ $$2) s=g(r,t)$$ $$3) t=h(s,r)$$ Thus, when ##s \neq t##, it cannot be true that ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## is a perfect cube, which implies that in this case ##r\in \mathbb{I}## regardless of whether ##2^k## is a perfect cube or not. This would then imply that ##a \in \mathbb{I}## or ##c \in \mathbb{I}##. Therefore, for any ##s,t##, there cannot exist a possible rational solution. Similar cases can be proven with ##g(r,t)## and ##h(s,r)## based on the 1-1 correspondence. There only exists a trivial rational solution, ##x=y=z=0## since $$2(0)^6+3(0)^6=(0)^3$$ $$\implies 0=0$$
Mentor
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My attempt at 14 (I'm a novice so bare with me!)
Sure. You put much effort in it, I appreciate that, and it is not an easy problem. Nevertheless I'll have to be strict.
Consider a solution ##(a,b,c)## to the equation $$2x^6+3y^6=z^3$$ where ##a,b,c \neq 0## and ##a,b,c \in \mathbb{Q}##.
And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases ##c=0## or ##a=0##?
This yields $$2a^6+3b^6=c^3$$ $$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$ $$\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$ We search for the possibility of a rational solution.
Now I get lost. What do you need ##r,s,t## for?
If $$\left( \frac {a^2}{c} \right)^3 \neq \frac {r^3}{t^3}$$ for ##r,t \in \mathbb {Z}##, ...
This is never true. We can always write a quotient of two quotients as one quotient.
... then ## \sqrt[3] \frac {r}{t} \in \mathbb {I}## ## \forall r,t## with no common factor, due to the prime factorization theorem. This would imply that either ##a \in \mathbb {I}## or ##c \in \mathbb {I}##. The same logic follows for ##\left( \frac {b^2}{c} \right)^3##. So for a rational solution to exist, it must be that $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ where ##s \in \mathbb {Z}##, and we have reduced ##\frac{r}{t}## and ##\frac{s}{t}## to simplest form, as can be done with any rational number. Then $$2 \left( \frac{r^3}{t^3} \right)+3 \left( \frac {s^3}{t^3} \right) = 1$$ $$\implies 2r^3 + 3s^3=t^3$$
Now we have a solution ##r=-1, s=t=1## but no solution for the original question!
With algebra, we can show that $$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$ If ##t,s## are both even, then ##\frac {s}{t}## is not fully simplified, but we assumed that we did simplify ##\frac {s}{t}## previously, so this cannot be the case. If ##t,s## are even and odd respectively, or vice versa, then the numerator ##t^3-3s^3## is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of ##f(s,t)## will be uniquely irrational (irrational with no common factor) and so ##r## will be irrational in this case, implying ##a\in \mathbb{I}## or ##c\in \mathbb{I}##. If ##s,t## are both odd, then ##t^3-3s^3## is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write $$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$ where ##k,n_i \in \mathbb {N}## and ##P_i## is a prime number. Here, ##2^k## can be a perfect cube, but how do we know that ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## is never a perfect cube (which may imply a possible rational solution)? Consider the case where ##s=t##. Then $$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}$$ $$\implies r^3=-t^3$$ From this we see that in the case where ##s=t##, ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## must be a perfect cube, since ##2r^3=t^3-3s^3## (and it must be odd as well, since any factor of 2 from the prime factorization is contained within ##2^k##). However, if ##r=-t## then we get a complex set of solutions from ## \frac {a^2}{c}## and ##\frac {b^2}{c}##. It can be shown ...
Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.
You were on the right track. Divisibility is the key. You should consider the remainders of a division by ##7##. There are not many remainders possible for cubes.
And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases ##c=0## or ##a=0##?
Understood. I am quite unsure exactly of the nature of a "rational solution," whether each ##a,b,c## have to all be rational or if a rational solution could be constituted by something like ##(a,b,c)## where only ##a,b \in \mathbb{Q}## and maybe ##c\in \mathbb{R}## for example. Either way, I understand that 0 is indeed rational, and so I have left out those cases.
Now I get lost. What do you need ##r,s,t## for?
My goal was to reduce the problem down to integers instead of working with rational numbers, since then I could apply the parity arguments.
This is never true. We can always write a quotient of two quotients as one quotient.
I believe that was actually the intention of my argument. If ##a,c## are rational numbers (each an integer divided by an integer), then ##\frac{a^2}{c}## can be written as a single fraction in reduced form, ##\frac{r}{t}## where ##r,t## are integers.
Now we have a solution ##r=-1, s=t=1## but no solution for the original question!
This is a solution, but it would be a complex solution to the original equation and so I was under the presumption that a solution of this form would not be accepted as a rational solution to the original question.
Never, ever use this platitude! You bet it is in nine of ten cases wrong and thus needlessly embarrassing.
Would you mind pointing out exactly where my logic fails within this argument? I value this problem as a learning experience as well as a challenge. Thank you for the haste reply!
Mentor
2021 Award
Most of what you wrote can be directly achieved. If we have an equation with quotients, we can always make it an integer equation, just multiply the quotients to get the least common multiple in the denominator.
Using the definition of a prime, namely ##p## is prime, if it is not ##\pm 1## and ##p\,|\,ab## requires ##p\,|\,a## or ##p\,|\,b## makes the cancellations you need. But the reduction to ##2r^3+3s^3=t^3## isn't the solution, since ##(r,s,t)=(-1,1,1)## is a counterexample.
I read your proof very carefully and achieved your conclusions with slightly different arguments. So maybe the proof can be corrected by the additional assumption, that all ##r,s,t## are positive. I got at least as far as ##2r^3+3s^3=t^3## with coprime pairs ##(r,t)## and ##(s,t)##. But then the solution with ##r=-1## came to mind.
You could really try to pass all equations to their remainders modulo ##7## and look for common divisors.
If ##2a^6+3b^6=c^3 ## then the same equation holds if we substitute ##a,b,c## by their remainders of any division, esp. ##7##.
An example of what I mean: ##13^2=12^2+5^2##. Division by ##5## yields as remainders ##9=3^2\equiv 2^2+0^2 = 4 \mod 5##, since ##9## and ##4## have the same remainder modulo a division by ##5##. This works for any number. The above problem can be tackled by remainders modulo ##7##. The remainders modulo seven are ##\{\,0,1,2,3,4,5,6\,\}## which is cubed ##\{\,0,1,8,27,64,125,216\,\}## with remainders ##\{\,0,1,1,6,1,6,6\,\}=\{\,0,1,6\,\}##. This is a very limited selection of possibilities.
Kyle Nemeth
Understood, the substitutions I made were clearly not necessary and the fact that any equation with quotients can be written as an integer equation by multiplying the least common denominator is an approach I did not think of.
But the reduction to ##2r^3+3s^3=t^3## isn't the solution, since ##(r,s,t)=(−1,1,1)## is a counterexample.
I do not assert that this equation has no solutions, but rather any potential solution will ultimately give a solution to the original equation $$2x^6+3y^6=z^3$$ that is not rational.
So maybe the proof can be corrected by the additional assumption, that all ##r,s,t## are positive.
In order for ##2x^6+3y^6=z^3## to have a solution that is rational, isn't there a certain restriction on which values may be negative in a solution ##(r,s,t)##? Since $$\frac {a^2}{c} = \frac {r}{t}, \frac {b^2}{c} = \frac {s}{t}$$ In order that ##a## or ##b## not be complex, for ##c>0##, mustn't it be the case that a potential solution must exist of the forms ##(r,s,t)## where ##r,s,t<0## or ##r,s,t>0##? or if ##c<0##, then we must have ##r,s<0##,##t>0## or ##r,s>0##, ##t<0##? But if ##r,s<0## and ##t>0##, we can let ##r=-\alpha##, ##s=-\beta## and ##t=\gamma##, for integers ##\alpha,\beta,\gamma>0##, then (using ##f(s,t)## in post 113) $$-\alpha = \sqrt[3] \frac{\gamma^3-3(-\beta)^3}{2}$$ $$\implies -\alpha=\sqrt[3] \frac{\gamma^3+3\beta^3}{2}$$ and this equation is never true, since ##\alpha,\beta,\gamma>0##, and this is similarly the case for where ##r,s>0##, ##t<0##.
And what in the other cases? We may assume that not all three of them, or even two equal zero, since then we are done. But what is in the cases ##c=0## or ##a=0##?
If ##a=0##, then $$3b^6=c^3$$ $$\implies \frac {c}{b^2}=\sqrt[3] 3$$ which implies that ##c## or ##b## must be irrational, since a number ##\frac {p}{q}\in \mathbb{I}## if either ##p## or ##q## is irrational. If ##b=0##, then $$2a^6=c^3$$ $$\implies \frac{c}{a^2}=\sqrt[3] 2$$ and so ##c## or ##a## must be irrational. If ##c=0##, then $$2a^6+3b^6=0$$ $$\implies \frac {a}{b}=\sqrt[6] \frac {-3}{2}$$ which also implies ##a## or ##b## must be irrational (and complex).
You could really try to pass all equations to their remainders modulo ##7## and look for common divisors.
Ahh I see, I currently do not have much knowledge of modular arithmetic so that is another approach I was not aware of and I can see how this approach is far more simple than mine.
Mentor
2021 Award
I will check later. I'm too tired now to say something reasonable, or to read something correctly.
We can assume ##a.b,c > 0## from the start, after the cases ##c=0## (easy) and ##a=0## (prime definition with ##p=2## if ##b=0## or symmetrically ##p=3## if ##a=0##) are done. A solution with a negative number immediately also gets a solution with a positive number, and ##c<0## is impossible.
This leads to ##2r^3+3s^3=t^3## with positive integers ##r,s,t##. That they may be assumed pairwise coprime is again a consequence of primality.
E.g. let us assume a prime ##p## divides ##r## and ##t##. Then
\begin{align*}
p \,|\, 3s^3=t^3-2r^3 & \Longrightarrow p\,|\,3 \text{ or } p\,|\,s^3\\
\text{ (a) }p\,|\,s^3 &\Longrightarrow p\,|\,s\\
&\Longrightarrow p^3\,|\,s^3\\
&\Longrightarrow p^3 \text{ divides every term and can be cancelled }\\
\text{ (b) }p\,|\,3&\Longrightarrow p=3\\
&\Longrightarrow s^3= 9\cdot t'^3 - 2\cdot 9\cdot r'^3\\
&\Longrightarrow 3\,|\,s^3\\
&\Longrightarrow 3\,|\,s\\
&\Longrightarrow 3^3\,|\,s^3\\
&\Longrightarrow 3^3 \text{ divides every term and can be cancelled }
\end{align*}
This is the way primality is normally used. Here it allows us to assume that ##r,s,t## are pairwise coprime, since otherwise we would get a common factor cubed, which we can cancel and still have the same equation structure ##2r'^3+3s'^3=t'^3##. So we continue to cancel all common factors. Now we have
• ##2r^3+3s^3=t^3##
• ##r,s,t > 0##
• ##r,s,t## are pairwise coprime
If ##c=0##, then ##2a^6+3b^6=0 \quad \implies \frac {a}{b}=\sqrt[6] \frac {-3}{2}## which also implies ##a## or ##b## must be irrational (and complex).
You need ##b\neq 0## here, but this is not necessary. Both terms of the right hand side are always non negative. If their sum equals zero, then both summands have to be zero, i.e. ##a=b=c=0##.
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Mentor
2021 Award
Consider a solution ##(a,b,c)## to the equation
$$2x^6+3y^6=z^3$$
where ##a,b,c \neq 0## and ##a,b,c \in \mathbb{Q}##. This yields
$$2a^6+3b^6=c^3$$
$$\implies 2\frac {a^6}{c^3}+3 \frac{b^6}{c^3}=1$$
$$\implies 2 \left( \frac {a^2}{c} \right)^3 +3 \left( \frac {b^2}{c} \right)^3=1$$
I skip the part here that is a bit confusing and unnecessary. We can always transform ##2a^6+3b^6=c^3## into ...
$$\implies 2r^3 + 3s^3=t^3$$
where we may assume w.l.o.g. ##r,s,t > 0## and ##r,s,t## are pairwise coprime integers (cp. post #118).
With algebra, we can show that
$$f(s,t)=\sqrt[3] \frac{t^3-3s^3}{2}$$
If ##t,s## are both even...
Obsolete, if we first guarantee that ##s## and ##t## are coprime.
... then ##\frac {s}{t}## is not fully simplified, but we assumed that we did simplify ##\frac {s}{t}## previously, so this cannot be the case. If ##t,s## are even and odd respectively, or vice versa, then the numerator ##t^3-3s^3## is odd. By definition, an odd number does not contain a factor of 2 and so the numerator and denominator of ##f(s,t)## will be uniquely irrational (irrational with no common factor) and so ##r## will be irrational in this case, implying ##a\in \mathbb{I}## or ##c\in \mathbb{I}##.
What we really have in this case is, that a number ##z+\frac{1}{2}## is under the root. We need to show that there is no integer ##r## with ##r^3=z+\frac{1}{2}##. But this is true and no argument involving irrationality is needed.
Keep it simple and only use what you have, step by step.
If ##s,t## are both odd, then ##t^3-3s^3## is even and the numerator may contain a factor of 2. By the prime factorization theorem, we may write
$$t^3-3s^3=2^k(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})$$
where ##k,n_i \in \mathbb {N}## and ##P_i## is a prime number. Here, ##2^k## can be a perfect cube, but how do we know that ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## is never a perfect cube (which may imply a possible rational solution)? Consider the case where ##s=t##. Then
$$r = f(s,t)=\sqrt[3] \frac {t^3-3t^3}{2}=\sqrt[3] \frac {-2t^3}{2}=-\sqrt[3] {t^3}$$
$$\implies r^3=-t^3$$
From this we see that in the case where ##s=t##...
... which again is obsolete, since ##s## and ##t## can be assumed coprime.
...##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## must be a perfect cube, since ##2r^3=t^3-3s^3## (and it must be odd as well, since any factor of 2 from the prime factorization is contained within ##2^k##). However, if ##r=-t## then we get a complex set of solutions from ## \frac {a^2}{c}## and ##\frac {b^2}{c}##. It can be shown for the equation ##2r^3+3s^3=t^3## that there is a 1-1 correspondence over the rationals for three separate cases: $$1) r = f(s,t)$$
$$2) s=g(r,t)$$
$$3) t=h(s,r)$$
Summary: We are left with ##s\neq t##, both odd, and
$$\dfrac{t^3-3s^3}{2} = \left( 2^k P_1^{n_1}\cdot P_2^{n_2}\cdot \ldots \cdot P_{n_k}^{n_k}\right)^{1/3} \in \mathbb{Z} \;\;\text{ i.e. }\;\; 3\,|\, k,n_j$$
Thus, when ##s \neq t##, it cannot be true that ##(P_1^{n_1}\cdot P_2^{n_2}\cdot P_3^{n_3}\cdot {\dots} \cdot P_i^{n_i} \cdot {\dots})## is a perfect cube...
Sorry, but I do not see this. Why is it impossible that all powers are divisible by three? What happens if they were? E.g. all zero is a possibility.
...which implies that in this case ##r\in \mathbb{I}## regardless of whether ##2^k## is a perfect cube or not. This would then imply that ##a \in \mathbb{I}## or ##c \in \mathbb{I}##. Therefore, for any ##s,t##, there cannot exist a possible rational solution. Similar cases can be proven with ##g(r,t)## and ##h(s,r)## based on the 1-1 correspondence. There only exists a trivial rational solution, ##x=y=z=0## since
$$2(0)^6+3(0)^6=(0)^3$$
$$\implies 0=0$$
Last edited:
sage gogenkil
You are hard to follow.
Done in post #89, although the additional variables make it unnecessarily complicated (see post #90).
We do not "see" it. What we see is that you invented out of the blue a certain number ##t##. What you should have done is telling us, that you set ##t:= \sqrt[3]{\frac{3}{2}y} \in \mathbb{R}## such that ##y=\frac{2}{3}t^3##. But now ##x^3\neq \frac{1}{2}-t^3##.
To correct this, we have to set ##t:=\sqrt[3]{\frac{3}{2}}y##. Now ##x^3= \frac{1}{2}-t^3##.
How that?
Sorry for late reply. I was not able to come online a lot since a long time. Anyway, I tried to prove it, but I couldn't. Hence my proof is wrong. Sorry.
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# Ray Optics and Optical Instruments Class 12 notes Physics Chapter 9
## Introduction
In the present chapter, we shall study the phenomena like reflection, refraction, and dispersion of light. We then go on to describe the construction and working of some important optical instruments, including the human eye.
We can see objects because of the light reflected or emitted by them. Nature has given us eyes, which can detect a small region of the electromagnetic radiation spectrum. This region (visible to us) is called light. From common experience, we can say that light travels with a great speed and that it travels in a straight line.
## Reflection of Light
The phenomenon in which a light ray is sent back into the same medium from which it is coming, on interaction with a boundary, is called reflection. The boundary can be a rigid surface or just an interface between two media.
### Law of reflection
1. The angle of reflection equals the angle of incidence ∠i = ∠r.
2. The incident ray reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane.
## Spherical Mirrors
A spherical mirror is a part of a sphere. If one of the surfaces is silvered, the other surface acts as the reflecting surface. When the convex face is silvered, and the reflecting surface is concave, the mirror is called a concave mirror. When its concave face is silvered and a convex face is a reflecting face, the mirror is called a convex mirror.
1. Centre of curvature: Centre of curvature is the center of a sphere of which, the mirror is a part.
2. The radius of curvature: The radius of curvature is the radius of the sphere of which, the mirror is a part.
3. Pole of the mirror: Pole is the geometric center of the mirror.
4. Principal axis: Principal axis is the line passing through the pole and center of curvature.
5. Normal: Any line joining the mirror to its center of curvature is normal.
## Sign Convention
1. According to the Cartesian sign convention, all distances are measured from the pole of the mirror or the optical center of the lens.
2. The distances measured in the same direction as the incident light are taken as positive and those measured in the direction opposite to the direction of incident light are taken as negative.
3. The heights measured upwards with respect to the x-axis and normal to the principal axis of the mirror/lens are taken as positive. The heights measured downwards are taken as negative.
According to this, the focal length of a convex mirror/lens is the positive and the focal length of a concave mirror/lens is negative.
## Focal Length of Spherical Mirrors
When a parallel beam of light is incident on a concave mirror, and a convex mirror. The rays are incident at points close to the pole P of the mirror and make small angles with the principal axis. The reflected rays converge at a point F on the principal axis of a concave mirror. For a convex mirror, the reflected rays appear to diverge from a point F on its principal axis.
Point F is called the principal focus of the mirror. The distance between the focus F and the pole P of the mirror is called the focal length of the mirror, denoted by f.
If R be the radius of curvature of the mirror then the relation between R and f is given by
f=\frac{R}{2}
### Number of images formed by two plane mirrors
Consider two plane mirrors inclined at an angle θ (in degrees) with each other. Multiple reflections take place from the surfaces of these mirrors and more than one image is formed. To find out a number of images, we compute n=[\frac{360°}{\theta}]. Now, if
1. n is even, number of images = n – 1
2. n is odd, the number of images equals
## Refraction
When light travels from one transparent medium to another, it deviates from its original path. The phenomenon of change in the path of light at the medium of separation as it goes from one medium to another medium is called refraction.
### Snell’s Law
1. The incident ray, the refracted ray, and the normal to the interface at the point of incidence, all lie in the same plane.
2. Normal at the medium of separation and ‘i’ and ‘r’ are the angles of incidence and refraction respectively then for pair of media and for a given color \frac{sin i}{sin r}=\mu_{21}= Constant
\mu_{21} is called the refractive index of the second medium w.r.t. first medium.
### Absolute Refractive index
The absolute refractive index of a medium is defined by the ratio of the speed of light in a vacuum to the speed of light in the medium \mu=\frac{c}{v}, where c is the speed of light in a vacuum and v is the speed of light in the medium.
## Total Internal Reflection
When light travels from an optically denser medium to a rarer medium at the interface, it is partly reflected back into the same medium and partly refracted to the second medium. This reflection is called internal reflection.
When a ray of light enters from a denser medium to a rarer medium, it bends away from the normal, the angle of refraction (r) being larger than the angle of incidence (i). As the angle of incidence increases, so does the angle of refraction, till, for the ray, the angle of refraction is 90º. The angle of incidence corresponding to an angle of refraction 90º is called the critical angle (ic). If the angle of incidence is increased still further, refraction is not possible, and the incident ray is totally reflected. This is called total internal reflection.
sin i_{c}=\frac{\mu_1}{\mu_2}
### Application of Total Internal Reflection
#### (i). Mirage
On hot summer days, the air near the ground becomes hotter than the air at higher levels. The refractive index of air increases with its density. Hotter air is less dense and has a smaller refractive index than cooler air. If the air currents are small, that is, the air is still, and the optical density at different layers of air increases with height.
As a result, light from a tall object such as a tree passes through a medium whose refractive index decreases towards the ground. Thus, a ray of light from such an object successively bends away from the normal and undergoes total internal reflection, if the angle of incidence for the air near the ground exceeds the critical angle. To a distant observer, the light appears to be coming from somewhere below the ground.
The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object. Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts.
#### (ii). Diamond
The critical angle for the diamond-air interface (≈ 24.4°) is very small, therefore once light enters in diamond, it undergoes total internal reflections inside it, as a result, it sparkles brilliantly. By cutting the diamond suitable multiple total internal reflections can be made to occur.
#### (iii) .Prism
Prisms are designed to bend light by 90° or 180° the phenomenon of total internal reflection, such prisms are used to invert images without changing their size.
#### (iv). Optical fibres
Optical fiber is a transmission medium to carry the optical signal without any appreciable loss. The structure of optical fiber consists of a core surrounded by cladding. The core is the denser medium of refractive index μ1 and cladding is the relatively rarer medium of refractive index μ2 such that μ1 > μ2.
When a light incident at one end of the core gets refracted and will be incident on the interface between the core and cladding at an angle greater than the critical angle and the light continues reflecting in the core.
## Lens Formula
The lens formula relates the distance of the object from the lens with the distance of the image from the lens. It is given by
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
Where u = object distance
v = image distance
f = focal length
## Mirror Formula
The mirror formula relates the distance of the object from the mirror with the distance of the image from the mirror. It is given by
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
Where u = object distance
v = image distance
f = focal length
## Lens Maker’s Formula
Lens Maker’s formula gives the focal length of a lens in terms of the nature of the surfaces by which the lens is bounded and the nature of the material of the lens.
Let us consider the situation shown in the figure. C1 and C2 are the centers of curvature of two spherical surfaces of the thin lens. O is the object and I' is the image due to the first refraction. Let radii of curvature be R1 and R2.
For the first refraction at image distance is v1. From the formula for refraction at a curved surface, we get
\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_{2}-n_{1}}{R_1} .....(i)
Final image position is I, which is also the image due to second refraction. Let this image distance be v. For the second refraction, v1 becomes the object distance. Hence we get,
\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_{1}-n_{2}}{R_2} .....(ii)
Adding (i) and (ii), we get
(\frac{n_{1}}{v}-\frac{n_{1}}{u})=(n_{2}-n_1)(\frac{1}{R_1}-\frac{1}{R_2})
\frac{1}{v}-\frac{1}{u}=(\frac{n_2}{n_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})
\frac{1}{v}-\frac{1}{u}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})
According to the definition of the focal length f
\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})
This is called the “Lens Maker’s formula”.
## Combination of thin lenses in contact
Consider two lenses A and B of focal length f1 and f2 placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens A. The first lens produces an image at I1. Since image I1 is real, it serves as a virtual object for the second lens B, producing the final image at I.
For the image formed by the first lens A, we get
\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1} .....(i)
For the image formed by the second lens B, we get
\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2} .....(ii)
Adding Eqs. (i) and (ii), we get
\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}
If the two lens-system is regarded as equivalent to a single lens of focal length f, we have
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
so that we get
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
The derivation is valid for any number of thin lenses in contact.
## Power of a Lens
The power of a lens is defined as the reciprocal of the focal length in meters. SI unit of power of lens is Diopter (D).
P=\frac{1}{f}
If several thin lenses of power P1, P2, P3,... are in contact, the effective power of their combination is given by
P=P_{1}+P_{2}+P_{3}+...
## Dispersion
When white light is incident on a prism, different colors having different wavelengths suffer different deviations. The phenomenon of splitting light into its component colours is known as dispersion. The pattern of colour components of light (VIBGYOR) is called the spectrum of light. The deviation produced by a thin prism depends on the refractive index.
### Angular Dispersion
Angular dispersion produced by a prism for white light is the difference in the angles of deviation for two extreme colours i.e. violet and red. It is given by
θ = δV - δR
θ = (nV - nR)A
### Dispersive Power
The dispersive power of a prism is defined as the ratio of angular dispersion to the mean deviation produced by the prism.
Dispersive power \omega=\frac{\delta_{V}-\delta_{R}}{\delta_{Y}}
## Some Natural Phenomena Due to Sunlight
### (i). Scattering of Light
The change in direction of light by the particles of the medium through which light passes is known as scattering. As sunlight travels through the earth's atmosphere it gets scattered by the atmospheric particles.
### (ii). Rayleigh Law of Scattering
The amount of scattering is inversely proportional to the fourth power of the wavelength. Therefore, the light of shorter wavelengths is scattered much more than light of longer wavelengths.
scattering\propto\frac{1}{\lambda^4}
### (iii). Blue Sky
Blue has a shorter wavelength than red and is scattered much more strongly. In fact, violet is scattered even more than blue having a shorter wavelength. But since our eyes are more sensitive to blue than violet, we see the sky blue.
### (iv). Reddish appearance of the sun in full moon near the horizon
At sunrise and sunset sun rays pass through a larger distance through the atmosphere so most of the blue and other shorter wavelengths are removed by scattering. The least scattered light reaches our eyes and the sun seems reddish.
## Optical Instruments
### (i). The Eye
Light enters the eye through the cornea a curved front surface. It passes through the pupil which is the central hole in the iris. The size of the pupil can change under the control of muscles. The light is further focussed by the eye lens on the retina. The retina is a film of nerve fibers covering the curved black surface of the eye. The retina contains rods and cones which sense light intensity and colour respectively and transmit electrical signals via the optic nerve to the brain.
The shape (curvature) and therefore the focal length of the lens can be modified somewhat by ciliary muscles. So images are formed at the retina for objects at all distances. This property of the eye is called accommodation.
The closest distance for which the eye lens can focus light on the retina is called the least distance of distinct vision or the near point. The standard value for normal vision is taken as 25 cm (Symbol D). If the object is too close to the eye; the lens cannot curve enough to focus the image on the retina, and the image is blurred.
#### # Near sightedness or Myopia
The light from a distant object arriving at the eye lens may get converged at a point in front of the retina. This defect is known as nearsightedness or myopia. A concave lens is used to correct the defect.
#### # Farsightedness or Hypermetropia
If the eye lens focuses the incoming light from a distant object at a point behind the retina, then the defect is called farsightedness or hypermetropia. A convergent lens is needed to compensate for the defect in vision.
#### # Astigmatism
This occurs when the cornea is not spherical in shape. It may have large curvature in the vertical plane than in the horizontal plane. It results in lines in one direction being well focussed while those in a perpendicular direction may appear distorted.
### (ii). The microscope
A simple magnifier or microscope is a converging lens of a small focal length. The lens nearest the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens, the eyepiece, which functions essentially as a simple microscope or magnifier, producing an enlarged virtual final image.
The first inverted image is thus near (at or within) the focal point of the eyepiece, at a distance appropriate for final image formation at infinity, or a little closer for image formation at the near point. Clearly, the final image is inverted with respect to the original object.
Magnification power is given by
m=\frac{v_0}{u_o}[\frac{D}{v}+\frac{D}{f_e}]
### (iii). Telescope
The telescope is used to provide angular magnification of distant objects. It also consists of an objective and an eyepiece. But here, the objective has a large focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at the second focal point of the convex objective lens. The eyepiece magnifies this image producing a final inverted image.
Magnifying power is given by
m=-f_{0}[\frac{1}{f_e}+\frac{1}{v}]
### (iv). Reflecting Telescope
One such arrangement using a convex secondary mirror to focus the incident light, which now passes through a hole in the objective primary mirror, is known as a reflecting telescope or a Cassegrain telescope.
Advantages of taking mirror objectives are
1. There is no chromatic aberration in a mirror.
2. If a parabolic reflecting surface is chosen, spherical aberration is also removed.
3. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality and can be supported over.
4. The entire back surface not just over rim, unlike lens.
## Summary
• Reflection is governed by the equation ㄥi = ㄥr and refraction by Snell’s law, sin i / sin r = n.
• The critical angle of incidence ic for a ray incident from a denser to a rarer medium is that angle for which the angle of refraction is 90°.
• Cartesian sign convention:
1. Distances measured in the same direction as the incident light are positive.
2. Those measured in the opposite direction are negative.
3. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis.
4. The heights measured upwards above the x-axis and normal to the principal axis of the mirror/lens are taken as positive.
5. The heights measured downwards are taken as negative.
• Mirror equation: where u and v are object and image distances, respectively and f is the focal length of the mirror.
• Lens makers formula.
\frac{1}{f}=(\mu-1)[\frac{1}{R_1}-\frac{1}{R_2}]
• Dispersion is the splitting of light into its constituent colors.
• The Eye: The eye has a convex lens with a focal length of about 2.5 cm. This focal length can be varied somewhat so that the image is always formed on the retina. This ability of the eye is called accommodation. In a defective eye,
• if the image is focussed before the retina (myopia), a diverging corrective lens is needed.
• if the image is focussed beyond the retina (hypermetropia), a converging corrective lens is needed.
• Astigmatism is corrected by using cylindrical lenses.
• Magnifying power m of a simple microscope is given by m = 1 + (D/f), where D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens.
• Magnifying power m of a telescope is the ratio of the β angle subtended at the eye by the image to the angle α subtended at the eye by the object. m = β / α.
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Cody
Problem 2226. Wayfinding 4 - Crossing, level 2
Solution 547725
Submitted on 18 Dec 2014
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Fail
%% AB = [ -10 10 ; -10 10 ]; F{1} = [ -5 5 -8 0 -6 2 7 1 ]; F{2} = [ 4 -2 4 6 6 -7 -3 -2 ]; F{3} = [ -5 6 3 -1 -6 -8 4 -2 ]; f = WayfindingPassed(AB,F); f_correct = [ 1 3 2 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
2 Fail
%% AB = [ 0 21 ; 0 0 ]; f_correct = randperm(20); F = arrayfun(@(n)[n+[0 1 1 0];-1 -1 1 1],f_correct,'uni',0); f = WayfindingPassed(AB,F); assert(isequal(f(f_correct),f_correct(f)));
Error: Improper assignment with rectangular empty matrix.
3 Fail
%% AB = [ -10 10 ; -10 10 ]; F{1} = [ -5 9 0 -4 5 2 8 -1 -9 -8 ]; F{2} = [ -2 -10 -4 0 8 7 -5 5 ]; F{3} = [ -6 2 10 10 -4 8 ]; F{4} = [ -10 8 -10 4 -8 2 ]; F{5} = [ 0 4 8 -3 1 -10 9 -8 -5 2 ]; F{6} = [ 6 6 -9 10 6 -3 4 -7 ]; F{7} = [ 9 7 -7 0 7 -5 ]; F{8} = [ 2 0 10 6 -10 0 ]; F{9} = [ -7 2 -7 -7 3 5 -3 7 ]; F{10} = [ -5 6 1 5 -10 0 8 4 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 7 8 10 7 3 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
4 Fail
%% AB = [ -10 10 ; -10 10 ]; F{1} = [ 2 5 8 5 2 -2 9 -5 ]; F{2} = [ -8 2 -2 8 -8 -2 0 4 5 -9 8 -2 ]; F{3} = [ 5 -6 -2 1 0 10 10 -8 0 10 -2 -5 ]; F{4} = [ 10 -4 -10 -2 9 4 1 8 -4 -1 ]; F{5} = [ -9 -7 2 -3 2 -9 -4 5 ]; F{6} = [ -3 10 6 9 4 -2 10 -6 2 2 5 -5 ]; F{7} = [ -1 -5 -5 3 0 -4 ]; F{8} = [ 8 -6 8 10 -7 8 -2 -5 3 7 ]; F{9} = [ 1 -10 -3 10 5 -5 -6 3 -6 8 ]; F{10} = [ -7 0 8 -8 7 -8 3 9 ]; f = WayfindingPassed(AB,F); f_correct = [ 5 9 10 9 3 1 8 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
5 Fail
%% AB = [ 4 -6 ; 0 0 ]; F{1} = [ -4 -4 2 2 -2 -3 -3 -2 2 2 -4 2 -4 -4 -2 2 2 -2 -2 2 4 4 ]; f = WayfindingPassed(AB,F); f_correct = [ 1 ]; assert(isequal(f,f_correct));
Error: In an assignment A(I) = B, the number of elements in B and I must be the same.
6 Fail
%% AB = [ 0 0 ; 15 -8 ]; F{1} = [ -4 4 4 -4 6 2 6 2 ]; F{2} = [ -2 -2 6 6 -2 -0 -4 -0 -4 -0 ]; F{3} = [ -1 -1 2 2 -6 -4 -6 -4 ]; F{4} = [ -1 1 -1 1 -7 -7 -9 -9 ]; F{5} = [ -2 2 -1 2 -1 1 14 10 6 6 10 14 ]; f = WayfindingPassed(AB,F); f_correct = [ 5 5 1 2 3 4 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
7 Fail
%% AB = [ 0 0 ; -6 6 ]; F{1} = [ -5 7 7 -5 -9 -9 9 9 ]; F{2} = [ -1 1 1 -1 -7 -7 -5 -5 ]; F{3} = [ -2 -2 2 2 4 2 2 4 ]; F{4} = [ 2 2 -2 -2 4 2 2 4 ]; F{5} = [ -1 1 1 -1 2 2 -2 -2 ]; F{6} = [ -2 0 -2 -2 -3 -4 ]; F{7} = [ 0 2 2 -3 -4 -2 ]; F{8} = [ -1 0 1 -8 -6 -8 ]; f = WayfindingPassed(AB,F); f_correct = [8 2 1 7 1 5 4 1]; assert(isequal(f,f_correct));
Error: In an assignment A(I) = B, the number of elements in B and I must be the same.
8 Fail
%% AB = [ 2 -2 ; 8 -6 ]; F{1} = [ -4 -4 4 4 -4 -0 -0 -4 ]; F{2} = [ -4 -4 4 4 2 6 6 2 ]; f = WayfindingPassed(AB,F); f_correct = [2 1]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
9 Fail
%% AB = [ 8 -4 ; 8 -8 ]; F{1} = [ -6 2 2 -4 -4 8 8 -6 -6 -6 -4 -4 2 2 4 4 ]; F{2} = [ -2 -2 4 4 -0 -2 -2 -0 ]; f = WayfindingPassed(AB,F); f_correct = [ 1 2 1 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
10 Fail
%% AB = [ -8 8 ; 8 -8 ]; F{1} = [ -2 -2 0 0 -0 2 2 -0 ]; F{2} = [ 2 4 4 -6 -6 -4 2 4 4 2 2 -4 -4 2 -0 -0 -6 -6 4 6 6 4 2 2 4 4 -4 -4 ]; F{3} = [ -3 -3 1 0 -1 -3 -3 -1 ]; F{4} = [ 5 9 9 5 -3 -3 -9 -9 ]; F{5} = [ -9 -10 -10 -9 9 9 10 10 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 1 2 4 ]; assert(isequal(f,f_correct)); AB = [ 0 0 ; -8 8 ]; F{1} = [ -4 -2 -2 -4 8 8 4 4 ]; F{2} = [ 2 4 4 2 -0 -0 -6 -6 ]; F{3} = [ -4 -2 -2 -6 -6 -4 -4 -6 -6 -4 ]; f = WayfindingPassed(AB,F); assert(isempty(f));
Error: Improper assignment with rectangular empty matrix.
11 Fail
%% AB = [ 7 -8 ; 0 0 ]; F{1} = [ 8 9 9 8 3 3 -2 -2 ]; F{2} = [ -9 -7 -7 -4 -4 -3 -3 0 0 1 1 4 4 5 5 -2 -8 -9 -2 -2 2 2 -2 -2 2 2 -2 -2 2 2 -2 -2 3 4 3 2 ]; F{3} = [ -2 -1 -1 -2 1 1 -4 -4 ]; F{4} = [ -6 -5 -5 -3 1 2 2 3 3 1 -4 -6 1 1 -3 -5 -5 -4 1 1 -5 -8 -7 -4 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 4 2 3 2 4 2 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
12 Fail
%% AB = [ 0 -2 ; 0 -4 ]; F{1} = [ -3 3 3 2 2 -2 -2 2 2 -3 -5 -5 3 3 -3 -3 2 2 3 3 ]; F{2} = [ -1 1 1 -1 1 1 -1 -1 ]; F{3} = [ -4 4 4 5 5 -5 -5 -4 4 4 -7 -7 5 5 -1 -1 ]; F{4} = [ -5 -4 -4 4 4 -5 -1 -1 -6 -6 -7 -7 ]; f = WayfindingPassed(AB,F); f_correct = [ 2 1 ]; assert(isequal(f,f_correct));
Error: Improper assignment with rectangular empty matrix.
13 Fail
%% AB = [ -2 0 ; 6 -6 ]; F{1} = [ 2 -4 -4 2 2 -2 0 -2 2 -4 -4 4 4 2 2 -0 -2 -2 ]; f = WayfindingPassed(AB,F); f_correct = [ 1 1 1 ]; assert(isequal(f,f_correct));
Error: In an assignment A(I) = B, the number of elements in B and I must be the same.
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# Tetrahedron
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Regular Tetrahedron
Type Platonic solid
Elements F = 4, E = 6
V = 4 (χ = 2)
Faces by sides 4{3}
Schläfli symbol {3,3} and s{2,2}
Wythoff symbol 3 | 2 3
| 2 2 2
Coxeter–Dynkin
Symmetry Td, A3, [3,3], (*332)
Rotation group T, [3,3]+, (332)
References U01, C15, W1
Properties Regular convex deltahedron
Dihedral angle 70.528779° = arccos(1/3)
3.3.3
( Vertex figure)
Self-dual
( dual polyhedron)
Net
A tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex. A regular tetrahedron is one in which the four triangles are regular, or "equilateral," and is one of the Platonic solids.
The tetrahedron is one kind of pyramid, the second most common type; a pyramid has a flat base, and triangular faces above it, but the base can be of any polygonal shape, not just square or triangular.
Like all convex polyhedra, a tetrahedron can be folded from a single sheet of paper.
## Formulas for regular tetrahedron
For a regular tetrahedron of edge length $a$:
Surface area $A=a^2\sqrt{3} \,$ Volume $V=\begin{matrix}{1\over12}\end{matrix}a^3\sqrt{2} \,$ Height $h=\sqrt{6}(a/3) \,$ Angle between an edge and a face $\arctan(\sqrt{2}) \,$ (approx. 55°) Angle between two faces $\arccos(1/3) = \arctan(2\sqrt{2}) \,$ (approx. 71°) Angle between the segments joining the centre and the vertices ${\pi \over 2} + \arcsin(1/3)\,$ (approx. 109.471°) Solid angle at a vertex subtended by a face $3 \arccos(1/3) - \pi \,$ (approx. 0.55129 steradians) Radius of circumsphere $R=\sqrt{6}(a/4) \,$ Radius of insphere that is tangent to faces $r=\sqrt{6}(a/12) \,$ Radius of midsphere that is tangent to edges $r_M=\sqrt{2}(a/4) \,$ Radius of exspheres $r_E=\sqrt{6}(a/6) \,$ Distance to exsphere centre from a vertex $\sqrt{6}(a/2) \,$
Note that with respect to the base plane the slope of a face ($2 \sqrt{2}$) is twice that of an edge ($\sqrt{2}$), corresponding to the fact that the horizontal distance covered from the base to the apex along an edge is twice that along the median of a face. In other words, if C is the centroid of the base, the distance from C to a vertex of the base is twice that from C to the midpoint of an edge of the base. This follows from the fact that the medians of a triangle intersect at its centroid, and this point divides each of them in two segments, one of which is twice as long as the other (see proof).
## Volume of any tetrahedron
The volume of any tetrahedron is given by the pyramid volume formula:
$V = \frac{1}{3} Ah \,$
where A is the area of the base and h the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apexes to the opposite faces are inversely proportional to the areas of these faces.
For a tetrahedron with vertices a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), and d = (d1, d2, d3), the volume is (1/6)·|det(ab, bc, cd)|, or any other combination of pairs of vertices that form a simply connected graph. This can be rewritten using a dot product and a cross product, yielding
$V = \frac { |(\mathbf{a}-\mathbf{d}) \cdot ((\mathbf{b}-\mathbf{d}) \times (\mathbf{c}-\mathbf{d}))| } {6}.$
If the origin of the coordinate system is chosen to coincide with vertex d, then d = 0, so
$V = \frac { |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| } {6},$
where a, b, and c represent three edges that meet at one vertex, and $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ is a scalar triple product. Comparing this formula with that used to compute the volume of a parallelepiped, we conclude that the volume of a tetrahedron is equal to 1/6 of the volume of any parallelepiped which shares with it three converging edges.
It should be noted that the triple scalar can be represented by the following determinants:
$6 \cdot \mathbf{V} =\begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix}$ or $6 \cdot \mathbf{V} =\begin{vmatrix} \mathbf{a} \\ \mathbf{b} \\ \mathbf{c} \end{vmatrix}$ where $\mathbf{a} = (a_1,a_2,a_3) \,$ is expressed as a row or column vector etc.
Hence
$36 \cdot \mathbf{V^2} =\begin{vmatrix} \mathbf{a^2} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{b} & \mathbf{b^2} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c} & \mathbf{c^2} \end{vmatrix}$ where $\mathbf{a} \cdot \mathbf{b} = ab\cos{C}$ etc.
which gives
$\mathbf{V}= \frac {abc} {6} \sqrt{1 + 2\cos{A}\cos{B}cos{C}-\cos^2{A}-\cos^2{B}-\cos^2{C}} \,$
If we are given only the distances between the vertices of any tetrahedron, then we can compute its volume using the formula:
$288 \cdot V^2 = \begin{vmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\ 1 & d_{21}^2 & 0 & d_{23}^2 & d_{24}^2 \\ 1 & d_{31}^2 & d_{32}^2 & 0 & d_{34}^2 \\ 1 & d_{41}^2 & d_{42}^2 & d_{43}^2 & 0 \end{vmatrix}.$
If the determinant's value is negative this means we can not construct a tetrahedron with the given distances between the vertices.
## Distance between the edges
Any two opposite edges of a tetrahedron lie on two skew lines. If the closest pair of points between these two lines are points in the edges, they define the distance between the edges; otherwise, the distance between the edges equals that between one of the endpoints and the opposite edge.
## Three dimensional properties of a generalized tetrahedron
As with triangle geometry, there is a similar set of three dimensional geometric properties for a tetrahedron. A generalised tetrahedron has an insphere, circumsphere, medial tetrahedron and exspheres. It has respective centers such as incenter, circumcenter, excenters, Spieker centre and points such as a centroid. However there is, generally, no orthocenter in the sense of intersecting altitudes. There is an equivalent sphere to the triangular nine point circle that is the circumsphere of the medial tetrahedron. However its circumsphere does not, generally, pass through the base points of the altitudes of the reference tetrahedron.
To resolve these inconsistencies, a substitute centre known as the Monge point that always exists for a generalized tetrahedron is introduced. This point was first identified by Gaspard Monge. For tetrahedra where the altitudes do intersect, the Monge point and the orthocenter coincide. The Monge point is define as the point where the six midplanes of a tetrahedron intersect. A midplane is defined as a plane that is orthogonal to an edge joining any two vertices that also contains the centroid of an opposite edge formed by joining the other two vertices.
An orthogonal line dropped from the Monge point to any face is coplanar with two other orthogonal lines to the same face. The first is an altitude dropped from a corresponding vertex to the chosen face. The second is an orthogonal line to the chosen face that passes through the orthocenter of that face. This orthogonal line through the Monge point lies mid way between the altitude and the orthocentric orthogonal line.
The Monge point, centroid and circumcenter of a tetrahedron are colinear and form the Euler line of the tetrahedron. However, unlike the triangle, the centroid of a tetrahedron lies at the midpoint of its Monge point and circumcenter.
There is an equivalent sphere to the triangular nine point circle for the generalized tetrahedron. It is the circumsphere of its medial tetrahedron. It is a twelve point sphere centered at the circumcenter of the medial tetrahedron. By definition it passes through the centroids of the four faces of the reference tetrahedron. It passes through four substitute Euler points that are located at a distance of 1/3 of the way from M, the Monge point, toward each of the four vertices. Finally it passes through the four base points of orthogonal lines dropped from each Euler point to the face not containing the vertex that generated the Euler point.
If T represents this twelve point center then it also lies on the Euler line, unlike its triangular counterpart, the center lies 1/3 of the way from M, the Monge point towards the circumcenter. Also an orthogonal line through T to a chosen face is coplanar with two other orthogonal lines to the same face. The first is an orthogonal line passing through the corresponding Euler point to the chosen face. The second is an orthogonal line passing through the centroid of the chosen face. This orthogonal line through the twelve point center lies mid way between the Euler point orthogonal line and the centroidal orthogonal line. Furthermore, for any face, the twelve point centre lies at the mid point of the corresponding Euler point and the orthocenter for that face.
The radius of the twelve point sphere is 1/3 of the circumradius of the reference tetrahedron.
If OABC forms a generalized tetrahedron with a vertex O as the origin and vectors $\mathbf{a}, \mathbf{b} \,$ and $\mathbf{c} \,$ represent the positions of the vertices A, B and C with respect to O, then the radius of the insphere is given by:
$r= \frac {6V} {|\mathbf{b} \times \mathbf{c}| + |\mathbf{c} \times \mathbf{a}| + |\mathbf{a} \times \mathbf{b}| + |(\mathbf{b} \times \mathbf{c}) + (\mathbf{c} \times \mathbf{a}) + (\mathbf{a} \times \mathbf{b})|} \,$
and the radius of the circumsphere is given by:
$R= \frac {|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})|} {12V} \,$
which gives the radius of the twelve point sphere:
$r_T= \frac {|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})|} {36V} \,$
where:
$6V= |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \,$
The vector position of various centers are given as follows:
The centroid
$\mathbf{G} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{4} \,$
The circumcenter
$\mathbf{O}= \frac {\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})} {12V} \,$
The Monge point
$\mathbf{M} = \frac {\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})(\mathbf{b} \times \mathbf{c}) + \mathbf{b}\cdot (\mathbf{c} + \mathbf{a})(\mathbf{c} \times \mathbf{a}) + \mathbf{c} \cdot (\mathbf{a} + \mathbf{b})(\mathbf{a} \times \mathbf{b})} {12V} \,$
The Euler line relationships are:
$\mathbf{G} = \mathbf{M} + \frac{1}{2} (\mathbf{O}-\mathbf{M})\,$
$\mathbf{T} = \mathbf{M} + \frac{1}{3} (\mathbf{O}-\mathbf{M})\,$
It should also be noted that:
$\mathbf{a} \cdot \mathbf{O} = \frac {\mathbf{a^2}}{2} \quad\quad \mathbf{b} \cdot \mathbf{O} = \frac {\mathbf{b^2}}{2} \quad\quad \mathbf{c} \cdot \mathbf{O} = \frac {\mathbf{c^2}}{2}\,$
and:
$\mathbf{a} \cdot \mathbf{M} = \frac {\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})}{2} \quad\quad \mathbf{b} \cdot \mathbf{M} = \frac {\mathbf{b} \cdot (\mathbf{c} + \mathbf{a})}{2} \quad\quad \mathbf{c} \cdot \mathbf{M} = \frac {\mathbf{c} \cdot (\mathbf{a} + \mathbf{b})}{2}\,$
## Geometric relations
A tetrahedron is a 3- simplex. Unlike in the case of other Platonic solids, all vertices of a regular tetrahedron are equidistant from each other (they are in the only possible arrangement of four equidistant points).
A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual.
A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are
(+1, +1, +1);
(−1, −1, +1);
(−1, +1, −1);
(+1, −1, −1).
For the other tetrahedron (which is dual to the first), reverse all the signs. The volume of this tetrahedron is 1/3 the volume of the cube. Combining both tetrahedra gives a regular polyhedral compound called the stella octangula, whose interior is an octahedron. Correspondingly, a regular octahedron is the result of cutting off, from a regular tetrahedron, four regular tetrahedra of half the linear size (i.e., rectifying the tetrahedron). The above embedding divides the cube into five tetrahedra, one of which is regular. In fact, 5 is the minimum number of tetrahedra required to compose a cube.
Inscribing tetrahedra inside the regular compound of five cubes gives two more regular compounds, containing five and ten tetrahedra.
Regular tetrahedra cannot tessellate space by themselves, although it seems likely enough that Aristotle reported it was possible. However, two regular tetrahedra can be combined with an octahedron, giving a rhombohedron which can tile space.
However, there is at least one irregular tetrahedron of which copies can tile space. If one relaxes the requirement that the tetrahedra be all the same shape, one can tile space using only tetrahedra in various ways. For example, one can divide an octahedron into four identical tetrahedra and combine them again with two regular ones. (As a side-note: these two kinds of tetrahedron have the same volume.)
The tetrahedron is unique among the uniform polyhedra in possessing no parallel faces.
## Intersecting tetrahedra
An interesting polyhedron can be constructed from five intersecting tetrahedra. This compound of five tetrahedra has been known for hundreds of years. It comes up regularly in the world of origami. Joining the twenty vertices would form a regular dodecahedron. There are both left-handed and right-handed forms which are mirror images of each other.
## The isometries of the regular tetrahedron
The proper rotations and reflections in the symmetry group of the regular tetrahedron
The vertices of a cube can be grouped into two groups of four, each forming a regular tetrahedron (see above, and also animation, showing one of the two tetrahedra in the cube). The symmetries of a regular tetrahedron correspond to half of those of a cube: those which map the tetrahedrons to themselves, and not to each other.
The tetrahedron is the only Platonic solid that is not mapped to itself by point inversion.
The regular tetrahedron has 24 isometries, forming the symmetry group Td, isomorphic to S4. They can be categorized as follows:
• T, isomorphic to alternating group A4 (the identity and 11 proper rotations) with the following conjugacy classes (in parentheses are given the permutations of the vertices, or correspondingly, the faces, and the unit quaternion representation):
• identity (identity; 1)
• rotation about an axis through a vertex, perpendicular to the opposite plane, by an angle of ±120°: 4 axes, 2 per axis, together 8 ((1 2 3), etc.; (1±i±j±k)/2)
• rotation by an angle of 180° such that an edge maps to the opposite edge: 3 ((1 2)(3 4), etc.; i,j,k)
• reflections in a plane perpendicular to an edge: 6
• reflections in a plane combined with 90° rotation about an axis perpendicular to the plane: 3 axes, 2 per axis, together 6; equivalently, they are 90° rotations combined with inversion (x is mapped to −x): the rotations correspond to those of the cube about face-to-face axes
## The isometries of irregular tetrahedra
The isometries of an irregular tetrahedron depend on the geometry of the tetrahedron, with 7 cases possible. In each case a 3-dimensional point group is formed.
• An equilateral triangle base and isosceles (and non-equilateral) triangle sides gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C3v, isomorphic to S3.
• Four congruent isosceles (non-equilateral) triangles gives 8 isometries. If edges (1,2) and (3,4) are of different length to the other 4 then the 8 isometries are the identity 1, reflections (12) and (34), and 180° rotations (12)(34), (13)(24), (14)(23) and improper 90° rotations (1234) and (1432) forming the symmetry group D2d.
• Four congruent scalene triangles gives 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein four-group V4Z22, present as the point group D2.
• Two pairs of isomorphic isosceles (non-equilateral) triangles. This gives two opposite edges (1,2) and (3,4) that are perpendicular but different lengths, and then the 4 isometries are 1, reflections (12) and (34) and the 180° rotation (12)(34). The symmetry group is C2v, isomorphic to V4.
• Two pairs of isomorphic scalene triangles. This has two pairs of equal edges (1,3), (2,4) and (1,4), (2,3) but otherwise no edges equal. The only two isometries are 1 and the rotation (12)(34), giving the group C2 isomorphic to Z2.
• Two unequal isosceles triangles with a common base. This has two pairs of equal edges (1,3), (1,4) and (2,3), (2,4) and otherwise no edges equal. The only two isometries are 1 and the reflection (34), giving the group Cs isomorphic to Z2.
• No edges equal, so that the only isometry is the identity, and the symmetry group is the trivial group.
## A law of sines for tetrahedra and the space of all shapes of tetrahedra
A corollary of the usual law of sines is that in a tetrahedron with vertices O, A, B, C, we have
$\sin\angle OAB\cdot\sin\angle OBC\cdot\sin\angle OCA = \sin\angle OAC\cdot\sin\angle OCB\cdot\sin\angle OBA.\,$
One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface.
Putting any of the four vertices in the role of O yields four such identities, but in a sense at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity. One reason to be interested in this "independence" relation is this: It is widely known that three angles are the angles of some triangle if and only if their sum is a half-circle. What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be a half-circle. Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by this sine law further reduce the number of degrees of freedom, not from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.
## Computational uses
Complex shapes are often broken down into a mesh of irregular tetrahedra in preparation for finite element analysis and computational fluid dynamics studies.
## Applications and uses
The ammonium ion is tetrahedral
Several of the Tetrahedron scoring pieces and goals from the 2005 FIRST Robotics Competition game
Packaging
• The company Tetrapak's Tetra Classic is in the shape of a tetrahedral.
Chemistry
• The tetrahedron shape is seen in nature in covalent bonds of molecules. For instance in a methane molecule (CH4) the four hydrogen atoms lie in each corner of a tetrahedron with the carbon atom in the centre. For this reason, one of the leading journals in organic chemistry is called Tetrahedron. Ammonium is another example.
• Angle from the centre to any two vertices is $\arccos{\left(-\tfrac{1}{3}\right)}$, or approximately 109.47°.
Electronics
• If each edge of a tetrahedron were to be replaced by a one ohm resistor, the resistance between any two vertices would be 1/2 ohm.
Symbolism
• The tetrahedron represents the classical element fire.
Games
• Especially in roleplaying, this solid is known as a d4, one of the more common polyhedral dice.
• Tetrahedrons constructed of 1 1/4" PVC pipe, which were known as 'tetras', were used as the main scoring object on the 2005 FIRST Robotics Competition game Triple Play. The object of the game was to stack these 'tetras' onto larger tetrahedron goals which here placed in a 3×3 matrix.
• Some Rubik's Cube-like puzzles are tetrahedral, such as the Pyraminx and Pyramorphix.
• Pyramid head from the Silent Hill games has a tetrahedral on top of his head.
• In the Xeelee Sequence of science fiction books by author Stephen Baxter, a blue-green tetrahedron is the symbol of free humanity.
• The Triforces of the Legend of Zelda cartoon series are green and red tetrahedrons. The triforce's representation in the actual game series (starting in A Link to the Past) is that of an unfolded tetrahedron.
• The story arc of the fan series "Star Trek: Hidden Frontier" evolves around gigantic ancient artifacts, which later become a centre part of the series. The artifacts are referred to as Tetrahedrons and have the shape of such a geometrical body. In the series, the Tetrahedrons possess the ability to produce a great deal of energy and are of unknown material and origin, however, they appear to be several million years old and it is outlined that these devices have been built by an ancient civilization.
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# Introduction to Graph Theory
## The Knigsberg Bridge Problem o
The city of Knigsberg was located on the Pregel river in Prussia. The river dio vided the city into four separate landmasses, including the island of Kneiphopf. These four regions were linked by seven bridges as shown in the diagram. Residents of the city wondered if it were possible to leave home, cross each of the seven bridges exactly once, and return home. The Swiss mathematician Leonhard Euler (1707-1783) thought about this problem and the method he used to solve it is considered by many to be the birth of graph theory.
x e1
1
X W
e2 w e5 e4 e3
e6 y e7 z
Y Z
7
Exercise 1.1. See if you can nd a round trip through the city crossing each bridge exactly once, or try to explain why such a trip is not possible. The key to Eulers solution was in a very simple abstraction of the puzzle. Let us redraw our diagram of the city of Knigsberg by representing each of the o land masses as a vertex and representing each bridge as an edge connecting the vertices corresponding to the land masses. We now have a graph that encodes the necessary information. The problem reduces to nding a closed walk in the graph which traverses each edge exactly once, this is called an Eulerian circuit. Does such a circuit exist?
Fundamental Denitions
We will make the ideas of graphs and circuits from the Knigsberg Bridge o problem more precise by providing rigorous mathematical denitions. A graph G is a triple consisting of a vertex set V (G), an edge set E(G), and a relation that associates with each edge, two vertices called its endpoints (not necessarily distinct). Graphically, we represent a graph by drawing a point for each vertex and representing each edge by a curve joining its endpoints. For our purposes all graphs will be nite graphs, i.e. graphs for which V (G) and E(G) are nite sets, unless specically stated otherwise. Note that in our denition, we do not exclude the possibility that the two endpoints of an edge are the same vertex. This is called a loop, for obvious reasons. Also, we may have multiple edges, which is when more than one edge shares the same set of endpoints, i.e. the edges of the graph are not uniquely determined by their endpoints. A simple graph is a graph having no loops or multiple edges. In this case, each edge e in E(G) can be specied by its endpoints u, v in V (G). Sometimes we write e = uv. When two vertices u, v in V (G) are endpoints of an edge, we say u and v are adjacent. A path is a simple graph whose vertices can be ordered so that two vertices are adjacent if and only if they are consecutive in the ordering. A path which begins at vertex u and ends at vertex v is called a u, v-path. A cycle is a simple graph whose vertices can be cyclically ordered so that two vertices are adjacent if and only if they are consecutive in the cyclic ordering. We usually think of paths and cycles as subgraphs within some larger graph. A subgraph H of a graph G, is a graph such that V (H) V (G) and E(H) E(G) satisfying the property that for every e E(H), where e has endpoints u, v V (G) in the graph G, then u, v V (H) and e has endpoints u, v in H, i.e. the edge relation in H is the same as in G. A graph G is connected if for every u, v V (G) there exists a u, v-path in G. Otherwise G is called disconnected. The maximal connected subgraphs of G are called its components. A walk is a list v0 , e1 , v1 , . . . , ek , vk of vertices and edges such that for 1 i k, the edge ei has endpoints vi1 and vi . A trail is a walk with no repeated edge. A u, v-walk or u, v-trail has rst vertex u and last vertex v. When the rst and last vertex of a walk or trail are the same, we say that they are closed. A closed trail is called a circuit. With this new terminology, we can consider paths and cycles not just as subgraphs, but also as ordered lists of vertices and edges. From this point of view, a path is a trail with no repeated vertex, and a cycle is a closed trail (circuit) with no repeated vertex other than the rst vertex equals the last vertex. Alternatively, we could consider the subgraph traced out by a walk or trail.
## Walks Trails (no edge is repeated) Paths
(no vertex is repeated)
Circuits
(closed trails)
Cycles
An Eulerian trail is a trail in the graph which contains all of the edges of the graph. An Eulerian circuit is a circuit in the graph which contains all of the edges of the graph. A graph is Eulerian if it has an Eulerian circuit. The degree of a vertex v in a graph G, denoted deg v, is the number of edges in G which have v as an endpoint.
Exercises
## Consider the following collection of graphs:
(a) (b) (c) (d)
(e)
(f)
(g)
(h)
1. Which graphs are simple? 2. Suppose that for any graph, we decide to add a loop to one of the vertices. Does this aect whether or not the graph is Eulerian? 3. Which graphs are connected? 4. Which graphs are Eulerian? Trace out an Eulerian circuit or explain why an Eulerian circuit is not possible. 5. Are there any graphs above that are not Eulerian, but have an Eulerian trail? 6. Give necessary conditions for a graph to be Eulerian. 7. Give necessary conditions for a graph to have an Eulerian trail.
8. Given that a graph has an Eulerian circuit beginning and ending at a vertex v, is it possible to construct an Eulerian circuit beginning and ending at any vertex in the graph? 9. Eulers House. Baby Euler has just learned to walk. He is curious to know if he can walk through every doorway in his house exactly once, and return to the room he started in. Will baby Euler succeed? Can baby Euler walk through every door exactly once and return to a dierent place than where he started? What if the front door is closed?
Front Door
Euler's House
Piano Rm Dining Rm Garage Kitchen Family Rm Master Bd Hall Living Rm Study
Conservatory Yard
## Characterization of Eulerian Circuits
We have seen that there are two obvious necessary conditions for a graph to be Eulerian: the graph must have at most one nontrivial component, and every vertex in the graph must have even degree. Now a more interesting question is, are these conditions sucient? That is, does every connected graph with vertices of even degree have an Eulerian circuit? This is the more dicult question which Euler was able to prove in the armative. Theorem 1. A graph G is Eulerian if and only if it has at most one nontrivial component and its vertices all have even degree. There are at least three dierent approaches to the proof of this theorem. We will use a constructive proof that provides the most insight to the problem. There is also a nonconstructive proof using maximality, and a proof that implements an algorithm. We will need the following result. Lemma 2. If every vertex of a graph G has degree at least 2, then G contains a cycle. Proof. Let P be a maximal path in G. Maximal means that the path P cannot be extended to form a larger path. Why does such a path exist? Now let u be an endpoint of P . Since P is maximal (cannot be extended), every vertex adjacent to u must already be in P . Since u has degree at least two, there is an edge e extending from u to some other vertex v in P , where e is not in P . The edge e together with the section of P from u to v completes a cycle in G. Proof of theorem. We have already seen that if G is Eulerian, then G has at most one nontrivial component and all of the vertices of G have even degree. We just need to prove the converse. Suppose G has at most one nontrivial component and that all of the vertices of G have even degree. We will use induction on the number of edges n. Basis step: When n = 0, a circuit consisting of just one vertex is an Eulerian circuit. Induction step: When n > 0, each vertex in the nontrivial component of G has degree at least 2. Why? By the lemma, there is a cycle C in the nontrivial component of G. Let G be the graph obtained from G by deleting E(C). Note that G is a subgraph of G which also has the property that all of its vertices have even degree. Why? Note also that G may have several nontrivial components. Each of these components of G must have number of edges < n. Why? By the induction hypothesis, each of these components has an Eulerian circuit. To construct an Eulerian circuit for G, we traverse the cycle C, but when a component of G is entered for the rst time (why must every component intersect C?), we detour along an Eulerian circuit of that component. This circuit ends at the vertex where we began the detour. When we complete the traversal of C, we have completed the Eulerian circuit of G.
deg v = 2 #E(G)
vV (G)
## Proposition 3. (Degree-Sum Formula) If G is a graph, then
where #E(G) is the number of edges in G. Proof. This is simply a matter of counting each edge twice. The details are left as an exercise. This formula is extremely useful in many applications where the number of vertices and number of edges are involved in calculations. For example, we will learn later about the graph invariants of Euler characteristic and genus; the degree-sum formula often allows us to prove inequalities bounding the values of these invariants. A fun corollary of the degree-sum formula is the following statement, also known as the handshaking lemma. Corollary 4. In any graph, the number of vertices of odd degree is even. Or equivalently, the number of people in the universe who have shaken hands with an odd number of people is even. Proof. Try to solve this one yourself. Hint: Split the sum on the left hand side of the degree-sum formula into two piecesone over vertices of even degree and one over vertices of odd degree.
Important Graphs
There are two special types of graphs which play a central role in graph theory, they are the complete graphs and the complete bipartite graphs. A complete graph is a simple graph whose vertices are pairwise adjacent. The complete graph with n vertices is denoted Kn .
K1
K2
K3
K4
K5
Before we can talk about complete bipartite graphs, we must understand bipartite graphs. An independent set in a graph is a set of vertices that are pairwise nonadjacent. A graph G is bipartite if V (G) is the union of two disjoint (possibly empty) independent sets, called partite sets of G. Similarly, a graph is k-partite if V (G) is the union of k disjoint independent sets. 6
## 3 dierent bipartite graphs
A 3-partite graph
A complete bipartite graph is a simple bipartite graph such that two vertices are adjacent if and only if they are in dierent partite sets. The complete bipartite graph with partite sets of size m and n is denoted Km,n .
K1,1
K2,2
K2,3
K3,3
Exercises
1. Determine the values of m and n such that Km,n is Eulerian 2. Prove or disprove: (a) Every Eulerian bipartite graph has an even number of edges. (b) Every Eulerian simple graph with an even number of vertices has an even number of edges. What if we also assume that the graph has only one component? 3. When is a cycle a bipartite graph? 4. Oh no! Baby Euler has gotten into the handpaints. His favorite colors are blue and yellow. Baby Euler wants to paint each room in the house (including the hall) either blue or yellow such that every time he walks from one room to an adjacent room, the color changes. Is this possible?
5. If we consider the graph corresponding to Eulers house, the previous problem is equivalent to assigning the color blue or yellow to each vertex of the graph so that no two vertices of the same color are adjacent. This is called a 2-coloring of the graph. What is the relationship between 2coloring vertices of a graph and bipartite graphs?
## k-partite and k-colorable
A k-coloring of a graph G, is a labeling of the vertices f : V (G) S, where S is some set such that |S| = k. Normally we think of the set S as a collection of k dierent colors, say S = {red, blue, green, etc.}, or more abstractly as the positive integers S = {1, 2, . . . , k}. The labels are called colors. A k-coloring is proper if adjacent vertices are dierent colors. A graph is k-colorable if it has a proper k-coloring. The chromatic number (G) is the least positive integer k such that G is k-colorable. You should notice that a graph is k-colorable if and only if it is k-partite. In other words, k-colorable and k-partite mean the same thing. You should convince yourself of this by determining the k dierent partite sets of a kcolorable graph and conversely determine a k-coloring of a k-partite graph. In general it is not easy to determine the chromatic number of a graph or even if a graph is k-colorable for a given k.
Exercises
1. If a graph G is k-partite, what do we know about (G)? 2. Show that (G) = 1 if and only if G is totally disconnected, i.e. all of the components of G contain only 1 vertex. 3. For a nite graph G, is (G) also nite? Find an upper bound, or nd a nite graph G which cannot be colored by nitely many colors. 4. Determine the chromatic number for each of the following graphs:
(a)
(b)
(c) Any cycle of odd length (the length of a cycle is the number of edges in the cycle). (d) Any cycle of even length.
(f)
(g)
## (j) This graph is called the hypercube, or 4-dimensional cube.
(k) This is an example of an innite graph. If the vertices of the graph are the integer coordinates, then this is also an example of a unit distance graph, since two vertices are adjacent if and only if they are distance one apart. 5. Suppose that the graph G is bipartite, i.e. 2-colorable, is it possible for G to contain a cycle of odd length?
10
10
## Characterization of Bipartite Graphs
We have just seen that if G is bipartite, then G contains no cycles of odd length. Equivalently, this means that if G does have a cycle of odd length, then G is not bipartite, hence not 2-colorable (this is the contra-positive of the previous statement). You should look back at the problem of coloring the rooms in Eulers house to determine if there is such an odd cycle. The obvious question now is whether or not the converse of the above statement is true. That is, if G contains no cycles of odd length, does it hold that G is bipartite? The answer is yes. Theorem 5. A graph G is bipartite if and only if G contains no cycles of odd length. Proof. You should have already proved the forward direction in the exercises, so we will prove the other direction. Suppose that G contains no cycles of odd length. We might as well assume that G contains only one component, since if it has more than one, we can form a bipartition of the graph from the bipartition of each of its components. Thus assume G has one component. Pick a vertex u V (G). For each v V (G), let f (v) be the minimum length of a u, v-path. Since G is connected, f (v) is dened for each v V (G). Let X = {v V (G) | f (v) is even} and Y = {v V (G) | f (v) is odd}. We wish to show that X and Y are independent sets of vertices. Indeed, if there are adjacent vertices v, v both in X (or both in Y ), then the closed walk consisting of the shortest u, v-path, plus the edge v to v , plus the reverse of the shortest u, v -path, is a closed walk of odd length. It can be shown by induction that every closed walk of odd length contains an odd cycle, but this contradicts our hypothesis that G contains no cycles of odd length. Therefore no two vertices in X (or Y ) are adjacent, i.e. X and Y are independent so that G is bipartite. Exercise 10.1. Show that every closed walk of odd length contains a cycle of odd length. Hint: Use induction on the length l of the closed walk. If the closed walk has no repeated vertex, then it is a cycle of odd length. If it does have a repeated vertex v, then break the closed walk into two shorter walks.
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## Upper and Lower Bounds for (G)
(G) #V (G).
We have already seen an upper bound for (G) in the exercises, namely
For the particular case of the complete graph we have (Kn ) = #V (Kn ) = n, so this is the best possible upper bound for the chromatic number in terms of the size of the vertex set. However, we may derive other upper bounds using other structural information about the graph. As an example, we will show that (G) (G) + 1, 11
where the number (G) is the maximum degree of all the vertices of G. Begin with an ordering v1 , v2 , . . . , vn of all of the vertices of G. The greedy coloring of G colors the vertices in order v1 , v2 , . . . vn and assigns to each vi the lowest-indexed color which has not already been assigned to any of the previous vertices in the ordering that are adjacent to vi . Note that in any vertex ordering of G, each vertex vi has at most (G) vertices which are adjacent to vi and have appeared earlier in the ordering. Thus as we color each vertex of G, we never need more than (G) + 1 colors. It follows that (G) (G) + 1.
To give a useful lower bound, we dene a set of vertices called a clique, which is complementary to the notion of an independent set dened earlier. A clique in a graph is a set of pairwise adjacent vertices. The clique number of a graph G, denoted (G), is the maximum size of a clique in G. In eect, a clique corresponds to a subgraph (whose vertices are the vertices of the clique) that is itself a complete graph. Thus if (G) = n, then there is a clique of size n corresponding to a subgraph of G that is equivalent to Kn . Since it will require at least n colors to color the vertices in this clique, we have that (G) (G).
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## Unit Distance GraphsAn Open Problem
A unit distance graph is quite simply a graph whose vertices are points in the plane (or more generally any Euclidean space), with an edge between two vertices u and v if and only if the distance between u and v is 1. Consider the unit distance graph whose set of vertices is the entire plane. Let us denote this graph by P . This is denitely not a nite graph, as there are uncountably many vertices, and for each vertex v there are uncountably many edges having v as an endpoint! In particular, for a given vertex v which corresponds to a point (x, y) R2 in the plane, the vertices which are adjacent to v are those which correspond to the points lying on the circle of radius 1 centered at (x, y). Exercise 12.1. What is (P )? This number is also known as the chromatic number of the plane. This question can be restated more simply as follows: How many colors are needed so that if each point in the plane is assigned one of the colors, no two points which are exactly distance 1 apart will be assigned the same color? This problem has been open since 1956 and it is known that the answer is either 4, 5, 6 or 7 (apparently it is not very dicult to prove that these are the only possibilities). I was able to show rather easily that the answer is 3, 4, 5, 6, or 7 but I did not spend enough time working on the problem 12
to determine how dicult it would be to eliminate 3 as a possibility. I encourage you to work on this yourself to get a feel for the subtleties of the problem. I dont think you will need much help eliminating 1 or 2. To prove 1 that (P ) 9, try tiling the plane with 2 1 squares and coloring the squares 2 in a clever pattern. To show (P ) 7, use a similar technique of tiling the plane into colored shapes. To eliminate 3 you will probably need to expand on the type of argument used to eliminate 2, but youre on your own here. Eliminate any of the remaining numbers and you can publish your results. To learn more about this and related open problems in graph theory, visit http://dimacs.rutgers.edu/hochberg/undopen/graphtheory/graphtheory.html .
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## The Four Color Theorem and Planar Graphs
Arguably the most famous theorem in the eld of graph theory is the Four Color Theorem. For an excellent history and explanation of the problem, see the article in Wikipedia at http://en.wikipedia.org/wiki/Four color theorem. Briey, this theorem states that 4 colors are sucient to color regions in the plane so that no two regions which border each other have the same color. It is trivial to verify that 3 colors is not sucient, and the proof that 5 colors is sufcient is not dicult. That 4 colors is indeed sucient to color any subdivision of the plane, proved to be an extremely dicult problem that was nally solved in 1976 with the aid of a computer. This computer-aided proof has proved to be quite unsatisfying to many mathematicians. The four color theorem can be stated quite simply in terms of graph theory. Just as with the Knigsberg bridge problem, or the exercise about Eulers house, o we abstract by representing the important information with a graph. Each region in the plane is represented as a vertex; two vertices are adjacent if and only if their corresponding regions border each other; and coloring the regions corresponds to a proper coloring of the vertices of the graph. You should notice that all possible graphs formed from such planar regions share an important property, namely they can be drawn in the plane without having to cross edges. This motivates the denition of planar graphs. A graph is planar if it can be drawn in the plane without crossings. (Examples of planar and nonplanar graphs.) Theorem 6. (Four color theoremoriginally stated by P.J. Heawood 1890) For any planar graph G, we have (G) 4. Proof. K. Appel and W. Haken 1976.
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Exercises
(a) K4 (b) K5 (c) K2,3 (d) K3,3
## 1. Determine if the following graphs are planar or nonplanar.
(e)
(f)
(g)
2. Find an example of a planar graph that is not 3-colorable. 3. Does the four color theorem imply that you need at most 4 colors to color a political map of the world so that each country is assigned a color and no two adjacent countries have the same color? Explain why or provide a counterexample. (This is a bit of a trick question) 4. Consider any planar graph G. Draw this graph in the plane so that there are no crossings. We refer to the regions of the plane bounded by the edges of the graph as faces, and denote the set of faces of G by F (G). Compute X(G) = #V (G) #E(G) + #F (G) for each of the graphs (number of vertices minus the number of edges plus the number of faces). This is called the Euler characteristic of the graph. What trend do you notice? 5. (Kuratowskis theorem) Kuratowski proved that a nite graph is planar if and only if it contains no subgraph that is isomorphic to or is a subdivision of K5 or K3,3 . In this sense, K5 and K3,3 are the basic building blocks of nonplanar graphs. Consider the graph from Exercise 1(f) above. Can you nd a subgraph of this graph which looks like K5 or K3,3 ? What about the graph of the hypercube shown in Section 9?
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## The Genus of a Graph
The ability to draw a graph in the plane without crossings is equivalent to being able to draw a graph on a sphere without crossings. For example, if you can draw a graph on a sphere, simply puncture the sphere in the middle of one of the faces formed by the edges of the graph and then stretch out this hole until you can lay the sphere at onto the plane. The result will be a drawing of the graph in the plane with no crossings. Conversely, if you can draw a graph in 14
the plane without crossings, take the outer face (the face containing ) and reverse the process above by in essence wrapping the plane around the sphere (the point at corresponds to the punctured hole in the sphere). The sphere is what we call a surface of genus zero. The genus of the surface tells you how many doughnut holes are in the surface. Thus a sphere has genus zero, a torus has genus one, a two-holed torus has genus 2, and so on. There are nonplanar graphs (hence cannot be drawn on a sphere without crossings), that can however be drawn on a torus without crossings. The graph K5 is such a graph. Similarly, there are graphs which cannot be drawn on a torus but can be drawn on a two-holed torus. The minimum genus of surface upon which a graph can be drawn without crossing edges is called the genus of a graph and is denoted (G). It can be shown that any nite graph can be drawn without crossings on a surface of large enough genus. Therefore the genus of a nite graph is well-dened. The genus of most important graphs has been calculated. For example (Kn ) = and (Km,n ) = (n 3) (n 4) 12 (m 2) (n 2) 4
where denotes the ceiling function (these calculations can be found in Hararys book on graph theory). So for example, (K4 ) = 0, (K5 ) = 1, (K7 ) = 1 and (K8 ) = 2. Thus K4 is the largest complete graph which can be drawn on the sphere and K7 is the largest complete graph which can be drawn on the torus. Exercise 15.1. The accompanying gure shows how to draw K5 on the torus without crossing edges. Try to draw K6 or K7 on the torus.
K5 on the torus
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# Find the area of a parallelogram with sides of 6 and 12 and an angle of 60°
106,286 results
maths
A parallelogram has two adjacent sides of length 4cm and 6cm reapectively. If the included angle measures 52 degrees, find the area of the parallelogram.
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I have six questions which will be on a test two weeks from now, and I do not understand. Help me find out these theorems for the blanks. 1. If one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a_______? 2. If the diagonals of a ...
geometry
abcd is a parallelogram with sides ab=12 cm,bc=10 cm and diagonal ac=16 cm .find the area of the parallelogram also find the distance between its shortest sides
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abcd is a parallelogram with sides ab=12 cm,bc=10 cm and diagonal ac=16 cm .find the area of the parallelogram also find the distance between its shortest sides
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If the perimeter of a parallelogram is 140 m, the distance between a pair of opposite sides is 7 meters and its area is 210 sq m, find the length of two adjacent sides of the parallelogram.
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If the perimeter of a parallelogram is 140 m, the distance between a pair of opposite sides is 7 meters and its area is 210 sq m, find the length of two adjacent sides of the parallelogram.
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THe area of a parallelogram is 594, and the lengths of its sides are 32 and 46. Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram
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Find the area of a parallelogram with sides of 6 and 12 and an angle of 60°.
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The sides of a parallelogram are 15 ft and 17 ft. One angle is 40° while another angle is 140°. Find the lengths of the diagonals of the parallelogram (to the nearest tenth of a foot)
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Given: QRST is a parallelogram. Prove: QRST is a square. Complete the proof below by choosing the reason for line number 2 and line number 6. Reason Statement 1. QRST is a parallelogram. Given 2. QRST is a rectangle 3. is a right angle Definition of a right angle. 4. ...
e
Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure \$45^\circ\$.
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Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees.
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Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees.
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1) Find correct to six decimal places root of the equation cos(x)= x for xE[0, pi/2] using Newton's Method. 2) A triangle has two constant lengths of 10 cm and 15 cm. The angle between two constant sides increases at a rate of 9 deg/min. Find the rate of increase of the third ...
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1.Determine the area of a parallelogram in which 2 adjacent sides are 10cm and 13cm and the angle between them is 55 degrees? 2.If the area of triangle ABC is 5000m squared with a=150m and b=70m what are the two possible sizes of angle C?
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1)If diagonals of a rhombus are 10 cm and 24 cm. find the area and perimeter of the rhombus. 2)A regular hexagon with a perimeter of 24 units is inscribed in a circle. Find the radius of the circle. 3)Find the altitude,perimeter and area of an isosceles trapezoid whose sides ...
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The area of a parallelogram with a side of 5.6 in is equal to the area of a rectangle with sides 7 in and 8 in. Find the altitude to the given side in the parallelogram.
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The area of a parallelogram with a side of 5.6 in is equal to the area of a rectangle with sides 7 in and 8 in. Find the altitude to the given side in the parallelogram.
Geometry
Can some one please check these for me.Thanks Angle DEF is similar to Angle HJK and the scale factor of angle DEF to anle HJK IS 5/2. If EF =15 FIND JK Here is what I came up with Am I correct 5/2 = EF/JK 5/2 =15/JK 5(JK)=30 JK=30/5 JK=6 Given: ABCD is a parallelogram; <1 ...
Maths
The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is 30cm. Find the area of the parallelogram by using heron's formula
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The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is 30cm. Find the area of parallelogram by using heron's formula
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The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is 30cm. Find the area of the parallelogram by using heron's formula
math
the measures of a pair of two adjacent sides of a parallelogram are 15cm and 20 cm.if the length of one of its diagonal is 25cm, find the area of the parallelogram
Math
the measure of pair of adjacent sides of a parallelogram are 15cm and 20cm.if the length of one diagonals is 25cm find the area of the parallelogram.
Trig-Algebra help asap
The adjacent sides of a parallelogram measure 8 cm and 12 cm and one angle measures 60 degrees. Find the area of the parellelogram.
trig
The lengths of 2 adjacent sides of a parallelogram are 42 cm and 36cm. an angle of the parallelogram is 4o degrees. Find the measure of the longest diagonal to the nearest tenth of a centimeter.
physics
Consider four vectors F1, F2, F3, and F4, with magnitudes are F1 = 41 N, F2 = 24 N, F3 =27 N, and F4 = 51 N, and angle 1 = 150 deg, angle 2 = −160 deg, angle 3 = 39 deg, and angle 4 = −57 deg, measured from the positive x axis with the counter clockwise angular ...
geometry
The longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. Find the area of the parallelogram if the diagonals intersect at angle of 70 degrees.
geometry
In parallelogram ABCD, BC = 8, DC = 12, and measure of angle D is 50 degrees. Find the area of the parallelogram to the nearest tenth.
math
Trigonometry - determine the area of a parallelogram in which 2 sides are 10 and 13 and the angle between them is 55
maths
two adjacent sides of a parallelogram measure 15 cm and 20 cm while the diagonal opposite their common vertex measure 25 cm.find the area of he parallelogram.
Math check
1. If LMNO is a parallelogram, and LM=2y-9 and NO=y-2, find the value of 'y'. Does y=7? 2. RSTV is a parallelogram. RT and SV intersect at Q. RQ=5x+1 and QT=3x+15. Find OT. Does QT=7? 3. RATS is a parallelogram. If angle s= (8x) and angle t=(7x) then find the value of 'x'. ...
math
The bases of a right prism are parallelograms with sides a=10 cm, b=6 cm, and altitude of the parallelogram towards side a, ha = 3 cm. Find the surface area of the parallelogram, if the height of the prism is h=12 cm.
Geometry
Given: ABCD is a parallelogram; <1 is congruent to <2 To Prove: ABCD is a Rhombus Plan: Show <2 is congruent to < CAB. hence CB is congruent to AB, making ABCD a parallelogram with consecutive sides congruent PROOF Statements Reasons 1. ABCD is a parallelogram 1....
Find Area of Parallelogram
Find the area of the parallelogram with one corner at P1 and adjacent sides P1P2 and P1P3. NOTE: There is an arrow over P1P2 and P1P3. What does that arrow mean? P1 = (0, 0, 0), P2 = (2, 3, 1), P3 = (-2, 4, 1)
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the perimeter of a parallelogram is 28 cm and ratio of the adjacent sides is 3:4 find the sides of a parallelogram?
Geometry
I have four questions: 1. The area of trapezoid ABCD is 60. One base is 4 units longer than the other, and the height of the trapezoid is 5. Find the length of the median of the trapezoid. 2. In trapezoid ABCD, line BC is parallel to line AD, angle ABD = 105 degrees, angle A...
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Find the area of the parallelogram. It's A parallelogram has a lower side measuring 24 centimeters and a right side measuring 9 centimeters. A dashed vertical segment with endpoints on the upper and lower sides is 8 centimeters and meets the lower left vertex at a right angle.
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Find the area of a regular pentagon if its apothem is approximately 4 ft long and each of its sides are 5.8 ft long. So I tried dividing the pentagon into triangles...but then I came up with the answer 24.41? I know that's not right. Can you hLep me at all? I will be happy to ...
Trig
The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. Find the lengths of the sides of the parallelogram. (Hint: The diagonals bisect each other.)
geometry
a parallelogram has the vertices (-1,2), (4,4), (2,-1), and (-3,-3). determine what type of parallelogram. find the perimeter and area. i found the parallelogram part and its a rhombus but i cant find the perimeter and area
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Find the EXACT VALUE of csc(-11pi/12) If that angle is in radians, it is easy. A full circle has 2PI radians...or 360 deg so 22PI/24 is the same as 22*15 deg, or -30 deg. The 30 deg triangle is well known. The sin is 1/2, so the csc is 2 Figure out the sign of the csc from ...
Math
Really stuck on this question not sure where im going wrong a trapezium with top and bottom sides paralell. With a diagonal line running from the top left to the bootom right hand corner splitting the trapez into two triangles. the top left corner angle is formed by 66 deg and...
math
parallelogram A is similar to Parallelogram B. If the area of parallelogram b is 162 square units, what is the area of parallelogram A?
SOLID
the altitude BE of parallelogram ABCD divides the side AD into segments in the ratio 1:3. Find the area of the parallelogram if the length of its shorter side is 14 cm, and one of its interior angle measures 60 degrees.
geometry
1. the opposite angles of a parallelogram measure(x=30) and (2x-50).Find the measure of each angle of the parallelogram. 2. the ratio of two consecutive angles of a parallelogram is 2:3. find the measurement of the angles of a parallelogram. 3.the difference between the ...
oblique triangle trigonometry
A parking lot has the shape of a parallelogram. The lengths of two adjacent sides are 70meters and 100meters. The angle between the two sides is 70o. What is the area of the parking lot?
Physics
Consider a piece of metal that is at 5 deg C. If it is heated until it has twice the internal energy, its temperature will be. a)556 deg C b)283 deg C c)273 deg C d)278 deg C e)10 deg C
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find the area of parallelogram whose adjacent sides are +3+3 and -3-2+.
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find the area of the parallelogram that has vectors as adjacent sides u = -2i +j +5k v = 4i -3j -3k
science (physics) Need to solve this today please
(a) Express the vectors A, B, and C in the figure below in terms of unit vectors. (b) Use unit vector notation to find the vectors R = A + B + C and S = C – A – B. (c) What are the magnitude and directions of vectors R and S? Well the picture goes and has a vector A ...
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In a parallelogram abcd,if angle a =(2x+25) and angle b = (3x_5),find the value of x and the measures of each angle of the parallelogram
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In a parallelogram abcd,if angle a =(2x+25) and angle b = (3x_5),find the value of x and the measures of each angle of the parallelogram
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In a parallelogram abcd,if angle a =(2x+25) and angle b = (3x_5),find the value of x and the measures of each angle of the parallelogram
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I really need some help please!!!!! Am I on the right tract? Given: ABCD is a parallelogram; <1 is congruent to <2 To Prove: ABCD is a Rhombus Plan: Show <2 is congruent to < CAB. hence CB is congruent to AB, making ABCD a parallelogram with consecutive sides ...
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The diagonal of a parallelogram 6cm and 8cm long and they intersect at an angle of 55 degrees. calculate the area of the parallelogram
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A parallelogram has an area of 8x^2 - 2x -45. The height of the parallelogram is 4x + 9. I know the formula for a parallelogram is A=bh. Please work and explain how I get the length of the base of the parallelogram. Thanks
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If one angle of parallelogram is twice of its adjacent angle. Find angle of parallelogram.
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The formula A=bh is used to find the area of a parallelogram. If the base of a parallelogram is double and its height is doubled, how does this affect the area.
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Find the area of the parallelogram that has the vectors as adjacent sides. u = i+2j+2k v = i+k I know that I have to use (-2,-2,2) sqrt (1-2)^2 + (2+2)^2 + (2-2)^2 = sqrt17 Area = sqrt 17. Is this correct?
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Two sides of a parallelogram are 9 and 13 with an included angle of 57 degree & 29 minutes.fine the distance between the longer sides.
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Two adjacent sides of a parallelogram are four cm and five cm find the perimeter of the parallelogram
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the sides of a quadrilateral abcd are 6cm,8cm,11cm and 12cm respectively,and the angle between the first two sides is a right angle.find its area.
geometry
solution of the questions The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.
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solution of the questions The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.
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if an angle of parallelogram is two third of its adjacent angle,find the angles of the parallelogram
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If there is a parallelogram which has four sides xy and x+25 ,4x+35 abc is parallelogram find x,y,z
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The sides of parallelogram 2x+y,3y-2x,5y-8,4x-3 are given in cm.find x and y and the perimeter of the parallelogram
Pre-calculus
Find the lengths of the diagonals of a parallelogram with adjacent sides of 12cm and 13 cm if the angle between the sides is 50 degrees. The shorter diagonal is what in terms of cm? The longer diagonal is what in terms of cm?
math
Two sides of a triangle have lengths 8 m and 24 m. The angle between them is increasing at a rate of 0.05 rad/s. Find the rate at which the area of the triangle is changing when the angle between the sides of fixed length is 135°. Round your answer to two decimal places. So ...
Math
A square and a parallelogram have the same area. If a side of the square is 40m and the height of the parallelogram is 20m, find the length of the corresponding base of the parallelogram.
Trigonometry
The lengths of the diagonals of a parallelogram are 20 inches and 30 inches. The diagonals intersect at an an angle 35 degrees. Find the lengths of the parallelogram's sides lengths to the nearest hundredth.
Algebra 1
In a right triangle the adjacent side = 12, the opposite side = 35, and the hypotenuse = 37. What does the opposite angle =? a-71,1 deg b-18.9 deg c-29.2 deg d-90 deg my answer is 18.89 degrees
Math
suppose that you know that a given parallelogram has at least one right angle and at least two adjacent sides that sre congruent. Can you make a conclusion about what kind of parallelogram this is? I say no because it can be a square or a rectangle.
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What is the value of x? image is on: tiny pic . com/r/nx1abn/8 I choose c 90 deg a 60 deg b 72 deg c 90 deg d 120 deg
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Find all missing sides/angles. Round each answer to the nearest unit. Angles/sides ABC (left to right) 1) Angle c= 70 deg side b= 28 yd side c= 26 yd 2) Side a= 18 cm Side b= 24 cm Side c= 28 cm
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Suppose you find the ratio of the lengths of adjacent sides in a parallelogram. This ratio is equivalent to the ratio of the adjacent sides in another parallelogram. Are the figures similar? Explain.
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The diagonal of a rectangle is 25 meters long makes an angle of 36 ° with one of the rectangle. Find the area and the perimeter of the parallelogram. how to find the b to arrive to the area?
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A parallelogram is 22 inches long and 14 inches wide. How do you find the height of the parallelogram? How do you find the area of the parallelogram ?
math , correction
theres a diagram which in the diagram the points are: (-3,1),(0,3),(2,0), (-1,-2). And the directions say: use the concept of slope to determine if the given figure is a parallelogram or a rectangle: so this is what i did please can someone check it to see if i am correct. M...
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the measure of an angle of a parallelogram is 12 degrees less than 3 times the measure of an adjacent angle. explain how to find the measures of all the interior angles of the parallelogram.
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suppose a parallelogram has an area of 84 square units. Describe a triangle related to this parallelogram, and find the triangle's area, base, and height.
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The vertices of a parallelogram are the origin and the points A(-1,4), B(3,6), and C(7,2). Write the vector equations of the lines that make up the sides of the parallelogram
Math - vector equation of a parallelogram
The vertices of a parallelogram are the origin and the points A(-1,4), B(3,6), and C(7,2). Write the vector equations of the lines that make up the sides of the parallelogram
Math
If A=<2,1> and B=<-1,1>, find the magnitude and direction angle for -2A +3B Find all specified roots: Cube roots of 8i A parallelogram has sides of length 32.7cm and 20.2cm. If the longer diagonal has a length of 35.1cm, what is the angle opposite this diagonal?
geometry
Quadrilateral QRST has two pairs of congruent sides, but it is not a parallelogram. What figure is it? What further condition would it have to satisfy to be a parallelogram? A. The figure is a square. To be a parallelogram, it would have to have pairs of opposite congruent ...
Geometry
a parallelogram has the vertices (0,3), (3,0), (o,-3), and (-3,0). determine what type of parallelogram and find the perimeter and area
maths
find the area of the parallelogram that has the vectors as adjacent sides u = -2i + j + 5k v = 4i - 3j - 3k a. 11 b. squareroot 26 c. 86 d. 2 squareroot 86 e. 1
maths
ABCD is a parallelogram. AB=9cm,BC=5cm and DE is perpendicular to AB.If the area of parallelogram is 36sqcm,find AE?
maths
ABCD is a parallelogram and AE perpendicular to DC if AB is 20cm. and the area of parallelogram is 80cm. then find the AE which is 90degree?
maths
The measures of two adjacent sides of a parallelogram are in the ratio 17:7. if the second side measures 3.5 cm, find the perimeter of the parallelogram.
maths
the measure of one angle of a parallelogram is 25 degree.find the measures of the other angles of the parallelogram
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A container of hot water at 80 deg C cools to 79 deg C in 15 seconds when it is placed in a room that is at 20 deg C. Use Newton's law of cooling to estimate the time it will take for the container to cool from 70 deg C to 69 deg C.
physics
A container of hot water at 80 deg C cools to 79 deg C in 15 seconds when it is placed in a room that is at 20 deg C. Use Newtons law of cooling to estimate the time it will take for the container to cool from 70 deg C to 69 deg C.
Math
Select the geometric figure that possesses all of the characteristics: (1) four equal sides (2) both pairs of opposite sides are parallel (3) it doesn't contain any right angle A) square B) parallelogram C) rhombus D) isosceles triangle I know it's not A or D
geometry
The measures of two consecutive angles of a parallelogram are in the Ratio 3:7. Find the measure of an acute angle of the parallelogram
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The measure of an angle of a parallelogram is 12 degrees less than 3 times the measure of an adjacent angle. find the measure of all the interior angels of the parallelogram.
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You are on page 1of 12
# MB – 1
BINOMIAL THEOREM
C1 Binomial Expression :
Any algebraic expression which contains two dissimilar terms is called binomial expression.
2 1 1
For example : x y , x y 2
, 3 x, x 2 1 3 etc.
xy ( x 1)1 / 3
C2 Statement of Binomial theorem :
If x, y R and n N, then :
(x + y)n = nC0 anb0 + nC1 an – 1b1 + nC2an – 2b2 + .... + nCran – rbr + ... + nCn a0bn
n
or (x y )n n
C ra nrb r
r0
## Now, putting y = 1 in the binomial theorem
or (1 + x)n = nC0 + nC1 + nC2x2 + .... + nCrxr + .... + nCnxn
n
(1 x )n n Cr x r
r0
Practice Problems :
1. Using binomial theorem, indicate which number is larger (1.1)10000 or 1000.
2. Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate (2 + 1)6 + (2 – 1)6.
3. Show that 9n + 1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
4. Using binomial theorem, prove that 6n – 5n always leaves remaining 1 when divided by 25.
## C3 Properties of Binomial Theorem :
(i) The number of terms in the expansion is n + 1.
(ii) The sum of the indices of x and y in each term is n.
(iii) The binomial coefficients (nC0, nC1.......nCn) of the terms equidistant from the
begining and the end are equal, i.e. nC0 = nCn, nC1 = nCn – 1 etc.
C4 Some important terms in the expansion of (x + y)n :
(i) General term :
(r + 1)th term of (x + y)n is Tr + 1 = nCrxn – ryr
(ii) Middle term/(s) :
n 2
(a) If n is even, there is only middle term, which is th term.
2
(b) if n is odd, there are two middle terms, which are
n 1 n1
th and 1 th terms.
2 2
(iii) Numerically greatest term in the expansion of (x + y)n, n N
Let Tr and Tr + 1 be the rth and (r + 1)th terms respectively
n
Tr = Cr – 1 xn–(r – 1) yr – 1
n
Tr + 1 = Cr xn – r yr
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB– 2
Tr 1 n
Cr x n r y r nr1 y
Now, r 1
.
Tr n Cr 1 x
n r 1
y r x
n r 1 y n1 x
Consider
Tr 1
1, 1, 1 , r n1
Tr r x r y x
1
y
Practice Problems :
2n 1 · 3 · 5 ... ( 2n 1)
1. Show that the middle term in the expansion of (1 x) is · 2n · xn, where n N.
n!
2. Show that the coefficient of the middle term in the expansion of (1 + x)2n is the sum of the coefficients
of two middle terms in the expansion of (1 + x)2n – 1.
3. Find the value of r, if the coefficients of (2r + 4)th and (r – 2)th terms in the expansion of (1 + x)18 are
equal.
4. If the coefficient of (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)n are in the ratio
1 : 3 : 5, find n and r
5. The 2nd, 3rd and 4th terms in the expansion of (x + y)n are 240, 720 and 1080 respectively. Find the
values of x, y and n.
6. Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
6
3 1
7. Find the term independent of x in the expansion of x 2 .
2 3 x
8. Find the coefficient of a4 in the product (1 + 2a)4 (2 – a)5 using binomial theorem.
m
3
9. The sum of the coefficients of the first three terms in the expansion of x 2 , x 0 , m being a
x
natural number, is 559. Find the term of the expansion containing x3.
2n
1 1.3.5....( 2n 1) . 2n
10. Show that the greatest coefficients in the expansion of x is .
x n!
## 11. Express ( x x 2 1 ) 6 ( x x 2 1 ) 6 as a polynomial in x.
12. If a1, a2, a3 and a4 be any four consecutive coefficients in the expansion of (1 + x)n, prove that
a1 a3 2a 2 .
a1 a 2 a 3 a 4 a 2 a 3
[Answers : (3) 6 (4) n = 7, r = 3 (5) x = 2, y = 3 and n = 5 (6) 171 (7) 5/12 (8) –438 (9) –5940 x3]
C5 Multinominal Theorem
n
n n!
As we know the Binomial Theorem ( x y ) n
n
Crx nr
y
r xnr y r
r0 r0 ( n r )! r!
putting n – r = r1, r = r2
n! r1 r2
therefore, ( x y )n x .y
r1 r2 n r1 ! r2 !
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB – 3
n
Total number of terms in the expansion of (x + y) is equal to number of non-negative integral solution of
r1 + r2 = n i.e. n + 2 – 1C2 – 1 = n + 1C1 = n + 1
In the same fashion we can write the multinominal theorem
n!
( x1 x 2 x 3 .....xk )n x1r1 .x r22 ....x rkk
r !
r1 r2 .... rk n 1 2r !...rk !
Here total number of terms in the expansion of (x1 + x2 + .... + xk)n is equal to number of non-negative
integral solution of r1 + r2 + ..... + rk = n i.e. n + k – 1Ck – 1
Practice Problems :
10
1
1. (i) the middle term in the expansion of x (ii) the coefficient of x 32 and x –17 in the
2y
15
1
expansion of x 4 3
x
2. Find the coefficient of x5 in the expansion of the product (1 + 2x)5 (1 – x)7.
63x 5
[Answers : (1) (i) (ii) 1365, – 1365 (2) 171]
8y 5
## C6 Properties of Binomial Coefficients :
(1 + x)n = C0 + C1x + C2x2 + .... + Crxr + .... + Cnxn ....(1)
(1) The sum of the binomial coefficients in the expansion of (1 + x)n is 2n
Putting x = 1 in (1)
n
C0 + nC1 + nC2 + .... + nCn = 2n ....(2)
n
n
or Cr 2 n
r 0
## (2) Again putting x = –1 in (1), we get
n
C0 – nC1 + nC2 – nC3 + .... + (–1)n nCn = 0 ....(3)
n
r n
or ( 1)
r 0
Cr 0
(3) The sum of the binomial coefficients at odd position is equal to the sum of the binomial
coefficients at even position and each is equal to 2n – 1 i.e.,
n
C0 + nC2 + nC4 + ................. = 2n – 1
n
C1 + nC3 + nC5 + ................. = 2n – 1
(4) Sum of two consecutive binomial coefficients
n
Cr + nCr – 1 = n + 1Cr
n
Cr nr1
(5) Ratio of two consecutive binomial coefficients n
Cr 1 r
n n n 1 n(n 1) n 2
(6) Cr C r 1 Cr 2
r r(r 1)
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB– 4
Practice Problems :
1. Prove the following identities :
n
(a) C0 + nC2 + nC4 + .... = 2n – 1
n
(b) C1 + nC3 + nC5 + .... = 2n – 1
n
(c) C0 + 3 nC1 + 5 nC2 + .... + (2n + 1)nCn = (n + 1)2n
n
(d) C1 – 2 nC2 + 3 nC3 – .... + (–1)n – 1 nnCn = 0
(e) C1 + 2 C2 + 3 C3 + .... + n Cn = n 2n – 1
(f) C0 + 2 C1 + 3 C2 + .... + (n + 1) Cn = 2n + n 2n – 1
C1 C 2 1
(g) C0 ....
2 3 n1
C1 C C 3n1 1
(h) 2C 0 2 2 2 3 2 .... 2n 1 n =
2 3 n1 n1
C 2 C4 2n
(i) C0 ....
3 5 n1
(j) 2C0 + 5 C1 + 8 C2 + .... + (3n + 2) Cn = (3n + 4) 2n – 1
If n R then,
## n(n 1) 2 n(n 1)(n 2) 3
(1 x)n 1 nx x x ..........
2! 3!
n(n 1)(n 2).....(n r 1) f
x ...........
r!
Remarks
(i) The above expansion is valid for any rational number other then a whole number if |x| < 1.
(ii) When the index is a negative integer or a fraction then number of terms in the expansion of (1 + x)n is
infinite. and the symbol nCr cannot be used to denote the coefficient of the general term.
(iii) The first terms must be unity in the expansion, when index ‘n’ is a negative integer or fraction.
## n(n 1)(n 2).....(n r 1) r
(iv) The general term in the expansion of (1 + x)n is Tr 1 x
r!
(v) When ‘n’ is any rational number other than whole number then approximate value of
(1 + x)n is 1 + nx (x2 and higher powers of x can be neglected)
(vi) Expansion to be remembered (|x| < 1)
(a) (1 + x)–1 = 1 – x + x2 – x3 + .......... + (–1)r xr + ..........
(b) (1 – x)–1 = 1 + x + x2 + x5 + .......... + xr + ..........
(c) (1 + x)–2 = 1 – 2x + 3x2 – 4x3 + ........... + (–1)r (r + 1) xr + ..........
(d) (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + .......... + (r + 1)xr + ..........
Practice Problems :
1. Find the coefficient of x6 in the expansion of (1 – 2x)–5/2.
(1 3x 2 )
2. Find the coefficient of x10 in the expansion of , mentioning the condition under which the
(1 x 2 ) 3
result holds.
15015
[Answers : (1) ]
16
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB – 5
SINGLE CORRECT CHOICE TYPE
## 1. The expansion 10. The coefficient of x 10 in the expansion of
[x + (x3 – 1)1/2]5 + [x – (x3 – 1)1/2]5 is a polynomial of (1 + x2 – x3)8 is
degree (a) 476 (b) 496
(a) 5 (b) 6 (c) 506 (d) 528
n
(c) 7 (d) 8 11. The coefficient of x in the expansion of
2. Given positive integers r > 1, n > 2 and the 2 3 n
2
coefficients of (3r)th and (r + 2)th terms in the 1 x x x .... x is
binomial expansion (1 + x) 2n are equal. Then n 1! 2! 3! n!
equals to
(a) 3r (b) 3r – 1 2n 2n1
(c) 3r + 1 (d) none of these (a) (b)
n! n!
3. If in the expansion of (1 + x)m (1 – x)n, the coeffi-
(c) n! 2n (d) n! 2n – 1
cient of x and x2 are 3 and –6 respectively, then m
is 12. The coefficient of x 4 in the expansion of
(1 + x + x2 + x3)n is
(a) 6 (b) 9
n
(c) 12 (d) 24 (a) C0nC4 + nC0nC2
n
(b) C0nC2 + nC2nC1
n r
( 1) (c) n
C0nC4 + nC2 · nC1
4. If n is an odd natural number then
r 0
n
Cr (d) n
C0nC4 + nC0nC2 + nC2 · nC1
equals 13. The coefficient of x 5 in the expansion of
(a) 0 (b) 1/n (1 + x2)5 · (1 + x)4 is
(c) n/2 n
(d) none of these (a) 60 (b) 61
5. The total number of dissimilar terms in the (c) 59 (d) none
expansion of (x1 + x2 + .... + xn)3 is 14. If (2x – 3x ) = a0 + a1x + a2x .... a12x12, then the a0
2 6 2
and a6 will be
n 3 3n 2 (a) a0 = 6C026 and a6 = 0
(a) n3 (b)
4 (b) a0 not possible and a6 = 6C026
(c) a6 not possible and a0 = 6C026
n ( n 1)(n 2) n 2 (n 1) 2
(c) (d) (d) a0 = 0 and a6 = 6C026
6 4
15. The term independent of x in the expansion of
6. The coefficient of x 5 in the expansion of 7
(1 + x)21 + (1 + x)22 + .... + (1 + x)30 is 1
(1 – x2) x is
(a) 51
C5 (b) 9
C5 x
31 21 30
(c) C6 – C6 (d) C5 + 20C5 (a) 4th term (b) 5th term
7. If x is positive, the first negative term in the expan- (c) 6th term (d) none
sion of (1 + x)27/5 is 16. The coefficient of x 4
in the expansion of
(a) 5th term (b) 8th term
(1 2x 3x 2 )
(c) 6th term (d) 7th term is
(1 x) 2
8. Sum of the coefficient of the terms of degree m in
the expansion of (1 + x)n (1 + y)n (1 + z)n is (a) 21 (b) 22
n 3 n
(a) ( Cm) (b) 3( Cm) (c) 23 (d) none
n 3n
(c) C3m (d) Cm 17. The term independent of x in the expansion of
9. The coefficient of xk in the expansion of 10
x1 x1
2
E = 1 + (1 + x) + (1 + x) .... (1 + x) is n 2/ 3 is
x x 1 x x1 / 2
1/ 3
n n+1
(a) Ck (b) Ck
10 10
n+1
(a) C5 (b) C6
(c) Ck + 1 (d) none
10 10
(c) C4 (d) C3
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB– 6
100 2 100
18. If (1 + x) = c0 + c1x + c2x .....c100x then the value
c1 c 3 c 5
of k1 and k2 will be 25. The value of .... is equal to
2 4 6
c2 c4
where c 0 ...... k 1 and 2n
3 5 (a) (b) n · 2n
n
c1 c 3 c 5
...... k 2
2 4 6 2n 1 2n 1
(c) (d)
n1 n1
2100 2100 2100 2100
(a) and (b) and n
100 101 101 100 n r
## 100 101 100 100
26. If (1 x) c x
r 0
r then
2 2 2 2 1
(c) and (d) and
100 101 101 101
c c c
19. If (1 + x)100 = c0 + c1x + c2x2.....c100x100 then the value 1 1 1 2 ....... 1 n
c0
c1 c n 1
of c02 + c12 + c22.....c1002 will be
200 200
(a) c101 (b) c99
(n 1)n
(c) 200
c100 (d) none (a) (b) (–1)n + 2 (–2)
n!
20. If (1 + x)101 = c0 + c1x + c2x2.....c101x101 then the value
of n n 1 (n 1)n1
(c) (d)
2 2 2
c – c + c – c .....c
0 1 2 3
2
101
2 (n 1)! n!
202
(a) 0 (b) c101
–202 201
1
(c) c101 (d) c100 27. The coefficient of in the expansion of
x
n
21. If (1 + x) n = c 0 + c 1 x + c 2 x 2 ......c nx n then 1
(1 x)n 1 is
c1 2c 2 3c 3 nc x
..... n will be
c0 c1 c2 c n 1
n!
(a)
n 2 (n 1) 3n(n 2) (n 1)!(n 1)!
(a) (b)
2 8
( 2n )!
(b)
n(n 1) (n 1) (n 2) (n 1)!(n 1)!
(c) (d)
2 2
( 2n)!
2
(c)
1 x ( 2n 1)!( 2n 1)!
22. In the expansion of the coefficient of xn
1 x (d) none
will be 28. If n is even positive integer, then the condition that
(a) 4n (b) 4n – 3 the greatest term in the expansion of (1 + x)n may
have the greatest coefficient also is
(c) 4n + 1 (d) none
n n2
c1 2c 2 3c 3 15c15 (a) x
23. ..... is equal to n2 n
c 0 c1 c2 14
(a) 100 (b) 120 n1 n
(b) x
(c) –120 (d) none n n1
## 24. If (1 + x) = c 0 + c 1 x + c 2 x .... c 15 x 15 then
15 2
n n4
c2 + 2c3 + 3c4....14 c15 equal to (c) x
n4 4
(a) 14 · 214 (b) 13 · 214 (d) none of these
(c) 13 · 214 – 1 (d) none
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB – 7
1/5 20
29. The sum of the rational terms of (2 + 3) is 38. If (1 – x + x ) = a 0 + a 1x + a 2x 2 ....a 2nx 2n then
2 n
## (a) 71 (b) 85 a0 + a2 + a4 + .... + a2n will be
(c) 97 (d) none 3n 1 3n 1
30. The number of terms which are free from radical (a) (b)
2 2
signs in the expansion of (y1/5 + x1/10)55 is
(a) 5 (b) 6 1 3n 1
(c) (d) 3n
(c) 4 (d) none 2 2
31. If sum of coefficient in the expansion of (x + y)n is 39. The sum of the last eight coefficients in the
4096 then the greatest binomial coefficient is expansion of (1 + x)15 is
(a) 12
C6 (b) 12
C7 (a) 216 (b) 215
(c) 12
C8 (d) none (c) 214 (d) none of these
1
1
1
..... is equal to 2 2 C1 2 3 C 2 211 C10
32. 40. 2C 0 ..... is
1! (n 1)! 3! (n 3)! 5! (n 5)! 2 3 11
## 2n1 311 1 211 1
(a) (b) ( 1)n 1 (a) (b)
n! 11 11
2n 113 1 112 1
(c) (d) none (c) (d)
2(n! ) 11 11
33. The number of terms in the expansion of
(a + b + c)n will be
(a) 3 (b) n+2
(n 1)(n 2)
(c) (d) none
2
34. The coefficient of x n in the expansion of
(1 + x + x2.....)–n is
(a) 1 (b) (–1)n
(c) n (d) n+1 ANSWERS (SINGLE CORRECT
35. The coefficient of x5 in CHOICE TYPE)
(1 + 2x + 3x2 .... )–3/2 is
(a) 21 (b) 25 1. c 11. a 21. c 31. a
(c) 26 (d) none of these 2. d 12. d 22. a 32. a
2 2 2
36. The coefficient of x y , yzt and xyzt in the 3. c 13. a 23. b 33. c
expansion of (x + y + z + t)4 are in the ratio.
4. a 14. b 24. d 34. b
(a) 1:2:4 (b) 4:2:1
5. c 15. b 25. d 35. d
(c) 2:1:4 (d) 4:1:2
37. If (1 + x) = c0 + c1x + c2x + c2x .....c100x100 then the
100 2 3 6. c 16. b 26. a 36. a
value of c 0 + c 2 + c 4 + ...... = k 1 and 7. b 17. c 27. b 37. a
c1 + c3 + c4 + ....... = k2 will be
8. d 18. d 28. a 38. a
(a) k1 = k2 = 299 (b) k1 = k2 = 298
(c) k1 = k2 = 2101 (d) k1 k2 9. c 19. c 29. d 39. c
10. a 20. a 30. b 40. a
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB– 8
## EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE Comprehension-3
Comprehension-1 Let P be a product given by
Consider the following expansion of P = (x + a1) (x + a2).....(x + an)
2 n 2 3 2n
(1 + x + x ) = a0 + a1x + a2x + a3x + ...... + a2nx
n
1. Let S1 = a0 + a3 + a6....... and let S1 = ai 1
i
S2 = a1 + a4 + a7.......
S3 = a2 + a5 + a8.......
then S1 : S2 : S3 will be S2 a a , S a a a
i j
i j 3
i j k
i j k
(a) 1:2:3 (b) 3:2:1
(c) 1:1:1 (d) 3:2:1 and so on, then it can be shown that
2. The value of P = xn + S1xn – 1 + S2xn – 2 + ... + Sn
a 0 2 – a 1 2 + a 2 2 – a 3 2 ..... (–1) n – 1
a 2n – 1
is 7. The coefficient of x 99 in the expression of
(x – 1) (x – 2) (x – 3) .... (x – 100) must be
k a n [1 ( 1)n a n ] . The value of ‘k’ is (a) –99 (b) –4950
(c) –5050 (d) 5050
(a) ½ (b) 2
203
8. The coefficient of x in the expression
(c) 1 (d) –½
(x – 1) (x – 2) (x – 3) .... (x20 – 20) must be
2 3
3. Consider the following statement :
(a) 11 (b) 12
(i) The value of
n (c) 13 (d) 15
C0 . ar – nCl . ar – 1 + nC2 ar – 2....(–1)r nCr . a0
is zero when r is not multiple of 3. 9. The total number of terms in P must be
(ii) The value of (a) 22n (b) 2n – 1
n
C0 . ar – nCl . ar – 1 + nC2 ar – 2....(–1)r nCr . a0 (c) 2n + 1 (d) 2n
is (–1)m . nCm when r is multiple of 3. MATRIX MATCH TYPE
Then
Matching-1
(a) both are correct
Column - A Column - B
(b) only (i) is correct
n n
(c) only (ii) is correct Ck n .2 n 1 1
(d) both are incorrect
(A) (k 1)(k 2)
k 0
(p)
(n 1)(n 2)
Comprehension-2
n n
If a, b, c, d be four consecutive coefficients in the Cr 2n 1
binomial expansion of (1 + x)n then (B)
r 0
( 1)r r3
Cr
(q)
n1
ab bc cd
4. , , are in n n
C 2k 1 3
a b c
(C) 2k
(r)
(n 3)
(a) A.P. (b) G.P. k 1
## (c) H.P. (d) none n n
Ck 2n 2 n 3
5.
a
,
b
,
c
are in
(D) k2
k 0
(s)
(n 1) (n 2)
ab bc cd
Matching-2
(a) A.P. (b) G.P.
Match the column A with their suitable term
(c) H.P. (d) none
independent of x given in column B
6. (bc + ad), 2(ac2 – b2d), (b – c) are in
Column - A Column - B
(a) A.P. (b) G.P. –2 –3 10
(A) (1 + x + x + x ) (p) –336
(c) H.P. (d) none
10
2 1
(B) x 2 2 (q) 168
x
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB – 9
2/3 1/3 5
(C) (x + 4x + 4) · (r) 184756 3. The term independent of x in the expansion of
n
9 1
1 1 (1 + x)n 1 is
1/ 3 2/ 3 x
x 1 x x 11/ 3
(a) 0, if n is odd
10
1
(D) ( 2 3x 4x 2 ) x (s) 11851 n 1
x
(b) ( 1) 2 · n C n 1 if n is odd
Matching-3 2
(1 x x 2 )10 n
If 2
= a0 + a1x + a2x2 + .... + anxn + .... (c) (1) 2 · n C n if n is even
1 x
2
then find
Column - A Column - B (d) none of these
2n
## (B) a0 – a1 + a2 – .... (q) 2 1
(a) a0 + a2 + a4 .... = a + a1 + a2....
(C) a0 + a2 + a4 + .... (r) 1 2 0
(D) a1 + a3 + a5 + .... (s) 0 (b) an + 1 < an
Matching-4 (c) an – 3 = an + 3
Column - A Column - B (d) none of these
4
(A) The coefficient of x in (p) 62640 5. Let
the expansion of 6
6
2
(1 + x + x2)3 f ( x) x 2 1 x 2 1
x2 1 x2 1
(B) The coefficient of x10 in (q) 6
(7 + x + x2 + x3 + x4 + x5)8 then
(C) The coefficient of x in 7
(r) 60 (a) f(x) is a polynomial of the sixth degree
the expansion of in x
## (xy + yz + zx)6 6. The value of x in the expression (x x
log10 x 5
) if the
MULTIPLE CORRECT CHOICE TYPE third term in the coefficient is (10) is 6
p m
n (a) 10 (b) 10–5/2
1. If k = a n4 + a n3 + a n2 + a n + a
0 1 2 3 4
(c) 20 (d) none
p 1 m 1 k 1 7. The numerically greatest term in the expansion of
then (3 – 2x)9 when x = 1 is
## 1 1 (a) 3rd term (b) 4th term
(a) a0 (b) a1 (c) 5th term (d) 6th term
24 4
n
11 1 3 x 1
(c) a2 (d) a3 8. In the expansion of when x = , it is
24 4 2 3 2
2. Let (1 + x + x2 + x3 + x4)10 = a0 + a1x + a2x2 + ... + known that the 6th term is the greatest term. The
a40x40. Then possible positive integral values of n is
(a) a1 = 20 (b) a2 = 110 (a) 49 (b) 52
(c) a2 = 55 (d) a1 + a2 = 65 (c) 57 (d) 59
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB– 10
2 n 2 2n
9. If (1 + x + x ) = a0 + a1x + a2x + .... + a2nx then the
2 35
value of 1. STATEMENT-1 : a0 + a3 + a6 + .... a36 = (2 + 1)
3
(a) a0 + a1 + a2 + ... + a2n is 3n
STATEMENT-2 : a0 + a1 + a2 + .... + a36 = 236 and
(b) a0 – a1 + a2 – ... + a2n is 1 a0 + a2 + a4 + .... + a36 = 235
1 3n 2. STATEMENT-1 : The coefficient of the middle
(c) a0 + a2 + a4 + ... + a2n is term in (1 + x) 2n is equal to the sum of the
2
coefficient of the two middle terms in (1 + x)2n – 1.
3n 1 STATEMENT-2 : nCr – 1 + nCr = n + 1Cr.
(d) a1 + a3 + a5 + ... + a2n – 1 is
2 3. STATEMENT-1 : If n is a positive integer then
32n – 1 + 24n – 32n2 is divisible by 512 if n > 2.
10. If (1 + x + x2)n = a0 + a1x + a2x2 + .... + a2nx2n then
choose the correct statement STATEMENT-2 : It can be shown by using the
2 2 2 2 2 binomial theorem.
(a) a – a + a – a +.... + a
0 1 2 3 2n
= an
4. STATEMENT-1 : If n be an even positive integer
(b) a0a2 – a1a3 + a2a4 – .... + a2n – 2 · a2n = an + 1
then
(c) ai = a2n – i
(d) none of these 1 1 1 1 2n 1
....
1! n 1! 3 ! n 3 ! 5 ! n 5 ! n 1 !1 ! n!
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion) STATEMENT-2 : If n is even then
and STATEMENT-2 (Reason). Each question has n
C1 + nC3 + nC5 + .... + nCn – 1 = 2n – 1
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
5. STATEMENT-1 : (101)50 > (100)50 + (99)50
(A) Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation STATEMENT-2 : It can be shown by using the
for Statement-1 binomial theorem.
(B) Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Let (1 + x)36 = a0 + a1x + a2x2 + ... + a36x36
## (Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1. c 2. a 3. a 4. c 5. a 6. b
7. c 8. c 9. d
MATRIX MATCH TYPE
1. [A-s, B-r, C-q, D-s] 2. [A-s; B-r; C-q; D-p] 3. [A-p; B-p; C-p; D-s]
4. [A-q; B-s; C-p; D-r]
MULTIPLE CORRECT CHOICE TYPE
1. a, b, c, d 2. c, d 3. a, c 4. a, b, c 5. a, b, c, d
6. a, b 7. b, c 8. a, b, c, d 9. a, b, c, d
10. a, b, c
ASSERTION-REASON TYPE
1. B 2. A 3. A 4. A 5. A
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB – 11
INITIAL STEP EXERCISE
(SUBJECTIVE)
1. Given that the 4th term in the expansion of
C 0 C1 C 2 Cn
10 (h) .... ( 1) n
3x 1 4 7 3n 1
2 has a maximum numerical value, find
8
the range of values of x for which this will be true. 3n .n!
2. If a, b, c, d be four consecutive coefficients in the 1.4.7...( 3n 1)
binomial expansion of (1 + x)n then prove that 5. If a and b are two complex numbers, then find the
2 sum of (n + 1) terms of the series
b ac
if x > 0. aC0 – (a + d)C1 + (a + 2d)C2 – (a + 3d)C3 + ...
bc (a b )(c d )
6. If (2 + 3)n = I + f where I and n are positive
2n
3. If the greatest term in the expansion of (1 + x) has integers and 0 < f < 1, show that I is an odd integer
and (1 – f) (I + f) = 1.
10 11
the greatest coefficient if and only if x , 7. If (9 + 45)n = I + f, n and I being positive integers
11 10
and f is a proper fraction, show that (I – 1)f + f2 is
m an even integer.
1
and the fourth term in the expansion of kx 8. Show that [(3 + 1)2n] + 1 is divisible by 2n + 1 for all
x
n N.
n 9. If 2nC r = C r, prove that : C 12 – 2C 22 + 3C 3 2 –
is , then find the value of mk.
4 + .....................– 2n C2n2 = (–1)n – 1 nCn.
4. If C0, C1, C2,.........Cn denote the coefficients in the 10. If (66 + 14)2n + 1 = P, prove that the integral part of
expansion of (1 + x)n, prove that P is an even integer and P f = 202n + 1 where f is the
fractional part of P.
(a) (C0 + C1) (C1 + C2).....(Cn – 1 + Cn) =
11. If (1 + x + 2x2)20 = a0 + a1x + a2x2 +.... + a40x40 then
n
C1C2 ..........Cn (n 1) find the value of a0 + a2 + a4 + ... + a38.
.
n! n
2
12. Show that r ( n r )C r n 2 ( 2n 2 C n ) .
r 0
C1 C C n(n 1)
(b) 2 2 .... n n 13. Show that
C0 C1 C n 1 2
n
n
2n! . C k sin( kx ) cos[(n – k ) x ] 2 n 1 sin( nx )
(c) C 02 C12 C 22 ........... C 2n k 0
n!n!
(d) C0Cr + C1Cr+1 + C2Cr+2 + ............. + Cn – rCn 14. If (1 + x)n = C0 + C1x + C2x2 + .... + Cnxn, prove that
=
2n!
.
(C C )
0 i j n
i j
2
(n 1) · 2n Cn 2 2n .
n r!n r!
n
(e) 12C1 + 22C2 + 32C3 + ....... + n2Cn = n
(n + 1)2n – 2
15. Prove that r r 0
2
· nC r p r q n – r = npq + n 2p 2 if
## (f) C0 . 2nCn – C1 . 2n – 2Cn + C2 . 2n–4Cn – ... = 2n p + q = 1.
(g) 16. Show that the roots of the equation
C0 C C C n! (n 1)! ax2 + 2bx + c = 0 are real and unequal, where a, b, c
1 2 .... ( 1) n n are the three consecutive coefficients in any
n n1 n 2 2n ( 2n )! binomial expansion with positive integral index.
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
MB– 12
FINAL STEP EXERCISE
(SUBJECTIVE)
1. If (1 + x)n = C0 + C1x + C2x2 + ............... + Cnxn then 7. If (1 + x)n = C0 + C1x + C2x2 + .... + Cnxn (n N) then
prove that : 2
n
3
C 1
(a) (i j) C C i j
1
n 2 2n 1 · 2n C n show that
k 1
k k
C k 1 12
n(n 1) 2 (n 2) .
0 i j n 2
8. Show that
2 n
i i n 1 C 0 C1 C 2 C 3 Cn
(b) C i
Cj
2 C r
.... ( 1)n
0 i j n r0 1 5 9 13 4n 1
1
(c) i.j.C C i j n 2 2 2n 3 . 2n 2 C n 1 4n .n!
0 i j n 2
1.5.9....(4n 1)
2n 1
2. Show that the HM of 2n + 1Cr and 2n + 1Cr + 1 is 9. If (1 + 2x + 2x2)n = a0 + a1x + a2x2 + .... + a2nx2n, n N
n1
times of 2nCr. Also show that Show that a4 equals to
2n 1
r n . 1 3 n n 2 n 2
( 1)r 1 · .2 n(n 1)(1.3 3.5 C1 5.7 C2
r 1
2n
Cr
n1 3
3. Find the sum of the series
.... (2n 3)(2n 1))
n
r n 1 3r 7 r 15r 10. Show that
( 1) . C r r
2r
3r
4 r ....upto m terms
r 0 2 2 2 2 C 0 C1
1 x
C2
1 2x
C3
1 3x
.... 0
2
4. Show that 1 nx (1 nx ) (1 nx ) 3
## 1 1 1 11. Show that
C1 C 2 C 3 ...... (1) n 1 C n
2 3 n 1 n n(n 1)
C0 C1 C2
m! (m 1)! (m 2)!
1 1 1
1 ....
2 3 n n(n 1)....2.1
2 n .... Cn
5. Given that S n = 1 + q + q + .... + q and (m n)!
2 n
q 1 q 1 q1 (m n 1)(m n 2)...(m 2n )
n 1 ....
2 2 2 (m n )!
where q 1 prove that 12. If n > 3, then
n+1
C1 + n + 1C2 . S1 + n + 1C3 . S2 + .... + n + 1Cn + 1Sn = 2n.n (i) C0ab – C1(a – 1)(b – 1) + C2(a – 2)(b – 2)
6. Let k and n be positive integers and put – . . . + (–1)nCn(a – n)(b – n) = 0
Sk = 1k + 2k + 3k + .... + nk show that (ii) C0abc – C1(a – 1)(b – 1)(c – 1) + C2
m+1
C1S1 + m + 1C2S2 + ... + m + 1CmSm = (a – 2)(b – 2)(c – 2) – ... + (–1)nCn
(n + 1)m + 1 – (n + 1) (a – n)(b – n)(c – n) = 0
## ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
64 64
1. x , 2 2, 3. 3 5. 0 11. 219(220 – 1)
21 21
## ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
( 2mn 1)
3.
2mn (2n 1)
Einstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
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This HTML version of "Think Stats 2e" is provided for convenience, but it is not the best format for the book. In particular, some of the math symbols are not rendered correctly.
# Chapter 3 Probability mass functions
The code for this chapter is in `probability.py`. For information about downloading and working with this code, see Section ??.
## 3.1 Pmfs
Another way to represent a distribution is a probability mass function (PMF), which maps from each value to its probability. A probability is a frequency expressed as a fraction of the sample size, `n`. To get from frequencies to probabilities, we divide through by `n`, which is called normalization.
Given a Hist, we can make a dictionary that maps from each value to its probability:
```n = hist.Total()
d = {}
for x, freq in hist.Items():
d[x] = freq / n
```
Or we can use the Pmf class provided by `thinkstats2`. Like Hist, the Pmf constructor can take a list, pandas Series, dictionary, Hist, or another Pmf object. Here’s an example with a simple list:
```>>> import thinkstats2
>>> pmf = thinkstats2.Pmf([1, 2, 2, 3, 5])
>>> pmf
Pmf({1: 0.2, 2: 0.4, 3: 0.2, 5: 0.2})
```
The Pmf is normalized so total probability is 1.
Pmf and Hist objects are similar in many ways; in fact, they inherit many of their methods from a common parent class. For example, the methods `Values` and `Items` work the same way for both. The biggest difference is that a Hist maps from values to integer counters; a Pmf maps from values to floating-point probabilities.
To look up the probability associated with a value, use `Prob`:
```>>> pmf.Prob(2)
0.4
```
The bracket operator is equivalent:
```>>> pmf[2]
0.4
```
You can modify an existing Pmf by incrementing the probability associated with a value:
```>>> pmf.Incr(2, 0.2)
>>> pmf.Prob(2)
0.6
```
Or you can multiply a probability by a factor:
```>>> pmf.Mult(2, 0.5)
>>> pmf.Prob(2)
0.3
```
If you modify a Pmf, the result may not be normalized; that is, the probabilities may no longer add up to 1. To check, you can call `Total`, which returns the sum of the probabilities:
```>>> pmf.Total()
0.9
```
To renormalize, call `Normalize`:
```>>> pmf.Normalize()
>>> pmf.Total()
1.0
```
Pmf objects provide a `Copy` method so you can make and modify a copy without affecting the original.
My notation in this section might seem inconsistent, but there is a system: I use Pmf for the name of the class, `pmf` for an instance of the class, and PMF for the mathematical concept of a probability mass function.
## 3.2 Plotting PMFs
`thinkplot` provides two ways to plot Pmfs:
• To plot a Pmf as a bar graph, you can use `thinkplot.Hist`. Bar graphs are most useful if the number of values in the Pmf is small.
• To plot a Pmf as a step function, you can use `thinkplot.Pmf`. This option is most useful if there are a large number of values and the Pmf is smooth. This function also works with Hist objects.
In addition, `pyplot` provides a function called `hist` that takes a sequence of values, computes a histogram, and plots it. Since I use Hist objects, I usually don’t use `pyplot.hist`.
Figure ?? shows PMFs of pregnancy length for first babies and others using bar graphs (left) and step functions (right).
By plotting the PMF instead of the histogram, we can compare the two distributions without being mislead by the difference in sample size. Based on this figure, first babies seem to be less likely than others to arrive on time (week 39) and more likely to be a late (weeks 41 and 42).
Here’s the code that generates Figure ??:
``` thinkplot.PrePlot(2, cols=2)
thinkplot.Hist(first_pmf, align='right', width=width)
thinkplot.Hist(other_pmf, align='left', width=width)
thinkplot.Config(xlabel='weeks',
ylabel='probability',
axis=[27, 46, 0, 0.6])
thinkplot.PrePlot(2)
thinkplot.SubPlot(2)
thinkplot.Pmfs([first_pmf, other_pmf])
thinkplot.Show(xlabel='weeks',
axis=[27, 46, 0, 0.6])
```
`PrePlot` takes optional parameters `rows` and `cols` to make a grid of figures, in this case one row of two figures. The first figure (on the left) displays the Pmfs using `thinkplot.Hist`, as we have seen before.
The second call to `PrePlot` resets the color generator. Then `SubPlot` switches to the second figure (on the right) and displays the Pmfs using `thinkplot.Pmfs`. I used the `axis` option to ensure that the two figures are on the same axes, which is generally a good idea if you intend to compare two figures.
## 3.3 Other visualizations
Histograms and PMFs are useful while you are exploring data and trying to identify patterns and relationships. Once you have an idea what is going on, a good next step is to design a visualization that makes the patterns you have identified as clear as possible.
In the NSFG data, the biggest differences in the distributions are near the mode. So it makes sense to zoom in on that part of the graph, and to transform the data to emphasize differences:
``` weeks = range(35, 46)
diffs = []
for week in weeks:
p1 = first_pmf.Prob(week)
p2 = other_pmf.Prob(week)
diff = 100 * (p1 - p2)
diffs.append(diff)
thinkplot.Bar(weeks, diffs)
```
In this code, `weeks` is the range of weeks; `diffs` is the difference between the two PMFs in percentage points. Figure ?? shows the result as a bar chart. This figure makes the pattern clearer: first babies are less likely to be born in week 39, and somewhat more likely to be born in weeks 41 and 42.
For now we should hold this conclusion only tentatively. We used the same dataset to identify an apparent difference and then chose a visualization that makes the difference apparent. We can’t be sure this effect is real; it might be due to random variation. We’ll address this concern later.
## 3.4 The class size paradox
Before we go on, I want to demonstrate one kind of computation you can do with Pmf objects; I call this example the “class size paradox.”
At many American colleges and universities, the student-to-faculty ratio is about 10:1. But students are often surprised to discover that their average class size is bigger than 10. There are two reasons for the discrepancy:
• Students typically take 4–5 classes per semester, but professors often teach 1 or 2.
• The number of students who enjoy a small class is small, but the number of students in a large class is (ahem!) large.
The first effect is obvious, at least once it is pointed out; the second is more subtle. Let’s look at an example. Suppose that a college offers 65 classes in a given semester, with the following distribution of sizes:
``` size count
5- 9 8
10-14 8
15-19 14
20-24 4
25-29 6
30-34 12
35-39 8
40-44 3
45-49 2
```
If you ask the Dean for the average class size, he would construct a PMF, compute the mean, and report that the average class size is 23.7. Here’s the code:
``` d = { 7: 8, 12: 8, 17: 14, 22: 4,
27: 6, 32: 12, 37: 8, 42: 3, 47: 2 }
pmf = thinkstats2.Pmf(d, label='actual')
print('mean', pmf.Mean())
```
But if you survey a group of students, ask them how many students are in their classes, and compute the mean, you would think the average class was bigger. Let’s see how much bigger.
First, I compute the distribution as observed by students, where the probability associated with each class size is “biased” by the number of students in the class.
```def BiasPmf(pmf, label):
new_pmf = pmf.Copy(label=label)
for x, p in pmf.Items():
new_pmf.Mult(x, x)
new_pmf.Normalize()
return new_pmf
```
For each class size, `x`, we multiply the probability by `x`, the number of students who observe that class size. The result is a new Pmf that represents the biased distribution.
Now we can plot the actual and observed distributions:
``` biased_pmf = BiasPmf(pmf, label='observed')
thinkplot.PrePlot(2)
thinkplot.Pmfs([pmf, biased_pmf])
thinkplot.Show(xlabel='class size', ylabel='PMF')
```
Figure ?? shows the result. In the biased distribution there are fewer small classes and more large ones. The mean of the biased distribution is 29.1, almost 25% higher than the actual mean.
It is also possible to invert this operation. Suppose you want to find the distribution of class sizes at a college, but you can’t get reliable data from the Dean. An alternative is to choose a random sample of students and ask how many students are in their classes.
The result would be biased for the reasons we’ve just seen, but you can use it to estimate the actual distribution. Here’s the function that unbiases a Pmf:
```def UnbiasPmf(pmf, label):
new_pmf = pmf.Copy(label=label)
for x, p in pmf.Items():
new_pmf.Mult(x, 1.0/x)
new_pmf.Normalize()
return new_pmf
```
It’s similar to `BiasPmf`; the only difference is that it divides each probability by `x` instead of multiplying.
## 3.5 DataFrame indexing
In Section ?? we read a pandas DataFrame and used it to select and modify data columns. Now let’s look at row selection. To start, I create a NumPy array of random numbers and use it to initialize a DataFrame:
```>>> import numpy as np
>>> import pandas
>>> array = np.random.randn(4, 2)
>>> df = pandas.DataFrame(array)
>>> df
0 1
0 -0.143510 0.616050
1 -1.489647 0.300774
2 -0.074350 0.039621
3 -1.369968 0.545897
```
By default, the rows and columns are numbered starting at zero, but you can provide column names:
```>>> columns = ['A', 'B']
>>> df = pandas.DataFrame(array, columns=columns)
>>> df
A B
0 -0.143510 0.616050
1 -1.489647 0.300774
2 -0.074350 0.039621
3 -1.369968 0.545897
```
You can also provide row names. The set of row names is called the index; the row names themselves are called labels.
```>>> index = ['a', 'b', 'c', 'd']
>>> df = pandas.DataFrame(array, columns=columns, index=index)
>>> df
A B
a -0.143510 0.616050
b -1.489647 0.300774
c -0.074350 0.039621
d -1.369968 0.545897
```
As we saw in the previous chapter, simple indexing selects a column, returning a Series:
```>>> df['A']
a -0.143510
b -1.489647
c -0.074350
d -1.369968
Name: A, dtype: float64
```
To select a row by label, you can use the `loc` attribute, which returns a Series:
```>>> df.loc['a']
A -0.14351
B 0.61605
Name: a, dtype: float64
```
If you know the integer position of a row, rather than its label, you can use the `iloc` attribute, which also returns a Series.
```>>> df.iloc[0]
A -0.14351
B 0.61605
Name: a, dtype: float64
```
`loc` can also take a list of labels; in that case, the result is a DataFrame.
```>>> indices = ['a', 'c']
>>> df.loc[indices]
A B
a -0.14351 0.616050
c -0.07435 0.039621
```
Finally, you can use a slice to select a range of rows by label:
```>>> df['a':'c']
A B
a -0.143510 0.616050
b -1.489647 0.300774
c -0.074350 0.039621
```
Or by integer position:
```>>> df[0:2]
A B
a -0.143510 0.616050
b -1.489647 0.300774
```
The result in either case is a DataFrame, but notice that the first result includes the end of the slice; the second doesn’t.
My advice: if your rows have labels that are not simple integers, use the labels consistently and avoid using integer positions.
## 3.6 Exercises
Solutions to these exercises are in `chap03soln.ipynb` and `chap03soln.py`
Exercise 1 Something like the class size paradox appears if you survey children and ask how many children are in their family. Families with many children are more likely to appear in your sample, and families with no children have no chance to be in the sample.
Use the NSFG respondent variable `NUMKDHH` to construct the actual distribution for the number of children under 18 in the household.
Now compute the biased distribution we would see if we surveyed the children and asked them how many children under 18 (including themselves) are in their household.
Plot the actual and biased distributions, and compute their means. As a starting place, you can use `chap03ex.ipynb`.
Exercise 2
In Section ?? we computed the mean of a sample by adding up the elements and dividing by n. If you are given a PMF, you can still compute the mean, but the process is slightly different:
x =
∑ i
pi xi
where the xi are the unique values in the PMF and pi=PMF(xi). Similarly, you can compute variance like this:
S2 =
∑ i
pi (xi − x)2
Write functions called `PmfMean` and `PmfVar` that take a Pmf object and compute the mean and variance. To test these methods, check that they are consistent with the methods `Mean` and `Var` provided by Pmf.
Exercise 3 I started with the question, “Are first babies more likely to be late?” To address it, I computed the difference in means between groups of babies, but I ignored the possibility that there might be a difference between first babies and others for the same woman.
To address this version of the question, select respondents who have at least two babies and compute pairwise differences. Does this formulation of the question yield a different result?
Hint: use `nsfg.MakePregMap`.
Exercise 4
In most foot races, everyone starts at the same time. If you are a fast runner, you usually pass a lot of people at the beginning of the race, but after a few miles everyone around you is going at the same speed.
When I ran a long-distance (209 miles) relay race for the first time, I noticed an odd phenomenon: when I overtook another runner, I was usually much faster, and when another runner overtook me, he was usually much faster.
At first I thought that the distribution of speeds might be bimodal; that is, there were many slow runners and many fast runners, but few at my speed.
Then I realized that I was the victim of a bias similar to the effect of class size. The race was unusual in two ways: it used a staggered start, so teams started at different times; also, many teams included runners at different levels of ability.
As a result, runners were spread out along the course with little relationship between speed and location. When I joined the race, the runners near me were (pretty much) a random sample of the runners in the race.
So where does the bias come from? During my time on the course, the chance of overtaking a runner, or being overtaken, is proportional to the difference in our speeds. I am more likely to catch a slow runner, and more likely to be caught by a fast runner. But runners at the same speed are unlikely to see each other.
Write a function called `ObservedPmf` that takes a Pmf representing the actual distribution of runners’ speeds, and the speed of a running observer, and returns a new Pmf representing the distribution of runners’ speeds as seen by the observer.
To test your function, you can use `relay.py`, which reads the results from the James Joyce Ramble 10K in Dedham MA and converts the pace of each runner to mph.
Compute the distribution of speeds you would observe if you ran a relay race at 7.5 mph with this group of runners. A solution to this exercise is in `relay_soln.py`.
## 3.7 Glossary
• Probability mass function (PMF): a representation of a distribution as a function that maps from values to probabilities.
• probability: A frequency expressed as a fraction of the sample size.
• normalization: The process of dividing a frequency by a sample size to get a probability.
• index: In a pandas DataFrame, the index is a special column that contains the row labels.
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# Solving Equations with Logarithms – A Level Maths Revision
So, we hear you’re looking to boost your skills in solving equations with logarithms…
Stick around with Beyond Revision and we’ll ensure you meet the same high performance levels on both sides of the classroom – in revision practice at home and in formal examinations!
### Where to Start?
Once you have a grip on the basics of logarithms, you need to be capable of solving equations with logarithms.
If you want to check your understanding of logarithms before you start, try this multiple-choice quiz. The content of the blog below, along with a more extensive set of questions, can be downloaded in PDF and PowerPoint form here. Once you’re happy you’ve fully grasped the AS Level content on exponentials and logarithms, try these exam-style questions.
A major part of solving equations is ‘doing the same to both sides’. This can include taking logs of both sides. You can use this information to solve equations with logarithms and exponentials:
\begin{aligned} &\log_a{b} = c \text{ can be written as } a^c=b \\ &\text{(where }a>0, a \neq 1 \text{ and }a\text{ is real)} \\ \\ &\log_a{b}+\log_a{c}=\log_a{bc} \\ &\log_a{b}-\log_a{c}=\log_a{\frac{b}{c}} \\ &\log_a{b^c}=c\log_a{b} \end{aligned}
### Example Question 1
$3^x = 7$
To solve this equation without knowing about logarithms, you would probably use trial and improvement. x must lie between 1 and 2 because 31 = 3 and 32 = 9, so you could use a calculator to evaluate 31.5 and so on.
However, if the unknown in an equation appears as a power, you can get straight to the answer by taking logs of both sides. It’s up to you what base you use, but if you use the base given (3 in this example), the solution will be simpler.
$3^x = 7$
Taking logs (to base 3) of each side gives:
$\log_3{3^x} = \log_3{7}$
Using the log law for powers gives:
$x\log_3{3}=log_3{7}$
The unknown no longer features as a power and log33=1:
\begin{aligned} &x = \log_3{7} \\ &x=1.77\text{ (to 2d.p.)} \end{aligned}
When you take logs of both sides, that log can be to any base. Although it can be easier to use a base number from the question, log to the base 10 is often used instead, as it is a single button on your calculator (usually just labelled log) which does not require you to enter a base. This is helpful if the base number in the question is awkward, as even slight rounding in a log question can lead to considerable inaccuracy in your answer. Let’s try the same question using log10:
\begin{aligned} &3^x = 7 \\ &\log_{10}{3^x} = \log_{10}{7} \\ & x\log_{10}{3}=log_{10}{7} \end{aligned}
The unknown no longer features as a power. log103 can’t be cancelled out in the way log33 can be.
\begin{aligned} &x = \log_{10}{7} \div \log_{10}{3} \\ &x=1.77\text{ (to 2d.p.)} \end{aligned}
As a final alternative:
\begin{aligned} &\log_a{b} = c \text{ can be written as } a^c=b \\ &3^x = 7 \text { can be written as } \log_3{7}=x \\ &x=1.77\text{ (to 2d.p.)} \end{aligned}
### Example Question 2
$5^x = 3^{x-1}$
Using the base number given in the equation makes finding a solution easier – here there are two to pick from. Let’s choose base 5.
\begin{aligned} &\log_5{5^x} = \log_5{3^{x-1}} \\ & x\log_5{5}=(x-1)\log_5{3} \\ &x = x\log_5{3}-\log_5{3} \\ &\log_5{3} = x\log_5{3} - x \\ &\log_5{3} = x(\log_5{3}-1) \\ &x = \frac{\log_5{3}}{\log_5{3}-1} \\ &x = -2.15 \text{ (to 2d.p.)} \end{aligned}
Again, substitute it back in to check that it gives correct values.
As well as using logs to solve equations involving exponentials, you’ll need to be able to solve equations which already contain logs. You’ll usually use log laws to do so.
### Example Question 3
$\log_3{(x+2)}+\log_3{5}=\log_3{(9x)}$
Using the first law of logs gives:
$\log_3{(5x+10)}=\log_3{(9x)}$
Therefore:
\begin{aligned} &5x + 10 = 9x \\ &x = 2.5 \end{aligned}
Substitute it back into the original equation to check.
### Example Question 4
$\log_7{(x-2)}+\log_7{x}=\log_7{(x+10)}$
Using the first law of logs gives:
$\log_7{(x^2-2x)}=\log_7{(x+10)}$
In this case, removing the logs gives a quadratic. This can be solved as normal:
\begin{aligned} & x^2-2x=x + 10 \\ &x^2-3x-10=0 \\ &(x-5)(x+2)=0 \\& x = 5 \text{ or } x=-2 \end{aligned}
Again, substitute it back into the original equation to check. This time, you will find that -2 is not a valid solution because there is no power which acts upon 7 to give a negative result – the log of a negative number gives a mathematical error on a calculator.
### Example Question 5
$\log_4{(x^2+10x)}=2+\log_4{(x+1)}$
You may also see equations with a mix of logarithmic and non-logarithmic terms:
\begin{aligned}&\log_4{(x^2+10x)}=2+\log_4{(x+1)} \\& \log_4{(x^2+10x)}-\log_4{(x+1)}=2 \\ & \log_4{\frac{x^2+10x}{x+1}} = 2\end{aligned}
The inverse of taking the logarithm to base 4 of a term is raising 4 to the power of that term. Therefore, if both sides are taken as a power of 4, you get:
\begin{aligned} &\frac{x^2+10x}{x+1}=4^2 \\ \\ &x^2 + 10x = 16(x+1) \\ &x^2 + 10x -16x - 16 = 0 \\ &(x-8)(x+2)=0 \\&x = 8 \text{ or } x=-2 \end{aligned}
As before, you can reject x = -2 as a possible answer, leaving x = 8.
## Solving Equations with Logarithms – Practice Questions
1. $4^x = 18$
2. $4^x - 6 = 11$
3. $7^{3x} = 5$
4. $8^{x-1} = 4^x$
5. $4^x \times 4^{x+5} = 4^{x+11}$
6. $3 \times 2^x = 9$
7. $5^{2x-3} = 6^{150}$
8. $\log_4{p} + log_4{(p-8)} = \log_4{(2p-10)} + \log_4{2}$
9. $\log_{10}{s} +\log_{10}{(s-1)} = \log_{10}{(3s + 12)}$
10. $\log_6{(m^2-41m)} = 2 + \log_6{(1-m)}$
1. $\boldsymbol{ 4^x = 18 }$
\begin{aligned} &\log_4{4^x}=\log_4{18} \\&x=\log_4{18}\\&x=2.08 \end{aligned}
2. $\boldsymbol{ 4^x - 6 = 11 }$
\begin{aligned} &4^x = 17 \\ &\log_4{4^x} = \log_4{17} \\ &x\log_4{4}=\log_4{17} \\ & x = \log_4{17} \\ &x=2.04 \end{aligned}
3. $\boldsymbol{ 7^{3x} = 5 }$
\begin{aligned} &\log_7{7^{3x}}=\log_7{5} \\&3x\log_7{7}=\log_7{5} \\ &3x = \log_7{5} \\ &x=0.276 \end{aligned}
4. $\boldsymbol{ 8^{x-1} = 4^x }$
\begin{aligned} &\log_4{8^{x-1}}=\log_4{4^x} \\ &(x-1)\log_4{8}=x\log_4{4} \\ &x\log_4{8}-\log_4{8}=x \\ &\log_4{8} = x\log_4{8} - x \\ &\log_4{8}=x(\log_4{8}-1) \\&x=\frac{\log_4{8}}{\log_4{8}-1} \\ &x=3 \end{aligned}
5. $\boldsymbol{ 4^x \times 4^{x+5} = 4^{x+11} }$
\begin{aligned} &4^{2x+5} = 4^{x+11} \\ &2x + 5 = x + 11 \\&x=6 \end{aligned}
6. $\boldsymbol{ 3 \times 2^x = 9 }$
\begin{aligned} &2^x=3\\&\log_2{2^x}=\log_2{3} \\&x\log_2{2}=\log_2{3} \\&x = \log_2{3} \\ &x=1.58 \end{aligned}
7. $\boldsymbol{ 5^{2x-3} = 6^{150} }$
\begin{aligned} &\log_5{5^{2x-3}}=\log_5{6^{150}} \\ &(2x-3)\log_5{5} = \log_5{6^{150}} \\ &2x-3=\log_5{6^{150}} \end{aligned}
If you now try to use your calculator to evaluate the right-hand side, you’ll find you can’t: 6150 is too large to work with. But, if you rearrange the right-hand side, this can be circumvented:
\begin{aligned} &2x-3=150\log_5{6} \\ &x=0.5 \times (150\log_5{6}+3) \\ &x = 85.0 \end{aligned}
8. $\boldsymbol{ \log_4{p} + log_4{(p-8)} = \log_4{(2p-10)} + \log_4{2} }$
\begin{aligned} &log_4(p^2-8p)=\log_4{(4p-20)} \\ &p^2 -8p = 4p-20 \\ &p^2-12p+20=0 \\&(p-10)(p-2) = 0 \\ &p=10 \text{ or } p = 2 \end{aligned}
9. $\boldsymbol{ \log_{10}{s} +\log_{10}{(s-1)} = \log_{10}{(3s + 12)} }$
\begin{aligned} &\log_{10}{(s(s-1))} = \log_{10}{(3s+12)} \\ &s^2-s=3s+12 \\&s^2-4s-12=0 \\ &(s-6)(s+2) = 0 \\ &s=6 \text{ or } s=-2 \text{ (reject)} \end{aligned}
10. $\boldsymbol{ \log_6{(m^2-41m)} = 2 + \log_6{(1-m)} }$
\begin{aligned} &\log_6{(m^2-41m)}-\log_6{(1-m)}=2 \\ & \log_6{\frac{m^2-41m}{1-m}}=2 \\ &6^2=\frac{m^2-41m}{1-m} \\ &36(1-m)=m^2-41m \\ & 36-36m = m^2-41m \\&m^2-5m-36=0\\&(m-9)(m+4)=0 \\&m=9 \text{ or } m=-4 \text{ (reject)} \end{aligned}
Do you feel confident now in solving equations with logarithms? If so, feel free to move on to more of our blogs and practice content here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.
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# 6.3: Using the Central Limit Theorem
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## jkesler
[latexpage]
Before we begin this chapter, it is important for you to understand when to use the central limit theorem (CLT). If you are being asked to find the probability of the mean, use the CLT for the mean. If you are being asked to find the probability of a sum or total, use the CLT for sums. This also applies to percentiles for means and sums.
### NOTE
If you are being asked to find the probability of an individual value, do not use the Central Limit Theorem. Use the distribution of its random variable.
### Examples of the Central Limit Theorem
#### Law of Large Numbers
The law of large numbers says that if you take samples of larger and larger size from any population, then the mean $\bar x$ of the sample tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for $\bar X$ is $\frac{\sigma}{\sqrt{n}}$. ) This means that the sample mean $\bar x$ must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers.
### Example 6.8
A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:
1. The probability that the mean stress score for the 75 students is less than two.
2. The 90th percentile for the mean stress score for the 75 students.
3. The probability that the total of the 75 stress scores is less than 200.
4. The 90th percentile for the total stress score for the 75 students.
Let X = one stress score.
Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75.
Since the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (In other words, the individual stress scores are uniformly distributed between 1 and 5. See Continuous Random Variables for an explanation on the uniform distribution).
$\mu_x = \frac{a+b}{2}=\frac{1+5}{2}=3$
$\sigma_x = \sqrt{\frac{(b-a)^2}{12}}=\sqrt{\frac{(5-1)^2}{12}}=1.15$
For problems a. and b., let $\bar X =$ the mean stress score for the 75 students. Then,
$\bar X \sim N\left(3, \frac{1.15}{\sqrt{75}} \right)$
Find $P(\bar x <2)$. Draw the graph.
### Try It 6.8
Use the information in Example 6.8, but use a sample size of 55 to answer the following questions.
1. Find $P(\bar x<7)$.
2. Find P(Σx > 170).
3. Find the 80th percentile for the mean of 55 scores.
4. Find the 85th percentile for the sum of 55 scores.
### Example 6.9
Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. (Note: you do not need to know anything about the exponential distribution to solve this problem, but if you are curious, you can read about it at Wikipedia).
Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.
Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.
$X\sim Exp\left( \frac{1}{22} \right)$. From previous chapters, we know that μ = 22 and σ = 22.
Let $\bar X=$ the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance.
$\bar X \sim N\left( 22, \frac{22}{\sqrt{80}} \right)$ by the central limit theorem for sample means
Using the CLT to find probability
1. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find $P(\bar x > 20)$. Draw the graph.
2. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer’s excess time is longer than 20 minutes. This is asking us to find P(x > 20).
3. Explain why the probabilities in parts a and b are different.
#### Using the CLT to find percentiles
Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph.
### Try It 6.9
Use the information in Example 6.9, but change the sample size to 144.
1. Find $P(20<\bar x < 30)$.
2. Find P(Σx is at least 3,000).
3. Find the 75th percentile for the sample mean excess time of 144 customers.
4. Find the 85th percentile for the sum of 144 excess times used by customers.
### Example 6.10
In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.
1. Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.
2. Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.
3. Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.
4. Find the value that is two standard deviations above the sample mean.
5. Find the IQR for the sum of the sample times.
### Try It 6.10
Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.
1. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.
2. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
3. If the sample were four women between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used?
### Example 6.11
A study was done about violence against prostitutes and the symptoms of the posttraumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.
1. In a sample of 25 prostitutes, what is the probability that the mean age of the prostitutes is less than 35?
2. Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.
3. In a sample of 49 prostitutes, what is the probability that the sum of the ages is no less than 1,600?
4. Is it likely that the sum of the ages of the 49 prostitutes is at most 1,595? Interpret the results.
5. Find the 95th percentile for the sample mean age of 65 prostitutes. Interpret the results.
6. Find the 90th percentile for the sum of the ages of 65 prostitutes. Interpret the results.
### Try It 6.11
According to Boeing data, the 757 airliner carries 200 passengers and has doors with a height of 72 inches. Assume for a certain population of men we have a mean height of 69.0 inches and a standard deviation of 2.8 inches.
1. What doorway height would allow 95% of men to enter the aircraft without bending?
2. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?
3. For engineers designing the 757, which result is more relevant: the height from part a or part b? Why?
### HISTORICAL NOTE
: Normal Approximation to the Binomial
Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution:
• there are a certain number n of independent trials
• the outcomes of any trial are success or failure
• each trial has the same probability of a success p
Recall that if X is the binomial random variable, then X ~ B(n, p). The shape of the binomial distribution needs to be
similar to the shape of the normal distribution. To ensure this, the quantities np
and nq must both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation $\sigma = \sqrt{npq}$. Remember that q = 1 – p. In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example.
### Example 6.12
Suppose in a local Kindergarten through 12th grade (K – 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.
1. Find the probability that at least 150 favor a charter school.
2. Find the probability that at most 160 favor a charter school.
3. Find the probability that more than 155 favor a charter school.
4. Find the probability that fewer than 147 favor a charter school.
5. Find the probability that exactly 175 favor a charter school.
Let X = the number that favor a charter school for grades K trough 5. X ~ B(n, p) where n = 300 and p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and $\sigma = \sqrt{npq}$. The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y. Y ~ N(159, 8.6447).
For part a, you include 150 so P(X ≥ 150) has normal approximation P(Y ≥ 149.5) = 0.8641.
1-NORM.DIST(149.5,159,8.6447,TRUE) = 0.8641.
For part b, you include 160 so P(X ≤ 160) has normal appraximation P(Y ≤ 160.5) = 0.5689.
NORM.DIST(160.5,159,8.6447,TRUE) = 0.5689
For part c, you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572.
1-NORM.DIST(155.5,159,8.6447,TRUE) = 0.6572.
For part d, you exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741.
NORM.DIST(146.5,159,8.6447,TRUE) = 0.0741
For part e,P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083.
NORM.DIST(175.5,159,8.6447,TRUE)- NORM.DIST(174.5,159,8.6447,TRUE)= 0.0083
Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. In a spreadsheet application (like Google Sheets), you can use the BINOM.DIST() function to calculate binomial distribution probabilities.
For Example 6.10, the probabilities are calculated using the following binomial distribution: (n = 300 and p = 0.53). Compare the binomial and normal distribution answers.
P(X ≥ 150) :1 - BINOM.DIST(149, 300,0.53,TRUE) = 0.8641
P(X ≤ 160) :B INOM.DIST(160,300,0.53,TRUE) = 0.5684
P(X > 155) :1 - BINOM.DIST(155,300,0.53,TRUE) = 0.6576
P(X < 147) : BINOM.DIST(146,300,0.53,TRUE) = 0.0742
P(X = 175) : BINOM.DIST(175,300,0.53,FALSE) = 0.0083 (You use the FALSE argument since we don’t want the cumulative probabilities)
### Try It 6.12
In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor.
6.3: Using the Central Limit Theorem is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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# Phi Digit Calculator
## World's Simplest Number Tool
Quickly calculate golden ratio digits in your browser. To get your φ digits, just specify how many of them you need in the options below, and this utility will generate a sequence of that many digits. Created by developers from team Browserling.
Tool Options
#### Golden Ratio Digits Options
Calculate this amount of φ digits.
Write prefix "1." in front of the positive φ digits and prefix "-0." in front of the negative φ digits.
#### Which φ value to output?
Calculate golden ratio digits by using formula (1+√5)/2.
Calculate golden ratio digits by using formula (1-√5)/2.
#### Golden Ratio Digits Separator
Separate φ digits with this symbol.
By default, digits are joined together. Enter the "\n" symbol to separate digits by the line break.
## What Is a Phi Digit Calculator?
This is an online browser-based utility for generating a list of digits of the golden ratio. It's said that two objects are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two objects. Roughly speaking, this proportion can be expressed as 62% to 38%. The golden ratio is denoted by the Greek letter phi (Φ or φ) and often appears in science and nature. Many other names are used for the phi constant, in particular, golden mean, golden number, golden section, and lesser-known – extreme and mean ratio. The golden section can be calculated by solving the quadratic equation φ² - φ - 1 = 0. It has two solutions: φ = (1+√5)/2 (positive constant) and φ = (1-√5)/2 (negative constant). In this tool, you can calculate both the positive and negative golden proportion and get any number of φ digits, up to one million values. You can also get only digits of the ratio, without the whole number part (either "1." for positive phi or "-0." for negative), and optionally set a separator between φ digits. That's numberwang!
## Phi Digit Calculator Examples
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### 1000 Golden Ratio Digits
In this example, we find the first one thousand digits of the positive golden number, that is, of the constant φ = (1+√5)/2.
1.6180339887498948482045868343656381177203091798057628621354486227052604628189024497072072041893911374847540880753868917521266338622235369317931800607667263544333890865959395829056383226613199282902678806752087668925017116962070322210432162695486262963136144381497587012203408058879544547492461856953648644492410443207713449470495658467885098743394422125448770664780915884607499887124007652170575179788341662562494075890697040002812104276217711177780531531714101170466659914669798731761356006708748071013179523689427521948435305678300228785699782977834784587822891109762500302696156170025046433824377648610283831268330372429267526311653392473167111211588186385133162038400522216579128667529465490681131715993432359734949850904094762132229810172610705961164562990981629055520852479035240602017279974717534277759277862561943208275051312181562855122248093947123414517022373580577278616008688382952304592647878017889921990270776903895321968198615143780314997411069260886742962267575605231727775203536139362
Required options
These options will be used automatically if you select this example.
Calculate this amount of φ digits.
Write prefix "1." in front of the positive φ digits and prefix "-0." in front of the negative φ digits.
Calculate golden ratio digits by using formula (1+√5)/2.
Separate φ digits with this symbol.
click me
### 200 Negative φ Digits
In this example, we calculate 200 digits of the second golden mean root. This root is negative and is calculated using the formula φ = (1-√5)/2.
-0.61803398874989484820458683436563811772030917980576286213544862270526046281890244970720720418939113748475408807538689175212663386222353693179318006076672635443338908659593958290563832266131992829026788
Required options
These options will be used automatically if you select this example.
Calculate this amount of φ digits.
Write prefix "1." in front of the positive φ digits and prefix "-0." in front of the negative φ digits.
Calculate golden ratio digits by using formula (1-√5)/2.
Separate φ digits with this symbol.
click me
### Dash-separated Phi Digits
In this example, we exclude the whole part from the golden section (that is, we don't print the "1." at the start), and generate 100 digits, separating them with the dash symbol.
6-1-8-0-3-3-9-8-8-7-4-9-8-9-4-8-4-8-2-0-4-5-8-6-8-3-4-3-6-5-6-3-8-1-1-7-7-2-0-3-0-9-1-7-9-8-0-5-7-6-2-8-6-2-1-3-5-4-4-8-6-2-2-7-0-5-2-6-0-4-6-2-8-1-8-9-0-2-4-4-9-7-0-7-2-0-7-2-0-4-1-8-9-3-9-1-1-3-7-4
Required options
These options will be used automatically if you select this example.
Calculate this amount of φ digits.
Write prefix "1." in front of the positive φ digits and prefix "-0." in front of the negative φ digits.
Calculate golden ratio digits by using formula (1+√5)/2.
Separate φ digits with this symbol.
### Pro tipsMaster online number tools
You can pass options to this tool using their codes as query arguments and it will automatically compute output. To get the code of an option, just hover over its icon. Here's how to type it in your browser's address bar. Click to try!
https://onlinetools.com/number/calculate-phi-digits?count=1000&include-base=true&positive-phi=true&separator=
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## CTRL+A, CTRL+C, CTRL+V
January 4, 2011
Imagine you have a special keyboard with the following keys:
1. A
2. Ctrl+A
3. Ctrl+C
4. Ctrl+V
where CTRL+A, CTRL+C, CTRL+V each acts as one function key for “Select All”, “Copy”, and “Paste” operations respectively.
If you can only press the keyboard for N times (with the above four keys), please write a program to produce maximum numbers of A. If possible, please also print out the sequence of keys.
That is to say, the input parameter is N (No. of keys that you can press), the output is M (No. of As that you can produce).
This question seemed like a brand new interview question from Google. Someone posted this problem on a discussion board, and it generated a lot of discussions, which unfortunately none of them is really helpful in solving this problem. Here, I discuss my approach and solution to this problem, and just like those other tricky questions from a typical Google interview — the common pitfalls you might fall into.
One common strategy in problem solving is to always begin with small examples.
It is trivial to notice that for N <= 6, M = N. But how about the case where N = 7? Here is where the most common pitfall that most people would fall into (myself included).
Most people reason that for N = 7, the answer is M = 8, because the sequence S = { A, A, A, A, CTRL+A, CTRL+C, CTRL+V } produces a total of 8 A’s.
Wait, the copied text is still in the buffer after a paste operation. We could have applied CTRL+V twice to double the previous text, sweet!
How about S = { A, A, A, CTRL+A, CTRL+C, CTRL+V, CTRL+V } which produces a total of 9 A’s?
Unfortunately, both of the above answers are incorrect, as the correct answer for N = 7 is M = 7. This is simply because the sequence of { CTRL+A, CTRL+C, CTRL+V } does not double the previous text. Why? Take a moment to let this to sink into your brain.
Answers for N up to 7 is easy, which is M = N. But how about N = 8?
For N = 8 the answer is M = 9, where S = { A, A, A, CTRL+A, CTRL+C, CTRL+V, CTRL+V, CTRL+V }.
For N = 9 the answer is M = 12, where S = { A, A, A, CTRL+A, CTRL+C, CTRL+V, CTRL+V, CTRL+V, CTRL+V }.
You might ask why all A’s are typed before the sequence of { CTRL+A, CTRL+C, CTRL+V } operations.
Assume that we could insert A’s at the back of some sequence of { CTRL+A, CTRL+C, CTRL+V } and yield a maximum sequence. If we take all the A’s from the back and insert it at the front, this modified sequence must yield a larger number of A’s, since the number of A’s is multipled from the beginning. Therefore, by contradiction, all A’s must always be inserted at the front to yield the maximum number of A’s. Similar for the case where A’s are inserted in the middle of the sequence.
Before we proceed further, we introduce the following notation:
• Define 4A as a sequence of { A, A, A, A }. Therefore, 5A would then mean { A, A, A, A, A }.
• Define 2D as a sequence of CTRL+A, CTRL+C, CTRL+V, CTRL+V, which simply means double the previous text. Note that 3D does not double the previous text, it actually triples the previous text.
With this notation in place, it is much easier to work with this problem. Using the above notation, we rewrite our answer for N = 8 and N = 9.
N = 8, S = { 3A3D }, M = 9.
N = 9, S = { 3A4D }, M = 12.
The value of M could be obtained simply by multiplying the numbers, isn’t that neat?
Working our way up:
N = 10, S = { 4A4D }, M = 16.
N = 11, S = { 5A4D }, M = 20.
As you can see, the pattern here is pretty obvious, let’s summarize as follow:
• The solution so far for N > 7 is to find integers a and b such that ab yields the largest product, subjected to the condition where a+b = N-2.
• Both a and b are easy to find, as the largest product is found when the difference of a and b is less than or equal to one.
Similarly,
N = 12, S = { 5A5D }, M = 25.
N = 13, S = { 5A6D }, M = 30.
N = 14, S = { 6A6D }, M = 36.
Be extra cautious for N = 15.
When N = 15, does the sequence { 6A7D } yields the maximum where M = 42?
Imagine if you have a very large number of keystrokes to enter, does pressing CTRL+V forever gives you the maximum sequence? Remember, you can redo the entire { CTRL+A, CTRL+C, CTRL+V } operations again and potentially maximizes the sequence.
For N = 15, the maximum sequence should be:
{ 3A4D4D }, which yields M = 48.
Similarly,
N = 16, S = { 4A4D4D }, M = 64.
N = 21, S = { 3A4D4D4D }, M = 192.
N = 25, S = { 4A5D5D5D }, M = 500.
N = 26, S = { 5A5D5D5D }, M = 625.
N = 27, S = { 3A4D4D4D4D }, M = 768.
Let’s generalize the above:
M = MAX (a1 . a2ak),
where
a
1 + a2 + … + ak = n – 2(k-1)
To obtain M = MAX (a1 . a2ak),
it is necessary that the below condition must be met:
i, j ∈{ 1, 2, … , k } : MAX ( | aiaj | ) = 1.
To obtain M, we can first divide a1 + a2 + … + ak by k to obtain the average as a reference, and the rest should be straightforward.
Now the final problem lies in how to obtain the value of k efficiently. I am pretty sure this could be solved easily using Number Theory, but so far, my best solution is to use a brute force method to obtain k.
Below is the C++ code for my solution. It is pretty straightforward to output the sequence S. Given N, the function f() returns the maximum value of M.
VN:F [1.9.22_1171]
Rating: 4.8/5 (25 votes cast)
CTRL+A, CTRL+C, CTRL+V, 4.8 out of 5 based on 25 ratings
### 86 responses to CTRL+A, CTRL+C, CTRL+V
1. Is it possible to use DP to solve the problem?
Let M(i, j) denote the maximum number of A we can get by pressing A for i times and using totally j keys (i <= j). Then the recursion is as follows:
M(i, j) = max_{k<=j}{i*M(i,k-3)*(j-k+1)}, 1<=i<=N, N-i<=j<=N.
Correct me if I'm wrong.
VA:F [1.9.22_1171]
+1
• could someone please explain why does i*M(i,k-3)*(j-k+1) mean?
i stands for A is pressed i times, M(i, k-3) means max number of a by pressing A for i times and using totally k-3 keys, . then what the heck is j-k+1? and why do you need to multiple them together?
VA:F [1.9.22_1171]
0
• A DP solution is not correct because you have to take into account how many A’s are in the clipboard
VN:F [1.9.22_1171]
0
• I really doubt the DP works for this problem, since DP only works when current decision is based on the next decision. But when current decision affect the previous ones, I don’t know whether there is a way for that.
VN:F [1.9.22_1171]
0
2. My solution using DP. The O(n^3) complexity is not so good though.
int max_key_A(int n) {
if (n <= 7) return n;
int *M = new int[n+1];
for (int i = 0; i <= n; i++)
M[i] = 0;
int maxa = 0;
for (int i = 1; i <= n; i++) {
M[i] = i;
for (int j = i; j <= n; j++) {
for (int k = i+3; k <= j; k++) {
if (M[j] < (j-k+1)*M[k-3])
M[j] = (j-k+1)*M[k-3];
}
}
if (M[n] > maxa) maxa = M[n];
}
delete[] M;
return maxa;
}
VA:F [1.9.22_1171]
0
3. @Anonymous:
Awesome!
I haven't thought of using DP, will study your solution later.
I have tested your code and so far it matches my output for N=1 to N=87 (just change your return type from int to unsigned int to avoid overflow).
For N=88, my output is different than yours. What I'm suspecting is my code might have overflow problem in the line:
pow(t, (double)power);
caused by findMaxK() function.
VA:F [1.9.22_1171]
0
4. How about using { Ctrl+A, Ctrl+C, A, Ctrl+V } instead of { Ctrl+A, Ctrl+C, Ctrl+V, Ctrl+V }
VA:F [1.9.22_1171]
0
5. This definitely does not work:
{ Ctrl+A, Ctrl+C, A, Ctrl+V }
Because when you do CTRL+A, you are doing a select all. After the copy operation, all text are still being selected. When you type A, all text would then be replaced with a single A.
VA:F [1.9.22_1171]
0
6. A DP running in O(n^2):
int kcount(int n)
{
int* s = new int[n+1];
for(int i = 1; i <= n; ++i)
{
s[i] = i;
}
for(int i =1; i <= n-4; ++i)
{
int val = 2*s[i];
if(s[i+4] < val)
{
s[i+4] = val;
}
int delta = s[i];
for(int j = i+5; j <= n; ++j)
{
val += delta;
if(val > s[j])
{
s[j] = val;
}
}
}
int result = s[n];
delete[] s;
return result;
}
VA:F [1.9.22_1171]
0
7. Interesting post!
Because longer sequences of #D are dominated by repeated sequences of smaller counts (8D < 3D3D, but has same number of keystrokes) it is possible to use O(N) DP to solve it efficiently.
In fact, I suspect that because for longer runs 4D repeated is the most efficient, that it is possible to run in O(1) time, with a fixed amount of slop at the beginning and end to have sequences that are more efficient for a particular non multiple length of 4D4D4D… However I'm not sure how big that fixed slop is (it is probably pretty large).
VA:F [1.9.22_1171]
0
8. Can you please explain in more details why for N =7, {A, A, A,CTRL+A, CTRL+C, CTRL+V,CTRL+V } do not work to produce 9'A but for N=8, {A, A, A,CTRL+A, CTRL+C, CTRL+V,CTRL+V , CTRL+V} it works and produces 9 A's. I am confused why repeated CTRL+V works for N=8 but for not N=7.
VA:F [1.9.22_1171]
+1
9. @Anonymous1:
Yup, you are right. It is possible to solve using DP in O(N). And you are also right about the possibility to run in O(1) time, and someone from the mitbbs forum had discussed about this upper bound you mentioned. I will update my post later with the overall conclusions.
VA:F [1.9.22_1171]
0
10. @Anonymous2:
Think carefully. To double the previous text, you need to do {CTRL+A, CTRL+C, CTRL+V, CTRL+V}. Two CTRL+V's are needed. Therefore, for the example you gave in N=7, it only produces 6A's.
VA:F [1.9.22_1171]
+2
• I still do not get why ctrlA + ctrlC + ctrlV does not double the existing text.
What does the following set of operations result in ?
{ A, CTRL+A, CTRL+C, CTRL+V}
What does the following set of operations result in ?
{ A, A, A, CTRL+A, CTRL+C, CTRL+V}
VA:F [1.9.22_1171]
+2
• Try that out in a text editor like notepad. { A, CTRL+A, CTRL+C, CTRL+V} should yield “A”. { A, A, A, CTRL+A, CTRL+C, CTRL+V} should yield “AAA”. This is because CTRL+V replaces the selected text by CTLR+A. Anyway, this is not the heart of discussion. The main idea should not be affected by this rule.
VN:F [1.9.22_1171]
+2
• Oops, it indeed replaces the selected text. Yes, it does not affect the main idea. It was however disturbing me. Thanks for the clarification.
VA:F [1.9.22_1171]
0
• then It is better to highlight and clarify this confusing assumption at the beginning of post , because the original question doesn’t mention you can not move cursor.
VA:F [1.9.22_1171]
0
• You only have a keyboard that has CTRL, A, C, V. You definitely don’t have arrows.
VA:F [1.9.22_1171]
0
11. Here is my solution for reference:
using DP:
M[n] = max{ 1+ M[n-1], (n-k-3)*M[k] + M[k] }
Code:
#include
#include
#include
using namespace std;
int main (int argc, char const* argv[])
{
int n = atoi(argv[1]);
int *M = new int[n];
M[0] = 1;
for( int i = 1; i < n; i += 1)
{
int maxi = 0;
maxi = maxi > (1+M[i-1])? maxi:(1 + M[i-1]);
for(int k = 0; k <= i-4; k++)
{
int temp = M[k] + (i-k-3) * M[k];
maxi = maxi > temp? maxi: temp;
}
M[i] = maxi;
}
cout << "result:" << M[n-1] << endl;
delete[] M;
return 0;
}
VA:F [1.9.22_1171]
0
12. I think N = 8, S = { A, A, A, CTLA, CTLC, CTLV,, CTLV, CTLV }, M = 12
SO HOW ABOVE DESCRIBE ALGORITHM IS CORRECT.
VA:F [1.9.22_1171]
0
• Yes, I agree with you. I found many mistakes with the given solution.. for N=7, I got M=8.(A, A, A, ctrlA, ctrlC, ctrlV, crtlV) => gives M=9; I think this solution is wrong.
VA:F [1.9.22_1171]
0
• this gives 9 i think
VN:F [1.9.22_1171]
0
13. I solved it using DP taking O(n^2). I put my code here : http://ideone.com/tucp4. It shows the sequence of keys to get the max number of As printed.
VA:F [1.9.22_1171]
0
14. @buried.shopno – could you explain the M[i-j] * (j-2)?
VA:F [1.9.22_1171]
0
• @dwight: To double the sequence it needs 4 key press (CTRL+A, CTRL+C, CTRL+V, CTRL+V), to triple the sequence it needs 5 key press (CTRL+A, CTRL+C, CTRL+V, CTRL+V, CTRL+V), so on.
So, when j = 4, M [i-4] has the value of max. number of A’s sequence using (i-4) chars, and (j-2) which is 2 gives “number of times” M[i-4] will be repeated by a sequence of CTRL+A, CTRL+C, CTRL+V, CTRL+V.
For next iteration, j = 5, hence (j-2) gives 3 times of M[i-5]. Hope you get it now.
VA:F [1.9.22_1171]
0
15. There is a concise DP solution. FYI.
1: 1
2: 2
3: 3
4: 4
5: 5
6: 6
7: 7
8: 9
9: 12
10: 16
11: 20
12: 25
13: 30
14: 36
15: 48
16: 64
17: 80
18: 100
19: 125
20: 150
21: 192
22: 256
23: 320
24: 400
25: 500
26: 625
27: 768
28: 1024
29: 1280
30: 1600
31: 2000
32: 2500
33: 3125
34: 4096
35: 5120
36: 6400
37: 8000
38: 10000
39: 12500
40: 16384
41: 20480
42: 25600
43: 32000
44: 40000
45: 50000
46: 65536
47: 81920
48: 102400
49: 128000
50: 160000
51: 200000
52: 262144
53: 327680
54: 409600
55: 512000
56: 640000
57: 800000
58: 1048576
59: 1310720
60: 1638400
61: 2048000
62: 2560000
63: 3200000
64: 4194304
65: 5242880
66: 6553600
67: 8192000
68: 10240000
69: 12800000
70: 16777216
71: 20971520
72: 26214400
73: 32768000
74: 40960000
75: 51200000
76: 67108864
77: 83886080
78: 104857600
79: 131072000
80: 163840000
81: 204800000
82: 268435456
83: 335544320
84: 419430400
85: 524288000
86: 655360000
87: 819200000
88: 1073741824
89: 1342177280
90: 1677721600
91: 2097152000
92: 2621440000
93: 3276800000
94: 4294967296
95: 5368709120
96: 6710886400
97: 8388608000
98: 10485760000
99: 13107200000
100: 17179869184
VA:F [1.9.22_1171]
0
• I don’t think this output is right:
For example N=10: A A A A A CTRL+A CTRL+C CTRL+V CTRL+V CTRL+V the output should be 20!
VN:F [1.9.22_1171]
-2
• A A A A A CTRL+A CTRL+C CTRL+V CTRL+V the result is 10,you may have a try.
VN:F [1.9.22_1171]
0
• Wrong, your sequence only gets you 15 As. The answer of N=10, M=16 is correct. The CORRECT sequence is A A A A CTRL+A CTRL+C CTRL+V CTRL+V CTRL+V CTRL+V
VA:F [1.9.22_1171]
0
16. Suppose the last time you press Ctrl+A is after n-k, then F(n) = F(n-k) * (k-2).
F(n) = max(F(n-4)*2, F(n-5)*3, F(n-6)*4, F(n-7)*5, F(n-8)*6, F(n-9)*7).
This can be easily done in O(n) using DP and the code should be very simple.
There’s no need to consider k >= 10, because F(n-4)*2 = F(n-4-(k-4)) * (k-4 – 2) = F(n-k) * 2 * (k-6) already >= F(n-k) * (k-2) for k >= 10.
As mentioned earlier, O(1) might be possible but it will be harder to prove.
VA:F [1.9.22_1171]
0
17. The concise DP solution quoted by tanliboy is O(n^2). The DP solution using Haitao’s formula is O(n) time and O(1) space (using a ring buffer). They seem to give the same results.
However the code in the original post gives different results for some inputs. For example f(33) = 3072 while both DP solution give 3125. 3125 = 5*f(26) is clearly more correct than 3072. Is it a logic error or just FP rounding error in the original post?
VA:F [1.9.22_1171]
0
• I think you got the second DP solution wrong. It also gives 3072 as the result for N = 33. So 3072 is the right result.
VA:F [1.9.22_1171]
0
• Ignore my last reply. Just found a small glitch in my code. Anonymous is right, it does give 3125 instead of 3072.
VA:F [1.9.22_1171]
+1
18. N – Number of A’s
1 – 1
2 – 2
3 – 3
4 – 4
5 – 5
6 – 6 (Use 4th,5th,6th key for Ctrl-A,C,V)
7 – 8 (Use 5th,6th,7th key for Ctrl-A,C,V)
8 – 10(Use 6th,7th,8th key for Ctrl-A,C,V)
9 – 12 (Use 4t,5th,6th AND 7th,8th,9th key for Ctrl-A,C,V)
It can be solved by recursion:
if (n<6)
T(n) = n;
else
T(n) = (num – 3)*2
int findAs(int numKeys)
{
if(numKeys < 6)
return numKeys;
else
return( 2* findAs(numKeys-3) );
}
VA:F [1.9.22_1171]
-1
• Correction in above solution:
When n is 7 we can also press ctr-V key and get number of A’s to be 9 instead of 8.
So little modification in recursion function:
if(n= T2(n) )
….{ T(n) = T1(n)
………copiedAs = T(n-3)
….}
….else
….{ T(n) = T2(n)
….}
}
1 – 1
2 – 2
3 – 3
4 – 4
5 – 5
6 – Max(6,5) = 6, CopiedAs= 3
7 – Max(8,9) = 9, CopiedAs= 3
8 – Max(10,12) = 12, CopiedAs= 3
9 – Max(12,15) = 15, CopiedAs= 3
10- Max(18,18) = 18, CopiedAs= 9 (Chose T1 over T2 when equal and change copiedAs)
11- Max(24,27) = 27, CopiedAs= 9
12- Max(30,36) = 36, CopiedAs= 9
13- Max(36,45) = 45, CopiedAs= 9
14- Max(54,54) = 54, CopiedAs= 27 (CopiedAs changed)
.
.
.
.
int findAs(int nKeys)
{ static int copiedAs = 0;
….if(num= result2)
……..{ copiedAs = temp;
………….return result1;
……..}
……..else
…………return result2;
….}
}
VA:F [1.9.22_1171]
-1
• int findAs(int nKeys)
{ static int copiedAs = 0;
….if(num= result2)
……..{ copiedAs = temp;
………….return result1;
……..}
……..else
…………return result2;
….}
}
VA:F [1.9.22_1171]
0
• Not able to post complete code…
int findAs(int nKeys)
{ static int copiedAs = 0;
….if(num= result2)
……..{ copiedAs = temp;
………….return result1;
……..}
……..else
…………return result2;
….}
}
VA:F [1.9.22_1171]
0
• Correction: N=7 M=7 is correct. Open a notepad and try it. For N=7, it’s impossible to get M=8 or 9.
VN:F [1.9.22_1171]
0
19. I have the following idea, correct me if I am wrong
================================================================
Since we have a chain of keystrokes like this:
k “A”s ->(3 keystrokes, i.e., Ctrl-ACP)->2k “A”s ->(3 keystrokes)->4k “A”s ->(3 keystrokes)->….->2^n “A”s
we use totally (k+3*n = N) keystrokes and produce k*2^n “A”s.
Define the function f(k) = k*2^{(N-k)/3} as the number of consecutive “A”s produced, where 1 <= k <= N is a variable — by knowing the exact value of k, we will be able to see the maximum.
It is clear that f(k) is monotonically increase before it reaches the maximum and monotonically decrease after that. View this function as an array with index [0,...,N-1]. Hence, we could use a (modified) binary search algorithm to identify the maximum of f(k), which is the solution to this problem.
If the above is true, the algorithm works with O(log N) time.
VA:F [1.9.22_1171]
0
20. this question can be done in O(1).
the strategy should be like this:
for N, we type ‘A’ k times, and then we do “ctrl-A, ctrl-C,ctrl-V,ctrl-V,…ctrl-V”
now we can see M = max { max{ k(N-2-k) for k = 1,…,N-3 } , N }
to make k(N-2-k) max, we can choose k that nearest to N-2-k, assume k = N-2-k, we get k=(N-2)/2; adopt this k, we calculate max k(N-2-k), then we get max M.
no iterative, no recursive, no other space!
correct me, if i am wrong~!
VA:F [1.9.22_1171]
+1
21. I have an dp method with the same results as someone post before. O(n^2) time complexity and O(n) space complexity
find a position for the last c+a, c+c, c+v, then after that all c+v.
for(i = 8; i 0; k–){
long long len = s[k] * ( i – (k-2));
if(len > s[i]) s[i] = len;
}
VA:F [1.9.22_1171]
0
• long long len = s[k] * ( i – (k-2));
should be
long long len = s[k] * ( i – (k+2)); check the code
VA:F [1.9.22_1171]
0
22. #include
#include
using namespace std;
int main()
{
int N;
while(cin>>N)
{
long long int arr[N+1];
arr[0]=0;
for(int i=1;i<8;i++)
{
arr[i]=i;
}
for(int i=8;i<=N;i++)
{
long long int max = -1;
for(int k=1;kmax)max = mul;
}
arr[i] = max;
}
cout<<arr[N]<<endl;
}
}
VA:F [1.9.22_1171]
0
• copy paste didnt work.. Retrying.
#include
#include
using namespace std;
int main()
{
int N;
while(cin>>N)
{
long long int arr[N+1];
arr[0]=0;
for(int i=1;i<8;i++)
{
arr[i]=i;
}
for(int i=8;i<=N;i++)
{
long long int max = -1;
for(int k=1;kmax)max = mul;
}
arr[i] = max;
}
cout<<arr[N]<<endl;
}
}
VA:F [1.9.22_1171]
0
23. Hi,
I was trying to understand the solution proposed by 1337c0d3r for this problem
He has mentioned a statement :
Let’s generalize the above:
M = MAX (a1 . a2 … ak),
where
a1 + a2 + … + ak = n – 2(k-1)
To obtain M = MAX (a1 . a2 … ak),
Can sb please, let me know :
1- What are a1, a2…ak ?
2- What is ’k’ and how a1 + a2 + … + ak = n – 2(k-1) ?
VN:F [1.9.22_1171]
0
24. Attached my code in Java, Time O(n), Space O(1)
VA:F [1.9.22_1171]
0
• correction:
it’s not “//aAxD”, it’s //aAaD–(a+1)D(a+1)D, there are (k-x) times a, (k) times a+1
VA:F [1.9.22_1171]
+1
25. Hi guys:
I think this problem can be converted to
find the max product of the sequnence a1*a2*…*an, with ai takes (ai+2)step and the total step is n.
Here is my DP code :
int maxCtrlDp(int n){
//number of factor can get from x1Dx2D…xkD in n step
vector xDtable(n+1,1);
for(int i=1;i+2<=n;i++){
xDtable[i+2]=i;
}
//find the max factor in n step
for(int x=2;x+2<=n;x++){
for(int i=2+x;i<=n;i++){
xDtable[i]=max(xDtable[i],xDtable[i-2-x]*x);
}
}
int maxA=0;
for(int i=1;i<=n;i++){//number of A
maxA=max(maxA,i*xDtable[n-i]);
}
return maxA;
}
VA:F [1.9.22_1171]
0
26. Using greedy strategy
#include
#include
using namespace std;
int findMax(int N)
{
int i=0;
int M=0;
int copy=0;
while (i<N)
{
if (i<3)
{
M++; i++; cout<=N)
{
if (!copy)
{ M++; i++; cout<<"A ";}
else
{ M+=copy; i++; cout< M)
{ M += 3*copy; i+=3; cout<= 3)
{ copy=M; M+=copy; i+=3; cout<<"ctrl+A ctrl+C ctrl+V ";}
else
{ M++; i++; cout<<"A ";}
}
}
cout<<endl;
return M;
}
main(int argc, char **argv)
{
int m = findMax(atoi(argv[1]));
cout<<m<<endl;
}
VN:F [1.9.22_1171]
0
27. DP O(n^2)
int kcount(int n)
{
int* s = new int[n+1];
for(int i = 1; i <= n; ++i)
{
s[i] = i;
}
for(int i =4; i <= n; ++i)
{
for(int j = 0; j s[i])
{
s[i] = val;
}
}
}
int result = s[n];
delete[] s;
return result;
}
VA:F [1.9.22_1171]
0
• something missing due to display error, retype it
for(int i =4; i <= n; ++i)
{
for(int j = 0; js[i])
s[i] = val;
}
}
VA:F [1.9.22_1171]
0
28. I come up with a O(n) solution with O(1) space.
The key points are:
1) it can be proved that “A” sould be entered at the beginning (proof by controdiction)
2) it can be proved that for X<N, if X-1 yeilds smaller number of A than X does, then all numbers previous to X can no longer be helpful yeilding more A's.
So I use a deque to maintain the number of A yeilded from X to N-1 to compute that value for N. I do not know how to prove the O(1) space, but I print out the size of my deque and it is always <10.
VN:F [1.9.22_1171]
0
29. Hey, the solution is really easy if you found the rule here.
It’s simple O(n) and O(1) dp
4D is the most efficient because it takes 6 strokes, and get 4 times of the original text, which is the best you can get from 6 strokes after any certain stroke. there’s no other possibility for 6 strokes, because it can’t be 2D2D since it takes 8 strokes.
so you get max[n]=4*max[n-6] // if (4*max[n-6]>n), really simple right?!
so simply traverse the array and apply the rule.
you can check the result post by @tanliboy in the comment above
VA:F [1.9.22_1171]
0
30. static int f(int keys, int copies)
{
if (copies == 1)
return keys;
else
{
int x = (keys – 2 * (copies – 1)) / copies;
return x * f(keys – x – 2, copies – 1);
}
}
static int f(int keys)
{
int oldCount = 0;
int count = keys;
int split = 1;
while (oldCount < count)
{
split++;
oldCount = count;
count = f(keys, split);
}
return oldCount;
}
VN:F [1.9.22_1171]
0
31. mark,very interesting. most early answers seem to be uncorrect
VN:F [1.9.22_1171]
0
32. Hi
I found a little bug for your algorithm. For example:
N = 33, S = { 5A5D5D5D5D }, M = 3125. But your output is : S = { 3A4D4D4D4D4D }, m = 3072.
here is my code:
VN:F [1.9.22_1171]
0
33. Below is the recursive algo for simplicity.
int f(int n)
{
return re(n, 0, 0);
}
// n1: the strokes left
// cCopied: copied As
// c: the count of As already generated
int re(int n1, int cCopied, int c)
{
if(n1==0)
{
return c;
}
int t1 = re(n1-1, cCopied, c+1); // if the next stroke is A
int t2 = re(n1-1, cCopied, c+cCopied); // if the next stroke is ctrl-v
int t3 = INT_MIN;
if(n1>3) // if the next strokes are ctrl-a, ctrl-c, ctrl-v, ctrl-v
{
t3 = re(n1-4, c, 2c);
}
return max(t1, t2, t3);
}
VA:F [1.9.22_1171]
0
34. The solution is wrong. For N=7, it should be 8. The key sequence is:
[A, A, A, A, CTRL+A, CTRL+C, CTRL+V]
VA:F [1.9.22_1171]
0
35. Yet another DP in O(n^2)
VA:F [1.9.22_1171]
0
• Hmm, some parts were not copied correctly:
VA:F [1.9.22_1171]
+1
36. Ah, duh, it was the xml tag, sorry >.<"
` int max_keys(int strokes) { assert(strokes >= 0); if (strokes < 3) return strokes;`
` int* solutions = new int[strokes]; solutions[0] = 1; solutions[1] = 2; solutions[2] = 3;`
` for (int stroke_index = 3; stroke_index < strokes; stroke_index++) { solutions[stroke_index] = 1 + solutions[stroke_index - 1];`
` for (int copied_index = stroke_index - 4; copied_index >= 1; copied_index--) solutions[stroke_index] = max( solutions[stroke_index], (stroke_index - (copied_index + 2)) * solutions[copied_index]); }`
` int max = solutions[strokes - 1]; delete[] solutions;`
` return max; }`
VA:F [1.9.22_1171]
+1
37. hum… 9 maps to 16 as my result as code below:
private static int calculateMaxAs(int n) {
if (n 3) {
return (int) Math.pow(2, condition) * (1 + (n – 1) % 3);
} else if (condition > 1 && condition <= 3) {
return (n – 7) * 4 + 8;
} else {
return n;
}
}
VA:F [1.9.22_1171]
+1
• The paste is wrong:( see below and hope that it is going to be paste right:
private static int calculateMaxAs(int n) {
if (n 3) {
return (int) Math.pow(2, condition) * (1 + (n – 1) % 3);
} else if (condition > 1 && condition <= 3) {
return (n – 7) * 4 + 8;
} else {
return n;
}
}
8-12
9-16
VA:F [1.9.22_1171]
0
38. VA:F [1.9.22_1171]
0
39. VA:F [1.9.22_1171]
0
• second half….
VA:F [1.9.22_1171]
0
40. >>>>>>>>>>>>>>>>
Actually there is a tricky answer for this question, (still need to code as above eventually:)). First the answer will be only click on A key…N times… N is the max … Reason is that if there is no mouse involved, which I mean after Ctrl+A the focus doesn’t move to next space, whatever you paste or click would always replace what you have….
VA:F [1.9.22_1171]
0
41. paths:
VA:F [1.9.22_1171]
0
• VA:F [1.9.22_1171]
0
• if n>7 && n<13, let's say 9… A A A A Ctrl+A Ctrl+C Ctrl+V Ctrl+V Ctrl+V
VA:F [1.9.22_1171]
0
• if n>=13, let’s say 14… A Ctrl+A Ctrl+C Ctrl+V Ctrl+A Ctrl+C Ctrl+V Ctrl+A Ctrl+C Ctrl+V Ctrl+A Ctrl+C Ctrl+V Ctrl+V
VA:F [1.9.22_1171]
0
42. I think this is a indirect DP problem with two parts. The first part is calculate the times array, the second part is using the times array to compute the maximum.
The times array is that: given a string with length k, how many times longer we can get using index operations. For example times[i] = t indicates that using i operations we can get t*k length strings.
Then this problem transform to find the times array and using it to compute maximum.
How to get times array? Actually, I found that C-A C-C C-V get 1*original string, C-A C-C C-V C-V get 2*original string etc. So we need to use these sequence to cover at most n slots. So my formula is times[i] = max{ 1*times[i-3], 2 *time[i-4], 3*times[i-5] ….(j-2)*times[i-j].}
Then we just make continuous x As at the beginning with following times. Of course, n <= 7, best = 7.
Totally, time complexity is O(n^2)
VN:F [1.9.22_1171]
0
43. 12345678910
VA:F [1.9.22_1171]
0
44. Here is my solution with code at stackoverflow.com. It also prints the sequence of keystrokes.
http://stackoverflow.com/questions/4606984/maximum-number-of-characters-using-keystrokes-a-ctrla-ctrlc-and-ctrlv/22542395#22542395
VA:F [1.9.22_1171]
0
45. ctrl+ v
VA:F [1.9.22_1171]
0
46. ctrl+A,CTRL+C,CTRL+V
VA:F [1.9.22_1171]
0
47. xyz said on May 25, 2014
int MaxCopy(int n){
int *table=(int *)malloc(sizeof(int)*(n+1));
memset(table,0,sizeof(int)*(n+1));
for(int i=0;i<=n;i++){
table[i]=i;
}
for(int i=0;i<=n;i++){
for(int j=i+4;j<=n;j++){
table[j]=max(table[j],table[i]*(j-i-2));
}
}
int res=table[n];
free(table);
return res;
}
VA:F [1.9.22_1171]
0
48. O(n) solution in C including printing of key sequence:
http://gatecse.in/wiki/Printing_A
VA:F [1.9.22_1171]
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49. I think there should be something wrong with the answer…
E.g. n=7, m =9; n =8, m =12(aaa, aaa(ctrl a, c, v), aaa(ctrl v), aaa(ctrl v))
VN:F [1.9.22_1171]
0
50. VA:F [1.9.22_1171]
0
51. The formula to have the max A is f(x) = (N – 3x)*2^x , x in (0,N/3).
VN:F [1.9.22_1171]
0
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# MA322-001
Wed Jan 15: First day of class. Went over the syllabus. Worked out part of example 1 in the textbook. You are responsible for reading all of 1.1 and solving #5,#9,#17+18(just explain how to do it), 23, 29, 31, 33, 34. I handed out the quiz but we didn't have time to do it. You can check your work against the answer key.
Thu: Make sure to reread 1.1, work on the homework, and skim 1.2.
Fri Jan 17: Second day of class. We worked a full example, and carefully described what the solutions look like, and how we managed to solve it. The quiz covered HW 1.1 and what we did in class. You can check your answers; most people did extremely well, and everybody did well enough. Bold words we covered were: free variables, pivots, echelon form, reduced echelon form. We technically just learned Gaussian elimination as well. Before class started, we played and analyzed a silly game:
Game: Here is a silly game based on magic squares (add up the rows and get the same number for each row, add up the columns and get the same number of each column; we play on easy difficulty: the row sum is not required to be the same as the column sum). The first player gets two moves in a row; then the second player has to finish the magic square, if they can! Can they? Who has the advantage?
Sat-Mon: Reread 1.2, work on the homework, and skim 1.3. Here is a worked version of HW 1.2 #7. You may also find page 51 useful, as we'll use that idea (stoichiometry) for our example.
Wed Jan 22: Third day of class: 1.3 and stoichiometry. We covered vector geometry (walking on the lattice, example 4) and some stoichiometry (section 1.6 page 51). We took notes on a worksheet (the quiz is on the last page). You can check your work here. The “harder stoichiometry” problem is for you to try if like to test whether you've figured it out. Quiz question #3 was too soon, we'll handle it Friday (also it had a typo! Fixed on the online version, and here is the updated key).
HW 1.3 changed to HW 1.3 #5, 7, 9, 11, 13*, 22, 29*. For 13, convince yourself it is solved the same way as 11. Read 29, and convince yourself that it makes sense. You don't need to do the calculations on these two if you are pressed for time.
Fri Jan 24: Fourth day of class: 1.4 and stoichiometry. We went over the answers to Wednesday's quiz carefully, including a stoichiometry worksheet. We discussed how vectors are just lists of numbers, and answers are just lists of numbers, so answers are vectors. In our stoichiometry example, we wrote the answer as a line in vector form, and then covered how to find the (vector) equation of a line between two points. Importantly, slope is now a vector rather than a ratio. This is also called parametric form. We didn't have time for the quiz, but I'll collect it at the beginning of class on Monday.
Extra homework: Finish the quiz from Friday.
Mon Jan 27:Fifth day of class: 1.4 and 1.5. We'll finish up 1.4 and 1.5, making sure to cover all the different ways of writing down both the questions and the answers. Particular and homogeneous solutions. Vector form of the solution. The quiz from Friday is due at the beginning of class. You can check your answer here. We also had a quiz today; answers will be posted soon.
Tue Jan 28:9:30am to 10:30am special MA322 office hours.
Wed Jan 29:Sixth day of class: (1) Equation of a plane, (2) when can we find a particular solution, (3) Review all the ways to ask the same question (systems of equations, vector equation=span=linear combination, matrix equation), all the ways to write down the answer (basic and free variables, particular and homogeneous, vector form). Review the only way we know to solve them (guassian elimination). Review stoichiometry if time.
Fri Jan 31: Seventh day of class: EXAM. People did pretty well. If you made B- or less, make sure to seriously restudy this material. These sorts of questions will keep coming up in this class. If you made an A or less it doesn't hurt to review the parts you missed, but much of it should be come clear as we keep working. Solution key available end of today.
Mon Feb 3: Class cancelled by UK. It was going to be awesome too. With sledding. Since I couldn't return your exams, I put the grade on blackboard.
Wed Feb 5: Exams 1.7 Linear dependence AND exam return. You can check your work on the take home quiz. HW1.7 #1,9,15*,17*,19* (the star means no calculations needed).
Fri Feb 7: 1.8 linear transformations and 1.9 their matrices. These two topics go together wonderfully. You can check your work on the quiz . HW1.8 #1,13,14,15,16,19. HW1.9 #1*,3*,5*,13*,15*
Mon Feb 10: 2.1 Matrix operations (adding matrices, multiplying by a scalar, transpose, matrix multiplication). You can check you work on the quiz. HW2.1 #1,5,7*,10,11,12
Wed Feb 12: 2.2 Matrix inversion (when does a matrix have an inverse, how to find it, elementary matrices). The quiz is take-home, but you can check your work on this partial solution (full solution). HW2.2 #1,5,7,33
Fri Feb 14: 2.3 Matrix invertibility (when does a matrix have an inverse) The quiz was take home, and you can check your work. HW2.3 #1*,3*,5*,7*,13*
Mon Feb 17: 3.2/3.3 Matrix determinants (relationship to volume, how to actually compute them) The quiz was take home, and check your work. HW3.3#19,21,23
Wed Feb 19: Review.
Fri Feb 21: Exam
Mon Feb 24: 4.1 vector spaces: know the (10) axioms of a vector space, and the (3 part) test for a subspace. Be able to give examples of vector spaces and subspaces. HW 4.1 #1,2,3,5,6,7,8,9,11,19,21,(32,33,37). None involve computation. The last three will be used in this course, but won't be on the quiz.
Wed Feb 26: 4.2 Null spaces, column spaces, and linear transformations: Know the definition of null space, and how it relates to page 2 of exam 1 and column dependence. Know the definition of column space and how it relates to row dependence. Know the definition of linear transformation and how to use section 1.8/1.9 to think of them as matrices. The quiz is TAKE-HOME (write up solutions to the first two in your own words. I'll put up solutions as well). HW 4.1 #1,2,3,4,5,6,7,8,9,11,15,19,21 (again) and HW 4.2 #1,3,7,9,11,13,17*.
Fri Feb 28: 4.1 / 4.2 Vector spaces, subspaces, null spaces, column spaces (images). We'll go over more examples and make sure we get this. This is a very hard section, because it is all about perspective. This is the part that will actually let you use MA322 to get your job done, not just your MA322 homework. The quiz for Friday is take-home.
Mon Mar 3: Snow Day.
Wed Mar 5: 4.3. Writing down subspaces nicely. We talked about how subspaces are spans (this is like Col(A)). In 4.3, we talk about how to make sure none of the spanning vectors are redundant (this is like Nul(A)). The quiz will cover HW 4.2 #1,3,7,9,11,13,17 and the new stuff. Friday's quiz is due at the beginning of class.
Fri Mar 7: 4.4 and 4.5. Coordinate systems and dimension, especially for column spaces. The quiz.
Mon Mar 10: Review with answers.
Make sure you can answer questions like:
• Is this a subspace? Explain why or why not.
• What is the dimension of this subspace. Give a basis.
• Is this vector in this subspace? Give its coordinates with respect to this basis.
You'll need to be able to handle “subspaces” expressed as images of functions, and as solutions to equations (function value equals blah). The functions will either have domain and range column vectors (and then functions are just called matrices, like “A”), or domain and range contained in the set of real-valued functions with domain the real numbers, (and then the functions are called linear transformations, like the derivative, the definite integral, the evaluation at a point).
Wed Mar 12: Exam on 4.1 through 4.5.
Fri Mar 14: Work on the project due Fri Mar 28, 2014
Mon Mar 24: 5.1: Eigenvectors and eigenvalues. Some matrices would be diagonal if they were only given the chance to change coordinates. quiz with answers. HW5.1 #1,3,9,13,17,33 (and try #33 with A=[0,1;1,1]).
Wed Mar 26: 5.2: One way to find eigenvalues. Probably some review of diagonal matrices so that we realize how nice they are. quiz with with answers.
Fri Mar 28: Quickly go over project (Ch 4). Work on the Fibonacci fun (Ch 5). quiz it up with with answers.
Mon Mar 31: Review. I'll cover this population model (which is a very primitive version of section 5.6).
Wed Apr 2: Exam over chapter 5, with answers.
Fri Apr 4: No class. NCUR conference is using our classroom.
Mon Apr 7: 6.1 and 6.2: Inner products and orthogonality. quiz with answers.
Wed Apr 9: 6.3: Projections. quiz with answers.
Fri Apr 11: 6.4 and 6.5: Gram-Schmidt and least squares solutions. quiz with answers.
Mon Apr 14: Ch 6 review with answers.
Wed Apr 16: Ch6 exam with answers.
Fri Apr 18: Chapter 7.1: Emphasizing the so-called “spectral decomposition” of a symmetric matrix as a sum of rank one matrices uuT where the u are orthogonal column vectors. Such a sum has (uTu,u) as its eigenpairs. The quiz for today with answers. HOMEWORK: Find the spectral decomposition of the diagonal matrix with entries 1,3,5.
Mon Apr 21: Chapter 7.2: Quadratic forms, emphasizing so-called “principal axes”. Quadratic forms are basically “energy” and the principal axes allow us to find the principal contributors to both maximum and minimum energy. We'll talk about how this can be used to understand statistical energy, also known as “variance.” The worksheet and quiz with answers.
Wed Apr 23: Chapter 7.3: Constrained optimization. Actually find those minimum and maximum energies. quiz with answers.
Fri Apr 25: Chapter 7.4: SVD. We will write ANY matrix (even rectangular ones) as a sum of rank one matrices uvT where the u are orthogonal column vectors of a certain length, and the v are orthogonal column vectors of a certain (possibly different) length. Principal axes, energy levels, etc. will still basically work even for rectangular matrices. We worked the quiz together. The HOMEWORK is described on a webpage. You'll need a matrix calculator, but the homework includes a link to a free online interface to the free matrix calculator mentioned in class.
Mon Apr 28: Chapter 7.5: Image processing and statistical analysis. We reduce the dimension of statistical observations to something more managable. quiz.
Wed Apr 30: Chapter 7 review. Symmetric and rectangular; eigen and singular.
Fri May 2: Chapter 7 review. Practice exam with brief answers.
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# Statistical analysis: Wikis
Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.
# Encyclopedia
(Redirected to Statistics article)
More probability density will be found the closer one gets to the expected (mean) value in a normal distribution. Statistics used in standardized testing assessment are shown. The scales include standard deviations, cumulative percentages, percentile equivalents, Z-scores, T-scores, standard nines, and percentages in standard nines.
Statistics is the science of making effective use of numerical data relating to groups of individuals or experiments. It deals with all aspects of this, including not only the collection, analysis and interpretation of such data, but also the planning of the collection of data, in terms of the design of surveys and experiments.[1]
A statistician is someone who is particularly versed in the ways of thinking necessary for the successful application of statistical analysis. Often such people have gained this experience after starting work in any of a number of fields. There is also a discipline called mathematical statistics, which is concerned with the theoretical basis of the subject.
The word statistics can either be singular or plural.[2] In its singular form, a statistic is a quantity (such as a mean) calculated from a set of data,[3] whereas statistics is the mathematical science discussed in this article.
## Scope
Statistics is considered by some to be a mathematical science pertaining to the collection, analysis, interpretation or explanation, and presentation of data,[4] while others consider it to be a branch of mathematics[5] concerned with collecting and interpreting data.[6] Because of its empirical roots and its focus on applications, statistics is usually considered to be a distinct mathematical science rather than a branch of mathematics.[7][8]
Statisticians improve the quality of data with the design of experiments and survey sampling. Statistics also provides tools for prediction and forecasting using data and statistical models. Statistics is applicable to a wide variety of academic disciplines, including natural and social sciences, government, and business.
Statistical methods can be used to summarize or describe a collection of data; this is called descriptive statistics. This is useful in research, when communicating the results of experiments. In addition, patterns in the data may be modeled in a way that accounts for randomness and uncertainty in the observations, and are then used to draw inferences about the process or population being studied; this is called inferential statistics. Inference is a vital element of scientific advance, since it provides a prediction (based in data) for where a theory logically leads. To further prove the guiding theory, these predictions are tested as well, as part of the scientific method. If the inference holds true, then the descriptive statistics of the new data increase the soundness of that hypothesis. Descriptive statistics and inferential statistics (a.k.a., predictive statistics) together comprise applied statistics.[9]
## History
Some scholars pinpoint the origin of statistics to 1663, with the publication of Natural and Political Observations upon the Bills of Mortality by John Graunt.[10] Early applications of statistical thinking revolved around the needs of states to base policy on demographic and economic data, hence its stat- etymology. The scope of the discipline of statistics broadened in the early 19th century to include the collection and analysis of data in general. Today, statistics is widely employed in government, business, and the natural and social sciences.
Its mathematical foundations were laid in the 17th century with the development of probability theory by Blaise Pascal and Pierre de Fermat. Probability theory arose from the study of games of chance. The method of least squares was first described by Carl Friedrich Gauss around 1794. The use of modern computers has expedited large-scale statistical computation, and has also made possible new methods that are impractical to perform manually.
The American Statistical Association has ranked Deming, Fisher, and Rao as the greatest statisticians of all time.[citation needed]
## Overview
In applying statistics to a scientific, industrial, or societal problem, it is necessary to begin with a population or process to be studied. Populations can be diverse topics such as "all persons living in a country" or "every atom composing a crystal". A population can also be composed of observations of a process at various times, with the data from each observation serving as a different member of the overall group. Data collected about this kind of "population" constitutes what is called a time series.
For practical reasons, a chosen subset of the population called a sample is studied — as opposed to compiling data about the entire group (an operation called census). Once a sample that is representative of the population is determined, data is collected for the sample members in an observational or experimental setting. This data can then be subjected to statistical analysis, serving two related purposes: description and inference.
“... it is only the manipulation of uncertainty that interests us. We are not concerned with the matter that is uncertain. Thus we do not study the mechanism of rain; only whether it will rain.”
Dennis Lindley, "The Philosophy of Statistics", The Statistician (2000).
The concept of correlation is particularly noteworthy for the potential confusion it can cause. Statistical analysis of a data set often reveals that two variables (properties) of the population under consideration tend to vary together, as if they are connected. For example, a study of annual income that also looks at age of death might find that poor people tend to have shorter lives than affluent people. The two variables are said to be correlated; however, they may or may not be the cause of one another. The correlation phenomena could be caused by a third, previously unconsidered phenomenon, called a lurking variable or confounding variable. For this reason, there is no way to immediately infer the existence of a causal relationship between the two variables. (See Correlation does not imply causation.)
For a sample to be used as a guide to an entire population, it is important that it is truly a representative of that overall population. Representative sampling assured, inferences and conclusions can be safely extended from the sample to the population as a whole. A major problem lies in determining the extent to which the sample chosen is actually representative. Statistics offers methods to estimate and correct for any random trending within the sample and data collection procedures. There are also methods for designing experiments that can lessen these issues at the outset of a study, strengthening its capability to discern truths about the population. Statisticians describe stronger methods as more "robust".(See experimental design.)
The fundamental mathematical concept employed in understanding potential randomness is probability. Mathematical statistics (also called statistical theory) is the branch of applied mathematics that uses probability theory and analysis to examine the theoretical basis of statistics. The use of any statistical method is valid only when the system or population under consideration satisfies the basic mathematical assumptions of the method.
Misuse of statistics can produce subtle, but serious errors in description and interpretation — subtle in the sense that even experienced professionals make such errors, and serious in the sense that they can lead to devastating decision errors. For instance, social policy, medical practice, and the reliability of structures like bridges all rely on the proper use of statistics. Even when statistics are correctly applied, the results can be difficult to interpret for those lacking expertise. The statistical significance of a trend in the data — which measures the extent to which a trend could be caused by random variation in the sample — may or may not agree with an intuitive sense of its significance. The set of basic statistical skills (and skepticism) that people need to deal with information in their everyday lives properly is referred to as statistical literacy.
## Statistical methods
### Experimental and observational studies
A common goal for a statistical research project is to investigate causality, and in particular to draw a conclusion on the effect of changes in the values of predictors or independent variables on dependent variables or response. There are two major types of causal statistical studies: experimental studies and observational studies. In both types of studies, the effect of differences of an independent variable (or variables) on the behavior of the dependent variable are observed. The difference between the two types lies in how the study is actually conducted. Each can be very effective.
An experimental study involves taking measurements of the system under study, manipulating the system, and then taking additional measurements using the same procedure to determine if the manipulation has modified the values of the measurements. In contrast, an observational study does not involve experimental manipulation. Instead, data are gathered and correlations between predictors and response are investigated.
An example of an experimental study is the famous Hawthorne study, which attempted to test changes to the working environment at the Hawthorne plant of the Western Electric Company. The researchers were interested in determining whether increased illumination would increase the productivity of the assembly line workers. The researchers first measured the productivity in the plant, then modified the illumination in an area of the plant and checked if the changes in illumination affected productivity. It turned out that productivity indeed improved (under the experimental conditions). However, the study is heavily criticized today for errors in experimental procedures, specifically for the lack of a control group and blindness. The Hawthorne effect refers to finding that an outcome (in this case, worker productivity) changed due to observation itself. Those in the Hawthorne study became more productive not because the lighting was changed but because they were being observed.[citation needed]
An example of an observational study is one that explores the correlation between smoking and lung cancer. This type of study typically uses a survey to collect observations about the area of interest and then performs statistical analysis. In this case, the researchers would collect observations of both smokers and non-smokers, perhaps through a case-control study, and then look for the number of cases of lung cancer in each group.
The basic steps of an experiment are:
1. Planning the research, including determining information sources, research subject selection, and ethical considerations for the proposed research and method.
2. Design of experiments, concentrating on the system model and the interaction of independent and dependent variables.
3. Summarizing a collection of observations to feature their commonality by suppressing details. (Descriptive statistics)
4. Reaching consensus about what the observations tell about the world being observed. (Statistical inference)
5. Documenting / presenting the results of the study.
### Levels of measurement
There are four types of measurements or levels of measurement or measurement scales used in statistics:
• nominal,
• ordinal,
• interval, and
• ratio.
They have different degrees of usefulness in statistical research. Ratio measurements have both a zero value defined and the distances between different measurements defined; they provide the greatest flexibility in statistical methods that can be used for analyzing the data. Interval measurements have meaningful distances between measurements defined, but have no meaningful zero value defined (as in the case with IQ measurements or with temperature measurements in Fahrenheit). Ordinal measurements have imprecise differences between consecutive values, but have a meaningful order to those values. Nominal measurements have no meaningful rank order among values.
Since variables conforming only to nominal or ordinal measurements cannot be reasonably measured numerically, sometimes they are called together as categorical variables, whereas ratio and interval measurements are grouped together as quantitative or continuous variables due to their numerical nature.
### Key terms used in statistics
#### Null hypothesis
Interpretation of statistical information can often involve the development of a null hypothesis in that the assumption is that whatever is proposed as a cause has no effect on the variable being measured.
The best illustration for a novice is the predicament encountered by a jury trial. The null hypothesis, H0, asserts that the defendant is innocent, whereas the alternative hypothesis, H1, asserts that the defendant is guilty.
The indictment comes because of suspicion of the guilt. The H0 (status quo) stands in opposition to H1 and is maintained unless H1 is supported by evidence "beyond a reasonable doubt". However, "failure to reject H0" in this case does not imply innocence, but merely that the evidence was insufficient to convict. So the jury does not necessarily accept H0 but fails to reject H0.
#### Error
Working from a null hypothesis two basic forms of error are recognised:
• Type I errors where the null hypothesis is falsely rejected giving a "false positive".
• Type II errors where the null hypothesis fails to be rejected and an actual difference between populations is missed.
#### Confidence intervals
Most studies will only sample part of a population and then the result is used to interpret the null hypothesis in the context of the whole population. Any estimates obtained from the sample only approximate the population value. Confidence intervals allow statisticians to express how closely the sample estimate matches the true value in the whole population. Often they are expressed as 95% confidence intervals. Formally, a 95% confidence interval of a procedure is any range such that the interval covers the true population value 95% of the time given repeated sampling under the same conditions. If these intervals span a value (such as zero) where the null hypothesis would be confirmed then this can indicate that any observed value has been seen by chance. For example a drug that gives a mean increase in heart rate of 2 beats per minute but has 95% confidence intervals of -5 to 9 for its increase may well have no effect whatsoever.
The 95% confidence interval is often misinterpreted as the probability that the true value lies between the upper and lower limits given the observed sample. However this quantity is more a credible interval available only from Bayesian statistics.
#### Significance
Statistics rarely give a simple Yes/No type answer to the question asked of them. Interpretation often comes down to the level of statistical significance applied to the numbers and often refer to the probability of a value accurately rejecting the null hypothesis (sometimes referred to as the p-value).
When interpreting an academic paper reference to the significance of a result when referring to the statistical significance does not necessarily mean that the overall result means anything in real world terms. (For example in a large study of a drug it may be shown that the drug has a statistically significant but very small beneficial effect such that the drug will be unlikely to help anyone given it in a noticeable way.)
### Examples
Some well-known statistical tests and procedures are:
## Specialized disciplines
Some fields of inquiry use applied statistics so extensively that they have specialized terminology. These disciplines include:
In addition, there are particular types of statistical analysis that have also developed their own specialised terminology and methodology:
Statistics form a key basis tool in business and manufacturing as well. It is used to understand measurement systems variability, control processes (as in statistical process control or SPC), for summarizing data, and to make data-driven decisions. In these roles, it is a key tool, and perhaps the only reliable tool.
## Statistical computing
gretl, an example of an open source statistical package
The rapid and sustained increases in computing power starting from the second half of the 20th century have had a substantial impact on the practice of statistical science. Early statistical models were almost always from the class of linear models, but powerful computers, coupled with suitable numerical algorithms, caused an increased interest in nonlinear models (such as neural networks) as well as the creation of new types, such as generalized linear models and multilevel models.
Increased computing power has also led to the growing popularity of computationally-intensive methods based on resampling, such as permutation tests and the bootstrap, while techniques such as Gibbs sampling have made use of Bayesian models more feasible. The computer revolution has implications for the future of statistics with new emphasis on "experimental" and "empirical" statistics. A large number of both general and special purpose statistical software are now available.
## Misuse
There is a general perception that statistical knowledge is all-too-frequently intentionally misused by finding ways to interpret only the data that are favorable to the presenter. A famous saying attributed to Benjamin Disraeli is, "There are three kinds of lies: lies, damned lies, and statistics." Harvard President Lawrence Lowell wrote in 1909 that statistics, "...like veal pies, are good if you know the person that made them, and are sure of the ingredients."
If various studies appear to contradict one another, then the public may come to distrust such studies. For example, one study may suggest that a given diet or activity raises blood pressure, while another may suggest that it lowers blood pressure. The discrepancy can arise from subtle variations in experimental design, such as differences in the patient groups or research protocols, which are not easily understood by the non-expert. (Media reports usually omit this vital contextual information entirely, because of its complexity.)
By choosing (or rejecting, or modifying) a certain sample, results can be manipulated. Such manipulations need not be malicious or devious; they can arise from unintentional biases of the researcher. The graphs used to summarize data can also be misleading.
Deeper criticisms come from the fact that the hypothesis testing approach, widely used and in many cases required by law or regulation, forces one hypothesis (the null hypothesis) to be "favored," and can also seem to exaggerate the importance of minor differences in large studies. A difference that is highly statistically significant can still be of no practical significance. (See criticism of hypothesis testing and controversy over the null hypothesis.)
One response is by giving a greater emphasis on the p-value than simply reporting whether a hypothesis is rejected at the given level of significance. The p-value, however, does not indicate the size of the effect. Another increasingly common approach is to report confidence intervals. Although these are produced from the same calculations as those of hypothesis tests or p-values, they describe both the size of the effect and the uncertainty surrounding it.
## Statistics applied to mathematics or the arts
Traditionally, statistics was concerned with drawing inferences using a semi-standardized methodology that was "required learning" in most sciences. This has changed with use of statistics in non-inferential contexts. What was once considered a dry subject, taken in many fields as a degree-requirement, is now viewed enthusiastically. Initially derided by some mathematical purists, it is now considered essential methodology in certain areas.
• In number theory, scatter plots of data generated by a distribution function may be transformed with familiar tools used in statistics to reveal underlying patterns, which may then lead to hypotheses.
• Methods of statistics including predictive methods in forecasting, are combined with chaos theory and fractal geometry to create video works that are considered to have great beauty.
• The process art of Jackson Pollock relied on artistic experiments whereby underlying distributions in nature were artistically revealed. With the advent of computers, methods of statistics were applied to formalize such distribution driven natural processes, in order to make and analyze moving video art.
• Methods of statistics may be used predicatively in performance art, as in a card trick based on a Markov process that only works some of the time, the occasion of which can be predicted using statistical methodology.
• Statistics is used to predicatively create art, as in applications of statistical mechanics with the statistical or stochastic music invented by Iannis Xenakis, where the music is performance-specific. Though this type of artistry does not always come out as expected, it does behave within a range predictable using statistics.
## Notes
1. ^ Dodge, Y. (2003) The Oxford Dictionary of Statistical Terms, OUP. ISBN 0199206139
2. ^ "Statistics". Merriam-Webster Online Dictionary.
3. ^ "Statistic". Merriam-Webster Online Dictionary.
4. ^ Moses, Lincoln E. Think and Explain with statistics, pp. 1 - 3. Addison-Wesley, 1986.
5. ^ Hays, William Lee, Statistics for the social sciences, Holt, Rinehart and Winston, 1973, p.xii, ISBN 978-0030779459
6. ^ Statistics at Encyclopedia of Mathematics
7. ^ Moore, David (1992). "Teaching Statistics as a Respectable Subject". Statistics for the Twenty-First Century. Washington, DC: The Mathematical Association of America. pp. 14–25.
8. ^ Chance, Beth L.; Rossman, Allan J. (2005). "Preface". Investigating Statistical Concepts, Applications, and Methods. Duxbury Press. ISBN 978-0495050643.
9. ^ Anderson, , D.R.; Sweeney, D.J.; Williams, T.A.. Statistics: Concepts and Applications, pp. 5 - 9. West Publishing Company, 1986.
10. ^ Willcox, Walter (1938) The Founder of Statistics. Review of the International Statistical Institute 5(4):321-328.
## References
• Best, Joel (2001). Damned Lies and Statistics: Untangling Numbers from the Media, Politicians, and Activists. University of California Press. ISBN 0-520-21978-3.
• Desrosières, Alain (2004). The Politics of Large Numbers: A History of Statistical Reasoning. Trans. Camille Naish. Harvard University Press. ISBN 0-674-68932-1.
• Hacking, Ian (1990). The Taming of Chance. Cambridge University Press. ISBN 0-521-38884-8.
• Lindley, D.V. (1985). Making Decisions (2nd ed. ed.). John Wiley & Sons. ISBN 0-471-90808-8.
• Tijms, Henk (2004). Understanding Probability: Chance Rules in Everyday life. Cambridge University Press. ISBN 0-521-83329-9.
# Study guide
Up to date as of January 14, 2010
(Redirected to Statistics article)
### From Wikiversity
Correlation examples
Statistics is an applied branch of Mathematics.
## Category tree
### Online videos
#### Statistics rap
These (selected) student-made "statistics rap" videos can be used as fun introductions/reviews of concepts commonly taught in undergraduate statistics courses:
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# SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV
Probability Sampling
Probability Sampling; Simple, Systematic, Cluster, Stratified, Multistage
Mehdi Osooli
Knowledge Hub on HIV surveillance, Kerman University of Medical Sciences, Iran
The main objectives
1. Giving an introduction to basics of probability sampling 2. Making participants familiar with practical aspects of each probability sampling method 3. Comparing pros and cons of different probability sampling methods
Concept and basics of probability sampling methods
One of the most important issues in researches is selecting an appropriate sample. Among sampling methods, probability sample are of much importance since most statistical tests fit on to this type of sampling method. Representativeness and generalize-ability will be achieved well with probable samples from a population, although the matter of low feasibility of a probable sampling method or high cost, don’t allow us to use it and shift us to the other non-probable sampling methods. In probability sampling we give known chance to be selected to every unit of the population. We usually want to estimate some parameters of a population by a sample. These parameters estimates when we don’t observe whole population usually have some errors. Fortunately in probability sampling it is possible that we know how much our estimates are trustable or close to the parameter value from population by computing standard errors of estimates. This is not easily possible in non-probability sampling methods.
Types of probability sampling methods Simple Random Sampling
What is it? Simple random sampling is selecting randomly some units from a known and well defined population. In this method the sampling frame should be known and all units should have same chance for being selected. How is it down? (Example) In simple random sampling, from population of N, n units are selected randomly and the chance of being selected for all units is equal. Different methods and tools can be used for creating 27
000 from 76. 12 and 18 respectively. How is it down? (Example) First we should have the list of the population and according to the total number of sample needed we define a value of “k” to jump over population units and selecting units. The k was defined as 38 and a number between 1 and 38 was chosen. We decided to select a sample of 2. Criticisms The chance of selecting a non-representative sample is very high in this method of sampling especially when there is a correlation between the place of the unit in the population list and the 28 . Criticisms Although when the population is not very big it is possible to do simple random sampling. Example: You have been asked to perform a KAP survey in a prison.Probability Sampling SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV random numbers for sample selection. you can use a random number generator to generate 300 numbers between 1 and 2000. Suppose the random number is 3. third and fourth units will be 9. 2009. Standard random number tables and soft-wares with ability of generating random numbers like Open-Epi or Stata are available. Since. Choosing 12 then 38 was added to that and second person was the record number 50 and the next units were chosen adding each time the 38 to the previous selected record. Uses Simple random sampling is a good method for comparing the precision of different methods of sampling and also useful for teaching general probabilistic sampling rules. Systematic Random Sampling What is it? In systematic random sampling we use the order of the population list or the place of units in the population for choosing the sample. the k is 6 the second. we can define k=6 and draw a random number between 1 and 6. The precision of systematic random sampling is higher than simple random sampling. According to the participants names repeated units were excluded and replaced by new units with the same method. You think that a sample of 300 would be satisfactory for your work. Uses Systematic random sampling is very easy and less time consuming. If we want select 5 units over a population of 50. The list of blood donors was available on computer software and the order of patients was according to the date of their referral. In big population and wide geographical sampling areas it is not easy to take a list form all units and randomly selecting them.000. other methods of random sampling are preferable to it because they brought more precise estimates from population. The list of all 2000 prisoners has been given to you. Most of the time you would have some repeated numbers that should be replaced by new numbers. If you want choose 300 of them for interview randomly. Example: We want to estimate the prevalence of HIV infection among volunteer blood donors in Tehran.
hotel based and brothel based. Here it is possible that using the “k” you jump over some specific units and select in case different units of the population. In this case it is more reasonable to take random samples from these subdivisions. How is it down? (Example) Stratified sampling is done in two major steps. A schematic example for systematic sampling with k=10 Stratified Random Sampling What is it? In some situations the population can be divided in to sub population which share some characteristics internally. First we should define population strata’s and second select a sample form each stratum. The population of the strata’s should be known. Example: You want to estimate the prevalence of STI among female sex workers in a capital city. You have found from the formative assessment that there are three types of FSWs in the city.SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV Probability Sampling characteristics of the unit that should be observed. The level of shaded color in the below figure indicated the level of frequency of risky behaviors. They somehow differ from each other regarding percentage of high risk behavior and medical consultant and available health care services. The sub divisions here are called “strata”. The strata’s should be non-overlapping homogenous internally but heterogeneous externally. street based. 29 .
although in real world that is not easily achievable. you have access to the list of all the FSWs. How is it down? (Example) 30 . FSWs Hotel-based FSWs Brothel-based FSWs Streetbased Uses Four uses can be proposed for stratified random sampling: 1. Cluster Random Sampling What is it? In sampling from big population. the most common method is cluster sampling. When you want achieve certain precision or information for specific subdivision of the population stratified sampling is very useful. In this method we divide population in to sub divisions called clusters. 2. 3. you have three strata of FSWs and to have a precise estimate of STI prevalence of the target population. you should select sample from each group/strata.Probability Sampling SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV As all the FSWs in the city have registered in the health department. Clusters are representative sub-samples of population. In this case each area is a stratum and you will allocate your sample from each stratum separately. In some cases the sampling problems are different in different study fields again here dividing population into subdivisions will enable you to define specific methods and criteria for work in each division. Using this method you will need smaller sample size with higher precision. Criticisms The assumption of little variation and similarity within strata’s is very important to benefit from this method. In this case. Since the variability within strata’s is very small the smaller numbers will give you satisfactory precision and combining these estimates as well bring precise estimates too. this means that the distribution of population units in each cluster is heterogeneous. Using stratified sampling is more helpful in studies in multiple administrative areas. By using stratified sampling the overall precision of the estimates will be more exact. 4.
Big national surveys usual are done using multi stage sampling methods. How is it down? (Example) At first we assess the study population and its criteria to fit for different sampling methods. Then according to the population’s structure we define our sampling framework and take our sample. The inference about sample and final analysis is a bit complicated and should be based and adapted on the procedure of the sampling and different methods were used. Criticisms Multi stage method needs careful design. we include whole selected cluster’s unit in to our sample while in multi stage cluster sampling we will choose randomly just some units within clusters. Uses The benefits of each sampling type are achieved using multi stage method. In household surveys also each family can be considered as a cluster. In one step cluster sampling after selecting few clusters. in one step method all patients form each hospital should be included while in multistage cluster sampling we first select some hospitals then a number of patients within each hospital (cluster) will be selected.SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV Probability Sampling Clusters are parts of the population with almost same elements or units of the population but in smaller scale. Criticisms The precision of cluster sample is lower than stratified sampling and it needs bigger sample sizes to bring same precision. Uses In sampling from wide geographical area’s it is possible to define neighboring regions as a cluster. Multi stage sampling What is it? In this method you do several sampling steps and use different sampling methods to achieve your desired sample. It is possible to use both probabilistic and non probabilistic methods together but should keep in mind that non-probabilistic samples are not representative of the population. Tehran has around 120 hospitals. 31 . Example: In assessing the satisfaction of HIV positive patients from hospital based health care services in city of Kerman you can assume each hospital in the city and allocate a random sample size from each hospital to reach your desired sample size. Cluster sampling can be done in just one step or multiple steps.
Logistically difficult if sample geographically dispersed Systematic 1. Construct sample frame 1. Requires sample large enough to make precise estimates for each strata 4.Probability Sampling SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV Summary Table 1 provides a brief summary of various conventional sampling techniques including their advantages and disadvantages. Requires sample frame of entire survey population 2. Add SI to random start and select person. Using random number/lottery lottery draw time-consuming 1. 1. Can increase precision of indicator estimates 1. Define the strata and construct sample frame for each strata 2. Logistically difficult if sample geographically dispersed 3. Random numbers or lottery not required 2. Requires sample frame of entire for survey population target population 1. Sampling Simple random Steps Advantages Disadvantages 1. Select random start between 1 and SI & select that person 4.Summary of conventional sampling techniques. Select people randomly 2. Calculate indicator estimates for each strata and for population 1. Population estimates require weighting Stratified 32 . Create a list of the target population 2. Table 1 . Concept is easy 2. Logistically difficult if sample to understand and from sample frame using geographically dispersed analyse random number table or 3. Easy to analyse 1. Produces unbiased estimates of indicators for the strata 2. Take a simple/systematic sample from each strata 3. etc. Requires sample frame of entire target population 2. Calculate sampling interval (SI) 3.
Select clusters using simple/systematic sampling 3. Sample concentrated the random start in geographical areas 4. Only need sample frame of clusters and & SI individuals in selected 3. Sample equal numbers of people from selected clusters 1. Sample size. Only need sample frame of clusters and individuals in selected clusters 2. Decreases precision of estimates. fixed cluster size cluster size 1. Sample concentrated in geographical areas 3. thus. Sample equal proportions of people per cluster 1.SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV Probability Sampling Table 1 . select random start between 1 1. Size of clusters required prior to sampling Cluster: Cluster: Equal probability. Decreases precision of estimates. Size of clusters required for weighted analysis 1. thus. Construct sample frame of clusters 2. Construct sample frame of clusters 2. precision of estimates unpredictable 33 . thus. Add SI to random start & select cluster 5. Select cluster using simple/systematic sampling 3. Decreases precision of estimates. Select cluster whose clusters cumulative size contains 2. Construct sample frame of clusters 2. Sampling Cluster: Probability proportional to size (PPS) or equal probability sampling Steps Advantages Disadvantages 1. Sample concentrated in geographical areas 1. Sample equal numbers of people from selected clusters 1. Only need sample frame of clusters and individuals in selected clusters 2.Summary of conventional sampling techniques. Weighted analysis required for unbiased estimates 3. requires larger sample size 2. requires larger sample size 2. Size of clusters required for proportional sampling 3. Don’t need cluster sizes prior to sampling 1. continued. proportional Equal probability. thus. Calculate SI. requires larger sample size 2.
Tryfos P.Y. [in Persian]. Lemeshow. answer the following questions: a. John Willey & Sons.Probability Sampling SAMPLING METHODS FOR POPULATION AT INCREASED RISK OF HIV Question(s) to be discussed Looking at Table 1.Y. 2nd ed. 1992 Thompson M. London Chapman & Hall. 1983 Levy. 2nd ed.S. Sampling of Population methods and applications. et al. S. 2nd ed. 1996 34 . Sampling Techniques.G. John Willey & Sons. N. Sampling method for applied research. What are the steps to take when using a Cluster: equal probability. 3rd ed. 1999 Thompson S. 2007 Malek Afzali H et al. N.E. Applied Research Methodology in Medical Sciences. 1997 Chereii A. P. John Willey & Sons. John Willey & Sons Inc. Tehran University of medical Sciences. What is the disadvantage of using stratified sampling method when it comes to making population estimates? References • • • • • • • Cochrane W. Sampling. Theory of Sample survey. N.Y. What are the advantages to using a systematic sampling method? b.. 3rd ed. Sampling and estimating Sample Size in Medical Research [in Persian]. fixed cluster size sampling method? c.
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### Archive
Posts Tagged ‘programming’
## Sum of the Euler Totient function
Given a positive integer ${n}$, the Euler totient function ${\varphi(n)}$ is defined as the number of positive integers less than ${n}$ which are co-prime with ${n}$ (i.e. they have no common factors with ${n}$). There are formulas for computing ${\varphi(n)}$ starting from the factorization of ${n}$. One such formula is
$\displaystyle \varphi(n) = n \prod_{p|n} \left(1-\frac{1}{p}\right),$
where the product is made over all primes dividing ${n}$.
If you have to compute ${\varphi(n)}$ for all numbers less than a threshold then another property could be useful: ${\varphi}$ is arithmetic, that is, ${\varphi(mn) = \varphi(m)\varphi(n)}$ whenever ${\gcd(m,n)=1}$. Therefore you could store all values computed until ${k}$ and for computing the value ${\varphi(k+1)}$ there are two possibilities: ${k+1=p^\alpha}$ is a prime power and then ${\varphi(k+1) = p^{\alpha}-p^{\alpha-1}}$ or ${k+1}$ is composite and ${k+1 = mn}$ with ${m,n\leq k,\ \gcd(m,n)=1}$. Then use the stored values to compute ${\varphi(k+1)=\varphi(m)\varphi(n)}$.
I now come to the main point of this post: computing the sum of all values of the totient function up to a certain ${N}$:
$\displaystyle \text{Compute } S(N) = \sum_{i=1}^N \varphi(i).$
One approach is to compute each ${\varphi(i)}$ and sum them. I will call this the brute-force approach. For all numerical purposes I will use Pari-GP in this post. On my computer it takes less than a second to compute ${S(10^6)}$ and about ${12}$ seconds to compute ${S(10^7)}$. This is super linear in time, since the algorithm computes the factorization for each ${n}$ and then sums the values. Using the sieve approach could improve the timing a bit, but the algorithm is still super linear.
In some Project Euler problems it is not uncommon to have to compute something like ${S(10^{11})}$ or even larger. Therefore, there must be more efficient ways to compute ${S(N)}$ out there, so let’s study some of the properties of ${S(N)}$. In another post I dealt with the acceleration of the computation of the sum of the divisor function.
We have ${S(N) = \sum_{i=1}^N \varphi(i)}$ which is the number of pairs ${(a,b)}$ with ${1\leq a\leq b \leq N}$ such that ${\gcd(a,b) =1}$. It is not difficult to see that the total number of such pairs is ${n(n+1)/2}$. Moreover, the possible values of ${\gcd(a,b)}$ are ${1,2,...,N}$. Now, if for ${m \leq N}$ we search instead for pairs satisfying ${\gcd(a,b)=m}$ then we have ${a = ma',\ b = mb'}$ with ${\gcd(a',b')=1}$ and we get
$\displaystyle 1 \leq a' \leq b' \leq N/m,\ \gcd(a',b')=1.$
There fore the number of pairs with gcd equal to ${m}$ is ${S(\lfloor N/m\rfloor )}$. Now we arrive at an interesting recursive formula:
$\displaystyle S(N) = \frac{n(n+1)}{2} - \sum_{m=2}^N S(\lfloor N/m \rfloor ).$
At a first sight this looks more complicated, but there is a trick to keep in mind whenever you see a summation over ${m}$ of terms of the form ${\lfloor N/m \rfloor}$: these quantities are constant on large intervals. Indeed,
$\displaystyle \lfloor N/m \rfloor = d \Leftrightarrow md \leq N < m(d+1)\Leftrightarrow N/(d+1)
Therefore we can change the index of summation from ${m}$ to ${d=\lfloor N/m \rfloor}$. The range of ${d}$ for which the interval ${I_d = [N/(d+1),N/d]}$ contains more than one integer is of order ${\sqrt(N)}$. Indeed, ${N/d-N/(d+1) = N/(d(d+1))}$. Therefore for ${d\leq \sqrt(N)}$ we should have at least one integer in the interval ${I_d}$. The part where ${d}$ is larger than ${\sqrt{N}}$ corresponds to ${m}$ smaller than ${\sqrt{N}}$. Therefore, we can split ${S(N)}$ into two sums, each of order ${\sqrt{N}}$. and get that
$\displaystyle S(N) = \frac{n(n+1)}{2}- \sum_{m=2}^{\sqrt{N}} S(\lfloor N/m \rfloor)-\sum_{d=1}^{\sqrt{N}}\left(\lfloor N/d \rfloor-\lfloor N/(d+1)\rfloor\right)S(d),$
where in the last sum we must make sure that ${d \neq \lfloor N/d \rfloor}$ in order to avoid duplicating terms in the sum.
Therefore we replaced a sum until ${N}$ to two sums with upper bound ${\sqrt{N}}$. The complexity is not ${\sqrt{N}}$, but something like ${N^{2/3}}$ since we have a recursive computation. Nevertheless, with this new formula and using memoization, to keep track of the values of ${S}$ already computed, we can compute ${S(N)}$ very fast:
${S(10^6) = 303963552392}$ is computed instantly (vs ${1}$ second with brute force)
${S(10^7) = 30396356427242}$ takes ${1}$ second (vs ${12}$ seconds with brute force)
${S(10^8)}$ takes ${5}$ seconds (vs over ${3}$ minutes with brute force)
${S(10^9)}$ takes ${30}$ seconds
${S(10^{11})}$ takes about ${12}$ minutes
etc. Recall that these computations are done in Pari GP, which is not too fast. If you use C++ you can compute ${S(10^8)}$ in ${0.2}$ seconds, ${S(10^9)}$ in ${1}$ second and ${S(10^{10})}$ in ${6}$ seconds and ${S(10^{11})}$ in under a minute, if you manage to get past overflow errors.
Categories: Number theory, Programming
## FreeFem++ Tutorial – Part 2
Here are a few tricks, once you know the basics of FreeFem. If your plan is straightforward: define the domain, build the mesh, define the problem, solve the problem, plot the result… then things are rather easy. If you want to do stuff in a loop, like an optimization problem, things may get more complicated, but FreeFem still has lots of tricks up its sleeves. I’ll go through some of them.
1. Defining the geometry of the domain using border might be tricky at the beginning. Keep in mind that the domain should be on the left side of the curves definining the boundaries. This could be acheived in the parametrization chosen or you could reverse the parametrization of a particular part of the domain in the following way. Suppose you have the pieces of boundary C1,C2,C3,C4. You wish to build a mesh with the command mesh Th = buildmesh(C1(100)+C2(100)+C3(100)+C4(100));but you get an error concerning a bad sens on one of the boundaries. If you identify, for example, that you need to change the orientation of C3, you can acheive this by changing the sign of the integer defining the number of points on that part of the boundary: mesh Th = buildmesh(C1(100)+C2(100)+C3(-100)+C4(100));You could make sure that you have the right orientations and good connectivities for all the boundaries by running a plot command before trying mesh: something like
plot(C1(100)+C2(100)+C3(-100)+C4(100));
should produce a graph of all your boundaries with arrows showing the orientations. This is good as a debug tool when you don’t know where the error comes from when you define the domain.
2. Keep in mind that there are also other ways to build a mesh, like square, which meshes a rectangular domain with quite a few options and trunc which truncates or modifies a mesh following some criteria. Take a look in the documentation to see all the options.
3. Let’s say that you need to build a complex boundary, but with many components with similar properties. Two examples come to mind: polygons and domains with many circular holes. In order to do this keep in mind that it is possible to define a some kind of “vectorial boundary”. Let’s say that you want to mesh a polygon and you have the coordinates stored in the arrays xs,ys. Furthermore, you have another array of integers ind which point to the index of the next vertex. Then the boundary of the polygon could be defined with the following syntax:
border poly(t=0,1; i){
x=(1-t)*xx[i]+t*xx[ind(i)];
y=(1-t)*yy[i]+t*yy[ind(i)];
label=i;
}
Now the mesh could be constructed using a vector of integers NC containing the number of desired points on each of the sides of the polygon:
mesh Th = buildmesh (poly(NC));
4. You can change a 2D mesh using the command adaptmesh. There are various options and you’ll need to search in the documentation for a complete list. I use it in order to improve a mesh build with buildmesh. An example of command which gives good results in some of my codes is:
The parameters are related to the size of the triangles, the geometric properties of the griangles and the maximal number of vertices you want in your mesh. Experimenting a bit might give you a better idea of how this command works in practice.
5. There are two ways of defining the problems in FreeFem. One is with solve and the other one is with problem. If you use solve then FreeFem solves the problem where it is defined. If you use problem then FreeFem remembers the problem as a variable and will solve it whenever you call this variable. This is useful when solving the same problem multiple times. Note that you can modify the coefficients of the PDE and FreeFem will build the new problem with the updated coefficients.
6. In a following post I’ll talk about simplifying your code using macros and functions. What is good to keep in mind is that macros are verbatim code replacements, which are quite useful when dealing with complex formulas in your problem definition or elsewhere. Functions allow you to run a part of the code with various parameters (just like functions in other languages like Matlab).
I’ll finish with a code which computes the eigenvalue of the Laplace operator on a square domain with multiple holes. Try and figure out what the commands and parameters do.
int N=5; //number of holes
int k=1; //number of eigenvalue
int nb = 100; // parameter for mesh size
verbosity = 10; // parameter for the infos FreeFem gives back
real delta = 0.1;
int nbd = floor(1.0*nb/N*delta*2*pi);
int bsquare = 0;
int bdisk = 1;
// vertices of the squares
real[int] xs(N^2),ys(N^2);
real[int] initx = 0:(N^2-1);
for(int i=0;i<N^2;i++){
xs[i] = floor(i/N)+0.5;
ys[i] = (i%N)+0.5;
}
cout << xs << endl;
cout << ys << endl;
int[int] Nd(N^2);
Nd = 1;
Nd = -nbd*Nd;
cout << Nd << endl;
// sides of the square
border Cd(t=0,N){x = t;y=0;label=bsquare;}
border Cr(t=0,N){x = N;y=t;label=bsquare;}
border Cu(t=0,N){x = N-t;y=N;label=bsquare;}
border Cl(t=0,N){x = 0;y=N-t;label=bsquare;}
border disks(t=0,2*pi; i){
x=xs[i]+delta*cos(t);
y=ys[i]+delta*sin(t);
label=bdisk;
}
plot(Cd(nb)+Cr(nb)+Cu(nb)+Cl(nb)+disks(Nd));
mesh Th = buildmesh(Cd(nb)+Cr(nb)+Cu(nb)+Cl(nb)+disks(Nd));
plot(Th);
int[int] bc = [0,1]; // Dirichlet boundary conditions
fespace Vh(Th,P2);
// variables on the mesh
Vh u1,u2;
// Define the problem in weak form
varf a(u1,u2) = int2d(Th)
(dx(u1)*dx(u2) + dy(u1)*dy(u2))+on(1,u1=0)//on(C1,C2,C3,C4,u1=1)
+on(bc,u1=0);
varf b([u1],[u2]) = int2d(Th)( u1*u2 ) ;
// define matrices for the eigenvalue problem
matrix A= a(Vh,Vh,solver=Crout,factorize=1);
matrix B= b(Vh,Vh,solver=CG,eps=1e-20);
// we are interested only in the first eigenvalue
int eigCount = k;
real[int] ev(eigCount); // Holds eigenvalues
Vh[int] eV(eigCount); // holds eigenfunctions
// Solve Ax=lBx
int numEigs = EigenValue(A,B,sym=true,sigma=0,value=ev,vector=eV);
cout << ev[k-1] <<span id="mce_SELREST_start" style="overflow:hidden;line-height:0;"></span>< endl;
plot(eV[k-1],fill=1,nbiso=50,value=1);
Here is the mesh and the solution given by the above program:
## Using parfor in Matlab
We all know that loops don’t behave well in Matlab. Whenever it is possible to vectorize the code (i.e. use vectors and matrices to do simultaneous operations, instead of one at a time) significant speed-up is possible. However, there are complex tasks which cannot be vectorized and loops cannot be avoided. I recently needed to compute eigenvalues for some 10 million domains. Since the computations are independent, they could be run in parallel. Fortunately Matlab offers a simple way to do this, using parfor.
There are some basic rules one need to respect to use parfor efficiently:
1. Don’t use parfor if vectorization is possible. If the task is not vectorizable and computations are independent, then parfor is the way to go.
2. Variables used in each computation should not overlap between processors. This is for obvious reasons: if two processors try to change the same variable using different values, the computations will be meaningless in the end.
3. You can use an array or cell to store the results given by each processor, with the restriction that processors should work on disjoint parts of the array, so there is no overlap.
The most restrictive requirement is the fact that one cannot use the same variables in the computations for different processors. In order to do this, the simplest way I found was to use a function for the body of the loop. When using a matlab function, all variables are local, so when running the same function in parallel, the variables won’t overlap, since they are local to each function.
So instead of doing something like
parfor i = 1:N
commands ...
array(i) = result
end
you can do the following:
parfor i=1:N
array(i) = func(i);
end
function res = func(i)
commands...
This should work very well and no conflict between variables will appear. Make sure to initialize the array before running the parfor, a classical Matlab speedup trick: array = zeros(1,N). Of course, you could have multiple outputs and the output array could be a matrix.
There is another trick to remember if the parpool cannot initialize. It seems that the parallel cluster doesn’t like all the things present in the path sometimes. Before running parfor try the commands
c = parcluster('local');
c.parpool
If you recieve an error, then run
restoredefaultpath
c = parcluster('local');
c.parpool
and add to path just the right folders for your code to work.
## Metapost – or how to code an image
What software do you use when you need to draw a nice image? I often need to draw things related to research and I never managed to efficiently use a software which uses the mouse or touchpad to draw and modify things. Moreover, if you need to add some mathematical text to the figure things get even more complicated. For me it is more natural to use code to generate graphics, if this is possible. Using something like Matlab to draw is possible. The advantage is that once you have a working code which produces what you want, you can easily modify it. If you have an image where you need to repeat things, using loops can facilitate the job (programmers will understand…).
This is were Metapost comes into play. I found this a long time ago while searching for a tool to build nice graphics for my master thesis. Being used to LaTeX for typesetting math, it was a natural way to draw using code. I do not claim it is the best or the simplest way, but for me it works. Most importantly, it allows me to build high quality, vectorized graphics, which are memory efficient. The advantage of vector graphics is that you can zoom as much as you want and you’ll never see the pixels.
There are a bunch of places where you can learn Metapost. I’ll put here some references I use a lot. The easiest way to start drawing in Metapost is to take a look at some existing codes and modify them. You can find lots of examples in this tutorial. If you have a linux system, you can compile metapost .mp files to obtain high quality pdfs using the command mptopdf. If not, then you can use the online Metapost editor found here. In any case, here are some of my codes for some drawings in Metapost.
Here is the code for my website logo: If you want to see the lossless vectorized pdf click to see the following file: logo-0 You can see that it has a border which changes from one color to another. This was done using a loop and a parameter to vary the color as we go along the boundary.
prologues:=3;
verbatimtex
%&latex
\documentclass{minimal}
\begin{document}
etex
beginfig(0);
% copy from here to use the online previewer
u:=25; % 25 = 25bp = 25 PostScript points = 25/72 in
wi:=10; % width in units u
he:=7; % height in units u
hoehe:=he*u; % height
breite:=wi*u;
%for i=0 upto he:
%draw (0, i*u)--(breite, i*u) withcolor .7white;
%endfor
%for j=0 upto wi:
%draw (j*u, 0)--(j*u, hoehe) withcolor .7white;
%endfor;
path p,q;
p:=(6u,0.5u)--(6u,0)--(0,0)--(0,6u)--(6u,6u)--(6u,5.5u);
pickup pensquare scaled 20;
draw p;
h=length(p);
numeric c,d,detail;
color a,b,co;
a:=(1,0.45,0);
b:=.2white;
co:=.2white;
detail:=500;
for i=1 upto (detail/2):
q := subpath (h*(i-1)/detail,h*i/detail) of p;
draw q withcolor (2*i/detail)[a,b];
endfor;
for i=detail downto (1+detail/2):
q := subpath (h*(i-1)/detail,h*i/detail) of p;
draw q withcolor (2*(detail-i)/detail)[a,b];
endfor;
label ("B",(1/15)*(2.6u,2.7u)) scaled 15 withcolor co ;
label ("2" infont defaultfont scaled 8,(4.9u,4.1u))
withcolor co;
% end copy here for online previewer
endfig;
end;
Here’s another example of a figure where I needed a grid of disks aligned on top of another figure. Using loops in Metapost allowed me to get what I wanted:
For the high quality pdf click here: cioranescu-murat-0 and see the code below
prologues:=3;
verbatimtex
%&latex
\documentclass{minimal}
\begin{document}
etex
beginfig(0);
u:=25; % 25 = 25bp = 25 PostScript points = 25/72 in
wi:=10; % width in units u
he:=7; % height in units u
hoehe:=he*u; % height
breite:=wi*u;
path p,pa;
p:=(0,4u)..(0,2u)..(2u,2u)..(3u,u)..(3u,0)..(6u,u)..(7u,3u)..(9u,5u)..cycle;
pa:=(3u,5u)..(5u,6u)..(8u,6u)..(8u,3.8u)..(7.5u,3.4u)..(7u,3u)..cycle;
draw p;
fill p withcolor .8white;
pair a,b;
h=length(p);
i=5;
numeric c,d;
c:=0.65*h;
d:=0.7*h;
a:= point c of p;
b:= point d of p;
pickup pencircle scaled 1.5;
%draw a;
%draw b;
path q,r,s;
q:= subpath (c,d) of p;
%draw q withcolor red;
%r:= b..((a+b)/2 shifted (.5(a-b) rotated -90))..a;
%draw r;
%fill buildcycle(r,q) withcolor .5red;
%fill pa withcolor .9999blue;
%label.rt(btex $\Omega$ etex,.5(u,3u)) scaled 2;
draw (-u,9u)-- (10u,9u)--(10u,-2u)--(-u,-2u)--cycle;
for i=0 upto 9:
for j=-1 upto 8:
draw fullcircle scaled 0.3u shifted (i*u, j*u);
fill fullcircle scaled 0.3u shifted (i*u, j*u) withcolor white;
endfor
endfor;
endfig;
end;
There are many ways today to draw what you want. Metapost is one of the tools available, if you like coding.
Categories: Programming
## Project Euler – Problem 264
Today I managed to solve problem 264 from Project Euler. This is my highest rating problem until now: 85%. You can click the link for the full text of the problem. The main idea is to find all triangles ABC with vertices having integer coordinates such that
• the circumcenter O of each of the triangles is the origin
• the orthocenter H (the intersection of the heights) is the point of coordinates (0,5)
• the perimeter is lower than a certain bound
I will not give detailed advice or codes. You can already find a program online for this problem (I won’t tell you where) and it can serve to verify the final code, before going for the final result. Anyway, following the hints below may help you get to a solution.
The initial idea has to do with a geometric relation linking the points A, B, C, O and H. Anyone who did some problems with vectors and triangles should have come across the needed relation at some time. If not, just search for important lines in triangles, especially the line passing through O and H (and another important point).
Once you find this vectorial relation, it is possible to translate it in coordinates. The fact that points A, B, C are on a circle centered in O shows that their coordinates satisfy an equation of the form $x^2+y^2=n$, where $n$ is a positive integer, not necessarily a square… It is possible to enumerate all solutions to the following equation for fixed $n$, simply by looping over $x$ and $y$. This helps you find all lattice points on the circle of radius $\sqrt{n}$.
Once these lattice points are found one needs to check the orthocenter condition. The relations are pretty simple and in the end we have two conditions to check for the sum of the x and y coordinates. The testing procedure is a triple loop. We initially have a list of points on a circle, from the previous step. We loop over them such that we dont count triangles twice: i from 1 to m, j from i+1 to m, k from j+1 to m, etc. Once a suitable solution is found, we compute the perimeter using the classical distance formula between two points given in coordinates. Once the perimeter is computed we add it to the total.
Since the triple loop has cubic complexity, one could turn it in a double loop. Loop over pairs and construct the third point using the orthocenter condition. Then just check if the point is also on the circle. I didn’t manage to make this double loop without overcounting things, so I use it as a test: use double loops to check every family of points on a given circle. If you find something then use a triple loop to count it properly. It turns out that cases where the triple loop is needed are quite rare.
So now you have the ingredients to check if on a circle of given radius there are triangles with the desired properties. Now we just iterate over the square of the radius. The problem is to find the proper upper bound for this radius in order to get all the triangles with perimeter below the bound. It turns out that a simple observation can get you close to a near optimal bound. Since in the end the radii get really large and the size of the triangles gets really large, the segment OH becomes small, being of fixed length 5. When OH is very small, the triangle is almost equilateral. Just use the upper bound for the radius for an equilateral triangle of perimeter equal to the upper bound of 100000 given in the problem.
Using these ideas you can build a bruteforce algorithm. Plotting the values of the radii which give valid triangles will help you find that you only need to loop over a small part of the radii values. Factoring these values will help you reduce even more the search space. I managed to solve the problem in about 5 hours in Pari GP. This means things could be improved. However, having an algorithm which can give the result in “reasonable” time is fine by me.
## Project Euler 607
If you like solving Project Euler problems you should try Problem number 607. It’s not very hard, as it can be reduced to a small optimization problem. The idea is to find a path which minimizes time, knowing that certain regions correspond to different speeds. A precise statement of the result can be found on the official page. Here’s an image of the path which realizes the shortest time:
## FreeFem to Matlab – fast mesh import
October 14, 2016 1 comment
I recently wrote a brief introduction to FreeFem++ in this post. FreeFem is a software designed for the numerical study of partial differential equations. It has the advantage of being able to easily define the geometry of the domain, construct and modify meshes, finite element spaces and solve problems on these meshes.
I use Matlab very often for numerical computations. Most of the numerical stuff I’ve done (take a look here if you want) was based on finite differences methods, fundamental solutions and other classical techniques different from finite elements methods. Once I started using finite elements I quickly realized that Matlab is not that easy to work with if you want some automated quality meshing. PDEtool is good, but defining the geometry is not easy. There is also a simple tool: distmesh which performs a simple mesh construction for simple to state geometries. Nevertheless, once you have to define finite element spaces and solve problems things are not easy…
This brings us to the topic of this post: is it possible to interface Matlab and FreeFem? First, why would someone like to do this? Matlab is easier to code and use than FreeFem (for one who’s not a C expert…), but FreeFem deals better with meshes and solving PDE with finite elements. Since FreeFem can be called using system commands, it is possible to call a static program from Matlab. FreeFem can save meshes and variables to files. Let’s see how can we recover them in Matlab.
There is a tool called “FreeFem to Matlab” developed by Julien Dambrine (link on Mathworks). There’s also a longer explanation in this blog post. I recently tried to use the tool and I quickly found that it is not appropriate for large meshes. It probably scans the mesh file line by line which makes the loading process lengthy for high quality meshes. Fortunately there’s a way to speed up things and I present it below. I will not cover the import of the data (other than meshes) since the function importdata from the FreeFem to Matlab tool is fast enough for this.
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dan's math@home - problem of the week - archives
Problem Archives page 11
1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+ . prob index
101 -- The Fab Five !
102 - Similar but Diff
103 - Weird Numbers
104 Dan's Prime Code
105- Ez Come, Ez Go
106- New Year Phone
107- Geometry Gems
108 - Cover The Cube
109- Based On What?
110 - Family Of Four
Problem #101 - Posted Sunday, November 19, 2000
A set contains five integers. When distinct elements of this set are added together,
two at a time, the complete list of different possible sums that result is:
637, 669, 794, 915, 919, 951, 1040, 1072, 1197.
a) Figure out (if possible) the original five integers in the set.
b) What are the means; of the original set, and the set of sums ?
c) Given a set of 'sums', do five such numbers always exist ?
Solution: From Arthur Morris : Let the integers be x1, x2, x3, x4 and x5 in ascending order.
a) Assume the three smallest sums are { x1 + x2 = 637 , x1 + x3 = 669 , x2 + x3 = 794 } .
Solve the 3 simultaneous equations ; { x1 = 256 , x2 = 381 , x3 = 413 } .
There are only 9 different sums given, and the 5 elements in the set produce ten sums so there
are two summations with the same value. These two sums must involve four different elements
with the sum of the smallest and largest equal two the sum of the middle terms.
x1 + x4 = x2 + x3 and so x4 = 538. The largest sum must be the sum of the two largest elements,
x5 = 1197 - 538 = 659. The five elements are 256, 381, 413, 538 and 659.
b)
The mean of the original set is 449.4. The mean of the sum set is 910.44. (some of you put 898.8)
c) Five such numbers do not always exist, because I would have come up the sum set without
using values 4 through 8 in the original set. (Dan's note: 9 equations in 5 unknowns aren't always consistent!)
Arthur Morris . . . . 10 pts - Thanks for providing this week's solution!
Irina Kolesnik . . . . . . 7 pts - Nice use of remainders (794 = 2 mod 4 )
Ludwig Deruyck . . .5 pts - Good answer; I meant the actual 9 sums average.
Sudipta Das . . . . . . . 4 pts - Clever, let S = repeated sum; S + 8194 = 0 mod 4.
Slanky . . . . . . . . . . . . 3 pts - Start with the means, get info on the nos, nice strategy!
Lisa Schechner . . . . 3 pts - Nice idea on c), impossible if all 'sums' were odd!
Phil Sayre . . . . . . . . . 2 pts - Right on the Fab Five, part c) is indeed doubtful usually.
Al B . . . . . . . . . . . . . . 2 pts - c) is true if the sums come from actual numbers, yes.
Montta . . . . . . . . . . . . 2 pts - I like how you started with estimating a ~ 250, etc.
Kia Davidson . . . . . . 1 pt - Thanks for trying it; here they are, the mystery numbers!
Problem #102 - Posted Monday, November 27, 2000
Drew drew two similar triangles, both with integer sides. Two sides of one triangle
were the same as two in the other.The other (unmatching) sides differed by a prime.
a) What are the smallest triangles making this possible? (Give corresp. prime difference.)
b) What is the smallest prime for which this is possible? (Give corresponding triangles.)
Answers for a) and b) may or may not be different.
Solution (by frequent responder Lisa Schechner):
I'm not sure what the formal definition for "the smallest triangles" would be, so I can't answer exactly.
Let one triangle be given by its edges: a<b<c. Then there exists k>1 such that the second triangle's edges
are given by: ka<kb<kc, with ka=b, kb=c. Since k must be rational, we can assume that k=m/n, with
m, n are relatively prime, m < [(1+sqrt(5))/2] n and n>1. (Dan's note : this is the Golden Ratio.)
This means that the two triangles should have the form of {n^3,mn^2,nm^2} and {mn^2,nm^2,m^3}
and the prime number should have the form: m^3 - n^3.
The smallest prime p having the form p= m^3 - n^3, such that n>1 and n, m are rel. prime is p = 19
(m=3, n=2). So my conclusion is that the smallest prime = 19.
This corresponds to the triangles: {8,12,18} and {12,18,27}. (Dan's note: 27 - 8 = 19 = 3^3 - 2^3.)
Arthur Morris . . . . 10 pts - Good sys of eqns using k = i/j ; right: 1, 2, 4 not a triangle.
Lisa Schechner . . . . 7 pts - Also if m^3 - n^3 = (m-n)(m^2 + mn + n^2) = prime then m-n = 1.
Phil Sayre. . . . . . . . . 5 pts - Oh, my, yes, a cubic equation! (Devoured by the Python - language.)
Jesse Goldlink. . . . . 4 pts - Not 234, 468; not 124, 248, but yes! 8.12.18, 12.18.27 !
Al B . . . . . . . . . . . . . . 3 pts - I bumped you up a point because 3 < 19 and 2,3,4 ~ 3,4,5 almost.
Sudipta Das . . . . . . . 2 pts - Pretty good but 1, 2, 4 can't form a triangle because 1 + 2 < 4.
Ludwig Deruyck . . . 2 pts - Yes, if 2 _corresp_ sides are the same the triangles are congruent.
Irina Kolesnik . . . . . 2 pts - if a1 = p/(k^3 - 1) then no solution if k = integer, but k = 1.5.
Slanky . . . . . . . . . . . . 2 pts - 2,2,3 isn't similar to 2,3,4 , one being isosceles, but p=2 is small.
Les Billig (new) . . . . . 2 pts - Welcome to the contest! (Good but same prob with the 1,2,4 'triangle'.)
Montta . . . . . . . . . . . . 2 pts - For the triangles to be similar they must have the same shape.
Matt Albrecht (new) . . 2 pts - Welcome! Right, you can answer as long as the problem is still up!
Problem #103 - Posted Friday, December 8, 2000
You may know a 'perfect number' is one whose proper divisors add up to the number,
such as 6: 1+2+3 = 6. In an 'abundant number' the divisors add up to more than the
number, like 12: 1+2+3+4+6 = 16 > 12. A 'weird number' is an abundant number with
no subset of divisors adding to the number. The number 12 isn't weird 'cause 1+2+3+6 = 12.
What are the first two weird numbers?
Solution: The first 2 weird numbers are 70 and 836. You can try to find a slightly abundant
number with a divisor sum that exceeds the number by a non-divisor, then look more closely.
Thanks to Stanley Benkoski of West Valley College in Calif. for his 'weird' concept; see Lisa's score comment.
A few of you (including me) wrote programs or did a search; but I think Phil put it best when
describing his program that spit out abundant candidates to be submitted to manual recounts:
"This is a prize example of the difference between machine and human intelligence."
According to Lisa, the late great Paul Erdos once offered a (deficient) 10 dollars to anyone
finding any odd weird numbers, their existence is still in question! (Is the \$10 still on the table?)
(Dan's note: Same with odd perfect numbers; for odd abundant numbers look in my Problem Archives.)
Lisa also gave the longest list of weird nos (for us weirdos that appreciate them, she says!)
70, 836, 4030, 5830, 7192, 7912, 9272, 10430, 10570, 10792, 10990, 11410, 11690, 12110,
12530, 12670, 13370, 13510, 13790, 13930, 14770, 15610, 15890, 16030, 16310, 16730,...
Ludwig Deruyck . . . 10 pts - Good answer, thanks for the mathematica code!
Lisa Schechner. . . . . 7 pts - Thanks! I had dinner with someone with an 'Erdos number' of 1!
Irina Kolesnik . . . . . 5 pts - You got 'em! How did your program prove there was no subset?
Phil Sayre . . . . . . . . . 4 pts - Python strikes again, or at least paralyzes the problem.
Arthur Morris . . . . . 3 pts - Good try - 70 works, but 102 = 17 + 34 + 51.
Al B . . . . . . . . . . . . . . 2 pts - Right; 70 is first; the next one is after quite a gap.
Aimee Jacot (new) . . . 1 pt - Hi, thanks for entering; 2 and 3 are weird-esque but not abundant.
Problem #104 - Posted Sunday, December 17, 2000
The first 26 primes (2, 3, 5, ...) can be put in correspondence with the letters A through
Z, so the 'Prime Code' for the word CAB would be the product 5 * 2 * 3 = 30.
a) What's the Prime Code for the word BIKER ? b) Decode this message... 913 1511191 110618. c) What (English) word comes closest to a million?
Please submit all parts in one message. No proper nouns. Ranked by order received AND best c); results avgd.
One point penalty for each resubmission of 'improved' answers to c).
Solution: The 'alphabet primes' are 2, 3, 5, 7, 11, . . . 97, 101. (There are 25 primes under 100.)
a) BIKER makes a product of 3 * 23 * 31 * 11 * 61 = 1,435,269 (not all that close to a million).
b) 913 = 11 * 83 = E W ; 1511191 = 11 * 37 * 47 * 79 = E L O V ; 110618 = 2 * 19 * 41 * 71 = A H M T
The message decodes as WE LOVE MATH , which is pretty obvious by now, after over 100 problems!
c) This was a popular activity, thanks for the 'positive' feedback; you've 'primed' me for more hard 'code' facts.
The best entry (so far!) is COLIC, an acute abdominal pain: 5 * 47 * 37 * 23 * 5 = 999925 (within 75; ouch!)
FFFRY = 999973 (27) is more onomotopaeic than a real word. Next closest (thanks mostly to Phil) include:
JUNE = 1001341 (1341), BUOY = 998421 (1579), MODAL = 998186 (1814),
Also near a million are: SELL = 1008953, POLE = 1013837, AVOID = 1195586, PASTA = 1008484, DON'T = 1004437;
ERASE, WARY, CHIME, FINED; less English are ADIOS = 1013978, JOSE = 1004531, KOPF = 1003873.
You can try to make sentences of just these: (Neon door: "June & Jose sell pasta - adios, colic! Don't erase.)
WINNERS - Problem 104 . keep sending in good words!. (back to top) . leader board
Arthur Morris . . . . . 10 pts - COLIC - a good thing, in this case! (Used MATLAB)
Sue B . . . . . . . . . . . . . 8 pts - My dictionary has COLIC too - should I operate on it?
Phil Sayre . . . . . . . . . 6 pts - 'Python' says: JUNE, MODAL and BUOY are good 'n' close!
Ludwig Deruyck . . . 5 pts - Good words: DOOR = 943923 , NEON = 955923; first entry.
Irina Kolesnik . . . . . 5 pts - Yes, JUNE and BUOY; A-ZINC = 998890 is a vitamin slogan!
Slanky . . . . . . . . . . . . 4 pts - Got COLIC? (f.y.i. 9131511191110618 = 2*17*7643*60631*579569)
Martin Gritch (new) . .3 pts - Welcome to a Mathematica user! (FFFRY sounds like something frying!)
Al B . . . . . . . . . . . . . . 3 pts - Hey Al, I enjoy eating PASTA but I don't SELL or KILL it.
Robert Hussey. . . . . 2 pts - Hello again! AVOID was good; you slightly missed 'E' in BIKER.
Problem #105 - Posted Monday, December 25, 2000
At their traditional end-of-the-millenium poker game, Clifton and Lawrence agree on
the stakes for each hand: The loser pays 1/3 of the money he has remaining, to the winner.
After a while, Lawrence gives up: "You now have exactly three times the cash I have,
you've won the last few hands, and I've lost just about four bucks!"
"But you won every hand before that," Clifton replied, "in fact you've won the same
number of hands I have!" How many hands were played, how much money did they
each start with, and how much do they have now?
Solution: From Arthur Morris - Thanks again!!
Let Lawrence's holding at the end be x pennies, Clifton's at 3x and the length of each streak at N.
At the start of Lawrence's losing streak, he had (1.5 x)^N.
(1.5 = reciprocal of 2/3 = 1 - 1/3 ; Dan)
N<4 or else Lawrence would have had more than their combined holdings.
If "few" >1. then N=2 or N=3. For N=2, Lawrence has 2.25 x at the start of his losing streak,
and has 0.0625 at the start of the game and is, in fact a winner. Thus N=3.
Working backward, Larry had 3.375 x
(= (27/8) x ) at the start of his losing streak, and 1.890625 x
at the start of the game. He lost 0.890625 x ( = (399/448) x ). We must find x such that the loss is an
integer and about 400 pennies ( 448 x = 399 ). In working backward, I used a factor or 1/2 a total of
6 times, so will try some multiple of 2^6. I find x = (7 times 64) = 448 to solve the problem.
They played 6 hands. Clifton started with \$9.45 and ended with \$13.44.
Lawrence
(don't call me Larry) started with \$8.47 and ended with \$4.48.
Dan's Note: I made a table to show how much each player had at each stage, after 0, 1, . . . , 6 hands.
Their total always equals \$17.92, and Lawrence lost \$3.99, about four bucks. Check that 14.48 = 3 * 4.48.
Some of you sent in solutions in which L. had lost exactly 4 dollars, but the 1/3 payoff isn't exact then.
Some close answers: L: 8.49 -> 4.49, C: 9.47 -> 13.47 (Lisa and Phil); Lo: 8.50 (Sue), 8.55 (Al).
After --> 0 hands 1 hand 2 hands 3 hands 4 hands 5 hands 6 hands Clifton 9.45 6.30 4.20 2.80 7.84 11.20 13.44 Lawrence 8.47 11.62 13.72 15.12 10.08 6.72 4.48
WINNERS - Problem 105 . (besides Clifton!) . (back to top) . leader board
Arthur Morris . . . . . 10 pts - I worked through your solution, others can too! (Hints in brown)
Lisa Schechner. . . . . 5 pts - You were a 20th century entry but not exact pennies at each hand.
Phil Sayre . . . . . . . . . 4 pts - Good use of exponentials but I think you rounded to nearest 1c.
Ludwig Deruyck . . . 3 pts - I like your use of tables but Lawrence's loss is over last 3 games.
Al B . . . . . . . . . . . . . . 3 pts - Your total of \$18.20 is too rich for this penny poker game!
Sue B . . . . . . . . . . . . . 3 pts - Very good use of fractions to find divisors of x; it almost worked.
Slanky . . . . . . . . . . . . 2 pts - Yep, some of those n's give those answers but are they all \$4 losses?
Problem #106 - Posted Wednesday, January 3, 2001
After my recent move, I got a new phone number. I forgot to write it down, but it had a
3-digit prefix and the rest was another 4 digits, like this: xxx-xxxx. I did remember that
the prefix, subtracted from half the square of the rest of the number, gave me the whole
phone number as a result ! What was my new number ? (Explain steps fully for best ranking!)
Solution - It's fine to write a program to square all the 4-digit numbers and test 'em out with an exhaustive
search, and better to set up an equation. But let's try a numerical argument instead, for maximum style.
Let's call the number N = abc-defg ; put x = abc and y = defg; then N = 10000 x + y.
The condition is that y^2 - x = N = 10000 x + y , so y^2 - 2 y = 20002 x ; (y - 1)^2 = 20002 x + 1.
Thus y - 1 = +1 or -1 mod 20002 ; since 20002 = 73 * 274 we have y - 1 = + 1 or -1 mod 73 and mod 274.
The only interesting solution is y = 2 mod 73 and y = 0 mod 274 ; y = 2192 by the Chinese Remainder Theorem.
Then (y-1)^2 = 2191^2 = 4800481 = 20002 x + 1 ; x = 240 ; My phone number is 240-2192 (don't call!)
-- Nikita, Sue, and others noticed 000-0000 works; my sources say 000-0002 is OK2. -- ALL OF YOU GOT 240-2192 !!
WINNERS - Problem 106 . . (back to top) . leader board . . A few of you mentioned your 'language of choice'!
Arthur Morris. . . . . . . 10 pts - First in again, good use of equations and MATLAB search for y = 2192.
Ludwig Deruyck . . . . . 7 pts - Got y^2 / 2 - x = 1000x + y ; used BASIC program to search for whole # x.
Lisa Schechner. . . . . . . 6 pts - Good answer involving 137 k where 20002 = 2*73*137. (Bonus pt!)
Sue B . . . . . . . . . . . . . . . 5 pts - Did a great 'guess/check/adjust' approach, backed up by algebra!
Phil Sayre . . . . . . . . . . . 4 pts - Nice approach; b = 1 + sqrt(1 + 20002 a), let PYTHON do the search!
Sudipta Das . . . . . . . . . 4 pts - Excellent answer, good use of primes 10001 = 73 * 137. (Bonus pt!)
Al B . . . . . . . . . . . . . . . . 3 pts - Good to bound it : 1416 < xxxx < 4472 , cuts down the search!
Nikita Kuznetsov (new) 2 pts - Welcome! 000-0000 ok too but you get an annoyed operator! Visual C++
Paul Langford (new) . . . 1 pt - Got your partial answer as a YAHOO greeting card! Good unconventionality!
Problem #107 - Posted Friday, January 12, 2001
a) Given that two of the three sides of a right triangle are 3 and 4, what's the shortest possible
length for the third side? b) If (6,9) and (10,3) are the coordinates of two opposite vertices of
a square, what are the coords of the other two? c) One circle has radius 5 and center at (0, 5).
A second circle has radius 12, center (12,0). Find the radius and center of a third circle which
passes through the center of the 2nd circle and both intersection pts of the first two circles.
Solution: I'm bach after a short absence... This was a three-part problem, so let's bring in three experts!
a) Slanky sez: What fun remembering my Geometry for this weeks problem! Since it's a right triangle, the
sum of the squares of the two shorter sides equals the square of the longest side. If I want the shortest possible
third side, I will assume that 4 is the longest side and solve the equation a^2 + 3^2 = 4^2. a^2 = 7, so
a = square root of 7 =approx= 2.65. (Dan's note: Sqrt[7] is 'the' answer, but just for fun, Mathematica gives us
2.64575131106459059050161575363926042571025918308245 . . .)
b) Sue B (contest winner 1999/2000) sez: Find the midpoint of the two points (6, 9) and (10, 3) using the
midpoint formula. midpoint: x = (6 + 10) / 2 = 8 ; y = (9 + 3) / 2 = 6 ; midpoint = (8 , 6)
Find the slope of the line using the two given points. m = (9 - 3) / (6 - 10) = 6 / (-4) = - 3/2
The line perpendicular to this one has a slope of 2/3. Two points on this second line that are equidistant
from the midpoint are (5, 4) and (11, 8). 8 + 3 = 11 , 6 + 2 = 8 ; 8 - 3 = 5 , 6 - 2 = 4 .
Use the distance formula to check that the distances between all four outside points are the same, verifying
that the object is indeed a square. (Dan's note: The distance from any corner to the center (8, 6) is Sqrt[13].)
c) Phil Sayre sez: c) Let f(x,y,r)=x*x+y*y-r*r and let (x1,y1) be the intersection of f(x,y-5,5)=0 and
f(x-12,y,12)=0. The difference of these two equations yields the relation y=12*x/5, from which we obtain
x1=24*25/13*13, y1=10*12*12/13*13. Three chords of the desired circle are formed by the three points
(0,0), (12,0), and (x1,y1) . The perpendicular bisectors of each chord meet at the center of the circle.
In particular, the x-coordinate, x0=12/2=6. Then from f(x0,y0,r)=0 we obtain y0=sqrt(r*r-36).
From f(x0-x1,y0-y1,r)=0 we obtain, after substitutions for x0, y0,
r^2-36={[x1*(x1-12)+y1*y1]/2*y1}^2=(5/2)^2, hence r^2=42.25, r=36. From this result,
y0=sqrt(42.25-36)=2.5. Summary: radius = 6.5, center = (6, 2.5). (Dan's note: What can I add?!)
WINNERS - Problem 107 . (back to top) . leader board . The center and all intersec's were 'rational points' - Dan
Arthur Morris. . . . . . . 10 pts - The Art Morris juggernaut plows on! First and fastest yet thourough!
Ludwig Deruyck . . . . . 7 pts - You got the (600,169, 1440,169) intersec pt like I did. Good taste!
Phil Sayre . . . . . . . . . . . 5 pts - I'll let our new president check your equations for validiosity.
Lisa Schechner. . . . . . . 4 pts - Correct on all answers, even though rather light on the reasons...
Al B . . . . . . . . . . . . . . . . 4 pts - Wow, a 12-step square process and arctangents of 5/12 and 3/2 !
Sudipta Das . . . . . . . . . 3 pts - I liked your use of fractions and eqns of circles; part a) was a bit off.
Sue B . . . . . . . . . . . . . . . 3 pts - Your certificate for 1999/2000 has been authenticated and is on its way!
Slanky . . . . . . . . . . . . . . 3 pts - How's it feel to have your work quoted on a worldide basis!?
Conrad Tan (new) . . . . . . 2 pts - Yep, go sqrt[52]/2 out from (8, 6) with slope 2/3. That's the spot!
Daria Eiteneer (new) . . . 2 pts - Welcome to the contest, you two! The radius of the circle is smaller.
Problem #108 - Posted Sunday, January 21, 2001 Cover The Cube ! (back to top) The T-shape at the right can cover the six faces of a cube. How many other shapes can you find that cover the cube? Please give your answers as lists of the six squares used. The T-shape in the picture would be called {a5, b2, b3, b4, b5, c5}. Rotations and reflections (flips) don't count as different. Shapes must be connected; squares must touch along a whole edge. Solution: These are the eleven (including the T-shape) that I'm aware of: Arthur sent them all in; Phil provided some personal names; other names and graphics courtesy dansmath.com (me). (Note: This is one of my all-time favorite problems; I've been saving it for you!)
1. "Tee-shape" . . . {a4, b1, b2, b3, b4, c4} (pictured above right) . . . There are ten others:
2. "High F" . . . . . {a4, b1, b2, b3, b4, c3} . . . . 3. "Low F". . . . {a4, b1, b2, b3, b4, c2}
4. "Backwds S" . .
{a4, b1, b2, b3, b4, c1} . . . . 5. "Cross" . . . . {a3, b1, b2, b3, b4, c3}
6. "Saguaro cactus"{a3, b1, b2, b3, b4, c2} . . . .7. "Skinny" . . . {a3, a4, a5, b1, b2, b3}
8. "Scary stairs" . . {a3, a4, b1, b2, b3, c1} . . . . 9. "Saguaro 2" .{a3, a4, b1, b2, b3, c2}
10."High five" . . . {a3, a4, b1, b2, b3, c3} . . . .11. "Staircase" . {a3, a4, b2, b3, c1, c2}
Arthur Morris. . . . . . . 10 pts - Thanks again, allow me to copy and paste your list to save typing!
Sue B . . . . . . . . . . . . . . . 6 pts - Got all 10, but one was an illegal L and another was a repeat.
Lisa Schechner. . . . . . . 5 pts - All yours were good; you just overlooked the staircase one.
Phil Sayre . . . . . . . . . . . 4 pts - Nice names, 9 shapes, original visual system, like chess positions.
Al B . . . . . . . . . . . . . . . . 4 pts - You found 8 of the extra 10; good going, no repeats.
Richard Clarke (new). . . 3 pts - Six shapes, two repeats, welcome to the contest!
Ludwig Deruyck . . . . . 3 pts - Four of your 5 were different; 2 were the same (hey, that makes 6 !?)
Bruce (new) . . . . . . . . . . 2 pts - Three shapes, all good & different ; welcome to the contest!
Edwin Wijono (new) . . . 2 pts - Welcome; also three (other) shapes; keep on entering, all you math-heads!
Problem #109 - Posted Tuesday, January 30, 2001
Based On What?! Here's a fun rhyme I found; see if you can answer it! (back to top)
The square of nine is 121 ; I know it looks quite weird.
But still I say it's really true ; the way we figure here.
And nine times ten is 132 ; the self-same rule, you see.
So whatcha say I'd have to write for five times twenty-three?
Solution(s): More than one of you said the title was a clue. I knew only method (1), for which
Phil S. added his own rhyme (see below) but Lisa, Po, and others showed me another (2), and
Chris North and Sue B convinced me (3) there was yet a third alternative!
(1) Let's explore different bases: Say the base is n. Place values are 1, n, n^2, etc going right to left.
Then "The square of nine is 121" means 9^2 = 81 base 10 = 1*n^2 + 2*n + 1*1. Solving for n ;
n^2 + 2n + 1 = 81 ==> (n + 1)^2 = 81 ==> n + 1 = 9 ==> n = 8. (Ah, octal, from my home country!)
Checking , "nine times ten" is 90 base 10 , but 90 = 1*64 + 3*8 + 2*1 , so it's 132 base 8. This
makes it clear: "five times twenty-three" is 115 base 10 = 1*64 + 6*8 + 3*1 = 163 written in octal.
(2) Lisa suggested that in Joe the Contractor's math (2) you add 2 feet to every dimension, so if you
want a 9 x 9 ft room, it takes up (9+2)(9+2) = 121 sq ft of space. Also 9 x 10 would 'equal up' to
(9+2)(10+2) = 11 * 12 = 132 ; therefore 5 x 23 would be done as 7 * 25 = 175 in Joe's Math.
(3) In (1) we multiplied decimal numbers, then converted to octal; let's convert first then view the
answer as decimal: 9 x 9 -> 11 * 11 = 121 ; 9 x 10 -> 11 * 12 = 132 ; 5 x 23 -> 5 * 27 = 135.
Arthur Morris. . . . . . 10 pts - Yes, it's 'doubly weird' all right, no 9 in base 8, just 'nine' = 11.
Lisa Schechner. . . . . . 8 pts - You got the base 8 answer & were first (bonus pt) to show me 175.
Ludwig Deruyck . . . . 5 pts - I appreciate how you explicitly solved for the base B. 163 rules!
Phil Sayre . . . . . . . . . . 4 pts - Your rhyme is self-explanatory; repeated below for posterity!
Sue B . . . . . . . . . . . . . . 4 pts - Thanks for makin' the octal/decimal vs decimal/octal distinction!
Chris North . . . . . . . . 3 pts - I didn't support your 135 at first but I view it as a ful partner now!
Po-Nien Chen . . . . . . 3 pts - Another supporter of the "add two and multiply" theory. Good!
Al B . . . . . . . . . . . . . . . 3 pts - "Add two to everything" except the final answer; it works!
Sudipta Das . . . . . . . . 3 pts - Good -175- you also solved for the base before proceeding.
Filipe Gonçalves. . . . 2 pts - Nice to have you back; you must have added twos to get the 175.
Tat Tsui (new). . . . . . . . 2 pts - That's the idea; add 2's , but 7 * 23 = 161 is only one two...
Slanky . . . . . . . . . . . . . 1 pt - Sneaked the answer in under the bell ; see my version of (1).
Problem #110 - Posted Saturday, February 10, 2001
"Hey, nice looking family!" said Fred, seeing the photo. "I met your younger boy today;
he said he was nine, and your wife reminded me his brother is older."
Frank agreed, saying, "It's odd about all our ages. If you total the squares of my age and
the boys' ages, you get my wife's age times the total of my age and the boys' ages."
If the ages are all whole numbers, can you figure them out?
Solution: From your entries I have thirteen answers, so Lisa's "No I can't and neither can you" is a winner.
Some of you did searches for ages (F, M, S) using Excel, Basic, Python, etc. I feel that mathematically the
only restrictions should be F, M, S >= 9 ; how old do parents have to be? Older could mean a few days...
Ludwig noticed the words 'odd about all our ages' and featured the all-odd solution (49, 37, 21).
Here's a list of submitted answers; Mo is mother's age at birth of oldest son. My source had only answer j).
Red is biologically dubious, brown is outside some 'average age' assumptions of Mo, purp older bro = 9.
a) (27, 21, 18) Mo=3 ; b) (29, 29, 14) Mo=15 ; c) (36, 27, 9) Mo=18 ; d) (36, 27, 18) Mo=9 ;
e) (39, 29, 14) Mo=15 ; f) (39, 29, 15) Mo=14 ; g) (49, 37, 16) Mo=19 ; h) (49, 37, 21) Mo=16 ;
i) (53, 41, 12) Mo=29 ; j) (54, 41, 18) Mo=23 ; k) (54, 41, 23) Mo=18 ; l) (63, 51, 9) Mo=42 ;
m) (64, 49, 24) Mo=25 ; n) (64, 49, 25) Mo=25 ; o) (66, 53, 11) Mo=42 ; p) (74, 59, 15) Mo=44 ;
q) (90, 75, 12) Mo = 63 (!)
(Dan's note: Some others I've found are (16, 13, 12), (18, 15, 15), (23, 17, 11), (25, 19, 15), (45, 39, 39),
(63, 51, 42), (72, 65, 65), (66, 57, 54), (81, 73, 72), (92, 83, 81), (99, 79, 54), etc.
Don't gimme flack about a zero-year-old mother, please; it's a math problem... My 'AppleWorks' spreadsheet for
M = (F^2+S^2+9^2)/(F+S+9) gave whole numbers M at a whopping 49 places for the domain 9 <= F,S<= 100 ;
who's to say how old these people are? We do assume F,M>=S>=9; this cuts it down to 45 'legal' answers.)
Lisa Schechner. . . . . 10 pts - Yep, answer the question that's asked! Two examples prove it's 'no'.
Slanky. . . . . . . . . . . . . 6 pts - Good use of algebra but soln i) not unique; your x < 60 missed (66,53,11)
Ludwig Deruyck . . . . 5 pts - Found four solns, featuring the one with the 11-year old. Good 'odd one'!
Arthur Morris . . . . . . 5 pts - Submitted ten answers assuming M,F>=S+18, M<60, F<75
Al B . . . . . . . . . . . . . . . 4 pts - Four answers, including the "biologically difficult" (90, 75, 12).
Phil Sayre . . . . . . . . . . 4 pts - Found the most (13) answers with 27<=F<=75, M>=S+18, 10<=S<=25.
Sudipta Das . . . . . . . . 3 pts - The answer you got, (39, 29, 14), was a correct (and popular) one.
Sue B . . . . . . . . . . . . . . 3 pts - Nine ways of solving your Diophantine equation, 'illogicals' are ok!
THANKS to all of you who have entered, or even just clicked and looked. My site is now in its fifth season - OVER 25,000 HITS so far! (Not factorial.) Help it grow by telling your friends, teachers, and family about it. YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 3*23*29 A.D.
Problem Archives Index
Probs & answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90
Probs & answers . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+
Problems only . . . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+
Browse the complete problem list, check out the weekly leader board,
| 9,428 | 27,711 |
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# diameter
## Definitions
• LARVA OF EPHEMERA, OR MAY-FLY, MAGNIFIED TWO DIAMETERS
• WordNet 3.6
• n diameter the length of a straight line passing through the center of a circle and connecting two points on the circumference
• n diameter a straight line connecting the center of a circle with two points on its perimeter (or the center of a sphere with two points on its surface)
• ***
Webster's Revised Unabridged Dictionary
• Interesting fact: It would take approximately twenty-four trees that are on average six to eight inches in diameter to produce one ton of newsprint for the Sunday edition of the New York Times
• Diameter (Geom) A diametral plane.
• Diameter (Geom) Any right line passing through the center of a figure or body, as a circle, conic section, sphere, cube, etc., and terminated by the opposite boundaries; a straight line which bisects a system of parallel chords drawn in a curve.
• Diameter (Arch) The distance through the lower part of the shaft of a column, used as a standard measure for all parts of the order. See Module.
• Diameter The length of a straight line through the center of an object from side to side; width; thickness; as, the diameter of a tree or rock.
• ***
Century Dictionary and Cyclopedia
• Interesting fact: A maple tree is usually tapped when the tree is at least 45 years old and has a diameter of 12 inches
• n diameter In geometry, a chord of a circle or a sphere which passes through its center; in general
• n diameter a chord of a conic cutting it at points tangents to which are parallel;
• n diameter a line intersecting a quadric surface at points where the tangent planes are parallel. The conception was extended by Newton to other algebraic curves by means of the following theorem: If on each of a system of parallel chords of a curve of the nth order there be taken the center of mean distances of the n points where the chord meets the curve, the locus of this center is a straight line, which may be called a diameter or the curve.
• n diameter The length of a diameter; the thickness of a cylindrical or spherical body as measured, in the former case on a diameter of a cross-section made perpendicular to the axis, and in the latter on a line passing through the center: as, a tree two feet in diameter; a ball three inches in diameter. In architecture, the diameter of the lower face of the shaft of a column, divided into 60 parts, forms a scale by which all the parts of a classical order are commonly measured. The 60th part of the diameter is called a minute, and 30 minutes make a module.
• n diameter The diameter (see def. 2) of the object observed, taken as a convenient measure of linear magnification used in micros-copy and in telescopic work.
• ***
Chambers's Twentieth Century Dictionary
• Interesting fact: An adult esophagus can range from 10 to 14 inches in length and is one inch in diameter
• n Diameter dī-am′e-tėr the measure through or across: a straight line passing through the centre of a circle or other figure, terminated at both ends by the circumference
• ***
## Quotations
• Samuel Johnson
“You teach your daughters the diameters of the planets and wonder when you are done that they do not delight in your company.”
## Etymology
Webster's Revised Unabridged Dictionary
F. diamètre, L. diametros, fr. Gr. ; dia` through + measure. See Meter
Chambers's Twentieth Century Dictionary
Through Fr. and L. from Gr. diametrosdia, through, metrein, to measure.
## Usage
### In literature:
They were composed of tubes, each about two feet long and eighteen inches in diameter.
"A History of Art in Chaldæa & Assyria, v. 1" by Georges Perrot
At length, in 1839, two specula, each three feet in diameter, were turned out in such perfection as to prompt a still bolder experiment.
"A Popular History of Astronomy During the Nineteenth Century" by Agnes M. (Agnes Mary) Clerke
Therefore let the diameter of the Sun be to the diameter of Venus as 30' to 1' 12''.
"The Astronomy of Milton's 'Paradise Lost'" by Thomas Orchard
Connect opposite points of the outer circle by drawing two more diameters.
"Construction Work for Rural and Elementary Schools" by Virginia McGaw
The actual diameter of the moon is about 2163 miles, which is somewhat more than one-quarter the diameter of the earth.
"Astronomy of To-day" by Cecil G. Dolmage
But in reckoning magnifying power, diameter, not area, is always considered.
"Pleasures of the telescope" by Garrett Serviss
First a block of wood is cut about the same diameter as the internal diameter of the stack.
"Boys' Book of Model Boats" by Raymond Francis Yates
The spores globose, 8-10u in diameter.
"The Mushroom, Edible and Otherwise" by M. E. Hard
They are about one foot in length and may be had from two to six inches inside diameter.
"Rural Hygiene" by Henry N. Ogden
The diameter of the semicircle is about 18 ft., and it is floored with mosaic.
"The Shores of the Adriatic" by F. Hamilton Jackson
The oval grave pits were from 4-1/2 to 6 feet deep and from 3 to 4 feet in greatest diameter.
"Ancient art of the province of Chiriqui, Colombia" by William Henry Holmes
As the diameter of the mat increases, the less often is it necessary to add.
"Philippine Mats" by Hugo H. Miller
Other things being equal, a bulb is valuable according to its vertical diameter.
"The Gladiolus" by Matthew Crawford
Masts of 35 inches diameter, 35 yards long, L110.
"Glimpses of the Past" by W. O. Raymond
Total length of animal 1.3; diameter of peduncle .4 of an inch.
"A Monograph on the Sub-class Cirripedia (Volume 1 of 2)" by Charles Darwin
Trees that I set out in the spring of 1885 measure six to ten inches in diameter.
"The Apple" by Various
In size they range from less than one inch in diameter and depth to more than twenty inches in diameter and a foot in depth.
"Ancient Pottery of the Mississippi Valley" by William H. Holmes
The largest Douglas firs are sometimes over 400 years old and 60 to 70 inches in diameter.
"The Forests of Mount Rainier National Park" by Grenville F. Allen
Cut a circle six to six and a half inches in diameter, using as guide a pie-tin or cardboard, if a regular cutter is not at hand.
"The Century Cook Book" by Mary Ronald
Can the same method be used to find the diameter of the sun?
"A Text-Book of Astronomy" by George C. Comstock
***
### In news:
Why is the diameter size of a pressure safety valve outlet line bigger than the inlet line.
The intersection of Sand Hill Road and Sharon Park Drive in Menlo Park was closed for several hours Aug 28 as crews repaired a large hole — about 12 feet in diameter — that opened up during a construction mishap.
Woll has introduced a Diamond Plus Induction 8-inch diameter sauté pan.
This end mill from Performance Micro Tool is 0.0004 inch in diameter.
In 2006, the company added a Kason recycle screening deck atop the existing screener , boosting its capacity from 100-150 gpm to 225-250 gpm without increasing its diameter.
Diameter Oscillation of Axonemes in Sea-Urchin Sperm Flagell.
It's a permanent seal and takes care of holes up to 6mm in diameter, which is probably a larger area than you think.
The diameter and position of the shackle prevent unauthorized entry from pulling the cylinder, core or plug.
Look past the OEM shift boot or cover that surrounds the main stick and check the diameter.
An optical instrument that scans and measures openings and wire diameters, the system supports sieves up to 18 in.
) in diameter and 100 mm long, in 14 hours in the case of C- or A-plane sapphire.
Specimens of plastic pipe and tubing with outside diameters greater than 4 in (101.6 mm) shall be mounted in accordance with Section 6.2.
(762-mm) diameter, a 220-grit square-edge diamond wheel, an adjustable squeegee -clamping system, a spring-steel platform sliding of squeegees for uniform sharpening, vacuum removal system, and more.
Kore expands large diameter handlebar offerings.
diameter in a variety of bright and happy glazes.
***
### In science:
The growth of patterns, i.e. vortices and density clusters, is controlled by diffusive modes , and typical diameters, Lv (τ ) of these patterns grow as √τ .
Asymptotic Energy Decay in Inelastic Fluids
In order to apply the Arzela-Ascoli theorem, it is enough to show that the diameters of the images of the mi are uniformly bounded.
A norm on homology of surfaces and counting simple geodesics
Since the lengths of mi approach lh , it follows that the diameters of all the components of mi , for i sufficiently great, are uniformly bounded by 2lh .
A norm on homology of surfaces and counting simple geodesics
Trumpler based this on his study of open clusters in which he compared the luminosities and distances of open clusters with the distances obtained by assuming that all their diameters were the same.
Cosmic Dust in the 21st Century
CMS solenoid magnet (inner diameter of 5.9 m).
Studies of the Response of the Prototype CMS Hadron Calorimeter, Including Magnetic Field Effects, to Pion, Electron, and Muon Beams
This structure has, however, the disadvantage of its size (outer diameter 430 mm). Since some structures at the upstream end will have to fit into focusing solenoids, we are now also studying a slotted iris structure with an outer diameter of only 174 mm as an alternative.
Design of a 3 GHz Accelerator Structure for the CLIC Test Facility (CTF 3) Drive Beam
In order for this filter to work effectively, substantial waveguide length is required. This lead to an outer diameter of the 3 GHz TDS of approximately 430 mm.
Design of a 3 GHz Accelerator Structure for the CLIC Test Facility (CTF 3) Drive Beam
By fitting the liquid and gas densities to the law of rectilinear diameters, we extrapolate the critical point at Tc = 0.418 and ρc = 0.134.
Phase Behavior of a Simple Model for Membrane Proteins
By fitting ρgas and ρliq to the law of rectilinear diameters, we extrapolate the critical point at βc = 2.392 and ρc = 0.134.
Phase Behavior of a Simple Model for Membrane Proteins
By the isoperimetric inequality for the square in the lattice Z2 (see ), A′ has l1 diameter at least cI √q tn for some constant cI > 0.
On the mixing time of simple random walk on the super critical percolation cluster
PP), mats of irregular PP fibres with various fiber diameters (between 15 and 33 µm) and foams of different material type: PP, polyethylene (PE) and Rohacell (RC).
Prototype tests for the ALICE TRD
Average pulse height as function of the drift time for pions and electrons for a radiator of 17 µm diameter fibres at the momentum of 1 GeV/c.
Prototype tests for the ALICE TRD
We show in Fig. 5 the drift time distribution of the average pulse height summed over the adjacent pads, hP H i, for pions and electrons in case of a fibre radiator with 17 µm fibre diameter (X=0.3 g/cm2).
Prototype tests for the ALICE TRD
In a separate study we have found that the fibre diameter influences very little the TR yield.
Prototype tests for the ALICE TRD
Radiators with fibres with 17 and 33 µm diameter show similar TR performance for the same density and thickness (see Fig. 7 below).
Prototype tests for the ALICE TRD
***
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# Sport Science: Table Tennis
http:// www.youtube.com/watch?v=rzz-NxGpQh4. Sport Science: Table Tennis . Chapter 7 Circular Motion and Gravitation. Why it matters
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## Sport Science: Table Tennis
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1. http://www.youtube.com/watch?v=rzz-NxGpQh4 Sport Science: Table Tennis
2. Chapter 7Circular Motion and Gravitation Why it matters Circular motion is present all around you – from a rotating Ferris wheel in an amusement park, to a space shuttle orbiting Earth, to Earth revolving around the sun.
3. Tangential Speed • Uniform Circular Motion • Centripetal Acceleration • Tangential Acceleration • Centripetal Force • Centrifugal “Force” CH7 – Sec 1 Circular Motion Vocabulary
4. Hammer Throw (King of Kings) - YouTube Hammer Throw – YouTubeCircular Motion Video – YouTube
5. The tangential speed (vt) of an object in circular motion is the object’s speed along an imaginary line drawn tangent to the circular path. • Tangential speed depends on the distance from the object to the center of the circular path. • When the tangential speed is constant, the motion is described as uniform circular motion. Tangential Speed
6. The acceleration of an object moving in a circular path and at constant speed is due to a change in direction. • An acceleration of this nature is called a centripetal acceleration. Centripetal Acceleration
7. (a) As the particle moves from A to B, the direction of the particle’s velocity vector changes. • (b) For short time intervals, ∆v is directed toward the center of the circle. • Centripetal acceleration is always directed toward the center of a circle. Centripetal Acceleration cont.
8. In circular motion, an acceleration due to a change in speed is called tangential acceleration. • To understand the difference between centripetal and tangential acceleration, consider a car traveling in a circular track. • Because the car is moving in a circle, the car has a centripetal component of acceleration. • If the car’s speed changes, the car also has a tangential component of acceleration. Centripetal Acceleration cont.
9. A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is the car’s tangential speed? • Is this an example of uniform circular motion? Why or why not? Practice Problem
10. Consider a ball of mass m that is being whirled in a horizontal circular path of radius r with constant speed. • The force exerted by the string has horizontal and vertical components. The vertical component is equal and opposite to the gravitational force. Thus, the horizontal component is the net force. • This net force, which is directed toward the center of the circle, is a centripetal force. Centripetal Force
11. Newton’s 2nd Law can be combined with the equation for centripetal acceleration to derive an equation for centripetal force: Centripetal Force cont.
12. If the centripetal force vanishes, the object stops moving in a circular path. • A ball that is on the end of a string is whirled in a vertical circular path. • If the string breaks at the position shown in (a), the ball will move vertically upward in free fall. • If the string breaks at the top of the ball’s path, as in (b), the ball will move along a parabolic path. Centripetal Force cont.
13. Consider a car traveling at high speed and approaching an exit ramp that curves to the left. As the driver makes the sharp left turn, the passenger slides to the right and hits the door. • What causes the passenger to move toward the door? Describing a Rotating System
14. As the car enters the ramp and travels along a curved path, the passenger, because of inertia, tends to move along the original straight path. • If a sufficiently large centripetal force acts on the passenger, the person will move along the same curved path that the car does. The origin of the centripetal force is the force of friction between the passenger and the car seat. • If this frictional force is not sufficient, the passenger slides across the seat as the car turns. Describing a Rotating System cont.
15. A pilot is flying a small plane at 56.6 m/s in a circular path with a radius of 188.5 m. The centripetal force needed to maintain the plane’s circular motion is 1.89 x 104 N. What is the plane’s mass? Practice Problem
16. Earth and many other planets in our solar system travel in nearly circular orbits around the sun. Thus, a centripetal force must keep them in orbit. One of Isaac Newton’s great achievements was the realization that the centripetal force that holds the planets in orbit is the very same force that pulls an apple toward the ground – GRAVITATIONAL FORCE. SEC2 - Newton’s Law of Universal Gravitation
17. Orbiting objects are in free fall. • To see how this idea is true, we can use a thought experiment that Newton developed. Consider a cannon sitting on a high mountaintop. Each successive cannonball has a greater initial speed, so the horizontal distance that the ball travels increases. If the initial speed is great enough, the curvature of Earth will cause the cannonball to continue falling without ever landing. Gravitational Force
18. Gravitational Force is the mutual force of attraction between particles of matter • Gravitational Force depends on the masses and the distances of those particles of matter • Gravitational force acts between all masses, regardless of size. Gravitational Force
19. G is called the Constant of Universal Gravitation G = 6.673 x 10-11 N∙m2/kg2 Note: When calculating the gravitational force between Earth and our sun, use the distance between their centers. Newton’s Law of Universal Gravitation
20. The gravitational forces that two masses exert on each other are always equal in magnitude and opposite in direction. • This is an example of Newton’s third law of motion. • One example is the Earth-moon system, shown on the next slide. • As a result of these forces, the moon and Earth each orbit the center of mass of the Earth-moon system. Because Earth has a much greater mass than the moon, this center of mass lies within Earth. Gravitational Force cont.
21. Newton’s Law of Universal Gravitation
22. Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force between them is 8.92 x 10-11 N. Practice Problem
23. Gravity is a field force (as opposed to a contact force) • Earth’s gravitational field can be described by the gravitational field strength (g). • The value of (g) is equal to the magnitude of the gravitational force exerted on a unit mass at that point: g=Fg/m • The gravitational field (g) is a vector quantity that points in the direction of the gravitational force. Law of Gravitation cont.
24. Free-fall acceleration at any location (ag) is always equal to the gravitational field strength (g) at that location. • On Earth, ag = g = 9.81 m/s2 Note: Although gravitational field strength (g) and free-fall acceleration (ag) are equivalent, they are not the same thing. For instance, when you hang an object from a spring scale, you are measuring gravitational field strength. Because the mass is at rest, there is no measurable acceleration. Gravitational field strength equals free-fall acceleration
25. Weight is the magnitude of the force due to gravity. w=m∙g • Weight equals mass times gravitational field strength. • Your weight changes with your location in the universe. • As your distance from Earth’s center increases, the value of g decreases, so your weight also decreases. Weight changes with location
26. Selling Gold: A scam artist hopes to make a profit by buying and selling gold at different altitudes for the same price per weight. Should the scam artist buy or sell at the higher altitude? Explain. Conceptual Challenge
27. Inertial mass – property of an object to resist acceleration • Gravitational mass – how objects attract one another How do we know that gravitational mass is equal to inertial mass? The fact that the acceleration of objects in free fall is always the same confirms that the two types of masses are equal. A more massive object experiences a greater gravitational force, but the object resists acceleration by just that same amount. For this reason, all masses fall with the same acceleration (disregarding air resistance, of course). Gravitational Mass equals Inertial Mass
28. Turn in for 5 Bonus Points!! • What is a Black Hole? • What causes Ocean Tides? (Explain in no less than 5 sentences) BONUS POINTS
29. OBJECTIVES: • Kepler’s 3 Laws of Planetary Motion • Solve problem involving orbital speed and period • Compare Weight and Apparent Weightlessness CH7 Sec 3 – Motion in Space
30. Kepler’s laws describe the motion of the planets. • First Law: Each planet travels in an elliptical orbit around the sun, and the sun is at one of the focal points. • Second Law: An imaginary line drawn from the sun to any planet sweeps out equal areas in equal time intervals. • Third Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun. Kepler’s 3 Laws of Plenetary Motion
31. Kepler’s 1st Law Illustrated
32. According to Kepler’s2nd Law, if the time a planet takes to travel the arc on the left (∆t1) is equal to the time the planet takes to cover the arc on the right (∆t2), then the area A1is equal to the area A2. Thus, the planet travels faster when it is closer to the sun and slower when it is fartheraway. Kepler’s 2nd Law Illustrated
33. Kepler’s 3rd Law states that T2r3 • T – Orbital Period is the time a planet takes to finish one full revolution • r – mean distance between the planet and the sun • So, Kepler’s third law can also be stated as follows: • The constant of proportionality is 4p2/Gm, where m is the mass of the object being orbited. Gis called the Constant of Universal Gravitation G = 6.673 x 10-11 N∙m2/kg2 Kepler’s 3rd Law
34. Kepler’s3rd Law leads to an equation for the period of an object in a circular orbit. The speed of an object in a circular orbit depends on the same factors: • Note that m is the mass of the central object that is being orbited. The mass of the planet or satellite that is in orbit does not affect its speed or period. Kepler’s 3rd Law cont.
35. Period and Speed of an Orbiting Object Magellan was the first planetary spacecraft to be launched from a space shuttle. During the spacecraft’s fifth orbit around Venus, Magellan traveled at a mean altitude of 361km. If the orbit had been circular, what would Magellan’s period and speed have been? Practice Problem
36. To learn about apparent weightlessness, imagine that you are in an elevator: • When the elevator is at rest, the magnitude of the normal force acting on you equals your weight. • If the elevator were to accelerate downward at 9.81 m/s2, you and the elevator would both be in free fall. You have the same weight, but there is no normal force acting on you. • This situation is called apparent weightlessness. • Astronauts in orbit experience apparent weightlessness. Weight and Weightlessness
37. Weight and Weightlessness
38. Objectives: • Examine the motion of a rotating rigid object. • Distinguishbetween torque and force. • Calculate the magnitude of a torque on an object. • Identify the six types of simple machines. • Calculate the mechanical advantage and efficiency of a simple machine. CH7 Sec 4 – Torque and Simple Machines
39. Rotational and translational motion can be analyzed separately. • For example, when a bowling ball strikes the pins, the pins may spin in the air as they fly backward. • These pins have both rotational and translational motion. • In this section, we will isolate rotational motion. • In particular, we will explore how to measure the ability of a force to rotate an object. Rotational Motion
40. Torque isa quantity that measures the ability of a force to rotate an object around some axis. • How easily an object rotates depends on two things: • how much force is applied (perpendicular to Lever Arm) – Force • wherethe force is applied – Lever Arm • Lever Arm - The perpendicular distance from the axis of rotation to a line drawn along the direction of the force is equal to d sin q. t = Fdsin q torque = force lever arm The S.I. Unit of Torque is the N·m Torque is a vector quantity The Magnitude of a Torque
41. The applied force may act at an angle. • However, the direction of the lever arm (d sin q) is always perpendicular to the direction of the applied force, as shown here. • d – distance to the axis • F – force • θ – angle between the force & axis • d sin θ– lever arm Magnitude of a Torque
42. In each example, the cat is pushing on the door at the same distance from the axis. To produce the same torque, the cat must apply greater force for smaller angles. Torque and the Lever Arm
43. A machine is any device that transmits or modifies force, usually by changing the force applied to an object. • All machines are combinations or modifications of six fundamental types of machines, called simple machines. • These six simple machines are the lever, pulley, inclined plane, wheel and axle, wedge, and screw Simple Machines
44. Simple Machines
45. Because the purpose of a simple machine is to change the direction or magnitude of an input force, a useful way of characterizing a simple machine is to compare the output and input force. • This ratio is called mechanical advantage. • If friction is disregarded, mechanical advantage can also be expressed in terms of input and output distance. Simple Machines cont.
46. The diagrams show two examples of a trunk being loaded onto a truck. • In the first example, a force (F1) of 360 N moves the trunk through a distance (d1) of 1.0 m. This requires 360 N•m of work. • In the second example, a lesser force (F2) of only 120 N would be needed (ignoring friction), but the trunk must be pushed a greater distance (d2) of 3.0 m. This also requires 360 N•m of work. Simple Machines cont.
47. The simple machines we have considered so far are ideal, frictionless machines. • Real machines, however, are not frictionless. Some of the input energy is dissipated as sound or heat. • The efficiency of a machine is the ratio of useful work output to work input. (Remember W = F·d) • The efficiency of an ideal (frictionless) machine is 1, or 100 percent. • The efficiency of real machines is always less than 1. Simple Machines cont.
48. The efficiency of a squeaky pulley system is 73%. The pulleys are used to raise a mass to a certain height. What force is exerted on the machine if a rope is pulled 18.0 m in order to raise a 58 kg mass a height of 3.0 m? Sample Problem
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# COMPOUND INTEREST WORD PROBLEMS WORKSHEET
## About "Compound Interest Word Problems Worksheet"
Compound Interest Word Problems Worksheet :
Worksheet given in this section is much useful to the students who would like to practice solving word problems on compound interest.
The formula given below can be used to find accumulated value in compound interest.
## Compound Interest Word Problems Worksheet - Problems
Problem 1 :
Mr. George invests \$800 in an account which pays 20% compound interest per year. If interest is compounded half yearly, find the the accumulated value and compound interest after 2 years.
Problem 2 :
A person invests certain amount in compound interest scheme. If the money invested is doubled itself in 3 years, how long will it take for the money to become 4 times of itself ?
Problem 3 :
The compound interest and simple interest on a certain sum for 2 years is \$1230 and \$1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principle ?
Problem 4 :
Mr. David borrowed \$15,000 at 12% per year compounded annually. He repaid \$7000 at the end of 1st year. What amount should he pay at the end of second year to completely discharge the load ?
Problem 5 :
There is 50% increase in an amount deposited in simple interest for 5 years. What will be the compound interest of \$10,000 after 3 years at the same rate ?
## Compound Interest Word Problems Worksheet - Solutions
Problem 1 :
Mr. George invests \$800 in an account which pays 20% compound interest per year. If interest is compounded half yearly, find the the accumulated value and compound interest after 2 years.
Solution :
The formula to find accumulated value in compound interest is
Given : P = 800, r = 20% or 0.2, n = 2 and t = 2.
Then, we have
r /n = 0.2 / 2 = 0.1
nt = 2 ⋅ 2 = 4
Finding accumulated value :
A = 800(1 + 0.1)
A = 800(1.1)
A = 800 x 1.4641
A = \$1171.28
Finding compound interest :
C. I = Accumulated value - Principal
C. I = 1171.28 - 800
C. I = 371.28
Hence, the accumulated value is \$1171.28 and the compound interest is \$371.28.
Let us look at the next problem on "Compound interest word problems worksheet".
Problem 2 :
A person invests certain amount in compound interest scheme. If the money invested is doubled itself in 3 years, how long will it take for the money to become 4 times of itself ?
Solution :
Let "P" be the amount invested initially.
It is given that the money is doubled in 3 years.
That is, P becomes 2P in 3 years.
Because the investment is in compound interest, the principal in the 4th year will be 2P.
And 2P becomes 4P (it doubles itself) in the next 3 years.
So, at the end of 6 years accumulated value will be 4P.
Hence, the amount deposited will become 4 times of itself in 6 years.
Let us look at the next problem on "Compound interest word problems worksheet".
Problem 3 :
The compound interest and simple interest on a certain sum for 2 years is \$1230 and \$1200 respectively. The rate of interest is same for both compound interest and simple interest and it is compounded annually. What is the principle ?
Solution :
Given : Simple interest for two years is \$1200.
So, simple interest for one year is \$600.
Given : Compound interest for two years is \$1230.
Fact : When it is compounded annually, interest earned in both compound interest and simple interest for one year on the same principal would be same.
So, compound interest for 1st year is \$600 and for 2nd year is \$630.
When we compare the compound interest for 1st year and 2nd year, it is clear that the interest earned in 2nd year is \$30 more than the first year.
Because, \$600 interest earned in the first year earned this additional \$30 interest in the second year.
Because it is compounded annually, it can be considered as simple interest for one year.
That is,
Principal = 600, Interest = 30
The formula to find simple interest is
I = P ⋅ ⋅ t
Plug I = 30, P = 600 and t = 1.
30 = 600 ⋅ r ⋅ 1
30 = 600r
Divide both sides by 600.
30 / 600 = r
0.05 = r
5% = r
Given : The rate of interest is same for both compound interest.
So, the rate of interest in simple interest is 5%.
In the given problem, simple interest earned in two years is 1200.
I = P ⋅ ⋅ t
Plug I = 1200, r = 5% or 0.05 and t = 2.
1200 = P ⋅ 0.05 2
1200 = P 0.1
Divide both sides by 0.1
1200 / 0.1 = P
12000 = P
Hence, the principal is \$12,000.
Let us look at the next problem on "Compound interest word problems worksheet".
Problem 4 :
Mr. David borrowed \$15,000 at 12% per year compounded annually. He repaid \$7000 at the end of 1st year. What amount should he pay at the end of second year to completely discharge the load ?
Solution :
The formula to find accumulated value in C.I is
To find the accumulated value for the first year,
Plug P = 15000, r = 12 % or 0.12, n = 1 in the above formula
A = 15000(1 + 0.12)1
A = 15000(1.12)
A = 16800
Given : Amount paid at the end of 1st year is \$7000.
So, the balance to be repaid is
= 16800 - 7000
= 9800
This \$9800 is going to be the principal for the 2nd year.
Now we need to calculate the accumulated value for the principal 9800 in one year.
That is, the amount to be paid at the end of 2nd year to completely discharge the loan.
A = 9800(1 + 0.12)1
A = 9800 ⋅ 1.12
A = 10976
Hence, he has to pay \$10,976 at the end of the second year to completely discharge the loan.
Let us look at the next problem on "Compound interest word problems worksheet".
Problem 5 :
There is 50% increase in an amount deposited in simple interest for 5 years. What will be the compound interest of \$10,000 after 3 years at the same rate ?
Solution :
Given : There is 50% increase in an amount in 5 years at simple interest.
Then , the increase in 1 year is
= 50% / 5
= 10%
The amount deposited in simple interest is increased by 10% in 1 year.
So, the rate of interest in simple interest is 10%.
Given : The rate of interest is same for both compound interest and simple interest
Because the rate of interest is same for both compound interest and simple interest, we can use the rate of interest 10% in compound interest.
To know compound interest for 3 years, plug
P = 12000, r = 0.1, n = 1 and t = 3
in the formula of compound interest.
So, we have
A = 10000(1 + 0.1)3
A = 10000(1.1)3
A = 10000 ⋅ 1.331
A = 13,310
Compound interest for 3 years :
C.I = A - P
C.I = 13,310 - 10000
C.I = 3,310
Hence the compound interest after 3 years at the same rate of interest is \$3310.
After having gone through the stuff given above, we hope that the students would have understood "Compound interest word problems worksheet".
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# 19.3: Work and Heat are not State Functions
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## Heat and work are path functions
Heat ($$q$$) and work ($$w$$) are path functions, not state functions:
1. They are path dependent.
2. They are energy transfer → they are not intrinsic to the system.
##### Path Functons
Functions that depend on the path taken, such as work ($$w$$) and heat ($$q$$), are referred to as path functions.
## Reversible versus irreversible
Let's consider a piston that is being compressed at constant temperature (isothermal) to half of its initial volume:
1. Start with cylinder 1 liter, both external and internal pressure 1 bar.
2. Peg the piston in a fixed position.
3. Put cylinder in a pressure chamber with $$P_{ext}=2$$ bar.
4. Suddenly pull the peg.
The piston will shoot down till the internal and external pressures balance out again and the volume is 1/2 L. Notice that the external pressure was maintained constant at 2 bar during the peg-pulling and that the internal and external pressures were not balanced at all time. In a $$P-V$$ diagram of an ideal gas, $$P$$ is a hyperbolic function of $$V$$ under constant temperature (isothermal), but this refers to the internal pressure of the gas. It is the external one that counts when computing work and they are not necessarily the same. As long as $$P_{external}$$ is constant, work is represented by a rectangle.
The amount of work being done is equal to the shaded region and in equation:
$w=-\int^{V_2}_{V_1}{PdV}=-P_{ext}(V_2-V_1)=-P\Delta V \nonumber$
This represents the maximum amount of work that can be done for an isothermal compression. Work is being done on the system, so the overall work being done is positive. Let's repeat the experiment, but this time the piston will compress reversibly over infinitesimally small steps where the $$P_{ext}=P_{system}$$:
For an ideal gas, the amount of work being done along the reversible compression is:
$w=-\int^{V_2}_{V_1}{PdV}=-nRT\int^{V_2}_{V_1}{\frac{1}{V}}=-nRT\ln{\left(\frac{V_2}{V_1}\right)} \nonumber$
The amount of work being done to the two systems are not the same in the two diagrams (see the gray areas). Work is not a state, but a path function, as it depends on the path taken. You may say, what's the big difference. In both cases, the system is compressed from state 1 to state 2. The key is the word suddenly. By pegging the position in place for the first compression, we have created a situation where the external pressure is higher than the internal pressure ($$P_{ext}>P$$). Because work is done suddenly by pulling the peg out, the internal pressure is struggling to catch up with the external one. During the second compression, we have $$P_{ext}=P$$ at all times. It's a bit like falling off a cliff versus gently sliding down a hill. Path one is called an irreversible path, the second a reversible path.
##### Reversible vs. Irreversible Processes
A reversible path is a path that follows a series of states at rest (i.e., the forces are allowed to balance at all times). In an irreversible one the forces only balance at the very end of the process.
Notice that less work is being done on the reversible isothermal compression than the one-step irreversible isothermal compression. In fact, the minimum amount of work that can be done during a compression always occurs along the reversible path.
### Isothermal Expansion
Let's consider a piston that is being expanded at constant temperature (isothermal) to twice of its initial volume:
1. Start with cylinder 1 liter in a pressure chamber with both an external and internal pressure of 2 bar.
2. Peg the piston in a fixed position.
3. Take the cylinder out of the pressure chamber with Pext= 1 bar.
4. Suddenly pull the peg.
The piston will shoot up till the internal and external pressures balance out again and the volume is 2 L. Notice that the external pressure was maintained constant at 1 bar during the peg-pulling and that the internal and external pressures were not balanced at all time.
The amount of irreversible work being done is again equal to the shaded region and the equation:
$w=-P\Delta V=-P_{ext}(V_2-V_1)=-P\Delta V \nonumber$
This represents the minimum amount of work that can be done for an isothermal expansion. Work is being done on the system, so the overall work being done is negative. Let's repeat the experiment, but this time the piston will compress reversibly over infinitesimally small steps where the $$P_{ext}=P_{system}$$:
Notice that not only is more work is being done than the one-step irreversible isothermal expansion, but it is the same amount of work being done as the reversible isothermal compression. This is the maximum amount of work that can be done during an expansion.
19.3: Work and Heat are not State Functions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
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# 11.1 Systems of linear equations: two variables (Page 5/20)
Page 5 / 20
Solve the system of equations by addition.
$\begin{array}{c}2x+3y=8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x+5y=10\end{array}$
$\left(10,-4\right)$
## Identifying inconsistent systems of equations containing two variables
Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different $\text{\hspace{0.17em}}y$ -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as $\text{\hspace{0.17em}}12=0.$
## Solving an inconsistent system of equations
Solve the following system of equations.
We can approach this problem in two ways. Because one equation is already solved for $\text{\hspace{0.17em}}x,$ the most obvious step is to use substitution.
$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+2y=13\hfill \\ \text{\hspace{0.17em}}\left(9-2y\right)+2y=13\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9+0y=13\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9=13\hfill \end{array}$
Clearly, this statement is a contradiction because $\text{\hspace{0.17em}}9\ne 13.\text{\hspace{0.17em}}$ Therefore, the system has no solution.
The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.
We then convert the second equation expressed to slope-intercept form.
Comparing the equations, we see that they have the same slope but different y -intercepts. Therefore, the lines are parallel and do not intersect.
$\begin{array}{l}\begin{array}{l}\\ y=-\frac{1}{2}x+\frac{9}{2}\end{array}\hfill \\ y=-\frac{1}{2}x+\frac{13}{2}\hfill \end{array}$
Solve the following system of equations in two variables.
$\begin{array}{l}2y-2x=2\\ 2y-2x=6\end{array}$
No solution. It is an inconsistent system.
## Expressing the solution of a system of dependent equations containing two variables
Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as $\text{\hspace{0.17em}}0=0.$
## Finding a solution to a dependent system of linear equations
Find a solution to the system of equations using the addition method .
$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+3y=2\\ 3x+9y=6\end{array}$
With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If we multiply both sides of the first equation by $\text{\hspace{0.17em}}-3,$ then we will be able to eliminate the $\text{\hspace{0.17em}}x$ -variable.
$\begin{array}{l}\underset{______________}{\begin{array}{ll}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3x-9y\hfill & =-6\hfill \\ +\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x+9y\hfill & =6\hfill \end{array}}\hfill \\ \begin{array}{ll}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\hfill & =0\hfill \end{array}\hfill \end{array}$
We can see that there will be an infinite number of solutions that satisfy both equations.
Solve the following system of equations in two variables.
The system is dependent so there are infinite solutions of the form $\text{\hspace{0.17em}}\left(x,2x+5\right).$
## Using systems of equations to investigate profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation $\text{\hspace{0.17em}}R=xp,$ where $\text{\hspace{0.17em}}x=$ quantity and $\text{\hspace{0.17em}}p=$ price. The revenue function is shown in orange in [link] .
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
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# 4.1.2: Points in the Coordinate Plane
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## Points in the Coordinate Plane
Megan is on vacation in Rome and is trying to find the restaurant where she will meet her friends for lunch. She got a map from her hotel that is formatted like a coordinate grid. The hotel is at the origin of the map and she was told that the restaurant is at the point (4,5) on the map. How can Megan find this location on the map so she can make it to the restaurant?
In this concept, you will learn how to name and graph ordered pairs of integer coordinates in a coordinate plane.
### Naming and Graphing Points in the Coordinate Plane
The coordinate plane is a grid created by a horizontal number line, the x-axis, intersecting a vertical number line, the y-axis. The point of intersection is called the origin.
The coordinate plane allows you to describe locations in two-dimensional space.
Each point on the coordinate plane can be named by a pair of numbers called an ordered pair in the form (x,y).
• The first number in an ordered pair identifies the x-coordinate of the point. This coordinate describes how far to the right or left of the y-axis a point is.
• The second number in an ordered pair identifies the y-coordinate of the point. This coordinate describes how far above or below the x-axis a point is.
Here is an example.
Plot the point (3,−4) on the coordinate plane.
This point has an x-coordinate of 3 and a y-coordinate of -4.
To plot this point, first start at the origin.
Then, find the location of the x-coordinate. Because the x-coordinate is positive, you will be moving to the right. Move to the right along the x-axis 3 units until you find 3.
Next, look at your y-coordinate. Because the y-coordinate is negative, you will be moving down. Move down from the 3 on the x-axis 4 units until you are lined up with the -4 on the y-axis.
The answer is shown plotted on the coordinate plane below.
Here is another example.
Give the ordered pair for the point plotted below.
To write the ordered pair you need both the x-coordinate and the y-coordinate.
Start at the origin. You need to figure out how far to the right/left you need to move and then how far up/down you need to move to reach your point.
First notice that you need to move 5 units to the right from the origin to reach the point. This means the x-coordinate is 5.
Next notice that you do not need to move up or down from the x-axis at all to reach the point. This means the y-coordinate is 0.
The answer is that the ordered pair is (5,0).
### Examples
Example 4.1.2.1
Earlier, you were given a problem about Megan, who is on vacation in Rome.
She's at her hotel and trying to use a map to find the location of the restaurant where she will be meeting her friends for lunch. The restaurant is at the point (4,5) on the map.
Solution
This point has an x-coordinate of 4 and a y-coordinate of 5.
To find this point, Megan should start at the origin on the map (her hotel).
Then, she should find the location of the x-coordinate. Because the x-coordinate is positive, she will be moving to the right. She needs to move to the right 4 units until she reaches the 4 on the x-axis.
Next, Megan should look at her y-coordinate. Because the y-coordinate is positive, she will be moving up. She should move up 5 units from the 4 on the x-axis.
The answer is shown plotted on the coordinate plane below.
Example 4.1.2.2
This coordinate grid shows locations in Jimmy’s city. Name the ordered pair that represents the location of the city park.
Solution
To write the ordered pair you need both the x-coordinate and the y-coordinate.
Start at the origin. You need to figure out how far to the right/left you need to move and then how far up/down you need to move to reach your point.
First notice that you need to move 2 units to the left from the origin in order to be exactly above the point for the city park. This means the x-coordinate is -2.
Next notice that you need to move down 6 units from the -2 on the x-axis to reach the point. This means that the y-coordinate is -6.
The arrows below show how you should have moved your finger to find the coordinates
The answer is that the ordered pair for the city park is (−2,6).
Example 4.1.2.3
Plot the point (0,3) on the coordinate plane.
Solution
This point has an x-coordinate of 0 and a y-coordinate of 3.
To plot this point, first start at the origin.
Then, find the location of the x-coordinate. Because the x-coordinate is 0, you do not need to move to the right or to the left from the origin. Stay at the origin.
Next, look at your y-coordinate. Because the y-coordinate is positive you will be moving up. Move up from the origin 3 units until you are lined up with the 3 on the y-axis.
The answer is shown plotted on the coordinate plane below.
Example 4.1.2.4
Give the ordered pair for the point plotted below.
Solution
To write the ordered pair you need both the x-coordinate and the y-coordinate.
Start at the origin. You need to figure out how far to the right/left you need to move and then how far up/down you need to move to reach your point.
First notice that you need to move 6 units to the right from the origin to be exactly above the point. This means the x-coordinate is 6.
Next notice that you need to move down 3 units from the 6 on the x-axis to reach your point. This means your y-coordinate is -3.
The answer is that the ordered pair is (6,−3).
Example 4.1.2.5
Plot the point (−2,−5) on the coordinate plane.
Solution
This point has an x-coordinate of -2 and a y-coordinate of -5.
To plot this point, first start at the origin.
Then, find the location of the x-coordinate. Because the x-coordinate is negative, you will be moving to the left. Move to the left 2 units until you reach the -2 on the x-axis.
Next, look at your y-coordinate. Because the y-coordinate is negative, you will be moving down. Move down 5 units from the -2 on the x-axis.
The answer is shown plotted on the coordinate plane below.
### Review
1. Name the ordered pair that represents each of these points on the coordinate plane.
1. roller coaster
2. Ferris wheel
3. carousel
4. log flume
Name the ordered pairs that represent the vertices of triangle FGH.
1. F
2. G
3. H
Name the ordered pairs that represent the vertices of pentagon ABCDE.
1. A
2. B
3. C
4. D
5. E
6. On the grid below, plot point V at (−6,4).
1. On the grid below, plot point a triangle with vertices R(4,−1),S(4,−4) and T(−3,−4).
### Vocabulary
Term Definition
x−axis The x−axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
x−coordinate The x−coordinate is the first term in a coordinate pair, commonly representing the value of the input or independent variable.
y−axis The y−axis is the vertical number line of the Cartesian plane.
Abscissa The abscissa is the x−coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 3 is the abscissa.
Cartesian Plane The Cartesian plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin.
Coordinate Plane The coordinate plane is a grid formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate plane is also called a Cartesian Plane.
Ordinate The ordinate is the y-coordinate of the ordered pair that represents a plotted point on a Cartesian plane. For the point (3, 7), 7 is the ordinate.
Origin The origin is the point of intersection of the x and y axes on the Cartesian plane. The coordinates of the origin are (0, 0).
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In mathematics, a superparticular ratio, also called a superparticular number or epimoric ratio, is the ratio of two consecutive integer numbers.
More particularly, the ratio takes the form:
${\displaystyle {\frac {n+1}{n))=1+{\frac {1}{n))}$ where n is a positive integer.
Thus:
A superparticular number is when a great number contains a lesser number, to which it is compared, and at the same time one part of it. For example, when 3 and 2 are compared, they contain 2, plus the 3 has another 1, which is half of two. When 3 and 4 are compared, they each contain a 3, and the 4 has another 1, which is a third part of 3. Again, when 5, and 4 are compared, they contain the number 4, and the 5 has another 1, which is the fourth part of the number 4, etc.
— Throop (2006), [1]
Superparticular ratios were written about by Nicomachus in his treatise Introduction to Arithmetic. Although these numbers have applications in modern pure mathematics, the areas of study that most frequently refer to the superparticular ratios by this name are music theory[2] and the history of mathematics.[3]
## Mathematical properties
As Leonhard Euler observed, the superparticular numbers (including also the multiply superparticular ratios, numbers formed by adding an integer other than one to a unit fraction) are exactly the rational numbers whose continued fraction terminates after two terms. The numbers whose continued fraction terminates in one term are the integers, while the remaining numbers, with three or more terms in their continued fractions, are superpartient.[4]
${\displaystyle \prod _{n=1}^{\infty }\left({\frac {2n}{2n-1))\cdot {\frac {2n}{2n+1))\right)={\frac {2}{1))\cdot {\frac {2}{3))\cdot {\frac {4}{3))\cdot {\frac {4}{5))\cdot {\frac {6}{5))\cdot {\frac {6}{7))\cdots ={\frac {4}{3))\cdot {\frac {16}{15))\cdot {\frac {36}{35))\cdots =2\cdot {\frac {8}{9))\cdot {\frac {24}{25))\cdot {\frac {48}{49))\cdots ={\frac {\pi }{2))}$
represents the irrational number π in several ways as a product of superparticular ratios and their inverses. It is also possible to convert the Leibniz formula for π into an Euler product of superparticular ratios in which each term has a prime number as its numerator and the nearest multiple of four as its denominator:[5]
${\displaystyle {\frac {\pi }{4))={\frac {3}{4))\cdot {\frac {5}{4))\cdot {\frac {7}{8))\cdot {\frac {11}{12))\cdot {\frac {13}{12))\cdot {\frac {17}{16))\cdots }$
In graph theory, superparticular numbers (or rather, their reciprocals, 1/2, 2/3, 3/4, etc.) arise via the Erdős–Stone theorem as the possible values of the upper density of an infinite graph.[6]
## Other applications
In the study of harmony, many musical intervals can be expressed as a superparticular ratio (for example, due to octave equivalency, the ninth harmonic, 9/1, may be expressed as a superparticular ratio, 9/8). Indeed, whether a ratio was superparticular was the most important criterion in Ptolemy's formulation of musical harmony.[7] In this application, Størmer's theorem can be used to list all possible superparticular numbers for a given limit; that is, all ratios of this type in which both the numerator and denominator are smooth numbers.[2]
These ratios are also important in visual harmony. Aspect ratios of 4:3 and 3:2 are common in digital photography,[8] and aspect ratios of 7:6 and 5:4 are used in medium format and large format photography respectively.[9]
Every pair of adjacent positive integers represent a superparticular ratio, and similarly every pair of adjacent harmonics in the harmonic series (music) represent a superparticular ratio. Many individual superparticular ratios have their own names, either in historical mathematics or in music theory. These include the following:
Examples
Ratio Cents Name/musical interval Ben Johnston
notation
above C
Audio
2:1 1200 duplex:[a] octave C'
3:2 701.96 sesquialterum:[a] perfect fifth G
4:3 498.04 sesquitertium:[a] perfect fourth F
5:4 386.31 sesquiquartum:[a] major third E
6:5 315.64 sesquiquintum:[a] minor third E
7:6 266.87 septimal minor third E
8:7 231.17 septimal major second D-
9:8 203.91 sesquioctavum:[a] major second D
10:9 182.40 sesquinona:[a] minor tone D-
11:10 165.00 greater undecimal neutral second D-
12:11 150.64 lesser undecimal neutral second D
15:14 119.44 septimal diatonic semitone C
16:15 111.73 just diatonic semitone D-
17:16 104.96 minor diatonic semitone C
21:20 84.47 septimal chromatic semitone D
25:24 70.67 just chromatic semitone C
28:27 62.96 septimal third-tone D-
32:31 54.96 31st subharmonic,
inferior quarter tone
D-
49:48 35.70 septimal diesis D
50:49 34.98 septimal sixth-tone B-
64:63 27.26 septimal comma,
63rd subharmonic
C-
81:80 21.51 syntonic comma C+
126:125 13.79 septimal semicomma D
128:127 13.58 127th subharmonic
225:224 7.71 septimal kleisma B
256:255 6.78 255th subharmonic D-
4375:4374 0.40 ragisma C-
The root of some of these terms comes from Latin sesqui- "one and a half" (from semis "a half" and -que "and") describing the ratio 3:2.
## Notes
1. Ancient name
## Citations
1. ^ Throop, Priscilla (2006). Isidore of Seville's Etymologies: Complete English Translation, Volume 1, p. III.6.12, n. 7. ISBN 978-1-4116-6523-1.
2. ^ a b Halsey, G. D.; Hewitt, Edwin (1972). "More on the superparticular ratios in music". American Mathematical Monthly. 79 (10): 1096–1100. doi:10.2307/2317424. JSTOR 2317424. MR 0313189.
3. ^ Robson, Eleanor; Stedall, Jacqueline (2008), The Oxford Handbook of the History of Mathematics, Oxford University Press, ISBN 9780191607448. On pp. 123–124 the book discusses the classification of ratios into various types including the superparticular ratios, and the tradition by which this classification was handed down from Nichomachus to Boethius, Campanus, Oresme, and Clavius.
4. ^ Leonhard Euler; translated into English by Myra F. Wyman and Bostwick F. Wyman (1985), "An essay on continued fractions" (PDF), Mathematical Systems Theory, 18: 295–328, doi:10.1007/bf01699475, hdl:1811/32133, S2CID 126941824((citation)): CS1 maint: multiple names: authors list (link). See in particular p. 304.
5. ^ Debnath, Lokenath (2010), The Legacy of Leonhard Euler: A Tricentennial Tribute, World Scientific, p. 214, ISBN 9781848165267.
6. ^ Erdős, P.; Stone, A. H. (1946). "On the structure of linear graphs". Bulletin of the American Mathematical Society. 52 (12): 1087–1091. doi:10.1090/S0002-9904-1946-08715-7.
7. ^ Barbour, James Murray (2004), Tuning and Temperament: A Historical Survey, Courier Dover Publications, p. 23, ISBN 9780486434063, The paramount principle in Ptolemy's tunings was the use of superparticular proportion..
8. ^ Ang, Tom (2011), Digital Photography Essentials, Penguin, p. 107, ISBN 9780756685263. Ang also notes the 16:9 (widescreen) aspect ratio as another common choice for digital photography, but unlike 4:3 and 3:2 this ratio is not superparticular.
9. ^ The 7:6 medium format aspect ratio is one of several ratios possible using medium-format 120 film, and the 5:4 ratio is achieved by two common sizes for large format film, 4×5 inches and 8×10 inches. See e.g. Schaub, George (1999), How to Photograph the Outdoors in Black and White, How to Photograph Series, vol. 9, Stackpole Books, p. 43, ISBN 9780811724500.
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Cody
Problem 174. Roll the Dice!
Solution 1407345
Submitted on 6 Jan 2018 by claus becker
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x1 = zeros(1,6000); x2 = zeros(1,6000); for ii = 1:6000 [x1(ii),x2(ii)] = rollDice(); end numCt = sum( bsxfun( @eq, x1, (1:6)' ), 2 ) + sum( bsxfun( @eq, x2, (1:6)' ), 2 ); assert(all(round(numCt/200) == 10) && sum(numCt) == 12000)
W = 6 2 W = 6 3 W = 6 4 W = 4 5 W = 3 1 W = 5 6 W = 3 3 W = 2 4 W = 1 1 W = 4 4 W = 5 1 W = 6 5 W = 2 2 W = 6 1 W = 5 6 W = 5 2 W = 1 5 W = 3 1 W = 4 4 W = 1 2 W = 5 5 W = 5 3 W = 2 6 W = 5 6 W = 6 6 W = 5 4 W = 6 1 W = 2 2 W = 2 6 W = 3 4 W = 3 3 W = 3 6 W = 4 5 W = 2 6 W = 1 4 W = 1 2 W = 6 4 W = 2 5 W = 5 4 W = 3 5 W = 6 6 W = 1 3 W = 5 5 W = 3 2 W = 1 3 W = 3 4 W = 6 3 W = 6 5 W = 1 2 W = 5 1 W = 1 2 W = 6 4 W = 2 5 W = 1 3 W = 5 5 W = 6 3 W = 5 2 W = 3 2 W = 3 6 W = 3 4 W = 4 2 W = 5 5 W = 6 3 W = 6 6 W = 4 5 W = 1 5 W = 4 2 W = 2 4 W = 6 5 W = 6 2 W = 1 3 W = 5 4 W = 3 2 W = 6 2 W = 2 1 W = 3 5 W = 1 1 W = 3 2 W = 3 3 W = 6 6 W = 2 1 W = 4 1 W = 3 1 W = 4 3 W = 6 5 W = 1 1 W = 3 2 W = 2 6 W = 5 5 W = 2 6 W = 6 5 W = 3 3 W = 2 1 W = 1 2 W = 1 6 W = 3 5 W = 4 3 W = 5 1 W = 5 4 W = 3 4 W = 4 1 W = 4 4 W = 1 3 W = 6 4 W = 1 1 W = 5 4 W = 2 6 W = 1 2 W = 1 1 W = 1 1 W = 3 6 W = 4 2 W = 2 5 W = 1 6 W = 3 5 W = 1 5 W = 6 2 W = 3 6 W = 3 2 W = 2 6 W = 3 5 W = 5 6 W = 2 1 W = 5 1 W = 4 1 W = 1 2 W = 3 6 W = 6 2 W = 1 1 W = 1 6 W = 3 4 W = 4 2 W = 6 2 W = 1 4 W = 6 4 W = 6 1 W = 6 5 W = 3 6 W = 6 1 W = 2 1 W = 2 2 W = 1 1 W = 4 3 W = 5 3 W = 4 2 W = 5 4 W = 4 4 W = 5 1 W = 5 1 W = 4 3 W = 2 6 W = 6 1 W = 4 1 W = 3 5 W = 1 3 W = 5 1 W = 4 5 W = 4 6 W = 4 1 W = 1 6 W = 6 5 W = 4 5 W = 2 3 W = 3 3 W = 1 6 W = 1 5 W = 2 2 W = 3 4 W = 6 4 W = 5 3 W = 5 6 W = 5 5 W = 4 6 W = 5 3 W = 6 1 W = 4 4 W = 4 2 W = 6 3 W = 1 2 W = 1 1 W = 5 5 W = 5 1 W = 1 2 W = 1 6 W = 6 6 W = 4 3 W = 4 1 W = 2 5 W = 6 1 W = 3 6 W = 2 2 W = 4 2 W = 6 3 W = 6 2 W = 1 5 W = 2 2 W = 4 1 W = 1 4 W = 5 4 W = 1 3 W = 5 3 W = 4 1 W = 5 6 W = 6 6 W = 4 4 W = 4 2 W = 1 1 W = 1 2 W = 2 4 W = 3 5 W = 4 2 W = 5 6 W = 2 2 W = 1 6 W = 1 2 W = 6 1 W = 4 4 W = 3 6 W = 3 2 W = 6 1 W = 5 6 W = 4 4 W = 5 2 W = 4 2 W = 1 3 W = 6 6 W = 4 6 W = 2 4 W = 6 6 W = 2 3 W = 3 1 W = 4 5 W = 1 6 W = 2 2 W = 1 5 W = 3 3 W = 3 2 W = 6 2 W = 3 2 W = 6 4 W = 5 6 W = 5 5 W = 2 1 W = 5 2 W = 2 3 W = 2 6 W = 2 5 W = 2 3 W = 4 1 W = 3 5 W = 2 3 W = 1 3 W = 6 5 W = 2 3 W = 5 3 W = 4 4 W = 4 5 W = 6 3 W = 2 3 W = 1 5 W = 1 2 W = 2 2 W = 1 2 W = 1 2 W = 2 3 W = 3 2 W = 1 4 W = 6 3 W = 4 2 W = 6 1 W = 6 2 W = 3 3 W = 2 5 W = 4 2 W = 1 4 W = 6 5 W = 1 3 W = 4 4 W = 6 2 W = 4 2 W = 5 4 W = 5 2 W = 6 1 W = 2 6 W = 6 6 W = 1 4 W = 6 6 W = 2 5 W = 6 3 W = 2 1 W = 6 6 W = 1 6 W = 1 6 W = 4 6 W = 2 2 W = 3 4 W = 5 4 W = 3 5 W = 6 2 W = 2 3 W = 3 3 W = 1 4 W = 2 2 W = 2 5 W = 1 1 W = 6 3 W = 6 3 W = 5 2 W = 6 3 W = 2 4 W = 5 3 W = 1 3 W = 4 2 W = 2 4 W = 6 3 W = 5 1 W = 3 1 W = 6 4 W = 3 5 W = 1 6 W = 6 4 W = 2 4 W = 2 6 W = 6 6 W = 4 4 W = 5 6 W = 4 1 W = 6 6 W = 2 1 W = 2 4 W = 3 5 W = 5 3 W = 5 2 W = 2 3 W = 4 5 W = 2 6 W = 4 2 W = 4 1 W = 1 3 W = 1 2 W = 3 6 W = 5 3 W = 6 6 W = 4 5 W = 2 6 W = 6 5 W = 1 2 W = 3 3 W = 6 6 W = 1 1 W = 5 2 W = 1 1 W = 2 6 W = 3 1 W = 4 1 W = 2 6 W = 1 2 W = 6 2 W = 5 6 W = 5 2 W = 2 2 W = 5 6 W = 3 4 W = 3 5 W = 5 6 W = 3 1 W = 6 5 W = 6 4 W = 4 4 W = 4 2 W = 2 4 W = 5 6 W = 2 3 W = 1 5 W = 2 5 W = 2 5 W = 1 4 W = 6 3 W = 2 1 W = 2 6 W = 2 5 W = 1 5 W = 4 1 W = 1 2 W = 2 4 W = 1 2 W = 2 5 W = 5 1 W = 4 6 W = 5 4 W = 5 2 W = 6 1 W = 5 5 W = 1 1 W = 5 6 W = 2 2 W = 4 5 W = 4 1 W = 5 5 W = 1 3 W = 2 2 W = 4 5 W = 3 6 W = 5 4 W = 1 6 W = 3 6 W = 4 3 W = 3 6 W = 4 4 W = 4 4 W = 2 5 W = 6 6 W = 3 5 W = 4 5 W = 4 2 W = 6 5 W = 6 2 W = 5 6 W = 6 2 W = 3 3 W = 4 6 W = 3 1 W = 2 5 W = 6 2 W = 4 6 W = 4 6 W = 4 4 W = 4 5 W = 4 5 W = 3 6 W = 2 6 W = 2 3 W = 1 3 W = 3 4 W = 5 3 W = 3 2 W = 5 3 W = 2 5 W = 6 2 W = 3 4 W = 1 4 W = 3 5 W = 2 1 W = 2 4 W = 6 4 W = 2 6 W = 3 6 W = 2 5 W = 3 3 W = 4 2 W = 1 6 W = 3 5 W = 4 2 W = 2 6 W = 6 3 W = 5 6 W = 2 2 W = 5 3 W = 2 2 W = 3 4 W = 5 6 W = 1 5 W = 3 2 W = 4 6 W = 4 4 W = 2 4 W = 5 4 W = 5 3 W = 3 3 W = 3 4 W = 4 1 W = 4 6 W = 2 6 W = 4 3 W = 2 4 W = 6 1 W = 2 4 W = 6 1 W = 1 6 W = 6 3 W = 3 3 W = 1 5 W = 1 4 W = 3 1 W = 2 5 W = 3 1 W = 1 6 W = 6 1 W = 2 6 W = 2 2 W = 5 6 W = 6 2 W = 5 2 W = 6 5 W = 4 4 W = 1 5 W = 2 4 W = 5 3 W = 3 6 W = 1 6 ...
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### Spoiled for choice
Hi! I'm Atul, in my last year of school and this is going to be my first blog post. For an introduction, I have a bronze at the IMO and a silver at the APMO this year. Apart from math, I also love reading, playing table tennis and making terrible jokes. Something I seem to miss out on saying often is that I love cats, especially colourful ones. Anyway, let's begin!
Like many things in math, the axiom of choice has a deceptively simple statement. All it says is - "Given a collection of nonempty sets, it is possible to pick an element from each of them". That... sounds pretty obvious and something that just should be true. And well, it is. At least for a finite collection of sets, it is definitely true. With some thought, it's also clear how to do it if the collection is countably infinite. But what if the number of sets is just extremely huge? When the sets are uncountably infinite, that's when the problems begin.
So far I've been rather handwavy, so let's get the formalities out of the way. A function $f$ defined on a collection of nonempty sets $X$, is called a choice function if $f(S) \in S$ for all $S \in X$. The axiom of choice states that every collection of sets $X$ has a choice function.
I should clarify though - although the axiom of choice leads to some very strange and crazy results, it is completely consistent and there's no "contradiction" that you can obtain from this.
There are many statements that can be shown to be equivalent to the axiom of choice, such as "The cartesian product of nonempty sets is nonempty" or "Every set of numbers can be well-ordered". The proof of the first one isn't too hard and you can try it as an exercise, the second one uses ordinals, which are too complicated for this blog post, so I won't do that either.
Perhaps one of the most famous consequences of the Axiom of Choice (Or AC for short) is the Banach-Tarski Paradox, which states that "given a solid ball in 3d space, it can be split into finitely many pieces and rearranged into two solid balls of the exact same radius", which sounds ridiculous. But, I'm getting repetitive, this is pretty complicated too, and uses group theory, so not that either.
So what will we do, you ask. Well let's beign with a simple sounding problem and then move on to more cool ones. They won't be boring, I promise.
Problem 1: Prove that every infinite set of numbers has a countably infinite subset.
Proof: The naive approach to this would probably be to just keep picking elements, don't we just get an infinite subset? The problem is that that only lets you get a subset with an arbitrarily large number of elements, but never infinite. And in fact, this problem cannot be solved without the axiom of choice.
Say $S$ is this infinite set of numbers and let $X$ be the power set of $S$. Let $f$ be a choice function on $X$. Define a function $g$ the following way. $g(1) = f(S)$ and $g(n) = f(S \setminus \{g(1), g(2), \cdots, g(n-1)\})$ for $n \geqslant 2$. This looks pretty similar to what we did above, but here there's no "choice" that needs to be made, the function is well-defined because the choice function does all of that for us. But fine that's still underwhelming.
Let's try an actually interesting (and counterintuitive) problem now:
Problem 2: There are countably many gnomes with hats in countably many colours. Each gnome can see the colours of every hat except the one that’s on his head. The gnomes can strategize before the hats are on their heads, but they cannot talk or communicate with each other once the hats are on. Show that they can figure out a strategy such that all, but finitely many of them will guess the colour of their hat correctly (all gnomes guess simultaneously).
Proof: This seems pretty messed up and no way it could be true. Unfortunately with the axiom of choice, it is possible. The idea is pretty clever and hard to think of if you haven't seen it before. Let $S$ be the set of all infinite sequences of natural numbers, FOR EXAMPLE: $1, 4, 5, 1, 2, \cdots$ Define an equivalence relation where $x \sim y$ if $x,y \in S$ and $x,y$ differ in at most finitely many places (you can also think of this as saying beyond a point $x$ and $y$ are the same sequence). Then, let $X$ be the set of all equivalence classes, and let $f$ be a choice function on $X$.
The gnomes then order themselves $g_1, g_2, \cdots$ and also order the colors $c_1, c_2, \cdots$ beforehand. Then, they consider the sequence of colours that the respective gnomes are wearing, clearly all gnomes see the entire sequence (call this sequence $s$) apart from one place (their own). But by the way the equivalence relation was defined, this is enough to determine the equivalence class (say $E$) that $s$ belongs to. Then, the $k$th gnome guesses the $k$th number in the sequence $f(E)$. Once again, by the way, the equivalence relation was defined, $f(E)$ and $s$ differ in finitely many places, and hence finitely many gnomes guess incorrectly, as desired. $\blacksquare$
If you thought gnomes were smart, well you'd be right, but ducks are even smarter, as we'll see in this next problem:
Problem 3: There are $100$ identical rooms with countably many boxes labelled with natural numbers. Inside each box, there is a real number. After discussing a strategy, $100$ ducks are sent to different rooms without any means of communication after that. While in the room the duck can open up boxes (perhaps countably many) and see the real number inside. Then each duck needs to guess the real number inside one of the unopened boxes. Show that there is a strategy that guarantees at least $99$ of those ducks makes a correct guess.
Proof: No, I promise the problem is true (if you actually believe that the problem statement even might be true, then you're either crazy or a genius). This seems even crazier than the previous one, there at least finitely many gnomes guessed wrong, but here at most one does?! Well look at the solution and see for yourself.
Define $X$ the same as the above solution, but with real numbers instead of natural numbers being elements of the infinite sequences. The fact that we're now using real numbers (which are uncountable) instead of natural numbers (which are reasonable and countable) actually does not change anything, we might as well have allowed uncountably many hats, that would have worked too, check it for yourself! But that's where the similarity ends, this time, the ducks are going to be a lot smarter.
As above, let $f$ be the choice function on $X$ and let $g(S)$ denote the last digit at which $S$ and $f(S)$ differ (for some infinite sequence of reals $S$).
Number the boxes $B_1, B_2, \cdots$ and create $100$ groups, $G_1, G_2, \cdots, G_{100}$ where $G_i$ contains all boxes numbered $i \pmod {100}$. Let $b_i$ (for $1 \leqslant i \leqslant 100$) denote the sequence of real numbers in the boxes $B_i, B_{i+100}, B_{i+200}, \cdots$.
Let's pick a duck, say duck numbered $k$. Then the duck follows a two-stage process: in the first stage, it opens every box not in $G_k$. Therefore, it knows the sequences $b_1, \cdots, b_{100}$ except for $g_k$. It then computes the following number $$N_k = \max(\{b_1, b_2, \cdots, b_{100}\} \setminus \{b_k\}) + 1$$Then in the second stage, the duck opens all boxes of the form $B_{100i + k}$ for $i > N$ and then obtains some number say $b_k'$, and then guesses the $N$th digit of $f(b_k') = f(b_k)$ (where the equality is due to the equivalence relation).
The reason this works is that suppose some duck, say duck $k$, guesses wrongly. Then that means that $f(b_k)$ and $b_k$ differ in the $N$th place, and so $g(b_k) \geqslant N > b_i$ for any $i \neq k$. Since this cannot happen for two ducks (as there can be at most one strict maximum), it follows that at most one duck guesses wrongly, and so at least $99$ guess correctly, as desired. $\blacksquare$
That... was quite something. Now you've seen first-hand the weirdness that such an innocent and simple-looking statement can lead to. Before you give up hope in math, there's a more sane alternative for all of this. It's called the axiom of countable choice, where it restricts $X$ to be a countable collection of sets. This avoids all of the crazy and weird results due to the AC but also leaves behind all the nice things you need to prove results in analysis.
Oh and one last thing, if you enjoy doing puzzles, you may want to re-read the blog post again, begin from the beginning, or so I've been told often. If you need some help, here's an extract from a nonexistent book:
"staring into the Abyss, i was staying with the Cinians when i heard a sound below. listening to songs by Sia, i was doing pretty much nothing when i saw a couple of Mice scurrying across the floor. i nearly dropped my Purse in fright; that wouldn't do, i was the Scion of a family known for bravery. i was curious though, so i followed them, over the Wire fence, nearly getting my Hair caught in it. i don't know why i felt so drawn to them, maybe it's because at the Core, i'm a Rat too."
To finish, you may want to look at places with numbers, which might help. Anyway, good luck!
~Atul
### LMAO Revenge
Continuing the tradition of past years, our seniors at the Indian IMO camp(an unofficial one happened this year) once again conducted LMAO, essentially ELMO but Indian. Sadly, only those who were in the unofficial IMOTC conducted by Pranav, Atul, Sunaina, Gunjan and others could participate in that. We all were super excited for the problems but I ended up not really trying the problems because of school things and stuff yet I solved problem 1 or so did I think. Problem 1: There is a grid of real numbers. In a move, you can pick any real number , and any row or column and replace every entry in it with . Is it possible to reach any grid from any other by a finite sequence of such moves? It turned out that I fakesolved and oh my god I was so disgusted, no way this proof could be false and then when I was asked Atul, it turns out that even my answer was wrong and he didn't even read the proof, this made me even more angry and guess what? I was not alone, Krutarth too fakesol
### The importance of "intuition" in geometry
Hii everyone! Today I will be discussing a few geometry problems in which once you "guess" or "claim" the important things, then the problem can easily be finished using not-so-fancy techniques (e.g. angle chasing, power-of-point etc. Sometimes you would want to use inversion or projective geometry but once you have figured out that some particular synthetic property should hold, the finish shouldn't be that non trivial) This post stresses more about intuition rather than being rigorous. When I did these problems myself, I used freehand diagrams (not geogebra or ruler/compass) because I feel that gives a lot more freedom to you. By freedom, I mean, the power to guess. To elaborate on this - Suppose you drew a perfect diagram on paper using ruler and compass, then you would be too rigid on what is true in the diagram which you drew. But sometimes that might just be a coincidence. e.g. Let's say a question says $D$ is a random point on segment $BC$, so maybe
### Edge querying in graph theory
In this post, I will present three graph theory problems in increasing difficulty, each with a common theme that one would determine a property of an edge in a complete graph through repeated iterations, and seek to achieve a greater objective. ESPR Summer Program Application: Alice and Bob play the following game on a $K_n$ ($n\ge 3$): initially all edges are uncolored, and each turn, Alice chooses an uncolored edge then Bob chooses to color it red or blue. The game ends when any vertex is adjacent to $n-1$ red edges, or when every edge is colored; Bob wins if and only if both condition holds at that time. Devise a winning strategy for Bob. This is more of a warm-up to the post, since it has a different flavor from the other two problems, and isn't as demanding in terms of experience with combinatorics. However, do note that when this problem was first presented, applicants did not know the winner ahead of time; it would be difficult to believe that Bob can hold such a strong
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# Complex Exponents
## Recommended Posts
How does one go about calculating complex exponents? To be more specific, what is the value of ai, abi, and ac + bi? While reading a book about prime numbers, I came across a section about Riemann's Hypothesis. From what I understand, Riemann's Hypothesis has to do with the graph of the zeta function when it is fed a complex number. The zeta function uses its argument as an exponent.
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1) Euler's Formula says that e^(ix) = cos x + i sin x. (you can read about this, here http://en.wikipedia.org/wiki/Euler's_formula for example) If a is a real number, by properties of logarithms a = e^(ln a). Therefore, a^i = (e^(ln a))^i. Since the normal rules of powers still apply, this = e^(i ln a), and by applying Euler's formula this is cos (ln a) + i sin (ln a).
2) a^(bi) = (e^(ln a))^(bi) = e^(i*b(ln a)). Therefore applying Euler's formula again, we get cos (b(ln a)) + i sin (b(ln a)).
3) a^(c+bi) = a^c times a^(bi) by laws of exponents, so multiplying a^c through number 2, we get a^c(cos (b(ln a)) + i sin (b(ln a))).
Now, this assumes that a is a real number. It gets a bit more complicated if it's not. If you want more information or if I was unclear, let me know.
(throughout, I have used ^ as notation for (to the power of))
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Thank you. I took a look at the proof on Wikipedia and it made good sense to me. Now I just have a few more questions. First of all, I want to be sure in my thinking that 22i is an irrational number in both dimensions. I will show the steps I took to calculate. If anything here is wrong, please tell me.
22i = cos(2(ln(2))) + i(sin(2(ln(2))))
22i = cos(2(.693)) + i(sin(2(.693)))
22i = cos(1.386) + i(sin(1.386))
22i = .183 + .983i
It just seems strange that an "integer" imaginary number used as an exponent of another integer would be irrational.
Also, where did mathematicians come up with those formulas for sines and cosines? I always wondered about that. I used to think that they had to draw triangles with a protractor and then measure sides to come up with those until I realised that a calculator would not have enough memory to store tables of those functions.
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Also, where did mathematicians come up with those formulas for sines and cosines? I always wondered about that. I used to think that they had to draw triangles with a protractor and then measure sides to come up with those until I realised that a calculator would not have enough memory to store tables of those functions.
A calculator will find trigonometric values using the power series expansion. A Maclaurin series (special case of the Taylor series) is an infinite series which approximates a function. You can get any desired degree of accuracy by taking as many terms as desired.
For example, sin x = x - x^3/3! + x^5/5! - x^7/7! + ...
This series converges extremely rapidly, and the error is easy to estimate because the signs alternate. The error is less than the next missing term -- for example, if I cut off the series after 4 terms, the error will be less than x^9/9!.
It just seems strange that an "integer" imaginary number used as an exponent of another integer would be irrational.
Your calculations are fine. Most numbers raised to the integral multiples of i will not be "nice" any more than cos 1 is nice. (Btw, for a nifty trick, you can check your calculations by googling 2^(2i). Google calculator will automatically find it for you.)
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Could you show me a proof for the validity of the series? I always feel guilty if I use a formula and don't know how or why it works.
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Could you show me a proof for the validity of the series? I always feel guilty if I use a formula and don't know how or why it works.
You simply do a Taylor series expansion of the sine function about zero. Just calculate the terms.
All the even terms go away because the even derivatives would reproduce the sine, with sin(0)=0 - the underlying reason, of course, being that the sine is an odd function.
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You simply do a Taylor series expansion of the sine function about zero. Just calculate the terms.
All the even terms go away because the even derivatives would reproduce the sine, with sin(0)=0 - the underlying reason, of course, being that the sine is an odd function.
This is correct, so I will not bother restating it :D. I will add that this is taught in calculus 2, and if you have not had calculus it probably won't make any sense.
However, if you have had calculus but it's been a while, here's a blog which includes two step-by-step derivations:
http://blogs.ubc.ca/...ansion-of-sinx/
(note that the next-previous links at the bottom lead to the expansions for e^x and cos x)
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I am a ninth-grader who has never taken calculus, so I often have a hard time understanding complicated proofs. I like math very much though, so if I wind up learning calculus from this thread, I won't be surprised. So I will ask a few questions here and maybe the proof will become clear to me. Is a derivative the result of a function? It looks like it is to me, because, as the ex page on that site says, when f(x) = ex and when x = 0, f(x) is always equal to 1 (because anything to the power of 0 is 1). So if x = 1, would the derivative be e? Or e2 when x = 2? And back at the first page, is d in the formula in step 1 the derivative?
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Not to discourage you, but I don't think this is going to work in a thread. If you are interested in learning calculus, get yourself a textbook - a used older edition of Stewart's Single Variable calculus will cost you a few \$ online- and start reading. It will really make sense, conceptually, calculus is not really difficult.
I, and probably some others on this board, will be happy to answer specific questions that arise.
All the best.
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I didn't mean that I was hoping to learn all of calculus, but I just hoped that someone might answer some questions for me. My mother, who is a poster here, told me that I might enjoy asking about these things on the self-education forum. If there is no way to demonstrate these formulas without using calculus, I'll come back to this once I've studied some basic concepts in calculus. I have an interest in math, and enjoy reading about and trying to figure out concepts ahead of my formal studies.
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Algorithm, since you seem to really enjoy thinking about mathematics, you might enjoy taking a look at What is Mathematics? by Courant, Robbins, Stewart. It's a survey of lots of different mathematical fields, including prime numbers, complex numbers, the Zeta function, calculus and much more. You wouldn't sit down and read this from cover to cover, but it might be handy to have on hand for those questions that seem to come up.
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OK, let me try to give you the two basic ideas of calc, so you have a starting point.
The Derivative is a concept that allows us to find the slope of a function at any given point. If you draw a straight line through any two points, you know that the slope is rise over run, or (f(x2)-f(x1)/(x2-x1). The main idea is to let the points slide closer and closer together, making the x2 and x1 closer together and the run basically zero. In this limit, the slope of that line becomes the slope of a tangent on the curve. this is called the derivative, and part of calculus explores how to calculate derivatives of functions. There are procedures to do that without having to go through the actual line/slope process.
Now, the Taylor series is a kind of approximation. Imagine you have a function and you want to know how big the value is approximately close to some point x_0 for which you know the functional value. So, you know that if your x is close to x_0, the functional value has to be close to f(x_0). Now of course that is not entirely correct if the function is not constant. So, you pretend the curve is straight (which is very good if you are close), and correct for the deviation. That works in the derivative. Now, since the curve is not exactly straight (unless it is linear), you have to correct for curvature, and get a second term which is related to the derivative of the derivative. If the curve is not a quadratic function, this, too, is not exact, so you keep correcting with higher and higher terms involving higher derivatives. The sum of these terms is called a Taylor Series and can be calculated for any function which is well behaved (it can't have holes or kinks or jumps).
That's where those formulas come from.
The other main idea of calculus is the integral, which has to do with finding the area under a curve. There is a surprising relationship between this and the derivative. Doing integrals is much more complicated, since there is no general rule how to find them for all functions. that's why the calculus text will have only one or two chapters on derivatives, but many more on integration.
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Algorithm, how much math have you had? What are you formally studying now? This would help with recommendations for further reading.
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Actually, I've read parts of that What Is Mathematics? book before. I never got to the part about calculus, because they were so in-depth that I thought it would be too hard to understand. I found it very useful for stuff like whole-number solutions to the Pythagorean Theorem and prime numbers.
I didn't know that mathematicians had ways of correcting the curved functions. Are more steps required the higher the exponent is? For example, would a quintic function require more steps for adjustment than a quartic function, and so forth?
I'm doing Geometry this year (grade 9) and I've already taken Algebra I. I think that Algebra II concepts would probably come easily for me though; I've read about things like logarithms, factorials, and the trig functions for interest and understand how to use them.
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Actually, I've read parts of that What Is Mathematics? book before. I never got to the part about calculus, because they were so in-depth that I thought it would be too hard to understand. I found it very useful for stuff like whole-number solutions to the Pythagorean Theorem and prime numbers.
You might want to take another look at it sometime soon, especially after you've mastered algebra and geometry. Another book that helped my daughter get an intuitive feel for calculus was Calculus for the Forgetful by Wojciech Kosek. It's a slim little book that gets to the heart of the subject without a lot of details. (it does take a mature reading level) After that she filled in the details of calculus with a more standard book.
I didn't know that mathematicians had ways of correcting the curved functions. Are more steps required the higher the exponent is? For example, would a quintic function require more steps for adjustment than a quartic function, and so forth.
Regentrude was talking above about approximating functions f(x), which can be a trig function (or exponential or rational or any kind of function) near a point x0 by using a series of polynomial functions. We like to do that in math because polynomials (a + bx + cx^2 + dx^3+ ...) are well-behaved and usually easier to work with.
It turns out that you can use calculus to find the values of the coefficients a, b, c, d, ... in that expansion. We can start by finding just the a+bx part of the series, which is a straight line approximation to f(x) [remembering how straight lines are written as y = bx +a].
Each new term in the series corrects the previous approximation a little bit more, introducing more curvature by adding a higher power of x. The further out you go in the series, the better the approximation. If you go infinitely far, and if the f(x) function is nice & smooth enough, the series sums exactly to f(x), at least in an interval around your point x0.
Now, the formulas that allow you to calculate a, b, c, etc, for any given function f(x) and point x0 involve the derivatives of f(x) (which are functions themselves!) evaluated at x0. As you go further out, you need higher order derivatives.
So, let f(x) = sin x. Suppose we want to approximate sin x near x=0:
We can use calculus to figure out that, near x=0, sinx is approximately equal to the straight line approximation
sin 0 + f '(0)*x
= 0 + cos(0) *x
= 0 + 1*x
= x
(here f '(x) is the first derivative function for sin x, which actually equals cos x)
So sin x is approximately equal to x near 0. Try it on your calculator. For small radian values, sin x and x are nearly equal.
We can go on and use more terms in our approximation of sin x. Each term involves calculating a higher derivative of sin x. Using more and more terms gains us more accuracy. If we go on forever, we get an exact expression for sinx made up of infinitely many polynomial terms, & it involves calculating all orders of derivatives of sin x. That creature is what we call a Taylor (or Maclaurin) series for sin x. It is:
sin x = x - x^3/3! + x^5/5! - x^7/7! + ...
This one happens to be valid for all real numbers x. That doesn't always happen; usually it's only valid in an interval around x0. Your calculator uses this series expansion to find values of sin x when you press the sin button.
Does that help at all? If not, feel free to keep asking questions. :001_smile:
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I didn't know that mathematicians had ways of correcting the curved functions. Are more steps required the higher the exponent is? For example, would a quintic function require more steps for adjustment than a quartic function, and so forth?
If you mean for the Taylor series approximation? Yes, that would be so. But the REALLY cool thing is that the technique is not restricted to polynomials themselves - you can approximate functions like sine and cosine and exponential functions too! The more terms you include, the more accurate the approximation, but if you are close to your known reference value, the higher terms become smaller and smaller because they go in powers of the distance from that position. Really neat stuff.
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Thank you for this insight. You're making me even more interested in calculus than I was before! I will certainly read the recommended books if I can get them from our library system. Is there a uniform system for finding how to calculate these derivative functions?
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Hi Algorithm!
Let's see. The derivative of f(x) is also a function, usually called f '(x).
The value of the derivative f '(x0) is defined to be the slope of the graph of y = f(x) at x0. What is the slope of a curved graph at a given point on the graph? It's defined to be the slope of the tangent line to the graph at that point. [A tangent line is a straight line that intersects the graph at exactly one point].
You can find a formula for f '(x) by realizing that for values of x very near x0, the ratio
[f(x) - f(x0)] / [ x-x0]
represents the slope [rise over run] of the secant line to the graph of y = f(x) that goes through the points
(x0, f(x0)) and (x, f(x)). [A secant line is a just a straight line going through two points on the graph]
Now suppose that x approaches x0. At the same time, that secant line approaches the tangent line to the graph of y = f(x) at the point (x0, f(x0)).
So as x --> x0, (read this "as x approaches x0"), the slope ratio above approaches the slope of the tangent line to y = f(x) at x = x0. But, hey! that's the same thing as the derivative of f(x) at x0 . So:
f '(x0) = limit [f(x) - f(x0)]
. ............ x-->x0 [ x-x0 ]
That's the basic formula for finding a derivative function. It takes a bit of practice to be able to calculate the limit above for a given function f(x). Don't worry, you'll get plenty of practice in calculus class someday!
And fortunately, you'll find a lot of patterns. For instance, if you apply this method to powers of x: f(x) = xn, (n not equal to 0), you can show that the derivative function is always f '(x) = n x n-1. Nice, huh?
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And so the derivative is written as function-name'(x)? Making the derivative of f(x) = f'(x) and so on?
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And so the derivative is written as function-name'(x)? Making the derivative of f(x) = f'(x) and so on?
Yes. That's one notation. It can also be written dy/dx.
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Ah, I was going to post, but forgot. Algorithm, if you're looking for more fun and interesting reading in this general area, you might try "e: the story of a number", by Eli Maor. I was given it for Christmas this year and enjoyed it. Parts of it require knowledge that you won't have yet (especially calculus), but since you clearly aren't thrown off by that, you may enjoy it.
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Fractional calculus and tetration - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Fractional calculus and tetration (/showthread.php?tid=937) Fractional calculus and tetration - JmsNxn - 11/17/2014 Well hello everybody. It's been a while since I posted here. I've been working vivaciously on iteration and fractional calculus and the ways the two intertwine and I've found a nice fact about tetration. I've been able to prove an analytic continuation of tetration for bases $1 < \alpha < e^{1/e}$ and I've been wondering about the base change formula, if this admits a solution for $\alpha > e^{1/e}$. The solution I've generated is periodic with period $2\pi i / \log(\beta)$ where $\beta$ is the attracting fixed point of $1 < \alpha < e^{1/e}$. The solution is culminated in two papers, where the first paper is purely fractional calculus and the second paper reduces problems in iteration to problems in fractional calculus. On the whole, I am able to iterate entire functions $\phi$ with fixed points $\xi_0$ such that $0<\phi'(\xi_0) < 1$. And then only able to iterate the function within the region where $\xi$ is such that $\lim_{n \to \infty}\phi^{\circ n}(\xi) = \xi_0$. For exponentiation this implies we can only iterate exponential functions with a fixed point such that its derivative is attracting and positive. This equates to functions with fixed points whose derivative is positive and less than one. From this I am wondering if it is possible to extend tetration to bases bigger than $\eta$. Since I have been able to generate tetration bases less than $\eta$ which are unique and determined by a single equation involving only naltural exponentiations of the number, $\alpha,\, \alpha^\alpha,\,\alpha^{\alpha^{\alpha}}, \,^4 \alpha,\,....$. As in we only need know the natural iterates of $\alpha$ exponentiated, to produce the complex iterates. I am mostly just wondering about the basechange formula to see if I can generate bases greater than $e^{1/e}$ using the method I've found. The benefit is that my method is unique, no other function is exponentially bounded which interpolates the values of tetration on the naturals. All I ask is if anyone can explain the base change formula clearly and if knowing tetration for bases < \eta can extend tetration for greater values. I have not written my formula for tetration as I am trying to write a paper which contains it in a simple proof. I wish to hide the simple proof ^_^. RE: Fractional calculus and tetration - sheldonison - 11/19/2014 (11/17/2014, 09:50 PM)JmsNxn Wrote: Well hello everybody. It's been a while since I posted here. I've been working vivaciously on iteration and fractional calculus and the ways the two intertwine and I've found a nice fact about tetration. I've been able to prove an analytic continuation of tetration for bases $1 < \alpha < e^{1/e}$ and I've been wondering about the base change formula, if this admits a solution for $\alpha > e^{1/e}$. The solution I've generated is periodic with period $2\pi i / \log(\beta)$ where $\beta$ is the attracting fixed point of $1 < \alpha < e^{1/e}$.... I just posted a similar comment on Mathstack. I like it enough to copy it here... Technically, $^x b$, for $b, The Kneser Tetration becomes ambiguous. For real bases $b>e^{1/e}$, Tetration is well defined, and analytic with singularities at negative integers<=-2. The base $b=e^{1/e}$ is the branch point, where iterating no longer grows arbitrarily large. I investigated what happens to Tetration when we extend it analytically to complex bases, and it turns out that for $be^{1/e}$, we are using both complex conjugate fixed points to generate Kneser's real valued at the real axis Tetration. And if we move the base in a circle around $e^{1/e}$ slowly using complex bases, from a real base greater than $e^{1/e}$ to one less than $e^{1/e}$, then we get to a function still uses both the attracting and repelling fixed points, but the function is no longer real valued at the real axis. Using the attracting fixed point is not the same function as Kneser's Tetration. Ok, now about the "base change" function. I could find a link on this forum, but here is a very short description: If you develop real valued Tetration by iterating the logarithm of another super-exponentially growing function, you get a function that is infinitely differentiable and looks a lot like Tetration, but it turns out to be nowhere analytic. Lets say $f(x)$ is a super-exponentially growing function, and we want to develop the "base change" Tetration for $b>e^{1/e}$ by iterating the logarithm of f(x) as follows: $\text{Tet}_b(x) \; = \lim_{n \to \infty} \log_b^{o n} \left(f(x+k_n+n)\right); \;\;\;\;\; f(k_n+n)\; = \; ^n b; \;\;\;\; k_n\;$ quickly converges to a constant as n increases If f is Tetration for another base, or many other super-exponentially growing functions, then it turns out that all of the derivatives at the real axis converge, but they eventually grow too fast for this base change Tetration function to be an analytic function. Also, the base change function is not defined in the complex plane. I haven't posted a rigorous proof of the nowhere analytic result. RE: Fractional calculus and tetration - JmsNxn - 11/19/2014 (11/19/2014, 06:00 PM)sheldonison Wrote: (11/17/2014, 09:50 PM)JmsNxn Wrote: Well hello everybody. It's been a while since I posted here. I've been working vivaciously on iteration and fractional calculus and the ways the two intertwine and I've found a nice fact about tetration. I've been able to prove an analytic continuation of tetration for bases $1 < \alpha < e^{1/e}$ and I've been wondering about the base change formula, if this admits a solution for $\alpha > e^{1/e}$. The solution I've generated is periodic with period $2\pi i / \log(\beta)$ where $\beta$ is the attracting fixed point of $1 < \alpha < e^{1/e}$.... I just posted a similar comment on Mathstack. I like it enough to copy it here... Technically, for $^x b$, even for $b, The Kneser Tetration becomes ambiguous. For real bases $b>e^{1/e}$, Tetration is well defined, and analytic with singularities at negative integers<=-2. The base $b=e^{1/e}$ is the branch point, where iterating no longer grows arbitrarily large. I investigated what happens to Tetration when we extend it analytically to complex bases, and it turns out that for$be^{1/e}$, we are using both complex conjugate fixed points to generate Kneser's real valued at the real axis Tetration. And if we move the base in a circle around $e^{1/e}$ slowly using complex bases, from a real base greater than $e^{1/e}$ to one less than $e^{1/e}$, then we get to a function still uses both the attracting and repelling fixed points, but the function is no longer real valued at the real axis. Using the attracting fixed point is not the same function as Kneser's Tetration. Ok, now about the "base change" function. I could find a link on this forum, but here is a very short description: If you develop real valued Tetration by iterating the logarithm of another super-exponentially growing function, you get a function that is infinitely differentiable and looks a lot like Tetration, but it turns out to be nowhere analytic. Lets say $f(x)$ is a super-exponentially growing function, and we want to develop the "base change" Tetration for $b>e^{1/e}$ by iterating the logarithm of f(x) as follows: $\text{Tet}_b(x) \; = \lim_{n \to \infty} \log_b^{o n} \left(f(x+k_n+n)\right); \;\;\;\;\; f(k_n+n)\; = \; ^n b; \;\;\;\; k_n\;$ quickly converges to a constant as n increases If f is Tetration for another base, or many other super-exponentially growing functions, then it turns out that all of the derivatives at the real axis converge, but they eventually grow too fast for this base change Tetration function to be an analytic function. Also, the base change function is not defined in the complex plane. I haven't posted a rigorous proof of the nowhere analytic result. AHA! Thank you for the clarification! So I have proven a unique periodic extension of tetration for bases between 1 and eta with an attracting fixed point. My extension $F$ is also the sole extension that is bounded by $|F(z)| < C e^{\alpha |\Im(z)| + \rho|\Re(z)|}$ where $\rho, \alpha, C \in \mathbb{R}^+$ and $\alpha < \pi/2$. What you said concurs with what I suspected however, that since my fractional calculus iteration only works on attracting fixed points it means its useless for bases greater than eta. Crud! Nonetheless I'll post my extension soon. I just need to iron out all the wrinkles in my article. I'm trying to tie this proof with all my other fractional calculus proofs (continuum sums, the differsum, and assorted recursion concepts). RE: Fractional calculus and tetration - fivexthethird - 11/20/2014 (11/19/2014, 10:54 PM)JmsNxn Wrote: My extension $F$ is also the sole extension that is bounded by $|F(z)| < C e^{\alpha |\Im(z)| + \rho|\Re(z)|}$ where $\rho, \alpha, C \in \mathbb{R}^+$ and $\alpha < \pi/2$. The regular iteration for bases $1 satisfies that, as it is periodic and bounded in the right halfplane. What complex bases does it work for? Does it work for base eta? RE: Fractional calculus and tetration - sheldonison - 11/20/2014 (11/20/2014, 02:56 AM)fivexthethird Wrote: (11/19/2014, 10:54 PM)JmsNxn Wrote: My extension $F$ is also the sole extension that is bounded by $|F(z)| < C e^{\alpha |\Im(z)| + \rho|\Re(z)|}$ where $\rho, \alpha, C \in \mathbb{R}^+$ and $\alpha < \pi/2$. The regular iteration for bases $1 satisfies that, as it is periodic and bounded in the right halfplane. What complex bases does it work for? Does it work for base eta?The Schroeder equations that give the formal solution for attracting (and repelling) fixed points do not work for the parabolic case, base eta=e^(1/e). See Will Jagy's post on mathstack. I have written pari-gp program that implements Jean Ecalle's formal Abel Series, Fatou Coordinate solution for parabolic points with multiplier=1; this is an asymptotic non-converging series, with an optimal number of terms to use. To get more accurate results, you may iterate f or $f^{ -1}$ a few times before using the Abel series. RE: Fractional calculus and tetration - JmsNxn - 11/20/2014 (11/20/2014, 02:56 AM)fivexthethird Wrote: (11/19/2014, 10:54 PM)JmsNxn Wrote: My extension $F$ is also the sole extension that is bounded by $|F(z)| < C e^{\alpha |\Im(z)| + \rho|\Re(z)|}$ where $\rho, \alpha, C \in \mathbb{R}^+$ and $\alpha < \pi/2$. The regular iteration for bases $1 satisfies that, as it is periodic and bounded in the right halfplane. What complex bases does it work for? Does it work for base eta? Quite literally only for those real bases so far. I'm thinking there might be a way to retrieve it for other bases, but that would require a lot of generalizing on the bare machinery I have now--making the fractional calculus techniques apply on repelling fixed points. All in all, the method I have only works for $1, and since the usual iteration is periodic and bounded like that, it must be mine as well. Currently I'm looking at how more regularly behaved functions look when they're iterated using FC--maybe that'll help me draw some more conclusions.
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```Standards Academy
Day 1
Objectives
• Understand the Critical Areas of our grade
levels.
• Examine the importance of vertical
• What is conceptual understanding?
• Cognitive Rigor Matrix
Who are we?
•
•
•
•
•
Name
School
District
Years spent teaching 3rd or 4th grade
– Keeping those previously mentioned in mind
Norms
Things we all need in order for our time
together to be productive……….
1. .
2. ..
3. …
4. ….
Vertical Alignment
Vertical Alignment
1. Form six groups.
a)
b)
c)
d)
e)
f)
Operations and Algebraic Thinking Grade 3
Operations and Algebraic Thinking Grade 4
Operations and Algebraic Thinking Grade 5
Number and Operation Fractions Grade 3
Number and Operations Fractions Grade 4
Number and Operations Fractions Grade 5
core and discuss it with your group to solidify
3. Re-group so that you are now composed of 3
members one from each grade level.
a)
b)
3, 4, 5 Numbers and Algebraic thinking
3, 4, 5 Numbers and Operations Fractions
5. Combine groups of three, still working within your
domain but you should now have 6 members with
notes.
6. Create a poster displaying the vertical alignment
Reflection
2. If so, why?
3. If not, what would have been more
4. How will this activity change your
teaching next year?
Critical
Areas
Critical Area Graphic
Organizer
• Identify at least one or two important mathematical
concepts within this critical area. What do students
need to learn prior to these concepts?
• What evidence would convince you that a student
understands these concepts?
• What common misconceptions do students have
when studying this critical area? What challenges
have you had in teaching these concepts?
• How do these concepts support learning in later
Reflection
• How do the critical areas help bring focus to the
• How will you use the critical areas to inform your
• What questions do you still have about the
critical areas?
• How has this activity increased your
understanding of the instructional core?
Conceptual
Understanding
What does understand really mean?
1.
2.
3.
4.
5.
6.
7.
8.
Form groups of 5-6 and sit together at a table.
The group discusses the question prompt until time is called
The leader’s role is to record the major points of the conversation that
take place at the table and to then summarize the conversation using
the recorded notes just before rotating to the next table.
The leader stays put; the rest of the group rotates to the next table.
The leader (the one who didn’t move) presents a summary of the
conversation recorded from the former group to the new group.
The new group discusses the new question prompt until time is called.
Again the leader takes notes and summarizes.
Round 1
Discuss the following:
The word "understanding" is used frequently
throughout the document. What do you
think it means and why is it important in a
child's ability to learn math concepts?
Round 2
Reflect on the quote:
"A significant indicator of conceptual
understanding is being able to represent
mathematical situations in different ways,
knowing how different representations can be
useful for different purposes."
Explain why you think it's important for a child
to be familiar with more than one way to solve
a problem.
Round 3
Discuss the following:
How can a teacher help students make
connections between concepts and
representations?
DOK & Cognitive Rigor
1. Develop a shared understanding of the
Cognitive Rigor Matrix.
2. Use the Cognitive Rigor Matrix to:
• Consider rigor expectations in the Common Core
• Examine learning expectations and critical thinking
• Evaluate sample assessments/tasks & rubrics
Before we begin…
Take a couple of minutes to write
“cognitive rigor” as it relates to
instruction, learning, and/or
assessment.
definition
fractions with like denominators.
• What is a basic computational question you might
• What is a more rigorous question you might ask?
Hess Article
Annotate:
• ? On parts of the article that need clarification
• ! For any “ahas” as you read
• Underline or highlight any key terms or
vocabulary
• * next to any key/main ideas from the article
The Hess Cognitive
Rigor Matrix: Integrates
Bloom’s & Webb’s
Blooms Taxonomy
What type of thinking
(verbs) is needed to
Webb’s Depth of
Knowledge
How deeply do you have to
understand the content to
successfully interact with it?
How complex is the
content?
Blooms Taxonomy
Draw
Identify
Define
List
Memorize
Calculate
Illustrate
Who, What, When, Where, Why
Arrange
Design
Tabulate
Repeat
Recall
Connect
Level
Four
Apply Concepts
(Extended
Thinking)
Critique
Analyze
Create
Match
Recognize
Prove
Categorize
Use
Infer
Level One
(Recall)
Synthesize
Webb’s
Depth of
Knowledge
Measure
Name
Graph
Organize
Classify
Describe
Explain
Interpret
Level Two
(Skill/
Concept)
Level Three
(Strategic Thinking)
Modify
Cause/Effect
Relate
Predict
Compare
Interpret
Estimate
Revise
Assess
Summarize
Develop a Logical Argument
Critique
Use Concepts to Solve
Non-Routine Problems
Apprise
Formulate
Investigate
Draw Conclusions
Hypothesize
Construct
Compare
Explain
Differentiate
Show
Webb’s Depth-of-Knowledge
Levels
• DOK 1- Recall & Reproduction-Recall a fact, term, principle,
concept, or perform a routine procedure.
• DOK 2- Basic Application of Skills/Concepts-Use of information,
conceptual knowledge, select appropriate procedures for a task, two
or more steps with decision points along the way, routine problems,
organize/display data, interpret/use simple graphs.
• DOK 3- Strategic Thinking-Requires reasoning, developing a plan
or sequence of steps to approach problem; requires some decision
making and justification; abstract, complex, or non-routine; often
more than one possible answer or approach.
• DOK 4- Extended Thinking-An original investigation or application
to real word; requires time to research, problem solve, and process
multiple conditions of the problem or task; non-routine
manipulations, across disciplines/content area/multiple sources.
difficulty!
The intended student learning outcome determines
the DOK level. What mental processing must occur?
While verbs may appear to point to a DOK level, it is
what comes after the verb that is the best indicator of
the rigor/DOK level.
Let’s see some examples…
It’s what after the verb…
• Describe how two characters are alike and different.
• Describe the information contained in graphics or
data tables in the text; or the rule for rounding of a
number
• Describe the data or text evidence that supports
• Describe varying perspectives on global climate
change using supporting scientific evidence, and
identify the most significant effects it might have on
the planet in 100 years
DOK 1
Recall and Reproduction
DOK 2
Recall, locate basic
facts, definitions,
details, events
Remember
Understand
Apply
Select appropriate
words for use when
intended meaning is
clearly evident.
Explain relationships
Summarize
State central idea
Use language
structure, word
relationships
to determine meaning
Analyze
Use context for word
meanings
Use information using
text features
Identify the kind of
information contained
in a graphic, table,
visual, etc.
Compare literary elements,
facts, terms and events
Analyze format,
organization and text
structures
DOK 4
Reasoning and
Thinking
Extended Thinking
-Explain how concepts
or ideas specifically
relate to other content
domains.
Explain, generalize or
connect ideas using
supporting evidence
(quote, text, evidence)
Devise an approach
among many
alternatives to
research a novel
problem
Use concepts to solve
non-routine problems
and justify
Analyze or interpret
author’s craft (e.g.,
literary devices,
viewpoint, or potential
bias) to critique a text
Analyze multiple
sources or multiple
text. Analyze
complex abstract
themes
Evaluate relevancy,
accuracy and
completeness of
information across
texts or sources
Cite evidence and develop a
logical argument for
conjectures based on one
text or problem
Evaluate
Create
DOK 3
Skills and
Concepts
-Brainstorm ideas,
concepts, problems,
or perspectives
related to a topic or
concept.
Selected Response
Generate conjectures
or hypotheses based
on observations or
prior knowledge
Constructed
Response
Develop a complex model or
approach for a given
situation
Develop an alternative
solution
Synthesize across
multiple sources/ texts
Articulate a new voice,
theme, or perspective
Performance
What DOK Level?
What DOK Level?
What DOK Level?
What DOK Level?
What DOK Level?
Find the next three terms in the pattern
and determine the rule for the
following pattern of numbers:
1, 4, 8, 11, 15, 18, 22, 25, 29, …
Revised Bloom’s
Taxonomy
Remember
Retrieve knowledge from longterm memory, recognize, recall,
locate, identify
Understand
Construct meaning, clarify,
paraphrase, represent, translate,
illustrate, give examples, classify,
categorize, summarize,
generalize, infer logical
conclusion (such as from
examples given), predict,
compare/contrast, match like
ideas, explain, construct models
Apply
Carry out or use a procedure in
a given situation; carry out
(apply to a familiar task), or use
DOK with Multiplication
Webb’s DOK Level 1
Recall & Reproduction
Webb’s DOK Level 2
Skills & Concepts
Draw a model to
represent the problem
5x3
Estimate a reasonable
23 x 8. Explain how you
why it is reasonable.
Find the area of this
shape.
Write a multiplication
statement comparing
brown eyes to blue.
Analyze
Break into constituent parts,
determine how parts relate,
differentiate between relevantirrelevant, distinguish, focus,
select, organize, outline, find,
coherence, deconstruct
Evaluate
Make judgments based on
criteria, check, detect,
inconsistencies or fallacies,
judge, critique
Create
Reorganize elements into new
patterns/structures, generate,
hypothesize, design, plan,
construct, produce
Webb’s DOK Level 3
Strategic Thinking/Reasoning
Compare the 2 strategies used. Explain
the process each student used and
explain who has the correct
Create a table displaying all
the factors of 48.
Webb’s DOK Level 4
Extended Thinking
Can the distributive
property be used with
division? Justify your
conclusions with data
generated from your
investigation.
A
Construct an area model
for the polygon from the
table
with
the
largest
area.
D
Describe
the trend
displayed
in the
data table
to a multiplication
problem called?
B
C
E
Make a conclusion based on
the data presented, use
mathematical relationships
F
DOK
Question
Stems
Questions
3.G.1-2
or
4.G.1-3
Wrap Up & Reflect
Questions or Concerns……..
Complete a 4 point evaluation:
•
•
•
•
What went well today?
What could be improved?
What do you need more support in?
What did you master?
```
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# Finite Mathematics and Calculus with Applications
## Educators
Problem 1
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 2
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 3
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 4
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 5
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 6
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 7
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 8
Find the locations of any absolute extrema for the functions with graphs as follows.
GRAPH
Check back soon!
Problem 9
What is the difference between a relative extremum and an absolute extremum?
Check back soon!
Problem 10
Can a relative extremum be an absolute extremum? Is a relative extremum necessarily an absolute extremum?
Check back soon!
Problem 11
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x^{3}-6 x^{2}+9 x-8 ; [0,5]$$
Check back soon!
Problem 12
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x^{3}-3 x^{2}-24 x+5 ; [-3,6]$$
Check back soon!
Problem 13
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{1}{3} x^{3}+\frac{3}{2} x^{2}-4 x+1 ; [-4,2]$$
Check back soon!
Problem 14
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{1}{3} x^{3}-\frac{1}{2} x^{2}-6 x+3 ; [-4,4]$$
Check back soon!
Problem 15
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x^{4}-18 x^{2}+1 ; [-4,4]$$
Check back soon!
Problem 16
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x^{4}-32 x^{2}-7 ; [-5,6]$$
Check back soon!
Problem 17
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{1-x}{3+x} ; [0,3]$$
Check back soon!
Problem 18
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{8+x}{8-x} ;[4,6]$$
Check back soon!
Problem 19
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{x-1}{x^{2}+1} ; [1,5]$$
Check back soon!
Problem 20
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{x}{x^{2}+2} ; [0,4]$$
Check back soon!
Problem 21
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\left(x^{2}-4\right)^{1 / 3} ; [-2,3]$$
Check back soon!
Problem 22
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\left(x^{2}-16\right)^{2 / 3} ; [-5,8]$$
Check back soon!
Problem 23
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=5 x^{2 / 3}+2 x^{5 / 3} ; d[-2,1]$$
Check back soon!
Problem 24
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x+3 x^{2 / 3} ; [-10,1]$$
Check back soon!
Problem 25
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x^{2}-8 \ln x ;[1,4]$$
Check back soon!
Problem 26
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{\ln x}{x^{2}} ;[1,4]$$
Check back soon!
Problem 27
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x+e^{-3 x} ;[-1,3]$$
Check back soon!
Problem 28
Find the absolute extrema if they exist, as well as all values of $x$ where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=x^{2} e^{-0.5 x} ; [2,5]$$
Check back soon!
Problem 29
Graph each function on the indicated domain, and use the capabilities of your calculator to find the location and value of the absolute extrema.
$$f(x)=\frac{-5 x^{4}+2 x^{3}+3 x^{2}+9}{x^{4}-x^{3}+x^{2}+7} ;[-1,1]$$
Check back soon!
Problem 30
Graph each function on the indicated domain, and use the capabilities of your calculator to find the location and value of the absolute extrema.
$$f(x)=\frac{x^{3}+2 x+5}{x^{4}+3 x^{3}+10} ;[-3,0]$$
Check back soon!
Problem 31
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=2 x+\frac{8}{x^{2}}+1, x>0$$
Check back soon!
Problem 32
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=12-x-\frac{9}{x}, x>0$$
Check back soon!
Problem 33
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=-3 x^{4}+8 x^{3}+18 x^{2}+2$$
Check back soon!
Problem 34
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=x^{4}-4 x^{3}+4 x^{2}+1$$
Check back soon!
Problem 35
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=\frac{x-1}{x^{2}+2 x+6}$$
Check back soon!
Problem 36
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=\frac{x}{x^{2}+1}$$
Check back soon!
Problem 37
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=\frac{\ln x}{x^{3}}$$
Check back soon!
Problem 38
Find the absolute extrema if they exist, as well as all values of $x$ where they occur.
$$f(x)=x \ln x$$
Check back soon!
Problem 39
Find the absolute maximum and minimum of $f(x)=$ $2 x-3 x^{2 / 3}(a)$ on the interval $[-1,0.5] ;(b)$ on the interval $[0.5,2] .$
Check back soon!
Problem 40
Let $f(x)=e^{-2 x} .$ For $x>0,$ let $P(x)$ be the perimeter of the rectangle with vertices $(0,0),(x, 0),(x, f(x))$ and $(0, f(x)) .$ Which of the following statements is true? Society of Actuaries.
a. The function $P$ has an absolute minimum but not an absolute maximum on the interval $(0, \infty) .$
b. The function $P$ has an absolute maximum but not an absolute minimum on the interval (0, $\infty$ ).
c. The function $P$ has both an absolute minimum and an absolute maximum on the interval $(0, \infty)$ .
d. The function $P$ has neither an absolute maximum nor an absolute minimum on the interval $(0, \infty),$ but the graph of the function $P$ does have an inflection point with positive $x$ -coordinate.
e. The function $P$ has neither an absolute maximum nor an absolute minimum on the interval $(0, \infty),$ and the graph of the function $P$ does not have an inflection point with positive $x$ -coordinate.
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Problem 41
Bank Robberies The number of bank robberies in the United States for the years 2000–2009 is given in the following figure. Consider the closed interval [2000, 2009]. Source: FBI.
a. Give all relative maxima and minima and when they occur on the interval.
b. Give the absolute maxima and minima and when they occur on the interval. Interpret your results.
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Problem 42
Bank Burglaries The number of bank burglaries (entry into or theft from a bank during nonbusiness hours) in the United States for the years 2000–2009 is given in the figure on the following page. Consider the closed interval [2000, 2009]. Source: FBI.
a. Give all relative maxima and minima and when they occur on the interval.
b. Give the absolute maxima and minima and when they occur on the interval. Interpret your results.
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Problem 43
Profit The total profit $P(x)$ (in thousands of dollars) from the sale of $x$ hundred thousand automobile tires is approximated by
$$P(x)=-x^{3}+9 x^{2}+120 x-400, \quad x \geq 5$$
Find the number of hundred thousands of tires that must be sold to maximize profit. Find the maximum profit.
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Problem 44
Profit A company has found that its weekly profit from the sale of $x$ units of an auto part is given by
$$P(x)=-0.02 x^{3}+600 x-20,000.$$
Production bottlenecks limit the number of units that can be made per week to no more than $150,$ while a long-term contract requires that at least 50 units be made each week. Find the maximum possible weekly profit that the firm can make.
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Problem 45
Average Cost Find the minimum value of the average cost for the given cost function on the given intervals.
$C(x)=x^{3}+37 x+250$ on the following intervals.
a. $1 \leq x \leq 10 \qquad$ b. $10 \leq x \leq 20$
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Problem 46
Average Cost Find the minimum value of the average cost for the given cost function on the given intervals.
$C(x)=81 x^{2}+17 x+324$ on the following intervals.
a. $1 \leq x \leq 10 \qquad$ b. $10 \leq x \leq 20$
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Problem 47
Each graph gives the cost as a function of production level. Use the method of graphical optimization to estimate the production level that results in the minimum cost per item produced.
GRAPH
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Problem 48
Each graph gives the cost as a function of production level. Use the method of graphical optimization to estimate the production level that results in the minimum cost per item produced.
GRAPH
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Problem 49
Each graph gives the profit as a function of production level. Use graphical optimization to estimate the production level that gives the maximum profit per item produced.
GRAPH
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Problem 50
Each graph gives the profit as a function of production level. Use graphical optimization to estimate the production level that gives the maximum profit per item produced.
GRAPH
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Problem 51
A marshy region used for agricultural drainage has become contaminated with selenium. It has been determined that flushing the area with clean water will reduce the selenium for a while, but it will then begin to build up again. A biologist has found that the percent of selenium in the soil $x$ months after the flushing begins is given by
$$f(x)=\frac{x^{2}+36}{2 x}, \quad 1 \leq x \leq 12$$
When will the selenium be reduced to a minimum? What is the minimum percent?
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Problem 52
Salmon Spawning The number of salmon swimming upstream to spawn is approximated by
$$S(x)=-x^{3}+3 x^{2}+360 x+5000, 6 \leq x \leq 20,$$
where $x$ represents the temperature of the water in degrees Celsius. Find the water temperature that produces the maximum number of salmon swimming upstream.
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Problem 53
Molars Researchers have determined that the crown length of first molars in fetuses is related to the postconception age of the tooth as
$$L(t)=-0.01 t^{2}+0.788 t-7.048$$
where $L(t)$ is the crown length (in millimeters) of the molar $t$ weeks after conception. Find the maximum length of the molar $t$ of first molars during weeks 22 through $28 .$ Source: American Journal of Physical Anthropology.
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Problem 54
Fungal Growth Because of the time that many people spend indoors, there is a concern about the health risk of being exposed to harmful fungi that thrive in buildings. The risk appears to increase in damp environments. Researchers have discovered that by controlling both the temperature and the relative humidity in a building, the growth of the fungus $A$ . versicolor can be limited. The relationship between temperature and relative humidity, which limits growth, can be described by
$$R(T)=-0.00007 T^{3}+0.0401 T^{2}-1.6572 T+97.086$$
$$15 \leq T \leq 46,$$
where $R(T)$ is the relative humidity (in percent) and $T$ is the temperature (in degrees Celsius). Find the temperature at which the relative humidity is minimized. Source: Applied and Environmental Microbiology.
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Problem 55
Gasoline Mileage From information given in a recent business publication, we constructed the mathematical model
$$M(x)=-\frac{1}{45} x^{2}+2 x-20, 30 \leq x \leq 65$$
to represent the miles per gallon used by a certain car at a speed of $x$ mph. Find the absolute maximum miles per gallon and the absolute minimum and the speeds at which they occur.
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Problem 56
Gasoline Mileage For a certain sports utility vehicle,
$$M(x)=-0.015 x^{2}+1.31 x-7.3,30 \leq x \leq 60$$
represents the miles per gallon obtained at a speed of $x$ mph. Find the absolute maximum miles per gallon and the absolute minimum and the speeds at which they occur.
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Problem 57
Area A piece of wire 12 $\mathrm{ft}$ long is cut into two pieces. (See the figure.) One piece is made into a circle and the other piece is made into a square. Let the piece of length $x$ be formed into a circle. We allow $x$ to equal 0 or $12,$ so all the wire may be used for the square or for the circle.
Radius of circle $=\frac{x}{2 \pi} \qquad$ Area of circle $=\pi\left(\frac{x}{2 \pi}\right)^{2}$
Side of square $=\frac{12-x}{4} \quad$ Area of square $=\left(\frac{12-x}{4}\right)^{2}$
Where should the cut be made in order to minimize the sum of the areas enclosed by both figures?
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Problem 58
Area A piece of wire 12 $\mathrm{ft}$ long is cut into two pieces. (See the figure.) One piece is made into a circle and the other piece is made into a square. Let the piece of length $x$ be formed into a circle. We allow $x$ to equal 0 or $12,$ so all the wire may be used for the square or for the circle.
Radius of circle $=\frac{x}{2 \pi} \qquad$ Area of circle $=\pi\left(\frac{x}{2 \pi}\right)^{2}$
Side of square $=\frac{12-x}{4} \quad$ Area of square $=\left(\frac{12-x}{4}\right)^{2}$
Where should the cut be made in order to make the sum of the areas maximum? (Hint: Remember to use the endpoints of a domain when looking for absolute maxima and minima.)
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Problem 59
Area A piece of wire 12 $\mathrm{ft}$ long is cut into two pieces. (See the figure.) One piece is made into a circle and the other piece is made into a square. Let the piece of length $x$ be formed into a circle. We allow $x$ to equal 0 or $12,$ so all the wire may be used for the square or for the circle.
Radius of circle $=\frac{x}{2 \pi} \qquad$ Area of circle $=\pi\left(\frac{x}{2 \pi}\right)^{2}$
Side of square $=\frac{12-x}{4} \quad$ Area of square $=\left(\frac{12-x}{4}\right)^{2}$
For the solution to Exercise 57, show that the side of the square equals the diameter of the circle, that is, that the circle can be inscribed in the square.*
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Problem 60
Area A piece of wire 12 $\mathrm{ft}$ long is cut into two pieces. (See the figure.) One piece is made into a circle and the other piece is made into a square. Let the piece of length $x$ be formed into a circle. We allow $x$ to equal 0 or $12,$ so all the wire may be used for the square or for the circle.
Radius of circle $=\frac{x}{2 \pi} \qquad$ Area of circle $=\pi\left(\frac{x}{2 \pi}\right)^{2}$
Side of square $=\frac{12-x}{4} \quad$ Area of square $=\left(\frac{12-x}{4}\right)^{2}$
Information Content Suppose dots and dashes are transmitted over a telegraph line so that dots occur a fraction $p$ of the time (where $0 < p < 1 )$ and dashes occur a fraction $1-p$ of the time. The information content of the telegraph line is given by $I(p),$ where
$$I(p)=-p \ln p-(1-p) \ln (1-p)$$
a. Show that $I^{\prime}(p)=-\ln p+\ln (1-p)$
b. Set $I^{\prime}(p)=0$ and find the value of $p$ that maximizes the information content.
c. How might the result in part b be used?
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10.2.1: Linear Systems of Differential Equations (Exercises)
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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ Q10.2.1 1. Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants $$c_1$$ and $$c_2$$. 1. $$\begin{array}{ccl}y'_1&=&2y_1 + 4y_2\\ y_2'&=&4y_1+2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{6t}+c_2\twocol1{-1}e^{-2t}$$ 2. $$\begin{array}{ccl}y'_1&=&-2y_1 - 2y_2\\ y_2'&=&-5y_1 + \phantom{2}y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{-4t}+c_2\twocol{-2}5e^{3t}$$ 3. $$\begin{array}{ccr}y'_1&=&-4y_1 -10y_2\\ y_2'&=&3y_1 + \phantom{1}7y_2;\end{array} \quad {\bf y}=c_1\twocol{-5}3e^{2t}+c_2\twocol2{-1}e^t$$ 4. $$\begin{array}{ccl}y'_1&=&2y_1 +\phantom{2}y_2 \\ y_2'&=&\phantom{2}y_1 + 2y_2;\end{array} \quad {\bf y}=c_1\twocol11e^{3t}+c_2\twocol1{-1}e^t$$ 2. Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants $$c_1$$, $$c_2$$, and $$c_3$$. 1. $$\begin{array}{ccr}y'_1&=&- y_1+2y_2 + 3y_3 \\ y_2'&=&y_2 + 6y_3\\y_3'&=&- 2y_3;\end{array}$$ $${\bf y}=c_1\threecol110e^t+c_2\threecol100e^{-t}+c_3\threecol1{-2}1e^{-2t}$$ 2. $$\begin{array}{ccc}y'_1&=&\phantom{2y_1+}2y_2 + 2y_3 \\ y_2'&=&2y_1\phantom{+2y_2} + 2y_3\\y_3'&=&2y_1 + 2y_2;\phantom{+2y_3}\end{array}$$ $${\bf y}=c_1\threecol{-1}01e^{-2t}+c_2\threecol0{-1}1e^{-2t}+c_3\threecol111e^{4t}$$ 3. $$\begin{array}{ccr}y'_1&=&-y_1 +2y_2 + 2y_3\\ y_2'&=&2y_1 -\phantom{2}y_2 +2y_3\\y_3'&=&2y_1 + 2y_2 -\phantom{2}y_3;\end{array}$$ $${\bf y}=c_1\threecol{-1}01e^{-3t}+c_2\threecol0{-1}1e^{-3t}+c_3\threecol111e^{3t}$$ 4. $$\begin{array}{ccr}y'_1&=&3y_1 - \phantom{2}y_2 -\phantom{2}y_3 \\ y_2'&=&-2y_1 + 3y_2 + 2y_3\\y_3'&=&\phantom{-}4y_1 -\phantom{3}y_2 - 2y_3;\end{array}$$ $${\bf y}=c_1\threecol101e^{2t}+c_2\threecol1{-1}1e^{3t}+c_3\threecol1{-3}7e^{-t}$$ 3. Rewrite the initial value problem in matrix form and verify that the given vector function is a solution. 1. $$\begin{array}{ccl}y'_1 &=&\phantom{-2}y_1+\phantom{4}y_2\\ y_2'&=&-2y_1 + 4y_2,\end{array} \begin{array}{ccr}y_1(0)&=&1\\y_2(0)&=&0;\end{array}$$ $${\bf y}=2\twocol11e^{2t}-\twocol12e^{3t}$$ 2. $$\begin{array}{ccl}y'_1 &=&5y_1 + 3y_2 \\ y_2'&=&- y_1 + y_2,\end{array} \begin{array}{ccr}y_1(0)&=&12\\y_2(0)&=&-6;\end{array}$$ $${\bf y}=3\twocol1{-1}e^{2t}+3\twocol3{-1}e^{4t}$$ 4. Rewrite the initial value problem in matrix form and verify that the given vector function is a solution. 1. $$\begin{array}{ccr}y'_1&=&6y_1 + 4y_2 + 4y_3 \\ y_2'&=&-7y_1 -2y_2 - y_3,\\y_3'&=&7y_1 + 4y_2 + 3y_3\end{array},\; \begin{array}{ccr}y_1(0)&=&3\\ y_2(0)&=&-6\\ y_3(0)&=&4\end{array}$$ $${\bf y}=\threecol1{-1}1e^{6t}+2\threecol1{-2}1e^{2t}+\threecol0{-1}1e^{-t}$$ 2. $$\begin{array}{ccr}y'_1&=& \phantom{-}8y_1 + 7y_2 +\phantom{1}7y_3 \\ y_2'&=&-5y_1 -6y_2 -\phantom{1}9y_3,\\y_3'&=& \phantom{-}5y_1 + 7y_2 +10y_3,\end{array}\ \begin{array}{ccr}y_1(0)&=&2\\ y_2(0)&=&-4\\ y_3(0)&=&3\end{array}$$ $${\bf y}=\threecol1{-1}1e^{8t}+\threecol0{-1}1e^{3t}+\threecol1{-2}1e^t$$ 5. Rewrite the system in matrix form and verify that the given vector function satisfies the system for any choice of the constants $$c_1$$ and $$c_2$$. 1. $$\begin{array}{ccc}y'_1&=&-3y_1+2y_2+3-2t \\ y_2'&=&-5y_1+3y_2+6-3t\end{array}$$ $${\bf y}=c_1\left[\begin{array}{c}2\cos t\\3\cos t-\sin t\end{array}\right]+c_2\left[\begin{array}{c}2\sin t\\3\sin t+\cos t \end{array}\right]+\twocol1t$$ 2. $$\begin{array}{ccc}y'_1&=&3y_1+y_2-5e^t \\ y_2'&=&-y_1+y_2+e^t\end{array}$$ $${\bf y}=c_1\twocol{-1}1e^{2t}+c_2\left[\begin{array}{c}1+t\\-t\end{array} \right]e^{2t}+\twocol13e^t$$ 3. $$\begin{array}{ccl}y'_1&=&-y_1-4y_2+4e^t+8te^t \\ y_2'&=&-y_1-\phantom{4}y_2+e^{3t}+(4t+2)e^t\end{array}$$ $${\bf y}=c_1\twocol21e^{-3t}+c_2\twocol{-2}1e^t+\left[\begin{array}{c} e^{3t}\\2te^t\end{array}\right]$$ 4. $$\begin{array}{ccc}y'_1&=&-6y_1-3y_2+14e^{2t}+12e^t \\ y_2'&=&\phantom{6}y_1-2y_2+7e^{2t}-12e^t\end{array}$$ $${\bf y}=c_1\twocol{-3}1e^{-5t}+c_2\twocol{-1}1e^{-3t}+ \left[\begin{array}{c}e^{2t}+3e^t\\2e^{2t}-3e^t\end{array}\right]$$ 6. Convert the linear scalar equation \[P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y(t)=F(t) \tag{A}$
into an equivalent $$n\times n$$ system
${\bf y'}=A(t){\bf y}+{\bf f}(t),\nonumber$
and show that $$A$$ and $${\bf f}$$ are continuous on an interval $$(a,b)$$ if and only if (A) is normal on $$(a,b)$$.
7. A matrix function
$Q(t)=\left[\begin{array}{cccc}{q_{11}(t)}&{q_{12}(t)}&{\cdots }&{q_{1s}(t)} \\ {q_{21}(t)}&{q_{22}(t)}&{\cdots }&{q_{2s}(t)} \\ {\vdots }&{\vdots }&{\ddots }&{\vdots } \\ {q_{r1}(t)}&{q_{r2}(t)}&{\cdots }&{q_{rs}(t)} \end{array} \right] \nonumber$
is said to be differentiable if its entries $$\{q_{ij}\}$$ are differentiable. Then the derivative $$Q'$$ is defined by
$Q(t)=\left[\begin{array}{cccc}{q'_{11}(t)}&{q'_{12}(t)}&{\cdots }&{q'_{1s}(t)} \\ {q'_{21}(t)}&{q'_{22}(t)}&{\cdots }&{q'_{2s}(t)} \\ {\vdots }&{\vdots }&{\ddots }&{\vdots } \\ {q'_{r1}(t)}&{q'_{r2}(t)}&{\cdots }&{q'_{rs}(t)} \end{array} \right] \nonumber$
1. Prove: If $$P$$ and $$Q$$ are differentiable matrices such that $$P+Q$$ is defined and if $$c_1$$ and $$c_2$$ are constants, then $(c_1P+c_2Q)'=c_1P'+c_2Q'.\nonumber$
2. Prove: If $$P$$ and $$Q$$ are differentiable matrices such that $$PQ$$ is defined, then $(PQ)'=P'Q+PQ'.\nonumber$
8. Verify that $$Y' = AY$$.
1. $$Y=\left[\begin{array}{cc}{e^{6t}}&{e^{-2t}}\\{e^{6t}}&{-e^{-2t}} \end{array} \right],\quad A=\left[\begin{array}{cc}{2}&{4}\\{4}&{2} \end{array} \right]$$
2. $$Y=\left[\begin{array}{cc}{e^{-4t}}&{-2e^{3t}}\\{e^{-4t}}&{5e^{3t}} \end{array} \right],\quad A=\left[\begin{array}{cc}{-2}&{-2}\\{-5}&{1} \end{array} \right]$$
3. $$Y=\left[\begin{array}{cc}{-5e^{2t}}&{2e^{t}}\\{3e^{2t}}&{-e^{t}} \end{array} \right],\quad A=\left[\begin{array}{cc}{-4}&{-10}\\{3}&{7} \end{array} \right]$$
4. $$Y=\left[\begin{array}{cc}{e^{3t}}&{e^{t}}\\{e^{3t}}&{-e^{t}} \end{array} \right],\quad A=\left[\begin{array}{cc}{2}&{1}\\{1}&{2} \end{array} \right]$$
5. $$Y = \left[\begin{array}{ccc} e^t&e^{-t}& e^{-2t}\\ e^t&0&-2e^{-2t}\\ 0&0&e^{-2t}\end{array}\right], \quad A = \left[\begin{array}{ccc}{-1}&{2}&{3}\\{0}&{1}&{6}\\{0}&{0}&{-2} \end{array} \right]$$
6. $$Y = \left[\begin{array}{ccc} {-e^{-2t}}&{-e^{-2t}}& {e^{4t}}\\ {0}&{e^{-2t}}&{e^{4t}}\\ {e^{-2t}}&{0}&{e^{4t}}\end{array}\right], \quad A = \left[\begin{array}{ccc}{0}&{2}&{2}\\{2}&{0}&{2}\\{2}&{2}&{0} \end{array} \right]$$
7. $$Y = \left[\begin{array}{ccc} {e^{3t}}&{e^{-3t}}& {0}\\ {e^{3t}}&{0}&{-e^{-3t}}\\ {e^{3t}}&{e^{-3t}}&{e^{-3t}}\end{array}\right], \quad A = \left[\begin{array}{ccc}{-9}&{6}&{6}\\{-6}&{3}&{6}\\{-6}&{6}&{3} \end{array} \right]$$
8. $$Y = \left[\begin{array}{ccc} {e^{2t}}&{e^{3t}}& {e^{-t}}\\ {0}&{-e^{3t}}&{-3e^{-t}}\\ {e^{2t}}&{e^{3t}}&{7e^{-t}}\end{array}\right], \quad A = \left[\begin{array}{ccc}{3}&{-1}&{-1}\\{-2}&{3}&{2}\\{4}&{-1}&{-2} \end{array} \right]$$
9. Suppose
${\bf y}_1=\twocol{y_{11}}{y_{21}}\quad \text{and} \quad{\bf y}_2=\twocol{y_{12}}{y_{22}}\nonumber$
are solutions of the homogeneous system
${\bf y}'=A(t){\bf y}, \tag{A}$
and define
$Y= \left[\begin{array}{cc}{y_{11}}&{y_{12}}\\{y_{21}}&{y_{22}}\end{array}\right].\nonumber$
1. Show that $$Y'=AY$$.
2. Show that if $${\bf c}$$ is a constant vector then $${\bf y}= Y{\bf c}$$ is a solution of (A).
3. State generalizations of (a) and (b) for $$n\times n$$ systems.
10. Suppose $$Y$$ is a differentiable square matrix.
1. Find a formula for the derivative of $$Y^2$$.
2. Find a formula for the derivative of $$Y^n$$, where $$n$$ is any positive integer.
3. State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions.
11. It can be shown that if $$Y$$ is a differentiable and invertible square matrix function, then $$Y^{-1}$$ is differentiable.
1. Show that ($$Y^{-1})'= -Y^{-1}Y'Y^{-1}$$. (Hint: Differentiate the identity $$Y^{-1}Y=I$$.)
2. Find the derivative of $$Y^{-n}=\left(Y^{-1}\right)^n$$, where $$n$$ is a positive integer.
3. State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions.
12. Show that Theorem 10.2.1 implies Theorem 9.1.1. HINT: Write the scalar function $P_{0}(x)y^{(n)}+P_{1}(x)y^{(n-1)}+\cdots +P_{n}(x)y=F(x)\nonumber$ as an $$n\times n$$ system of linear equations.
13. Suppose $${\bf y}$$ is a solution of the $$n\times n$$ system $${\bf y}'=A(t){\bf y}$$ on $$(a,b)$$, and that the $$n\times n$$ matrix $$P$$ is invertible and differentiable on $$(a,b)$$. Find a matrix $$B$$ such that the function $${\bf x}=P{\bf y}$$ is a solution of $${\bf x}'=B{\bf x}$$ on $$(a,b)$$.
This page titled 10.2.1: Linear Systems of Differential Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.
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# 2016 AMC 10A Problems/Problem 24
The following problem is from both the 2016 AMC 10A #24 and 2016 AMC 12A #21, so both problems redirect to this page.
## Problem
A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$
## Solution 1
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,W); label("B",B,NW); label("C",C,NE); label("D",D,ENE); label("O",O,S); label("\theta",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("E",E,NE); F=extension(C,O,A,D); label("F",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$
Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$. Since $\frac{OA}{AB} = \frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$.
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$
## Solution 2 (Algebra)
To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,W); label("B",B,NW); label("C",C,NE); label("D",D,E); label("E",E,WSW); label("O",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); [/asy]$
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Let the intersection of $BD$ and $OC$ be point $E$. Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem, $$\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}$$
We solve for $x$: $$1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2$$ $$2\sqrt{(1-x^2)(2-x^2)}=2x^2-1$$ $$4(1-x^2)(2-x^2)=(2x^2-1)^2$$ $$8-12x^2+4x^4=4x^4-4x^2+1$$ $$8x^2=7$$ $$x=\frac{\sqrt{14}}{4}$$
By Ptolemy's Theorem, $$AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2$$
Substituting values, $$1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2$$ $$1+AD=\frac{7}{2}$$ $$AD=\frac{5}{2}$$
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{\textbf{(E) } 500}$.
## Solution 3 (HARD Algebra)
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,W); label("B",B,NW); label("C",C,NE); label("D",D,E); label("O",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); [/asy]$
Let quadrilateral $ABCD$ be inscribed in circle $O$, where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$.
By the Pythagorean Theorem, the length of $OH$ is \begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore, \begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$, we get: $$\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}$$ $$\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}$$ $$60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)$$ $$40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}$$ $$400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}$$ $$(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}$$ $$7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)$$ $$7x^2 - 5600x + 1120000 = 320000 - x^2$$ $$8x^2 - 5600x + 800000 = 0$$ $$x^2 - 700x + 100000 = 0$$
The roots of this quadratic are found by using the quadratic formula: \begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$, then $ABCD$ would be a square. Thus, the radius of the circle would be $$\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}$$ Which is a contradiction. Therefore, our answer is $\boxed{500}.$
## Solution 4 (Trigonometry Bash)
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,W); label("B",B,NW); label("C",C,NE); label("D",D,E); label("O",O,S); label("\theta",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]$
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get the following equation: $$(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta$$ Substituting the values in, we get $$(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta$$ Canceling out, we get $$\cos\theta=\frac{3}{4}$$ Because $\angle AOB$, $\angle BOC$, and $\angle COD$ are congruent, $\angle AOD = 3\theta$. To find the remaining side ($AD$), we simply have to apply the law of cosines to $\Delta AOD$. Now, to find $\cos 3\theta$, we can derive a formula that only uses $\cos\theta$: $$\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta$$ $$\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)$$ $$\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta$$ It is useful to memorize the triple angle formulas ($\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$). Plugging in $\cos\theta=\frac{3}{4}$, we get $\cos 3\theta= -\frac{9}{16}$. Now, applying law of cosines on triangle $OAD$, we get $$(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}$$ $$\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}$$ $$AD=200 \cdot \frac{5}{2}=\boxed{500}$$
## Solution 5 (Easier Trigonometry)
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,W); label("B",B,NW); label("C",C,NE); label("D",D,ENE); label("O",O,S); label("\theta",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("E",E,NW); label("F",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); [/asy]$
Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \frac{3}{4}$.
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$, we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$. In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$, so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$, so $\angle DCF = \theta$.
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \boxed{500}$.
## Solution 6 (Ptolemy's Theorem)
$[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]$
Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: $$\boxed{\textbf{(E) } 500.}$$
## Solution 7 (Trigonometry)
Since all three sides equal $200$, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\sqrt{7},200\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\frac{100}{200\sqrt{2}}=\frac{\sqrt{2}}{4}$. Similarly, the cosine is $\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}$. Since there are three sides, and since $\sin\theta=\sin\left(180-\theta\right)$,we seek to find $2r\sin 3\theta$. First, $\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}$ and $\cos 2\theta=\frac{3}{4}$ by Pythagorean. $$\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac{\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}$$ $$2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac{5\sqrt{2}}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(E)}\text{ 500}}$$
## Solution 8 (Area By Brahmagupta's Formula)
For simplicity, scale everything down by a factor of 100. Let the inscribed trapezoid be $ABCD$, where $AB=BC=CD=2$ and $DA$ is the missing side length. Let $DA=2x$. If $M$ and $N$ are the midpoints of $BC$ and $AD$, respectively, the height of the trapezoid is $OM-ON$. By the pythagorean theorem, $OM=\sqrt{OB^2-BM^2}=\sqrt7$ and $ON=\sqrt{OA^2-AN^2}=\sqrt{8-x^2}$. Thus the height of the trapezoid is $\sqrt7-\sqrt{8-x^2}$, so the area is $\frac{(2+2x)(\sqrt7-\sqrt{8-x^2})}{2}=(x+1)(\sqrt7-\sqrt{8-x^2})$. By Brahmagupta's formula, the area is $\sqrt{(x+1)(x+1)(x+1)(3-x)}$. Setting these two equal, we get $(x+1)(\sqrt7-\sqrt{8-x^2})=\sqrt{(x+1)(x+1)(x+1)(3-x)}$. Dividing both sides by $x+1$ and then squaring, we get $7-2(\sqrt7)(\sqrt{8-x^2})+8-x^2=(x+1)(3-x)$. Expanding the right hand side and canceling the $x^2$ terms gives us $15-2(\sqrt7)(\sqrt{8-x^2})=2x+3$. Rearranging and dividing by two, we get $(\sqrt7)(\sqrt{8-x^2})=6-x$. Squaring both sides, we get $56-7x^2=x^2-12x+36$. Rearranging, we get $8x^2-12x-20=0$. Dividing by 4 we get $2x^2-3x-5=0$. Factoring we get, $(2x-5)(x+1)=0$, and since $x$ cannot be negative, we get $x=2.5$. Since $DA=2x$, $DA=5$. Scaling up by 100, we get $\boxed{\textbf{(E)}\text{ 500}}$.
## Solution 9 (Similar Triangles)
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,NW); label("B",B,NW); label("C",C,NE); label("D",D,NE); label("E",E,SW); label("F",F,SE); label("O",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("\alpha = 90-\frac{\theta}{2}",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("\theta",O,3N); draw(anglemark(E,F,O)); label("\alpha",F,3SW); draw(anglemark(D,F,C)); label("\alpha",F,3NE); draw(anglemark(F,C,D)); label("\alpha",C,3SSE); draw(anglemark(C,D,F)); label("\theta",(RADIUS-0.04)*dir(31.586),3WNW); [/asy]$ Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity. From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\triangle{OCB} \cong \triangle{ODC}$, $\angle{OCD} = \alpha$. This means that $\angle{CDF} = 180-2\alpha = \theta$, which leads to $\triangle{OCB} \sim \triangle{DCF}$. Since we know that $\overline{CD} = 200$, $\overline{DF} = 200$, and by similar reasoning $\overline{AE} = 200$. Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$, which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$. We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$, and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}$ - ColtsFan10
## Solution 10 (Complex Numbers)
We first scale down by a factor of $200\sqrt{2}$. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$, so that $AD$ is the length of the fourth side. We draw this in the complex plane so that $D$ corresponds to the complex number $1$, and we let $C$ correspond to the complex number $z$. Then, $A$ corresponds to $z^3$ and $B$ corresponds to $z^2$. We are given that $\lvert z \rvert = 1$ and $\lvert z-1 \rvert = 1/\sqrt{2}$, and we wish to find $\lvert z^3 - 1 \rvert=\lvert z^2+z+1\rvert \cdot \lvert z-1 \rvert=\lvert (z^2+z+1)/\sqrt{2} \rvert$. Let $z=a+bi$, where $a$ and $b$ are real numbers. Then, $a^2+b^2=1$ and $a^2-2a+1+b^2=1/2$; solving for $a$ and $b$ yields $a=3/4$ and $b=\sqrt{7}/4$. Thus, $AD = \lvert z^3 - 1 \rvert = \lvert (z^2+z+1)/\sqrt{2} \rvert = \lvert (15/8 + 5\sqrt{7}/8 \cdot i)/\sqrt{2} \rvert = \frac{5\sqrt{2}}{4}$. Scaling back up gives us a final answer of $\frac{5\sqrt{2}}{4} \cdot 200\sqrt{2} = \boxed{\textbf{(E)} 500}$.
~ Leo.Euler
## Solution 11 (Trignometry + Ptolemy’s)
Let angle $C$ be $2a$. This way $BD$ will be $400sin(a)$. Now we can trig bash. As the circumradius of triangle $BCD$ is $200\sqrt{2}$, we can use the formula $$R=\frac{abc}{4A}$$ and $$A=\frac{absin(C)}{2}$$ and plug in all the values we got to get $$200\sqrt{2}=\frac{200^2 \cdot 400sin(a)}{4 \cdot (\frac{200^2 sin(2a)}{2})}$$. This boils down to $$\sqrt{2}=\frac{sin(a)}{sin{2a}}$$. This expression can further be simplified by the trig identity $$sin(2a)=2sin(a)cos(a)$$. This leads to the final simplified form $$2\sqrt{2}=\frac{1}{cos(a)}$$. Solving this expression gives us $$cos(a)=\frac{\sqrt{2}}{4}$$. However, as we want $sin(a)$, we use the identity $sin^2+cos^2=1$, and substitute to get that $sin(a)=\frac{\sqrt{14}}{4}$, and therefore BD is $100\sqrt{14}$.
Then, as $ABCD$ is a cyclic quadrilateral, we can use Ptolemy’s Theorem (with $AD=x$) to get $$14 \cdot 100^2=200x+200^2$$. Finally, we solve to get $\boxed{\textbf{(E) } 500}$.
-dragoon
## Solution 12 (Simple Trigonometry with Geometric Observations)
$[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("A",A,W); label("B",B,NW); label("C",C,NE); label("D",D,E); label("O",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); [/asy]$ Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$. Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$. Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$. Let $F$ the feet of the altitude from $O$ to $AD$. Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$, because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$. Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$. Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$, but $0<\alpha<180 \implies cos(\alpha)=3/4$. Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$
- [mathMagicOPS]
## Remark (Morley's Trisector Theorem)
This problem is related to M. T. Naraniengar's proof of Morley's Trisector Theorem. This problem is taken from the figure of the Lemma of M. T. Naraniengar's proof, as shown below.
If four points $Y'$, $Z$, $Y$, $Z'$ satisfy the conditions
$\quad$ $1.$ $Y'Z = ZY = YZ'$ and
$\quad$ $2.$ $\angle YZY'$ = $\angle Z'YZ$ = $180^{\circ} - 2a > 60^{\circ}$
then they lie on a circle.
The Lemma is used to prove Morley's Trisector Theorem by constructing an equilateral triangle at $YZ$ and extending $AY'$ and $AZ'$ as shown below.
~ pi_is_3.14
~IceMatrix
~ pi_is_3.14
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# The Proof is Trivial! Watch
Problem 44***
Let be a polynomial with coefficients in . Suppose that , where is a prime number. Suppose also that is irreducible over . Then there exists a prime number such that does not divide for any integer .
Problem 45*
Let be a prime number, . Given that the equation has an integer solution, then .
Since p>2, then all of p^(k,l,m) are odd. Since odd + odd + odd = odd, n^2 must be odd, which means n must be odd. If n is odd it must be congruent to either 1,3,5 or 7 mod 8. 1^2 = 1, 3^2 = 9 , 5^2 =25, 7^2 =49. Hence n^2 is congruent to 1 mod 8.
p^(k,l,m) are all odd so are congruent to 1,3,5 or 7 mod 8. If any of k,l,m are even p^(k,l,m) will be congruent to 1 mod 8, by similar reasoning to above. For odd powers they will be congruent to 1,3,5 or 7 mod8. If p is congruent to 3 mod 8 for example, p to an odd power will also be congruent to 3 mod 8.
The congruencies must 'add up' to 1 mod 8, as n^2 is congruent to 1 mod 8. So:
if p is congruent to 1 mod 8:
1mod8 +1mod8 +1mod8 does not equal 1mod8 so this isn't possible.
p is congruent to 3 mod 8:
3mod8 + 3mod8 + 3mod8 = 9mod8 = 1mod 8 so this is possible.
p is congruent to 5 mod8 :
5mod8 + 5mod8 + 5mod8 = 15mod8 = 7mod 8 not required.
5mod8 + 5mod8 + 1mod8 = 11mod8 = 3mod8 not required.
5mod8 + 1mod8 + 1mod8 = 7mod8 not required.
1mod8 + 1mod8 + 1mod8 = 3mod8 not required, so p can't be congruent to 5mod8.
For p is congruent to 7mod8:
7mod8 + 1mod8 + 1mod8 = 9mod8 = 1mod8, hence it is possible.
So either:
p is congruent to 3 mod 8 or -1mod8.
2. (Original post by shamika)
This feels like a challenge to come up with something. Now if only I knew any physics...
That's exactly what I've been thinking, but I can't come up with anything either.
3. (Original post by shamika)
This feels like a challenge to come up with something. Now if only I knew any physics...
(Original post by ukdragon37)
That's exactly what I've been thinking, but I can't come up with anything either.
Well all of particle physics is to do with symmetry groups, which slightly related to sets (isn't it?), so....there is potential.
Heck. The Higgs mechanism is to do with symmetry breaking.
4. Problem 46
This is a little Mechanics problem that I come up with (you can use a calculator or you can leave your answers exact) :
A car, in length, is travelling along a road at a constant speed of . The car takes up the whole of one half of a road of width . You want to cross the road at a constant speed of . You will cross the road at an angle degrees to the shortest path to the other side of the road, where theta is positive to the right of this line and negative to the left. The car is approaching you from a distance of to the left of this line, keeping to the opposite side of the road. This car is the only car on the road. Find all possible angles such that you can cross the road without being hit.
5. Solution 46:
If we are moving at a speed of 2.5, then the speed can be resolved into two components, and , with the cosx being the component travelling across the road. As such, the time take to cross the road is:
In this time, you will have travelled a distance of
The time taken for the car to reach this point is:
We want our time travelled to be less, so we cross the road first:
, where k is the square root of 100^2+17.5^2 and y is
Also, we can travel towards the car. In this case, the tan(x) becomes -tan(x), which means that our y changes signs, and so the angle must be less than:
This angle is measured to the left which means we multiply by -1, and so we have the inequality:
However, this is only half of the question. There is also the possibility the car passes by before we reach halfway:
This means that the time taken for us to reach halfway is greater than the time it takes the car to pass by:
Finally, if we choose to travel towards the car, tan(x) becomes -tan(x), and so we have an angle of:
We multiply by -1, and get the final solution of:
6. (Original post by DJMayes)
Solution 46:
If we are moving at a speed of 2.5, then the speed can be resolved into two components, and , with the cosx being the component travelling across the road. As such, the time take to cross the road is:
In this time, you will have travelled a distance of
The time taken for the car to reach this point is:
We want our time travelled to be less, so we cross the road first:
, where k is the square root of 100^2+17.5^2 and y is
However, this is only half of the question. There is also the possibility the car passes by before we reach halfway:
This means that the time taken for us to reach halfway is greater than the time it takes the car to pass by:
Very good You just need to find the negative values for x as well.
7. (Original post by metaltron)
x
I think the question should possibly read .
8. (Original post by DJMayes)
Solution 43
This needs a small correction before I can put it in the OP.
9. (Original post by und)
This needs a small correction before I can put it in the OP.
It doesn't need putting in; it's essentially identical to your solution anyway so there's no point having both.
10. (Original post by und)
I think the question should possibly read .
If it does, then I'm going to struggle to find an answer within a reasonable timeframe! Anyway what I posted previously was my current ramblings.
11. How do you guys latex so fast? That would have taken me an hour and would still be riddled with errors.
12. (Original post by metaltron)
Very good You just need to find the negative values for x as well.
Edited them in.
13. (Original post by bananarama2)
How do you guys latex so fast? That would have taken me an hour and would still be riddled with errors.
I have fast fingers
14. (Original post by Felix Felicis)
I have fast fingers
Use them to fill in the holes in your working do you?
15. (Original post by DJMayes)
Edited them in.
Looks good to me. What did you think of the question?
16. (Original post by bananarama2)
Use them to fill in the holes in your working do you?
I felt the heat from that burn all the way over here!
(Original post by metaltron)
Looks good to me. What did you think of the question?
I think that the idea of the question is a very good one. However, it might be an idea to try find some slightly nicer numbers.
17. (Original post by bananarama2)
Use them to fill in the holes in your working do you?
(Original post by Lord of the Flies)
So many to choose from as well
Do I really have that many holes in my solutions? Would you mind pointing one or two out?
18. (Original post by metaltron)
Looks good to me. What did you think of the question?
It's very similar to a STEP question I've seen, but I can't remember which paper. DJ will know obviously. I think it could have been posed as a general question, because numbers are yucky.
19. (Original post by Felix Felicis)
Do I really have that many holes in my solutions?
I was just trying to make a joke but it didn't quite work, as often.
20. (Original post by metaltron)
If it does, then I'm going to struggle to find an answer within a reasonable timeframe! Anyway what I posted previously was my current ramblings.
I'm not actually sure, it's just what I thought at first and then LotF's example with p=3 sort of cemented that belief.
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Question 24 (1 point) What is the percent concentration (mass/volume) of Nazo if a 834 mL aqueous solution contains 123 g of Na2O? a way to measure concentration. Example 2: mass percent = (molar mass of the element/total molecular mass of compound) x 100 = (72.0642/180.156) x 100 = 0.4000 x 100 = 40.00%. Our calculator will help you will all the conversions, so don't stress. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: Percent Composition by Mass (%) This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100. Mass per cent is a way of expressing a concentration or describing the component in a particular mixture. 2.0 L of an aqueous solution of potassium chloride contains 45.0 g of KCl. Mass Percent. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. 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# If a,b,c,d,e and f are distinct positive integers and
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If a,b,c,d,e and f are distinct positive integers and [#permalink] 27 Sep 2006, 02:20
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If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22
(2) b=4 and f=2
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Probably would have guessed A on this one on test day.
Spent over 5 mins trying to find a solution.
Basically, all I came up with is a, b, c, d, e, f should be {1, 2, 3, 4, 5 , 6} although not necessarily in that order.
No matter what combination I try to plug in the equation it won't balance.
(E)??
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It gets pretty easy when both the statements are used.
If we take the numbers as 1, 2 ,3 ,4 ,5 ,6. ( Dont know if 0 can be a possibility )
10ac+bc= 100d+10e+f
or , c ( 10a + b) = 100d + 10e + f { putting the values of b and f )
or , c ( 10a + 4) = 100d + 10e + 2 {LHS will have 2 in the units place.
In order to match this the value of c
has to be 3 , because 3 * 4 = 12 }
or, 3 (10a + 4) = 100d + 10e + 2
or, 30a + 12 = 100d + 10e + 2
or, 30a + 10 = 100d + 10e
or, 3a + 1 = 10d + e {Now put a = 5 , d = 1 , and e = 6 }
This can be solved. So a = 5
However, my doubt is can't we come to the same conclusion without knowing the values of b and f ?
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If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22
(2) b=4 and f=2
So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.
ok here is my attempt.
I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.
but a, d,e could be any numbers
no statement A: this condition gives many possibilities.
combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4
case 2: d=2
then a+e<11
no possible values
case3: d=3
a+e<10
no possible values.
Thus C.
Anyone else tried this? Any other ideas?
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If the first statement is a+b+c+d+e+f <=22 then A will be the answer.
But with the given conditions as it is the answer would be E.
Hey kevin where r u?
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Last edited by cicerone on 25 Sep 2008, 00:14, edited 1 time in total.
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2times wrote:
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22
(2) b=4 and f=2
So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.
ok here is my attempt.
I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.
but a, d,e could be any numbers
no statement A: this condition gives many possibilities.
combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4
case 2: d=2
then a+e<11
no possible values
case3: d=3
a+e<10
no possible values.
Thus C.
Anyone else tried this? Any other ideas?
Excellent approach, but were you too quick to dismiss A?
if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.
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kevincan wrote:
2times wrote:
If a,b,c,d,e and f are distinct positive integers and 10ac+bc= 100d+10e+f, what is the value of a?
(1) a+b+c+d+e+f < 22
(2) b=4 and f=2
So I got soooo bored of writing applications that I had to find something to do with my brain. That is how I am back here. btw.. hope everyone's prep is going on well.
ok here is my attempt.
I satrted with statement B (looks like it has more info than A)
so if (10a+4)c = 100d+10e+2
=> c has to be equal to 3 (you multiply a number ending in 4 with a number ending in 2 the multiple has to be 3.
this quick step tells me b = 4 ; c = 3 & f =2.
but a, d,e could be any numbers
no statement A: this condition gives many possibilities.
combining
a+d+e < 13 (22-4-3-2).....(i)
also 30a+12 = 100d+10e+2
=> 3a +1 = 10d +e.....(ii)
case 1: d= 1
then a+e< 12
only possible value of a that will give each number as a distinct integer is 4
case 2: d=2
then a+e<11
no possible values
case3: d=3
a+e<10
no possible values.
Thus C.
Anyone else tried this? Any other ideas?
Excellent approach, but were you too quick to dismiss A?
if a+b+c+...+f< 22, then (a,b,c,d,e,f) must be a permutation of the first six positive integers.
Intuitively, I know that (A) is sufficient, just can't seem to come up with the right combination in under two minutes. Gosh, the GMAT seems to be getting harder all the time
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Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.
If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).
We are told that 10ac+bc=100d+10e+f
Let's write it as:
.ab
x.c
___
def
Focussing on the units digits, it's clear that none of them can be 1 or 5.
Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?
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Hey kevin, i think the above equation will not have a solution if all of them are distinct and the solution set is 1,2,3,4,5,6
GMAT Instructor
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kevincan wrote:
Keep in mind that this is NOT a GMAT question, so don't estimate the difficulty of the exam by the questions I post. GMATPrep is by far the best indicator. These questions are just for fun and meant to help us to look for creative solutions.
If a+b+c+d+e+f<22 and each is a distinct positive integer, it follows that (a,b,c,d,e,f) is a permutation of (1,2,3,4,5,6).
We are told that 10ac+bc=100d+10e+f
Let's write it as:
.ab
x.c
___
def
Focussing on the units digits, it's clear that none of them can be 1 or 5.
Could b and c be 2 and 3 or vice versa? If that case, f would be 6, the two digit number de would simply be ac, which is not possible since the six numbers have to be distinct. This makes it clear that bc>9, since we NEED a carry-over. Only 3 and 4 will do! I have to start a class! Can somebody continue?
so if b and c are 3 and 4 (or vice versa) f would be 2
Let's see if (b,c)=(3,4)
.a3
X 4
----
de2
The two digit number de is 4*a+1- a would have to be 5 or 6, but in each case d would be 2- not permissible
.a4
X.3
----
de2
This time, the two digit number de is 3*a+1- a must be 5, so de is 16
54*3=162
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# Probability
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