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https://www.jiskha.com/search?query=random&page=32
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# random
4,716 results, page 32
1. ## algebra 1
From a shipment of 350 light-bulbs, a sample of 70 was selected at random and tested. If 21 light-bulbs in the sample were found to be defective, how many defective light-bulbs would be expected in the entire shipment?
2. ## LBCC
An apartment building has the following apartments: 2 bdrm 3 bdrm 4 bdrm 1st floor 7 3 4 2nd floor 9 0 11 3rd floor 8 4 7 If an apartment is selected at random, what is the probability that it is a 2 bedroom apartment on the 3rd floor?
3. ## maths-probabilty
Bag 1 contains 2 white and 3 red balls and bag 2 contains 4 white and 5 red balls. 1 ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag 2
4. ## Math
The odds are 3 to 5 in favor of drawing a blue marble in one random drawing from an urn containing only red and blue marbles. There are more than 100 marbles in the urn. what is the minimum number of red marbles that there could be in the urn?
5. ## statistics
A researcher wants to study the spending habits of customers of a local shopping mall. The mall manager claims that the average spending per customer is \$70, but the researcher believes that the average is less than \$70. A simple random sample of 350
6. ## Math
You use a line of best fit for a set of data to make a prediction about an unknown value. The correlation coefficient for your data is -0.993. How confident can you be that your predicted value will be reasonably close to the actual value? a. I can't be
7. ## Math (Probability)
When Lino submitted an essay for his English class, it was 9 pages long. Lino's essay had errors on pages 2 and 7. If the teacher chose 3 pages uniformly at random to read, then the probability that she didn't read any of the pages with errors can be
8. ## probability April-004
In a random sample of 95 sections of pipe in a chemical plant, 16 showed signs of serious corrosion. Construct a 98% confidence interval for the true proportion of pipe sections showing signs of serious corrosion, using the large sample confidence interval
9. ## Finite
A box contains 8 blue balls, 3 white balls, and 5 red balls, and that we choose two balls at random from the box. What is the probability of neither being blue given that neither is white?
10. ## Math-Probability
An urn contains four red balls and three white balls. A sample of 2 balls is selected at random from the urn. What is the probability that at least one of the balls is red?
11. ## statistics
An urn contains 10 balls of which 6 are white, 3 are red and 1 is green. Two balls are drawn at random from the urn. Find the probability distribution of number of white balls drawn. Hence state the expected number of white balls one can get in such a
12. ## statistics
EXERCISE: POSSIBLE VALUES OF THE ESTIMATES Suppose that the random variable È takes values in the interval [0,1]. a) Is it true that the LMS estimator is guaranteed to take values only in the interval [0,1]? b) Is it true that the LLMS estimator is
13. ## math
The number of immigrants living in a certain country with a large population makes up 35% of the population. In a random sample poll of 40 people what is: a)the expected number of non-immigrants will be polled? b)the probability that no immigrants will be
14. ## Math
Suppose that either of two instruments might be used for making a certain measurement. Instrument 1 yields a measurement whose p.d.f. is f1(x)=2x, 0 <x<1 Instrument 2 yields a measurement whose p.d.f. is f2(x)=3x^2, 0 <x<1 Suppose that one of
15. ## statistics
An urn contains 10 balls of which 6 are white, 3 are red and 1 is green. Two balls are drawn at random from the urn. Find the probability distribution of number of white balls drawn. Hence state the expected number of white balls one can get in such a
16. ## stats
A manufacturer of automobile batteries claims that the distribution of battery life is 54 months with a standard deviation of 6 months. We take a random sample of 50 batteries. a. Find the probability that their mean life is less than 52 months. b. Find
17. ## Maths
Suppose that either of two instruments might be used for making a certain measurement. Instrument 1 yields a measurement whose p.d.f. is f1(x)=2x, 0 <x<1 Instrument 2 yields a measurement whose p.d.f. is f2(x)=3x^2, 0 <x<1 Suppose that one of
18. ## PROBABILITY
R. H. Bruskin Associates Market Research found that 40% of Americans do not think that having a college education is important to succeed in the business world. If a random sample of five Americans is s selected, find these probabilities. (a) Exactly two
19. ## math
R. H. Bruskin Associates Market Research found that 30% of Americans do not think that having a college education is important to succeed in the business world. If a random sample of five Americans is s selected, find these probabilities. (a) Exactly two
20. ## statistics
1000 raffle tickets are sold and the winner is chosen at random there is only one prize \$500 in cash you buy one ticket(a) what is the probability you will win the prize of \$500 (b) your expected earnings can be found by multiplying the value of the prize
21. ## Statistics
A national survey of small businesses reports that 70% of all small businesses now have a web-site. Assume that for a randomly selected group of 30 small businesses, the number with a web-site has a binomial distribution. In a random sample of 30 small
22. ## Psychology
HELP !!! Which statistics concept is described? Choose from these answers: [(A):statistical test] [(B):Null Hypothesis] [(C):alpha] [(D):one-tailed test] [(E):two-tailed test] [(F):Type I error] [(G):Type II error] [(H):the statistic Z] 1)When a hypothesis
23. ## PLZ help!! Stats....
In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that: (A) exactly two will have this mild side effect (B) at least
24. ## Stats
According to a previous report, 15% of households had some type of high-speed Internet connection. Suppose 12 households are selected at random and the number of households with high-speed Internet is recorded. (a) Find the probability that
25. ## Did i do this right??? STATS
3. In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that: (A) exactly two will have this mild side effect (B) at least
26. ## statistics
Suppose a shipment of 500 machine parts contains 62 defective and 438 non-defective machine parts. From the shipment you take a random sample of 25. You are interested in the number of defective machine parts out of 25 trials and corresponding
27. ## Can u help me Ms. Sue...Stats
In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that: (A) exactly two will have this mild side effect (B) at least
28. ## Math
According to the U.S. Census Bureau's most recent data on the marital status of the 242 million Americans aged 15 years and older, 124.2 million are currently married and 74.5 million have never been married. If one person from these 242 million persons is
29. ## maths
Bibi wants to send invitations to some of her friends to come to her birthday party. She has contact information for 37 friends stored on her phone, but she only wants to invite 21 of those people to her party. She tries to send out a message to the 21
30. ## stats
In a straight 3 lottery you choose a 3 digit number. A 3 digit number is chosen at random, If your number mataches you win \$250. 1. How many diifrent three digit numbers are there? I thought you would do 3^3 but that only equaled 27 so i assumed im wrong
31. ## math
Bibi wants to send invitations to some of her friends to come to her birthday party. She has contact information for 37 friends stored on her phone, but she only wants to invite 21 of those people to her party. She tries to send out a message to the 21
32. ## Probabilities
Bibi wants to send invitations to some of her friends to come to her birthday party. She has contact information for 37 friends stored on her phone, but she only wants to invite 21 of those people to her party. She tries to send out a message to the 21
33. ## statistics
In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 12 patients using the drug is selected. If you wanted to find the probability that exactly three will have this mild side effect,
34. ## Statistics / Math
If a student is chosen at random, find the probability of getting someone who is a man or a non-smoker. Round your answer to three decimal places. ********non***regular heavy total man*****135*****64******5*****204 women***187****21*******9*****217
35. ## statistics
A study was conducted in order to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the
36. ## statistics
A study was conducted in order to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the
37. ## statistics
A study was conducted in order to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the
38. ## statistics
A study was conducted in order to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the
39. ## proportion test
In a manufacturing plant, the old machine produced widgets with a defect rate of 12%. The boss wants to check if the new machine has a different defect rate. So he took a random of 820 widgets, and found 85 of them were defective. Are these results
40. ## Math
A random sample was taken from Wyoming. Eighty percent of students reported that they like dogs, 64% reported they like cats, and 97% reported that they like dogs or cats. What percentage of students like both dogs and cats?
41. ## statistics
A survey among US adults of their favorite toppings on a cheese pizza reported that 43% favored pepperoni, 14% favored mushrooms, and 6% favored both pepperoni and mushrooms. What is the probability that a random adult favored pepperoni or mushrooms on
Susan has red, blue, and yellow sweaters. Joanne has green, red, and white sweaters. Diane's sweaters are red, blue, and mauve. Each girl has only one sweater of each color and will pick a sweater to wear at random. Find the probability that each girl
43. ## statistics
A survey among US adults of their favorite toppings on a cheese pizza reported that 43% favored pepperoni, 14% favored mushrooms, and 6% favored both pepperoni and mushrooms. What is the probability that a random adult favored pepperoni or mushrooms on
44. ## math
A box contains 300 colored paper clips. Tony pulled 25 paper clips out at random and found 6 pink, 4 white, 10 yellow, and 5 blue clips in the sample from the data. Tony predicts there are about 150 yellow paper clips in the entire box. Ia tony's
45. ## statistics
we developed a 95% z confidence interval for the mean BMI of women aged 20 to 29 years, based on a national random sample of 654 such women. we assumed there that the population standard deviation was known to be 7.5. in fact the sample data had mean BMI
46. ## STATISTICS
A random sample of clients at a weight loss center were given a dietary supplement to see if it would promote weight loss. The center reported that the 100 clients lost an average of 34 pounds, and that a 95% confidence interval for the mean weight loss
X, is assumed to be a random variable by placing letters of the word “YACHT” in a hat. If these letters are withdrawn then it gets replaced. A value of 1 is given when a vowel is retrieved, and a value of 2 is given when a consonant is retrieved while
48. ## math
There are 5 options on the dessert menu at a restaurant. Erin and Ellen like all of the choices equally, so they each choose a dessert at random from the menu. What is the probability that Erin will choose apple pie and Ellen will choose strawberry
49. ## statistical analysis
2. The Computer Systems Department has eight faculty, six of whom are tenured. Dr. Vonder, the chair, wants to establish a committee of three department faculty members to review the curriculum. If she selects the committee at random: a. What is the
50. ## statistics
Based on a random sample of 25 units of lawnmowers, the average weight is 102 lbs., and the sample standard deviation is 10 lbs. We would like to decide if there is enough evidence to establish that the average weight for the population of lawnmowers is
51. ## Statistics
Olive-Oil producers have expanded to approximately 120 producers statewide. Analysts estimate the average price consumers pay for a bottle of olive oil is \$21, with a standard deviation of \$2.40. What is the probability that the average price paid by a
52. ## math
In an extra-sensory-perception experiment, a blindfolded subject has two rows of blocks before him. Each row has blocks numbered 1 to 10 arranged in random order. The subject is to place one hand on a block in the first row and then try to place his other
53. ## mathematics
The numbers 1,2,…,17 are divided into 5 disjoint sets. One set has 5 elements, one set has 4 elements, two sets have 3 elements and the last set contains the 2 remaining elements. Two players each choose a number from 1 to 17 at random. The probability
54. ## maths
The numbers 1,2,…,17 are divided into 5 disjoint sets. One set has 5 elements, one set has 4 elements, two sets have 3 elements and the last set contains the 2 remaining elements. Two players each choose a number from 1 to 17 at random. The probability
55. ## Statictics
2. Herpetologists (snake specialists) found that a certain species of reticulated python have an average length of 20.5 feet with a standard deviation of 2.3 feet. The scientists collect a random sample of 30 adult pythons and measure their lengths. In
The Principal of SACRO Primary is concerned that the average speed of cars travelling along Princess Road exceeds the recommended speed of 60km/h. A random sample of 10 cars had their speeds measured by radar and the results (in km/h) are as follows: 81,
57. ## Statistics
Suppose a basketball player, Player A, made 80% of her free throw attempts last season and that she continues to shoot free throws at the same rate. Assume that free throw attempts are independent. Let the random variable X be the number of free throws
58. ## Math probability
Assume that 60% of community college students are female. 1) if we pick 100 students at random justify why we can or cannot model this using a normal model. 2) find the probability that between 50 and 70 of the 100 students selected are female. 3) fidn Q1
59. ## proportion test
How to solve this? In a manufacturing plant, the old machine produced widgets with a defect rate of 12%. The boss wants to check if the new machine has a different defect rate. So he took a random of 820 widgets, and found 85 of them were defective. Are
60. ## math probability
Laura and Adrian attend a dinner party. The 12 guests are seated at random around a round table, with all seating arrangements equally likely. (a) What is the probability that Laura and Adrian sit next to each other. (b) Explain your answer by giving a
61. ## Statistics
A packaging device is set to fill detergent power packets with a mean weight of 5 Kg. The standard deviation is known to be 0.01 Kg. These are known to drift upwards over a period of time due to machine fault, which is not tolerable. A random sample of 100
62. ## Probability and math statistics
We have two coins, C0 and C1, that looks identical but have different probabilities (0.5 and 0.7 respectively) of showing Heads. A coin is chosen at random and then is tossed X times until a Head comes up. Denote by H0 and H1 the hypotheses that the chosen
63. ## Math
the average length of time for students to register for fall classes at a certain college has been 50 minutes. a new registration procedure using modern computing machine is being tried. if a random sample of 12 students had an average registration time of
64. ## statistics/probability
in testing a new drug. researchers found that 20% of all patients using it will have a mild side effect. a random sample of 14 patients using the drug is selected. find the probability that: a) exactly three will have mild side effects. b) at least four
65. ## Statistics
The monthly incomes from a random sample of workers in a factory are shown below. Monthly Income (In \$1,000) 4.0 5.0 7.0 4.0 6.0 6.0 7.0 9.0 a. Compute the standard error of the mean (in dollars). b. Compute the margin of error (in dollars) at 95%
66. ## Statistics
A large furniture store has begun a new ad campaign on local television. Before the campaign, the long term average daily sales were \$24,819. A random sample of 40 days during the new ad campaign gave a sample mean daily sale of x(bar)=\$25,910 with sample
67. ## probability
A jar contains 3 red balls and 2 green balls. You draw one ball at random. Determine the probability of selecting a. red b. green c. blue d. red or green p.s: the reason I am asking this questions is because we never learned about it in class and I don't
68. ## Statistics
According to a recent survey, 75% of all customers will return to the same grocery store. Suppose 8 customers are selected at random: A) What is the probability that exactly five of the customers will return? B) What is the profability that all 8 will
69. ## Statistics
An automobile in surer has found that repair claims have a mean of \$1520 and a stardard deviation of \$770. Suppose that the next 100 claims can be ragarded as a random sample from the long-run claims process 1. What is the mean and standrd deviation of the
70. ## stats
The number of cars sold annually by used car salespeople is normally distributed with a standard deviation of 19.22. A random sample of salespeople was taken and the number of cars each sold is listed her. 79 43 58 66 101 63 79 33 58 71 60 101 74 55 88 The
71. ## Statistics
To make causal inferences from observed differences between groups based on varied levels of treatment, it is necessary to...? A. Find significant differences between groups at at-least the .05 level B. Use T-test or ANOVA procedures to conduct the
72. ## ap statistics
The manager at Air Cargo feels that the weights of packages shipped recently are less than in the past. Records show that in the past packages have had a mean weight of 36.7 lb. and a variance of 201.64 lb2. A random sample of last months shipping yielded
73. ## Stats
In testing a new drug, researchers found that 10% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that: (A) exactly two will have this mild side effect (B) at least
74. ## Statisics
I need help on this question i am confused. I don't think the answers i got are right The library checks out an average of 320 books per day, with a standard deviation of 75 books. Suppose a simple random sample of 30 days is selected for observation. What
75. ## Math: Probability
I need help. Here's the queston. Julia has 3 red marbles, 4 blues,3 yellow and 6 black in a bag. She takes one out at random. The probability its black ___ The probability its yellow ___ The probability it isnt red ___
Susan has red, blue, and yellow sweaters. Joanne has green, red, and white sweaters. Diane's sweaters are red, blue, and mauve. Each girl has only one sweater of each color and will pick a sweater to wear at random. Find the probability that each girl
77. ## Geometry
A circle of radius 1 is drawn in the plane. Four non-overlapping circles each of radius 1, are drawn (externally) tangential to the original circle. An angle ã is chosen uniformly at random in the interval [0,2ð). The probability that a half ray drawn
78. ## Statistics
a. Critical Thinking Critical Thinking Suppose you and a friend each take different random samples of data pairs (x,y) from the sample population. Assume the samples are the same size. Based on your samples, you compute r= 0.83. Based on her sample, your
79. ## Math : Probability and Standard Deviation
This is a word problem. "We know that for certain workers the mean wage is \$5/hr with a standard deviation of \$.50. We need to find out if a worker is chosen at random what is the probability that the workers' wage is between \$4.50 and \$5.50? Assume a
X, is assumed to be a random variable by placing letters of the word “YACHT” in a hat. If these letters are withdrawn then it gets replaced. A value of 1 is given when a vowel is retrieved, and a value of 2 is given when a consonant is retrieved while
81. ## statistics
Please help- A random sample of n=500 households where 150 have combined cable,phone, and internet service. Firrst, calculate p-hat, the sample proportion of households having such service. Then use a confidence level for pi, the proportion of all
82. ## Statistics
A student group claims that first year students at a university study 2.5 hours (150 minutes) per night during the school week. A skeptic suspects that they study less than that on average. He takes a random sample of 30 first year students and finds that
83. ## Finite Math
:
Four cards are drawn without replacement from a well shuffled standard deck of cards. Let X be the number of aces drawn. 12. What are the possible values of the random variable X? Write them in increasing order as a set using roster (or list)
84. ## math
Independent random samples, each containing 800 observations, were selected from two binomial populations. The samples from populations 1 and 2 produced 320 and 400 successes, respectively. Construct a 90% confidence interval for the true difference in
85. ## statistics
The population standard deviation (σ) for a standardized achievement test that is normally distributed is 8. a.) Calculate the standard error of the mean if you draw a random sample of 50 test scores. Show your work for full credit . . .
86. ## Cards
As shown above, a classic deck of cards is made up of 52 cards, 26 are black, 26 are red. Each color is split into two suits of 13 cards each (clubs and spades are black and hearts and diamonds are red). Each suit is split into 13 individual cards (Ace,
87. ## Statistics
Twelve different video games showing substance use were observed and the duration times of game play ( in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. use the data to
88. ## Stats.
Another one that I need help.... At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 520 ml. The filling process follows a normal distribution with a known process standard deviation of 4 ml. (a) Which sampling
89. ## statistics
twenty light bulbs are tested to see if they claims last long as the manufacturers claims they do. three light bulbs failed the test. two light bulbs are selected at random without replacement from this collection of bulbs find the probability that both
90. ## statistics
1. The scores on a standardized test for high school students are normally distributed with a population mean of 500 and a population standard deviation of 100. Use the empirical rule (0LI refers to this as the standard deviation rule.). Make sure you
Please, only formula and steps to do these exercises. Thank you For the following data from independent samples, could the null hypothesis that the population means are equal be rejected at the 0.05 level? Sample1 Sample2 Sample3 Sample4 15.2 10.2 14.9
92. ## Statistics (42)
A simple random sample of 60 items resulted in a sample mean of 96. The population standard deviation is 16. a. Compute the 95% confidence interval for the population mean (to 1 decimal). ( , ) b. Assume that the same sample mean was obtained from a sample
93. ## communication
X and Y are discrete jointly distributed discrete valued random variables. The relation between their joint entropy H(X,Y) and their individual entropies H(X),H(Y) is H(X,Y)≤H(X)+H(Y), equality holds when X,Y are independent H(X,Y)≤H(X)+H(Y), equality
94. ## conditional probability
Among a group of people, 10% are from NYC and 90% are not. All people from NYC are under 18 years old, and 60% of those not from NYC are also under 18 years old. One person is chosen at random, and he or she is under 18. What is the probability that the
95. ## Statistics
Jim's systolic blood pressure is a random variable with a mean of 145 mmHg and a standard deviation of 20 mmHg. For Jim's age group, 140 is the cutoff for high blood pressure. If Jim's systolic blood pressure is taken at a randomly chosen moment, what is
96. ## PHYSICS
can someone help me with physics lab homework please: load 300 g at 60 degree mass at 180 deg 150 g mass at 270 deg 260 g for step two resolution of 300-g at 60 degree f load= fx'= fy'= percent error for each component for setup II? WOULD THE HYPOTHETICAL
97. ## marketing research
A packaging device is set to fill detergent power packets with a mean weight of 5 Kg. The standard deviation is known to be 0.01 Kg. These are known to drift upwards over a period of time due to machine fault, which is not tolerable. A random sample of 100
98. ## Biostatitics
Based on data collected by the National Center for Health Statistics and made available to the public in the Sample Adult database, an estimate of the percentage of adults who indicated to have at some point of their life been tested for HIV is 32 percent
99. ## statistics
the manufacturing of jstcare automobile uses a random sample of 500 car owners to determeine if the proportion of satisfied caar owners is different for jetcare than for other cars a)perform a hypothesis test t the 5 percent level if the sample contained
100. ## Math
A bag contains 6 black balls,5 red balls and 4 white balls,if 3 are selected at random without replacement determine the probability that (1)all 3 are black (2)all 3 are red (3)2 are black and 1 white (4)at least 1 of each colour is selected.
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# Quant Question of the Day Chat
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Re: Quant Question of the Day Chat [#permalink]
gmatophobia wrote:
PS Question 1 - Mar 25 x and y are positive integers. When 16x is divided by y, the quotient is x, and the remainder is 4. What is the sum of all possible y-values? A) 7 B) 12 C) 19 D) 26 E) 41 Source: GMAT Prep Now | Difficulty: Hard
E
We can write it as -> 16x = yx
16x = yx + 4
x(16-y) = 4
we know x and y are +ve integers. Hence, y can be 15,14 or 12.
Sum = 41
Originally posted by ChandlerBong on 24 Mar 2023, 23:40.
Last edited by ChandlerBong on 24 Mar 2023, 23:42, edited 1 time in total.
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Re: Quant Question of the Day Chat [#permalink]
How do we get from 16x/y = x + 4. To 16x = yx +4?
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Re: Quant Question of the Day Chat [#permalink]
We’re putting in the values and checking. Y and X should always be positive integers and for the values of y = 15, 14 or 12 we’re getting +ve value of x. Any lower value will result in -ve value or fractional value of x.
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Re: Quant Question of the Day Chat [#permalink]
How did we get to the equation 16x = yx +4?
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Re: Quant Question of the Day Chat [#permalink]
joe123x wrote:
How did we get to the equation 16x = yx +4?
Dividend = Divisor x Quotient + Remainder
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Re: Quant Question of the Day Chat [#permalink]
gmatophobia wrote:
DS Question 1 - Mar 25 3 siblings - Alan, Betty and Carl - were the only ones to receive part of a \$300,000 inheritance. Did any of the three receive more than forty percent of the total inheritance? 1) Carl received a larger inheritance than Alan and a larger inheritance than Betty. 2) The combined total of Alan’s and Betty’s inheritances was equal to three times Betty’s inheritance. Source: EMPOWERGmat | Difficulty: Hard
anyone who solved this ? pls share an approach
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Re: Quant Question of the Day Chat [#permalink]
AbhinavKumar wrote:
anyone who solved this ? pls share an approach
You can refer to the official solution here- data-sufficiency-pack-4-question-4-3-siblings-alan-betty-and-carl-208874.html#p1606576
gmatophobia wrote:
DS Question 1 - Mar 25 3 siblings - Alan, Betty and Carl - were the only ones to receive part of a \$300,000 inheritance. Did any of the three receive more than forty percent of the total inheritance? 1) Carl received a larger inheritance than Alan and a larger inheritance than Betty. 2) The combined total of Alan’s and Betty’s inheritances was equal to three times Betty’s inheritance. Source: EMPOWERGmat | Difficulty: Hard
C
gmatophobia wrote:
PS Question 1 - Mar 25 x and y are positive integers. When 16x is divided by y, the quotient is x, and the remainder is 4. What is the sum of all possible y-values? A) 7 B) 12 C) 19 D) 26 E) 41 Source: GMAT Prep Now | Difficulty: Hard
E
DS Question 1 - Mar 26
If |a – b| = 6 and |b – c| = 15, then what is the value of |c|?
(1) |a – c| = 9
(2) |b| = 9
Source: EMPOWERGmat | Difficulty: Hard
PS Question 1 - Mar 26
In how many ways can the letters of the word "GRAPHITE" be rearranged such that the order in which the vowels appear does not change?
A. 8P3
B. 8C3 * 3!
C. 8C3
D. 8C3 * 5!
E. 8C3 * 3! * 5!
Source: Medium | Difficulty: Hard
Originally posted by gmatophobia on 26 Mar 2023, 03:10.
Last edited by gmatophobia on 26 Mar 2023, 03:16, edited 2 times in total.
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Re: Quant Question of the Day Chat [#permalink]
PS Question 1 - Mar 26
In how many ways can the letters of the word "GRAPHITE" be rearranged such that the order in which the vowels appear does not change?
A. 8P3
B. 8C3 * 3!
C. 8C3
D. 8C3 * 5!
E. 8C3 * 3! * 5!
Source: Others | Difficulty: Medium
PS Question 1 - Mar 26
In how many ways can the letters of the word "GRAPHITE" be rearranged such that the order in which the vowels appear does not change?
A. 8P3
B. 8C3 * 3!
C. 8C3
D. 8C3 * 5!
E. 8C3 * 3! * 5!
Source: Others | Difficulty: Medium
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Re: Quant Question of the Day Chat [#permalink]
gmatophobia wrote:
DS Question 1 - Mar 26 If |a – b| = 6 and |b – c| = 15, then what is the value of |c|? (1) |a – c| = 9 (2) |b| = 9 Source: EMPOWERGmat | Difficulty: Hard
E
gmatophobia wrote:
PS Question 1 - Mar 26 In how many ways can the letters of the word "GRAPHITE" be rearranged such that the order in which the vowels appear does not change? A. 8P3 B. 8C3 * 3! C. 8C3 D. 8C3 * 5! E. 8C3 * 3! * 5! Source: Others | Difficulty: Medium
D
PS Question 1 - Mar 28
A furniture manufacturer has two machines, but only one can be used at a time. Machine A is utilized during the first shift and Machine B during the second shift, while both work half of the third shift. If Machine A can do the job in 12 days working two shifts and Machine B can do the job in 15 days working two shifts, how many days will it take to do the job with the current work schedule?
A. 14
B. 13
C. 11
D. 9
E. 7
Source: GMAT Club Tests | Difficulty: Hard
DS Question 1 - Mar 28
If set S consists of the numbers n, -2, and 4, is the mean of set S greater than the median of set S ?
(1) n > 2
(2) n < 3
Source: Manhattan GMAT | Difficulty: Hard
Originally posted by gmatophobia on 28 Mar 2023, 02:36.
Last edited by gmatophobia on 28 Mar 2023, 02:37, edited 1 time in total.
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Re: Quant Question of the Day Chat [#permalink]
gmatophobia wrote:
E
Solution for this question ?
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Re: Quant Question of the Day Chat [#permalink]
rishabhg2712 wrote:
Solution for this question ?
You can refer the solution posted here - data-sufficiency-pack-3-question-4-if-a-b-208298.html#p2884446
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Re: Quant Question of the Day Chat [#permalink]
hello quick question: Julia purchased a car on an installment plan. She made a down payment of \$2,550 and then made n monthly payments of \$155 each. If Julia paid a total of \$9,060 for the car, how many monthly payments did she make?
I have 42 but answer is supposed to be 48
🤔
Julia purchased a car on an installment plan. She made a down payment of \$2,550 and then made n monthly payments of \$155 each. If Julia paid a total of \$9,060 for the car, how many monthly payments did she make?
A: 30
B: 36
C: 42
D: 48
E: 54
Originally posted by supernoob on 28 Mar 2023, 03:06.
Last edited by supernoob on 28 Mar 2023, 03:07, edited 1 time in total.
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Re: Quant Question of the Day Chat [#permalink]
supernoob wrote:
I have 42 but answer is supposed to be 48
It cannot be. It has to be C: 42
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Re: Quant Question of the Day Chat [#permalink]
Exactly because after down payment, she is left with 6510. Which divides evenly by 155 and gives 42. If there would have been some interest on the remaining amount - then 48 would have been a possibility. As per above question, 42 should be the right answer
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Re: Quant Question of the Day Chat [#permalink]
ShrikunjAgrawal wrote:
Exactly because after down payment, she is left with 6510. Which divides evenly by 155 and gives 42. If there would have been some interest on the remaining amount - then 48 would have been a possibility. As per above question, 42 should be the right answer
It is mentioned that she payed 9050 total for the car , answer must be 42
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Re: Quant Question of the Day Chat [#permalink]
I agree but the test disagrees
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did they provide any explanation??
Re: Quant Question of the Day Chat [#permalink]
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# Pigeonhole. Thomas Mildorf. September 15, What? Are those pigeons? And why are you stuffing them into holes?
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1 Pigeonhole Thomas Mildorf September 15, 2007 What? Are those pigeons? And why are you stuffing them into holes? Discussion The pigeonhole principle states that, given positive integers m and n with m > n, to stuff m pigeons into n hole requires that at least two pigeons are stuffed into the same hole. More specifically, some hole contains at least m/n pigeons. Furthermore, some hole contains at most m/n pigeons. Simple and intuitive, this concept arises frequently in Olympiad problems. Many characteristics can be assigned to the pigeons and holes; what is appropriate depends on the problem at hand. Some ideas include ordering within holes, forcing distinctness, and operations on pigeons. Even infinity plays an interesting, if infrequent, role. If an infinite set is partitioned into finitely many disjoint sets, then one of the subsets is itself infinite. In the opposite sense, a finite set cannot be partitioned into infinitely many disjoint nonempty subsets. The problems below have quick solutions involving pigeonhole. What aspects of each tell us this? Questions of existence suggest an averaging approach that considers many things at once only to extract a small bit of abstract information. Residues modulo n (likely holes) are relatively few in number compared to subsets of an n element set (likely pigeons), and choosing slightly more than half of a collection makes that extra item extremely interesting. It is worth observing that the fact that subsets are so numerous makes them good pigeons. Think about the following problems before reading their solutions: 1. A set S contains n integers. Prove that there exists a nonempty subset T S such that the sum of the elements of T is a multiple of n. 2. (RMO, India 1998) Let n be a positive integer. Show that if S is a subset of {1, 2, 3,..., 2n} containing n + 1 elements, then (a) there are two distinct coprime elements of S; (b) there are two distinct elements of S such that one divides the other. 3. Suppose that in a room of n 2 people, some pairs of people shake hands. Show that two people shook hands with the same number of other people. 1
2 4. (Zuming) The squares of an 8 8 checkerboard are filled with the numbers {1, 2,..., 64}. Prove that some two adjacent squares (sharing a side) contain numbers differing by at least 5. In the first problem, we can probably assume that residues modulo n will be the pigeons. With this in mind, we should need only about n pigeons. Given two subsets whose elements give equivalent remainders, we might adjust one set based on the other - we could use set subtraction if one set contained the other. Additionally, since the size of the subset T is not specified, it makes sense to consider the subsets of S as posets under inclusion. That is, set inclusion gives a partial ordering relation - an ordering where elements are not necessarily comparable - on the subsets. A natural inclusion structure is ascending chains, sequences of sets where one contains the next. So let the elements of S be s 1,..., s n, and define the subsets T k = {s 1, s 2,..., s k } for k = 1, 2,..., n. The solution is now clear. The partial sum corresponding to T k is then t k = s s k for 1 k n. If t k is a multiple of n for some k, we are done. Otherwise, t i t j (mod n) for some i < j, and in this case n divides t j t i = s i+1 + s i s j. The second problem clearly calls for two proofs based on pigeonhole, where n + 1 items are classified into n groups; pigeonhole then gives us a pair of items that grouped together. We must choose holes appropriately. For (a), the holes should be pairs of relatively prime numbers. Although there probably are many possibilities, the Euclidean algortihm quickly verifies that consecutive integers are relatively prime, so we can use {1, 2}, {3, 4},..., {2n 1, 2n}. For (b), we need n equivalence classes such that for any two equivalent elements, one divides the other. If the numbers m 1, m 2,..., m k, are equivalent, then without loss of generality we may assume that m 1 divides m 2, that m 2 divides m 3, and so on. It makes sense to generate m 2,..., m k from m 1 by adding prime factors, so as to pack as many numbers from {1, 2, 3,..., 2n} together. A natural candidate is the smallest prime, 2. Having reached this idea, it is actually easier to work in reverse, removing powers of 2. Starting from an integer i such that n < i 2n, remove the powers of 2 from i one at a time. The sequences of integers thus determined partition the set S into n equivalence classes based on common odd parts. We finally note that if x and y are two integers such that x < y while both have the same odd part, then x divides y, and we are done. In the third problem, the presumable holes are number of other people whose hand was shook. Since there are n possibilities and n people, there is exactly one, very uniform fail case. So we suppose for the sake of contradiction that no two people shook hands with the same number of others. Then the people shook hands with 0, 1, 2,..., n 1 others, one person per number. But this means that one person shook hands with nobody while another shook hands with everybody! Impossible. To approach the last problem, we might begin by noting that the largest possible difference is 64 1 = 63, much larger than we require. Indeed, as long as it is split into at most 15 2
3 differences, the maximal difference will give a solution via pigeonhole. It may happen that 1 and 64 are not adjacent. However, there we can clearly find a path between the squares labeled 1 and 64 using not more than 15 squares total. The 14 (or fewer) differences between the pairs of numbers in adjacent squares in the path sum to 63, so that, by pigeonhole, at least one such difference is at least 5. Problems 1. Five lattice points are chosen in the plane lattice. Prove that two of these points determine a line segment passing through a third lattice point. Outline. Look at the parities of the coordinates of each point, then check that one of the midpoints is a lattice point. 2. Let set S contain five integers greater each than 1 and less than 120. Show that S contains a prime or two elements of S share a prime divisor. Outline. Suppose that the set S = {s 1, s 2, s 3, s 4, s 5 } contains no primes and that no two elements of S share a prime divisor. Then s i = p i1 p i2 for i = 1,..., 5. There are four primes less than 11. Thus, there exists an j such that p j1, p j2 11. It follows that s j 121 > 120, a contradiction. 3. (Zuming) Let n be a positive integer that is not divisible by 2 or 5. Prove that there is a multiple of n consisting entirely of ones. Outline. Find two integers consisting entirely of 1 s that are equivalent mod n, then remove the zeros from their difference. 4. Show that for any positive real ɛ there exist positive integers m and n such that mπ n < ɛ. Outline. Fix a positive integer N such that 1 < ɛ. The fractional part of any positive N real lies in precisely one of the sets [0, 1/N), [1/N, 2/N),..., [(N 1)/N, 1). Some two of elements of {π, 2π,..., (N + 1)π}, have a fractional part in the same interval; their difference is what we need. 5. Show that any set of 27 different positive odd integers, all less than 100, contains a pair whose sum is 102. Solution. There are 24 pairs summing to 102: {3, 99}, {5, 97},..., {49, 53}. The above pairs exclude 1 and 51, but choosing 25 of the above numbers requires choosing both numbers from at least one pair. 3
4 6. Let n be a positive integer and P a set of n primes. Show that given any set S of n + 1 positive integers whose prime divisors are contained in P, we can find a subset T S such that the product of the elements of T is a perfect square. Outline. Consider the products of the elements of each subset. Use pigeonhole to find two subsets A and B that such that the products of the elements in each agree as far as the parities of the exponents of the primes in P. Then use the set (A B) \ (A B). 7. Does there exist an infinite sequence of positive integers, containing each positive integer exactly once, such that the sum of the first n terms is divisible by n for every n? Outline. Yes. Construct the sequence one element at a time, alternating between a large number and the least unused number. The large numbers are chosen so that their successors will satisfy the division property. They can be found with the help of the Chinese Remainder Theorem after noting that two consecutive integers are always relatively prime. 8. Let n > 1 be a positive integer. Suppose that 2n chess pieces are placed at distinct squares of an n n chessboard. Show that four of these pieces determine the vertices of a parallelogram. Outline. Imagine each pair of pieces as determining a vector in the first or fourth quadrant. There are n(2n 1) 1 possibilities, owing to the board dimensions, but there are n(2n 1) pairs. 9. (Zuming) Prove that every convex polygon with an even number of sides has a diagonal that is not parallel to any of its sides. Outline. For a 2n-gon, there are n(2n 3) diagonals and 2n sides, while each side is parallel to at most n 2 diagonals. 10. (Zuming) Let a 1,..., a 100 and b 1,..., b 100 be two permutations of the integers from 1 to 100. Prove that, among the products a 1 b 1, a 2 b 2,..., a 100 b 100, there are two with the same remainder upon division by 100. Outline. Suppose the remainders are all different. Then every residue appears once. Since there are 50 odd numbers, each pair {a i, b i } comprises numbers of the same parity. It follows that every even number is a product of two even numbers, so a multiple of 4. Then there can be no remainder of the form 4k The set {1, 2, 3,..., 16} is partitioned into three sets. Prove that one of the subsets contains some numbers x, y, z (not necessarily distinct) such that x + y = z. Outline. By pigeonhole, a subset, say A, contains at least six elements, a 1 < a 2 < a 3 < a 4 < a 5 < a 6. The differences a i a 1 as i = 2, 3,..., 6 must lie in the other two sets, say B or C. Then WLOG B contains three of the differences, b 1 < b 2 < b 3. Then b 2 b 1, b 3 b 1, b 3 b 2 lie in C yet b 3 b 1 = (b 3 b 2 ) + (b 2 b 1.) 4
5 12. A 6 6 rectangular grid is tiled with nonoverlapping 1 2 rectangles. Must there be a line through the interior of the grid that does not pass through a rectangle? Outline. Yes. Using an inductive parity argument, we can prove that any of the 10 lines must pass through an even number of 1 2 rectangles. Then because there are just 18 small rectangles, one of the 10 lines pierces none. Homework 1. Every edge and every diagonal of a regular hexagon is assigned one of two colors. Show that some three of these segments consititute a monochromatic triangle. Solution. Suppose the colors are red and blue. Label the hexagon A 1 A 2 A 3 A 4 A 5 A 6. By the pigeon hole principle, some three edges from the set {A 1 A i i = 2, 3,..., 6} are the same color. Without loss of generality, A 1 A 2, A 1 A 3, and A 1 A 4 are blue. Then either the edges of A 2 A 3 A 4 are all red, or a blue edge in this triangle completes a blue triangle with vertex A Let a 1, a 2,..., a 99 be a permutation of 1, 2, 3,..., 99. Prove that there exist two equal numbers from a 1 1, a 2 2,..., a Solution. Note that 99 i=1 99 a i i a i i = 0 (mod 2). ( ) i=1 On the other hand, a i i < 99. Thus, if no two such numbers are equal, then contrary to (*). 99 i=1 a i i = = (mod 2), 3. (RMO, India 1996) Suppose that A is a 50-element subset of {1, 2, 3,..., 100} such that no two numbers from A add up to 100. Show that A contains a square. Solution. Suppose for the sake of contradiction that A does not contain a square. Consider the pairs {1, 99}, {2, 98},..., {49, 51}. Observe that the intersection A {36, 64} is empty. It follows that regardless of whether 50 A, the set A contains at least 49 elements from the 48 other pairs. By the pigeon hole principle, this contradicts the condition that A contains no pair of numbers adding up to (St. Petersburg, 1997) Fifty numbers are chosen from the set {1,..., 99}, no two of which sum to 99 or 100. Prove that the chosen numbers are 50, 51,..., 99. Outline. Observe that no two of the chosen numbers can occur consecutively in the sequence 50, 49, 51, 48, 52,..., 1, 99. 5
6 5. (NMO, India 1994) Let S be a set of 181 square integers. Prove that there exists a 19-element subset T S such that the sum of the elements of T is divisble by 19. Outline. Note that there are just 10 quadratic residues in modulo 19; we can find 19 squares having the same remainder upon division by (Iran 1996) Suppose that a chessboard is made into a torus by identifying both pairs of opposite edges together. How many knights can be added to the resulting board such that no two knights attack each other? Outline. 32. It is possible to place the knights on all of the squares of one color. On the other hand, k knights occupy k squares and attack 8k squares, yet each of 64 k squares can be attacked by at most 8 knights. 7. (Zuming) In each of 3 classes there are n students. Every student has exactly n + 1 friends altogether in the other two classes (not his class.) Prove that in each class you can find one student, so that all three of them, taken in pairs, are friends. Outline. Look at a student having the least number of friends in a different class; suppose the student is in class A, has k friends in class B, and at least n k +1 friends in class C. Any friend of the student in class B must be friends with at least k students in class C. Then observe that the original student is unfamiliar with at most k 1 students in class C. 8. (IMO 1972) Prove that from a set of ten distinct two-digit natural numbers, it is possible to select two disjoint nonempty subsets whose members have the same sum. Outline. There are 2 10 = 1024 subsets of a ten element set, but the sum of the numbers in any subset is a nonnegative integer less than Given two sets A and B whose elements have the same sum, we can find sets A \ (A B) and B \ (A B). 6
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# Regents Recap — June, 2017: Trouble with Dilations (and Logic)
The emphasis on transformations in Common Core Geometry has proven to be a challenge for the creators of the New York State Regents. Here’s the latest example.
This is a tricky question. So tricky, in fact, that it tripped up those responsible for creating this exam.
Dilation is a similarity mapping (assuming, as we do, that the scale factor is non-zero), and translation is a congruence mapping. Thus, any composition of the two will be a similarity mapping, but not necessarily a congruence mapping. So in the above question, statement II will always be true, and statements I and IV are not always true.
Statement III requires closer attention. Under most circumstances, translations and dilations map lines to parallel lines, and so the same would be true of their compositions. However, if the center of dilation lies on a given line, or the translation is parallel to the given line, then that line will be mapped onto itself under the transformation.
This means that the answer to this test question hinges on the question, “Is a line parallel to itself?”
If the answer is yes, then statement III will always be true, and so (3) II and III will be the correct answer. If the answer is no, then statement III won’t always be true. and so (1) II only will be the correct answer.
So which is the correct answer? Well, that’s tricky, too. The answer key provided by New York state originally gave (3) as the correct answer. But several days later, the NYS Department of Education issued a memo instructing graders to accept both (1) and (3) as correct. Apparently, the state isn’t prepared to take a stance on this issue.
Their final decision is amusing, as these two answer choices are mutually exclusive: either statement III is always true or it isn’t always true. It can’t be both. Those responsible for this exam are trying to get away with quietly asserting that (P and not P) can be true!
Oddly enough, this wasn’t the only place on this very exam where this issue arose. Here’s question 6:Notice that this question directly acknowledges that the location of the center of dilation impacts whether or not a line is mapped to a parallel line. It’s not entirely correct (a center’s location on the line, not the segment, is what matters) but it demonstrates some of the knowledge that was lacking in question 14. How, then, did the problem with question 14 slip through?
As is typical, the state provided a meaningless and generic explanation for the error: this problem was a result of discrepancies in wording. But there are no discrepancies in wording here. This is simply a careless error, one that should have been caught early in the test production process, and one that would have been caught if production of these exams were taken more seriously.
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#### Amy Hogan · July 20, 2017 at 12:03 pm
I like your explanation of the “discrepancies” here, Patrick.
In my mind, this comes down to an issue of incomplete information regarding the transformation. To me, clear communication of a dilation includes a scale factor AND a center. Blanket statements about transformations like this make it so that the test-taker is forced to imagine what those things _might_ be.
As we know, dilations don’t act in a consistent manner in the plane. Negative scale factors, a scale of 1 (perhaps trivial, but still a possibility under this generic language), something greater than zero but less than one, etc. will produce different results. Likewise, varying centers of dilation (on vertex, on line segment, inside triangle, outside triangle,…) will give some interesting and very different results. And, most definitely, as you mentioned, where that center of dilation is will influence whether or not the image’s segments are parallel to its pre-image.
The lack of this information in this problem is equivalent to hand-waving on the part of the test committee. “Well, you know, a dilation.” The geometry student is going to have to fill in the blanks. A creative geometry student is going to fill in those blanks creatively. And in a timed test, why is this necessary? Adding more specific information about a scale and center wouldn’t inhibit the assessment of the question’s intent, as far as I can tell.
It shouldn’t be the test-takers job to both write the question and answer it. Perhaps in the future, students will have Mad Lib Regents questions: [Geometric Figure] is the image of [Geometric Figure] after a [Transformation] followed by a [Transformation]. Which statement(s) will [Always, Sometimes, Never] be true? 😛
#### MrHonner · July 20, 2017 at 5:17 pm
I think it’s appropriate, and important, to push students to think about abstract mathematical concepts (an arbitrary dilation, for example). When specified properly, these various transformations do have general properties that can be reasoned about. To me, the only problem with “incomplete information” is that of incomplete understanding in the minds of those who oversee this exam.
#### Jerome Dancis · July 20, 2017 at 2:03 pm
The real question is: “Is a line parallel to itself?”. As noted, this depends on the definition. Hopefully, there is a glossary or list of definitions on the state website.
If the definition excludes a line being parallel to itself, then being parallel would not be a transitive property. Boo. This would then require the awkward statement:
Theorem. If Line A is parallel is to Line B, which in turn is parallel to Line C, then Line A is parallel to Line C (except when Lines A and C are the same).
#### MrHonner · July 20, 2017 at 5:10 pm
Jerome-
There is no official list of definitions provided by the state. The closest approximation to that are the curricular materials provided by EngageNY (NY’s official unofficial curriculum), and those materials support the claim that a line is not parallel to itself.
#### Jerome Dancis · July 21, 2017 at 10:37 am
Mathematics is written literally and explicitly. Merely, supporting a claim is not Mathematics.
Emily DeSantis, a spokeswoman for the NYSED, mentioned that the Geometry Regents Exam is written and graded according to the concepts included in geometry, specifically cluster G.SRT.B, ( in her recent email to ABC News). Yes, the Geometry Regents Exam made national news; albeit just the grading of one problem.
Does this cluster G.SRT.B not provide definitions?
Does this cluster G.SRT.B or EngageNY include:
Theorem. If Line A is parallel is to Line B, which in turn is parallel to Line C, then Line A is parallel to Line C (except when Lines A and C are the same).
or
Theorem. If Line A is parallel is to Line B, which in turn is parallel to Line C, then Line A is parallel to Line C.
#### MrHonner · July 21, 2017 at 11:55 am
Jerome-
You are welcome to inspect for yourself.
The G.SRT standards are here, and I’ve create a Google search of Engage NY materials for “parallel line theorem” here.
#### Max McMahon · July 23, 2017 at 8:38 pm
Though it’s not part of the geometry standards, they do touch on the idea that parallel lines and the same line are distinct entities through standard form systems of linear equations (ax + by = c and a’x + b’y = c’). They define a system of linear equations that has no solution as parallel lines (a/a’ = b/b’ only) and a system of linear equations that has infinite solutions as the same line (a/a’ = b/b’ = c/c’).
#### MrHonner · July 24, 2017 at 8:48 am
Yes, good point. This is another situation where consequences of the definition of parallel pop up: are the graphs of two linear equations with no common solution “parallel”, or “parallel and distinct”?
#### Max McMahon · July 23, 2017 at 8:18 pm
At east it isn’t the 2008 8th grade exam where they gave points for doing a question wrong and not for doing it right…
#### MrHonner · July 24, 2017 at 8:34 am
Max-
I’m not familiar with that problem. Do you have a link?
But it reminds me of this Regents question from 2011, where the student’s score decreased for being less incorrect.
#### Max McMahon · July 24, 2017 at 4:39 pm
I can’t find the link (though I kept the scoring guide, it’s in storage somewhere) but the problem was “solve and check: 3(p + 6) = 5p + 4” with separate scorings for the ‘solve’ and the ‘check.’ The ‘solve’ part was fine, but the check was horrific (the most horrific part being how difficult it was to find someone at the site or headquarters who even understood why it was a problem).
On the rubric, they completely ignored the substitution of 3(7 + 6) = 5(7) + 4.
The first line was 21 + 18 = 35 + 4. The graders at my site were almost universally not giving any points for putting 3(13) = 35 + 4. It would have been one thing (still frustrating but logistically understandable) if they gave credit for doing it both ways, but, for over two days, literally nobody i spoke to understood that using the distributive property (rather than the order of operations) to check this problem was flat out wrong. Finally I ended up FAXING an explanation as to why it was wrong, demonstrating that if you mess up the distribution by only multiplying the variable, you’ll get that the answer is 1 and not 7, and a subsequent check (making the same mistake) would confirm that one was the correct answer. In the afternoon of the 3rd (and final) day of grading, somebody at headquarters finally conceded that it was flat out wrong and that they were giving points to kids who got it wrong but not those who got it right, but told me that it was too late to do anything about it. I was a first year teacher so didn’t want to piss anybody off any more than I had already, but my one of my biggest regrets from now 10 years of teaching is that I didn’t fight even harder (or just gone to the News or Post). But the worst part wasn’t that the rubric was wrong to begin with, it was how few people, including fellow teachers, who even partially understood that there was even a problem in the first place…
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## how to find delta v in physics
Although, I don't think I related exactly to internal energy. What we have is a quadratic equation in .In some situations you may be able to solve this equation by factoring, but, more likely, you will have to resort to using the quadratic formula: He holds a BS in Physics from the University of California, Santa Barbara and an MS in Theoretical Physics from San Francisco State University. The initial and final potential energy of the charge is (q V1) and (q V2), respectively. Solve for v, u, a or s; final velocity, initial velocity, acceleration ar displacement. But since fuel mass grows exponentially with delta-v, it’s easier to work with delta-v instead of fuel mass directly. To get from the ground to low Earth orbit (LEO) requires a delta-V of about 8600 m/sec. change in velocity (∆v) is measured in metres per second (m/s) (∆ is the Greek letter delta, representing 'change in') time taken ( t ) is measured in seconds (s) Energy is so important to so many subjects that there is a tendency to define a special energy unit for each major topic. Sign up to join this community. The find_peaks function returns the indices where the peaks are located and also some information about them. Find the potential difference V2 - V3. The delta values are taken at the beginning and end. (Note that formulas are not given on the test.) And something tells me that I might have already done this pseudo-proof in the physics play list. Velocity Definition Physics The time rate of change of displacement of an object in a particular direction is called its velocity. Kinematics v ave = ∆x ∆t v ave =averagevelocity ∆x =displacement ∆t =elapsedtime The definition of average ve-locity. That is a number that lies between 10^8428 and 10^13359. This will be particularly noticeable in the chapters on modern physics. Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Home Questions Tags Users Unanswered Rocket Propulsion, delta V, acceleration and time. Let's say my system is this cube. The mass is distributed over the two velocities on line three. This article has been viewed 1,218,165 times. v ave = (v i + v f) 2 v ave … I've done it a few times in career mode when I want to feel like a true rocket scientist but I recommend just getting like KER and just using the delta-v map … Boundless Physics. For example, to find the delta between 1/3 and 1/2, you must first find a common denominator. Once the uncertainty in momentum Δp is found, the uncertainty in velocity can be found from Δp = mΔv. $$m\frac{\Delta v}{r_f} = \frac{GMm}{r_f^2},$$ which results in the ultra-classic $$\Delta v = \sqrt{\frac{GM}{r_f}}.$$ But now you deal with a rocket here and there is an essential point your forgot: the rocket will loose mass as it burns fuel! Using the equals sign in the uncertainty principle to express the minimum uncertainty, we have $\displaystyle\Delta{x}\Delta{p}=\frac{h}{4\pi}\\$ Solving for Δp and substituting known values gives You actually want to find minima so the trick is to put flip the signs on the currents. Feb 2015 1 0. Solution for Part 1. Question: Find The Potential Difference Delta V = VB - VA In The Following Circuit Elements: In The Circuit Shown, Find (a) The Current Through The Resistance R, (b) The Potential Difference Delta V = VB - VA , And (c) The Value Of R. This problem has been … Learning Objectives. A variation would be W = V(delta P), where V is volume, and delta P is the change in pressure. Formulas used in physics are presented and explained. The mass of the vehicle is 3300 kg. $\Delta v$ is usually given in kilometers per second. Lifecycle Services Our Global Service Centers support all your service requirements or technical queries. And to figure out the ct prime coordinate, we would just go parallel to the x prime axis. Change in velocity, or delta-V, is the most important measure of "distance" in space flight. It's the clearest measure of how hard a rocket has to work to change position. Then to find the mass of the fuel you need just use the other equations in this thread (the one with the natural log). Delta-V ( ∆v) is a measure ... Delta-V ( ∆v) is commonplace notation used in mathematics and particularly in physics to denote a change or difference in velocity. problem with potential difference (delta V) Thread starter nakenjiex; Start date Feb 24, 2015; Tags delta difference potential problem; Home. Velocity Formula in Physics Its SI unit is m/s. for the SAT Subject test in physics. It doesn’t matter if you slow down or speed up, the fuel consumed is the same for a given delta-v. The radius of the orbit is 2RE. To get from LEO to the Moon requires a different delta-V. Free online physics calculators for velocity and displacement. a * t = 10 * 12 =120; Subtract the product from the final velocity. $\begingroup$ Ultimately, we use delta-v to determine the amount of fuel needed to change the trajectory to a desired one using impulse thrusts (sudden change in velocity). We will only use the indices. So I'll do that here. Search for: Quantities of Rotational Kinematics. According to the Wikipedia article you mention the $\Delta v$ for Mars is 10.7 km/sec rather than the value of 6.2 you mention. Quickly find device installation kits for Emerson products. Using the contour plot you can determined the potential difference [Delta]V between the initial and final position of charge q (make sure you get the sign correct). Thus, the right side obviously ends up as the difference between two momenta. The momentum shown as the mass times the final velocity, i.e., mv', is the final momentum, and the momentum shown … Forums. Well, Delta U can be used to calculate the change of Delta F with temperature (van't Hoff equation). And let's say the dimensions of the cube are x in every direction. Rotational Kinematics, Angular Momentum, and Energy. 306 0. Online physics calculators for motion in a straight line under constant acceleration: Average velocity, Velocity as a function of initial velocity, acceleration and time, Displacement as a function of velocity and time, Displacement as a function of initial velocity, acceleration and time Feb 24, 2015 #1 I'm asked to: b. We want to do this exponentially. So let's say-- let me just draw it-- I have a cube. The V i = ?, V f = 200 m/s, a = 10 m/s 2, t = 12 s; Multiply the acceleration and time. Chemical rockets for relativistic travel are beyond … High School Physics Help. Electricity / Magnetism N. nakenjiex. In this case, it looks like this: 1/3 x 2/2 = 2/6 and 1/2 x 3/3 = 3/6. To figure out the total work done on, or by, a gas system, they use the formula W = P (delta)V. W stands for work, P is the pressure of the system (for gases), and delta V is the change in volume for the system. For comparison, the entire mass of the Universe is estimated to be 10^53 kg. The Determination of Vehicle Speeds from Delta-V in Two Vehicle Planar Collisions J Neades AiTS, A5 Lakeside Business Park, South Cerney, Gloucestershire, UK R Smith Faculty of Technology, De Montfort University, Leicester, UK. But you are right it is not so usefull for an adiabatical reaction. The angle of rotation is a measurement of the amount (the angle) that a figure is rotated about a fixed point— often the center of a circle. Delta-v (more known as "change in velocity"), symbolized as ∆v and pronounced delta-vee, as used in spacecraft flight dynamics, is a measure of the impulse per unit of spacecraft mass that is needed to perform a maneuver such as launching from or landing on a planet or moon, or an in-space orbital maneuver.It is a scalar that has the units of speed. I checked it's definition in Prigogine, Defay, Chemical Thermodynamics and it is in deed defined at fixed T and V, as I said. Relative Delta. Homework Statement A space vehicle is in circular orbit about the earth. Introductory Physics Homework Help. A good explanation of delta-V is given in Chapter 2 of Space … Now how do you actually go in between, transform, from x to x prime and from ct to ct prime? In the context of a motor vehicle crash, ∆v specifically refers to the change in velocity between pre-collision and post-collision trajectories of a vehicle. So this would be the ct prime coordinate. In line two we change delta v to the quantity of the final velocity minus the original velocity, as one can do with any delta quantity. Angular Position, Theta . The kinematics equation that relates , vo, a, and is:. ∆v = vafter −vbefore Delta-V emerged in the 1970s in the … But I think you'll find it OK, or reasonably satisfying. Services & Consulting Project Services Our performance services offer expertise and support to help take your operation to the next level. The $\Delta v$ is useful because it tells you how much work you need to get to your target i.e. Substituting this into the expression for $$v$$ gives \[v = \dfrac{r \Delta \theta}{\Delta t} = r\omega. Point 2 is just below the sphere, and point 3 is right beside the … To do this, multiply the denominators together, then multiply the numerator in each fraction by the denominator of the other fraction. Depending on the situation, delta-v can be either a spatial vector (Δv) or scalar (Δv).In either case it is equal to the acceleration (vector or scalar) integrated over time: = − = ∫ (vector version) And to do that, … Calculate final velocity as a function of initial velocity, acceleration and displacement using v^2 = u^2 + 2as. V i = V f – (a * t) = 200 – 120 = 80 V i = 80 m/s east; Write your answer correctly. It only takes a minute to sign up. Find delta-v; Hohmann transfer orbit Thread starter oddjobmj; Start date Nov 19, 2013; Nov 19, 2013 #1 oddjobmj. So that would be the x prime coordinate in her frame of reference. Now the minima are maxima and the peak finding algorithm will find them. We find that for a chemical rocket to achieve a deltaV of 42% of the speed of light, we would need e^28000 kilograms of fuel for each kilogram of equipment, structure, engines and payload. How do … Although this problem looks a lot like example 7, the v o t term does not disappear, which makes this a whole different ball game. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Assess the relationship between radians the the … ‘Delta v’ is engineering speak for change in velocity. ABSTRACT The change of a vehicle’s velocity, delta-V ( v), due to an impact is often calculated and used in the scientific investigation of road … It is desired to transfer the vehicle to a circular orbit of radius … Its dimensional formula is [M°LT-1]. Each formula row contains a description of the variables or constants that make up the formula, along with a brief explanation of the formula. With over 15 years of experience, Sean has worked as a physics and math instructor and tutor for Stanford University, San Francisco State University, and Stanbridge Academy. v i is the initial velocity (vector) v ix is the component of the initial velocity along the horizontal direction x (scalar) v iy is the component of the initial velocity along the vertical direction y (scalar) θ is the initial angle that v … To find her x prime coordinate, we would just go parallel to the ct prime axis. Subtract 2/6 from 3/6 to arrive at the delta, which is 1/6. In general physics, delta-v is a change in velocity.The Greek uppercase letter Δ (delta) is the standard mathematical symbol to represent change in some quantity.. Pairing that with a delta-v map that you can find through the wiki you can plan missions down to a few kg of fuel. 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# The business’s headquarters was moved first from Atlanta to Macon, and
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21 Mar 2018, 06:23
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The business’s headquarters was moved first from Atlanta to Macon, and then, after three more years of declining profits, from Macon to Omaha, during which the business succeeded in recapturing its dominant presence in the market.
(A) from Macon to Omaha, during which the business
(B) from Macon to Omaha, where it
(C) from Macon to Omaha, where they
(D) it was then moved from Macon to Omaha, where it
(E) from Macon to Omaha, where the business
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21 Mar 2018, 09:32
I go with A. Which modifies from Macon to Omaha.
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21 Mar 2018, 09:34
It is going to bring a lot of learning to posters and readers if the entire analysis or query is posted. Just posting the favorite choices solves no purpose. Hope it makes sense. Let's discuss
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21 Mar 2018, 10:06
arvind910619 wrote:
HKD1710 wrote:
The business’s headquarters was moved first from Atlanta to Macon, and then, after three more years of declining profits, from Macon to Omaha, during which the business succeeded in recapturing its dominant presence in the market.
(A) from Macon to Omaha, during which the business
(B) from Macon to Omaha, where it
(C) from Macon to Omaha, where they
(D) it was then moved from Macon to Omaha, where it
(E) from Macon to Omaha, where the business
Imo E
A wrong we can not use during here as it refers to time not place .
B is incorrect as it is wrong here .Headquarter can not capture market but business can so out .it is ambiguous
C Again they is wrong
D use of it ambiguous here
Think E is more clear and only business can succeed in capturing market .
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22 Mar 2018, 01:03
I think E is the right answer because the use of possession in the beginning(business's) needs a direct mention of the noun later in the sentence(business) that takes place only in option E. Hence the answer
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24 Mar 2018, 23:28
imho D,
Since there are only two elements in the list: from Atlanta to Macon and from Macon to Omaha, we don't need a comma before and. Comma before and signifies that as per FANBOYS there has to be another subject after "and". D is the only such choice, where both it refers to headquarters.
what is wrong in this approach?
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25 Mar 2018, 00:16
anil0548 wrote:
imho D,
Since there are only two elements in the list: from Atlanta to Macon and from Macon to Omaha, we don't need a comma before and. Comma before and signifies that as per FANBOYS there has to be another subject after "and". D is the only such choice, where both it refers to headquarters.
what is wrong in this approach?
from X to Y and then after sometime
it was then moved from Y to Z
how many "then" do we need?
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04 Jun 2018, 07:39
HKD1710 wrote:
The business’s headquarters was moved first from Atlanta to Macon, and then, after three more years of declining profits, from Macon to Omaha, during which the business succeeded in recapturing its dominant presence in the market.
(A) from Macon to Omaha, during which the business
(B) from Macon to Omaha, where it
(C) from Macon to Omaha, where they
(D) it was then moved from Macon to Omaha, where it
(E) from Macon to Omaha, where the business
VERITAS PREP OFFICIAL SOLUTION:
The most actionable decision point on this problem is the pronoun choice - you have decisions at the end of the sentence between "it," "they," and "the business." And note that (as in this case) when answer choices replace a pronoun with an "actual" noun, you should always be a bit skeptical of whether a pronoun is the right choice at all.
Here the subject of the sentence is "the business's headquarters," or more succinctly "headquarters." This is singular, so "they" would be an incorrect pronoun for singular/plural agreement reasons. But even more importantly the headquarters isn't what "succeeded in recapturing its dominant presence in the market." The business is what succeeded, after its headquarters had been moved. So none of the pronouns are correct: "they" commits an agreement error and "it" commits a logical reference error. Therefore (B), (C), and (D) are all incorrect.
Between (A) and (E), the remaining choices, the difference is one of modifier "during which" vs. "where." Because that modifier describes locations and not time periods, you must use "where," making choice (E) correct.
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Re: The business’s headquarters was moved first from Atlanta to Macon, and &nbs [#permalink] 04 Jun 2018, 07:39
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## Friday, February 29, 2008
### Odd and Even Functions
Hello, fellow classmates. My name is Paul, and I will be your scribe for today, and I will be writing about "odd" and "even" functions in an overview of Friday's class.
So our class today started with a review of yesterday's class (slide 2), which basically stated that:
In the formula y = af[b(x-c)]+d:
- If a > 1, the graph is stretched vertically
- If 0 < |a| <> 1, the graph is compressed horizontally
- The y-coordinates of f are multiplied by a
- If b > 1, the graph is compressed horizontally ("speeds up")
- If 0 < |b| < 1, the graph is stretched horizontally ("slows down")
- The x-coordinates of f are multiplied by (1/b)
Following the review, we spent some time practicing graphing functions, which can be seen on slide 3.
REMEMBER!
Stretches before translations!
Here's a quick rundown of the formula and what each variable does:
y = af[b(x-c)]+d
a -> Vertical stretch/compression
d -> Vertical shift
b -> Horizontal stretch/compression
c -> Horizontal shift
Note: Remember that you have to stretch the graph first, but it doesn't have to be both stretches first (abcd/abdc)! You can stretch the x axis and then shift it before stretching the y axis(adbc), the important part is that the stretch for each axis must happen before the shift for its respective axis. This was discussed on slide 5, where we wrote the possible orders you can apply transformations.
After the practice with drawing graphs, we were introduced to images, which is the graph after it has been shifted/stretched (slides 4 and 6).
Also, when talking about an image, the "formula" is:
(image) is the image of (original coordinates) under (function).
So an example would be:
(0,7) is the image of (-2,-2) under the function y=-3f[1/2(x-4)]+1
We then proceeded to learn all about the online bookmark service and website Del.icio.us.
I won't really go into detail about this, since it's pretty self-explanatory. However in the interest of thoroughness, I've made a quick image here:
And with that, we proceeded to our next topic of reflections (slide 7):
- Basically, a vertical reflection is a reflection along the x-axis which occurs when the y-coordinates of any function f(x) are multiplied by (-1).
- Conversely, a horizontal reflection is a reflection along the y-axis which occurs when the x-coordinates of any function f(x) are multiplied by (-1).
Slide 8 contains a picture of a reflected sine wave, neato.
Inverses are also introduced on slide 7:
As stated by the slide, the inverse of f(x) is f^-1(x) [f to the power (-1) then/applied to (x)]. Which looks like this (image follows):
Thank you, Paint, for that excellent demonstration.
And in BIG SCARY LETTERS, I WILL MAKE SURE YOU DO NOT FORGET THAT f^-1(x) IS NOT EQUAL TO [(1/f(x)]!!!
The truth is that
(can you handle this?)... [f(x)]^-1 (note how the power is on the outside), is the one that is equal to [1/f(x)].
So in summary:
[f(x)]^-1 = [1/f(x)] √ Correct
f^-1(x) = [1/f(x)] X Wrong
Its also pronounced "eff inverse." And finally, f^-1(x) undoes what f(x) does.
Lastly, we started talking about "odd" and "even" functions (slides 10 and 9 respectively):
A function is "even" if it is symmetrical about the y-axis, which occurs only when f(x) = f(-x). If you look, a cosine curve is a nice example of an "even" function.
A function is "odd" if it is symmetrical about the origin, which occurs only when f(-x) = -f(x) (the negative of the whole function). If you look, a sine curve is a nice example of an "odd" function.
And that concludes my scribe post summarizing the topics we covered during our class last Friday. Now about the scribe...
Since some people actually want to be scribe (unlike myself who was hoping to be scribe for pi day), I hereby declare that benofschool has the honour of being the next scribe for Monday's class.
And don't forget to get a Delicious (mmm, delicious) account if you haven't already! Here's a direct link for you who are too lazy to type it (shame on you), or like convenient things.
Del.icio.us
So I bid you adieu and offer my sincerest apologies for the lateness of my post. Unfortunately, I had work Sunday and was busy the majority of Saturday. I did infact start this post on Friday, but did not manage to finish it until today. I am highly aware of the time I am posting this at, especially considering I had resolved to post it first thing on Friday.
Good night, and I'll see you tomorrow.
P.S. I would like to offer the following advice to the following scribes:
Blogger's draft system is not without its flaws, and I have had to rewrite the second paragraph (involving the overview of the a and b variables) numerous times because the draft system confuses my use of the arrow signs (<>) as parts of html tags and deletes everything inbetween. Thus, if you think you're not going to finish your post, just save yourself some hassle and put it in a notepad document instead of using the draft system.
### Today's Slides: February 29
Homework is: Exercise #8.
Here are today's slides...
## Thursday, February 28, 2008
### Transformations
I am deeply sorry I posted so late, and I apologize to whomever was waiting for a scribe post. Unfortunately I didn't check the blog yesterday and hadn't realized that I was the next scribe as I was up working on a project for another class. Anyways, I'll recall as much as I can remember.
We started off the class waiting outside... but then he let us back in and let us write the second half of the test where we drew the graph of a function and wrote two functions for a graph. As you can see in the second slide, we went over the question just in case anyone u, anxieties, questions and any good jokes. In slide three we quickly went over what the answer was for the second question. He showed us a quick process to easily write the function. He found 'D' was 3 because the sinusoidal axis had moved UP 3 units. He found this by looking at the distance from crest to trough and found that distance was 6, so then he discovered D was 3. From that he also concluded that the amplitude was also 3. Finding the period, he saw that one full wave had occurred in the time of 'pi' so then he took B = 2pi/period, and found B = 2, as you can see in the slide. For Sine, there was no phase shift as the graph already started at the origin of the sinusoidal axis. For Cosine, he took the closest maximum value to the Y axis and found the phase shift is pi/4. So now we have the Functions. But don't do what I did and REMEMBER to put either y = 'function' or for whatever function its asking you to find (ex. f(x) = ...) You'll LOSE a WHOLE MARK as it is an EXPRESSION and not a FUNCTION.
We then looked at translations and how 'a' and 'b' respectively affect the function and it's position where 'a' affects the x coordinate and where 'b' affects the y coordinates as you can see in slide 4.
'A' is like the phase shift, and shifts the 'x' position of the function left or right, and b is like the sinusoidal axis, but really it just shifts the function up or down (depending on the sign).
We then looked at stretches. In f(x) = ... x is what is used to graph, so if in the case there is a 2 in the f(x) (ex. f(2x) or you can just look in the slides), the 2 would affect how the function looks. If the two happened to be in front of the whole function, the whole function would be multiplied by two. We then headed over to http://fooplot.com to see how they would get affected by an integer in front of the 'x'. The larger the integer, the smaller the function looks, because in a sense, it takes less time, as to where if it were an integer, that would mean it would take longer, thus stretching the graph out.
Unfortunately we didn't make it to Compressions.
It's probably no-use saying this at this time of the night but homework is exercise 8. Again, I'm truly sorry for posting so late, especially to the people who were looking for the scribe to post to look for some answers.
~Rence_Out
Oh yeah, Next scribe will be... *checks list*
DARTH PAUL! I CHOOSE YOU! QUICK ATTACK, NOW!
To simplify that unnecessary outburst of randomness, Paul's the next scribe.
### Today's Slides: February 28
Here they are ...
## Wednesday, February 27, 2008
### New Unit: Transformations
Hi everyone! Its finally my turn to scribe, and today we started a new unit. But first I must talk about what little happened during the morning class...
In the morning, we began by looking over the review questions that were on yesterday's slides. Mr. K told us not to worry about the second part of those review questions as much since we didn't really go over how to do them, but we did in the afternoon to some extent.
After we had finished briefly looking over those questions, we were given the first part of our Circular Functions test. Notice how I say "first part". The reason why I say this is because tomorrow morning, we will be getting a second part to the test, which should only take us about 5 minutes to do, according to Mr. K. The second part will consist of graphing work. That was all that happened in the morning class.
Onto the afternoon class, which we had a late start on because of Mr. K talking to the principal or something. Anyways, Mr. K first started off by showing off some trippy spiral themed pictures from www.flickr.com. Then he took it one step further by showing us his newly downloaded firefox add on, PicLens, which gave him the ability to look through all the pictures in a cool way. Its hard to explain for me so if you're interested in this, go to www.piclens.com!
Okay back to math! So, we started on a new unit called Transformations. The first thing we did was look at the graph for f(x) = x^2, which looks like this:
We then looked at what would happen when 2 was added to the function to make f(x) = x^2 + 2. The graph ended up looking like this:
As you can see, by adding 2, the parabola was shifted upwards by two units. We then looked back at the sine graph and did the same thing to it, except we subtracted 1, therefore causing the graph to shift downwards by one unit.
Now, if you're looking at slide 3 of today's class, you'll notice that we tried graphing y(x) = 2^x. Don't fret if you don't understand what this is. The reason why the graph looks like that is because since x is equal to a number, when 2 is to the exponent of that number, it will increase as so, making the graph look like that. To help you understand it better, here is a table that shows you a simple pattern of the relationship between exponents and a number, which in this case is 2:
See the pattern? Starting from exponent 4 all the way down to negative exponent 4, the values keep halving themselves. So if x is equal to 4, that would mean the y coordinate would be 16. If x was equal to -4, that would mean the y coordinate would be at 1/16.
Okay, so now that you know why it looks like that, what would the graph look like if I were to do make it f(x) = 2^(x+3)? Well, if you're still looking on that slide, then you'd see that the graph moved 3 units to the left. This is pretty much like the phase shift in those trig function graphs we did for the past few units. The only difference is that it isn't called the phase shift, since phase shift is only used when dealing with trig functions, instead, we call it the horizontal shift.
So in conclusion to this part of the afternoon class, we can say that we write these functions in the form of y = f(x-b) + a, where "b" is the horizontal shift and "a" is the vertical shift.
By the way, the weird shaped thing below the f(x) = 2^x graphs are called "piece wise functions". They are called this because of they are split into pieces. As you can see in that specific piece wise function, there are three parts to it, thus why its a piece wise function.
For the rest of the class, we were given questions to work on based on the things I just talked about. The first question, as shown in slide 6, asked us to to write f(x) = sin(x-2) + 5 in terms of g(x). Since f(x) is basically g(x) shifted 4 units to the left, the only possible answer was g(x) = sin(x+2) + 5. As you can see, the only thing that was changed was "b". To get +2, we basically just added 4 since f(x) is units to the left of g(x), meaning that we'd have to move right 4 units from f(x) to get to g(x). Simple, right?
Okay, so onto slide 9. Remember those review questions from yesterday's slide? Well, we learned how to solve those now. I'll use the same example as shown in slide 9, which was sin(2x) = 1/2. To make this easier to work with, you can Let
θ = 2x and then work from there for now so that you've got sin θ = 1/2. Now, you should already know how to solve for this by now, and if you don't, look over the circular function notes again, mkay?
Yeah, so if you did it right, you should have gotten the answers in the black. If you're confused with the 13
π/6 and the 7π/6, we just got those by adding 12π/6 which is equal to 2π to π/6 and the reason why we did this is because of the original 2x. Since its 2x, when you look at it on the graph, you'll see that there are two complete waves, meaning that there's going to be double the solutions. The same can be said if it was 4x instead of 2x. In the case of 4x, there'd be 8 solutions, and if it was 6x instead of 2x, there'd be 12 solutions. You see?
You think you're done with the question now eh? Well, you're not. Since you'd switched out 2x for
θ, that means that you're going to have to put back the 2x. Then from there, you should be able to work with it, so if you did everything correct, you should get the answers in red. Voila, wasn't that oh so simple? :)
Um, so, I think I'm done now! Yay! Oh wait, just forgot to mention that tonight's homework is Exercise 7, questions 1-10, since you should have already finished 11-20 and if you haven't, you are a naughty child.
Now for the moment of truth, since I'm the last scribe in this cycle, that means I have a whole list to choose from now, ehehe. Therefore, I choose .....LAWRENCE.
Yeah, back to the beginning we go! Anyways, remember that there is part 2 of the Circular Functions test tomorrow, so good luck everyone!
~kristina
### Today's Slides: February 27
Here they are ...
## Tuesday, February 26, 2008
### Tuesday, February 26
My Turn to Scribe 2 ~ HOOOOOOOOOO!
It was my turn to scribe so...
At the start of class, Mr. K started by showing us a few websites with varying purposes. First he showed us skrbl.com which is basically an online whiteboard. Then he presented twitter.com which you can use to communicate with people but you have to sign up with an account.
After viewing the features of those sites, we finally got to buckle down and write our Pre-test. It was basically a scaled down version of a normal test. There were 2 multiple choice, 2 short answer, and 2 long answers. We were alloted 15 minutes. After our time limit was up, we assembled into groups of about 3 people. Each group would have to hand in the a paper with the best possible solutions. Following that we went over the answers to the test. Bell Rings*
ＴＨＥ END。
The next scribe is kristina.
### Circular Functions BOB
Hello, hello again. I'm back to do some.. BOB-ing. Does anyone else find it strange that you call it BOB? Or maybe that's just me.. Anyways..
I think this unit, for someone who hasn't touched any math for a YEAR, was a pretty good unit to start with. I'm not saying it was easy, but it was simple enough that I was able to wrap my math-challenged mind around it, and that was more than awesome. I found that having the unit circle forever drawn in my head really helped me; to me, having that down is understanding everything else.
But what I'm really worrying about is the graphing, because I missed the classes that focused on it. But I am getting it, just reaally slowly, kind of like Jamie, but.. probably slower.
Soo, that's my BOB for Circular Functions. Good luck on the test tomorrow you guys. =)
### Today's Slides: February 26
Here they are ...
## Monday, February 25, 2008
### BOB, Circular Functions
Well the unit Circular Functions was a bit confusing for me in the beginning because of that whole converting degrees to radians it took me a week to understand it but now i get it. another thing that also confused me was the functions secant and co secant it confusted me because the words do not match haha but i get it now. Now the only thing that is troubling me is the graghing for some odd reason i get confused with the period part (B) in DABC another thing that i need to remmember is to put label my graph and also put arrows on my graph. The part that i liked about this unit is that you wont lose any marks for forgeting to put the untits for radian, because there are no units ahaha.
Well now i guess its time to study for the test on wednesday.
Richard
### BOB on Circular Functions
Now that there's a test, I'm slightly nervous, but anxious to get it over with. This unit at first was quite simple and straight forward because converting degrees to radians in respect to proportions is quite easy.
The unit circle however, was really annoying because of all the memorizing we had to do. We had to memorize each point for every radian measurement on the unit circle, but everything in math is a pattern, and as soon as you have the pattern etched in memory then it should be no problem memorizing things such as the unit circle. I still sometimes have trouble with the points because I have yet to memorize it all. In time I will though. I guess I should draw out the complete unit circle a couple times without any outside help such as notes.
The next part I found worth mentioning is the trigonometry equations, we did these in grade 11, and I remember these quite well. I believe these equations were introduced to us in preparation for the sine and cosine waves that we are currently working on. Found from the equation f(x) = AsinB(x-C)+D or f(x) = AcosB(x-C)+D. Sure, it may look hard at first but with all great puzzles, you must put it together piece by piece. Plotting each value in order. Preferably in the order DABC. I found that plotting these were not too much of a hassle but I have yet to perfect it, due to simple mistakes. In time I will get wiser though.
Well that just about sums up my understandings on the unit of circular functions. All in all, this unit was a little rough around the edges, but still understandable.
Until next time!
-Francis
### Jamie's First BOB
Hello universe, and hello there BOB...I think everyone's introduced. I see from previous posts, everyone knows BOB.
There's an upcoming test this week and the normal thing is to be nervous. But I'm not normal, I'm just something else..and if you're something else, you can have different feelings like anxiety and quivering in fear. But what way better to review than to talk about what was learned in the past month or so?
The first unit was about Circular Functions. Functions about Circles...that's just plain cool. Radical, really. and round.. I just like making the "R" sound....TORRRRRRRRRRRRRRRRRRRRRTILLA nailed it! But yes, no more tangents. This unit was a great introductory unit, and it made me feel better about myself because I felt so smart coming up with answers out of no where but that was the beginning. It was basically a review of grade 11 pre-cal, from where i just freshly sprung from last semester...We took "rules" like CAST and disregarded them-- REBELS!! and we took theories apart and properly determined the difference between an equation and a proportion.
BUT THERE IS ONE SLIGHT PROBLEM that might just lead to my demise in the near future. I am a SLOOOOOOW learner. Or as benofschool used to say, "as swift as a turtle"...actually slower than that. A sloth. I just have a problems with prioritizing [spelling?] thoughts in my head. I know the most random of things, like the lyrics to any Sweeney Todd song or how to talk in different accents, but I still can't grasp the whole idea of the unit circle and the DABC concept. Don't get me wrong, I know it and where the information comes from and understand it, it's just I can't register these values fast enough to finish a test!!!! URGH!!! I'm not making any excuses like a scapegoat, but I'm just saying I can do it...in about a century. The funny thing is, my magic number is not 6. It's more like 6, 295,048...I know I'm overexaggerating, but I'm getting the point across. I'm even trying a new thing [the idea came from the walkie talkie device on the blog] where I recorded myself singing the sincostan values in quadrant one and then making my way across all of the quadrants because I believe music is my TOOL. Yes, I try to be innovative you know....also modest, but I suppose I'm not doing a great job at that.
OKAY YES...Mr. K. did say to not MEMORIZE but REMEMBER...but sometimes I just the provided example to the idea of the paradox, FORGET TO REMEMBER. Here I was last year being told that the "VERTEX WAS EVERYTHING", but now it's more like "IF YOU DON'T KNOW THE UNIT CIRCLE, YOU WILL FAIL." How encouraging. and the PARAMETERS!!! for graphing these functions. But I promise to you world, BOB, and myself that I will master the art of circle-- THE ROUND SENSEI.
By the way, good luck on your tests everyone! Let's count down to it like it's another New Years DAY! Don't mind my being nutty, it's just I'm trying to think of the glass being half full right now. hahah
### Monday, February 25
MY TURN TO SCRIBE ~ HOOOOOOO!
It was my turn to scribe so...
First of all, we started out by going over the test about maximum area of a triangle. So we have point P on the unit circle. B on the x-axis and it has to be a right angle triangle. What's the maximum area? Well we know that the area of a triangle is base x height divided by 2 and the base and height is equal to the x and y components of the triangle. We also know that the x and y components are equal to the sine and cosine of the central angle. We tested out the values and found that as the angle goes up the area would also increase but after π/4 radians the area would go down again. That means the sine and cosine of π/4 radians will give up the largest area.
After that, we went over question N on the sheet. Turns it out it was not a typo and we actually had to factor out the 2 pi. This will give us a period of 1. After that we went over some more transformations and Ｉ a few important things we should know. For example, parameter B is not the period itself, but actually helps us calculate the period. If we substitute B in the formula: period = 2π/b. Another important tip we should know is if we are given parameter A = -1/2 and we are asked for the amplitude, then the answer would be 1/2 and not -1/2. This is because the amplitude is the distance from the sinusoidal axis to the maximum or minimum points. Therefore, distance cannot be negative because this would destroy all logic. In other words, read your questions carefully, if it asks for amplitude, then discard the negative sign but if it asks for the value of parameter A then you include the value of A.
A few minutes later, we were given a wave graphed on the Cartesian Plane and were asked to write the equation of the graph in sine and cosine. Turns out there are INFINITE ways of writing the equation so... there's nothing to complain about there and plus we will only be required to only write one or two (max) equations on tests and the provincial exam.
Somewhere in the class, we looked at the infamous tangent function. Finally after a few of us were dying to see it, it was revealed. The tangent function is kinda special. It appears to be like x³ but IS NOT. The curvature is slighty different. Another feature of the wonderful tangent functions is asymptotes. Asymptotes are areas where tangent cannot exist. Since tangent is a trigonometrical function it has to do with triangles. At certain angles, tangent does not form a triangle therefore tangent cannot exist. Tangent can also be infinitely large unlike sine and cosine which are confined to -1~1.
That pretty much wraps up the class. Until next time.
### Today's Slides and Homework: February 25
Here they are ...
And your homework is here. Have fun!!
## Sunday, February 24, 2008
### BOB; Circular Functions
So knowing there's a test this week. I'll be BOBing.. while bobbing my head to music. Okay, lame, I know.
To the point. We started the semester without even introducing ourselves and our names which was a definite change from other teacher's ice breaking activities. And of we started, learning about Radians and what not, converting from degrees, a measurement we've been using in pre-cal for how long now, and finding our measurement was 'x' number pi over 'x' number. At first I had a hard time grasping the concept but later became comfortable after studying it for a while.
Later on, we learned the Unit Circle and all the radian measurements around the circle. Yet another item to remember, and the words of consequence from Mr. K when we questioned the idea of failing to memorize such a thing and he put it out plainly "You'll fail." And that's how the Pi pie crumbles... or is it how the cookie crumbles. Okay whatever, I'll stop with the mediocre attempts at what's called a joke. After much practice, I became comfortable with the Unit circle. Then finding the arc length was something else as well as we found that no matter how stretched out from the origin, it still shares the same proportionality.
The waves I found I was having a little trouble but had an easier time understanding than the whole Unit Circle memorization task. The only trouble I really had was with the Period ('B') but after Ben showed me how to do it it became clearer. I'll definitely study for this bad boy... Lol
G'niight all and good luck on the Test. Remember, the apple doesn't fall from the tree... wait what does that have to do with anything...?
-Lawrence
### My BOB for Circular Functions
The test is coming up soon so I thought the time was ripe for a BOB. Our very first unit is circular functions. At first, it was relatively easy. Nothing too difficult, I would say it was exactly like Grade 11 but units in radians instead of degrees. Took me a while to adjust to radians since we have been using degrees for most of our lives but as Mr. K always proclaims "FRACTIONS ARE OUR FRIENDS". Casting away CAST took zero effort whatsoever. As so, the memorization of the unit circle. Since it always has to be a root of 1, 2 or 3 over 2. Converting radians to degrees and vice versa was cakewalk since setting up a proportion is easy for me. All thanks to my Grade 8 Math teacher: MR. TRAN!
As things progressed on, we were exposed to more things that I found familiar from Grade 11. For example, solving for trigonometry functions of X was one of them. The only difference is that the angles are in radians now. Another thing was sine and cosine waves and their transformations.
I liked a few things about this unit. The fact that radians have no units was a load off my mind (in case I forget to write the unit on a test and lose a 1/2 mark). Solving for exact values of stuff was also enjoyable.
That brings my BOB to an end. Good night and Good Luck!
### BOB: Circular Functions
Hello everyone, since the test for circular functions is supposed to be this week, I'm here doing my BOB.
Okay, so during the beginning of this unit, when we were just introduced to converting degrees to radians and whatnot, I found that this was really easy. Sure, it was a bit confusing at first since I wasn't used to working with radians but it eventually got better as the unit progressed. Now I actually like working in radians better than degrees!
One part I had trouble with AT FIRST was learning the unit circle. As soon as I saw all those numbers on that unit circle, I panicked. What made it even worse was when Mr. K told us that we had to memorize EVERYTHING that was on the darn thing, and at that point, I was ready to cry. It didn't help much when Roxanne asked him what would happen if we didn't memorize the thing by the time he told us, to which he answered "You'll fail." But now that I know the cursed thing that caused me so much pain, I feel much better.
Now onto the graphing bits. The basic graphing wasn't so bad, but the added stuff, such as DABC, was. Sure, DABC was fun to say, but it was hard to understand at first since the last part of that class was rushed and it was so much to absorb in just those last 2 minutes we had.
Um, yeah, that's all I've got to say for this unit. It had its hard and easy moments, and overall, it was pretty good. That is all.
### BOB: Circular Functions
Since our test will be sometime this week, I decided to make my BOB entry before I forget. =)
In the beginning of this unit, I wasn't so comfortable converting degrees to radians. Up and till this unit I've always been using degrees but later in this unit, it became easy. This unit wasn't as bad as I thought it would be because from last year's precal I had trouble with circular functions. Now, I understand how to figure out certain word problems and convert degrees into radians and vice versa. I also found that finding the sine, cosine, and tangent of angles are pretty easy and it's probably because of the mini quizzes we had. But finding cosecant, secant, and cotangent can be a bit confusing at times and may take up a bit of time for me to figure out. What helps me a lot is to look for patterns to solve different kinds of questions which makes it much easier to do.
On the other hand, considering the fact that I missed a couple classes last week, I was a bit confused with the graphing we had to do. At first I was like, "what is this?" But after a few explanations and practicing doing questions I got the hang of it. Remembering DABC actually helped me a lot to do the problems. Yet, I'm not totally comfortable with graphing because it probably takes me more time than others in the class.
In conclusion, this unit was "OK" and I hope I do well in the first test as well as everyone else. With that said, goodluck to everyone!
PS..were we taught the graph for tangent when I wasn't here? Because in excercise 6, I believe there is a question about it and I didn't know how to do it?! Help?
### bob for Circular functions
Since our test is approaching for this unit I thought it was a good time to bob!
I was able to understand most things in this unit but my "muddiest point" would be the graphing part. I need still need some more practice with graphing those kinds of questions where all the DABC changes are there. I'm really comfortable with the A,B,C,D transformations but just not all together.
The part of this unit I really liked was probably those kinds of questions where you just have to find the exact values and simplify. At first that was kind of hard since we've all been accustomed to using degrees. I remember converting all the radians to degrees for each question, which was really time consuming. As I learned the exact values on the unit circle, I've found those kinds of questions way easier and didn't need to put it to degrees.
There's my bob for this unit. Good luck to everyone on the test this week?!
### BOB.1 [Circular Functions]
Well now that I'm back from my Kenora basketball trip, and once again have computer access I thought it would be an oppourtune moment to do my BOB post for this unit. Alas, I find myself here doing exactly that. BOBbing!
Well I'll start with the bad, and work my way towards the good, seeing as how it also goes like that in terms of chronological order from the beginning of the unit to the end :)
The hardest part of this unit was definitely the very beginning. Now this had alot to do with the fact that I missed the first two days of class, having been sick with the flu (>_<). So by the time I had returned to the class on Wednesday (that being the third day of class) I had basically no idea of what was going on. However, with some help from my classmates, and a couple nights spent awake much later then I should have been, I managed to figure my way around the first couple days of work, and catch up to everyone else.
On the flip side, the easiest part of the unit, has probably been the memorization, of the patterns in the unit circle (which eventually lead to the memorization of the unit circles basic values for sine, cosine, tangent, etc.) I found this rather easy, because all I had to do, was draw the unit circle over the weekend, without referring to my notes. How simple is that? So after drawing the unit circle, probably 7 or 8 times over the weekend, taking about 3-5 mins each time (with that time frame getting smaller as I went along,) I managed to get a fairly good grasp of basic angles locations on the unit circle, and the trigonometric function values of those said angles.
Overall this unit wasn't very difficult to understand or grasp (at least for me.) I'd say the hardest parts of it all were related to the memorization of certain things (like what A,B,C,D mean in graphing f(x) = AsinB(x-C) + D). I also found it somewhat difficult at times to remember what to do when adding radicals or similar things that we learned in previous grades, and I haven't had to use in awhile. Besides that I found much of what we did within my abilities to understand and put to use in the exercises we've been assigned.
Well I think that about sums it up for this BOB. I hope everyone enjoyed their long weekend and I also hope everyone enjoys the rest of today :) I shall see everyone on Monday!
Ciao!
~Justus
## Friday, February 22, 2008
### BOB Version 1: Circular Functions
Nope, not done blogging for today. Just doing my BOB…
And so, Circular Functions, the unit, put a lot of emphasis on the unit circle. Memorizing (no, not memorize, but remember!) the values using mnemonics was the muddiest thing that I didn’t like about this unit. But remembering the unit circle did help me out as we progress through the course. Also just as equally as muddy was the group work assignment, The Unit Circle Triangle. I didn’t like that assignment ‘cause of its “challengingness.”
I remember in gr.11 we were studying how to graph trigonometric functions and Mr.K exposed us to the equation f(x) = AsinB(x-C) + D. He taught us what A, B, and D did to the sine graph but told us that we’ll learn what B does in gr.12 precal. Now, after over a year of waiting, I found out what B meant and that became my moment of clarity. B stretches/shrinks the graph horizontally.
I also liked how Mr.K did his 1st class on the 1st day of school because I remembered it as inspiring.
I also liked the helpful advice Mr.K said to one of his students: Remember folks, Mr.K says, “If you don’t memorized the unit circle, you will fail.”
By the way, does anyone know how to do Exercise 2 #15?
If f (x) = 2x + 3, find k so that f(k+2) = k + f(k).
2(k+2) + 3 = k + 2k + 3
2k + 7 = 3k + 3
4 = k
...which doesn't make sense to me.
Also, one of our questions in our homework asked us to graph the tangent function. We weren't taught that.
### Wrapping Up and Concluding Circular Functions
3 assignments, 2 periods, 1 class. This is PC40SW08 (Pre-Calculus 40S Winter 2008).
As of now, Mr.K is still out-of-province because he was selected by The Council at the Conference last week to help write a textbook, but will return from his quest hopefully by Monday. And so, we had the same substitute as previous class'.
Period 1:
Our first assignment was Quiz 2 on Circular Functions in which we were not allowed to use our calculator. It was not a hand-in because the substitute says "according to the instructions Mr.K gave me, it did not say to hand anything in." So we assumed that we are not to hand that in. The class worked quietly during that quiz.
After finishing Quiz 2 on Circular Functions, we were assigned Graphing Trigonometric Functions. We were also told not to use a calculator when doing that assignment, but were allowed to use the calculator to check our answers. That is for homework, folks.
Period 5:
In our afternoon session, we were told to continue working on this morning's pre-calculus assignments and that Exercise 7: Translations #11-20 is also for homework.
HOMEWORK:
* Graphing Trigonometric Functions
* Exercise 7: Translations #11-20
* Circular Functions Test on Tuesday 26 or Wednesday 27
Next scribe is AnhThi.
That is all.
## Wednesday, February 20, 2008
Hey everybody, Paul here, doing my scribe post (since I got "voluntold" to). I hope 8pm isnt too late for a scribe post, but I guess you guys don't really need it all that soon anyway as we didn't do a lesson.
So yeah, today in Pre-Calculus 40S we had a substitute (Mr. Rekrus?), and he gave us an assignment per Kuropatwa's orders. The assignment was on circular functions and it went something like this:
You have a unit circle, and on this circle you have a point P, which is along the circumfrence of the circle and in Quadrant I. You also have point A, which is at the center of the circle (thus making it the origin), and a point B, which is positive along the X axis. If triangle APB is a right angle triangle, find the value of P that creates the largest area of the triangle.
Since this was a group assignment, the class was split into three groups. There was also supposed to be a time limit of 20 minutes, but the question utterly stumped the groups for most of the class, so the time limit was disregarded. Some groups finished the question before the end of class, some spent the entire class trying to solve it and some groups made little paper cranes.
And that was what we did today in Pre-Calculus.
Today's homework is Exercise 6, Questions 1-20. I suggest you atleast try them since we do have a test coming up.
As for the next scribe, I have chosen Zeph since he asked so nicely.
Good night and farewell.
## Tuesday, February 19, 2008
### Graphing Sine and Cosine Functions
Hello, Francis here with your daily school scribe. Now I'll introduce you to the beginning of class. Let's bring it back.
Mr. Kuropatwa firstly showed us a new feature available on our blog. It's a translation feature that can translate this blog into one of many different languages. Brought to you by Google.
Mr. Kuropatwa also stated that the circular functions test will be sometime in the middle of next week. Which means that it would be a smart idea to start on your BOB posts.
Starting tomarrow there will be a substitute teacher and same goes for Thursday, because Mr. Kuropatwa won't be showing up for class on those days.
All of these functions that we graphed today are based on patterns and as said by Mr. Kuropatwa "Mathematics is the science of patterns". Okay now with today's lesson which was completely made up of graphing those sine and cosine functions, that are ever so interesting. As seen on slide 1 we started off class by graphing the sine function: y = sinx - 1 and on slide 2 the cosine function: y = 2cosx, then we stepped up to a slightly harder function found on slide 3 and I suggest you see how all that turned out, so check out that slide show. When graphing any of these functions you should remember a simple pattern: "1,2,3,4" where 1 is the maximum point, while 2 is the average point(between the max. and min. points) and 3 is the minimum point and 4 is the average once again. Remember when graphing one of these functions you should put arrows at the end, pointing in the direction the function is about to head to next, so if it's at the max point, and it has nowhere to go but down, you should end the line with an arrow pointing downward. Also remember not to put sharp points when you change direction from up to down or down to up, put curves, because if you put sharp peaks, then be prepared to lose your precious marks.
The general equation for these functions is: f(x) = Acos(x-C) + D or
f(x) = Asin(x-C) + D.
When graphing these functions from these equations remember the word: DABC. This work or phrase is like the BEDMAS of graphing. It's the order in which you should always graph.
Now I will expose what these letters actually mean.
Firstly with D: is the vertical movement of the curve itself, it determines the sinusoidal axis(the axis between the max. points and min. points, its the average value of the curve.
Second is A: This is the amplitude it determines the max. and min. points of the curve on either side of the sinusoidal axis. The graph is stretched if the value of A is: |A| > 1 and its compressed if the value of A is: 0 < |A| < 1.
Third is B: this determines the period of the graph with an equation found on slide 18. If the value of B is negative then the graph curve will start with the pattern being 1,2,3,4. Where 1 is the min. point and 2 is the average point, and 3 is the max. point and lastly 4 being the average point.
Last is C: This shifts the curve left of right (horizontally). This is also called the "Phase Shift". If the value of C is negative(C <> 0) the curve shifts to the right.
For the sake of summarizing I will say this:
D - Lay down the sinusoidal axis.
A - Stretch of the amplitude.
B - Label the x-axis.
C - Look at the scale and slide the curve accordingly.
All of the details are found on slides 17-19. I suggest you check it out.
The sine and cosine curves are quite similar with the difference being that when at zero, cosine = 1 and sine = 0. But if the sine curve is slid to the right by pi/2, then at 0 sine = 1. With this, you can state that cosx = sin(x +(pi/2)) or sinx = cos(x-(pi/2)).
Well thats the 411 of today's class, hope it was helpful. The next helpful person to be voluntold(forced to do it for free, hence "volunteer that is told") is Paul.
That's me, the one and only, Francis! Until next time, I'm out.
### Today's Slides: February 19
Here they are ...
### BOB for Circular Functions
Hello everybody,
I believe the test on Circular Functions is this week so I decided to BOB ahead of time. This is our first unit of the course and I am doing great so far. The thing that really helped me in this unit is that activity that Mr.K got us to do and it was the one with drawing the unit circle and getting us to memorize those certain sine/cosine/tangent/cosecant/secant/cotangent of the special radian values. One thing I would like to do before the test is a little more practice. It should just give me a firm hold on the unit before the test.
Well that was my BOB and until our next text or my next scribe (which ever comes first) see ya and good luck on your tests.
## Saturday, February 16, 2008
### Circular Functions (Feb. 15, 2008)
Hello everyone it's Roxanne here and I am your scribe for today or shall i say tonight. I know it's pretty late but I just got home and I decided to do this before I forget.
We first started off with reviewing two questions that we should know how to do already. The questions should be on slide 2 if anyone needs to go back and check. For the second question, where it says cos(pi/2-pi/6), remember that cosine is a function and not a value to distribute. Such as the question 2(x+7), where it will equal 2x+14.
On to slide 3, we looked at the sine graph. It shows that if you look at the black wave the origin is zero aka sine θ=0, and as it moves to sine π/2 = 1 the curve goes up by one. Then it goes back on the x axis as it changes to sine π. As it changes to sine 3π/2 = -1, the curve is now at the bottom at -1 on the x axis. Finally, as it changes to 2π, it curve moves back on the x axis where it equals zero. It then repeats itself over and over again. REMEMBER: 1, 2, 3, 4! Therefore, if you change the input of the sine function such as sine x+1, the line would be higher than the orginial which is the black curve. If you change it to sine x-2, it would be below the black curve.
After we looked at slide 4 which shows us the the curve of cosine. But before we started talking about the graph, one of the students suggested that you could look at the the curve as two half circles. One would be on the top and one would be at the bottom. However, this is not true because in fact the curve of cosine behaves differently compared to the sine curve.
In our graphing calculator we put Y1= sin(x) and Y2=√(1-(x-π/2)². It then showed us the the cosine graph with almost, not quite a half circle on the top curve. Therefore, it proves that the curves are NOT two half circles.
We also used our graphing calculator to graph the cosine curve. Remember when using the graphing calculator go to zoom 4 aka the Friendly window. The friendly window does not create distortion such as zoom 6 where it takes the shape of a square rather than a rectangle. Mr. K also said that there are 95 pixels on the x axis and 63 pixels on the y axis. He explained that if subtract 1 from either of them, you get 47 on the x axis and 31 on the y axis. On calculator, it should show for the Xmin=-4.7 and the Xmax=4.7. It should also show Ymin= -3.1 and Ymax= 3.1. These are also the values that we calculated after you subtract by one and divided by two. The reason why you subtract one is because both the 95th pixel and 63rd pixel is counted as the origin.
Well I suppose that is all! Well at least I hope so because I am quite tired and I want to go to bed! So.. see ya later! OH ! I almost forgot, next scribe will be .......... the one and only FRANCIS! (: alrighty, bye folks!
## Friday, February 15, 2008
### Today's Slides: February 15
Here they are ...
Working with the Unit Circle Exercises - Answers
1. (i) 7π/3
(ii) 1
(iii) I
(iv) (1/2, √(3)/2)
2. (a) (i) 8π/3 (ii) 1 (iii) II (iv) (-1/2, √(3)/2)
(b) (i) 11π/3 (ii) 1 (iii) IV (iv) (1/2, -√(3)/2)
(c) (i) -17π/3 (ii) 2 (iii) I (iv) (-1/2, √(3)/2)
3. (i) 510°
(ii) 1
(iii) II
(iv) (-√(3)/2, 1/2)
4. (a) (i) 1350° (ii) 3 (iii) on negative y-axis (iv) (0, -1)
(b) (i) -2250° (ii) 6 (iii) IV (iv) (√(2)/2, -√(3)/2)
(c) (i) 1590° (ii) 4 (iii) II (iv) (-√(3)/2, 1/2)
5. (a) 1 (b) 1 (c) 1 (d) 1
6. ???
7. (a) yes, because (3/5)2 + (4/5)2 = 1 (b) no (c) yes(d) yes
### The Scribe List
This is The Scribe List. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.
This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.
IMPORTANT: Make sure you label all your Scribe Posts properly or they will not be counted.
Cycle 3
FrancisJoyceElevenbenofschoolroxanne JamieNeRd123CzephAnhThiRichardHi I'm Justus nelsaRencekristinaPaul
3.23
## Thursday, February 14, 2008
Hello everyone I am Richard and I am your scribe for Valentines day
Today we learned that there are four different ways to write a function but they all mean exactly the same thing, but they are just 4 different points of view. All the examples have to do with Mr. K's pet otter and that he eats two fish a day
1) In words
Example My pet Otter eats two fish a day.
2) In A table , you would draw it out in a table
Example below
3) in a Graph, You would graph out the information
example below
4) In an Equation you would make an equation for your data.
Example : y = 2x
Slide two
we did some problems
jsut a reminder dont when solving for the cos or sin or tan of a number and you find it you dont have to put the sin, cos ,tan in front of it anymore.
example cos² ( 5π/6) → cos² (-√3/2 ) which is wrong
the correct way is cos² ( 5π/6) → (-√3/2 )
After we did ten mental math questions
on slide 6 we also did some questions we didn't have enough time to do all of them
and the questions that you did not do is for homework.
☺☺☺☺☺☺☺☺☺☺☺
In the afternoon Mr K was away because he had to go to a confrence to sspeak about leadership. the sub's name was Ms. Hallson. she gave us an assignment and that it was due tommorow.
well that was my scribe i hope it was usefull.... the nest scribe is going to be roxanne
### Today's Slides: February 14
Here they are ...
## Wednesday, February 13, 2008
### Circular Functions (February 13, 2008)
Well guys it's me, Justus here, and as the scribe for today I'm writing this blog post :)
To begin I'd like to say that we had a substitute today, a very nice lady by the name of, Ms. Cheekie. Anyways after introducing herself, she informed us that we had a quiz on circular functions, which we had the whole class to do, but must be handed in at the end of the class. If you missed the class I'd sugges talking to Mr. Kuropatwa and seeing if you'd be able to write the quiz, as I assume its for marks.
After that, we were handed out a worksheet on circular functions, entitled, "Working With the Unit Circle Exercises." Which was to be worked on once completed the quiz. If we didnt finish the worksheet during regular class time, Ms. Cheekie said that it would be due during TOMORROWS CLASS (as in February 14th, 2008, at 9:00am.)
That basically sums up what happened during todays class, and thus, sums up my scribe post. However, because this scribe was significantly shorter then most posts, and I'm not entirely sure if this material counted as a legitimate scribe post (regardless of the fact that we do need to know of the hand in sheet), I decided to just post anyways. Should it be unsatisfactory as a scribe post, I will help Richard (whom I've selected as the next scribe) out with tomorrows scribe (since it is a two period day.)
That is all, until next time.
## Tuesday, February 12, 2008
### SOLVING TRIGONOMETRIC EQUATIONS
Hello world. I am Jamie and I come in peace. I’m stranded in a lodge somewhere, looking out through a window, perhaps the one that you see on this blog page and I’m feeling rather…snog. Haha. The wonders of the English language— everyday there is a word of the day in the room. Yesterday was picayune and today’s was petard. I was just wondering if he was doing this in descending alphabetical order. Sometimes I just go into TANGENTS and look for random things around the room and there I see…a word…for the day…it just magically changes.
Anyhow, in all seriousness [kind of], today’s first SECTOR of math notes was all about SOLVING TRIGONOMETRIC EQUATIONS. Half of the class time was spent reviewing content that was from the grade 11 curriculum in regards to finding the smallest angle formed by the terminal arm [the arm which extends from the origin point of a Cartesian plane to a point on the grid] and made with the x-axis. This is known as the RELATED ANGLE.
The class did many problems similar to that, but before advancing, we habitual humans had to do our third mental math exercise on our unit circles. THINK IN RADIANS everyone!! To celebrate, just do the clock dance.
Immediately after the refresher course, the easy related angle concept slowly adapted into a more complex one. Again, everyone had to look back and dwell on their factoring skills. But in alternative of factoring variables and isolating x in binomials and trinomials, we instead isolate sine, cosine or tangent of x.
ex. Solve for x on the interval [0, 2π] in 2 sin2x = sin x. The first thing to do is transpose sin x to the other side so that one side equals zero. Then factor sin x out of the equation to simplify, making it easier to find the sine value of x. Finally, find the measure of the angles using these values and in RADIANS!!
2 sin2x = sin x à 2 sin2x - sin x = 0 à sin x (2 sin x – 1) à sin x = 0, ½
Therefore x = 0, π, 2π and π/6 and 5π/6.
**NOTE if you cannot see how to factor this, you can also substitute sin x with another value like “a” [or Helena Bonham Carter, Queen Elizabeth, Johnny Depp, Jamie, etc…whatever floats your luxury yacht.] just as long as it is not the same variable like x because x does not EQUAL sin x. sin x is a function of x, therefore a manipulated x cannot equal x. Just remember to substitute back what you substituted in the beginning to get full marks and clarify what values are.
~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
In the afternoon, we continued going over the same material in order to receive a thorough understanding on the idea. As usual though, we took it a step further.
Do you ever open up a calculus book and wonder what those symbols mean? I do, but I’m just the big nerd. Hence, Jamie123C was born. That’s what you do when you don’t have wonders like…I don’t know…the internet around you. Flabberjacks, I still don’t have internet. How am I blogging then, you say? It’s all magic. I’m not lying. I went to Hogwarts and everything. Everything was swell there, I had an owl, people were chipper waving around their magic wands. Mine was the root of a phoenix feather. Not kidding. Here I am again— my mind going astray off to Fleet Street or some place like that. Oh, Sweeney Todd.
Anyhow, I think I’m finished, I’m no longer derailed on my train of thought. Where was I? Ah, yes. Symbols— they just look like the DaVinci code or something. As each of us grow mathematically, we come across the meaning of some of them. Starting off with “x Ɛ R”. This is familiar in terms of finding the domain. So instead of finding the values of sine of x in between [0, 2π], which happens to be the distance of revolution around the circle once, the domain will be x Ɛ R which circulates the circle multiple times, infinite times because it has no restriction in domain. But having limited time which should be spent like gold, we can’t count all of the values which apply. So we have to find a pattern and make an equation just as shown in the following example:
cos2 x – 2 cos x = 0 à cos x (cos x – 2) = 0 à cos x = 0 and 2.
***cos x = 2 is EXTRANEOUS because the cosine of 2 is undefined since cosine value cannot be more than one. The same thing applies for sine. Tangent on the other hand can be more than one. [sinθ <> 1] Further explanation of this is on the toy, patent pending— erm, math tool: Dave’s Unit Circle Applet, link is found in slides for today.
Therefore x = π/6 and 5π/6
But since there is no restriction on the domain, possible values could be 13π/6, etc…
x = π/6 + 2kπ; k Ɛ I OR in this situation, x = 5π/6 + 2kπ; k Ɛ I.
where k represents a variable that is always a whole number and an integer to find out how many times you can go around the unit circle and find the same value and k is an element of an integer…
This scribe post is a little lengthy and droning, especially with the jokes and lessons merged together. You know, I try. I think I’ve summed up today. My bloggerish mood has begun to fade, so who shall take on these powers tomorrow? The answer to that is in this question…[sort of an inside joke to the math room] Let’s just pretend I’m situated in quadrant two and there are three other people around me. In which quadrant is both tanθ > 0 and sinθ < 0? ANSWER? QUADRANT III!! So that means Justus is next. Sorry man, it’s just that you’re adjacent to me and it worked, math-wise… I try to make every experience educational. Hey, who knows, it could’ve been Queen Elizabeth sitting behind me and scribing tomorrow.
oh yeah. pps I adore you [sequel to the ps I love you, going to be a greater movie] I BElieve homework tonight is Exercise 5
### Today's Slides: February 12
Here they are ...
## Monday, February 11, 2008
### Trigonometric Equations
Alrightey I finally got home. Sorry for the late post had some technical difficulties with the computer also. Anyways here's the recap of what we did on todays slides.
Slide 2:
Here we see the reciprocal trig functions we learned about yesterday. Just for review, the new trig functions we learned were cosecant, secant, and cotangent. Mr. K emphasized that most kids go wrong when they think that cosecant is with cos and secant is with sin. So make sure to not do that! Anyways as it says on the slide, we get cosecant by 1/sin(theta), secant by 1/cos(theta), and cotangent by 1/tan(theta). Or we can just change the numerator with the denominator to save time. As I learned today, don't switch any signs. If it's negative in the denominator it will still be negative when it becomes your numerator when you flip the fraction.
Slide 3:
Ok the question here was a bit of a review from the questions we've been doing for the past few days. First we must find the hypotenuse of the big triangle. We do that by plotting the point (9,-40), which are your x and y values. Then we must find the cos and sin values. We do this by finding the angles adjacent to theta over the hypotenuse (for the cos), and for sin we put the angle opposite of theta over the hypotenuse. The cos and sin value you get will be the coordinates for theta! As for the second question on the slide, just list the cos, sin, tan, secant, cosecant, and cotangent values. Remember for the reciprocal trig functions, just flip the fraction.
Slide 5:
We've been practicing how to find the exact values with questions like these on this slide. Doing these questions are a lot easier if you know the values around the unit circle. K i'll make this brief, all you have to do is substitue the exact values in and simplify!
Slide 6:
Oh yes I remember this slide from this morning. Mr. K challenged our mathematical minds to do these questions. But like he said there's a method to his madness or something like that. I think we all know how to do these kinds of questions. We just isolate x or factor.
Slide 7:
We now see what Mr. K was trying to do by starting us off easy. We do exactly what we did on the last slide. We simply isolate sinx and get the value of 1/2. Now what on the unit circle do we know has a y value of 1/2? pi/6, and 5pi/6 of course.
Slide 8:
1+2cosx = 5cosx
The question gets a bit tougher but we still do the same things. Isolate cosx and it will equal 1/3. Now we take our trusty calculators and press 2nd fnc cos then 1/3. Make sure to be on radians. We then get the value 1.2309. But there is another value because cos is also positive in quadrant 4. How do we get the value in quadrant 4? Just put in your calculator 2pi - 1.2309 to get the other value of x, which is 5.0522 the related angle. K a quick summary, if looking for related angle in q2 you take the reference angle and subtract it from pi. If looking for related angle in q3 you take the reference angle and add it to pi. And that's where we ended!
Hope you guys understood my rambling. Oh yeah next scriiibe is...........................JAMIE since I took your turn today. KK goodniight everyone *faints*.
### Today's Slides: February 11
Here they are ...
## Sunday, February 10, 2008
### The Great Almighty Unit Circle. Plus other stuff .. That can be almighty if you want it to. ^^
Hey everybody! Sorry for the two days and a half late post. Anyways, what I'm going to be discussing in my blog is what we've done that day. First, we started it off by finding that there is a super Pi on the Chinese coin. After that we discussed the points on the Unit Circle. What was Sinθ, Cosθ, and Tanθ were on each point of the Unit Circle. Then we discovered patterns and methods of remembering the whole unit circle.
The unit circle is called the unit circle because the radius is 1. The angles we used in the unit circle was 30°, 45° and 60°. We converted degrees into radians. 30° = Pi/6, 45° = Pi/4 and 60° = Pi/3.
**
- In a right angle triangle, we know that Sinθ = OPP / HYP, Cosθ = ADJ / HYP, and Tanθ = OPP / ADJ.
SOH CAH TOA
- We then find that Tanθ = Sinθ / Cosθ
** In the unit circle, Pi/6, if you drop straight down from P(θ) to form a right angle triangle, you can find the coordinates of P(θ). As you can see in my professional diagram below.
The coordinates of P(θ) is (Cosθ, Sinθ) .. Because we know the unit circle's radius is 1, and the hypotenuse is 1, then we can figure out Sinθ, and Cosθ, using Pi/6 (30 Degrees) ..
- Sin Pi/6 = 1/2
- Cos Pi/6 = v3/2
** Keep in mind that it will always be over 2
So P(θ) = P(v3/2, 1/2) Knowing that these are the points that represent P(θ) then that must mean that Cosθ is represented by the X axis and Sinθ is represented by the Y axis. -Tanθ = y/x
**
That is how we found the points for the unit circle.
- A short lesson we learned was complementary angles.
- It is the sum of 2 angles that add up to 90°
Ex. 30° + 60° = 90°
Sin30° = 1/2
Cos60° = 1/2
** The unit circle can be found on the last lesson's slide (The one with Happy New Year!!). Slide #6/8 .. It is a very good unit circle.
-The method of remembering each point is you have to remember 3 numbers.
1, 2, and 3.
Here is a link to my amazing, incredibly, outstanding paint skills for a clear example of one way of remembering the unit circle.
The explaining will be here.
-The black numbers represent Sin
θ
-The red numbers represent Cos
θ
-The green numbers represent Tan
θ
What do you notice?
-RIGHT! 1, 2 and 3. (You're probably thinking in your head .. "uh.. no")
**Anyways
The numbers you see will always be square rooted over 2.
Yes, Square root of 3, over 2
-Square root of 2, over 2
-Square root of 1, over 2.
**Now how do I know when the number is 1, 2 or 3?
The purple lines in the circle represent the Y axis, also Sin
θ.
Ditto for the blue lines, it represents the X axis, Cos
θ.
**What do you notice?
-RIGHT! That Pi/4 or 45° will always be square root of 2 over 2. (Not including Tangent)
As you see the longer the lines are extended, it is square root of 3 over 2.
The shorter it is, it is 1 half.
Between that is square root of 2 over 2.
Yea the clock dance becomes useful. Unfortunately.
**Now that we have figured out Sin
θ and Cosθ .. How about Tanθ?
Well .. You should have already picked up that the middle numbers are always 1.
**The closer tangent is to the X axis, it is under 1. Yes 1 over square root of 3.
The farther it is, (over the X axis) it is square root of 3 over 1. Or just square root of 3.
**I hope this helped. I was in a rush.
**The next scribe will be the person who sits the closest to me.
Yup, you got it right again! Joyce will be the next scribe =)
This is Agent Eleven, out.
Better days.
## Friday, February 8, 2008
### Today's Slides: February 8
Here they are ...
## Thursday, February 7, 2008
### Exact Values In The Unit Circle
Hello everyone, I'm Nelsa, today's scribe. I'm so sorry for the late post, Thursday is a really busy day for me.
Mr. Kuropatwa began the class by asking everyone if we have read the 'Digital Ethics' post and taking everyone's oath (by shaking our hands) that we'll abide by those rules. He also explained 'Blogging On Blogging' - which you can learn more about by reading that post - and how to properly label your posts. But of course, this is a math class, and that's probably what you all are waiting for. Most of the things we studied today, reviewed a lot of what we learned yesterday, so I'm sorry if it seems like I'm repeating the things Ben has detailed on the last scribe post.
The first slide required us to find which quadrant P(5) can be found in. Most of us were more than slightly confused because of the lack of a y-coordinate, but Mr. K simply told us to think in 'radians'. With this in mind, we reviewed what we already knew, which was, 180°=π and 360°=2π. π is approximately 3, which means that 2π would be approximately 6. In order to get 5, we subtract one from six, and whenever we subtract, we move in a clockwise direction, which would mean that P(5) is in quadrant IV.
We were also asked to find which quadrant contained sinθ <> 0. To do this, Mr. K told us to forget all about 'CAST', which I'm sure we all learned last year, and instead, understand what's going on. Cosine, as we know from 'SOHCAHTOA', is the x-axis or coordinate, and sine, is the y-axis or coordinate. Keeping this in mind, we know that in quadrant I, both cosine and sine is positive, as both the x-axis and the y-axis is positive. This also means that tangent is positive, because when you divide two positives (which is really what tangent is, sine/cosine), it's a positive. In the second quadrant, cosine is negative, because it's before the zero, but sine is still positive, which means that tangent is negative. In the third quadrant, both cosine and sine is negative, as both are before the zero, which means tangent is positive. Finally, in the fourth quadrant, cosine is positive, and sine is negative, making tangent negative as well. Using this information, we were able to answer the first question on the second slide.
The equation of a unit circle is x(squared) + y(squared) = 1. The third slide asked us to determine whether the point (1/root of 5, 2/root of five) is on the circumference of the unit circle. So using the above formula, we figured out that the point (1/root of 5, 2/root of five) is on the circumference of the unit circle. But a simpler way to find the answer is to understand, again, that cosine is the x-coordinate, and sine is the y-coordinate. The radius of a unit circle is 1 (hence 'unit'), so the cosine squared, plus the sine squared should equal to 1, which is the hypotenuse.
Mr. K then went on to talk about the person, who died because of one of the triangles included in your basic geometry set. I forgot the person's name, but basically, a group of people who call themselves the Pythagoreans believed that the world was made up of (only) rational numbers. One of the triangles in a geometry set, besides having a 90° angle, also have 45° angles. The hypotenuse of this triangle, is the root of 2, which is an in irrational number. The Pythagoreans were aghast, and kept this fact hidden from the rest of the world, as it would ruin everything that they believed in. But one person from the inner circle (who were the only people who knew) told everyone else, and, to make a long story short, was thrown off a cliff.
So that's basically what we learned in class today. Again, I'm so, so sorry for the late post. The next scribe is twenty-seven hundred, ninety-eight, twenty, ELEVEN, nine. =)
### Today's Slides: February 7
Here they are ...
### Blogging On Blogging (BOB)
We were talking about exactly what sort of post you're supposed to make to get that one blogging mark on your test. The kind of post I'd like you to make should have one or more of these characteristics:
• Write about what you understand the least in the unit so far; your personal "Muddiest Point."
• A reflection on a particular class.
• A reflective comment on your progress in the course.
• A comment on something that you've learned that you thought was "cool".
• A comment about something that you found very hard to understand but now you get it! Describe what sparked that "moment of clarity" and what it felt like.
• Have you come across something we discussed in class out there in the "real world" or another class? Describe the connection you made.
Your posts do not have to be long. I'm far more interested in the quality of what you write rather than the quantity.
Make certain you always use 3 labels on your post: [your name], [unit tag], BOB
When you share where you are in your learning a few days before the unit test I can address those issues in class so, hopefully, you will get much more than one extra mark on the test. ;-)
Happy Blogging!
## Wednesday, February 6, 2008
### Circular Funtions and Circles On a Cartesian Plain
Okay I hello everybody, this is benofschool with today's scribe on what we did. We started with talking about Digital Ethics or things to be careful of on the internet. Watch the video on the blog to find out more.
Mr.K also told us what new things to look for on the blog in the near future. Such features include a walkie-talkie sort of thing found on the blog which allows people with a microphone on the computer to send voice messages to other people who are online on the blog. Another cool feature is a little number in the corner of the blog which shows the number of people online which helps with using the walkie-talkie feature. The last but not least feature is a chat/shout box which allows us to chat. But with these features also come trust. If we use these features we must follow the Digital Ethics Rules and keep things appropriate. Remember don't post anything that you don't want anybody to know.
Okay onto the math. We did a small review on what we did yesterday. One reminder about units. There are no units for radian answers. All of the units used in finding the radian answer have been reduced to 1 which doesn't change any answer so it is as if they are not there. One more thing, the equation to convert degrees to radians and vice versa is not a formula but a proportion. It shows how degree values are proportional to radian values. So don't memorize that exact equation but remember how degree values are proportional to radian values.
The first question we worked on is on the second slide on the Feb.6 slide post. One solution is showed on the slide but there was one trick that could be used. We know that a full circle is 2π and we can find out that how the hands are found on a clock at 4:00 that it is one-third of the clock or the full circle. So just divide 2π by 3 which gives us our answer with little work.
The second question we worked on is found on the third slide. (Ignore that little message at the bottom at the slide it is irrelevant). That is the correct solution to the problem and I will now explain the process. To answer the problem you must find the area of the circle using the area of a circle equation. Then make a proportional equation like the one on the slide. It shows that 30° out of 360° is equal to the unknown area out of the total area of the circle. Now all that you have to do is solve for the unknown or on the slide, A. Remember to include units where necessary.
That ended our first class for today. Notice that I said first. There was another one.
In our second class we continued with a new problem and the first question required us to know what the word "coterminal" means. Coterminal means end together. So in the question we need to find out what other radian ends where -π/3 (I don't know how to write fractions on the blog) ends or the positive coterminal of -π/3. So we know that a full circle in radians is 2π. we need to subtract π/3 to 2π to find where it ends. If you are wondering what a negative angle is, a negative angle is the same as going backwards from 360° or 2π. One example would be -50°. That is 360°-50°= 310°. Same goes with radians in this problem. We need to subtract π/3 from 2π. By using basic fraction subtracting skills we can find the answer. To answer the problem we need to multiply 2π by 1 or a number that equals to 1 to change 2π into a number that can help us. We need to multiply 2π by 3/3 and it becomes 6π/3 and now we can subtract π/3.
If you are wondering why we multiplied 2π by 1, we did that to turn the number's appearance in a way that would help us without changing the value of the number. 1 is the identity element of multiplication which means that if we multiply any number by 1 or a number that is equivalent to 1, the value or "identity" of the number doesn't change.
The final problem the class worked with is found on the fifth (5th) slide. P(θ) means the point where a ray intersects the circumference of a circle or in this case the unit circle. The unit circle is a circle that has a radius of one on a Cartesian Plain. To find that angle we can use the SOHCAHTOA rule but we need to find the x and y values of the triangle made shown on the slide. The height or opposite side of the triangle is found by looking at the y-coordinate. The x or adjacent length of the triangle can be found by looking at the x-coordinate the given point. Now that we know two sides we can use the Pythagorean Theorem to find the hypotenuse. But wait, if the sides of the triangle that are not the hypotenuse is 8 units long and 6 units long the hypotenuse must be 10 units long because of the Pythagorean triples. Pythagorean triples are side lengths of triangles are whole numbers like 3, 4, 5 and 65, 72, 97. We can now find the unknown angle made by the ray and the x-axis. Now it doesn't matter where we connect the ray to the x-axis creating a right angle the unknown angle is always that angle found. Even if one of the coordinates were to change integers as in 8 into -8. The angle created by the new created ray with the x-axis will always be that angle. The sine, cosine, and tangent of that angle will be the same as the angle's related angle but in some cases the integer might change depending on which quadrant the angle is found.
Related angles are angles that are the same distance in angles from the x-axis. One example would be 30°. Its related angles would be 150°, 210°, 330° because these values are all 30° or 1/12 away from the x-axis π or 2π.
We can find determine the integer of the sine, cosine, and tangent of an angle by the quadrant in which the angle is located. Sine is a measurement of the y-axis which means in the y-value is the 1st and 2nd quadrant it is positive because the y-value is positive and in the 3rd and 4th quadrant the value of sine is negative as with the negative y-value. For cosine it depends on the integer of the x-value. Tangent is sine/cosine which means that if the sine value is positive and the cosine is negative both according to what was said earlier the tangent would be negative because + / - = -. So cast CAST away and remember the sine=y, cosine=x, and tangent=y/x.
That was all we learned today in class. Homework is to watch the Digital Ethics video, read the Digital Ethics post, Exercise #2, and send Mr.K an email to be invited if you haven't yet.
The next scribe will be nelsa!!!
Bye Bye for now and see you in tomorrow's class. Remember "Beans have souls."
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### 07-Waves and Sound
```Physics
Unit 7
This Slideshow was developed to accompany the textbook
OpenStax Physics
By OpenStax College and Rice University
2013 edition
Some examples and diagrams are taken from the textbook.
Slides created by
[email protected]
16.9 Waves
16.2 Period and Frequency in Oscillations
Waves
A traveling disturbance
Carries energy from place to place
When a boat makes a wave,
the water itself does not get up and move
the water pushes a little, then moves back
energy is transferred in the wave and is what you feel
16.9 Waves
16.2 Period and Frequency in Oscillations
Transverse
Up and down disturbance
Wave travels left or right
Disturbance is perpendicular to direction of travel
Examples:
Radio waves, light waves, microwaves, stringed
instruments
16.9 Waves
16.2 Period and Frequency in Oscillations
Longitudinal Waves
Disturbance is left and right
Direction of travel is left or right
Disturbance and direction of travel are parallel
Series of compressed and stretched regions
Example:
Sound
16.9 Waves
16.2 Period and Frequency in Oscillations
Other
Water waves are a combination
Water at the surface of a water wave travels in small
circles
16.9 Waves
16.2 Period and Frequency in Oscillations
Periodic pattern is
regularly repeated
Cycle one unit of pattern
Wavelength () Distance of
one cycle
Amplitude (A) height from
equilibrium to crest
16.9 Waves
16.2 Period and Frequency in Oscillations
Period (T) time it takes for one
cycle
Unit: s
Frequency (f) # of cycles per
second
Unit: 1/s = 1 hertz (Hz)
f=1/T
v=/T=f
16.9 Waves
16.2 Period and Frequency in Oscillations
WAUS operates at a frequency of 90.7 MHz. These
waves travel at 2.99x108 m/s. What is the wavelength
and period of these radio waves?
= 3.30 m
T = 1.10 x 10-8 s
16.9 Waves
16.2 Period and Frequency in Oscillations
You are sitting on the beach and notice that a seagull
floating on the water moves up and down 15 times in 1
minute. What is the frequency of the water waves?
f = 0.25 Hz
Day 68 Homework
Wave hello to some exercises.
16P7-10, 47-50, 53-55
16CQ2, 5, 6
7) 16.7 ms
8) 0.400 s/beat
9) 400 Hz
10) 12500 Hz
47) 9.26 days
48) 11.3 m
49) 40.0 Hz
50) 7.50 times
53) 700 m
54) 2.50 × 109 Hz
55) 34.0 cm
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
A mass is hung from a
spring
If it just hangs, it is at
equilibrium position
If stretched and released,
it bounced up and down
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
Hooke’s Law
= −
F = restoring force
x = distance displaced
k = spring constant
Force will pull the mass back
toward equilibrium
As mass gets to equilibrium, it has
momentum, so it continues past
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
Energy in Hooke’s Law
Since a force acts over a distance, work is done
1 2
=
2
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
A Nerf dart gun uses a spring to launch a dart. If it takes
24 N of force to compress the spring 6 cm, what is the
spring constant? How much potential energy does it
contain?
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
Simple harmonic motion
Frequency independent
of amplitude
= 2
1
=
2
If a graph of position versus
time of simple harmonic
motion is made, a wave is
formed
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
Think of a point on a string some distance (x) from the origin
We want to know the vertical displacement (y) of the particle at
any given time
If the wave repeats, then it will look like a sine (or cosine) graph
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
2
= cos
2
= −max sin
2
=
=
2
=−
cos
Notice a is proportional to and
opposite direction of x
16.1 Hooke’s Law
16.3 Simple Harmonic Motion
A wave has an amplitude of 1.5 cm, a speed of 20 m/s,
and a frequency of 100 Hz. Write the equation of the
wave position of the wave.
= 0.015 cos(200)
Day 69 Homework
These problems harmonize with
the lesson
16P1-5, 13-15, 18-19
17CQ1-2
1) 1230 N/m, 6.88 kg, 4.00 mm
2) 1.57 × 105 N/m, 77 kg
3) 889 N/m, 133 N
4) 7.54 cm, 3.25 × 105 N/m
5) 6.53 × 103 N/m, Yes
13) 2.37 N/m
14) 2m
15) 0.389 kg
18) 94.7 kg
19) 1.37 Hz
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
Some vibrating object like a speaker moves and compresses
the air
Air pressure rises called Condensation
Condensation moves away at speed of sound
Object moves back creating less air pressure called
Rarefaction
Rarefaction moves away at speed of sound
Particles move back and forth
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
Distance between consecutive condensations or
rarefactions is wavelength
String or speaker makes air molecule vibrate
That molecule pushes the next one to vibrate and
so on
When it hits the ear, the vibrations are interpreted
as sound
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
1 cycle = 1 condensation + 1 rarefaction
Frequency = cycles / second
1000 Hz = 1000 cycles / second
Each frequency has own tone
Sounds with 1 frequency called Pure Tone
Healthy young people can hear frequencies of 20 to 20,000 Hz
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
Brain can interpret frequency as pitch
High freq = high pitch
Subjective because most people don’t have perfect
pitch
Some electronic devices can produce and detect exact
frequencies
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
A telephone uses pure
tones
Each column and row is
assigned a different tone
As a button is pushed,
two tones are produced
The computer at the
routing center “hears” the
two tones and routes the
call appropriately
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
The condensations have more pressure than the
rarefactions
Amplitude = highest pressure
Typical conversation, Amp = 0.03 Pa
Atmospheric air pressure = 101,000 Pa
Loudness is ear’s interpretation of pressure amplitude
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
For all waves
=
Sound travels slowest in gases, faster in liquids, and fastest
in solids
Air 343 m/s
Fresh Water 1482 m/s
Steel 5960 m/s
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
Speed of sound depends on properties of medium (like
speed of wave on string did)
In gases
Sound is transmitted only when molecules collide
So we derive formula from speed of molecules
And speed changes with temperature
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
For air
= 331
where T is in Kelvin
273
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
What wavelength corresponds to a frequency of concert
A which is 440 Hz if the air is 25 °C?
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
Sound is emitted from the hull of a ship.
It bounces off some object.
The echo returns to a receiver on the hull of the ship
How far away is a ship if it takes 3.4 s to receive a return
signal in seawater?
d = 2618 m
Day 70 Homework
These problems sound like
you could speed right through
them.
17P1-9
17CQ4
1) 0.288 m
2) 3400 Hz
3) 332 m/s
4) 5.96 × 103 m/s, steel
5) 343 m/s
6) 363 m/s
7) 0.223
8) 924 m
9) 7.70 m, Can only find big
stuff like ships
17.3 Sound Intensity and Sound Level
Sound waves carry energy that can do work
Amount of energy transported per second = power
Units: J/s = W
17.3 Sound Intensity and Sound Level
As sound moves away
out over a larger and
larger area
As the areas get bigger,
intensity at any 1 point is
less
Units:
W/m2
17.3 Sound Intensity and Sound Level
17.3 Sound Intensity and Sound Level
If sound is transmitted uniformly in all directions, the
areas are the surfaces of spheres.
ℎ = 4r 2
=
4r2
17.3 Sound Intensity and Sound Level
Intensity is proportional to amplitude2
Δ 2
=
2
where
Δ = pressure amplitude
= density of the medium
= speed of the wave
17.3 Sound Intensity and Sound Level
You and a friend are watching fireworks that are launching
from the observatory. You are standing right in front of
Berman Hall (150 m) and your friend is across campus at
AA (700 m). The sound intensity at AA is 0.2 W/m2. What
is the sound intensity at your location, and how much
power is the firework emitting?
P = 1.23 x 106 W
I = 4.36 W/m2
17.3 Sound Intensity and Sound Level
Unit of measure to compare two sound intensities.
Based on how human ear perceives loudness.
If you double the intensity, I, the sound isn’t twice
as loud.
Use a logarithmic scale
17.3 Sound Intensity and Sound Level
Intensity Level
= 10 log
0
Where
= intensity level
I and I0 are intensities of two sounds
-12 W/m2
I0 is usually 1.0 x 10
Unit: dB (decibel)
An intensity level of zero only means that I = I0 since log (1) = 0
17.3 Sound Intensity and Sound Level
Intensity can be measured
Loudness is simply how ear perceives
Doubling intensity does not double loudness
17.3 Sound Intensity and Sound Level
You double the intensity of sound coming from a stereo.
What is the change in loudness?
= 3 dB
Experiment shows that if the intensity level increases by
10 dB, the sound will seem twice as loud.
See Table 17.2
17.3 Sound Intensity and Sound Level
What is the intensity of a 20 dB sound?
Day 71 Homework
This is intense!
17P12-16, 19, 21-22, 26
17CQ5-7
12) 3.16 × 10−4 /2
13) 1.26 × 10−3 /2
14) 3.04 × 10−4 /2
15) 85 dB
16) 106 dB
19) 8.00 × 10−10 /2 ,
8.00 × 10−9 /2
21) 1.58 × 10−13 /2
22) 70.0 dB
26) 1.45 × 10−3
17.4 Doppler Effect
Have you ever listened to a ambulance drive by quickly with
their lights and sirens going?
What did it sound like?
High pitch as they were coming, low pitch as they were
leaving.
Called Doppler effect after Christian Doppler who first
labeled it.
17.4 Doppler Effect
Stationary Source
17.4 Doppler Effect
Moving Source
17.4 Doppler Effect
Deriving the formula
Moving toward object
’ = - vsT
Where
= wavelength of wave
’ = perceived wavelength
vs = velocity of source
T = Period of wave
17.4 Doppler Effect
fo = frequency observed
vw = speed of wave
’ = perceived wavelength
fo = frequency observed
fs = frequency of source
vw = speed of wave
vs = speed of source
17.4 Doppler Effect
Moving Observer
Encounters more
condensations than if
standing still
17.4 Doppler Effect
General Case
Combine the two formulas
Both observer and source can
be moving
WARNING!
vw, vs, and vo are signless
Use the top signs when that
object is moving towards the
other object
±
=
∓
17.4 Doppler Effect
You are driving down the
road at 20 m/s when you
approach a car going the
other direction at 15 m/s
loudly. If you hear a certain
note at 600 Hz, what is the
original frequency?
(Assume speed of sound is
343 m/s)
542 Hz
17.4 Doppler Effect
Wave bounce off water drops in storms
Computer checks to compare the frequencies
Can compute to see how fast the clouds are spinning
Day 72 Homework
Move yourselves to do
these exercises
17P30-35
16CQ16
30) 878 Hz, 735 Hz
31) 138000 Hz, 1770 Hz
32) 3790 Hz
33) 3.05 m/s
34) 12.9 m/s, 193 Hz
35) 1.030, Yes
16.10 Superposition and Interference
Often two or more wave pulses move through the same
space at once
When two or more waves are present simultaneously at
the same place, the resultant disturbance is the sum of
the disturbances from individual waves
16.10 Superposition and Interference
16.10 Superposition and Interference
After 2 seconds, what is the height of the resultant
pulse at x = 2, 4, and 6 cm?
0, -2, 2
16.10 Superposition and Interference
Imagine that there are 2 speakers facing each
other. Both speakers produce the same sound at
the same time. = 1 m
16.10 Superposition and Interference
One of the speakers is moved back half a wavelength
16.10 Superposition and Interference
16.10 Superposition and Interference
16.10 Superposition and Interference
Beats
When two frequencies are the same
Constructive and Destructive Interference give twice
the amplitude or no amplitude
What if the two frequencies are just slightly different?
16.10 Superposition and Interference
16.10 Superposition and Interference
16.10 Superposition and Interference
Beat Frequency = difference of the two source
frequencies
Beats = | f1 – f2 |
16.10 Superposition and Interference
A simple way to tune musical instruments is with beats
If the notes are out of tune, you hear beats
Adjust the tuning and try again
If the frequency of the beats is higher, adjust the other
way
Keep adjusting until there are no more beats
16.10 Superposition and Interference
Two car horns have an
average frequency of 420
Hz and a beat frequency
of 40 Hz. What are the
frequencies of both
horns?
440 Hz, 400 Hz
Day 73 Homework
Don’t beat around the bush,
start the problems now!
16P57-62
17CQ8-10
57) 4 Hz
58) 261.5 Hz or 258.5 Hz
59) 462 Hz, 4 Hz
60) 4099.75 Hz, 4100.25 Hz
61) 3.33 m/s, 1.25 Hz
62) 21 Hz, 22 Hz, 43 Hz
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
One end of a string is
attached to a fixed point.
The other end is
vibrated up and down.
The standing wave is
formed.
Nodes – No move
Antinodes – most
movement
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
The wave travels along the string until it hits the other
end
The wave reflects off the other end and travels in the
opposite direction, but upside down
The returning wave hits the vibrating end and reflects
again (this side the wave is right side up)
Unless the timing is just right the reflecting wave and
the new wave will not coincide
When they do coincide, the waves add due to
constructive interference
When they don’t coincide; destructive interference
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
Harmonics
When you vibrate the string faster, you can get
standing waves with more nodes and antinodes
Standing waves are named by number of antinodes
1 antinode 1st harmonic (fundamental freq)
2 antinodes 2nd harmonic (1st overtone)
3 antinodes 3rd harmonic (2nd overtone)
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
f1 = fundamental frequency (1st harmonic)
f2 = 2f1 (2nd harmonic)
f3 = 3f1 (3rd harmonic)
Harmonics Example
If the fundamental is 440 Hz (concert A)
nd harmonic = 2(440 Hz) = 880 Hz (High A)
2
rd harmonic = 3(440 Hz) = 1320 Hz
3
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
To find the fundamental frequencies and harmonics of a
string fixed at both ends
=
2
Where
th harmonic
fn = frequency of the n
n = integer (harmonic #)
vw = speed of wave
L = length of string
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
Just like stringed instruments rely on standing
transverse waves on strings
Wind instruments rely on standing longitudinal sound
waves in tubes
The waves reflect off the open ends of tubes
One difference at the ends are antinodes instead of
nodes
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
Tube open at both ends
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
Formula for Tube Open at Both Ends
Distance between antinodes = ½
Tube must be integer number of ½
=
=
1
2
or =
2
=
2
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
What is the lowest frequency playable by a flute that is
0.60 m long if that air is 20 °C.
f = 285.8 Hz
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
Tube open at one end
17.5 Sound Interference and Resonance:
Standing Waves in Air Columns
Tube Open at One End
Node at the closed end
Antinode at the open end
At fundamental frequency L = ¼
nd harmonic adds one more node or ½
The 2
Thus the lengths are odd integer multiples of ¼
=
4
Day 74 Homework
these problems
17P38-48
17CQ11-13, 15
38) 0.7 Hz
39) 44 Hz, 55 Hz, 132 Hz, 88 Hz,
33 Hz, 77 Hz
40) 0.2 Hz, 0.5 Hz, 0.3 Hz
41) 263.5 Hz or 264.5 Hz
42) 256 Hz, 512 Hz
43) 96 Hz, 160 Hz, 224 Hz
44) 180 Hz, 270 Hz, 360 Hz
45) 65.4 cm
46) 1.56 m
47) 0.974 m
48) 0.334 m, 259 Hz
17.6 Hearing
17.7 Ultrasound
Hearing
Pitch
Perception of
frequency
20 Hz – 20000 Hz
Most sensitive to 2000
– 5000 Hz
Can distinguish
between pitches that
vary by at least 0.3 %
Loudness
Perception of intensity
-12 W/m2 – 1012
Range 10
W/m2
Most people can discern a
intensity level difference of
3 dB
17.6 Hearing
17.7 Ultrasound
Ultrasound
Used in obstetrics to examine a
fetus, used to examine some organs,
and blood flow
High frequency sound aimed at
target
Sound reflects at boundary of
tissues with different acoustic
impedances
Computer compiles picture from
where echoes come from
Acoustic impedance
=
See table 17.5
Intensity reflection coefficient
2 − 1 2
=
1 + 2 2
Higher coefficient, more
reflection
17.6 Hearing
17.7 Ultrasound
Calculate the intensity reflection coefficient of
ultrasound when going from water to fat tissue (like a
baby in the womb).
a = 0.00317
This means 0.317% of the sound is reflected.
17.6 Hearing
17.7 Ultrasound
Cavitron Ultra Surgical Aspirator
Used to remove inoperable brain tumors
Tip of instrument vibrates at 23 kHz
Shatters tumor tissue that comes in contact
Better precision than a knife
17.6 Hearing
17.7 Ultrasound
High-Intensity Focused Ultrasound
Sound is focused on a region of the body.
The waves entering the body don’t do damage
Only damage done where focused (like sun and magnifying glass)
The focused energy at target causes heating which kills abnormal
cells
17.6 Hearing
17.7 Ultrasound
Doppler Flow Meter
Transmitter and receiver placed on skin
High frequency sound emitted
Sound reflects off of blood cells
Since cells are moving, Doppler effect exists
Computer can find rate of flow by counting the returned
frequency
Used to find areas of narrowed blood vessels
Narrowest area fastest flow
Day 75
Applying science is called
engineering.
17P57-61, 72, 74-76, 80, 83
57) 498.5 Hz, 501.5 Hz
58) No
59) 82 dB
60) 3 dB, yes
61) 48 dB, 9 dB, 0 dB, -7 dB,
20 dB
72) 170 dB
74) 103 dB
75) 10 cm
76) 1.00, 0.823
80) 5.78 × 10−4 m, 2.67 × 106
Hz
83) 0.192 m/s
```
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# Network Layer
Question 1
Assume that source S and destination D are connected through two intermediate routers labeled R. Determine how many times each packet has to visit the network layer and the data link layer during a transmission from S to D.
A Network layer – 4 times and Data link layer – 4 times B Network layer – 4 times and Data link layer – 3 times C Network layer – 4 times and Data link layer – 6 times D Network layer – 2 times and Data link layer – 6 times
GATE CS 2013 Network Layer
Discuss it
Question 1 Explanation:
Router is a network layer device. See the following diagram from http://www.oreillynet.com/network/2001/04/13/net_2nd_lang.html So every packet passes twice through data link layer of every intermediate router.
Question 2
In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are
A Last fragment, 2400 and 2789 B First fragment, 2400 and 2759 C Last fragment, 2400 and 2759 D Middle fragment, 300 and 689
GATE CS 2013 Network Layer
Discuss it
Question 2 Explanation:
M = 0 indicates that this packet is the last packet among all fragments of original packet. So the answer is either A or C. It is given that HLEN field is 10. Header length is number of 32 bit words. So header length = 10 * 4 = 40 Also, given that total length = 400. Total length indicates total length of the packet including header. So, packet length excluding header = 400 - 40 = 360 Last byte address = 2400 + 360 - 1 = 2759 (Because numbering starts from 0)
Question 3
Consider a source computer(S) transmitting a file of size 106 bits to a destination computer(D)over a network of two routers (R1 and R2) and three links(L1, L2, and L3). L1connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D.Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second.Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?
A 1005 ms B 1010 ms C 3000 ms D 3003 ms
GATE CS 2012 Network Layer
Discuss it
Question 3 Explanation:
Question 4
Consider a network with five nodes, N1 to N5, as shown below. The network uses a Distance Vector Routing protocol. Once the routes have stabilized, the distance vectors at different nodes are as following. N1: (0, 1, 7, 8, 4) N2: (1, 0, 6, 7, 3) N3: (7, 6, 0, 2, 6) N4: (8, 7, 2, 0, 4) N5: (4, 3, 6, 4, 0) Each distance vector is the distance of the best known path at the instance to nodes, N1 to N5, where the distance to itself is 0. Also, all links are symmetric and the cost is identical in both directions. In each round, all nodes exchange their distance vectors with their respective neighbors. Then all nodes update their distance vectors. In between two rounds, any change in cost of a link will cause the two incident nodes to change only that entry in their distance vectors. 52. The cost of link N2-N3 reduces to 2(in both directions). After the next round of updates, what will be the new distance vector at node, N3.
A (3, 2, 0, 2, 5) B (3, 2, 0, 2, 6) C (7, 2, 0, 2, 5) D (7, 2, 0, 2, 6)
GATE CS 2011 Network Layer
Discuss it
Question 4 Explanation:
Question 5
Consider the same data as given in previous question. After the update in the previous question, the link N1-N2 goes down. N2 will reflect this change immediately in its distance vector as cost, infinite. After the NEXT ROUND of update, what will be cost to N1 in the distance vector of N3?
A 3 B 9 C 10 D Infinite
GATE CS 2011 Network Layer
Discuss it
Question 5 Explanation:
Question 6
One of the header fields in an IP datagram is the Time to Live(TTL)field.Which of the following statements best explains the need for this field?
A It can be used to prioritize packets B It can be used to reduce delays C It can be used to optimize throughput D It can be used to prevent packet looping
GATE CS 2010 Network Layer
Discuss it
Question 6 Explanation:
Time to Live can be thought as an upper bound on the time that an IP datagram can exist in the network. The purpose of the TTL field is to avoid a situation in which an undeliverable datagram keeps circulating.
Question 7
Consider a network with 6 routers R1 to R6 connected with links having weights as shown in the following diagram All the routers use the distance vector based routing algorithm to update their routing tables. Each router starts with its routing table initialized to contain an entry for each neighbour with the weight of the respective connecting link. After all the routing tables stabilize, how many links in the network will never be used for carrying any data?
A 4 B 3 C 2 D 1
GATE CS 2010 Network Layer
Discuss it
Question 7 Explanation:
Question 8
Consider the data given in above question. Suppose the weights of all unused links in the previous question are changed to 2 and the distance vector algorithm is used again until all routing tables stabilize. How many links will now remain unused?
A 0 B 1 C 2 D 3
GATE CS 2010 Network Layer
Discuss it
Question 8 Explanation:
Question 9
For which one of the following reasons does Internet Protocol (IP) use the timeto- live (TTL) field in the IP datagram header
A Ensure packets reach destination within that time B Discard packets that reach later than that time C Prevent packets from looping indefinitely D Limit the time for which a packet gets queued in intermediate routers.
Network Layer GATE-CS-2006
Discuss it
Question 9 Explanation:
following are lines from wikipedia Time to live (TTL) or hop limit is a mechanism that limits the lifespan or lifetime of data in a computer or network. TTL may be implemented as a counter or timestamp attached to or embedded in the data. Once the prescribed event count or timespan has elapsed, data is discarded. In computer networking, TTL prevents a data packet from circulating indefinitely.
Question 10
Consider the following three statements about link state and distance vector routing protocols, for a large network with 500 network nodes and 4000 links.
```[S1] The computational overhead in link state protocols
is higher than in distance vector protocols.
[S2] A distance vector protocol (with split horizon)
avoids persistent routing loops, but not a link
state protocol.
[S3] After a topology change, a link state protocol
will converge faster than a distance vector
protocol.```
Which one of the following is correct about S1, S2, and S3 ?
A S1, S2, and S3 are all true. B S1, S2, and S3 are all false. C S1 and S2 are true, but S3 is false D S1 and S3 are true, but S2 is false
Network Layer GATE-CS-2014-(Set-1)
Discuss it
Question 10 Explanation:
```Link-state:
Every node collects complete graph structure
Each computes shortest paths from it
Each generates own routing table
Distance-vector
No one has copy of graph
Nodes construct their own tables iteratively
Each sends information about its table to neighbors ```
Source: http://www.cs.cmu.edu/~srini/15-441/S05/lectures/10-Routing.ppt
```[S1] The computational overhead in link state protocols
is higher than in distance vector protocols.
[S2] A distance vector protocol (with split horizon)
avoids persistent routing loops, but not a link
state protocol.
[S3] After a topology change, a link state protocol
will converge faster than a distance vector
protocol.```
S1 is clearly true as in Link State all nodes compute shortest path for whole network graph. S3 is also true as Distance Vector protocol has count to infinity problem and converges slower. S2 is false. In distance vector protocol, split horizon with poison reverse reduces the chance of forming loops and uses a maximum number of hops to counter the 'count-to-infinity' problem. These measures avoid the formation of routing loops in some, but not all, cases
There are 45 questions to complete.
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# Smallest subarray with sum greater than a given value
Given an array arr[] of integers and a number x, the task is to find the smallest subarray with a sum greater than the given value.
Examples:
```arr[] = {1, 4, 45, 6, 0, 19}
x = 51
Output: 3
Minimum length subarray is {4, 45, 6}
arr[] = {1, 10, 5, 2, 7}
x = 9
Output: 1
Minimum length subarray is {10}
arr[] = {1, 11, 100, 1, 0, 200, 3, 2, 1, 250}
x = 280
Output: 4
Minimum length subarray is {100, 1, 0, 200}
arr[] = {1, 2, 4}
x = 8
Output : Not Possible
Whole array sum is smaller than 8.```
Naive approach: A simple solution is to use two nested loops. The outer loop picks a starting element, the inner loop considers all elements (on right side of current start) as ending element. Whenever sum of elements between current start and end becomes more than the given number, update the result if current length is smaller than the smallest length so far.
Below is the implementation of the above approach:
## C++
`# include ``using` `namespace` `std;` `// Returns length of smallest subarray with sum greater than x.``// If there is no subarray with given sum, then returns n+1``int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)``{`` ``// Initialize length of smallest subarray as n+1`` ``int` `min_len = n + 1;` ` ``// Pick every element as starting point`` ``for` `(``int` `start=0; start x) ``return` `1;` ` ``// Try different ending points for current start`` ``for` `(``int` `end=start+1; end x && (end - start + 1) < min_len)`` ``min_len = (end - start + 1);`` ``}`` ``}`` ``return` `min_len;``}` `/* Driver program to test above function */``int` `main()``{`` ``int` `arr1[] = {1, 4, 45, 6, 10, 19};`` ``int` `x = 51;`` ``int` `n1 = ``sizeof``(arr1)/``sizeof``(arr1[0]);`` ``int` `res1 = smallestSubWithSum(arr1, n1, x);`` ``(res1 == n1+1)? cout << ``"Not possible\n"` `:`` ``cout << res1 << endl;` ` ``int` `arr2[] = {1, 10, 5, 2, 7};`` ``int` `n2 = ``sizeof``(arr2)/``sizeof``(arr2[0]);`` ``x = 9;`` ``int` `res2 = smallestSubWithSum(arr2, n2, x);`` ``(res2 == n2+1)? cout << ``"Not possible\n"` `:`` ``cout << res2 << endl;` ` ``int` `arr3[] = {1, 11, 100, 1, 0, 200, 3, 2, 1, 250};`` ``int` `n3 = ``sizeof``(arr3)/``sizeof``(arr3[0]);`` ``x = 280;`` ``int` `res3 = smallestSubWithSum(arr3, n3, x);`` ``(res3 == n3+1)? cout << ``"Not possible\n"` `:`` ``cout << res3 << endl;` ` ``return` `0;``}`
## Java
`import` `java.io.*;``class` `SmallestSubArraySum``{`` ``// Returns length of smallest subarray with sum greater than x.`` ``// If there is no subarray with given sum, then returns n+1`` ``static` `int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)`` ``{`` ``// Initialize length of smallest subarray as n+1`` ``int` `min_len = n + ``1``;` ` ``// Pick every element as starting point`` ``for` `(``int` `start = ``0``; start < n; start++)`` ``{`` ``// Initialize sum starting with current start`` ``int` `curr_sum = arr[start];` ` ``// If first element itself is greater`` ``if` `(curr_sum > x)`` ``return` `1``;` ` ``// Try different ending points for current start`` ``for` `(``int` `end = start + ``1``; end < n; end++)`` ``{`` ``// add last element to current sum`` ``curr_sum += arr[end];` ` ``// If sum becomes more than x and length of`` ``// this subarray is smaller than current smallest`` ``// length, update the smallest length (or result)`` ``if` `(curr_sum > x && (end - start + ``1``) < min_len)`` ``min_len = (end - start + ``1``);`` ``}`` ``}`` ``return` `min_len;`` ``}` ` ``// Driver program to test above functions`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr1[] = {``1``, ``4``, ``45``, ``6``, ``10``, ``19``};`` ``int` `x = ``51``;`` ``int` `n1 = arr1.length;`` ``int` `res1 = smallestSubWithSum(arr1, n1, x);`` ``if` `(res1 == n1+``1``)`` ``System.out.println(``"Not Possible"``);`` ``else`` ``System.out.println(res1);` ` ``int` `arr2[] = {``1``, ``10``, ``5``, ``2``, ``7``};`` ``int` `n2 = arr2.length;`` ``x = ``9``;`` ``int` `res2 = smallestSubWithSum(arr2, n2, x);`` ``if` `(res2 == n2+``1``)`` ``System.out.println(``"Not Possible"``);`` ``else`` ``System.out.println(res2);` ` ``int` `arr3[] = {``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250``};`` ``int` `n3 = arr3.length;`` ``x = ``280``;`` ``int` `res3 = smallestSubWithSum(arr3, n3, x);`` ``if` `(res3 == n3+``1``)`` ``System.out.println(``"Not Possible"``);`` ``else`` ``System.out.println(res3);`` ``}``}` `// This code has been contributed by Mayank Jaiswal`
## Python3
`# Python3 program to find Smallest``# subarray with sum greater``# than a given value` `# Returns length of smallest subarray``# with sum greater than x. If there``# is no subarray with given sum,``# then returns n+1``def` `smallestSubWithSum(arr, n, x):` ` ``# Initialize length of smallest`` ``# subarray as n+1`` ``min_len ``=` `n ``+` `1` ` ``# Pick every element as starting point`` ``for` `start ``in` `range``(``0``,n):`` ` ` ``# Initialize sum starting`` ``# with current start`` ``curr_sum ``=` `arr[start]` ` ``# If first element itself is greater`` ``if` `(curr_sum > x):`` ``return` `1` ` ``# Try different ending points`` ``# for current start`` ``for` `end ``in` `range``(start``+``1``,n):`` ` ` ``# add last element to current sum`` ``curr_sum ``+``=` `arr[end]` ` ``# If sum becomes more than x`` ``# and length of this subarray`` ``# is smaller than current smallest`` ``# length, update the smallest`` ``# length (or result)`` ``if` `curr_sum > x ``and` `(end ``-` `start ``+` `1``) < min_len:`` ``min_len ``=` `(end ``-` `start ``+` `1``)`` ` ` ``return` `min_len;` `# Driver program to test above function */``arr1 ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``19``]``x ``=` `51``n1 ``=` `len``(arr1)``res1 ``=` `smallestSubWithSum(arr1, n1, x);``if` `res1 ``=``=` `n1``+``1``:`` ``print``(``"Not possible"``)``else``:`` ``print``(res1)` `arr2 ``=` `[``1``, ``10``, ``5``, ``2``, ``7``]``n2 ``=` `len``(arr2)``x ``=` `9``res2 ``=` `smallestSubWithSum(arr2, n2, x);``if` `res2 ``=``=` `n2``+``1``:`` ``print``(``"Not possible"``)``else``:`` ``print``(res2)` `arr3 ``=` `[``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250``]``n3 ``=` `len``(arr3)``x ``=` `280``res3 ``=` `smallestSubWithSum(arr3, n3, x)``if` `res3 ``=``=` `n3``+``1``:`` ``print``(``"Not possible"``)``else``:`` ``print``(res3)`` ` `# This code is contributed by Smitha Dinesh Semwal`
## C#
`// C# program to find Smallest``// subarray with sum greater``// than a given value``using` `System;` `class` `GFG``{`` ` ` ``// Returns length of smallest`` ``// subarray with sum greater`` ``// than x. If there is no`` ``// subarray with given sum,`` ``// then returns n+1`` ``static` `int` `smallestSubWithSum(``int` `[]arr,`` ``int` `n, ``int` `x)`` ``{`` ``// Initialize length of`` ``// smallest subarray as n+1`` ``int` `min_len = n + 1;` ` ``// Pick every element`` ``// as starting point`` ``for` `(``int` `start = 0; start < n; start++)`` ``{`` ``// Initialize sum starting`` ``// with current start`` ``int` `curr_sum = arr[start];` ` ``// If first element`` ``// itself is greater`` ``if` `(curr_sum > x)`` ``return` `1;` ` ``// Try different ending`` ``// points for current start`` ``for` `(``int` `end = start + 1;`` ``end < n; end++)`` ``{`` ``// add last element`` ``// to current sum`` ``curr_sum += arr[end];` ` ``// If sum becomes more than`` ``// x and length of this`` ``// subarray is smaller than`` ``// current smallest length,`` ``// update the smallest`` ``// length (or result)`` ``if` `(curr_sum > x &&`` ``(end - start + 1) < min_len)`` ``min_len = (end - start + 1);`` ``}`` ``}`` ``return` `min_len;`` ``}` ` ``// Driver Code`` ``static` `public` `void` `Main ()`` ``{`` ``int` `[]arr1 = {1, 4, 45,`` ``6, 10, 19};`` ``int` `x = 51;`` ``int` `n1 = arr1.Length;`` ``int` `res1 = smallestSubWithSum(arr1,`` ``n1, x);`` ``if` `(res1 == n1 + 1)`` ``Console.WriteLine(``"Not Possible"``);`` ``else`` ``Console.WriteLine(res1);` ` ``int` `[]arr2 = {1, 10, 5, 2, 7};`` ``int` `n2 = arr2.Length;`` ``x = 9;`` ``int` `res2 = smallestSubWithSum(arr2,`` ``n2, x);`` ``if` `(res2 == n2 + 1)`` ``Console.WriteLine(``"Not Possible"``);`` ``else`` ``Console.WriteLine(res2);` ` ``int` `[]arr3 = {1, 11, 100, 1, 0,`` ``200, 3, 2, 1, 250};`` ``int` `n3 = arr3.Length;`` ``x = 280;`` ``int` `res3 = smallestSubWithSum(arr3,`` ``n3, x);`` ``if` `(res3 == n3 + 1)`` ``Console.WriteLine(``"Not Possible"``);`` ``else`` ``Console.WriteLine(res3);`` ``}``}` `// This code is contributed by ajit`
## PHP
` ``\$x``) ``return` `1;` ` ``// Try different ending`` ``// points for current start`` ``for` `(``\$end``= ``\$start` `+ 1; ``\$end` `< ``\$n``; ``\$end``++)`` ``{`` ``// add last element`` ``// to current sum`` ``\$curr_sum` `+= ``\$arr``[``\$end``];` ` ``// If sum becomes more than`` ``// x and length of this subarray`` ``// is smaller than current`` ``// smallest length, update the`` ``// smallest length (or result)`` ``if` `(``\$curr_sum` `> ``\$x` `&&`` ``(``\$end` `- ``\$start` `+ 1) < ``\$min_len``)`` ``\$min_len` `= (``\$end` `- ``\$start` `+ 1);`` ``}`` ``}`` ``return` `\$min_len``;``}` `// Driver Code``\$arr1` `= ``array` `(1, 4, 45,`` ``6, 10, 19);``\$x` `= 51;``\$n1` `= sizeof(``\$arr1``);``\$res1` `= smallestSubWithSum(``\$arr1``, ``\$n1``, ``\$x``);` `if` `((``\$res1` `== ``\$n1` `+ 1) == true)`` ``echo` `"Not possible\n"` `;``else`` ``echo` `\$res1` `, ``"\n"``;` `\$arr2` `= ``array``(1, 10, 5, 2, 7);``\$n2` `= sizeof(``\$arr2``);``\$x` `= 9;``\$res2` `= smallestSubWithSum(``\$arr2``, ``\$n2``, ``\$x``);` `if` `((``\$res2` `== ``\$n2` `+ 1) == true)`` ``echo` `"Not possible\n"` `;``else`` ``echo` `\$res2` `, ``"\n"``;` `\$arr3` `= ``array` `(1, 11, 100, 1, 0,`` ``200, 3, 2, 1, 250);``\$n3` `= sizeof(``\$arr3``);``\$x` `= 280;``\$res3` `= smallestSubWithSum(``\$arr3``, ``\$n3``, ``\$x``);``if` `((``\$res3` `== ``\$n3` `+ 1) == true)`` ``echo` `"Not possible\n"` `;``else`` ``echo` `\$res3` `, ``"\n"``;` `// This code is contributed by ajit``?>`
## Javascript
``
Output
```3
1
4
```
Time Complexity: O(n2).
Auxiliary Space: O(1)
Efficient Solution: This problem can be solved in O(n) time using the idea used in this post.
## C++14
`// O(n) solution for finding smallest subarray with sum``// greater than x``#include ``using` `namespace` `std;` `// Returns length of smallest subarray with sum greater than``// x. If there is no subarray with given sum, then returns``// n+1``int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)``{`` ``// Initialize current sum and minimum length`` ``int` `curr_sum = 0, min_len = n + 1;` ` ``// Initialize starting and ending indexes`` ``int` `start = 0, end = 0;`` ``while` `(end < n) {`` ``// Keep adding array elements while current sum`` ``// is smaller than or equal to x`` ``while` `(curr_sum <= x && end < n)`` ``curr_sum += arr[end++];` ` ``// If current sum becomes greater than x.`` ``while` `(curr_sum > x && start < n) {`` ``// Update minimum length if needed`` ``if` `(end - start < min_len)`` ``min_len = end - start;` ` ``// remove starting elements`` ``curr_sum -= arr[start++];`` ``}`` ``}`` ``return` `min_len;``}` `/* Driver program to test above function */``int` `main()``{`` ``int` `arr1[] = { 1, 4, 45, 6, 10, 19 };`` ``int` `x = 51;`` ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]);`` ``int` `res1 = smallestSubWithSum(arr1, n1, x);`` ``(res1 == n1 + 1) ? cout << ``"Not possible\n"`` ``: cout << res1 << endl;` ` ``int` `arr2[] = { 1, 10, 5, 2, 7 };`` ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2[0]);`` ``x = 9;`` ``int` `res2 = smallestSubWithSum(arr2, n2, x);`` ``(res2 == n2 + 1) ? cout << ``"Not possible\n"`` ``: cout << res2 << endl;` ` ``int` `arr3[] = { 1, 11, 100, 1, 0, 200, 3, 2, 1, 250 };`` ``int` `n3 = ``sizeof``(arr3) / ``sizeof``(arr3[0]);`` ``x = 280;`` ``int` `res3 = smallestSubWithSum(arr3, n3, x);`` ``(res3 == n3 + 1) ? cout << ``"Not possible\n"`` ``: cout << res3 << endl;` ` ``return` `0;``}`
## Java
`// O(n) solution for finding smallest subarray with sum``// greater than x``import` `java.io.*;``class` `SmallestSubArraySum {`` ``// Returns length of smallest subarray with sum greater`` ``// than x. If there is no subarray with given sum, then`` ``// returns n+1`` ``static` `int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)`` ``{`` ``// Initialize current sum and minimum length`` ``int` `curr_sum = ``0``, min_len = n + ``1``;` ` ``// Initialize starting and ending indexes`` ``int` `start = ``0``, end = ``0``;`` ``while` `(end < n) {`` ``// Keep adding array elements while current sum`` ``// is smaller than or equal to x`` ``while` `(curr_sum <= x && end < n)`` ``curr_sum += arr[end++];` ` ``// If current sum becomes greater than x.`` ``while` `(curr_sum > x && start < n) {`` ``// Update minimum length if needed`` ``if` `(end - start < min_len)`` ``min_len = end - start;` ` ``// remove starting elements`` ``curr_sum -= arr[start++];`` ``}`` ``}`` ``return` `min_len;`` ``}`` ``// Driver program to test above functions`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr1[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``19` `};`` ``int` `x = ``51``;`` ``int` `n1 = arr1.length;`` ``int` `res1 = smallestSubWithSum(arr1, n1, x);`` ``if` `(res1 == n1 + ``1``)`` ``System.out.println(``"Not Possible"``);`` ``else`` ``System.out.println(res1);` ` ``int` `arr2[] = { ``1``, ``10``, ``5``, ``2``, ``7` `};`` ``int` `n2 = arr2.length;`` ``x = ``9``;`` ``int` `res2 = smallestSubWithSum(arr2, n2, x);`` ``if` `(res2 == n2 + ``1``)`` ``System.out.println(``"Not Possible"``);`` ``else`` ``System.out.println(res2);` ` ``int` `arr3[]`` ``= { ``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250` `};`` ``int` `n3 = arr3.length;`` ``x = ``280``;`` ``int` `res3 = smallestSubWithSum(arr3, n3, x);`` ``if` `(res3 == n3 + ``1``)`` ``System.out.println(``"Not Possible"``);`` ``else`` ``System.out.println(res3);`` ``}``}` `// This code has been contributed by Mayank Jaiswal`
## Python3
`# O(n) solution for finding smallest``# subarray with sum greater than x` `# Returns length of smallest subarray``# with sum greater than x. If there``# is no subarray with given sum, then``# returns n + 1` `def` `smallestSubWithSum(arr, n, x):` ` ``# Initialize current sum and minimum length`` ``curr_sum ``=` `0`` ``min_len ``=` `n ``+` `1` ` ``# Initialize starting and ending indexes`` ``start ``=` `0`` ``end ``=` `0`` ``while` `(end < n):` ` ``# Keep adding array elements while current`` ``# sum is smaller than or equal to x`` ``while` `(curr_sum <``=` `x ``and` `end < n):`` ``curr_sum ``+``=` `arr[end]`` ``end ``+``=` `1` ` ``# If current sum becomes greater than x.`` ``while` `(curr_sum > x ``and` `start < n):` ` ``# Update minimum length if needed`` ``if` `(end ``-` `start < min_len):`` ``min_len ``=` `end ``-` `start` ` ``# remove starting elements`` ``curr_sum ``-``=` `arr[start]`` ``start ``+``=` `1` ` ``return` `min_len` `# Driver program``arr1 ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``19``]``x ``=` `51``n1 ``=` `len``(arr1)``res1 ``=` `smallestSubWithSum(arr1, n1, x)``print``(``"Not possible"``) ``if` `(res1 ``=``=` `n1 ``+` `1``) ``else` `print``(res1)` `arr2 ``=` `[``1``, ``10``, ``5``, ``2``, ``7``]``n2 ``=` `len``(arr2)``x ``=` `9``res2 ``=` `smallestSubWithSum(arr2, n2, x)``print``(``"Not possible"``) ``if` `(res2 ``=``=` `n2 ``+` `1``) ``else` `print``(res2)` `arr3 ``=` `[``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250``]``n3 ``=` `len``(arr3)``x ``=` `280``res3 ``=` `smallestSubWithSum(arr3, n3, x)``print``(``"Not possible"``) ``if` `(res3 ``=``=` `n3 ``+` `1``) ``else` `print``(res3)` `# This code is contributed by``# Smitha Dinesh Semwal`
## C#
`// O(n) solution for finding``// smallest subarray with sum``// greater than x``using` `System;` `class` `GFG {` ` ``// Returns length of smallest`` ``// subarray with sum greater`` ``// than x. If there is no`` ``// subarray with given sum,`` ``// then returns n+1`` ``static` `int` `smallestSubWithSum(``int``[] arr, ``int` `n, ``int` `x)`` ``{`` ``// Initialize current`` ``// sum and minimum length`` ``int` `curr_sum = 0, min_len = n + 1;` ` ``// Initialize starting`` ``// and ending indexes`` ``int` `start = 0, end = 0;`` ``while` `(end < n) {`` ``// Keep adding array elements`` ``// while current sum is smaller`` ``// than or equal to x`` ``while` `(curr_sum <= x && end < n)`` ``curr_sum += arr[end++];` ` ``// If current sum becomes`` ``// greater than x.`` ``while` `(curr_sum > x && start < n) {`` ``// Update minimum`` ``// length if needed`` ``if` `(end - start < min_len)`` ``min_len = end - start;` ` ``// remove starting elements`` ``curr_sum -= arr[start++];`` ``}`` ``}`` ``return` `min_len;`` ``}` ` ``// Driver Code`` ``static` `public` `void` `Main()`` ``{`` ``int``[] arr1 = { 1, 4, 45, 6, 10, 19 };`` ``int` `x = 51;`` ``int` `n1 = arr1.Length;`` ``int` `res1 = smallestSubWithSum(arr1, n1, x);`` ``if` `(res1 == n1 + 1)`` ``Console.WriteLine(``"Not Possible"``);`` ``else`` ``Console.WriteLine(res1);` ` ``int``[] arr2 = { 1, 10, 5, 2, 7 };`` ``int` `n2 = arr2.Length;`` ``x = 9;`` ``int` `res2 = smallestSubWithSum(arr2, n2, x);`` ``if` `(res2 == n2 + 1)`` ``Console.WriteLine(``"Not Possible"``);`` ``else`` ``Console.WriteLine(res2);` ` ``int``[] arr3`` ``= { 1, 11, 100, 1, 0, 200, 3, 2, 1, 250 };`` ``int` `n3 = arr3.Length;`` ``x = 280;`` ``int` `res3 = smallestSubWithSum(arr3, n3, x);`` ``if` `(res3 == n3 + 1)`` ``Console.WriteLine(``"Not Possible"``);`` ``else`` ``Console.WriteLine(res3);`` ``}``}` `// This code is contributed by akt_mit`
## PHP
` ``\$x` `&&`` ``\$start` `< ``\$n``)`` ``{`` ``// Update minimum`` ``// length if needed`` ``if` `(``\$end` `- ``\$start` `< ``\$min_len``)`` ``\$min_len` `= ``\$end` `- ``\$start``;` ` ``// remove starting elements`` ``\$curr_sum` `-= ``\$arr``[``\$start``++];`` ``}`` ``}`` ``return` `\$min_len``;``}` `// Driver Code``\$arr1` `= ``array``(1, 4, 45,`` ``6, 10, 19);``\$x` `= 51;``\$n1` `= sizeof(``\$arr1``);``\$res1` `= smallestSubWithSum(``\$arr1``,`` ` ` ``\$n1``, ``\$x``);``if``(``\$res1` `== ``\$n1` `+ 1)``echo` `"Not possible\n"` `;``else``echo` `\$res1` `,``"\n"``;` `\$arr2` `= ``array``(1, 10, 5, 2, 7);``\$n2` `= sizeof(``\$arr2``);``\$x` `= 9;``\$res2` `= smallestSubWithSum(``\$arr2``,`` ``\$n2``, ``\$x``);``if``(``\$res2` `== ``\$n2` `+ 1)``echo` `"Not possible\n"` `;``else``echo` `\$res2``,``"\n"``;` `\$arr3` `= ``array``(1, 11, 100, 1, 0,`` ``200, 3, 2, 1, 250);``\$n3` `= sizeof(``\$arr3``);``\$x` `= 280;``\$res3` `= smallestSubWithSum(``\$arr3``,`` ``\$n3``, ``\$x``);`` ` `if``(``\$res3` `== ``\$n3` `+ 1)``echo` `"Not possible\n"` `;``else``echo` `\$res3``, ``"\n"``;` `// This code is contributed by ajit``?>`
## Javascript
``
Output
```3
1
4
```
Time Complexity: O(n).
Auxiliary Space: O(1)
Another Approach: Binary Search
1. First calculates the cumulative sum of the vector elements and stores them in the sums vector.
2. Then iterates through the sums vector and finds the lower bound of the target sum for each possible subarray.
3. If the lower bound is found and it’s not equal to the target sum (i.e., the subarray sum is greater than the target),
4. Calculates the length of the subarray and updates the ans variable if the length is smaller than the current value.
5. Finally, returns the ans value or 0 if ans was not updated.
## C++
`// O(n log(n) solution for finding smallest subarray with``// sum greater than x``#include ``using` `namespace` `std;` `int` `smallestSubArrayLen(``int` `target, vector<``int``>& nums)``{`` ``// Get the length of the input vector`` ``int` `n = nums.size();`` ``// If the vector is empty, return 0`` ``if` `(n == 0)`` ``return` `0;`` ``// Initialize the minimum subarray length to INT_MAX-1`` ``int` `ans = INT_MAX - 1;`` ``// Create a new vector "sums" with size n+1, initialized`` ``// to all zeros`` ``vector<``int``> sums(n + 1, 0);`` ``// Compute the running sum of nums and store it in`` ``// "sums"`` ``for` `(``int` `i = 1; i <= n; i++)`` ``sums[i] = sums[i - 1] + nums[i - 1];`` ``// Iterate through each starting index i`` ``for` `(``int` `i = 1; i <= n; i++) {`` ``// Calculate the target sum for the subarray`` ``// starting at index i`` ``int` `to_find = target + sums[i - 1];`` ``// Find the first element in "sums" that is >=`` ``// to_find`` ``auto` `bound = lower_bound(sums.begin(), sums.end(),`` ``to_find);`` ``// If such an element is found and it is not equal`` ``// to to_find itself`` ``if` `(bound != sums.end() && *bound != to_find) {`` ``// Compute the length of the subarray and update`` ``// ans if necessary`` ``int` `len = bound - (sums.begin() + i - 1);`` ``ans = min(ans, len);`` ``}`` ``}`` ``// Return ans if it was updated, otherwise return 0`` ``return` `(ans != INT_MAX - 1) ? ans : 0;``}` `/* Driver program to test above function */``int` `main()``{`` ``vector<``int``> arr1 = { 1, 4, 45, 6, 10, 19 };`` ``int` `target1 = 51;`` ``cout << ``"Length of Smallest Subarray :"`` ``<< smallestSubArrayLen(target1, arr1) << endl;` ` ``vector<``int``> arr2 = { 1, 10, 5, 2, 7 };`` ``int` `target2 = 9;`` ``cout << ``"Length of Smallest Subarray :"`` ``<< smallestSubArrayLen(target2, arr2) << endl;` ` ``vector<``int``> arr3 = { 1, 1, 1, 1, 1, 1, 1, 1 };`` ``int` `target3 = 11;`` ``cout << ``"Length of Smallest Subarray :"`` ``<< smallestSubArrayLen(target3, arr3) << endl;` ` ``vector<``int``> arr4`` ``= { 1, 11, 100, 1, 0, 200, 3, 2, 1, 250 };`` ``int` `target4 = 280;`` ``cout << ``"Length of Smallest Subarray :"`` ``<< smallestSubArrayLen(target4, arr4) << endl;` ` ``return` `0;``}`
Output
```Length of Smallest Subarray :3
Length of Smallest Subarray :1
Length of Smallest Subarray :0
Length of Smallest Subarray :4
```
Time Complexity: O (n log(n)).
Auxiliary Space: O(n)
Thanks to Ankit and Nitin for suggesting this optimized solution.
How to handle negative numbers?
The above solution may not work if input array contains negative numbers. For example arr[] = {- 8, 1, 4, 2, -6}. To handle negative numbers, add a condition to ignore subarrays with negative sums. We can use the solution discussed in Find subarray with given sum with negatives allowed in constant space
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# Cube
Cube
TypePlatonic solid
Regular polyhedron
Parallelohedron
Zonohedron
Plesiohedron
Hanner polytope
Faces6
Edges12
Vertices8
Symmetry groupoctahedral symmetry ${\displaystyle \mathrm {O} _{\mathrm {h} }}$
Dihedral angle (degrees)90°
Dual polyhedronregular octahedron
Propertiesconvex,
face-transitive,
edge-transitive,
vertex-transitive
In geometry, a cube is a three-dimensional solid object bounded by six square faces. It has twelve edges and eight vertices. It can be represented as a rectangular cuboid with six square faces, or a parallelepiped with equal edges. It is an example of many type of solids: Platonic solid, regular polyhedron, parallelohedron, zonohedron, and plesiohedron. The dual polyhedron of a cube is the regular octahedron.
The cube can be represented in many ways, one of which is the graph known as the cubical graph. It can be constructed by using the Cartesian product of graphs. The cube was discovered in antiquity. It was associated with the nature of earth by Plato, the founder of Platonic solid. It was used as the part of the Solar System, proposed by Johannes Kepler. It can be derived differently to create more polyhedrons, and it has applications to construct a new polyhedron by attaching others. It can be generalized as tesseract in four-dimensional space.
## Properties
A cube is a special case of rectangular cuboid in which the edges are equal in length.[1] Like other cuboids, every face of a cube has four vertices, each of which connects with three congruent lines. These edges form square faces, making the dihedral angle of a cube between every two adjacent squares being the interior angle of a square, 90°. Hence, the cube has six faces, twelve edges, and eight vertices. [2] Because of such properties, it is categorized as one of the five Platonic solids, a polyhedron in which all the regular polygons are congruent and the same number of faces meet at each vertex.[3]
### Measurement and other metric properties
A face diagonal in red and space diagonal in blue.
Given that a cube with edge length ${\displaystyle a}$. The face diagonal of a cube is the diagonal of a square ${\displaystyle a{\sqrt {2}}}$, and the space diagonal of a cube is a line connecting two vertices that is not in the same face, formulated as ${\displaystyle a{\sqrt {3}}}$. Both formulas can be determined by using Pythagorean theorem. The surface area of a cube ${\displaystyle A}$ is six times the area of a square:[4] ${\displaystyle A=6a^{2}.}$ The volume of a cuboid is the product of length, width, and height. Because the edges of a cube are all equal in length, it is:[4] ${\displaystyle V=a^{3}.}$
A unit cube is a special case where each cube's edge is 1 unit length. The surface area and the volume of a unit cube is 1.[5][6] It has Rupert property, meaning a polyhedron with the same or larger size can pass through into the hole of the unit cube. The Prince Rupert's cube, named after Prince Rupert of the Rhine, is the largest cube that can fit inside, with the size being 6% larger.[7]
A geometric problem of doubling the cube—alternatively known as the Delian problem—requires the construction of a cube with a volume twice the original by using a compass and straightedge solely. Ancient mathematicians could not solve this old problem until French mathematician Pierre Wantzel in 1837 proved it was impossible.[8]
### Relation to the spheres
With edge length ${\displaystyle a}$, the inscribed sphere of a cube is the sphere tangent to the faces of a cube at their centroids, with radius ${\textstyle {\frac {1}{2}}a}$. The midsphere of a cube is the sphere tangent to the edges of a cube, with radius ${\textstyle {\frac {1}{\sqrt {2}}}a}$. The circumscribed sphere of a cube is the sphere tangent to the vertices of a cube, with radius ${\textstyle {\frac {\sqrt {3}}{2}}a}$.[9]
For a cube whose circumscribed sphere has radius ${\displaystyle R}$, and for a given point in its three-dimensional space with distances ${\displaystyle d_{i}}$ from the cube's eight vertices, it is:[10] ${\displaystyle {\frac {1}{8}}\sum _{i=1}^{8}d_{i}^{4}+{\frac {16R^{4}}{9}}=\left({\frac {1}{8}}\sum _{i=1}^{8}d_{i}^{2}+{\frac {2R^{2}}{3}}\right)^{2}.}$
### Symmetry
The cube has octahedral symmetry ${\displaystyle \mathrm {O} _{\mathrm {h} }}$. It is composed of reflection symmetry, a symmetry by cutting into two halves by a plane. There are nine reflection symmetries: the five are cut the cube from the midpoints of its edges, and the four are cut diagonally. It is also composed of rotational symmetry, a symmetry by rotating it around the axis, from which the appearance is interchangeable. It has octahedral rotation symmetry ${\displaystyle \mathrm {O} }$: three axes pass through the cube's opposite faces centroid, six through the cube's opposite edges midpoints, and four through the cube's opposite vertices; each of these axes is respectively four-fold rotational symmetry (0°, 90°, 180°, and 270°), two-fold rotational symmetry (0° and 180°), and three-fold rotational symmetry (0°, 120°, and 240°).[11][12][13]
The dual polyhedron can be obtained from each of the polyhedron's vertices tangent to a plane by the process known as polar reciprocation.[14] One property of dual polyhedrons generally is that the polyhedron and its dual share their three-dimensional symmetry point group. In this case, the dual polyhedron of a cube is the regular octahedron, and both of these polyhedron has the same symmetry, the octahedral symmetry.[15]
The cube is face-transitive, meaning its two squares are alike and can be mapped by rotation and reflection.[16] It is vertex-transitive, meaning all of its vertices are equivalent and can be mapped isometrically under its symmetry.[17] It is also edge-transitive, meaning the same kind of faces surround each of its vertices in the same or reverse order, all two adjacent faces have the same dihedral angle. Therefore, the cube is regular polyhedron because it requires those properties.[18]
### Classifications
The cube is a special case among every cuboids. As mentioned above, the cube can be represented as the rectangular cuboid with edges equal in length and all of its faces are all squares.[1] The cube may be considered as the parallelepiped in which all of its edges are equal edges.[19]
The cube is a plesiohedron, a special kind of space-filling polyhedron that can be defined as the Voronoi cell of a symmetric Delone set.[20] The plesiohedra include the parallelohedrons, which can be translated without rotating to fill a space—called honeycomb—in which each face of any of its copies is attached to a like face of another copy. There are five kinds of parallelohedra, one of which is the cuboid.[21] Every three-dimensional parallelohedron is zonohedron, a centrally symmetric polyhedron whose faces are centrally symmetric polygons,[22]
## Construction
An elementary way to construct a cube is using the net. A net is an arrangement of edge-joining polygons constructing a polyhedron by connecting along the edges of those polygons. Here, there are eleven different cube's net.[23]
In analytic geometry, a cube may be constructed using the Cartesian coordinate systems. For a cube centered at the origin, with edges parallel to the axes and with an edge length of 2, the Cartesian coordinates of the vertices are ${\displaystyle (\pm 1,\pm 1,\pm 1)}$.[24] Its interior consists of all points ${\displaystyle (x_{0},x_{1},x_{2})}$ with ${\displaystyle -1 for all ${\displaystyle i}$. A cube's surface with center ${\displaystyle (x_{0},y_{0},z_{0})}$ and edge length of ${\displaystyle 2a}$ is the locus of all points ${\displaystyle (x,y,z)}$ such that ${\displaystyle \max\{|x-x_{0}|,|y-y_{0}|,|z-z_{0}|\}=a.}$
The cube is Hanner polytope, because it can be constructed by using Cartesian product of three line segments. Its dual polyhedron, the regular octahedron, is constructed by direct sum of three line segments.[25]
## Representation
### As a graph
The graph of a cube, and its construction
According to Steinitz's theorem, the graph can be represented as the skeleton of a polyhedron; roughly speaking, a framework of a polyhedron. Such a graph has two properties. It is planar, meaning the edges of a graph are connected to every vertex without crossing other edges. It is also 3-connected graph, meaning that, whenever a graph with more than three vertices, and two of the vertices are removed, the edges remain connected.[26][27] The skeleton of a cube can be represented as the graph, and it is called the cubical graph, a Platonic graph. It has the same number of vertices and edges as the cube, twelve vertices and eight edges.[28]
The cubical graph is a special case of hypercube graph or ${\displaystyle n}$-cube—denoted as ${\displaystyle Q_{n}}$—because it can be constructed by using the operation known as the Cartesian product of graphs. To put it in a plain, its construction involves two graphs connecting the pair of vertices with an edge to form a new graph.[29] In the case of the cubical graph, it is the product of two ${\displaystyle Q_{2}}$; roughly speaking, it is a graph resembling a square. In other words, the cubical graph is constructed by connecting each vertex of two squares with an edge. Notationally, the cubical graph can be denoted as ${\displaystyle Q_{3}}$.[30] It is a unit distance graph.[31]
Like other graphs of cuboids, the cubical graph is also classified as the prism graph.[32]
### In orthogonal projection
An object illuminated by parallel rays of light casts a shadow on a plane perpendicular to those rays, called an orthogonal projection. A polyhedron is considered equiprojective if, for some position of the light, its orthogonal projection is a regular polygon. The cube is equiprojective because, if the light is parallel to one of the four lines joining a vertex to the opposite vertex, its projection is a regular hexagon. Conventionally, the cube is 6-equiprojective.[33]
### As a configuration matrix
The cube can be represented as configuration matrix. A configuration matrix is a matrix in which the rows and columns correspond to the elements of a polyhedron as in the vertices, edges, and faces. The diagonal of a matrix denotes the number of each element that appears in a polyhedron, whereas the non-diagonal of a matrix denotes the number of the column's elements that occur in or at the row's element. As mentioned above, the cube has eight vertices, twelve edges, and six faces; each element in a matrix's diagonal is denoted as 8, 12, and 6. The first column of the middle row indicates that there are two vertices in (i.e., at the extremes of) each edge, denoted as 2; the middle column of the first row indicates that three edges meet at each vertex, denoted as 3. The following matrix is:[34] ${\displaystyle {\begin{bmatrix}{\begin{matrix}8&3&3\\2&12&2\\4&4&6\end{matrix}}\end{bmatrix}}}$
## Appearances
### In antiquity
Sketch of a cube by Johannes Kepler
Kepler's Platonic solid model of the Solar System
The Platonic solid is a set of polyhedrons known since antiquity. It was named after Plato in his Timaeus dialogue, who attributed these solids with nature. One of them, the cube, represented the classical element of earth because of its stability.[35] Euclid's Elements defined the Platonic solids, including the cube, and using these solids with the problem involving to find the ratio of the circumscribed sphere's diameter to the edge length.[36]
Following its attribution with nature by Plato, Johannes Kepler in his Harmonices Mundi sketched each of the Platonic solids, one of them is a cube in which Kepler decorated a tree on it.[35] In his Mysterium Cosmographicum, Kepler also proposed the Solar System by using the Platonic solids setting into another one and separating them with six spheres resembling the six planets. The ordered solids started from the innermost to the outermost: regular octahedron, regular icosahedron, regular dodecahedron, regular tetrahedron, and cube.[37]
### Polyhedron, honeycombs, and polytopes
Some of the derived cube, the stellated octahedron and tetrakis hexahedron.
The cube can appear in the construction of a polyhedron, and some of its types can be derived differently in the following:
• When faceting a cube, meaning removing part of the polygonal faces without creating new vertices of a cube, the resulting polyhedron is the stellated octahedron.[38]
• Attaching a square pyramid to each square face of a cube produces its Kleetope, a polyhedron known as the tetrakis hexahedron.[39] Suppose one and two equilateral square pyramids are attached to their square faces. In that case, they are the construction of an elongated square pyramid and elongated square bipyramid respectively, the Johnson solid's examples.[40]
• Each of the cube's vertices can be truncated, and the resulting polyhedron is the Archimedean solid, the truncated cube.[41] When its edges are truncated, it is a rhombicuboctahedron.[42] Relatedly, the rhombicuboctahedron can also be constructed by separating the cube's faces and then spreading away, after which adding other triangular and square faces between them; this is known as the "expanded cube". It also can be constructed similarly by the cube's dual, the regular octahedron.[43]
• The snub cube is an Archimedean solid that can be constructed by separating away the cube square's face, and filling their gaps with twisted angle equilateral triangles;a process known as snub.[44]
The honeycomb is the space-filling or tessellation in three-dimensional space, meaning it is an object in which the construction begins by attaching any polyhedrons onto their faces without leaving a gap. The cube can be represented as the cell, and examples of a honeycomb are cubic honeycomb, order-5 cubic honeycomb, order-6 cubic honeycomb, and order-7 cubic honeycomb.[45] The cube can be constructed with six square pyramids, tiling space by attaching their apices.[46]
Polycube is a polyhedron in which the faces of many cubes are attached. Analogously, it can be interpreted as the polyominoes in three-dimensional space.[47] When four cubes are stacked vertically, and the other four are attached to the second-from-top cube of the stack, the resulting polycube is Dali cross, after Salvador Dali. The Dali cross is a tile space polyhedron,[48][49] which can be represented as the net of a tesseract. A tesseract is a cube analogous' four-dimensional space bounded by twenty-four squares, and it is bounded by the eight cubes known as its cells.[50]
## References
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22. ^ In higher dimensions, however, there exist parallelopes that are not zonotopes. See e.g. Shephard, G. C. (1974). "Space-filling zonotopes". Mathematika. 21 (2): 261–269. doi:10.1112/S0025579300008652. MR 0365332.
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sure to describe on which tick marks each point is plotted and how many tick marks are between each integer. It’s still true that we’re depending on an interpretation of the integral … and obviously tru practice problems solutions hw week select (by induction) ≥ 4 5 The fundamental theorem of arithmetic says that every integer larger than 1 can be written as a product of one or more prime numbers in a way that is unique, except for the order of the prime factors. Propositions 30 and 32 together are essentially equivalent to the fundamental theorem of arithmetic. Every positive integer has a unique factorization into a square-free number and a square number rs 2. mitgliedd1 and 110 more users found this answer helpful. The most important maths theorems are listed here. Use sigma notation to write the sum: 9 14 6 8 5 6 4 4 3 2 5. Please be Fundamental Theorem of Arithmetic The Basic Idea. Do you remember doing division in Arithmetic? It states that any integer greater than 1 can be expressed as the product of prime number s in only one way. According to Fundamental theorem of Arithmetic, every composite number can be written (factorised) as the product of primes and this factorization is Unique, apart from the order in which prime factors occur. …. If you are considering these as subjects or concepts of Mathematics and not from a biology perspective, then arithmetic represents a constant growth and a geometric growth represents an exponential growth. * The Fundamental Theorem of Arithmetic states that every positive integer/number greater than 1 is either a prime or a composite, i.e. Basic math operations include four basic operations: Addition (+) Subtraction (-) Multiplication (* or x) and Division ( : or /) These operations are commonly called arithmetic operations.Arithmetic is the oldest and most elementary branch of mathematics. Prime numbers are thus the basic building blocks of all numbers. Video transcript. In the case of C [ x], this fact, together with the fundamental theorem of Algebra, means what you wrote: every p (x) ∈ C [ x] can be written as the product of a non-zero complex number and first degree polynomials. Take $$\pi = 22/7$$ Pls dont spam. of 25152 and 12156 by using the fundamental theorem of Arithmetic 9873444080 (a) 24457576 (b) 25478976 (c) 25478679 (d) 24456567 (Q.49) Find the largest number which divides 245 and 1029 leaving remainder 5 in each case. Converted file can differ from the original. home / study / math / applied mathematics / applied mathematics solutions manuals / Technology Manual / 10th edition / chapter 5.4 / problem 8A. Proving with the use of contradiction p/q = square root of 6. (Q.48) Find the H.C.F and L.C.M. The square roots of unity are 1 and –1. Theorem 2: The perpendicular to a chord, bisects the chord if drawn from the centre of the circle. If is a differentiable function of and if is a differentiable function, then . Fundamental theorem of algebra (complex analysis) Fundamental theorem of arbitrage-free pricing (financial mathematics) Fundamental theorem of arithmetic (number theory) Fundamental theorem of calculus ; Fundamental theorem on homomorphisms (abstract algebra) Fundamental theorems of welfare economics The Fundamental Theorem of Arithmetic An integer greater than 1 whose only positive integer divisors… 2 positive integers a and b, GCD (a,b) is the largest positive… can be expressed as a unique product of primes and their exponents, in only one way. Theorem 6.3.2. For example, 1200 = 2 4 ⋅ 3 ⋅ 5 2 = ⋅ 3 ⋅ = 5 ⋅ … The history of the Fundamental Theorem of Arithmetic is somewhat murky. 1 $\begingroup$ I understand how to prove the Fundamental Theory of Arithmetic, but I do not understand how to further articulate it to the point where it applies for $\mathbb Z[I]$ (the Gaussian integers). There are systems where unique factorization fails to hold. This is called the Fundamental Theorem of Arithmetic. The fundamental theorem of arithmetic states that any integer greater than 1 has a unique prime factorization (a representation of a number as the product of prime factors), excluding the order of the factors. Fundamental principle of counting. In general, by the Fundamental Theorem of Algebra, the number of n-th roots of unity is n, since there are n roots of the n-th degree equation z u – 1 = 0. Remainder Theorem and Factor Theorem. Proof: To prove Quotient Remainder theorem, we have to prove two things: For any integer a … See answer hifsashehzadi123 is waiting for your help. Write the first 5 terms of the sequence whose nth term is ( 3)!! Play media. (By uniqueness of the Fundamental Theorem of Arithmetic). Every such factorization of a given $$n$$ is the same if you put the prime factors in nondecreasing order (uniqueness). "7 divided by 2 equals 3 with a remainder of 1" Each part of the division has names: Which can be rewritten as a sum like this: Polynomials. For example, 75,600 = 2 4 3 3 5 2 7 1 = 21 ⋅ 60 2. The Fundamental Theorem of Arithmetic for $\mathbb Z[i]$ Ask Question Asked 2 days ago. The Basic Idea is that any integer above 1 is either a Prime Number, or can be made by multiplying Prime Numbers together. Free Rational Roots Calculator - find roots of polynomials using the rational roots theorem step-by-step This website uses cookies to ensure you get the best experience. Or another way of thinking about it, there's exactly 2 values for X that will make F of X equal 0. Within abstract algebra, the result is the statement that the ring of integers Zis a unique factorization domain. Fundamental theorem of arithmetic, Fundamental principle of number theory proved by Carl Friedrich Gauss in 1801. Euclid anticipated the result. The course covers several variable calculus, optimization theory and the selected topics drawn from the That course is aimed at teaching students to master comparative statics problems, optimization Fundamental Methods of Mathematical Economics, 3rd edition, McGrow-Hill, 1984. Problem 8A from Chapter 5.4: a. What is the height of the cylinder. So the Assumptions states that : (1) $\sqrt{3}=\frac{a}{b}$ Where a and b are 2 integers Quotient remainder theorem is the fundamental theorem in modular arithmetic. The Fundamental theorem of Arithmetic, states that, “Every natural number except 1 can be factorized as a product of primes and this factorization is unique except for the order in which the prime factors are written.” This theorem is also called the unique factorization theorem. Also, the important theorems for class 10 maths are given here with proofs. The fundamental theorem of arithmetic: For each positive integer n> 1 there is a unique set of primes whose product is n. Which assumption would be a component of a proof by mathematical induction or strong mathematical induction of this theorem? Functions in this section derive their properties from the fundamental theorem of arithmetic, which states that every integer n > 1 can be represented uniquely as a product of prime powers, … (See Gauss ( 1863 , Band II, pp. It may help for you to draw this number line by hand on a sheet of paper first. Play media. Or: how to avoid Polynomial Long Division when finding factors. 8.ОТА начало.ogv 9 min 47 s, 854 × 480; 173.24 MB. . The two legs meet at a 90° angle and the hypotenuse is the longest side of the right triangle and is the side opposite the right angle. Mathway: Scan Photos, Solve Problems (9 Similar Apps, 6 Review Highlights & 480,834 Reviews) vs Cymath - Math Problem Solver (10 Similar Apps, 4 Review Highlights & 40,238 Reviews). Thefundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorizationtheorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. Elements of the theorem can be found in the works of Euclid (c. 330–270 BCE), the Persian Kamal al-Din al-Farisi (1267-1319 CE), and others, but the first time it was clearly stated in its entirety, and proved, was in 1801 by Carl Friedrich Gauss (1777–1855). Within abstract algebra, the result is the statement that the Technology Manual (10th Edition) Edit edition. The file will be sent to your Kindle account. The Fundamental Theorem of Arithmetic is one of the most important results in this chapter. The fundamental theorem of algebra tells us that because this is a second degree polynomial we are going to have exactly 2 roots. Use the Fundamental Theorem of Arithmetic to justify that... Get solutions . The unique factorization is needed to establish much of what comes later. n n a n. 2. This site is using cookies under cookie policy. Other readers will always be interested in your opinion of the books you've read. n n 3. Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a 2 + b 2 = c 2.Although the theorem has long been associated with Greek mathematician-philosopher Pythagoras (c. 570–500/490 bce), it is actually far older. Book 7 deals strictly with elementary number theory: divisibility, prime numbers, Euclid's algorithm for finding the greatest common divisor, least common multiple. Applications of the Fundamental Theorem of Arithmetic are finding the LCM and HCF of positive integers. ОООО If the proposition was false, then no iterative algorithm would produce a counterexample. In this and other related lessons, we will briefly explain basic math operations. You can specify conditions of storing and accessing cookies in your browser. Simplify: ( 2)! Also, the relationship between LCM and HCF is understood in the RD Sharma Solutions Class 10 Exercise 1.4. Factorial n. Permutations and combinations, derivation of formulae and their connections, simple applications. 11. Carl Friedrich Gauss gave in 1798 the first proof in his monograph “Disquisitiones Arithmeticae”. The Fundamental Theorem of Arithmetic | L. A. Kaluzhnin | download | Z-Library. The fundamental theorem of arithmetic is Theorem: Every n∈ N,n>1 has a unique prime factorization. For example, 252 only has one prime factorization: 252 = 2 2 × 3 2 × 7 1 The same thing applies to any algebraically closed field, … function, F: in other words, that dF = f dx. More formally, we can say the following. It may take up to 1-5 minutes before you receive it. If 1 were a prime, then the prime factor decomposition would lose its uniqueness. Exercise 1.2 Class 10 Maths NCERT Solutions were prepared according to … The following are true: Every integer $$n\gt 1$$ has a prime factorization. 2 Addition and Subtraction of Polynomials. Converse of Theorem 1: If two angles subtended at the centre, by two chords are equal, then the chords are of equal length. By the choice of F, dF / dx = f(x).In the parlance of differential forms, this is saying that f(x) dx is the exterior derivative of the 0-form, i.e. It is used to prove Modular Addition, Modular Multiplication and many more principles in modular arithmetic. By … Paul Erdős gave a proof that also relies on the fundamental theorem of arithmetic. thefundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorizationtheorem, states that every integer greater than 1 either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. A Startling Fact about Brainly Mathematics Uncovered Once the previous reference to interpretation was removed from the proofs of these facts, we’ll have a true proof of the Fundamental Theorem. Mathematics College Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainder, all rational roots, all possible roots, and actual roots of the given function. The fundamental theorem of arithmetic is truly important and a building block of number theory. 437–477) and Legendre ( 1808 , p. 394) .) Here is a set of practice problems to accompany the Rational Functions section of the Common Graphs chapter of the notes for Paul Dawkins Algebra course at Lamar University. You can write a book review and share your experiences. The divergence theorem part of the integral: Here div F = y + z + x. Stokes' theorem is a vast generalization of this theorem in the following sense. Viewed 59 times 1. Suppose f is a polynomial function of degree four, and $f\left(x\right)=0$. In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to the order of the factors. It also contains the seeds of the demise of prospects for proving arithmetic is complete and self-consistent because any system rich enough to allow for unique prime factorization is subject to the classical proof by Godel of incompleteness. Answer: 1 question What type of business organization is owned by a single person, has limited life and unlimited liability? Well, we can also divide polynomials. A right triangle consists of two legs and a hypotenuse. The fundamental theorem of calculus and accumulation functions. This is because we could multiply by 1 as many times as we like in the decomposition. This means p belongs to p 1 , p 2 , p 3 , . It simply says that every positive integer can be written uniquely as a product of primes. Mathematics College Use the Fundamental Theorem of Calculus to find the "area under curve" of f (x) = 6 x + 19 between x = 12 and x = 15. All exercise questions, examples and optional exercise questions have been solved with video of each and every question.Topics of each chapter includeChapter 1 Real Numbers- Euclid's Division Lemma, Finding HCF using Euclid' Find a formula for the nth term of the sequence: , 24 10, 6 8, 2 6, 1 4, 1 2 4. Which of the following is an arithmetic sequence? Of particular use in this section is the following. Can two numbers have 15 as their HCF and 175 … Mathematics C Standard Term 2 Lecture 4 Definite Integrals, Areas Under Curves, Fundamental Theorem of Calculus Syllabus Reference: 8-2 A definite integral is a real number found by substituting given values of the variable into the primitive function. corporation partnership sole proprietorship limited liability company - the answers to estudyassistant.com If possible, download the file in its original format. The fundamental theorem of arithmetic is Theorem: Every n∈ N,n>1 has a unique prime factorization. (9 Hours) Chapter 8 Binomial Theorem: History, statement and proof of the binomial theorem for positive integral indices. The fundamental theorem of arithmetic or the unique-prime-factorization theorem. Thank You for A2A, In a layman term, A rational number is that number that can be expressed in p/q form which makes every integer a rational number. Implicit differentiation. Download books for free. For example, 75,600 = 2 4 3 3 5 2 7 1 = 21 ⋅ 60 2. ivyong22 ivyong22 ... Get the Brainly App Download iOS App Definition 1.1 The number p2Nis said to be prime if phas just 2 divisors in N, namely 1 and itself. If A and B are two independent events, prove that A and B' are also independent. Carl Friedrich Gauss gave in 1798 the first proof in his monograph “Disquisitiones Arithmeticae”. Find books Real Numbers Class 10 Maths NCERT Solutions are extremely helpful while doing your homework or while preparing for the exam. Active 2 days ago. (・∀・). From Fundamental theorem of Arithmetic, we know that every composite number can be expressed as product of unique prime numbers. Paul Erdős gave a proof that also relies on the fundamental theorem of arithmetic. Get Free NCERT Solutions for Class 10 Maths Chapter 1 ex 1.2 PDF. So I encourage you to pause this video and try to … Theorem 1: Equal chords of a circle subtend equal angles, at the centre of the circle. Every positive integer has a unique factorization into a square-free number and a square number rs 2. Find the value of b for which the runk of matrix A=and runk is 2, 1=112=223=334=445=556=667=778=8811=?answer is 1 because if 1=11 then 11=1, Describe in detail how you would create a number line with the following points: 1, 3.25, the opposite of 2, and – (–4fraction of one-half). It may takes up to 1-5 minutes before you received it. Click now to get the complete list of theorems in mathematics. Thus 2 j0 but 0 -2. It provides us with a good reason for defining prime numbers so as to exclude 1. Euclid anticipated the result. The fourth roots are ±1, ±i, as noted earlier in the section on absolute value.
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# Case study Essay
Jane wants to setup a photo shop. The cost to rent an office is \$150 per week. The variable cost of making one photo is \$20 and she can sell it for \$50.
1. Jane has to sell _5_ photos per week to break even. (Please only enter an integer and include no units.)
Jane wants to setup a photo shop. The cost to rent an office is \$150 per week. The variable cost of making one photo is \$20 and she can sell it for \$50.
2. If Jane sells 10 units, her profits would be __150__dollars. (Please only enter an integer and include no units.)
Paul wants to choose one of the two investment opportunities over three possible scenarios. Investment 1 will yield a return of \$10,000 in Scenario 1, \$2,000 in Scenario 2, and a negative return of -\$5,000 in Scenario 3. Investment 2 will yield a return of \$6,000 in Scenario 1, \$4,000 in Scenario 2, and zero in Scenario 3. The probability for Scenario 1 is 0.2, for Scenario 2 is 0.3, and for Scenario 3 is 0.5.
3. If you were to choose the investment that maximizes Paul’s Expected Money Value (EMV), then you should choose __________.
A.Investment 1
B.Investment 2
C.Indifferent
Paul wants to choose one of the two investment opportunities over three possible scenarios. Investment 1 will yield a return of \$10,000 in Scenario 1, \$2,000 in Scenario 2, and a negative return of -\$5,000 in Scenario 3. Investment 2 will yield a return of \$6,000 in Scenario 1, \$4,000 in Scenario 2, and zero in Scenario 3. The probability for Scenario 1 is 0.2, for Scenario 2 is 0.3, and for Scenario 3 is 0.5.
4. If Paul is uncertain about the return for Investment 1 in Scenario 1, then this return has to be __21500___ dollars in order to make Paul indifferent between these two investments (i.e. the two investments would have the same EMV.) (Please only enter an integer and include no units.) Sam has a cleaning service. To better allocate his resources, he would like to forecast his weekly orders based on the order number he received in…
# My future employment opportunities Essay
That was the extreme consequence she could have faced and it shouldn’t eave been taken that far. They could have easily told her to take it off or Just tell her to make her profile private so that her students wouldn’t be able to see her profile. She should have gotten her degree because she did everything required to do so. Although she may not get a Job from an employer who sees that picture, she should still get her degree from the school. I think that employers should have the full extent of looking at peoples social networks.
Just because thieve have a couple pictures of them drinking or attending parties does not mean that they will not benefit your many. It only means that some people are more open about their social life and willing to share that with the world. As long as you are able to take care of the work at your Job and help your employers, what you do on your leisure time shouldn’t affect anything. There is nothing currently on my Backbone because I cleaned it all up last year when I started learning how it can affect my future employment opportunities. ) The webmasters of funerals is an interesting example of the flatness of the world. Is this an Invasion of privacy or do next-of-kin have the right to make such a decision? What other significant events in a person’s life might be suitable for webmasters? Identify at least three such events and then do some research to determine If webmasters of those events Is already taking place. I don’t think at all that this Is invasion of privacy if the family says that is okay for them to do so.
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Saying comments like that on social networks when you are followed by hundreds of thousands of people reflects you and any company you work for. Even if it’s Just your opinion, like it was in this case, he still reflected the company and people might think differently of his company. It may cause people to think that is what the company thinks of the situation. I do not however think its grounds for dismissal from a school though. 4) Very few people would question the service and commitment of military personnel to our country.
The two sergeants who created the Youth video openly criticized delta Airlines for its charging of baggage fees to military personnel. Is the open form of criticism of businesses and their practices acceptable? Will it help businesses be more accountable to customers? Yes it’s acceptable because people should be able to see how customers are criticizing a company. If they are not allowed to see some of the potential problems they could end up facing how would hey really know if they wanted to purchase from them. Plus with the bad criticism good will come.
Yes it will help businesses more accountable to their customers because they will have to fix the problems and know what they’re doing well so that they can continue to do so. 5) The use of Backbone (or any other social networking site) can truly make a person’s life transparent, available for the whole world to see. Should there be legislating regulation the openness of your life on the internet? Can we expect society somehow to regulate this without any laws? I don’t think that there would be any type of regulation on the internet because people should be able to share with everybody whatever they want to.
The more open somebody is, the more they will be willing to share. So who are we to put limits on what somebody can share to their friends, family, and the world. Not really because the only way that society will regulate what is put on the internet is if society was to be on the same page, which I feel we are not. Some people think there should be limits while others do not. Everybody is brought up on their own communities norms and values, which will determine what they think is socially fitting.
# Case study Essay
Case Problem 1: Young Professional Magazine
1.Descriptive Statistics for the quantitative variables follow:
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Variable N Mean SE Mean StDev Minimum Maximum Skewness Age 410 30.112 0.199 4.024 19.000 42.000 -0.03
Investments 410 28538 781 15811 0.000 133400 1.71
Transactions 410 5.973 0.153 3.101 0.000 21.000 1.21
Household Income 410 74460 1720 34818 16200 322500 2.01
Descriptive statistics for the qualitative variables follow:
GenderMale: 229Proportion male:.5585
Plan R.E. purchaseYes:181Proportion yes:.4415
Have ChildrenYes:219Proportion yes:.5341
2.95% Confidence Intervals
Variable N Mean StDev SE Mean 95% CI Age 410 30.1122 4.0240 0.1987 (29.7215, 30.5029)
Variable N Mean StDev SE Mean 95% CI Household Income 410 74459.5 34818.2 1719.5 (71079.3, 77839.8)
We can conclude with 95% confidence that the mean age of subscribers to
Young Professional is between 29.72 and 30.50 years of age. And, we can conclude with 95% confidence that the mean household income of subscribers is between \$71,079 and \$77,840.
3.95% Confidence Intervals
Variable X N Sample p 95% CI
Broadband Access 256 410 0.624390 (0.577514, 0.671266)
Variable X N Sample p 95% CI
Have Children? 219 410 0.534146 (0.485861, 0.582431)
We can conclude with 95% confidence that the proportion of subscribers with broadband access is between 57.75% and 67.13% and that the proportion of subscribers with children is between 48.59% and 58.24%.
4. Young Professional should be a good advertising outlet for online brokers. We see that most of the subscribers have financial investments exclusive of their home (the mean amount is \$28,538) and some of them have a substantial amount of investments. (Several have over \$100,000 of investments). Another factor to consider is the number of stock, bond, and mutual fund transactions. The mean number is approximately 6 per year and several subscribers make significantly more transactions than that. Finally a large proportion of subscribers have broadband access (the sample proportion is .6244) and this makes them more likely to do business with an online broker. 5.The survey results allow us to estimate that the mean age of subscribers is 30.12 years and that 53.41% of subscribers have children. Given the age of subscribers, it is reasonable to assume that their children are young. Thus, we conclude that subscribers to Young Professional would be a good target market for companies selling educational software and computer games for young children.
6.A variety of answers are possible here. But, from the survey results, it seems clear that articles about investing would have appeal to many readers. Articles about real estate and architecture would probably appeal to those subscribers planning to make a real estate purchase. Technology related articles would probably appeal to readers as well as an occasional article on parenting and child care.
Case Problem 2: Gulf Real Estate Properties
The variables are as follows:
GV ListThe list price of a Gulf View condominium
GV SaleThe sale price of the Gulf View condominium
GV DaysThe number of days to sell the Gulf View condominium
NGV ListThe list price of a No Gulf View condominium
NGV SaleThe sale price of the No Gulf View condominium
NGV DaysThe number of days to sell the No Gulf View condominium
1/2.The results obtained using Minitab are as follows:
Descriptive Statistics: GV List, GV Sale, GV Days, NGV List, NGV Sale, NGV Days
Variable N Mean Median TrMean StDev SE Mean GV List 40 474.0 437.0 462.0 197.3 31.2 GV Sale 40 454.2 417.5 441.2 192.5 30.4 GV Days 40 106.00 96.00 102.64 52.22 8.26 NGV List 18 212.8 212.5 210.0 48.9 11.5 NGV Sale 18 203.2 203.5 201.8 43.9 10.3 NGV Days 18 135.0 126.0 127.8 76.3 18.0
Variable Minimum Maximum Q1 Q3
GV List 169.9 975.0 332.7 551.9
GV Sale 165.0 975.0 314.3 530.6
GV Days 28.00 282.00 71.25 138.75
NGV List 148.0 322.0 174.9 241.0
NGV Sale 135.5 292.5 172.4 230.0
NGV Days 48.0 338.0 62.5 154.8
3.Gulf View condominiums are the more expensive as usually anticipated. A Gulf View condominium lists for a mean price of \$474,000 and a median price of \$437,000. A No Gulf View condominium lists for a mean price of \$212,800 and a median price of \$212,500. The Gulf View condominiums are over twice as expensive as the No Gulf View condominiums.
The standard deviation of the list price for Gulf View condominiums is \$197,300 and the standard deviation of the list price for No Gulf View condominiums is \$48,900. Thus, the Gulf View condominiums have a greater variation in list price.
The most expensive list-price is for a Gulf View condominium at \$975,000 while the least expensive list-price is a No Gulf View condominium at \$148,000.
A Box Plot shows that the No Gulf View condominiums do not have any outliers. A similar box plot shows the highest four Gulf View list prices are outliers: \$885,000, \$895,000 and two at \$975,000. These show 4/40 or 10% of the Gulf View condominiums have an unusually high list price.
The mean number of days to sell a condominium is slightly better for Gulf View condominiums than for No Gulf View condominiums (106 days vs. 135 days). On average, it appears to take a little over three months to sell a Gulf View condominium and a little over four months to sell a No Gulf View condominium. The quickest sale was a Gulf View condominium that sold in less than a month (28 days). The slowest sale was for a No Gulf View condominium that took almost a year to sell (338 days).
Gulf View condominiums with a mean list price of \$474,000 and a mean sale price of \$454,200 sell on average \$474,000 – \$454,200 = \$19,800 (4.2%) below list price. No Gulf View condominiums with a mean list price of \$212,800 and a mean sale price of \$203,200 appears to sell on average \$212,800 – \$203,200 = \$9,600 (4.5%) below list price. There is only a small percentage difference in list and sale price for the two types of condominiums.
In summary, Gulf View condos are substantially more expensive and tend to sell slightly faster. The discount off the list price is roughly 4.0 to 4.5% for both Gulf View and No Gulf View condos.
4.Using the Minitab 1-Sample t procedure
Variable N Mean StDev SE Mean 95.0% CI GV Sale 40 454.2 192.5 30.4 ( 392.7, 515.8) GV Days 40 106.00 52.22 8.26 ( 89.30, 122.70)
95% Confidence Intervals:
Gulf View Mean Sales Price\$392,700 to \$515,800
Gulf View Mean Days to Sell 89 to 123
5.Using the Minitab 1-Sample t procedure
Variable N Mean StDev SE Mean 95.0% CI NGV Sale 18 203.2 43.9 10.3 ( 181.4, 225.0) NGV Days 18 135.0 76.3 18.0 ( 97.1, 172.9)
95% Confidence Intervals:
No Gulf View Mean Sales Price\$181,400 to \$225,000
No Gulf View Mean Days to Sell 97 to 173
6.Gulf View condominiums
With n = 40, the margin of error is \$59,600. To reduce the margin of error to \$40,000 the recommended sample size.
No Gulf View condominiums
With n = 18, the margin of error is \$21,800. To reduce the margin of error to \$15,000 the recommended sample size.
7.From part 3, a Gulf View condominium sells on average 4.2% below its list price.
The estimated selling price is \$589,000 x .958 = \$564,262. Using the mean days, it is estimated to sell in about 106 days.
From part 3, a No Gulf View condominium sells on average 4.5% below its list price.
The estimated selling price is \$285,000 x .955 = \$272,175. Using the mean days, it is estimated to sell in about 135 days.
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# Search by Topic
#### Resources tagged with Visualising similar to Simple Train Journeys:
Filter by: Content type:
Stage:
Challenge level:
### There are 189 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Visualising
### Map Folding
##### Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
### Counting Cards
##### Stage: 2 Challenge Level:
A magician took a suit of thirteen cards and held them in his hand face down. Every card he revealed had the same value as the one he had just finished spelling. How did this work?
### Shunting Puzzle
##### Stage: 2 Challenge Level:
Can you shunt the trucks so that the Cattle truck and the Sheep truck change places and the Engine is back on the main line?
### Celtic Knot
##### Stage: 2 Challenge Level:
Building up a simple Celtic knot. Try the interactivity or download the cards or have a go on squared paper.
### Putting Two and Two Together
##### Stage: 2 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
##### 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.
### Counters
##### Stage: 2 Challenge Level:
Hover your mouse over the counters to see which ones will be removed. Click to remover them. The winner is the last one to remove a counter. How you can make sure you win?
### Tangram Tangle
##### Stage: 1 Challenge Level:
If you split the square into these two pieces, it is possible to fit the pieces together again to make a new shape. How many new shapes can you make?
### Display Boards
##### Stage: 2 Challenge Level:
Design an arrangement of display boards in the school hall which fits the requirements of different people.
### 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?
### Single Track
##### Stage: 2 Challenge Level:
What is the best way to shunt these carriages so that each train can continue its journey?
##### Stage: 2 Challenge Level:
How many DIFFERENT quadrilaterals can be made by joining the dots on the 8-point circle?
### Waiting for Blast Off
##### Stage: 2 Challenge Level:
10 space travellers are waiting to board their spaceships. There are two rows of seats in the waiting room. Using the rules, where are they all sitting? Can you find all the possible ways?
### Paw Prints
##### Stage: 2 Challenge Level:
A dog is looking for a good place to bury his bone. Can you work out where he started and ended in each case? What possible routes could he have taken?
### Dodecamagic
##### Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
### Four Triangles Puzzle
##### Stage: 1 and 2 Challenge Level:
Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?
### Cover the Camel
##### Stage: 1 Challenge Level:
Can you cover the camel with these pieces?
### Routes 1 and 5
##### Stage: 1 Challenge Level:
Find your way through the grid starting at 2 and following these operations. What number do you end on?
### Open Boxes
##### Stage: 2 Challenge Level:
Can you work out how many cubes were used to make this open box? What size of open box could you make if you had 112 cubes?
### Knight's Swap
##### Stage: 2 Challenge Level:
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### Tessellate the Triominoes
##### Stage: 1 Challenge Level:
What happens when you try and fit the triomino pieces into these two grids?
### Painting Possibilities
##### Stage: 2 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### 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?
### Nine-pin Triangles
##### Stage: 2 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
### Domino Numbers
##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
### Tricky Triangles
##### Stage: 1 Challenge Level:
Use the three triangles to fill these outline shapes. Perhaps you can create some of your own shapes for a friend to fill?
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Tetrafit
##### Stage: 2 Challenge Level:
A tetromino is made up of four squares joined edge to edge. Can this tetromino, together with 15 copies of itself, be used to cover an eight by eight chessboard?
### Square Corners
##### Stage: 2 Challenge Level:
What is the greatest number of counters you can place on the grid below without four of them lying at the corners of a square?
### Coded Hundred Square
##### Stage: 2 Challenge Level:
This 100 square jigsaw is written in code. It starts with 1 and ends with 100. Can you build it up?
### 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?
### One Big Triangle
##### Stage: 1 Challenge Level:
Make one big triangle so the numbers that touch on the small triangles add to 10. You could use the interactivity to help you.
### Teddy Bear Line-up
##### Stage: 1 Challenge Level:
What is the least number of moves you can take to rearrange the bears so that no bear is next to a bear of the same colour?
### 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?
### A City of Towers
##### Stage: 1 Challenge Level:
In this town, houses are built with one room for each person. There are some families of seven people living in the town. In how many different ways can they build their houses?
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### Put Yourself in a Box
##### Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### World of Tan 18 - Soup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?
### World of Tan 17 - Weather
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the watering can and man in a boat?
### Right or Left?
##### Stage: 2 Challenge Level:
Which of these dice are right-handed and which are left-handed?
### Cubes Within Cubes
##### Stage: 2 and 3 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
### Twice as Big?
##### Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
##### Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
### World of Tan 20 - Fractions
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the chairs?
### World of Tan 19 - Working Men
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this shape. How would you describe it?
### 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?
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### World of Tan 21 - Almost There Now
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the lobster, yacht and cyclist?
### Endless Noughts and Crosses
##### Stage: 2 Challenge Level:
An extension of noughts and crosses in which the grid is enlarged and the length of the winning line can to altered to 3, 4 or 5.
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## UnkleRhaukus 2 years ago Show that the function satisfies the DE
1. UnkleRhaukus
2. UnkleRhaukus
\begin{align*} I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\mathrm du\\ \frac{\mathrm dI(a)}{\mathrm da}&=\int\limits_0^\infty \frac{\partial}{\partial a}e^{-u^2}\cos(au)\mathrm du\\ &=-a\int\limits_0^\infty e^{-u^2}\sin(au)\mathrm du\\ \\ \end{align*}
3. UnkleRhaukus
now what should i do ?
firstly $$\frac{dI}{da}=\int\limits_{0}^{\infty}-ue^{-u^2}\sin(au)du$$ when we consider the left side of the differential equation, \begin{align*}2\frac{dI}{da}+aI &= \int\limits_{0}^{\infty}(-2u)e^{-u^2}\sin(au)du+\int \limits_{0}^{\infty}e^{-u^2}a\cos(au)du\\ &=\int\limits_0^\infty\frac{de^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\sin(au)}{du}du\\ &=\int \limits_0^\infty \frac{d\,e^{-u^2}\sin(au)}{du}\,du\\ &=\int\limits_0^\infty d\,e^{-u^2}\sin(au)\\ &=\left[e^{-u^2}\sin(au)\right]_0^{\infty}=0-0=0 \end{align*}
5. UnkleRhaukus
some kinda reverse integration by parts?
6. UnkleRhaukus
im not quite sure how you got your third line
7. UnkleRhaukus
can you explain how you got from \begin{align*} \int\limits_{0}^{\infty}(-2u)e^{-u^2}\sin(au)du+\int \limits_{0}^{\infty}e^{-u^2}a\cos(au)du\\ \text {to}\\ =\int\limits_0^\infty\frac{de^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\sin(au)}{du}du\end{align*} @BAdhi
we know that $$\frac{d\,e^{-u^2}}{du}=(-2u)\,e^{-u^2}$$ and $$\frac{d\,\sin(au)}{du}=a\cos(au)$$ so, \begin{align*}\int\limits_0^\infty (-2u)\,e^{-u^2}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}a\cos(au)\,du&=\int\limits_0^\infty \frac{d\,e^{-u^2}}{du}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}\frac{d\,\sin(au)}{du}\,du\\ &=\int\limits_0^\infty \underbrace{\frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}}_{(1)}\,du\end{align*} (1) can be reduced from the differentiation of a multiplication, property (i.e.) $$\frac{d\,uv}{dx}=u\frac{du}{dx}+v\frac{dv}{dx}$$ $$\frac{d\,\left(e^{-u^2}\sin(au)\right)}{du}=\frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}$$ (1) can be replaced with this, \begin{align*}\int\limits_0^\infty (-2u)\,e^{-u^2}\sin(au)\,du+\int\limits_0^\infty e^{-u^2}a\cos(au)\,du&=\int\limits_0^\infty \frac{d\,e^{-u^2}}{du}\sin(au)+ e^{-u^2}\frac{d\,\sin(au)}{du}\,du \\ &=\int\limits_0^\infty\frac{d\,\left(e^{-u^2}\sin(au)\right)}{du}du\end{align*} Is it clear enough for you?
9. UnkleRhaukus
yes
10. UnkleRhaukus
\begin{align*} 2\frac{\mathrm dI(a)}{\mathrm da}+aI(a) &=\int\limits_0^\infty -2ue^{-u^2}\sin(au)\mathrm du+\int\limits_0^\infty ae^{-u^2}\cos(au)\mathrm du\\ &=\int\limits_0^\infty\frac{\mathrm d }{\mathrm d u}\left(e^{-u^2}\sin(au)\right)\mathrm du\qquad\text{(using the product rule)}\\ &=\left.e^{-u^2}\sin(au)\right|_0^\infty\\ &=0 \end{align*}
11. UnkleRhaukus
\begin{align*} I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du\\ I(0)&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\ \frac{\sqrt \pi}2&=\int\limits_0^\infty e^{-u^2}\,\mathrm du\\ \text{let }u=v^{1/2}\\ \mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\ \frac{\sqrt \pi}2&=\int\limits_0^\infty e^{-v}\frac{v^{-1/2}\,\mathrm dv}2\\ {\sqrt \pi}&=\int\limits_0^\infty {v^{-1/2}e^{-v}\,\mathrm dv}\\ {\sqrt \pi}&=\Gamma(\tfrac12)\\ \end{align*}
12. UnkleRhaukus
im not sure what they mean by an expression for $$I(a)$$
13. Sepeario
14. UnkleRhaukus
@Sepeario yes i have tried differentiating $$I(a)$$
15. rahul91
@UnkleRhaukus integrate the expression using integration by parts technique.then subst for I(0).
16. UnkleRhaukus
\begin{align*} I(a)&=\int\limits_0^\infty e^{-u^2}\cos(au)\,\mathrm du\\ \text{let }u=v^{1/2}\\ \mathrm du=\frac{v^{-1/2}}2\,\mathrm dv\\ I(a)&=\int\limits_0^\infty e^{-v}\cos(a\sqrt v)\frac{v^{-1/2}\,\mathrm dv}2\\ \end{align*} how do i integrate by parts, there are three terms? do i break up the cos into e^... bits?
17. sirm3d
$2\frac{ dI }{ da }+aI=0\\2 \frac{ dI }{ I }+a\space da=0\\2lnI+\frac{ a^2 }{ 2 }=C$when $$\displaystyle a=0,\;I=\frac{\sqrt{\pi}}{2}$$$2\ln \frac{ \sqrt{\pi} }{ 2 }+\frac{ 0^2 }{ 2 }=C$$2\ln I+\frac{a^2}{2}=2\ln\frac{\sqrt{\pi}}{2}$$\vdots\\\ln \left[\frac{ 4 }{ \pi }I^2(a)\right]=-\frac{ a^2 }{ 2 }\\I^2(a)=\frac{\pi}{4}\exp(-a^2/2)$
18. UnkleRhaukus
Thankyou @sirm3d & @BAdhi \begin{align*} 2\frac{\mathrm dI(a)}{\mathrm da}+aI(a)&=0\\ 2\frac{\mathrm dI(a)}{I(a)}+a\,\mathrm da&=0\\ 2\int\frac{\mathrm dI(a)}{I(a)}+\int a\,\mathrm da&=0\\ 2\ln |I(a)|+\frac{a^2}2&=c \end{align*} \begin{align*} 2\ln |I(0)|&=c\\ 2\ln\frac{\sqrt\pi}2&=c\\ \ln\frac{\pi}4&=c\end{align*} \begin{align*} 2\ln |I(a)|+\frac{a^2}2&=\ln\frac{\pi}4\\ \ln |I(a)|&=\ln{\frac{\sqrt\pi}2}-\frac{a^2}4\\ I(a)&={\frac{\sqrt\pi}2}e^{-\frac{a^2}4} \end{align*}
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# RD Sharma Solutions For Class 12 Maths Exercise 7.1 Chapter 7 Adjoint and Inverse of a Matrix
RD Sharma Solutions for Class 12 Maths Exercise 7.1 Chapter 7 Adjoint and Inverse of a Matrix is provided here. The solutions for the exercise wise answers are prepared by the experts at BYJU’S in the best possible way which are easily understandable by students.
The PDF of RD Sharma Solutions for Class 12, Exercise 7.1 of Chapter 7 Adjoint and Inverse of a Matrix can be downloaded from the given links. This exercise consists of two levels according to the increasing order of difficulties. Let us have a look at the important topics covered in this exercise.
• Definition and meaning of adjoint of a square matrix
• The inverse of a matrix
• Some useful results on invertible matrices
• Determining the adjoint and inverse of a matrix
• Determining the inverse of a matrix when it satisfies the matrix equation
• Finding the inverse of a matrix by using the definition of inverse
• Finding a non – singular matrix when adjoint is given
## RD Sharma Solutions For Class 12 Maths Chapter 7 Adjoint and Inverse of a Matrix Exercise 7.1:-
### Access another exercise of RD Sharma Solutions For Class 12 Maths Chapter 7 – Adjoint and Inverse of a Matrix
Exercise 7.2 Solutions
### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 7 – Adjoint and Inverse of a Matrix Exercise 7.1
1. Find the adjoint of each of the following matrices:
Verify that (adj A) A = |A| I = A (adj A) for the above matrices.
Solution:
(i) Let
A =
Cofactors of A are
C11Â = 4
C12 = – 2
C21 = – 5
C22 = – 3
(ii) Let
A =
Therefore cofactors of A are
C11Â = d
C12 = – c
C21 = – b
C22Â = a
(iii) Let
A =
Therefore cofactors of A are
C11 = cos α
C12 = – sin α
C21 = – sin α
C22 = cos α
(iv) Let
A =
Therefore cofactors of A are
C11Â = 1
C12 = tan α/2
C21 = – tan α/2
C22Â = 1
2. Compute the adjoint of each of the following matrices.
Solution:
(i) Let
A =
Therefore cofactors of A are
C11 = – 3
C21Â = 2
C31Â = 2
C12Â = 2
C22 = – 3
C23Â = 2
C13Â = 2
C23Â = 2
C33 = – 3
(ii) Let
A =
Cofactors of A
C11Â = 2
C21Â = 3
C31 = – 13
C12 = – 3
C22Â = 6
C32Â = 9
C13Â = 5
C23 = – 3
C33 = – 1
(iii) Let
A =
Therefore cofactors of A
C11 = – 22
C21Â = 11
C31 = – 11
C12Â = 4
C22 = – 2
C32Â = 2
C13Â = 16
C23 = – 8
C33Â = 8
(iv) Let
A =
Therefore cofactors of A
C11Â = 3
C21 = – 1
C31Â = 1
C12 = – 15
C22Â = 7
C32 = – 5
C13Â = 4
C23 = – 2
C33Â = 2
Solution:
Given
A =
Therefore cofactors of A
C11Â = 30
C21Â = 12
C31 = – 3
C12 = – 20
C22 = – 8
C32Â = 2
C13 = – 50
C23 = – 20
C33 = 5
Solution:
Given
A =
Cofactors of A
C11 = – 4
C21 = – 3
C31 = – 3
C12Â = 1
C22Â = 0
C32Â = 1
C13Â = 4
C23Â = 4
C33Â = 3
Solution:
Given
A =
Cofactors of A are
C11 = – 3
C21Â = 6
C31Â = 6
C12 = – 6
C22Â = 3
C32 = – 6
C13 = – 6
C23 = – 6
C33Â = 3
Solution:
Given
A =
Cofactors of A are
C11Â = 9
C21Â = 19
C31 = – 4
C12Â = 4
C22Â = 14
C32Â = 1
C13Â = 8
C23Â = 3
C33Â = 2
7. Find the inverse of each of the following matrices:
Solution:
(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
Now, |A| = cos θ (cos θ) + sin θ (sin θ)
= 1
Hence, A – 1 exists.
Cofactors of A are
C11 = cos θ
C12 = sin θ
C21 = – sin θ
C22 = cos θ
(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
Now, |A| = – 1 ≠0
Hence, A – 1 exists.
Cofactors of A are
C11Â =Â 0
C12 = – 1
C21 = – 1
C22Â = 0
(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
Now, |A| = 2 + 15 = 17 ≠0
Hence, A – 1 exists.
Cofactors of A are
C11Â = 1
C12Â = 3
C21 = – 5
C22Â = 2
8. Find the inverse of each of the following matrices.
Solution:
(i) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 1(6 – 1) – 2(4 – 3) + 3(2 – 9)
= 5 – 2 – 21
= – 18≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 5
C21 = – 1
C31 = – 7
C12 = – 1
C22 = – 7
C32Â = 5
C13 = – 7
C23Â = 5
C33 = – 1
(ii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 1 (1 + 3) – 2 (– 1 + 2) + 5 (3 + 2)
= 4 – 2 + 25
= 27≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 4
C21Â = 17
C31Â = 3
C12 = – 1
C22 = – 11
C32Â = 6
C13Â = 5
C23Â = 1
C33 = – 3
(iii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 2(4 – 1) + 1(– 2 + 1) + 1(1 – 2)
= 6 – 2
= – 4≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 3
C21Â = 1
C31 = – 1
C12Â = + 1
C22Â = 3
C32Â = 1
C13 = – 1
C23Â = 1
C33Â = 3
(iv) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 2(3 – 0) – 0 – 1(5)
= 6 – 5
= 1≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 3
C21 = – 1
C31Â = 1
C12 = – 15
C22Â = 6
C32 = – 5
C13Â = 5
C23 = – 2
C33Â = 2
(v) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 0 – 1 (16 – 12) – 1 (– 12 + 9)
= – 4 + 3
= – 1≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 0
C21 = – 1
C31Â = 1
C12 = – 4
C22Â = 3
C32 = – 4
C13 = – 3
C23Â = 3
C33 = – 4
(vi) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
= 0 – 0 – 1(– 12 + 8)
= 4≠0
Hence, A – 1 exists
Cofactors of A are
C11 = – 8
C21Â = 4
C31Â = 4
C12Â = 11
C22 = – 2
C32 = – 3
C13 = – 4
C23Â = 0
C33Â = 0
(vii) The criteria of existence of inverse matrix is the determinant of a given matrix should not equal to zero.
|A| =
 – 0 + 0
= – (cos2 α – sin2 α)
= – 1≠0
Hence, A – 1 exists
Cofactors of A are
C11 = – 1
C21Â = 0
C31Â = 0
C12Â = 0
C22 = – cos α
C32 = – sin α
C13Â = 0
C23 = – sin α
C33 = cos α
9. Find the inverse of each of the following matrices and verify that A-1A = I3.
Solution:
(i) We have
|A| =
= 1(16 – 9) – 3(4 – 3) + 3(3 – 4)
= 7 – 3 – 3
= 1≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 7
C21 = – 3
C31 = – 3
C12 = – 1
C22Â = 1
C32Â = 0
C13 = – 1
C23Â = 0
C33Â = 1
(ii) We have
|A| =
= 2(8 – 7) – 3(6 – 3) + 1(21 – 12)
= 2 – 9 + 9
= 2≠0
Hence, A – 1 exists
Cofactors of A are
C11Â = 1
C21Â = 1
C31 = – 1
C12 = – 3
C22Â = 1
C32Â = 1
C13Â = 9
C23 = – 5
C33 = – 1
10. For the following pair of matrices verify that (AB)-1 = B-1A-1.
Solution:
(i) Given
Hence, (AB)-1 = B-1A-1
(ii) Given
Hence, (AB)-1 = B-1A-1
Solution:
Given
Solution:
Given
Solution:
Given
Solution:
Solution:
Given
A =
 and B – 1 =
Here, (AB) – 1 = B – 1 A – 1
|A| = – 5 + 4 = – 1
Cofactors of A are
C11 = – 1
C21Â = 8
C31 = – 12
C12Â = 0
C22Â = 1
C32 = – 2
C13Â = 1
C23 = – 10
C33Â = 15
(i) [F (α)]-1 = F (-α)
(ii) [G (β)]-1 = G (-β)
(iii) [F (α) G (β)]-1 = G (-β) F (-α)
Solution:
(i) Given
F (α) =
|F (α)| = cos2 α + sin2 α = 1≠0
Cofactors of A are
C11 = cos α
C21 = sin α
C31Â = 0
C12 = – sin α
C22 = cos α
C32Â = 0
C13Â = 0
C23Â = 0
C33Â = 1
(ii) We have
|G (β)| = cos2 β + sin2 β = 1
Cofactors of A are
C11 = cos β
C21Â = 0
C31 = -sin β
C12Â = 0
C22Â = 1
C32Â = 0
C13 = sin β
C23Â = 0
C33 = cos β
(iii) Now we have to show that
[F (α) G (β)] – 1 = G (– β) F (– α)
[G (β)] – 1 = G (– β)
[F (α)] – 1 = F (– α)
And LHS = [F (α) G (β)] – 1
= [G (β)] – 1 [F (α)] – 1
= G (– β) F (– α)
Hence = RHS
Solution:
Consider,
Solution:
Given
Solution:
Given
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# Integration by Substitution and Separable Differential Equations
Contributed by:
The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.
1. 6.2
Integration by
Substitution &
Separable
Differential
Equations
Greg Kelly
M.L.King Jr. Birthplace, Atlanta, GA Hanford High School
Photo by Vickie Kelly, 2002 Richland, Washington
2. The chain rule allows us to differentiate a wide variety
of functions, but we are able to find antiderivatives for
only a limited range of functions. We can sometimes
use substitution to rewrite functions in a form that we
can integrate.
3. Example 1:
5
x 2 dx Let u x 2
du dx
5
u du The variable of integration
must match the variable in
1 6 the expression.
u C
6
6 Don’t forget to substitute the value
x 2 C
for u back into the problem!
6
4. (Exploration 1 in the book) One of the clues that we look for is
2 if we can find a function and its
1 x 2 x dx derivative in the integrand.
2
The derivative of 1 x is 2 x dx .
1
2
u 2
du Let u 1 x
3 du 2 x dx
2
u C
2
3
Note that this only worked because
3 of the 2x in the original.
2
3
1 x 2 2
C Many integrals can not be done by
substitution.
5. Example 2:
4 x 1 dx Let u 4 x 1
du 4 dx
1
1
Solve for dx.
u 2
du
4
1
du dx
3
4
2 1
u C
2
3 4
3
1
u C 2
6
1 3
4 x 1 2 C
6
6. Example 3:
cos 7 x 5 dx Let u 7 x 5
du 7 dx
1
cos u 7 du 1
du dx
7
1
sin u C
7
1
sin 7 x 5 C
7
7. Example: (Not in book)
x 2
sin x 3
dx Let u x 3
du 3 x 2 dx
1
sin u du 1
3 2
du x dx
3
1 2
We solve for x dx
cos u C
3 because we can find it
in the integrand.
1 3
cos x C
3
8. Example 7:
4
x cos x dx
sin
4
sin x cos x dx Let u sin x
4
du cos x dx
u du
1 5
u C
5
1 5
sin x C
5
9. Example 8:
The technique is a little different
for definite integrals.
2
tan
4
x sec x dx
0
new limit Let u tan x
1
u du
0
du sec 2 x dx
new limit We can find
1
u 0 tan 0 0 new limits,
1 2 and then we
u don’t have
2 0 u tan 1 to substitute
4 4
back.
1
2
We could have substituted back and
used the original limits.
10. Example 8:
Using the original limits:
2
tan
4
x sec x dx
0
Let u tan x
u du
0
4
du sec 2 x dx
Leave the
u du limits out until
you substitute Wrong!
1 1
2
tan tan 0
2
back.
The
2 limits
4 don’t
2 match!
1 2
u
2 This is
1 usually
1 1 2 1 2
tan x
2 4 1 0 more work
2 2 2 than finding
2 0 new limits
11. Example: (Exploration 2 in the book)
1
3x
2 3
x 1 dx Let u x 3 1 u 1 0
1
2
du 3x dx u 1 2
1
2
u
0
2
du
3 2
2 Don’t forget to use the new limits.
u 2
3 0
2 3
2 4 2
2 2 2 2
3 3 3
12. Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
g x h y h y 0
dx
Example:
dy Multiply both sides by dx and divide
2 xy 2
dx both sides by y2 to separate the
variables. (Assume y2 is never zero.)
dy
2
2 x dx
y
y 2 dy 2 x dx
13. Separable Differential Equations
A separable differential equation can be expressed as
the product of a function of x and a function of y.
dy
g x h y h y 0
dx
Example:
2
dy
dy 2 x dx
y
2 xy 2 1 2 Combined
dx y C1 x C2 constants of
dy 1 integration
2
2 x dx x 2 C
y y
1 1
y 2 dy 2 x dx 2 y y 2
x C x C
14. Example 9:
dy x2
2 x 1 y e
2
Separable differential equation
dx
1 x2
2
dy 2 x e dx
1 y
1 x2 u x 2
1 y 2 dy 2 x e dx du 2 x dx
1 u
1 y 2 dy du
e
tan 1 y C1 eu C2
1 x2
tan y C1 e C2
1 x2
tan y e C Combined constants of integration
15. Example 9:
dy x2
2 x 1 y e
2
dx
1 x2
tan y e C We now have y as an implicit
function of x.
x2
tan tan y tan e C We can find y as an explicit function
1
of x by taking the tangent of both
sides.
y tan e C x2
Notice that we can not factor out the constant C, because
the distributive property does not work with tangent.
16. In another generation or so, we might be able to use
the calculator to find all integrals.
Until then, remember that half the AP exam and half the
nation’s college professors do not allow calculators.
You must practice finding integrals by hand until you are
good at it!
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# 7.1 Coordinate geometry, equation of tangent, transformations (Page 2/2)
Page 2 / 2
${m}_{f}=\frac{{y}_{1}-{y}_{0}}{{x}_{1}-{x}_{0}}$
The tangent (line $g$ ) is perpendicular to this line. Therefore,
${m}_{f}×{m}_{g}=-1$
So,
${m}_{g}=-\frac{1}{{m}_{f}}$
Now, we know that the tangent passes through $\left({x}_{1},{y}_{1}\right)$ so the equation is given by:
$\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-{y}_{1}& =& -\frac{1}{{m}_{f}}\left(x-{x}_{1}\right)\hfill \\ \hfill y-{y}_{1}& =& -\frac{1}{\frac{{y}_{1}-{y}_{0}}{{x}_{1}-{x}_{0}}}\left(x-{x}_{1}\right)\hfill \\ \hfill y-{y}_{1}& =& -\frac{{x}_{1}-{x}_{0}}{{y}_{1}-{y}_{0}}\left(x-{x}_{1}\right)\hfill \end{array}$
For example, find the equation of the tangent to the circle at point $\left(1,1\right)$ . The centre of the circle is at $\left(0,0\right)$ . The equation of the circle is ${x}^{2}+{y}^{2}=2$ .
Use
$y-{y}_{1}=-\frac{{x}_{1}-{x}_{0}}{{y}_{1}-{y}_{0}}\left(x-{x}_{1}\right)$
with $\left({x}_{0},{y}_{0}\right)=\left(0,0\right)$ and $\left({x}_{1},{y}_{1}\right)=\left(1,1\right)$ .
$\begin{array}{ccc}\hfill y-{y}_{1}& =& -\frac{{x}_{1}-{x}_{0}}{{y}_{1}-{y}_{0}}\left(x-{x}_{1}\right)\hfill \\ \hfill y-1& =& -\frac{1-0}{1-0}\left(x-1\right)\hfill \\ \hfill y-1& =& -\frac{1}{1}\left(x-1\right)\hfill \\ \hfill y& =& -\left(x-1\right)+1\hfill \\ \hfill y& =& -x+1+1\hfill \\ \hfill y& =& -x+2\hfill \end{array}$
## Co-ordinate geometry
1. Find the equation of the cicle:
1. with centre $\left(0;5\right)$ and radius 5
2. with centre $\left(2;0\right)$ and radius 4
3. with centre $\left(5;7\right)$ and radius 18
4. with centre $\left(-2;0\right)$ and radius 6
5. with centre $\left(-5;-3\right)$ and radius $\sqrt{3}$
1. Find the equation of the circle with centre $\left(2;1\right)$ which passes through $\left(4;1\right)$ .
2. Where does it cut the line $y=x+1$ ?
3. Draw a sketch to illustrate your answers.
1. Find the equation of the circle with center $\left(-3;-2\right)$ which passes through $\left(1;-4\right)$ .
2. Find the equation of the circle with center $\left(3;1\right)$ which passes through $\left(2;5\right)$ .
3. Find the point where these two circles cut each other.
2. Find the center and radius of the following circles:
1. ${\left(x-9\right)}^{2}+{\left(y-6\right)}^{2}=36$
2. ${\left(x-2\right)}^{2}+{\left(y-9\right)}^{2}=1$
3. ${\left(x+5\right)}^{2}+{\left(y+7\right)}^{2}=12$
4. ${\left(x+4\right)}^{2}+{\left(y+4\right)}^{2}=23$
5. $3{\left(x-2\right)}^{2}+3{\left(y+3\right)}^{2}=12$
6. ${x}^{2}-3x+9={y}^{2}+5y+25=17$
3. Find the $x-$ and $y-$ intercepts of the following graphs and draw a sketch to illustrate your answer:
1. ${\left(x+7\right)}^{2}+{\left(y-2\right)}^{2}=8$
2. ${x}^{2}+{\left(y-6\right)}^{2}=100$
3. ${\left(x+4\right)}^{2}+{y}^{2}=16$
4. ${\left(x-5\right)}^{2}+{\left(y+1\right)}^{2}=25$
4. Find the center and radius of the following circles:
1. ${x}^{2}+6x+{y}^{2}-12y=-20$
2. ${x}^{2}+4x+{y}^{2}-8y=0$
3. ${x}^{2}+{y}^{2}+8y=7$
4. ${x}^{2}-6x+{y}^{2}=16$
5. ${x}^{2}-5x+{y}^{2}+3y=-\frac{3}{4}$
6. ${x}^{2}-6nx+{y}^{2}+10ny=9{n}^{2}$
5. Find the equations to the tangent to the circle:
1. ${x}^{2}+{y}^{2}=17$ at the point $\left(1;4\right)$
2. ${x}^{2}+{y}^{2}=25$ at the point $\left(3;4\right)$
3. ${\left(x+1\right)}^{2}+{\left(y-2\right)}^{2}=25$ at the point $\left(3;5\right)$
4. ${\left(x-2\right)}^{2}+{\left(y-1\right)}^{2}=13$ at the point $\left(5;3\right)$
## Rotation of a point about an angle $\theta$
First we will find a formula for the co-ordinates of $P$ after a rotation of $\theta$ .
We need to know something about polar co-ordinates and compound angles before we start.
## Polar co-ordinates
Notice that : $sin\alpha =\frac{y}{r}$ $\therefore y=rsin\alpha$ and $cos\alpha =\frac{x}{r}$ $\therefore x=rcos\alpha$
so $P$ can be expressed in two ways:
1. $P\left(x;y\right)$ rectangular co-ordinates
2. $P\left(rcos\alpha ;rsin\alpha \right)$ polar co-ordinates.
## Compound angles
(See derivation of formulae in Ch. 12)
$\begin{array}{ccc}\hfill sin\left(\alpha +\beta \right)& =& sin\alpha cos\beta +sin\beta cos\alpha \hfill \\ \hfill cos\left(\alpha +\beta \right)& =& cos\alpha cos\beta -sin\alpha sin\beta \hfill \end{array}$
## Now consider ${P}^{\text{'}}$ After a rotation of $\theta$
$\begin{array}{ccc}& & P\left(x;y\right)=P\left(rcos\alpha ;rsin\alpha \right)\hfill \\ & & {P}^{\text{'}}\left(rcos\left(\alpha +\theta \right);rsin\left(\alpha +\theta \right)\right)\hfill \end{array}$
Expand the co-ordinates of ${P}^{\text{'}}$
$\begin{array}{ccc}\hfill x-\mathrm{co-ordinate of}{P}^{\text{'}}& =& rcos\left(\alpha +\theta \right)\hfill \\ & =& r\left[cos,\alpha ,cos,\theta ,-,sin,\alpha ,sin,\theta \right]\hfill \\ & =& rcos\alpha cos\theta -rsin\alpha sin\theta \hfill \\ & =& xcos\theta -ysin\theta \hfill \end{array}$
$\begin{array}{ccc}\hfill y-\mathrm{co-ordinate of P\text{'}}& =& rsin\left(\alpha +\theta \right)\hfill \\ & =& r\left[sin,\alpha ,cos,\theta ,+,sin,\theta ,cos,\alpha \right]\hfill \\ & =& rsin\alpha cos\theta +rcos\alpha sin\theta \hfill \\ & =& ycos\theta +xsin\theta \hfill \end{array}$
which gives the formula ${P}^{\text{'}}=\left[\left(,x,cos,\theta ,-,y,sin,\theta ,;,y,cos,\theta ,+,x,sin,\theta ,\right)\right]$ .
So to find the co-ordinates of $P\left(1;\sqrt{3}\right)$ after a rotation of 45 ${}^{\circ }$ , we arrive at:
$\begin{array}{ccc}\hfill {P}^{\text{'}}& =& \left[\left(,x,cos,\theta ,-,y,sin,\theta ,\right),;,\left(,y,cos,\theta ,+,x,sin,\theta ,\right)\right]\hfill \\ & =& \left[\left(1cos{45}^{\circ }-\sqrt{3}sin{45}^{\circ }\right),;,\left(\sqrt{3}cos{45}^{\circ }+1sin{45}^{\circ }\right)\right]\hfill \\ & =& \left[\left(\frac{1}{\sqrt{2}},-,\frac{\sqrt{3}}{\sqrt{2}}\right),;,\left(\frac{\sqrt{3}}{\sqrt{2}},+,\frac{1}{\sqrt{2}}\right)\right]\hfill \\ & =& \left(\frac{1-\sqrt{3}}{\sqrt{2}},;,\frac{\sqrt{3}+1}{\sqrt{2}}\right)\hfill \end{array}$
## Rotations
Any line $OP$ is drawn (not necessarily in the first quadrant), making an angle of $\theta$ degrees with the $x$ -axis. Using the co-ordinates of $P$ and the angle $\alpha$ , calculate the co-ordinates of ${P}^{\text{'}}$ , if the line $OP$ is rotated about the origin through $\alpha$ degrees.
$P$ $\alpha$ 1. (2, 6) 60 ${}^{\circ }$ 2. (4, 2) 30 ${}^{\circ }$ 3. (5, -1) 45 ${}^{\circ }$ 4. (-3, 2) 120 ${}^{\circ }$ 5. (-4, -1) 225 ${}^{\circ }$ 6. (2, 5) -150 ${}^{\circ }$
## Characteristics of transformations
Rigid transformations like translations, reflections, rotations and glide reflections preserve shape and size, and that enlargement preserves shape but not size.
## Geometric transformations:
Draw a large 15 $×$ 15 grid and plot $▵ABC$ with $A\left(2;6\right)$ , $B\left(5;6\right)$ and $C\left(5;1\right)$ . Fill in the lines $y=x$ and $y=-x$ . Complete the table below , by drawing the images of $▵ABC$ under the given transformations. The first one has been done for you.
Description Transformation (translation, reflection, Co-ordinates Lengths Angles rotation, enlargement) ${A}^{\text{'}}\left(2;-6\right)$ ${A}^{\text{'}}{B}^{\text{'}}=3$ ${\stackrel{^}{B}}^{\text{'}}={90}^{\circ }$ $\left(x;y\right)\to \left(x;-y\right)$ reflection about the $x$ -axis ${B}^{\text{'}}\left(5;-6\right)$ ${B}^{\text{'}}{C}^{\text{'}}=4$ $tan\stackrel{^}{A}=4/3$ ${C}^{\text{'}}\left(5;-2\right)$ ${A}^{\text{'}}{C}^{\text{'}}=5$ $\therefore \stackrel{^}{A}={53}^{\circ },\stackrel{^}{C}={37}^{\circ }$ $\left(x;y\right)\to \left(x+1;y-2\right)$ $\left(x;y\right)\to \left(-x;y\right)$ $\left(x;y\right)\to \left(-y;x\right)$ $\left(x;y\right)\to \left(-x;-y\right)$ $\left(x;y\right)\to \left(2x;2y\right)$ $\left(x;y\right)\to \left(y;x\right)$ $\left(x;y\right)\to \left(y;x+1\right)$
A transformation that leaves lengths and angles unchanged is called a rigid transformation. Which of the above transformations are rigid?
## Exercises
1. $\Delta ABC$ undergoes several transformations forming $\Delta {A}^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ . Describe the relationship between the angles and sides of $\Delta ABC$ and $\Delta {A}^{\text{'}}{B}^{\text{'}}{C}^{\text{'}}$ (e.g., they are twice as large, the same, etc.)
Transformation Sides Angles Area Reflect Reduce by a scale factor of 3 Rotate by 90 ${}^{\circ }$ Translate 4 units right Enlarge by a scale factor of 2
2. $\Delta DEF$ has $\stackrel{^}{E}={30}^{\circ }$ , $DE=4\phantom{\rule{0.166667em}{0ex}}\mathrm{cm}$ , $EF=5\phantom{\rule{0.166667em}{0ex}}\mathrm{cm}$ . $\Delta DEF$ is enlarged by a scale factor of 6 to form $\Delta {D}^{\text{'}}{E}^{\text{'}}{F}^{\text{'}}$ .
1. Solve $\Delta DEF$
2. Hence, solve $\Delta {D}^{\text{'}}{E}^{\text{'}}{F}^{\text{'}}$
3. $\Delta XYZ$ has an area of $6\phantom{\rule{0.166667em}{0ex}}{\mathrm{cm}}^{2}$ . Find the area of $\Delta {X}^{\text{'}}{Y}^{\text{'}}{Z}^{\text{'}}$ if the points have been transformed as follows:
1. $\left(x,y\right)\to \left(x+2;y+3\right)$
2. $\left(x,y\right)\to \left(y;x\right)$
3. $\left(x,y\right)\to \left(4x;y\right)$
4. $\left(x,y\right)\to \left(3x;y+2\right)$
5. $\left(x,y\right)\to \left(-x;-y\right)$
6. $\left(x,y\right)\to \left(x;-y+3\right)$
7. $\left(x,y\right)\to \left(4x;4y\right)$
8. $\left(x,y\right)\to \left(-3x;4y\right)$
#### Questions & Answers
encoding is the last step in memory stracture
crow and angle have feathers and can fly " it is a connective concept
FARWA
my mother always take decision based on reasoning ; she is practicng realistic thinking .
FARWA
all human are mortal is a formal concept
FARWA
sparrow is a best prototype of reptile
FARWA
echoic memory reflect humans visual system
getting the information back from stored area of brain is called retrieval
FARWA
no , echoic memory reflect upon things you hear , that's Iconic memory which deals with visual information
Nancy
I'll be studying psychology soon. Where do I start?
I would start by learning basic terminology and definitions cause there's a lot of it and learning the different concepts behind psychology and there meanings and there use then go onto introduction into psychology
why is psychology a science?
it's a social science
Alex
what we define as science is a systematic arrangement of knowledge based on experimental and observational data.so psychology is a science. as for what kind of science it is depends on how and what information are you using in your approach towards understanding the said data and to what conclusion.
imirror
depending on your approach and understanding Psychology can classify as medical or social science like anthropology
imirror
your friend is under stress because she has not prepared well for the examination what strategies would you suggest her? express in 100 or 150 words
hello
Masi
yess
dcs
where on the disorders does paranoid thought come in?
hot shots will give you paranoid thoughts
Kimberly
Kimberly 😂😆😂 yes hot shots will give you paranoid thoughts
Brandon
what compound motivation is attention extension
what is psychology
The scientific study of the mind
Gina
scientific study of mind and behaviour
Siva
...
Saeed
Scientific study of mind soul behaviour and experiences
Ayush
scientific srtudy of behaviour and cognitive processes
Reshmi
Apply social psychology on real life in Nigerian universities campus
what is the meaning of an idiosyncratic pattern
I am here for the first time just here to learn...
hi I'm new on here first time
Lisa
hello
GOPAL
hi,am new here
jennifer
what is this group all about
jennifer
Suppose an individual with OCD experiences obsessive thoughts about germs, contamination, and disease whenever she encounters a doorknob. What might have constituted a viable unconditioned stimulus?
What are factors that influence learning?
Environment Heredity(I am not sure about heredity)
Tusita
Enos
Sure, anytime
Tusita
I have other questions also
Enos
Based on the factors affecting learning, how do we improve learning
Enos
I think that addressing that everyone learns in their own time
James
Peer group influence can also be another reason
Sorie
Also knowing what's going on at home. what pressure are the parents putting on them.
James
am I close or no
James
also is there a lack of care? going one more step. with peer groups do an activity that shows how much they have in common
James
I think one's society has a big role to play in determining what/when & how he/she learns.
BECHEM
multiple reasons at different stages. at the earliest stages of development in a child is behaviour of people around. A baby starts learning basic social expressions and actions by mimicking the people around it. further social and cultural practices enforces learning and behaviour
imirror
from need to nutrition everything affect development and learning when choosing on a single factor of influence it is necessary to keep these other factors in the mind as environment around people is very plastic and flexible and there are many variables playing active role
imirror
social interaction is surely a big influence hence can be used to improve learning.
imirror
to encourage learning it is important to establish curiosity and notion of possibility for more perspectives habit of questioning possibilities and cause and effect in a great way to encourage learning in my opinion. language is another important factor
imirror
under what schedule of reinforcement do animals learn from ?
the role of emotional intelligence in a courtship behavior
1
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Transcript angle modulation
```FREQUENCY MODULATION(FM )
CHAPTER 3
FREQUENCY MODULATION
(FM)
Part 1
FM :Introduction
FM is the process of varying the
frequency of a carrier wave in
proportion to a modulating signal.
The amplitude of the carrier wave is
kept constant while its frequency and a
rate of change are varied by the
modulating signal.
FM :Introduction (cont…)
Fig 3.1 : Frequency
Modulated signal
FM :Introduction (cont…)
i.
ii.
iii.
iv.
waveforms are :
The frequency varies.
The rate of change of carrier frequency changes is
the same as the frequency of the information
signal.
The amount of carrier frequency changes is
proportional to the amplitude of the information
signal.
The amplitude is constant.
FM :Introduction (cont…)
The FM modulator receives two signals,the information signal
from an external source and the carrier signal from a built in
oscillator.
The modulator circuit combines the two signals producing a FM
signal which passed on to the transmission medium.
The demodulator receives the FM signal and separates it,
passing the information signal on and eliminating the carrier
signal.
Federal Communication Coporation (FCC) allocation for a
standard broadcast FM station is as shown in Fig.3.2
Figure 3.2: FM frequency allocation by FCC
Analysis of FM
Mathematical analysis:
Let message signal:
m t Vm cos mt
(3.1)
And carrier signal:
c t Vc cos[ ct ]
(3.2)
Where carrier frequency is very much higher
than message frequency.
Analysis of FM (cont’d)
In FM, frequency changes with the change of the
amplitude of the information signal. So the
instantenous frequency of the FM wave is;
(3.3)
K is constant of proportionality
i c Kvm t C KVm cos mt
Analysis of FM(cont’d)
Thus, we get the FM wave as:
vFM (t ) Vc cos 1 VC cos(C t
KVm
m
sin mt )
vFM (t ) VC cos(C t m f sin mt )
3.4
Where modulation index for FM is given by
mf
KVm
m
Analysis of FM(cont’d)
Frequency deviation: ∆f is the relative
placement of carrier frequency (Hz) w.r.t its
unmodulated value. Given as:
max C KVm
min C KVm
d max C C min KVm
d KVm
f
2
2
FM(cont’d)
Therefore:
KVm
f
;
2
f
mf
fm
f Vm ;
Example 3.1
FM broadcast station is allowed to have
a frequency deviation of 75 kHz. If a 4
kHz (highest voice frequency) audio
signal causes full deviation (i.e. at
maximun amplitude of information
signal) , calculate the modulation index.
Example 3.2
Determine the peak frequency
deviation, f , and the modulation
index, mf, for an FM modulator with a
deviation Kf = 10 kHz/V. The
modulating signal to be transmitted is
Vm(t) = 5 cos ( cos 10kπt).
Equations for Phase- and Frequency-Modulated Carriers
Tomasi
Electronic Communications Systems, 5e
Inc.
Upper Saddle River, New Jersey 07458
FM&PM (Bessel function)
Thus, for general equation:
vFM (t ) VC cos(C t m f cos mt )
vFM (t ) VC sin( C t m f sin mt )
VC sin C t cos( m f sin mt ) cos c t sin( m f sin mt )
Bessel function
vt FM VC [ J 0 (m f ) sin C t J1 (m f )sin( C m )t sin( C m )t
VC [ J 2 (m f ) sin( C 2m )t sin( C 2m )t
VC [ J 3 (m f ) sin( C 3m )t sin( C 3m )t ] ..
B.F. (cont’d)
It is seen that each pair of side band is preceded by J
coefficients. The order of the coefficient is denoted by
subscript m. The Bessel function can be written as
mf
J m m f
2
n
1 m f / 22 m f / 24
....
n! 1!n 1! 2!n 2!
N=number of the side frequency
M=modulation index
Bessel Functions of the First Kind, Jn(m)
for some value of modulation index
B.F. (cont’d)
Representation of frequency spectrum
Angle Modulation
Part 2
FM Bandwidth
Power distribution of FM
Generation & Detection of FM
Application of FM
FM Bandwidth
Theoretically, the generation and transmission of FM requires
infinite bandwidth. Practically, FM system have finite bandwidth
and they perform well.
The value of modulation index determine the number of sidebands
that have the significant relative amplitudes
If n is the number of sideband pairs, and line of frequency
spectrum are spaced by fm, thus,, the bandwidth is:
B fm 2nf m
For n=>1
FM Bandwidth (cont’d)
Estimation of transmission b/w;
Assume mf is large and n is approximate mf +2; thus
Bfm=2(mf +2)fm
=2(
f
2) f m
fm
B fm 2(f 2 f m )........(1)
(1) is called Carson’s rule
Deviation Ratio (DR)
The worse case modulation index which produces the widest
output frequency spectrum.
DR
f (max )
f m (max )
Where
∆f(max) = max. peak frequency deviation
fm(max) = max. modulating signal frequency
Example 3.3
An FM modulator is operating with a peak
frequency deviation f =20 kHz.The
modulating signal frequency, fm is 10 kHz,
and the 100 kHz carrier signal has an
amplitude of 10 V. Determine :
a) The minimum bandwidth using Bessel
Function table.
b)The minimum bandwidth using Carson’s
Rule.
Sketch the frequency spectrum for (a), with
actual amplitudes.
Example 3.4
For an FM modulator with a modulation index
m=1, a modulating signal Vm(t)=Vm
sin(2π1000t), and an unmodulated carrier
Vc(t) = 10sin(2π500kt), determine :
a) Number of sets of significant side
frequencies
b) Their amplitudes
c) Draw the frequency spectrum showing
their relative amplitudes.
Example 3.4 (solution)
a) From table of Bessel function, a modulation index
of 1 yields a reduced carrier component and three
sets of significant side frequencies.
b) The relative amplitude of the carrier and side
frequencies are
Jo = 0.77 (10) = 7.7 V
J1 = 0.44 (10) = 4.4 V
J2 = 0.11 (10) = 1.1 V
J3 = 0.02 (10) = 0.2 V
Example 3.4 (solution) :
FREQUENCY SPECTRUM
7.7 V
4.4V
4.4V
1.1V
1.1V
0..2
497
J3
0..2
498
J2
499
500
501
502
503
J1
JO
J1
J2
J3
FM Power Distribution
As seen in Bessel function table, it shows that as the
sideband relative amplitude increases, the carrier
amplitude,J0 decreases.
This is because, in FM, the total transmitted power is
always constant and the total average power is equal to
the unmodulated carrier power, that is the amplitude of
the FM remains constant whether or not it is
modulated.
FM Power Distribution
(cont’d)
In effect, in FM, the total power that is originally in the
carrier is redistributed between all components of the
spectrum, in an amount determined by the modulation
index, mf, and the corresponding Bessel functions.
At certain value of modulation index, the carrier
component goes to zero, where in this condition, the
power is carried by the sidebands only.
Average Power
Vc2
Pc
R
The average power in unmodulated carrier
The total instantaneous power in the angle modulated carrier.
m(t ) 2 Vc2
Pt
cos 2 [ct (t )]
R
R
2
Vc2 1 1
Vc
Pt
cos[ 2ct 2 (t )]
R 2 2
2R
The total modulated power
Vc2 2(V1 ) 2 2(V2 ) 2
2(Vn ) 2
Pt P0 P1 P2 .. Pn
..
2R
2R
2R
2R
Example 3.5
a) Determine the unmodulated carrier
power for the FM modulator and
condition given in example 3.4,
(assume a load resistance RL = 50 Ώ)
b) Determine the total power in the
angle modulated wave.
Example 3.5 : solution
a) Pc = (10)(10)/(2)(50) =1 W
b)
7.7 2
2(4.4) 2
2(1.1) 2
2(0.2) 2
Pt
2(50)
2(50)
2(50)
2(50)
Pt 0.5929 0.3872 0.0242 0.0008 1.0051W
Generation of FM
Two major FM generation:
i)
Direct method:
straight forward, requires a VCO whose oscillation frequency
has linear dependence on applied voltage.
Disadvantage: the carrier frequency tends to drift and must
be stabilized.
example circuit:
i)
ii)
iii)
iv)
i)
ii)
Reactance modulator
Varactor diode
Generation of FM (cont’d)
ii) Indirect method:
Frequency-up conversion.
Two ways:
i.
ii.
a.
b.
iii.
Heterodyne method
Multiplication method
One most popular indirect method is the Armstrong
modulator
Armstrong modulator
Vm(t)
fm
Integrator
Balanced
modulator
Phase
shifter
Vc(t)
fc
Frequency
multiplier
(x n)
Down
converter
Crystal oscillator
Armstrong modulator
For example:
Let fm =15Hz and fc= 200kHz
At frequency deviation= 75kHz,it need a frequency
multiplication by a factor, n,
n=75000/15=5000;
So it need a chain of four triplers (34) and six
ie:n= (34) x (26)=5184,
But,
n x fc=5000 x 200kHz=1000MHz
So, down converter with oscillating
needed to put fc in the
FM band of 88MHz-108MHz
doublers (26),
frequency=900MHz is
FM Detection/Demodulation
FM demodulation
is a process of getting back or regenerate the
original modulating signal from the modulated FM
signal.
It can be achieved by converting the frequency
deviation of FM signal to the variation of
equivalent voltage.
The demodulator will produce an output where its
instantaneous amplitude is proportional to the
instantaneous frequency of the input FM signal.
FM detection (cont’d)
To detect an FM signal, it is necessary to have
a circuit whose output voltage varies linearly
with the frequency of the input signal.
The most commonly used demodulator is the
PLL demodulator. Can be use to detect either
NBFM or WBFM.
PLL Demodulator
V0(t)
FM input
Phase
detector
Low pass
filter
Amplifier
fVc0
VCO
Vc(t)
PLL Demodulator
The phase detector produces an average output voltage that is
linear function of the phase difference between the two input
signals. This low frequency component is selected by LPF.
After amplification, part of the signal is fed back through VCO
where it results in frequency modulation of the VCO frequency.
When the loop is in lock, the VCO frequency follows or tracks
the incoming frequency.
PLL Demodulator
Let instantaneous freq of FM Input,
fi(t)=fc +k1vm(t),
and the VCO output frequency,
f VCO(t)=f0 + k2Vc(t);
f0 is the free running frequency.
For the VCO frequency to track the
instantaneous incoming frequency,
fvco = fi; or
PLL Demodulator
f0 + k2Vc(t)= fc +k1vm(t), so,
Vc (t ) f c f 0 k1vm (t )
If VCO can be tuned so that fc=f0, then
Vc (t ) k1vm (t )
Where Vc(t) is also taken as the output voltage,
which therefore is the demodulated output
Comparison AM and FM
Its the SNR can be increased without increasing transmitted
power about 25dB higher than in AM
Certain forms of interference at the receiver are more easily to
suppressed, as FM receiver has a limiter which eliminates the
amplitude variations and fluctuations.
The modulation process can take place at a low level power
stage in the transmitter, thus a low modulating power is needed.
Power content is constant and fixed, and there is no waste of
power transmitted
There are guard bands in FM systems allocated by the
standardization body, which can reduce interference between
Application of FM
used by most of the field VHF portable,
mobile and base radios in exploration
use today. It is preferred because of its
immunity to noise or interference and at
the frequencies used the antennas are
of a reasonable size.
Summary of angle modulation
-what you need to be familiar with
Summary (cont’d)
Summary (cont’d)
a)
Bandwidth:
Actual minimum bandwidth from Bessel
table:
B 2(n f m )
b) Approximate minimum bandwidth using
Carson’s rule:
B 2(f f m )
Summary (cont’d)
Multitone modulation (equation in general):
i c Kvm1 Kvm 2
i c 2f1 cos 1t 2f 2 cos 2t....
f1
f 2
i C t
sin 1t
sin 2t......
f1
f2
Summary (cont’d)
v fm t VC sin i
f1
f 2
v fm t VC sin[ C t
sin 1t
sin 2t ]
f1
f2
VC sin[ C t m f 1 sin 1t m f 2 sin 2t ]...........
Summary (cont’d)Comparison NBFM&WBFM
ANGLE MODULATION
Part 3
Wideband FM gives significant improvement in the SNR at the output
of the RX which proportional to the square of modulation index.
Angle modulation is resistant to propagation-induced selective fading
since amplitude variations are unimportant and are removed at the
Angle modulation is very effective in rejecting interference. (minimizes
the effect of noise).
Angle modulation allows the use of more efficient transmitter power in
information.
Angle modulation is capable of handing a greater dynamic range of
modulating signal without distortion than AM.
Angle modulation requires a transmission
bandwidth much larger than the message
signal bandwidth.
The capture effect where the wanted signal
may be captured by an unwanted signal or
noise voltage.
Angle modulation requires more complex and
inexpensive circuits than AM.
END OF ANGLE
MODULATION
```
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| 3.59375 | 4 |
CC-MAIN-2021-04
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latest
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| 0.813268 |
https://www.prepbytes.com/blog/linked-list/length-of-longest-palindrome-list-in-a-linked-list-using-o1-extra-space/
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# Length of longest palindrome list in a linked list using O(1) extra space
Last Updated on March 10, 2022 by Ria Pathak
### Introduction
The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.
### Problem Statement
According to the problem statement, our task is to find the length of the longest palindrome list in a linked list.
### Problem Statement Understanding
We have been given a linked list. We have to find the length of the longest palindrome list in the given linked list.
Let’s understand this problem with the help of examples.
• In the above given linked list, the sublist 2→6→1→6→2 is a palindrome, and also it is the longest palindrome in the list.
• As the length of 2→6→1→6→2 is 5, so our output will be 5.
Let’s take another example if the input list is: head→9→7→9→11→5→7→4→4→7.
• In this example, there are two sublists that are palindrome, one is 9→7→9, and another is 7→4→4→7. But we have to print the length of the longest palindrome list, and that is 4.
##### Some more examples
Sample Input 1: 1→2→3→3→2→7
Sample Output 1: 4
Sample Input 2: 3→6→9→4→7→7→4→9→6→11
Sample Output 2: 8
Now I think from the above examples, the problem statement is clear. So let’s see how we will approach it.
Before moving to the approach section, try to think about how you can approach this problem.
• If stuck, no problem, we will thoroughly see how we can approach the problem in the next section.
Let’s move to the approach section.
### Approach
The first thing that we need to focus on is that a palindrome is a sequence that reads the same backward and forwards.
• Let’s say if we have a sequence “abcba” and we want to check whether it is a palindromic sequence or not. So, for a sequence to be a palindromic sequence, it should follow the property that when we reverse the first half of the sequence, then the first half should become the same as the second half of the sequence.
• “abcba” is an odd-length palindromic sequence. So, when we reverse the first two characters of the sequence, i.e., ab to ba, then it becomes the same as the last two characters of the sequence, i.e. ba.
### Algorithm
• Start traversing through the linked list.
• Declare pointers curr and prv of type Node.
• Initialize curr with head and prv with NULL.
• curr points to the current node of the linked list.
• In every step, the linked list from head to curr is reversed, and prv points to this reversed linked list.
• For every current node, We will call function countCommon(). This function will return the number of common elements in the two linked lists.
• We will call this function two times to check for odd length palindrome and even length palindrome.
• In the case of odd length palindrome, our answer will be (2* No. of common elements )+1.
• In the case of even length palindrome, our answer will be (2* No. of common elements).
• When we call countCommon() function, in the case of odd length palindrome, we will exclude the current node and, in the case of even length palindrome, we will include the current node.
• We are including and excluding current node in the above step because we have to care about the length of the palindrome linked list.
• It is already known that in odd length palindrome, the middle element doesn’t have any match. Ex. “aba” here aba is a palindrome but ‘b’ doesn’t have any match.
• Store the length of longest palindrome sublist in a variable named result.
• Update result, whenever we found a sublist having length greater than the value of the result.
### Code Implementation
```#include<stdio.h>
#include<stdlib.h>
#define max(x, y) (((x) > (y)) ? (x) : (y))
#define min(x, y) (((x) < (y)) ? (x) : (y))
struct Node
{
int data;
struct Node* next;
};
// function for counting the common elements
int countCommon(struct Node *a,struct Node *b)
{
int count = 0;
// loop to count common in the list starting
// from node a and b
for (; a && b; a = a->next, b = b->next)
// increment the count for same values
if (a->data == b->data)
++count;
else
break;
return count;
}
// Returns length of the longest palindrome
// sublist in given list
{
int result = 0;
struct Node *prev = NULL;
// loop till the end of the linked list
while (curr)
{
// The sublist from head to current
// reversed.
struct Node *next = curr->next;
curr->next = prev;
// check for odd length palindrome
// by finding longest common list elements
// beginning from prev and from next (We
// exclude curr)
result = max(result,
2*countCommon(prev, next)+1);
// check for even length palindrome
// by finding longest common list elements
// beginning from curr and from next
result = max(result,
2*countCommon(curr, next));
// update prev and curr for next iteration
prev = curr;
curr = next;
}
return result;
}
// Utility function to create a new list node
struct Node *newNode(int key)
{
struct Node* temp = (struct Node*)malloc(sizeof(struct Node));
temp->data = key;
temp->next = NULL;
return temp;
}
/* Driver program to test above functions*/
int main()
{
/* Let us create a linked lists to test
the functions
Created list is a: 2->4->3->4->2->15 */
printf("%d \n",ans);
return 0;
}
```
```#include<bits stdc++.h="">
using namespace std;
/* Node structure of the linked list node */
class Node{
public:
int data;
Node* next;
};
/* Using this function we will be counting the number of common elements in the given two linked lists */
int countCommon(Node *a, Node *b){
int count = 0;
for (; (a!=NULL && b!=NULL); a = a->next, b = b->next)
if (a->data == b->data){
++count;
}
else{
break;
}
return count;
}
/* Using this function we will be finding the length of the longest palindrome sublist in the given linked list */
int result = 0;
Node *prev = NULL, *curr = head;
while (curr){
Node *next = curr->next;
curr->next = prev;
result = max(result,
2*countCommon(prev, next)+1);
result = max(result,
2*countCommon(curr, next));
prev = curr;
curr = next;
}
return result;
}
/* Using this function we will be creating a new list node */
Node *newNode(int key){
Node *temp = new Node;
temp->data = key;
temp->next = NULL;
return temp;
}
int main(){
cout<<"Length of longest palindrome list is: ";
return 0;
}
```
```class LargestPalindrome
{
static class Node
{
int data;
Node next;
}
static int countCommon(Node a, Node b)
{
int count = 0;
for (; a != null && b != null;
a = a.next, b = b.next)
if (a.data == b.data)
++count;
else
break;
return count;
}
{
int result = 0;
Node prev = null, curr = head;
while (curr != null)
{
// The sublist from head to current reversed
Node next = curr.next;
curr.next = prev;
/* check for odd length palindrome by finding longest common list elements
beginning from prev and from next (We exclude curr) */
result = Math.max(result,2 * countCommon(prev, next)+1);
/* check for even length palindrome by finding longest common list elements
beginning from curr and from next */
result = Math.max(result,2*countCommon(curr, next));
// update prev and curr for next iteration
prev = curr;
curr = next;
}
return result;
}
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}
/* Driver code*/
public static void main(String[] args)
{
}
}
```
```# Linked List node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# function for counting the common elements
def countCommon(a, b) :
count = 0
while ( a != None and b != None ) :
if (a.data == b.data) :
count = count + 1
else:
break
a = a.next
b = b.next
return count
# Returns length of the longest palindrome
# sublist in given list
result = 0
prev = None
while (curr != None) :
next = curr.next
curr.next = prev
result = max(result, 2 * countCommon(prev, next) + 1)
result = max(result, 2 * countCommon(curr, next))
prev = curr
curr = next
return result
# Utility function to create a new list node
def newNode(key) :
temp = Node(0)
temp.data = key
temp.next = None
return temp
print("Length of longest palindrome list is:", maxPalindrome(head))
```
#### Output
Length of longest palindrome list is: 3
Time Complexity: O(N2), For every node of the linked list, We have traversed through the remaining part of the linked list for comparison of the common elements.
[forminator_quiz id=”5052″]
So, In this blog, we have learned how to find the length of the longest palindrome list in a linked list. This is a basic problem and is good for strengthening your concepts in LinkedList and if you want to practice more such problems, you can checkout Prepbytes (Linked List).
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| 4.34375 | 4 |
CC-MAIN-2024-18
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en
| 0.894127 |
idt.carbonaraadheera.fun
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crawl-data/CC-MAIN-2021-31/segments/1627046154320.56/warc/CC-MAIN-20210802110046-20210802140046-00076.warc.gz
| 325,725,805 | 12,734 |
# Stivaletti caprice felicia nero donna in linea :
By | 04.10.2020
Now we can relate the odds for males and females and the output from the logistic regression. The intercept of -1. Using the odds we calculated above for males, we can confirm this: log(. The coefficient for female is the log of odds ratio between the female group and male group: log(1. So we can get the odds ratio by exponentiating the coefficient for female.
Most statistical packages display both the raw regression coefficients and the exponentiated coefficients for logistic regression models. The table below is created by Stata. In other words, the odds of being in an honors class when the math score is zero is exp(-9.
These odds are very low, but if we look at the distribution of the variable math, we will see that no one in the sample has math score lower than 30. In fact, all the test scores in the data set were standardized around mean of 50 and standard deviation of 10. So the intercept in this model corresponds to the log odds of being in an honors class when math is at the hypothetical value of zero. How do we interpret the coefficient for math.
We will use 54. Then the conditional logit of being in an honors class when the math score is held at 54 isWe can examine the effect of a one-unit increase in math score. When the math score is held at 55, the conditional logit of being in an honors class isWe can say now that the coefficient for math is the difference in the log odds. In other words, for a one-unit increase in the math score, the expected change in log odds is. Can we translate this change in log odds to the change in odds.
Recall that logarithm converts multiplication and division to addition and subtraction. Its inverse, the exponentiation converts addition and subtraction back to multiplication and division. Applying such a model to our example dataset, each estimated coefficient is the expected change in the log odds of being in an honors class for a unit increase in the corresponding predictor variable holding the other predictor variables constant at certain value.
Each exponentiated coefficient is the ratio of two odds, or the change in odds in the multiplicative scale for a unit increase in the corresponding predictor variable holding other variables at certain value. Here is an example. In all the previous examples, we have said that the regression coefficient of a variable corresponds to the change in log odds and its exponentiated form corresponds to the odds ratio. This is only true when our model does not have any interaction terms.
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The Act has been referred to the European Commission and the mandatory standstill period expires on March 31. A number of online gambling operators have since announced that they would quit the Polish market rather than apply for a new Polish license.
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Category: idt
| 3,858 | 18,398 |
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| 3.65625 | 4 |
CC-MAIN-2021-31
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latest
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en
| 0.915527 |
https://gmatclub.com/forum/either-perry-s-faction-or-tucker-s-faction-but-not-both-will-win-con-90306.html
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crawl-data/CC-MAIN-2019-09/segments/1550247484648.28/warc/CC-MAIN-20190218033722-20190218055722-00177.warc.gz
| 570,761,359 | 157,263 |
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# Either Perry’s faction or Tucker’s faction, but not both, will win con
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Either Perry’s faction or Tucker’s faction, but not both, will win con [#permalink]
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Updated on: 10 Oct 2018, 21:00
6
28
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Difficulty:
55% (hard)
Question Stats:
62% (01:34) correct 38% (01:33) wrong based on 908 sessions
### HideShow timer Statistics
Either Perry’s faction or Tucker’s faction, but not both, will win control of the government. If Perry’s faction wins, the nation will suffer economically. If Tucker’s faction wins, the nation will suffer militarily.
Given the statements in the passage, which one of the following statements must be true?
(A) It is possible, but not certain, that the nation will neither suffer economically nor suffer militarily.
(B) If the nation suffers economically, it is certain that Perry’s faction has won control of the government.
(C) It is certain that the nation will suffer either economically or militarily, and also certain that it will not suffer both.
(D) If the nation suffers militarily, it is possible, but not certain, that Tucker’s faction has won control of the government.
(E) If the nation suffers both economically and militarily, it is certain that neither Perry’s faction nor Tucker’s has won control of the government.
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Originally posted by SudiptoGmat on 11 Feb 2010, 06:53.
Last edited by Bunuel on 10 Oct 2018, 21:00, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Either Perry’s faction or Tucker’s faction, but not both, will win con [#permalink]
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26 Apr 2013, 19:51
16
3
A very good question.
Either Perry’s faction or Tucker’s faction, but not both, will win control of the government. - It means that someone (Perry or Tucker) will definitely win the election & it's not possible that BOTH will win.
If Perry’s faction wins, the nation will suffer economically. -
This is tricky part - we can restate in other term.
(ST-1)- If Perry’s faction wins, IT'S CERTAIN THAT the nation will suffer economically & IT'S ALSO POSSIBLE THAT the nation will suffer militarily.
If Tucker’s faction wins, the nation will suffer militarily.
(ST-2)If Tucker’s faction wins, IT'S CERTAIN THAT the nation will suffer militarily & IT'S ALSO POSSIBLE THAT the nation will suffer economically.
(A) It is possible, but not certain, that the nation will neither suffer economically nor suffer militarily. - It's not possible that the nation will not suffer neither way because it's CERTAIN that either PERRY or TUCKER will win the election. This means that the country will suffer either militarily or economically at least. Thus INCORRECT
(B) If the nation suffers economically, it is certain that Perry’s faction has won control of the government. - In Powerscore CR terms MISTAKEN REVERSAL is incorrect. Incorrect because country can suffer economically under Tucker's control as well as per (ST-2). INCORRECT
(C) It is certain that the nation will suffer either economically or militarily, and also certain that it will not suffer both. - Both (ST-1, 2) indicate that the government can suffer both ways. Thus INCORRECT
(D) If the nation suffers militarily, it is possible, but not certain, that Tucker’s faction has won control of the government. - Restated (ST-2). Thus CORRECT
(E) If the nation suffers both economically and militarily, it is certain that neither Perry’s faction nor Tucker’s has won control of the government. (ST-2) indicate that the government can suffer both ways under the control of TUCKER. Thus INCORRECT
I hope this detailed explanation will help many aspirants.
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##### General Discussion
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Re: Either Perry s faction or Tucker s faction, but not both, [#permalink]
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22 Apr 2010, 13:03
What is the OA? IMO it has to be D.
(A) It is possible, but not certain, that the nation will neither suffer economically nor suffer militarily.
(B) If the nation suffers economically, it is certain that Perry’s faction has won control of the government.
(C) It is certain that the nation will suffer either economically or militarily, and also certain that it will not suffer both.The original statement only says what will happen, but not what won't. It never says that either faction could will only suffer in one way. Either faction could suffer both economically and militarily.
(D) If the nation suffers militarily, it is possible, but not certain, that Tucker’s faction has won control of the government.This is the best answer. Again, the only certainty that we have is that one of the two factions will win, and that each faction will cause the nation to suffer in at least one way. No matter how the nation suffers, it is possible but not certain that either faction won.
(E) If the nation suffers both economically and militarily, it is certain that neither Perry’s faction nor Tucker’s has won control of the government
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Re: Either Perry s faction or Tucker s faction, but not both, [#permalink]
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08 Jul 2010, 21:07
Initially I selected c , but looking at the explanation by Vannbj, D seems to be correct answer.
@Vannbj,
Please advise what are the other different type of CR questions where we need to consider this approach.
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Re: Either Perry s faction or Tucker s faction, but not both, [#permalink]
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09 Jul 2010, 10:02
2
Initially I selected c , but looking at the explanation by Vannbj, D seems to be correct answer.
@Vannbj,
Please advise what are the other different type of CR questions where we need to consider this approach.
I'm not sure I understand your question. But I’d take this approach with just about every critical reasoning question. You usually can break them down into logical equations like math. For instance, “If Bob jumps then his hat will fall off” can be written as “j => h” where “=>” means “then”, “j” means “Bob Jumps” and “h” means “Bob’s hat will fall off.” If we weren’t on a computer then I’d just draw “=>” as an arrow pointing to the right. We can then write the contrapositive “-h => -j” which means in “If Bob’s hat didn’t fall off then he did not jump.” The contrapositive of a logical statement is ALWAYS true and is the first thing I tend to infer from a question like this. But notice that we didn’t say “-j => -h” this says “If Bob didn’t jump then his hat didn’t fall off.” This is a common logical flaw but it’s not substantiated from the original statement. Something else could have caused the hat to fall off. You can generally make up your own equations for any if-then statements. I hope this helps and that I answered your question but like I said, I’m not entirely sure that I understood it. If it does help then I accept kudos
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Re: Either Perry’s faction or Tucker’s faction, but not both, will win con [#permalink]
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03 Dec 2013, 10:51
1
SudiptoGmat wrote:
Either Perry’s faction or Tucker’s faction, but not both, will win control of the government. If Perry’s faction wins, the nation will suffer economically. If Tucker’s faction wins, the nation will suffer militarily.
(1) Either Perry’s faction or Tucker’s faction will win the election , both can't win / loose.
(2) Perry’s faction wins = suffer economically
(3) Tucker’s faction wins = suffer militarily
Given the statements in the passage, which one of the following statements must be true?
(A) It is possible, but not certain, that the nation will neither suffer economically nor suffer militarily.
Violates the condition (1) , either of the 2 contestants must win and the country will either have to suffer Either Militarily or Economically.
(B) If the nation suffers economically, it is certain that Perry’s faction has won control of the government.
One can't be certain - There are multitude of other factors.
(C) It is certain that the nation will suffer either economically or militarily, and also certain that it will not suffer both.
The Country will either suffer economically or militarily , the underlined part distorts the picture.
(D) If the nation suffers militarily, it is possible, but not certain, that Tucker’s faction has won control of the government.
True , it is just a possibility , nothing confirm can be stated.
(E) If the nation suffers both economically and militarily, it is certain that neither Perry’s faction nor Tucker’s has won control of the government.
Not true , might be that the conditions were favourable and there were efficient management techniques to suppress the same...
(D) is the best option among the rest hence ( D ) is the answer.
The golden rule here is : We can't be certain about something here , there is Just a possibility / Probability of occurrence / Non Occurrence of an Event.
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Re: Either Perry’s faction or Tucker’s faction, but not both, will win con [#permalink]
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14 Dec 2013, 08:20
3
Just to encapsulate what some of the earlier posters have been suggesting, remember that on a "must be true" question, when in doubt, generally* go with the weaker answer.
It is counterintuitive for some at first, but "maybe" and "possibly" allow for more options and are likely to hold true in more situations (which is exactly what you want for a "must be true" question).
*Of course, if the weak statement is false outright, this won't apply, but you probably aren't struggling with those choices in the first place.
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Re: Either Perry’s faction or Tucker’s faction, but not both, will win con [#permalink]
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Re: Either Perry’s faction or Tucker’s faction, but not both, will win con [#permalink] 10 Oct 2018, 21:05
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Posted on by Kalkicode
Code Bit Logic
# Remove nth bit of a number
The given problem is about removing the nth bit of a given number. We are provided with a positive integer `num` and an integer `n` representing the position of the bit to be removed (1-indexed, from right to left). The task is to remove the nth bit from the binary representation of the number and return the resulting number.
## Problem Statement
Given a positive integer `num` and an integer `n`, remove the nth bit from the binary representation of `num` and return the resulting number.
## Example
Let's take an example to illustrate the problem. Consider `num = 21` and `n = 3`. The binary representation of `21` is `10101`. After removing the 3rd bit (from right to left), we get `1001`, which is `9` in decimal.
## Pseudocode
Before diving into the algorithm, let's start with a pseudocode representation of the approach to remove the nth bit:
``````FUNCTION removeBit(num, n):
IF n < 1 OR num < 0:
RETURN
value = 1 << (n - 1)
IF value > num:
PRINT "Output:", num
RETURN
result = num XOR value
value = value - 1
value = result AND value
result = result - value
value = value << 1
result = result OR value
result = result >> 1
PRINT "Output:", result
``````
## Algorithm Explanation
1. The function `removeBit(num, n)` takes the number `num` and the position `n` of the bit to be removed as inputs.
2. We first check if the value of `n` is less than 1 or if `num` is negative. If either condition is true, we return from the function since it's not possible to remove a bit in these cases.
3. We calculate the value of `1` left-shifted by `(n - 1)` positions, which will set the nth bit to `1` and all other bits to `0`.
4. If the calculated value is greater than `num`, then removing the bit will not affect the given number, so we print the output as the original number `num`.
5. If the calculated value is not greater than `num`, we proceed to remove the nth bit.
6. We use bitwise XOR operation (`^`) to deactivate the nth bit, effectively setting it to `0` while keeping other bits unchanged.
7. We subtract `1` from the calculated value, which will result in all `1`s from the nth position (inclusive) to the rightmost bit (inclusive).
8. We use bitwise AND operation (`&`) with the result to unset all the bits from the nth position to the rightmost bit, effectively removing the nth bit.
9. We update the `result` by subtracting the calculated value, effectively removing the nth bit.
10. We shift the value one position to the left using bitwise left shift (`<<`) to prepare it to be added to the result in the next step.
11. We use bitwise OR operation (`|`) with the result to add the new left portion obtained from the previous step.
12. Finally, we shift the result one position to the right using bitwise right shift (`>>`) to remove the extra zero bit added in the previous step.
13. We print the final output, which is the number obtained after removing the nth bit.
## Resultant Output Explanation
The code uses the algorithm described above to remove the nth bit from the given numbers and prints the output.
• For `num = 21` and `n = 3`, the output is `9`. The binary representation of `21` is `10101`, and after removing the 3rd bit, we get `1001`, which is `9` in decimal.
• For `num = 73` and `n = 5`, the output is `41`. The binary representation of `73` is `1001001`, and after removing the 5th bit, we get `101001`, which is `41` in decimal.
• For `num = 80` and `n = 7`, the output is `16`. The binary representation of `80` is `1010000`, and after removing the 7th bit, we get `10000`, which is `16` in decimal.
• For `num = 11` and `n = 5`, the output is `11`. The binary representation of `11` is `01011`, and after removing the 5th bit, we get `1011`, which is `11` in decimal.
• For `num = 6` and `n = 1`, the output is `3`. The binary representation of `6` is `110`, and after removing the 1st bit, we get `11`, which is `3` in decimal.
## Code Solution
Here given code implementation process.
``````// C Program
// Remove nth bit of a number
#include <stdio.h>
// Remove a Bit at given position of a number
void removeBit(int num, int n)
{
if (n < 1 || num < 0)
{
return;
}
int value = (1 << (n - 1));
// Display result
printf("\n Given Number : %d", num);
printf("\n Remove Bit : %d-th", n);
if (value > num)
{
// When after remove bit is not effect to given number
printf("\n Output : %d\n", num);
return;
}
int result = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result & value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
printf("\n Output : %d\n", result);
}
int main(int argc, char const *argv[])
{
// num = 21, n = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
removeBit(21, 3);
// num = 73 n = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
removeBit(73, 5);
// num = 80 n = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
removeBit(80, 7);
// num = 11 n = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
removeBit(11, 5);
// num = 6 n = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
removeBit(6, 1);
return 0;
}``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````/*
Java program
Remove nth bit of a number
*/
public class BitManipulation
{
// Remove a Bit at given position of a number
public void removeBit(int num, int n)
{
if (n < 1 || num < 0)
{
return;
}
int value = (1 << (n - 1));
// Display result
System.out.print("\n Given Number : " + num);
System.out.print("\n Remove Bit : " + n + "-th");
if (value > num)
{
// When after remove bit is not effect to given number
System.out.print("\n Output : " + num + "\n");
return;
}
int result = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result & value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
System.out.print("\n Output : " + result + "\n");
}
public static void main(String[] args)
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}
}``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````// Include header file
#include <iostream>
using namespace std;
/*
C++ program
Remove nth bit of a number
*/
class BitManipulation
{
public:
// Remove a Bit at given position of a number
void removeBit(int num, int n)
{
if (n < 1 || num < 0)
{
return;
}
int value = (1 << (n - 1));
// Display result
cout << "\n Given Number : " << num;
cout << "\n Remove Bit : " << n << "-th";
if (value > num)
{
// When after remove bit is not effect to given number
cout << "\n Output : " << num << "\n";
return;
}
int result = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result &value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
cout << "\n Output : " << result << "\n";
}
};
int main()
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
return 0;
}``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````// Include namespace system
using System;
/*
C# program
Remove nth bit of a number
*/
public class BitManipulation
{
// Remove a Bit at given position of a number
public void removeBit(int num, int n)
{
if (n < 1 || num < 0)
{
return;
}
int value = (1 << (n - 1));
// Display result
Console.Write("\n Given Number : " + num);
Console.Write("\n Remove Bit : " + n + "-th");
if (value > num)
{
// When after remove bit is not effect to given number
Console.Write("\n Output : " + num + "\n");
return;
}
int result = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result & value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
Console.Write("\n Output : " + result + "\n");
}
public static void Main(String[] args)
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}
}``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````<?php
/*
Php program
Remove nth bit of a number
*/
class BitManipulation
{
// Remove a Bit at given position of a number
public function removeBit(\$num, \$n)
{
if (\$n < 1 || \$num < 0)
{
return;
}
\$value = (1 << (\$n - 1));
// Display result
echo "\n Given Number : ". \$num;
echo "\n Remove Bit : ". \$n ."-th";
if (\$value > \$num)
{
// When after remove bit is not effect to given number
echo "\n Output : ". \$num ."\n";
return;
}
\$result = \$num;
if ((\$result ^ \$value) != 0)
{
// Deactivation of a removal bit
\$result = \$result ^ \$value;
}
\$value = (\$value - 1);
// Get the left portion
\$value = \$result & \$value;
// unset all bit in left part
\$result = \$result - \$value;
// Shift by one of left part
\$value = \$value << 1;
// Add new left part into result
\$result = \$result | \$value;
// Shift right by one bit
\$result = \$result >> 1;
echo "\n Output : ". \$result ."\n";
}
}
function main()
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}
main();``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````/*
Node Js program
Remove nth bit of a number
*/
class BitManipulation
{
// Remove a Bit at given position of a number
removeBit(num, n)
{
if (n < 1 || num < 0)
{
return;
}
var value = (1 << (n - 1));
// Display result
process.stdout.write("\n Given Number : " + num);
process.stdout.write("\n Remove Bit : " + n + "-th");
if (value > num)
{
// When after remove bit is not effect to given number
process.stdout.write("\n Output : " + num + "\n");
return;
}
var result = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result & value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
process.stdout.write("\n Output : " + result + "\n");
}
}
function main()
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}
main();``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````# Python 3 program
# Remove nth bit of a number
class BitManipulation :
# Remove a Bit at given position of a number
def removeBit(self, num, n) :
if (n < 1 or num < 0) :
return
value = (1 << (n - 1))
# Display result
print("\n Given Number : ", num, end = "")
print("\n Remove Bit : ", n ,"-th", end = "")
if (value > num) :
# When after remove bit is not effect to given number
print("\n Output : ", num )
return
result = num
if ((result ^ value) != 0) :
# Deactivation of a removal bit
result = result ^ value
value = (value - 1)
# Get the left portion
value = result & value
# unset all bit in left part
result = result - value
# Shift by one of left part
value = value << 1
# Add new left part into result
result = result | value
# Shift right by one bit
result = result >> 1
print("\n Output : ", result )
def main() :
# num = 21, position = 3
# 21 = (10101)
# After remove bit at 3rd position
# 9 = (1001)
# num = 73 position = 5
# 73 => 1001001
# After remove 5-th bit
# (41) 101001
# num = 80 position = 7
# 73 => 1010000
# After remove 7-th bit
# (16) 10000
# num = 11 position = 5
# 73 => 01011
# After remove 5-th bit
# (11) 1011
# num = 6 position = 1
# 6 => 110
# After remove 1-st bit
# (3) 11
if __name__ == "__main__": main()``````
#### Output
`````` Given Number : 21
Remove Bit : 3 -th
Output : 9
Given Number : 73
Remove Bit : 5 -th
Output : 41
Given Number : 80
Remove Bit : 7 -th
Output : 16
Given Number : 11
Remove Bit : 5 -th
Output : 11
Given Number : 6
Remove Bit : 1 -th
Output : 3``````
``````# Ruby program
# Remove nth bit of a number
class BitManipulation
# Remove a Bit at given position of a number
def removeBit(num, n)
if (n < 1 || num < 0)
return
end
value = (1 << (n - 1))
# Display result
print("\n Given Number : ", num)
print("\n Remove Bit : ", n ,"-th")
if (value > num)
# When after remove bit is not effect to given number
print("\n Output : ", num ,"\n")
return
end
result = num
if ((result ^ value) != 0)
# Deactivation of a removal bit
result = result ^ value
end
value = (value - 1)
# Get the left portion
value = result & value
# unset all bit in left part
result = result - value
# Shift by one of left part
value = value << 1
# Add new left part into result
result = result | value
# Shift right by one bit
result = result >> 1
print("\n Output : ", result ,"\n")
end
end
def main()
# num = 21, position = 3
# 21 = (10101)
# After remove bit at 3rd position
# 9 = (1001)
# num = 73 position = 5
# 73 => 1001001
# After remove 5-th bit
# (41) 101001
# num = 80 position = 7
# 73 => 1010000
# After remove 7-th bit
# (16) 10000
# num = 11 position = 5
# 73 => 01011
# After remove 5-th bit
# (11) 1011
# num = 6 position = 1
# 6 => 110
# After remove 1-st bit
# (3) 11
end
main()``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3
``````
``````/*
Scala program
Remove nth bit of a number
*/
class BitManipulation
{
// Remove a Bit at given position of a number
def removeBit(num: Int, n: Int): Unit = {
if (n < 1 || num < 0)
{
return;
}
var value: Int = (1 << (n - 1));
// Display result
print("\n Given Number : " + num);
print("\n Remove Bit : " + n + "-th");
if (value > num)
{
// When after remove bit is not effect to given number
print("\n Output : " + num + "\n");
return;
}
var result: Int = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result & value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
print("\n Output : " + result + "\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: BitManipulation = new BitManipulation();
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}
}``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````/*
Swift 4 program
Remove nth bit of a number
*/
class BitManipulation
{
// Remove a Bit at given position of a number
func removeBit(_ num: Int, _ n: Int)
{
if (n < 1 || num < 0)
{
return;
}
var value: Int = (1 << (n - 1));
// Display result
print("\n Given Number : ", num, terminator: "");
print("\n Remove Bit : \(n)-th", terminator: "");
if (value > num)
{
// When after remove bit is not effect to given number
print("\n Output : ", num );
return;
}
var result: Int = num;
if ((result ^ value) != 0)
{
// Deactivation of a removal bit
result = result ^ value;
}
value = (value - 1);
// Get the left portion
value = result & value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value << 1;
// Add new left part into result
result = result | value;
// Shift right by one bit
result = result >> 1;
print("\n Output : ", result );
}
}
func main()
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}
main();``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
``````/*
Kotlin program
Remove nth bit of a number
*/
class BitManipulation
{
// Remove a Bit at given position of a number
fun removeBit(num: Int, n: Int): Unit
{
if (n < 1 || num < 0)
{
return;
}
var value: Int = (1 shl(n - 1));
// Display result
print("\n Given Number : " + num);
print("\n Remove Bit : " + n + "-th");
if (value > num)
{
// When after remove bit is not effect to given number
print("\n Output : " + num + "\n");
return;
}
var result: Int = num;
if ((result xor value) != 0)
{
// Deactivation of a removal bit
result = result xor value;
}
value = (value - 1);
// Get the left portion
value = result and value;
// unset all bit in left part
result = result - value;
// Shift by one of left part
value = value shl 1;
// Add new left part into result
result = result or value;
// Shift right by one bit
result = result shr 1;
print("\n Output : " + result + "\n");
}
}
fun main(args: Array < String > ): Unit
{
// num = 21, position = 3
// 21 = (10101)
// After remove bit at 3rd position
// 9 = (1001)
// num = 73 position = 5
// 73 => 1001001
// After remove 5-th bit
// (41) 101001
// num = 80 position = 7
// 73 => 1010000
// After remove 7-th bit
// (16) 10000
// num = 11 position = 5
// 73 => 01011
// After remove 5-th bit
// (11) 1011
// num = 6 position = 1
// 6 => 110
// After remove 1-st bit
// (3) 11
}``````
#### Output
`````` Given Number : 21
Remove Bit : 3-th
Output : 9
Given Number : 73
Remove Bit : 5-th
Output : 41
Given Number : 80
Remove Bit : 7-th
Output : 16
Given Number : 11
Remove Bit : 5-th
Output : 11
Given Number : 6
Remove Bit : 1-th
Output : 3``````
## Time Complexity
The time complexity of the given code is O(1) because the algorithm performs a fixed number of bitwise operations, regardless of the size of the input number `num`. The number of iterations or operations in the algorithm is constant, making it a constant-time operation.
## Comment
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# Semi-major axis
Semi-major axis
The semi-major axis of an ellipse
The major axis of an ellipse is its longest diameter, a line that runs through the centre and both foci, its ends being at the widest points of the shape. The semi-major axis is one half of the major axis, and thus runs from the centre, through a focus, and to the edge of the ellipse; essentially, it is the measure of the radius of an orbit taken at the orbit's two most distant points. For the special case of a circle, the semi-major axis is the radius. One can think of the semi-major axis as an ellipse's long radius.
The length of the semi-major axis a of an ellipse is related to the semi-minor axis' length b through the eccentricity e and the semi-latus rectum , as follows:
$b = a \sqrt{1-e^2},\,$
$\ell=a(1-e^2),\,$
$a\ell=b^2.\,$
The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches. Thus it is the distance from the center to either vertex (turning point) of the hyperbola.
A parabola can be obtained as the limit of a sequence of ellipses where one focus is kept fixed as the other is allowed to move arbitrarily far away in one direction, keeping fixed. Thus $a\,\!$ and $b\,\!$ tend to infinity, a faster than b.
## Ellipse
The semi-major axis is the mean value of the smallest and largest distances from one focus to the points on the ellipse. Now consider the equation in polar coordinates, with one focus at the origin and the other on the positive x-axis,
$r(1-e\cos\theta)=\ell.\,$
The mean value of $r={\ell\over{1+e}}\,\!$ and $r={\ell\over{1-e}}\,\!$, (for $\theta = \pi \, \text{and} \, \theta = 0$) is
$a={\ell\over 1-e^2}.\,$
In an ellipse, the semimajor axis is the geometric mean of the distance from the center to either focus and the distance from the center to either directrix.
## Hyperbola
The semi-major axis of a hyperbola is, depending on the convention, plus or minus one half of the distance between the two branches; if this is a in the x-direction the equation is:
$\frac{\left( x-h \right)^2}{a^2} - \frac{\left( y-k \right)^2}{b^2} = 1.$
In terms of the semi-latus rectum and the eccentricity we have
$a={\ell \over e^2-1 }.$
The transverse axis of a hyperbola coincides with the semi-major axis.[1]
## Astronomy
### Orbital period
In astrodynamics the orbital period T of a small body orbiting a central body in a circular or elliptical orbit is:
$T = 2\pi\sqrt{a^3 \over \mu}$
where:
a is the length of the orbit's semi-major axis
μ is the standard gravitational parameter
Note that for all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity.
The angular momentum H of a small body orbiting a central body in a circular or elliptical orbit is:
$H = \sqrt{a \cdot \mu \over (1-e^2)}$
where:
a and μ are as defined above
e is the eccentricity of the orbit
In astronomy, the semi-major axis is one of the most important orbital elements of an orbit, along with its orbital period. For solar system objects, the semi-major axis is related to the period of the orbit by Kepler's third law (originally empirically derived),
$T^2 \propto a^3 \,$
where T is the period in years, and a is the semimajor axis in astronomical units. This form turns out to be a simplification of the general form for the two-body problem, as determined by Newton:
$T^2= \frac{4\pi^2}{G(M+m)}a^3\,$
where G is the gravitational constant, and M is the mass of the central body, and m is the mass of the orbiting body. Typically, the central body's mass is so much greater than the orbiting body's, that m may be ignored. Making that assumption and using typical astronomy units results in the simpler form Kepler discovered.
The orbiting body's path around the barycentre and its path relative to its primary are both ellipses. The semi-major axis used in astronomy is always the primary-to-secondary distance; thus, the orbital parameters of the planets are given in heliocentric terms. The difference between the primocentric and "absolute" orbits may best be illustrated by looking at the Earth-Moon system. The mass ratio in this case is 81.30059. The Earth-Moon characteristic distance, the semi-major axis of the geocentric lunar orbit, is 384,400 km. The barycentric lunar orbit, on the other hand, has a semi-major axis of 379,700 km, the Earth's counter-orbit taking up the difference, 4,700 km. The Moon's average barycentric orbital speed is 1.010 km/s, whilst the Earth's is 0.012 km/s. The total of these speeds gives the geocentric lunar average orbital speed, 1.022 km/s; the same value may be obtained by considering just the geocentric semi-major axis value.
### Average distance
It is often said that the semi-major axis is the "average" distance between the primary focus of the ellipse and the orbiting body. This is not quite precise, as it depends on what the average is taken over.
• averaging the distance over the eccentric anomaly (q.v.) indeed results in the semi-major axis.
• averaging over the true anomaly (the true orbital angle, measured at the focus) results, oddly enough, in the semi-minor axis $b = a \sqrt{1-e^2}\,\!$.
• averaging over the mean anomaly (the fraction of the orbital period that has elapsed since pericentre, expressed as an angle), finally, gives the time-average
$a \left(1 + \frac{e^2}{2}\right).\,$
The time-average of the inverse of the radius[clarification needed], r −1, is a −1.
### Energy; calculation of semi-major axis from state vectors
In astrodynamics semi-major axis a can be calculated from orbital state vectors:
$a = { - \mu \over {2\varepsilon}}\,$
for an elliptical orbit and, depending on the convention, the same or
$a = {\mu \over {2\varepsilon}}\,$
and
$\varepsilon = { v^2 \over {2} } - {\mu \over \left | \mathbf{r} \right |}$
and
$\mu = G(M+m ) \,$
(standard gravitational parameter), where:
• v is orbital velocity from velocity vector of an orbiting object,
• $\mathbf{r }\,$ is cartesian position vector of an orbiting object in coordinates of a reference frame with respect to which the elements of the orbit are to be calculated (e.g. geocentric equatorial for an orbit around Earth, or heliocentric ecliptic for an orbit around the Sun),
• G is the gravitational constant,
• M and m are the masses of the bodies.
Note that for a given amount of total mass, the specific energy and the semi-major axis are always the same, regardless of eccentricity or the ratio of the masses. Conversely, for a given total mass and semi-major axis, the total specific energy is always the same. This statement will always be true under any given conditions.
## References
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• semi-major axis — didysis pusašis statusas T sritis fizika atitikmenys: angl. semi major axis vok. große Halbachse, f rus. большая полуось, f pranc. demi grand axe, m … Fizikos terminų žodynas
• Semi-minor axis — The semi minor axis of an ellipse In geometry, the semi minor axis (also semiminor axis) is a line segment associated with most conic sections (that is, with ellipses and hyperbolas). One end of the segment is the center of the conic section, and … Wikipedia
• semi — semi·abstract; semi·abstraction; semi·aerial; semi·amphibious; semi·annual; semi·annually; semi·anthracite; semi·ape; semi·aquatic; semi·arboreal; semi·arch; semi·arid; semi·auto; semi·autonomous; semi·basement; semi·beam; semi·bejan;… … English syllables
• semi-ellipse — semi ellipseˈ noun Half of an ellipse, bounded by a diameter, esp the major axis • • • Main Entry: ↑semi … Useful english dictionary
• Semi-rigid airship — Semi rigid airships are airships with a partial framework. These often consist of a rigid, occasionally flexible, keel frame along the long axis under the aerodynamic hull envelope. The partial framework can also be inside the hull. Semi rigids… … Wikipedia
• Transverse axis — refers to an axis which is (side to side, relative to some defined forward direction). In particular:* Transverse axis (aircraft) * For a hyperbola the transverse axis is in the same direction as the semi major axis … Wikipedia
• Major depressive disorder — For other depressive disorders, see Mood disorder. Major Depressive Disorder Classification and external resources … Wikipedia
• semimajor axis — /sem ee may jeuhr, sem uy , sem ee , sem uy / 1. Geom. one half the major axis of an ellipse. 2. Astron. one half the major axis of the ellipse that one celestial body describes around another, as a planet around the sun or a satellite around a… … Universalium
• Hyperbola — This article is about a geometrical curve, a conic section. For the term used in rhetoric, see Hyperbole … Wikipedia
• Ellipse — Elliptical redirects here. For the exercise machine, see Elliptical trainer. This article is about the geometric figure. For other uses, see Ellipse (disambiguation). Not to be confused with ellipsis. An ellipse obtained as the intersection of a… … Wikipedia
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You are on page 1of 21
# CALCULUS RESUME
Physics Department of Education Faculty of Mathematic and Science Ganesha University of Education 2011
## By: Gde Parie Perdana Class A, Semester I 1113021059
CALCULUS SUMMARY
I. FUNCTION a. Definition of Function A function f is a matching rule that links each element x in a set, called the region of origin (domain), with a unique value f(x) of the second set, the set obtained in this way is called the outcome function (the codomain).
Example of Function If f is a function from A to B we write: f: A B Which means that f maps A to B. A is the area of origin (domain) of f and B is called the results (codomain) of f. A is a collection of things, such as numbers.
Here are some examples: Set of even numbers: {..., -4, -2, 0, 2, 4 ...} Set of odd numbers: {..., -3, -1, 1, 3, ...} Set of prime numbers: {2, 3, 5, 7, 11, 13, ...} Positive multiples of 3 that are less than 10: {3, 6, 9}
## b. Domain, Codomain and Range
In this illustration:
## The set "A" is the Domain,
The set "B" is the Codomain, And the set of elements that get pointed to in B (the actual values produced by the function) are the Range, also called the Image.
c. Sort of Function Elementary Function o Constant function Is the simplest function with the general form is y = a. Identity function is the function that the general form y = x. o Linear functions The general form is y = a + bx. Where a is a constant and b is coefficients. o Quadratic equation The general form of the quadratic equation is y = ax2 + bx + c. With a0. The letters of a, b and c are called coefficients: the quadratic coefficient a is the coefficient of x2, the linear coefficient b is the coefficient of x, and c is a constant coefficient also called interest-free. o Trigonometry functions Trigonometric functions are functions in the form of sine and cosine (trigonometric parameters). A simple example such as y= sin ax. o Polynomial function Polynomial function is a function that contains a lot of interest in the independent variables, has the form ( ) integer called the power of the polynomial. Rational Function The definition of a rational function is a quotient polynomial function. ( ) Implicit Function Implicit Function is a function of independent and the nondependent variables are placed on the same segment. . Where, n is a positive
Explicit Function Explicit Function is a function where the independent and the nondependent variables are at a different segment.
Parametric Function Parametric function is functions of the independent variables are bounded to other variables.
d. Two Special Functions Absolute Function The absolute function is even function. This function is defined by:
| | { }
Function of the largest integer The greatest integer is neither even function nor odd function. = largest integer, smaller or equal to x.
## The graph of absolute function and function of the largest integer:
e. Operation of Functions A function is not same as a number but same as two numbers a and b can be add to get a new number a + b likewise to function f and g can be add to find new function f +g. If you have two functions f and g by the formula ( ) , ( )
. We can
make a new function (f + g), (f g), (f . g), and ( ) by giving value x to each of function. Suppose that f and g have natural domain so that each operation we can be defined:
Operation
( ( ( )( ) )( ) )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )
## Domain [0,) [0,) [0,) (0,)
( ) ( )
f. Translation By observing how the function is formed from the simple, can help us to drawing the graph.
## | |, then we can draw the graph of | by apply the concept of translation.
g. Trigonometry Function Trigonometric functions are functions in the form of sine and cosine (trigonometric parameters). A simple example such as The graph of sine and cosine .
Base on the graph above, we can get some points. 1. Both function and have the value in interval -1 until 1.
2. Both of the graphs repeated in contiguous interval as long as 2. 3. The graph of is symmetry to origin point, (0.0), meanwhile
is symmetry to the y axis (so, sine function is odd function and cosine function is even function). 4. The graph of the right. is same as , but has translation units to
h. Period and amplitude of trigonometric function A function is called to be periodic if there is number p so that: ( ) ( )
(
For all the members of real number x in domain of f. the smallest p the number like that called the period of f. the function of sine called be periodic because
)
## for all of x. Its true if: ( )
( )
If the periodic functions of f get the maximum and minimum value we defined that the amplitude A is half of the distance between highest and lowest point.
i. Relation to Angle Trigonometry Angles are commonly measured either in degrees or in radius. One radian is by definition the angle corresponding to an arc of length 1 on the unit circle.
## This leads to the results
j. Trigonometric Identity
II.
( )
( )
( )
( )
| ( ) |
## (no matter how small) provided that
there is a corresponding .
d. Theorem of Limit Let n be a positive integer, k be a constant, and f and g be functions that have limits at c. 1. 2. 3. 4. 5. 6. 7. 8. 9. ( )
[ ( ) [ ( ) [ ( )
( ) ( )
( ) ( )] ( )] ( )]
( ) provided ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
[ ( )]
( )
( )] ( ), provided ( )
when n is even.
Example, if
*
( ) ( ) ( )+ ( )
and
( )
, find
( ) ( )+
= =[
( )
( )] .
( )
= [4]2. = 32
e. Trigonometry Limit
1. lim it
sin ax ax a lim it x 0 bx x 0 sin bx b tan ax ax a 2. lim it lim it x 0 bx x 0 tan bx b sin ax tan ax a 3. lim it lim it x 0 sin bx x 0 tan bx b sin ax tan ax a 4. lim it lim it tan bx x 0 sin bx b x0
f. Limit at Infinity Limit For example f defined at [c,) for some numbers of c. We say that for each there is number related to M so that | ( ) | ( ) if
Limit For example f defined at [c,) for some numbers of c. We say that if for each there is number related to M so that: ( ) ( )
g. Infinite Limit We say that corresponding ( ) such that ( ) if for every positive number M, there exists a
Example:
and
## the graph is,
h. Asymptote We say that the line of x = c is the vertical asymptote of from the four formula is true. 1. 2. 3. 4.
( ) ( ) ( ) ( )
( ) if one or more
( )
## is the horizontal asymptote of the graph of
( )
( ) if
atau
i. Continuity at one point If f has a definition in an open interval which is contain c. We say that f continue at c if
( )
( )
1. There is
## 2. There is ( )(where c consisted in the domain of f). 3.
( ) ( )
j. Continuity on an interval We say that f continue at an open interval (a,b) if f continue at every points in the interval. f continue at an close interval [a,b] if f : 1. Continue at (a,b) 2. 3.
( ) ( ) ( ) ( )
III. DERIVATIVE a. Gradients Tangent line is a line that touches a curve at only one point. The slope of the tangent line is a derivative of the curve that is in contact with the tangent line.
( )
( )
Tangent line the curve y = f (x) at point P (c, f (c)) is the line through P with slope.
( )
( )
b. Derivative The derivative of a function f (x) is another function f '(x) whose value on any number c is
( )
( )
( )
or
( )
( )
( )
( )
)( )
( )
( )( ) ( )
( )
( )( ) ( )
( )
)( )
( ) ( ) ( )
( ) ,
## Quotient rule, if f and g is functions which in differentiation with ( ) then
( ) ( ) ( ) ( ) ( ) ( )
( )
d. Derivative of trigonometry function There are several formulas derived in the sine and cosine functions. ( ( ( ( ( ( e. Leibniz Notation for the Derivative Suppose now that the independent variable change from x to x + x. the corresponding change in the dependent variable, y, will be ( ) ( ) And the ratio
( ) ( )
) ) ) ) ) )
## In addition, there is also a more specializes.
Represents the slope of a secant line though (x,f(x)), as show in graph below. y f(x+ ) (x, ( ))
F(x)
(x,f(x))
x+
as
0, the slope of this second line approached that of the tangent line, and
for this letter slope of this secant line approaches that of the tangent line, and for this letter slope we use the symbol . Thus, ( f. The Chain Rule Facilitate the functioning of the chain rule decrease the degree of the polynomial function. Suppose if differentiable in ( ) dan ( ). If g differentiable in x and f , defined by (
)( )
( )
( )
## ( ), then composite function
)( )
( ( )) differentiable in x and ( g. Higher Derivatives Notation for derivatives Derivative First Second Third Fourth n-th Notation f ( ) ( ) ( )
( )
( ( )) ( )
## ( ) Notation y Notation D Leibniz Notation
( ) ( )
( )
( )
( )
( )
h. Implicit differentiation To derivative the implicit function we have to use implicit differentiation, as an example of a function
( )
( )
## IV. APPLICATION OF DERIVATIVE a. Maxima and Minima
Suppose that S is the domain of f, which have within c point. We can say that: 1. 2. 3. ( ) is maximum value f on S, if ( ) is minimum value f on S, if ( ) is extreme value f on S, if ( ) ( ) ( ) for all x in S; ( ) for all x in S;
## ( ) is maximum or minimum value
4. The function we want maximize or minimize is objective function. If f continue at close interval [ ], then f achieve maximum or minimum value in
there. Suppose that f is an interval I which contain c point. If ( ) is extreme value, then c must be critical point; that is c must be one of: 1. Tip point of I; 2. Stationery point of f, that is a point where 3. Singular point of f, that is a point where
( )
; or
## ( ) does not exist.
b. The Monotony and Concavity Suppose f is defined on interval I (open, close or neither). We say that: 1. f is increasing I if, for ever pair of number x1 and x2 in I. ( ) ( ) ( ) ( ) 2. f is decreasing I if, for ever pair of number x1 and x2 in I. 3. f is strictly monotonic on I if f is either increasing on I or decreasing on I.
c. Monotonicity Theorem Suppose f continuous on an interval I and differentiable at every interior point of I. 1. If 2. If
( ) ( )
for all x interior to I, then f is increasing on I. for all x interior to I, then f decreasing on I.
d. The Second Law and Concavity Suppose f differentiable on an open interval I. We say that f (as well as its graph) is concave up on I if f is increasing on I, and we say that f is concave down on I if f is decreasing on I.
( ) ( )
## for all x in I, then f is concave up on I. for all x in I, then f is concave down on I.
f. Inflection Points Let f be continuous at c. we call (c,f(c)) an inflection point of the graph of f if f is concave up on one side and concave down on the other side. Below is the graph.
g. Local Extreme Global maximum value is the simply the largest of the local maximum values. Similarly, the global minimum value is the smallest of the local minimum values. Let S, the domain of f, contains the point c. We say that: 1. ( ) is a local maximum value of f if there is an interval (a,b) containing c such that ( ) is the maximum value of f on ( 2. such that ( ) is the minimum value of f on ( 3. ) ) ; ( ) is a local minimum value of f if there is an interval (a,b) containing c ; ( ) is a local extreme value of f if it is either a local maximum or a local minimum value.
h. First and Second Derivative Test To proof the graph we can use the first derivative of function. Let f be continuous on an open interval (a,b) that contains a critical point c. 1. If
( )
( )
## for all x in (c,b), then ( ) is a local
maximum value of f. 2. If
( )
( )
## for all x in (c,b), then ( ) is a local
minimum value of f. 3. If
( ) has the same sign on both sides of c, then
## ( ) is not a local extreme
value of f. Besides the first derivative of function, there is another test for local maxima and minima that is sometimes easier to apply. It is the second derivative at the
and
( )
( ) ( )
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Live Chat
# KMP algorithm Assignment Help
## 1. Knuth-Morris-Pratt Introduction
Knuth-Morris-Pratt also called KMP is a pattern matching algorithm.
There are various other algorithms which are used for pattern matching. For example, Rabin Karp algorithm, Boyer Moor etc. KMP is one of them.
First, we need to understand, what is pattern matching algorithm?
### 1.1 Pattern matching algorithm
Let’s say we have a large text T containing words [1…..n] and we need to find the occurrence of pattern P [1..m] in this text T. To accomplish this task, we need some algorithms. Those algorithms are called pattern matching algorithms and KMP is one of them. But here two questions arise:
• Why do we need a separate algorithm for this task?
• What is the application of pattern matching? Where do we need pattern matching?
## KMP algorithm Assignment Help By Online Tutoring and Guided Sessions at AssignmentHelp.Net
Firstly, we’ll see the role of pattern matching.
#### 1.1.1 Why do we need Pattern matching?
In this section, we’ll talk about the applications of pattern matching. In daily life, whenever we search a particular string in notepad or word file, this search utility is the result of pattern searching algorithms. And there are unlimited areas where we are using the concept of pattern matching, we have listed some of them in the following figure.
#### 1.1.2 Why do we need different algorithms?
As we saw in above figure, the importance of pattern matching. There is also an important detail about pattern matching which we should know. It is not necessary that every time people will search only text. They may want to search different types of data like audio, video, image etc. So tackle with different data types, we need better algorithms. After finding the location of matching pattern, it can also be used to substitute some component of that pattern with another string sequence that is first search then replace.
## 2. The Naïve Method
Let’s first study the naïve method for doing this. The idea is to take each word and compare it with the pattern.
```{`
Pseudocode for this method
Function match(text[], pattern[])
{
// n is size of text
// m is size of pattern
for(i = 0; i < n; i++)
{
for(j = 0; j < m && i + j < n; j++)
if(text[i + j] != pattern[j])
break; //doesn’t match, break the inner loop
if(j == m) // match found
}
}
`}```
As we know, the problem of pattern matching is like a needle in haystack and If the length of the text is n and that of pattern is m. Then, it will take O(n*m) iterations to complete the task. In that case, the time complexity for worst case is O(mn). That is why, this method is impractical.
## 3. KMP algorithm
This algorithm was developed by Donald Knuth, Vaughan Pratt and James H. Morris in 1977. The idea was to search occurrences of a word within a text by using the result of previously matched characters. It uses the observation through which if a mismatch occurs, the word itself will provide sufficient information to determine where we’ll find the next match. At high level, there is a similarity between KMP and the naïve algorithm. In both the cases, the scanning is done in the same order that is from 1 to n-m, where n is the length of the text and m is the length of the pattern. But in KMP algorithm, it uses the information obtained by the previous occurrences of the pattern in the whole text. By using this information, it is able to skip those parts of the text which cannot contain the pattern we are searching. If we can calculate that how much we can shift in advance, then we can reduce the number of iterations. We have explained this in the following example:
```{`
Position: 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Text (t): a b a b a a b b a b a b b a
Pattern (p): a b a b b
a b a b b
a b a b b
a b a b b
`}```
Start from the very first position, here the pattern is matched up to position 4th. At position 5 mismatch occurred, so shifting the pattern one place will not do any good. We also need to match the pattern at position 3 with the text at position 5, so as to make sure we don’t miss any matches. We’ll continue matching from position 5. Then, we encounter another mismatch at position 6. This shifting does not work. This is because, we did not consider the fact that ab is repeating itself in the pattern, thus only two places shift resulted in mismatch. Since the condition pattern at position 1 pattern at position 4 is satisfied. Thus, we take a shift of three places and resume matching at position 6 and here we find mismatch at position 8. Then again, we shifted the pattern three places. And finally, we find the pattern at position 9 of the text.
The naïve algorithm would have taken 23 iterations, but this algorithm will give result in 15 iterations.
Now, we’ll mathematically deduce the above explained shift:
Let A[1….m] be the array which gives the information that how much shift do we need. This array is called Prefix array, it stores the information of previous matches in such a way that will exactly tell us the position where we have to jump for finding the match. The length of this array is equal to the length of the pattern. For example, If this array at any column contains the value 1. That means the sequence up to which the text matched contain the same prefix and suffix of length 1. The length of the longest proper prefix in the pattern that matches a proper suffix in the same pattern. So, we can shift accordingly. In other words, we do not need to backtrack on the text.
A[1]=0
A[i] = max (s| (s=0) or (p[1… s-1] = p[i-s+1 .. i-1] and p[s] p[i])), the maximisation of s is needed that means minimised shift. So, that, no occurrence of pattern will be missed
### 3.1 Matching algorithm
```{`
1. i=1; j=1;
2. while (i<=m) and (j<=n) do begin
3. while (i>=0) and (p[i]<>[j])
4. do i=h[i];
5. i=i+1; j=j+1
6. end
7. if i <= m then print(“pattern is not found in the text”)
8. else print(“found at”, j-i+1)
`}```
Let the function f(i) be defined by
```{`
f(1) = 0
f(i) = max{ s | (1 <= s < i) and (p[1 .. s-1] = p[i-s+1 .. i-1]) }
`}```
Both the definition of A[i] and f(i) are almost same. But in the case of A[i], we need to satisfy the condition of p[s] p[i] which is not needed in the case of f(i). So, in case of the prefix array, the pattern matching of p is on pattern itself. We’ll use this function f(i) in next section.
### 3.2 Prefix array
In this section, we’ll write algorithm for computing prefix array that is A[i].
```{`
1. t=0; A[1]:=0;
2. for i=2 to m do begin
3. while (t>0) and (p[i-1]<>p[t]) //t=f(i-1)
4. do t=A[t];
5. t:=t+1;
6. if p[i]<>p[t] //t=f(i)
7. then A[i]:=t else A[i]:=A[t]
8. end
`}```
Explanation of the above algorithm:
We are proceeding from left to right. In algorithm, we mentioned that t=f(i-1) in line 3. Then, in that case, we can extend the matched portion by 1 (line 5).
If p[t] p[i-1], we slide the pattern over itself until we find p[t]=p[i-1] and we can then extend the matching portion.
In line 7, the condition satisfied, so A[i] is set to t.
## 4. Example illustrating KMP algorithm
```{`Pattern p = a b a b b
i = 2, at line 7 t=1,
p[1]<>p[2],
f(2) = 1, A[2]=1
i
a b a b b
a b a b b
t
i = 3, t=1 at line 4,
p[1]<>p[2],
t=A[1]=0, at line 7, t=1, f(3)=1, p[1]=p[3], A[3]=0
a b a b b
a b a b b
i = 4 t=1 at line 4. p[3]=p[1], t:=t+1, t=2.
At line 7, p[4]=p[2], A[4]:=A[2]=1
i = 5 t=2, f(4)=2. At line 4, p[4]=p[2], t=t+1=3. f(5)=3.
At line 7, p[5]<>p[3],
A[5]=t=3
Finally we have
i | 1 2 3 4 5
------------------------------
pat | a b a b b
f | 0 1 1 2 3
A | 0 1 0 1 3
`}```
## 5. Analysis of KMP algorithm
Creating prefix array is a pre-processing task and if the length of pattern is m. Then to execute that algorithm, the running time will be O(m), the space complexity will also be O(m) and in second algorithm, the running time will be O(n) where n is the string length in the text.
So, the total time complexity for this algorithm is O(m+n).
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# Banked Curve Train Condominium on Mars
• Yachtsman
In summary, the writer is researching ways to simulate Earth's gravity on Mars and has worked on the math to determine the necessary speed and bank angle. He is interested in what would make for safe and reasonable height and width dimensions for the condos and other design considerations.
Yachtsman
I am researching for a sci-fi screenplay that I am writing for fun. To keep bone density from losing the estimated 1.5% per month on Mars I want to simulate Earth gravity by having the colonists live aboard condominiums that are fitted with a rail car suspension and travel in a perfect circle in a banked curve. I envision two pairs (4 total) of rails for each track system to provide safety with the redundancy of a second set of rails and wheels as backup for any failure.
This is sci-fi and so I am able to make certain reasonable assumptions in this pretend world for our future Martian colonists; I envision that we are able to easily power the needs of this future city on rails.
I've worked on the math and can easily determine the speed necessary for a frictionless banked curve (I chose frictionless because I want gravity to feel like it's falling straight down through the floor) at the various radius' that I choose for each track, and I calculate the necessary bank angle to be approximately 67.21 =DEGREES(ACOS(3.8/9.81)). Wow, a really steep bank--the reason is because the Martian gravity is so weak compared to the target gravity of 9.81m/s2. There is great debate regarding the Martian gravity, I'm most impressed with the latest estimates of 3.8/s2, so I chose that for my math.
OK, all that to finally ask my question:
What are reasonable width and height dimensions for these railed condos? I want to know what formula's and rules of thumb I can use to determine what is safe and reasonable design.
Does length of the "rail car-condo" have any effect, or is it simply width and height based upon the angle, speed and gravity?
Obviously a very tall and narrow car will be more dangerous, especially during a wind storm, so what is the math to show how wind will effect the cars--remember, they have some strength of a bicycle wheel because all cars are connected together in a circle; and the connection points can be stronger than a typical rail car because the angle is typically fixed--except for the rare occurrence where we'll have cars change tracks so that repairs or maintenance can be done to a particular track. Because of this I think that part of the math will depend upon how much I want to rely upon this strength in the linkage points; because of they were truly very solid and strong, then it would be impossible for wind to blow over a 1km diameter ring regardless of the height being a few stories tall; but failure does happen and so it's safer to not rely heavily upon this strength.
A rail with a radius of 500m traveling along the banked rail will require a velocity of approximately 67.246m/s; I derive this using =SQRT(500 * SQRT(9.812 - 3.82))
I'm interested to know what would make for safe and reasonable height and width for the condos--I am especially interested in the formulas used.
Other design consideration are, what to do if power fails. Let's say that we have a 3-story tall condo banked at over 67°, it would be best to have a way to automatically adjust the angle of the train car and/or the track so that they don't fall over, and especially so that people don't fall out of bed in the middle of the night during such a disaster.
Good movies think through things like this, I want to create a quality design for my script.
Thank you in advance for any ideas or thoughts that you may have on the subject.
Edit: I had a thought, rather than having the wheels under the cars like a normal train, I could place them in front and back of the cars, make the rails wider than the car for safety, attach each car to the wheels via a central pivot point that is located above the center of gravity for the car; in this way the cars would automatically tilt to the perfect angle based upon the acceleration of the car. With this design the pivot point must be located high above the rail or the rails are wide enough for the bottom of the car to swing in between the rails; this actually becomes possible when the rails are about 1.6 times the width of the car; in this design a rounded trench would be dug between the banked rails or outer rail would be raised above ground. It would be cheaper and safer to dig the trench.
Last edited:
Yachtsman, In the days before man went into space, elaborate schemes were devised to simulate Earth gravity. Rotating donut-shaped space stations were popular. The station in the movie "2001" is an example. It shows how an incoming ship would dock in the center. The crew would then experience changing gravity as they moved from the center to the rim, where they would live. They would keep in shape by doing laps around the circumference.
But the idea of a rotating station was never put into practice. Why not? Because living on a merry-go-round would drive people crazy! For example one thing they forgot to consider was the Coriolis force. Every time you moved your hand it would be deflected sideways. Many things like antennas and solar panels need to be kept pointing in the same direction. Eventually they decided the drawbacks were too great, and it was better to just put up with the zero g.
Bill_K said:
Yachtsman, In the days before man went into space, elaborate schemes were devised to simulate Earth gravity. Rotating donut-shaped space stations were popular. The station in the movie "2001" is an example. It shows how an incoming ship would dock in the center. The crew would then experience changing gravity as they moved from the center to the rim, where they would live. They would keep in shape by doing laps around the circumference.
But the idea of a rotating station was never put into practice. Why not? Because living on a merry-go-round would drive people crazy! For example one thing they forgot to consider was the Coriolis force. Every time you moved your hand it would be deflected sideways. Many things like antennas and solar panels need to be kept pointing in the same direction. Eventually they decided the drawbacks were too great, and it was better to just put up with the zero g.
I appreciate any feedback, even kindly worded critique as you've done. Thanks for that. From what my research has shown, the sickness that you speak of greatly reduceds as the diameter increases; I've read many studies that show life from the "Coriolis effect" is fine with diameters of 1km or larger--and the larger the better.
To terraform Mars we need some form of artificial gravity because studies also show that the colonists would lose their bone density at 1.5% per month there; in just a few short years Mars would "feel" like Earth gravity to them and they could never come back to Earth if they wanted. I suppose they could just go there and over time our bodies would adjust after a few centuries of breeding. But, I'd like to think that we would prefer being a people who want to stay strong, not weak, and so we would devise ways to keep the people there strong. So, with that premise in mind, I thought through options, and it seems that the only 3 ways for artificial gravity would be:
1. A train condominium concept similar to what I'm describing
2. frequent trips to space where you have the tethered space station concept (tethered in excess of 1km span apart)
3. or some futuristic artificial gravity that seems impossible to imagine at this point since we're not made of metal
If we don't have added gravity then our people there will eventually become weakened Matians who can never leave the planet; this thought is not my preference and so I'd like to imagine a solution.
One thing that I thought of is that if i dig into the soil (I wanted to say dig into the mars, since we say dig into the Earth LOL), and create a rounded trench, this allows the condo cars to travel partially subterranian, and that would greatly reduce the wind effect; so as long as my condos are not over 4 stories high and I can bury at least 1 of the stories, this has the "feeling" of being reasonbly safe assuming a wide base--as mentioned, the rail base would have to be 1.6 times with width, and so some simply math shows that 3 stories is incredibly secure with a base of 1.6 the width; the rails simply need to be strong enough to support the strain. It would be nice if I could find some formulas that help me determine if I can go with 5 stories, or 6, or 10; what is the safe height limit for a tran car that is 10 to 15 meters wide with a rail base 1.6x of that?
I forgot to mention one important point: I was not planning for people to live on these condo's all the time--simply as a place to sleep and get some of their day spent with Earth gravity; it seemed that getting some Earth G time on Mars was cheaper and better than getting it in space because at least on Mars you could get some added gravity every day, whereas the tethered space station idea requires a person to spend extended stays at the station and then rotate down back down to Mars on periodic crew schedules; at this point our colonist seem to be living more like remote oil riggers who travel to the remote site during crew changes; it's not nearly as fun of a lifestyle as getting some Earth G time while sleeping and then spending some time there during the early morning and late evening to get some walking time.
I'm curious if any studies have thought about simply adding weight to the clothing people wear while walking around on Mars. If I am wearing clothing that has a heavy metal lined into the clothing so that my body is close to an Earth weight. This added weight would force my bones and muscles to work harder; an interesting thought.
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Apparently bones and muscles are straight up "programmed" to deteriorate in reduced gravity or with reduced action. It's an actual evolutionary survival mechanism (albeit obsolete one) that humans have, and it seems far from impossible to turn it off with some genetic engineering, or even perhaps with pills. I understand that there is active research in this field as it's obviously a potentially huge market.
Point is, by the time we'll be vacationing/ living on Mars, reduced gravity could very well be a rather minor problem. If this reduced gravity problem is just a nuisance in your story then that might be one way to make it go away. If it's the fundamental part of your story though, then, well...please ignore my post :)
Lsos said:
Apparently bones and muscles are straight up "programmed" to deteriorate in reduced gravity or with reduced action. It's an actual evolutionary survival mechanism (albeit obsolete one) that humans have, and it seems far from impossible to turn it off with some genetic engineering, or even perhaps with pills. I understand that there is active research in this field as it's obviously a potentially huge market.
Point is, by the time we'll be vacationing/ living on Mars, reduced gravity could very well be a rather minor problem. If this reduced gravity problem is just a nuisance in your story then that might be one way to make it go away. If it's the fundamental part of your story though, then, well...please ignore my post :)
I know we've all watched sci-fi and laughed about how fake it was; I would never want to do such a thing, so while the focus is not on the gravity, I do need to have some explanation about how we're dealing with the gravity--this is typically done in side passing conversations where it's not so obvious that a gravity lesson is being given; e.g. a Martian school teacher or parent is teaching a child about something, two scientists joking about how it was in the old days before it was solved, a TV program talks about a landmark anniversary and the news anchor explains how it was in the old days, ect.
If a pill is the solution then I need to watch someone take a pill and then have some explanation as to what the pill does and give credibility to the movie.
Movies are very visual, in fact the rule is to show it rather than say it if you can; and having a huge train of condo's is certainly visual and can be incorporated into a story-line in a captivating manner. My movie does not rely upon a train.
Thanks for idea about the pill, I had not seen or thought of that. The math used for doing the train idea has been fun. Maybe I'll have the trains and show them being phased out because the pill is now perfected. That gives me my visual plus shows how technology improves. I would imagine people would still like the trains as a training gravity in preparation of traveling back to earth--even if our muscles are artificially stronger than normal based upon Mars gravity, a person would need to come off of the drugs when traveling to Earth if they behave more like a steroid rather than simply halting decay--my premise for this is that the drug are steroid like, and under greater gravity stress it seems plausible that the muscles grow even more under the greater stress.
However this brings another interesting idea with such a pill, which is what happens when "some" people on Earth want superhuman strength (we always have bodybuilders and those who seek the ideal body to be impressive, plus athletes, etc)? And the same for our Martian brothers and sisters?
With all "advances" we seem to give up something--there seems to be a dark side with the benefits no matter what the human race seems to invent.
Last edited:
Yachtsman said:
However this brings another interesting idea with such a pill, which is what happens when "some" people on Earth want superhuman strength (we always have bodybuilders and those who seek the ideal body to be impressive, plus athletes, etc)? And the same for our Martian brothers and sisters?
This is a large driving force behind such a solution. People spend long hours at the gym fighting an ultimately uphill battle against the muscle degradation which their own body is commanding. If you could reduce the rate of muscle degradation (which is regulated internally) the time spent at the gyum would be more effective and long lasting.
It's not so much about "superhuman" strength as it is about efficiency. After all, if you want superhuman strength you can already get it simply by spending many hours a day at the gym with a personal trianer. Yet, there's no ethical issues about this...in fact, these people are generally praised and admired. If there's a way to streamline the whole process without any side effects, it might be very beneficial to everybody...especially now when there are more fat people in the world then there are hungry people.
Lsos said:
This is a large driving force behind such a solution. People spend long hours at the gym fighting an ultimately uphill battle against the muscle degradation which their own body is commanding. If you could reduce the rate of muscle degradation (which is regulated internally) the time spent at the gyum would be more effective and long lasting.
It's not so much about "superhuman" strength as it is about efficiency. After all, if you want superhuman strength you can already get it simply by spending many hours a day at the gym with a personal trianer. Yet, there's no ethical issues about this...in fact, these people are generally praised and admired. If there's a way to streamline the whole process without any side effects, it might be very beneficial to everybody...especially now when there are more fat people in the world then there are hungry people.
I agree with so much of what you are saying here; although I tend to be more of a person of "balance" and feel that anything can be taken to excess that is eventually bad, it seems there is a lot of gray between where that "excess" line falls--we are all unique and should be applauded for pursuing our love and ambitions.
My pondering of the super-human idea (both physical and mental IQ) is very close to the concept presented in the movie Gattaca where designer babies become a reality. Here is my favorite quote from the movie from the Geneticist who is trying to talk the parents into going with a designer baby rather than a natural birth, "We want to give your child the best possible start. Believe me, we have enough imperfection built in already. Your child doesn't need any more additional burdens. Keep in mind, this child is still you. Simply, the best, of you. You could conceive naturally a thousand times and never get such a result."
What is the world like when it's impossible to "rise above your station in life" without being augmented with technology? Who received the "gift" of augmentation? it's an interesting thought.
Lemme ask this. Is your reason that the Martian people want to stay strong plausible? If they are living most of their lives there, then I don't see any big issue with weakening, especially if you have to go out of your way to exercise or spend money on drugs to keep you strong. If you have to go to Earth, I could easily see bulking back up or taking the drugs for a period of time before heading back, or even on the trip itself.
Drakkith said:
Lemme ask this. Is your reason that the Martian people want to stay strong plausible? If they are living most of their lives there, then I don't see any big issue with weakening, especially if you have to go out of your way to exercise or spend money on drugs to keep you strong. If you have to go to Earth, I could easily see bulking back up or taking the drugs for a period of time before heading back, or even on the trip itself.
Yeah, I know what you mean... you're right and I'd speculated on that too from time to time; each time I think about it the two thoughts that run through my mind are:
1. people tend to always want to be strong not weak, and if a simple pill (assuming that it's not grossly expensive or has side effects) solves the problem that I can't imagine many choosing to become weak if they had the option not to.
2. people from Earth will visit and they will be so much stronger; Martian dwellers would be weak and vulnerable to such people if they were truly 1.5, 2 or even up to 3-times stronger.
An interesting questions is whether Earthlings will always be stronger because whatever drugs help the Martians will also be used by Earthlings and "assuming" gravity allows a person to grow stronger than their current state, it's possible that Earthlings would always be stronger; perhaps there is a limit to how much advantage the extra gravity provides, and so it seems there is a law of diminishing return where the gap decreases; so it has the highest potential of an Earthling being 3-times stronger and then through diminishing returns could decrease to 2x and finally down to potentially 1x (no advantage).
Part of my movie is going to deal with the issues of augmentation and so this is an interesting thought; it's certainly going to be our future at some point.
Yachtsman said:
[*]people tend to always want to be strong not weak, and if a simple pill (assuming that it's not grossly expensive or has side effects) solves the problem that I can't imagine many choosing to become weak if they had the option not to.
I'd say there are many things to consider, not just whether someone chooses to be weak or strong. Personal views on medication being necessary or not, how rich/poor someone is, how physical a persons job is, their hobbies, etc. Many people probably wouldn't even care if the loss of bone and muscle didn't have a negative impact on their day to day lives. Hell, I could easily see the existence of activists that promote adapting to Mars and to not take pills.
[*]people from Earth will visit and they will be so much stronger; Martian dwellers would be weak and vulnerable to such people if they were truly 1.5, 2 or even up to 3-times stronger.
Maybe. Remember the trip to Mars probably isn't going to be a short one, so unless they are taking medications themselves then they will lose muscle and bone mass over time as well. I can imagine some interesting plot options when you have an influx of people that are physically stronger than the natives. And that probably also influences peoples view on augmentations.
Drakkith said:
I'd say there are many things to consider, not just whether someone chooses to be weak or strong. Personal views on medication being necessary or not, how rich/poor someone is, how physical a persons job is, their hobbies, etc. Many people probably wouldn't even care if the loss of bone and muscle didn't have a negative impact on their day to day lives. Hell, I could easily see the existence of activists that promote adapting to Mars and to not take pills.
Maybe. Remember the trip to Mars probably isn't going to be a short one, so unless they are taking medications themselves then they will lose muscle and bone mass over time as well. I can imagine some interesting plot options when you have an influx of people that are physically stronger than the natives. And that probably also influences peoples view on augmentations.
I agree with you; I think the activist idea is pretty funny, it made me laugh, and has some truth to it--also the fact that people choose so many different paths for their life's journey.
Regarding the trip over there, it would either be done with bone density drugs and/or a rocket design that provides for artificial gravity with centripetal force; studies have shown that it does not matter what direction the rotation is in relationship to the rocket's travel toward Mars.
My favorite designs are where you use the massive weight of the rocket booster as a counterweight and then extend the smaller habitation station at the end of a long tether that is about 500 to 1000m long; the longer the better for comfort and yet the longer the more risk for having the tether break; you would want to use multiple tethers and have enough fuel aboard the station to get back to the counterweight should massive tether failure occur where all of the lines are cut. Getting back to the counterweight assumes that you can repair and start rotating again, and if repairs are not possible you still need to have fuel to reach this general location so that you are still back on course; the counterweight would also be "off course" but not as much since it's larger and I assume it would have some fuel left over to get it back on course as well.
I think that it's safer to have the rotation as perpendicular to the line of travel rather than in parallel because if such a failure did occur close to Mars or Earth with a parallel rotation they have a 10 to 20% chance of being catapulted straight into the gravitational field of these planets with little time to react. During such an event it seems that it would be best to use the fuel to obtain the best possible planetary orbit safely rather than trying to get back to the counterweight. You'd try to stabilize your spin so that you can point your rockets toward the planet and then start slowing down your speed in a direction and manner to achieve orbit.
The science of terraforming Mars is quite interesting as well; what an incredible feat of engineering that would be. It seems that it would take at least 150 Earth years or more to achieve. I've read things that show 3 stages of warming the planet, planting trees, and then finally breathable atmosphere. I read speculation that the oxygen left the planet because of the weaker gravity; if this is true then we may need to plant extra trees to provide for the gradual oxygen loss. It's certainly an interesting problem.
Thanks for the ideas.
## 1. What is a Banked Curve Train Condominium on Mars?
A Banked Curve Train Condominium on Mars is a proposed concept for a futuristic living space on the planet Mars. It is essentially a train that travels along a curved track, providing housing for its passengers.
## 2. How will the Banked Curve Train Condominium on Mars work?
The train will be powered by renewable energy sources, such as solar panels, and will use magnetic levitation technology to travel along a curved track. The train will be self-sustaining, with living quarters, food production, and waste management all built into its design.
## 3. Who will be able to live in the Banked Curve Train Condominium on Mars?
The train will be open to anyone who is able to afford a ticket. It is envisioned as a luxury living space for those who want to experience life on Mars in a unique and innovative way.
## 4. What are the benefits of living in a Banked Curve Train Condominium on Mars?
Living in a Banked Curve Train Condominium on Mars would provide a one-of-a-kind experience, with the opportunity to live and travel on the red planet. It would also allow for a self-sufficient and sustainable lifestyle, with access to resources and amenities through the train's design.
## 5. What are the challenges of implementing a Banked Curve Train Condominium on Mars?
There are several challenges that would need to be addressed in order to make this concept a reality. These include the cost of building and maintaining the train, the technological advancements needed for its operation, and the potential effects on the environment and other aspects of Mars.
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# Search by Topic
#### Resources tagged with Mathematical reasoning & proof similar to Incentre Angle:
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### There are 179 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof
### Pent
##### Stage: 4 and 5 Challenge Level:
The diagram shows a regular pentagon with sides of unit length. Find all the angles in the diagram. Prove that the quadrilateral shown in red is a rhombus.
### No Right Angle Here
##### Stage: 4 Challenge Level:
Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.
### Picturing Pythagorean Triples
##### Stage: 4 and 5
This article discusses how every Pythagorean triple (a, b, c) can be illustrated by a square and an L shape within another square. You are invited to find some triples for yourself.
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
### Picture Story
##### Stage: 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### The Golden Ratio, Fibonacci Numbers and Continued Fractions.
##### Stage: 4
An iterative method for finding the value of the Golden Ratio with explanations of how this involves the ratios of Fibonacci numbers and continued fractions.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### Iffy Logic
##### Stage: 4 and 5 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### The Great Weights Puzzle
##### Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Proof: A Brief Historical Survey
##### Stage: 4 and 5
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Matter of Scale
##### Stage: 4 Challenge Level:
Prove Pythagoras' Theorem using enlargements and scale factors.
### To Prove or Not to Prove
##### Stage: 4 and 5
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Proof Sorter - Quadratic Equation
##### Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### Convex Polygons
##### Stage: 3 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### Pareq Exists
##### Stage: 4 Challenge Level:
Prove that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.
### Unit Interval
##### Stage: 4 and 5 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
### Zig Zag
##### Stage: 4 Challenge Level:
Four identical right angled triangles are drawn on the sides of a square. Two face out, two face in. Why do the four vertices marked with dots lie on one line?
### Magic Squares II
##### Stage: 4 and 5
An article which gives an account of some properties of magic squares.
### Impossible Sandwiches
##### Stage: 3, 4 and 5
In this 7-sandwich: 7 1 3 1 6 4 3 5 7 2 4 6 2 5 there are 7 numbers between the 7s, 6 between the 6s etc. The article shows which values of n can make n-sandwiches and which cannot.
### Mouhefanggai
##### Stage: 4
Imagine two identical cylindrical pipes meeting at right angles and think about the shape of the space which belongs to both pipes. Early Chinese mathematicians call this shape the mouhefanggai.
### Tessellating Hexagons
##### Stage: 3 Challenge Level:
Which hexagons tessellate?
### Folding Squares
##### Stage: 4 Challenge Level:
The diagonal of a square intersects the line joining one of the unused corners to the midpoint of the opposite side. What do you notice about the line segments produced?
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
##### Stage: 3 Challenge Level:
What can you say about the angles on opposite vertices of any cyclic quadrilateral? Working on the building blocks will give you insights that may help you to explain what is special about them.
### More Marbles
##### Stage: 3 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
### Tree Graphs
##### Stage: 4 Challenge Level:
A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree. . . .
### Converse
##### Stage: 4 Challenge Level:
Clearly if a, b and c are the lengths of the sides of a triangle and the triangle is equilateral then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true, and if so can you prove it? That is if. . . .
### Calculating with Cosines
##### Stage: 4 and 5 Challenge Level:
If I tell you two sides of a right-angled triangle, you can easily work out the third. But what if the angle between the two sides is not a right angle?
### Advent Calendar 2011 - Secondary
##### Stage: 3, 4 and 5 Challenge Level:
Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
### Diophantine N-tuples
##### Stage: 4 Challenge Level:
Take any whole number q. Calculate q^2 - 1. Factorize q^2-1 to give two factors a and b (not necessarily q+1 and q-1). Put c = a + b + 2q . Then you will find that ab+1 , bc+1 and ca+1 are all. . . .
### Marbles
##### Stage: 3 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### A Long Time at the Till
##### Stage: 4 and 5 Challenge Level:
Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Folding Fractions
##### Stage: 4 Challenge Level:
What fractions can you divide the diagonal of a square into by simple folding?
### Whole Number Dynamics II
##### Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
### Long Short
##### Stage: 4 Challenge Level:
A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4. Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot. . . .
### Whole Number Dynamics III
##### Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Find the Fake
##### Stage: 4 Challenge Level:
There are 12 identical looking coins, one of which is a fake. The counterfeit coin is of a different weight to the rest. What is the minimum number of weighings needed to locate the fake coin?
### Composite Notions
##### Stage: 4 Challenge Level:
A composite number is one that is neither prime nor 1. Show that 10201 is composite in any base.
### Königsberg
##### Stage: 3 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Pattern of Islands
##### Stage: 3 Challenge Level:
In how many distinct ways can six islands be joined by bridges so that each island can be reached from every other island...
### Children at Large
##### Stage: 3 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
### Take Three from Five
##### Stage: 3 and 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Rhombus in Rectangle
##### Stage: 4 Challenge Level:
Take any rectangle ABCD such that AB > BC. The point P is on AB and Q is on CD. Show that there is exactly one position of P and Q such that APCQ is a rhombus.
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Angle Trisection
##### Stage: 4 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
| 2,476 | 10,435 |
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| 4.375 | 4 |
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https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/6_Hyperbolic_functions/6.7_Miscellaneous/6.7.1_Hyperbolic_functions/rese725.htm
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| 677,110,098 | 5,170 |
### 3.725 $$\int \frac{A+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx$$
Optimal. Leaf size=82 $-\frac{-A b \sinh (x)-A c \cosh (x)+b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}-\frac{c C \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}$
[Out]
-((c*C*ArcTan[(c*Cosh[x] + b*Sinh[x])/Sqrt[b^2 - c^2]])/(b^2 - c^2)^(3/2)) - (b*C - A*c*Cosh[x] - A*b*Sinh[x])
/((b^2 - c^2)*(b*Cosh[x] + c*Sinh[x]))
________________________________________________________________________________________
Rubi [A] time = 0.0790252, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {3154, 3074, 206} $-\frac{-A b \sinh (x)-A c \cosh (x)+b C}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}-\frac{c C \tan ^{-1}\left (\frac{b \sinh (x)+c \cosh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}$
Antiderivative was successfully verified.
[In]
Int[(A + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^2,x]
[Out]
-((c*C*ArcTan[(c*Cosh[x] + b*Sinh[x])/Sqrt[b^2 - c^2]])/(b^2 - c^2)^(3/2)) - (b*C - A*c*Cosh[x] - A*b*Sinh[x])
/((b^2 - c^2)*(b*Cosh[x] + c*Sinh[x]))
Rule 3154
Int[((A_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> -Simp[(b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Co
s[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - c*C)/(a^2 - b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d +
e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - c*C, 0]
Rule 3074
Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{A+C \sinh (x)}{(b \cosh (x)+c \sinh (x))^2} \, dx &=-\frac{b C-A c \cosh (x)-A b \sinh (x)}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}-\frac{(c C) \int \frac{1}{b \cosh (x)+c \sinh (x)} \, dx}{b^2-c^2}\\ &=-\frac{b C-A c \cosh (x)-A b \sinh (x)}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}-\frac{(i c C) \operatorname{Subst}\left (\int \frac{1}{b^2-c^2-x^2} \, dx,x,-i c \cosh (x)-i b \sinh (x)\right )}{b^2-c^2}\\ &=-\frac{c C \tan ^{-1}\left (\frac{c \cosh (x)+b \sinh (x)}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}-\frac{b C-A c \cosh (x)-A b \sinh (x)}{\left (b^2-c^2\right ) (b \cosh (x)+c \sinh (x))}\\ \end{align*}
Mathematica [A] time = 0.439, size = 155, normalized size = 1.89 $-\frac{\sinh (x) \left (2 b c^2 C \sqrt{b+c} \tan ^{-1}\left (\frac{b \tanh \left (\frac{x}{2}\right )+c}{\sqrt{b-c} \sqrt{b+c}}\right )-A (b-c)^{3/2} (b+c)^2\right )+2 b^2 c C \sqrt{b+c} \cosh (x) \tan ^{-1}\left (\frac{b \tanh \left (\frac{x}{2}\right )+c}{\sqrt{b-c} \sqrt{b+c}}\right )+b^2 C \sqrt{b-c} (b+c)}{b (b-c)^{3/2} (b+c)^2 (b \cosh (x)+c \sinh (x))}$
Antiderivative was successfully verified.
[In]
Integrate[(A + C*Sinh[x])/(b*Cosh[x] + c*Sinh[x])^2,x]
[Out]
-((b^2*Sqrt[b - c]*(b + c)*C + 2*b^2*c*Sqrt[b + c]*C*ArcTan[(c + b*Tanh[x/2])/(Sqrt[b - c]*Sqrt[b + c])]*Cosh[
x] + (-(A*(b - c)^(3/2)*(b + c)^2) + 2*b*c^2*Sqrt[b + c]*C*ArcTan[(c + b*Tanh[x/2])/(Sqrt[b - c]*Sqrt[b + c])]
)*Sinh[x])/(b*(b - c)^(3/2)*(b + c)^2*(b*Cosh[x] + c*Sinh[x])))
________________________________________________________________________________________
Maple [A] time = 0.067, size = 115, normalized size = 1.4 \begin{align*} -2\,{\frac{1}{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}b+2\,c\tanh \left ( x/2 \right ) +b} \left ( -{\frac{ \left ( A{b}^{2}-A{c}^{2}-Ccb \right ) \tanh \left ( x/2 \right ) }{ \left ({b}^{2}-{c}^{2} \right ) b}}+{\frac{bC}{{b}^{2}-{c}^{2}}} \right ) }-2\,{\frac{Cc}{ \left ({b}^{2}-{c}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,\tanh \left ( x/2 \right ) b+2\,c}{\sqrt{{b}^{2}-{c}^{2}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x)
[Out]
-2*(-(A*b^2-A*c^2-C*b*c)/(b^2-c^2)/b*tanh(1/2*x)+b*C/(b^2-c^2))/(tanh(1/2*x)^2*b+2*c*tanh(1/2*x)+b)-2*C*c/(b^2
-c^2)^(3/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*c)/(b^2-c^2)^(1/2))
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 2.42513, size = 1648, normalized size = 20.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="fricas")
[Out]
[-(2*A*b^3 - 2*A*b^2*c - 2*A*b*c^2 + 2*A*c^3 - (C*b*c - C*c^2 + (C*b*c + C*c^2)*cosh(x)^2 + 2*(C*b*c + C*c^2)*
cosh(x)*sinh(x) + (C*b*c + C*c^2)*sinh(x)^2)*sqrt(-b^2 + c^2)*log(((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x)*sinh(
x) + (b + c)*sinh(x)^2 - 2*sqrt(-b^2 + c^2)*(cosh(x) + sinh(x)) - b + c)/((b + c)*cosh(x)^2 + 2*(b + c)*cosh(x
)*sinh(x) + (b + c)*sinh(x)^2 + b - c)) + 2*(C*b^3 - C*b*c^2)*cosh(x) + 2*(C*b^3 - C*b*c^2)*sinh(x))/(b^5 - b^
4*c - 2*b^3*c^2 + 2*b^2*c^3 + b*c^4 - c^5 + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)^2 + 2*
(b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)*sinh(x) + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b
*c^4 + c^5)*sinh(x)^2), -2*(A*b^3 - A*b^2*c - A*b*c^2 + A*c^3 - (C*b*c - C*c^2 + (C*b*c + C*c^2)*cosh(x)^2 + 2
*(C*b*c + C*c^2)*cosh(x)*sinh(x) + (C*b*c + C*c^2)*sinh(x)^2)*sqrt(b^2 - c^2)*arctan(sqrt(b^2 - c^2)/((b + c)*
cosh(x) + (b + c)*sinh(x))) + (C*b^3 - C*b*c^2)*cosh(x) + (C*b^3 - C*b*c^2)*sinh(x))/(b^5 - b^4*c - 2*b^3*c^2
+ 2*b^2*c^3 + b*c^4 - c^5 + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)^2 + 2*(b^5 + b^4*c - 2
*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*cosh(x)*sinh(x) + (b^5 + b^4*c - 2*b^3*c^2 - 2*b^2*c^3 + b*c^4 + c^5)*sinh
(x)^2)]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))**2,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [A] time = 1.14731, size = 112, normalized size = 1.37 \begin{align*} -\frac{2 \, C c \arctan \left (\frac{b e^{x} + c e^{x}}{\sqrt{b^{2} - c^{2}}}\right )}{{\left (b^{2} - c^{2}\right )}^{\frac{3}{2}}} - \frac{2 \,{\left (C b e^{x} + A b - A c\right )}}{{\left (b^{2} - c^{2}\right )}{\left (b e^{\left (2 \, x\right )} + c e^{\left (2 \, x\right )} + b - c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+C*sinh(x))/(b*cosh(x)+c*sinh(x))^2,x, algorithm="giac")
[Out]
-2*C*c*arctan((b*e^x + c*e^x)/sqrt(b^2 - c^2))/(b^2 - c^2)^(3/2) - 2*(C*b*e^x + A*b - A*c)/((b^2 - c^2)*(b*e^(
2*x) + c*e^(2*x) + b - c))
| 3,889 | 7,685 |
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# Wien's Displacement Law frequency vs wavelength
1. Jan 16, 2009
### buttersrocks
I hope this is the right place. Since the law is derived from statistical mechanics, I assumed this was the place to post. I understand how the Wien's Displacement law is derived. What I don't understand, other than mathematically, is why there is a difference between the peak intensity as measured by wavelength and frequency. Wikipedia does as good a job as anyone in providing the background: http://en.wikipedia.org/wiki/Wien's_displacement_law
Seeing as each wavelength has a one-to-one correspondence with a particular frequency, I would expect that the peak would occur at the same place when plotted against either. Again, I understand how, from the math, we find that it doesn't. What is wrong with my thinking?
If I were to go do an experiment and, let's say, filter out all light except for a particular frequency/wavelength, and do this repeatedly across the spectrum, which plot would I expect and why?
Thanks.
2. Jan 16, 2009
### mgb_phys
There isn't a difference, you can just divide C by the curve to get the curve for frequency.
The only reason it's quoted in terms of wavelength is that for reasonable temperature objects you are in a region where you normally work in wavelengths (UV-IR) rather than energy (x-ray) or frequency (radio)
3. Jan 16, 2009
### buttersrocks
Can you please elaborate? I'm certain that the formula $$c=\lambda\nu$$, does not relate the peak of the frequency curve to the peak of the wavelength curve. You can check by attempting to calculate the temperature of the sun given the form found using wavelength and the form using frequency. If that formula worked, the temperatures would agree, but they do not.
Start with any arbitrary temperature and you will find that they don't relate by that equation.
For peak intensity: (excuse the units)
$$\nu_{max} = 5.879\times10^{10} T$$
and
$$\lambda_{max} T = 2.898\times10^{6}$$
Maybe I'm misinterpreting what you mean by "divide C by the curve."
4. Jan 16, 2009
### mgb_phys
Whatever peak wavelength you get from wien's law, you must get the corresponding peak frequency from $$c=\lambda\nu$$
(sorry lambda isn't showing up for me ?)
5. Jan 16, 2009
### bartek2009
maximum of the blackbody spectrum
I agree with buttersrocks, I also derived the two displacement laws with exactly the same constants of proportionality. Apparently these are not the only relations and if you look at the bottom of this page
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/wien3.html" [Broken]
you should know what to look for in other sources. Heald is mentioned but I just looked through his "Classical Electromagnetic Radiation" and I couldn't find anything related to this problem. His other book, "Physics of Waves", also doesn't seem to cover it but I only look at the table of contents so I'm not sure.
Last edited by a moderator: May 3, 2017
6. Jan 16, 2009
### Cthugha
No, you don't. The scales are pretty nonlinear. You know that $$\nu=\frac{c}{\lambda}$$ so you can easily see that the range of lambdas you span by going one unit step in $$\nu$$ further depends strongly on which $$\nu$$ you are at. The number of lambdas will be very large for small $$\nu$$ and will be rather small for large $$\nu$$. If you go one unit step in $$\lambda$$ you will see something similar. This is just an application of the chain rule and has already been discussed here: https://www.physicsforums.com/showthread.php?t=132258.
7. Jan 16, 2009
### mgb_phys
Sorry am I misunderstanding the question?
If Wein's law predicts a peak wavelength for a blackbody at a certain temperature then a photon of that wavelength has a frequncy given by c=\lambda \nu
The relationship between the wavelength and frequency of a photon doesn't depend on what created it.
Similairly a peak wavelength - temperature plot and a peak frequency - temperature plot are going to have different shapes ( one is the reciprocal of the other) but you can convert between them with the above equation.
8. Jan 16, 2009
### Cthugha
Well, you can convert the peak wavelength of Wien's law to a frequency, but the peak frequency of Wien's law will NOT give the same result. This is a result of the chain rule.
Just consider you have two distributions and want to change variables $$\nu$$=$$\nu(\lambda)$$:
f($$\lambda$$) d$$\lambda$$=g($$\nu$$) d$$\nu$$
Now g($$\nu$$)=f($$\lambda$$) $$\frac{d \lambda}{d \nu}$$
Differentiate this with respect to $$\lambda$$ to find the peak for $$\lambda$$:
$$\frac{d g(\nu)}{d \lambda}=\frac{d f(\lambda)}{d \lambda} \frac{d \lambda}{d \nu} + f(\lambda) \frac{d^2 \lambda}{d \nu^2}$$
At the peak wavelength one of the terms vanishes:
$$\frac{d g(\nu(\lambda_{PEAK}))}{d \lambda}=f(\lambda_{PEAK}) \frac{d^2 \lambda}{d \nu^2}$$
So if both distributions are supposed to have the peak at the same wavelength, the term $$\frac{d^2 \lambda}{d \nu^2 }$$ must vanish. This is clearly not the case here, so both distributions will peak at different wavelengths.
9. Jan 17, 2009
### cmos
To get back to the OP's question:
First the part I think you already know:
The blackbody spectrum is a plot of power distribution throughout wavelengths/frequency. So if you want to find the power for some band of wavelengths $$\Delta \lambda$$, then you would integrate. Say we measure wavelength bandwidths of 100 nm; e.g. 600-700 nm, 700-800 nm, 800-900 nm, etc. In all cases, we have kept $$\Delta \lambda=100 nm$$ constant. Now convert each of these numbers into frequency. You will find now that $$\Delta \nu$$ does not remain constant. This is the reason for the different curves depending if you are working in the wavelength domain or the frequency domain.
It all comes down to conservation of energy. An integration over a certain range in the wavelength domain must give the same value for the corresponding range in the frequency domain. However, holding the wavelength range constant and moving throughout the spectrum has the effect of varying the frequency range and vice-versa. This is due to the inverse relation between frequency and wavelength.
Suppose I gave you a bag of marbles with different colors/weight: 2 x red/2 g, 5 x orange/2 g, 3 x yellow/4 g, 3 x green/4 g, and 2 x blue/6 g. Which type of marble do we have the most off?
Orange marbles!
No!
4 g marbles!
um....
Because we don't have a direct relationship between color and weight, we have different peak positions.
Depends on your experiment. Are you measuring wavelength (e.g. using diffraction gratings) or frequency/energy (e.g. using atomic transitions)?
10. Jan 17, 2009
### buttersrocks
Bear with me for a moment...
Okay, I understand the combinatorial example, but does it apply? For each particular frequency there is one and only one wavelength that it relates to. You don't have 100hz and 1000hz both corresponding to the same wavelength. Hence, my confusion.
Also, how do you distinguish whether you are measuring wavelength or frequency, since they are directly related. You can typically express atomic transitions in terms of wavelength as well. Don't we know that there IS a direct relationship between frequency and wavelength? ($$c=\nu\lambda$$) Or is that relationship not as fundamental as I thought? (I can't see how this would be the case.) Perhaps I just need a bit more elaboration.
11. Jan 17, 2009
### Cthugha
Just try something: Take some random pairs of equally spaced frequencies, for example 100 Hz and 1100 Hz, 3100 Hz and 4100 Hz and 1000000 Hz and 1001000 Hz. Now calculate the corresponding wavelengths. You will see that although the differences in frequency are constant, the differences in wavelength will be very different. You can also convert some constant wavelength differences into frequencies to see that it works the other way round, too. Note, that this works for ANY difference, no matter, whether you choose 1000 Hz, 100000000Hz or 0.1 Hz so any linear scale in frequency will give you a different peak than using a linear scale in wavelength.
12. Jan 17, 2009
### buttersrocks
Okay, I see where it is coming from now. Thanks.
13. Jan 17, 2009
### cmos
True, the combinatorial example isn't exactly what's going on with the blackbody example, the point I was trying to make is that in both cases, you do not have a direct relationship between the two domains you are examining. Note that $$\nu=c/\lambda$$ is not a "direct" relationship. A direct relation would be energy and frequency. In that case the Wien peaks match up.
As for how you know what you are measuring; you have to think about it. You will never measure precisely x, but rather some $$x\pm\Delta x$$. So in the laboratory, it is wise to make your experiment hold $$\Delta x$$ constant. So, for example a diffraction grating, the diffraction properties are directly related to wavelength ($$m\lambda=d sin \theta$$), therefore you hold $$\Delta \lambda$$ constant and get the blackbody spectrum in the wavelength domain.
14. Jan 17, 2009
### buttersrocks
Thanks cmos. That last part of the last post really solidified it for me. You are correct, I was not using "direct relationship" in the mathematically rigorous sense; I should have seen that. I think that this discussion really helped to destroy a few subconscious misconceptions. Much appreciated.
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Skip to Content
# How to Measure 4 Quarts of Water Full Guide of 2023
This site is supported by our readers. We may earn a commission, at no cost to you, if you purchase through links.
Are you feeling confused about how to measure 4 quarts of water? If so, don’t worry. You’re not alone! Understanding measurements can be tricky for a lot of people, but with the right information and guidance, it’s totally doable.
In this article, we’ll give you everything you need to know about measuring 4 quarts of water (that’s 16 cups!). We’ll break down all the conversions from quart to cup, pint, gallon, and liter in easy-to-follow steps.
## Key Takeaways
• 4 quarts of water are equivalent to 16 cups.
• 4 quarts of water are also equal to 8 pints.
• 4 quarts of water is equal to 1 gallon.
• 4 quarts of water is approximately equal to 3.79 liters.
## What is the Equivalent Measure for 4 Quarts?
You’re asking how to measure 4 quarts of water? Here’s a simple trick: since 1 quart equals 4 cups, 4 quarts would be 16 cups. So grab your 4-cup measuring cup and fill it up 4 times to get those 4 quarts measured out lickety-split.
How much easier could it be? With this quick quart to cup conversion, you can precisely measure any liquid volume. Understanding these fundamental imperial conversions empowers you to tackle complex measurement word problems.
Equipped with the knowledge that 4 quarts equals 16 cups, you can excel at test prep and excel in the kitchen.
Converting between quarts, cups, pints, and gallons may seem tricky, but grasping the relationships between imperial units is actually quite straightforward.
## How Many Cups is 4 Quarts?
4 quarts equals 16 cups, girl.
• 4 quarts = 1 gallon
• 16 cups = 1 gallon
• 1 quart = 4 cups
• 4 cups = 1 quart
To measure 4 quarts of water:
• Get a 1-gallon container
• Fill it up with water
• That’s 4 quarts or 16 cups of water
Measuring water for recipes or other purposes is made easy when you know the basic conversions. With a quart-to-cup ratio of 1:4, measuring out 4 quarts just requires filling a 1-gallon jug.
## How Many Pints is 4 Quarts?
Since 4 quarts equals 2 pints, imagine the refreshing taste of 2 pints of crisp water quenching your thirst on a hot summer day. Knowing how to convert between quarts and pints allows you to easily measure the right amount when cooking or staying hydrated.
Here are 3 key points:
• 4 quarts = 8 pints
• 2 quarts = 4 pints
• 1 quart = 2 pints
When making a recipe that calls for 4 quarts of water, you can measure out 2 pints instead.
Converting between quarts and pints gives you measurement flexibility. So whether you’re following a recipe or trying to hit your daily water intake goal, you can use either units interchangeably. Understanding quart to pint conversions ensures you accurately measure the right liquid volume.
## How Many Gallons is 4 Quarts?
Well now, wouldn’t you know, those 4 quarts equal one whole gallon of water right there.
• 1 quart = 0.25 gallons
• 4 quarts = 4 * 0.25 gallons = 1 gallon
• The formula is: Quarts * 0.25 = Gallons
When dealing with imperial volume conversions, it’s key to remember the relationships between units. A quart is a quarter of a gallon. So, four quarts together make up one full gallon. Keeping conversion formulas like quarts to gallons handy helps convert between units.
Practice converting units like cups, pints, quarts, gallons, ounces, and liters.
## How Many Liters Are in 4 Quarts?
Feel the smooth 3.79 liters of cool water in your hands as you grasp the container holding 4 quarts.
1. Picture a 1-liter soda bottle from the grocery store.
2. Now envision holding almost 4 of those bottles in your hands.
3. See the condensation dripping down as you firmly grasp the mass of liquid.
When converting between imperial and metric volume, remember that 1 quart equals 0.95 liters. Applying this quick conversion calculation reveals that 4 quarts is equivalent to 3.
So holding 4 quarts of any liquid is like carrying around 4 pounds of bottled water – you can sense the heft and substantial quantity in an intuitive way.
Grasping the reality of different volume units allows you to effortlessly convert in either direction.
## Quart to Cup Conversion: Step-by-Step
To measure 4 quarts of water in cups, you must first refer to the conversion chart to see that 1 quart equals 4 cups. Then, set up the expression 4 quarts x 4 cups/quart, multiply to find the result of 16 cups.
### Using the Conversion Chart
You’ll catch a glimpse of the chart’s insight and trust its logical aid by transforming quarts to cups. The conversion chart displays imperial fluid measures in orderly columns and rows, instantly revealing equivalent volumes.
Scanning its grid, you can match the position of 1 quart to the position of 4 cups, confirming their equal value. Relying on its data, you can convert any quart amount to cups or other ratios, such as quarts to fluid ounces.
Through its mapping, the chart instills confidence in calculating conversions for recipes, experiments, and quantitative tasks.
### Setting Up the Conversion Expression
Now, set up the conversion as an expression equating quarts and cups. Using the chart, build a conversion expression that shows the ratio of quarts to cups. This critical mathematical step sets you up to correctly multiply and obtain the conversion.
Although measurement conversions can seem challenging, carefully setting up the expression ensures that you apply the correct quart-to-cup ratio for accurate real-world measurements. With practice, you will soon master this essential calculation skill for kitchen and test word problems.
### Multiplying to Find the Result
After setting up the conversion expression, multiply to find the resulting cups in four quarts of water.
1. Set up the conversion as: 4 quarts x 4 cups / 1 quart
2. Multiply across: 4 x 4 = 16
3. The units cancel, leaving cups.
So, 4 quarts equals 16 cups.
Performing the multiplication finds the volume outcome.
To determine cup volume from quarts, set up a conversion expression, then multiply to get the resulting cups. This fundamental technique underlies all imperial conversions. Mastering it equips you to solve SAT and ACT word problems involving quart to cup conversions.
Ashley Sufflé Robinson’s guides teach essential skills for excelling on standardized tests.
## Quart to Pint Conversion: Step-by-Step
Since 1 quart equals 2 pints, you’d measure 4 quarts of water as 8 pints. For example, when baking bread, you’d need 4 quarts of water, which is the same as 8 pints based on the conversion.
To convert, just multiply the quarts by 2 to get the number of pints. So for a recipe calling for 4 quarts, you’d grab 8 pints when shopping. This pint conversion comes in handy at the store when shopping for ingredients like vegetable broth or coconut milk.
Double-checking the conversion helps ensure accuracy. For a full vegan grocery list, knowing quart to pint conversions allows proper measuring during meal prep and cooking. Whether following a recipe requiring 2 quarts of almond milk (4 pints) or grabbing vegetables like carrots, conversions provide a deeper understanding of amounts.
Taking time to study measurement conversions leads to greater kitchen confidence when cooking.
## Quart to Gallon Conversion: Step-by-Step
Convert your 4 quarts by dividing by 4; that’s 1 gallon of water for you.
To recap quart to gallon conversions:
• There are 4 quarts in 1 gallon. Divide the quarts by 4 to get gallons.
• 1 gallon equals 128 fluid ounces. Useful for measuring large volumes.
• A gallon is about 3.8 liters. Great for metric conversions.
• Gallons work for tangible examples like gas, milk, paint.
Without conversion tools, dividing and multiplying is key. For gallons, think of groups of 4 quarts.
## Quart to Liter Conversion: Step-by-Step
Cause analogies rock, converting 4 quarts to liters is like translating a book to reach more readers. When facing the complex imperial system, utilize a handy conversion formula. To convert 4 quarts to liters, begin with the conversion ratio: 1 quart equals 0.
946 liters. Then multiply: 4 quarts x 0.946 liters/quart = 3.784 liters. VoilĂ ! Like translating a novel, converting between metric and imperial requires finesse. Yet with the right tools, we can build bridges across cultural divides. Through mindful effort, even convoluted conversions become simpler, bringing us closer to liberation from archaic systems.
## How to Measure 4 Quarts of Water?
After converting quarts to cups, you now know that 4 quarts equals 16 cups. To accurately measure 4 quarts of water, use a liquid measuring cup with cup markings. Fill the 4-cup mark four separate times, pouring each measured 4 cups into a large container.
To check your work, fill the 1-cup mark 16 times, pouring into the same container. The water level should reach the 4-quart mark. For faster measuring, use a 4-quart pitcher. Fill it once from the tap, checking the accuracy with a separate quart-measuring container.
Utilizing the proper tools and techniques prevents spills and ensures precision when portioning liquids for recipes.
## Conclusion
Measuring 4 quarts of water doesn’t need to be a challenge. With a basic understanding of the imperial volume conversion system, you can use a chart to calculate the equivalent measurements in cups, pints, gallons, and liters.
By breaking down each conversion step by step, you can easily measure 4 quarts of water and understand how it relates to other measurements. So, next time you’re presented with a volume conversion challenge, you can confidently measure 4 quarts of water.
References
• whoatwherewhy.com
## Mutasim Sweileh
Mutasim is an author and software engineer from the United States, I and a group of experts made this blog with the aim of answering all the unanswered questions to help as many people as possible.
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# Integration operation, and its relation to differentials
Tags:
1. Nov 30, 2015
### Mr Davis 97
I need to get a few things straight about the integration operation (as an intro calc student). I understand that integration is a process that takes a function and returns its antiderivative. We can think of it as an operator, where $\displaystyle \int...dx$ is kind of like an opening and a closing bracket for an input function. This is how I interpret integration: the summa and the differential (the closing "bracket") are inseparable, since they are part of the same notation. However, my professor confused me with his derivation of velocity under constant acceleration:
$$\displaystyle \frac{dv}{dt} = a$$
$$\displaystyle v\frac{dv}{dt} = va$$
$$\displaystyle v\frac{dv}{dt} = a\frac{dx}{dt}$$
Then he "cancels" the $dt$
$$vdv = adx$$
The next part is what confuses me:
$$\int_{v_1}^{v_2}vdv = a\int_{x_1}^{x_2}dx$$
which comes out to be
$$\frac{1}{2}v_2^2 - \frac{1}{2}v_1^2 = a(x_2 - x_1)$$
My primary question is, how did the two summas appear, if there were no corresponding differentials at the time of application? This is disconcerting because the same operation is supposed to be applied to both sides of the equation, and those look like two different operations in terms of two different variables. Shouldn't he have done something like $\displaystyle \int_{t_1}^{t_2}vdv~dt = \int_{t_1}^{t_2}adx~dt$?
If it's no trouble, I have two additional questions. Why does he use $\displaystyle \int_{v_1}^{v_2}...dv$ rather than $\displaystyle \int...dv$? Also, what justifies that he "cancels" the dt differentials?
2. Nov 30, 2015
### andrewkirk
That last formula $vdv = adx$ is, strictly speaking, not meaningful. It's just a shorthand for a formula that is meaningful, which is :
$$\int_{v_1}^{v_2}vdv = \int_{x_1}^{x_2}adx$$
provided that $v_1\equiv v(t_1); v_t\equiv v(t_2);x_1\equiv x(t_1); x_2\equiv x(t_2)$.
The justification for this is that, since $v\frac{dv}{dt} = a\frac{dx}{dt}$, we have
$$\int_{t_1}^{t_2}v\frac{dv}{dt}\,dt= \int_{t_1}^{t_2}a\frac{dx}{dt}\,dt$$
We then apply the rule for change of integration variable, $t\to v$ on the LHS and $t\to x$ on the RHS, to obtain
$$\int_{v_1}^{v_2}vdv = \int_{x_1}^{x_2}adx$$
He then pulls the $a$ outside the integral, which is valid if $a$ is constant. The post doesn't say whether that is the case.
3. Nov 30, 2015
### Mr Davis 97
So is it better to do it like he did, since it is faster, or do it the rigorous way, since it acts as a sanity check?
4. Nov 30, 2015
### andrewkirk
The former - the 'definite integral' - is a real number. The latter - the 'indefinite integral' - is an equivalence class of functions. In physics one generally needs to get to a number sooner or later rather than just a function.
The sense in which integration is an 'inverse' of differentiation is that:
$$\frac{d}{dx}\bigg(\int_a^x f(u)\,du\bigg)=f(x)$$
provided that $a\leq x$. Note that we needed to use a definite integral to write this.
5. Nov 30, 2015
### andrewkirk
It depends on the context. Personally I think it is poor practice to do it like that when teaching, because it sows confusion and encourages sloppy thinking. On the other hand, when doing your own derivations it does no harm, and can speed things up, as long as you keep awareness of what you're about. You can always go back and make such steps rigorous later on, if the branch you are exploring turns out to be fruitful.
6. Nov 30, 2015
### Mr Davis 97
Okay, that makes sense. I have another question. In introductory calculus we are told that dy/dx is one entity, a derivative, and not a quotient of differentials. However, in practice, why does turn out that letting the differentials act as separate entities in a quotient lead to correct answers, like in the above case?
7. Nov 30, 2015
### andrewkirk
I think it's because, while $\frac{dy}{dx}$ is not a ratio, it is the limit of a ratio (as the denominator tends to zero).
So any manipulation that could be performed on the ratio inside the limit and then validly moved outside the limit - using the many useful properties of limits - ends up making things look like the $\frac{dy}{dx}$ is a ratio.
A nice example of this is the chain rule: $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ where $y$ is a function of $x$, which is a function of $t$.
We write
$$\frac{dy}{dt}=\lim_{h\to 0}\left(\frac{y(x(t+h))-y(x(t))}{h}\right) = \lim_{h\to 0}\left(\frac{y(x(t+h))-y(x(t)}{x(t+h)-x(t)}\cdot \frac{x(t+h)-x(t)}{h}\right)\\ = \lim_{h\to 0}\left(\frac{y(x(t+h))-y(x(t)}{x(t+h)-x(t)}\right)\cdot \lim_{h\to 0}\left(\frac{x(t+h)-x(t)}{h}\right)\\ =\frac{dy}{dx}\frac{dx}{dt}$$
We wrote the derivatives as limits of ratios, manipulated the ratios inside the limit, then moved the multiplication outside the limit, using the 'product of limits' theorem.
8. Nov 30, 2015
### Mr Davis 97
9. Nov 30, 2015
### Mr Davis 97
I actually have one last question. If we can consider the differential at the end of an integral as simply a closing bracket that is convenient because it tells with respect to which variable one is integrating, why is it so important when it comes to u-substitution? Why does it just seem as a placeholder with simple integrals, and becomes a part of teh mathematics with more complex ones?
10. Nov 30, 2015
### SteamKing
Staff Emeritus
I think you're assuming that just because the variable of integration comes at the end of the expression, it doesn't serve any real purpose. That's no so.
When the integrand (that's the stuff following the integral sign) is developed, the differential serves to indicate the variable with respect to which the integration is being performed. Now, when using u-substitution to simplify finding an antiderivative, you want the original integral ∫ f(x) dx = ∫ g(u) du. Since integration is the limit of a sum, you're pretty much stuck with ensuring that f(x) dx = g(u) du, also.
11. Nov 30, 2015
### andrewkirk
It's not just a closing bracket. That becomes really clear when you do Riemann-Stieltjes integrals, which are important in measure theory and especially in probability theory.
Again it helps to think in terms of limits (of sums in this case, rather than ratios), to which end recall that the integral sign is just an elongated 'S' for 'sum'. A simplified version of the Riemann integral can be written as:
$$\int_a^b f(x)dx\equiv \lim_{n\to\infty}\sum_{k=1}^n f(x(k))\delta x$$
where $\delta x\equiv \frac{b-a}{n}$ and $x(k)\equiv a+(b-a)\cdot\frac{k}{n}$.
The $dx$ in the integral corresponds to the $\delta x$ in the sum inside the limit.
If we substitute $u=2x$ in these then, for a given value of $n$, we have $\delta u= \frac{2b-2a}{n}=2\frac{b-a}{n}=2\delta x$. This is reflected in the integral as follows:
$$\int_a^b f(x)dx\equiv \lim_{n\to\infty}\sum_{k=1}^n f(x(k))\delta x\\ =\lim_{n\to\infty}\sum_{k=1}^n f(\frac{u(k)}{2})\frac{\delta u}{2}$$
[where $u(k)\equiv 2x(k)$]
and this is equal to
$$\int_{2a}^{2b} \frac{1}{2}f(\frac{u}{2})du$$
12. Dec 1, 2015
### micromass
OP: you reeally want to do nonstandard calculus! Check out Keisler's free book: https://www.math.wisc.edu/~keisler/calc.html
It will explain what $dx$ really is and why it behaves so damn nice. It will definitely open your eyes and convince you that the nonrigorous proof in your OP actually is pretty rigorous!
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# Regents Review #5 What else do I need to know?.
## Presentation on theme: "Regents Review #5 What else do I need to know?."— Presentation transcript:
Regents Review #5 What else do I need to know?
Pythagorean Theorem a2 + b2 = c2 The two missing sides are 8 and 10.
For any right triangle with legs (shorter sides) a and b and hypotenuse (the longest side) c, the square of the hypotenuse is equal to the sum of the squares of the other two sides. a2 + b2 = c2 The length of the shortest side of a right triangle is 6 inches. The lengths of the other two sides are consecutive even integers. Write an equation that can be used to find the missing sides. b) Solve the equation for the two missing sides. a2 + b2 = c2 62 + x2 = (x + 2)2 x: length of one leg x + 2: length of hypotenuse 8 8 + 2 = 10 62 + x2 = (x + 2)2 36 + x2 = x2 + 4x + 4 36 = 4x + 4 32 = 4x 8 = x The two missing sides are 8 and 10.
Shaded Area In the diagram, the dimensions of the large rectangle are 3x – 1 by 3x + 7 units. The dimensions of the cut-out rectangle are x by 2x + 5 units. Represent the area of the shaded region as a simplified polynomial expression. Area of Shaded Region = Area of Big Rectangle – Area of Small Rectangle “Whole Shape” – “Inside Shape” Area = (3x – 1)(3x + 7) – (2x + 5)(x) Multiply each pair of polynomials A = (9x2 + 18x – 7) – (2x2 + 5x) Distribute the - sign A = 9x2 + 18x – 7 – 2x2 – 5x Combine like terms A = 7x2 + 13x – 7 The area of the shaded region is 7x2 + 13x – 7 square units.
Interval Notation [ means to include < or > ( means do not include < or > For Example: [12,16) means…. All Real Numbers 12 through 16, including 12 but not 16 Represent the domain of f(x) = 2x + 3 graphed over the interval -4 < x < 6. [-4,6] (2) (-4,6) (3) [-4,6) (4) (-4,6] 2) Represent the domain and range of the function Domain: [-3, ∞) Range: [0, ∞) Remember: Infinity ∞ always uses )
Linear Functions Equation: y – 5 = 3(x – 2)
Linear Functions written in slope-intercept form identify the slope (rate of change) and y-intercept of the function. Linear Functions written in point-slope form identify the slope (rate of change) and one point that lies on the function. In order to write a linear function in point-slope form, follow these steps… Calculate the slope Identify a point on the line Replace m with the slope and x1 and y1 with the coordinates of a point on the function. Equation: y – 5 = 3(x – 2)
Linear Functions Parallel lines have the same slope and different y-intercepts. Perpendicular lines intersect at a 90⁰ angle. These lines have opposite reciprocal slopes. Write the equation of a line in point slope form that is perpendicular to 8x – 2y = 20 and passes through the point (1, -6). 8x – 2y = 20 y = 4x – 10, m = 4 Slope of perpendicular line = m = point: (1, -6) y – y1 = m(x – x1) y + 6 = - ¼(x – 1)
Quadratic Functions y = – (x – 2) 2 + 4 y = a(x – h) 2 + k
Consider the parent function y = x2 with the vertex (0,0). The function y = (x – 4)2 shifts the parent function to the right 4 units. The vertex of the new function is (4, 0). The function y = (x – 4)2 – 5 shifts the parent function to the right 4 units and down 5 units. The vertex of the new function is (4, -5). The graph pictured to the left represents a transformation of y = x2. Write an equation to represent this graph. y = a(x – h) 2 + k Vertex: (2, 4) Parabola opens down, a = -1 y = – (x – 2) 2 + 4
Sequences Explicit vs. Recursive
An explicit formula that defines a sequence can tell you any term of the sequence. A recursive formula that defines a sequence can tell you the next term provided that you know the previous term. Consider the sequence 8, 24, 72, 216, … Explicit Formula: s(n) = n-1 Find the 5th term s(5) = s(5) = 648 The 5th term is 648 Recursive Formula: s(1) = 8 s(n) = s(n – 1) 3 s(5) = s(5 – 1) 3 s(5) = s(4) 3 s(5) = S(5) = 648
Sequences Explicit vs. Recursive Explicit Formulas:
Arithmetic: an = a1 + d(n – 1) Geometric: an = a1 rn – 1 Recursive Formulas: Arithmetic: an = an-1 + d Geometric: an = an-1 r Sometimes Recursive Formulas are more complicated. Find the 4th term given the recursive formula: f(1) = 12 and f(n) = f(n – 1) + 3n f(n) = f(n – 1) + 3n f(2) = f(2 – 1) + 3(2) f(2) = f(1) + 3(2) f(2) = f(2) = 18 f(n) = f(n – 1) + 3n f(3) = f(3 – 1) + 3(3) f(3) = f(2) + 3(3) f(3) = f(3) = 27 f(n) = f(n – 1) + 3n f(4) = f(4 – 1) + 3(4) f(4) = f(3) + 3(4) f(4) = f(4) = 39 The 4th term of the sequence is 39.
Rational Equations 8x(x + 12) = 2x(2x + 4) 8x2 + 96x = 4x2 + 8x
When solving rational equations (equations with algebraic fractions), combine fractions and set up a proportion. Remember: A common denominator is needed to add or subtract fractions. 8x(x + 12) = 2x(2x + 4) 8x2 + 96x = 4x2 + 8x 4x2 + 88x = 0 4x(x + 11) = 0 4x = x + 11 = 0 x = x = -11 FOO FOO Reject 0 because it makes the equation undefined. Solution: x = -11
Work Word Problems Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours. How long would it take the two painters together to paint the house? Hours to Complete the Job Job Completed per Hour (rate) Combined Labor 1st painter: 12 hours 1st painter: per hour Equation: 2nd painter: 8 hours 2nd painter: per hour Together: x hours Together: per hour Together, the painters can complete the job in 4.8 hours (just under 5 hours).
Work Word Problems One pipe can fill a pool 1.25 times faster than a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used? Hours to Complete the Job Job Completed per Hour (rate) Combined Labor Pipe A: x hours Pipe A: per hour Equation: Pipe B: 1.25x Pipe B: per hour Together: 5 hours Together: per hour It takes the fast pipe 9 hours. It takes the slow pipe hours (9 X 1.25). It would take 11 hours and 15 minutes to fill the pool if only the slow pipe is used.
Profit Word Problems Profit = Income – Expenses (P = I – E)
Net Profit Revenue Cost Gross Profit The cost of operating Hannah’s Biscotti Company is \$750 per week plus \$0.05 to make each biscotti cookie. Write a function, C(b), to model the company’s weekly costs for producing b biscotti cookies. What is the total weekly cost in dollars if the company produces 5,000 biscotti cookies. Hannah’s company makes a gross profit of \$0.40 for each biscotti cookie they sell. If they sold all 5000 biscotti cookies, would they make money or lose money? C(b) = b C(b) = (5000) = \$ It costs the company \$1000 to make 5000 cookies Gross Profit = 0.40(5000) = \$2000 \$ earned before expenses are subtracted Net Profit = Income – Expenses = 2000 – 1000 = 1000 The company will earn \$1000
Age Word Problems Sue is 5 years older than Ann. In 6 years, Sue’s age will be 11 years less than twice Ann’s age then. How old is each person now? Person Age Now Age In 6 Years Ann x x + 6 Sue x + 5 (x + 5) + 6 = x + 11 Future Sue will be 11 years less than twice Future Ann x = (x + 6) – 11 x + 11 = 2(x + 6) – 11 x + 11 = 2x + 12 – 11 x + 11 = 2x + 1 11 = x + 1 10 = x Right now, Ann is 10 years old and Sue is 15 years old. Remember: It is helpful to organize information in a table prior to creating an equation.
Coin Word Problems Joe has \$ He has 7 more dimes than nickels. How many of each does he have? Coin Value Quantity Total Value Nickels .05 x .05x Dimes .10 7 + x .10(7 + x) .05x + .10(7 + x) = or 5x + 10(7 + x) = 250 5x x = 250 15x + 70 = 250 15x = 180 x = 12 Joe has 12 nickels and 19 dimes. Check: 12 nickels = 60 cents 19 dimes = \$1.90 Total: \$ \$0.60 = \$2.50 Remember: (Value)(Quantity) = Total value of Coins \$ per coin x how many = total \$
Ratio Word Problems The measures of two supplementary angles are in the ratio of 3:7. What is the measure of the larger angle? Let 3x = the measure of the smaller angle Let 7x = the measure of the larger angle smaller angle = 54 degrees (3)(18) larger angle = 126 degrees (7)(18) Remember: Include an x in each part of the ratio. Ex: Donna wants to make 4lbs of trail mix made up of almonds, walnuts and raisins. She wants to mix one part almonds, two parts walnuts, and three parts raisins. Ratio 1:2:3 Let x = the amount of almonds Let 2x = the amount of walnuts Let 3x = the amount of raisins x + 2x + 3x = 4 3x + 7x = 180 10x = 180 x = 18 The larger angle measures 126○
Consecutive Integer Word Problems
Find two consecutive integers whose sum is -35. x: 1st consecutive integer x + 1: 2nd consecutive integer -18 -17 ( ) Remember: Consecutive integers count by 1’s Ex: x, x+1, x+2, x+3…. Consecutive odd or even integers count by 2’s Ex: x, x+2, x+4, x+6… Negative integers doesn’t change anything
Statistics Quantitative Data is numerical, meaning it can be counted or measured. Ex: height of a flagpole, weight of a backpack Categorical Data (Qualitative) is not numerical, meaning it can be observed. Ex: type of toppings on a pizza, favorite ice cream flavor Univariate refers to single variable data. Ex: the number of pets each student owns Bivariate refers to two variable data. Ex: a person’s shoe size compared to their height
Statistics A population is a group that you want information about.
A sample is part of a population that is used to make estimates about the population. In a random sample, each member has an equal chance of being selected, and the sample is representative of the entire population. A biased sample favors one or more parts of the population over others. Ex: You want to conduct a survey to find out what type of music people listen to. Determine which of these scenarios is biased. You ask every fifth person leaving a Taylor Swift concert about the type of music they listen to. You ask every fifth person leaving the local mall about the type of music they listen to. Biased Unbiased
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# Walker's guide to Slide Presentation
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10 Dec 2009, 11:00
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Cool Job Walker...Nice info...
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10 Dec 2009, 13:42
A very helpful post, I'm just starting on my presentation.
Kudos to you Walker!
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08 Apr 2010, 14:01
thanks for this excellent information.
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14 May 2010, 18:36
Walker this is an excellent discussion on slide presentations. It is thoughtful and practical. I plan to use these techniques in my next presentation at work.
Thanks
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02 Aug 2010, 15:58
Nice post!
Thanks!!
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26 Aug 2010, 20:04
Thanks Walker! Great guide, very helpful for those of us in the grinding through the applications. Thanks!
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26 Aug 2010, 23:45
interesting.. nice
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03 Oct 2010, 07:04
thanks!
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03 Oct 2010, 07:04
thanks!
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05 Oct 2010, 21:28
I have a question for you - is that you in the picture?
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06 Oct 2010, 05:18
yes, it's me.
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06 Oct 2010, 17:04
wow, where were you [thread] this whole time!?!
thanks for pointers... helping me perfect some of the final touches on my first two slides.
+1
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05 Nov 2010, 14:01
Wow! amazing stuff. Keep on rocking
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Re: Walker's guide to Slide Presentation [#permalink]
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08 Dec 2010, 14:30
I have been sitting in front of my PowerPoint with absolutely no clue about what to say in those four slides. Now with your guide I'm starting to have ideas bouncing in my head and to understand the kind of information Booth is looking for.
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Re: Walker's guide to Slide Presentation [#permalink]
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24 Dec 2010, 12:39
great job!!
Thanks..
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Re: Walker's guide to Slide Presentation [#permalink]
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22 Feb 2012, 09:30
didn't realize that booth presentation allowed imagery...i thought i had read somewhere that you had to keep to a certain size...and images/videos weren't allowed. i guess i now know not to trust that source...
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Re: Walker's guide to Slide Presentation [#permalink]
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22 Feb 2012, 09:40
fxsunny wrote:
didn't realize that booth presentation allowed imagery...i thought i had read somewhere that you had to keep to a certain size...and images/videos weren't allowed. i guess i now know not to trust that source...
The slides are printed so video and flash images aren't allowed, but you can def use static pictures. In fact I'd say 80% of applicants use pictures in their Booth PPT.
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Re: Walker's guide to Slide Presentation [#permalink]
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23 Dec 2012, 12:05
Wonderful article walker...
+1 from me...
Wish you merry Xmas and a very Happy 2013 ahead....
Cheers!
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Re: Walker's guide to Slide Presentation [#permalink]
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01 Jan 2014, 22:42
Hello from the GMAT Club MBAbot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: Walker's guide to Slide Presentation [#permalink]
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09 May 2015, 10:35
Hello from the GMAT Club MBAbot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: Walker's guide to Slide Presentation [#permalink] 09 May 2015, 10:35
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THERMO Spoken Here! ~ J. Pohl © ( C7905) ( C8100 - )
# Gravity Work-Rate:
This is simply a discussion of terms, ideas and if what we think we know we know. A student having given some thought to a quite simple event of a Block, used definitions from physics along with Newton's 2nd Law. The student arrived at a contradiction. At the very lowest level... a contradiction? Fortunate for us (or maybe not), the student wrote down the diellema. No answer will follow this. The answer depends upon how you see it.
A physics student imagined an event of a Block. She modeled the Block as a BODY then analyzed the physical situation in two manners and obtained results that seemed contradictory. The student's wrtten statement is below. Beneath that a discussion: What was said and what does it mean.
======
Student statement: Suppose a block is being raised at constant velocity by a tension in a cord against the force of gravity. Is there any work done on the block?
My first answer was "no," since Work = Force*Distance, Force = Mass*Acceleration, and if velocity is constant then acceleration is zero and force equals zero. So then work is zero.
However work also comprises mgΔH, and if the height is changing then there is an increase in potential energy. So my second answer is yes there is work. Please resolve this contradiction.
======
First let's try to agree on our assumptions of what the student's statement means. I suggest the following:
Scenario: The physical scenario, the reality in mind, the matter in consideration, is the Block as it experiences its event. The Block, a model of physical reality, is being raised by a force acting through action of "tension through an attached cord" while the "force of gravity" acts oppositely on the Block, in opposition to the raising event. Nonetheless, the Block moves upward at constant velocity.
System: The defined system is the Block (only). Earth as acknowledged contributes to the event by way of its mass (at a distance), via its "gravity force." From this perspective Earth and the Block do not form a "potential energy" pair.
Question: "Is there any work done on the Block?" This request is specific. There is "work," there is "work-net," "work done on" and "work done by..." The request is "work done on."
If agreed that the above is what the discussion is about, below we discuss perspectives. The student analyses the system from two vantages (1) and (2)
- - - - - -
Student Analysis 1: "... since the following equations apply as shown...
... one concludes "work" is zero.
==========
Author 1: This argument by intersplice of definitions concludes "The work of a Block that moves at constant velocity equals zero" appears sound but how is it relevant to the question asked?
(i) "Force equals mass times acceleration" is true in a dimensional sense. That is [F] = [m][L][t]-2. If "work equals zero" is then "work done on equal zero," or what?
This argument is imprecise. Does it address the Block? For the Block: F ≠ mA but ΣF = mA. ==========
- - - -
Student 2: "... but work comprises mgoΔH,... if height is changing... conclude "work done on" is NOT zero.
==========
Author 2: "work" does not "comprises mgoΔH." Work and To indtoduce potential energy (being an Earth-and-Block as system idea) is to return to change the system from Block (only) to be Earth and Block (combined as system). What has change of potential energy to do with "work done on?" Please show effort, show steps of your logic.
Also The interest is "work done on." To head this way, please show what "work done on" has to do with is, and how the "work" as in the above equations being zero parses to conclude "work done on" is zero.
==========
Author's Comments: The student is clear in asking for "work done on."
## Explanation:
The student's statement is unclear. Work is not germane for this "steady-continuing" event. "Work-rate" is the proper work-related form. Work (and work-rate) as ideas arrived to engineering physics about the time Watt and Newcomen developed the first steam engines. The student does not define the system for either analysis. What system, the reader must deduce.
Conclusion 1: The student is correct. The work-rate of the Block is zero as a sum of two equal and opposite work-rates in action.
Conclusion 2. The student, by mention of potential energy, is using "Block and Earth" as the system. A work-rate occurs, it is the tension of the cord times its vertical velocity. This work-rate equals the rate of increase of potential energy of the Earth/Body system.
Solution 1: For the first analysis, the student uses the Block as system. Also he makes the classic algebra-based-physics mistake. His use of "F" in Newton's 2nd Law causes him to overlook a force. One should use ΣF in the 2nd Law, as Newton intended.
So our task: Block as system, multiply Newton's 2nd Law by Block velocity. Obtain relation for work-rate.
Newton's 2nd Law addresses momentum of the Block in our "0Z" coordinate system. Best ALWAYS to use ΣF. (1)
Mass of the Block and its vertical velocity are constant so left-of-equality equals zero. (2)
Expand ΣF to establish gravity force as equal and opposite to the cord force. This is obvious but needed later. (3)
Return to Eqn-2, multiply the equation by the vertical velocity of the Block. (4)
Now expand the sum of forces. (5)
Write the forces as magnitudes and directions. (6)
The vector multiplications determine the signs: (7)
Force times velocity equals work-rate. The sum of these work-rates is zero. (8)
Conclusion 1: The student is correct. The work-rate of the Block is zero as a sum of two equal and opposite work-rates in action.
######################
Solution 2. The student's statement "work equals mgoH" is not applicable from the aspect work. Work-rate as "work-rate = d(mgoH)/dt" is relevant. The mention of increasing potential energy reveals the system to be the Block and Earth taken together. This development below addresses the Block initially with Earth participating via gravity force. Block and Earth are brought together near the end.
Newton's 2nd Law addresses momentum of the Block in our "0Z" coordinate system. Best ALWAYS to use ΣF. (9)
In this and a great many instances, gravity force is relevant. Remove the force of gravity from the sum of forces. (10)
As before, multiply Eqn-9 by the velocity of the Block. Three terms are obtained. (11)
By basic definitions of calculus, the term left-of-equality Eqn-11 is altered as follows: (12)
Right-of-equality, the first term changes as follows: (13)
Since zo is a constant, dzo/dt = 0. The term is "(z - zo)" as a convenience. Place the results, Eqn-12 and Eqn-13 into Eqn-11. Thus with a little grass-roots calculus, Newton's 2nd Law multiplied by velocity, can be represented as Eqn-14.
Important: Although Newton's 2nd Law has been transformed, the system it addressed initially has not changed. That system is the Block. The terms of Eqn-14, left to right are: A modification of momentum, the work-rate of the gravity force and the work-rate of all other forces. This form, known to many persons in the years after Newton's death, displays a new idea, energy of a BODY.
(14)
A Body has characteristics (position and velocity) and properties (mass, momentum for now - energy soon). Supposing the mass of a Body is known, for any event its position and velocity can be made quantitative, is observable, and measurable from an external non-intefering reference (called extrinsic, by some). The terms of Eqn-14 are specified simply in terms of velocity and elevation relative to Earth.
(15)
When we move the gravity work-rate term from right-of-equality to left-of-equality, the effect is change our system from "Block" to "Block and Earth."
(16)
As a last step we define kinetic energy as KE, and potential energy as PE. Substitute these definitions into Eqn-17.
(17)
We can group the terms within the differentiation.
(18)
The above concludes the development.
Conclusion 2. The student, by mention of potential energy, is using "Block and Earth" as the system. Consequently Eqn-16 applies. Since the vertical speed is constant, the first term is zero - Eqn-19. A work-rate occurs, it is the tension of the cord times its vertical velocity. This work-rate equals the rate of increase of potential energy of the Earth/Body system.
(19)
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S-1: describes a scenario: ... a block moves upward at constant speed by pull of a cord. Gravity force opposes the motion. We assume some manner structure with a "lifting motor" to power the event. Analysis generally begins from the vantage of a physical secnario approximated by a sketch - The scenario - Figure 1.
By Newton's method, the next task is to select (from the scenario) a system (that matter to be analyzed) then represent it as a Free-Body-Diagram (FBD). An obvious system is the Block. To draw a FBD requires experience. The completed FBD is Figure 2.
To draw a FBD requires a number of decisions. The FBD is a sketch of the system once it is isolated from the remaining matter of the scenario. Newton's idea was to "isolated" the matter otf the system from all else - the all else he called, the surroundings. That is a first step. So though the system was isolated, its events were still subject to influence of the surroundings. The complexity of these ideas, physical truths and their approximations are depicted graphically (and by notations) on the FBD.
1. Model: Physical reality is always studied approximated as a model. The Block is modeled as a Body.
2. Coordinates-Basis/Characteristics: Position, and velocity are system characteristics. A coordinate reference 0Z with origin and unit vector: K is notated.
3. Properties/Property Equations: Our Block has three properties: one property, its mass which is constant.
4. Event: The abstract event is the relation of time to physics is No event, permanance, batch and continuing are common choices. Event is the abstract of time.. as it continues. Physically real
Body and Boundary Forces: Since the system is imagined a part apart from reality, forces are applied to approximated.
5. ... and others not needed here.
Question 1: Is there work? Strictly speaking, No. Work is associated with a start/stop or commence/end type of event. Events for which a force is displaced an increment without regard for the time required. Our event the time-wise nature "continuing." For our event (continuing) "work rate" is germane.
Newton's 2nd Law has no work-related term. However, were the 2nd Law multiplied by the vector velocity, two work-rate terms appear. The case we study is very special. Since vertical speed is constant, left-of-equality is zero, which greatly reduces this discussion. , we avoid
The terms right-of-equanity in the 2nd Law (multiplied by velocity are work-rate terms. Work is energy or momentum change of a system. Work-rate is energy rate of change or momentum rate of change. to or from a system. The tension of the cord times its vertival velocity
from surroundings, needs a manner to recognize that "surroundings being there" might influence the event. System selection, isolation, selection of forces with notation and a new sketch... these steps yield a free-body-diagram.
## Gravity Work Rate:
This is being developed!
Premise presently unwritted!
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• Math Tips, Tricks, and Games
## Improve Critical Thinking with 3rd Grade Multiplication Word Problems
Last Updated on June 3, 2022 by Thinkster
When done well, 3rd grade multiplication word problems present an opportunity for teachers and parents to help kids improve their critical thinking skills. These problems require a significant amount of thought in order to decide which operation to use, figure out the answer, and then determine whether or not the answer makes sense.
Here are some strategies to help children practice the problem-solving process. They work well to fine-tune those critical thinking skills and help with 3rd grade multiplication word problems.
## Read the Problem to Find Meaningful Information
In order to solve 3rd grade multiplication word problems, students must learn in order to read the problem to locate critical information. The critical information includes the question asked, the clue word or words that indicate which operation to use, and the numbers that will be used to find that information. They might underline clue words and circle the question or highlight the numbers to help with this process.
3rd grade is an excellent time to introduce this process or to fine-tune it if it has been introduced earlier. Most third-graders have strong enough reading skills to be able to decode the words and start looking a little deeper into their meaning. As a result, they are better able to focus on honing their critical thinking skills.
Asking questions is one of the key components of critical thinking. After reading a word problem students need to ask themselves several questions, including:
• What am I trying to find?
• Will the answer be larger or smaller than the numbers given?
• What operation would work best to find this answer?
• What is my estimated answer?
• Does my estimated answer make sense?
These critical thinking processes will help them arrive at the right answer with less struggle.
## Visualizing the Problem
Many 3rd grade multiplication word problems work with numbers too large to draw a picture. Students must therefore solve the problem mentally, yet visualizing the process is still an important critical thinking skill. Teach children to draw mental pictures to visualize the word problems they are solving.
As you can see, 3rd grade multiplication word problems are an excellent way to begin developing those critical thinking skills. If you want to work on these skills at home, but lack the time or knowledge to sit down and work through word problems with your child, Thinkster Math can help. Thinkster’s innovative tablet-based math learning program will help your child focus on those critical thinking skills, then give them the tools they need to solve math word problems accurately with the help of knowledgeable math teachers. It’s fun to play and also rewards children for their hard work , so your child will be motivated to learn.
If you’re looking for online math tutoring to help your child, you can try Thinkster risk-free .
Thinkster provides a full-fledged online tutoring platform (driven by AI, behavioral, and data science), as well as supplemental math worksheets , math homework help , test prep , and more. Our Parent Insights App allows you to monitor your student ‘s work and learning improvements at any time.
An elite, expert math tutor and online teaching system work together to help your student go beyond just learning math – we want them to master it.
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## Mastering Math: How to Crack Challenging Math Problems in Under a Minute!
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## Ditch the Timed Test! Alternative Ideas for Assessing Multiplication Facts
• September 18, 2023
Assessing multiplication facts gives some of us chills…and not the good kind! Do you have a childhood trauma story that starts with, “When I was in third grade we had to memorize our multiplication facts….” We all have experienced timed tests where we had to race against the clock to complete as many problems as possible or get a perfect score to climb our way up Multiplication Mountain. But as educators, it’s time to shift our focus away from these anxiety-inducing assessments and towards more engaging and effective ways of assessing multiplication facts.
First and foremost, make sure you’ve done a thorough job giving your students a solid foundation for understanding multiplication and strategies for different types of multiplication facts BEFORE you start assessing whether students have the facts memorized or not. The lessons in this multiplication unit are highly engaging and offer tons of strategies and practice activities.
Once you’ve taught all the factors, you can start assessing multiplication facts. Here are some fun and creative ideas to assess multiplication fluency in your classroom without resorting to the stress of timed tests.
## Goal Setting & Progress Tracking
Have your students set realistic goals for themselves and track their progress on a chart in the classroom. For example, their measurements could be to improve by one fact per week–if they got 5 out of 20 correct last week, celebrate when they get 6 out of 20 correct this week. As a teacher, you could use colored stickers or stars each time they achieve their weekly goal, leading to a sense of accomplishment and progress. Also, continuously communicating with each student about their progress helps to promote accountability and focus on specific areas of improvement.
## Technology-Based Assessment
Incorporating technology into students’ learning has become an increasingly common practice in recent years. Multiplication applications such as Math Duel, Prodigy, and Times Tables Rockstars are readily available on devices like mobile phones, tablets, and desktop computers. Websites like xtramath , Freckle , and 99Math are growth-promoting tools that students could use to take self-assessments and track progress in a gamified form of learning.
## Create a Multiplication Museum
What better way to show off multiplication mastery than to create a museum? Ask students to display their knowledge of multiplication by creating posters, models, and dioramas of their favorite multiplication facts or strategies. This activity encourages creativity, collaboration, and critical thinking, and students will take ownership of their learning.
## Peer Quizzing
Peer quizzing is a fantastic method of assessing multiplication facts without causing stress to the student. As students of a similar age group, they are bound to be more comfortable sharing with each other while still receiving valuable feedback. Take advantage of this by pairing students to answer multiplication questions for each other using these free flashcards and quiz pages . This method allows both students to work equally in a safe and relaxed learning environment.
## Create a Multiplication Book
Another fun way to assess multiplication facts is to have them create a multiplication facts pattern book. Students can show you they know their multiplication facts and explore the patterns that the facts create on a 0-100 number chart at the same time. This activity encourages creativity and critical thinking while also assessing multiplication facts in a subtle manner.
## Two-Tone Tables
Here’s a colorful idea for assessing multiplication facts! Give each student a blank multiplication chart and have them use two different colors of markers or colored pencils. Set a timer for using color #1 and have students fill in as much of the multiplication chart as possible. At the end of the timer, have students switch to color #2. Have students set an improvement goal for themselves to get more and more of the multiplication table done in color #1 each time you do the activity. Have students write the date on each table and keep them so students can easily see how much they have improved over time.
There are so many fun and engaging ways for assessing multiplication facts that don’t include timed tests. By incorporating creative projects and games you can keep your students motivated and engaged while ensuring they are mastering the essential skill of multiplication. Remember, education is not just about memorization but also about creativity, problem-solving, and critical thinking skills. By incorporating these fun ideas for assessing multiplication facts into your teaching, you can both challenge and engage your students while keeping the learning process enjoyable and stress-free.
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## Multiplication Word Problems 4th Grade
Welcome to our Multiplication Word Problems for 4th Grade. Here you will find our range of printable multiplication problems which will help your child apply and practice their multiplication and times tables skills to solve a range of 'real life' problems at a 4th grade level.
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• 4th Grade Multiplication Problems Worksheets
• Easier & Harder Multiplication Worksheets
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## Multiplication Word Problems 4th Grade Online Quiz
Multiplication word problems, 4th grade multiplication problems.
Here you will find a range of problem solving worksheets involving multiplication.
Each sheet involves solving a range of written multiplication problems.
There are 3 levels of difficulty for each worksheet below: A,B and C.
Worksheet A is the easiest level, suitable for children at the beginning of their grade.
Worksheet B is a medium level worksheets for children who are working at the expected level in their grade.
Worksheet C is set at a harder level, suitable for children who are more able mathematicians.
The problems in each worksheet are similar in wording, but the numbers involved become trickier as the level gets harder.
To encourage careful checking and thinking skills, each sheet includes one 'trick' question which is not a multiplication problem. Children need to spot this word problem, and work out which operation they need to solve it.
• apply their multiplication and times tables skills at a 4th grade level;
• apply their times table knowledge to work out related facts;
• recognise multiplication problems, and try to spot 'trick' problems;
• solve a range of 'real life' problems.
Some of the sheets have a UK version with spelling and currency symbols set for the UK.
## 4th Grade Multiplication Word Problem Sheets
Series 4 sheet 1 set.
• Series 4 Sheet 1A (easier)
• Series 4 Sheet 1B (medium)
• Series 4 Sheet 1C (hard)
• PDF Series 4.1 (6 sheets)
• PDF Series 4.1 UK version (6 sheets)
## Series 4 Sheet 2 Set
• Series 4 Sheet 2A (easier)
• Series 4 Sheet 2B (medium)
• Series 4 Sheet 2C (hard)
• PDF Series 4.2 (6 sheets)
• PDF Series 4.2 UK version (6 sheets)
## Series 4 Sheet 3 Set
• Series 4 Sheet 3A (easier)
• Series 4 Sheet 3B (medium)
• Series 4 Sheet 3C (hard)
## Multiplication Word Problems Walkthrough Video
This short video walkthrough shows several problems from our Multiplication Problems Worksheet 4.3A being solved and has been produced by the West Explains Best math channel.
If you would like some support in solving the problems on these sheets, check out the video!
Looking for some easier Multiplication Problems?
In our 3rd Grade Multiplication word problem area, you will find a range of multiplication problems aimed at 3rd graders.
The following areas are covered:
• basic multiplication fact sheets;
• multiplication facts to 10x10;
• 2 digits x 1 digit
• Multiplication Word Problem Worksheets 3rd Grade
Looking for some harder Multiplication Problems?
In our 5th Grade Multiplication word problem area, you will find a range of multiplication problems aimed at 5th graders.
• multiplication fact sheets;
• multiplication related facts to 10x10 e.g. 6 x 70, 8 x 0.6, etc;
• problems needing written multiplication methods to solve e.g. 2 digits x 2 digits, decimal multiplication
• Multiplication Problems Printable 5th Grade
## More Recommended Math Worksheets
Take a look at some more of our worksheets similar to these.
Looking for more 4th Grade Word Problems?
Each problem sheet comes complete with answers, and is available in both standard and metric units where applicable.
Many of the problems are based around 'real-life' problems and data such as the world's heaviest animals.
• apply their addition, subtraction and problem solving skills;
• apply their knowledge of rounding and place value;
• solve a range of 'real life' problems;
• attempt more challenging longer problems.
Using the problems in this section will help your child develop their problem solving and reasoning skills.
• 4th Grade Math Word Problems
## Multiplication Times Table Charts
Here you will find a selection of Multiplication Times Table Charts to 10x10 or 12x12 to support your child in learning their multiplication facts.
There is a wide selection of multiplication charts including both color and black and white, smaller charts, filled charts and blank charts.
• Learn their multiplication facts to 10x10 or 12x12;
• Practice their multiplication table.
All the free printable Math charts in this section are informed by the Elementary Math Benchmarks.
• Large Multiplication Chart
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• Multiplication Times Tables Chart to 10x10
• Times Table Grid to 12x12
• Blank Multiplication Charts to 10x10
• Blank Printable Charts to 12x12
• Multiplication Math Games
Here you will find a range of Free Printable Multiplication Games.
The following games develop the Math skill of multiplying in a fun and motivating way.
• learn their multiplication facts;
• practice and improve their multiplication table recall;
• develop their strategic thinking skills.
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This quick quiz tests your knowledge and skill at solving multiplication word problems by tens and hundreds.
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## Multiplication Worksheets
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## 20 Math Critical Thinking Questions to Ask in Class Tomorrow
• November 20, 2023
The level of apathy towards math is only increasing as each year passes and it’s up to us as teachers to make math class more meaningful . This list of math critical thinking questions will give you a quick starting point for getting your students to think deeper about any concept or problem.
Since artificial intelligence has basically changed schooling as we once knew it, I’ve seen a lot of districts and teachers looking for ways to lean into AI rather than run from it.
The idea of memorizing formulas and regurgitating information for a test is becoming more obsolete. We can now teach our students how to use their resources to make educated decisions and solve more complex problems.
With that in mind, teachers have more opportunities to get their students thinking about the why rather than the how.
## Looking for more about critical thinking skills? Check out these blog posts:
• Why You Need to Be Teaching Writing in Math Class Today
• How to Teach Problem Solving for Mathematics
• Turn the Bloom’s Taxonomy Verbs into Engaging Math Activities
## What skills do we actually want to teach our students?
As professionals, we talk a lot about transferable skills that can be valuable in multiple jobs, such as leadership, event planning, or effective communication. The same can be said for high school students.
It’s important to think about the skills that we want them to have before they are catapulted into the adult world.
Do you want them to be able to collaborate and communicate effectively with their peers? Maybe you would prefer that they can articulate their thoughts in a way that makes sense to someone who knows nothing about the topic.
Whatever you decide are the most essential skills your students should learn, make sure to add them into your lesson objectives.
## When should I ask these math critical thinking questions?
Critical thinking doesn’t have to be complex or fill an entire lesson. There are simple ways that you can start adding these types of questions into your lessons daily!
## Start small
Add specific math critical thinking questions to your warm up or exit ticket routine. This is a great way to start or end your class because your students will be able to quickly show you what they understand.
## Add critical thinking questions to word problems
Word problems and real-life applications are the perfect place to add in critical thinking questions. Real-world applications offer a more choose-your-own-adventure style assignment where your students can expand on their thought processes.
They also allow your students to get creative and think outside of the box. These problem-solving skills play a critical role in helping your students develop critical thinking abilities.
## Keep reading for math critical thinking questions that can be applied to any subject or topic!
• Explain the steps you took to solve this problem
• Draw a diagram to prove your solution.
• Is there a different way to solve this problem besides the one you used?
• How would you explain _______________ to a student in the grade below you?
• Why does this strategy work?
• Use evidence from the problem/data to defend your answer in complete sentences.
## When you want your students to justify their opinions
• What do you think will happen when ______?
• Do you agree/disagree with _______?
• What are the similarities and differences between ________ and __________?
• What suggestions would you give to this student?
• What is the most efficient way to solve this problem?
• How did you decide on your first step for solving this problem?
## When you want your students to think outside of the box
• How can ______________ be used in the real world?
• What might be a common error that a student could make when solving this problem?
• How is _____________ topic similar to _______________ (previous topic)?
• What examples can you think of that would not work with this problem solving method?
• What would happen if __________ changed?
• Create your own problem that would give a solution of ______________.
• What other math skills did you need to use to solve this problem?
## Let’s Recap:
• Rather than running from AI, help your students use it as a tool to expand their thinking.
• Identify a few transferable skills that you want your students to learn and make a goal for how you can help them develop these skills.
• Add critical thinking questions to your daily warm ups or exit tickets.
• Ask your students to explain their thinking when solving a word problem.
• Get a free sample of my Algebra 1 critical thinking questions ↓
7 thoughts on “20 math critical thinking questions to ask in class tomorrow”.
I would love to see your free math writing prompts, but there is no place for me to sign up. thank you
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## 5 Ways to Stop Thinking for Your Students
Too often math students lean on teachers to think for them, but there are some simple ways to guide them to think for themselves.
Who is doing the thinking in your classroom? If you asked me that question a few years ago, I would have replied, “My kids are doing the thinking, of course!” But I was wrong. As I reflect back to my teaching style before I read Building Thinking Classrooms by Peter Liljedahl (an era in my career I like to call “pre-thinking classroom”), I now see that I was encouraging my students to mimic rather than think .
My lessons followed a formula that I knew from my own school experience as a student and what I had learned in college as a pre-service teacher. It looked like this: Students faced me stationed at the board; I demonstrated a few problems while students copied what I wrote in their notes. I would throw out a few questions to the class to assess understanding. If a few kids answered correctly, I felt confident that the lesson had gone well. Some educators might call this “ I do, we do, you do .”
What’s wrong with this formula? When it was time for them to work independently, which usually meant a homework assignment because I used most of class time for direct instruction, the students would come back to class and say, “The homework was so hard. I don’t get it. Can you go over questions 1–20?” Exhausted and frustrated, I would wonder, “But I taught it—why didn’t they get it?”
Now in the “peri-thinking classroom” era of my career, my students are often working at the whiteboards in random groups as outlined in Liljedahl’s book. The pendulum has shifted from the teacher doing the thinking to the students doing the thinking. Do they still say, “I don’t get it!”? Yes, of course! But I use the following strategies to put the thinking back onto them.
## 5 Ways to Get Your Students to Think
1. Answer questions with a refocus on the students’ point of view. Liljedahl found in his research that students ask three types of questions: “(1) proximity questions—asked when the teacher is close; (2) stop thinking questions—most often of the form ‘is this right’ or ‘will this be on the test’; and (3) keep thinking questions—questions that students ask so they can get back to work.” He suggests that teachers acknowledge “proximity” and “stop thinking questions” but not answer them.
Try these responses to questions that students ask to keep working:
• “What have you done so far?”
• “Where did you get that number?”
• “What information is given in the problem?”
• “Does that number seem reasonable in this situation?”
2. Don’t carry a pencil or marker. This is a hard rule to follow; however, if you hold the writing utensil, you’ll be tempted to write for them . Use verbal nudges and hints, but avoid writing out an explanation. If you need to refer to a visual, find a group that has worked out the problem, and point out their steps. Hearing and viewing other students’ work is more powerful .
3. We instead of I . When I assign a handful of problems for groups to work on at the whiteboards, they are tempted to divvy up the task. “You do #30, and I’ll do #31.” This becomes an issue when they get stuck. I inevitably hear, “Can you help me with #30? I forgot how to start.”
I now require questions to use “we” instead of “I.” This works wonders. As soon as they start to ask a question with “I,” they pause and ask their group mates. Then they can legitimately say, “ We tried #30, and we are stumped.” But, in reality, once they loop in their group mates, the struggling student becomes unstuck, and everyone in the group has to engage with the problem.
4. Stall your answer. If I hear a basic computation question such as, “What is 3 divided by 5?” I act like I am busy helping another student: “Hold on, I need to help Marisela. I’ll be right back.” By the time I return to them, they are way past their question. They will ask a classmate, work it out, or look it up. If the teacher is not available to think for them, they learn to find alternative resources.
5. Set boundaries. As mentioned before, students ask “proximity” questions because I am close to them. I might reply with “Are you asking me a thinking question? I’m glad to give you a hint or nudge, but I cannot take away your opportunity to think.” This type of response acknowledges that you are there to help them but not to do their thinking for them.
When you set boundaries of what questions will be answered, the students begin to more carefully craft their questions. At this point of the year, I am starting to hear questions such as, “We have tried solving this system by substitution, but we are getting an unreasonable solution. Can you look at our steps?” Yes!
Shifting the focus to students doing the thinking not only enhances their learning but can also have the effect of less frustration and fatigue for the teacher. As the class becomes student-centered, the teacher role shifts to guide or facilitator and away from “sage on the stage.”
As another added benefit, when you serve as guide or facilitator, the students are getting differentiated instruction and assessment. Maybe only a few students need assistance with adding fractions, while a few students need assistance on an entirely different concept. At first, you might feel like your head is spinning trying to address so many different requests; however, as you carefully sift through the types of questions you hear, you will soon be comfortable only answering the “keep thinking” questions.
## Practicing multiplication tables
Common Core Standards: Grade 3 Operations & Algebraic Thinking
CCSS.Math.Content.3.OA.C.7, CCSS.Math.Content.3.OA.D.8
This worksheet originally published in Math Made Easy for 3rd Grade by © Dorling Kindersley Limited .
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## Related Resources
• Chinook’s Edge Website
## Strategies vs Models
Strategy: How students solve the problem. (Examples: Decomposition, Compensation, Traditional Algorithm)
Model: How students notate their thinking. (Examples: Number line, Ten Frames, Base Ten Blocks)
## Important to Know:
One strategy can be explained using multiple models. One model can be used to explain multiple strategies.
Fluency with strategies evolves over time.
## The Development of Mathematical Reasoning
Pamela Harris' Development of Mathematical Reasoning: Counting Strategies ⇒ Additive Thinking Strategies ⇒ Multiplicative Reasoning Strategies ⇒ Proportional Reasoning Strategies ⇒ Functional Reasoning Strategies
Students develop mathematical reasoning as they develop more sophisticated thinking. A student who solves 4 x 5 by using counting strategies or additive thinking strategies is not engaging in multiplicative reasoning. Although, this student will correctly determine the answer to 4 x 5 using other strategies, if they do not develop multiplicative reasoning, exploring more complex questions will be extremely challenging. For example, a student may use counting or additive thinking strategies as part of their process for solving 1.2 x 2.3 or (x + 2)(x - 1), but will be unable to solve it without using multiplicative thinking.
When a student develops fluency in multiple strategies and more sophisticated strategies, the student can confidently select the most appropriate strategy based on the numbers.
## Naming Strategies:
Mathematical strategies should be named using the math utilized in the strategy. It might be cute to name the strategy after the student who first demonstrates it in class but when a student moves to a different class or a different school, non-conventional naming is not helpful. A grade level team and, preferably, a school, should agree upon names for the strategies that will be explored in math class. There aren't that many of them. Decomposing, compensating, partitioning by place value, give and take, and the traditional algorithm will cover almost all possible strategies.
## Strategies for this concept:
Below you will find a short description for each of the most typically used strategies for multiplicative thinking. Each strategy will include one or two models to help notate the thinking within the strategy. When solving a question, a student might use a single strategy or apply two or more strategies. Note that, although the sample question provided below could be solved using other reasoning, only the reasoning appropriate to this concept has been included.
Examples have been included for each strategy.
Multiplication: 5 x 18
## Compensation
Adjust one of the numbers to make it easier to work with while keeping the other number the same. Solve the new question. Adjust the answer to account for (compensate for) the change made to the first number.
## Decomposition
Decompose (break apart) one or both of the numbers into more manageable values in order to make it easier to solve.
The Distributive Property: Building Arrays and the Area Model: Explore the distributive property through arrays and the area model as it can be developed from Kindergarten to Grade 8.
Multiplication – Decomposition Strategy: These videos explore multiplication through decomposition.
Division- Decomposition Strategy : These videos explore division through decomposition while making connections to the traditional algorithm.
## “Doubles / Halving”
The name of this strategy is a misnomer as it is not limited to knowing doubles or halves. You might use triples, thirds, fourths, etc. The first example below focuses on “give and take” where you double the 5 at the same time that you are halving the 18. The second example could relate to compensation as it first doubles the 10, and then halves the result.
The traditional algorithm (often called the standard algorithm) is a standardized process for solving multiplication and division questions. It focuses on solving a question digit by digit. Caution: The Traditional Algorithm most often described by Canadians is the traditional algorithm for Canada. Other countries have their own traditional algorithms. Keep this in mind if using this phrasing with students. ( Source )
## Formative Assessment
Ways to show 4 x 3 – 4 questions. Students determine if the example shows a way to think about 4 x 3 and provide an explanation. Student exemplars included. <Original Website>
Encyclopedia of the Sciences of Learning pp 2389–2392 Cite as
## Multiplicative Thinking and Learning
• Parmjit Singh 2
• Reference work entry
578 Accesses
Multiplicative reasoning ; Multiplicative structures ; Multiplicative thinking
Multiplication (and division) is an arithmetic operator used on numbers while thinking is a cognitive process involving the mind of learners reacting to incoming information. Multiplicative thinking represents the learner’s mental adaptive processing of multiplication concepts by using different methods and approaches in various mathematical problem contexts. Considering the level of complexity inherent in the nature of multiplication, one requires a more complex approach when thinking about numbers and operations. Multiplicative thinkers are those who have understood the concept of multiplication and are able to apply the concepts and solve problems relationally.
## Theoretical Background
Multiplicative thinking has gained more recognition and interest in recent years, following the early work of Vergnaud in 1983 . The growth of multiplicative thinking is critical for a learner’s...
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Confrey, J. (1994). Splitting, similarity and rate of change: new approaches to multiplication and exponential functions. In G. Harel & J. Confrey (Eds.), The development of multiplicative reasoning in the learning of mathematics (pp. 293–332). Albany, NY: State University of New York Press.
Dienes, Z. P. (1967). Building up mathematics . London: Hutchinson Educational Publishers.
Jacob, L., & Willis, S. (2001). Recognising the difference between additive and multiplicative thinking in young children. In J. Bobis, B. Perry & M. Mitchelmore. (Eds.), Numeracy and beyond (Proceedings of the 24th Annual Conference of the Mathematics Education Research Group of Australasia, Sydney pp. 306–313). Sydney: MERGA.
Steffe, L. (1994). Children’s multiplying schemes. In G. Harel & J. Confrey (Eds.), The development of multiplicative reasoning in the learning of mathematics (pp. 3–40). Albany: State University of New York Press.
Susan, B. E., & Erin, T. (2006). The emergence of multiplicative thinking in children’s solutions to paper folding tasks. Journal of Mathematical Behavior, 25 , 46–56.
Vergnaud, G. (1988). Multiplicative structures. In J. Hiebert & M. Behr (Eds.), Number concepts and operations in the middle grades (pp. 141–161). Reston: National council of Teachers of Mathematics.
Vergnaud, G. (1983). Multiplicative structures. In R. Lesh & M. Landau (Eds.), Acquisition of mathematics concepts and processes (pp. 127–174). New York: Academic.
## Author information
Authors and affiliations.
University Technology MARA, Shah Alam, Selangor, Malaysia
Parmjit Singh ( Faculty of Education )
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## Corresponding author
Correspondence to Parmjit Singh .
## Editor information
Editors and affiliations.
Faculty of Economics and Behavioral Sciences, Department of Education, University of Freiburg, 79085, Freiburg, Germany
Norbert M. Seel
## Rights and permissions
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Cite this entry.
Singh, P. (2012). Multiplicative Thinking and Learning. In: Seel, N.M. (eds) Encyclopedia of the Sciences of Learning. Springer, Boston, MA. https://doi.org/10.1007/978-1-4419-1428-6_531
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## Critical Thinking: Test-taking Practice for Math Grade 3
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## Mathematics
Domain - operations and algebraic thinking, grade 3.
Represent and solve problems involving multiplication and division.
Math.3.OA.A.3 : Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
Understand properties of multiplication and the relationship between multiplication and division.
Math.3.OA.B.5 : Apply properties of operations as strategies to multiply and divide. Examples: If 6 × 4 = 24 is known, then 4 × 6 = 24 is also known. (Commutative property of multiplication.) 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30. (Associative property of multiplication.) Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. (Distributive property.)
Solve problems involving the four operations, and identify and explain patterns in arithmetic.
Math.3.OA.D.8 : Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.(This standard is limited to problems posed with whole numbers and having whole-number answers; students should know how to perform operations in the conventional order when there are no parentheses to specify a particular order).
## Domain - Number and Operations in Base Ten, Grade 3
Use place value understanding and properties of operations to perform multi-digit arithmetic.4
Math.3.NBT.A.2 : Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.
Math.3.NBT.A.1 : Use place value understanding to round whole numbers to the nearest 10 or 100.
Math.3.NBT.A.3 : Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., 9 × 80, 5 × 60) using strategies based on place value and properties of operations.
## Domain - Number and Operations-Fractions, Grade 3
Develop understanding of fractions as numbers.
Math.3.NF.A.2 : Understand a fraction as a number on the number line; represent fractions on a number line diagram.
Math.3.NF.A.3 : Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
## Domain - Develop understanding of fractions as numbers., Grade 3
Understand a fraction as a number on the number line; represent fractions on a number line diagram.
Math.3.NF.A.2a : Represent a fraction 1/ b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1/ b and that the endpoint of the part based at 0 locates the number 1/ b on the number line.
Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
Math.3.NF.A.3b : Recognize and generate simple equivalent fractions, e.g., 1/2 = 2/4, 4/6 = 2/3). Explain why the fractions are equivalent, e.g., by using a visual fraction model.
Math.3.NF.A.3a : Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
Math.3.NF.A.3c : Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Examples: Express 3 in the form 3 = 3/1; recognize that 6/1 = 6; locate 4/4 and 1 at the same point of a number line diagram .
## Domain - Measurement and Data, Grade 3
Solve problems involving measurement and estimation of intervals of time, liquid volumes, and masses of objects.
Math.3.MD.A.2 : Measure and estimate liquid volumes and masses of objects using standard units of grams (g), kilograms (kg), and liters (l).(Excludes compound units such as cm3 and finding the geometric volume of a container) Add, subtract, multiply, or divide to solve one-step word problems involving masses or volumes that are given in the same units, e.g., by using drawings (such as a beaker with a measurement scale) to represent the problem.(Excludes multiplicative comparison problems (problems involving notions of "times as much").
Math.3.MD.A.1 : Tell and write time to the nearest minute and measure time intervals in minutes. Solve word problems involving addition and subtraction of time intervals in minutes, e.g., by representing the problem on a number line diagram.
Represent and interpret data.
Math.3.MD.B.3 : Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets .
Geometric measurement: understand concepts of area and relate area to multiplication and to addition.
Math.3.MD.C.7 : Relate area to the operations of multiplication and addition.
Math.3.MD.C.5 : Recognize area as an attribute of plane figures and understand concepts of area measurement.
Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures.
Math.3.MD.D.8 : Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters.
## Domain - Geometry, Grade 3
Reason with shapes and their attributes.
Math.3.G.A.1 : Understand that shapes in different categories (e.g., rhombuses, rectangles, and others) may share attributes (e.g., having four sides), and that the shared attributes can define a larger category (e.g., quadrilaterals). Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.
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1. PDF Learning to Think Mathematically About Multiplication
This book is designed to help students develop a rich understanding of multiplication and division through a variety of problem contexts, models, and methods that elicit multiplicative thinking. Elementary level math textbooks have historically presented only one construct for multiplication: repeated addition.
2. Improve Critical Thinking with 3rd Grade Multiplication Word Problems
Visualizing the Problem. Many 3rd grade multiplication word problems work with numbers too large to draw a picture. Students must therefore solve the problem mentally, yet visualizing the process is still an important critical thinking skill. Teach children to draw mental pictures to visualize the word problems they are solving.
3. Ditch the Timed Test! Alternative Ideas for Assessing Multiplication
Take advantage of this by pairing students to answer multiplication questions for each other using these free flashcards and quiz pages. This method allows both students to work equally in a safe and relaxed learning environment. ... This activity encourages creativity and critical thinking while also assessing multiplication facts in a subtle ...
4. Multiplication Word Problems 4th Grade
4th Grade Multiplication Problems. Here you will find a range of problem solving worksheets involving multiplication. Each sheet involves solving a range of written multiplication problems. There are 3 levels of difficulty for each worksheet below: A,B and C. Worksheet A is the easiest level, suitable for children at the beginning of their grade.
5. 101 Great Higher-Order Thinking Questions for Math
The answer is by utilizing higher-order thinking questions for math. Higher-order thinking questions are critical thinking questions that require students to infer, apply, predict, connect, evaluate, and judge knowledge in new ways. The answers to these questions require prior knowledge and an expansive schema so that readers can see beyond the ...
6. Multiplication Worksheets
It encourages critical thinking in solving questions like '10 is 2 times as many as 5.' Customizable for specific learning needs, it can be converted into flash cards for interactive study or used in distance learning settings to accommodate different teaching modalities. ... Given the large number of problems in the worksheet, students are ...
7. Multiplication Critical Thinking
30 seconds. 1 pt. Which problems show the commutative property or the "flip-flop property" for multiplication? 4 x 5 = 20 5 x 4 = 20. 7 x 3 = 21 7 x 3 = 21. 6 x 6 = 36 6 x 6 = 36. 6 x 2 = 12 12 x 2 = 6.
8. What is multiplicative thinking?
Additive thinking is the ability to join and separate numbers. Subitizing, using manipulatives, story problems, balance equation activities, building numbers, and more are needed for additive thinking. Counting, even skip counting, is still additive thinking. Both early numeracy and additive thinking must be strong for students to build up to ...
9. Multiplication and Division Facts: Critical Thinking (Gr. 4)
Provided by Scott Foresman, an imprint of Pearson, the world's leading elementary educational publisher. Its line of educational resources supports teachers and helps schools and districts meet demands for adequate yearly progress and reporting.
10. Critical Thinking Math Problems: Examples and Activities
Cite this lesson. Critical thinking is an important factor in understanding math. Discover how critical thinking can help with real-world problem solving, using examples and activities like asking ...
11. Multiplication Critical Thinking
Multiplication Critical Thinking - Worksheets, Lessons, and Printables. Multiplication Worksheets. 2 Digit Multiplication. 2nd Grade Multiplication. 3rd Grade Multiplication. 4th Grade Multiplication. 5th Grade Multiplication. 6th Grade Multiplication.
12. 20 Math Critical Thinking Questions to Ask in Class Tomorrow
Start small. Add critical thinking questions to word problems. Keep reading for math critical thinking questions that can be applied to any subject or topic! When you want your students to defend their answers. When you want your students to justify their opinions. When you want your students to think outside of the box.
13. PDF Multiplicative thinking practice questions
Multiplicative Thinking Practice Questions. 1. Estimate the answer to 58790 ÷ 29, explaining your reasoning. 2. How would you demonstrate the multiplication 3 X 4 using arrays model. Use this model to show the commutative property of multiplication. 3. Solve 132 × 15 using two different methods. These two methods must be distinctly different.
14. Multiplication: Critical Thinking (Gr. 4)
Provided by Scott Foresman, an imprint of Pearson, the world's leading elementary educational publisher. Its line of educational resources supports teachers and helps schools and districts meet demands for adequate yearly progress and reporting.
15. Promoting Independent Critical Thinking in Math
5 Ways to Get Your Students to Think. 1. Answer questions with a refocus on the students' point of view. Liljedahl found in his research that students ask three types of questions: " (1) proximity questions—asked when the teacher is close; (2) stop thinking questions—most often of the form 'is this right' or 'will this be on the ...
16. Practicing multiplication tables
This math worksheet gives your third grader multiplication practice with equations and word problems involving shapes, volume, money, and logic. MATH | GRADE: 3rd. Print full size.
17. Multiplication and Division Concepts: Critical Thinking (Gr. 2)
Multiplication and Division Concepts: Critical Thinking (Gr. 2) Use these practice worksheets to accompany the initial teaching of multiplication and division. They provide varied strategies for solving problems with this topic that you can use them to differentiate your lessons, student assessments, and supplemental material.
18. Multiplicative Thinking
Below you will find a short description for each of the most typically used strategies for multiplicative thinking. Each strategy will include one or two models to help notate the thinking within the strategy. When solving a question, a student might use a single strategy or apply two or more strategies. Note that, although the sample question ...
19. PDF Critical Thinking Worksheet Grades 6-8: Mathematical Concepts
Engage student's critical thinking skills with the use of these worksheets in the classroom. Click here: critical_thinking_034-download.pdf to download the document. ... Students might create a special question mark symbol to post next to any item for which contradictory sources can be found Note: The Food Timeline is a resource that documents ...
20. Multiplication and Critical Thinking: Lesson Plan
For instance, they will need to represent "2+2+2+2" as "2∙4.". The second set of problems will include a set of multiplications for which the learners will have to provide answers (e.g., 3∙2=6). Finally, the third set will involve problems requiring the application of critical thinking.
21. Multiplicative Thinking and Learning
Definition. Multiplication (and division) is an arithmetic operator used on numbers while thinking is a cognitive process involving the mind of learners reacting to incoming information. Multiplicative thinking represents the learner's mental adaptive processing of multiplication concepts by using different methods and approaches in various ...
22. Critical Thinking: Test-taking Practice for Math Grade 3
Mathematics Domain - Operations and Algebraic Thinking, Grade 3. Represent and solve problems involving multiplication and division. Math.3.OA.A.3: Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
23. Worksheet Library: Critical Thinking: Grades 3-5
Scratch Your Brain. Use addition and subtraction to figure out solutions to these brain benders. (Grades 3-5) From One Word to the Next. Change a letter in the previous word to make the word that completes each phrase. (Grades 3-5) Root Words. Complete this activity about words that have /capt/ or /tact/ as a root.
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Vous êtes sur la page 1sur 9
# Network Flow
## The network flow problem is as follows:
Given a connected directed graph G with non-negative integer
weights, (where each edge stands for the capacity of that edge),
and two distinguished vertices, s and t, called the source and
the sink, such that the source only has out-edges and the sink
only has in-edges, find the maximum amount of some
commodity that can flow through the network from source to
sink.
One way to imagine the situation is imagining each edge as a
pipe that allows a certain flow of a liquid per second. The
source is where the liquid is pouring from, and the sink is
where it ends up. Each edge weight specifies the maximal
amount of liquid that can flow through that pipe per second.
Given that information, what is the most liquid that can flow
from source to sink per second, in the steady state?
Here is an example:
12
a -----> b
16 / ^
^ \ 20
/ 4|
| \
s
|
|7 t
\ | 10
| /
13 \ v
| /4
c ------>d
14
capacity 9
11
/
s
\
8 \
12
a -----> b
/ ^
^ \ 15
1|
| \
|
|7 t
|0
| /
v
| /4
c ------>d
11
flow of 4.
## In order for the assignment of flows to be valid, we must have
the sum of flow coming into a vertex equal to the flow coming
out of a vertex, for each vertex in the graph except the source
and the sink. This is the conservation rule. Also, each flow must
be less than or equal to the capacity of the edge. This is the
capacity rule.
The flow of the network is defined as the flow from the source,
or into the sink. For the situation above, the network flow is 19.
Cuts
A cut is a partition of vertices (V s,Vt) such that the s Vsand
t Vt.Anedgethatgoesfromutovisaforwardedgeifu Vs
andv Vt.Iftheoppositeistrue,thenitisabackwardedge.
AlsoletFequaltheflowofanetwork.
Foranycut,definetheflowacrossthecuttobethesumofthe
flowsoftheforwardedgesminusthesumoftheflowsofthe
backwardsedges.
Theflowacrossanycutequalstheflowofanetwork.
Inourpreviousexample,theflowwas19.Considertheflow
acrossthecutVs={s,a,b}.Wehavethreeforwardedgeswith
flowsof8,4,and15andtwobackwardedgeswithflowsof1
and7.Summing,wehave8+4+1517=19,asdesired.
I will supplement the proof in the book by making the
followingobservations:
1)ThesumoftheflowsofalltheverticesinVsistheflowofthe
networkbecausethissumis0forallnonsourceverticesinVs,
andequaltotheflowatthesource.
2)WeneedonlytoshowthattheflowNOTacrossVsis 0. This
toeachother.Thus,thenetflowofalltheedgeswithinVs is 0,
because for two different vertices in Vsweaddandsubtractthe
sameflowforanedge.
Wecanusethistoshowthattheflowofanetworkcannot
exceedthecapacityofanycut.
Simplyput,ourbestcaseisifwedon'thaveanybackward
edgeswithflow.Inthiscase,wearesimplyleftwithforward
edges, each with a particular capacity. The sum of these
capacitiesisthecapacityofthecut,andanupperboundonthe
flowofthatcut.
MaximumFlow
Wewillusetwoideastohelpusdeterminewhetherornota
flowismaximum.Namely,wemustshowthatabsolutelyno
Theideaswewilllookare residualcapacity and augmenting
paths.
Theresidualcapacityofanedgefromvertexutovertexvis
simplyitsunusedcapacitythedifferencebetweenitscapacity
and its flow in the direction of the edge. In the opposite
direction(fromvtou),thisvalueisdefinedastheflowofthe
edge.
Also,theresidualcapacityofapathisdefinedastheminimum
of the residual capacities of the edges on that path. This
particularvalueisthemaximumexcessflowwecanpushdown
thatparticularpath.Thus,anaugmentingpathisdefinedas
onewhereyouhaveapathfromthesourcetothesinkwhere
everyedgehasanonzeroresidualcapacity.Namely,
flowislessthancapacityforeachforwardedge
flowisgreaterthan0foreachbackwardsedge
Inourexamplebelow,considerthefollowingpath:
s>a>c>b>t
Theforwardedgesonthispatharesa,ac,andbt
Thebackwardedgeiscb.
11
/
s
\
8 \
12
a -----> b
/ ^
^ \ 15
1|
| \
|
|7 t
|0
| /
v
| /4
c ------>d
11
## , with an edge from b->c with
flow of 4.
Forsawehave11/16,ac0/10,bt15/20
Finallyforcb,wehave4/9.
Thus,theminimumresidualcapacityofthepathis4,whichis
thelimitgiventousbycb.Now,let'saugmentthispath:
15
/
s
\
8 \
12
a -----> b
/ ^
^ \ 19
1|
| \
|
|7 t
|4
| /
v
| /4
c ------>d
11
## Hopefully it should be clear that if we find an augmenting
path, adding the minimal residual capacity of that path to each
edge on that path yields another valid flow. In essence, what we
are doing is simply adding some fluid to a particular path in
the network. On a backwards edge, we are actually taking
away fluid.
Now, we will show that if a flow network has no augmenting
path, then it has a cut of maximum capacity.
If no augmenting path exists, then that means for all paths
from source to sink, there is at least one edge without any
residual capacity. Now, consider partitioning the vertices
according to this rule: Place in Vs all vertices v that DO have
augmenting paths from the source s. Then place all other
vertices in Vt. (Note that the sink t must be in this set since
there is no augmenting path to t.)
Now, consider all the edges with respect to this cut. Clearly,
NONE of these edges have any residual capacity. If they did,
then we would have added the vertex that they reach to the set
Vs. Thus, for this particular cut, the flow at each forward edge
is to its capacity and the flow at each backwards edge is 0. As
## we showed before, this situation represents the maximal flow
possible.
Now, based upon the fact that augmenting paths can always be
used to increase flow, and that the minimum capacity of a cut
constrains the maximal flow, we can now claim the following:
The Max-Flow, Min-Cut Theorem: The value of the maximal
flow in a flow network equals the value of the minimum cut.
Withourexample,wecanseethatthereisnoaugmentingpath
fromstot.Iftherewere,itwouldhavetogothroughband
utilize edge bt. But all three edges leading to bt have no
residualcapacity.
Now, consider forming the set of vertices that do have
augmenting paths from s. This set includes {s, a, c, d} As
mentionedbefore,itisimpossibletoreachbonanaugmenting
15
/
s
\
8 \
12
a -----> b
/ ^
^ \ 19
1|
| \
|
|7 t
|4
| /
v
| /4
c ------>d
11
## Now, if we look at this cut, we find that the maximal value of
this cut is the sum of the capacities of the edges ab, db, and dt
which is 12+7+4 = 23. (Notice that we did not add or subtract
anything for edge bc with capacity 9, since it is a backwards
edge.)
Ford-Fulkerson Algorithm
The Ford-Fulkerson Algorithm is a basic consequent of the
work above. In its simplest form, we do the following:
While there exists an augmenting path
Add the appropriate flow to that augmenting path
We can check the existence of an augmenting path by doing a
graph traversal on the network (with all full capacity edges
removed.) This graph, a subgraph with all edges of full
capacity removed is called a residual graph.
It is difficult to analyze the true running time of this algorithm
because it is unclear exactly how many augmenting paths can
be found in an arbitrary flow network. In the worst case, each
augmenting path adds 1 to the flow of a network, and each
search for an augmenting path takes O(E) time, where E is the
number of edges in the graph. Thus, at worst-case, the
algorithm takes O(|f|E) time, where |f| is the maximal flow of
the network.
## The Edmonds-Karp Algorithm
This algorithm is a variation on the Ford-Fulkerson method
which is intended to increase the speed of the first algorithm.
The idea is to try to choose good augmenting paths. In this
algorithm, the augmenting path suggested is the augmenting
path with the minimal number of edges. (We can find this
using BFS, since this finds all paths of a certain length before
moving on to longer paths.) The total number of iterations of
the algorithm using this strategy is O(VE). Thus, its total
running time is O(VE2).
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# Math Quiz: Problems On Time And Work!
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What we have here is some math trivia questions quiz, which is made up of problems on time and work! Do you think you have the necessary skills to solve it? How about you give it a shot and get to find out if you may have a hard time when it comes to word problems. All the best!
• 1.
### Two pipes A and B can fill a cistern in 37 1/ 2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour , if the B is turned off after…
• A.
5 min
• B.
8 min
• C.
7 min
• D.
9 min
D. 9 min
Explanation
LCM of (75/2 and 45) = 225, assume this is total unit of work. Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min. Since A works for 30 mins, he will finish = 6 x 30 = 180 units. Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
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• 2.
### P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
• A.
8/15
• B.
7/15
• C.
9/15
• D.
6/15
A. 8/15
Explanation
Amount of work P can do in 1 day = 1/15 Amount of work Q can do in 1 day = 1/20 Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60 Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15 Fraction of work left = 1 – 7/15= 8/15
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• 3.
### A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?Which one do you like?
• A.
32 1/2
• B.
37 1/2
• C.
17 1/2
• D.
35 1/2
B. 37 1/2
Explanation
Work done by A in 20 days = 80/100 = 8/10 = 4/5 Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1) Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B) Work done by A and B in 1 day = 1/15 ---(2) Work done by B in 1 day = 1/15 – 1/25 = 2/75 => B can complete the work in 75/2 days = 37 ½ days
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• 4.
### P can lay railway track between two stations in 16 days. Q can do the same job in 12 days. With the help of R, they complete the job in 4 days. How many days does it take for R alone to complete the work?
• A.
9 2/5
• B.
9 1/5
• C.
9 3/5
• D.
None of these
C. 9 3/5
Explanation
Amount of work P can do in 1 day = 1/16 Amount of work Q can do in 1 day = 1/12 Amount of work P, Q and R can together do in 1 day = 1/4 Amount of work R can do in 1 day = 1/4 - (1/16 + 1/12) = 3/16 – 1/12 = 5/48 => Hence R can do the job on 48/5 days = 9 (3/5) days
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• 5.
### X, Y, and Z contract a work for Rs. 550. Together X and Y are supposed to do 7/11 of the work. How much does Z get?
• A.
110
• B.
330
• C.
200
• D.
150
C. 200
Explanation
(A + B)'s work = 7/11 So, C's work = 1 - (7/11) = (11 - 7) / 11 = 4/11 Thus, ratio of (A + B) and C's share = (7/11) : (4/11) = 7 : 4 Hence, C's share in a contract of Rs. 550 = Rs. 550 x (4/11) = Rs. 200.
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• 6.
### In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes and there are no invalid votes, by how many votes did it lose the election?
• A.
36000
• B.
48000
• C.
24000
• D.
15000
A. 36000
Explanation
Let the percentage of the total votes secured by Party D be x% Then the percentage of total votes secured by Party R = (x - 12)% As there are only two parties contesting in the election, the sum total of the votes secured by the two parties should total up to 100% i.e., x + x - 12 = 100 2x - 12 = 100 or 2x = 112 or x = 56%. If Party D got 56% of the votes, then Party got (56 - 12) = 44% of the total votes. 44% of the total votes = 132,000 i.e.,44/100*T = 132,000 T = 132000*100/44 = 300,000 votes. The margin by which Party R lost the election = 12% of the total votes = 12% of 300,000 = 36,000.
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• 7.
### A shepherd has 1 million sheep at the beginning of the Year 2000. The numbers grow by x% (x > 0) during the year. A famine hits his village in the next year and many of his sheep die. The sheep population decreases by y% during 2001 and at the beginning of 2002, the shepherd finds that he is left with 1 million sheep. Which of the following is correct?
• A.
X > y
• B.
Y > x
• C.
X = y
• D.
Cannot be determined
A. X > y
Explanation
Let us assume the value of x to be 10%. Therefore, the number of sheep in the herd at the beginning of year 2001 (end of 2000) will be 1 million + 10% of 1 million = 1.1 million In 2001, the numbers decrease by y% and at the end of the year the number sheep in the herd = 1 million. i.e., 0.1 million sheep have died in 2001. In terms of the percentage of the number of sheep alive at the beginning of 2001, it will be (0.1/1.1)*100 % = 9.09%. From the above illustration it is clear that x > y.
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• 8.
### A vendor sells 60 percent of apples he had and throws away 15 percent of the remainder. Next day he sells 50 percent of the remainder and throws away the rest. What percent of his apples does the vendor throw?
• A.
20
• B.
17
• C.
23
• D.
21
C. 23
Explanation
Let the number of apples be 100. On the first day he sells 60% apples ie.,60 apples. Remaining apples =40. He throws 15% of the remaining i.e., 15% of 40 = 6.Now he has 40 - 6 = 34 apples The next day he throws 50% of the remaining 34 apples i.e., 17. Therefore, in all, he throws 6 + 17 =23 apples.
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• 9.
### Tina, Mina, Gina, Lina and Bina are 5 sisters, aged in that order, with Tina being the eldest. Each of them had to carry a bucket of water from a well to their house. Their buckets' capacities were proportional to their ages. While returning, equal amount of water got splashed out of their buckets. Who lost maximum amount of water as a percentage of the bucket capacity?
• A.
Bina
• B.
Tina
• C.
Gina
• D.
Lina
A. Bina
Explanation
Tina is the older and Bina is the youngest. So, Bina's bucket would have been the smallest. Each sister lost an equal amount of water. As a proportion of the capacity of their buckets, Bina would have lost the most.
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• 10.
### If S is 150 percent of T, then T is what percent of S + T?
• A.
20%
• B.
25%
• C.
40%
• D.
10%
C. 40%
Explanation
The easiest way to solve this question is by assuming a value for T. Take T to be 100. Therefore, S = 150% of T = 150% of 100 = 150. So, S + T = 150 + 100 = 250. We need to find out T as a percent of S + T i.e., (100/250) * 100 = 40%
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• 11.
### A school sold drama tickets for 100 each for donating to an orphanage. One member sold 75% of his tickets and had 80 tickets left. How much money did the member collect?
• A.
12000
• B.
18000
• C.
24000
• D.
36000
C. 24000
Explanation
25 % = 80
75% = 240
240 X 100 = 2,40,000
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• 12.
### When processing flower - nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower - nectar must be processed to yield 1kg of honey, if nectar contains 50% water, and the honey obtained from this nectar contains 15% water?
• A.
1.5 Kg
• B.
1.7 Kg
• C.
1.25 Kg
• D.
2 Kg
B. 1.7 Kg
Explanation
Flower-nectar contains 50% of non-water part. In honey this non-water part constitutes 85% (100-15). Therefore, 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg Therefore, amount of flower-nectar needed = (0.85/0.5) * 1kg = 1.7 kg.
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• 13.
### Four men and three women can do a job in 6 days. When five men and six women work on the same job, the work gets completed in 4 days. How long will a woman take to do the job, if she works alone on it?
• A.
54
• B.
36
• C.
18
• D.
25
A. 54
Explanation
Let the amount of work done by a man in a day be 'm' and the amount of work done by a woman in a day be 'w'. Therefore, 4 men and 3 women will do 4m + 3w amount of work in a day. If 4 men and 3 women complete the entire work in 6 days, they will complete 16th of the work in a day. Hence, 4m + 3w = 16 ----- eqn (1) From statement (2), we know 5 men and 6 women take 4 days to complete the job. i.e., 5 men and 4 women working together will complete 14th of the job in a day. So, 5m + 6w = 14 ----- eqn (2) m = 1/36. w = 1/54. Hence, she will take 54 days to do the entire work.
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• 14.
### Shyam can do a job in 20 days, Ram in 30 days and Singhal in 60 days. If Shyam is helped by Ram and Singhal every 3rd day, how long will it take for them to complete the job?
• A.
10
• B.
15
• C.
20
• D.
25
B. 15
Explanation
Shyam can complete 1/20th of the job in a day, Ram can complete 1/30th of the job in a day, and Singhal can complete 1/60th of the job in a day. Every 3rd day, Shyam is helped by Ram and Singhal, so the work done on those days is 1/20 + 1/30 + 1/60 = 1/10th of the job. Therefore, in 3 days, they complete 1/10th of the job. To complete the entire job, they will need 3 * 10 = 30 days. However, since they started on the first day, the total time taken will be 30 - 3 = 27 days. Therefore, the correct answer is 15.
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• 15.
### A father can do a certain job in x hours. His son takes twice as long to do the job. Working together, they can do the job in 6 hours. How many hours does the father take to do the job?
• A.
9
• B.
15
• C.
20
• D.
12
A. 9
Explanation
If the father takes x hours to do the job and the son takes twice as long, then the son takes 2x hours to do the job. When they work together, their combined work rate is 1/x + 1/2x = 1/6. Simplifying this equation, we get 3/2x = 1/6. Cross-multiplying, we find that 2x = 18, so x = 9. Therefore, the father takes 9 hours to do the job.
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How much time did it take a certain car to travel 400 : GMAT Data Sufficiency (DS)
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# How much time did it take a certain car to travel 400
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How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
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Re: Difficult DS Problem- Need Help! [#permalink]
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06 Sep 2010, 11:41
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jjewkes wrote:
Can you help me with this problem??
How much time did it take a car to travel 400km?
(1) The car traveled the first 200 km in 2.5 hours.
(2) If the car's average speed had been 20km per hour greater than it was, it would have traveled the 400km in 1 hour less than it did.
I know that statement 2 is sufficient, but I cannot figure out why. I am getting a difficult quadratic equation when I try to put it into formulas. Obviously statement 1 is insufficient. Please help me understand this logic/algebra. Thanks!
Jeremiah
Let the average speed of the car be $$s$$ km/h and the time it spent to cover 400 km be $$t$$ hours, then $$st=400$$ ($$s=\frac{400}{t}$$). Question: $$t=?$$
(1) The car traveled the first 200 km in 2.5 hours --> we don't know the time the car needed to travel the second 200 km, so this statement is clearly insufficient.
(2) If the car's average speed had been 20km per hour greater than it was, it would have traveled the 400km in 1 hour less than it did --> $$(s+20)(t-1)=400$$ --> $$st-s+20t-20=400$$ --> as $$st=400$$ and $$s=\frac{400}{t}$$ --> $$400-\frac{400}{t}+20t-20=400$$ --> 400 cancels out and after simplification we'll get: $$20t^2-20t-400=0$$ --> $$t^2-t-20=0$$ --> $$(t-5)(t+4)=0$$ --> $$t=5$$ or $$t=-4$$ (not a valid solution as time can not be negative), so $$t=5$$. Sufficient.
Hope it helps.
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Re: Difficult DS Problem- Need Help! [#permalink]
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07 Sep 2010, 21:11
It can be solved in an easier way.
Say it traveled initially with 80km per hour. So it covered 400 km in 5 hours.
Now adding 20km per hour . i.e 100 km per hour, it would have covered 400 km in 4 hours . Which is 1 hour less as per statement 2.
Makes life easy when u plugin numbers.
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Re: Difficult DS Problem- Need Help! [#permalink]
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08 Sep 2010, 09:09
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anticipation wrote:
It can be solved in an easier way.
Say it traveled initially with 80km per hour. So it covered 400 km in 5 hours.
Now adding 20km per hour . i.e 100 km per hour, it would have covered 400 km in 4 hours . Which is 1 hour less as per statement 2.
Makes life easy when u plugin numbers.
totally agree. while algebra is nice sometimes number plugin is easier to understand
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Re: Difficult DS Problem- Need Help! [#permalink]
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20 Oct 2010, 05:46
Do we always have to solve the quadratic in such cases? You have 1 equation and you will obviously get 1 right answer as these are real speeds and we know for sure a solution exists. Given that this is a DS problem, is it safe to arrive at the equation and move on?
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Re: Data Sufficiency - Rate & Time problem [#permalink]
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03 Dec 2010, 18:02
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Yellow22 wrote:
How much time did it take a certain car to travel 400 km?
1) The car travelled the first 200 KM in 2.5 hrs?
2) If the Car's average speed was 20 KM per hour greater than it was, it would have travelled the 400 KM in 1 hour less time than it did.
I guess the solution is already clear to you.
Just for intellectual purposes, look at an alternative method:
If the car actually took t hours to cover 400 km, in the last 1 hour, the car travels a distance which is equal to 20*(t - 1) km
This must be the distance it covered in each hour since we are considering average speed.
400 = 20*(t - 1)*t
t(t - 1) = 20
t = 5 hrs
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 02 Jun 2011 Posts: 4 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: Data Sufficiency - Rate & Time problem [#permalink] ### Show Tags 19 Jun 2011, 06:40 1 This post was BOOKMARKED VeritasPrepKarishma wrote: Yellow22 wrote: How much time did it take a certain car to travel 400 km? 1) The car travelled the first 200 KM in 2.5 hrs? 2) If the Car's average speed was 20 KM per hour greater than it was, it would have travelled the 400 KM in 1 hour less time than it did. I guess the solution is already clear to you. Just for intellectual purposes, look at an alternative method: If the car actually took t hours to cover 400 km, in the last 1 hour, the car travels a distance which is equal to 20*(t - 1) km This must be the distance it covered in each hour since we are considering average speed. 400 = 20*(t - 1)*t t(t - 1) = 20 t = 5 hrs Nice one. Could't understand completely at first glance until I broke the equation down : ( Let original Avg Speed = X km/hr, Time taken = t hr., Distance = 400 km) From statement 2, => (X+20) km/hr * (t-1) hr = 400 km => [ X km/hr * (t-1)hr ]+ [20 km/hr * (t-1)hr] = 400 km => [ X km/hr * (t-1)hr ]+ [20 *(t-1) km/hr * 1hr] = 400 km ---> Equation (1) From the original problem statement, the car travels 400 km at X km/hr in t hrs So if the car travels at X km/ hr , the distance covered in the first (t-1hrs) is given by [ X km/hr * (t-1)hr ] and the distance covered in the last 1 hr is given by [20 *(t-1) km/hr * 1hr] Hence speed in the last 1 hr = [20 *(t-1)] km/hr ( and this would be the avg speed for the entire distance of 400km) [20 *(t-1)] km/hr * t hr = 400 km Upon solving we get t= 5hrs. Thanks Karishma. Manager Status: SC SC SC SC SC.... Concentrating on SC alone. Joined: 20 Dec 2010 Posts: 240 Location: India Concentration: General Management GMAT Date: 12-30-2011 Followers: 3 Kudos [?]: 60 [1] , given: 47 Re: Car and 400 Kms Distance - Ivy 26 [#permalink] ### Show Tags 13 Nov 2011, 02:10 1 This post received KUDOS Hey.. The answer is B. It is given that the distance to be covered is 400kms in the question. Taking speed as s km/hr and time as t hrs, we have the eqn 400 = st ----> (1) In B, it i given as the same distance would be covered by less than an hour if the car had traveled in a speed greater than 20km/hr that it was. so this can be equated as, 400 = (s+20)(t-1) ---> (2) Equating (1) and (2) we have, st = (s+20)(t-1) st = st + 20t - s - 20 0 = 20t - s - 20 Again substitute the value of s from eqn (1) 0 = 20t - 400/t - 20 divide it by 20 0 = t -20/t - 1 Multiply by t throughout, 0 = t^2 - 20 - t so, t^2 - t - 20 = 0 Solving for t, we get t=5 , -4 T=-4 is not possible, hence t = 5 hrs Hope it helps. _________________ D- Day December 30 2011. Hoping for the happiest new year celebrations ! Aiming for 700+ Kudo me if the post is worth it Manager Status: Enjoying the MBA journey :) Joined: 09 Sep 2011 Posts: 137 Location: United States (DC) Concentration: General Management, Entrepreneurship GMAT 1: 710 Q49 V38 WE: Corporate Finance (Other) Followers: 19 Kudos [?]: 120 [1] , given: 16 Re: Car and 400 Kms Distance - Ivy 26 [#permalink] ### Show Tags 13 Nov 2011, 02:21 1 This post received KUDOS The only information given in the question is that the total distance travelled by the car is 400kms. We have to find the time taken by the car to do so. Now, let's take a look at the given options. Statement 1 says that the car traveled the first 200 kilometers in 2.5 hours. This does not give us any information about the remaining 200 km that the car travelled. Hence this is clearly Insufficient. Statement 2 says that If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hours less time than it did. This seems to be quite a lengthy statement, so let's break this down into simpler sentences to make the given information clear. Let's assume that the car's average speed for the 400 km travel was 'x' km per hour. Also, let the time taken to travel be 't' hours. So, we have t = 400/x -----> (A) Statement 2 says that if the car's average speed had been 20 km per hour greater than it was (earlier). This indicates that the new speed is 'x+20' km per hour The car would have travelled 400 km in 1 hour less time than it did (earlier). This indicates that the time taken later (in the new scenario) is 't-1' hours We know that the car still travelled 400 km, so the new equation can be written as t-1 = 400/(x+20) -----> (B) But we know from 'A' that t = 400/x Substituting this value in the equation 'B', we get 400/x - 1 = 400/(x+20) Simplifying, we get, 400x = 400x - x^2 + 8000 -20x i.e. x^2 + 20x - 8000 = 0 Solving the above quadratic equation yields the solutions x = 80 OR x = -100 As the speed of the car cannot be negative, the required speed is 80 km per hour Using this, we can determine the time taken by the car to travel 400 km. This comes to 5 hours. Hence Statement 2 alone is sufficient to answer the question. Hope this helps. Cheers! _________________ MBA Candidate 2015 | Georgetown University McDonough School of Business VP Status: Top MBA Admissions Consultant Joined: 24 Jul 2011 Posts: 1134 GMAT 1: 780 Q51 V48 GRE 1: 1540 Q800 V740 Followers: 125 Kudos [?]: 550 [0], given: 19 Re: Car and 400 Kms Distance - Ivy 26 [#permalink] ### Show Tags 13 Nov 2011, 02:30 Using statement 2, if v is the speed of the car then: 400/v = 400/(v+20) + 1 Solve this to get v and then compute 400/v. Sufficient. So (B). _________________ GyanOne | Top MBA Rankings and MBA Admissions Blog Top MBA Admissions Consulting | Top MiM Admissions Consulting Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738 Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 64 Kudos [?]: 605 [1] , given: 355 Re: Difficult DS Problem- Need Help! [#permalink] ### Show Tags 12 Oct 2013, 10:27 1 This post received KUDOS hemanthp wrote: Do we always have to solve the quadratic in such cases? You have 1 equation and you will obviously get 1 right answer as these are real speeds and we know for sure a solution exists. Given that this is a DS problem, is it safe to arrive at the equation and move on? I guess something i've noticed in this kind of questions is the following. Once you have that a increase in rate will obviously lead to a decrease in time, say in the form (r+x)(t-y), where r and t are rate and time respectively, and asuming you know the value for rt as well, or the distance, then you know for sure you are going to have a quadratic equation with two roots one positive and one negative. Therefore, I usually stop solving right there and move on. Would like to here the opinion of the Math experts in regards to this. It could save us a nice 40 seconds at least if it is usually correct. I know we must be careful though cause there might be some exceptions on problems where this might not work, although I haven't seen one where it has not worked yet. Hope it helps Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 64 Kudos [?]: 605 [1] , given: 355 Re: How much time did it take a certain car to travel 400 [#permalink] ### Show Tags 31 Dec 2013, 06:59 1 This post received KUDOS jjewkes wrote: How much time did it take a certain car to travel 400 kilometers? (1) The car traveled the first 200 kilometers in 2.5 hours. (2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did. Statement 1 Clearly insuff Statement 2 Now here's a trick, when I get something like this (r+20)(t-1) = 400, where rt=400 Then I know two things: First, I will be able to eliminate the first 'rt' with the 400 at the other side and I will also multiply all the expression again by 'r' to get 400 again Second, I will end up with a quadratic with negative sign and two different values that will give me a positive solution and a negative solution Since time can only be positive then I know that this statement is going to be sufficient without even solving With not much more to add, this answer is a clear B Is this all clear? Cheers! J GMAT Club Legend Joined: 09 Sep 2013 Posts: 13959 Followers: 591 Kudos [?]: 167 [0], given: 0 Re: How much time did it take a certain car to travel 400 [#permalink] ### Show Tags 12 Jun 2015, 10:51 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Senior Manager Joined: 10 Mar 2013 Posts: 291 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 10 Kudos [?]: 100 [0], given: 2405 Re: How much time did it take a certain car to travel 400 [#permalink] ### Show Tags 31 Oct 2015, 18:21 I'm surprised that no one realized that we (2) allows us to solve for both s and t, since we now have a system of two equations with two unknowns. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7187 Location: Pune, India Followers: 2169 Kudos [?]: 14029 [0], given: 222 Re: How much time did it take a certain car to travel 400 [#permalink] ### Show Tags 02 Nov 2015, 20:26 TooLong150 wrote: I'm surprised that no one realized that we (2) allows us to solve for both s and t, since we now have a system of two equations with two unknowns. Note that it is not necessary that 2 equations in 2 variables will give you a unique solution. The lines depicted by the equations might be parallel or the same line. Similarly, it is not necessary that a quadratic will give two solutions - it might give a unique solution. Hence, these situations warrant further inspection if you go the algebra way. Check out these two posts for more on this topic: http://www.veritasprep.com/blog/2011/06 ... -of-thumb/ http://www.veritasprep.com/blog/2011/08 ... nd-points/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: How much time did it take a certain car to travel 400 [#permalink]
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08 Nov 2015, 04:57
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
How much time did it take a certain car to travel 400 kilometers?
(1) The car traveled the first 200 kilometers in 2.5 hours.
(2) If the car's average speed had been 20 kilometers per hour greater than it was, it would have traveled the 400 kilometers in 1 hour less time than it did.
There are 2 variables (v=velocity, t=time), one equation(vt=400) and 2 further equations from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, the fact that the car traveled the first 200km in 2.5hrs is not helpful; there is no explanation that the velocity is constant, so we do not know anything about the next 200km, so this is insufficient.
From condition 2, (v+20)(t-1)=400, vt=400. This is sufficient, so the answer becomes (B).
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: How much time did it take a certain car to travel 400 [#permalink]
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17 Dec 2015, 11:03
Statement 1: What about the remaining 200 kms. Did it take 10,00 hours or 10 hours or 2.356 hours? Insufficient.
Statement 2: Let initial speed be x and time taken be y. Distance(i.e. 400) = xy. According to information in Statement 2, (x+20)(y-1) = 400.
Eliminating laziness and solving the equations we would get xy = xy + 20y - x - 20
20y - x = 20
Hence B
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Re: How much time did it take a certain car to travel 400 [#permalink]
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01 Aug 2016, 09:39
jlgdr wrote:
jjewkes wrote:
Statement 2
Now here's a trick, when I get something like this (r+20)(t-1) = 400, where rt=400
Then I know two things:
First, I will be able to eliminate the first 'rt' with the 400 at the other side and I will also multiply all the expression again by 'r' to get 400 again
Second, I will end up with a quadratic with negative sign and two different values that will give me a positive solution and a negative solution
Since time can only be positive then I know that this statement is going to be sufficient without even solving
J
I have one questions to your approach. How do you know that there won't be two positive solutions? If I only think of quadratic equations with negative sign there also can be two positive solutions (y=x^2-3x+2).
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Re: How much time did it take a certain car to travel 400 [#permalink]
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01 Aug 2016, 09:59
22gmat wrote:
jlgdr wrote:
jjewkes wrote:
Statement 2
Now here's a trick, when I get something like this (r+20)(t-1) = 400, where rt=400
Then I know two things:
First, I will be able to eliminate the first 'rt' with the 400 at the other side and I will also multiply all the expression again by 'r' to get 400 again
Second, I will end up with a quadratic with negative sign and two different values that will give me a positive solution and a negative solution
Since time can only be positive then I know that this statement is going to be sufficient without even solving
J
I have one questions to your approach. How do you know that there won't be two positive solutions? If I only think of quadratic equations with negative sign there also can be two positive solutions (y=x^2-3x+2).
The equation above is of the form (x-a)(x+b) which will always give one positive and one negative solution.
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Re: How much time did it take a certain car to travel 400 [#permalink]
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04 Aug 2016, 07:36
Got it. Thank you very much abhimahna.
Kind regards.
Re: How much time did it take a certain car to travel 400 [#permalink] 04 Aug 2016, 07:36
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70.82 kg to lbs - 70.82 kilograms to pounds
Before we move on to the practice - that is 70.82 kg how much lbs conversion - we are going to tell you few theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 70.82 kg to lbs? 70.82 kilograms it is equal 156.1313739484 pounds, so 70.82 kg is equal 156.1313739484 lbs.
70.82 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, that is International System of Units (in short form SI).
From time to time the kilogram is written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. First definition was simply but hard to use.
Then, in 1889 the kilogram was described using the International Prototype of the Kilogram (in short form IPK). The IPK was prepared of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was in use until 2019, when it was substituted by a new definition.
Today the definition of the kilogram is build on physical constants, especially Planck constant. Here is the official definition: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.
70.82 kilogram to pounds
You learned a little bit about kilogram, so now we can move on to the pound. The pound is also a unit of mass. We want to emphasize that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is in use in the British and United States customary systems of measurements. Naturally, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is defined as 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains.
The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
How many lbs is 70.82 kg?
70.82 kilogram is equal to 156.1313739484 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
70.82 kg in lbs
The most theoretical section is already behind us. In this part we will tell you how much is 70.82 kg to lbs. Now you learned that 70.82 kg = x lbs. So it is high time to get the answer. Just see:
70.82 kilogram = 156.1313739484 pounds.
This is a correct outcome of how much 70.82 kg to pound. You may also round off the result. After it your result is exactly: 70.82 kg = 155.804 lbs.
You know 70.82 kg is how many lbs, so look how many kg 70.82 lbs: 70.82 pound = 0.45359237 kilograms.
Obviously, this time it is possible to also round off the result. After rounding off your result is as following: 70.82 lb = 0.45 kgs.
We also want to show you 70.82 kg to how many pounds and 70.82 pound how many kg outcomes in tables. Let’s see:
We are going to start with a table for how much is 70.82 kg equal to pound.
70.82 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
70.82 156.1313739484 155.8040
Now see a table for how many kilograms 70.82 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
70.82 0.45359237 0.45
Now you know how many 70.82 kg to lbs and how many kilograms 70.82 pound, so it is time to go to the 70.82 kg to lbs formula.
70.82 kg to pounds
To convert 70.82 kg to us lbs you need a formula. We are going to show you a formula in two different versions. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 156.1313739484 outcome in pounds
The first formula will give you the most accurate result. Sometimes even the smallest difference can be significant. So if you need an exact result - this formula will be the best for you/option to calculate how many pounds are equivalent to 70.82 kilogram.
So let’s go to the second formula, which also enables calculations to learn how much 70.82 kilogram in pounds.
The shorter formula is as following, have a look:
Number of kilograms * 2.2 = the outcome in pounds
As you can see, the second formula is simpler. It could be the best option if you need to make a conversion of 70.82 kilogram to pounds in easy way, for example, during shopping. You only need to remember that final outcome will be not so correct.
Now we are going to show you these two versions of a formula in practice. But before we will make a conversion of 70.82 kg to lbs we want to show you easier way to know 70.82 kg to how many lbs without any effort.
70.82 kg to lbs converter
An easier way to learn what is 70.82 kilogram equal to in pounds is to use 70.82 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. It is based on longer formula which we showed you in the previous part of this article. Thanks to 70.82 kg pound calculator you can effortless convert 70.82 kg to lbs. Just enter number of kilograms which you need to convert and click ‘convert’ button. You will get the result in a second.
So let’s try to convert 70.82 kg into lbs using 70.82 kg vs pound calculator. We entered 70.82 as a number of kilograms. Here is the outcome: 70.82 kilogram = 156.1313739484 pounds.
As you can see, our 70.82 kg vs lbs calculator is intuitive.
Now we can move on to our main issue - how to convert 70.82 kilograms to pounds on your own.
70.82 kg to lbs conversion
We will begin 70.82 kilogram equals to how many pounds calculation with the first formula to get the most accurate result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 156.1313739484 the result in pounds
So what need you do to know how many pounds equal to 70.82 kilogram? Just multiply amount of kilograms, this time 70.82, by 2.20462262. It is equal 156.1313739484. So 70.82 kilogram is 156.1313739484.
It is also possible to round off this result, for example, to two decimal places. It is equal 2.20. So 70.82 kilogram = 155.8040 pounds.
It is high time for an example from everyday life. Let’s calculate 70.82 kg gold in pounds. So 70.82 kg equal to how many lbs? And again - multiply 70.82 by 2.20462262. It gives 156.1313739484. So equivalent of 70.82 kilograms to pounds, if it comes to gold, is equal 156.1313739484.
In this example you can also round off the result. Here is the result after rounding off, in this case to one decimal place - 70.82 kilogram 155.804 pounds.
Now we can go to examples converted using a short version of a formula.
How many 70.82 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 155.804 the outcome in pounds
So 70.82 kg equal to how much lbs? And again, you need to multiply amount of kilogram, this time 70.82, by 2.2. Look: 70.82 * 2.2 = 155.804. So 70.82 kilogram is 2.2 pounds.
Let’s do another calculation using shorer version of a formula. Now convert something from everyday life, for example, 70.82 kg to lbs weight of strawberries.
So calculate - 70.82 kilogram of strawberries * 2.2 = 155.804 pounds of strawberries. So 70.82 kg to pound mass is equal 155.804.
If you learned how much is 70.82 kilogram weight in pounds and are able to calculate it with use of two different versions of a formula, let’s move on. Now we want to show you all outcomes in charts.
Convert 70.82 kilogram to pounds
We realize that results presented in charts are so much clearer for most of you. We understand it, so we gathered all these results in tables for your convenience. Thanks to this you can easily compare 70.82 kg equivalent to lbs outcomes.
Start with a 70.82 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
70.82 156.1313739484 155.8040
And now see 70.82 kg equal pound chart for the second formula:
Kilograms Pounds
70.82 155.804
As you see, after rounding off, when it comes to how much 70.82 kilogram equals pounds, the results are not different. The bigger amount the more considerable difference. Remember it when you want to make bigger amount than 70.82 kilograms pounds conversion.
How many kilograms 70.82 pound
Now you learned how to calculate 70.82 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What do you say about 70.82 kilogram to pounds and ounces calculation?
We want to show you how you can calculate it little by little. Let’s start. How much is 70.82 kg in lbs and oz?
First things first - you need to multiply number of kilograms, in this case 70.82, by 2.20462262. So 70.82 * 2.20462262 = 156.1313739484. One kilogram is equal 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 70.82 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So your result is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then final outcome will be exactly 2 pounds and 33 ounces.
As you can see, conversion 70.82 kilogram in pounds and ounces simply.
The last calculation which we are going to show you is conversion of 70.82 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.
To convert it it is needed another formula. Before we show you this formula, let’s see:
• 70.82 kilograms meters = 7.23301385 foot pounds,
• 70.82 foot pounds = 0.13825495 kilograms meters.
Now let’s see a formula:
Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 70.82 foot pounds to kilograms meters you have to multiply 70.82 by 0.13825495. It gives 0.13825495. So 70.82 foot pounds is equal 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 70.82 foot pounds will be exactly 0.14 kilogram meters.
We hope that this calculation was as easy as 70.82 kilogram into pounds conversions.
We showed you not only how to do a conversion 70.82 kilogram to metric pounds but also two other conversions - to know how many 70.82 kg in pounds and ounces and how many 70.82 foot pounds to kilograms meters.
We showed you also other way to make 70.82 kilogram how many pounds conversions, that is using 70.82 kg en pound calculator. It is the best option for those of you who do not like calculating on your own at all or this time do not want to make @baseAmountStr kg how lbs calculations on your own.
We hope that now all of you can do 70.82 kilogram equal to how many pounds conversion - on your own or using our 70.82 kgs to pounds converter.
Don’t wait! Convert 70.82 kilogram mass to pounds in the way you like.
Do you want to make other than 70.82 kilogram as pounds conversion? For instance, for 5 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so simply as for 70.82 kilogram equal many pounds.
How much is 70.82 kg in pounds
At the end, we are going to summarize the topic of this article, that is how much is 70.82 kg in pounds , we prepared for you an additional section. Here you can see all you need to remember about how much is 70.82 kg equal to lbs and how to convert 70.82 kg to lbs . Let’s see.
How does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 70.82 kg to pound conversion formula . Check it down below:
The number of kilograms * 2.20462262 = the result in pounds
Now you can see the result of the conversion of 70.82 kilogram to pounds. The exact answer is 156.1313739484 pounds.
There is also another way to calculate how much 70.82 kilogram is equal to pounds with second, easier version of the formula. Have a look.
The number of kilograms * 2.2 = the result in pounds
So now, 70.82 kg equal to how much lbs ? The answer is 156.1313739484 lb.
How to convert 70.82 kg to lbs in an easier way? You can also use the 70.82 kg to lbs converter , which will make whole mathematical operation for you and give you a correct result .
Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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An Example of Undetermined Coefficients
Flash and JavaScript are required for this feature.
Description: This method is also successful for forces and solutions equal to polynomials times exponentials. Substitute into the equation!
Related section in textbook: 2.6b
Instructor: Prof. Gilbert Strang
GILBERT STRANG: OK. So can I begin with a few words about the big picture of solving differential equations? So if that was a nonlinear equation, we would go to computer solutions. And Cleve Moler is making a parallel video series about the Matlab suite of codes for solving differential equations.
Then when that equation is linear, as it is here, with constant coefficients, as like the 1, minus 3, and the 2, we can always get a formula for the answer. Involves an integral. There's still one integral to do involving the impulse response. And you'll see that.
But there are few, the most beautiful, the most simple equations when the right-hand side has a special form. And that's one possibility-- t, or t squared we could deal with, or e to the t. We'll see all those. Then we know what the solution looks like.
For example-- first of all, we know the null solutions, of course. That's when this is 0. And then I just wrote for e to the st, and I find s could be 1 or s could be 2. Those are two solutions with right side 0. So we can match initial conditions. But we need a particular solution.
And that's where this one is especially simple. So the idea is to try the form we know the solution will have. So I'm going to try a particular solution. When I see a t there, I'll want a t in the particular solution. But I also need the constant term. So I'll try a plus bt.
What I mean by try is put that into the equation, match the left side and the right side, and we have a solution. And I'll just go ahead and do that. So there's several-- this video is mostly about the list of possible nice functions. And that's one of them.
OK. So if I put that into the equation, the second derivative is 0 for that. So I have minus 3. The first derivative is just b. Then I have plus 2 times y itself, which is a plus bt. So the left side of the equation is just this much. And it has to equal 4t.
And that we can make happen. I see 2bt and 4t here, so I want b equal to 2. And so when b is 2, then I have 4t matching 4t. And then I have minus 3b plus 2a should be 0. Minus 3b plus 2a, the constant part there we don't want, so that should be 0. We already know b is 2, so that's minus 6 plus 2a. a is 3. a is 3.
Now, that's perfectly satisfied by b equal 2a equal 3, and that's the answer. Correct answer is b was 2, a was 3, and we don't have to say try anymore. We got it. Done.
One other right-hand side. Once you get a nice one like this, you look for more. If it was t squared, we would assume-- what would we assume if it was a 4t squared there? We would assume a plus bt plus ct squared. We want to match the right-hand side.
Now, a different type of right-hand side we know already. What if this was e to the-- say, e to the 3t? Or e to the st. Let me put any exponent there for the moment. So now we have a different right-hand side. A very nice function. The best function of differential equations.
And now what we will try with this right-hand side, e to the st. We've seen it before. The particular solution is just y e to the st. The undetermined coefficients were a and b in the first time. Here the undetermined coefficient is capital Y. I'm just going to plug that into the equation and match the left side and right side. And I'll determine this coefficient, capital Y.
So what happens when I put that into the equation? I get second derivative brings down an s squared, and first derivative, we have a minus 3. The first derivative brings down an s, and I have plus 2. All that is Y e to the st, right? I put in Y e to the st. The derivatives brought down this familiar polynomial. And it's all supposed to match e to the st.
We can do that. That's a perfect match. The left side has the same form as the right side. I can cancel the e to the st's, and I discover that Y is 1 over s squared minus 3s plus 2. Good. That's the coefficient, 1 over that.
Let's see. I have to make two comments. One comment is that s-- we're totally golden if that s is imaginary. If s is i omega, well, minus i omega, or both, both possibilities, those will give us sine and cosine. So we'll add those to the nice function.
So the nice functions are t, polynomials. E to the st, exponentials. Sines and cosines. And you see this worked perfectly. Well, perfectly except we have to be sure that we don't have 1/0.
When would we have 1/0? If s is 1, this would be 1 minus 3 plus 2, that would be 1/0. I can't deal with s equal 1 with that assumption. Doesn't work. Also, s equal 2. What's special about s equal 1 and s equal 2? Those make that 0. They give the null solutions.
And I know that if this happened to be e to the t, and a null solution was also e to the t, I have to fix the particular solution by giving it an extra factor t. Let me do that at the end of the lecture. That's the resonance idea, where my null solution is the same as what I hope for the particular solution. So I have to change that particular solution with an extra factor t.
Let me keep going here to be sure we get all the possibilities. Again, we're getting a very small set of nice functions. But fortunately, they appear quite often in applications. Constants, linear like 4t, exponentials like e to the 5t, oscillation like e to the i omega t. So let me make a list.
So polynomials, like the forcing function could be t, could be 1. Constant, that would be like a ramp. That would be like a step function. They could be 3 to the st, but not s equals 1 and 2 in this problem. They could be e to the i omega t, and e to the minus i omega t, which leads me directly to cosine of omega t and sine of omega t.
I'm creating here a list of the nice functions-- polynomials, exponentials, cosine and sine, because those come from exponentials, and finally, I can multiply these. I could have, for example, t e to the st. I could allow that. And now, all I have to do is tell you, what form do I try to plug-in?
The form will have undetermined coefficients. I'll substitute in the equation, and I'll determine the coefficients. So here we saw a plus bt. That was good. That was the Y. Here we saw Y e to the st. That was good.
Here, what will I have? If I have cosine, I need the sine there also. So I'll have to allow a combination of those. Say M on this plus N of that. If I tried to do cosines alone, I would be in danger of taking the derivative, getting a sine, and having nothing to match. So I'll take that.
Now, final question, or next question is, if I multiply two of these, the product rule is still going to tell me that derivatives of that have the same form as that. Derivatives of this t e to the st have the same-- they also involve t or e to the st, with factors s, from the product rule.
So what do I assume here? Well, when I see t there, I have to include, as I did up there, also constants. And if I saw a t squared, I would go up to t squared. I'd have three coefficients. Now, that e to the st, I can keep.
So what I've put on the right-hand side is the right form to assume. It's just like good advice. Put that into the differential equation when this is the right-hand side. Match left side with the right side. That will tell you a and b. And you have the answer. You have the particular solution. You have a particular solution, and that's what you wanted.
OK. And if I had t cosine omega t, do you want to see in the whole business? If I had t cosine omega t-- oh dear, it's getting a little messy. I'd need an a plus bt to deal with the t. And I'd need a cosine to deal with the cosine.
And then, just to make the problem a little messier, can't be helped, I'd need the sine. So I need a sine omega t. And I'm going to need a c plus dt there. OK. I'm up to four coefficients to determine by plugging in. It uses more ink, but it doesn't use more thinking. You just put it in, match all terms, and you discover A, B, C, and D.
Finally, I have to say that word about resonance that I mentioned earlier. The possibility that if s happened to be 1 or 2 on the right-hand side, that's also a null solution. This method would give me 1/0, infinite answer, no good. And we know what to do.
We know how to deal with resonance. So with resonance-- so could I just finally complete the whole story. Resonance. In this example, s equal 1 or 2. What would I do with f of t equals e to the t? If that was the right-hand side, that would give me resonance.
It's got the exponent s equal 1, which is also in the null solution, e to the t. So what do I try? So I try-- everybody knows what happens when there's resonance. When you have this doubling up. You need an extra factor t to rescue. So you would try y of t. y, this is the particular solution I'm looking for.
You would try t e to the t, with a Y. And you would plug that in. You would find the right number for capital Y. And you'd have the particular solution. Only, I think, doing a few examples of this, you get the knack of assume the right form, put it into the equation, match left side with right side, and that reveals the undetermined coefficients. It tells you what a and b and capital Y and c and d, tells you what they all have to be.
So this is a really good method that applies to really nice right-hand sides. Good. Thanks.
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# E12 Assignment 12
Description:
Problems:
1. Sinusoidal Steady State of First Order System
A system has transfer function
Determine the sinusoidal steady state output if the input is:
1. x(t)=sin(0.1·t)
2. x(t)=cos(t+45°)
3. x(t)=3·sin(10·t-15°)
4. x(t)=4·sin(100·t)
1. Does the system exhibit high pass or low pass behavior (i.e., does it preferentially transmit high or low frequency inputs)?
Note: it isn't necessary to draw the Bode plot for this problem, you can simply calculate the magnitude and phase of H(jω) at each frequency.
Solution:
a) H=0.01∠89° (see below). Output is 0.01·sin(0.1·t+89°).
```>> H=0.1j/(10+0.1j)
H = 0.0001 + 0.0100i
>> abs(H)
ans = 0.0100
>> angle(H)*180/pi
ans = 89.4271```
b) H=0.1∠84° (see below). Output is 0.1·cos(t+129°).
```>> H=1j/(10+1j)
H = 0.0099 + 0.0990i
>> abs(H)
ans = 0.0995
>> angle(H)*180/pi
ans = 84.2894```
c) H=0.71∠45° (see below). Output is 2.1·sin(10·t+30°).
```>> H=10j/(10+10j)
H = 0.5000 + 0.5000i
>> abs(H)
ans = 0.7071
>> angle(H)*180/pi
ans = 45```
d) H=1∠6° (see below). Output is 4·sin(100·t+6°).
```>> H=100j/(10+100j)
H = 0.9901 + 0.0990i
>> abs(H)
ans = 0.9950
>> angle(H)*180/pi
ans = 5.7106```
e) The high frequency inputs are transmitted without attenuation. This is a high pass system.
2. Bode Plot, Distinct, Real Poles and Zeros
Consider the transfer function .
1. Plot the asymptotic (straight line approximations) magnitude and phase plots for a Bode diagram. Assume 3.2 is halfway between 1 and 10 on a log scale. Suitable paper is here.
2. Using the Bode plot estimate the sinusoidal steady state output if the input is cos(5t).
3. Verify by calculating H(j5). Note: Matlab's "freqresp" command does this, or you can evaluate with your calculator or by hand).
4. Use Matlab to plot the Bode diagram using the "bode" command. Use the "grid" command to add a grid (useful for the next part).
5. Transfer your asymptotic approximation onto the Matlab plot.
Solution:
a)
Constant=31 Magnitude=31≈30dB, phase = 0°
Zero at the s=-10 (i.e., break frequency is 10 rad/sec). Slope is +20 dB/dec, phase goes from 0 to 90°
Pole at s=-1, and at s=-3.2 (i.e., break frequencies at 1 and 3.2 rad/sec). Slope from each is -20 db/dec, and phase goes from 0 to -90°
b) At ω=5, H(jω)≈12 dB→1012/20=4.0, ∠H(jω)≈-100° Output=4.0cos(5t-100°).
c) Output=3.7cos(5t-109°);
```>> H=tf(10*[1 10],[1 4.2 3.2]);
ans = 3.6936
ans = -109.5058```
d) >> bode(H); grid
e) See part a.
3. Complex Bode Plot
1. Sketch (by hand) the Bode plot for . Suitable paper is here. You may assume that the phase drops from 0° to -180° instantaneously at the break frequency of the underdamped poles.
2. Verify with Matlab.
Solution:
a)
s2+2s+25 yields ω0=5, ζ=0.2
Constant = 3.2 ≈ 10 dB;
Double zero at the origin yields +40 dB/dec with phase of +180° (or -180°)
Pole at s=-100 yields slope of -20 dB/dec and phase from 0 to -90°
Underdamped poles with ω0=5, ζ=0.2 gives a peak of about 1/2ζ=2.5=8 dB.
b) >> H=tf(8000*[1 0 0],conv([1 100],[1 2 25])); bode(H); grid
4. Find System From Bode - Real poles
a) Find the differential equation for the system with Bode plot shown.
b) Show a system of your own design that has the given Bode plot (e.g., mass-spring-dashpot, resistor-capacitor-inductor. For this problem the system can even be resistor-capacitor (two capacitors) or resistor-inductor).
c) What are ζ and ω0?
Solution:
a,c) DC gain=1 (0 dB) and there is a double pole at w0=10 rad/sec (mag drops 40 dB/dec, and phase goes from 0 to -180°.
b)
5. Height of Underdamped Peak
Consider the transfer function .
• Show that the peak of |G(jω)| occurs at
Hint: if X is positive, then the extrema of X occurs at the same place as the extrema of X2 (or, the extrema of √X occurs at the same place as that of X), and the peak of X occurs at the same place as the minimum of 1/X.
• Show that the peak has amplitude .
Solution:
a) . Peak of |G(jω)| is at same frequency as minimum of 1/|G(jω)|^2. Find this by differentiation:
b)
We can pull some terms out of the square root
6. Find zeta for which Bode Approximations are valid
The peak due to a second order underdamped system, , occurs at frequency and has amplitude . We often use approximations when sketching by hand.
a) For the approximation, for what value of ζ is the approximation accurate to within 10%?
b) For the approximation , for what value of ζ is the approximation accurate to within 10%?
c) If we say that, for hand plots, we only need to plot a resonant peak if it is bigger than 6 dB (recall that 6 dB is about a factor of two), what is the corresponding value of ζ?
The results of this problem indicate that if we want to make a hand plot, we can use the given approximations for resonant frequency and peak amplitude.
Solution:
a)
b)
c)
Since we only have a peak for ζ<1/√2=0.71, we use the lower value of ζ=0.26 (so we don't need to plot for ζ>0.26).
In other words, if the peak is large enough to plot in a hand plot, the value of ζ is small enough that we can use approximations for the height of the peak and the frequency at which the peak occurs.
7. Transfer Function Properties (1)
Consider a system transfer function with Transfer Function
Assume none of denominator coefficients are zero, and that m≤n. Show that
1. In response to a step input, the steady state value of the system is zero if a0=0, otherwise it is a0/b0.
2. That the value of the transfer function goes to zero at low frequencies (s→0) if a0=0, otherwise it is a0/b0.
3. In response to a step input, the value at t=0+ is zero if m<n, otherwise it is an/bn.
4. That the value of the transfer function goes to zero at high frequencies (s→∞) if m<n, otherwise it is an/bn.
5. (optional) If m<n then the first n-m-1 derivatives of the step response are zero at t=0+.
Solution:
a)
b)
c)
d)
e)
lower order derivatives will also be zero, but any derivative of higher order will not be equal to zero.
8. Transfer Function Properties (2)
Verify that each of the five results of the previous problem is correct by plotting the step response and Bode plots of the four Transfer Functions given below. Use ζ=0.25, ωn=10. You needn't do any calculations. Use Matlabs "tf," "step" and "bode" command. Note that each command can plot multiple system responses simultaneously (i.e., step(sys1, sys2,....)).
Solution:
```% Set values.
w=10; z=0.25;
% Now define transfer functions
GLP=tf([0 0 w*w],[1 2*z*w w*w]);
GBP=tf([0 2*z*w 0],[1 2*z*w w*w]);
GHP=tf([1 0 0],[1 2*z*w w*w]);
GN=tf([1 0 w*w],[1 2*z*w w*w]);
%% Step response
step(GLP,GBP,GHP,GN);
legend('G_{LP}','G_{BP}','G_{HP}','G_N')
%% Bode
bode(GLP,GBP,GHP,GN);
legend('G_{LP}','G_{BP}','G_{HP}','G_N')```
a) GLP. Step response has final value of 1, initial value of 0, and initial slope of zero. Bode=1 at low frequencies and 0 at high frequencies.
b) GBP. Step response has final value of 0, initial value of 0, and initial slope not equal to zero. Bode=0 at low frequencies and 0 at high frequencies.
c) GHP. Step response has final value of 0, initial value of 1, and initial slope not equal to zero. Bode=0 at low frequencies and 1 at high frequencies.
d) GN. Step response has final value of 1, initial value of 1, and initial slope not equal to zero. Bode=1 at low frequencies and 1 at high frequencies.
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## Project Euler — Problem 87 Solution
#### Problem
The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:
28 = 22 + 23 + 24
33 = 32 + 23 + 24
49 = 52 + 23 + 24
47 = 22 + 33 + 24
How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?
#### Solution
```// generate all prime numbers under <= this max
let max = int64(sqrt(double(50000000L)))
// initialise the list with 2 which is the only even number in the sequence
// only check the prime numbers which are <= the square root of the number n
let hasDivisor n =
|> Seq.takeWhile (fun n' -> n' <= int64(sqrt(double(n))))
|> Seq.exists (fun n' -> n % n' = 0L)
// only check odd numbers <= max
let potentialPrimes = Seq.unfold (fun n -> if n > max then None else Some(n, n+2L)) 3L
// populate the prime numbers list
for n in potentialPrimes do if not(hasDivisor n) then primeNumbers <- primeNumbers @ [n]
// use the same hasDivisor function instead of the prime numbers list as it offers
// far greater coverage as the number n is square rooted so this function can
// provide a valid test up to max*max
let isPrime n = if n = 1L then false else not(hasDivisor(n))
|> Seq.collect (fun n ->
|> Seq.map (fun n' -> pown n 2 + pown n' 3)
|> Seq.takeWhile (fun sum -> sum < 50000000L))
|> Seq.collect (fun sum ->
|> Seq.map (fun n -> sum + pown n 4)
|> Seq.takeWhile (fun sum' -> sum' < 50000000L))
|> Seq.distinct
|> Seq.length
```
The biggest prime that can appear in the equation is equals to the square root of 50000000 – 8 – 16, which, incidentally is just over 7000, so the numbers of primes involved is reasonably small which bodes well for a fast solution!
The logic in this solution is otherwise simple, using the cached prime numbers list to first generate numbers (< 50 million) that can be written as the sum of a prime square and prime cube; then for each number see if we can add a prime fourth power and get a sum less than 50 million.
One thing that caught me out initially was the need to search for distinct numbers as some of these numbers do overlap, but otherwise this is a solution which runs happily under 3 seconds.
## Project Euler — Problem 92 Solution
#### Problem
A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
For example,
Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
How many starting numbers below ten million will arrive at 89?
#### Solution
```let max = 10000000
// build up a cache for all the numbers from 1 to 10 million
let cache = Array.init (max+1) (fun n ->
match n with
| 0 | 1 -> Some(false)
| 89 -> Some(true)
| _ -> None)
// define function to add the square of the digits in a number
n.ToString().ToCharArray()
|> Array.sumBy (fun c -> pown (int(c.ToString())) 2)
// define function to take an initial number n and generate its number chain until
// it gets to a number whose subsequent chain ends with 1 or 89, which means that
// all previous numbers will also end in the same number
let processChain n =
let rec processChainRec n (list: int list) =
if cache.[n] = None then processChainRec (addSquaredDigits n) (list@[n])
else list |> List.iter (fun n' -> cache.[n'] <- cache.[n])
processChainRec n []
// go through all the numbers from 2 to 10 million using the above function
[2..10000000] |> List.iter processChain
// how many numbers whose number chain ends with 89?
let answer = cache |> Array.filter (fun n -> n = Some(true)) |> Array.length
```
This is actually a fairly simple problem, with the main challenge being how to make it run fast enough to obey with the one minute rule which is where the cache comes in.
Using the property that if a chain starting with the number n, i.e. n, n1, n2, …, ends in 89 then the chain starting with any of the numbers n1, n2, … will also end in 89 we can minimise the amount of computation required by caching previous results. Note I didn’t have to mark the cache as mutable in this case because the Array construct in F# is a fixed-sized, zero-index, mutable collections of elements.
## Project Euler — Problem 50 Solution
#### Problem
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most consecutive primes?
#### Solution
```// generate all prime numbers under <= this max
let max = 10000
// initialise the list with 2 which is the only even number in the sequence
// only check the prime numbers which are <= the square root of the number n
let hasDivisor n =
|> Seq.takeWhile (fun n' -> n' <= int(sqrt(double(n))))
|> Seq.exists (fun n' -> n % n' = 0)
// only check odd numbers <= max
let potentialPrimes = Seq.unfold (fun n -> if n > max then None else Some(n, n+2)) 3
// populate the prime numbers list
for n in potentialPrimes do
// use the same hasDivisor function instead of the prime numbers list as it offers
// far greater coverage as the number n is square rooted so this function can
// provide a valid test up to max*max
let isPrime n = if n = 1 then false else not(hasDivisor(n))
// define function which computes the sum of the all consecutive primes starting
// from n, and returns the longest sequence which sums to a prime
let getPrimeSequence n =
|> Seq.filter (fun n' -> n' > n)
|> Seq.scan (fun (sum, count) n' -> (sum+n', count+1)) (n, 1)
|> Seq.takeWhile (fun (sum, count) -> sum < 1000000)
|> Seq.filter (fun (sum, count) -> isPrime sum)
|> Seq.maxBy (fun (sum, count) -> count)
// for all numbers, find the longest sequence
let answer = primeNumbers |> Seq.map getPrimeSequence |> Seq.maxBy (fun (sum, count) -> count)
```
Again, utilizing the PGsimple 3 algorithm from here, this solution runs in just over a second as I’ve restricted it to check sequences starting with a prime < 10000 which perhaps unsurprisingly, has provided me with the answer.
The only tricky part in this problem is how to build up a list of sequences for each starting prime and find the longest sequence starting from that number which results in a prime < 1000000. Fortunately you can do this rather easily with Seq.scan in F# which works similarly to Seq.fold but gives you all the intermediate results as well as the final result. Used in conjunction with Seq.takeWhile it allows me to get all the sequences which sums up to a number less than 1000000. The rest is pretty straight forward, filter out the sums that aren’t prime numbers and find the longest sequence, and do this for all the primes we have generated to find the answer to the problem.
## Project Euler — Problem 57 Solution
#### Problem
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
By expanding this for the first four iterations, we get:
1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666…
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379…
The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
#### Solution
```// define function to return all the numerator-denominator pairs for the first n expand
let expand n =
Seq.unfold (fun (num, denom) -> Some((num, denom), (denom*2I+num, denom+num))) (3I, 2I)
|> Seq.take n
expand 1000
|> Seq.filter (fun (num, denom) -> num.ToString().Length > denom.ToString().Length)
|> Seq.length
```
If you look at the patterns 3/2, 7/5, 17/12, 41/29, and so on, it’s easy to spot a pattern where the numerator and denominator of iteration n can be derived from the iteration n-1:
Numerator(n) = Numerator(n-1) + 2 * Denominator(n-1)
Denominator(n) = Numerator(n-1) + Denominator(n-1)
## Project Euler — Problem 58 Solution
#### Problem
Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 62%.
If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
#### Solution
```let hasDivisor(n) =
let upperBound = int(sqrt(double(n)))
[2..upperBound] |> Seq.exists (fun x -> n % x = 0)
let isPrime(n) = if n = 1 then false else not(hasDivisor(n))
// define function that returns the number on the corners of a spiral of length n
let getCornerNumbers n =
match n with
| 1 -> [1]
| _ when n % 2 = 0 -> []
| _ -> [3..-1..0] |> List.map (fun n' -> n*n - n'*(n-1))
let mutable cornerNumbers, primeNumbers, size = 0, 0, 1
let mutable continueLoop = true
while continueLoop do
// get the numbers that appear at the corners of a spiral of the given size
let newNumbers = getCornerNumbers size
// increment the totals
cornerNumbers <- cornerNumbers + newNumbers.Length
let ratio = double(primeNumbers) / double(cornerNumbers)
if ratio < 0.1 && size > 1 then continueLoop <- false else size <- size + 2
size
```
UPDATE: Having stumbled upon some very good algorithms for generating prime number sequence here, I decided to revisit my solution here and using the PGSimple3 algorithm the new code now runs in seconds!
```// generate all prime numbers under <= this max
let max = 100000
// initialise the list with 2 which is the only even number in the sequence
// only check the prime numbers which are <= the square root of the number n
let hasDivisor n =
|> Seq.takeWhile (fun n' -> n' <= int(sqrt(double(n))))
|> Seq.exists (fun n' -> n % n' = 0)
// only check odd numbers <= max
let potentialPrimes = Seq.unfold (fun n -> if n > max then None else Some(n, n+2)) 3
// populate the prime numbers list
for n in potentialPrimes do
// use the same hasDivisor function instead of the prime numbers list as it offers
// far greater coverage as the number n is square rooted so this function can
// provide a valid test up to max*max
let isPrime n = if n = 1 then false else not(hasDivisor(n))
// define function that returns the number on the corners of a spiral of length n
let getCornerNumbers n =
match n with
| 1 -> [1]
| _ when n % 2 = 0 -> []
| _ -> [3..-1..0] |> List.map (fun n' -> n*n - n'*(n-1))
let mutable cornerNumbers, primeNumbers, size = 0, 0, 1
let mutable continueLoop = true
while continueLoop do
// get the numbers that appear at the corners of a spiral of the given size
let newNumbers = getCornerNumbers size
// increment the totals
cornerNumbers <- cornerNumbers + newNumbers.Length
| 3,412 | 11,583 |
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For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the : GMAT Data Sufficiency (DS)
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For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the
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For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
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aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given: $$a$$ and $$b$$ are integers, also $$\sqrt{a^3-a^2-b}=7$$ --> $$a^3-a^2-b=49$$
(1) a^2 - a = 12 --> $$a=-3$$ or $$a=4$$. Now, both values of $$a$$ give an integer solution for $$b$$ ($$b=85$$ or $$b=-1$$), so both values are valid. Not sufficient.
(2) b^2 - b = 2 --> $$b=-1$$ or $$b=2$$ --> if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$ BUT if if $$b=2$$ then $$a^3-a^2=51$$ --> $$a^2(a-1)=51=3*17$$ --> this equation has no integer solution for $$a$$, hence only the first case is valid: $$b=-1$$ and $$a=4=integer$$. Sufficient.
Answer: B.
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08 Dec 2010, 09:13
can you elaborate on how you solved $$a^2(a-1) = 48$$ to $$a = 4$$?
Similarly how did you conclude that $$a^2(a-1) = 51$$ will not have an interger as a solution?
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aalriy wrote:
can you elaborate on how you solved $$a^2(a-1) = 48$$ to $$a = 4$$?
Similarly how did you conclude that $$a^2(a-1) = 51$$ will not have an interger as a solution?
$$a^2(a-1)=48$$: as $$a$$ is an integer than we have that 48 is a product of a perfect square (a^2) and another positive integer (a-1): 48=1*48=4*12=16*3, so after some trial and error you'll get $$a=4$$;
$$a^2(a-1)=51=1*51$$: the same here. But 51 can be represented as a product of perfect square and another integer only in one way 51=1*51, which doesn't fit for a^2(a-1), so this equation doesn't have integer solution for $$a$$.
Hope it's clear.
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aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given in the question stem that $$(a^3-a^2-b) = 49$$$$\to a^2(a-1) = 49+b$$
From F.S 1, we know that a(a-1) = 12, thus, $$a*12 = 49+b \to$$ a is an integer for both b = -1 or b = -13. Thus, we get two different values of a, Insufficient.
From F. S 2, we know upon solving for the quadratic $$b^2-b-2$$, the integral roots are -1 or 2.
Now,$$a^2(a-1) = 49+b$$
For b = 2, $$a^2(a-1) = 51$$. Assuming that a is odd/even, the given product is an arrangement like odd*odd*even = even, and 51 is not even.
Similarly, for a is even, the arrangement will be like even*even*odd = even, and just as above 51 is not even. Thus, $$b\neq{2}$$
For b = -1,$$a^2(a-1)$$ = 48. Now all the integral roots of this polynomial can only be factors of 48, including both negative and positive factors.
However, any negative factor will never satisfy the given polynomial as because (a-1) will become a negative expression, which can never equal 48.Hence, the given polynomial has no integral solutions in -48,-24,-12,-8,-6,-4,-3,-2,-1.
By the same logic, we know that any integral solution,if present will be one of the positive factors of 48.It fails for 1,2,3 and we find that a=4 is a root.If for a=4, the expression equals 48, then for a value above 4, the expression $$a^2(a-1)$$is bound to be greater than 48.
Thus, the only solution possible for the given polynomial is a=4, a unique value,Sufficient.
B.
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aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Tough one...Took more than 3 min 50 sec and ended up getting it wrong...
We can change the given equation to (a^3-a^2-b) = 49 (squaring both sides)
From St 1 we have a^2-a=12 ----> substituting in above given eqn we get
{a(a^2-a)- b} =49 ------> 12a-b=49 -----> a = (49+b)/12
Now a and b are integers therefore (49+b)/12 should be an integers
Possible values b=11, a=5
b=23, a =6, b=-1, a=4
So St 1 alone is not sufficient
St 2 says b^2-b =2 -----> b(b-1) =2 possible values of b are
b=2 or b=-1
Substituting in given expression we get
(a^3-a^2-b) = 49
a^3-a^2= 51 or------> a^2(a-1)= 51 (17*3) we see that 51 even after reducing to prime factors gives us no Integer value of a
a^3-a^2= 48 -----> a^2(a-1)= 48 (4^2)*3 and hence we get value of a= 4 or -4
Substituting values of b=-1 and a=4 or -4, we can see that only for a=4 the above given equation is proven.
Hence a =4
I did highlight the value of b=-1, a=4 from statement and can be taken as a hint without solving completely eqn 2
Ans B
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03 Jul 2013, 09:00
mridulparashar1 wrote:
aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
a^3-a^2= 48 -----> a^2(a-1)= 48 (4^2)*3 and hence we get value of a= 4 or -4
Substituting values of b=-1 and a=4 or -4, we can see that only for a=4 the above given equation is proven.
Hence a =4
I did highlight the value of b=-1, a=4 from statement and can be taken as a hint without solving completely eqn 2
Ans B
Minor mistake. The given polynomial will not yield a = -4 as a root.
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10 Nov 2013, 08:31
Bunuel wrote:
aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given: $$a$$ and $$b$$ are integers, also $$\sqrt{a^3-a^2-b}=7$$ --> $$a^3-a^2-b=49$$
(1) a^2 - a = 12 --> $$a=-3$$ or $$a=4$$. Now, both values of $$a$$ give an integer solution for $$b$$ ($$b=85$$ or $$b=-1$$), so both values are valid. Not sufficient.
(2) b^2 - b = 2 --> $$b=-1$$ or $$b=2$$ --> if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$ BUT if if $$b=2$$ then $$a^3-a^2=51$$ --> $$a^2(a-1)=51=3*17$$ --> this equation has no integer solution for $$a$$, hence only the first case is valid: $$b=-1$$ and $$a=4=integer$$. Sufficient.
Answer: B.
if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$
how do you solve something like that, is it just quick trial & error since you know a & b have to be integers and there's only a few values of a that would give results somewhat near 48?
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10 Nov 2013, 11:10
AccipiterQ wrote:
Bunuel wrote:
aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given: $$a$$ and $$b$$ are integers, also $$\sqrt{a^3-a^2-b}=7$$ --> $$a^3-a^2-b=49$$
(1) a^2 - a = 12 --> $$a=-3$$ or $$a=4$$. Now, both values of $$a$$ give an integer solution for $$b$$ ($$b=85$$ or $$b=-1$$), so both values are valid. Not sufficient.
(2) b^2 - b = 2 --> $$b=-1$$ or $$b=2$$ --> if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$ BUT if if $$b=2$$ then $$a^3-a^2=51$$ --> $$a^2(a-1)=51=3*17$$ --> this equation has no integer solution for $$a$$, hence only the first case is valid: $$b=-1$$ and $$a=4=integer$$. Sufficient.
Answer: B.
if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$
how do you solve something like that, is it just quick trial & error since you know a & b have to be integers and there's only a few values of a that would give results somewhat near 48?
Yes, that's correct. We know that $$a$$ is an integer, thus $$a^2(a-1)=(perfect \ square)*(positive \ integer)=48$$. From here you can use trial and error and find that $$a=4$$.
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Re: For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the [#permalink]
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01 Aug 2014, 01:34
Bunuel wrote:
aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given: $$a$$ and $$b$$ are integers, also $$\sqrt{a^3-a^2-b}=7$$ --> $$a^3-a^2-b=49$$
(1) a^2 - a = 12 --> $$a=-3$$ or $$a=4$$. Now, both values of $$a$$ give an integer solution for $$b$$ ($$b=85$$ or $$b=-1$$), so both values are valid. Not sufficient.
(2) b^2 - b = 2 --> $$b=-1$$ or $$b=2$$ --> if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$ BUT if if $$b=2$$ then $$a^3-a^2=51$$ --> $$a^2(a-1)=51=3*17$$ --> this equation has no integer solution for $$a$$, hence only the first case is valid: $$b=-1$$ and $$a=4=integer$$. Sufficient.
Answer: B.
Thanks for the solution
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Re: For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the [#permalink]
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08 Sep 2014, 00:03
I have just one question, why isnt the absolute value used after the expression is squared?
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Re: For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the [#permalink]
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08 Sep 2014, 03:10
sunaimshadmani wrote:
I have just one question, why isnt the absolute value used after the expression is squared?
$$\sqrt{x}=2$$ --> $$x = 4$$. Where should there be an absolute value?
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Re: For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the [#permalink]
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02 Sep 2016, 00:51
Bunuel wrote:
aalriy wrote:
For integers a and b, if (a^3 – a^2 – b)^1/2 = 7, what is the value of a?
(1) a^2 - a = 12
(2) b^2 - b = 2
Given: $$a$$ and $$b$$ are integers, also $$\sqrt{a^3-a^2-b}=7$$ --> $$a^3-a^2-b=49$$
(1) a^2 - a = 12 --> $$a=-3$$ or $$a=4$$. Now, both values of $$a$$ give an integer solution for $$b$$ ($$b=85$$ or $$b=-1$$), so both values are valid. Not sufficient.
(2) b^2 - b = 2 --> $$b=-1$$ or $$b=2$$ --> if $$b=-1$$ then $$a^3-a^2=48$$ --> $$a^2(a-1)=48$$ --> $$a=4=integer$$ BUT if if $$b=2$$ then $$a^3-a^2=51$$ --> $$a^2(a-1)=51=3*17$$ --> this equation has no integer solution for $$a$$, hence only the first case is valid: $$b=-1$$ and $$a=4=integer$$. Sufficient.
Answer: B.
The value of b in (1) should be b=-85 not 85!
Re: For integers a and b, if (a^3 a^2 b)^1/2 = 7, what is the [#permalink] 02 Sep 2016, 00:51
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4.4 Pascal's Triangle
The array of numbers below is called Pascal's Triangle, it was discovered by a French mathematician, Blaise Pascal. In the triangle each term is equal to the sum of both numbers directly above it. The first and last number of each row is 1 considering the number before it always 1.
In the formula for Pascal's triangle rows r represented by n, and positions r. Both rows and positioning start with 0.
If tn,r represents the term in row n, position r, then tn,r= tn-1,r-1 + tn-1, r.
row0 1
row1 1 1
row2 1 2 1
row3 1 3 3 1
row4 1 4 6 4 1
row5 1 5 10 10 5 1
positions: t0,0
t1,0 t1,1
t2,0 t2,1 t2,2
t3,0 t3,1 t3,2 t3,3
t4,0 t4,1 t4,2 t4,3 t4,4
t5,0 t5,1 t5,2 t5,3 t5,4 t5,5
t6,0 t6,1 t6,2 t6,3 t6,4 t6,5 t6,6
Example:
tn,r=tn-1,r-1 +tn-1,r
=6,2
=,1+ t5,2
You can use the following chart to help you count position on rows
number of rows number of #'s in total term in Pascal's triangle
1 1 t2,2
2 3 t3,2
3 6 t4,2
4 10 t5,2
5 15 t6,2
6 21 t7,2
There are several number patterns in Pascal's Triangle
sum row (n)
1 1 1
1 + 1 * 2 2
1 + 2 + 1 4 3
1 + 3 + 3 + 1 8 4
1 + 4 + 6 + 4 + 1 16 5
1 + 5 + 10 + 10 + 5 + 1 32 6
1 + 6 + 15 + 20 + 15 + 6 + 1 64 7
The sum of the terms in row n of the triangle is n
2
Example A: 13
2 = 8192
Example B: which row has a sum of 128
using your calculator, 128= 2 to the power of 7.
Therefore the row which has the sum of 128 is row 7
log128 divided by log2 equals 7
Homework pg. 251 # 1-5, 7a, 9-11
Check My Work: (OSMAN OSMAN)
Click on the link below & open the file, it's waaaaay esier..!!
http://mdm4u1.wikidot.com/local--files/4-4-pascal-s-triangle/Chapter4.doc
Osman Osman
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Chapter 4 ( 4.4 Pascal's Triangle)
4.4 Pascal's Triangle:
Pascal's Triangle is an array of numbers arranged in staggered rows.
Check the diagram below:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
Pascal's method of building his triangle is simple , where each term equals the sum of the two terms immediately above it.
The first and the last row are both equal to 1 since the only term immediately above them t is always a 1.
This process or method of arranging these numbers is represented by a formula:
tn,r = t n-1, r-1 + tn-1, r
Where t n,r represents the term in a row n , position r .
For example: t6,2 = t5,1 + t5,2
Note that both the row and position labeling begin with 0 . (check diagram below)
1 Row 0 t 0,0
1 1 Row 1 t 1,0 t1,1
1 2 1 Row 2 t2,0 t2,1 t2,2
1 3 3 1 Row 3 t 3,0 t3,1 t3,2 t3,3
1 4 6 4 1 Row 4 t4,0 t4,1 t4,2 t4,3 t4,4
1 5 10 10 5 1 Row 5 t5,0 t5,1 t5,2 t5,3 t5,4 t5,5
1 6 15 20 15 6 1 Row 6 t6,0 t6,1 t6,2 t6,3 t6,4 t6,5 t6,6
Applications on Pascal’s Triangle:
Example 1: (Pascal’s Method)
a) The first 6 terms in row 25 of Pascal’s triangle are 1, 25 , 300 , 2300 , 12650 , and 53130. Determine the first 6 terms in row 26.
b) Use Pascal’s method to write a formula a for each of the following terms
i) t12,5
ii) t40,32
iii) tn+,r+1
Solution:
a) t26,1 = 1 t26,2 = 1+25 t26,3 = 25 +300 = 325
t26,4 = 300 + 2300 t26,5 = 2300 + 12650 t26,6 = 12650 + 53130
= 2600 = 14950 = 65780
b) i) t12,5 = t11,4 + t11,5
ii) t40,32 = t39,31 + t39,32
iii) tn+1,r+1 = tn,r + tn,r+1
Example 2: (Row Sums)
Which row in Pascal’s Triangle has the sum of its terms equal to 32768?
Solution:
The terms in any row n is 2^n . Dividing 32768 by 2 repeatedly, you find that 32768 = 2^15. Thus, it is row 15 of Pascal’s Triangle that has terms totalling 32768.
Example 3: (Divisibility)
Determine whether tn,2 is divisible by tn,1 in each row of Pascal’s Triangle
Solution:
Row tn,2 / tn,1 Divisible
0 and 1 n/a n/a
2 0.5 no
3 1 yes
4 1.5 no
5 2 yes
6 2.5 no
7 3 yes
It appears that tn,2 is divisible by tn,1 only in odd-numbered rows.
However, 2tn,2 is divisible by tn,1 in all rows that have three or more terms.
Example 4: (Triangular Numbers)
Coins can be arranged in the shape of an equilateral triangle as shown:
a) Continue the pattern to determine the number of coins in triangle with four, five , and six rows.
b) Locate these numbers in Pascal’s triangle.
c) Relate Pascal’s triangle to the number of coins in a triangle with n rows.
d) How many coins are in a triangle with 12 rows?
Solution:
a) The numbers of coins in the triangles follows the pattern 1+2+3+…… as shown in the table below.
b) The number of coins in the triangles matches the entries on the third diagonal of Pascal’s triangle.
1. Row # of coins Term in Pascal’s Triangle
1 1 t2,2
2 3 t3,2
3 6 t4,2
4 10 t5,2
5 15 t5,2
6 21 t6,2
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
c) If we compare the entries in the first and third columns of the table, we observe, the the number of the term from Pascal’s triangle is always one greater than the number of rows in the equilateral triangle. The position of the term in the row, r, is always 2. Thus, the number of coins in a triangle with n rows is equal to the term tn+1, 2 in Pascal’s triangle.
d) t12+1,2 = t13,2 = 78
A triangle with 12 rows contains 78 coins.
Hint:
Triangular Numbers : are numbers that correspond to the number of items stacked in a triangular array.
Example 5: (Perfect Squares)
Find a relationship between perfect squares and the sum of pairs of entries in Pascal’s triangle?
Solution:
Looking at the third diagonal in Pascal’s triangles.
n N^2 Entries in Pascal’s Triangle Term in Pascal’s Triangle
1 1 1 t2,2
2 4 1+3 t2,2 + t3,2
3 9 3+6 t3,2 + t4,2
4 16 6+10 t4,2 + t5,2
Key Concepts:
• Each term in Pascal’s Triangle is equal to the sumof two adjacent terms in the row immediately above tn,r = tn-1,r-1 + tn-1,r where tn,r represents the rth term in a row n.
• The sum of terms in Pascal’s triangle in row n in Pascal’s Triangle is 2^n.
• The terms in the third diagonal of Pascal’s triangle are triangular numbers.
Many other number patterns occur in Pascal’s triangle.
Usefull Sites for students who are interested in Pascal’s Traingle:
http://mathforum.org/workshops/usi/pascal/pascal/sierpinski.html
Applications of Pascal's Triangle to Sierpinski's triangle and fractals
http://mathforum.org/workshops/usi/pascal/pascal_hsdisc.html
Explorations in Pascal's triangle using number patterns
http://mathforum.org/workshops/usi/pascal/mid.color_pascal.html
Explorations in Pascal's Triangle involving Coloring Multiples
http://hsb.iitm.ernet.in/~jm/mar_april02/articles/pascal.htm
Background in Pascal's Triangle
http://www.roma.unisa.edu.au/07305/pascal.htm
Binomial Theorem and the Pascal Triangle
http://www.pbs.org/mathline
The Smithville Families - applications to probability
By: Osman Osman
page revision: 11, last edited: 12 Apr 2007 03:25
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• https://me.yahoo.com
Cantilever Beams
Formulae for the shear and deflection of Cantilever Beams under a selection of differing loadings.
Overview
This section covers Beams used as Cantilever. The examples include Beams which are "Built-in" at one end and either supported or guided at the other.
Fixed At One End With A Uniform Load.
The stress is given by: $\inline&space;S&space;=&space;\displaystyle\frac{W}{2\;Z\;l}\;\;(l&space;-&space;x)^2$
The Stress at the Support: $\inline&space;S_s&space;=&space;\displaystyle\frac{W\;l}{2\;Z}$
If the cross section is Constant then this is the maximum stress.
The Deflection is given by: $\inline&space;y&space;=&space;\displaystyle\frac{W\;x^2}{24\;E\;I\;l}\;\;[2l^2&space;+&space;(2l&space;-&space;x)^2]$
The Maximum deflection is at the end and is: $\inline&space;\hat{y}&space;=&space;\displaystyle\frac{W\;l^3}{8\;E\;I}$
A beam is a horizontal structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment.
Fixed At One End. Load At The Other
The Stress is given by: $\inline&space;S&space;=&space;\displaystyle\frac{W}{Z}\;\;(l&space;-&space;x)$
The Stress at the Support: $\inline&space;S&space;=&space;\displaystyle\frac{W\;l}{Z}$
If the Cross-Section is Constant, then this is the Maximum Stress.
The Deflection at any point is given by: $\inline&space;y&space;=&space;\displaystyle\frac{W\;x^2}{6\;E\;I}\;\;(3\,l&space;-&space;x)$
The Maximum Deflection is at the end and is: $\inline&space;\hat{y}&space;=&space;\displaystyle\frac{W\;l^3}{3\;E\;I}$
Deflection is a term that is used to describe the degree to which a structural element is displaced under a load.
Fixed At One End. Load Intermediate.
Between the Fixed End and the Load: $\inline&space;\displaystyle&space;S&space;=&space;\frac{W}{Z}\;\;(l&space;-&space;x)$
Beyond the Load the stress is zero.
At the Fixed End: $\inline&space;\displaystyle&space;S&space;=&space;\frac{W\;l}{Z}$
If the Cross-section is Constant, then this is the Maximum Stress.
The Deflection of any point between the Fixed end and the Load is: $\inline&space;\displaystyle&space;y&space;=&space;\frac{W\;x^2}{6\;E\;I}\;\;(3\;l&space;-&space;x)$
Beyond the load the Deflection is: $\inline&space;\displaystyle&space;y&space;=&space;\frac{W\;l^2}{6\;E\;I}\;\;(3\;x-&space;l)$
The Maximum Deflection at the "Free" end is: $\inline&space;\displaystyle&space;y_{max}&space;=&space;\frac{W\;l^2}{6\;E\;I}\;\;(2\;l&space;+&space;3\;b)$
Deflection at the Load: $\inline&space;\displaystyle&space;y&space;=&space;\frac{W\;l^3}{3\;E\;I}$
Fixed At One End. Supported At The Other. Uniform Load.
The Stress at any point is : $\inline&space;S&space;=&space;\displaystyle\frac{W\;(l&space;-&space;x)}{2\;Z\;l}\;\;\left(\displaystyle\frac{1}{4}\;l&space;-&space;x&space;\right)$
The Maximum Stress at the Fixed End: $\inline&space;y&space;=&space;\displaystyle\frac{W\;l}{8\;Z}$
The Stesss is zero at $\inline&space;\displaystyle&space;x&space;=&space;\frac{1}{4}\;l$. The greatest negative Stres is at $\inline&space;\displaystyle&space;x&space;=&space;\frac{5}{8}\;l$ and is:
$S\;=&space;-&space;\frac{9}{128}\;\;\frac{W\'l}{Z}$
The deflection is given by: $\inline&space;y&space;=&space;\displaystyle\frac{W\;x^2(l&space;-&space;x)}{48\;E\;I\;l}\;\;(3\;l&space;-&space;2\;x)$
The maximum Deflection is at x = 0.5785 l and is: $\inline&space;\hat{y}&space;=&space;\displaystyle\frac{W\;l^3}{185\;E\;I}$
The Deflection at the centre is: $\inline&space;y_c&space;=&space;\displaystyle\frac{W\;l^3}{192\;E\;I}$
The Deflection at the point of greatest negative Stress, i.e. at $\inline&space;x&space;=&space;\displaystyle\frac{5}{8}\;l$, is:
$y&space;=&space;\frac{W\;l^3}{187\;E\;I}$
Fixed At One End, Supported At The Other With A Central Point Load.
The Stress between the Fixed Point and the Load: $\inline&space;S&space;=&space;\displaystyle\frac{W}{16\;Z}\left(3\;l&space;-&space;11\;x&space;\right)$
Between the Support and the Load: $\inline&space;S\;=&space;-&space;\displaystyle\frac{5}{16}\;\displaystyle\frac{W\;v}{Z}$
Stress at the Fixed end. This is the maximum value $\inline&space;S&space;=&space;\displaystyle\frac{3}{16}\;\displaystyle\frac{W\;l}{Z}$
The Stress is Zero at $\inline&space;\displaystyle&space;x&space;=&space;\frac{3}{11}\;l$
The Greatest negative Stress is at the centre and is $\inline&space;\displaystyle&space;-&space;\frac{5}{32}\;\;\frac{W\;l}{z}$
The Deflection of any Point between the Fixed End and the Load:
$y&space;=&space;\frac{W\;x^2}{96\;E\;I}\;\;(9\;l&space;-&space;11\;x)$
The Deflection of any Point between the Support and the Load
$y&space;=&space;\frac{W\;v}{96\;E\;I}\;\;(3\;l^2&space;-&space;5\;v^2)$
The Maximum Deflection is at $\inline&space;\displaystyle&space;v&space;=&space;0.4472\;l$
$\hat{y}&space;=&space;\frac{W\;l^3}{107.33\;E\;I}$
The Deflection of the Load is: $\inline&space;\displaystyle\frac{7}{768}\;\;\displaystyle\frac{W\;l^3}{E\;I}$
Fixed At One End And Free But Guided At The Other. Uniform Load.
The Stress at any Point $\inline&space;S&space;=&space;\displaystyle\frac{W\;l}{Z}\;\;\left\{\displaystyle\frac{l}{3}&space;-&space;\displaystyle\frac{x}{l}\;+\displaystyle\frac{1}{2}\;\left(\displaystyle\frac{x}{l}&space;\right)^2&space;\right\}$
The Maximum Stress is at the support and is $\inline&space;\displaystyle&space;\frac{W\;l}{3\;Z}$
The Stress is zero when $\inline&space;\displaystyle&space;x&space;=&space;0.4227\;l$
The Greatest negative stress is at the free end and is $\inline&space;\displaystyle&space;-&space;\frac{W\;l}{3\;Z}$
The Deflection at any Point is given by: $\inline&space;y&space;=&space;\displaystyle\frac{W\;x^2}{24\;E\;I\;l}\;\;(2l&space;-&space;x)^2$
The Maximum Deflection is at the free end and is: $\inline&space;\hat{y}&space;=&space;\displaystyle\frac{W\;l^3}{12\;E\;I}$
Fixed At One End. Free But Guided At The Other. Point Load.
Stress at any Point: $\inline&space;S&space;=&space;\displaystyle\frac{W}{Z}\;\;\;\left(\displaystyle\frac{1}{2}l&space;-&space;x&space;\right)$
The Stress at the Support $\inline&space;S_s&space;=&space;\displaystyle\frac{W\;l}{2\;Z}$
The Stress at the Free End: $\inline&space;S_f\;=&space;-&space;\displaystyle\frac{W\;l}{2\;Z}$
These are the Maximum Stresses and are equal and opposite. The Stress is zero at the Centre.
The Deflection at any Point is: $\inline&space;y&space;=&space;\displaystyle\frac{W\;x^2}{12\;E\;I}\;\;(3l&space;-&space;2x)$
The Maximum Deflection is at the Free End and is: $\inline&space;\hat{y}&space;=&space;\displaystyle\frac{W\;l^3}{12\;E\;I}$
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### Times Right
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Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?
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##### Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
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When I type a sequence of letters my calculator gives the product of all the numbers in the corresponding memories. What numbers should I store so that when I type 'ONE' it returns 1, and when I type. . . .
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Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
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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!?
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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?
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Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
### Diggits
##### Stage: 3 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
### I'm Eight
##### Stage: 1, 2, 3 and 4 Challenge Level:
Find a great variety of ways of asking questions which make 8.
### More Mods
##### Stage: 4 Challenge Level:
What is the units digit for the number 123^(456) ?
### Long Multiplication
##### Stage: 3 Challenge Level:
A 3 digit number is multiplied by a 2 digit number and the calculation is written out as shown with a digit in place of each of the *'s. Complete the whole multiplication sum.
### Two Many
##### Stage: 3 Challenge Level:
What is the least square number which commences with six two's?
### Expenses
##### Stage: 4 Challenge Level:
What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?
### One O Five
##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### Repeaters
##### Stage: 3 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.
### Number Rules - OK
##### Stage: 4 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...
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Product Doubles Sudoku
##### Stage: 3 and 4 Challenge Level:
Each clue number in this sudoku is the product of the two numbers in adjacent cells.
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
##### Stage: 3 and 4 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Magic Potting Sheds
##### Stage: 3 Challenge Level:
Mr McGregor has a magic potting shed. Overnight, the number of plants in it doubles. He'd like to put the same number of plants in each of three gardens, planting one garden each day. Can he do it?
##### Stage: 3 Challenge Level:
If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why?
##### Stage: 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Round and Round and Round
##### Stage: 3 Challenge Level:
Where will the point stop after it has turned through 30 000 degrees? I took out my calculator and typed 30 000 ÷ 360. How did this help?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Integrated Product Sudoku
##### Stage: 3 and 4 Challenge Level:
This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.
### Going Round in Circles
##### Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Ones Only
##### Stage: 3 Challenge Level:
Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
### Skeleton
##### Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### So It's Times!
##### Stage: 2 and 3 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### Missing Multipliers
##### Stage: 3 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### A Chance to Win?
##### Stage: 3 Challenge Level:
Imagine you were given the chance to win some money... and imagine you had nothing to lose...
### 3388
##### Stage: 3 Challenge Level:
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
### The Unmultiply Game
##### Stage: 2, 3 and 4 Challenge Level:
Unmultiply is a game of quick estimation. You need to find two numbers that multiply together to something close to the given target - fast! 10 levels with a high scores table.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Vedic Sutra - All from 9 and Last from 10
##### Stage: 4 Challenge Level:
Vedic Sutra is one of many ancient Indian sutras which involves a cross subtraction method. Can you give a good explanation of WHY it works?
### Galley Division
##### Stage: 4 Challenge Level:
Can you explain how Galley Division works?
### What Is the Question?
##### Stage: 3 and 4 Challenge Level:
These pictures and answers leave the viewer with the problem "What is the Question". Can you give the question and how the answer follows?
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Home - Theorems_And_Conjectures - Perfect And Prime Numbers
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Perfect And Prime Numbers: more detail
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1. Perfect-Key
Prime numbers (blue background). Perfect numbers (yellow background). Continue thesequences indefinitely in the chart to reveal all perfect and prime numbers.
http://www.borderschess.org/Perfect-Key.htm
Extractions: n:=1 n:=2 n:=3 n:=4 n:=5 n:=6 n:=7 n:=8 n:=9 a(n) b(n) c(n) f(n) g(n) Prime numbers (blue background). Perfect numbers (yellow background). The product sequence algorithm of c(n) is the solution key to unlocking all Prime numbers (shaded blue) and Perfect numbers (shaded yellow) for numbers greater than 6 in our number system. The idea for the algorithm comes from my article: "Prime Consideration for the Perfect Number" shown below. The n sequence (n:=1, n:=2, .. , n:=9) can be easily increased and iterated in a software program. Results for the n sequence, shown in the chart above, were derived from MathCad Professional and Mathematica computer programs. If f(n) is prime, then b(n) is perfect. Also, if f(n) is prime, it will be a Mersenne prime. There are various ways to check a number for primeness. Factoring f(n) can be very slow on extremely large numbers. Using the IsPrime function (Mathcad) or PrimeQ function (Mathematica) takes much less time but gives a probability that a number is Prime. Use of IsPrime or PrimeQ may need to be verified on very large numbers by an independent source. Check out GIMPS (The Great Internet Mersenne Prime Search) at http://www.mersenne.org/prime.htm
2. Large Prime Numbers
Mersenne primes. The new perfect number generated with the new Mersenneprime is the 34th known perfect number and has 757,263 digits.
http://www.isthe.com/chongo/tech/math/prime/prime_press.html
Extractions: Mersenne Prime Digits and Names EAGAN, Minn., September 3, 1996 Computer scientists at SGI 's former Cray Research unit, have discovered a large prime number while conducting tests on a CRAY T90 series supercomputer. The prime number has 378,632 digits. Printed in newspaper-sized type, the number would fill approximately 12 newspaper pages. In mathematical notation, the new prime number is expressed as , which denotes two, multiplied by itself 1,257,787 times, minus one. Numbers expressed in this form are called Mersenne prime numbers after Marin Mersenne, a 17th century French monk who spent years searching for prime numbers of this type. See Chris Callwell's prime page for more information on prime numbers. Prime numbers can be divided evenly only by themselves and one. Examples include 2, 3, 5, 7, 11 and so on. The Greek mathematician
3. A Prime Way To Find Perfect Numbers
SINE WAVE. A prime Way to Find perfect numbers. return. prime numbers are numbers that cannot be expressed as the product of 2 whole numbers, other than 1 and the number itself. is 28 (1+2+4+7+14=28). These are called perfect numbers and they are very rare is an interesting relationship between even perfect numbers and certain prime numbers
http://users.andara.com/~brsears/primperf.htm
Extractions: A Prime Way to Find Perfect Numbers Prime numbers are numbers that cannot be expressed as the product of 2 whole numbers, other than 1 and the number itself. They are called prime because they cannot be factored into smaller numbers. The number 1 is not defined as prime. Here is a list of all the prime numbers less than 1000: How can we determine whether a number is prime or not? One obvious way is to divide the number by each smaller number, except 1. If any divide evenly, then the number is not prime. Actually we only need to try the numbers less than or equal to the square root of the number we are testing. For very large numbers, this method can take a long time. Even using modern computers, there is a practical limit to how many prime numbers we can find. Numbers that are not prime, except for 1, are called composite. Any composite number can be expressed as the product of a unique set of prime numbers. It is possible to prove that there are infinitely many prime numbers. We can also make an observation about the relative frequency of prime numbers. The prime number theorem tells us: If P(N) is the number of primes less than or equal to N, then the ratio N/P(N) approaches the natural logarithm of N as N approaches infinity.
4. Perfect, Amicable And Sociable Numbers
perfect, amicable and sociable numbers. ( how to have fun perfect numbers, and, apart from 1 and 2, no augmented modified exponential perfect numbers. Also, any prime factor of
http://xraysgi.ims.uconn.edu:8080/amicable.html
Extractions: HTTP 200 Document follows Date: Sun, 06 Jun 2004 07:20:00 GMT Server: NCSA/1.5.2 Last-modified: Tue, 30 Dec 2003 02:14:15 GMT Content-type: text/html Content-length: 29165 Introduction Perfect numbers Amicable numbers Sociable numbers ... Technical appendix For a number n , we define s(n) to be the sum of the aliquot parts of n, i.e., the sum of the positive divisors of n, excluding n itself: so, for example, s(8)=1+2+4=7, and s(12)=1+2+3+4+6=16. If we start at some number and apply s repeatedly, we will form a sequence: s(15)=1+3+5=9, s(9)=1+3=4, s(4)=1+2=3, s(3)=1, s(1)=0. If we ever reach 0, we must stop, since all integers divide 0. There are three obvious possibilities for the behavior of this aliquot sequence It can terminate at like the example above. It can fall into an aliquot cycle , of length 1 (a fixed point of s) , or greater It can grow without bound and approach infinity A perfect number is a cycle of length 1 of s , i.e., a number whose positive divisors (except for itself) sum to itself. For example, 6 is perfect (1+2+3=6), and in fact 6 is the smallest perfect number. The next two perfect numbers are 28 (1+2+4+7+14=28) and 496 (1+2+4+8+16+31+62+124+248=496).
5. Mersenne Primes: History, Theorems And Lists
We know that all even perfect numbers are a Mersenne prime times a powerof two (theorem one above), but what about odd perfect numbers?
http://www.utm.edu/research/primes/mersenne/
Extractions: History, Theorems and Lists A forty first Mersenne found May 2004: 2 Early History Perfect Numbers and a Few Theorems Table of Known Mersenne Primes The Lucas-Lehmer Test and Recent History ... Conjectures and Unsolved Problems See also Where is the next larger Mersenne prime? and Mersenne heuristics For remote pages on Mersennes see the Prime Links' Mersenne directory Primes: Home Largest Proving How Many? ... Mailing List Many early writers felt that the numbers of the form 2 n -1 were prime for all primes n , but in 1536 Hudalricus Regius showed that 2 -1 = 2047 was not prime (it is 23 89). By 1603 Pietro Cataldi had correctly verified that 2 -1 and 2 -1 were both prime, but then incorrectly stated 2 n -1 was also prime for 23, 29, 31 and 37. In 1640 Fermat showed Cataldi was wrong about 23 and 37; then Euler in 1738 showed Cataldi was also wrong about 29. Sometime later Euler showed Cataldi's assertion about 31 was correct. Enter French monk Marin Mersenne (1588-1648). Mersenne stated in the preface to his Cogitata Physica-Mathematica (1644) that the numbers 2 n -1 were prime for n 31, 67, 127 and 257
6. Mersenne Primes: History, Theorems And Lists
The definitive pages on the Mersenne primes and the related mathematics! We know that all even perfect numbers are a Mersenne prime times a power of two (theorem one above), but what about odd perfect numbers? If there is one
http://www.utm.edu/research/primes/mersenne.shtml
Extractions: History, Theorems and Lists A forty first Mersenne found May 2004: 2 Early History Perfect Numbers and a Few Theorems Table of Known Mersenne Primes The Lucas-Lehmer Test and Recent History ... Conjectures and Unsolved Problems See also Where is the next larger Mersenne prime? and Mersenne heuristics For remote pages on Mersennes see the Prime Links' Mersenne directory Primes: Home Largest Proving How Many? ... Mailing List Many early writers felt that the numbers of the form 2 n -1 were prime for all primes n , but in 1536 Hudalricus Regius showed that 2 -1 = 2047 was not prime (it is 23 89). By 1603 Pietro Cataldi had correctly verified that 2 -1 and 2 -1 were both prime, but then incorrectly stated 2 n -1 was also prime for 23, 29, 31 and 37. In 1640 Fermat showed Cataldi was wrong about 23 and 37; then Euler in 1738 showed Cataldi was also wrong about 29. Sometime later Euler showed Cataldi's assertion about 31 was correct. Enter French monk Marin Mersenne (1588-1648). Mersenne stated in the preface to his Cogitata Physica-Mathematica (1644) that the numbers 2 n -1 were prime for n 31, 67, 127 and 257
7. Prime Numbers
Euclid also showed that if the number 2 n 1 is prime then the number2 n-1 (2 n - 1) is a perfect number. The mathematician Euler
http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Prime_numbers.html
Extractions: Prime numbers and their properties were first studied extensively by the ancient Greek mathematicians. The mathematicians of Pythagoras 's school (500 BC to 300 BC) were interested in numbers for their mystical and numerological properties. They understood the idea of primality and were interested in perfect and amicable numbers. You can see more about these numbers in the History topics article Perfect numbers By the time Euclid 's Elements appeared in about 300 BC, several important results about primes had been proved. In Book IX of the Elements Euclid proves that there are infinitely many prime numbers. This is one of the first proofs known which uses the method of contradiction to establish a result. Euclid also gives a proof of the Fundamental Theorem of Arithmetic: Every integer can be written as a product of primes in an essentially unique way.
8. The Prime Glossary: Perfect Number
Welcome to the prime Glossary a collection of definitions, information and facts all related to prime numbers. This pages contains the entry titled 'perfect number.' Come explore a new prime term
http://www.utm.edu/research/primes/glossary/PerfectNumber.html
Extractions: (another Prime Pages ' Glossary entries) Glossary: Prime Pages: Many ancient cultures endowed certain integers with special religious and magical significance. One example is the perfect numbers, those integers which are the sum of their positive proper divisors . The first three perfect numbers are The ancient Christian scholar Augustine explained that God could have created the world in an instant but chose to do it in a perfect number of days, 6. Early Jewish commentators felt that the perfection of the universe was shown by the moons period of 28 days. Whatever significance ascribed to them, these three perfect numbers above, and 8128, were known to be "perfect" by the ancient Greeks, and the search for perfect numbers was behind some of the greatest discoveries in number theory. For example, in Book IX of Euclid 's elements we find the first part of the following theorem (completed by Euler some 2000 years later).
9. Perfect Numbers
11 12 13 1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 let them be called the radicalsof perfect numbers, since whenever they are prime, they produce them.
http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/Perfect_numbers.html
Extractions: It is not known when perfect numbers were first studied and indeed the first studies may go back to the earliest times when numbers first aroused curiosity. It is quite likely, although not certain, that the Egyptians would have come across such numbers naturally given the way their methods of calculation worked, see for example [17] where detailed justification for this idea is given. Perfect numbers were studied by Pythagoras and his followers, more for their mystical properties than for their number theoretic properties. Before we begin to look at the history of the study of perfect numbers, we define the concepts which are involved. Today the usual definition of a perfect number is in terms of its divisors, but early definitions were in terms of the 'aliquot parts' of a number. An aliquot part of a number is a proper quotient of the number. So for example the aliquot parts of 10 are 1, 2 and 5. These occur since 1 = , and 5 = . Note that 10 is not an aliquot part of 10 since it is not a proper quotient, i.e. a quotient different from the number itself. A perfect number is defined to be one which is equal to the sum of its aliquot parts. The four perfect numbers 6, 28, 496 and 8128 seem to have been known from ancient times and there is no record of these discoveries.
10. "Computing Perfect(prime) Numbers" By SAMIEL@FASTLANE.NET
perfect numbers A perfect number is a number whose divisors not including the original number add up to where p is prime and 2^p1 is prime, so 2^2
http://www.bsdg.org/swag/MATH/0108.PAS.html
Extractions: Back to MATH SWAG index Back to Main SWAG index Original PROGRAM Perfect VAR tmp num longint j k byte Function IsPrime num longint boolean Var tmp boolean j longint Begin tmp true if num mod then tmp false for j to round sqrt num do if odd j then if num mod j then tmp false if num then tmp true IsPrime tmp End BEGIN tmp writeln 'Perfect numbers...' for j to do begin tmp tmp if IsPrime j then begin num tmp if IsPrime num then begin num num tmp div writeln num ' is perfect' end end end END Samiel samiel fastlane net http //www.fastlane.net/~samiel Considering that is prime and is the largest known prime there are obviously better methods Use the Lucas Lehmer test B n B n mod p where B if B n then p is prime The division is a special case and is done easily because of the all ONE 's when 2^p-1 is written in binary. All that is necessary is a fast implementation of multiplication and this can be done with FFT 's. See http //www.utm.edu/research/primes/mersenne.shtml or for software http //ourworld.compuserve.com/homepages/justforfun/prime.htm Well considering we are looking at numbers under bits your code would probably not get done as fast as mine though in the long run say over bits or so it would All the FFT 's and rather long multiplication would slow it down considerably Even if you could get a really fast implementation of FFT 's and multiplications, we'
11. ONJava.com: Web Services And The Search For Really Big Prime Numbers [Aug. 29, 2
Web services should open up new avenues of computing. Such as? This article shows how Web services are an ideal model for computing Mersenne prime numbers, some of the largest primes yet prime numbers, Mersenne numbers, Mersenne prime numbers, and perfect numbers. A prime number Also, a relationship between prime numbers and perfect numbers has been known
http://www.onjava.com/pub/a/onjava/2002/08/28/wsdc.html
Extractions: What do searching for extraterrestrials, curing cancer, and finding big prime numbers all have in common? These problems are all being attacked with grid computing, a a technique of breaking a large problem into small tasks that can be computed independently. While projects like Seti@home and The Greatest Internet Mersenne Prime Search have received plenty of press for using the Internet to distribute tasks to end users around the globe, grid computing also takes place in more controlled environments, such as research and financial settings. But it is by using the power of the Internet and the ability to discover and access idle processes on users' machines that grid computing (once called
12. SS > Factoids > Perfect Number
is prime then 2 perfect. Euler all even perfect numbers are of the form 2(p1)(2p-1), where 2p-1 is a Mersenne prime ( and so p is prime
http://public.logica.com/~stepneys/cyc/p/perfect.htm
13. The Prime Glossary: Multiply Perfect
Welcome to the prime Glossary a collection of definitions, information and facts all related to prime numbers. This pages contains the entry titled 'multiply perfect.' Come explore a new prime
http://primes.utm.edu/glossary/page.php?sort=MultiplyPerfect
14. Landon Curt Noll's Prime Pages
Landon Curt Noll Landon Curt Noll s picture. prime numbers, Mersenneprimes, perfect numbers, etc. Mersenne prime Digits and Names
http://www.isthe.com/chongo/tech/math/prime/
15. Math Forum - Ask Dr. Math Archives: Elementary Prime Numbers
and why are they prime numbers? prime numbers vs. prime Factors perfect numbers Basics, History 11/3/1996 What is the next perfect number after 28? prime and Composite numbers
http://mathforum.com/library/drmath/sets/elem_prime_numbers.html
16. Math Forum - Ask Dr. Math Archives: Elementary Prime Numbers
perfect numbers Basics, History 11/3/1996 What is the next perfect number after28? prime and Composite numbers, Sieve of Eratosthenes 01/28/1997 I need
http://mathforum.org/library/drmath/sets/elem_prime_numbers.html
17. Computing Perfect(prime) Numbers
28=14*7*4*2*1=14+7+4+2+1. The definition of a perfect number can also be definedas, 2^(p1)*(2^p-1) where p is prime and 2^p-1 is a Mersenne prime
18. Trailpost 2: Properties Of Prime Numbers
Mersenne primes and perfect numbers. You can find perfect numbers by using theformula (2 n 1) * 2 n-1 , where n in 2 n - 1 makes a Mersenne prime.
Extractions: Prime Numbers Natural numbers are either prime or composite numbers. A prime number is a natural number that can only be divided by one and itself. In other words, it has exactly two factors. For example the number can only be divided by and , so is a prime number. has the factors and so is a composite number. Numbers that have more than two factors are composite numbers. Marin Mersenne 1588 - 1648 A special type of prime is called a Mersenne prime. Mersenne primes are calculated using the formula n . Any prime number calculated using the formula is a Mersenne prime. For example, the number is a Mersenne prime since and is a prime number. Note that the formula does not always produce prime numbers. For example, , which is not a prime number. Mersenne primes are very rare. In fact there are only known Mersenne primes as of January 1998. Mersenne primes can be used to calculate a special type of number called Perfect numbers. These are numbers whose factors when added together equal the number. For example the number
19. Ancient Greek Number Theory And Prime Numbers
was able to find that each of these numbers is of the form 2 n1 (2 n -1), where2 n -1 is prime. Euclid proved that all numbers of this form were perfect.
http://www.mlahanas.de/Greeks/Primes.htm
Extractions: Pythagoras discovered the relation between harmony and numbers. The Pythagoreans saw the number one as the primordial unity from which all else is created. Two was the symbol for the female, three for the male and therefore five (two + three) symbolized marriage. The number four was symbolic of harmony, because two is even, so four (two times two) is "evenly even". Four symbolized the four elements out of which everything in the universe was made (earth, air, fire, and water). Ten that was the sum from one to four was a very special number. The ancient Greeks believed that all numbers had to be rational numbers. 2500 years ago Greeks discovered that if all the common prime numbers were removed from the top and bottom of the ratio then one of the two numbers had to be odd. This we can term reduced form . Obviously, if top and bottom were both even, then both could be divide by the number two and this could be eliminated from both. The Greeks then went on to show that for a right triangle with sides [1:1:square root of two] that the hypotenuse of the triangle, the square root of two, in reduced form could not have either top or bottom number odd. Consequently, it cannot be a rational number.
20. Mathematics Enrichment Workshop: The Perfect Number Journey
Beginning with the number 1, and keep adding the powers of 2 (ie doubling the numbers),until you get a sum which is a prime number. A perfect number is then
http://home1.pacific.net.sg/~novelway/MEW2/lesson1.html
Extractions: by Heng O.K. What are perfect numbers? Mathematicians and nonmathematicians have been fascinated for centuries by the properties and patterns of numbers. They have noticed that some numbers are equal to the sum of all of their factors (not including the number itself). The smallest such example is , since = 1 + 2 + 3. Such numbers are called perfect numbers The search for perfect numbers began in ancient times. The first three perfect numbers: and were known to the ancient mathematicians since the time of Pythagoras (circa 500 BC). How to find perfect numbers? Euclid (circa 300 BC), the famous Greek mathematician, devised a simple method for computing perfect numbers. Beginning with the number 1, and keep adding the powers of 2 (i.e. doubling the numbers), until you get a sum which is a prime number . A perfect number is then obtained by multiplying this sum to the last power of 2. In the exercise that follows, you are going to use this method to determine the next two perfect numbers. The first few rows in the table demonstrate the calculations being carried out to compute the first three perfect numbers. Apply this technique now, and let's see how fast you can find the fourth perfect number.
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# Range and Update Sum Queries with Factorial
Given an array arr[] of N integers and number of queries Q. The task is to answer three types of queries.
1. Update [l, r] – for every i in range [l, r] increment arr[i] by 1.
2. Update [l, val] – change the value of arr[l] to val.
3. Query [l, r] – calculate the sum of arr[i]! % 109 for all i in range [l, r] where arr[i]! is the factorial of arr[i].
Prerequisite :Binary Indexed Trees | Segment Trees
Examples:
Input: Q = 6, arr[] = { 1, 2, 1, 4, 5 }
3 1 5
1 1 3
2 2 4
3 2 4
1 2 5
3 1 5
Output:
148
50
968
1st query, the required sum is (1! + 2! + 1! + 4! + 5!) % 109 = 148
2nd query, the array becomes arr[] = { 2, 3, 2, 4, 5 }
3rd query, array becomes arr[] = { 2, 4, 2, 4, 5 }
4th query, the required sum is (4! + 2! + 4!) % 109 = 50
5th query, the array becomes arr[] = { 2, 5, 3, 5, 6 }
6th query, the required sum is (2! + 5! + 3! + 5! + 6!) % 109 = 968
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Naive Approach: A simple solution is to run a loop from l to r and calculate sum of factorial of elements (pre-computed) in the given range for the 3rd query. For the 2nd query, to update a value, simply replace arr[i] with the given value i.e. arr[i] = val. For the 1st type query, increment the value of arr[i] i.e. arr[i] = arr[i] + 1.
Efficient Approach: It can be observed from careful analysis that 40! is divisible by 109, that means factorial of every number greater than 40 will be divisible by 109. Hence, that adds zero to our answer for the 3rd query. The idea is to reduce the time complexity for each query and update operation to O(logN). Use Binary Indexed Trees (BIT) or Segment Trees. Construct a BIT[] array and have two functions for query and update operation.
• Now, for each update operation of the 1st type, the key observation is that the number in given range can at max be updated to 40, since after that it won’t matter as it will add zero to our final answer. We will use a set to store the index of only those numbers which are lesser than 10 and use binary search to find the l index of the update query and increment the l index until every element is updated in range of that update query. If the arr[i] becomes greater than or equal to 40 after incrementing by 1, remove it from the set as it will not affect our answer of sum query even after any next update query.
• For the update operation of the 2nd type, call the update function with the given value. Also, the given value is < 40, insert the index of the element to be replaced with into the set and if the given value is ≥ 40, remove it from the set since it will have no importance in sum query.
• For the sum query of the 3rd type, simply do query(r) – query(l – 1).
Below is the implementation of the above approach:
`// CPP program to calculate sum of ` `// factorials in an interval and update ` `// with two types of operations ` `#include ` `using` `namespace` `std; ` ` ` `// Modulus ` `const` `int` `MOD = 1e9; ` ` ` `// Maximum size of input array ` `const` `int` `MAX = 100; ` ` ` `// Size for factorial array ` `const` `int` `SZ = 40; ` ` ` `int` `BIT[MAX + 1], fact[SZ + 1]; ` ` ` `// structure for queries with members type, ` `// leftIndex, rightIndex of the query ` `struct` `queries { ` ` ``int` `type, l, r; ` `}; ` ` ` `// function for updating the value ` `void` `update(``int` `x, ``int` `val, ``int` `n) ` `{ ` ` ``for` `(x; x <= n; x += x & -x) ` ` ``BIT[x] += val; ` `} ` ` ` `// function for calculating the required ` `// sum between two indexes ` `int` `sum(``int` `x) ` `{ ` ` ``int` `s = 0; ` ` ``for` `(x; x > 0; x -= x & -x) ` ` ``s += BIT[x]; ` ` ``return` `s; ` `} ` ` ` `// function to return answer to queries ` `void` `answerQueries(``int` `arr[], queries que[], ` ` ``int` `n, ``int` `q) ` `{ ` ` ``// Precomputing factorials ` ` ``fact[0] = 1; ` ` ``for` `(``int` `i = 1; i < 41; i++) ` ` ``fact[i] = (fact[i - 1] * i) % MOD; ` ` ` ` ``// Declaring a Set ` ` ``set<``int``> s; ` ` ``for` `(``int` `i = 1; i < n; i++) { ` ` ` ` ``// inserting indexes of those ` ` ``// numbers which are lesser ` ` ``// than 40 ` ` ``if` `(arr[i] < 40) { ` ` ``s.insert(i); ` ` ``update(i, fact[arr[i]], n); ` ` ``} ` ` ``else` ` ``update(i, 0, n); ` ` ``} ` ` ` ` ``for` `(``int` `i = 0; i < q; i++) { ` ` ` ` ``// update query of the 1st type ` ` ``if` `(que[i].type == 1) { ` ` ``while` `(``true``) { ` ` ` ` ``// find the left index of query in ` ` ``// the set using binary search ` ` ``auto` `it = s.lower_bound(que[i].l); ` ` ` ` ``// if it crosses the right index of ` ` ``// query or end of set, then break ` ` ``if` `(it == s.end() || *it > que[i].r) ` ` ``break``; ` ` ` ` ``que[i].l = *it; ` ` ``int` `val = arr[*it] + 1; ` ` ` ` ``// update the value of arr[i] to ` ` ``// its new value ` ` ``update(*it, fact[val] - fact[arr[*it]], n); ` ` ` ` ``arr[*it]++; ` ` ` ` ``// if updated value becomes greater ` ` ``// than or equal to 40 remove it from ` ` ``// the set ` ` ``if` `(arr[*it] >= 40) ` ` ``s.erase(*it); ` ` ` ` ``// increment the index ` ` ``que[i].l++; ` ` ``} ` ` ``} ` ` ` ` ``// update query of the 2nd type ` ` ``else` `if` `(que[i].type == 2) { ` ` ``int` `idx = que[i].l; ` ` ``int` `val = que[i].r; ` ` ` ` ``// update the value to its new value ` ` ``update(idx, fact[val] - fact[arr[idx]], n); ` ` ` ` ``arr[idx] = val; ` ` ` ` ``// If the value is less than 40, insert ` ` ``// it into set, otherwise remove it ` ` ``if` `(val < 40) ` ` ``s.insert(idx); ` ` ``else` ` ``s.erase(idx); ` ` ``} ` ` ` ` ``// sum query of the 3rd type ` ` ``else` ` ``cout << (sum(que[i].r) - sum(que[i].l - 1)) ` ` ``<< endl; ` ` ``} ` `} ` ` ` `// Driver Code to test above functions ` `int` `main() ` `{ ` ` ``int` `q = 6; ` ` ` ` ``// input array using 1-based indexing ` ` ``int` `arr[] = { 0, 1, 2, 1, 4, 5 }; ` ` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` ` ` ``// declaring array of structure of type queries ` ` ``queries que[q + 1]; ` ` ` ` ``que[0].type = 3, que[0].l = 1, que[0].r = 5; ` ` ``que[1].type = 1, que[1].l = 1, que[1].r = 3; ` ` ``que[2].type = 2, que[2].l = 2, que[2].r = 4; ` ` ``que[3].type = 3, que[3].l = 2, que[3].r = 4; ` ` ``que[4].type = 1, que[4].l = 2, que[4].r = 5; ` ` ``que[5].type = 3, que[5].l = 1, que[5].r = 5; ` ` ` ` ``// answer the Queries ` ` ``answerQueries(arr, que, n, q); ` ` ``return` `0; ` `} `
Output:
```148
50
968
```
My Personal Notes arrow_drop_up
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The U.S. presidential election is now only a few weeks away. The politics of this election are of course interesting and important, but I do not want to discuss these topics here (there is not exactly a shortage of other venues for such a discussion), and would request that readers refrain from doing so in the comments to this post. However, I thought it would be apropos to talk about some of the basic mathematics underlying electoral polling, and specifically to explain the fact, which can be highly unintuitive to those not well versed in statistics, that polls can be accurate even when sampling only a tiny fraction of the entire population.
Take for instance a nationwide poll of U.S. voters on which presidential candidate they intend to vote for. A typical poll will ask a number $n$ of randomly selected voters for their opinion; a typical value here is $n = 1000$. In contrast, the total voting-eligible population of the U.S. – let’s call this set $X$ – is about 200 million. (The actual turnout in the election is likely to be closer to 100 million, but let’s ignore this fact for the sake of discussion.) Thus, such a poll would sample about 0.0005% of the total population $X$ – an incredibly tiny fraction. Nevertheless, the margin of error (at the 95% confidence level) for such a poll, if conducted under idealised conditions (see below), is about 3%. In other words, if we let $p$ denote the proportion of the entire population $X$ that will vote for a given candidate $A$, and let $\overline{p}$ denote the proportion of the polled voters that will vote for $A$, then the event $\overline{p}-0.03 \leq p \leq \overline{p}+0.03$ will occur with probability at least 0.95. Thus, for instance (and oversimplifying a little – see below), if the poll reports that 55% of respondents would vote for A, then the true percentage of the electorate that would vote for A has at least a 95% chance of lying between 52% and 58%. Larger polls will of course give a smaller margin of error; for instance the margin of error for an (idealised) poll of 2,000 voters is about 2%.
I’ll give a rigorous proof of a weaker version of the above statement (giving a margin of error of about 7%, rather than 3%) in an appendix at the end of this post. But the main point of my post here is a little different, namely to address the common misconception that the accuracy of a poll is a function of the relative sample size rather than the absolute sample size, which would suggest that a poll involving only 0.0005% of the population could not possibly have a margin of error as low as 3%. I also want to point out some limitations of the mathematical analysis; depending on the methodology and the context, some polls involving 1000 respondents may have a much higher margin of error than the idealised rate of 3%.
– Assumptions and conclusion –
Not all polls are created equal; there are a certain number of hypotheses on the methodology and effectiveness of the poll that we have to assume in order to make our mathematical conclusions valid. We will make the following idealised assumptions:
1. Simple question. Voters polled can only offer one of two responses, which I will call A and not-A; thus we ignore the effect of third-party candidates, undecided voters, or refusals to respond. In particular, we do not try to combine this data with other questions about the polled voters, such as demographic data. We also assume that the question is unambiguous and cannot be misinterpreted by respondents (see Hypothesis 3 below).
2. Perfect response rate. All voters polled offer a response; there are no refusals to respond to the poll, or failures to make contact with the voter being polled. (This is a special case of 1., but deserves to be emphasised.) In particular, this excludes polls that are self-selected, such as internet polls (since in most cases, a large fraction of viewers of a web page with a poll will refuse to respond to that poll).
3. Honest responses. The response given by a voter to the poll is an accurate representation whether that voter intends to vote for $A$ or not; thus we ignore response-distorting effects such as the Bradley effect or push-polling, as well as tactical voting, frivolous responses, misunderstanding of the question, or attempts to “game” a poll by the respondents.
4. Fixed poll size. The number $n$ of polled voters is fixed in advance; in particular, one cannot keep polling until one has achieved some desired outcome, and then stop.
5. Simple random sampling (without replacement). Each one of the $n$ voters polled is selected uniformly at random among the entire population $X$, thus each voter is equally likely to be selected by the poll, and no non-voter can be selected by the poll. (In particular, we make the important assumption that there is no selection bias.) Furthermore, each polled voter is chosen independently of all the others, except for the one condition that we do not poll any given voter more than once. (Thus, once a voter is polled, that voter is “crossed off the list” of the pool $X$ of voters that one randomly selects from to determine the next voter polled.) In particular, we assume that the poll is not clustered.
6. Honest reporting. The results of the poll are always reported, with no inaccuracies; one cannot cancel, modify, or ignore a poll once it has begun. In particular, one cannot conduct multiple polls and only report the “best” results (thus running the risk of confirmation bias).
Polls which deviate significantly from these hypotheses (e.g. due to complex questions, self-selection or other selection bias, confirmation bias, inaccurate responses, a high refusal rate, variable poll size, or clustering) will generally be less accurate than an idealised poll with the same sample size. Of course, there is a substantial literature in statistics (and polling methodology) devoted to measuring, mitigating, avoiding, or compensating for these less ideal situations, but we will not discuss those (important) issues here. We will remark though that in practice it is difficult to make the poll selection truly uniform. For instance, if one is conducting a telephone poll, then the sample will of course be heavily biased towards those voters who actually own phones; a little more subtly, it will also be biased toward those voters who are near their phones at the time the poll was conducted, and have the time and inclination to answer phone calls. As long as these factors are not strongly correlated with the poll question (i.e. whether the voter will vote for A), this is not a major concern, but in some cases, the poll methodology will need to be adjusted (e.g. by reweighting the sample) to compensate for the non-uniformity.
As stated in the introduction, we let $p$ be the proportion of the entire population $X$ that will vote for $A$, and $\overline{p}$ be the proportion of the polled voters that will vote for $A$ (which, by Hypotheses 2 and 3, is exactly equal to the proportion of polled voters that say that they will vote for $A$). Under the above idealised conditions, if the number $n$ of polled voters is 1,000, and the size of the population $X$ is 200 million, then the margin of error is about 3%, thus ${\Bbb P}( \overline{p}-0.03 \leq p \leq \overline{p} + 0.03 ) \geq 0.95$. (See this margin of error calculator for what happens with different choices of parameters.)
There is an important subtlety here: it is only the unconditional probability of the event $\overline{p}-0.03 \leq p \leq \overline{p} + 0.03$ that is guaranteed to be greater than 0.95. If one has additional prior information about $p$ and $\overline{p}$, then the conditional probability of this event, relative to this information, may be very different. For instance, if one had, prior to the poll, a very good reason to believe that $p$ is almost certainly between 0.4 and 0.6, and then the poll reports $\overline{p}$ to be 0.1, then the conditional probability that $\overline{p}-0.03 \leq p \leq \overline{p}+0.03$ occurs should be lower than the unconditional probability. [Note though that having priori information just about $p$, and not $\overline{p}$, will not cause the probability to drop below 95%, as this bound on the confidence level is uniform in $p$.] The question of how to account for prior information is a very delicate one in Bayesian probability, and will not be discussed here.
One special case of the above point is worth emphasising: the statement that $\overline{p}-0.03 \leq p \leq \overline{p} + 0.03$ is true with at least 95% probability is only valid before one actually conducts the poll and finds out the value of $\overline{p}$. Once $\overline{p}$ is computed, the statement $\overline{p}-0.03 \leq p \leq \overline{p} + 0.03$ is either true or false, i.e. occurs with probability 1 or 0 (unless one takes a Bayesian approach, as mentioned above). [This phenomenon of course occurs all the time in probability. For instance, if x denotes the outcome of rolling a fair six-sided die, then before one performs this roll, the probability that x equals 1 will be 1/6, but after one has seen what the value of this die is, the probability that x equals 1 will be either 1 or 0.]
– Nobody asked for my opinion! –
One intuitive argument against a poll of small relative size being accurate goes something like this: a poll of just 1,000 people among a population of 200,000,000 is almost certainly not going to poll myself, or any of my friends or acquaintances. If the opinions of myself, and everyone that I know, is not being considered at all in this poll, how could this poll possibly be accurate?
It is true that if you know, say, 5,000 voting-eligible people, then chances are that none of them (or maybe one of them, at best) will be contacted by the above poll. However, even though the opinions of all these people are not being directly polled, there will be many other people with equivalent opinions that will be contacted by the poll. Through those people, the views of yourself and your friends are being represented. [This may seem like a very weak form of representation, but recall that you and your 5,000 friends and acquaintances still only represent 0.0025% of the total electorate.]
Now one may argue that no two voters are identical, and that each voter arrives at a decision of who to vote for their own unique reasons. True enough – but recall that this poll is asking only a simple question: whether one is going to vote for A or not. Once one narrowly focuses on this question alone, any two voters who both decide to vote for A, or to not vote for A, are considered equivalent, even if they arrive at this decision for totally different reasons. So, for the purposes of this poll, there are only two types of voters in the world – A-voters, and not-A-voters – with all voters in one of these two types considered equivalent. In particular, any given voter is going to have millions of other equivalent voters distributed throughout the population $X$, and a representative fraction of those equivalent voters is likely to be picked up by the poll.
As mentioned before, polls which offer complex questions (for instance, trying to discern the motivation behind one’s voting choices) will inherently be less accurate; there are now fewer equivalent voters for each individual, and it is harder for a poll to pick up each equivalence class in a representative manner. (In particular, the more questions that are asked, the more likely it becomes that the responses to at least one of these questions will be inaccurate by an amount exceeding its margin of error. This provides a limit as to how much information one can confidently extract from data mining any given data set.)
– Is there enough information? –
Another common objection to the accuracy of polls argues that there is not enough information (or “degrees of freedom”) present in the poll sample to accurately describe the much larger amount of data present in the full population; 1,000 bits of data cannot possibly contain 200,000,000 bits of information. However, we are not asking to find out so much information; the purpose of the poll is to estimate just a single piece of information, namely the number $p$. If one is willing to accept an error of up to 3%, then one can represent this piece of information in about five bits rather than 200,000,000. So, in principle at least, there is more than enough information present in the poll to recover this information; one does not need to sample the entire population to get a good reading. (The same general philosophy underlies compressed sensing, but that’s another story.)
As before, the accuracy degrades as one asks more and more complicated questions. For instance, if one were to poll 1,000 voters for their opinions on two unrelated questions A and B, each of the answers to A and B would be accurate to within 3% with probability 95%, but the probability that the answers to A and B were simultaneously accurate to within 3% would be lower (around 90% or so), and so any data analysis that relies on the responses to both A and B may not have as high a confidence level as data analysis that relies on A and B separately. This is consistent with the information-theoretic perspective: we are demanding more and more bits of information on our population, and it is harder for our fixed data set to supply so much information accurately and confidently.
– Swings –
One intuitive way to gauge the margin of error of a poll is to see how likely such a poll is to accurately detect a swing in the electorate. Suppose for instance that over the course of a given time period (e.g. a week), 7% of the voters switch their vote from not-A to A, while another 2% of the voters switch their vote from A to not-A, leading to a net increase of 5% in the proportion $p$ of voters voting for A. How does would this swing in the vote affect the proportion $\overline{p}$ of the voters being polled, if one imagines the same voters being polled at both the start of the week and at the end of the week?
If the poll was conducted by simple random sampling, then each of the 1,000 voters polled would have a 7% probability of switching from not-A to A, and and a 2% probability of switching from A to not-A. Thus, one would expect about 70 of the 1,000 voters polled to switch to A, and about 20 to switch to not-A, leading to a net swing of 50 voters, that would increase $\overline{p}$ by 5%, thus matching the increase in p. Now, in practice, there will be some variability here; due to the luck of the draw, the poll may pick up more or less than 70 of the voters switching to A, and more or less than 20 of the voters switching to not-A. But having 1,000 voters to sample is just about large enough for the law of large numbers to kick in and ensure that the number of voters switching to A picked up by the poll will be significantly larger than the number of voters switching to not-A. Thus, this poll will have a good chance of detecting a swing of size 5% or more, which is consistent with the assertion of a margin of error of about 3%. [In appealing to the law of large numbers, we are implicitly exploiting the uniformity and independence assumptions in Hypothesis 5.]
It is worth noting that this swing of 5% in an electorate of 200,000,000 voters represents quite a large shift in absolute terms: fourteen million voters switching to A and four million switching away from A. Quite a few of these shifting voters will be picked up by the poll (in contrast to one’s sphere of friends and acquaintances, which is likely to be missed completely).
– Irregularity –
Another intuitive objection to polling accuracy is that the voting population is far from homogeneous. For instance, it is clear that voting preferences for the U.S. presidential election vary widely among the 50 states – shouldn’t one need to multiply the poll size by 50 just to accomodate this fact? Similarly for distinctions in voting patterns based on gender, race, party affiliation, etc.
Again, these irregularities in voter distribution do not affect the final accuracy of the poll, for two reasons. Firstly, we are asking only the simple question of whether a voter votes for A or not-A, and are not breaking down the answers to this question by state, gender, race, or any other factor; as stated before, two voters are considered equivalent as long as they have the same preference for A, even if they are in different states, have different genders, etc. Secondly, while it is conceivable that the poll will cluster its sample in one particular state (or one particular gender, etc.), thus potentially skewing the poll, the fact that the voters are selected uniformly and independently of each other prevents this from happening very often. (And in any event, clustering in a demographic or geographic category is not what is of direct importance to the accuracy of the poll; the only thing that really matters in the end is whether there is clustering in the category of A-voters or not-A-voters.) The independence hypothesis is rather important. If for instance one were to poll by picking one particular location in the U.S. at random, and polling 1,000 people from that location, then the responses would be highly correlated (as one could have picked a location which happens to highly favour A, or highly favour not-A) and would have a much larger margin of error than if one polled 1,000 people at random across the U.S..
[Incidentally, in the specific case of the U.S. presidential election, statewide polls are in fact more relevant to the outcome of the election than nationwide polls, due to the mechanics of the U.S. Electoral College, but this does not detract from the above points.]
– Analogies –
Some analogies may help explain why the relative size of a sample is largely irrelevant to the accuracy of a poll.
Suppose one is in front of a large body of water (e.g. a sea or ocean), and wants to determine whether it is a freshwater or saltwater body. This can be done very easily: dip one’s finger into the body of water and taste a single drop. This gives an extremely accurate result, even though the relative proportion of the sample size to the population size is, literally, a drop in the ocean; the quintillions of water molecues and salt molecues present in that drop are more than sufficient to give a good reading of the salinity of the water body.
[To be fair, in order for this reading to be accurate, one needs to assume that the salinity is uniformly distributed across the body of water; if for instance the body happened to be nearly fresh on one side and much saltier on the other, then dipping one’s finger in just one of these two sides would lead to an inaccurate measurement of average salinity. But if one were to stir the body of water vigorously, this irregularity of distribution disappears. The procedure of taking a random sample, with each sample point being independent of all the others, is analogous to this stirring procedure.]
Another analogy comes from digital imaging. As we all know, a digital camera takes a picture of a real-world object (e.g. a human face) and converts it into an array of pixels; an image with a larger number of pixels will generally lead to a more accurate image than one with fewer. But even with just a handful of pixels, say 1,000 pixels, one is already able to make crude distinctions between different images, for instance to distinguish a light-skinned face from a dark-skinned face (despite the fact that skin colour is determined by millions of cells and quintillions of pigment molecues). See for instance this well-known (and very low resolution) image of a US president, by Leon Harmon:
– Appendix: Mathematical justification –
One can compute the margin of error for this simple sampling problem very precisely using the binomial distribution; however I would like to present here a cruder but more robust estimate, based on the second moment method, that works in much greater generality than the setting discussed here. (It is closely related to the arguments in my previous post on the law of large numbers.) The main mathematical result we need is
Theorem. Let X be a finite set, let A be a subset of X, and let $p := |A|/|X|$ be the proportion of elements of X that lie in A. Let $x_1, \ldots, x_n$ be sampled independently and uniformly at random from X (in particular, we allow repetitions). Let $\overline{p} := |\{1 \leq i \leq n: x_i \in A \}|/n$ be the proportion of the $x_1,\ldots,x_n$ (counting repetition) that lie in A. Then for any $r > 0$, one has
$\displaystyle {\Bbb P}( |\overline{p}-p| \leq r ) \geq 1 - \frac{1}{4 n r^2}$. (1)
Proof. We use the second moment method. For each $1 \leq i \leq n$, let $I_i$ be the indicator of the event $x_i \in A$, thus $I_i := 1$ when $x_i \in A$ and $I_i = 0$ otherwise. Observe that each $I_i$ has a probability of p of equaling 1, thus
$p = {\Bbb E} I_i.$
On the other hand, we have
$\overline{p} = \frac{1}{n} \sum_{i=1}^n I_i$.
Thus
$\overline{p}-p = \frac{1}{n} \sum_{i=1}^n I_i - {\Bbb E}(I_i)$;
squaring this and taking expectations, we obtain
${\Bbb E} |\overline{p}-p|^2 = \frac{1}{n^2} \sum_{i=1}^n {\bf Var}(I_i) + \frac{2}{n} \sum_{1 \leq i < j \leq n} {\bf Cov}(I_i,I_j)$
where ${\bf Var}(I_i) := {\Bbb E} (I_i-{\Bbb E} I_i)^2$ is variance of $I_i$, and ${\bf Cov}(I_i,I_j) := {\Bbb E}( (I_i-p) (I_j-p))$ is the covariance of $I_i, I_j$.
By assumption, the random variable $I_i, I_j$ for $i \neq j$ are independent, and so the covariances ${\bf Cov}(I_i, I_j)$ vanish. On the other hand, a direct computation shows that
${\bf Var}(I_i) = p - p^2 = \frac{1}{4} - (p-\frac{1}{2})^2 \leq \frac{1}{4}$
for each i. Putting all this together we conclude that
${\Bbb E} |\overline{p}-p|^2 \leq \frac{1}{4n}$
and the claim (1) follows from Markov’s inequality. $\Box$
Applying this theorem with n=1000 and $r=1/\sqrt{200} \approx 0.07$, we conclude that p and $\overline{p}$ lie within about 7% of each other with probability at least 95%, regardless of how large the population X is. In the context of an election poll, this means that if one samples 1000 voters independently at random (with replacement) whether they would vote for A, the margin of error for the answer would be at most 7% at the 95% confidence level.
Remark 1. Observe that the proof of the above theorem did not really need the $x_i$ to be fully independent of each other; the key thing was that each $x_i$ was close to uniformly distributed, and that the covariances between the indicators $I_i, I_j$ were small. (Thus one only needs pairwise independence rather than joint independence for the theorem to hold.) Because of this, one can also obtain variants of the above theorem when one selects $x_1,\ldots,x_n$ for random sampling without replacement (known as simple random sampling); now there is a slight correlation between $I_i, I_j$, but it turns out to be negligible when X is large, for instance when n=1000 and $|X| \sim 10^8$. (For this range of parameters, there is a non-trivial probability of a birthday paradox occurring, so the two sampling methods are genuinely different from each other; but they turn out to have almost the same margin of error anyway.) $\diamond$
Remark 2. If one assumes joint independence instead of pairwise independence, one can obtain slightly sharper inequalities than (1) (e.g. by using the Chernoff inequality), but at the 95% confidence level, this gives a relatively modest improvement only in the margin of error (in our specific example, the optimal margin of error is about 3% rather than 7%). $\diamond$
Remark 3. An inspection of the argument shows that if p is known to be very small or very large, then the margin of error is better than what (1) predicts. (In the most extreme case, if p=0 or p=1, then it is easy to see that the margin of error is zero.) But in the case of election polls, p is generally expected to be close to 1/2, and so one does not expect to be able to improve the margin of error much from this effect. And in any case, we don’t know the value of p exactly in practice (otherwise why would we be doing the poll in the first place?). $\diamond$
Remark 4. In real world situations, it can be difficult or impractical to get the $x_i$ to be close to uniformly distributed (because of sampling bias), and to keep the correlations low (because of effects such as clustering). Because of this, one often needs to perform a more complicated sampling procedure than simple random sampling, which requires more sophisticated statistical analysis than given by the above theorem. This is beyond the scope of this post, though. $\diamond$
[Updated, October 13: added emphasis that the confidence level only applies before one performs the poll, not afterwards.]
[Updated, October 17: Minor corrections; thanks to Tom Verhoeff for pointing them out.]
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Kinematics in 2-D (and 3-D) From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin, [email protected] The primary objective of the SOFiSTiK VERiFiCATiON Benchmarks is to verify the capabilities of SOFiSTiK by means of nontrivial problems which are bound to reference solutions. Inverse Kinematics for swerve with 3 or more wheels. This is the inverse kinematics problem. 348 m/s 2D Motion 2D Kinematics. zUnit 10 - Electrostatics. Shed the societal and cultural narratives holding you back and let step-by-step Physics textbook solutions reorient your old paradigms. 4 sec Vwest =. Last edited: Mar 23, 2011. The acceleration of P at time t seconds, t ≥ 0, is (3t + 5) m s–2 in the positive x-direction. ezyEXAMs > Jee mains > physics > kinematics one/two Questions and Answers of NEET Kinematics - TopperLearning KINEMATICS IN 2 DIMENSION notes and MCQ for IIT-JEE, NEET Motion in a Plane NCERT Solutions Class11 physics. The bullet (mass m = 0. Question TitleKinematics Problems I A car begins driving from a stationary position. AP1 Kinematics Page 2 2. How fast is Georgia running now? 2. Find the north and east components of the displacement for the hikers shown in Figure. When the object in the image may be one of many known. A proper understanding of kinematics require an understanding of vector calculus too. What is the initial velocity ? 1. Includes a starter on Speed = Distance / Time, diagnostic questions using RAG&White, examples and practice, then Nando's themed differentiated questions (with answers) and a literacy plenary to write to their future selves what mistakes to avoid and to explain in their own words - "metacognition" or something! Please let me know what you think :). Find the following for path B in Figure: (a) the total distance traveled, and (b) the magnitude and direction of the displacement from start to finish. Consider a simple flow situation, such as an airfoil in a. " Considers only motion " Determined by positions, velocities, accelerations. The kinematic equations are simplifications of object motion. An inverse kinematic solution was studied by training the neural network with the robot's end effector Cartesian coordinates and its corresponding joint configurations. In the present work, a new simultaneous multiple-model method for 2D elements with node-dependent kinematics is developed, for the analysis of electro-mechanical problems. " Considers only motion " Determined by positions, velocities, accelerations. Substituting v i and v f into equation 1 gives vsinT vsinT gt. 7 km at 59° west of north. 1 Kinematic Chains. Real life applications are also included as they show how these concepts in physics are used in engineering systems for example. Read through the , problem ,. Kinematics 2D Skill Building Problems Notebook. Smith's Website Below you will find answer keys to the worksheet at the end of each lesson. (2) Includes 350 figures to help students. Kinematics, Dynamicspdf. a = v2/r Challenge Problem Solutions Circular Motion Dynamics Challenge Problem. 4 km [S] before getting to the 7-Eleven. Newton-Raphson, Newmark Numerical Integration Method, etc. Show Step-by-step Solutions Try the free Mathway calculator and problem solver below to practice various math topics. Position and speed at any time can be calculated from the motion equations. It is possible that the object stops. 2d-kinematics-problems-with-solutions-pdf 2/6 Downloaded from dominiqueanselbook. Choosing for the X-axis the vertical line of motion OY with the origin at O , we see from the figure that the displacement x of the Projectile Motion Problems. I'm not a chemist, I'm a 3D geometry problem-solver, so I don't know what the implications are. your dog takes many short side trips to chase squirrels, examine fire hydrants. In the absence of frictional drag, an object near the surface of the earth will fall with the constant acceleration of gravity g. The primary objective of the SOFiSTiK VERiFiCATiON Benchmarks is to verify the capabilities of SOFiSTiK by means of nontrivial problems which are bound to reference solutions. In the diagrams below, a ball is on a flat horizontal surface. Kinematic equations relate the variables of motion to one another. Equation of motion of a chain. Kinematics in 2D The instantaneous acceleration is given by: Kinematics in 2D Example. Kinematics Test Review (Answers). 235 m/s VNorth =. When t = 0, the velocity of P is 2 ms–1 in the positive x-direction. 4 sec Vwest =. Godage et al. Kinematic Curve Analysis Consider the following velocity vs. Review Notes available in PDF format @ Dynamics Problem Decomposition Dynamics! Kinematics! Kinetics! Solution: – 2D projectile motion. 23 1 Homework Solutions 27 2 Kinematics of Particles 29 2. It contains distance vs time graphs, velocity vs time graphs, and kinematics problems with both constant speed and accelerated motion. Acces PDF 2d Kinematics Answer Key 2d Kinematics Answer Key Yeah, reviewing a books 2d kinematics answer key could be credited with your near associates listings. 2D kinematics estimation of postural sway using single-axis accelerometers Fabio Bagalà Abstracts / Gait & Posture 30S (2009) S26–S74 S59 ON/OFF burst regions are depicted. impulse, momentum, and conservation answers pdf. This is the Readme file for a simple 2D kinematic vehicle's steering motion and visualzation implemented in Matlab's Simulink. H C Verma Kinematics Objective Solution is helpful for students aspiring for IIT JEE Mains/Advanced and other engineering/medical exams. Solution north component 3. This is not a problem for animation software as it has a lot of resources to calculate the animations. The speed was 6. countered in solving kinematics problems (though, some of these ideas are more universal, and can be applied to some problems of other fields of physics). Uncertainty Propagation PHYS 1120 Error Propagation Solutions. Show the formula(ae) used to solve the problem with only numbers D. Problem 1: (30 points) – Basic kinematics in 2D Motion A target is placed a horizontal distance R = 65. Choosing for the X-axis the vertical line of motion OY with the origin at O , we see from the figure that the displacement x of the Projectile Motion Problems. they are constants), we treat them like we would any other. Your answer seems reasonable. Kumar When closed loops are present in the kinematic chain (that is, the chain is no longer serial, or even open), it is more difficult to determine the number of degrees of freedom or the mobility of the robot. Position and speed at any time can be calculated from the motion equations. KinematicsSolver objects allow users to formulate and numerically solve kinematics problems for their Simscape™ Multibody™ models. Kinematics in 2-D (and 3-D) From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin, [email protected] Download File PDF 2d Kinematics Answer Key 2d Kinematics Answer Key|dejavusanscondensedbi font size 10 format Thank you for reading 2d kinematics answer key. Vector Problems Kinematics Problems Newton's Laws Problems Types of Forces Inclined Planes Pulleys Springs Work, Energy, Power Conservation Laws, Momentum Circular Motion Gravity Thermodynamics Solution: e) The initial speed of the hat is v1 = 20m/s and the final speed is 0m/s. AP Physics Practice Test Solutions: Vectors; 2-D Motion ©2011, Richard White www. kinematics exam problems with answers relative motion, problems solutions exam kinematics Kinematics + exams + problems wwwphysicstitorial in kinematics short cuts for kinematics physics exam kinematics shortcuts on kinematics njit kinematics exam kinamatics exam solution. Three of the equations assume constant acceleration (equations 1, 2, and 4), and the other equation assumes zero acceleration and constant velocity (equation 3). fixed reference system. Tips for. Regarding model-based methods, they mathematically formulate the system’s behavior using kine-matics and dynamics equations [5][6]. Answers have been provided below for the review. Modular kinematics of planar mechanisms. SOLVED PROBLEMSThe steps for solving the problem will be the following: Identify the problem as a 2D one. Extra Practice with answers. This means rz=2(7-4)2 Solving for the total time yields T= 13. 2d-kinematics-problems-with-solutions-pdf 2/6 Downloaded from dominiqueanselbook. Physics Calculators Online. (2) Includes 350 figures to help students. YES! Now is the time to redefine your true self using Slader’s Physics answers. Kinematics Test Review To get ready for the unit test, try problems #1-17, and #19. Homework Solutions: Chapter 3 Homework Solutions. 5] Problem Set 1 (PDF). Analytical Solutions:. Additional details: (1) Features 150 multiple-choice questions and nearly 250 free-response problems, all with detailed solutions. 8: Written solution to lamp problem. Kinematics does not usually provide solutions to problems:to be a solu-tion,a flow configuration should also satisfy Newton’s second law,the object of the next Chapter. V I would not claim any novelty or originality of this note, since almost all of problems in the book belong to standard material of quantum eld theory. But kinematics establishes constraints that a dynamical solution must satisfy. Also fits well with a discussion of dimensions, handling large numbers and, possibly, dimensional analysis. The equation for Q1, in fact, depends on the solution for Q2. 4 sec Vwest =. 2d Kinematics Answer Key View 2D_Kinematics_Problem_Set_Key. Hc Verma Class 11 Physics Part 1 Solutions For Chapter 3 Rest And Showing top 8 worksheets in the category kinematics. 20 m/s 2 , and = 842 s. Sample Problems Chapter 0: A Mathematics Review. 1 Chapter 2 - Kinematics 2. We either know the velocity or acceleration, or the dependence of velocity on time or acceleration on time. Free solved physics problems on kinematics. 01 kg) leaves the gun at a speed of v0 = 145 m/s. Horizontal Motion V x= (V o) x x= x o+ (V o) xt Vertical Motion V y= (V o) y gt y= y o+. I'm guessing you are looking for 3D kinematic freedom that amounts to more than the trivial case I've cited? You're looking for flexibility where at least one joint is moving in relation to the others? That, too, spawns a simplistic solution. Show any necessary math E. worksheet_-_3_-_graphing_problems_answer_key. Kinematics Test Review (Answers). 2 Transformations between BODY and NED 2. Start a free trial now!. This is facilitated by resolving both the path tracking task and the optimal inverse kinematics problem simultaneously. 348 m/s 2D Motion 2D Kinematics. Qualitatively derive both the position (x) vs. 23 1 Homework Solutions 27 2 Kinematics of Particles 29 2. A single inverse solution branch consists of a set of configurations which have a manifold structure in the joint space of dimension equal to the number of redundant degrees of freedom. Vector Problems Kinematics Problems Newton's Laws Problems Types of Forces Inclined Planes Pulleys Springs Work, Energy, Power Conservation Laws, Momentum Circular Motion Gravity Thermodynamics Solution: e) The initial speed of the hat is v1 = 20m/s and the final speed is 0m/s. But there is a simple formula that one can derive for this purpose. Inverse Kinematics Issues • While FK is relatively easy to evaluate. This is the Readme file for a simple 2D kinematic vehicle's steering motion and visualzation implemented in Matlab's Simulink. 2d-kinematics-problems-with-solutions-pdf 2/6 Downloaded from dominiqueanselbook. The following projectiles are launched on horizontal ground with the same initial speed. This is a simple 1D , Kinematics , acceleration example , problem ,. f 2 available phase space. pdf (252 KB). Physics problems: kinematics Problem 5. Physics 12 Stuff. The math-level of the problems is for students wh. H C Verma Kinematics Objective Solution is helpful for students aspiring for IIT JEE Mains/Advanced and other engineering/medical exams. Determine the magnitude of the acceleration of the car. The solutions are prepared by best teachers from Kota. The time between the ball leaving the cliff and hitting the ground is: (A) 2 3 2 s (B) 2 3 s (C) 2 s (D) 4 s (E) 5 s 2. The kinematics quantities can be related as: Position - the location of an object. Two Dimensional Rotational Steps to Solving 2D Kinematics Problems. (a) Robot kinematics and alternative solutions (b)Tool path computation. There are some points given below that Vedantu recommends to JEE aspirants: In the chapter’s PDF, it contains numerical problems which are answered precisely by explaining each question. Additional details: (1) Features 150 multiple-choice questions and nearly 250 free-response problems, all with detailed solutions. The solution of the inverse kinematics problem leads to an important and useful concept of the workspace of a serial manipulator and the approaches to obtain the workspace and determine its properties are also presented. MeKin2D is a collection of subroutines for kinematic simulation of planar linkages using a modular approach, for synthesis and analysis of disk-cam Figure 2: Simulation of the animated character in Figure 1 done. We will now study kinematics in two dimensions. Example 1: On a hot summer day a person goes for a walk to see if they can find a 7-Eleven to buy a Slurpee. Show a plot of the states (x(t) and/or y(t)). Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. We show results on a 7-joint arm grasping objects in a cluttered environment, an 18 DoF quadruped standing on stepping stones, and a. Chapter 4: Kinematics in 2D Motion in a plane, vertical or horizontal But, the motion in the x- and y-directions are independent, except that they are coupled by the time Therefore, we can break the problem into x and y parts’’ We must use vectors: displacement r = x + y velocity v = v x+ v y acceleration a = a x + a y Usually, y x r x y. 3) Find the equation in the table that contains all 4 involved quantities. Get Started. This will include conditioning and QC of interpretations, producing valid 2D geological models and, importantly, quantifying and minimizing geological uncertainties. 1 Kinematic Chains. NOW is the time to make today the first day of the rest of your life. 1D and 2D Kinematics Skills and Practices Taught/Emphasized in This Investigation Science Practices Activities 1. The PDF uploaded on our website comprises the kinematics notes which are well systematized and supportive for learning with ease. Download Kinematics Problems Part 2 Solutions. SOLUTIONS FOR PRACTICE PROBLEMS 81 Now we know that the stone falls the last half of its distance in 4 seconds. Class 11 Physics Notes Chapter 2 Kinematics. 3 Transformations between ECEF and NED 2. Since this example problem deals with a two-joint robotic arm whose inverse kinematics formulae can be derived, it is possible to test the answers that the ANFIS networks produce with the answers from the derived formulae. (2) Includes 350 figures to help students. Kinematics Exams and Problem Solutions Kinematics Exam1 and Answers (Distance, Velocity, Acceleration, Graphs of Motion) Kinematics Exam2 and Answers(Free Fall) Kinematics Exam3 and Answers (Projectile Motion) Kinematics Exam4 and Answers (Relative Motion, Riverboat Problems). Approximately how far will the package travel in the horizontal direction from the. Homework Solutions: Chapter 3 Homework Solutions. using MeKin2D subroutines. From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin, [email protected] Pay attention to significant figures! (6490 m/s) 8. Show the answer with proper units VIII. Unit Extra Practice w/ Answers: The attached file has lots of extra practice questions (answers are at the end). 235 m/s VNorth =. Latest Articles. Students use data from the different parts of the investigation to create graphs of the. Uncertainty Propagation PHYS 1120 Error Propagation Solutions. 15 Systems Key Honors Worksheets W3. In this survey, we present a. time and the acceleration vs. 12H "Hamburger Problems" with Key W3. Kinematics, Dynamicspdf. The solution of the inverse kinematics problem leads to an important and useful concept of the workspace of a serial manipulator and the approaches to obtain the workspace and determine its properties are also presented. Typically, the geometrical solutions of physics problems rep- that it represents surface areas under graphs. PROJECTILE MOTION We see one dimensional motion in previous topics Examples of rotational kinematics. Your answer seems reasonable. Problem 3 A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure. 4 KB Views: 151. For example, the average velocity vector is v = ( d f − d o )/ t , where d o and d f are the initial and final displacement vectors and t is the time elapsed. This is the inverse kinematics problem. Determine the displacement from t = 0s to t = 4 s. H C Verma Kinematics Objective Solution is helpful for students aspiring for IIT JEE Mains/Advanced and other engineering/medical exams. This node-variable capability enables one to vary the kinematic assumptions within the same finite plate element. 1 Kinematic Chains. File Type PDF Kinematics Sample Problems And Solutions Kinematics Sample Problems And Solutions|freesansbi font size 12 format When somebody should go to the books stores, search inauguration by shop, shelf by shelf, it is in point of fact problematic. It helps students to attempt objective. 1D Kinematics - Position and Velocity: Problem Set 1 (PDF) Derivatives in Kinematics. f 2 available phase space. 1 Introduction In this chapter, as in the previous chapter, we won’t be concerned with the actual forces that cause an object to move the way it is moving. Unit 2 kinematics in 2d. Any problem that asks you to describe the motion of an object without worrying about the cause of that motion is a kinematics problem, no matter what was given or requested in the problem. Source of problems • Non-linear equations (sin, cos in rotation matrices). Furthermore, the author presents methods for fast computation of approximate optimal solutions to planning problems with changing parameters. Giancoli Answers is your best source for the 7th and 6th Edition Giancoli physics solutions. PHYS 1120 1D Kinematics Solutions. He discusses the key concepts and formulae of. In the present work, a new simultaneous multiple-model method for 2D elements with node-dependent kinematics is developed, for the analysis of electro-mechanical problems. 13 Newton’s Law Quick Hitters 2 Key W3. Kinematics does not usually provide solutions to problems:to be a solu-tion,a flow configuration should also satisfy Newton’s second law,the object of the next Chapter. The time between the ball leaving the cliff and hitting the ground is: (A) 2 3 2 s (B) 2 3 s (C) 2 s (D) 4 s (E) 5 s 2. See full list on physics. 2d-kinematics-problems-with-solutions-pdf 2/6 Downloaded from dominiqueanselbook. Kinematics (2D) Laws, Principles (so-called formulae) Solution A Solution B Solution C Problem Answer Critical ThinkerCritical Thinker One would just plug in the numbers and if it didn't come out to be a correct answer then he/she would just change the positive to negative and so on. Assume that it is important for the ANFIS networks to have low errors within the operating range 0. _electric_field_on_a_single_charge_teachers_notes. This problem starts at t = 0. Vision is replete with such problems, as is limb control. ©1996-99 by K. 2 m from the base. Motion Problems with 2 Objects. PHYSICS Kinematics Objectives. Newton-Raphson, Newmark Numerical Integration Method, etc. impulse, momentum, and conservation answers pdf. Godage et al. For this type of robot a solution to the inverse kinematics problem, needed for generating desired trajectories in the Cartesian space (2D) is found by using a feed-forward neural network. Kinematics Derivations (cont. Get Started. The Wooley Bear 2. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. The goal of kinematics is to mathematically describe the trajectory of an object over time. Each equation contains four variables. Students use data from the different parts of the investigation to create graphs of the. Question TitleKinematics Problems I A car begins driving from a stationary position. INVERSE KINEMATICS For a kinematic mechanism, the inverse kinematic problem is difficult to solve. SOLIDWORKS Motion takes the designer beyond the free drag movement available in a CAD environment and into a true physical calculation of the forces and motions of an assembly as it would move under environmental loads (external forces) and/or internal loads (motors, springs, and dampers). The kinematic equations relate position and velocity at any two points in the motion. " System of equations is usually under-defined " Multiple solutions. This is not to be passed in, but is intended to give you some more problems to get your ready to complete the unit test on Kinematics. Answer the following based on the velocity vs. Once you break it down, this is a reasonably simple geometry problem. Read through the , problem ,. ezyEXAMs > Jee mains > physics > kinematics one/two Questions and Answers of NEET Kinematics - TopperLearning KINEMATICS IN 2 DIMENSION notes and MCQ for IIT-JEE, NEET Motion in a Plane NCERT Solutions Class11 physics. Answers have been provided below for the review. These approaches are mainly divided into two types. I'm guessing you are looking for 3D kinematic freedom that amounts to more than the trivial case I've cited? You're looking for flexibility where at least one joint is moving in relation to the others? That, too, spawns a simplistic solution. Physics problems: kinematics Problem 5. More emphasis on the topics of physics included in the SAT physics subject with hundreds of problems with detailed solutions. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. 2d practice questions Overview of Energy Problems. If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old. Solution t v v v v at d v t at v v ad d i f f i i f i 2 2 2 1 2 Using what we know from the problem, we can eliminate the third and fourth equations of motion. com on January 30, 2021 by guest allowing students to steer clear of those if they wish. Lesson 1) to download a PDF version of the answer key. This OER repository is a collection of free resources provided by Equella. Kinematics in 2-D (and 3-D) From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin, [email protected] In the absence of frictional drag, an object near the surface of the earth will fall with the constant acceleration of gravity g. It consists of h c verma short answer solution, h c verma exercise solution, h c verma objective solutions. Stop wasting time looking for files and revisions. 1 Straight-Line The minimum time solution is to use maximum braking and acceleration. time and the acceleration vs. The ball, even as it moves upwards and sideways through the air, experiences a force of gravity acting on it, which causes it to accelerate downwards at g. -3 3 Euler angles -3 Remaining d. The negative solutions are not relavent to this problem. 8Read through this section of your text after watching the FlipIt pre-lecture video. Read through the , problem ,. The math-level of the problems is for students wh. 2d: 2 2 v , , , dr dv d r a r vdt v adt dt dt dt ³³ E9 Use trigonometric functions to solve problems in context, including problems involving vectors, kinematics and forces Commentary The extension of the techniques to 2D greatly increases the scope, sophistication and general interest of the scenarios that can be investigated; these may involve. Practice Problems: Kinematics Click here to see the solutions. ) We can use the initial angular velocity and the time to find the angular acceleration, here assumed to be constant. Answers have been provided below for the review. The solutions to the problems are initially hidden, and can be shown in gray boxes or hidden again by clicking. In this page we have 1D Kinematics Sample Problems And. PHYSICS Kinematics Objectives. com on January 30, 2021 by guest allowing students to steer clear of those if they wish. Classwork: Students will work in groups to practice solving problems involving projectiles launched at an angle from a height. time graph. time curve. Any problem that asks you to describe the motion of an object without worrying about the cause of that motion is a kinematics problem, no matter what was given or requested in the problem. Describe qualitatively how motion the motion of the ball will change. Kinematics Eqn #3 Proof solution pdf Kinematics 3rd eqn intro ques solution pdf Ch 5 page 180 tractor trailer problem solution pdf momentum 2D #1 solution pdf. Robot Geometry and Kinematics -7- V. Braking Car. First you should try to solve the problems while keeping in mind those ideas which are suggested for the given problem. work problems. Projectiles and vectors Online practice problems with answers: http://cstephenmurray. Powered by Create your own unique website with customizable templates. Designed for civil engineering & mining. 5] Problem Set 1 (PDF). Unit 2 Vectors and Projectiles (2D Kinematics) Assignment Sheet Trigonometry for Physics. Furthermore, the author presents methods for fast computation of approximate optimal solutions to planning problems with changing parameters. Show a plot of the states (x(t) and/or y(t)). Kinematics Practice Problems Answers. PROJECTILE MOTION We see one dimensional motion in previous topics Examples of rotational kinematics. The acceleration of P at time t seconds, t ≥ 0, is (3t + 5) m s–2 in the positive x-direction. What is his current position (in yards)? 3. The problem wi th this solution is that the computational cost is too high for complex bone-structures in games. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. A diagram is indispensable. Using the relations introduced in Problem 2. impulse, momentum, and conservation answers pdf. Read PDF 2d Kinematics Problems With Solutions 2d Kinematics Problems With Solutions Getting the books 2d kinematics problems with solutions now is not type of challenging means. Get Started. Additional details: (1) Features 150 multiple-choice questions and nearly 250 free-response problems, all with detailed solutions. Lesson 8 Solving Friction Problems. semester review problems pdf. A program to help physics students understand Kinematics and Dynamics. fixed reference system. For the sketch, recall that on a d versus t curve an object moving forward with a. (a) How long did it take the rock land on the ground (b) Find the range of the rock. (a) Robot kinematics and alternative solutions (b)Tool path computation. Sketch a possible x-t graph for the motion of the. 8Read through this section of your text after watching the FlipIt pre-lecture video. Planned features The Caliko library is an implementation of the FABRIK (Forward And Backward Reaching Inverse Kinemat - ics. (The solution is presented on the next page. 0 m/s as it rolls up an incline. The Wooley Bear 2. 0m/s at an altitude of 125m. Automatically solve the set of nonlinear equations of motion using appropriate numerical solution algorithms: e. The required equations and background reading to solve these problems is given on the kinematics page. time graph. This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on "Kinematics of Motion". Therefore it would be desired to adopt optimization techniques. The errors contained in the DEM data are corrected with ANUDEM method (Hutchinson, 2004; Zhang et al. Choose an origin and a coordinate system. 5 The student can re-express key elements of natural phenomena across multiple representations in the domain. Kinematics in 2-D (and 3-D) From Problems and Solutions in Introductory Mechanics (Draft version, August 2014) David Morin, [email protected] We offer 2d kinematics problems with solutions pdf and numerous books collections These are enough to solve problems given in this book. The first part is a station model in which students will work in groups of 3-4 to solve problems in 6 stations. solution: (a) 480 m (b) 379 m, 18. Solution to Problem 2. Sixty Baseball Physics Problems © page 1 !! Sixty Baseball Physics Problems (many with correct solutions!) !. It helps students to attempt objective. " System of equations is usually under-defined " Multiple solutions. Last edited: Mar 23, 2011. In this page we have 1D Kinematics Sample Problems And. Hc Verma Class 11 Physics Part 1 Solutions For Chapter 3 Rest And Showing top 8 worksheets in the category kinematics. Physics 12 Stuff. they are constants), we treat them like we would any other. See Appendix 3 for a complete list of problem types, solutions, and additional suggestions. Inverse Kinematics Issues IK is challenging because while f() may be relatively easy to evaluate, f-1() usually isn’t For one thing, there may be several possible solutions for Φ, or there may be no solutions Even if there is a solution, it may require complex and expensive computations to find it. A 50 g ball rolls off a table and lands 2 m from the base of the table. Problem 2: This problem is exactly analogous to a one-dimensional linear kinematics problem: a block has an initial velocity v = 30 m/s and slides to a halt in 10 seconds; what is the distance it slid through? a. Remember the lab is also due on Monday the 26th. Chapter 4: Kinematics in 2D Motion in a plane, vertical or horizontal But, the motion in the x- and y-directions are independent, except that they are coupled by the time Therefore, we can break the problem into x and y parts’’ We must use vectors: displacement r = x + y velocity v = v x+ v y acceleration a = a x + a y Usually, y x r x y. txt) or read online for free. kinematics in 2-D (and 3-D). It extends backward and forward in time. semester review problems solved pdf. • The existence of multiple solutions. For JEE Main other Engineering Entrance Exam Preparation, JEE Main Physics Kinematics Previous Year Questions with Solutions is given below. 2d Kinematics Answer Key View 2D_Kinematics_Problem_Set_Key. This set of Machine Kinematics Multiple Choice Questions & Answers (MCQs) focuses on "Kinematics of Motion". Download File PDF 2d Kinematics Answer Key 2d Kinematics Answer Key|dejavusanscondensedbi font size 10 format Thank you for reading 2d kinematics answer key. Then click the button to check the answer or use the link to view the solution. Report the final value of each state as t \to \infty. SOLVED PROBLEMSThe steps for solving the problem will be the following: Identify the problem as a 2D one. Show the formula(ae) used to solve the problem with only numbers D. Pulley in Physics is one of the most interesting topics in mechanics. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration. 1 Reference frames 2. Download kinematics for free. To see how the kinematic equations generate motion graphs for the ball in Figure 2-19a, work through Example 2-2-6 Solving Kinematics Problems I: Uniform Acceleration Example 2-8 Example 2-7 Revisited Repeat Example 2-7 using the equations of motion. 01 kg) leaves the gun at a speed of v0 = 145 m/s. Peskin and D. 1, show that given the components ij of a 2d strain tensor in a basis e i: 1. 23 1 Homework Solutions 27 2 Kinematics of Particles 29 2. accepted solution. • Solution (Inverse Kinematics)- A “solution” is the set of joint variables associated with an end effector’s desired position and orientation. Show any necessary math E. The kinematics quantities can be related as: Position - the location of an object. Detailed solutions. Identify the Problem. and so on When you arrive at the park, do you and your dog have the same displacement?. kinematics-2d. 3) Find the equation in the table that contains all 4 involved quantities. • Solution Strategies – Closed form Solutions - An analytic expression includes all solution sets. The second part is a 5 question practice worksheet. Give a written description of the motion. Download File PDF 2d Kinematics Answer Key 2d Kinematics Answer Key ANSWERS: 1. A 50 g ball rolls off a table and lands 2 m from the base of the table. Physics 12 Stuff. Remember the lab is also due on Monday the 26th. txt) or read online for free. Pulley in Physics is one of the most interesting topics in mechanics. File Type PDF Physics Kinematics Problems And Solutions Physics Problems: kinematics r = 11. • Kinematics. And the math can be simplified down to that very point. Of course you also need to review basic algebra, trigonometry and calculus. But there is a simple formula that one can derive for this purpose. Example 1: On a hot summer day a person goes for a walk to see if they can find a 7-Eleven to buy a Slurpee. •2D approximations used for analysis Two-dimensional kinematics of common The second part of the solution begins by expressing the length of the. Northern Highlands Regional HS / Overview. Source of problems • Non-linear equations (sin, cos in rotation matrices). 4 sec Vwest =. If we treat the allowed contact points on the beam as an open inerval, then the space is topologically equivalent to R. Physics 1120: 2D Kinematics Solutions File Type PDF Physics Kinematics Problems And Solutions speed of 5. A technique for solving the inverse kinematics problem using artificial neural networks was introduced for a PUMA 560 robot. But there is a simple formula that one can derive for this purpose. We show results on a 7-joint arm grasping objects in a cluttered environment, an 18 DoF quadruped standing on stepping stones, and a. Kinematics does not usually provide solutions to problems:to be a solu-tion,a flow configuration should also satisfy Newton’s second law,the object of the next Chapter. The PDF uploaded on our website comprises the kinematics notes which are well systematized and supportive for learning with ease. Physics concepts are clearly discussed and highlighted. Chapter 2 One-Dimensional Kinematics Q. UNIT 1 Kinematics 10 2D Motion NAME DATE Scenario An engineer is testing the design of a. t x Involved Unneeded. Move your mouse over the "Answer" to reveal the answer or click on the "Complete Solution" link to reveal all of the steps required for solving projectile motion problems. Chapter 4: Kinematics in 2D Motion in a plane, vertical or horizontal But, the motion in the x- and y-directions are independent, except that they are coupled by the time Therefore, we can break the problem into x and y parts’’ We must use vectors: displacement r = x + y velocity v = v x+ v y acceleration a = a x + a y Usually, y x r x y. com on January 30, 2021 by guest allowing students to steer clear of those if they wish. Determine the displacement from t = 0s to t = 4 s. 5 (a) I: the point of contact on the beam (which determines the angle of the wheel, since rolling is enforced). Pay attention to significant figures! (6490 m/s) 8. 14 Systems Key W3. 3) Find the equation in the table that contains all 4 involved quantities. PSI AP Physics C – Kinematics 2D Multiple Choice Questions 1. We will simply take the motion as. A program to help physics students understand Kinematics and Dynamics. Because many problems provide or ask about information at more than two points, I identify each point with its own subscript to keep them straight. ME 230 Kinematics and Dynamics and apply it in order to solve problems that involve force, velocity and displacement Solution: (continued) T 1 + U 1-2 = T 2. (moderate) A racecar, moving at a constant tangential speed of 60 m/s, takes one lap around a circular track in 50 seconds. f 2 available phase space. We show results on a 7-joint arm grasping objects in a cluttered environment, an 18 DoF quadruped standing on stepping stones, and a. 235 m/s VNorth =. Monkey problem exploration Using the simulation and the problem presented on the worksheet (see file below), students are asked to calculate and plot the scenario of launching the banana staight at the falling monkey. e a = v/t and unit of velocity is m/s so, unit of. 1D Kinematics - Position and Velocity: Problem Set 1 (PDF) Derivatives in Kinematics. A proper understanding of kinematics require an understanding of vector calculus too. • Solution Strategies – Closed form Solutions - An analytic expression includes all solution sets. When the light turns green, the driver accelerates so that the cars speedometer reads 10 m/s after 5 s. A useful problem-solving strategy was presented for use with these equations and two examples were given that illustrated the use of the strategy. Warning - it has 46 pages! So only print what you need. Kinematics 2D Skill Building Problems Notebook. Powered by Create your own unique website with customizable templates. kinematics quantity - to inappropriately interchange quantities such as position, velocity, and acceleration. UNIT 1 Kinematics 10 2D Motion NAME DATE Scenario An engineer is testing the design of a. ©1996-99 by K. The speed was 6. CH6: 21, 22, 42. Answers and Replies. Solving a projectile motion problem α A basketball launched on a level surface travels 15 m and reaches a maximum height of 6. Therefore it would be desired to adopt optimization techniques. (a) Robot kinematics and alternative solutions (b)Tool path computation. Unit Extra Practice w/ Answers: The attached file has lots of extra practice questions (answers are at the end). 5 The student can re-express key elements of natural phenomena across multiple representations in the domain. File Type PDF Physics Kinematics Problems And Solutions Physics Problems: kinematics r = 11. 1-D Kinematics Problems 1. Topic 3: Kinematics – Displacement, Velocity, Acceleration, 1- and 2-Dimensional Motion Source: Conceptual Physics textbook (Chapter 2 - second edition, laboratory book and concept-development practice book; CPO physics textbook and laboratory book Types of Materials: Textbooks, laboratory manuals, demonstrations, worksheets and activities. This paper presents a novel inverse kinematics solution for robotic arm based on artificial neural network (ANN) architecture. PDF download free. Unit 1 - Kinematics in 1D. Determine t1 , t2 and the car's maximum 2. Kinematics (2D) Laws, Principles (so-called formulae) Solution A Solution B Solution C Problem Answer Critical ThinkerCritical Thinker One would just plug in the numbers and if it didn't come out to be a correct answer then he/she would just change the positive to negative and so on. Position and speed at any time can be calculated from the motion equations. Unit 2 kinematics in 2d. Kinematic Character Controller a relatively low-level character controller solution that is not tied to any specific game genre and is made to be cleanly integrated into any project/architecture with as little friction or bloat as possible. Solution t v v v v at d v t at v v ad d i f f i i f i 2 2 2 1 2 Using what we know from the problem, we can eliminate the third and fourth equations of motion. We pose the problem so that it is suited for solution using a computer: 1. Warning - it has 46 pages! So only print what you need. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration. This node-variable capability enables one to vary the kinematic assumptions within the same finite plate element. 2 km [E] , and finally 1. This OER repository is a collection of free resources provided by Equella. Oh, thank heaven! Determine the displacement of the person. Introduction to S TATICS D and YNAMICS Chapters 1-10 Rudra Pratap and Andy Ruina Spring 2001 °c Rudra Pratap and Andy Ruina, 1994-2001. Problem 3 A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure. 2 m from the base. Maybe you have knowledge that, people have search numerous times for their chosen novels like this 2d kinematics answer key, but end up in infectious downloads. Derive the forward kinematics equations using the DH convention. 1 Kinematic Chains. To do that, we use three main equations. pdf - Inverse Kinematics ¥End-effector postions specified by spline curves!1!2 X = (x,y) l2 l1 (0,0) y x t Inverse Kinematics ¥Problem for more complex structures "System of equations is usually under-defined "Multiple solutions!1!2 l2 l1 (0,0) X =. Any problem that asks you to describe the motion of an object without worrying about the cause of that motion is a kinematics problem, no matter what was given or requested in the problem. We will simply take the motion as. " Kinematics Questions and Answers " pdf provides problems and solutions for class 9 competitive exams. Motion Problems with 2 Objects. Pedestrian and Bike at Intersection. less than 2 m from the base. The concepts for solving this problem are encapsu-lated in the kinematics formula: What this tells you is how the velocity changes over time due to a constant acceleration. Show the formula(ae) used to solve the problem with only variables C. Kinematics Q3 Understand, use and derive the formulae for constant acceleration for motion in a straight line using vectors in 2d Q4 Use calculus in kinematics for motion in a straight line using vectors in 2d: 2 2 v , , , dr dv d r a r vdt v adt dt dt dt ³³ E9 Use trigonometric functions to solve problems in. Then click the button to check the answer or use the link to view the solution. We either know the velocity or acceleration, or the dependence of velocity on time or acceleration on time. A box sits on a horizontal wooden board. A particle P moves on the x-axis. Kinematics: Dimensions and estimates Introduces students to the idea of making estimates, a ubiquitous skill in physics. time curves and draw them on the blank graphs below. 01 kg) leaves the gun at a speed of v0 = 145 m/s. 19, 2019, 7:19 p. Review of Units : Feb 3 : 1D Kinematics Review : Feb 4 : Motion Problems with 2 Objects. Download kinematics for free. Be able to convert between accelerations in m/s2 and g’s. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Real fluids: Slightly viscous shear and pressure will cause fluid to deform and move Objectives: The kinematics of the fluid motion —the velocity —acceleration, and —the description and visualization of its motion. Second, the inverse kinematics problem for a manipulator with redundant DoF is locally ill-posed in that each solution branch contains an infinite number of solutions. Chapter 2 One-Dimensional Kinematics Q. pdf from MATH 76 at Rockland College International. inverse kinematic solution adopted to consider robot movements in tool axis selection is explained. Since this example problem deals with a two-joint robotic arm whose inverse kinematics formulae can be derived, it is possible to test the answers that the ANFIS networks produce with the answers from the derived formulae. Determine the displacement from t = 0s to t = 4 s. Solution to Problem 2. Last edited: Mar 23, 2011. Pedestrian and Bike at Intersection. Physics 1120: 2D Kinematics Solutions 1. Horizontal Motion V x= (V o) x x= x o+ (V o) xt Vertical Motion V y= (V o) y gt y= y o+. For motion in two dimensions, the earlier kinematics equations must be expressed in vector form. Kinematics PYQP. Looking for free solution. Inverse Kinematics Issues • While FK is relatively easy to evaluate. ezyEXAMs > Jee mains > physics > kinematics one/two Questions and Answers of NEET Kinematics - TopperLearning KINEMATICS IN 2 DIMENSION notes and MCQ for IIT-JEE, NEET Motion in a Plane NCERT Solutions Class11 physics. 2d Kinematics Answer Key View 2D_Kinematics_Problem_Set_Key. Unit 2 Vectors and Projectiles (2D Kinematics) Assignment Sheet Trigonometry for Physics. The inverse kinematics (IK) algorithm is implemented in both 2D EE106A Discussion 4: Inverse Kinematics famaysaxena, isabella. In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 seconds at 8 yd/s. Kinematic Curve Analysis Consider the following velocity vs. Kinematics does not usually provide solutions to problems:to be a solu-tion,a flow configuration should also satisfy Newton’s second law,the object of the next Chapter. More emphasis on the topics of physics included in the SAT physics subject with hundreds of problems with detailed solutions. The displacement time graph for Y end is (a) Ellipse (b) Circle (c) Parabola (d) Straight line Solution. This is the inverse kinematics problem. -3 3 Euler angles -3 Remaining d. The wheels can be located anywhere on the robot; they don’t have to be at the corners of a rectangle. (2) Includes 350 figures to help students. A small rock was thrown at 300 with horizontal direction and speed of 15. Kinematics Exams and Problem Solutions Kinematics Exam1 and Answers (Distance, Velocity, Acceleration, Graphs of Motion) Kinematics Exam2 and Answers(Free Fall) Kinematics Exam3 and Answers (Projectile Motion) Kinematics Exam4 and Answers (Relative Motion, Riverboat Problems). 26 MODULE 2. 5 km [N], then 4. (moderate) A racecar, moving at a constant tangential speed of 60 m/s, takes one lap around a circular track in 50 seconds. Determine the displacement from t = 2 s to t = 6 s. SOLUTIONS FOR PRACTICE PROBLEMS 81 Now we know that the stone falls the last half of its distance in 4 seconds. The first part is a station model in which students will work in groups of 3-4 to solve problems in 6 stations. Kinematics: Dimensions and estimates Introduces students to the idea of making estimates, a ubiquitous skill in physics. Mechanical Physics lesson # 9 Kinematics in 2D part2 Solución de problemas relacionados con el movimiento parabólico. 4º east of north. Approximate Solutions. This is a two part activity. The complexity of inverse kinematic solution arises with the increment of degrees of freedom. 15 Systems Key Honors Worksheets W3. 2-D Kinematics The problem we run into with 1-D kinematics, is that well…it’s one dimensional. Unit 2 kinematics in 2d. Any problem that asks you to describe the motion of an object without worrying about the cause of that motion is a kinematics problem, no matter what was given or requested in the problem. Also fits well with a discussion of dimensions, handling large numbers and, possibly, dimensional analysis. Your answer seems reasonable. But there is a simple formula that one can derive for this purpose. Of course you also need to review basic algebra, trigonometry and calculus. Download ME8492 Kinematics of Machinery Lecture Notes, Books, Syllabus, Part-A 2 marks with answers and ME8492 Kinematics of Machinery Important Part-B 13 & 15 marks Questions, PDF Book, Question Bank with answers Key. A diagram is indispensable. Braking Car. The kinematic equations relate position and velocity at any two points in the motion. Now, we will try to explain motion in two dimensions that is exactly called “projectile motion”. 7 km at 59° west of north. To see how the kinematic equations generate motion graphs for the ball in Figure 2-19a, work through Example 2-2-6 Solving Kinematics Problems I: Uniform Acceleration Example 2-8 Example 2-7 Revisited Repeat Example 2-7 using the equations of motion. 12 Dynamics Word Problems Key W3. Solutions of Kinematics Practice Questions 3. Expand the requested time horizon until the solution reaches a steady state. Kinematics worksheet with answers. Projectiles and vectors Online practice problems with answers: http://cstephenmurray. Dalitz Plot – 2D Phase Space How can resonances be studied in multi-body decays? Consider 3 body decay M →m 1m 2m 3 (all spin 0) Degrees of freedom Complete dynamics described by two variables! Usual choice 15 3 Lorentz-vectors 12 3 Masses -3 Energy conserv. 5 The student can re-express key elements of natural phenomena across multiple representations in the domain. You should work out the algebra to prove the final result on the. final velocity with any displacement practice with answers: File Size: 139 kb: File Type: pdf. 15P Solution: Chapter 4 Two-Dimensional Kinematics Q. Start a free trial now!. Author by : Waldron Language : en Publisher by : John Wiley & Sons Format Available : PDF, ePub, Mobi Total Read : 63 Total Download : 455 File Size : 49,6 Mb GET BOOK. Kinematics worksheet 2 Kinematics worksheet 2. Lesson 1) to download a PDF version of the answer key. Figure out what we need: To get v 0x we need the flight time to use Can now use to get t. The solutions are prepared by best teachers from Kota. Chapter 3: 2D Kinematics Thursday January 22nd Reading: up to page 44 in the text book (Ch. Smith's Website Below you will find answer keys to the worksheet at the end of each lesson. A new artificial neural network approach for inverse kinematics is proposed. PHYSICS Kinematics Objectives. Oh, thank heaven! Determine the displacement of the person. kinematics-2d. Physics is an interesting and important branch of Science, that studies about matter. Note: Assume x = 0 when t = 0. The complex matrix calculations grow exponentially with every bone that is added. Determine t1 , t2 and the car's maximum 2. • RiRequire ClComplex and EiExpensive computations to find a solution.
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College Physics for AP® Courses 2e
# 17.4Doppler Effect and Sonic Booms
College Physics for AP® Courses 2e17.4 Doppler Effect and Sonic Booms
## Learning Objectives
By the end of this section, you will be able to:
• Define Doppler effect, Doppler shift, and sonic boom.
• Calculate the frequency of a sound heard by someone observing Doppler shift.
• Describe the sounds produced by objects moving faster than the speed of sound.
The characteristic sound of a motorcycle buzzing by is an example of the Doppler effect. The high-pitch scream shifts dramatically to a lower-pitch roar as the motorcycle passes by a stationary observer. The closer the motorcycle brushes by, the more abrupt the shift. The faster the motorcycle moves, the greater the shift. We also hear this characteristic shift in frequency for passing race cars, airplanes, and trains. It is so familiar that it is used to imply motion and children often mimic it in play.
The Doppler effect is an alteration in the observed frequency of a sound due to motion of either the source or the observer. Although less familiar, this effect is easily noticed for a stationary source and moving observer. For example, if you ride a train past a stationary warning bell, you will hear the bell’s frequency shift from high to low as you pass by. The actual change in frequency due to relative motion of source and observer is called a Doppler shift. The Doppler effect and Doppler shift are named for the Austrian physicist and mathematician Christian Johann Doppler (1803–1853), who did experiments with both moving sources and moving observers. Doppler, for example, had musicians play on a moving open train car and also play standing next to the train tracks as a train passed by. Their music was observed both on and off the train, and changes in frequency were measured.
What causes the Doppler shift? Figure 17.13, Figure 17.14, and Figure 17.15 compare sound waves emitted by stationary and moving sources in a stationary air mass. Each disturbance spreads out spherically from the point where the sound was emitted. If the source is stationary, then all of the spheres representing the air compressions in the sound wave centered on the same point, and the stationary observers on either side see the same wavelength and frequency as emitted by the source, as in Figure 17.13. If the source is moving, as in Figure 17.14, then the situation is different. Each compression of the air moves out in a sphere from the point where it was emitted, but the point of emission moves. This moving emission point causes the air compressions to be closer together on one side and farther apart on the other. Thus, the wavelength is shorter in the direction the source is moving (on the right in Figure 17.14), and longer in the opposite direction (on the left in Figure 17.14). Finally, if the observers move, as in Figure 17.15, the frequency at which they receive the compressions changes. The observer moving toward the source receives them at a higher frequency, and the person moving away from the source receives them at a lower frequency.
Figure 17.13 Sounds emitted by a source spread out in spherical waves. Because the source, observers, and air are stationary, the wavelength and frequency are the same in all directions and to all observers.
Figure 17.14 Sounds emitted by a source moving to the right spread out from the points at which they were emitted. The wavelength is reduced and, consequently, the frequency is increased in the direction of motion, so that the observer on the right hears a higher-pitch sound. The opposite is true for the observer on the left, where the wavelength is increased and the frequency is reduced.
Figure 17.15 The same effect is produced when the observers move relative to the source. Motion toward the source increases frequency as the observer on the right passes through more wave crests than she would if stationary. Motion away from the source decreases frequency as the observer on the left passes through fewer wave crests than he would if stationary.
We know that wavelength and frequency are related by $vw=fλvw=fλ$, where $vwvw$ is the fixed speed of sound. The sound moves in a medium and has the same speed $vwvw$ in that medium whether the source is moving or not. Thus $ff$ multiplied by $λλ$ is a constant. Because the observer on the right in Figure 17.14 receives a shorter wavelength, the frequency she receives must be higher. Similarly, the observer on the left receives a longer wavelength, and hence he hears a lower frequency. The same thing happens in Figure 17.15. A higher frequency is received by the observer moving toward the source, and a lower frequency is received by an observer moving away from the source. In general, then, relative motion of source and observer toward one another increases the received frequency. Relative motion apart decreases frequency. The greater the relative speed is, the greater the effect.
## The Doppler Effect
The Doppler effect occurs not only for sound but for any wave when there is relative motion between the observer and the source. There are Doppler shifts in the frequency of sound, light, and water waves, for example. Doppler shifts can be used to determine velocity, such as when ultrasound is reflected from blood in a medical diagnostic. The recession of galaxies is determined by the shift in the frequencies of light received from them and has implied much about the origins of the universe. Modern physics has been profoundly affected by observations of Doppler shifts.
For a stationary observer and a moving source, the frequency fobs received by the observer can be shown to be
$fobs=fsvwvw±vs,fobs=fsvwvw±vs,$
17.20
where $fsfs$ is the frequency of the source, $vsvs$ is the speed of the source along a line joining the source and observer, and $vwvw$ is the speed of sound. The minus sign is used for motion toward the observer and the plus sign for motion away from the observer, producing the appropriate shifts up and down in frequency. Note that the greater the speed of the source, the greater the effect. Similarly, for a stationary source and moving observer, the frequency received by the observer $fobsfobs$ is given by
$fobs=fsvw±vobsvw,fobs=fsvw±vobsvw,$
17.21
where $vobsvobs$ is the speed of the observer along a line joining the source and observer. Here the plus sign is for motion toward the source, and the minus is for motion away from the source.
## Example 17.4
### Calculate Doppler Shift: A Train Horn
Suppose a train that has a 150-Hz horn is moving at 35.0 m/s in still air on a day when the speed of sound is 340 m/s.
(a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes?
(b) What frequency is observed by the train’s engineer traveling on the train?
### Strategy
To find the observed frequency in (a), $fobs=fsvwvw±vs,fobs=fsvwvw±vs,$ must be used because the source is moving. The minus sign is used for the approaching train, and the plus sign for the receding train. In (b), there are two Doppler shifts—one for a moving source and the other for a moving observer.
### Solution for (a)
(1) Enter known values into $fobs=fsvwvw–vs.fobs=fsvwvw–vs.$
$f obs = f s v w v w − v s = 150 Hz 340 m/s 340 m/s – 35.0 m/s f obs = f s v w v w − v s = 150 Hz 340 m/s 340 m/s – 35.0 m/s$
17.22
(2) Calculate the frequency observed by a stationary person as the train approaches.
$fobs = ( 150 Hz ) ( 1.11 ) = 167 Hz fobs = ( 150 Hz ) ( 1.11 ) = 167 Hz$
17.23
(3) Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes.
$f obs = f s v w v w + v s = 150 Hz 340 m/s 340 m/s + 35.0 m/s f obs = f s v w v w + v s = 150 Hz 340 m/s 340 m/s + 35.0 m/s$
17.24
(4) Calculate the second frequency.
$fobs = ( 150 Hz ) ( 0.907 ) = 136 Hz fobs = ( 150 Hz ) ( 0.907 ) = 136 Hz$
17.25
### Discussion on (a)
The numbers calculated are valid when the train is far enough away that the motion is nearly along the line joining train and observer. In both cases, the shift is significant and easily noticed. Note that the shift is 17.0 Hz for motion toward and 14.0 Hz for motion away. The shifts are not symmetric.
### Solution for (b)
(1) Identify knowns:
• It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero.
• Relative to the medium (air), the speeds are $vs=vobs=35.0 m/s.vs=vobs=35.0 m/s.$
• The first Doppler shift is for the moving observer; the second is for the moving source.
(2) Use the following equation:
$fobs=[ fs vw±vobsvw ]vwvw±vs.fobs=[ fs vw±vobsvw ]vwvw±vs.$
17.26
The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source.
(3) Because the train engineer is moving in the direction toward the horn, we must use the plus sign for $vobs;vobs;$ however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for $vsvs$. But the train is carrying both the engineer and the horn at the same velocity, so $vs=vobsvs=vobs$. As a result, everything but $fsfs$ cancels, yielding
$fobs = fs.fobs = fs.$
17.27
### Discussion for (b)
We may expect that there is no change in frequency when source and observer move together because it fits your experience. For example, there is no Doppler shift in the frequency of conversations between driver and passenger on a motorcycle. People talking when a wind moves the air between them also observe no Doppler shift in their conversation. The crucial point is that source and observer are not moving relative to each other.
## Sonic Booms to Bow Wakes
What happens to the sound produced by a moving source, such as a jet airplane, that approaches or even exceeds the speed of sound? The answer to this question applies not only to sound but to all other waves as well.
Suppose a jet airplane is coming nearly straight at you, emitting a sound of frequency $fsfs$. The greater the plane’s speed $vsvs$, the greater the Doppler shift and the greater the value observed for $fobsfobs$. Now, as $vsvs$ approaches the speed of sound, $fobsfobs$ approaches infinity, because the denominator in $fobs=fsvwvw±vsfobs=fsvwvw±vs$ approaches zero. At the speed of sound, this result means that in front of the source, each successive wave is superimposed on the previous one because the source moves forward at the speed of sound. The observer gets them all at the same instant, and so the frequency is infinite. (Before airplanes exceeded the speed of sound, some people argued it would be impossible because such constructive superposition would produce pressures great enough to destroy the airplane.) If the source exceeds the speed of sound, no sound is received by the observer until the source has passed, so that the sounds from the approaching source are mixed with those from it when receding. This mixing appears messy, but something interesting happens—a sonic boom is created. (See Figure 17.16.)
Figure 17.16 Sound waves from a source that moves faster than the speed of sound spread spherically from the point where they are emitted, but the source moves ahead of each. Constructive interference along the lines shown (actually a cone in three dimensions) creates a shock wave called a sonic boom. The faster the speed of the source, the smaller the angle $θθ$.
There is constructive interference along the lines shown (a cone in three dimensions) from similar sound waves arriving there simultaneously. This superposition forms a disturbance called a sonic boom, a constructive interference of sound created by an object moving faster than sound. Inside the cone, the interference is mostly destructive, and so the sound intensity there is much less than on the shock wave. An aircraft creates two sonic booms, one from its nose and one from its tail. (See Figure 17.17.) During television coverage of space shuttle landings, two distinct booms could often be heard. These were separated by exactly the time it would take the shuttle to pass by a point. Observers on the ground often do not see the aircraft creating the sonic boom, because it has passed by before the shock wave reaches them, as seen in Figure 17.17. If the aircraft flies close by at low altitude, pressures in the sonic boom can be destructive and break windows as well as rattle nerves. Because of how destructive sonic booms can be, supersonic flights are banned over populated areas of the United States.
Figure 17.17 Two sonic booms, created by the nose and tail of an aircraft, are observed on the ground after the plane has passed by.
Sonic booms are one example of a broader phenomenon called bow wakes. A bow wake, such as the one in Figure 17.18, is created when the wave source moves faster than the wave propagation speed. Water waves spread out in circles from the point where created, and the bow wake is the familiar V-shaped wake trailing the source. A more exotic bow wake is created when a subatomic particle travels through a medium faster than the speed of light travels in that medium. (In a vacuum, the maximum speed of light will be $c=3.00×108 m/sc=3.00×108 m/s$; in the medium of water, the speed of light is closer to $0.75c0.75c$. If the particle creates light in its passage, that light spreads on a cone with an angle indicative of the speed of the particle, as illustrated in Figure 17.19. Such a bow wake is called Cerenkov radiation and is commonly observed in particle physics.
Figure 17.18 Bow wake created by a duck. Constructive interference produces the rather structured wake, while there is relatively little wave action inside the wake, where interference is mostly destructive. (credit: Horia Varlan, Flickr)
Figure 17.19 The blue glow in this research reactor pool is Cerenkov radiation caused by subatomic particles traveling faster than the speed of light in water. (credit: U.S. Nuclear Regulatory Commission)
Doppler shifts and sonic booms are interesting sound phenomena that occur in all types of waves. They can be of considerable use. For example, the Doppler shift in ultrasound can be used to measure blood velocity, while police use the Doppler shift in radar (a microwave) to measure car velocities. In meteorology, the Doppler shift is used to track the motion of storm clouds; such “Doppler Radar” can give velocity and direction and rain or snow potential of imposing weather fronts. In astronomy, we can examine the light emitted from distant galaxies and determine their speed relative to ours. As galaxies move away from us, their light is shifted to a lower frequency, and so to a longer wavelength—the so-called red shift. Such information from galaxies far, far away has allowed us to estimate the age of the universe (from the Big Bang) as about 14 billion years.
## Check Your Understanding
Why did scientist Christian Doppler observe musicians both on a moving train and also from a stationary point not on the train?
## Check Your Understanding
Describe a situation in your life when you might rely on the Doppler shift to help you either while driving a car or walking near traffic.
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# Why does spin determine particle statistics?
I understand that "spin" refers to the intrinsic angular momentum of a particle, which relates to the magnetic moment of a particle. I mostly follow the "Background" section of the spin-statistics theorem Wikipedia page, so I understand that the math works out such that half-integer spin particles are fermions and integer-spin particles are bosons.
But I do not understand why the intrinsic angular momentum of a particle has anything to do with whether it can coexist with another identical particle. And I really don't understand why the distinction would be half-integer vs integer values. Is there any intuitive way of understanding this?
• In general there is no relation between spin and statistics. Only in systems with Lorentz invariance (and in more than 2 dimensions) there is a sort of relation, but it is indirect: the spin-statistics theorem (cf. this PSE post). You might also want to read Baez. Nov 26, 2019 at 0:59
## 1 Answer
"Simple English":
I do not understand why the intrinsic angular momentum of a particle has anything to do with whether it can coexist with another identical particle
The coexistence of a particle with another (identical) particle is quantified by the value of the commutator. The commutator depends on the particular representation (how the particle transforms under Poincaré tranformations, that is Lorentz boosts, rotations and translations). The representation of a particle is uniquely labelled by its spin and mass.
So this is how the spin enters the picture.
And I really don't understand why the distinction would be half-integer vs integer values
The integer and half-integer values comes out from (simple) angular momentum algebra, see later.
How does it enter the particle statistics business? If you do the maths (see later), you'll see that the spin enters as $$2j$$ and not just $$j$$.
$$\newcommand{\mb}{\mathbf}$$ $$\newcommand{\im}{\mathrm{i}}$$
Let's work in 3D and assume you know that there are only two particle statistics, such that: $$[a, a^\dagger ] = 0 \quad \text{for bosons}, \tag{1}$$ $$\{a, a^\dagger \} = 0 \quad \text{for fermions}, \tag{2}$$ where $$a^\dagger$$ ($$a$$) is the creation (annihilation) operator for the quanta of the field.
There is a mathematical (topological) way of proving this, and an intuitive picture for visualising the topologically distinct paths in $$n$$ dimensions. But that's another matter.
How does spin come in?
The above distribution statistics for quantum particles stems from the requirement of indistinguishability between identical particles. In the true sense o the word, identical particles transform in the same irreducible representation of the Poincaré group, uniquely identified by their mass and spin. Hence, quantum mechanics is not enough and we need to do quantum field theory.
Quantum field theory's main addition to the pot is that it allows for the theory to be causal.
For a theory to be causal, the time ordering of physical events affecting the system's evolution cannot be reversed. This is especially problematic for space-like separations where a Lorentz boost may reverse the chronological order $$t_{\mathrm{final}} - t_{\mathrm{initial}} < 0$$. For causality to hold, any two space-like separated operators are required to commute: $$$$[\mathcal{O}_1(x), \mathcal{O}_2(y)] = 0 \quad \text{if} \quad (x-y)^2 < 0, \quad g_{\mu \nu} = (+,-,-,-),$$$$ to ensure their time ordering is irrelevant and not resulting in any physical consequence.
The relativistic counterparts of the commutation relations in eqs. $$1$$ and $$2$$ are: $$$$\begin{gathered} \text{bosons: }\qquad [a(\mb{p}, s), a^{\dagger}(\mb{p}', s')] = 2E_{\mb{p}} (2\pi)^3 \delta^{(3)}(\mb{p} - \mb{p}') \delta_{s, s'}, \\ \text{fermions: } \qquad \{a(\mb{p}, s), a^{\dagger}(\mb{p}', s') \} = 2E_{\mb{p}} (2\pi)^3 \delta^{(3)}(\mb{p} - \mb{p}') \delta_{s, s'} . \end{gathered}$$$$ A general field operator $$\phi$$ for non-interacting particles $$A$$ and $$B$$ is given by: $$$$\label{eq:fieldeq} \begin{gathered} \phi_{A, B}(x) = \int \frac{\mathrm{d}^3 \mb{p}}{(2\pi)^3} \frac{1}{2 E_{\mb{p}}} \sum_s \left [ \mathrm{e}^{-\im px}f_{A, B}(\mb{p},s)a(\mb{p},s) + \mathrm{e}^{\im px}h_{A, B}(\mb{p},s)a^{\dagger}(\mb{p},s) \right ]_{p^0 = E_{\mb{p}}}. \end{gathered}$$$$ Because operators $$\mathcal{O}(x)$$ are usually just a product of $$\prod_i \phi_i(x)$$, requiring $$[\mathcal{O}_1(x), \mathcal{O}_2(y)]=0$$ is the same as requiring $$\left [ \phi_A(x), \phi_B(y)\right ] = 0$$.
This results in: $$$$\label{eq:commutators} \begin{gathered} \text{for bosons: } \quad \left [ \phi_A(x), \phi_B(y)\right ] \propto 1 - (-1)^{2j} , \\ \text{for fermions: } \quad \left [ \phi_A(x), \phi_B(y)\right ] \propto 1 + (-1)^{2j}, \end{gathered}$$$$ where $$j$$, in the absence of orbital angular momentum, is the particle's spin. See later for a bit of a proof.
The ladder operators used in angular momentum algebra result in a quantised spectrum bounded by $$j$$ and $$-j$$, so that $$-j + n = j$$ with $$n \in \mathbb{N}$$. This requires $$j = n/2$$, i.e. either an integer or a half integer.
Hence, in conclusion:
$$$$\begin{gathered} \text{bosons} \Leftrightarrow \left [ a,a^\dagger \right ] = 0 \Leftrightarrow \text{integer spin}, \\ \text{fermions} \Leftrightarrow \left \{ a,a^\dagger \right \} = 0 \Leftrightarrow \text{half-integer spin}. \end{gathered}$$$$
Sketch of a proof
The foundation of the proof lies in the commutator $$[\phi_A, \phi_B]$$. The full, rigorous and index-full proof is given in The Quantum Field of Fields I by Weinberg, in the section about General causal fields. I am just going to sketch the proof here, as a guide for the argument presented in the reference.
The commutator $$[\phi_A, \phi_B]$$ depends on the commutators $$[f_A, f_B]$$ and $$[h_A, h_B]$$ of the functions that define the field operator. Hence we will turn our attention to the coefficients $$f$$ and $$h$$.
The function $$f(\mb{p})$$ for a generic momentum $$\mb{p}$$ can be generated from one at rest $$f(\mb{0})$$ by a Lorentz boost $$\Lambda$$: $$f(\mb{p}) = D(\Lambda)f(\mb{0})$$, with $$D$$ being some irreducible representation of the Lorentz group. The latter affects both space and time, and therefore obeys $$SO(4)$$ algebra -- technically $$SO(1,3)$$ -- which is isomorphic to $$SO(3)\times SO(3)$$. This can be easily understood by considering the same isomorphism in one fewer dimension: $$SO(3)$$ is isomorphic to $$SO(2)\times SO(2)$$, because any position on a 3D sphere can be written as a linear superposition of two orthogonal 2D circles. Hence, we can replace the four-dimensional vector $$\mathcal{J}\in SO(4)$$ by two three-dimensional vectors $$\mb{J}, \mb{K}\in SO(3)$$, with quantum numbers $$j$$ and $$k$$ obeying the usual angular momentum algebra and hence $$2j+1, 2k+1 \in \mathbb{N}$$.
Luckily we already know that the spherical harmonics $$Y^m_{\ell}$$ are the basis functions of the irreducible $$2\ell+1$$ space, and they transform according to $$Y^m_{\ell}(-\mathbf{r}) = (-1)^{\ell} Y^m_{\ell}(\mathbf{r})$$. Extending this to the generic representation $$D(\Lambda)$$, we expect a transformation $$D(\Lambda) \rightarrow (-1)^j (-1)^k = (-1)^{j+k}$$.
Identical particles transform according to the same irreducible representation of the Poincaré group, uniquely labelled by their mass $$m = \sqrt{\frac{1}{c^2}\,p^\mu p_\mu}$$ and spin. If $$A$$ and $$B$$ are identical, that is with equal mass and spin, then: $$$$[f_A, f_B] \propto D(\Lambda)D(\Lambda) \propto (-1)^{j+k} (-1)^{j+k} = (-1)^{2(j+k)}.$$$$ Hence, the total angular momentum $$j+k$$ determines the sign of the commutator and the statistics obeyed by the particles, resulting in the expressions above.
The treatment in this section was simplified by only considering real scalar fields, though it can be easily extended to complex fields representing charged particles. More importantly, the fields were assumed to be non-interacting and hence expressed in terms of single mode $$\mb{p}$$ creation and annihilation operators. Proofs of the spin-statistics theorem for interacting fields are much harder, and to be quite frank I do not understand them. However, it is argued in Weinberg that the cluster decomposition theorem ensures interacting fields can always be constructed from the non-interacting components considered here.
• Personnally, I did not understand your answer at all. And I seriously doubt this is what the OP meant by "intuitive way" either. Even Dirac's introduction to QM seems more readable to a novice than your answer. I am certain you know what you are talking about but you assume such a large number of prerequisites that I believe if the OP knew them, they would not ask this question in the first place. Any way you could start with a simple English approach first before diving into the algebra? Nov 26, 2019 at 18:25
• Added a paragraph at the beginning. Nov 26, 2019 at 18:36
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# Solving Equations Algebraically
#### Scatter Plot
Contents: This page corresponds to § 2.4 (p. 200) of the text.
Suggested Problems from Text:
p. 212 #7, 8, 11, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97
### Equations Involving Fractional Expressions or Absolute Values
A quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0.
### Factoring
This approach to solving equations is based on the fact that if the product of two quantities is zero, then at least one of the quantities must be zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text.
Example 1.
2x2 - 5x - 12 = 0.
(2x + 3)(x - 4) = 0.
2x + 3 = 0 or x - 4 = 0.
x = -3/2, or x = 4.
### Square Root Principle
If x2 = k, then x = ± sqrt(k).
Example 2.
x2 - 9 = 0.
x2 = 9.
x = 3, or x = -3.
Example 3.
Example 4.
x2 + 7 = 0.
x2 = -7.
x = ± .
Note that = = , so the solutions are
x = ± , two complex numbers.
### Completing the Square
The idea behind completing the square is to rewrite the equation in a form that allows us to apply the square root principle.
Example 5.
x2 +6x - 1 = 0.
x2 +6x = 1.
x2 +6x + 9 = 1 + 9.
The 9 added to both sides came from squaring half the coefficient of x, (6/2)2 = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = x2 + 2ax + a2. To get "a" one need only divide the x-coefficient by 2. Thus, to complete the square for x2 + 2ax, one has to add a2.]
(x + 3)2 = 10.
Now we may apply the square root principle and then solve for x.
x = -3 ± sqrt(10).
Example 6.
2x2 + 6x - 5 = 0.
2x2 + 6x = 5.
The method of completing the square demonstrated in the previous example only works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is 2, but we can change that by dividing both sides of the equation by 2.
x2 + 3x = 5/2.
Now that the leading coefficient is 1, we take the coefficient of x, which is now 3, divide it by 2 and square, (3/2)2 = 9/4. This is the constant that we add to both sides to complete the square.
x2 + 3x + 9/4 = 5/2 + 9/4.
The left hand side is the square of (x + 3/2). [ Verify this!]
(x + 3/2)2 = 19/4.
Now we use the square root principle and solve for x.
x + 3/2 = ± sqrt(19/4) = ± sqrt(19)/2.
x = -3/2 ± sqrt(19)/2 = (-3 ± sqrt(19))/2
So far we have discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right form to apply the square root principle may be rearranged and solved by factoring as we see in the next example.
Example 7.
x2 = 16.
x2 - 16 = 0.
(x + 4)(x - 4) = 0.
x = -4, or x = 4.
In some cases the equation can be solved by factoring, but the factorization is not obvious.
The method of completing the square will always work, even if the solutions are complex numbers, in which case we will take the square root of a negative number. Furthermore, the steps necessary to complete the square are always the same, so they can be applied to the general quadratic equation
ax2 + bx + c = 0.
The result of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula.
The solutions for the equation ax2 + bx + c = 0 are
We are saying that completing the square always works, and we have completed the square in the general case, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, we write it in the standard form to find the values of a, b, and c, then substitute these values into the Quadratic Formula.
One consequence is that you never have to complete the square to find the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so you still need to know how to do it!
Examples using the Quadratic Formula:
Example 8.
2x2 + 6x - 5 = 0.
In this case, a = 2, b = 6, c = -5. Substituting these values in the Quadratic Formula yields
Notice that we solved this equation earlier by completing the square.
Note: There are two real solutions. In terms of graphs, there are two intercepts for the graph of the function f(x) = 2x2 + 6x - 5.
Example 9.
4x2 + 4x + 1 = 0
In this example a = 4, b = 4, and c = 1.
• There is only one solution. In terms of graphs, this means there is only one x-intercept.
• The solution simplified so that there is no square root involved. This means that the equation could have been solved by factoring. (All quadratic equations can be solved by factoring! What I mean is it could have been solved easily by factoring.)
4x2 + 4x + 1 = 0.
(2x + 1)2 = 0.
x = -1/2.
Example 10.
x2 + x + 1 = 0
a = 1, b = 1, c = 1
Note: There are no real solutions. In terms of graphs, there are no intercepts for the graph of the function f(x) = x2 + x + 1. Thus, the solutions are complex because the graph of y = x2 + x + 1 has no x-intercepts.
The expression under the radical in the Quadratic Formula, b2 - 4ac, is called the discriminant of the equation. The last three examples illustrate the three possibilities for quadratic equations.
1. Discriminant > 0. Two real solutions.
2. Discriminant = 0. One real solution.
3. Discriminant < 0. Two complex solutions.
Notes on checking solutions
None of the techniques introduced so far in this section can introduce extraneous solutions. (See example 3 from the Linear Equations and Modeling section.) However, it is still a good idea to check your solutions, because it is very easy to make careless errors while solving equations.
The algebraic method, which consists of substituting the number back into the equation and checking that the resulting statement is true, works well when the solution is "simple", but it is not very practical when the solution involves a radical.
For instance, in our next to last example, 4x2 + 4x + 1 = 0, we found one solution x = -1/2.
The algebraic check looks like
4(-1/2)2 +4(-1/2) + 1 = 0.
4(1/4) - 2 + 1 = 0.
1 - 2 + 1 = 0.
0 = 0. The solution checks.
In the example before that, 2x2 + 6x - 5 = 0, we found two real solutions, x = (-3 ± sqrt(19))/2. It is certainly possible to check this algebraically, but it is not very easy. In this case either a graphical check, or using a calculator for the algebraic check are faster.
First, find decimal approximations for the two proposed solutions.
(-3 + sqrt(19))/2 = 0.679449.
(-3 - sqrt(19))/2 = -3.679449.
Now use a graphing utility to graph y = 2x2 + 6x - 5, and trace the graph to find approximately where the x-intercepts are. If they are close to the values above, then you can be pretty sure you have the correct solutions. You can also insert the approximate solution into the equation to see if both sides of the equation give approximately the same values. However, you still need to be careful in your claim that your solution is correct, since it is not the exact solution.
Note that if you had started with the equation 2x2 + 6x - 5 = 0 and gone directly to the graphing utility to solve it, then you would not get the exact solutions, because they are irrational. However, having found (algebraically) two numbers that you think are solutions, if the graphing utility shows that intercepts are very close to the numbers you found, then you are probably right!
Exercise 1:
Solve the following quadratic equations.
(a) 3x2 -5x - 2 = 0. Answer
(b) (x + 1)2 = 3. Answer
(c) x2 = 3x + 2. Answer
## Equations Involving Radicals
Equations with radicals can often be simplified by raising to the appropriate power, squaring if the radical is a square root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked.
If there is only one radical in the equation, then before raising to a power, you should arrange to have the radical term by itself on one side of the equation.
Example 11.
Now that we have isolated the radical term on the right side, we square both sides and solve the resulting equation for x.
Check:
x = 0
When we substitute x = 0 into the original equation we get the statement 0 = 2, which is not true!
So, x = 0 is not a solution.
x = 3
When we substitute x = 3 into the original equation, we get the statement 3 = 3. This is true, so x = 3 is a solution.
Solution: x = 3.
Note: The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+1)+1.
Look at what would have happened if we had squared both sides of the equation before isolating the radical term.
This is worse than what we started with!
If there is more than one radical term in the equation, then in general, we cannot eliminate all radicals by raising to a power one time. However, we can decrease the number of radical terms by raising to a power.
If the equation involves more than one radical term, then we still want to isolate one radical on one side and raise to a power. Then we repeat that process.
Example 12.
Now square both sides of the equation.
This equation has only one radical term, so we have made progress! Now isolate the radical term and then square both sides again.
Check:
Substituting x = 5/4 into the original equation yields
sqrt(9/4) + sqrt(1/4) = 2.
3/2 + 1/2 = 2.
This statement is true, so x = 5/4 is a solution.
Note on checking solutions:
The algebraic check was easy to do in this case. However, the graphical check has the advantage of showing that there are no solutions that we have not found, at least within the scope of the viewing rectangle. The solution is the x-coordinate of the intersection point of the graphs of y = 2 and y = sqrt(x+1)+sqrt(x-1).
Exercise 2:
Solve the equation sqrt(x+2) + 2 = 2x. Answer
## Polynomial Equations of Higher Degree
We have seen that any degree two polynomial equation (quadratic equation) in one variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a problem, then either the polynomial is of a special form which allows us to factor it, or we must approximate the solutions with a graphing utility.
### Zero Constant
One common special case is where there is no constant term. In this case we may factor out one or more powers of x to begin the problem.
Example 13.
2x3 + 3x2 -5x = 0.
x (2x2 + 3x -5) = 0.
Now we have a product of x and a quadratic polynomial equal to 0, so we have two simpler equations.
x = 0, or 2x2 + 3x -5 = 0.
The first equation is trivial to solve. x = 0 is the only solution. The second equation may be solved by factoring. Note: If we were unable to factor the quadratic in the second equation, then we could have resorted to using the Quadratic Formula. [Verify that you get the same results as below.]
x = 0, or (2x + 5)(x - 1) = 0.
So there are three solutions: x = 0, x = -5/2, x = 1.
Note: The solution is found from the intercepts of the graphs of f(x) = 2x3 + 3x2 -5x.
### Factor by Grouping
Example 14.
x3 -2x2 -9x +18 = 0.
The coefficient of x2 is -2 times that of x3, and the same relationship exists between the coefficients of the third and fourth terms. Group terms one and two, and also terms three and four.
x2 (x - 2) - 9 (x - 2) = 0.
These groups share the common factor (x - 2), so we can factor the left hand side of the equation.
(x - 2)(x2 - 9) = 0.
Whenever we find a product equal to zero, we obtain two simpler equations.
x - 2 = 0, or x2 - 9 = 0.
x = 2, or (x + 3)(x - 3) = 0.
So, there are three solutions, x = 2, x = -3, x = 3.
Note: These solutions are found from the intercepts of the graph of f(x) = x3 -2x2 -9x +18.
### Quadratic in Form
Example 15.
x4 - x2 - 12 = 0.
This polynomial is not quadratic, it has degree four. However, it can be thought of as quadratic in x2.
(x2) 2 -(x2) - 12 = 0.
It might help you to actually substitute z for x2.
z2 - z - 12 = 0 This is a quadratic equation in z.
(z - 4)(z + 3) = 0.
z = 4 or z = -3.
We are not done, because we need to find values of x that make the original equation true. Now replace z by x2 and solve the resulting equations.
x2 = 4.
x = 2, x = -2.
x2 = -3.
x = i , or x = -i.
So there are four solutions, two real and two complex.
Note: These solutions are found from the intercepts of the graph of f(x) = x4 - x2 - 12.
A graph of f(x) = x4 - x2 - 12 and a zoom showing its local extrema.
Exercise 3:
Solve the equation x4 - 5x2 + 4 = 0. Answer
## Equations Involving Fractional Expressions or Absolute Values
Example 16.
The least common denominator is x(x + 2), so we multiply both sides by this product.
This equation is quadratic. The Quadratic Formula yields the solutions
Checking is necessary because we multiplied both sides by a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the x-coordinate of the intersection point of the graphs of y = 1 and y = 2/x-1/(x+2).
Example 17.
5 | x - 1 | = x + 11.
The key to solving an equation with absolute values is to remember that the quantity inside the absolute value bars could be positive or negative. We will have two separate equations representing the different possibilities, and all solutions must be checked.
Case 1. Suppose x - 1 >= 0. Then | x - 1 | = x - 1, so we have the equation
5(x - 1) = x + 11.
5x - 5 = x + 11.
4x = 16.
x = 4, and this solution checks because 5*3 = 4 + 11.
Case 2. Suppose x - 1 < 0. Then x - 1 is negative, so | x - 1 | = -(x - 1). This point often confuses students, because it looks as if we are saying that the absolute value of an expression is negative, but we are not. The expression (x - 1) is already negative, so -(x - 1) is positive.
Now our equation becomes
-5(x - 1) = x + 11.
-5x + 5 = x + 11.
-6x = 6.
x = -1, and this solution checks because 5*2 = -1 + 11.
If you use the Java Grapher to check graphically, note that abs() is absolute value, so you would graph
5*abs(x - 1) - x - 11 and look at x-intercepts, or you can find the solution as the x-coordinates of the intersection points of the graphs of y = x+11 and y = 5*abs(x-1).
Exercise 4:
(a) Solve the equation Answer
(b) Solve the equation | x - 2 | = 2 - x/3 Answer
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Preface
This section provides an introduction to a very important class of ordinary differential equations and their solutions.
Introduction to Linear Algebra with Mathematica
Hypergeometric Functions
The differential equation $$\label{E66.1} x(1-x) \,y'' + [\gamma - (\alpha + \beta +1)x]\,y' - \alpha\beta\,y =0, %\eqno{(6.1)}$$ \index{Hypergeometric equation}% in which $\alpha$, $\beta$, and $\gamma$ are parameters, is called the {\bf hypergeometric equation}. Its singular points $x=0$, $x=1$, and $x=\infty$ are all regular. For Eq.\;\eqref{E66.1}, suppose we attempt a power series solution of the Frobenius form \index{Frobenius series}% $y(x) = x^m \,\sum_{k=0}^\infty \,a_k x^k = \sum_{k\geq 0} \,a_k x^{k+m} .$ We substitute the power-series form into the differential equation \eqref{E66.1} to obtain$$\sum_{k\geq 0} \,a_k (k+m)(k+m-1)\, x^{k+m-1} - \sum_{k\geq 0} \,a_k (k+m)(k+m-1)\, x^{k+m}$$$$+\gamma \,\sum_{k\geq 0} \,a_k (k+m)\, x^{k+m-1} - (\alpha + \beta + 1)\,\sum_{k\geq 0} \,a_k (k+m)\, x^{k+m} -\alpha \beta \,\sum_{k\geq 0} \,a_k x^{k+m} =0.$$ In two series that contain powers $x^{k+m-1}$, we shift the index by setting $k-1=j$, then $k=j+1$. When $k=0$, the index $j$ is -1, and we keep this term out of the sum. This yields $$a_0 m(m-1) x^{m-1} + \gamma a_0 m x^{m-1} + \sum_{j\geq 0} \,a_{j+1} (j+1+m)(j+m) \,x^{j+m}$$$$+ \gamma \,\sum_{j\geq 0} \,a_{j+1} (j+1+m) \,x^{j+m} - \sum_{k\geq 0} \,a_k (k+m)(k+m-1)\, x^{k+m}$$$$- (\alpha + \beta + 1)\,\sum_{k\geq 0} \,a_k (k+m)\, x^{k+m} -\alpha \beta \,\sum_{k\geq 0} \,a_k x^{k+m} =0.$$ We equate the coefficients of the power $m-1$ to zero. This leads to $a_0 m\, [m-1 +\gamma ] =0$. Assuming that $a_0 \neq 0$, we get $$m(m-1+\gamma )=0.$$ This equation has two roots $$m_1 =0 \quad\mbox{and}\quad m_2 = \gamma -1 .$$ We denote the corresponding solution of Eq.\;\eqref{E66.1} by $y_1 (x)$ and $y_2 (x)$. For $m= m_1 =0$ we have $$y(x) = \sum_{k\geq 0} \,a_k \, x^k , \quad y' (x) = \sum_{k\geq 1} \,k a_k \, x^{k-1} , \quad y'' (x) = \sum_{k\geq 2} \,k(k-1) a_k \, x^{k-2} .$$We substitute these power series into Eq.\;\eqref{E66.1} to obtain $$\sum_{k\geq 1} \,a_{k+1} \, k(k+1)\, x^k - \sum_{k\geq 2} \,a_k \, k(k-1)\, x^k + \gamma \,\sum_{k\geq 0} \,a_{k+1} \, (k+1)\, x^k$$$$-(\alpha + \beta +1)\,\sum_{k\geq 1} \,a_k \, k\, x^k - \alpha \beta\, \sum_{k\geq 0} \,a_k \, x^k =0 .$$Next, we equate the coefficients of each power of $x$ to zero. This yields $$\gamma a_1 - \alpha\beta \,a_0 =0 ,$$$$2 a_2 + \gamma 2 a_2 - (\alpha + \beta +1) a_1 - \alpha \beta a_1 =0\quad\mbox{or}\quad 2 (\gamma +1) a_2 - (\alpha +1)(\beta +1) a_1 =0 ,$$$$3(\gamma +2) \,a_3 - (\alpha +2)(\beta +2)\,a_2 =0 ,$$$$\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots$$$$(k+1)(\gamma +k) \,a_{k+1} - (\alpha +k)(\beta +k) a_k =0 .$$ If $\gamma \neq 0, -1, -2, \ldots ,-k, \ldots$ and $a_0 =1$ we obtain the following relations: \begin{eqnarray*} a_1 &=& \frac{\alpha\beta}{\gamma} , \\ a_2 &=& \frac{(\alpha +1)(\beta +1)}{2(\gamma +1)} \,a_1 = \frac{\alpha (\alpha +1)\beta (\beta +1)}{1\cdot 2 \gamma (\gamma +1)} , \\ a_3 &=& \frac{(\alpha +2)(\beta +2)}{2(\gamma +2)} \, a_2 = \frac{\alpha (\alpha +1)(\alpha +2)\beta (\beta +1)(\beta +2)}{1\cdot 2 \cdot 3 \gamma (\gamma +1)(\gamma +2)} , \\ &&\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ a_{k+1} &=& \frac{(\alpha +k)(\beta +k)}{(k+1)(\gamma +k)} \,a_k \\ &=& \frac{\alpha (\alpha +1)(\alpha +2)\cdots (\alpha +k)\beta (\beta +1)(\beta +2)\cdots (\beta +k)}{1\cdot 2 \cdot 3 \cdots (k+1)\gamma (\gamma +1)(\gamma +2)\cdots (\gamma +k)} . \end{eqnarray*} The power series with these coefficients, when $a_0$ is assigned the value 1, is denoted by $F(\alpha , \beta ; \gamma ; x)$ and is called the {\bf hypergeometric function}: \index{Hypergeometric function}% \index{Function!hypergeometric}% $$\label{E66.4} F(\alpha , \beta ; \gamma ; x) = 1 + \sum_{k\geq 0} \, \frac{\alpha (\alpha +1)(\alpha +2)\cdots (\alpha +k)\beta (\beta +1)(\beta +2)\cdots (\beta +k)}{(k+1)! \,\gamma (\gamma +1)(\gamma +2)\cdots (\gamma +k)} \, x^{k+1} . %\eqno{(6.4)}$$ Sometimes the hypergeometric function is denoted by $_2F_1 (\alpha , \beta ; \gamma ; x)$ or $F\left(\left.\pile{\alpha ,\beta }{\gamma }\right| x\right)$. To determine the domain of convergence of the power series \eqref{E66.4}, we apply the ratio test to obtain $$\lim_{k\rightarrow \infty} \,\left\vert \frac{u_{k+1}}{u_k} \right\vert = \lim_{k\rightarrow \infty} \,\left\vert \frac{(\alpha +k)(\beta +k)}{(1+k)(\gamma +k)} \, x\right\vert = |x| .$$ Therefore the series \eqref{E66.4} converges\footnote{It is proved in advanced books that the power series \eqref{E66.4} converges absolutely when $x=1$ if $\gamma - \alpha -\beta >0$, and diverges if $\gamma - \alpha -\beta \leq 0$. When $x=-1$ it converges absolutely when $\gamma - \alpha -\beta >0$, it converges when $-1 < \gamma - \alpha -\beta \leq 0$, and it diverges when $\gamma - \alpha -\beta \leq -1$.} (absolutely) when $-1< x < 1$ and diverges for $|x| >1$. \qed To find another solution $y_2 (x)$, we make the substitution $y_2 = x^{1-\gamma}\,u$. Then $u$ is a solution of the differential equation $$x(1-x)\, u'' + [(2-\gamma ) -(\alpha + \beta - 2\gamma +3) x]\,u' - (\alpha +1 -\gamma )(\beta +1 -\gamma )\,u =0$$which is a hypergeometric equation. Hence $u= F(\alpha +1 -\gamma , \beta +1 -\gamma ; 2-\gamma ; x)$ and $$y_2 (x) = x^{1-\gamma} \, F(\alpha +1 -\gamma , \beta +1 -\gamma ; 2-\gamma ; x) .$$ The functions $y_1 (x)$ and $y_2 (x)$ are linearly independent because the corresponding power series contain different powers. Therefore the general solution of Eq.\;\eqref{E66.1} is their linear combination: $$\label{E66.5} y(x) = %C_1 \,y_1 (x) + C_2 \,y_2 (x) = C_1 \,F(\alpha , \beta ; \gamma ; x) + C_2 \,x^{1-\gamma} \, F(\alpha +1 -\gamma , \beta +1 -\gamma ; 2-\gamma ; x) .$$ The coefficients of the hypergeometric series \eqref{E66.4} can be expressed in terms of the gamma-function as \index{Gamma function}% \index{Function!Gamma}% $F(\alpha , \beta , \gamma , x) = \frac{\Gamma (\gamma )}{\Gamma (\alpha )\Gamma (\beta )} \,\sum_{k=0}^\infty \,\frac{\Gamma (\alpha +k) \Gamma (\beta +k)}{k! \,\Gamma (\gamma +k)} \, x^k , \qquad \Gamma (\nu ) = \int_0^\infty t^{\nu -1} \,e^{-t} \,dt . %\eqno{(6.5)}$ Many elementary and special functions are of the hypergeometric type, as the following list shows \begin{align*} F(\alpha , \beta ; \beta ; x) &= (1-x)^{-\alpha} , &F(-n , \beta ; \beta ; -x) &= F(-n , 1; 1; -x ) = (1+x)^n , \\ F(-n , \beta ; \beta ; 1-x) &= x^n , &F(1,1;2;-x) & = \frac{1}{x} \,\ln (1+x) , \\ F\left( \frac{1}{2} , 1; \frac{3}{2} ; x^2 \right) &= \frac{1}{2x} \,\ln \frac{1+x}{1-x} , &F\left( \frac{1}{2} , \frac{1}{2} ; \frac{3}{2} ; x^2 \right) &= \frac{1}{x} \,\ln \arcsin x , \\ F\left( \frac{1}{2} , 1; \frac{3}{2} ; - x^2 \right) &= \frac{1}{x} \,\arctan x , &F\left( \frac{1}{2} , \frac{1}{2} ; \frac{1}{2} ; x^2 \right) &= (1- x^2 )^{1/2} = \sqrt{1- x^2} , \\ F\left( \frac{k}{2} , -\frac{k}{2} ; \frac{1}{2} ; x^2 \right) &= \cos (k\arcsin x) , &\lim_{\beta \rightarrow \infty} \, F\left( 1, \beta ; 1 ; \frac{x}{\beta} \right) &= e^x , \\ \lim_{\alpha \rightarrow \infty} \, F\left( \alpha , \beta ; \beta ; \frac{x}{\alpha} \right) &= e^x , &\lim_{\alpha \rightarrow \infty} \, F\left( \alpha, \alpha ; \frac{1}{2} ; -\frac{x^2}{4\alpha^2} \right) &= \cos x , \\ \lim_{\alpha \rightarrow \infty} \, F\left( \alpha, \alpha ; \frac{3}{2} ; -\frac{x^2}{4\alpha^2} \right) &= \frac{1}{x}\,\sin x , &\lim_{\alpha ,\beta\rightarrow \infty} \, F\left( \alpha, \beta ; \frac{1}{2} ; \frac{x^2}{4\alpha\beta} \right) &= \cosh x . \end{align*}
The special functions are extremely useful tools for obtaining closed form as well as series solutions to a variety of problems arising in science and engineering we tryed to reobtain the known results by the new method.
Although the Adomian’s goal is to find a method to unify linear and nonlinear, ordinary or partial differential equations for solving initial and boundary value problems, we shall deal in the following only with linear second order differential equations.
Because the solutions to many mathematical physics problems are expressed by the hypergeometric functions, we consider first this equation whose standard form is
$x \left( 1 - x \right) f'' (x) + \left[ \gamma - \left( \alpha + \beta + 1 \right) x \right] f' (x) - \alpha\beta\, f(x) =0$
and rewrite it as
\begin{align*} L\,f(x) &= x\, f'' (x) + \gamma \,f' (x) \equiv x^{1-\gamma} \frac{\text d}{{\text d}x} \left( x^{\gamma} \frac{{\text d}f}{{\text d}x} \right) \\ &= x^2 f'' (x) + \left[ \left( \alpha + \beta + 1 \right) x \right] f' (x) - \alpha\beta\, f(x) . \end{align*}
Applying Adomian's decomposition method, we seek its solution in the form:
$f(x) = u_0 (x) + u_1 (x) + u_2 (x) + \cdots ,$
with the initial condition u0 = 1. Then applying the inverse operator
$\left( L^{-1} f\right) (x) = \int_0^x x^{-\gamma} {\text d}x \int_0^x x^{\gamma -1} f(x) \,{\text d}x - f(0) ,$
we find
\begin{align*} u_n (x) &= L^{-1} \left[ x^2 u''_{n-1} (x) + \left( \alpha + \beta + 1 \right) x \,u'_{n-1} (x) + \alpha\beta\, u_{n-1} (x) \right] \\ &= \int_0^x x^{-\gamma} {\text d}x \int_0^x {\text d}x \,x^{\gamma -1} \left[ x^2 u''_{n-1} (x) + \left( \alpha + \beta + 1 \right) x \,u'_{n-1} (x) + \alpha\beta\, u_{n-1} (x) \right] {\text d}x . \end{align*}
So we get u1 = αβx/γ, and by induction
$u_n (x) = \frac{\alpha^{\overline{n}} \,\beta^{\overline{n}}{\gamma^{\overline{n}} \, \frac{x^n}{n!} , \qquad n=1,2,\ldots ;$
where
$\alpha^{\overline{n}} = \alpha \left( \alpha +1 \right) \cdots \left( \alpha + n -1 \right)$
is rising factorial of α. In this way, we get the well known solution
$f(x) = _2F_1 \left( \alpha , \beta; \gamma; x \right) = 1 + \sum_{n\ge 1} \frac{\alpha^{\overline{n}} \,\beta^{\overline{n}}{\gamma^{\overline{n}} \, \frac{x^n}{n!} .$
Example:
1. Dită, P. and Grama, N., On Adamian's decomposition method for solving differential equations, Institute of Atomic Physics, Bucharest, 2008.
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### Common Core State Standards (CCSS) Edition Lesson List
Expand All | Collapse All
Unit 1: Naming and Constructing Geometric Figures
Lesson
1-1 Introduction to the Student Reference Book
1-2 Points, Line Segments, Lines, and Rays
1-4 Parallelograms
1-5 Polygons
1-6 Drawing Circles with a Compass
1-7 Circle Constructions
1-8 Hexagon and Triangle Constructions
1-9 Progress Check 1
Unit 2: Using Numbers and Organizing Data
Lesson
2-1 A Visit to Washington D.C.
2-2 Many Names for Numbers
2-3 Place Value in Whole Numbers
2-4 Place Value with a Calculator
2-5 Organizing and Displaying Data
2-6 The Median
2-8 Displaying Data with Graphs
2-9 Subtraction of Multidigit Numbers
2-10 Progress Check 2
Unit 3: Multiplication and Division; Number Sentences and Algebra
Lesson
3-1 "What's My Rule?"
3-2 Multiplication Facts
3-3 Multiplication Facts Practice
3-4 More Multiplication Facts Practice
3-5 Multiplication and Division
3-6 World Tour: Flying to Africa
3-7 Finding Air Distances
3-8 A Guide for Solving Number Stories
3-9 True or False Number Sentences
3-10 Parentheses in Number Sentences
3-11 Open Sentences
3-12 Progress Check 3
Unit 4: Decimals and Their Uses
Lesson
4-1 Decimal Place Value
4-2 Review of Basic Decimal Concepts
4-3 Comparing and Ordering Decimals
4-4 Estimating with Decimals
4-6 Decimals in Money
4-7 Thousandths
4-8 Metric Units of Length
4-9 Personal References for Metric Length
4-10 Measuring in Millimeters
4-11 Progress Check 4
Unit 5: Big Numbers, Estimation, and Computation
Lesson
5-1 Extended Multiplication Facts
5-2 Multiplication Wrestling
5-3 Estimating Sums
5-4 Estimating Products
5-5 Partial-Products Multiplication (Part 1)
5-6 Partial-Products Multiplication (Part 2)
5-7 Lattice Multiplication
5-8 Big Numbers
5-9 Powers of 10
5-10 Rounding and Reporting Large Numbers
5-11 Comparing Data
5-12 Progress Check 5
Unit 6: Division; Map Reference Frames; Measures of Angles
Lesson
6-1 Multiplication and Division Number Stories
6-2 Strategies for Division
6-3 The Partial-Quotients Division Algorithm, Part 1
6-4 Expressing and Interpreting Remainders
6-5 Rotations and Angles
6-6 Using a Full-Circle Protractor
6-7 The Half-Circle Protractor
6-8 Rectangular Coordinate Grids for Maps
6-9 Global Coordinate Grid System
6-10 The Partial-Quotients Division Algorithm, Part 2
6-11 Progress Check 6
Unit 7: Fractions and Their Uses; Chance and Probability
Lesson
7-1 Review of Basic Fraction Concepts
7-2 Fractions of Sets
7-3 Probabilities When Outcomes Are Equally Likely
7-4 Pattern-Block Fractions
7-5 Fraction and Mixed-Number Addition and Subtraction
7-6 Many Names for Fractions
7-7 Equivalent Fractions
7-8 Fractions and Decimals
7-9 Comparing Fractions
7-10 The ONE for Fractions
7-11 Probability, Fractions, and Spinners
7-12 A Cube-Drop Experiment
7-12a Multiplying Fractions by Whole Numbers
7-13 Progress Check 7
Unit 8: Perimeter and Area
Lesson
8-1 Kitchen Layouts and Perimeter
8-2 Scale Drawings
8-3 Area
8-4 What Is the Area of My Skin?
8-5 Formula for the Area of a Rectangle
8-6 Formula for the Area of a Parallelogram
8-7 Formula for the Area of a Triangle
8-8 Geographical Area Measurements
8-9 Progress Check 8
Unit 9: Fractions, Decimals, and Percents
Lesson
9-1 Fractions, Decimals, and Percents
9-2 Converting "Easy" Fractions to Decimals and Percents
9-3 Using a Calculator to Convert Fractions to Decimals
9-4 Using a Calculator to Rename Fractions as Percents
9-5 Conversions among Fractions, Decimals, and Percents
9-6 Comparing the Results of a Survey
9-7 Comparing Population Data
9-8 Multiplication of Decimals
9-9 Division of Decimals
9-10 Progress Check 9
Unit 10: Reflections and Symmetry
Lesson
10-1 Explorations with a Transparent Mirror
10-2 Finding Lines of Reflection
10-3 Properties of Reflections
10-4 Line Symmetry
10-5 Frieze Patterns
10-6 Positive and Negative Numbers
10-7 Progress Check 10
Unit 11: 3-D Shapes, Weight, Volume, and Capacity
Lesson
11-1 Weight
11-2 Geometric Solids
11-3 Constructing Geometric Solids
11-4 A Volume Exploration
11-5 A Formula for the Volume of Rectangular Prisms
11-6 Subtraction of Positive and Negative Numbers
11-7 Capacity
11-8 Progress Check 11
Unit 12: Rates
Lesson
12-1 Introducing Rates
12-2 Solving Rate Problems
12-3 Converting Between Rates
12-4 Comparison Shopping: Part 1
12-5 Comparison Shopping: Part 2
12-6 World Tour Wrap-Up
12-7 Progress Check 12
Projects
Project
1 Making a Cutaway Globe
2 Using a Magnetic Compass
3 A Carnival Game
4 Making a Quilt
5 Which Soft Drink is the Best Buy?
6 Building and Viewing Structures
7 Numbers, Mayan Style
Algorithm Projects
Algorithm
7 U.S. Traditional Long Division, Part 1
8 U.S. Traditional Long Division, Part 2
### 3rd Edition Lesson List
Expand All | Collapse All
Unit 1: Naming and Constructing Geometric Figures
Lesson
1-1 Introduction to the Student Reference Book
1-2 Points, Line Segments, Lines, and Rays
1-4 Parallelograms
1-5 Polygons
1-6 Drawing Circles with a Compass
1-7 Circle Constructions
1-8 Hexagon and Triangle Constructions
1-9 Progress Check 1
Unit 2: Using Numbers and Organizing Data
Lesson
2-1 A Visit to Washington, D.C.
2-2 Many Names for Numbers
2-3 Place Value in Whole Numbers
2-4 Place Value with a Calculator
2-5 Organizing and Displaying Data
2-6 The Median
2-8 Displaying Data with a Bar Graph
2-9 Subtraction of Multidigit Numbers
2-10 Progress Check 2
Unit 3: Multiplication and Division; Number Sentences and Algebra
Lesson
3-1 "What's My Rule?"
3-2 Multiplication Facts
3-3 Multiplication Facts Practice
3-4 More Multiplication Facts Practice
3-5 Multiplication and Division
3-6 World Tour: Flying to Africa
3-7 Finding Air Distances
3-8 A Guide for Solving Number Stories
3-9 True or False Number Sentences
3-10 Parentheses in Number Sentences
3-11 Open Sentences
3-12 Progress Check 3
Unit 4: Decimals and Their Uses
Lesson
4-1 Decimal Place Value
4-2 Review of Basic Decimal Concepts
4-3 Comparing and Ordering Decimals
4-4 Estimating with Decimals
4-6 Decimals in Money
4-7 Thousandths
4-8 Metric Units of Length
4-9 Personal References for Metric Length
4-10 Measuring in Millimeters
4-11 Progress Check 4
Unit 5: Big Numbers, Estimation, and Computation
Lesson
5-1 Extended Multiplication Facts
5-2 Multiplication Wrestling
5-3 Estimating Sums
5-4 Estimating Products
5-5 Partial-Products Multiplication (Part 1)
5-6 Partial-Products Multiplication (Part 2)
5-7 Lattice Multiplication
5-8 Big Numbers
5-9 Powers of 10
5-10 Rounding and Reporting Large Numbers
5-11 Comparing Data
5-12 Progress Check 5
Unit 6: Division; Map Reference Frames; Measures of Angles
Lesson
6-1 Multiplication and Division Number Stories
6-2 Strategies for Division
6-3 The Partial-Quotients Division Algorithm, Part 1
6-4 Expressing and Interpreting Remainders
6-5 Rotations and Angles
6-6 Using a Full-Circle Protractor
6-7 The Half-Circle Protractor
6-8 Rectangular Coordinate Grids for Maps
6-9 Global Coordinate Grid System
6-10 The Partial-Quotients Division Algorithm, Part 2
6-11 Progress Check 6
Unit 7: Fractions and Their Uses; Chance and Probability
Lesson
7-1 Review of Basic Fraction Concepts
7-2 Fractions of Sets
7-3 Probability When Outcomes Are Equally Likely
7-4 Pattern-Block Fractions
7-6 Many Names for Fractions
7-7 Equivalent Fractions
7-8 Fractions and Decimals
7-9 Comparing Fractions
7-10 The ONE for Fractions
7-11 Probability, Fractions, and Spinners
7-12 A Cube-Drop Experiment
7-13 Progress Check 7
Unit 8: Perimeter and Area
Lesson
8-1 Kitchen Layouts and Perimeter
8-2 Scale Drawings
8-3 Area
8-4 What Is the Area of My Skin?
8-5 Formula for the Area of a Rectangle
8-6 Formula for the Area of a Parallelogram
8-7 Formula for the Area of a Triangle
8-8 Geographical Area Measurements
8-9 Progress Check 8
Unit 9: Fractions, Decimals, and Percents
Lesson
9-1 Fractions, Decimals, and Percents
9-2 Converting "Easy" Fractions to Decimals and Percents
9-3 Using a Calculator to Convert Fractions to Decimals
9-4 Using a Calculator to Rename Fractions as Percents
9-5 Conversions among Fractions, Decimals, and Percents
9-6 Comparing the Results of a Survey
9-7 Comparing Population Data
9-8 Multiplication of Decimals
9-9 Division of Decimals
9-10 Progress Check 9
Unit 10: Reflections and Symmetry
Lesson
10-1 Explorations with a Transparent Mirror
10-2 Finding Lines of Reflection
10-3 Properties of Reflections
10-4 Line Symmetry
10-5 Frieze Patterns
10-6 Positive and Negative Numbers
10-7 Progress Check 10
Unit 11: 3-D Shapes, Weights, Volume, and Capacity
Lesson
11-1 Weight
11-2 Geometric Solids
11-3 Constructing Geometric Solids
11-4 A Volume Exploration
11-5 A Formula for the Volume of Rectangular Prism
11-6 Subtraction of Positive and Negative Numbers
11-7 Capacity and Weight
11-8 Progress Check 11
Unit 12: Rates
Lesson
12-1 Introducing Rates
12-2 Solving Rate Problems
12-3 Converting Between Rates
12-4 Comparison Shopping: Part 1
12-5 Comparison Shopping: Part 2
12-6 World Tour and 50-Facts Test Wrap Ups
12-7 Progress Check 12
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# 1-0 Knapsack Problem – A Hands-on Guide (C++)
The 1-0 knapsack problem; an optimization puzzle famously solved with dynamic programming (dp). This post is merely my take on the problem, which I hope to provide a more hands-on approach.
To get started, try and attempt The Knapsack Problem (KNAPSACK) from SPOJ. Study the problem closely as I will referring to it throughout this guide. I will then explain how the general solution is derived and how dp is applied.
I assume you have known a bit of dp as a prerequisite, though if you haven’t you can check out my beginner friendly hands-on intro: 445A – Boredom – CodeForces Tutorial.
## The General Solution
A dp solution is usually derived from a recursive solution. So let’s start with that.
We define 2 containers: v and c, that contains the all the values of each item and the capacity they consume respectively, starting from index 1. So for example, v[2] and c[2] returns the value and size of the 2nd item.
What are we trying to optimize here? We are trying to maximize the value; the combination of items that yields the highest value. So let us define a function B(i, w) that returns the maximum value given a scope of items (i) and the capacity of the knapsack (w). The solution to KNAPSACK will therefore be B(N, S).
In solving it recursively, imagine we have all N items with us and our knapsack completely empty (current capacity is S), and we consider the items one by one from the last one (the N-th item).
What are the base cases of this problem? When there are no items to put in (i = 0), or when the knapsack cannot contain any items (w = 0).
Before we consider some i-th item, we first need to make sure that it can fit into the knapsack given its capacity, in other words, an i-th item should not be considered if c[i] > w. If this is so, you will consider the maximum value of the the scope of items excluding the i-th item, or B(i-1, w).
So what happens when you can put the item in the knapsack? You have 2 choices: To put it in (take), or not put it in (keep).
1. Keep: you exclude it from the scope of items in which you consider the maximum value, which is again B(i-1, w).
2. Take: you will get the value of the i-th item you select (v[i]), BUT, we should also consider the remaining space after adding the i-th item inside (w-c[i]). When considering items to add here, we need to exclude the item we already added, so the scope of items we consider is i-1. With this remaining space and this scope of items, we also want to get the maximum value. We can do this recursively via B(i-1, w-c[i]).
Choosing between keep and take is as simple as taking the maximum of the 2.
If you piece all this together, you will get the general solution:
$B(i,j) = \begin{cases} 0 \text{ if } i = 0\text{ or }w = 0\\ B(i-1, w) \text{ if } c[i] > w\\ max\Big( B(i-1, w), v[i] + B(i-1, w-c[i]) \Big) \end{cases}$
## Recursive Solution
Short and simple:
Plug it in the judge and you will get a TLE (Time Limit Exceeded).
## Dynamic Programming Solution
We use a 2D array, DP, containing N+1 rows and S+1 columns. The rows map to the scope of i items we consider (the top considers 0 items and the bottom considers all N items). The columns map to capacity (an individual column is denoted as the w-th column) of knapsack left to right from 0 to S. A cell DP[i][w] means “this is the maximum value given i items and capacity of w“. The maximum value for N items and capacity of S is therefore DP[N][S].
We first fill DP with base cases: where i = 0 and where w = 0. This means that the first row and first column are all 0. The order in which we solve this is simple: starting where i = 1 and w = 1, for each row from top to bottom, we fill the cells from left to right.
I will continue on with an example using inputs from Tushar Roy’s YouTube video (you should check it out), but I jumbled the order to prove that ordering of items is not important. Here is the DP array:
Cells shaded in pale blue is when item does not fit in capacity w, or where c[i] > w. In pale green cells we have a choice: keep or take. Notice that with every row, we are adding one more item to consideration, and with every column we increase the capacity by 1.
Here is the code:
## Getting the Selected Items
Some people use an auxiliary Boolean array (often called Keep) that keeps track of whether an item is selected or not. This seems to be an unnecessary occupation of space to me, since you can deduce the selected items from DP itself.
You start from DP[N][S], and from there:
• An item is selected, if the value of the cell directly above it is not equal to the current cell. When this happens, the capacity of the knapsack reduces by the weight of the selected item. With that new capacity you select the next item.
• An item is not selected, if the value of the cell directly above it is equal to the current cell. So we consider the next item by moving up one row; capacity remains unchanged.
This process continues until either there are no more items remaining, or the knapsack is full.
The table below shows the trail of the algorithm as it selects items (item 1 and 4) from the DP array we constructed before:
Below is the recursive function:
void pick(int i, int w)
{
if (i <= 0 || w <= 0) return;
int k = DP[i][w];
if (k != DP[i - 1][w]) {
cout << i << " "; // select!
pick(i - 1, w - c[i]); // capacity decreases
} else {
// move on to next item; capacity no change
pick(i - 1, w);
}
}
See the full implementation of this function in this gist.
## Application
Enough spoon feeding! It is time for you to try out some puzzles on your own. Conveniently UVa grouped a series of 3 questions that are slight variations of the 1-0 knapsack problem. I sort them here in order of difficulty:
1. 10130 – SuperSale (my solution)
2. 990 – Diving for Gold (my solution)
You will need to list down the items you select in this one.
– Beware of the tricky formatting! There shouldn’t be a blank line at the end of your output.
3. 562 – Dividing coins (my solution)
In my solution, there is a tip (short comment block before the main function) you can check out if you just want some pointers to get started.
# Matrix Chain Multiplication with C++ code – PART 3: Extracting the Sequence
In my previous post, I wrote about finding the minimum cost of multiplying a matrix chain using dynamic programming. In this post I build on top of what we have covered to extract the sequence of multiplying the chain itself.
This is part of a 3 part series:
1. Analysis and Design – analyze the problem and deduce a solution.
2. Implementation – implement the algorithm to find the minimum number of operations.
3. Extracting the Sequence – extract the sequence in which the chain is multiplied.
## Extract the Sequence
To do this we keep track of the point at which we split up the chain as prefix and suffix: the point (we define this from the previous post). We do this by storing it in another 2D array of the same size as DP, which we call splits:
int ops = DP[i][k] + DP[k + 1][j] + rc[i] * rc[k + 1] * rc[j + 1];
if (ops < DP[i][j]) {
DP[i][j] = ops;
splits[i][j] = k;
}
Now let us print out the elements of splits in dark blue. I will be reusing the example input from the previous post:
How do we interpret this? Let us start from where we got the main solution; where i = 1 and j = 6. This cell splits[1][6] signifies the last operation of multiplying the matrix chain – in the above table this maps to 3. This means in the last operation, a prefix from a chain $A_1 \ldots A_3$ multiplies with a suffix from a chain $A_{4} \ldots A_6$.
Let us now consider just the prefix. How then is prefix $A_1 \ldots A_3$ multiplied? We find that from splits[1][3], which in this case is 1. Therefore the it is formed by multiplying a prefix $A_1 \ldots A_1$ and a suffix $A_2 \ldots A_3$. In this prefix $A_1 \ldots A_1$ we hit a base case where i = j. Should this happen we return the matrix $A_1$. The suffix $A_2 \ldots A_3$ is split from from splits[2][3], which is 2. This in turns hits 2 base cases and returns the matrix $A_2$ and $A_3$.
If you repeat this recursive process with the top level suffix, you parenthesize the chain as such:
We can generalize a function mult that returns the resultant matrix that has been multiplied in the most optimal way from the matrix chain:
$mult(i, j)=\begin{cases}A_i \text{ if } i = j\\mult(i, k) \cdot mult(k+1, j)\text{ where } k = splits[i][j]\end{cases}$
The main solution is therefore mult(1, 6).
## Implementation
Below is an implementation of this:
Because we don’t have the contents of the matrices, I have the program output the parenthesis instead, along with the order in which it parenthesize them:
./mat-chain-mult-dp-trace < input.txt
multiply 2 and 3
multiply 1 and (2*3)
multiply 4 and 5
multiply (4*5) and 6
multiply (1*(2*3)) and ((4*5)*6)
((1*(2*3))*((4*5)*6))
Note that the numbers here are not integers but indices that map to its respective matrix in the matrix chain.
It is not that hard to convert the above code to multiply matrices. I will leave it as an exercise, should you be interested.
## Application
Why is this matrix chain multiplication problem so important that most computer science undergraduate programs must include it in their syllabus? The same reason applies for most algorithms you learn: some problems are just slight variations of another problem. Understanding one solution to a problem in the marrow of its bones may help you solve another similar problem.
Therefore, I encourage you to challenge yourself on these 2 problems to help solidify your understanding:
You can find my solutions to these problems in my github SPOJ repository.
# Matrix Chain Multiplication with C++ code – PART 2: Implementation
In the last post I explained how we got the general solution:
$B(i,j) = \begin{cases} 0 \text{ if } i = j \\ \displaystyle\min_{k=i}^{j-1} \begin{cases} B(i, k) + B(k+1, j) + r_i \cdot c_k \cdot c_j \end{cases} \end{cases}$
In this post we plug the formula in and watch it run! I will go through the recursive solution ($O(2^n)$), and then the more optimal dynamic programming solution ($O(n^3)$).
This is part of a 3 part series:
1. Analysis and Design – analyze the problem and deduce a solution.
2. Implementation – implement the algorithm to find the minimum number of operations.
3. Extracting the Sequence – extract the sequence in which the chain is multiplied.
## Input
Our input (input.txt), therefore, is the rows and columns of all n matrices:
7
30 35 15 5 10 20 25
The above input meant that there are 7 integers that correspond to the dimensions of 6 matrices:
The input only contains the rows from matrix $A_1$ until $A_6$, followed by the column of matrix $A_6$ . This is because a matrix A and only be multiplied with a matrix B if they are compatible (if $r_A = c_B$). So for brevity sake we do not need to put rows and columns of all matrices (otherwise the sequence will look like 30 35 35 15 15 5 5 10 10 20 20 25).
Notice, that if we place the sequence of integers in an array rcthe row of a matrix $A_i$ is rc[i] and the column is rc[i+1].
## Recursive Implementation
As surprising though it may be, the amount of code is surprisingly little:
I have written the code to follow the formula as close as possible, as such I start the index with 1 instead of 0. You can always change it to save that bit of space, but the code just serves to show that it works:
./mat-chain-mult-recursive < input.txt
15125
Now, if you print the inputs of the function B each time it is being called, you will see a lot of the same inputs are being called again and again. In geeksforgeeks.org’s article on matrix chain multiplication, they took the time to draw out the recursion tree.
## Dynamic Programming: Ordering
So now we have broken the main problem to small recurring subproblems (Overlapping Subproblems), which we can piece together to solve the main problem (Optimal Substructure). With this 2 properties, we can use dynamic programming to drastically reduce the running time of an exponential solution.
However, this problem is tricky to solve since it involves 2 variables i and j. We need a 2D array to keep track of this. In a single dimensional array ordering is trivial; we simply go from left to right, building the solution from the bottom up, but with 2 dimensions ordering becomes tricky. Which cell do we solve first?
Well, we know that:
• the base case is when i = j, then B(i, j) = 0.
• i cannot exceed j, so those areas will need to be grayed out.
• B(1, 6) is the solution.
So with that, let us fill up an array DP based on the inputs we have above:
So we need to find B(1, 6) yes? Well, what does B(1, 6) need? Well, if we unfold the general solution plugging in i as 1 and j as 6 we have 5 selections in which we choose the smallest:
$\begin{cases}k=1: B(1, 1) + B(2, 6) + r_1 \cdot c_1 \cdot c_6 \\k=2: B(1, 2) + B(3, 6) + r_1 \cdot c_2 \cdot c_6 \\k=3: B(1, 3) + B(4, 6) + r_1 \cdot c_3 \cdot c_6 \\k=4: B(1, 4) + B(5, 6) + r_1 \cdot c_4 \cdot c_6 \\k=5: B(1, 5) + B(6, 6) + r_1 \cdot c_5 \cdot c_6 \end{cases}$
To make it easier to visualize what B(1, 6) requires, I distinguish the selection by colour:
What you can see, is that B(1, 6) requires cells to the left and to the bottom of it. Well, we have the cells that represent the base cases solved. Perhaps the next logical step is to solve the next diagonal. This is possible, because all the values of all cells where j i = 1 depends on all cells where ji = 0. Consequently, all values of all cells where ji = 2 depends on all cells where ji = 1 and ji = 0; you can validate this by plugging in the numbers into the formula. Because all cells aside the base case requires cells below and to its left up until it hits the base case, we can continue filling up the cells of the array DP diagonally until we hit the top right cell.
From another perspective, we are finding all 5 pairs of the matrices in the chain and find their minimum cost (choose from 1 each), then we find all 4 triplets of matrices in the chain and find their minimum cost (choose from 2), then we find all 3 quadruplets of matrices in the chain and find their minimum cost (choose from 3), and this repeats until we hit the top right cell; each subsequent pass depending on the pass before it.
So we solve in diagonals in order ji = 0, 1, 2 … n – 1. Once you can sequentially iterate through the array DP in a diagonal fashion, you have everything you need to piece together the solution:
It helps that you calculate the cells by hand to grasp the essence of this algorithm.
## Dynamic Programming: Implementation
So what I did, was design a nested for loop to print the indices in the order I want. Once I do that I simply plug the formula inside, and I have my final solution:
Note that again, the index starts at 1.
Although code count is near indistinguishable, the performance is drastically improved at a negligible expense of space. We generalize that the improved implementation is $O(n^3)$ by counting the number of nested for loops.
## Conclusion
In the 3rd and final part, I will guide you through extracting the actual order in which the chain is multiplied. Check it out: Matrix Chain Multiplication with C++ code – PART 3: Extracting the Sequence.
# Matrix Chain Multiplication with C++ Code – PART 1: Analysis and Design
Ah. The matrix chain multiplication problem. A classic dynamic programming example in the undergraduate computer science world. I will be doing a series consisting of 3 parts:
1. Analysis and Design – analyze the problem and deduce a solution.
2. Implementation – implement the algorithm to find the minimum number of operations.
3. Extracting the Sequence – extract the sequence in which the chain is multiplied.
There’s practically a ton of articles out there that already writes about this problem. I intend to distinguish myself by expressing it in the simplest terms possible with examples, and season it with working C++11 source code. In other words, if you had problem understanding everywhere else, my hopes is that your search ends with this series.
In this first part, there will be no code. At all. Not even pseudocode. I promise some C++11 code in the coming 2 parts, but understanding this section will be important for implementation later.
## Prerequisites
I will assume you should have a bit of background in matrix math. It helps that you do a bit of refreshing if you have forgotten.
I will also assume you have known a bit of what dynamic programming is, as matrix chain multiplication is a slightly more complex example; you keep track of a 2D array instead of 1D array. I’ve written a beginner level hands on introduction to dynamic programming in a form of solving a problem which you can check out: 445A – Boredom – CodeForces Tutorial.
## Understanding the Problem
Given a chain of matrices, we want to multiply them. We cannot change the order of the matrices as that would change the result or make it incompatible, therefore we say matrix multiplication is not commutative. In addition, regardless of how you parenthesize (e.g. whether the sequence as multiplied as $A\cdot \left ( B\cdot C \right )$ or $\left ( A\cdot B \right )\cdot C$ ) the chain it will give you the same answer, thus we say it is associative.
How we parenthesize the matrices also determine how many operations will be performed. For a multiplication chain of integers, parenthesizing them does not affect the number of operations performed, but for matrices the change is significant. For example, given the matrices with the following dimensions:
If we define a row of a matrix as $r_A$ and the column as $c_A$, then the total number of multiplications required to multiply A and B is $r_A \cdot c_A \cdot c_{B}$.
• $A\cdot \left ( B\cdot C \right )$ -> 2*20*1 + 2*1*10 = 60 multiplications
• $\left ( A\cdot B \right )\cdot C$ -> 20*1*10 + 2*20*10 = 600 multiplications
The 1st case is 10 times faster!
Therefore, we want to figure out how to parenthesize the matrix chain such that we minimize the required number of operations.
However, to keep things simple, we will focus on the minimum number of operations. We will eventually find the sequence itself by building on top on this.
## Formulation
To put in formal terms, given a sequence of n matrices $A_1, A_2, A_3 \ldots A_n$ we wish to parenthesize the product of this sequence so as to minimize the number of operations required.
## Finding the Minimum Number of Operations
Suppose we have a function B(i, j) that computes the minimum number of required operations for multiplying a chain of matrices from matrix i to matrix j. The solution to the main problem would be B(1, n).
A standard technique to in solving a dynamic programming problem is usually to ask, “What is the last operation?”. Here, we ask what is the minimum total cost of the last operation? Regardless how the chain is parenthesized, there will always be 2 matrices remaining. The cost, is the minimum cost of some prefix + minimum cost of some suffix + the number of operations to multiply them. The cost of prefix and suffix in turn, is derived from the minimum number of operations that deduces them.
Because each matrix can only multiply with its adjacent matrix, a prefix can only start from $A_1$ to some matrix $A_k$, and a suffix can only start from $A_{k+1}$ to $A_n$, split at some index k. The cost in the last operation for the prefix is then B(1, k) and the cost of the suffix is B(k+1, n)k therefore ranges from 1 until n-1 in the last operation, and within a range i to j, k ranges from i to j-1. We choose k by selecting the minimum cost of all the possible costs within a range i to j.
The resultant dimensions from multiplying 2 matrices are important to finding the cost. In a sequence of matrices $A_i \ldots A_j$ If we split at a point k, the resultant dimensions of the prefix is $r_i \times c_k$ and the suffix is $r_{k+1} \times c_j$. The cost of multiplying these 2 matrices are therefore $r_i \cdot c_k \cdot c_j$.
## Running Through Possibilities
Let us consider the base case: When there is only 1 matrix. The prefix = suffix, and there are no operations performed, so the cost = 0.
Now what if there are only 2 matrices? There are no other possible combinations to compare with, so the minimum number of operations is the number of operations to multiply them. There is no cost to the prefix or suffix, because the cost for 1 matrix alone is 0.
3 matrices onwards is gets a little more challenging. When the chain has only 3 matrices, you have to choose between $\left ( A_1\cdot A_2 \right )\cdot A_3$ and $A_1\cdot \left ( A_2\cdot A_3\right )$. How we choose between the 2 is determined by cost of either multiplying $A_1$ and $A_2$ first, then take the resulting matrix (prefix) and multiply by $A_3$ (suffix), OR, multiplying $A_2$ and $A_3$ first, then multiplying that (suffix) by $A_1$ (prefix). Where we decide to make that split, either at $A_1$ or $A_2$, we determined by their cost. The matrix in which we choose we call it $A_k$, at a point k.
The minimum cost with 3 matrices, B(1, 3) is therefore $B(1, k) + B(k+1, 3) + r_1 \cdot c_k \cdot c_3$, and we choose between $B(1, 1) + B(2, 3) + r_1 \cdot c_1 \cdot c_3$ and $B(1, 2) + B(3, 3) + r_1 \cdot c_2 \cdot c_3$. You can continue doing this for 4 or 5 or 6 matrices until you start to see a pattern. Notice you can apply the same procedure to find B(1, 3) to find B(1, 2) – there is only 1 possible value for k for B(1, 2).
## The General Solution
So in a range i to j, we select from j – i possibilities, from i until j – 1. Those possibilities in turn call B recursively until it hits the base case where i = j. The general solution is therefore:
$B(i,j) = \begin{cases} 0 \text{ if } i = j \\ \displaystyle\min_{k=i}^{j-1} \begin{cases} B(i, k) + B(k+1, j) + r_i \cdot c_k \cdot c_j \end{cases} \end{cases}$
And the solution to the main problem would be B(1, n).
## Conclusion
This is all for the first part. Next post I will cover implementation: Matrix Chain Multiplication with C++ code – PART 2: Implementation.
# 445A – Boredom – CodeForces Tutorial
Inasmuch as this is a CodeForces tutorial, I’d also want to use Boredom as an introduction to dynamic programming (dp). I will also compare the dp solution to its alternate recursive implementation. This tutorial also caters to people who can’t seem to wrap their head around the solution or tutorial from CodeForces.
I won’t be explaining the problem itself. Should be pretty straightforward.
## Gotchas
The maximum points can exceed the limit for the 32-bit data type int. In some of the test cases that value can be as high as 9999899998 (the maximum capacity for 32-bit int is 2147483647). The maximum value for the n integers may be up to 10^5, but they can total up to a lot higher than that.
That is why you will see C/C++ solutions out there the data type long long int:
typedef long long int bigInt;
## Solution Overview
You can try to understand the succinct tutorial from CodeForces, if you haven’t already looked into it.
The sequence of n numbers needed to be grouped together and kept count, because when selecting a single number, it makes sense to repeatedly select it (thereby adding it to your points) until it doesn’t exist in the sequence. We will keep track of this count in array c, where c[i] contains the number of occurrences of i.
Let f be a function that calculates the maximum number of points to a value up to i. Now there will be 2 base cases:
1. If i = 0, then f(0) = 0
2. If i = 1, then f(0) = c[1]
I hope the 2nd point is clear. Up to 1 the maximum number of points you can get by selecting 1 is the number of occurrences of 1 in the array c.
When i >= 2, you have 2 choices in deciding which will be f(i):
1. Select f(– 2), in which you can also add with i * c[i].
2. Select f(i – 1), but in doing so you can’t choose to add i * c[i] because of the condition that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence.
Choosing between 1 and 2 is as simple as choosing which is largest.
Now let m be the largest number in the sequence. The solution is therefore f(m). The CodeForces tutorial writes f(n), but I doubt n here implies the number of elements in the sequence. To avoid confusion, I used m instead.
Most implementations I see online simply put f(100000).
It helps that you trace the algorithm step by step on paper, should it not be apparent to you how it works. For example for input:
10
10 5 8 9 5 6 8 7 2 8
You can track the values of i, c[i], and f(i):
i c[i] i*c[i] f(i)
0 0 0 0
1 0 0 0
2 1 2 2
3 0 0 2
4 0 0 2
5 2 10 12
6 1 6 12
7 1 7 19
8 3 24 36
9 1 9 36
10 1 10 46
Do not undermine this method; learning how to trace algorithms in a systematic way can be very helpful to help you understand it.
## Recursive Implementation
Based on the solution as detailed above, here the recursive implementation.
#include <algorithm>
#include <iostream>
#include <vector>
typedef long long int bigInt;
using namespace std;
vector<bigInt> c(100001, 0);
bigInt f(const int i)
{
if (i == 0) return 0;
if (i == 1) return c[1];
return max(f(i - 1), f(i - 2) + i*c[i]);
}
int main()
{
int n, x, m = 0;
cin >> n;
while(n--) {
cin >> x;
c[x]++;
m = max(m, x);
}
cout << f(m);
}
Don’t submit this solution; it will fail at test 9 (time limit exceeded).
That’s it! In 30 lines of code you can solve this problem! The issue? Well, try and increase the values in the sequence (not the number of elements in the sequence. This doesn’t affect the growth of the function). Even a value as small as 30 and I had to wait a few seconds for it to compute.
## Analysis of Recursive Implementation
We can define the performance of this recursive implementation by the number of recursive calls.
Let us trace the recursive calls of the function f in a recursion tree. I used paler colors to signify the depth of the recursion tree; the deeper it gets, the paler it gets:
That is a total of 9 recursive calls. How about f(5)?
5+9+1 (including itself) = 15 function calls.
If you observe for awhile, you will be able to deduce a pattern. Let q be a function that finds the number of function calls for f given i. Then we write q as: q(i)=q(i-1) + q(i-2) + 1, for all i >= 2; q(0)=1; q(1)=1. If you plot this function as a graph, you will get this:
So time complexity is O(2^n); Implementation scales exponentially.
By the time i reaches 18, q(18) = 8361. When i = 30, the number of function calls becomes 2692537! This would explain that bit of lag earlier. Just imagine i going as big as 10000. It is no surprise then, that when this happens, the program crashes due to stack overflow:
## Dynamic Programming Implementation
Dynamic programming, in essence, is an optimization technique that dramatically cuts down computational time but taking out a bit of space to store calculated answers to overlapping subproblems. The overlapping subproblems here are the calls to f, which for example in the recursion tree f(4) we can see that f(1) is called 3 times and f(2) and f(0) is called 2 times.
Furthermore, we can take the solutions the subproblems to compute the solution of the main problem. For example f(5) can be determined by f(4) and f(3), and f(4) can be determined by f(3) and f(2), and so on. This property is called Optimal Substructure.
To recap, the 2 properties that must exist for a problem to be made solvable by dp:
Let’s put theory to practice.
We can reduce the time complexity from O(2^n) to O(n) by keeping an array DP of the same size as cDP will store the calculated values of f for all i. The solution, therefore, is DP[m], since it contains f(m). Below is the dynamic programming solution:
#include <algorithm>
#include <iostream>
#include <vector>
typedef long long int bigInt;
using namespace std;
int main()
{
const int MAX_N = 100001;
int n, x;
vector<bigInt> dp(MAX_N, 0);
vector<bigInt> c(MAX_N, 0);
cin >> n;
while(n--) {
cin >> x;
c[x]++;
}
dp[0] = 0;
dp[1] = c[1];
for (int i = 2; i < MAX_N; i++) {
dp[i] = max(dp[i - 1], dp[i - 2] + i*c[i]);
}
cout << dp[MAX_N - 1];
}
In the dp solution, f(i) is computed from 0 to 10000 (bottom to top), whereas in the recursive solution f(i) is computed from m to 0 (top to bottom).
Notice I didn’t bother keep track the largest value in the sequence (variable m). You can have that bit of optimization if you use it, but it isn’t necessary for this problem. In essence we have f (the block of code inside that for loop) called a constant 10000 times each time this program is called, and this works granted the largest value in the sequence doesn’t exceed 10000.
What is fascinating about this implementation is that it solves the problem with the same amount of code! Ladies and gentlemen, the power of algorithms!
# 103 – Stacking Boxes (Dynamic Programming Method) – UVa Tutorial
In my previous post I wrote about solving the Stacking Boxes problem using graphs. In this post I will write about a simpler method that utilizes dynamic programming that solves the same problem with half the amount of code.
I will assume you have read my previous post on using graphs, though you haven’t you can check it out here: 103 – Stacking Boxes (Graph Method) – UVa Tutorial. I will refer longest increasing subsequence as LIS and dynamic programming as dp.
## On the O(n^2) LIS Dynamic Programming Solution
In using dynamic programming there are 2 variants of the algorithm: one solves in O(n^2) and another in O(n*log n), the later of which replaces the inner loop in the former algorithm with a binary search using an algorithm called Patience Sorting. I will be using the simpler, quadratic time variant. I won’t attempt to explain the algorithm in detail, but there are plenty of resources out there to help you wrap your head round it:
1. Tushar Roy’s step by step of the algorithm (youtube video)
2. Geeksforgeeks article
3. Bo Qian’s youtube video (with C++ code)
4. Article on Algorithmist (succinct, though can be tough to understand).
The first 2 resource will only find the length of the LIS, though that process is the heart of the algorithm. Once you can find that, getting the sequence itself is not that hard.
## A Generic LIS Implementation
The code here is based off a stackoverflow answer.
template<typename T>
deque<T> find_lis(const vector<T> &a,
function<bool(const T &, const T &)> comp = [&](const T &a, const T &b)->bool { return a < b; })
{
int maxLength = 1, bestEnd = 0;
vector<int> DP(a.size(), 1);
vector<int> prev(a.size(), -1);
for (int i = 1; i < a.size(); i++) { for (int j = i - 1; j >= 0; j--) {
if (DP[j] + 1 > DP[i] && comp(a[j], a[i])) {
DP[i] = DP[j] + 1;
prev[i] = j;
}
}
if (DP[i] > maxLength) {
bestEnd = i;
maxLength = DP[i];
}
}
deque<T> lis;
for (int i = bestEnd; i != -1; i = prev[i]) {
lis.push_front(a[i]);
}
return lis;
}
Using the function itself is pretty straightforward. You pass a vector that contains your sequence and it spits out the LIS as a deque. Here’s a github gist that test drives this function. You also have an option to modify the comparator, which in the gist is used to change the LIS algorithm to find the longest decreasing subsequence.
## Solution Explanation
Once you understood what the LIS algorithm is doing, the rest is pretty much a no brainer. Once you sort the boxes and its dimensions, you can find the LIS is about 5 lines of code:
auto comp = [&](const Box *a, const Box *b)->bool { return canFit(a, b); };
auto lis = find_lis<Box*>(boxes, comp);
cout << lis.size() << endl;
for (auto b : lis) cout << b->id << " ";
cout << endl;
The only thing you need to customize here is the comparator; you will get the LIS based on whether the boxes will fit into each other. I handle this will a simple lambda function comp which I input inside find_lis. The canFit function is the exact same function in the graph method solution.
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# Sphere : How Come 1/3 Ratio Applied?
• 11 Replies
• 771 Views
#### Danang
• 4271
• Everything will be "Phew" in its time :')
##### Sphere : How Come 1/3 Ratio Applied?
« on: April 16, 2020, 02:33:35 PM »
Is it that easy to figure out the volume of a sphere by the constant 1/3 ?
It surely fit to flat base area. But a fraction of sphere by definition is a ROUND base area with certain height.
If you treat the same way between flat base area and round base area, there will be miscalculation. Because the volume will be less 1/3 base area times height.
As you know, a pyramid has multiple heights depending on where the position is. Those are the minimum height, say, h=1 to h>1 as the pointer goes to the side points.
If transforemed to be a cone it will be like this:
Still, the height has multiple sizes.
Now let's compare both objects with a fraction of a sphere:
The heights are Constant Everywhere, which means the calculation of final volume will be LESS than 1/3 ratio of base area times height.
The multiple heights of pyramid will be substracted to be the flat minimum height >> h=1 or r=1.
Pyramid's multiple heights becomes sphere's flat height AKA radius which all becomes minimum.
How come you still use the constant of 1/3 ? Isn't a fraction of rounded sphere is ensmalled version of a pyramid?
Now we work on pi, but you can prove me wrong if you have evidence. Can you?
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#### Danang
• 4271
• Everything will be "Phew" in its time :')
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #1 on: April 17, 2020, 04:04:33 AM »
0:51
Use your ruler. It's more than 1/3.
Even the cylinder is not full after getting poured with sphere's water.
To apply phew legacy, i.e. Sphere Volume = (1/2 phew) ^3 × r^3
Now >> Sphere Volume = (1/2 pi) ^3 × r^3 = 3.87578 r^3.
The volume ratio of Sphere : Cylinder = 3.87578 : 6.28318 = 0.61685 : 1.
Not 0.66666 : 1
« Last Edit: April 17, 2020, 08:52:20 AM by Danang »
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#### Danang
• 4271
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##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #2 on: April 17, 2020, 08:47:09 AM »
No reply since yesterday...
case closed. 👌
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#### Danang
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##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #3 on: April 17, 2020, 08:50:20 AM »
With just ONE PUNCH, Rabinoz, JackBlack, MicroBeta, Sokarul, CuriouserAndCuriouser got KO!!! 😅😅😅
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?
#### JackBlack
• 15256
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #4 on: April 22, 2020, 03:17:25 PM »
No reply since yesterday...
case closed. 👌
Sure, case closed.
All you have are pathetic ravings of a mad man.
How about you try to actually provide the mathematically derivation of the volume of a sphere, instead of baseless claims like "it must be less than this".
I thought I had already provided it, showing how if the area of a circle is pi*r^2 (which it is) then the volume of a sphere MUST be (4/3)*pi*r^3
#### Danang
• 4271
• Everything will be "Phew" in its time :')
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #5 on: April 23, 2020, 02:47:48 AM »
your math should match with real experiment.
So far no experiment indicate 1/3 thing.
go to 0:40
use a ruler to measure the height of the empty space of the cylinder after it's poured with the 1st cone's water.
If you expect the empty space will be left a half after the second cone's water getting poured, you'll be wrong.
And the way the 1st and the 2nd red lines be drawn is not even. The 1st line is under the surface, the second is above the surface.
This experiment is not transparent by the way. They didn't show the entire cylinder.
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#### Danang
• 4271
• Everything will be "Phew" in its time :')
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #6 on: April 23, 2020, 02:50:34 AM »
"Maths Without Experiment is Blind" -- Albert Einstein (2020)
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#### JackBlack
• 15256
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #7 on: April 23, 2020, 05:14:17 AM »
your math should match with real experiment.
We have been over this before.
With pure math, you can elliminate errors.
With real experiments, there will always be some error.
If your expected result was 0.5, and your actual result was 0.51, do they disagree? That depends upon the level of error in the experiment.
Is your observed value actually 0.51, or do the errors make it 0.49?
It is bad enough to rely upon measurements like from a ruler, but then you start pouring water around...
Have you made sure every single molecule of water was transferred? Did you make sure none evaporated?
Minor errors in experiment cannot refute the pure math.
#### Danang
• 4271
• Everything will be "Phew" in its time :')
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #8 on: April 23, 2020, 05:34:56 PM »
Unfortunately the nature of RECTANGLE is DIFFERENT from the nature of CURVE.
1/3 constant is related with rectangled objects.
And for the sphere volume there must be three arc constants in multiplication, i.e. anything related to pi pi, or pi/2, or pi/4.
The three dimentions of a sphere are definitely not straight lines. Those are curved.
r=1, if a cube has the volume of 2r × 2r × 2r, a sphere will be [(pi/4)×2r] × [(pi/4)×2r] × [(pi/4)×2r] = [(pi/4)×2r]^3 = [(pi/2)×r]^3 = (pi/2)^3 × r^3
So if compared to a cube's volume which is 2^3 = 8, sphere volume is a little bit under 50%.
Goto the video
Use a ruler.
The blue water's volume/height is a bit more than 50% of the cube's volume/height.
Such (repeated) evidence confims that the sphere's volume is NOT a bit more than 50% of cube's volume.
It's a bit less than 50% of cube's volume.
« Last Edit: April 23, 2020, 05:37:13 PM by Danang »
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#### Bullwinkle
• The Elder Ones
• 19546
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##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #9 on: April 23, 2020, 05:52:06 PM »
1+1 not always 2
?
#### JackBlack
• 15256
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #10 on: April 24, 2020, 05:05:34 AM »
1/3 constant is related with rectangled objects.
Nope. 1/3 is for any object where:
There is a line.
Perpindicular to this line at any given point there is a shape.
This shape remains similar throughout the length of the line.
This shapes changes by a constant linear scale as it moves along the line, reaching a size of 0 at one end of the line.
This produces a 3D shape known as a pyramid or cone.
For a sphere you have (4/3)*pi, i.e. the volume is (4/3)*r^3. If you compare that to a cube, with a side length of 2*r, you will have a volume of 8*r^3 for the cube. This is a ratio of pi/6, or roughly 52%.
Just because a sphere is 3D doesn't mean you have to have pi in it 3 times.
This is just like area and length. They all only have it in there 1 time.
If you want more, then try higher dimensional spheres.
Such (repeated) evidence confims that the sphere's volume is NOT a bit more than 50% of cube's volume.
It's a bit less than 50% of cube's volume.
No, it doesn't. Firstly, you don't have repeated experiments, you have 1.
Even if you did, you are still completely ignoring any possible experiment errors involved.
You are yet to show your value contradicts known math.
In order to do so you actually need to know the experimental errors and get them low enough such that the known value lies outside.
This also includes any errors in the measurement.
The best section of the video has the cube (the relevant section of it) a mere ~155 px high, with each edge stretching across a few px.
That means each px is roughly 0.64 pp and thus each edge would be rouglhy 1.3 pp.
So when you combine 2 edges you are at an error of over 2 pp.
Far too much for you to tell the difference between 50% and pi/6.
You then have the systematic errors of the experiment (which do act against each other).
The ball must always be smaller than the cube or it wont fit. This is also seen from it wobbling. If the ball is just 4.9 cm instead of 5, your percentage drops to 0.49%, just below 50.
When you remove the ball, some water may come with it, making the ball appear larger.
Also, the ball has a mount on top which also displaces water making it larger.
You then have the question of if it is actually a cube or if it is just a rectangular prism, and if it is a sphere or an ellipsoid.
So what you actually have is an experiment with many significant limitations allowing you to see that the volume of a sphere is roughly 52%, in accordance with known math.
Meanwhile, like I said before, the pure math doesn't have this problem. It doesn't suffer from experimental error and instead provides the exact value (expressed in terms of pi) of the ratio.
#### MicroBeta
• 2460
##### Re: Sphere : How Come 1/3 Ratio Applied?
« Reply #11 on: May 09, 2020, 02:57:23 AM »
Here. I've derived the equation for the volume of a sphere for you. It shows where it all comes from.
It seems my original link didn't work so here is another one.
https://www.dropbox.com/s/nh2ly1g49fbgzk2/sphere.pdf?dl=0
Mike
Since it costs 1.82¢ to produce a penny, putting in your 2¢ if really worth 3.64¢.
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Prealgebra
# 10.6Introduction to Factoring Polynomials
Prealgebra10.6 Introduction to Factoring Polynomials
## Learning Objectives
By the end of this section, you will be able to:
• Find the greatest common factor of two or more expressions
• Factor the greatest common factor from a polynomial
## Be Prepared 10.6
Before you get started, take this readiness quiz.
1. Factor $5656$ into primes.
If you missed this problem, review Example 2.48.
2. Multiply: $−3(6a+11).−3(6a+11).$
If you missed this problem, review Example 7.25.
3. Multiply: $4x2(x2+3x−1).4x2(x2+3x−1).$
If you missed this problem, review Example 10.32.
## Find the Greatest Common Factor of Two or More Expressions
Earlier we multiplied factors together to get a product. Now, we will be reversing this process; we will start with a product and then break it down into its factors. Splitting a product into factors is called factoring.
In The Language of Algebra we factored numbers to find the least common multiple (LCM) of two or more numbers. Now we will factor expressions and find the greatest common factor of two or more expressions. The method we use is similar to what we used to find the LCM.
## Greatest Common Factor
The greatest common factor (GCF) of two or more expressions is the largest expression that is a factor of all the expressions.
First we will find the greatest common factor of two numbers.
## Example 10.80
Find the greatest common factor of $2424$ and $36.36.$
## Try It 10.159
Find the greatest common factor: $54,36.54,36.$
## Try It 10.160
Find the greatest common factor: $48,80.48,80.$
In the previous example, we found the greatest common factor of constants. The greatest common factor of an algebraic expression can contain variables raised to powers along with coefficients. We summarize the steps we use to find the greatest common factor.
## How To
### Find the greatest common factor.
1. Step 1. Factor each coefficient into primes. Write all variables with exponents in expanded form.
2. Step 2. List all factors—matching common factors in a column. In each column, circle the common factors.
3. Step 3. Bring down the common factors that all expressions share.
4. Step 4. Multiply the factors.
## Example 10.81
Find the greatest common factor of $5xand15.5xand15.$
## Try It 10.161
Find the greatest common factor: $7y,14.7y,14.$
## Try It 10.162
Find the greatest common factor: $22,11m.22,11m.$
In the examples so far, the greatest common factor was a constant. In the next two examples we will get variables in the greatest common factor.
## Example 10.82
Find the greatest common factor of $12x212x2$ and $18x3.18x3.$
## Try It 10.163
Find the greatest common factor: $16x2,24x3.16x2,24x3.$
## Try It 10.164
Find the greatest common factor: $27y3,18y4.27y3,18y4.$
## Example 10.83
Find the greatest common factor of $14x3,8x2,10x.14x3,8x2,10x.$
## Try It 10.165
Find the greatest common factor: $21x3,9x2,15x.21x3,9x2,15x.$
## Try It 10.166
Find the greatest common factor: $25m4,35m3,20m2.25m4,35m3,20m2.$
## Factor the Greatest Common Factor from a Polynomial
Just like in arithmetic, where it is sometimes useful to represent a number in factored form (for example, $1212$ as $2·6or3·4),2·6or3·4),$ in algebra it can be useful to represent a polynomial in factored form. One way to do this is by finding the greatest common factor of all the terms. Remember that you can multiply a polynomial by a monomial as follows:
$2(x + 7)factors 2·x + 2·7 2x + 14product 2(x + 7)factors 2·x + 2·7 2x + 14product$
Here, we will start with a product, like $2x+14,2x+14,$ and end with its factors, $2(x+7).2(x+7).$ To do this we apply the Distributive Property “in reverse”.
## Distributive Property
If $a,b,ca,b,c$ are real numbers, then
$a(b+c)=ab+acandab+ac=a(b+c)a(b+c)=ab+acandab+ac=a(b+c)$
The form on the left is used to multiply. The form on the right is used to factor.
So how do we use the Distributive Property to factor a polynomial? We find the GCF of all the terms and write the polynomial as a product!
## Example 10.84
Factor: $2x+14.2x+14.$
## Try It 10.167
Factor: $4x+12.4x+12.$
## Try It 10.168
Factor: $6a+24.6a+24.$
Notice that in Example 10.84, we used the word factor as both a noun and a verb:
$Noun7is a factor of14Verbfactor2from2x+14Noun7is a factor of14Verbfactor2from2x+14$
## How To
### Factor the greatest common factor from a polynomial.
1. Step 1. Find the GCF of all the terms of the polynomial.
2. Step 2. Rewrite each term as a product using the GCF.
3. Step 3. Use the Distributive Property ‘in reverse’ to factor the expression.
4. Step 4. Check by multiplying the factors.
## Example 10.85
Factor: $3a+3.3a+3.$
## Try It 10.169
Factor: $9a+9.9a+9.$
## Try It 10.170
Factor: $11x+11.11x+11.$
The expressions in the next example have several factors in common. Remember to write the GCF as the product of all the common factors.
## Example 10.86
Factor: $12x−60.12x−60.$
## Try It 10.171
Factor: $11x−44.11x−44.$
## Try It 10.172
Factor: $13y−52.13y−52.$
Now we’ll factor the greatest common factor from a trinomial. We start by finding the GCF of all three terms.
## Example 10.87
Factor: $3y2+6y+9.3y2+6y+9.$
## Try It 10.173
Factor: $4y2+8y+12.4y2+8y+12.$
## Try It 10.174
Factor: $6x2+42x−12.6x2+42x−12.$
In the next example, we factor a variable from a binomial.
## Example 10.88
Factor: $6x2+5x.6x2+5x.$
## Try It 10.175
Factor: $9x2+7x.9x2+7x.$
## Try It 10.176
Factor: $5a2−12a.5a2−12a.$
When there are several common factors, as we’ll see in the next two examples, good organization and neat work helps!
## Example 10.89
Factor: $4x3−20x2.4x3−20x2.$
## Try It 10.177
Factor: $2x3+12x2.2x3+12x2.$
## Try It 10.178
Factor: $6y3−15y2.6y3−15y2.$
## Example 10.90
Factor: $21y2+35y.21y2+35y.$
## Try It 10.179
Factor: $18y2+63y.18y2+63y.$
## Try It 10.180
Factor: $32k2+56k.32k2+56k.$
## Example 10.91
Factor: $14x3+8x2−10x.14x3+8x2−10x.$
## Try It 10.181
Factor: $18y3−6y2−24y.18y3−6y2−24y.$
## Try It 10.182
Factor: $16x3+8x2−12x.16x3+8x2−12x.$
When the leading coefficient, the coefficient of the first term, is negative, we factor the negative out as part of the GCF.
## Example 10.92
Factor: $−9y−27.−9y−27.$
## Try It 10.183
Factor: $−5y−35.−5y−35.$
## Try It 10.184
Factor: $−16z−56.−16z−56.$
Pay close attention to the signs of the terms in the next example.
## Example 10.93
Factor: $−4a2+16a.−4a2+16a.$
## Try It 10.185
Factor: $−7a2+21a.−7a2+21a.$
## Try It 10.186
Factor: $−6x2+x.−6x2+x.$
## Section 10.6 Exercises
### Practice Makes Perfect
Find the Greatest Common Factor of Two or More Expressions
In the following exercises, find the greatest common factor.
422.
$40 , 56 40 , 56$
423.
$45 , 75 45 , 75$
424.
$72 , 162 72 , 162$
425.
$150 , 275 150 , 275$
426.
$3 x , 12 3 x , 12$
427.
$4 y , 28 4 y , 28$
428.
$10 a , 50 10 a , 50$
429.
$5 b , 30 5 b , 30$
430.
$16 y , 24 y 2 16 y , 24 y 2$
431.
$9 x , 15 x 2 9 x , 15 x 2$
432.
$18 m 3 , 36 m 2 18 m 3 , 36 m 2$
433.
$12 p 4 , 48 p 3 12 p 4 , 48 p 3$
434.
$10 x , 25 x 2 , 15 x 3 10 x , 25 x 2 , 15 x 3$
435.
$18 a , 6 a 2 , 22 a 3 18 a , 6 a 2 , 22 a 3$
436.
$24 u , 6 u 2 , 30 u 3 24 u , 6 u 2 , 30 u 3$
437.
$40 y , 10 y 2 , 90 y 3 40 y , 10 y 2 , 90 y 3$
438.
$15 a 4 , 9 a 5 , 21 a 6 15 a 4 , 9 a 5 , 21 a 6$
439.
$35 x 3 , 10 x 4 , 5 x 5 35 x 3 , 10 x 4 , 5 x 5$
440.
$27 y 2 , 45 y 3 , 9 y 4 27 y 2 , 45 y 3 , 9 y 4$
441.
$14 b 2 , 35 b 3 , 63 b 4 14 b 2 , 35 b 3 , 63 b 4$
Factor the Greatest Common Factor from a Polynomial
In the following exercises, factor the greatest common factor from each polynomial.
442.
$2 x + 8 2 x + 8$
443.
$5 y + 15 5 y + 15$
444.
$3 a − 24 3 a − 24$
445.
$4 b − 20 4 b − 20$
446.
$9 y − 9 9 y − 9$
447.
$7 x − 7 7 x − 7$
448.
$5 m 2 + 20 m + 35 5 m 2 + 20 m + 35$
449.
$3 n 2 + 21 n + 12 3 n 2 + 21 n + 12$
450.
$8 p 2 + 32 p + 48 8 p 2 + 32 p + 48$
451.
$6 q 2 + 30 q + 42 6 q 2 + 30 q + 42$
452.
$8 q 2 + 15 q 8 q 2 + 15 q$
453.
$9 c 2 + 22 c 9 c 2 + 22 c$
454.
$13 k 2 + 5 k 13 k 2 + 5 k$
455.
$17 x 2 + 7 x 17 x 2 + 7 x$
456.
$5 c 2 + 9 c 5 c 2 + 9 c$
457.
$4 q 2 + 7 q 4 q 2 + 7 q$
458.
$5 p 2 + 25 p 5 p 2 + 25 p$
459.
$3 r 2 + 27 r 3 r 2 + 27 r$
460.
$24 q 2 − 12 q 24 q 2 − 12 q$
461.
$30 u 2 − 10 u 30 u 2 − 10 u$
462.
$y z + 4 z y z + 4 z$
463.
$a b + 8 b a b + 8 b$
464.
$60 x − 6 x 3 60 x − 6 x 3$
465.
$55 y − 11 y 4 55 y − 11 y 4$
466.
$48 r 4 − 12 r 3 48 r 4 − 12 r 3$
467.
$45 c 3 − 15 c 2 45 c 3 − 15 c 2$
468.
$4 a 3 − 4 a b 2 4 a 3 − 4 a b 2$
469.
$6 c 3 − 6 c d 2 6 c 3 − 6 c d 2$
470.
$30 u 3 + 80 u 2 30 u 3 + 80 u 2$
471.
$48 x 3 + 72 x 2 48 x 3 + 72 x 2$
472.
$120 y 6 + 48 y 4 120 y 6 + 48 y 4$
473.
$144 a 6 + 90 a 3 144 a 6 + 90 a 3$
474.
$4 q 2 + 24 q + 28 4 q 2 + 24 q + 28$
475.
$10 y 2 + 50 y + 40 10 y 2 + 50 y + 40$
476.
$15 z 2 − 30 z − 90 15 z 2 − 30 z − 90$
477.
$12 u 2 − 36 u − 108 12 u 2 − 36 u − 108$
478.
$3 a 4 − 24 a 3 + 18 a 2 3 a 4 − 24 a 3 + 18 a 2$
479.
$5 p 4 − 20 p 3 − 15 p 2 5 p 4 − 20 p 3 − 15 p 2$
480.
$11 x 6 + 44 x 5 − 121 x 4 11 x 6 + 44 x 5 − 121 x 4$
481.
$8 c 5 + 40 c 4 − 56 c 3 8 c 5 + 40 c 4 − 56 c 3$
482.
$−3 n − 24 −3 n − 24$
483.
$−7 p − 84 −7 p − 84$
484.
$−15 a 2 − 40 a −15 a 2 − 40 a$
485.
$−18 b 2 − 66 b −18 b 2 − 66 b$
486.
$−10 y 3 + 60 y 2 −10 y 3 + 60 y 2$
487.
$−8 a 3 + 32 a 2 −8 a 3 + 32 a 2$
488.
$−4 u 5 + 56 u 3 −4 u 5 + 56 u 3$
489.
$−9 b 5 + 63 b 3 −9 b 5 + 63 b 3$
### Everyday Math
490.
Revenue A manufacturer of microwave ovens has found that the revenue received from selling microwaves a cost of $pp$ dollars each is given by the polynomial $−5p2+150p.−5p2+150p.$ Factor the greatest common factor from this polynomial.
491.
Height of a baseball The height of a baseball hit with velocity $8080$ feet/second at $44$ feet above ground level is $−16t2+80t+4,−16t2+80t+4,$ with $t=t=$ the number of seconds since it was hit. Factor the greatest common factor from this polynomial.
### Writing Exercises
492.
The greatest common factor of $3636$ and $6060$ is $12.12.$ Explain what this means.
493.
What is the GCF of $y4y4$, $y5y5$, and $y10y10$? Write a general rule that tells how to find the GCF of $yaya$, $ybyb$, and $ycyc$.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?
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The inverse function exists only for the bijective function that means the function should be one-one and onto. Explore this in the following exercise. Then by the definition we take such that and we would like to show that , right? denotes composition).. l is a left inverse of f if l . The point $\left(3,1\right)$ tells us that $g\left(3\right)=1$. Find the inverse of the function $f\left(x\right)=\frac{2}{x - 3}+4\\$. Not all functions have inverse functions. Make sure $f$ is a one-to-one function. Example 2: Find the inverse function of f\left( x \right) = {x^2} + 2,\,\,x \ge 0, if it exists.State its domain and range. Show Instructions. For example, in our example above, is both a right and left inverse to on the real numbers. r is an identity function (where . By this definition, if we are given ${f}^{-1}\left(70\right)=a$, then we are looking for a value $a$ so that $f\left(a\right)=70$. With y = 5x − 7 we have that f = y and g = x. So a left inverse is epimorphic, like the left shift or the derivative? Usually, to find the Inverse Laplace Transform of a function, we use the property of linearity of the Laplace Transform. For instance, the map given by has the two-sided inverse I keep saying "inverse function," which is not always accurate.Many functions have inverses that are not functions, or a function may have more than one inverse. So if there are only finitely many right inverses, it's because there is a 2-sided inverse. This domain of ${f}^{-1}$ is exactly the range of $f$. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. To evaluate $g\left(3\right)$, we find 3 on the x-axis and find the corresponding output value on the y-axis. Inverse Function Calculator. This proof is invalid, because just because it has a left- and a right inverse does not imply that they are actually the same function. So in the expression ${f}^{-1}\left(70\right)$, 70 is an output value of the original function, representing 70 miles. r is an identity function (where . Note that the does not indicate an exponent. This discussion of how and when matrices have inverses improves our understanding of the four fundamental subspaces and of many other key topics in the course. An inverse function goes the other way! Use the horizontal line test. In this case, we are looking for a $t$ so that $f\left(t\right)=70$, which is when $t=90$. Solution. If the function is one-to-one, there will be a unique inverse. Inverse Function Calculator. One also says that a left (or right) unit is an invertible element, i.e. The function $$y\left( x \right) = \arcsin x$$ is defined on the open interval $$\left( { – 1,1} \right).$$ The sine of the inverse sine is equal [/latex] State the domains of both the function and the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. JavaScript is not enabled. Often the inverse of a function is denoted by . The domain of $f$ is $\left[4,\infty \right)$. Help us out by expanding it. This property ensures that a function g: Y → X exists with the necessary relationship with f. The inverse of a function is a function that "undoes" the action of a given function. In other words, we show the following: Let $$A, N \in \mathbb{F}^{n\times n}$$ where … Note that the -1 use to denote an inverse function is not an exponent. The calculator will find the inverse of the given function, with steps shown. This same quadratic function, as seen in Example 1, has a restriction on its domain which is x \ge 0.After plotting the function in xy-axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. The inverse will return the corresponding input of the original function $f$, 90 minutes, so ${f}^{-1}\left(70\right)=90$. If $$AN= I_n$$, then $$N$$ is called a right inverse of $$A$$. What is the inverse of the function $f\left(x\right)=2-\sqrt{x}? We now prove that a left inverse of a square matrix is also a right inverse. When the domain and range are subsets of the real numbers, one way to test this is the horizontal line test: if every horizontal line drawn in the plane intersects the graph of the function in at most one point, the function is injective. Often the inverse of a function is denoted by . Up Main page Main result. Here, he is abusing the naming a little, because the function combine does not take as input the pair of lists, but is curried into taking each separately.. Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. (An example of a function with no inverse on either side is the zero transformation on .) The inverse function takes an output of [latex]f$ and returns an input for $f$. Let us start with an example: Here we have the function f(x) = 2x+3, written as a flow diagram: The Inverse Function goes the other way: So the inverse of: 2x+3 is: (y-3)/2 . Let f : A → B be a function with a left inverse h : B → A and a right inverse g : B → A. So ${f}^{-1}\left(y\right)=\frac{2}{y - 4}+3\\$ or ${f}^{-1}\left(x\right)=\frac{2}{x - 4}+3\\$. If the original function is given as a formula— for example, $y$ as a function of $x\text{-\hspace{0.17em}}$ we can often find the inverse function by solving to obtain $x$ as a function of $y$. Perform function composition. This is called the two-sided inverse, or usually just the inverse f –1 of the function f http://www.cs.cornell.edu/courses/cs2800/2015sp/handouts/jonpak_function_notes.pdf For example, we can make a restricted version of the square function $$f(x)=x^2$$ with its range limited to $$\left[0,\infty\right)$$, which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function). This might be easier to visualize with a concrete example: Take the real function . Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right.For instance, the map given by → ↦ ⋅ → has the two-sided inverse → ↦ (/) ⋅ →.In this subsection we will focus on two-sided inverses. $\begin{cases}\hfill{ C }=\frac{5}{9}\left(F - 32\right)\hfill \\ C\cdot \frac{9}{5}=F - 32\hfill \\ F=\frac{9}{5}C+32\hfill \end{cases}$, By solving in general, we have uncovered the inverse function. 2.Prove that if f has a right inverse… In other words, in a monoid (an associative unital magma) every element has at most one inverse (as defined in this section). Solution. By using this website, you agree to our Cookie Policy. This article is a stub. Inverse Functions. Y, and g is a left inverse of f if g f = 1 X. The calculator will find the inverse of the given function, with steps shown. 1.Prove that f has a left inverse if and only if f is injective (one-to-one). However, ${f}^{-1}$ itself must have an inverse (namely, $f$ ) so we have to restrict the domain of ${f}^{-1}$ to $\left[2,\infty \right)$ in order to make ${f}^{-1}$ a one-to-one function. Here, he is abusing the naming a little, because the function combine does not take as input the pair of lists, but is curried into taking each separately.. So, supposedly there can not be a number R such that (n + 1) * R = 1, and I'm supposed to prove that. 8.We will choose the right half, so that the output of the inverse sine function is always between 90 and 90 . Using the table below, find and interpret (a) $\text{ }f\left(60\right)$, and (b) $\text{ }{f}^{-1}\left(60\right)$. r is a right inverse of f if f . 9.Here are the common values with which you should be familiar. Suppose we want to find the inverse of a function represented in table form. For example, the inverse of f(x) = sin x is f-1 (x) = arcsin x, which is not a function, because it for a given value of x, there is more than one (in fact an infinite number) of possible values of arcsin x. If an element a has both a left inverse L and a right inverse R, i.e., La = 1 and aR = 1, then L = R, a is invertible, R is its inverse. Learning Objectives. {\displaystyle *} is associative then if an element has both a left inverse and a right inverse, they are equal. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. By using this website, you agree to our Cookie Policy. So ${f}^{-1}\left(x\right)={\left(x - 2\right)}^{2}+4$. Say we want to find out if is injective. Those that do are called invertible. 10.Of course, we could also give the answers in radians, rather than degrees. By above, we know that f has a Determine whether or not given functions are inverses. A function $f\left(t\right)$ is given below, showing distance in miles that a car has traveled in $t$ minutes. $C=h\left(F\right)=\frac{5}{9}\left(F - 32\right)\\$. Recall, that $$\mathcal{L}^{-1}\left(F(s)\right)$$$is such a function f(t) that $$\mathcal{L}\left(f(t)\right)=F(s)$$$. For example, in our example above, is both a right and left inverse to on the real numbers. $\begin{cases}y=\frac{2}{x - 3}+4\hfill & \text{Set up an equation}.\hfill \\ y - 4=\frac{2}{x - 3}\hfill & \text{Subtract 4 from both sides}.\hfill \\ x - 3=\frac{2}{y - 4}\hfill & \text{Multiply both sides by }x - 3\text{ and divide by }y - 4.\hfill \\ x=\frac{2}{y - 4}+3\hfill & \text{Add 3 to both sides}.\hfill \end{cases}\\$, $\begin{cases}y=2+\sqrt{x - 4}\hfill \\ {\left(y - 2\right)}^{2}=x - 4\hfill \\ x={\left(y - 2\right)}^{2}+4\hfill \end{cases}$, Evaluating the Inverse of a Function, Given a Graph of the Original Function, Finding Inverses of Functions Represented by Formulas, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175, $f\left(t\right)\text{ (miles)}$, Read the inverse function’s output from the. Now for the inverses: is called a right inverse of if it cancels by the right side: for all . $F={h}^{-1}\left(C\right)=\frac{9}{5}C+32\\$. Find and interpret ${f}^{-1}\left(70\right)$. Notice that the range of $f$ is $\left[2,\infty \right)$, so this means that the domain of the inverse function ${f}^{-1}$ is also $\left[2,\infty \right)$. (ii) $$sin\left ( sin^{-1}\frac{1}{2}+sec^{-1}2 \right )+cos\left ( tan^{-1}\frac{1}{3}+tan^{-1}3 \right )$$ (iii) $$sin\;cos^{-1}\left ( \frac{3}{5} \right )$$ Solution: Although problem (iii) can be solved using the formula, but I would like to show you another way to solve this type of Inverse trigonometric function … Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. https://artofproblemsolving.com/wiki/index.php?title=Inverse_of_a_function&oldid=13692. This same quadratic function, as seen in Example 1, has a restriction on its domain which is x \ge 0.After plotting the function in xy-axis, I can see that the graph is a parabola cut in half for all x values equal to or greater than zero. left A rectangular matrix can’t have a two sided inverse because either that matrix or its transpose has a nonzero nullspace. The interpretation of this is that, to drive 70 miles, it took 90 minutes. A left unit that is also a right unit is simply called a unit. In this case, is called the (right) inverse function of . The Moore-Penrose inverse is equal to left inverse A+ = Aleft−1, when ρ (A) = n < m and equals the right inverse A+ = Aright−1, when ρ (A) = m < n. The Moore-Penrose inverse is equal to the matrix inverse A+ = A−1, when ρ (A) = m = n. View chapter Purchase book On … If the function is one-to-one, there will be a unique inverse. Informally, this means that inverse functions “undo” each other. Interchange $x$ and $y$. (One direction of this is easy; the other is slightly tricky.) f is an identity function.. For a function f: X → Y to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that f = y. Sometimes we will need to know an inverse function for all elements of its domain, not just a few. Inverse Laplace Transform: The inverse Laplace transform of the function {eq}Y(s) {/eq} is the unique function {eq}y(t) {/eq} that is continuous and satisfies {eq}L[y(t)](s)=Y(s). Checkpoint 7.3.16. By above, we know that f has a left inverse and a right inverse. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. To evaluate ${g}^{-1}\left(3\right)$, recall that by definition ${g}^{-1}\left(3\right)$ means the value of x for which $g\left(x\right)=3$. Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. denotes composition).. l is a left inverse of f if l . Free functions inverse calculator - find functions inverse step-by-step This website uses cookies to ensure you get the best experience. If $$MA = I_n$$, then $$M$$ is called a left inverse of $$A$$. If no two points on the graph of an inverse function can sit above or below each other, then reflecting these points over $$y = x$$ means that no two points on the graph of $$f(x)$$ can sit directly left or right of each other. In this case, we introduced a function $h$ to represent the conversion because the input and output variables are descriptive, and writing ${C}^{-1}$ could get confusing. $f$ and ${f}^{-1}$ are equal at two points but are not the same function, as we can see by creating the table below. Similarly, a function such that is called the left inverse function of . Typically, the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function. Therefore, to find the inverse of f\left( x \right) = \left| {x - 3} \right| + 2 for x \ge 3 is the same as finding the inverse of the line f\left( x \right) = \left( {x - 3} \right) + 2 for x \ge 3. Example 2: Find the inverse function of f\left( x \right) = {x^2} + 2,\,\,x \ge 0, if it exists.State its domain and range. Find the inverse of a one-to-one function … A function $g\left(x\right)$ is given in Figure 5. Therefore it has a two-sided inverse. Of course, for a commutative unitary ring, a left unit is a right unit too and vice versa. By looking for the output value 3 on the vertical axis, we find the point $\left(5,3\right)$ on the graph, which means $g\left(5\right)=3$, so by definition, ${g}^{-1}\left(3\right)=5$. {eq}f\left( x \right) = y \Leftrightarrow g\left( y \right) = x{/eq}. This holds for all $x$ in the domain of $f$. Find the inverse of the function $f\left(x\right)=2+\sqrt{x - 4}$. Show Instructions. Similarly, a function such that is called the left inverse function of . So in the expression ${f}^{-1}\left(70\right)$, 70 is an output value of the original function, representing 70 miles. Thus an inverse of f is merely a function g that is both a right inverse and a left inverse simultaneously. Valid Proof ( ⇒ ): Suppose f is bijective. Free Inverse Laplace Transform calculator - Find the inverse Laplace transforms of functions step-by-step This website uses cookies to ensure you get the best experience. If. Free functions inverse calculator - find functions inverse step-by-step This website uses cookies to ensure you get the best experience. The inverse of a function with range is a function if and only if is injective, so that every element in the range is mapped from a distinct element in the domain. Alternatively, recall that the definition of the inverse was that if $f\left(a\right)=b$, then ${f}^{-1}\left(b\right)=a$. f is an identity function.. Using the graph in Example 6, (a) find ${g}^{-1}\left(1\right)$, and (b) estimate ${g}^{-1}\left(4\right)$. Find $g\left(3\right)$ and ${g}^{-1}\left(3\right)$. 7.For the inverse sine, we have to choose between the right half of the circle, or the left half. The equation Ax = b always has at least one solution; the nullspace of A has dimension n − m, so there will be The calculator will find the Inverse Laplace Transform of the given function. Some functions have a two-sided inverse map, another function that is the inverse of the first, both from the left and from the right. If a function has both a left inverse and a right inverse, then the two inverses are identical, and this common inverse is unique (Prove!) In Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events.How do logarithmic graphs give us insight into situations? Typically, the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function. Then h = g and in fact any other left or right inverse for f also equals h. 3. $\endgroup$ – Peter LeFanu Lumsdaine Oct 15 '10 at 16:29 $\begingroup$ @Peter: yes, it looks we are using left/right inverse in different senses when the ring operation is function composition. We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. The Attempt at a Solution My first time doing senior-level algebra. So we need to interchange the domain and range. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. We'd like to be able to "invert A" to solve Ax = b, but A may have only a left inverse or right inverse (or no inverse). an element that admits a right (or left) inverse … $C=\frac{5}{9}\left(F - 32\right)$. JavaScript is required to fully utilize the site. By using this website, you agree to our Cookie Policy. The function has the property that . Note that the does not indicate an exponent. You also need to observe the range of the given function which is y \ge 2 because this will be the domain of the inverse function. Subtract [b], and then multiply on the right by b^j; from ab=1 (and thus (1-ba)b = 0) we conclude 1 - ba = 0. Intro to Finding the Inverse of a Function Before you work on a find the inverse of a function examples, let’s quickly review some important information: Notation: The following notation is used to denote a function (left) and it’s inverse (right). r is a right inverse of f if f . Solve for $x$ in terms of $y$ given $y=\frac{1}{3}\left(x - 5\right)\\$. And similarly a left inverse if for all . Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. The inverse function takes an output of $f$ and returns an input for $f$. However we will now see that when a function has both a left inverse and a right inverse, then all inverses for the function must agree: Lemma 1.11. However, just as zero does not have a reciprocal, some functions do not have inverses.. The domain and range of $f$ exclude the values 3 and 4, respectively. For example, consider the function given by the rule . In a monoid, the set of (left and right) invertible elements is … The formula we found for ${f}^{-1}\left(x\right)$ looks like it would be valid for all real $x$. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Let us return to the quadratic function $f\left(x\right)={x}^{2}$ restricted to the domain $\left[0,\infty \right)$, on which this function is one-to-one, and graph it as in Figure 7. Right inverse If A has full row rank, then r = m. The nullspace of AT contains only the zero vector; the rows of A are independent. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e.
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Split-quaternion
Coquaternion multiplication
× 1 i j k
1 1 i j k
i i −1 k −j
j j −k 1 −i
k k j i 1
In abstract algebra, the split-quaternions or coquaternions are elements of a 4-dimensional associative algebra introduced by James Cockle in 1849 under the latter name. Like the quaternions introduced by Hamilton in 1843, they form a four dimensional real vector space equipped with a multiplicative operation. Unlike the quaternion algebra, the split-quaternions contain zero divisors, nilpotent elements, and nontrivial idempotents. (For example, 12(1 + j) is an idempotent zero-divisor, and ij is nilpotent.) As a mathematical structure, they form an algebra over the real numbers, which is isomorphic to the algebra of 2 × 2 real matrices. For other names for split-quaternions see the Synonyms section below.
The set {1, i, j, k} forms a basis. The products of these elements are
ij = k = −ji,
jk = −i = −kj,
ki = j = −ik,
i2 = −1,
j2 = +1,
k2 = +1,
and hence ijk = 1. It follows from the defining relations that the set {1, i, j, k, −1, −i, −j, −k} is a group under coquaternion multiplication; it is isomorphic to the dihedral group of a square.
A coquaternion
q = w + xi + yj + zk, has a conjugate q* = wxi − yj − zk.
Due to the anti-commutative property of its basis vectors, the product of a coquaternion with its conjugate is given by an isotropic quadratic form:
Given two coquaterions p and q, one has N(p q) = N(p) N(q), showing that N is a quadratic form admitting composition. This algebra is a composition algebra and N is its norm. Any q ≠ 0 such that N(q) = 0 is a null vector, and its presence means that coquaternions form a "split composition algebra", and hence a coquaternion is also called a split quaternion.
When the norm is non-zero, then q has a multiplicative inverse, namely q*/N(q). The set
U = {q : qq* ≠ 0}
is the set of units. The set P of all coquaternions forms a ring (P, +, •) with group of units (U, •). The coquaternions with N(q) = 1 form a non-compact topological group SU(1,1), shown below to be isomorphic to SL(2, R).
Historically coquaternions preceded Cayley's matrix algebra; coquaternions (along with quaternions and tessarines) evoked the broader linear algebra.
Matrix representations
Let
q = w + xi + yj + zk,
and consider u = w + xi, and v = y + zi as ordinary complex numbers with complex conjugates denoted by u* = wxi, v* = yzi. Then the complex matrix
,
represents q in the ring of matrices, i.e. the multiplication of split-quaternions behaves the same way as the matrix multiplication. For example, the determinant of this matrix is
uu* − vv* = qq*.
The appearance of the minus sign, where there is a plus in H, distinguishes coquaternions from quaternions. The use of the split-quaternions of norm one (qq* = 1) for hyperbolic motions of the Poincaré disk model of hyperbolic geometry is one of the great utilities of the algebra.
Besides the complex matrix representation, another linear representation associates coquaternions with 2 × 2 real matrices. This isomorphism can be made explicit as follows: Note first the product
and that the square of each factor on the left is the identity matrix, while the square of the right hand side is the negative of the identity matrix. Furthermore, note that these three matrices, together with the identity matrix, form a basis for M(2, R). One can make the matrix product above correspond to jk = −i in the coquaternion ring. Then for an arbitrary matrix there is the bijection
which is in fact a ring isomorphism. Furthermore, computing squares of components and gathering terms shows that qq* = adbc, which is the determinant of the matrix. Consequently there is a group isomorphism between the unit quasi-sphere of coquaternions and SL(2, R) = {g ∈ M(2, R) : det g = 1}, and hence also with SU(1, 1): the latter can be seen in the complex representation above.
For instance, see Karzel and Kist[1] for the hyperbolic motion group representation with 2 × 2 real matrices.
In both of these linear representations the norm is given by the determinant function. Since the determinant is a multiplicative mapping, the norm of the product of two coquaternions is equal to the product of the two separate norms. Thus coquaternions form a composition algebra. As an algebra over the field of real numbers, it is one of only seven such algebras.
Generation from split-complex numbers
Kevin McCrimmon [2] has shown how all composition algebras can be constructed after the manner promulgated by L. E. Dickson and Adrian Albert for the division algebras ℂ, ℍ, & O. Indeed, he presents the multiplication rule
to be used when producing the doubled product in the real-split cases. As before, the doubled conjugate so that
If a and b are split-complex numbers and split-quaternion
then
Profile
The circle E lies in the plane z = 0.
Elements of J are square roots of +1.
Elements of I are square roots of −1
The subalgebras of P may be seen by first noting the nature of the subspace {zi + xj + yk : x, y, zR}. Let
r(θ) = j cos(θ) + k sin(θ)
The parameters z and r(θ) are the basis of a cylindrical coordinate system in the subspace. Parameter θ denotes azimuth. Next let a denote any real number and consider the coquaternions
p(a, r) = i sinh a + r cosh a
v(a, r) = i cosh a + r sinh a.
These are the equilateral-hyperboloidal coordinates described by Alexander Macfarlane and Carmody.[3]
Next, form three foundational sets in the vector-subspace of the ring:
E = {r P: r = r(θ), 0 θ < 2π}
J = {p(a, r) P: a R, r E}, hyperboloid of one sheet
I = {v(a, r) P: a R, r E}, hyperboloid of two sheets.
Now it is easy to verify that
{q P: q2 = 1} = J {1, −1}
and that
{q P: q2 = −1} = I.
These set equalities mean that when pJ then the plane
{x + yp: x, y R} = Dp
is a subring of P that is isomorphic to the plane of split-complex numbers just as when v is in I then
{x + yv: x, y R} = Cv
is a planar subring of P that is isomorphic to the ordinary complex plane C.
Note that for every rE, (r + i)2 = 0 = (r − i)2 so that r + i and r − i are nilpotents. The plane N = {x + y(r + i): x, yR} is a subring of P that is isomorphic to the dual numbers. Since every coquaternion must lie in a Dp, a Cv, or an N plane, these planes profile P. For example, the unit quasi-sphere
SU(1, 1) = {q P: qq* = 1}
consists of the "unit circles" in the constituent planes of P: In Dp it is a unit hyperbola, in N the "unit circle" is a pair of parallel lines, while in Cv it is indeed a circle (though it appears elliptical due to v-stretching).These ellipse/circles found in each Cv are like the illusion of the Rubin vase which "presents the viewer with a mental choice of two interpretations, each of which is valid".
Pan-orthogonality
When coquaternion q = w + xi + yj + zk, then the scalar part of q is w.
Definition. For non-zero coquaternions q and t we write q ⊥ t when the scalar part of the product q(t*) is zero.
• For every vI, if q, tCv, then qt means the rays from 0 to q and t are perpendicular.
• For every pJ, if q, tDp, then qt means these two points are hyperbolic-orthogonal.
• For every rE and every aR, p = p(a, r) and v = v(a, r) satisfy pv.
• If u is a unit in the coquaternion ring, then qt implies qutu.
Proof: (qu)(tu)* = (uu*)q(t*) follows from (tu)* = u*t*, which can be established using the anticommutativity property of vector cross products.
Counter-sphere geometry
The quadratic form qq* is positive definite on the planes Cv and N. Consider the counter-sphere {q: qq* = −1}.
Take m = x + yi + zr where r = j cos(θ) + k sin(θ). Fix θ and suppose
mm* = −1 = x2 + y2 − z2.
Since points on the counter-sphere must line on the conjugate of the unit hyperbola in some plane DpP, m can be written, for some pJ
.
Let φ be the angle between the hyperbolas from r to p and m. This angle can be viewed, in the plane tangent to the counter-sphere at r, by projection:
. Then
as in the expression of angle of parallelism in the hyperbolic plane H2 . The parameter θ determining the meridian varies over the S1. Thus the counter-sphere appears as the manifold S1 × H2.
Application to kinematics
By using the foundations given above, one can show that the mapping
is an ordinary or hyperbolic rotation according as
.
The collection of these mappings bears some relation to the Lorentz group since it is also composed of ordinary and hyperbolic rotations. Among the peculiarities of this approach to relativistic kinematic is the anisotropic profile, say as compared to hyperbolic quaternions.
Reluctance to use coquaternions for kinematic models may stem from the (2, 2) signature when spacetime is presumed to have signature (1, 3) or (3, 1). Nevertheless, a transparently relativistic kinematics appears when a point of the counter-sphere is used to represent an inertial frame of reference. Indeed, if tt* = −1, then there is a p = i sinh(a) + r cosh(a) ∈ J such that tDp, and an bR such that t = p exp(bp). Then if u = exp(bp), v = i cosh(a) + r sinh(a), and s = ir, the set {t, u, v, s} is a pan-orthogonal basis stemming from t, and the orthogonalities persist through applications of the ordinary or hyperbolic rotations.
Historical notes
The coquaternions were initially introduced (under that name)[4] in 1849 by James Cockle in the LondonEdinburghDublin Philosophical Magazine. The introductory papers by Cockle were recalled in the 1904 Bibliography[5] of the Quaternion Society. Alexander Macfarlane called the structure of coquaternion vectors an exspherical system when he was speaking at the International Congress of Mathematicians in Paris in 1900.[6]
The unit sphere was considered in 1910 by Hans Beck.[7] For example, the dihedral group appears on page 419. The coquaternion structure has also been mentioned briefly in the Annals of Mathematics.[8][9]
Synonyms
• Para-quaternions (Ivanov and Zamkovoy 2005, Mohaupt 2006) Manifolds with para-quaternionic structures are studied in differential geometry and string theory. In the para-quaternionic literature k is replaced with k.
• Musean hyperbolic quaternions
• Exspherical system (Macfarlane 1900)
• Split-quaternions (Rosenfeld 1988)[10]
• Antiquaternions (Rosenfeld 1988)
• Pseudoquaternions (Yaglom 1968[11] Rosenfeld 1988)
Notes
1. Karzel, Helmut & Günter Kist (1985) "Kinematic Algebras and their Geometries", in Rings and Geometry, R. Kaya, P. Plaumann, and K. Strambach editors, pp 437509, esp 449,50, D. Reidel ISBN 90-277-2112-2
2. Kevin McCrimmon (2004) A Taste of Jordan Algebras, page 64, Universitext, Springer ISBN 0-387-95447-3 MR 2014924
3. Carmody, Kevin (1997) "Circular and hyperbolic quaternions, octonions, sedionions", Applied Mathematics and Computation 84(1):2747, esp. 38
4. James Cockle (1849), On Systems of Algebra involving more than one Imaginary, Philosophical Magazine (series 3) 35: 434,5, link from Biodiversity Heritage Library
5. A. Macfarlane (1904) Bibliography of Quaternions and Allied Systems of Mathematics, from Cornell University Historical Math Monographs, entries for James Cockle, pp. 17–18
6. Alexander Macfarlane (1900) Application of space analysis to curvilinear coordinates, Proceedings of the International Congress of Mathematicians, Paris, page 306, from International Mathematical Union
7. A. A. Albert (1942), "Quadratic Forms permitting Composition", Annals of Mathematics 43:161 to 77
8. Rosenfeld, B.A. (1988) A History of Non-Euclidean Geometry, page 389, Springer-Verlag ISBN 0-387-96458-4
9. Isaak Yaglom (1968) Complex Numbers in Geometry, page 24, Academic Press
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Trigonometric Functions
# Wave Mathematics: Trigonometric Functions
by Gary Leo Welz, M.A./M.S.
Waves are familiar to us from the ocean, the study of sound, earthquakes, and other natural phenomenon. But as any surfer can tell you, ocean waves come in very different sizes, as can all waves. To fully understand waves, we need to understand measurements associated with these waves, such as how often they repeat (their frequency), and how long they are (their wavelength), and their vertical size (amplitude).
While these measurements help describe waves, they do not help us make predictions about wave behavior. In order to do that, we need to look at waves more abstractly, which we can do using a mathematical formula.
## Waves and circles
It is possible to look at waves mathematically because a wave's shape repeats itself over a consistent interval of time and distance. This behavior mirrors the repetition of the circle. Imagine drawing a circle on a piece of paper. Now imagine drawing that same shape while your friend slowly pulled the piece of paper out from under your pencil – the line you would have drawn traces out the shape of a wave. One rotation around the circle completes one cycle of rising and falling in the wave, as seen in the picture below.
Mathematicians use the sine function (Sin) to express the shape of a wave. The mathematical equation representing the simplest wave looks like this:
y = Sin(x)
This equation describes how a wave would be plotted on a graph, stating that y (the value of the vertical coordinate on the graph) is a function of the sine of the number x (the horizontal coordinate).
The sine function is one of the trigonometric ratios originally calculated by the astronomer Hipparchus of Nicaea in the second century B.C. when he was trying to make sense of the movement of the stars and moon in the night sky. More than 2000 years ago, when Hipparchus began to study astronomy, the movement of objects in the sky was a mystery. Hipparchus knew that the stars and moon tended to move through the night sky in a semi-circular fashion. Thus he felt that understanding the shape of a circle was important to understanding astronomy. Hipparchus began to observe that there was a relationship between the radius of a circle, the center angle made by a pie slice of that circle, and the length of the arc of that pie piece. If one knew any two of these values, the third value could be calculated. It was later realized that this relationship also applies to right triangles. Knowing one angle measure of a right triangle, you can calculate the ratio of the sides of the triangle. The exact size of the triangle can vary, but the ratio of the lengths of the sides is defined by the angle size. The specific relationships between the angle measure and the sides of the triangle are what we call the trigonometric functions, the three main functions consisting of:
• Sine A = opposite/hypotenuse
The word trigonometry means "measurement of triangles" and sine along with the cosine and tangent are called the trigonometric ratios because they originated with the ancient study of triangles.
Comprehension Checkpoint
The ancient astronomer Hipparchus discovered that knowing one angle measure of a right triangle allows you to calculate
## Trigonometric ratios become wave functions
But how do triangles relate to waves? In the early 17th century, two Frenchmen named Rene Descartes and Pierre Fermat co-developed what would become known as the Cartesian coordinate plane, more commonly known as the (x,y)-graphing plane. This invention was an extraordinary advance in the history of mathematics because it brought together, for the first time, the integration of the two great, but distinct branches of mathematics: geometry, the science of space and form, and algebra, the science of numbers. The invention of the Cartesian coordinate system soon led to the graphing of many mathematical relations including the sine and cosine ratios.
As it turns out, the trigonometric functions can also be defined in relation to the "unit circle," i.e. a circle with a radius equal to 1. When we put the unit circle on the Cartesian plane, we can begin to see how this works if we draw a triangle within the circle, as seen in the diagram below. According to our earlier discussion, the sine of angle A in the diagram equals the ratio of the opposite side over the hypotenuse. However, remember that we are working with a unit circle and the length of the hypotenuse is equal to the radius of the circle, or 1. Therefore,
Sin(A) = opp/1 = opp
So the sine of A gives the length of the opposite side of the triangle, or the y-coordinate on our Cartesian plane. Similarly, the cosine of angle A equals the ratio of the adjacent side over the hypotenuse. Since the length of the hypotenuse equals 1, the cosine of A gives the length of the adjacent side, or the x-coordinate on the Cartesian plane.
If we redraw this triangle as we move counterclockwise on the circle, we can begin to see that the trigonometric functions, in this case sine and cosine, take on a periodic quality. This means that sine, for example, increases to a maximum at the top of the circle, decreases to zero as we sweep left, and begins to take on negative values as we continue around the circle. At the bottom of the circle the sine function reaches a minimum value and the process begins again as we reach the right side of the circle. To better appreciate this idea, review the animation Sine, Cosine, and the Unit Circle linked below.
Sine, Cosine, and the Unit Circle
This animation illustrates how the values of the sine and cosine change as we sweep around the unit circle.
As you saw in the animation above, as angle A increases, the values of the trigonometric functions of A undergo a periodic cycle from 0, to a maximum of 1, down to a minimum of -1, and back to 0. There are several ways to express the measure of the angle A. One way is in degrees, where 360 degrees defines a complete circle. Another way to measure angles is in a unit called the radian, where 2π radians defines a complete circle. Angles smaller than 360 can be defined as fractions of this unit, for example 90° can be written as π/2 or 1.57 radians, 180° equals π or 3.14 radians.
Comprehension Checkpoint
The measure of angles can be expressed in degrees or in
## Plotting a basic sine wave
If we now plot the sine of the angle measured in radians along the Cartesian coordinate system, we see that we again get the characteristic rise and fall. However, since the angle measure is plotted along the x-axis (instead of the cosine of the angle), the graph that results is a continuous curve on the coordinate plane that resembles a physical wave, as seen below.
If you look closely at this graph you will see that the wave crosses the x-axis at multiples of 3.1416 - the value of pi. One full wave is completed at the value 6.2832, or 2π, exactly the circumference of the unit circle.
Understanding the origin of the sine function makes it easier to understand how it operates in relation to waves. As we saw earlier, the basic formula representing the sine function is:
y = Sin(x)
In this formula, y is the value on the y-axis obtained when one carries out the function Sin(x) for points on the x-axis. This results in the graph of the basic sine wave. But how can we represent other forms of waves, especially ones that are larger or longer? To graph waves of different sizes we need to add other terms to our formula. The first we will look at is amplitude.
y = ASin(x)
In this modification of the formula, A gives us the value of the amplitude of the wave – the distance it moves above and below the x-axis, or the height of the wave. In essence, what the modifier A does is increase (or amplify) the result of the function Sin(x), thus leading to larger resulting y values.
To modify the wavelength of a wave, or the distance from one point on a wave to an equal point on the following wave, the modifier k is used, as seen in the formula below.
y = ASin(k*x)
The multiplier k extends the length of the wave. Remember from our earlier discussion that the wavelength of our most simple wave is 2π, therefore wavelength in the final formula is determined simply be dividing 2π by the multiplier k, so wavelength (λ) = 2π/k.
Comprehension Checkpoint
The multiplier k is used to modify the _____ of a wave.
## Wave periods
Since waves always are moving, one more important term to describe a wave is the time it takes for one wavelength to pass a specific point in space. This term, referred to as the period, T, is equivalent to the wavelength, T = Period = 2π/k, however it is given in units of time (sec) rather than distance.
Understanding the mathematics behind wave functions allows us to better understand the natural world around us. For example, the differences between the colors you see on this page have to do with different wavelengths of light perceived by your eyes. Similarly, the difference between a bird?s song and the roar of a locomotive is due to the size of the sound waves emitted. Waves, and thus the mathematics of waves, constantly surround us.
### Summary
Waves, circles, and triangles are closely related. In fact, this relatedness forms the basis of trigonometry. Basic trigonometric functions are explained in this module and applied to describe wave behavior. The module presents Cartesian coordinate (x, y) graphing, and shows how the sine function is used to plot a wave on a graph.
### Key Concepts
• The sine function is one of many trigonometric ratios calculated by astronomer Hipparchus over 2,000 years ago.
• Understanding trigonometric functions allows for the understanding and prediction of an object’s movement.
• ##### NGSS
• HS-C3.5, HS-PS4.A1
Gary Leo Welz, M.A./M.S. “Wave Mathematics” Visionlearning Vol. MAT-1 (1), 2006.
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Connect with us
# Is sin 1 the same as arcsin
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## Should you write arcsin or sin 1?
If you are looking for a function, arcsin(x), makes more sense to me. This always returns a value in a specified range. I always interpret sin^{-1}(x) to be a pre-image: that is, any real number y such that sin(y) = x (this is not a function, since y is not unique).
## Why is sin inverse called arcsin?
If you have a numerical value and you want the size of the angle whose sine has this value, you get something like this, where the value is a number and the arcsin is expressed in degrees of arc. It essentially reverses the process of the sine function. It is called “arcsin” because it gives you a measure of the arc.
## What is the same as arcsin?
sin1x means the same as arcsin x, i.e., the arc whose sine is x.
## What is the difference between arcsin and sin?
Sine is a basic trigonometric function, and the arcsine is the inverse function of the sine.
## Is sin 1 the same as 1 sin?
The notation sin1(x) has been misunderstood to mean 1/sin(x). … So sin1(x) means the inverse sine of x, that is, the function that undoes the sine function. It is not equal to 1/sin(x).
## How do I know if I have SOH CAH TOA?
SOHCAHTOA is a mnemonic device helpful for remembering what ratio goes with which function.
1. SOH = Sine is Opposite over Hypotenuse.
2. CAH = Cosine is Adjacent over Hypotenuse.
3. TOA = Tangent is Opposite over Adjacent.
## How do you convert sin to arcsin?
The inverse sine function, arcsine, will take the ratio of the opposite/hypotenuse (x) and return the angle, θ. So, knowing that, for our triangle, arcsin(x) = θ we can also write that: Sine: sin(arcsin(x)) = x. Cosine: cos(arcsin(x)) = √(1-x²)
READ ALSO Who from macedonia conquered greece
## Is arcsin the same as cosecant?
As nouns the difference between arcsine and cosecant
is that arcsine is (trigonometry) any of several single-valued or multivalued functions that are inverses of the sine function symbol: arcsin, sin1 while cosecant is (trigonometry) in a right triangle, the reciprocal of the sine of an angle symbols: cosec, csc.
## Does arcsin cancel out sin?
The arcsine function is the inverse function for the sine function on the interval . So they “cancel” each other under composition of functions, as follows.
## How do you write arcsin?
y = arcsine of x = arcsin(x) = sin1(x). Another way to write x = sin(y). y = arctangent of x = arctan(x) = tan1(x). Another way to write x = tan(y).
## How do you get arcsin?
To calculate arcsin, press the “2nd” button and then the “sin” button. This will produce the “sin^-1” button. Enter the value you wish to calculate and press enter. The answer will appear.
## How do you calculate arcsin manually?
to compute x from sin(x). arcsin is defined to be the inverse of sin but restricted to a certain range. Hence arcsin(sin(x))=x if x is within this range (generally either 0 to 2π or −π to π) or a value y such that sin(y)=sin(x) i.e. y=x+2πn or y=π−x+2πm for some n∈Z or m∈Z and y is in this range.
## What is the value of arcsin?
Explanation: Arcsin can be thought of as the ANGLE with the specified value of sin. The range of arcsin or sin−1 is π2 to −π2 .
## Where is arcsin defined?
Definition: The arcsine of x, denoted arcsin(x), is defined as is defined as ‘the one angle between −π/2 and +π/2 radians (or between −90° and +90°) whose sine is x’. It is a one-to-one function.
READ ALSO Cost of Building an App in Nigeria 2023
## What does Arcsin look like?
The arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number.
arcsin.
## How do you do Arccos arcsin?
Arcsin, Arccos, Arctan : Example Question #2
For the above triangle, what is if and ? Explanation: We need to use a trigonometric function to find . We are given the opposite and adjacent sides, so we can use the and functions.
## What is the inverse of sin-1?
90°
Hence, sin1 (1) is equal to the angle whose sine is 1. Since the inverse sin1 (1) is 90° or Π/2.
## What is the domain of arcsin?
The domain of arcsinx is the interval [−1,1] and it is undefined elsewhere. Within this domain it has range [−π2,π2], and these values as input to sinx produce values in the range [−1,1].
## How do you convert arcsin to degrees?
Use this arcsine calculator to easily calculate the arcsine of a number. Online arcsine calculation tool with output in degrees or radians.
How to calculate the arcsine of a number?
## What does ARC mean in trig?
They switch around what’s the input and what’s the output. Each trig function has its associated inverse function. One way to denote this inverse is by writing the prefix “Arc” in front of the function. So Arc sine is the inverse of sine. Arc cosine is the inverse of cosine.
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# How to know if airpod case is charging
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By
## What Colour should AirPods be when charging?
Green means fully charged, and amber means less than one full charge remains. When you connect your Wireless Charging Case to a charger, or place it on a Qi-certified charging mat, the status light will stay on for eight seconds. If the light flashes white, your AirPods are ready to set up with one of your devices.
## Why are my AirPods case not charging?
Change Lightning Cables
Some people find out that the lightning cable is to blame for their Airpods case not charging. Your lightning cable could have a short or another issue that is preventing it from working properly. Getting a new lightning cable and hooking things up with that could make the difference.
## How long do AirPods case take to charge?
To turn a deceased battery into a fully charged battery through an AirPods case charge will take approximately twenty to thirty minutes. That means if you’re short of time, then you can quickly get your AirPods back to full fitness and ready to be used again.
## What does green light on AirPod case mean?
A green light indicates the AirPods or charging case are fully charged, while an amber light means there’s less than one full charge left.
## Why is my AirPod charging case blinking green?
Green means charged, and amber means less than one full charge remains. If the light flashes white, your AirPods are ready to set up with one of your devices. If the light flashes amber, you might need to set up your AirPods again.
## Do AirPods have a green light on the front?
The status light normally tells you if AirPods are charging or ready to pair. The second-generation case and Airpods Pro case has it on the front of the case. Normally the status light shows if your AirPods or case are charged, charging, or ready to pair. A flashing green light seems to show a problem.
## Why is my AirPod Pro case flashing orange?
A blinking orange light means that your AirPods are not pairing properly with your iPhone or the firmware is different on each AirPod and that they need to be reset and then paired again. It could also mean that you’ve got a fake pair of AirPods.
## How can you tell if AirPods are fake?
To put it shortly, the quickest way to spot fake AirPods is to scan the serial number found on the inside of the case (see pictures below on how to find that serial number). Once you get that code, pop it through checkcoverage.apple.com and see whether Apple confirms it for you.
## Why is my AirPods light flashing orange?
If they are solid orange, then you have less than one full charge on them. If the light is flashing white, that means that the device is ready to set up. On the other hand, if it blinks orange, it means that you have to set up your AirPods again.
## Why do my AirPods keep flashing amber?
Amber: your AirPods or AirPods Pro are not fully charged. White: your AirPods or AirPods Pro are ready to be set up. Flashing amber: there is a problem with your AirPods or AirPods Pro. You need to factory reset them and then re-pair them to your device.
## What to do if AirPods are not connecting?
If you’re having trouble getting your AirPods to connect, make sure your AirPods are charged, Bluetooth is turned on for the device you want to connect, and reset the device before trying again. If none of those steps work, you should un pair your AirPods from your device, reset the AirPods, and try to reconnect them.
READ ALSO Did the dust bowl affect all 50 states
## What does resetting my AirPods do?
Note that now the AirPods are reset they will no longer automatically recognize any of the devices linked to your iCloud account. Opening the AirPods case near to an iOS device will initiate the setup process, just like the first time you used them.
## Why wont my AirPods automatically connect?
Go to Settings > Bluetooth and tap the “i” icon next to your AirPods. 2 Then tap Forget This Device, and tap again to confirm. With the lid open, press and hold the setup button on the back of the case for about 15 seconds, until you see the status light flashing amber.
## What do I do if my AirPods won’t reset?
You may need to clean the contacts inside the case with a cotton swab. Then put the other one back in, and confirm you still see an amber light. *Then* you can do a hard reset by holding down the button for 15 seconds until you get yet another amber light and then flashing white.
## What do I do if my AirPods wont flash white?
Press and hold the setup button on the case for up to 10 seconds. The status light should flash white, which means that your AirPods are ready to connect. Hold the case, with your AirPods inside and the lid open, next to your iOS device.
## Are AirPods waterproof?
They are not waterproof but they do have sweat and dust resistance meaning they won’t be ruined by rain or falling in a puddle. That being said don’t like throw them in a pool or shower with them. They’re rated to be IPX4, so only sweat and splash proof.
## Can I use AirPods pro in the shower?
In spite of their improved protection, Apple also recommends not wearing AirPods Pro in the shower either. The reasons for this is the level of water resistance that’s on offer. While they can easily survive rain or a vigorous workout, AirPods Pro are not designed for use under running water.
## Will sweat ruin AirPods?
One word: sweat. After all, AirPods are neither sweat nor water-resistant. “Simply put, sweat—and any kind of water—will damage headphones permanently if it gets inside,” our executive editor, TJ Donegan, explains.
## Can I wear my Airpod pros in the shower?
Apple has two truly wireless earbud offerings. The AirPods Pro are water and sweat resistant, meaning they should survive heavy perspiration or a splash, though Apple tells users not to place them “under running water, such as a shower or faucet.”
## Can you overcharge AirPods?
The short answer is yes, it’s safe. Your AirPods can‘t overcharge and doing so overnight won’t damage their battery. In this article, we‘ll discuss new batteries but also about potential problems that could damage your battery and how to avoid them.
## Can you use AirPods on PS4?
Unfortunately, the PlayStation 4 doesn’t natively support AirPods. To connect AirPods to your PS4, you‘ll need to use a third-party Bluetooth. ‘: A beginner’s guide to the wireless technology Bluetooth is a wireless technology that allows the exchange of data between different devices.
## Why is AirPods so expensive?
There are several factors that combine to make the Airpods expensive. The first is that they’re an Apple product and the brand does not manufacture cheap products. There’s a fair amount of overhead that goes into the design, materials, and construction of each product manufactured.
READ ALSO Why is the marginal productivity of labor mpl likely to increase and then decline in the short run
## What Colour should AirPods be when charging?
Green means fully charged, and amber means less than one full charge remains. When you connect your Wireless Charging Case to a charger, or place it on a Qi-certified charging mat, the status light will stay on for eight seconds. If the light flashes white, your AirPods are ready to set up with one of your devices.
## Why are my AirPods case not charging?
Change Lightning Cables
Some people find out that the lightning cable is to blame for their Airpods case not charging. Your lightning cable could have a short or another issue that is preventing it from working properly. Getting a new lightning cable and hooking things up with that could make the difference.
## How long do AirPods case take to charge?
To turn a deceased battery into a fully charged battery through an AirPods case charge will take approximately twenty to thirty minutes. That means if you’re short of time, then you can quickly get your AirPods back to full fitness and ready to be used again.
## What does green light on AirPod case mean?
A green light indicates the AirPods or charging case are fully charged, while an amber light means there’s less than one full charge left.
## Why is my AirPod charging case blinking green?
Green means charged, and amber means less than one full charge remains. If the light flashes white, your AirPods are ready to set up with one of your devices. If the light flashes amber, you might need to set up your AirPods again.
## Do AirPods have a green light on the front?
The status light normally tells you if AirPods are charging or ready to pair. The second-generation case and Airpods Pro case has it on the front of the case. Normally the status light shows if your AirPods or case are charged, charging, or ready to pair. A flashing green light seems to show a problem.
## Why is my AirPod Pro case flashing orange?
A blinking orange light means that your AirPods are not pairing properly with your iPhone or the firmware is different on each AirPod and that they need to be reset and then paired again. It could also mean that you’ve got a fake pair of AirPods.
## How can you tell if AirPods are fake?
To put it shortly, the quickest way to spot fake AirPods is to scan the serial number found on the inside of the case (see pictures below on how to find that serial number). Once you get that code, pop it through checkcoverage.apple.com and see whether Apple confirms it for you.
## Why is my AirPods light flashing orange?
If they are solid orange, then you have less than one full charge on them. If the light is flashing white, that means that the device is ready to set up. On the other hand, if it blinks orange, it means that you have to set up your AirPods again.
## Why do my AirPods keep flashing amber?
Amber: your AirPods or AirPods Pro are not fully charged. White: your AirPods or AirPods Pro are ready to be set up. Flashing amber: there is a problem with your AirPods or AirPods Pro. You need to factory reset them and then re-pair them to your device.
## What to do if AirPods are not connecting?
If you’re having trouble getting your AirPods to connect, make sure your AirPods are charged, Bluetooth is turned on for the device you want to connect, and reset the device before trying again. If none of those steps work, you should un pair your AirPods from your device, reset the AirPods, and try to reconnect them.
READ ALSO How to clean old rusty tools
## What does resetting my AirPods do?
Note that now the AirPods are reset they will no longer automatically recognize any of the devices linked to your iCloud account. Opening the AirPods case near to an iOS device will initiate the setup process, just like the first time you used them.
## Why wont my AirPods automatically connect?
Go to Settings > Bluetooth and tap the “i” icon next to your AirPods. 2 Then tap Forget This Device, and tap again to confirm. With the lid open, press and hold the setup button on the back of the case for about 15 seconds, until you see the status light flashing amber.
## What do I do if my AirPods won’t reset?
You may need to clean the contacts inside the case with a cotton swab. Then put the other one back in, and confirm you still see an amber light. *Then* you can do a hard reset by holding down the button for 15 seconds until you get yet another amber light and then flashing white.
## What do I do if my AirPods wont flash white?
Press and hold the setup button on the case for up to 10 seconds. The status light should flash white, which means that your AirPods are ready to connect. Hold the case, with your AirPods inside and the lid open, next to your iOS device.
## Are AirPods waterproof?
They are not waterproof but they do have sweat and dust resistance meaning they won’t be ruined by rain or falling in a puddle. That being said don’t like throw them in a pool or shower with them. They’re rated to be IPX4, so only sweat and splash proof.
## Can I use AirPods pro in the shower?
In spite of their improved protection, Apple also recommends not wearing AirPods Pro in the shower either. The reasons for this is the level of water resistance that’s on offer. While they can easily survive rain or a vigorous workout, AirPods Pro are not designed for use under running water.
## Will sweat ruin AirPods?
One word: sweat. After all, AirPods are neither sweat nor water-resistant. “Simply put, sweat—and any kind of water—will damage headphones permanently if it gets inside,” our executive editor, TJ Donegan, explains.
## Can I wear my Airpod pros in the shower?
Apple has two truly wireless earbud offerings. The AirPods Pro are water and sweat resistant, meaning they should survive heavy perspiration or a splash, though Apple tells users not to place them “under running water, such as a shower or faucet.”
## Can you overcharge AirPods?
The short answer is yes, it’s safe. Your AirPods can‘t overcharge and doing so overnight won’t damage their battery. In this article, we‘ll discuss new batteries but also about potential problems that could damage your battery and how to avoid them.
## Can you use AirPods on PS4?
Unfortunately, the PlayStation 4 doesn’t natively support AirPods. To connect AirPods to your PS4, you‘ll need to use a third-party Bluetooth. ‘: A beginner’s guide to the wireless technology Bluetooth is a wireless technology that allows the exchange of data between different devices.
## Why is AirPods so expensive?
There are several factors that combine to make the Airpods expensive. The first is that they’re an Apple product and the brand does not manufacture cheap products. There’s a fair amount of overhead that goes into the design, materials, and construction of each product manufactured.
# How to make yourself puke
Published
on
By
## Can you throw up your guts?
While it sounds unpleasant and unusual, it’s possible to vomit up your own fecal matter. Known in medical literature as “feculent vomiting,” throwing up poop is usually due to some type of blockage in the intestines. Learn what causes someone to throw up poop, and how to treat this condition.
## What Herb makes you vomit?
After decades of ipecac use for poisoning, researchers looked at all of the evidence about ipecac syrup. They agreed that ipecac syrup reliably caused vomiting.
## What is in ipecac that makes you vomit?
The actions of ipecac are mainly those of major alkaloids, emetine (methylcephaeline) and cephaeline. They both act locally by irritating the gastric mucosa and centrally by stimulating the medullary chemoreceptor trigger zone to induce vomiting.
## Can you throw up your guts?
While it sounds unpleasant and unusual, it’s possible to vomit up your own fecal matter. Known in medical literature as “feculent vomiting,” throwing up poop is usually due to some type of blockage in the intestines. Learn what causes someone to throw up poop, and how to treat this condition.
## What Herb makes you vomit?
After decades of ipecac use for poisoning, researchers looked at all of the evidence about ipecac syrup. They agreed that ipecac syrup reliably caused vomiting.
## What is in ipecac that makes you vomit?
The actions of ipecac are mainly those of major alkaloids, emetine (methylcephaeline) and cephaeline. They both act locally by irritating the gastric mucosa and centrally by stimulating the medullary chemoreceptor trigger zone to induce vomiting.
# How to get rid of home button on screen
Published
on
By
## How do I get rid of the circle on my iPhone screen?
Answer: A: Go to settings, general, accessibilty, assistive touch, turn off.
## How do I get rid of the home button on my iPhone?
Of course you can also turn off the onscreen Home button in iOS too if you decide you don’t like it, or don’t need it: Open the “Settings” app in iOS. Go to “General” and then choose “Accessibility” and then tap on “AssistiveTouch:” Toggle the “AssistiveTouch” switch to OFF position.
## How do I get rid of the floating home button on my iPad?
Answer: A: Settings > General Accessibility > Assistive Touch — off.
## How do I get rid of the floating home button on my iPhone 7?
Go to Settings > General > Accessibility > AssistiveTouch and move the switch to off.
## Why did Apple get rid of home button?
According to the report, Apple is ditching the button to make room for a larger screen in one version of the next iPhone. The home button has been a staple in iPhone design since the beginning. It’s used to wake a device, return to the app grid, and, more recently, for Touch ID fingerprint recognition.
## Where is my add to home screen button?
Tap the menu icon (3 dots in upper right-hand corner) and tap Add to homescreen. You’ll be able to enter a name for the shortcut and then Chrome will add it to your home screen.
READ ALSO Who from macedonia conquered greece
## How do I put the home button on my screen?
To add the home button function on-screen, turn on AssistiveTouch in the Accessibility section of Settings. To use the home button, tap the AssistiveTouch button and then tap the home button when it appears in the pop-out menu.
## Where is the home button on s20?
By default, you get the standard Android navigation buttons on the bottom of the screen, including the multi-task button, the home button, and the back button.
## Why is my home button not clicking?
Your iPhone may just need Calibrating and Restoring! This is the most common problem with the home button lagging and can be caused by a software flaw or just because your phone needs a reset. Before restoring your device and upgrading the iOS software, make sure all data is backed up to either iCloud or iTunes.
## How do I get my home button back?
Swipe the screen to the left to get to the All tab. Scroll down until you locate the currently running home screen. Scroll down until you see the Clear Defaults button (Figure A). Tap Clear Defaults.
To do this, follow these steps:
1. Tap the home button.
2. Select the home screen you want to use.
3. Tap Always (Figure B).
## What do you do when your iPhone Home button doesnt work?
iOS has a neat feature that helps when hardware buttons stop working. To activate this, go to the Settings app, and select General, then Accessibility. Scroll down to the Interaction section, and tap AssistiveTouch. Now tap the button next to AssistiveTouch so that it turns green and slides to the on position.
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https://notebooks.ai/rmotr-curriculum/estimates-of-location-a38351b2
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# Estimates of Location
Last updated: January 13th, 2020
# Estimates of Location¶
Variables with measured or count data might have thousands of distinct values. A basic step in exploring your data is getting a "typical value" for each feature (variable): an estimate of where most of the data is located (i.e., its central tendency).
👉 External resource: Jake Vanderplas' keynote about Statistics for Hackers at PyCon 2016: https://www.youtube.com/watch?v=Iq9DzN6mvYA
## Hands on!¶
In [1]:
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
We'll use the following Google Play Store Apps dataset in this lesson:
In [2]:
df = pd.read_csv('data/googleplaystore.csv')
Out[2]:
App Category Rating Reviews Installs Price Content Rating Genres Android Ver
0 Photo Editor & Candy Camera & Grid & ScrapBook ART_AND_DESIGN 4.1 159 10,000+ 0 Everyone Art & Design 4.0.3 and up
1 Coloring book moana ART_AND_DESIGN 3.9 967 500,000+ 0 Everyone Art & Design;Pretend Play 4.0.3 and up
2 U Launcher Lite – FREE Live Cool Themes, Hide ... ART_AND_DESIGN 4.7 87510 5,000,000+ 0 Everyone Art & Design 4.0.3 and up
3 Sketch - Draw & Paint ART_AND_DESIGN 4.5 215644 50,000,000+ 0 Teen Art & Design 4.2 and up
4 Pixel Draw - Number Art Coloring Book ART_AND_DESIGN 4.3 967 100,000+ 0 Everyone Art & Design;Creativity 4.4 and up
## Measuring Central Tendency¶
Mathematically, central tendency means measuring the center or distribution of location of values of a data set. It gives an idea of the average value of the data in the data set and also an indication of how widely the values are spread in the data set.
That in turn helps in evaluating the chances of a new input fitting into the existing data set and hence probability of success.
#### Frequency distribution of a variable¶
The first thing we're going to do is plot a line of every sample value of the Rating column:
In [3]:
df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10840 entries, 0 to 10839
Data columns (total 9 columns):
App 10840 non-null object
Category 10840 non-null object
Rating 9366 non-null float64
Reviews 10840 non-null int64
Installs 10840 non-null object
Price 10840 non-null object
Content Rating 10840 non-null object
Genres 10840 non-null object
Android Ver 10838 non-null object
dtypes: float64(1), int64(1), object(7)
memory usage: 762.3+ KB
In [4]:
df['Rating'].head(10)
Out[4]:
0 4.1
1 3.9
2 4.7
3 4.5
4 4.3
5 4.4
6 3.8
7 4.1
8 4.4
9 4.7
Name: Rating, dtype: float64
In [5]:
df['Rating'].plot(color='#3498db', figsize=(12,6))
Out[5]:
<matplotlib.axes._subplots.AxesSubplot at 0x7f3865945c18>
### Histogram¶
It's a mess, so we're going to make an accurate representation of the distribution of the sample values by getting the frequency of each value.
In [6]:
df['Rating'].value_counts().head(10)
Out[6]:
4.4 1109
4.3 1076
4.5 1038
4.2 952
4.6 823
4.1 708
4.0 568
4.7 499
3.9 386
3.8 303
Name: Rating, dtype: int64
In [7]:
freq = df['Rating'].value_counts().sort_index()
freq_frame = freq.to_frame()
freq_frame.plot.bar(color='#3498db', figsize=(12,6))
Out[7]:
<matplotlib.axes._subplots.AxesSubplot at 0x7f38638a4978>
This plot of the frequency (count) of the values is known as a Histogram:
In [8]:
df['Rating'].plot.hist(bins=20, color='#3498db', figsize=(12,6))
Out[8]:
<matplotlib.axes._subplots.AxesSubplot at 0x7f38637fd208>
In [9]:
df['Rating'].plot.hist(bins=10, color='#3498db', figsize=(12,6))
Out[9]:
<matplotlib.axes._subplots.AxesSubplot at 0x7f386370ff28>
## Density estimates¶
Related to the histogram is a Density plot, which shows the distribution of data values as a continuous line.
This density plot can be thought of as a smoothed version of a histogram, although it is typically computed directly from the data through a kernel density estimate.
We'll use Seaborn library to make our plots:
In [10]:
plt.figure(figsize=(12,6))
sns.distplot(df['Rating'].dropna())
Out[10]:
<matplotlib.axes._subplots.AxesSubplot at 0x7f38637d99b0>
## Mean¶
The sum of all values divided by the number of values. Also known as average. This is the most basic estimate of location.
The formula to compute the mean for a set of $n$ values $x_1, x_2, ..., x_n$ is:
$$Mean = \mu = \overline{x} = \frac{\sum\limits_{i}^n x_i }{n}$$
Let's calculate the mean of the Rating of Google Play Store Apps.
In [11]:
list(df['Rating'])[0:15]
Out[11]:
[4.1, 3.9, 4.7, 4.5, 4.3, 4.4, 3.8, 4.1, 4.4, 4.7, 4.4, 4.4, 4.2, 4.6, 4.4]
In [12]:
mean_rating = df['Rating'].mean()
mean_rating
Out[12]:
4.191757420456972
In [13]:
plt.figure(figsize=(12,6))
sns.distplot(df['Rating'].dropna())
# Mean line
plt.axvline(mean_rating, color='#e74c3c', linestyle='dashed', linewidth=2)
Out[13]:
<matplotlib.lines.Line2D at 0x7f38635e0a90>
#### Pros¶
• It works well for lists that are simply combined (added) together.
• Easy to calculate: just add and divide.
• It’s intuitive — it’s the number "in the middle", pulled up by large values and brought down by smaller ones.
#### Cons¶
• The average can be skewed by outliers — it doesn't deal well with wildly varying samples. The average of 100, 200 and -300 is 0, which is misleading.
## Median¶
The value such that one-half of the data lies above and below. If there is an even number of data values, the middle value is one that is not actually in the data set, but rather the average of the two values that divide the sorted data into upper and lower halves.
$$6, 2, 9, 3, 13, 4, 9, 7, 12, 8, 10, 5, 13, 9, 3$$
Compared to the mean, which uses all observations, the median depends only on the values in the center of the sorted data.
The median will be a robust estimator of location since it is not influenced by outliers that could sked the results.
In [14]:
numbers = [6,2,9,3,13,4,9,7,12,8,10,5,13,9,3]
np.median(numbers)
Out[14]:
8.0
Now lets calculate the median for the Rating column of Google Play Store dataset:
In [15]:
median_rating = df['Rating'].median()
median_rating
Out[15]:
4.3
In [16]:
plt.figure(figsize=(12,6))
sns.distplot(df['Rating'].dropna())
# Mean line
plt.axvline(mean_rating, color='#e74c3c', linestyle='dashed', linewidth=2)
# Median line
plt.axvline(median_rating, color='yellow', linestyle='dashed', linewidth=2)
Out[16]:
<matplotlib.lines.Line2D at 0x7f38634b2978>
## Mode¶
The most commonly occurring category or value in a data set.
In [17]:
df['Rating'].value_counts().head()
Out[17]:
4.4 1109
4.3 1076
4.5 1038
4.2 952
4.6 823
Name: Rating, dtype: int64
In [18]:
mode_rating = df['Rating'].mode()[0]
mode_rating
Out[18]:
4.4
In [19]:
plt.figure(figsize=(12,6))
sns.distplot(df['Rating'].dropna())
# Mean line
plt.axvline(mean_rating, color='#e74c3c', linestyle='dashed', linewidth=2)
# Median line
plt.axvline(median_rating, color='yellow', linestyle='dashed', linewidth=2)
# Mode line
plt.axvline(mode_rating, color='green', linestyle='dashed', linewidth=2)
Out[19]:
<matplotlib.lines.Line2D at 0x7f3863332748>
👉 External resource: Khan Academy has a video showing an example of how to calculate Mean, Median and Mode: https://youtu.be/k3aKKasOmIw
## Range and Mid Range¶
### Range (max - min)¶
In [20]:
dist_range = df['Rating'].max() - df['Rating'].min()
dist_range
Out[20]:
4.0
In [21]:
plt.figure(figsize=(12,6))
sns.distplot(df['Rating'].dropna())
# Mean line
plt.axvline(mean_rating, color='#e74c3c', linestyle='dashed', linewidth=2)
# Range line
plt.axvline(dist_range, color='green', linestyle='dashed', linewidth=2)
Out[21]:
<matplotlib.lines.Line2D at 0x7f38633a9ef0>
### Mid range (range / 2)¶
In [22]:
plt.figure(figsize=(12,6))
sns.distplot(df['Rating'].dropna())
# Mean line
plt.axvline(mean_rating, color='#e74c3c', linestyle='dashed', linewidth=2)
# Range line
plt.axvline(dist_range / 2.0, color='green', linestyle='dashed', linewidth=2)
Out[22]:
<matplotlib.lines.Line2D at 0x7f3862d7b400>
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| 3.6875 | 4 |
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https://developmentality.wordpress.com/tag/car-talk/
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### Archive
Posts Tagged ‘car talk’
## Car Talk Puzzler #5: The Perfect Square Dance
PUZZLER: The Perfect Square Dance!
Sally invited 17 guests to a dance party. She assigned each guest a number from 2 to 18, keeping 1 for herself. The sum of each couple’s numbers was a perfect square. What was the number of Sally’s partner?
Puzzler
The fifth in my ongoing series of solving Car Talk Puzzlers with my programming language of choice. I’m using Python again, just like last time.
There are a few pieces to this. The first is, how do we generate a list of all possible pairs that match the perfect square constraint? With list comprehensions and a helper function this is easy.
```def perfect_square(n):
return math.sqrt(n) == int(math.sqrt(n))
# range creates a list that's exclusive of last number. Thus to go from 1 to n,
# use range(1, n+1)
guests = range(1, num_guests + 1)
# By enforcing the x# a whole bunch of equivalent pairs (e.g. (3,6) and (6,3)).
pairs = [(x,y) for x in guests for y in guests if x<y and perfect_square(x+y)]
>>> guests
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
>>> pairs
[(1, 3), (1, 8), (1, 15), (2, 7), (2, 14), (3, 6), (3, 13), (4, 5), (4, 12), (5, 11), (6, 10), (7, 9), (7, 18), (8, 17), (9, 16), (10, 15), (11, 14), (12, 13)]```
OK great. At this point I have 18 pairs; 9 of these pairs together form the correct setup. Now, I could try to be very clever and use logic to deduce the correct pairs. But I’m lazy and it’s a small problem, so I’m going to use brute force. I’m just going to take all combinations of 9 pairs and test to see which one uses each guest exactly once. Not only am I so lazy as to do it this way, I’m not even going to write code to calculate the combinations. Instead, I’ll use the excellent itertools package.
```>>> itertools.combinations([1,2,3,4,5], 2)
# This is a sort of generator object which lazily returns the values as needed.
# To force them to all be evaluated at once, to see how this works,
# wrap the call in a list function.
>>>list(itertools.combinations([1,2,3,4,5],2))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]
```
This performs exactly as how we would expect. Note that I want combinations rather than permutations because the order does not matter.
As I mentioned earlier, I assume that a brute force solution is going to be fast enough. Let me check my math here. How many 9-element combinations taken from an 18-element set are there?
N choose K formula - number of k element combinations from set of n elements
Yeah, I think we can handle checking 48620 combinations.
OK, but how do we know whether a given combination (9 pairs of numbers) match our criteria? The easiest way is to check if the number of unique elements is the same as the number of guests; this indicates that each guest is used only once. First we flatten our nested list of tuples into a single list, and then use a set to check uniqueness:
```
def flatten(nested):
""" Flatten one level of nesting. Returns a generator object
For instance:
list(flatten([(1,3),(5,6)])) --> [1,3,5,6] """
return itertools.chain.from_iterable(nested)
def all_guests_present_once(combination):
""" Returns whether each guest is present once
Combination is a list of tuples, e.g. [(1,5),(7,8)]
"""
flattened = list(flatten(combination))
return len(set(flattened)) == len(flattened)
>>> all_guests_present_once([(1,3),(4,5)])
True
>>> all_guests_present_once([(1,3),(3,6)])
False
```
As you can see, I’m shortcutting a little bit and not checking for the perfect square aspect, since we already did that when constructing our set of pairs.
OK we’re ready to throw it all together.
```
def dance_arrangement(num_guests):
"""
Returns a valid pairing for all guests if possible, else an empty set
"""
# Clearly you need an even number of guests to have everyone paired
if num_guests % 2 == 1:
return []
else:
# range creates a list that's exclusive of last number. Thus to go from 1 to n,
# use range(1, n+1)
guests = range(1, num_guests + 1)
# By enforcing the x # a whole bunch of equivalent pairs (e.g. (3,6) and (6,3)).
pairs = [(x,y) for x in guests for y in guests if x # brute force search
all_arrangements = itertools.combinations(pairs, num_guests / 2)
return filter(all_guests_present_once, all_arrangements)
```
Running the program with num_guests = 18, we get
```[((1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 18), (8, 17), (9, 16))]
```
Tada! Thus Sally’s partner was number 15.
I stumbled onto this page which also sought to solve this problem using programming; it ends with the question
“I wonder if 18 is unique as the total number of dancers which has a solution. It should be easy to modify the program to check”
It is very easy to modify, and I checked. I ran the program on all number of guests less than 20, and these are the results:
```8 [((1, 8), (2, 7), (3, 6), (4, 5))]
14 [((1, 8), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9))]
16 [((1, 8), (2, 7), (3, 6), (4, 5), (9, 16), (10, 15), (11, 14), (12, 13))]
18 [((1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 18), (8, 17), (9, 16))]
```
As you can see, 8, 14, and 16 guests can also be paired up in this way. Something to keep in mind the next time you are going to have a party.
Full sourcecode can be found on Github.
Categories: Python, Uncategorized
## Car Talk Puzzler #4: Flipping Ages
RAY: This was sent in many weeks ago by Wendy Gladstone, and as usual I tweaked it a little bit.
She writes: “Recently I had a visit with my mom and we realized that the two digits that make up my age when reversed resulted in her age. For example, if she’s 73, I’m 37. We wondered how often this has happened over the years but we got sidetracked with other topics and we never came up with an answer.
“When I got home I figured out that the digits of our ages have been reversible six times so far. I also figured out that if we’re lucky it would happen again in a few years, and if we’re really lucky it would happen one more time after that. In other words, it would have happened 8 times over all. So the question is, how old am I now?”
Here’s the fourth in my Car Talk Puzzler series; today I’m going to be using Python because it’s my current favorite language, and because it’s well suited to filtering, mapping, etc. I won’t put too much commentary here.
```# Find all the ages such that the second age is the reverse of the first age. Don't worry that there are a lot of impossibilities; we'll fix it through filtering
# Note that [::-1] is the slice operator that says iterate backwards through the string; this effectively reverses the list.
matching_ages = map(lambda x:(x, int(str(x)[::-1])), range(0,100))
matching_ages
# OUT: [(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 1), (11, 11), (12, 21), (13, 31), (14, 41), (15, 51), (16, 61), (17, 71), (18, 81), (19, 91), (20, 2), (21, 12), (22, 22), (23, 32), (24, 42), (25, 52), (26, 62), (27, 72), (28, 82), (29, 92), (30, 3), (31, 13), (32, 23), (33, 33), (34, 43), (35, 53), (36, 63), (37, 73), (38, 83), (39, 93), (40, 4), (41, 14), (42, 24), (43, 34), (44, 44), (45, 54), (46, 64), (47, 74), (48, 84), (49, 94), (50, 5), (51, 15), (52, 25), (53, 35), (54, 45), (55, 55), (56, 65), (57, 75), (58, 85), (59, 95), (60, 6), (61, 16), (62, 26), (63, 36), (64, 46), (65, 56), (66, 66), (67, 76), (68, 86), (69, 96), (70, 7), (71, 17), (72, 27), (73, 37), (74, 47), (75, 57), (76, 67), (77, 77), (78, 87), (79, 97), (80, 8), (81, 18), (82, 28), (83, 38), (84, 48), (85, 58), (86, 68), (87, 78), (88, 88), (89, 98), (90, 9), (91, 19), (92, 29), (93, 39), (94, 49), (95, 59), (96, 69), (97, 79), (98, 89), (99, 99)]
# Here we filter by only allowing matches in which the mother's age is greater than that of the child. Note the use of a lambda expression, basically an anonymous function.
filtered1 = filter(lambda (mother,child):mother > child, matching_ages)
filtered1
# OUT: [(10, 1), (20, 2), (21, 12), (30, 3), (31, 13), (32, 23), (40, 4), (41, 14), (42, 24), (43, 34), (50, 5), (51, 15), (52, 25), (53, 35), (54, 45), (60, 6), (61, 16), (62, 26), (63, 36), (64, 46), (65, 56), (70, 7), (71, 17), (72, 27), (73, 37), (74, 47), (75, 57), (76, 67), (80, 8), (81, 18), (82, 28), (83, 38), (84, 48), (85, 58), (86, 68), (87, 78), (90, 9), (91, 19), (92, 29), (93, 39), (94, 49), (95, 59), (96, 69), (97, 79), (98, 89)]
# Assume that the mother was at least 15 when she had the kid, and no more than 60
filtered2 = filter(lambda(mother, child):mother-child >= 15 and mother-child < 60, filtered1)
filtered2
# OUT: [(20, 2), (30, 3), (31, 13), (40, 4), (41, 14), (42, 24), (50, 5), (51, 15), (52, 25), (53, 35), (60, 6), (61, 16), (62, 26), (63, 36), (64, 46), (71, 17), (72, 27), (73, 37), (74, 47), (75, 57), (82, 28), (83, 38), (84, 48), (85, 58), (86, 68), (93, 39), (94, 49), (95, 59), (96, 69), (97, 79)]
len(filtered2)
# OUT: 30
# Create a new list comprised of the differences in age between mother and child
age_diff = map(lambda(mother,child):mother-child, filtered2)
age_diff
# OUT: [18, 27, 18, 36, 27, 18, 45, 36, 27, 18, 54, 45, 36, 27, 18, 54, 45, 36, 27, 18, 54, 45, 36, 27, 18]
sorted(age_diff)
# OUT: [18, 18, 18, 18, 18, 18, 18, 27, 27, 27, 27, 27, 27, 36, 36, 36, 36, 36, 45, 45, 45, 45, 54, 54, 54]
# The puzzler states that it's will happen a total of 8 times; that matches the age difference of 18 years
filter(lambda(mother,child):mother-child == 18, filtered3)
# OUT: [(20, 2), (31, 13), (42, 24), (53, 35), (64, 46), (75, 57), (86, 68), (97, 79)]
```
Thus the mother is currently 75 years old and the daughter is 57. Tada
Categories: Python
## Car Talk Puzzler #3: The Palindromic Odometer
The Palindromic Odometer
Driving along, Terry notices that the last four digits on the odometer are palindromic. A mile later, the last five digits are palindromic. [A mile later, the middle four digits are palindromic.] [One mile after that], all six are palindromic. What was the odometer reading when Terry first looked at it?
This is the third in my series of Car Talk puzzler solutions using whatever programming language strikes my fancy. Just like the last one I wrote, this one uses Scala. This is a much shorter post, as I am assuming some familiarity with Scala; go back and read the last post if you haven’t already.
```object PalindromicPuzzlerSolver {
// Pad out numbers to 6 places
// Note that I can cal Java library functions seamlessly
val df = new java.text.DecimalFormat("000000")
// A string is a palindrome if it reversed equals itself
def isPalindrome(word: String): Boolean = {
// Reverse is a method of a RichString, which a String can be implicitly converted
// into. Necessary to convert the RichString back into String after the reversal
word == word.reverse.toString()
}
def fitsDescription(miles: Int): Boolean = {
// Last 4 are palindome (dropping first 2)
isPalindrome(df.format(miles).drop(2)) &&
// Add one, last 5 are palindrome
isPalindrome(df.format(miles + 1).drop(1)) &&
// Add two more, all six are a palindrome
isPalindrome(df.format(miles + 3))
}
def main(args: Array[String]) {
val max = 1000000
println( 0 until max filter fitsDescription )
// Prints out
// RangeF(198888, 199999)
// This is the same as the java style:
// println( (0.until(max)).filter(fitsDescription) )
// This syntax only works when you are dealing with
// methods that take one argument.
}
}
```
Thus the solution is 198888, just as the answer proclaims. Another win for brute force.
There are three noteworthy things in this example, which didn’t come up in the last post (or weren’t explained fully)
1. Interoperability with Java
2. Use of drop
3. Convenience methods for specifying a range
## Interoperability with Java
Any Java class can be imported and used with Scala; that is one of its strongest selling points. Scala runs in the same JVM as Java, so it makes sense that it should have access to Java classes. In fact many Scala classes are thin wrappers around Java objects, with syntactic sugar built in. For instance, Scala has a BigInt class which wraps a Java BigInteger, while providing overloaded methods to allow them to be used more like standard numbers. For instance,
```scala> BigInt("103058235287305927350") + BigInt("288888203959230529352")
res16: BigInt = 391946439246536456702
res21: java.math.BigInteger = 391946439246536456702
```
Note that the + is actually a method call; this is exactly the same thing as calling
```scala> BigInt("103058235287305927350").+(BigInt("288888203959230529352"))
res22: BigInt = 391946439246536456702
```
In Scala, method names don’t have to be alphanumeric, which is how it gets away with what looks like operator overloading. This makes it a lot nicer to work with classes like BigDecimal or BigInteger, as you can use the more commonly used addition and subtraction symbols rather than explicitly calling “add” and “subtract” methods. Just realize that you are really calling methods like any other under the hood. When you understand this, you’ll also understand why Scala uses an underscore instead of * for wildcard – * can be used in method names, as it is used for multiplication!
## Drop
Scala has full support for slicing iterable objects; see my previous discussion on slices if this term is unfamiliar.
```scala> List(1,2,3,4).slice(1,3)
res27: List[Int] = List(2, 3)
scala> List(1,2,3,4).slice(1)
warning: there were deprecation warnings; re-run with -deprecation for details
res28: Seq[Int] = List(2, 3, 4)
scala> List(1,2,3,4).drop(1)
res29: List[Int] = List(2, 3, 4)
```
So drop(n) is exactly the same as slice(n) – it provides a new list with the first n elements removed. I use drop instead of slice due to the deprecation of the single argument form of slice.
Specifying a range
If you need to iterate over a range of numbers in Java, you’d do something like
for (int i = 0; i < max; i++) {
}
whereas in Python you can do something like
for x in range(max):
Scala similarly provides methods to create a range of numbers to iterate over.
```scala> 0 until 10
res9: Range = Range(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
scala> 0 to 10
res10: Range.Inclusive = Range(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
```
Again, I want to stress that until is a method like any other; you are just free to leave off the . and the parens when it is unambiguous what you mean. Note that these commands are actually creating a sequence in memory. If you only need to access each element one at a time, and you are dealing with a large range, you might instead create an Iterator rather than a sequence. The Iterator will create each subsequent number as necessary, rather than building up the whole list ahead of time. This is much more space efficient.
Note that Python makes this distinction as well:
```>>> range(10)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> xrange(10)
xrange(10)
```
xrange returns a generator object, as opposed to range which returns the actual list.
Returning to Scala, if I wanted an iterator instead I would do
```scala> Iterator.range(0,10)
res11: Iterator[Int] = non-empty iterator
scala> Iterator.range(0,10).foreach(println)
0
1
2
3
4
5
6
7
8
9
```
Conclusion
I have shown another solution using Scala, while illustrating a few more features. In particular one needs to understand that Scala’s syntactic sugar is actually masking standard function calls under the hood; x + y is the same as x.+(y), where “+” is a valid function name.
## Car Talk Puzzler #2: What do these words have in common? An imperative and functional approach using Java and Scala
May 15, 2010 1 comment
Here’s the second in my series of Car Talk puzzler solutions. Today I’m going to be illustrating a solution first in Java, and then in Scala in order to show how functional programming features can drastically reduce the number of lines of code, while still expressing the developer’s intent.
Puzzler:
What do these words have in common: pig, table, cab, real, yet, and ride?
RAY: Here’s the answer. If you notice, the first letter in each of the words is from farther down in the alphabet than all the subsequent letters. And if we assigned a number to every letter in the alphabet, so that A is one, B is two, C is three and so on…
TOM: Yeah, yeah, yeah.
RAY: Then the numerical place in the alphabet of the first letter is equal to the sum of the numerical places of all the other letters. For example, Table: T is 20, A is 1, B is 2, L is 12, E is 5, so 1 plus 2 plus 12 plus 5 –
TOM: Is 20.
So, we have the answer. But as the guys mention on the show, there are probably other words as well. Let’s do a brute force search through the dictionary to find all such words. The dictionary I”ll be using is available here, as a new line separated list of words, one per line. This makes it very easy to deal with.
Here’s the rather straightforward solution in Java:
``` import java.io.FileReader;
import java.io.FileNotFoundException;
import java.io.IOException;
/**
* Determines all the words such that the first letter's ordinal position
* equals the sum of the remaining letters' ordinal positions. For instance,
* "table"'s ordinal numbers are [20,1,2,12,5], and 20 = 1+2+12+5. Thus
* "table" has this property.
*/
public class OrdinalPositionPuzzlerSolver {
/**
* Assumes the string consists just of A-Z, a-z.
*/
public static boolean hasProperty(String s) {
if (s.length() < 2) {
return false;
}
char[] letters = s.toCharArray();
int sum = 0;
for (int i = 1; i < letters.length; i++) {
int ordinal = ordinal(letters[i]) ;
sum += ordinal;
}
return sum == ordinal(letters[0]);
}
/**
* @return the 1 based ordinal position of the letter. For instance,
* a is 1, b is 2, etc.
*/
public static int ordinal(char letter) {
return Character.toLowerCase(letter) - 'a' + 1;
}
public static void main(String[] args) {
String dictionaryPath = "dict.txt";
/**
http://www.javapractices.com/topic/TopicAction.do?Id=42
*/
try {
try {
String line = null;
while (( line = input.readLine()) != null){
if (hasProperty(line)) {
System.out.println(line);
}
}
}
catch (FileNotFoundException e) {
System.err.println("Error, file " + dictionaryPath + " didn't exist.");
}
finally {
input.close();
}
}
catch (IOException e) {
System.err.println("IO Error:" + e);
}
}
}
```
The only place that bears some explanation might be the ordinal function:
```/**
* @return the 1 based ordinal position of the letter. For instance,
* a is 1, b is 2, etc.
*/
public static int ordinal(char letter) {
return Character.toLowerCase(letter) - 'a' + 1;
}
```
Basically, letters (characters) in Java are really just numbers under the hood which get mapped back into letters when printed to the screen. So, we can do arithmetic using letters. ‘a’ + 1 = ‘b’, ‘b’ + 2 = ‘d’, and so on. (This only works because the letters a-z and A-Z are mapped to a continuous range of integers). So if we want to convert a->1, first we figure out a’s position from the start of the alphabet (‘a’ -‘a’ == 0, it’s the first), and add 1 to get the 1-based number.
The other thing to notice is how much junk code there is just to read the lines from our file:
```try {
try {
String line = null;
while (( line = input.readLine()) != null){
if (hasProperty(line)) {
System.out.println(line);
}
}
}
catch (FileNotFoundException e) {
System.err.println("Error, file " + dictionaryPath + " didn't exist.");
}
finally {
input.close();
}
}
catch (IOException e) {
System.err.println("IO Error:" + e);
}
```
Believe it or not, this is relatively idiomatic Java code to accomplish this task. First we wrap a file reader in a buffered reader so that we do not continuously access the file; rather we read chunks into the file and buffer them for in-memory access. That’s 20 lines of code just to iterate over the lines of a file, determine if they match a pattern, and print it to the terminal.
It really should not be this hard.
As I’ve said before, Java’s verbosity is one of my biggest complaints about it. The program works, and it’s relatively straightforward, but it can’t really be described as elegant. In that same post I mentioned my interest in Scala as an alternative to Java. It’s a compiled language that runs in the JVM, but through its type inferencing and functional language aspects, it is much more concise; the creator Martin Odersky claims a typical Source Line of Code (SLOC) reduction of 2x between equivalent Scala and Java programs. Let’s reimplement the program in Scala, and see if this claim rings true. (Note: I’m not an expert in scala so there’s probably even more concise, idiomatic ways to accomplish what I’m trying to do. But let’s start somewhere).
The syntax of scala is very similar to that of Java. One of the main differences is that type declarations occur after the variable rather than before.
``` int x = 0; // Java
val x: Int = 0 // Scala
```
Another syntactic difference is that generic type arguments occur within square brackets rather than angle brackets
``` List<Integer> intList = ...; // java
val intList: List[Int] = ..// scala
```
Because of this, array indexing is done through use of parentheses rather than square brackets.
``` // Java
int[] x = {1,2,3,4};
x[0] // 1
```
``` // Scala
val x: List[Int] = List(1,2,3,4)
x(0) // 1
```
Due to type inferencing, you often don’t need to declare the type of your variables; the compiler can figure it out. Thus the last declaration could be rewritten as
``` val x = List(1,2,3,4)
```
I highly recommend you watch Martin Odersky’s introduction to Scala video; it’s about an hour long and goes over the design considerations and differences from Java.
There are no static functions in Scala; rather you create a singleton object through the keyword ‘object’ and access the method in the same way.
Finally, there is a read/eval/execute interpreter built-in, accessible by calling ‘scala‘ in the command line. This allows you to experiment without having to explicitly compile any code; it allowed me to create the pieces iteratively before plunking them into a file for later compilation.
``` scala
Welcome to Scala version 2.7.3.final (Java HotSpot(TM) Client VM, Java 1.5.0_22).
Type in expressions to have them evaluated.
scala> 1+1
res0: Int = 2
scala> print("hello world")
hello world
scala>
```
OK, enough beating around the bush. Let’s solve this puzzler!
We’ll tackle this in 3 parts.
1) Summing a list of numbers
2) Completing the ordinal number function
3) Iterating over all the lines in the dictionary file
## Summing a list of numbers
A first pass at a sum function might be as follows:
```def sum(nums: Iterable[Int]): Int = {
var sum = 0
for (num <- nums) {
sum += num
}
sum
}
```
Note the declaration of sum as a var as opposed to a val; a val is equivalent to a final value while a var can be reassigned. Why did I declare the nums argument to be of type Iterable rather than List? Well, a List is iterable. So is an array. So are any number of things. We want to keep this function as generic as possible. Pasting this into our interpreter, we can test this out.
```scala> def sum(nums: Iterable[Int]): Int = {
| var sum = 0
| for (num <- nums) {
| sum += num
| }
| sum
| }
sum: (Iterable[Int])Int
scala> sum(List())
res4: Int = 0
scala> sum(List(1,2,3,4,5))
res5: Int = 15
scala> sum(Array(1,2,3,4))
res6: Int = 10
```
This works, but let’s look at another way to solve this problem – a functional rather than imperative solution. Java is inherently imperative, meaning you’ll see a lot of explicit while and for loops. Here’s just a quick excerpt of each the wikipedia articles to define each programming paradigm:
In computer science, functional programming is a programming paradigm that treats computation as the evaluation of mathematical functions and avoids state and mutable data. It emphasizes the application of functions, in contrast to the imperative programming style, which emphasizes changes in state.
In computer science, imperative programming is a programming paradigm that describes computation in terms of statements that change a program state. In much the same way that imperative mood in natural languages expresses commands to take action, imperative programs define sequences of commands for the computer to perform.
So whereas the imperative manner is completely explicit in how the sum is created, using a mutable state variable (sum), a more functional approach will eliminate the use of mutable state altogether.
The reduce function is absolutely crucial to understand in order to program in a functional manner; let’s examine Python’s definition of it before applying it to the problem at hand.
Here is the Python definition of reduce function:
reduce(function, iterable[, initializer])
Apply function of two arguments cumulatively to the items of iterable, from left to right, so as to reduce the iterable to a single value. For example, reduce(lambda x, y:x+y, [1, 2, 3, 4, 5]) calculates ((((1+2)+3)+4)+5). The left argument, x, is the accumulated value and the right argument, y, is the update value from the iterable. If the optional initializer is present, it is placed before the items of the iterable in the calculation, and serves as a default when the iterable is empty. If initializer is not given and iterable contains only one item, the first item is returned.
And the description of the built-in sum function is as follows:
sum(iterable[, start])
Sums start and the items of an iterable from left to right and returns the total. start defaults to 0. The iterable‘s items are normally numbers, and are not allowed to be strings. The fast, correct way to concatenate a sequence of strings is by calling ”.join(sequence). Note that sum(range(n), m) is equivalent to reduce(operator.add,range(n), m) To add floating point values with extended precision, see math.fsum().
While Scala doesn’t have a built-in sum function, it does have support for reducing a list through use of a function. It has both a reduce left (left to right reduction) and reduce right (right to left reduction). Here is Scala’s definition of reduceLeft:
def reduceLeft[B >: A](op : (B, A) => B) : B
Combines the elements of this iterable object together using the binary operator op, from left to right
Notes Will not terminate for infinite-sized collections.
Parameters
op – The operator to apply
Returns op(… op(a0,a1), …, an) if the iterable object has elements a0, a1, …, an.
Throws
Predef.UnsupportedOperationException – if the iterable object is empty.
This might be a little confusing but bear with me.
First, we need to write a binary function for our sum operator – after all, we want to pairwise reduce the list by using an add operator. A first pass is as follows:
```scala> def add(x: Int, y: Int): Int = { x + y }
res12: Int = 10
java.lang.UnsupportedOperationException: Nil.reduceLeft
at scala.List.reduceLeft(List.scala:1093)
at .(:6)
at .()
at RequestResult\$.(:3)
at RequestResult\$.()
at RequestResult\$result()
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethod...
res14: Int = 0
```
Note that you cannot call sum on an empty list any more; doing so results in an UnsupportedOperationException. This is arguably more correct than returning 0, but we will have to ensure that we don’t attempt to sum an empty iterable object.
So a working sum definition would be
```def sum(nums: Iterable[Int]): Int = {
def add(x: Int, y: Int): Int = { x + y }
}
```
Note that you can nest function definitions, as functions are objects in Scala, and thus can be defined wherever a variable could. Note too that I never have to explicitly return a value; the last evaluated value is implicitly returned by the function.
Finally, if you want to be even more terse, you can replace the explicitly defined add function with an anonymous one:
```def sum(nums: Iterable[Int]): Int = {
nums.reduceLeft(_+_)
}
```
Read the Daniel Spiewak’s excellent blog post for a better explanation for what’s going on here. Here’s a quick explanation:
(_+_) is actually a Scala shorthand for an anonymous function taking two parameters and returning the result obtained by invoking the + operator on the first passing the second. Less precisely: it’s addition. Likewise, (_*_) would define multiplication.
## Completing the ordinal number functional
Recall the function that we had in Java for computing the ordinal position of each letter:
``` public static boolean hasProperty(String s) {
if (s.length() < 2) {
return false;
}
char[] letters = s.toCharArray();
int sum = 0;
for (int i = 1; i < letters.length; i++) {
int ordinal = ordinal(letters[i]) ;
sum += ordinal;
}
return sum == ordinal(letters[0]);
}
/**
* @return the 1 based ordinal position of the letter. For instance,
* a is 1, b is 2, etc.
*/
public static int ordinal(char letter) {
return Character.toLowerCase(letter) - 'a' + 1;
}
```
Again, we could directly translate this code into Scala but doing so would not be making full use of Scala’s functional underpinnings. We are converting from characters of the string to their ordinal positions in the alphabet. While the Java code does this one letter at a time, keeping a sum, it’s just as correct to do the entire mapping process and then sum over the new list. Pictorially:
Just as there is a reduce function to convert an iterable object into a single entry, so too is there a function to convert an Iterable of one type into an Interable of another; it is the map function.
```// The underscore in an anonymous function like this is a placeholder for the entry in the iterable collection currently being processed
scala> List(1,3,4).map(_+5)
res19: List[Int] = List(6, 8, 9)
scala> List("hi","there").map(_.toUpperCase())
res22: List[java.lang.String] = List(HI, THERE)
```
The above examples map from Int-> Int and String->String but there’s nothing stopping us from converting types.
```scala> List(100,300,400).map(_.toString())
res23: List[java.lang.String] = List(100, 300, 400)
```
Finally, let’s apply the ordinal function to a string:
```scala> "cat".map(Character.toLowerCase(_) - 'a' + 1)
res24: Seq[Int] = ArrayBufferRO(3, 1, 20)
```
Thus the entire function for determining if the word is a match is
[sourecode language=”scala”]
/**
* Determines if the word has the desired property of the ordinal position
* of first letter equalling the sum of the ordinal positions of the other
* letters
* Expects letters in the range [a-zA-Z];
*/
def matches(word: String): Boolean = {
if (word.length() < 2) {
false
}
else {
val ordinals = word.map(Character.toLowerCase(_) – 'a' + 1)
val ordinalSum = sum(ordinals.drop(1))
ordinals(0) == ordinalSum
}
}
[/sourecode]
## Iterating over the lines in a file
Remember how it took 20 lines to iterate over all the lines in a file? Fortunately Scala makes it incredibly easy:
```val lines = scala.io.Source.fromFile("dict.txt").getLines
```
The variable is actually not an array or a list; it is an iterator over the lines in the file. But iterators are still iterable, so all of our previously mentioned functions for dealing with iterable objects still work. Again, rather than looping over each element, explicitly testing, and printing whether it matched, we will use a functional approach, namely the ‘filter’ function.
```scala> List(1,2,3,4).filter(_%2==0)
res25: List[Int] = List(2, 4)
```
The function passed into filter must return Boolean; since our matches() function does that, we can insert it right in:
```val lines = scala.io.Source.fromFile("dict.txt").getLines
val matching = lines.filter(matches)
matching.foreach(print(_))
```
When we run the complete code, we get the following list of matches:
• acknowledgements
approximation
bumptiously
capitalizations
circuitously
conceptualizing
consciousness
consistently
credulousness
crystallizing
disconcertingly
discriminatory
electrocutions
excommunication
existentialism
experimentally
exploitations
expressively
extraneously
governmentally
impersonations
incomprehensibly
indispensability
initializations
intercommunicated
interpolations
notwithstanding
organizationally
oversimplifying
phenomenologically
sportswriting
Wait a minute… what’s going on here? These obviously don’t follow the pattern. Let’s open a terminal and do some testing.
First we paste our two helper method definitions in
``` Welcome to Scala version 2.7.3.final (Java HotSpot(TM) Client VM, Java 1.5.0_22).
Type in expressions to have them evaluated.
scala> def sum(nums: Iterable[Int]): Int = {
| nums.reduceLeft[Int](_+_)
| }
sum: (Iterable[Int])Int
scala>
scala> /**
| * Determines if the word has the desired property of the ordinal position
| * of first letter equalling the sum of the ordinal positions of the other
| * letters
| * Expects letters in the range [a-zA-Z];
| */
scala> def matches(word: String): Boolean = {
| val ordinals = word.map(Character.toLowerCase(_) - 'a' + 1)
| val ordinalSum = sum(ordinals.drop(1))
| ordinals(0) == ordinalSum
| }
matches: (String)Boolean
scala> matches("table")
res0: Boolean = true
scala> matches("chair")
res1: Boolean = false
scala> matches("consistently")
res2: Boolean = false
```
This is very strange. “consistently” is one of the words on the list, but it doesn’t pass the test on its own. In fact, none of them do.
What’s the problem? After googling and debugging for a bit, I realized that the precondition I was assuming in my matches method was being violated; the letters were not in the range [A-Za-z]. How could that be? Well, it turns out that the getLines() method returns the lines in the file – with new lines included!
```scala> matches("consistently\r\n")
res5: Boolean = true
```
(The file had been created on Windows, and thus has a carriage feed as well as a new line marker).
Fortunately there is a method to eliminate whitespace (including control characters like these); it’s known as trim.
We can put the trim in one of two places, either in the matches() method itself,
```def matches(word: String): Boolean = {
val ordinals = word.trim().map(Character.toLowerCase(_) - 'a' + 1)
val ordinalSum = sum(ordinals.drop(1))
ordinals(0) == ordinalSum
}
scala> matches("consistently\r\n")
res6: Boolean = false
scala> matches("consistently")
res7: Boolean = false
```
Or we can introduce the trim while iterating over the lines in the collection. I’ve opted for this second approach; I’m making it a precondition of calling the method matches() that the input is already sanitized. We introduce an implicit mapping from the string to the trimmed version of the string:
```// An iterator of lines in the file; note that they have newline control characters
val lines = scala.io.Source.fromFile("dict.txt").getLines
// Filter the lines, eliminating the new lines through trim mapping
val matching = lines.filter(i => matches(i.trim()))
matching.foreach(print(_))
```
After making the change, the result is
• cab
• hag
• leaf
• leg
• maced
• mica
• mid
• need
• pig
• real
• ride
• same
• sand
• seam
• sod
• table
• toe
• vial
• ward
• whack
• who
• wick
• win
• yeas
• yet
• zebra
• zinc
The final version of the program is
```object OrdinalPositionSolver {
/**
* Determines if the word has the desired property of the ordinal position
* of first letter equalling the sum of the ordinal positions of the other
* letters
* Expects letters in the range [a-zA-Z];
*/
def matches(word: String): Boolean = {
if (word.length() < 2) {
false
}
else {
val ordinals = word.map(Character.toLowerCase(_) - 'a' + 1)
val ordinalSum = sum(ordinals.drop(1))
ordinals(0) == ordinalSum
}
}
def sum(nums: Iterable[Int]): Int = {
nums.reduceLeft[Int](_+_)
}
def main(args: Array[String]) {
// An iterator of lines in the file; note that they have newline control characters
val lines = scala.io.Source.fromFile("dict.txt").getLines
// Filter the lines, eliminating the new lines through trim mapping
val matching = lines.filter(i => matches(i.trim()))
matching.foreach(print(_))
}
}
```
## Conclusion
I have solved the puzzler in two different ways, once in an imperative Java style and once in a more functional Scala style. The functional way is extremely concise and expressive, doing in 30 lines of code what it takes the Java code to do in over 60, confirming the claim that a typical Scala program is at least twice as compact as the corresponding Java program. I have introduced some of the salient syntactic differences between Java and Scala. I have also introduced some of the mainstays of the functional programming paradigm, namely map, reduce, and filter. While there are efforts to make Java more functional, the fact that you cannot pass functions as objects severely inhibits the use of functional programming paradigms.
## Car Talk Puzzler #1: The Bank Temperature Sign
May 1, 2010 1 comment
One of my favorite podcasts is NPR’s Car Talk, and one of my favorite segments of the show is the weekly puzzler, in which listeners have a week to solve a puzzle that is usually automotive in nature. It struck me that this would be an ideal way to illustrate some beginning programming techniques, while also illustrating some real world problem solving uses of programming. Initially I will be posting solutions in Python, but I will probably show how to do them in other ways as well.
You can see the whole archive of 2010 puzzles, but the specific one I’m starting with can be found here.
In short:
PREVIOUS PUZZLER: Stevie and His Moto
Stevie’s riding his motorcycle to work when he sees a big sign displaying the temperature in Fahrenheit and in centigrade. The digits are exactly reversed. He notices the same thing on the way home. What were the temperatures?
This is an ideal choice for a programming solution because it can be brute forced; a computer can try all possible solutions near instantaneously, something they are very good at. Not all of the puzzlers are so conducive to a programmatic solution.
For those who already know the basics of programming, I’ll post my program first. Afterwards I post explanations more geared towards beginners, but there still is something that more advanced programmers might not know in Python.
Here’s my solution:
```def fahrenheitToCelsius(temp):
return (temp - 32) / 1.8
def reverseString(s):
# Step size is -1, so start at the end and work backwards
return s[::-1]
# Presumably it was above freezing if it's a spring day
LOW_TEMP_FAHRENHEIT = 32
# Presumably it was below 100 degrees
HIGH_TEMP_FAHRENHEIT = 100
def main():
for temp in range(LOW_TEMP_FAHRENHEIT, HIGH_TEMP_FAHRENHEIT):
# Round the float to nearest whole number, cast to integer
celsius = int(round(fahrenheitToCelsius(temp)))
# Convert the integers to strings so as to be able to reverse the digits
fahrenheitString = str(temp)
celsiusString = str(celsius)
# See if the reversed digits match
if fahrenheitString == reverseString(celsiusString):
print temp, celsius
if __name__ == '__main__':
main()
```
Spoiler:
```python Temperature.py
61 16
82 28
```
Tada! The program calculated the correct solution.
Let’s break the problem into a few logical steps:
1) Convert a Fahrenheit temperature to Celsius (we’re going to be able to do this in order to determine what the temperature on the signs must have been)
2) Round a number to the nearest whole number (converting Fahrenheit to Celsius will often leave a fractional part, but most signs don’t display temperature in increments smaller than 1 degree)
3) Reverse the digits of a number (in order to tell whether the temperatures are the reverse of each other)
4) Test a whole bunch of Fahrenheit temperatures to discover the matching Celsius ones, using steps 1 – 3.
Let’s explain each part in order.
## Convert Fahrenheit to Celsius
The first part is the most straightforward:
```def fahrenheitToCelsius(temp):
return (temp - 32) / 1.8
```
There’s not a lot to say about this part; this is the formula to convert from Fahrenheit to Celsius. If you’re new to Python or programming as a whole, this whole block is known as a method declaration; basically methods take in data, do some calculation with the data, and more often than not, return a new piece of data. In our case, we are passing in the temperature in Fahrenheit, and what comes out of it is a temperature in Celsius. Just to show that this indeed works:
```Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on win32
>>> def fahrenheitToCelsius(temp):
... return (temp - 32) / 1.8
...
>>> fahrenheitToCelsius(212)
100.0
>>> fahrenheitToCelsius(32)
```
## Round a number to nearest whole number
OK, how about rounding a fractional number to the nearest whole number?
Python has a function called, easily enough, round which does exactly that. You choose how many decimal places you want; by default it rounds to 0 (i.e. the nearest whole number)
```>>> round (2.252, 2)
2.25
>>> round (2.252 )
2.0
```
Notice how even though we round to 0 decimal places, the answer is “2.0”. This is because round returns a double rather than an integer. Integers are what we would think of as “whole numbers” – -1023, 2035, 733, etc. etc. Doubles are an (inexact) representation of a number with a fractional part. It’s a little beyond the scope of this tutorial, but suffice to say there are numbers that a computer cannot represent exactly
```>>> round (2.1, 1)
2.1000000000000001
```
The reason I make the distinction is that we want to treat our numbers as integers; after all we’re rounding them to 0 decimal places, so they are in fact integers. We need the computer to treat them that way. We do this with the int() method
```>>> round(2.23501)
2.0
>>> int(2.0)
2
```
Note too that we can combine these two expressions into one; you’ll often see computer programs written this way, as it allows us to express thoughts more concisely. So the above could be replaced by
```>>> int(round(2.23501))
2
```
This expression can be read as “after rounding the number to 0 decimal places, convert it to an integer”.
## Reverse the digits of a number
How do we reverse the digits of a number? Well, first, we’re going to treat the number as if it were an arbitrary string of letters (characters) rather than numbers. We are going to cast the integer into a string. Strings are a programming concept used to express textual information; they are usually enclosed in quotes.
```>>> "hello"
'hello'
```
The string representation of a number is different than the number itself. For instance, when two strings are “added” together, they become joined (concatenated). When two numbers are added together, they are added in the mathematical sense. For instance
```>>> "2" + "2"
'22'
>>> 2 + 2
4
```
We are going to treat the numbers as strings because it is fairly easy to reverse the characters of a string, but hard to do so for a number. With a string, you can access each individual character; you cannot (easily) do that with a number. To understand how to reverse the characters of a string, you need to understand a bit about iterating over collections in Python, as well as a concept known as “slice”.
Let’s start with the basic Python data structure known as a list. You can put a heterogenous (mixed) set of objects into a list:
```a = [2,3,5,6,"hello","world"]
```
In this case we’ve put a bunch of numbers, and then two strings into the list data structure. We can get out each element in the list by ‘indexing into’ the array. Unlike in the normal world, in Python you must count starting at 0. So instead of counting 1,2,3,…, you count 0,1,2,… . In other words, if we want to get out the first item we put into the list (the 2), we have to ask for the element at position 0.
```>>> a[0]
2
```
The number within brackets is the index of the element you are accessing from the list.
Now, what if we want the first 3 elements of the list instead of just the first one? To do that, we need to use what’s known as a “slice” in Python. The syntax is [startIndex:endIndex+1]. In other words, if we want the first three elements, those at position 0, 1, and 2, we would ask for
```>>> a[0:3]
[2, 3, 5]
```
This can be read as, “start at the element with index 0 and stop before you get to the element with index 3”. Yes, this is a little counter-intuitive – but just stick with me.
What if you want all of the items after a certain point? For instance, I want all the elements including and after the 2nd element. To do that, you omit the second number, but keep the colon. This is saying “start and keep going till you reach the end of the collection”
```>>> a[1:]
[3, 5, 6, 'hello', 'world']
```
You can also omit the first number, in which case it is assumed that you want to start at the beginning of the collection
```>>> a[:5]
[2, 3, 5, 6, 'hello']
>>> a[0:5]
[2, 3, 5, 6, 'hello']
```
What happens if we omit both numbers?
```>>> a[:]
[2, 3, 5, 6, 'hello', 'world']
>>> a
[2, 3, 5, 6, 'hello', 'world']
```
We get back all the items of the original list! This makes sense if you think about it; we assume we’re starting at the beginning, and we assume we’re ending at the end. Why do I bring this up? Because there’s one last element you can add to take a new slice out of array: the stride argument.
Whereas we’ve been taking some contiguous subset of the list, there’s nothing preventing us from skipping more than one number between every element, in effect taking every second, or every nth number.
```>>> a[::2]
[2, 5, 'hello']
```
Interestingly, this number doesn’t have to be positive either; if it’s negative it means we start at the end of the collection and work our way towards the front of the list:
```>>> a[::-2]
['world', 6, 3]
```
Putting this all together then, one way to get the reverse of a list is to take a new slice out of it, taking the whole thing but backwards:
```>>> a[::-1]
['world', 'hello', 6, 5, 3, 2]
```
(More advanced readers: an alternate way, and arguably easier, is to use a list comprehension and the reversed method:
```>>> [i for i in reversed(a)]
['world', 'hello', 6, 5, 3, 2]
```
)
Hopefully you understand how to take slices out of a list now; you can read a whole lot more about slices at any number of sites, including this great introduction to Python.
But why did I start talking about lists? Weren’t we trying to reverse a string, not a list? Well, it turns out you can view a string as a list of individual letters (characters), and treat it exactly the same as we did for lists, including indexing and slicing.
```>>> "hello"[0]
'h'
>>> "hello"[:3]
'hel'
>>> "hello"[::-1]
'olleh'
```
So, the method for reversing a string (or really any subscriptable object, but that’s another issue) is
```def reverseString(s):
# Step size is -1, so start at the end and work backwards
return s[::-1]
```
The only thing left to say in this regard is that if we want to convert a number to a string, this is accomplished by the str() method.
## Iterate over a range of numbers
We want to test a whole range of temperatures, looking for those that yield the properties desired in the puzzle. Fortunately there is an easy way to generate a list of numbers in a given range in Python: the range command.
```</div>
>>> range(5)
[0, 1, 2, 3, 4]
>>> range(5,10)
[5, 6, 7, 8, 9]
<div>```
Once we have our list of numbers, we want to iterate over them, pulling out each temperature. We do that with a simple for loop:
```>>> for x in range(5):
... print x
...
0
1
2
3
4```
Or, for our purposes
```for temp in range(LOW_TEMP_FAHRENHEIT, HIGH_TEMP_FAHRENHEIT):
```
Where LOW_TEMP_FAHRENHEIT and HIGH_TEMP_FARENHEIT are two constants defined to limit the range of numbers we are looking at.
## Conclusion
I’ve illustrated a brute-force solution to a recent puzzler. Despite being a fairly simple challenge, it illustrates some important features of Python, including range generation, list comprehension, slicing to reverse selections, and converting between different data types.
I’d love to see some other solutions to this problem, so post in the comments if you have ideas.
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+1(978)310-4246 credencewriters@gmail.com
Select Page
Label and color the layers of the Earth’s interior.
Use google Earth to examine the shape of the continents and the trend of the Hawaiian islands to see how these relate to the Theory of Plate Tectonics.
Calculate the rates of plate motion and density of continental and oceanic crustal rocks.
Create a concept sketch of the different types of plate boundaries and use Google Earth to locate plate boundaries and surface feature.
Name _________________________
Partner Name_____________________________
LAB 3: PLATE TECTONICS
Activity 3A: Rates of Change
One of the most important equations we will use in this course is the “dirt” equation:
Distance = Rate x Time (read as distance equals rate times time)
When you know two of the variables in this equation, you can figure out the third. We can
move the variables around in the equation to suit our needs (but then we couldn’t call it the
“dirt” equation):
Rate = Distance/Time (read as rate equals distance divided by time)
or
Time = Distance/Rate (time equals distance divided by rate)
Example: Car Speed
It’s summertime: you decide to go on a road trip. You want to figure out how long it will take
you to drive to the Grand Canyon. You look on your map app and see that you are 380 miles
away… when suddenly, your phone dies! Assuming you will drive at an average rate of 60 mph,
how many hours will it take you to drive to the Grand Canyon?
First, write down your known variables:
Distance =
Rate =
Then, calculate your unknown variable using one of the equations above. Show your work:
Time =
Geology Example: Rate of Sedimentation
While you’re out hiking, you encounter an interesting-looking rock bed. You can tell by the
pebbles incorporated throughout the rock bed that it was laid down by a river over a long time.
You want to find out how fast the bed was deposited by the river (this is called rate of
sedimentation).
You measure how thick the bed is (this is its distance, a measurement of length) and discover it
is 1.8 meters from the base to the top. Luckily, this bed is sandwiched by two ash beds laid
down by volcanic eruptions: ash can be dated by radiometric dating!
After analyzing the samples in a lab, you learn the lower ash bed is dated at 10.1278 Ma (read
as million years old) and the upper ash bed is dated at 10.1251 Ma.
Figure 3.1: Outcrop of layers of sedimentary rock and ash. The ash layers have been
What is the average rate of sediment deposition for your rock bed?
Distance (Length, width, or thickness) =
Time =
Rate =
2
Activity 3B: Plate Motion and Evidence
One of the most striking things about the geography of the continents today is how they appear
to fit together like puzzle pieces. The reason for this is clear: they once were connected in the
past and have since separated shifting into their current positions.
Use GoogleEarth Pro or open the browser version of Google Earth and zoom out to an eye
altitude (camera) of ~10,000 miles. Examine the coastlines of eastern South America and
Western Africa and notice how well they match in shape. There are scientifically important rock
deposits in southern Brazil, South America and Angola, Africa that show the northernmost
glacial deposits on the ancient continent of Pangaea, which indicates these two areas were
once connected.
1. Based on the shape of the two coastlines, give the present-day latitude and longitude of
two sites along the coasts of these countries that used to be connected when the two
continents were joined as a part of Pangaea (note: there are multiple correct answers):
a. Brazil:
i.
Latitude:
ii.
Longitude:
b. Angola
i.
Latitude:
ii.
Longitude:
2. Measure in centimeters the distance (Map Length) between the two points you
recorded in the previous question. Hint: use the ruler icon (left or top toolbar).
a. Distance (in centimeters) =
3. This portion of Pangaea broke apart 200,000,000 years ago. Using the distance you
measured above, calculate how fast (the rate) South America and Africa are separating
in cm/year. (Hint: the formula to use here is Rate= Distance/Time). SHOW YOUR WORK.
3
Examine the Western Coast of South America, the Eastern Coast of Asia, and the Pacific Ocean.
If South America and Africa are separating and the Atlantic Ocean is growing, then the opposite
must be occurring on the other side of the earth (the Americas are getting closer to Asia and
the Pacific Ocean is shrinking). It begs the question, when will the next supercontinent form? To
4. Measure the distance between North America and Mainland Asia in centimeters? (Hint:
measure across the Pacific at 40° N latitude, between Northern California and North
Korea)
a. Distance (in centimeters) =?
5. Using the distance above, and the rate calculated in question 3, determine the time it
will take to develop a new supercontinent. (Hint: the formula Rate= Distance/Time, can
be reworked to Time = Distance/Rate). SHOW YOUR WORK.
4
Activity 3C: The Structure of Internal Earth
Label each layer at their correct letter, then color the layers using the following colors:
â—‹
â—‹
â—‹
â—‹
Asthenosphere – Light green
Continental crust – yellow
Inner core – red
Lithosphere – gray
â—‹ Mantle – label, but do not color
â—‹ Mesosphere – dark green
â—‹ Oceanic crust – black
â—‹ Outer core – orange
Figure 3.2: A cross-section of Earth to label.
5
Activity 3D: Hot Spots
What Are Hot Spots?
In addition to GPS technology, geologists can also track plate motion using the location of hot
spots, volcanically active areas on the Earth’s surface that are caused by anomalously hot
mantle rocks underneath (Figure 3.3). This heat is the result of a mantle plume that rises from
deep in the mantle toward the surface resulting in the production of magma and volcanoes.
These mantle plumes occur within the asthenosphere or deeper, such that they are unaffected
by the movement of the continental or oceanic plates. Mantle plumes appear to be stationary
through time; therefore, as the tectonic plate moves over the hot spot, a linear chain of
volcanoes is produced. This gives geologists a wonderful view of the movement of a plate
through time with the distribution of volcanoes indicating the direction of motion and their
ages revealing the rate at which the plate was moving. Interested in hot spots? Want to learn
Figure 3.3: The life of an oceanic hot spot.
The Hawaiian Hot Spot
One of the most striking examples of a hot spot is underneath Hawaii. The mantle plume
generates magma that results in an active volcano on the seafloor, referred to as a seamount.
Each eruption causes the volcano to grow until it eventually breaks the surface of the ocean
and forms a volcanic island. As the oceanic plate shifts the volcano off the stationary hot spot,
the volcano loses its source for magma and becomes inactive. The volcano then cools down,
contracts, erodes, sinks slowly beneath the ocean surface, and is carried by the tectonic plate as
it moves over time. As each island moves away from the mantle plume, a new island will then
be formed at the hot spot in a continual conveyor belt of islands. Therefore, the scars of ancient
volcanic islands near Hawaii give a wonderful view of the movement of the tectonic plate
beneath the Pacific Ocean.
6
Use GoogleEarth Pro or open the browser version of Google Earth and type “Hawaii†into the
search bar and zoom out to an eye altitude (Camera) of 700 miles. Examine the chain of
Hawaiian Islands.
1. On the map of the Hawaiian Islands (Figure 3.4), include the following:
â— A North arrow
â— Label the following islands: Big Island of Hawaii, Kauai, Maui, Molokai, Oahu
Figure 3.4: Map view of the Hawaiian Islands.
2. Next, label on the map the ages of each of the islands. These ages were determined
through radiometric dating of the lava flows on the islands.
â— Big Island of Hawaii: 0 years old (Active)
â— Kauai: 5.1 million years old
â— Maui: 1.3 million years old
â— Molokai: 1.8 million years old
â— Oahu: 3.7 million years old
7
3. In Google Earth, measure the distance between the islands. To do this, measure from
the center of each island and its adjacent island in centimeters. (Hint: numbers will be
large)
a. Distance between the Big Island and Maui (in cm):
b. Distance between Maui and Molokai (in cm):
c. Distance between Molokai and Oahu (in cm):
d. Distance between Oahu and Kauai (in cm):
4. Look closely at each island in Google Earth and record the maximum elevation in
centimeters. (Hint: elevation can be determined by placing your cursor over a point and
reading the elevation on the lower right of the image by the latitude and longitude. The
elevation units can be changed in your settings. To locate the highest point on the
islands, tilt the image or use the 3D button. Convert meters to centimeters by multiplying
by 100.)
a. Big Island of Hawaii, max elevation (in cm):
b. Kauai, max elevation (in cm):
c. Maui, max elevation (in cm):
d. Molokai, max elevation (in cm):
e. Oahu, max elevation (in cm):
5. Consider the ages and positions of the islands listed above along with what you know
about plate tectonics and hotspots. In what general direction is the Pacific Plate
moving?
a. Northwest
b. Southeast
c. Northeast
d. Southwest
6. How fast was the Pacific plate moving during the last 1.3 million years between the
formation of the Big Island and Maui? Calculate your answer in cm/year. (Hint: Rate =
8
7. How fast was the Pacific plate moving between the formation of Oahu and Kauai?
8. Zoom out and examine the dozens of sunken volcanoes, referred to as extinct
seamounts, to the northwest of Niihau, these are named the Emperor Seamounts. As
one of these volcanic islands on the Pacific Plate moves off the hotspot it becomes
inactive, or extinct, and the island begins to sink as it and the surrounding tectonic plate
cool down. The speed the islands are sinking can be estimated by measuring the
difference in elevation between two islands and dividing by the difference in their ages.
Note, this method assumes the islands were a similar size when they were active.
a. Calculate how fast the Hawaiian Islands are sinking, by using the ages and
elevations of Maui and Kauai. (Hint: Rate = Distance/Time). SHOW YOUR WORK.
b. When will the Big Island of Hawaii sink below the surface of the ocean? (Hint:
use the rate calculated above, ignoring possible changes in sea level, and the
max elevation of the Big Island in centimeters; Time = Distance/ Rate). SHOW
9
Activity 3E: Plate Densities
An important property of geological plates is their density. Remember the asthenosphere has
fluid-like properties, such that tectonic plates ‘float’ relative to their density. This property is
called isostasy and represents the equilibrium between crustal height and relative density. This
is similar to buoyancy in water. For example, if a cargo ship has a full load of goods it will
appear lower than if it were empty because the density of the ship is on average higher.
Therefore, the relative density of two plates can control how they interact at a boundary and
the types of geological features found along the border between the two plates.
Recall the Earth’s crust is divided into two main types:
â— Continental crust, which is composed of granite, is relatively older and thicker than oceanic
crust. The thickness of the continental crust is between 25-70 km with an average thickness
around 30 km. The average density of granite is 2.75 g/cm3 (read as 2.75 grams per cubic
centimeter).
â— Oceanic crust, which is composed of basalt, is relatively younger and thinner than
continental crust. The thickness of the oceanic crust is between 5-10 km with an average
thickness of 7 km. The average density of basalt is 3.0 g/cm3 (read as 3.0 grams per cubic
centimeter).
1. Which crustal type is thicker?
2. Which crustal type is denser?
a. How did you determine which crustal type was denser?
b. Why do you think this is the case?
3. Which crustal type is more buoyant?
a. How did you determine this?
10
Measuring the density of rocks is fairly easy and can be done by first weighing the rocks and
then calculating their volume. The latter is best done by a method called fluid displacement
using a graduated cylinder. Water is added to the cylinder and the level is recorded, a rock is
then added to the cylinder and the difference in water levels equals the volume of the rock.
Density is then calculated as the mass divided by the volume (Density = Mass/Volume).
Figure 3.5 contains the information needed to calculate density. There are four rocks which
have weight (in grams) as well as the volume of water recorded by a graduated cylinder (in
milliliters) before and after the rock was added.
Notes:
â— Each line on the graduated cylinder represents 10 milliliter (ml).
â— When measuring volume, round to the nearest 10 ml line on the graduated cylinder.
â— Surface tension will often cause the water level to curve up near the edges of the
graduated cylinder creating a feature called a meniscus. To accurately measure the
volume, use the lowest level the water looks to occupy.
Figure 3.5: Density Experiment: Calculated volumes (in ml) and masses (in g) of rock samples A, B, C, and D.
11
4. The rock that most closely resembles the composition of continental crust is:
a. A
b. B
c. C
d. D
5. Based on your selection above, what is the density of this rock? Note: Density =
6. The rock that most closely resembles the composition of oceanic crust is:
a. A
b. B
c. C
d. D
7. Based on your selection above, what is the density of this rock? Note: Density =
8. Based on their densities, when oceanic and continental crust collide, the _____ crust would
sink below the ______ crust.
a. continental; oceanic
b. oceanic; continental
9. What type of boundary is represented in the question above?
a. Convergent, subduction
b. Convergent, Continental-Continental
c. Divergent
d. Transform
12
Activity 3F: Plate Boundaries
Earthquakes are great indicators of plate boundaries and are associated with all three boundary
types.
1. The Wadati-Benioff zone is associated with which type of plate boundary?
a. Divergent
b. Convergent collision zone (Continent-Continent)
c. Convergent subduction zone (Continent-Ocean or Ocean-Ocean)
d. Transform
2. Download the front portion of the The Dynamic Planet map from the USGS. Which of
the following locations represent a Wadati-Benioff zone?
a. 10°S, 110°W
b. 0°, 0°
c. 15°S, 180°
d. 30°N, 75°E
3. Use GoogleEarth Pro or open the browser version of Google Earth and type
34°46’16.2″N 118°44’58.2″W into the search bar. Zoom out to an eye altitude (camera)
of 10 miles. This is Quail Lake, a dammed river that sits directly on top of the San
Andreas Fault (SAF). The SAF is a well-known transform boundary with the North
American Plate on the northern side and the Pacific Plate on the southern side. This
boundary is running East-West in this area (dashed line in Figure 3.6). Zoom out more in
Google Earth and you should be able to see it better.
4. Examine the path of the waterway that feeds into and flows out of Quail Lake. What
direction is the North American plate moving in comparison to the Pacific Plate at this
location? Draw arrows indicating the motions on Figure 3.6.
a. East
b. West
5. Given that San Francisco is located on the North American Plate and Los Angeles is
located on the Pacific Plate, are these two cities getting closer together or farther apart
over time?
a. Closer
b. Farther
13
Figure 3.6: Map view of the North American and Pacific Plates.
6. Type 41°47’22.68″N, 124°15’0.51″W into the Google Earth Search bar. What type of
tectonic plates are present?
a. Ocean-Ocean
b. Ocean-Continent
c. Continent-Continent
7. What type of plate tectonic boundary is present?
a. Transform
b. Convergent, subduction zone
c. Convergent, collision zone (continent-continent)
d. Divergent
8. What features would you expect to occur at this type of boundary?
a. Volcanos, earthquakes and a trench
b. Volcanoes and a linear valley
c. Mountains and landslides
d. Earthquakes and offset rivers
14
9. Type 23°26’04â€N, 108°29’25â€W into the Google Earth Search bar. What type of process
is going on at this location?
b. Continental rifting
c. Subduction
10. What features would you expect to occur at this type of boundary?
a. Earthquakes and a trench
b. Submarine volcanic activity and earthquakes
c. Mountains and landslides
d. Earthquakes and offset rivers
11. Type 27°58’42.06″N, 86°55’11.53″E into the Google Earth Search bar. What type of
tectonic plates are present?
a. Ocean-Ocean
b. Ocean-Continent
c. Continent-Continent
12. What type of plate tectonic boundary is present?
a. Transform
b. Convergent, subduction zone
c. Convergent, collision zone
d. Divergent
13. Type 43°29’9.14″N, 128° 7’27.37″W into the Google Earth Search bar. This is known as
the Blanco Fracture Zone. What type of tectonic plates are present?
a. Ocean-Ocean
b. Ocean-Continent
c. Continent-Continent
14. What type of plate tectonic boundary does the Blanco Fracture Zone represent?
a. Transform
c. Convergent, collision zone
b. Convergent, subduction zone
d. Divergent
15. This plate boundary isn’t as simple as the previous examples, meaning another nearby
plate boundary directly influences it. Zoom out and examine the area, what other type
of boundary is nearby?
a. Divergent, continental rift
c. Convergent, collision zone
b. Convergent, subduction zone
d. Divergent, mid-ocean ridge
15
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# Operational Amplifier, Op Amp Parameters and calculations
## Operational amplifier:
The basic job of an amplifier is to amplify the input signal. In early days when digital computers were not evolved, at that time the different mathematical functions like addition, subtraction, integration, and differentiation were performed using this operational amplifier. So, just by connecting few resistors and capacitors, it is possible to perform the different mathematical operations.
It consists of two inputs and one output. Most of the operational amplifiers consist of two power supplies. The positive and the negative power supply. But there are many op-amp IC’s which runs on the single power supply. So, now in this operational amplifier, the input terminal which is marked by this positive sign is known as the non-inverting input terminal and another input terminal which is marked by this negative sign is the known as the inverting input terminal.
So, now if you see this operational amplifier, it is one kind of differential amplifier with the singal output. It means that this amplifier amplifies the difference between the two input signals. So, let’s say V1 and V2 are the input signals which is being applied to this operational amplifier and let’s say the gain of this operational amplifier is A, then the output will be equal:
Vout=A (V1– V2)
if we have applied the single input to this operational amplifier and we have grounded another input terminal then at the output you will get A times V1. Where A is the open loop gain of this operational Amplifier.
When there is no feedback from the output to the input side. So, suppose if you are applying the sinusoidal signal over here, then at the output that sinusoidal signal should be get multiplied
by the factor of this gain and at the output, you should get the amplified sinusoidal signal. Now, here the phase of this output voltage will be the same as the input voltage.
Likewise, whenever we are applying input to this negative terminal, and we are grounding another terminal then the output of this amplifier will be equal to minus A times the V2 because the difference between these two input terminals will be equal to 0 minus V2, that is equal to minus V2.
## Inverting terminal:
So, suppose let’s say if we are applying the sinusoidal signal at the input then at the output we will get the amplified sinusoidal signal which is having a 180-degree phase with respect to the input signal. That means the output will be get inverted by 180 degrees. And that is why this input terminal is known
as the inverting terminal. Because the output will be get inverted with respect to the input.
So, now here suppose if we apply the input signal between these two positive and the negative terminals then at the output we will get A times this differential input signal. Where here this A represents the open-loop gain of this operational amplifier.
Now, this operational amplifier is a very high gain amplifier. The value of gain used to be in the range
Of 105 to 106 So, let’s say, even if we apply the 1 mV of a signal between these two terminals, and let’s say if the gain of this op-amp is then at the output theoretically we should get 1 mV signal that is multiplied by 105 that is equal to 100V. Or let’s say if we apply 1V of a signal, then theoretically, we should get the output as 105 volts but that is not possible.
The output of this op-amp is restricted by the biasing voltages that are being applied to this op-amp. So, the output voltage will be between these biasing voltages.
## Positive and negative Feedback of operational amplifier:
We know that any circuit has two major parameters such as input and output. A condition in which some part of the output is fed back to the input is called as a feedback. In case of an op-amp we have two types of feedbacks such as positive feedback and a negative feedback. When some part of an output is fed back to a non-inverting terminal of an op-amp it is called as a positive feedback and when some part of an output is fed back to the inverting terminal of an op-amp it is called as a negative feedback resistor RF is called as a feedback resistor.
## Virtual ground:
A virtual ground concept the input impedance of an op-amp is very high hence an op-amp never draws any current at its input an input current is always zero amperes for current to be 0 the voltage must be 0. Let’s assume that some input is applied to an inverting terminal keeping a non-inverting terminal at the ground even though the input is applied an inverting terminal also behaves as a ground terminal at node a this concept is called as a virtual ground concept.
## Op Amp Parameters:
We will learn different parameters of op-amp as voltage gain, input impedance, output impedance, input offset voltage, input offset current, input bias current and bandwidth.
## Voltage gain:
Let’s start with the voltage gain it is defined as the ratio of output voltage to input voltage.
## Input impedance:
The second term is an input impedance the resistance offered by the input terminals of an op-amp is called as an input impedance. The voltage drop at the input of an op-amp must be very high hence the input impedance of an op-amp is always very high due to equation V equals I into R.
## Output impedance:
The third term is an output impedance the resistance offered by the output of an op-amp is called as an output impedance generally an output device like a speaker is connected next to an op-amp hence it is necessary that all the output of an op-amp must be passed to the next device in other words the voltage drop at output must be zero hence output impedance must be as low as possible.
## Input offset voltage:
The fourth term that we learn is an input offset voltage when input to an op-amp is zero the output should be zero ideally but if it’s not zero we need to apply some DC voltage at the input terminal to force the output voltage to be zero this applied voltage is called as an input offset voltage.
## Input offset current:
The next term we study is an input offset current. The difference between the currents into the two input terminals when the output is held at zero is called as an input offset current.
## Bias current:
The sixth term is an input bias current the average of the currents into the two input terminals with the output at zero volts is called as an input bias current.
## Bandwidth:
The last term is a bandwidth the range of frequencies for which an op-amp can be used is called as a bandwidth of an op-amp.
## Operational Amplifier as wave form generator:
An op-amp can also be used as a waveform generator. Here it generates different waveforms such as a square wave, triangular wave etc. First we will see an op-amp as a square wave generator. The schematic diagram for this application is as shown as we can see a capacitor C is connected to an inverting terminal and resistances and are connected to a non-inverting terminal.
Resistor R is connected as a negative feedback to the inverting terminal forming the RC circuit. As soon as the op-amp is supplied with the supply voltages +V and –V. We get some output as without any input applied the output should be zero but practically we get some nonzero output. R_a and R_b form a voltage divider network. Thus if initial V_out is nonzero. We get voltage across V_b also as nonzero thus we get a positive input at the non-inverting and inverting terminals and the output gets amplified by its gain say AV and reaches its maximum value Vout max thus we get the positive half of the square wave as we have a nonzero input at the inverting terminal. Now a capacitor also starts charging it will charge continuously till its voltage becomes greater than Vb as soon as voltage Vc is greater than Vb the inverting input becomes greater than the non-inverting input and hence op-amp output switches to negative voltage and gets amplified till minus vo max. Thus we get the negative half of the square wave this is the application of an op-amp as a square wave generator.
Now we see an op-amp as a triangular wave generator we have already seen that the output of the integrator is a triangular wave if the input given to it is a square wave thus to construct the triangular wave generator we combine two circuits such as a square wave generator followed by an integrator as shown and at the output of an integrator we get a triangular signal.
## Operational amplifier as differentiator:
An op-amp is a differentiator for this we replace the input resistor with the capacitor as shown applying KCl at node “A”
if If equals to Ic from the diagram the current If is equals to:
If= (Va– Vout )/R
The current flowing through a capacitor:
Ic= (CdVc)/dt
(Va– Vout )/R= (CdVc)/dt
(Va– Vout )/R= (Cd(Vin-Va ))/dt
But Va=0 virtual ground concept.
(- Vout )/R= (Cd(Vin ))/dt
Vout = (RC*d(Vin ))/dt
If RC is equal to gain Av.
Vout = (Av*dVin)/dt
We can see if we take R into C as a gain of an amplifier then the output is the differentiation of an input hence it is called as a differentiator.
## Operational amplifier as Integrator:
The next use of an op-amp is an integrator. If we interchange the position of the capacitor and the resistor of a differentiator circuit. We get the circuit of an op-amp as an integrator applying KCl at node “A”.
## Operational amplifier as difference amplifier:
The next application of an op-amp is a difference amplifier. Here we apply KCl at both the nodes node a and node B applying KCl at node B.
If we consider Z as a gain of op-amp then the output is an amplified version of the difference between two inputs hence it is called as a difference amplifier.
## Operational amplifier as summing amplifier:
An op-amp is also used for mathematical operations we will start with an op-amp as a summing amplifier we have three inputs as v1 v2 and v3 given to an inverting terminal of an op-amp with the currents as i1 i2 and i3. Applying KCl at node “X”:
Vx = 0 virtual ground concept.
Rearranging the equation in terms of V out we get V out equals:
We can see an output is an amplified version of the sum of all the input signals it is called as a summing amplifier.
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# How does one know what statements in Coq require Induction?
I was trying to learn Coq using the famous book Software Foundations. In it I found the following:
Theorem mult_0_r : forall n:nat,
n * 0 = 0.
Proof.
induction n as [| n IHn].
- simpl. reflexivity.
- simpl. rewrite -> IHn. reflexivity.
Qed.
which I understand perfectly but find rather unintuitive. I understand perfectly how each step works but it would have never occurred to me to use induction to prove such a trivial fact. In fact in the mathematical proof I had in mind that would have been a fact/property (or I guess an axiom) of 0. i.e. $$\forall n \in N, n \cdot 0 = 0$$ is true by definition. I guess in Coq (or the way we set up numbers? thats not true).
My biggest complaint or worry is that if such a trivial thing requires induction I feel now I am unable to recognize what needs induction (at least in Coq). I know it needs it here because I am in the induction chapter. But in normal maths its usually quite obvious because the problem is obviously recursive. But I wouldn't have really thought of that proposition as recursive. For example it goes on to prove more things as exercises:
Theorem plus_n_Sm : ∀n m : nat,
S (n + m) = n + (S m).
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_comm : ∀n m : nat,
n + m = m + n.
Proof.
(* FILL IN HERE *) Admitted.
Theorem plus_assoc : ∀n m p : nat,
n + (m + p) = (n + m) + p.
Proof.
(* FILL IN HERE *) Admitted.
which I am sure are not too difficult, but my worry is that in isolation I would have never thought such trivial statements required something as sophisticated as induction. I didn't even learn induction until last 2 years of highschool and didn't really do it seriously until college. So now I see trivial statements requiring what seems to me, sophisticated mathematics.
I just feel my intuition got really lost. When I do Coq proofs (in isolation), how do I know when to use induction and on what? I doubt there is a general procedure (of course) but proofs do exist. So there must be something guiding us to use induction in Coq.
• Every mathematical book that actually bothers to prove $n \cdot 0 = 0$ from Peano axioms does so by induction on $n$. This all looks unintuitive to you because you have never seen any proofs of such facts, and now you're looking at their formal incarnations. (Note: the relevant Peano axiom $0 \cdot n = 0$ does not imply $n \cdot 0 = 0$, you first need to prove $n \cdot m = m \cdot n$. However, most proofs of $n \cdot m = m \cdot n$ rely on $n \cdot 0 = 0$.) Jan 21, 2019 at 7:42
• @AndrejBauer seems theorem provers lack a severe case of common sense mathematics! As do computers in all AI fields. Jan 22, 2019 at 23:31
Coq allows one to prove mathematical theorems in a completely formal way. At first, this copes with our experience of doing maths, which is far more informal.
Most of the time, people doing maths are not exposed to mathematical logic. They do not know, for instance, how to define the set of real numbers. Or even the set of natural numbers. Numeric sets are simply taken for granted, together with their operations and related properties. On top of these "basic facts" people build more complex theorems. Essentially, the "basic facts" are used as if they were axioms.
Note that one could work at a similar level in Coq, if one wishes. One can import the Arith Coq library, which provides the basic facts of arithmetics, and exploit those.
However, the underlying logic of Coq does not take those properties for granted. The type of natural numbers, for instance, is not a primitive concept in the logic, but rather something which is defined using induction. Since that makes nat an inductive type, virtually every proof about naturals will require induction. (This also holds in set theory, e.g. Zermelo-Fraenkel, where naturals are also defined in a sort-of inductive way.)
Arithmetical operators are also defined using induction. Using a simplified syntax, one could define multiplication as
x * 0 = 0
x * (S y) = x + (x * y)
or, alternatively,
0 * x = 0
(S y) * x = x + (y * x)
Both definitions are equivalent. However, using the first one will result in us being able to prove x * 0 = 0 "without induction" -- it follows by definition. In Coq, we can simply write reflexivity and let the system expand the definition for us.
By comparison, 0 * x = 0 is not a trivial equation which follows from the definition: to expand the definition we need to know what is the second argument of the multiplication, but here it's merely an unknown x. Hence, for this we do need induction over x. There's no way around that.
Note that, if we used the second definition instead, we would prove 0 * x = 0 immediately, and instead require induction for x * 0 = 0. So, no definition is strictly better than the other one.
• just to study this specific example more in depth and hopefully gain an intuition when to use induction (in contrast to destruct for example). I understand that in either case x is unknown. But what seems to be a coincidence or even lucky that works is induction on x. Why exactly did we use induction on the unknown? How did it help? this remains a bit mysterious to me. It's still unclear to me when induction would be used or what the intuition for it is (especially contrasted with destruct for example). Jan 20, 2019 at 20:02
• @Pinocchio Well, if * was defined by induction on the first argument, and our goal to be proved is property (x * something), inducing on x looks promising, since that will make x * something simplify, unrolling the definition. Actually, destruct also does that, but it does not provide any induction hypothesis. Very roughly, you might look at destruct as a weak form of induction which does not provide induction hypotheses.
– chi
Jan 20, 2019 at 22:06
• is that why we use induction on n when dealing with something like (n + m) * p = (n * p) + (m * p). I tried p but the proof didn't move forward when I got to the inductive case. Jan 20, 2019 at 22:24
• @Pinocchio Yes, + and * are defined by induction on their first argument in the standard library of Coq.
– chi
Jan 20, 2019 at 22:36
As the others have mentioned, in Coq's standard library (or typical presentations of naturals in Coq), naturals are defined inductively, usually a la Peano. We could make other choices, e.g. one could imagine defining naturals as the free semiring on one generator which would give you many of the "axiomatic" facts for "free". (Well, you'd have to prove them about general rings...)
For any inductively defined type, T, whenever you want to prove a statement of the form forall x:T, P(x) you will need to use induction on T unless the proof doesn't actually depend on T, i.e. it is parametric, e.g. forall x:T, x = x. In fact, all proofs of this form can be written as an induction, they may just be a trivial induction. Often the induction will be hidden away in some lemma.
However, the key concept it seems you are unfamiliar/uncomfortable with is the notion of definitional equality in intensional dependent type theories like Coq. Let's say you state that $$2$$ means $$1+1$$. If you then ask whether $$2=1+1$$ there are no steps of proof required. You simply substituted the meaning of $$2$$ and see that $$1+1=1+1$$ which is true by reflexivity. More generally, (closed) terms in Coq have normal forms, and Coq treats any two terms with the same normal form as identical. This is definitional equality. Reducing to that normal form may involve large amounts of computation, but there is no proof rule that to tell Coq to reduce a term to normal form. It just does it in the background as necessary to perform type checking. Definitions, such as the one for multiplication reproduced below:
Fixpoint mul n m :=
match n with
| 0 => 0
| S p => m + p * m
end
where "n * m" := (mul n m) : nat_scope.
introduce new (non-normal) terms and new computational rules for normalizing them. In this case, it adds a rule that says 0 * x = 0. These follow from the generic rules for Fixpoint and match for the Coq type theory. If you were to ask whether forall x:nat, 0 * x = 0, Coq sees forall x:nat, 0 = 0 since 0 is the normal form of 0 * x. There is no computational rule saying how x * 0 should be reduced when x is a variable, so the "normal" form of x * 0 is x * 0. With this understanding, it is much easier to see why and how induction helps in this particular case. When we do induction, we get two new problems: 0 * 0 = 0 and (S p) * 0 = 0 which allows steps of computation to be performed. In the latter case, we get 0 + p * 0 which then leans on the computation rules induced by the definition of natural number addition, but also leaves us with n * 0 which requires the induction hypothesis to resolve.
At more advanced levels of proof engineering, it is common to try to set up definitions so that a decent amount of work can be done by this implicit computation mechanism. Extreme forms of this are reflection principles where we prove an implementation of a decision procedure correct after which we can simply run it and appeal to this result to prove other results. Techniques like this are behind things like the ring solver and ssreflect.
To see how you can prove something about natural numbers, you must know what you know about natural numbers. Usually, we learn all sorts of 'basic facts' about numbers in high school that are pretty much given without further proof, which we can use later to prove more interesting statements about numbers.
In most mathematical disciplines, this 'naive' approach to numbers is fine. However, when we care about the formalization of numbers, we want a 'minimal definition' of what a number (and related concepts such as successors, predecessors and arithmetic) is and prove all 'basic facts' from them.
So, in the book you're reading, all we know about numbers is the following definition:
Inductive nat : Type :=
| O
| S (n : nat).
That is, natural numbers are an inductive type, defined as either the natural number O or the successor of a natural number. This is the reason why we do proofs by induction: it is all we know about the natural numbers in this setting.
So, where did your intuition let you get lost? When learning mathematics, you get taught that 'basic facts' about numbers early on, which therefore seem easy and get taught induction fairly late (and when you actually have to prove things!) which makes it seem hard.
What is actually the case is that, from a viewpoint of formalisation, induction is the one fundamental property of natural numbers! The best way to see this is that, in fact, all other properties of natural numbers follow from it.
My final advice is not to worry. The fact that all 'basic facts' are suddenly not as basic as they may have seemed so far can be intimidating, but this is simply what formalisation is about: having rigid constructions for 'basic facts'. Once you have these in place, you can prove more complicated statements as usual, by relying on the 'basic facts' just proven.
• the thing that still concerns me is if you see the remaining 3 facts/theorems I wrote, I know they should use induction cuz they are in the induction chapter. But how would you have suspected to use induction in the first place? This is what still remains a but mysterious to me. Jan 20, 2019 at 20:04
• As I said, the fact that nat is defined as an inductive type (or recursively, if you like) and the fact that we know nearly nothing else about nat is something that indicates induction is a useful approach here. Be sure that you understand how inductive types related to proofs by induction! If you do, and still feel lost, then I'm afraid that your question goes into the "how should I prove things" territory, which is far to general to answer. Consider delaying this question until you actually get stuck proving something in a later chapter. Jan 20, 2019 at 20:54
• Ok if I get stuck on later chapters I shall come back. Though, my current strategy seems to try destruct and induction sort of randomly. Though, my biggest confusion is that in highschool things that needed induction were so obvious. But say, the commutativity property doesn't seem like an recursive statement (except for the fact that nat is inductive in Coq). It's just I never had though of commutativity as recursive before which is bothering me, but I will follow your advice and see what happens alter... Jan 20, 2019 at 21:08
I want to share my own experience of learning Coq and theorem proving in general.
Most of the time, the proof of a statement largely depends on the recursive structure of the function or operation at hand. IMHO it is even more important than the recursive structure of data types.
Again, let's look at multiplication of natural numbers.
Fixpoint mul n m :=
match n with
| 0 => 0
| S p => m + p * m
end
where "n * m" := (mul n m) : nat_scope.
Theorem mult_0_r : forall n, mul n 0 = 0.
(Before you write a proof, just focus on the definition of the function, and ignore what it means as a number operation.)
The definition of mul matches on the structure of n, then, if n = S p, recursively calls itself with p. The structure is the strong sign that you'll need induction on the first argument of mul. On the other hand, you won't need to do induction on the second argument, or even destruct, simply because it's not pattern-matched inside mul.
This applies to more complex statements as well. Say we were to prove distributivity of multiplication over addition:
Theorem mult_distr_plus_r : forall n m p, (n + m) * p = n * p + m * p.
And suppose that we know both plus and mult are recursive on the first argument. In this case, the natural choice to do induction on is n because, by its positions, destructing n gives the largest amount of expansion (three times in total):
(S n + m) * p = S (n + m) * p = p + (n + m) * p
S n * p + m * p = (p + n * p) + m * p
It's easy to see that you can complete this proof by IH and plus_assoc.
In contrast, destructing m gives just one, and p gives none, so you'll need more lemmas to complete the proof (e.g. plus_n_Sm, plus_comm if you destruct m, mult_n_Sm in addition if you destruct p).
(n + S m) * p = ?
n * p + S m * p = n * p + (p + m * p)
(n + m) * S p = ?
n * S p + m * S p = ?
Here's another example to show that proof by induction depends on the function structure. Suppose we have the Fibonacci function:
Fixpoint fib n :=
match n with
| 0 => 0
| 1 => 1
| S (S p) => fib p + fib (S p)
end.
Then we want to prove some property on it, say it gives the same result as another definition of Fibonacci.
Theorem fib_is_fib' : forall n, fib n = fib' n.
Now, the recursive structure of fib is more complicated than mul above. It has two base cases, and the result of fib (S (S p)) depends on both fib p and fib (S p). If you try good ol' induction (with tactic induction n.) on it, you immediately fail.
In order to prove a property of fib, you actually need a stronger induction principle:
• Given a property P on natural numbers, if all of the following holds:
• P holds for 0.
• P holds for 1.
• If P holds on n and S n, it holds on S (S n).
• Then P holds on all natural numbers.
In Coq:
Theorem nat_ind_fib :
forall P : nat -> Prop, (* For a given property P, *)
P 0 -> (* if P holds for 0 *)
P 1 -> (* and 1, *)
(forall n : nat, P n -> P (S n) -> P (S (S n))) ->
(* and P n and P (S n) implies P (S (S n)) *)
forall n : nat, P n. (* then P holds for all n. *)
Now we can prove a property on fib by induction using tactic induction n using nat_ind_fib.
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# Cycle (graph theory)
(Redirected from Directed cycle)
A graph with edges colored to illustrate path H-A-B (green), closed path or walk with a repeated vertex B-D-E-F-D-C-B (blue) and a cycle with no repeated edge or vertex H-D-G-H (red)
In graph theory, a cycle is a path of edges and vertices wherein a vertex is reachable from itself. There are several different types of cycles, principally a closed walk and a simple cycle; also, e.g., an element of the cycle space of the graph.
## Definitions
A closed walk consists of a sequence of vertices starting and ending at the same vertex, with each two consecutive vertices in the sequence adjacent to each other in the graph. In a directed graph, each edge must be traversed by the walk consistently with its direction: the edge must be oriented from the earlier of two consecutive vertices to the later of the two vertices in the sequence. The choice of starting vertex is not important: traversing the same cyclic sequence of edges from different starting vertices produces the same closed walk.
A simple cycle may be defined either as a closed walk with no repetitions of vertices and edges allowed, other than the repetition of the starting and ending vertex, or as the set of edges in such a walk. The two definitions are equivalent in directed graphs, where simple cycles are also called directed cycles: the cyclic sequence of vertices and edges in a walk is completely determined by the set of edges that it uses. In undirected graphs the set of edges of a cycle can be traversed by a walk in either of two directions, giving two possible directed cycles for every undirected cycle. (For closed walks more generally, in directed or undirected graphs, the multiset of edges does not unambiguously determine the vertex ordering.) A circuit can be a closed walk allowing repetitions of vertices but not edges; however, it can also be a simple cycle, so explicit definition is recommended when it is used.[1]
In order to maintain a consistent terminology, for the rest of this article, "cycle" means a simple cycle, except where otherwise stated.
## Chordless cycles
In this graph the green cycle (A-B-C-D-E-F-A) is chordless whereas the red cycle (G-H-I-J-K-L-G) is not. This is because the edge K-I is a chord.
A chordless cycle in a graph, also called a hole or an induced cycle, is a cycle such that no two vertices of the cycle are connected by an edge that does not itself belong to the cycle. An antihole is the complement of a graph hole. Chordless cycles may be used to characterize perfect graphs: by the strong perfect graph theorem, a graph is perfect if and only if none of its holes or antiholes have an odd number of vertices that is greater than three. A chordal graph, a special type of perfect graph, has no holes of any size greater than three.
The girth of a graph is the length of its shortest cycle; this cycle is necessarily chordless. Cages are defined as the smallest regular graphs with given combinations of degree and girth.
A peripheral cycle is a cycle in a graph with the property that every two edges not on the cycle can be connected by a path whose interior vertices avoid the cycle. In a graph that is not formed by adding one edge to a cycle, a peripheral cycle must be an induced cycle.
## Cycle space
The term cycle may also refer to an element of the cycle space of a graph. There are many cycle spaces, one for each coefficient field or ring. The most common is the binary cycle space (usually called simply the cycle space), which consists of the edge sets that have even degree at every vertex; it forms a vector space over the two-element field. By Veblen's theorem, every element of the cycle space may be formed as an edge-disjoint union of simple cycles. A cycle basis of the graph is a set of simple cycles that forms a basis of the cycle space.[2]
Using ideas from algebraic topology, the binary cycle space generalizes to vector spaces or modules over other rings such as the integers, rational or real numbers, etc.[3]
## Cycle detection
The existence of a cycle in directed and undirected graphs can be determined by whether depth-first search (DFS) finds an edge that points to an ancestor of the current vertex (it contains a back edge).[4] In an undirected graph, finding any already visited vertex will indicate a back edge. All the back edges which DFS skips over are part of cycles.[5] In the case of undirected graphs, only O(n) time is required to find a cycle in an n-vertex graph, since at most n − 1 edges can be tree edges.
Many topological sorting algorithms will detect cycles too, since those are obstacles for topological order to exist. Also, if a directed graph has been divided into strongly connected components, cycles only exist within the components and not between them, since cycles are strongly connected.[5]
For directed graphs, Rocha–Thatte Algorithm[6] is a distributed cycle detection algorithm. Distributed cycle detection algorithms are useful for processing large-scale graphs using a distributed graph processing system on a computer cluster (or supercomputer).
Applications of cycle detection include the use of wait-for graphs to detect deadlocks in concurrent systems.[7]
## Covering graphs by cycles
In his 1736 paper on the Seven Bridges of Königsberg, widely considered to be the birth of graph theory, Leonhard Euler proved that, for a finite undirected graph to have a closed walk that visits each edge exactly once, it is necessary and sufficient that it be connected except for isolated vertices (that is, all edges are contained in one component) and have even degree at each vertex. The corresponding characterization for the existence of a closed walk visiting each edge exactly once in a directed graph is that the graph be strongly connected and have equal numbers of incoming and outgoing edges at each vertex. In either case, the resulting walk is known as an Euler cycle or Euler tour. If a finite undirected graph has even degree at each of its vertices, regardless of whether it is connected, then it is possible to find a set of simple cycles that together cover each edge exactly once: this is Veblen's theorem.[8]When a connected graph does not meet the conditions of Euler's theorem, a closed walk of minimum length covering each edge at least once can nevertheless be found in polynomial time by solving the route inspection problem.
The problem of finding a single simple cycle that covers each vertex exactly once, rather than covering the edges, is much harder. Such a cycle is known as a Hamiltonian cycle, and determining whether it exists is NP-complete.[9] Much research has been published concerning classes of graphs that can be guaranteed to contain Hamiltonian cycles; one example is Ore's theorem that a Hamiltonian cycle can always be found in a graph for which every non-adjacent pair of vertices have degrees summing to at least the total number of vertices in the graph.[10]
The cycle double cover conjecture states that, for every bridgeless graph, there exists a multiset of simple cycles that covers each edge of the graph exactly twice. Proving that this is true (or finding a counterexample) remains an open problem.[11]
## Graph classes defined by cycles
Several important classes of graphs can be defined by or characterized by their cycles. These include:
## References
1. ^ Balakrishnan, V.K. (2005). Schaum's outline of theory and problems of graph theory ([Nachdr.]. ed.). McGraw–Hill. ISBN 978-0070054899.
2. ^ Gross, Jonathan L.; Yellen, Jay (2005), "4.6 Graphs and Vector Spaces", Graph Theory and Its Applications (2nd ed.), CRC Press, pp. 197–207, ISBN 9781584885054.
3. ^ Diestel, Reinhard (2012), "1.9 Some linear algebra", Graph Theory, Graduate Texts in Mathematics, 173, Springer, pp. 23–28.
4. ^ Tucker, Alan (2006). "Chapter 2: Covering Circuits and Graph Colorings". Applied Combinatorics (5th ed.). Hoboken: John Wiley & sons. p. 49. ISBN 978-0-471-73507-6.
5. ^ a b Sedgewick, Robert (1983), "Graph algorithms", Algorithms, Addison–Wesley, ISBN 0-201-06672-6
6. ^ Rocha, Rodrigo Caetano; Thatte, Bhalchandra (2015). "Distributed cycle detection in large-scale sparse graphs". Simpósio Brasileiro de Pesquisa Operacional (SBPO). doi:10.13140/RG.2.1.1233.8640.
7. ^ Silberschatz, Abraham; Peter Galvin; Greg Gagne (2003). Operating System Concepts. John Wiley & Sons, INC. p. 260. ISBN 0-471-25060-0.
8. ^ Veblen, Oswald (1912), "An Application of Modular Equations in Analysis Situs", Annals of Mathematics, Second Series, 14 (1): 86–94, doi:10.2307/1967604, JSTOR 1967604.
9. ^ Richard M. Karp (1972), "Reducibility Among Combinatorial Problems" (PDF), in R. E. Miller and J. W. Thatcher, Complexity of Computer Computations, New York: Plenum, pp. 85–103.
10. ^ Ore, Ø. (1960), "Note on Hamilton circuits", American Mathematical Monthly, 67 (1): 55, doi:10.2307/2308928, JSTOR 2308928.
11. ^ Jaeger, F. (1985), "A survey of the cycle double cover conjecture", Annals of Discrete Mathematics 27 – Cycles in Graphs, North-Holland Mathematics Studies, 27, pp. 1–12, doi:10.1016/S0304-0208(08)72993-1..
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Past Year Questions: Water Requirements of Crops
# Past Year Questions: Water Requirements of Crops | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) PDF Download
Q. 1 The command area of a canal grows only one crop, i.e., wheat. The base period of wheat is 120 days and its total water requirement, A, is 40 cm. If the canal discharge is 2 m3/s, the area, in hectares, rounded off to the nearest integer, which could be irrigated (neglecting all losses) is______ [2019 : 1 Mark, Set-II]
Ans: 5184 ha
Given data:
Base period,
B = 120 days
Delta of crop,
Δ = 40 cm
Discharge, Q= 2 m3/s
Area to be irrigated,
A = ?
∴ Duty of water,
and Area to be irrigated;
A = Q x D
Q. 2 The intensity of irrigation for the Kharif season is 50% for an irrigation project with culturable command area of 50,000 hectares. The duty for the Kharif season is 1000 hectare cumec. Assuming transmission loss of 10%, the required discharge (in cumec, up to two decimal places) at the head of the canal is __________ . [2018 : 1 Mark, Set-II]
Ans: 27.78cumec
Culturable command area = 50000 ha
Intensity of irrigation for kharif season = 50%
∴Area under kharif = 25000 ha
Duty for kharif season = 1000 ha/cumec
∴ Discharge at the head of field
Transmission/conveyance loss = 10%
Discharge at the head of canal
Q. 3 The culturable command area of a canal is 10000 ha. The area grows only two crops-rice in the Kharif season and wheat in the Rabi season. The design discharge of the canal is based on the rice requirements, which has an irrigated area of 2500 ha, base period of 150 days and delta of 130 cm. The maximum permissible irrigated area (in ha) for wheat, with a base period of 120 days and delta of 50 cm, i s _________ . [2017 : 2 Marks, Set-II]
Ans: 5199.966 ha ≈ 5200 ha
Method I
CCA = 10000 ha
For rice, Δr = 130 cm = 1.3 m
Ar = 2500ha
Br = 150 days
Duty,
For wheat also, discharge is 2.5077 m3/s,
Bw = 120days
Δ = 50 cm
Aw = Qx Dw
∴ Aw = 2.5077 x 2073.6
= 5199.966 ha ≈ 5200 ha
Method II
Dischage,
Now, equate Q for ricve and wheat
Hence, Qr = Qw
Q. 4 Field channel has cultivable commanded area of 2000 hectares. The intensities of irrigation for gram and wheat are 30% and 50% respectively. Gram has a kor period of 18 days, kor depth of 12 cm, while wheat has a kor period of 18 days and a kor depth of 15 cm. The discharge (in m3/s) required in the field channel to supply water to the commanded area during the kor period is ___________. [2015 : 2 Marks, Set-II]
Ans: 1.4275 m3/s
Gram and wheat are Rabi crops.
So both are grown in the same period.
Discharge required for gram,
Discharge required for wheat,
Q. 5 The two columns below show some parameters and their possible values.
Parameter:
P. Gross Command Area
Q. Permanent Wilting Point
R. Duty of canal water
S. Delta of wheat
Value:
I. 100 hectares/cumec
II. 6°C
III. 1000 hectares
IV. 1000 cm
V. 40 cm
VI. 0.12
Which of the following options matches the parameters and the values correctly?
(a) P-I, Q-ll, R-lll, S-IV
(b) P-lll, Q-VI, R-I, S-V
(c) P-I, Q-V, R-VI, S-ll
(d) P-lll, Q-ll, R-V, S-IV
[2015 : 1 Mark, Set-I]
Ans: (B)
Gross Command Area is measured in hectares. (P-III)
Duty has a unit of ha/cumec. (R-I)
Delta of wheat can be practically 40 cm ( S - V )
Q. 6 Irrigation water is to be provided to a crop in a field to bring the moisture content of the soil from the existing 18% to the field capacity of the soil at 28%. The effective root zone of the crop is 70 cm. It the densities of the soil and water are 1.3 g/cm3 and 1.0 g/cm3 respectively, the depth of irrigation water (in mm) required for irrigating the crop i s ________ . [2014 : 2 Marks, Set-II]
Ans: 91mm
Given,
Root zone depth,
d = 70 cm
Field capacity,
Fc = 28%
Existing moisture content,
w = 18%
Density of soil,
γ = 1.3 gm/cm3
Density of water,
γw = 1.0 gm/cm3
Depth of irrigation water required,
Note: It is to be understood that , but as it is not mentioned clearly in the question that the density of 1.3 g/cc is computed at which moisture content, hence, assuming the given density of 1.3 g/cc to be the dry density.
If it would have been given that density of 1.3 g/cc is at moisture content 18%, then we would have computed the dry density from the formula
Q. 7 The transplantation of rice requires 10 days and total depth of water required during transplantation is 48 cm. During trans-plantation there is an effective rainfall (useful for irrigation) of 8 cm. The duty of irrigation water in hectare/cumecs is
(a) 612
(b) 216
(c) 300
(d) 108 [2013 : 2 Marks]
Ans: (b)
Δ = Total water depth required - Effective rainfall = 48 - 8 = 40 cm
Q. 8 Wheat crop requires 55 cm of water during 120 days of base period. The total rainfall during this period is 100 mm. Assume the irrigation efficiency to be 60%. The area (in ha) of the land which can be irrigated with a canal flow of 0.01 m3/s is
(a) 13.82
(b) 18.85
(c) 23.04
(d) 230.40 [2012 : 2 Marks]
Ans: (a)
Base period, 6 = 120 days
Δ = Net water required by the crop during base period
= Total water requirement - Rainfall
= 55 - 10 = 45 cm = 0.45 m
Now,
⇒ D - 2304 ha/cumec
and Q= 0.01 m3/s
∴ Area = A = D x Q
= 2304 x 0.01 ha
= 23.04 ha
Efficiencyof irrigation = 60%
∴ Area which can be irrigated
Q. 9 The moisture holding capacity of the soil in a 100 hectare farm is 18 cm/m. The field is to be irrigated when 50 percent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75 percent. Details of crops planned for cultivation are as follows:
The area of crop 'Y’ that can be irrigated when the available capacity of irrigation system is 40 litres/sec is
(a) 40 hectares
(b) 36 hectares
(c) 30 hectares
(d) 27 hectares [2010 : 2 Marks]
Ans: (d)
Moisture holding capacity of soil for crop
Y = 18 x 0.8 = 14.4 cm
Allowable depletion of moisture
Consumptive use of water for crop,
Y = 0.4 cm/day
∴ Frequency of irrigation
Thus 7.2 cm depth of water is to be applied to the crop Y in the next 18 days.
Irrigation requirement =
Quantity of water required = Discharge through pump when pump works @ 10 hrs per day
Q. 10
The moisture holding capacity of the soil in a 100 hectare farm is 18 cm/m. The field is to be irrigated when 50 percent of the available moisture in the root zone is depleted. The irrigation water is to be supplied by a pump working for 10 hours a day, and water application efficiency is 75 per cent. Details of crops planned for cultivation are as follows:
The capacity of irrigation system required to irrigate crop ‘X in 36 hectares is
(a) 83 litres/sec
(b) 67 litres/sec
(c) 57 litres/sec
(d) 53 litres/sec [2010 : 2 Marks]
Ans: (b)
Moisture holding capacity of soil = 18 cm /m
Root zone depth of crop X = 1.0 m
Moisture holding capacity of soil for crop
X = 18 x 1 = 18cm
Allowable depletion of moisture
Consumptive use of water for crop
X = 0.5 cm/day
∴ Frequency of irrigation
Thus 9 cm depth of water is to be applied to the crop Xin the next 18 days.
Irrigation requirement
Quantity of water required
∴ Discharge through pump (if pump works for 10 hrs per day)
The document Past Year Questions: Water Requirements of Crops | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.
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## FAQs on Past Year Questions: Water Requirements of Crops - Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)
1. What are the water requirements of crops?
Ans. The water requirements of crops refer to the amount of water needed for optimal growth and development. It includes factors such as the crop type, stage of growth, climate, soil conditions, and management practices. Water requirements vary among different crops, with some requiring more water than others to achieve maximum yield.
2. How can I determine the water requirements of a specific crop?
Ans. The water requirements of a specific crop can be determined through various methods. One common approach is to use crop coefficient values, which are multiplied by reference evapotranspiration (ET0) to estimate crop water use. ET0 is calculated based on weather data, and the crop coefficient takes into account the crop's growth stage and other factors. Another method is to use soil moisture sensors or other irrigation scheduling tools to monitor the crop's water needs.
3. What are the consequences of under-watering crops?
Ans. Under-watering crops can have several negative consequences. It can lead to reduced crop growth and development, lower yields, and poor quality produce. Insufficient water can also make crops more susceptible to pests and diseases. Additionally, under-watering can cause stress to the plants, resulting in wilting, leaf curling, and ultimately plant death in severe cases.
4. What are the consequences of over-watering crops?
Ans. Over-watering crops can also have detrimental effects. Excessive water can lead to waterlogging, which deprives the roots of oxygen and hinders nutrient uptake. This can cause root rot and other root-related diseases. Over-watering can also leach nutrients from the soil, leading to nutrient deficiencies. Furthermore, it can contribute to soil erosion and wastage of water resources.
5. How can I efficiently manage water for crop irrigation?
Ans. Efficient water management for crop irrigation involves several strategies. It is essential to use irrigation methods that minimize water loss, such as drip irrigation or precision sprinklers. Monitoring soil moisture levels and using irrigation scheduling tools can help ensure that crops receive adequate water without over-watering. Implementing water-saving practices such as mulching, using cover crops, and optimizing irrigation timing can also contribute to efficient water management. Additionally, incorporating water-saving technologies like rainwater harvesting or using recycled water can further enhance water efficiency in crop irrigation.
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# Concepts For today:¶
• (delve briefly into) Solving for systems of linear equations
• Reduced Row Echelon Form as a Matrix Transformation
• Invertability
In [14]:
import numpy as np
# Solving for a System of Linear Equations¶
$$x + y = 7$$ $$x + 2y = 11$$
To solve for a system of equations, we must have as many equations as unknowns. The idea is to leverage the equations to isolate values for the unknowns.
Three potential outcomes:
1. We'll find many solutions (potentially an infinite number of solutions for potential values of $x$ and $y$
2. We'll find one solution.
3. We'll find no solution.
Essentially, we are trying to locate the point (or set of points) where these two lines intersect.
Note: The far deeper discussion on how solve a system of linear equations is covered in the reading.
Let's solve the above system.
$$x + y = 7\\ x + 2y = 11$$
$$x = 7 - y \\ x + 2y = 11$$
$$x = 7 - y \\ (7 - y) + 2y = 11$$
$$x = 7 - y \\ y = 4$$
$$x = 7 - (4) \\ y = 4$$
$$x = 3 \\ y = 4$$
Let's plug these values back in to get see if they work.
$$(3) + (4) = 7\\ (3) + 2(4) = 11$$
Looks good!
## System of Linear Equations in Matrix Form¶
$$1x + 1y = 7\\ 1x + 2y = 11$$
$$\begin{bmatrix} 1 & 1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 11 \end{bmatrix}$$
$$\textbf{A} \textbf{b} = \textbf{y}$$
Where:
• $\textbf{A}$ can be thought of as our data
• $\textbf{b}$ can be thought of as our unknown coefficients
• $\textbf{y}$ can be thought of as our outcomes for which we are trying to solve for
We can express this matrix as an "augmented matrix" (which will help as we perform row-wise operations)
$$\left| \begin{array}{cc|c} 1 & 1 & 7 \\ 1 & 2 & 11 \\ \end{array} \right|$$
## Reduced Row Echelon Form (rref)¶
The goal of RREF is to use row-wise addition/subtraction and scaling to reduce each column so that 1 of the row entries equals one and the rest of the entries equal 0.
In essence, we want $\textbf{A}$ to resemble $\textbf{I}$.
$$\begin{bmatrix} 1 & 1\\ 1 & 2 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$
Why might we want to do this?
Recall, if we think of matrix multiplication as a linear transformation (as we did last time), we'll remember that we're really fundamentally changing those unit vectors into a new coordinate system (e.g. $c_1\textbf{A}\hat{i} + c_2\textbf{A}\hat{j}$). The aim here is to reverse that process. That is, what steps do we need to take to go back to our original unit vectors ($\hat{i},\hat{j}$).
We can do this as we did before (i.e. when writing a function that would do the transformation for us) by performing the row-wise operations on both sides of the augmented matrix. The result will be the solution to our linear equation (if a solution exists).
Again, this is a simplification of rref and solving for systems of equations. Check out the reading for a more involved discussion. The point is to get the intuition of what is going on here.
$$\left| \begin{array}{cc|c} 1 & 1 & 7 \\ 1 & 2 & 11 \\ \end{array} \right|$$
Hold the first row fixed. How do we get position (2,1) in the matrix to be zero? Subtract the first row from the second.
$$\left| \begin{array}{cc|c} 1 & 1 & 7 \\ 1 - 1 & 2 - 1 & 11 - 7 \\ \end{array} \right|$$
$$\left| \begin{array}{cc|c} 1 & 1 & 7 \\ 0 & 1 & 4 \\ \end{array} \right|$$
Holding the second row fixed. How do we get position (1,2) in the matrix to be zero? Subtract the second row from the first.
$$\left| \begin{array}{cc|c} 1-0 & 1-1 & 7-4 \\ 0 & 1 & 4 \\ \end{array} \right|$$
$$\left| \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 4 \\ \end{array} \right|$$
We've found our solution!
Let's check it.
$$\begin{bmatrix} 1 & 1\\ 1 & 2 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 1(3) + 1(4)\\ 1(3) + 2(4) \end{bmatrix} = \begin{bmatrix} 7 \\ 11 \end{bmatrix}$$
In [2]:
A = np.array([[1,1],[1,2]])
x = np.array([3,4])
b = A.dot(x)
b
Out[2]:
array([ 7, 11])
### RREF as a matrix transformation¶
We can encode these instructions to reduce the data down as a matrix transformation (just as we did when we first introduced matrices). We'll perform our instructions on the identity matrix, $\textbf{I}$, like we did before.
Again, our instructions (i.e. the steps we took above):
1. Hold the first row constant and subtract the first row from the second row
$$f_1(x) = \begin{bmatrix} x_1\\x_2 - x_1\end{bmatrix}$$
1. Hold the second row constant and subtract the second row from the first row
$$f_2(x) = \begin{bmatrix} x_1 - x_2 \\x_2\end{bmatrix}$$
Let's perform these operations on our identity matrix.
$$\textbf{I} = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$
$$f_1(\textbf{I}) = \begin{bmatrix} 1 & 0\\0-1 & 1-0\end{bmatrix} = \begin{bmatrix} 1 & 0\\-1 & 1\end{bmatrix}$$
$$f_2(\begin{bmatrix} 1 & 0\\-1 & 1\end{bmatrix}) = \begin{bmatrix} 1-(-1) & 0-1 \\-1 & 1\end{bmatrix}= \begin{bmatrix} 2 & -1\\-1 & 1\end{bmatrix}$$
$$\begin{bmatrix} 2 & -1\\-1 & 1\end{bmatrix}$$
What is this matrix? It's the inverse of matrix $\textbf{A}$!
$$\textbf{A} = \begin{bmatrix}1 & 1\\ 1 & 2\end{bmatrix}$$
$$\textbf{A}^{-1} = \begin{bmatrix} 2 & -1\\-1 & 1\end{bmatrix}$$
We can use this inverse transformation to take our vector back to where we started.
$$\textbf{A}^{-1}\textbf{y} = \begin{bmatrix} 2 & -1\\-1 & 1\end{bmatrix} \begin{bmatrix}7\\11\end{bmatrix} = \begin{bmatrix}3\\4\end{bmatrix} = \textbf{b}$$
$$\textbf{A}\textbf{x} = \begin{bmatrix}1 & 1\\ 1 & 2\end{bmatrix} \begin{bmatrix}3\\4\end{bmatrix} = \begin{bmatrix}7\\11\end{bmatrix} = \textbf{y}$$
We can think of the inverse transformation as numerical instructions to solve for a system of linear equations!
In [3]:
T1 = np.array([[1,0],[-1,1]])
print("Our first tranformation matrix\n")
print(T1)
T2 = np.array([[1,-1],[0,1]])
print("\nOur second tranformation matrix\n")
print(T2)
Our first tranformation matrix
[[ 1 0]
[-1 1]]
Our second tranformation matrix
[[ 1 -1]
[ 0 1]]
In [4]:
# together
A_inv = T2.dot(T1)
In [5]:
b
Out[5]:
array([ 7, 11])
In [6]:
A_inv.dot(b)
Out[6]:
array([3, 4])
And where $\textbf{A}$ dotted with its inverse $\textbf{A}^{-1}$ transforms us back to $\textbf{I}$
In [7]:
A_inv.dot(A)
Out[7]:
array([[1, 0],
[0, 1]])
# Invertible Functions¶
Note (or recall) that we can only solve for a system where there is as many equations as there are unknowns.
We cannot solve this... $$x + y - 3z = -10\\ x - y + 2z = 3$$
But we could potentially solve this...
$$x + y - 3z = -10\\ x - y + 2z = 3 \\ 2x + y - z = -6$$
What does this mean for us in linear algebra land?
$$\begin{bmatrix} 1 & 1 & -3 \\1 & -1 & 2 \\ 2 & 1 & -1\end{bmatrix} \begin{bmatrix} x\\y\\z\end{bmatrix} = \begin{bmatrix} -10\\2\\-6\end{bmatrix}$$
The matrix we are inverting must always be a square matrix. That is, the rank of the column space must be equal to the rank of the row space.
This is just a fancy way of saying that none of the column vectors are a linear combination of another column vector, i.e. they are linearly independent. And likewise with the row vectors.
$$Rank(colspace(\textbf{A})) = Rank(rowspace(\textbf{A}))$$
$$N~Cols = N~Rows$$
Another way to think about this, if a matrix transformation reduces a vector or matrix into a lower dimension (dimension reduction), then we can't walk back to where we started. Dimension reduction always results in a loss of information.
More generally, let's thinking about what it means to invert a function.
$$f: x \mapsto y$$
An inverse function takes us back from our codomain $y$ to our original domain $x$
$$f^{-1}: y \mapsto x$$
But we can only do this for a function that is surjective and injective.
• surjective: there exists a mapping for every value of set $\textbf{x}$ "onto" set $\textbf{y}$. That is, every value from one set maps onto a value in the other set.
Surjective!
X Y
- -
a => z
b => y
c => x
d => x
Not Surjective
X Y
- -
a => z
b => y
c => x
d
• injective: there exists a "one-to-one" mapping of values in set $\textbf{x}$ onto set $\textbf{y}$. That is. there exists a unique mapping for each $\textbf{x}$ onto $\textbf{y}$
Injective!
X Y
- -
a => z
b => y
c => x
d => w
Not injective
X Y
- -
a => z
b => y
c => x
d => x
The idea is that every value of $x$ maps onto a unique value of $y$. If we don't have sufficient information, that is, if we don't have equal number of dimensions, then the function wouldn't be surjective and we can't invert.
### What to do if our matrix isn't square?¶
Recall from the last lecture that we can always generate a square matrix by projecting it back onto itself by squaring it.
In [8]:
B = np.random.randn(5,3)
B.shape
Out[8]:
(5, 3)
In [9]:
B.dot(B.T).shape
Out[9]:
(5, 5)
In [10]:
B.T.dot(B).shape
Out[10]:
(3, 3)
## How to determine if a matrix is invertible¶
Recall above that we encoded instructions regarding how to convert a $2 \times 2$ matrix into RREF.
Let's follow those steps again but this time on a more general representation of the matrix.
$$\textbf{A} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} R_1 \\ R_2 \end{bmatrix}$$
Let's put it in augmented matrix form, and perform our row-wise manipulations simultaneously on $\textbf{I}$
$$\left| \begin{array}{cc|cc} a & b & 1 & 0 \\ c & d & 0 & 1 \\ \end{array} \right|$$
Transformation 1:
$$\textbf{T}_1 = \begin{bmatrix} R_1 \\ aR_2 - cR_1 \end{bmatrix}$$
$$\left| \begin{array}{cc|cc} a & b & 1 & 0 \\ 0 & ad - bc & -c & a \\ \end{array} \right|$$
Transformation 2:
$$\textbf{T}_2 = \begin{bmatrix} (ad-bc)R_1 - bR_2 \\ R_2 \end{bmatrix}$$
$$\left| \begin{array}{cc|cc} a(ad-bc) & 0 & ad & -ab \\ 0 & ad - bc & -c & a \\ \end{array} \right|$$
Transformation 3: ensure the diagonals equal 1
$$\textbf{T}_3 = \begin{bmatrix} \frac{R_1}{a(ad-bc)} \\ \frac{R_2}{(ad-bc)} \end{bmatrix}$$
$$\left| \begin{array}{cc|cc} 1 & 0 & \frac{d}{(ad-bc)} & \frac{-b}{(ad-bc)} \\ 0 & 1 & \frac{-c}{(ad-bc)} & \frac{a}{(ad-bc)} \\ \end{array} \right|$$
This yields the formula for the inverse of a $2 \times 2$ matrix.
$$\frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
What is $ad-bc$?
#### The Determinant¶
$$det(\textbf{A}) = |\textbf{A}| = ad-bc$$
The determinant of matrix $\textbf{A}$ tells us if a (square) matrix is invertible. When we examine the above equation, it's obvious why this is. A fraction with a denominator of $0$ is undefined, meaning we can't solve it.
But what does this actually mean?
It means that the vectors composing the square matrix are not linearly independent.
In [11]:
A = np.array([[1,2],[2,4]])
A
Out[11]:
array([[1, 2],
[2, 4]])
In [12]:
np.linalg.det(A)
Out[12]:
0.0
To give a better intuition of what is going on, think of the determinant as the area of a square (when in $\Re^2$) generated by our two basis vectors. When we transform these vectors, that area grows and shrinks. When that area goes to zero, it means that we've collapsed to a lower dimension (i.e. down to a line if we were in $\Re^2$). Thus, the determinant tells us if there is sufficient information in the matrix to take us back to where we started (i.e. if the column and row space of our square matrix are actually linearly independent).
Here is a great video that outlines how we can think of the determinant of a matrix in $\Re^2$ as the area of a square.
Note that when the determinant of a matrix is 0 we call it singular.
NOTE: that finding the determinant for an $n \times n$ matrix is more involved! See the reading for a deeper discussion on this. We'll be relying on our computers to compute these values, but it's useful to have a deeper understanding of the steps.
### Which of these matrices are invertible?¶
In [15]:
X = np.array([[14,-7],[2,-1]])
Y = np.random.randn(4,5)
Z = np.array([[0,0,0,0],
[1,0,0,1],
[0,1,0,0]])
print(X,"\n")
print(Y,"\n")
print(Z,"\n")
[[14 -7]
[ 2 -1]]
[[-1.31692334 -0.44108299 0.56100078 0.93984351 2.24174342]
[-0.62457184 -0.16588639 0.14610647 -1.78707646 -0.05134602]
[ 0.57006418 -0.31545966 0.03206903 1.0371645 0.23118344]
[-1.00540481 1.65994245 -0.26802636 0.90013057 -0.13122613]]
[[0 0 0 0]
[1 0 0 1]
[0 1 0 0]]
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# Systems of Linear Equations
## Presentation on theme: "Systems of Linear Equations"— Presentation transcript:
Systems of Linear Equations
This is a Grade 11 Applied Math presentation for ‘Systems of Linear Equations’ The equation for a line has two variables: x and y If there are two unknowns (variables) you need two different facts (or equations) to find the unknowns MA30SA_C_SysLinEq.ppt Revised:
How much does the cow weigh?
6 Diagram A y x 7 Diagram B x y ? C x x is the weight of the Cow. y is the weight of the Beaver. 1 Cow and 1 Beaver weigh 6, 1 Cow and 2 Beavers weigh 7. How much does 1 Cow weigh by itself? How much does a beaver weigh?
‘Systems of Linear Equations’
Solution Did you say the cow weighs 5? So how much does a beaver weigh? Did you say 1? How did you get that answer?? That is all that this unit is about: ‘Systems of Linear Equations’
Definitions Systems of Linear Equations
A set of two or more linear equations with the same variables. Solution to a System of Linear Equations The set of all ordered pairs that satisfy all equations (make them true) In Grade 11 Applied we will only look at two equations with two variables Systems of Linear Equations are also sometimes called ‘Simultaneous Linear Equations’ if you google the subject
Other examples of typical problems for Systems of Linear Equations
Jason is twice as old as Scott. Scott is 3 years younger than Jason. How old is Jason? Two unknowns; ages of Scott and Jason The sum of two numbers equals 10, their difference equals 4. What are the two numbers? Two unknowns: Number A and Number B It takes you two hours to paddle upstream a distance of two km, and one hour to come back to the start. What is the speed of the canoe and of the current? Two unknowns: your paddling speed and the speed of the current
Graphing Tools This unit uses the TI-83 Graphing tool primarily for graphing There are lots of other graphing tools also, many are on-line All of the problems can be done manually using graph paper if necessary, but you will find graphing tools much more useful. This presentation will demonstrate the manual method
Graph the two equations
Graph the 2 equations: Line 1: x + y = 4 Line 2: 4x – 4y = 8 Notice the tables have one point, and only one, in common. In the graph, it is where the lines meet
Graph two more equations
Graph the 2 equations: Line 1: y = 3x + 2 Line 2: y = (1/2)x – 3 Notice the tables have one point, and only one, in common. In the graph, it is where the lines meet
Graph even more equations
Graph the 2 equations: Line 1: 2x + 2y = 4 Line 2: x + y = 2 Notice the table has all points in common for both lines. They are really the same lines.
Graph a final two equations
Graph the 2 equations: Line 1: y = 3x + 2 Line 2: y = 3x – 2 Notice the table has no points in common The lines never meet, they are parallel!
Types of Linear Systems
Consistent Have solution(s) Independent and consistent has one solution Lines have different slopes. Lines meet at one point Dependent and consistent has all the points on the line as a solution. The lines have same slopes and same intercepts They are the same line! Inconsistent The lines are parallel, lines never meet, they have no points in common.
Consistent (Dependent or Independent) or Inconsistent?
Solving Systems of Equations Using Algebra
There are two ways to solve Systems of Linear Equations using Algebra Using Algebra is best because you don’t need to graph lines The two methods: Elimination by Addition or Subtraction Elimination by Substitution
Given two equations 2x + y = 8 2x – y = 0 Label the equations 2x + y = 8 2x – y = 0 Eliminate one variable by adding or subtracting the equations. (1)+(2) 2x + y = 8 +(2x – y = 0) 4x + 0y = 8 x = 2 We know that x =2 now So just substitute it back into either equation (1) or (2) to find y: 2(2)+y=8 y = 4 The solution to the System is when x = 2 and y = 4. It is the point (2, 4)
Sometimes you need to change one of the equations, or both, so that when you add or subtract them one of the variables will be eliminated
Eliminate one variable by adding or subtracting the equations after multiplying one equation by a suitable constant. (1)+2*(2) (1) 4x + 2y = 16 (2) +(4x – 2y = 8) 8x + 0y = 24 x = 3 Given two equations 4x + 2y = 16 2x – y = 4 Label the equations 4x + 2y = 16 2x – y = 4 We know that x =3 now So just substitute it back into either equation (1) or (2) to find y: 4(3)+2y=16 y = 2 The solution to the System is when x = 3 and y = 2. It is the point (3, 2)
More examples Elimination
Solve: (1) 5x + 2y = 3 (2) 2x + 3y = –1 3*(1) x + 6y = 9 -2*(2) – (4x + 6y = –2) Subtract: 11x – 0y = 11 11x = x = 1 Substitute value x = 1 into either equation to find y 5(1) + 2y = 3 y = –1 Solution: (1, –1) Check the answer!! Solve: (1) 2x + 3y = 3 (2) –6x + 6y = 6 3*(1) x + 9y = 9 Add (2) +(–6x + 6y = 6) Add: x + 15y = 15 y = 1 Substitute value y = 1 into either equation to find x 2x + 3(1) = 3 2x = 0; x = 0 Solution: (0, 1) Check the answer!!
Review of Steps to Solve by Elimination
Put equations into a ‘standard’ order (ax + by = c) Label the equations Multiply either, or both, equations by some number so that one variable will be eliminated when the equations are added or subtracted Add or subtract the equations to eliminate one variable and find the remaining variable Find the eliminated variable by substituting the the value of the solved variable back into either equation of the system Check your answer (x, y) in both equations
Solving by Substitution
The steps are: Label the equations (1), (2) Isolate one variable and express it in terms of the other Substitute that expression into the other equation Solve for the now single variable Substitute the value of the found variable back into either equation to find the other unknown Check the answer in both equations Given: (1): 2x – y = 4 (2): x + 2y = – 3 From (2): x = – 3 – 2y Substitute into (1) 2(– 3 – 2y) – y = 4 – 6 – 4y – y = 4 – 6 – 5y = 4 – 5y = 10 So: y = – 2 Substitute into either eqn 2x – (– 2) = 4 2x + 2 = 4 ; 2x = 2 So x = 1 Now Check the Answer!
Sample Problem - Digits
Two numbers add to 12 The same two numbers have a difference of 4 Find the two numbers Give the unknowns labels: like a and b Write the two facts or equations: (1) a + b = 12 (2) a – b = 4 Add eqns (1) and (2) 2a + 0 =16 2a = 16 a = 8. So if a = 8 then b = 4 from either equation So the numbers are 8 and 4 Check to see if these numbers work in both equations
Sample Problems - Motion
An airplane covers a distance of 1500 miles in 3 hours when it flies with the wind, and in 3 and 1/3 hours when it flies against the wind. What is the speed of the plane in still air . What is the speed of the wind? Unknowns; a: speed of airplane; b: speed of wind a + b=(1500/3) = 500 mph with the wind a – b = 1500/3.333= 450 mph against the wind Solve from there! ??
Sample Problems - Age A father is 24 years older than his son.
In 8 years he will be twice as old as his son. Determine their current ages Let: f = father’s current age. s = son’s current age. f – s = 24 (f + 8) = 2*(s + 8) f + 8 = 2s + 16 f – 2s = 8 Eqn (1) – Eqn (2) – (f – 2s = 8) s = 16 f = 40 Check: (1): (40) – (16) = 24 (2): (40 + 8) = 2*(16 + 8)
Coin Problems Edgar has 20 coins in dimes and nickels, which together total \$1.40. How many of each does he have? Let n = nbr of nickels, let d = nbr of dimes (1): n(0.05)+d(0.10) = 1.40 (2): n + d = 20 Use substitution (2): n = 20 – d Substitute eqn (2) into eqn (1) (20 - d)*(0.05) + d*(0.10)= 1.40 d = 1.4 d=8, n=12 Check the answer in both equations It works!!
Word Problems 1000 tickets were sold. Adult tickets cost \$8.50, children's cost \$4.50, and a total of \$7300 was collected. How many tickets of each kind were sold Let a = nbr of adult tickets sold; let c = nbr of children tickets sold (1): a + c = 1000 (2): 8.5a + 4.5c = 7300 Substitute: a = 1000 – c 8.5(1000 – c) + 4.5c = 7300 8500 – 4c = 7300 c = and a = 700 700 adult tickets were sold, 300 children’s tickets were sold Now check answer in both equations!
The End Now you know how to solve Systems of Linear Equations with two unknowns If you take Pre-Calculus, you will learn how to solve for 3 unknowns! Just a few extra steps for three unknowns
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# Check if given number is Emirp Number or not
An Emirp Number (prime spelled backwards) is a prime number that results in a different prime when its decimal digits are reversed. This definition excludes the related palindromic primes.
Examples :
```Input : n = 13
Output : 13 is Emirp!
Explanation :
13 and 31 are both prime numbers.
Thus, 13 is an Emirp number.
Input : n = 27
Output : 27 is not Emirp.```
Objective: Input a number and find whether the number is an emirp number or not.
Approach: Input a number and firstly check if its a prime number or not. If the number is a prime number, then we find the reverse of the original number and check the reversed number for being prime or not. If the reversed number is also prime, then the original number is an Emirp Number otherwise it is not.
Below is the implementation of the above approach :.
## C++
`// C++ program to check if given``// number is Emirp or not.``#include ``using` `namespace` `std;` `// Returns true if n is prime.``// Else false.``bool` `isPrime(``int` `n)``{`` ``// Corner case`` ``if` `(n <= 1)`` ``return` `false``;` ` ``// Check from 2 to n-1`` ``for` `(``int` `i = 2; i < n; i++)`` ``if` `(n % i == 0)`` ``return` `false``;` ` ``return` `true``;``}` `// Function will check whether number``// is Emirp or not``bool` `isEmirp(``int` `n)``{`` ``// Check if n is prime`` ``if` `(isPrime(n) == ``false``)`` ``return` `false``;` ` ``// Find reverse of n`` ``int` `rev = 0;`` ``while` `(n != 0) {`` ``int` `d = n % 10;`` ``rev = rev * 10 + d;`` ``n /= 10;`` ``}` ` ``// If both Original and Reverse are Prime,`` ``// then it is an Emirp number`` ``return` `isPrime(rev);``}` `// Driver code``int` `main()``{`` ``int` `n = 13; ``// Input number`` ``if` `(isEmirp(n) == ``true``)`` ``cout << ``"Yes"``;`` ``else`` ``cout << ``"No"``;``}` `// This code is contributed by Anant Agarwal.`
## Java
`// Java program to check if given number is``// Emirp or not.``import` `java.io.*;``class` `Emirp {`` ``// Returns true if n is prime. Else`` ``// false.`` ``public` `static` `boolean` `isPrime(``int` `n)`` ``{`` ``// Corner case`` ``if` `(n <= ``1``)`` ``return` `false``;` ` ``// Check from 2 to n-1`` ``for` `(``int` `i = ``2``; i < n; i++)`` ``if` `(n % i == ``0``)`` ``return` `false``;` ` ``return` `true``;`` ``}` ` ``// Function will check whether number`` ``// is Emirp or not`` ``public` `static` `boolean` `isEmirp(``int` `n)`` ``{`` ``// Check if n is prime`` ``if` `(isPrime(n) == ``false``)`` ``return` `false``;` ` ``// Find reverse of n`` ``int` `rev = ``0``;`` ``while` `(n != ``0``) {`` ``int` `d = n % ``10``;`` ``rev = rev * ``10` `+ d;`` ``n /= ``10``;`` ``}` ` ``// If both Original and Reverse are Prime,`` ``// then it is an Emirp number`` ``return` `isPrime(rev);`` ``}` ` ``// Driver Function`` ``public` `static` `void` `main(String args[]) ``throws` `IOException`` ``{`` ``int` `n = ``13``; ``// Input number`` ``if` `(isEmirp(n) == ``true``)`` ``System.out.println(``"Yes"``);`` ``else`` ``System.out.println(``"No"``);`` ``}``}`
## Python3
`# Python3 code to check if``# given number is Emirp or not.` `# Returns true if n is prime.``# Else false.``def` `isPrime( n ):`` ` ` ``# Corner case`` ``if` `n <``=` `1``:`` ``return` `False`` ` ` ``# Check from 2 to n-1`` ``for` `i ``in` `range``(``2``, n):`` ``if` `n ``%` `i ``=``=` `0``:`` ``return` `False`` ` ` ``return` `True` `# Function will check whether``# number is Emirp or not``def` `isEmirp( n):`` ` ` ``# Check if n is prime`` ``n ``=` `int``(n)`` ``if` `isPrime(n) ``=``=` `False``:`` ``return` `False`` ` ` ``# Find reverse of n`` ``rev ``=` `0`` ``while` `n !``=` `0``:`` ``d ``=` `n ``%` `10`` ``rev ``=` `rev ``*` `10` `+` `d`` ``n ``=` `int``(n ``/` `10``)`` ` ` ` ` ``# If both Original and Reverse`` ``# are Prime, then it is an`` ``# Emirp number`` ``return` `isPrime(rev)` `# Driver Function``n ``=` `13` `# Input number``if` `isEmirp(n):`` ``print``(``"Yes"``)``else``:`` ``print``(``"No"``)`` ` `# This code is contributed by "Sharad_Bhardwaj".`
## C#
`// C# program to check if given``// number is Emirp or not.``using` `System;` `class` `Emirp {`` ``// Returns true if n is prime`` ``// Else false.`` ``public` `static` `bool` `isPrime(``int` `n)`` ``{`` ``// Corner case`` ``if` `(n <= 1)`` ``return` `false``;` ` ``// Check from 2 to n-1`` ``for` `(``int` `i = 2; i < n; i++)`` ``if` `(n % i == 0)`` ``return` `false``;` ` ``return` `true``;`` ``}` ` ``// Function will check whether number`` ``// is Emirp or not`` ``public` `static` `bool` `isEmirp(``int` `n)`` ``{`` ``// Check if n is prime`` ``if` `(isPrime(n) == ``false``)`` ``return` `false``;` ` ``// Find reverse of n`` ``int` `rev = 0;`` ``while` `(n != 0) {`` ``int` `d = n % 10;`` ``rev = rev * 10 + d;`` ``n /= 10;`` ``}` ` ``// If both Original and Reverse are Prime,`` ``// then it is an Emirp number`` ``return` `isPrime(rev);`` ``}` ` ``// Driver Function`` ``public` `static` `void` `Main()`` ``{`` ``int` `n = 13; ``// Input number`` ``if` `(isEmirp(n) == ``true``)`` ``Console.WriteLine(``"Yes"``);`` ``else`` ``Console.WriteLine(``"No"``);`` ``}``}` `// This code is contributed by vt_m.`
## PHP
``
## Javascript
``
Output :
`Yes`
Time complexity: O(n)
Auxiliary Space:O(1)
#### Approach#2: Using recursion
Check if the given number is prime or not If it is prime, reverse the number and check if the reversed number is also prime or not If both the original and reversed numbers are prime, then the given number is an Emirp number
#### Algorithm
1. Define a function is_prime() to check if a number is prime or not
2. Define a function reverse_number() to reverse a given number
3. Define a function is_emirp() which takes a number as input
4. Check if the given number is prime or not using is_prime() function
5. If it is not prime, return “Not Emirp”
6. Reverse the given number using reverse_number() function
7. Check if the reversed number is prime using is_prime() function
8. If both the original and reversed numbers are prime, return “Emirp”, else return “Not Emirp”
## C++
`#include ``#include ``#include ` `using` `namespace` `std;` `// Function to check if a number is prime``bool` `is_prime(``int` `num)``{`` ``if` `(num < 2) {`` ``return` `false``;`` ``}`` ``for` `(``int` `i = 2; i <= ``sqrt``(num); i++) {`` ``if` `(num % i == 0) {`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;``}` `// Function to reverse a number``int` `reverse_number(``int` `num)``{`` ``string str_num = to_string(num);`` ``string rev_str_num = ``""``;`` ``for` `(``int` `i = str_num.length() - 1; i >= 0; i--) {`` ``rev_str_num += str_num[i];`` ``}`` ``return` `stoi(rev_str_num);``}` `// Function to check if a number is an emirp``string is_emirp(``int` `num)``{`` ``if` `(!is_prime(num)) {`` ``return` `"Not Emirp"``;`` ``}`` ``int` `rev_num = reverse_number(num);`` ``if` `(is_prime(rev_num) && num != rev_num) {`` ``return` `"Emirp"``;`` ``}`` ``else` `{`` ``return` `"Not Emirp"``;`` ``}``}` `int` `main()``{`` ``int` `num = 27;`` ``cout << is_emirp(num) << endl;``}`
## Python3
`def` `is_prime(num):`` ``if` `num < ``2``:`` ``return` `False`` ``for` `i ``in` `range``(``2``, ``int``(num``*``*``0.5``)``+``1``):`` ``if` `num ``%` `i ``=``=` `0``:`` ``return` `False`` ``return` `True` `def` `reverse_number(num):`` ``return` `int``(``str``(num)[::``-``1``])` `def` `is_emirp(num):`` ``if` `not` `is_prime(num):`` ``return` `"Not Emirp"`` ``rev_num ``=` `reverse_number(num)`` ``if` `is_prime(rev_num) ``and` `num !``=` `rev_num:`` ``return` `"Emirp"`` ``else``:`` ``return` `"Not Emirp"` `num ``=` `27``print``(is_emirp(num))`
## Java
`import` `java.lang.Math;``import` `java.util.Scanner;` `public` `class` `Emirp {`` ``// Function to check if a number is prime`` ``public` `static` `boolean` `isPrime(``int` `num) {`` ``if` `(num < ``2``) {`` ``return` `false``;`` ``}`` ``for` `(``int` `i = ``2``; i <= Math.sqrt(num); i++) {`` ``if` `(num % i == ``0``) {`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;`` ``}` ` ``// Function to reverse a number`` ``public` `static` `int` `reverseNumber(``int` `num) {`` ``String strNum = Integer.toString(num);`` ``String revStrNum = ``""``;`` ``for` `(``int` `i = strNum.length() - ``1``; i >= ``0``; i--) {`` ``revStrNum += strNum.charAt(i);`` ``}`` ``return` `Integer.parseInt(revStrNum);`` ``}` ` ``// Function to check if a number is an emirp`` ``public` `static` `String isEmirp(``int` `num) {`` ``if` `(!isPrime(num)) {`` ``return` `"Not Emirp"``;`` ``}`` ``int` `revNum = reverseNumber(num);`` ``if` `(isPrime(revNum) && num != revNum) {`` ``return` `"Emirp"``;`` ``} ``else` `{`` ``return` `"Not Emirp"``;`` ``}`` ``}` ` ``public` `static` `void` `main(String[] args) {`` ``int` `num = ``27``;`` ``System.out.println(isEmirp(num));`` ``}``}`
## Javascript
`function` `isPrime(num) {`` ``if` `(num < 2) {`` ``return` `false``;`` ``}`` ``for` `(let i = 2; i <= Math.sqrt(num); i++) {`` ``if` `(num % i === 0) {`` ``return` `false``;`` ``}`` ``}`` ``return` `true``;``}` `function` `reverseNumber(num) {`` ``return` `parseInt(num.toString().split(``''``).reverse().join(``''``));``}` `function` `isEmirp(num) {`` ``if` `(!isPrime(num)) {`` ``return` `"Not Emirp"``;`` ``}`` ``const revNum = reverseNumber(num);`` ``if` `(isPrime(revNum) && num !== revNum) {`` ``return` `"Emirp"``;`` ``} ``else` `{`` ``return` `"Not Emirp"``;`` ``}``}` `const num = 27;``console.log(isEmirp(num));`
Output
`Not Emirp`
Time Complexity: O(sqrt(n)), n is the input number for which we are checking if it is an Emirp number or not.
Auxiliary Space: O(log n), n is the input number for which we are checking if it is an Emirp number or not.
My Personal Notes arrow_drop_up
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## ICSE Solutions for Class 6 Geography Voyage – Elements of a Map
APlusTopper.com provides step by step solutions for ICSE Solutions for Class 6 Geography Voyage. You can download the Voyage Geography ICSE Solutions for Class 6 with Free PDF download option. Geography Voyage for Class 6 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.
1. This is not one of the five basic elements of a map
1. a title
2. a north line
3. conventional symbols
4. a twine
2. This is one of the intermediate directions.
1. north
2. south
3. north-east
4. east
3. Places lying halfway between south and south-west will have a
1. south direction
2. west direction
3. south-south-west direction
4. west-south-west direction
4. This is the ratio of the distance between two places on a map to the actual distance between the same two places on the ground.
1. scale
2. plan
3. symbol
4. sketch
5. The numerator of a representative fraction is always
1. 1
2. 2
3. o
4. 100
B. Answer the following questions in brief.
Question 1.
What are the cardinal directions ?
N may or maynot be shown the top of the map or plan indicates the North direction. If no such symbol is shown, then the top portion of the map is taken as the North and the bottom, right and left of the map as the South, the East and the West respectively. These directions—North, South, East and West — are called the cardinal directions.
Question 2.
What is meant by the scale of a map ?
Scale on a map is the distance shown in the map. The scale is given just below the map.
Scale helps us to find out the correct distance between various points on a map. In a scale there is always a proportion between the dimension of the map and the actual area they represent.
Question 3.
Which three ways are used to represent the scale of a map ?
The proportion which exists between the map and actual surface of the Earth is called the scale. A scale can be expressed in three ways
1. by a statement
2. by representative fraction
3. linear scale or graph
Question 4.
Convert the statement 1 cm = 100 km into an R.F. scale.
1 km = 1000 m, 1 m = 100 cm
100 km = 100 × 1000 × 100 = 1,00,00,000 cm
Since RF = Distance on the map in cm
RFscale = 1 / 1,00,00,000 = 1:100,00,000
Question 5.
What method would you use to measure the length of a river ?
We use twine method to measure the length of a river. In the twine method, a twine is placed along the feature to be measured from one end to the other, carefully following all the curves and bends. The length of the twine is then measured in centimetres or inches using a ruler or linear scale. Thereafter, this length is converted into kilometres or miles using the scale of the given map.
Question 6.
What is a legend or key ?
A legend or key is an essential feature of any map. It explains the colours, signs, and symbols used in the map. It uses different colours to show the height or depth of an area above or below sea level respectively. It is provided near the top or the bottom of the map, either on the left-hand or right-hand side.
Question 7.
What are conventional signs and symbols ?
A variety of colourful signs and symbols are used on maps to show natural and man-made features on maps. These signs and symbols give plenty of information and are simple to draw and understand.
Some of these symbols are internationally recognized as they have been determined by convention, i.e., these symbols have been agreed upon and accepted internationally. Therefore, they are also called conventional signs and symbols.
Question 8.
Name the features that are shown in blue, green, and brown colours on a map.
• Blue : water bodies such as oceans, seas, bays, gulfs, lakes, rivers, streams.
• Green : lowlands or plains, both coastal and those built by rivers.
• Brown : hills, mountains, plateaux.
• Yellow : hot deserts.
• White : cold deserts, snow-covered regions.
Question 9.
Name the line of latitude that divides the earth into the northern and southern hemispheres.
The line of latitude that divides the earth into the northern and southern hemispheres is the equator (0°).
A network of horizontal and vertical lines or latitudes and latitudes drawn on a map or globe is called the grid system. Latitudes are the horizontal east-west lines. The longest latitude is the equator (0°).
Question 10.
What are the latitudinal and longitudinal extents of India? Use your atlas to find the answer.
The latitudinal extent of India is 20.59° N and the longitudinal extent of India is 79° E.
C. Answer the following questions in one or two paragraphs.
Question 1.
What are the five different elements of a map ? Explain them briefly ?
The following are the five essential features of a map.
1. Title – It indicate the subject of the map. e.g, distribution of natural vegetation, water bodies, climate, etc.
2. Scale – It denotes the relationship between map distances and actual ground distances. _
3. Direction – It refers to the cardinal direction, i.e., North, South, East and West, Conventionally, a map is aligned with the North towards the top.
4. Key or Legend – It explains the meaning of the symbols that are used in the map.
5. Grid system, usually formed by the cross-conjugation of the lines of longitudes and latitudes.
Question 2.
Explain the functions of the north line and legend on a map.
The direction on a map is measured with the help ofthe north line.
All maps have a key or a legend that explains the colours, signs and symbols used in them.
The equator (0° latitude) divides the earth into the hemispheres — the northern and the southern hemispheres. All lines of latitude are measured from the equator.
Question 3.
What is the difference between the cardinal directions and the intermediate directions ? Draw a diagram to shthe intermediate directions.
The four major directions i.e. North, South, East and West are called cardinal directions. The directions in between these four cardinal directions are called intermediate directions.
E.g North, East, North-North east, South west etc.
Question 4.
Describe any two ways of representing a map scale.
Map scales can be represented as
(a) Verbal or statement scale—i.e. the scale is stated in words as 1 cm = 5 km or 1 cm to 5 km. It means 1 cm on the map is equal to 5 km on ground.
(b) Representative fractions — In this system, the numerator expresses the distance on map and denominator represents the actual distance on ground. Both should have same units i.e. 1 cm on map represents 50,000 cm on ground.
R.F = distance on map cm / distance or ground cm
Question 5.
Why is the key or legend an important element of a map?
A legend or key is an essential feature of any map. It explainsthe colours, signs, and symbols used in the map. It uses different colours to show the height or depth of an area above or below sea level respectively. It is provided near the top or the bottom of the map, either on the left-hand or right-hand side.
Question 6.
Explain why colours are used in maps?
Colours play a significant role in maps. They make maps attractive to look at and easy to understand and interpret. On physical maps, the different colours are conventionally used in this way:
• Blue : water bodies such as oceans, seas, bays, gulfs, lakes, rivers, streams.
• Green : lowlands or plains, both coastal and those built by rivers.
• Brown : hills, mountains, plateaux.
• Yellow : hot deserts.
• White : cold deserts, snow-covered regions.
Question 7.
What are lines of latitude ? Give examples ?
Latitudes are the horizontal east-west lines. The longest latitudes is the equator (0°). All the other latitudes are parallel to the equator and thier length decreases towards the poles. Examples:
1. The Tropic of Cancer, at 2314°, is located north of the equator. It is the northernmost limit to which the direct rays of the sun are received.
2. The Tropic of Capricorn, at 23 V20, is located south of the equator. It is the southernmost limit to which the direct rays of the sun are received.
Question 8.
What is the grid system ? Why is it important ?
A network of horizontal and vertical lines or latitudes and longitudes drawn on a map or globe is called the grid system.
The grid system is an important feature of maps. It helps in locating places on the surface of the earth. For example, if you wanted to locate a place, you would look for its latitude and longtitude. The location of the place would be at the intersection of its latitude and longitude.
D. Look closely at the political map of India in your atlas and make a note of your observations and calculations in your notebook.
Question 1.
What is the scale of the map ?
A scale is the ratio between the distance on a map and the actual distance on the ground.
Question 2.
Calculate the N-S and E-W extents of India.
N-37°6\ S-8° 4′ and E 97°25\ W 68° T
Question 3.
Using the scale, calculate the distance between Delhi and Kolkata, Chennai and Thiruvananthapuram, and Mumbai and Panaji in kilometres.
1 cm = 61 km
Delhi and Kolkata 24 cm = 24 × 61 = 1464 km
Chennai and Thiruvananthapuram 10.3 cm = 628 km appropriate
Mumbai and Panaji 6.50 cm = 6.5 × 61 = 397 km
E. Complete the following chart
F. Practical work.
Question 1.
Draw a linear scale to show two men standing 30 m apart. Use the scale 1 cm = 3m.
Question 2.
A rectangular plot of length 200 m and width 125 m was bought by a businessman to set up a cosmetic factory. Draw this plot reduced to size using the scale 1cm = 25m. Show your work in class.
G Picture study
Study the diagram alongside and fill in the blanks with appropriate directions:
Example: A lies to the north-west of B.
1. T lies to the – of U.
2. P lies to the – of 0.
3. 0 lies to the – of P.
4. Y lies to the – of X.
5. U lies to the – of T.
1. T lies to the south of U.
2. P lies to the E of O.
3. O lies to the W of P
4. Y lies to the SW of X.
5. U lies to the N of T.
H. Complete the following table with the help of the list of conventional signs and symbols given in this chapter
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# randomized randomized algorithms 2 randomized algorithms 3 •so far we dealt with deterministic...
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• Randomized algorithms Inge Li Gørtz
�1
• Today
• What are randomized algorithms?
• Properties of randomized algorithms
• Three examples:
• Median/Select.
• Quick-sort
• Closest pair of points
Randomized algorithms
�2
Randomized Algorithms
�3
• So far we dealt with deterministic algorithms:
• Giving the same input to the algorithm repeatedly results in:
• The same running time.
• The same output.
• Randomized algorithm:
• Can make random choices (using a random number generator)
Randomized Algorithms
�4
• • A randomized algorithm has access to a random number generator rand(a, b), where a, b are integers, a < b.
• rand(a, b) returns a number x ∈ {a, (a + 1), . . . , (b − 1), b}.
• Each of the b − a + 1 numbers is equally likely.
• Calling rand(a, b) again results in a new, possibly different number.
• rand(a, b) “has no memory”. The current number does not depend on the results of previous calls.
• Statistically speaking: rand(a, b) generates numbers independently according to uniform distribution.
Random number generator
�5
• The algorithm can make random decisions using rand
Randomized algorithms
Running while (rand(0,1) = 0) do
print(“running”); end while print(“stopped”);
Julefrokost while (rand(1,4) = rand(1,4)) do have another Christmas beer;
end while go home;
�6
• A randomized algorithm might never stop.
• Example: Find the position of 7 in a three-element array containing the numbers 4, 3 and 7.
Properties of randomized algorithms
Find7(A[1…3]) goon := true; while (goon) do i := rand(1,3); if (A[i] = 7) then print(Bingo: 7 found at position i); goon := false; end if
end while
�7
• A randomized algorithm might find the wrong solution.
• Example: Find the position of the minimum element in a three-element array.
• If A = [5,6,4] and i = 2 and j = 1 then the algorithm will wrongly return 5.
Properties of randomized algorithms
MinOfThree(A[1…3]) i := rand(1,3); j := rand(1,3); return(min(A[i],A[j]);
�8
• • Both examples are not so bad as they look.
• It is highly unlikely that Find7 runs a long time.
• It is more likely that MinOfThree returns a correct result than a wrong one.
Properties of randomized algorithms
�9
Analysis of Randomized Algorithms
�10
• Running time: O(1).
• What is the chance that the answer of MinOfThree is correct?
• 9 possible pairs all with probability 1/9:
• (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)
• Assume wlog that A[1] is the minimum:
• 5 of the 9 pairs contain the index 1 and give the correct minimum.
• P[correct] = 5/9 > 1/2.
Analysis of MinOfThree MinOfThree(A[1…3])
i := rand(1,3); j := rand(1,3); return(min(A[i],A[j]);
�11
• Running MinOfThree once gives the correct answer with probaility 5/9 and the incorrect one with probability 4/9.
• Idea: Run MinOfThree many times and pick the smallest value returned.
• The probability that MinOfThree fails every time in k runs is:
• To be 99% sure to find the minimum, choose k such that
Analysis of MinOfThree MinOfThree(A[1…3])
i := rand(1,3); j := rand(1,3); return(min(A[i],A[j]);
�12
• • Running time is variable. What is the expected running time?
• Assume wlog that A[1] = 7.
• If rand(1, 3) = 1 then the while-loop is terminated.
• P[rand(1,3) = 1] = 1/3
• If rand(1, 3) ≠ 1 then the while-loop is not terminated.
• P[rand(1,3) ≠ 1] = 2/3
• The while-loop is terminated after one iteration with probability
P[rand(1,3) = 1] = 1/3.
Analysis of Find7 Find7(A[1…3])
goon := true; while (goon) do i := rand(1,3); if (A[i] = 7) then print(Bingo: 7 found at position i); goon := false; end if
end while
�13
• The while-loop stops after exactly two iterations if
• rand(1, 3) ≠ 1 in the first iteration.
• rand(1, 3) = 1 in the second iteration.
• This happens with probability 2/3 and 1/3, resp.
• The probability for both to happen is - by independence -
Analysis of Find7 Find7(A[1…3])
goon := true; while (goon) do i := rand(1,3); if (A[i] = 7) then print(Bingo: 7 found at position i); goon := false; end if
end while
�14
• The while-loop stops after exactly three iterations if
• rand(1, 3) ≠ 1 in the first two iterations.
• rand(1, 3) = 1 in the third iteration.
• This happens with probability 2/3, 2/3 and 1/3, resp.
• The probability for all three to happen is - by independence -
• The while-loop stops after exactly k iterations if
• rand(1, 3) ≠ 1 in the first k-1 iterations.
• rand(1, 3) = 1 in the kth iteration.
• This happens with probability 2/3, 2/3 and 1/3, resp.
• The probability for all three to happen is - by independence -
Analysis of Find7
�15
• We have
• The expected running time of Find7 is
• Idea: Stop Las Vegas algorithm if it runs “much longer” than the expected time and restart it.
Expected running time
�16
∑ k=0
k ⋅ xk = x
(1 − x)2 for |x | < 1.
• • We have seen two kinds of algorithms:
• Monte Carlo algorithms: stop after a fixed (polynomial) time and give
the correct answer with probability greater 50%.
• Las Vegas algorithms: have variable running time but always give the correct answer.
Types of randomized algorithms
�17
• Analyse the expected number of times running is printed:
• Analyse the expected number of beers you get if you follow the algorithm:
Randomized algorithms
Running while (rand(0,1) = 0) do
print(“running”); end while print(“stopped”);
Julefrokost while (rand(1,4) = rand(1,4)) do have another Christmas beer;
end while go home;
�18
Summations
∑ k=0
k ⋅ xk = x
(1 − x)2 for |x | < 1.
∑ k=0
xk = 1
(1 − x) for |x | < 1.
∑ k=0
k ⋅ xk−1 = 1
(1 − x)2 for |x | < 1.
Median/Select
�20
• • Given n numbers S = {a1, a2, …, an}.
• Median: number that is in the middle position if in sorted order.
• Select(S,k): Return the kth smallest number in S.
• Min(S) = Select(S,1), Max(S)= Select(S,n), Median = Select(S,n/2).
• Assume the numbers are distinct.
Select
Select(S, k) {
Choose a pivot s ∈ S uniformly at random.
For each element e in S if e < s put e in S’ if e > s put e in S’’
if |S’| = k-1 then return s
if |S’| ≥ k then call Select(S’, k)
if |S’| < k then call Select(S’’, k - |S’| - 1) }
�21
• Worst case running time:
• If there is at least an fraction of elements both larger and smaller than s:
• Limit number of bad pivots.
• Intuition: A fairly large fraction of elements are “well-centered” => random pivot
likely to be good.
Select Select(S, k) {
Choose a pivot s ∈ S uniformly at random.
For each element e in S if e < s put e in S’ if e > s put e in S’’
if |S’| = k-1 then return s
if |S’| ≥ k then call Select(S’, k)
if |S’| < k then call Select(S’’, k - |S’| - 1) }
T (n) = cn+ c(n� 1) + c(n� 2) + · · · = ⇥(n2). "
T (n) = cn+ (1� ")cn+ (1� ")2cn+ · · · =
� 1 + (1� ") + (1� ")2 + · · ·
� cn
cn/".
�22
• Phase j: Size of set at most and at least .
• Element is central if at least a quarter of the elements in the current call are smaller
and at least a quarter are larger.
• At least half the elements are central.
• Pivot central => size of set shrinks with by at least a factor 3/4 => current phase
ends.
• Pr[s is central] = 1/2.
• Expected number of iterations before a central pivot is found = 2 =>
expected number of iterations in phase j at most 2.
• X: random variable equal to number of steps taken by algorithm.
• Xj: expected number of steps in phase j.
• X = X1+ X2 + .…
• Number of steps in one iteration in phase j is at most .
• E[Xj] = .
• Expected running time:
Select n(3/4)j n(3/4)j+1
E[X] = X
j
E[Xj ] X
j
2cn
✓ 3
4
◆j = 2cn
X
j
✓ 3
4
◆j 8cn.
2cn(3/4)j cn(3/4)j
�23
Quicksort
�24
• • Given n numbers S = {a1, a2, …, an} return the sorted list.
• Assume the numbers are dis
Recommended
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# slides-6 by DaronMackey
VIEWS: 26 PAGES: 16
• pg 1
``` Variable Definition Notes & comments
Extended base dimension
system
Pi-theorem (also definition of
physical quantities,…)
Physical similarity Physical similarity means that
all Pi-parameters are equal
Galileo-number (solid
mechanics)
Reynolds number (fluid
mechanics)
Lectures 1-3 and PS2
Important concepts include the extended base dimension system, distinction
between units and dimensions, the formal Pi-theorem based procedure and the
concept of physical similarity.
Applications include calculation of physical processes like atomic explosion, drag
force on buildings etc.
1
r r r r r
x x = x1e1 + x2 e2 + x3e3 Position vector
r r r
v v = dx / dt Velocity vector
r r r
a a = dv / dt Acceleration vector
r r r r r r
p p = mv = m(v1e1 + v2 e2 + v3e3 ) Linear momentum
r r r Unit vectors that
e1 , e2 , e3 define coordinate
system = basis
Normal vector
r Always points outwards of
n
domain considered
r r r r r r
xi × pi xi × pi = xi × mi vi Angular momentum
r r r r r Force vector (force that acts
F F = Fx ex + Fy e y + Fz ez
on a material point)
Covered in lecture 4 and PS1
Basic definitions of linear momentum, angular momentum, normal vector of domain
boundaries
2
1. Every body continues in its state of
rest, or of uniform motion in a right
line, unless it is compelled to change
that state by forces impressed upon
it.
2. The change of motion is proportional
to the motive force impresses, and is
made in the direction of the right line Newton’s three laws
in which that force is impressed.
3. To every action there is always
opposed an equal reaction: or, the
mutual action of two bodies upon
each other are always equal, and
directed to contrary parts.
Dynamic resultant theorem
r def r
v Change of linear momentum
d ( p) / dt = d (mv) / dt = F
is equal to sum of external
forces
Dynamic moment theorem
Change of the angular motion
of a discrete system of i = 1,N
(
r def N r r
)
N r
d N r particles is equal to the sum of
∑ (xi × mi vi ) = ∑ xi × Fi ext = ∑ M iext
dt i=1 i=1 i=1 the moments (or torque)
generated by external forces
Static EQ (solve truss problems)
Lecture 4: These laws and concepts form the basis of almost everything we’ll do in
1.050.
The dynamic resultant theorem and dynamic moment theorem are important
concepts that simplify for the static equilibrium. This can be used to solve truss
problems, for instance.
3
Atomic bonds
O(Angstrom=1
E-10m)
Grains,
crystals,…
REV=
Representative volume element
‘d’=differential element
REV
dΩ Must be:
Continuum
representative (1) Greater than any in
volume element homogeneity (grains,
REV
molecules, atoms,..)
(2) Much smaller than size of the
system
Surface of domain Ω Note the difference between
∂Ω
‘d’ and ' ∂' operator
Skyscraper photograph courtesy of jochemberends on Flickr.
Lecture 5
The definition of REV is an essential concept of continuum mechanics: Separation
of scales, i.e., the three relevant scales are separated sufficiently. There are three
relevant scales in the continuum model. Note: The beam model adds another
scale to the continuum problem – therefore the beam is a four scale continuum
model.
4
Force density that acts on a
r
material plane with normal n
r
at point x
⎛T ⎞
r r r ⎜ x⎟ Stress vector
T (n, x) = ⎜ Ty ⎟
⎜ ⎟ (note: normal always points out
⎝ Tz ⎠
of domain)
Stress matrix
r r r r r r r
σ = σ ij ei ⊗ e j T (n , x ) = σ ( x ) ⋅ n Stress tensor
p Pressure (normal force per area
that compresses a medium)
Lecture 5, 6, 7
These concepts are very important. We started with the definition of the stress
vector that describes the force density on a particular surface cut.
The stress tensor (introduced by assembling the stress matrix) provides the stress
vector for an arbitrary plane (characterized by the normal vector). This requirement
represents the definition of the stress tensor; by associating each entry with two
vectors (this is a characteristic of a second order tensor).
The pressure is a scalar quantity; for a liquid the pressure and stress tensor are
linked by a simple equation (see next slide).
5
Divergence theorem (turn
surface integral into a volume
integral)
Differential equilibrium (solved
by integration)
on S :
d
on ∂Ω : T = T (n)
r r
div σ + ρ (g − a ) = 0 Differential E.Q. written out
for cartesian C.S.
In cartesian C.S.
EQ for liquid (no shear
stress=material law)
Lecture 5, 6, 7
We expressed the dynamic resultant theorem for an arbitrary domain and
transformed the resulting expression into a pure volume integral by applying the
divergence theorem. This led to the differential EQ expression; each REV must
satisfy this expression. The integration of this partial differential equation provides
us with the solution of the stress tensor as a function of all spatial coordinates.
6
Divergence of stress tensor in cylindrical C.S.
Divergence of stress tensor in spherical C.S.
PS 4 (cylindrical C.S.)
This slide quickly summarizes the differential EQ expressions for different
coordinate systems.
7
z
z
y h,b << l Beam geometry
Section
Section quantities - forces
=
Nx
=
Section quantities - moments
σ Stress tensor beam geometry
Lecture 8
Introduction of the beam geometry. The beam is a ‘special case’ of the continuum
theory. It introduces another scale: the beam section size (b,l) which are much
smaller than the overall beam dimensions, but much larger than the size of the
REV.
8
Beam EQ equations
+BCs
+BCs
2D planar beam EQ
z equations
x
Lectures 8, 9
The beam EQ conditions enable us to solve for the distribution of moments and
normal/shear forces.
The equations are simplified for a 2D beam geometry.
9
EQ for truss structures
(S.A.)
Strength criterion for
truss structures (S.C.)
Fmax
σ0 σ0 = Tensile strength limit
A0
P P
Concept: Visualization of the
‘strength’
x
x Number of atomic bonds per
x: marks bonds that break at max force area constant due to fixed
Fbond
A0
lattice parameter of crystal
Fmax = Fbond N A0 cell
# bonds per area A0 Therefore finite force per
Strength per bond area that can be sustained
Lecture 10, PS 5 (strength calculation)
10
Mohr plane (τ and σ)
r r r r r
T (x , n ) = σn + τt Mohr circle
(Significance: Display 3D
stress tensor in 2D)
σ ,τ
Basis in Mohr plane
σ I , σ II , σ III Principal stresses
r r r
uI ,uII , uIII Principal stress directions
Principal stresses and
directions obtained through
eigenvector analysis
Principal stresses
=Eigenvalues
Principal stress directions
=Eigenvectors
Lecture 11
11
At any point, σ must be:
Two pillars of
strength and
approach
(2) Strength compatible (S.C.)
• Equilibrium conditions “only” specify statically admissible
stress field, without worrying about if the stresses can
actually be sustained by the material – S.A.
From EQ condition for a REV we can integrate up
(upscale) to the structural scale
Examples: Many integrations in homework and in class;
Hoover dam etc.
to S.A., the stress field must be compatible with the
strength capacity of the material – S.C.
In other words, at no point in the domain can the stress
vector exceed the strength capacity of the material
Examples: Sand pile, foundation etc. – Mohr circle
Lecture 10, 11, 12 (application to beams in lectures 13-15)
12
Strength domain (general
Dk definition)
Equivalent to condition for S.C.
Max. shear stress
c
Dk ,Tresca Tresca criterion
v v
∀n : f (T ) = σ − c ≤ 0
Max. tensile stress
Dk ,Tension−cutoff Tension cutoff criterion
c
Lecture 11
13
N
Friction force
F frict = µN = tan ϕ ⋅ N F frict
F frict
Shear resistance increases
with increasing normal force
µ
τ =µ
Max. shear stress
Dk ,Mohr−Coulomb function of σ Mohr-Coulomb
σ
c cohesion
c=0 dry sand
τ
Angle of repose
σ
Lecture 12 (Mohr-Coulomb criterion)
The definition of friction is included here for completeness
14
DS Strength domain for beams
Moment capacity for beams
For rectangular cross-section
b,h
N x lim = N 0 N x lim = N 0 = bhσ 0 Strength capacity for beams
My Nx
f (M y , N x ) = + −1 ≤ 0
M0 Nx
2
My ⎛N ⎞
f (M y , N x ) = + ⎜ x ⎟ −1 ≤ 0
⎜
M0 ⎝ Nx ⎟
⎠ M-N interaction (linear)
f (M y , N x ) ≤ 0
1
M-N interaction (actual);
convexity
1
Lecture 13 and 14
15
Safe
strength
Linear combination is safe
domain
(convexity)
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# Maths Challenge
## 39 thoughts on “Year 6 Maths Challenge – Summer 1”
1. Arshdeep D. says:
Square-9
Triangle-5
2. Sahrah K. says:
1) 8x8x6=384squared
2) 9x3x2=54
3)4×3=12divided by 2 =6
4)3mx3mx3m=27m cubed
5) 4cmx7cm plus ecm 84cm squared
6)16x12x30 divided by 2=2880 cubed
Shapes 2
I can see 6 shapes.
I can see 12 shapes.
4. Lyba G. says:
Challenge for everyone
There are 5 triangles
I all so think that there are 9 squares
Gold
8x8x8=512cm 2
9x3x2=54cm 2
4x7x3=84cm 2
3x3x3=27cm 2
By Lyba
5. Sami M. says:
Gold challenge
1.8x8x6=384 squared
2.9x3x2=54
3.4Γ3=12 divided by 2=6
4.3mx3mx3m=27m cubed
5.4cmx7cm+ecm=84cm cubed
6.16x12x30 divided by 2=2880m cubedL
6. Alisha R. says:
There are 5 triangles.
There are 9 squares.
Gold
I think the formula to calculating surface area is to times the length and width then times that by the amount of faces there are.
So I think these are the answers:
8×8=64×6=384cm squared
9×3=12×6=72cm squared
4×11=44×5=220cm squared
7. Ruqayyah K. says:
Bronze challenge:
Triangles=5
Squares=9
Gold challenge;
8x8x8=512cm2
9x3x2=54cm2
3x3x3=27cm2
4x7x3=84cm2
16x12x30x20=792cm2
You needed a bit of work to solve it
8. Maleeha M. says:
ON PAPER
9. Yousaf K. says:
6 triangles
10 squares
Silver
36 cans of coke
10. Demi-Lea D. says:
C)4times3=12 divided by 2=6
11times4=44times3=132msquared
D)3mtimes3mtimes3mtimes=27m cubed
E)4cmtimes7cm+ecm=84cmcubed
F)16times12times30divided by 2 =2880m cubed
1.7 triangles
2. 9 squares
Gold challenge
1. 64 centimetres squared
2. 27 centimetres squared
3. 50 centimetre squared
4. 27 centimetre cubed
5. 84 centimetres cubed
6.
• Fatima S. says:
Challenge for everyone:πππ
A)3 B)8
Gold not volume:
1)8
2)3
3)3
From Fatima,6W
• Fatima S. says:
Challenge for everyone:πππ
A)3 B)8
C)Canβt do it on blogπ©βπ«ππ»I will do it on a paper
Gold not volume:
1)8
2)3
3)3
From Fatima,6W
I will do the
12. Demi-Lea D. says:
Starter:
1)9
2)11
3) iβm Going to do it on a Peice of paper
Iβm going to do gold
Hereβs the start
A)8 times 8 times 6=384 squared
B)9 times 3 times 2=54
9 times two times two=36
3 times 2 times 2=12
54+36+12=102 squared
Iβm going to do (c)in another post
13. Nidhi P. says:
Gold challenge:
surface area
1) 8 times 8=64
64 times by 6=384 cm squared
2) 3 times 9=27
27 times by 2=54
54 times by 6=324cm squared
3) 4 times 11=44
44 times 4= 176
176 times by 5=880cm squared
Volume:
1)3 cubed= 27 cm cubed
2)4 times 7 times 3=84 cm cubed
3)
14. Hassan L. says:
5 triangles and 9 squares.
Gold
A= 8×8=64
B=9×3=27
C=44
D=27
E=84
F=960
15. Reece S. says:
Normal challenge
There is 5 triangles in the picture βΌοΈ
And 8 squares
Silver challenge
330 cans of cola 33*10=330 33
330 *10=330
Gold challenge (A)
8*8=64. 64
6
64*6=384 384
(B)
ββββββββββββββββββββββββββββββββββββ
(C)
There are 5 rectangles and 9 squares.
Triangles I meant.
17. Safa M. says:
5 triangles and 9 squares.
Gold: 64cm squared
54cm squared
27cm squared
84cm squared
(Worked out on whiteboard)
18. Takwa S. says:
There are 6 triangle and 9 squares.
• Takwa S. says:
There are 5 squares
19. Nidhi P. says:
Bronze challenge
There are 5 triangles.
There are 11 squares.
Bronze:
5 triangles
9 squares
21. Moman L. says:
Starter
7 triangles
11 squares
Area
A=64
B=27
C=44
D=27
E=84
F=2880
6 triangles
8 squares
Gold
Square 64cm2
Rectangle 27cm2
C=44cm2
D=27cm2
E=84cm2
F=292cm2
23. Rayan O. says:
Bronze challenge:
πΊ=5
βΌοΈ=9
Gold challenge;
8x8x8=512cm2
9x3x2=54cm2
3x3x3=27cm2
4x7x3=84cm2
16x12x30x20=792cm2
24. Damien D. says:
Bronze challenge:
Triangles=5
Squares=9
Gold challenge;
8x8x8=512cm2
9x3x2=54cm2
3x3x3=27cm2
4x7x3=84cm2
16x12x30x20=792cm2
25. Sharazat H. says:
Gold challenge:
8x8x8=512cm2
9x3x2=54cm2
3x3x3=27cm2
4x7x3=84cm2
26. Umayr K. says:
Bronze
There are 5 Triangle.
There are 9 square.
Gold
512cm2
54cm2
27cm2
84cm2
792cm2
27. Abdullah S. says:
You will need 243 cans .
28. Razvan L. says:
Bronze
5 squares
9 squares
Gold
8 x 8 x 8 = 512cm2
9 x 3 x 2 = 54cm2
3 x 3 x 3 = 27cm2
4 x 7 x 3 = 84cm2
16 x 12 x 20 x 30 = 115200
29. Tameem A. says:
Bronze
There are 5 triangles
There are 9 squares
30. Priya M. says:
Bronze
There are 5 triangles.
There are 9 squares .
Gold
8x8x8=512cm2
9x3x2=54cm2
3x3x3=27cm2
7x4x3=84cm2
31. Rudaba W. says:
Bronze challenge:
Triangles=5
Squares=9
Gold challenge;
8x8x8=512cm2
9x3x2=54cm2
3x3x3=27cm2
4x7x3=84cm2
16x12x30x20=792cm2
Doing the challenges on paper and I will bring it in for Mrs Matharni.
33. Ibrahim R. says:
I have done bronze challenge on paper.
34. Ibrahim R. says:
Challenge for everyone
___________________
1) There are 7 triangles.
2) The are 11 squares.
Silver challenge
______________
1) You will need 243 cans of cola to fill the bath. How I worked this out is I saw how much a bath can hold which 80 litres which is 8000 centilitres and I divided it by 33 but it was 242.424 recurring so I added an extra can so I got 243.
Gold challenge
_____________
A)8x8x6=384 squared
B)9x3x2=54
9x2x2=36
3x2x2=12
54+36+12=102 squared
C)4×3=12 divided by 2=6
11×4=44×3=132+6=138m squared
D)3mx3mx3m=27m cubed
E)4cmx7cm+ecm=84cm cubed
F)16x12x30 divided by 2=2880m cubed
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### Lecture 17 - Block Diagrams
```Negative Feedback
R(s)
+-
E(s)
H(s)
Y(s)
B(s)
G(s)
Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Block Diagrams
Outline of Today’s Lecture
Review
A new way of representing systems
Coordinate transformation effects
hint: there are none!
Development of the Transfer Function from an ODE
Gain, Poles and Zeros
The Block Diagram
Components
Block Algebra
Loop Analysis
Block Reductions
Caveats
Alternative Method of Analysis
Up to this point in the course, we have been concerned about the
structure of the system and discribed that structure with a state
space formulation
Now we are going to analyze the system by an alternative method
that focuses on the inputs, the outputs and the linkages between
system components.
The starting point are the system differential equations or
difference equations.
However this method will characterize the process of a system
block by its gain, G(s), and the ratio of the block output to its
input.
Formally, the transfer function is defined as the ratio of the
Laplace transforms of the Input to the Output:
TF ( S )
Output ( s )
for s a complex number and the Laplace Variable
Input ( s )
Coordination Transformations
x1
d
d
x Ax Bu z Az Bu
dt
dt
y Cz Du
y Cx Du
z2
x2
for A TAT 1, B TB and C CT 1
Since the ouput y is unchanged by the transformation
z1
y C ( sI A) 1 B D
and
y C ( sI A) 1 B D
y G ( s )u(t ) G ( s )u(t ) G ( s ) G s
Thus the Transfer function is invariant under coordinate
transformation
Linear System Transfer Functions
General form of linear time invariant (LTI) system is expressed:
(n)
n1
n 2
m
m1
m2
y a1 y a2 y ... an2 y an1 y an y b0 u b1 u b2 u ... bn2u bn1u bnu
For an input of u(t)=est such that the output is y(t)=y(0)est
s
n
a1s n 1 a2 s n 2 ... an 2 s 2 an 1 2 an y0e st b0 s m b1s m 1 b2 s m 2 ... bn 2 s 2 bn 1s bn e st
b s bs
s a s
m
y (t ) y (0)e
st
0
m 1
1
n 1
n
1
b2 s m 2 ... bn 2 s 2 bn 1s bn
a2 s n 2 ... an 2 s 2 an 1 2 an
e st
The transfer function form is then
b0 s m b1s m1 b2 s m2 ... bn 2 s 2 bn 1s bn b( s )
y ( t ) G ( s )u ( t ) G ( s ) n
s a1s n 1 a2 s n 2 ... an 2 s 2 an 1 2 an
a(s)
Note that the transfer function for a simple ODE can be written as the ratio
of the coefficients between the left and right sides multiplied by powers of s
The order of the system is the highest exponent of s in the denominator.
Simple Transfer Functions
Differential
Equation
Transfer
Function
Name
yu
s
Differentiator
yu
1
s
1
s2
1
sa
Integrator
yu
y ay u
y 2n y n 2 y u
y k pu kd u ki udt
2nd order Integrator
1st order system
1
s 2n s n
2
k p kd s
ki
s
2
Damped Oscillator
PID Controller
Gain, Poles and Zeros
G ( s ) C ( sI A) 1 B D
G (0) D CA1B
b( s )
b( s )
K
a( s)
a( s )
yo bm
K
uo am
The roots of the polynomial in the denominator, a(s), are called
the “poles” of the system
The poles are associated with the modes of the system and these are
the eigenvalues of the dynamics matrix in a state space representation
The roots of the polynomial in the numerator, b(s) are called the
“zeros” of the system
The zeros counteract the effect of a pole at a location
The value of G(s) is the zero frequency or steady state gain of the
system
Actuate
Sense
Block Diagrams
Compute
Throughout this course, we have used block diagrams to show
different properties
Here, we will formalize the meaning of block diagrams
Controlle
r
Disturbance
Controller
Plant/Process
Input
r
S
kr
S
u
Output
y
d
x Ax Bu
dt
y Cx Du
Prefilter
Sensor
x
-K
Plant
State Feedback
State Controller
D
u
S
-1
S
S
c1
c2
z1
…
…
z2
a1
a2
S
S
S
S
cn-1
cn
zn-1
an-1
…
S
zn
an
y
Components
x
x
The paths represent variable values which
are passed within the system
xG(s)
G(s)
x
x+y
Blocks represent System components which
are represented by transfer functions and multiply
their input signal to produce an output
Addition and subtraction of signals are represented
by a summer block with the operation indicated
on the arrow
++
y
x
x
x
Branch points occur when a value is placed on two
lines: no modification is made to the signal
Block Algebra
x
x-y
+-
y
-
y
x
x-y
+-
x
G
+
z
x
-
z-x+y
x
z-x
-+
+
z
y
x-y
+
+
xGH
H
x
y
GH
z-x+y
++
x
y
z
xG
z-x+y
x
H
xGH
xH
G
xGH
Block Algebra
x
x
Gx
G
Gx
x
Gx
G
H
+-
(G-H)x
G
+-
z
z
x
+-
G
G(x-z)
x
x
Gx-z
1
G
z
G
+-
z
G
Gx
x
G
z
z
(G-H)x
G-H
Hx
Gx
Gx
G
x
x
Gx
G
Gz
G
G
G(x-z)
+-
Gx-z
Block Algebra
x
x
Gx
G
Gx
G
x
1
G
x
y
x-y
x
x-y
+-
x
y
y
+-
x-y
+-
y
x
x-y
+
G
H
x
1
H
y
+-
H
G
Closed Loop Systems
r
y
++
H
A positive feedback system
r
y
+-
H
A negative feedback system
Loop Analysis
(Very important slide!)
Negative Feedback
R(s)
E(s)
+-
H(s)
Y(s)
B(s)
G(s)
E ( s) R( s) B( s)
B( s ) G ( s )Y ( s )
Y ( s) H ( s) E ( s) H ( s) R( s) H ( s) B( s)
Y s H ( s ) R ( s ) H ( s )G ( s )Y ( s )
Y s H ( s )G ( s )Y ( s ) H ( s ) R ( s )
H ( s) R( s)
Y (s)
1 H ( s )(G ( s )
Y (s)
H (s)
T .F .
R( s ) 1 H ( s )G ( s )
Loop Analysis
Negative Feedback
Positive Feedback
R(s)
++
E(s)
H(s)
R(s)
Y(s)
+-
E(s)
H(s)
Y(s)
B(s)
B(s)
G(s)
E ( s) R( s) B( s )
B( s) Y ( s)
E ( s) R( s) B( s )
B ( s ) G ( s )Y ( s )
Y ( s) H ( s) E ( s) H ( s) R( s) H ( s) B( s)
Y ( s) H ( s) E ( s) H ( s) R( s) H ( s) B( s)
Y H ( s ) R ( s ) H ( s )Y ( s )
H ( s) R( s)
Y ( s)
1 H ( s)
Y ( s)
H ( s)
T .F .
R( s) 1 H ( s)
Y H ( s ) R ( s ) H ( s )G ( s )Y ( s )
H ( s) R( s)
Y ( s)
1 H ( s )(G ( s )
Y ( s)
H ( s)
T .F .
R ( s ) 1 H ( s )G ( s )
Block Reduction
Example
x
y
++
D
C
B
A
+-
+-
+-
E
F
First, uncross signals where possible
G
E
x
+-
+-
A
+-
G
F
B
C
D
++
y
Block Reduction
Example
x
D
C
B
A
+-
+-
+-
E
++-
y
G
F
E
BC
Next: Reduce Feed Forward Loops where possible
x
+-
+-
A
+-
G
F
B
C
D
++-
y
Block Reduction
Example
x
+-
+-
A
+-
E
BC
B
C
D
++-
y
G
F
Next: Reduce Feedback Loops starting with the inner most
x
+-
A
1 AG
+F
B
C
BCD E
BC
y
Block Reduction
Example
x
A
1 AG
+-
+-
B
C
C
BCD E
BC
BCD E
BC
y
F
AB
AB
1 AG
FAB
1 AG ABF
1
1 AG
x
+-
AB
1 AG ABF
ABC
ABC
1 AG ABF
ABC
1 AG ABF ABC
1
1 AG ABF
x
ABC
1 AG ABF ABC
BCD E
BC
y
y
Block Reduction
Example
x
ABC
1 AG ABF ABC
x
BCD E
BC
AB 2C 2 D ABCE
BC ABCG AB 2CF AB 2C 2
y
y
Loop Nomenclature
Reference
Input
R(s)
Prefilter
F(s)
+-
Error
signal
E(s)
Disturbance/Noise
Controller
C(s)
Open Loop
Signal
B(s)
+-
Plant
G(s)
Output
y(s)
Sensor
H(s)
The plant is that which is to be controlled with transfer function G(s)
The prefilter and the controller define the control laws of the system.
The open loop signal is the signal that results from the actions of the
prefilter, the controller, the plant and the sensor and has the transfer function
F(s)C(s)G(s)H(s)
The closed loop signal is the output of the system and has the transfer function
F ( s )C ( s )G ( s )
1 C ( s )G ( s ) H ( s )
Caveats: Pole Zero Cancellations
Assume there were two systems that were connected as such
R(s)
1
C ( s) 3
s 8s 2 17 s 10
s 1
G( s) 3
s 12 s 2 47 s 60
An astute student might note that C( s)
Y(s)
1
1
s 8s 17s 10 s 1 s 2 s 5
3
2
and then want to cancel the (s+1) term
This would be problematic: if the (s+1) represents a true
system dynamic, the dynamic would be lost as a result of the
cancellation. It would also cause problems for controllability and
observability. In actual practice, cancelling a pole with a zero
usually leads to problems as small deviations in pole or zero
location lead to unpredictable dynamics under the cancellation.
Caveats: Algebraic Loops
The system of block diagrams is based on the presence of
differential equation and difference equation
A system built such the output is directly connected to the
input of a loop without intervening differential or time
difference terms leads to improper block interpretations and
an inability to simulate the model.
+
-
2
When this occurs, it is called an Algebraic Loop. Such loops
are often meaningless and errors in logic.
x
Summary
x
The Block Diagram
xG(s)
G(s)
Components
Block Algebra
x
Loop Analysis
Block Reductions
x+y
++
y
Caveats
x
x
Negative Feedback
R(s)
+-
E(s)
H(s)
x
Y(s)
B(s)
G(s)
Next: Bode Plots
```
| 3,909 | 8,578 |
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# Search by Topic
#### Resources tagged with Investigations similar to Doplication:
Filter by: Content type:
Stage:
Challenge level:
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### 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?
##### 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?
### 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!
### 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.
### Month Mania
##### Stage: 1 and 2 Challenge Level:
Can you design a new shape for the twenty-eight squares and arrange the numbers in a logical way? What patterns do you notice?
### 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?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Sometimes We Lose Things
##### Stage: 2 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### 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?
### 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.
### 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?
### 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.
### Train Carriages
##### Stage: 1 and 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
### 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?
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Sending Cards
##### Stage: 2 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
### Division Rules
##### Stage: 2 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
##### Stage: 2 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Times
##### Stage: 2 Challenge Level:
Which times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!
### Building with Rods
##### Stage: 2 Challenge Level:
In how many ways can you stack these rods, following the rules?
### Sorting the Numbers
##### Stage: 1 and 2 Challenge Level:
Complete these two jigsaws then put one on top of the other. What happens when you add the 'touching' numbers? What happens when you change the position of the jigsaws?
### Stairs
##### Stage: 1 and 2 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
##### 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?
### Street Sequences
##### Stage: 1 and 2 Challenge Level:
Investigate what happens when you add house numbers along a street in different ways.
### Exploring Wild & Wonderful Number Patterns
##### Stage: 2 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### Mobile Numbers
##### Stage: 1 and 2 Challenge Level:
In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time?
### 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?
### New House
##### Stage: 2 Challenge Level:
In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
### Become Maths Detectives
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? Don't forget to keep visiting NRICH projects site for the latest developments and questions.
### 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?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Sweets in a Box
##### Stage: 2 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
### Round and Round the Circle
##### Stage: 2 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
### Room Doubling
##### Stage: 2 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### Magazines
##### Stage: 2 Challenge Level:
Let's suppose that you are going to have a magazine which has 16 pages of A5 size. Can you find some different ways to make these pages? Investigate the pattern for each if you number the pages.
### Worms
##### Stage: 2 Challenge Level:
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
### Calendar Patterns
##### Stage: 2 Challenge Level:
In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers?
### Five Coins
##### Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### The Great Tiling Count
##### Stage: 2 Challenge Level:
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
### Plants
##### Stage: 1 and 2 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### Number Squares
##### Stage: 1 and 2 Challenge Level:
Start with four numbers at the corners of a square and put the total of two corners in the middle of that side. Keep going... Can you estimate what the size of the last four numbers will be?
### Balance of Halves
##### Stage: 2 Challenge Level:
Investigate this balance which is marked in halves. If you had a weight on the left-hand 7, where could you hang two weights on the right to make it balance?
### 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.
### 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?
| 2,232 | 10,028 |
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| 4.03125 | 4 |
CC-MAIN-2017-39
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latest
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| 0.928856 |
https://schoolbag.info/sat/sat_3/16.html
| 1,566,251,423,000,000,000 |
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crawl-data/CC-MAIN-2019-35/segments/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00399.warc.gz
| 628,109,706 | 9,124 |
Math Test - Calculator - DIAGNOSTIC SAT - SAT 2016
## DIAGNOSTIC SAT
### Math Test – Calculator
55 MINUTES, 38 QUESTIONS
DIRECTIONS
For questions 1–30, solve each problem, choose the best answer from the choices provided, and fill in the corresponding circle on your answer sheet. For questions 31–38, solve the problem and enter your answer in the grid on the answer sheet. Please refer to the directions before question 31 on how to enter your answers in the grid. You may use any available space in your test booklet for scratch work.
NOTES
1. The use of a calculator is permitted.
2. All variables and expressions used represent real numbers unless otherwise indicated.
3. Figures provided in this test are drawn to scale unless otherwise indicated.
4. All figures lie in a plane unless otherwise indicated.
5. Unless otherwise indicated, the domain of a given function f is the set of all real numbers for which f(x) is a real number.
REFERENCE
The number of degrees of arc in a circle is 360.
The number of radians of arc in a circle is 2π.
The sum of the measures in degrees of the angles of a triangle is 180.
1
a – b = 10
a – 2b = 8
Based on the system of equations above, what is the value of b?
A) −2
B) −1
C) 1
D) 2
2
The average (arithmetic mean) of three numbers is 50. If two of the numbers have a sum of 85, what is the third number?
A) 75
B) 70
C) 65
D) 55
3
What number is the same percent of 225 as 9 is of 25?
A) 27
B) 54
C) 64
D) 81
4
RESULTS OF FAVORABILITY POLL
The table above shows the partial results of a favorability poll for a local politician. If the data shown are correct, how many of the women who were polled viewed the politician unfavorably?
A) 33
B) 43
C) 61
D) It cannot be determined by the information given.
5
If 22n-2 = 32, what is the value of n?
A) 2.0
B) 2.5
C) 3.0
D) 3.5
6
A bag of Nellie’s Nut Mix contains x ounces of walnuts, 15 ounces of peanuts, and 20 ounces of pecans. Which of the following expresses the fraction of the mix, by weight, that is walnuts?
A)
B)
C)
D)
7
In the triangle above, what is the value of k? (sin 35° = 0.574, cos 35° = 0.819, tan 35° = 0.700)
A) 6.00
B) 6.88
C) 8.40
D) 9.83
8
The figure above shows a polygon with five sides. What is the average (arithmetic mean) of the measures, in degrees, of the five angles shown?
A) 108°
B) 110°
C) 112°
D) 114°
Questions 9 and 10 are based on the graph below.
ANNUAL REVENUE PER STORE
9
The scatterplot above shows the annual revenue for each of the individual retail stores operated by a clothing company for each year from 2004 through 2012. Based on the line of best fit to the data shown, which of the following is closest to the average annual increase in revenue per store?
A) \$5,000
B) \$50,000
C) \$100,000
D) \$500,000
10
Which of the following statements is most directly justified by the data shown in the scatterplot above?
A) The average revenue per store increased by over 100% from 2005 to 2009.
B) The total number of retail stores increased by 50% from 2005 to 2012.
C) The total revenue from all stores in 2012 was more than three times the total revenue from all stores in 2004.
D) The total revenue from all stores in 2008 was over \$1 million.
11
Which of the following statements expresses the fact that the product of two numbers, a and b, is 6 greater than their sum?
A) ab + 6 > a + b
B) ab = a + b + 6
C) ab + 6 = a + b
D) ab > a + b + 6
12
Note: Figure not drawn to scale.
In the figure above, if m || l, what is the area, in square units, of the shaded rectangle?
A) 156
B) 168
C) 180
D) 192
13
The Glenville Giants have played a total of 120 games and have a win-to-loss ratio of 2 to 3. How many more games have they lost than won?
A) 24
B) 30
C) 40
D) 48
14
A culture of bacteria initially contained p cells, where p > 100. After one hour, this population decreased by . In the second and third hours, however, the population increased by 40% and 50%, respectively. At the end of those first three hours, what was the population of the culture?
A) 1.3p
B) 1.4p
C) 1.5p
D) 1.6p
15
If , what is the value of m2?
A)
B)
C)
D)
16
A jar contains only red, white, and blue marbles. It contains twice as many red marbles as white marbles and three times as many white marbles as blue marbles. If a marble is chosen at random, what is the probability that it is not red?
A)
B)
C)
D)
17
y = -3(x − 2)2 + 2
In the xy-plane, line l passes through the point (−1, 3) and the vertex of the parabola with equation above. What is the slope of line l?
A)
B)
C)
D)
18
A certain function takes an input value and transforms it into an output value according to the following three-step procedure:
Step 1: Multiply the input value by 6.
Step 2: Add x to this result.
Step 3: Divide this result by 4.
If an input of 7 to this function yields an output of 15, what is the value of x?
A) 12
B) 16
C) 18
D) 24
19
The variables x and y are believed to correlate according to the equation y = ax2 + bx + c, where ab, and c are constants. Which of the following scatterplots would provide the strongest evidence in support of the hypothesis that a < 0?
A)
B)
C)
D)
20
On a number line, the coordinates of points P and R are p and r, respectively, and p < r. If the point with coordinate x is closer to p than to r, then which of the following statements must be true?
A)
B)
C) |x − p|< r
D) |x + p|< r − p
21
Let function f(x) be defined by the equation . If m is a positive integer, then
A)
B)
C)
D) 2 − m
22
The value of y varies with x according to the equation y = a(x − 2)(x + 1), where a < 0. As the value of x increases from 0 to 5, which of the following best describes the behavior of y?
A) It increases and then decreases.
B) It decreases and then increases.
C) It increases only.
D) It decreases only.
23
If the expression is equivalent to the expression for all values of n, what is the value of k?
A) −12
B) −6
C) 6
D) 12
24
An online trading company charges a 3% commission for all stock purchases. If a trader purchases 200 shares of a stock through this company and is charged \$3,399 including commission, what is the cost per share for this stock?
A) \$16.45
B) \$16.48
C) \$16.50
D) \$16.52
25
For nonzero numbers w and y, if w is 50% greater than y, then what is the ratio of w−2 to y−2?
A) 4 to 9
B) 2 to 3
C) 9 to 4
D) 4 to 1
26
Every athlete in a group of 60 female varsity athletes at Greenwich High School either runs track, plays soccer, or does both. If one-third of the athletes in this group who play on the soccer team also run on the track team, and one-half of the athletes in this group who run on the track team also play on the soccer team, which of the following statements must be true?
A) This group contains 40 soccer players.
B) This group contains 20 athletes who play soccer but do not run track.
C) This group contains 20 athletes who play both track and soccer.
D) The number of soccer players in this group is 15 greater than the number of track team members in this group.
27
A portion of the graph of the quadratic function y = f(x) is shown in the xy-plane above. The function g is defined by the equation g(x) = f(x) + b. If the equation g(x) = 0 has exactly one solution, what is the value of b?
A) −2
B) −1
C) 1
D) 2
28
If cos x = a, where , and cos y = − a, then which of the following could be the value of y?
A) x + 2π
B) x + π
C)
D) − x + 2π
Questions 29 and 30 refer to the following table.
OPINION POLL ON PROPOSAL 81A
29
Of those surveyed who expressed an opinion on Proposal 81a, approximately what percentage are under 40 years of age?
A) 30%
B) 38%
C) 68%
D) 72%
30
If the data in the table above are assumed to be representative of the general voting population, which of the following statements is most directly justified by these data?
A) The approval rate for Proposal 81a generally decreases with the age of the voter.
B) The disapproval rate for Proposal 81a generally increases with the age of the voter.
C) Those who express an opinion on Proposal 81a are more likely to be over 64 than they are to be under 40.
D) In all three age categories, voters are more than twice as likely to approve of Proposal 18a than to have no opinion about it.
Student-Produced Response Questions
DIRECTIONS
For questions 31–38, solve the problem and enter your answer in the grid, as described below, on the answer sheet.
1. Although not required, it is suggested that you write your answer in the boxes at the top of the columns to help you fill in the circles accurately. You will receive credit only if the circles are filled in correctly.
2. Mark no more than one circle in any column.
3. No question has a negative answer.
4. Some problems may have more than one correct answer. In such cases, grid only one answer.
5. Mixed numbers such as must be gridded as 3.5 or .
(If is entered into the grid as , it will be interpreted as , not .)
6. Decimal answers: If you obtain a decimal answer with more digits than the grid can accommodate, it may be either rounded or truncated, but it must fill the entire grid.
31
If y varies inversely as x, and y = ½ when x = 10, then for what value of x does y = 25?
32
If x2 + 12x = 13, and x < 0, what is the value of x2?
33
Four triangles are to be cut and removed from a square piece of sheet metal to create an octagonal sign with eight equal sides, as shown in the figure above. If the total area of the removed material is 196 square centimeters, what is the perimeter, in centimeters, of the octagon?
34
If m and n are integers such that m2 + n2 = 40 and < 0 < n, what is the value of (m + n)2?
35
If (cos x)(sin x) = 0.2, what is the value of (cos x + sin x)2?
36
MONTHLY SALES (FEBRUARY)
The table above shows information about the February sales for five different cell phone models at a local store. What was the median price, to the nearest dollar, of the 240 phones sold in February?
Questions 37 and 38 are based on the scenario described below. Enter your responses on the corresponding grids on your answer sheet.
Performance Banner Company creates promotional banners that include company logos. The Zypz Running Shoe Company would like a 4-foot high and 20-foot long banner that includes its logo, which has a height-to-length ratio of 5:8.
37
If the logo were scaled so that its height matched the height of the banner and then were placed in the center of the banner, then what would be the width, in feet, of each margin on either side of the logo?
38
Performance Banner Company charges its customers \$1.20 per square foot for the banner material, \$2.50 per square foot of any printed logo, and \$32 in fixed costs per banner. The Zypz Running Shoe Company is considering two options for the banner: one with a single logo, and another with two logos. If these logos are all to be the same size as described in Part 1, what percent of the banner costs would the company save by choosing the single-logo option instead of the two-logo option? (Ignore the % symbol when entering into the grid. For example, enter 27% as 27.)
STOP
If you finish before time is called, you may check your work on this section only. Do not turn to any other section of the test.
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https://www.geeksforgeeks.org/array-range-queries-elements-frequency-value/?ref=lbp
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# Array range queries for elements with frequency same as value
• Difficulty Level : Expert
• Last Updated : 14 Sep, 2022
Given an array of N numbers, the task is to answer Q queries of the following type:
query(start, end) = Number of times a
number x occurs exactly x times in a
subarray from start to end
Examples:
Input : arr = {1, 2, 2, 3, 3, 3}
Query 1: start = 0, end = 1,
Query 2: start = 1, end = 1,
Query 3: start = 0, end = 2,
Query 4: start = 1, end = 3,
Query 5: start = 3, end = 5,
Query 6: start = 0, end = 5
Output : 1 0 2 1 1 3
Explanation:
In Query 1, Element 1 occurs once in subarray [1, 2];
In Query 2, No Element satisfies the required condition is subarray [2];
In Query 3, Element 1 occurs once and 2 occurs twice in subarray [1, 2, 2];
In Query 4, Element 2 occurs twice in subarray [2, 2, 3];
In Query 5, Element 3 occurs thrice in subarray [3, 3, 3];
In Query 6, Element 1 occurs once, 2 occurs twice and 3 occurs thrice in subarray [1, 2, 2, 3, 3, 3]
Method 1 (Brute Force): Calculate frequency of every element in the subarray under each query. If any number x has frequency x in the subarray covered under each query, we increment the counter.
Implementation:
## C++
/* C++ Program to answer Q queries to find number of times an element x appears x times in a Query subarray */#include using namespace std; /* Returns the count of number x with frequency x in the subarray from start to end */int solveQuery(int start, int end, int arr[]){ // map for frequency of elements unordered_map frequency; // store frequency of each element // in arr[start; end] for (int i = start; i <= end; i++) frequency[arr[i]]++; // Count elements with same frequency // as value int count = 0; for (auto x : frequency) if (x.first == x.second) count++; return count;} int main(){ int A[] = { 1, 2, 2, 3, 3, 3 }; int n = sizeof(A) / sizeof(A[0]); // 2D array of queries with 2 columns int queries[][3] = { { 0, 1 }, { 1, 1 }, { 0, 2 }, { 1, 3 }, { 3, 5 }, { 0, 5 } }; // calculating number of queries int q = sizeof(queries) / sizeof(queries[0]); for (int i = 0; i < q; i++) { int start = queries[i][0]; int end = queries[i][1]; cout << "Answer for Query " << (i + 1) << " = " << solveQuery(start, end, A) << endl; } return 0;}
## Java
/* Java Program to answer Q queries tofind number of times an element xappears x times in a Query subarray */import java.util.HashMap;import java.util.Map; class GFG{ /* Returns the count of number x withfrequency x in the subarray fromstart to end */static int solveQuery(int start, int end, int arr[]){ // map for frequency of elements Map mp = new HashMap<>(); // store frequency of each element // in arr[start; end] for (int i = start; i <= end; i++) mp.put(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1); // Count elements with same frequency // as value int count = 0; for (Map.Entry entry : mp.entrySet()) if (entry.getKey() == entry.getValue()) count++; return count;} // Driver codepublic static void main(String[] args){ int A[] = { 1, 2, 2, 3, 3, 3 }; int n = A.length; // 2D array of queries with 2 columns int [][]queries = { { 0, 1 }, { 1, 1 }, { 0, 2 }, { 1, 3 }, { 3, 5 }, { 0, 5 } }; // calculating number of queries int q = queries.length; for (int i = 0; i < q; i++) { int start = queries[i][0]; int end = queries[i][1]; System.out.println("Answer for Query " + (i + 1) + " = " + solveQuery(start, end, A)); }}} // This code is contributed by Rajput-Ji
## Python3
# Python 3 Program to answer Q queries# to find number of times an element x# appears x times in a Query subarrayimport math as mt # Returns the count of number x with# frequency x in the subarray from# start to enddef solveQuery(start, end, arr): # map for frequency of elements frequency = dict() # store frequency of each element # in arr[start end] for i in range(start, end + 1): if arr[i] in frequency.keys(): frequency[arr[i]] += 1 else: frequency[arr[i]] = 1 # Count elements with same # frequency as value count = 0 for x in frequency: if x == frequency[x]: count += 1 return count # Driver codeA = [1, 2, 2, 3, 3, 3 ]n = len(A) # 2D array of queries with 2 columnsqueries = [[ 0, 1 ], [ 1, 1 ], [ 0, 2 ], [ 1, 3 ], [ 3, 5 ], [ 0, 5 ]] # calculating number of queriesq = len(queries) for i in range(q): start = queries[i][0] end = queries[i][1] print("Answer for Query ", (i + 1), " = ", solveQuery(start,end, A)) # This code is contributed# by Mohit kumar 29
## C#
// C# Program to answer Q queries to// find number of times an element x// appears x times in a Query subarrayusing System;using System.Collections.Generic; class GFG{ /* Returns the count of number x with frequency x in the subarray from start to end */ public static int solveQuery(int start, int end, int[] arr) { // map for frequency of elements Dictionary mp = new Dictionary(); // store frequency of each element // in arr[start; end] for (int i = start; i <= end; i++) { if (mp.ContainsKey(arr[i])) mp[arr[i]]++; else mp.Add(arr[i], 1); } // Count elements with same frequency // as value int count = 0; foreach (KeyValuePair entry in mp) { if (entry.Key == entry.Value) count++; } return count; } // Driver code public static void Main(String[] args) { int[] A = { 1, 2, 2, 3, 3, 3 }; int n = A.Length; // 2D array of queries with 2 columns int[,] queries = {{ 0, 1 }, { 1, 1 }, { 0, 2 }, { 1, 3 }, { 3, 5 }, { 0, 5 }}; // calculating number of queries int q = queries.Length; for (int i = 0; i < q; i++) { int start = queries[i, 0]; int end = queries[i, 1]; Console.WriteLine("Answer for Query " + (i + 1) + " = " + solveQuery(start, end, A)); } }} // This code is contributed by// sanjeev2552
## Javascript
Output
Answer for Query 1 = 1
Answer for Query 2 = 0
Answer for Query 3 = 2
Answer for Query 4 = 1
Answer for Query 5 = 1
Answer for Query 6 = 3
Time Complexity: O(Q * N)
Method 2 (Efficient):
We can solve this problem using the MO’s Algorithm
We assign starting index, ending index and query number to each query, Each query takes the following form-
Starting Index(L): Starting Index of the subarray covered under the query;
Ending Index(R) : Ending Index of the subarray covered under the query;
Query Number(Index) : Since queries are sorted, this tells us original position of the query so that we answer the queries in the original order
Firstly, we divide the queries into blocks and sort the queries using a custom comparator.
Now we process the queries offline where we keep two pointers i.e. MO_RIGHT and MO_LEFT with each incoming query, we move these pointers forward and backward and insert and delete elements according to the starting and ending indices of the current query.
Let the current running answer be current_ans.
Whenever we insert an element we increment the frequency of the included element, if this frequency is equal to the element we just included, we increment the current_ans.If the frequency of this element becomes (current element + 1) this means that earlier this element was counted in the current_ans when it was equal to its frequency, thus we need to decrement current_ans in this case.
Whenever we delete/remove an element we decrement the frequency of the excluded element, if this frequency is equal to the element we just excluded, we increment the current_ans.If the frequency of this element becomes (current element – 1) this means that earlier this element was counted in the current_ans when it was equal to its frequency, thus we need to decrement current_ans in this case.
Implementation:
## C++
Output
Answer for Query 1 = 1
Answer for Query 2 = 0
Answer for Query 3 = 2
Answer for Query 4 = 1
Answer for Query 5 = 1
Answer for Query 6 = 3
Time Complexity of this approach using MO’s Algorithm is O(Q * sqrt(N) * logA) where logA is the complexity to insert an element A into the unordered_map for each query.
My Personal Notes arrow_drop_up
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Courses
# CBSE TEST PAPER-02 CLASS - XII MATHEMATICS (Linear Programming) Notes | EduRev
## : CBSE TEST PAPER-02 CLASS - XII MATHEMATICS (Linear Programming) Notes | EduRev
``` Page 1
CBSE TEST PAPER-02
CLASS - XII MATHEMATICS (Linear Programming)
Topic: - Linear Programming [ANSWERS]
Ans 01. Let the two tailors work for x days and y days respectively
Z = 150x + 200y
6x + 10y = 60
3 5 30
4 4 32
8 and x 0, y 0
x y
x y
x y
? + =
+ =
? + = = =
Z is minimum = 1350
When A work for 5 days B work for 3 days
Ans 02. Let P = x
Q = y
Z = 250x + 200y
3x + 1.5y = 18
2.5x + 11.25y = 45
2x + 3y = 24
x =0, y = 0
Z = Rs 1950
When x = 3 y = 6
Number of bags of brand P = 3 bags of brand Q = 6
Ans 03. Let the dietician mix x Kg of food X and y Kg of food Y.
Z = 16x + 20y
x + 2y = 10
2x + 2y =12
6
3 8
x y
x y
? + =
+ =
Cost is minimum = 112 When 2 Kg of food X and 4 Kg of food Y are mixed.
Ans 04. Let x toys of type A and y toys of type B
15
5
2
12 6 360
2 60
18 360
Z x y
x y
x y
x
= +
+ =
? + =
=
20
6 9 360
x
x y
? =
+ =
Page 2
CBSE TEST PAPER-02
CLASS - XII MATHEMATICS (Linear Programming)
Topic: - Linear Programming [ANSWERS]
Ans 01. Let the two tailors work for x days and y days respectively
Z = 150x + 200y
6x + 10y = 60
3 5 30
4 4 32
8 and x 0, y 0
x y
x y
x y
? + =
+ =
? + = = =
Z is minimum = 1350
When A work for 5 days B work for 3 days
Ans 02. Let P = x
Q = y
Z = 250x + 200y
3x + 1.5y = 18
2.5x + 11.25y = 45
2x + 3y = 24
x =0, y = 0
Z = Rs 1950
When x = 3 y = 6
Number of bags of brand P = 3 bags of brand Q = 6
Ans 03. Let the dietician mix x Kg of food X and y Kg of food Y.
Z = 16x + 20y
x + 2y = 10
2x + 2y =12
6
3 8
x y
x y
? + =
+ =
Cost is minimum = 112 When 2 Kg of food X and 4 Kg of food Y are mixed.
Ans 04. Let x toys of type A and y toys of type B
15
5
2
12 6 360
2 60
18 360
Z x y
x y
x y
x
= +
+ =
? + =
=
20
6 9 360
x
x y
? =
+ =
2 3 120
0, y 0
x y
x
? + =
= =
profit is maximum = 262.5 at A = 15 B = 30
Ans 05. X passengers travel by executive class and y passengers travel by economy class. L
Z = 1000x + 600y
x + y = 200
x = 20
y = 80
x = 0, y = 0
profit is maximum = 168000
When x = 120, y = 80
Ans 06.
5 3
410
2 2
Z x y = + +
60 0
50 0
100 ( ) 0
60 0
, 0
x
y
x y
x y
x y
- =
- =
- + =
+ - =
=
minimum = 510
when D ? 10
E ? 50
F ? 40
Ans 07.
3
3950
10 10
x y
Z = + +
4500 0
3000 0
3500 0
7000 ( ) 0
0, 0
x
y
x y
x y
x y
- =
- =
+ - =
- + =
= =
Minimum = 4400
A
B
E F D
100
100-(x+y)
40
x+y-60
50
50-y
50
Y
60
60-x
X
A
B
E F D
Y
X
7 0 0 0 L
7000 - (x+ y)
3 5 0 0
x+ y-3500
4 0 0 0 L
3 0 0 0 -y
3 0 0 0
4500-x
4 5 0 0
Page 3
CBSE TEST PAPER-02
CLASS - XII MATHEMATICS (Linear Programming)
Topic: - Linear Programming [ANSWERS]
Ans 01. Let the two tailors work for x days and y days respectively
Z = 150x + 200y
6x + 10y = 60
3 5 30
4 4 32
8 and x 0, y 0
x y
x y
x y
? + =
+ =
? + = = =
Z is minimum = 1350
When A work for 5 days B work for 3 days
Ans 02. Let P = x
Q = y
Z = 250x + 200y
3x + 1.5y = 18
2.5x + 11.25y = 45
2x + 3y = 24
x =0, y = 0
Z = Rs 1950
When x = 3 y = 6
Number of bags of brand P = 3 bags of brand Q = 6
Ans 03. Let the dietician mix x Kg of food X and y Kg of food Y.
Z = 16x + 20y
x + 2y = 10
2x + 2y =12
6
3 8
x y
x y
? + =
+ =
Cost is minimum = 112 When 2 Kg of food X and 4 Kg of food Y are mixed.
Ans 04. Let x toys of type A and y toys of type B
15
5
2
12 6 360
2 60
18 360
Z x y
x y
x y
x
= +
+ =
? + =
=
20
6 9 360
x
x y
? =
+ =
2 3 120
0, y 0
x y
x
? + =
= =
profit is maximum = 262.5 at A = 15 B = 30
Ans 05. X passengers travel by executive class and y passengers travel by economy class. L
Z = 1000x + 600y
x + y = 200
x = 20
y = 80
x = 0, y = 0
profit is maximum = 168000
When x = 120, y = 80
Ans 06.
5 3
410
2 2
Z x y = + +
60 0
50 0
100 ( ) 0
60 0
, 0
x
y
x y
x y
x y
- =
- =
- + =
+ - =
=
minimum = 510
when D ? 10
E ? 50
F ? 40
Ans 07.
3
3950
10 10
x y
Z = + +
4500 0
3000 0
3500 0
7000 ( ) 0
0, 0
x
y
x y
x y
x y
- =
- =
+ - =
- + =
= =
Minimum = 4400
A
B
E F D
100
100-(x+y)
40
x+y-60
50
50-y
50
Y
60
60-x
X
A
B
E F D
Y
X
7 0 0 0 L
7000 - (x+ y)
3 5 0 0
x+ y-3500
4 0 0 0 L
3 0 0 0 -y
3 0 0 0
4500-x
4 5 0 0
Ans 08. Let the fruit grower mix x bags of brand P and Y bags of brand Q
7
3
2
Z x y = +
2 240
3
3 270
2
3
2 310
2
0, 0
x y
x y
x y
x y
+ =
+ =
+ =
= =
minimum = 470 Kg
P = 40
Q = 100
Ans 09. Let Anil invest x in bond A and Y in bond B,
8 10
100 100
x y
P = +
x + y = 12000
x = 2000
y = 4000
x = 0, y = 0
P is maximum = 1160
x = 2000
y = 10,000
Ans 10. X dolls of type A and y dolls of type B.
Z = 12x + 16y
X + y = 1200
2
3 600
x
y
x y
=
= +
Profit is maximum = 16000
A = 800
B = 400
```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
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Вы находитесь на странице: 1из 12
# QUANT TECHNIQUES STRAIGHT FROM SERIAL CAT TOPPER
BYJU
INDEX
ss
es
## 1) CONSTANT PRODUCT RULE
2) POWER CYCLE
3) MINIMUM OF ALL REGIONS IN VENN DIAGRAMS
4) LAST 2 DIGITS TECHNIQUE
5) SIMLIAR TO DIFFERENT GROUPING ( P&C)
6) APPLICATION OF FACTORIALS
7) GRAPHICAL DIVISION FOR GEOMETRY
8) ASSUMPTION METHOD
la
This rule can be applied when we have two parameters whose product is constant, or in other words, when they are
inversely proportional to each other.
eg)Time x Speed = Distance, Price x consumption = Expenditure, Length x breadth= Area
A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other
Parameter if the parameters are inversely proportional.
Lets see the application with the following examples
AT
1) A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6
km/hr he would have reached 10 min early. Find the distance from his home to office?
Solution: Assume original speed= 4km/hr. Percentage increase in speed from 4 to 6= 50% or . 1/2 increase in speed
will result in 1/3 decrease in original time=30 minutes.(from given data). Original time= 90 minutes= 1.5 hours
2) A 20% increase in price of sugar. Find the % decrease in consumption a family should adopt so that the
expenditure remains constant
Solution: Here price x consumption= expenditure (constant)
Using the technique, 20% (1/5) increase results in 16.66%(1/6) decrease in consumption. Answer=16.66%
By
j
u'
s
2) POWER CYCLE
The last digit of a number of the form ab falls in a particular sequence or order depending on the unit digit of the
number (a) and the power the number is raised to (b). The power cycle of a number thus depends on its unit digit.
Consider the power cycle of 2
2 1=2,
2 5=32
2
2 =4
2 6=64
3
2 =8
2 7=128
4
2 =16
2 8=256
As it can be observed, the unit digit gets repeated after every 4 th power of 2. Hence, we can say that 2 has a power
cycle of 2,4,8,6 with frequency 4.
This means that, a number of the form
2 4k+1 will have the last digit as 2
2 4k+2 will have the last digit as 4
2 4k+3 will have the last digit as 8
2 4k+4 will have the last digit as 6 (where k=0, 1, 2, 3)
This is applicable not only for 2, but for all numbers ending in 2.
Therefore to find the last digit of a number raised to any power, we just need to know the power cycle of digits from 0
to 9, which are given below
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
## PDF created with pdfFactory Pro trial version www.pdffactory.com
Power cycle
0
1
2,4,8,6
3,9,7,1
4,6
5
6
7,9,3,1
8,4,2,6
9,1
Frequency
1
1
4
4
2
1
1
4
4
2
es
Unit digit
0
1
2
3
4
5
6
7
8
9
la
ss
For example
3)
Find the remainder when 375 is divided by 5.
1) Express power in the form, 4k+x where x=1, 2, 3, 4. In this case 75 = 4k+3.
2) Take the power cycle of 3 which is 3,9,7,1. Since the form is 4k+3, take the third digit in the cycle, which is 7
Any number divided by 5, the remainder will be that of the unit digit divided by 5. Hence the remainder is 2.
Sometimes, you may get a question in the term of variables, where you need to substitute values to get the answer in
the fastest way possible.
For example,
AT
4)
Find the unit digit of 73^4n
Put n=1, the problem reduces to 7 3^4, which is 781.
Since 81=4k+1, take the first digit in the power cycle of 7, which is 7.
What is the first non zero integer from the right in 83301957 + 8370 1982?
a) 3
b) 1
c) 9
d) none of these
8370 1982 will end with more number of zeroes so we need to consider only the first part. Rightmost non-zero integer of
the expression will be = unit digit of 8331957
= unit digit of 31957
Since 1957=4k+1, take the first digit in the power cycle of 3, which is 3.
5)
If N = (13)1! + 2! + 3! + ..+ 13! + (28)1! + 2! + 3!..+ 28! + (32)1! + 2! + 3! + ...+ 32!+ (67) 1! + 2! + 3! +
then the unit digit of N is
(a) 4
(b) 8
(c) 2
(d) none of these
Based on Power Cycle
After 4! Every number is of the form 4k+4 , here in all four terms power becomes
4k+1. So taking the first digit from the power cycles of 3,8,2, and 7 we will get the unit
digit as i.e 3+8+2+7 = 4. Ans = 0
6)
By
j
u'
s
...+ 67!
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
## PDF created with pdfFactory Pro trial version www.pdffactory.com
3) Useful technique to find the last 2 digits of any expression of the form ab
Depending on the last digit of the number in question, we can find the last two digits of that number.
We can classify the technique to be applied into 4 categories.
TYPE
METHOD
EXAMPLES
67
1) Numbers ending in 1
The last digit is always 1.
1) 21 =__41(2 * 7=__4)
The 2nd last digit = product of
tens digit of base * unit digit of
the power.
In 2167; 2 is the tens digit of
base and 7 is the unit digit of
power
## Last two digits: always 25
e.g.) 1555 34 = __ 25
3) Numbers ending in 3, 7, 9
## Change the power so that the
base ends with 1 and then use
the same technique as for those
numbers ending with 1.
## e.g.) 17288 =(174)72 (taking power 4 as
74 will end in 1.
(17 2* 172) 72
es
la
ss
## =( _89*_89)72 (as last 2 digits of
172=89)
= ( _21)72 (as last 2 digits of
89*89=21)
## Use the pattern of the number
1024 =2 10 i.e.
*2 10 raised to even power ends
with 76 and
* 210raised to odd power ends
with 24.
u'
s
AT
## It is also important to note that,
By
j
1. 76 multiplied by the last 2 digits of any power of 2 will end in the same last 2 digits
E.g. 76*04 = 04, 76*08 = 08, 76*16 = 16, 76*32 = 32
2. The last two digits of , x2, (50-x)2 , (50+x)2 , (100-x)2 will always be the same. For example last 2 digits of
122,382,622,882,112 2. will all be the same.
Also, last two digits of 11 2=392=612=89 2 =111 2=139 2=161 2=1892 and so on
3.To find the squares of numbers from 30-70 we can use the following method
7) To find 41 2
Step1 : Difference from 25 will be first 2 digits = 16
Step 2 : Square of the difference from 50 will be last 2 digits = 81 Answer = 1681.
8): To find 432
Step1 : Difference from 25 will be first 2 digits = 18
Step 2 :Square of the difference from 50 will be last 2 digits = 49 Answer = 1849
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4. Combining all these techniques we can find the last 2 digits for any number because every even number can be
written as 2* an odd number
## 4) MINIMUM OF ALL REGIONS IN VENN DIAGRAMS
9)
In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B, 80 use brand
C, 90 use brand D & 60 use brand E of the same product. What is the minimum possible number of men using all
the 5 brands, if all the 100 men use at least one of these brands?
es
## Sum of the difference from 100 = (100-100) + (100-75)+(100-80)+(100-90)+(100-60) = 95
Again take the difference from 100 = 5 (answer)
This method will be explained in detail in the next booklet.
ss
## 5) SIMILAR TO DIFFERENT GROUPING IN PERMUTATION & COMBINATION
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All questions in Permutation and Combination fall into 4 categories, and if you master these 4 categories, you can
understand all concepts in P&C easily.
1) Similar to Different
2) Different to Similar
3) Similar to Similar
4) Different to Different
In this booklet, we will look at the first category; i.e. Similar to Different, which entails
the number of ways of dividing n identical (similar) things into r distinct (different) groups
AT
a) NO LIMIT QUESTIONS
Let me explain this with an example. Suppose I have 10 identical chocolates to be divided among 3 people.
The 10 chocolates need to be distributed into 3 parts where a part can have zero or more chocolates.
So let us represent chocolates by blue balls. The straight red lines are used to divide them into parts. So you can see
that for dividing into 3 parts, you need only two lines.
u'
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Suppose you want to give 1 st person 1 chocolate, 2nd 3 chocolates and 3rd 6 chocolates. Then you can show it as:
By
j
Suppose you want to give one person 1 chocolate, another person 6 chocolates and another one 3, then it can be
represented as:
Now if first person gets 0, second gets 1 and third gets 9 chocolates then it can be represented as:
Now suppose you want to give first person 0, second also 0 and third all of 10 then you can show it like:
## BYJUS CAT CLASSES
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es
So, for dividing 10 identical chocolates among 3 persons you can assume to have 12 (10 balls +2 sticks) things
among which ten are identical and rest 2 are same and of one kind. So number of ways in which you can distribute
ten chocolates among 3 people is the same in which you can arrange 12 things among which 10 are identical and of
one kind while 2 are identical and of one kind which can be done in 12!/(10!2!)
ss
The above situation is same as finding the number of positive integral solutions of a + b + c = 10. a, b, c is the
number of chocolates given to different persons.
Or else I can also say how many integral points lie on the line a + b + c = 10 in the 1st quadrant. In both the cases
## b) LOWER LIMIT QUESTIONS
la
Now suppose we have a restriction that the groups cannot be empty i.e. in the above example all 3 persons
should get at least 1.
You have to divide ten chocolates among 3 persons so that each gets at least one. So in the start only give them one
each. This you will do in just 1 way as all the chocolates are identical. Now, you are left with 7 chocolates and you
have to divide them among 3 people in such that way that each gets 0 or more. You can do this easily as explained
above using balls and sticks.
Number of ways = 9! /(7!2!).
AT
The above situation is same as finding the number of positive natural number solutions of a + b + c = 10. a, b, c
is the number of chocolates given to different persons.
Or else I can also say how many integral natural points lie on the line a + b + c = 10 in the 1 st quadrant.
In both the cases the answer is 9C2.
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Now suppose I change the question and say that now you have to divide 10 chocolates among 3 persons in such
a way that one gets at least 1, second at least 2 and third at least 3.
Its as simple as the last one. First full fill the required condition.
Give 1 st person 1, second 2 and third 3 and then divide the left 4 (1012-3) chocolates among those 3 in such a
way that each gets at least 1.
This is same as arranging 4 balls and 2 sticks which can be done in 6C2 ways.
By
j
The above situation is same as finding the number of positive integral solutions of
a + b + c = 10 such that a > = 1, b > = 2, c > = 3. a, b, c is the number of chocolates given to different persons.
In this case the answer is 9C2.
10)
Rajesh went to the market to buy 18 fruits in all. If there were mangoes, bananas, apples and oranges for sale
then in how many ways could Rajesh buy at least one fruit of each kind?
a) 17C3
b) 18C4 c) 21C3
d) 21C4
This is a Grouping type 1 Similar to Distinct question, with a lower limit condition.
M+B+A+O=18
Remove one from each group, therefore 4 is subtracted from both sides. The problem changes to
M+B+A+O=14
Using our shortcut, The answer is the arrangement of 14 balls and 3 sticks
i.e. 17C3
## BYJUS CAT CLASSES
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11)
There are four balls to be put in five boxes where each box can accommodate any number of balls. In how
many ways can one do this if:
a) Balls are similar and boxes are different
1) 275
2) 70
c) 120
d) 19
When the balls are similar and the boxes are different, its a grouping type 1 question
A+B+C+D+E=4, where A,B,C,D,E are the different boxes. The number of ways of selection and distribution=8C4=70
The number of non negative integral solutions of x1+x2+x3 10
a) 84
b) 286
c) 220
d) none of these
es
12)
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By non-negative integral solutions, the conditions imply that we can have 0 and natural number values for x1, x2, x3,
and x4
To remove the sign add another dummy variable x4. The problem changes to
x1+x2+x3+x4=10
This is an example of grouping type1 (Similar to Distinct)
It is the arrangement of 10 balls and 3 sticks.
Using the shortcut of balls and sticks, Therefore the answer is 13C3=286
6) APPLICATION OF FACTORIALS
## Definition of Factorial N! = 1*2*3**(n-1)*n
Eg 1) 5!= 1*2*3*4*5=120
eg 2) 3!=1*2*3=6
A thorough understanding of Factorials is important because they play a pivotal role not only in understanding
concepts in Numbers but also other important topics like Permutation and Combination
I)
AT
## Highest power in a factorial or in a product
a)
Questions based on highest power in a factorial are seen year after year in CAT. Questions based on this can be
categorized based on the nature of the number (prime or composite) whose highest power we are finding in the
factorial, i.e
## Highest power of a prime number in a factorial:
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To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the
quotients.
## 13) The highest power of 5 in 100!
By
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100/5=20; 20/5=4; Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24
ALTERNATIVE METHOD
100/5 + 100/52 =20+4=24 (We take upto 52 as it is the highest power of 5 which is less than 100)
b)
## Highest power of a composite number in factorial
Factorize the number into primes. Find the highest power of all the prime numbers in that factorial using the previous
method. Take the least power.
## 14) To find the highest power of 10 in 100!
Solution: Factorize 10=5*2.
1. Highest power of 5 in 100! =24
2. Highest power of 2 in 100! =97
Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available.
So take the lesser i.e. 24 is the answer.
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
## 15) Highest power of 12 in 100!
Solution: 12=22 *3. Find the highest power of 22 and 3 in 100!
First find out the highest power of 2.
Listing out the quotients:100/2 = 50; 50/2 = 25; 25/2 = 12; 12/2 = 6; 6/2 = 3; 3/2 = 1
Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97. So highest power of 2 2 = 48 (out of 97 2s only 48 can make 22)
Now for the highest power of 3. 100/3 = 33; 33/3 = 11; 11/3 = 3; 3/3 = 1;
Highest power of 3 = 48, Highest power of 12 = 48
II)
## Number of zeros in the end of a factorial or a product
es
Finding the number of zeroes forms the base concept for a number of application questions. In base 10, number of
zeros in the end depends on the number of 10s; i.e. effectively, on the number of 5s
In base N, number of zeroes in the end highest power of N in that product
## 16) Find the number of zeroes in 13! In base 10
ss
Solution: We need to effectively find the highest power of 10 in 13! = Highest power of 5 in 13! As this power will be
lesser. 13/5=2
17) Find the number of zeroes in the end of 15! in base 12.
## Number of factors of any factorial
III)
la
Solution: Highest power of 12 in 15! =highest power of 22 *3 in 15! =Highest power of 3 in 15!= 5
There is a technique, which can be used to find the number of factors in a factorial
## 18) Find the factors of 12!
IV)
AT
STEP 1: Prime factorize 12! i.e. find out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! =
2 10*3 5*52*7*11
STEP2: Use the formula
N=am*bn (a, b are the prime factors). Then number of factors= (m+1)(n+1)
The number of factors= (10+1)(5+1)(2+1)(1+1)(1+1) =792. Answer=792
## Right most non zero integer in a factorial
By
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19) Find the right most non zero integer in 25! ? OR Find the remainder when 25! is divided by 107?( Both the
questions are conceptually the same)
There are 6 zeroes at the end of 25!.
Effectively we need to find the rightmost non-zero integer in 25!.
Number of 2s in 25! = 25/2+12/2+6/2+3/2 = 12+6+3+1=22,
Number of 3s in 25! = 25/3 + 8/3 = 8+2=10
Number of 5s in 25! = 25/5+5/1 =6,
Number of 7s in 25! =25/7=3
Number of 11s in 25! =2,
Number of 13s,17s,19s,23s in 25! =1
Remove 26x5 6. (form the 6 zeroes),
Right most non zero digit will be last digit of 2163107311 2.131.171.191.231
= 6.9.3.1.3.7.9.3 = __4. Answer is 4.
## If you know the unit digit in the product 1!*2!*3!*10! = 8 and
Unit digit in the product 11!*12*13*..20! = 8.then the answer is in one line as follows.
8*8*1*2*3 = Answer 4.This short cut method takes just 30 seconds.
## APPLICATION QUESTION BASED ON FACTORIAL
20) How many natural numbers are there such that their factorials are ending with 5 zeroes?
10! is 1*2*3*4*(5)*6*7*8*9*(2*5). From this we can see that highest power of 5 till 10! is 2. Continuing like this,
10!-14!, highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3).
Therefore, from 15! To 19! - The highest power of 5 will be 3.
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
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20!-24! Highest Power = 4, In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get
highest power of 5 as 6. The answer to the question is therefore, 0. There are no natural numbers whose factorials end
with 5 zeroes.
## 7) USE OF GRAPHICAL DIVISION IN GEOMETRY
la
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Lets look at a technique which will help you solve a geometry question in no time
21)
ABCD is a square and E and F are the midpoints of AB and BC respectively.
Find the ratio of Area( ABCD): Area(DEF)?
8) ASSUMPTION method
AT
FIGURE A
FIGURE B
Lets divide the figure using dotted lines as shown in Figure B. Area of ABCD=100%. Area AEDG=50%. Then Area in
shaded region 1(AED)= 25%. Similarly, DCFH=50%. Area in shaded region 2(DCF)=25%. Now Area of EOFB=
25%. .Area of shaded region 3(BEF)=12.5%. Total area outside triangle= 62.5%. Area inside triangle= 10062.5=37.5%. required ratio = 100/37.5 = 8:3
There are at least 10 questions out of 25 in CAT 2007 where you can apply this technique. This involves assuming
simple values for the variables in the questions, and substituting in answer options based on those values. Assumption
helps to tremendously speedup the process of evaluating the answer as shown below.
By
j
u'
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22) k & 2k 2 are the two roots of the equation x 2 px + q. Find q + 4q 2 + 6pq =
a) q2
b) p 3
c) 0
d) 2p3
Solution: Assume an equation with roots 1&2 (k=1)
=>p (sum of roots)= 3 and q(product of roots)=2.
Substitute in q + 4q 2 + 6pq = 54. Look in the answer options for 54 on substituting values of p=3 and q=2. we get 2p3
= 54.=> Ans = 2p3 .
23) Let x be the arithmetic mean and y,z be the two geometric means between any two positive numbers. The value
of (y3 + z3) / (xyz) is
a) 2
b) 3
c) 1/2
d) 3/2
Assume a GP 1
2
4
8 implies 2 GMs between 1 and 8, i.e. y=2 and z=4.
Arithmetic Mean, x =(8+1)/2 = 4.5 .Substitute in (y3 + z3) / (xyz) Answer = (2 3+43)/(2x4x4.5) = 2.
NOTE:- Assume a GP 1 1 1 1 then x=1, y=1 z=1. Answer on substitution=2, which will make the calculation
even faster, half of the problems in Algebra can be solved using assumption. This is not direct substitution. In the next
eg: see how you can use the same technique in an equation question.
## BYJUS CAT CLASSES
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## SOME ACTUAL QUESTIONS FROM CAT-2007 SOLVED USING THIS TECHNIQUE
c) q(n/2)(p+q)n/2
ss
25) Let a 1=p and b1=q, p and q are positive quantities. Define
a n=pbn-1, b n=qb n-1 for even n>1
a n=pan-1, b n=qa n-1 for odd n>1.
which of the following best describes a n+bn for even n.
a) qp (n/2)(p+q)
b) q(n/2)(p+q)n
(n/2)-1
n/2
d)q(pq)
(p+q)
e)q(pq)(n/2)-1(p+q)
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24) Consider the set S={2,3,42n+1), where n is a positive integer larger than 2007. Define X as the average of
odd integers in S and Y as the average of the even integers in S. What is the value of X-Y?
a) 1
b) n/2
c)(n+1)/2n
d) 2008
e) 0
The question is independent of n, which is shown below.
Take n=2. Then S= {2,3,4,5). X= 4 and Y=3. X-Y =1, Take n=3. then S={2,3,4,5,6,7}. X=5 and Y=4. X-Y=1
Hence you can directly mark the answer option (a) .You can solve the question in less than 60 seconds.
la
Since the answer options involve square roots, we will assume a1 =p=4 and b1 =q=4 (when n=1)
Now for n=2,we will get a2= 4*4=16 ; b 2=4*4=16
In our assumption, a2+b 2=32. Check where you are getting 8 among answer options. Answer is option (e) q(pq)(n/2)1
(p+q)= 4*8=32. This wont take more than more than two minutes.
AT
26) Let S be the set of all pairs (i,j) where 1i<jn and n4. Any two distinct members of
S are called friends, if they have one constituent of the pairs in common and enemies
otherwise. For example, if n=4, then S={(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}. Here (1,2),
(1,3) are friends but (1,4), (2,3) are enemies.
For general n, consider any two members of S that are friends. How many other members of S will be common
friends of both these members?
a) 2n-6
b) n/2(n-3)
c) n-2 d) (n2-7n+16)/2 e) (n2-5n+8)/2
We will start with n=5, then you will get two options (c) and (d).
Now take n =6, lets take two members (1,2) and (1,3). Then their common friends will be ( 1,4)(1,5),(1,6) and (2,3).i.e
Four common friends. So substitute n=6 among answer options and check where you get answer 4.Only option (c).
There were more questions which could be solved using similar strategies. The methods given above clearly show that
for someone with good conceptual knowledge and right strategies the quant section is a cakewalk.
By
j
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It is impossible for me to explain all the techniques I use in one booklet. A lot of
concepts and techniques like these helped me and my students to score 100 percentile in
CAT. To master these techniques and many more similarly unique ones, dont miss the
golden opportunity to attend the FREE workshop in your city.
BANGALORE, CHENNAI, MUMBAI, DELHI, PUNE, HYDERABAD, GURGAON,
MANIPAL, VELLORE, MANGALORE, MYSORE
<your name> to 9880031619 or 9986075402.
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
es
## By Serial CAT Topper BYJU.
Are you ready to crack Cat 2009? How confident are you that you will cross the 99%ile cutoff in
CAT 2009 Quant and DI Section?
ss
Check yourself ..
la
1)
Find the remainder when 123123123 300 digits is divided by 504?
2)
In an innings a team can lose a maximum of ten wickets. Find the probability that in a given
match the two teams will together lose more than 5 but less than 12 wickets?
3)
How many natural numbers less than 100 can be expressed as a difference of squares of whole
numbers exactly in one way?
AT
With around 3 lakh people battling for about 2000 seats you need to have an edge over others. Act smart and
attend the FREE workshop by Byju on strategies, methodologies and techniques to solve 50 such CAT level
questions with speed and accuracy.
u'
s
## What will you get in this free workshop?
Glimpse of Byju's unique and innovative techniques mastered by the 231 students (including quant &
DI toppers) who got IIM calls in 2008.
To explore mathematics through a prism you have never viewed before
Learn to tackle questions which are seemingly impossible to solve, in the best possible way with a
speed which will amaze you without compromising on accuracy.
A confidence to give your best in any competitive exam!
By
j
Milestones
Highest Success Ratio with 1 out of 8 Byjus students getting IIM Calls in CAT 2008.
231 students got calls from IIMs alone in 2008
212 Final Selections into IIMs in CAT 2007
Track Record of the Tutor
100 Percentiler in CAT 03 and CAT 05
Conducted Seminars in IITs, NITs & Foreign Universities in Speed Mathematics
Conducts Regular CAT Batches for top colleges like IIT Delhi, College of Engineering-Guindy
Chennai, VJTI- Mumbai, RVCE- Bangalore and exclusive batches for working engineers in
Bangalore and Pune.
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
## This is what his students have to say about him :
Byju sirs unique methods and easy to use techniques helped me a lot in achieving this. His classes &
workshops is the best way to prepare for CAT. You just need to see his class once to realize this.
Suhas Shetiya, Bangalore, 100 percentiler, Quant, CAT-08, (Calls: IIM A,B,C,L,I,K)
la
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## Well, let the numbers speak for themselves.
Before: My Quant score in CAT 2007 was 8. Percentile was in the range of 30-40.
After: My Quant score in CAT 2008 was 47. Percentile was 98.49
Byju has a very unique way of teaching. He tells you very, very practical and intuitive methods (like
assumption of values, Eulers method, Chinese Remainder Theorem, etc.) for solving problems. Using them
one can solve CAT-level (and higher level) problems insanely fast. But that alone is not the unique part. He
explains the logic behind using the shortcut and gives you a crystal-clear idea of the mechanism of the
method. The outcome is that you dont need to mug up shortcuts and you will never forget them.
The amount of practice he provides is phenomenal. And all of it is very relevant to CAT. Instead of solving
easy problems to brush up your basics, you solve the kind of problems which will make it into the CAT.
And save valuable time and effort.
I believe that joining Byjus classes is the best thing I did for my aim of cracking CAT. It changed the way I
approached the problem and the huge improvement in the results vouch for that.
Harshvarshan K, Pune, 99.8%ile (Calls- A,B,C,L,I,K)
AT
When I joined Byju classes, I had already tried my luck with CAT twice. One fine day I walked into a
demo class conducted by Byju and was completely blown away [I joined immediately]. The best part about
what I saw was a 'Cut the Crap' approach where important topics were given their due weightage and
concepts were discussed very effectively. The questions/techniques discussed were of the highest quality. He
himself takes all the classes and he took my preparation to the next level. I will advise all CAT aspirants to
attend one of his demo classes before joining any other institute
Ankit Bajaj, Bangalore, 99.91%ile (Calls- IIM B,C,L,I,K)
By
j
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Being from IIT, I thought there is nothing much to learn in quant and DI for CAT, but after attending
Byjus classes I learnt many shortcuts and improved reasoning skills that doubled my speed in quant and DI.
No one can teach quant and DI better than this serial CAT topper. I was confident that now whatever may be
the paper standard, I will get through and this played a very important role on D-Day. I attended one demo
class and decided immediately that this is the place that will push me into IIMs. I am not asking anyone to
join his classes, but I will suggest everyone to see any one class taken by him to realize that this is the best
place for cat preparation.
Lalit Kumar IIT Kharagpur, 99.66%ile (Calls: IIM A, B, C, L, I, K)
" I joined one premier institute for my preparation. But they are catering to the masses only covering the
basic things. Fortunately I came to know about Byju sir and his classes are very different, real value addition
and confidence booster. He discusses quality questions and gave direction to my random preparation. Full
credit to him for helping me realize my dream of getting into IIM A."
Anita Kishore, VJTI, Mumbai, 99.82%ile (Calls: IIM A,B,C,L,I,K)
"Few months of brain storming, unconventional yet professional training and I could see the difference.
.from a meager 90 odd% to a 99.10 all thanks to Byju."
Krishna Chaitanya Boga, Bangalore, 99.10%ile (Calls: IIM A,C,L,I,K)
## BYJUS CAT CLASSES
CONTACT: 09880031619/09986075402
## Anybody can claim anything!!
Don't take our word for it.
Experience the difference yourself
Attend a FREE SESSION in your city!
es
## BANGALORE, CHENNAI, MUMBAI, DELHI, PUNE, HYDERABAD, GURGAON,
MANIPAL, VELLORE, MANGALORE, MYSORE
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AT
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# How can we bound the probability that a random variable is maximal?
$\newcommand{\P}{\mathbb{P}}$Suppose we have $N$ independent random variables $X_1$, $\ldots$, $X_n$ with finite means $\mu_1 \leq \ldots \leq \mu_N$ and variances $\sigma_1^2$, $\ldots$, $\sigma_N^2$. I am looking for distribution-free bounds on the probability that any $X_i \neq X_N$ is larger than all other $X_j$, $j \neq i$.
In other words, if for simplicity we assume the distributions of $X_i$ are continuous (such that $\P(X_i = X_j) = 0$), I am looking for bounds on: $$\P( X_i = \max_j X_j ) \enspace.$$ If $N=2$, we can use Chebyshev's inequality to get: $$\P(X_1 = \max_j X_j) = \P(X_1 > X_2) \leq \frac{\sigma_1^2 + \sigma_2^2}{\sigma_1^2 + \sigma_2^2 + (\mu_1 - \mu_2)^2} \enspace.$$ I would like to find some simple (not necessarily tight) bounds for general $N$, but I have not been able to find (esthetically) pleasing results for general $N$.
Please note that the variables are not assumed to be i.i.d.. Any suggestions or references to related work are welcome.
Update: recall that by assumption, $\mu_j \geq \mu_i$. We can then use the above bound to arrive at: $$\P(X_i = \max_j X_j) \leq \min_{j > i} \frac{\sigma_i^2 + \sigma_j^2}{\sigma_i^2 + \sigma_j^2 + (\mu_j - \mu_i)^2} \leq \frac{\sigma_i^2 + \sigma_N^2}{\sigma_i^2 + \sigma_N^2 + (\mu_N - \mu_i)^2} \enspace.$$ This implies: $$( \mu_N - \mu_i ) \P( X_i = \max_j X_j ) \leq (\mu_N - \mu_i) \frac{\sigma_i^2 + \sigma_N^2}{\sigma_i^2 + \sigma_N^2 + (\mu_N - \mu_i)^2} \leq \frac{1}{2} \sqrt{ \sigma_i^2 + \sigma_N^2 } \enspace.$$ This, in turn, implies: $$\sum_{i=1}^N \mu_i \P( X_i = \max_j X_j ) \geq \mu_N - \frac{N}{2} \sqrt{ \sum_{i=1}^{N-1} (\sigma_i^2 + \sigma_N^2) } \enspace.$$ I am now wondering whether this bound can be improved to something that does not depend linearly on $N$. For instance, does the following hold: $$\sum_{i=1}^N \mu_i \P( X_i = \max_j X_j ) \geq \mu_N - \sqrt{ \sum_{i=1}^N \sigma_i^2 } \enspace?$$ And if not, what could be a counterexample?
• This bound can be tighter if you use the index $j$ that gives you the smaller upper bound instead of $N$. Note that this value depends on both the mean and the variance.
– user10525
Commented Jun 11, 2012 at 14:39
• @MichaelChernick : I don't believe that is correct. Suppose for instance we have three uniform distributions on $[0,1]$. Then, if I'm not mistaken, $P( X_1 < \max_j X_j ) = 2/3$, whereas $P(X_1 < X_2) = P(X_1 < X_3) = 1/2$. I do not know if you meant to write $P( X_i > \max_j X_j )$, but then the same example shows that it still isn't valid.
– MLS
Commented Jun 11, 2012 at 15:23
• @Michael: That is still not true, unfortunately. The events $A_j = \{X_i > X_j\}$ for fixed $i$ are not independent. Commented Jun 12, 2012 at 1:08
• @cardinal : Amongst other things, it's related to multi-armed bandits. If you pick an arm based on previous rewards, how big is the probability that you picked the best arm (that would be $P(X_N = \max_j X_j)$ in the notation above), and can we bound the expected loss for picking a sub-optimal arm?
– MLS
Commented Jun 12, 2012 at 15:41
• Crossposted to MathOverflow: mathoverflow.net/questions/99313 Commented Jun 13, 2012 at 0:09
# You can use the multivariate Chebyshev's inequality.
## Two variables case
For a single situation, $X_1$ vs $X_2$, I arrive at the same situation as Jochen's comment on Nov 4 2016
1) If $\mu_1 < \mu_2$ then $P(X_1>X_2) \leq (\sigma_1^2 + \sigma_2^2)/(\mu_1-\mu_2)^2$
### Derivation of equation 1
• using the new variable $X_1-X_2$
• transforming it such that it has the mean at zero
• taking the absolute value
• applying the Chebyshev's inequality
\begin{array} \\ P \left( X_1 > X_2 \right) &= P \left( X_1 - X_2 > 0 \right)\\ &= P\left( X_1 - X_2 - (\mu_1 - \mu_2) > - (\mu_1 - \mu_2)\right) \\ &\leq P\left( \vert X_1 - X_2 - (\mu_1 - \mu_2) \vert > \mu_2 - \mu_1\right) \\ &\leq \frac{\sigma_{(X_1-X_2- (\mu_1 - \mu_2))}^2}{(\mu_2 - \mu_1)^2} = \frac{\sigma_{X_1}^2+\sigma_{X_2}^2}{(\mu_2 - \mu_1)^2}\\ \end{array}
## Multivariate Case
The inequality in equation (1) can be changed into a multivariate case by applying it to multiple transformed variables $(X_n-X_i)$ for each $i<n$ (note that these are correlated).
A solution to this problem (multivariate and correlated) has been described by I. Olkin and J. W. Pratt. 'A Multivariate Tchebycheff Inequality' in the Annals of Mathematical Statistics, volume 29 pages 226-234 http://projecteuclid.org/euclid.aoms/1177706720
Note theorem 2.3
$P(\vert y_i \vert \geq k_i \sigma_i \text{ for some } i) = P(\vert x_i \vert \geq 1 \text{ for some } i) \leq \frac{(\sqrt{u} + \sqrt{(pt-u)(p-1)})^2}{p^2}$
in which $p$ the number of variables, $t=\sum k_i^{-2}$, and $u=\sum \rho_{ij}/(k_ik_j)$.
Theorem 3.6 provides a tighter bound, but is less easy to calculate.
## Edit
A sharper bound can be found using the multivariate Cantelli's inequality. That inequality is the type that you used before and provided you with the boundary $(\sigma_1^2 + \sigma_2^2)/(\sigma_1^2 + \sigma_2^2 + (\mu_1-\mu_2)^2)$ which is sharper than $(\sigma_1^2 + \sigma_2^2)/(\mu_1-\mu_2)^2$.
I haven't taken the time to study the entire article, but anyway, you can find a solution here:
A. W. Marshall and I. Olkin 'A One-Sided Inequality of the Chebyshev Type' in Annals of Mathematical Statistics volume 31 pp. 488-491 https://projecteuclid.org/euclid.aoms/1177705913
(later note: This inequality is for equal correlations and not sufficient help. But anyway your problem, to find the sharpest bound, is equal to the, more general, multivariate Cantelli inequality. I would be surprised if the solution does not exist)
• Could you provide a clear statement of the multivariate Chebyshev Inequality?
– whuber
Commented Jun 21, 2017 at 13:59
• I have edited the solution providing the entire theorem. Commented Jun 21, 2017 at 14:26
$$exp(t \cdot \mathbf{E}(\underset{1 \leq i \leq n}{max}X_{i}))$$
$$exp(t \cdot \mathbf{E}(\underset{1 \leq i \leq n}{max}X_{i})) \leq \mathbf{E}(exp( t \cdot \underset{1 \leq i \leq n}{max}X_{i})) = \mathbf{E}( \underset{1 \leq i \leq n}{max} \text{ }exp( t \cdot X_{i})) \leq \sum_{i=1}^{n} \mathbf{E} (exp(t \cdot X_{i})$$
Now for $exp(t \cdot X_{i}$ you have to plug in whatever the moment generating function of your random variable $X_{i}$ is (since it is just the definition of the mgf). Then, after doing so (and potentially simplifying your term), you take this term and take the log and divide by it by t so that you get a statement about the term $\mathbf{E}(\underset{1 \leq i \leq n}{max}X_{i})$. Then you can choose t with some arbitrary value (best so that the term is small so that the bound is tight).
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# CMPSCI 601: Theory of Computation
### David Mix Barrington
#### Solutions posted Mon 11 August 2003
Questions are in black, solutions in blue.
• Question 1 (30): Remember that the language 3-COLOR, the set of all undirected graphs that can be vertex-colored with three colors, is NP-complete. Here we consider some potential approximation schemes for this problem.
• (a,10) Suppose we had a poly-time algorithm that inputs an undirected graph and produces a coloring that used at most 5/4 times the optimal number of colors. Prove that in this case P=NP. (green text added 7 August 2003)
This algorithm can be used to decide the NP-complete language 3-COLOR. Given an undirected graph, we first apply the given algorithm and determine how many colors are used in its coloring. If there are three or fewer, we know that the input graph is 3-colorable and return "true". If it has four or more colors, we return "false". We are correct in returning "false" in this case because if the optmimal number were 3 or less, the algorithm would have used at most 3*(5/4) = 3.75 colors. Since it used four or more, there cannot have been a 3-coloring. If any NP-complete language can be decided in poly-time, then P=NP. (This decision procedure uses the poly-time given algorithm and then just counts the colors in the graph and answers, which is clear in poly-time overall.)
• (b,10) Suppose that we had a poly-time algorithm that inputs an undirected graph and produces a 3-coloring that is within one of optimal in terms of the number of correct vertices. That is, if there is a 3-coloring in which k of the n vertices are colored correctly, the algorithm produces a coloring with at least k-1 vertices colored correctly. (A vertex is colored correctly if all its neighbor's colors are different from its own color.) Prove that P=NP in this case.
Again, we will use the given algorithm to solve 3-COLOR in poly-time. Given an input graph G, let the graph H be two copies of G with no other edges. Give H to the given algorithm. (Since the size of H is only polynomially larger than that of G, the running time will still be polynomial in the size of G.) Suppose that G is 3-colorable. Then H is also 3-colorable. Since the coloring of H has only one vertex that is not colored correctly, at least one of the two copies of G must be entirely correct. We just check both of them and return "true" if either of then is correctly colored. Otherwise we return "false". It is clear that the extra work after the given algorithm is poly-time, and that we will return "true" iff G is 3-colorable. Since the NP-complete langauge 3-COLOR is in P under the assumption that this algorithm exists, P=NP in this case.
• (c,10) Suppose that we had a poly-time algorithm that inputs an n-vertex undirected graph and produces a coloring that has within sqrt(n) of the optimal number of correct vertices. That is, if there is a coloring with k vertices correct, it gets more than k - sqrt(n) vertices correct. Prove that P=NP in this case.
Let G be any input graph with n vertices and let H be the union of n separate copies of G. Clearly H is 3-colorable iff G is, and H has n2 vertices. Run the given algorithm on H to get a coloring. If G and H are 3-colorable, this coloring must have fewer than sqrt(n2) or fewer than n incorrect vertices, so at least one of the copies of G is correctly colored. Thus our decision procedure is to get this coloring, check each copy of G, and return "true" iff one of the copies is correctly colored. This is poly-time because the given algorithm takes time polynomial in n2 (hence also polynomial in n) and the extra work is clearly polynomial in the size of G. Since in this case the NP-complete language 3-COLOR is decidable in poly-time, P=NP in this case.
• Question 2 (30): Recall the ITERATED-PRODUCT problem from Question 2, HW#4:
• Input: a multiplication table for a group (a function from G times G to G obeying the group properties), and a sequence of elements of the group (a string w in G*). Note that if n is the input size, G has at most n elements and the length of w is at most n.
• Output: the product of the elements in the sequence.
Here are three questions concerning this problem:
• (a,5) Explain why, given results proved in HW#4 and in lecture, the ITERATED-PRODUCT problem is in the class AC1.
We proved on HW#4 that ITERATED-PRODUCT is in the class L. In lecture we showed that L is contained in NL (since a log-space DTM is a special case of a log-space NDTM) and that NL is contained in AC1 (because the Savitch argument puts the NL-complete problem REACH in AC1. So ITERATED-PRODUCT is in AC1.
• (b,15) Describe an AC1 circuit family for this problem directly.
Let m be the length of the string w. Make a binary tree of depth log(m) = O(log n), with m leaves. Each node of this tree will hold a partial product of some elements of w. For example, the leaves will hold w1, w2,..., wm. The next level will hold w1w2, w3w4, the next level's nodes will each hold a product of four elements of w, and so on until the root holds the final product.
At each node we must determine the required element, which is the product of the two elements at the node's children. We must thus build a circuit fragment that takes two group elements (coded as numbers from 1 to n) and gives their product. To do this we must use the multiplication table given in the read-only input, looking up the correct single entry of the table.
With a circuit, this means that we must take an OR of g2 subcircuit, where g is the size of the group. Each subcircuit returns the product of x and y, if the child nodes' elements are x and y, or returns 0 otherwise. The bitwise OR of these results thus gives us the product for the x and y that are actually the contents of the child nodes.
Since each binary multiplication is correct, the circuit gets the right answer for ITERATED-PRODUCT. Its size is polynomial because there are O(n) nodes, each with O(n) subcircuits of size O(log n). Its depth is O(log n) times the depth of the circuit at one node, and these circuits have depth O(1) each if we allow unbounded fan-in. So the whole circuit meets the conditions to put the problem in AC1.
• (c,10) Prove that the special case of ITERATED-PRODUCT, where the group has only O(1) elements, is in the class NC1.
The unbounded fan-in is used in the circuit of Problem 2-b only in taking the OR of g2 subcircuits in each node. But if g is only O(1), then g2 is also O(1) and the gates of fan-in O(1) may be replaced by O(1) gates of fan-in 2. Since the circuit still has depth O(log n) but now has fan-in two, the problem is now in NC1.
• Question 3 (25): Consider the following game played on an undirected graph G with n vertices. There are two players, and at any time each player occupies a vertex of G. On their move, a player may move from their current vertex to an adjacent vertex. There is a "victory table" V, where V is a set of pairs of vertices of G. Player 1 wins if at any time during the game, the pair (x,y) is in V where x is Player 1's position and y is player 2's position. Player 2 wins if the game continues for n moves without player 1 winning.
• (a,15) Describe a log-space alternating Turing machine M such that L(M) is the set {(G,V,x,y): Player 1 wins the game on G and V with starting position (x,y) under optimal play}.
The machine receives G, V, x, and y in its read-only input. It initializes a variable position1 to x and a variable position2 to y. It initializes a variable moveCount to 0, and then enters a loop.
Inside the loop, it checks whether (position1, position2) is in the set given by V and accepts if it does. Then if moveCount is even, player 1 names a node number z. The machine checks whether (position1, z) is in G. If it is, it changes position1 to z, if it isn't it rejects. If moveCount is odd, player2 names a node number z. Then if (position2, z) is in G the machine changes position2 to z, and if not it accepts. (Since player 2 has tried to cheat, she loses and player 1 wins in this case.)
If moveCount reaches n before the machine accepts or rejects, it then rejects (because player 1 wins in this case).
If either player has a winning strategy in the game as described above, that player can choose a series of nodes in the alternating Turing machine game, depending on her opponent's moves, that will guarantee that she wins. The ATM game thus simulates the original game.
• (b,10) Without using alternation, describe a poly-time decision procedure for this language. (Hint: make a table of all possible positions in the game.)
At any point of the game, the state is given by position1, position2, and moveCount. Following the hint, we make a table Winner indexed by all triples (position1,position2,count) of two positions and a move count. We first set Winner(x,y,n) to "player2" for all x and y. Then we set Winner(x,y,i) to "player1" for all triples where (x,y) is in V and i < n.
Now we have to fill in the other entries in the table. Starting with moveCount n-1 and working downward, we examine each position (x,y,i) that is not yet filled. If i is even, we look at all triples (z,y,i+1) where (x,z) is an edge in G. We set Winner(x,y,i) to "player1" if at least one of these is labeled "player1" or to "player2" if all the successors are labeled "player2". If i is odd, we set Winner(x,y,i) to "player2" if there is a triple (x,z,i+1) labeled "player2" and (y,z) an edge in G.
Once we have filled in the entire table, our output is Winner(x,y,0) for the particular x and y given by the input.
• Question 4 (15): Consider the following approximation algorithm for the VERTEX-COVER problem.
``````
Input: undirected graph G
for all nodes v
{mark(v) = false;}
edgeSet = empty set;
while there is an edge (x,y) with mark(x) = mark(y) = false
{edgeSet = edgeSet union {(x,y)}
mark(x) = true;
mark(y) = true;
for all nodes z adjacent to either x or y
{mark(z) = true;}
}
return edgeSet;
``````
Explain why this nondetermistic algorithm must produce a vertex cover. What is the approximation ratio of this algorithm for the MIN-VERTEX-COVER optimization problem? (It said "MAX-VERTEX-COVER" on the assignment, sorry.)
I'm afraid that I completely messed up this problem and I'll have to withdraw it. The algorithm above produces a set of edges, but a vertex cover is a set of vertices. It basically doesn't have too much to do with the vertex cover problem at all!
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29 January 2018
# Here is a somewhat trickier puzzle – Catching a mouse
There are five adjoining rooms – linearly arranged (from left to right). You know there is a mouse in one of those rooms. You also know that the mouse, every night moves to the adjoining room – either left or right. It does not stay in the same room two days in succession – meaning it will definitely move. However, it is completely random whether it will go left or right – unless, of course it is in one of the end rooms in which case, it has to go to the other side.
Now, you are allowed to open any one random room in the morning. If the mouse is there, you can trap it and catch it. But if you do not find it, you are not allowed to go to other rooms. You have to come back the next day and open any room you want again (including the one you opened to today – totally your choice).
What is the strategy you will use (of targeting rooms to open each morning) to eventually catch the mouse?
Posted January 29, 2018 by Rajib Roy in category "Puzzles
1. By Ambarish Mitra on
I think the best strategy is to choose one room and open that room everyday. The room in the middle can be chosen for the heck of choosing.
The probability that the mouse is in a given room is 0.2 everyday.
Works? or no?
1. By Ambarish Mitra on
The same problem if you choose any room. Here, in a probabilistic model, with zero previous knowledge, i think the strategy is to fix on a room is the best because the randomness is on the mouse movement and we reduce the randomness on our choice.
If it were a deterministic model, then the strategy would be entirely different. In a probablistic model, the expectation (E=Probability X occurance) is that the mouse will be caught in 3rd day. Can the mouse oscillate between 1 and 2 (or 4 and 5) till infinity? Sure, it is possible. But not probable.
2. By Ambarish Mitra on
Open door 2 on 2 consecutive mornings. If not found, then definitely the mouse is not in room1. Open door 4 next 2 mornings. If not found, then definitely not in 5. So, remaining rooms, 2, 3 and 4. Room 3 on 2 consecutive mornings will give the catch.
3. By Rajib Roy on
Ambarish not quite. The days might be getting mixed up. By hitting room 2 successively, you made sure that if the mouse was in Room 1 or 2 on DAY 1, you got it. Now doing the same with room 4 on day 3 and 4, you made sure that you caught it if it was in room 4 or 5 in DAY 3. Not DAY 1.
An example to show the algo not working
day 1 room 5 you open 2
Day 2 room 4 you open 2
Day 3 room 3 you open 4
Day 4 room 2 you open 4
Day 5 room 1 you open 3
Day 6 room 2 you open 3
2. By Ramesh Mantha on
if I am given very little time to make a decision, Id go with my gut instinct ,and choose the center room and keep opening it until I find the mouse. But the correct answer (and I admit I dont know yet) I think may be deduced by maximizing the probability of encountering the mouse which itself looks like sum of a series of decreasing probabilities. Reminds me that Bayes Theorem needs to be applied and I could never do it right. But this strikes close to home because we have a mice problem and I constantly trying to outsmart mice with various humane traps and strategies which the critters quickly learn to avoid.
3. By Debajyoti Roy on
Opening the doors in the the sequence of 2-3-4-2-3-4…. I can write up the dry run, am i in the right direction to catch it? 🙂
1. By Rajib Roy on
First I thought it won’t work. But actually it will. In fact, yours is a better solution than the one I had come up with. Yours has one step shorter than mine.
2. By Rajib Roy on
Debajyoti, I will post the generic solution for “n” rooms later. I am still cross-checking after applying your method which cuts out my first step.
3. By Rajib Roy on
Debajyoti, I am pretty sure it is 2,3,4…. n-1 and then repeat that with the possibility of having to have a gap day depending on whether n is odd or even.
4. By Rajib Roy on
Pradeep has come up with an alternate solution. Think about it as a symmetric problem for the second iteration- then there are multiple answers possible. In my case, I started with Room 1 and then 2..3… 4 and then iterate one more time after waiting out a day. Your solution is shorter by that extra step not required.
5. By Ramesh Mantha on
In my opinion, there is a chance the mouse is never caught no matter how many times you open the doors. Based on this discussion I see 2 variables that need to be optimized 1. number of days 2. probability of catching the mouse [since it is never going to be 1]. There is no one single solution I think. Thoughts?
6. By Rajib Roy on
I had come up with the strategy of 1,2,3,4… n-1 and then start over again (with the possibility of having to wait out one day depending on whether n is odd or even) and by the time you get to n-1 the second time, you would have caught the mouse.
But Debajyoti and Pradeep had a shorter solution. They realized that I do not need to do 1 in the beginning at all. So, I am going to explain their answer.
I will answer in the generic case “n” rooms and then show the specific case here with 5 rooms.
They said if you do 2,3,4,2,3,4 you will catch the mouse. (worst case scenario 6 days to catch the mouse). So, why does this solution work?
Realize that there is very little you can say about the mouse’s movement. We do not know where it started and which direction it would take any given day.
The ONLY KNOWN FACT is this – if it is in a ODD room today, tomorrow it will be in an EVEN room and vice versa. And actually, that is enough knowledge to outwit the mouse!!
On Day 1, the mouse can either be in an ODD room {1,3,5,7,9….} or in an EVEN room {2,4,6,8…}. (In our problem of course this is {1,3,5} and {2,4} )
EVEN CASE. Let’s start with the assumption the mouse was in an EVEN room. (We will do the ODD CASE later; there are no other cases possible)
On Day 1, we opened “2”. The mouse could have been in 2 or 4 or 6 or 8 (remember we have assumed it was in an even room). Well, if it was in 2 to begin with, we got it. If not, it is obviously in 4, 6, 8 …. .
On Day 2, the mouse has to be in one of the odd rooms 3, 5, 7, … (but not 1 since it cannot jump to 1 from any of the possible rooms 4, 6, 8 …..). On that day, we opened Room 3. If it was there, we got it. If not, obviously it was hiding in one of the rooms 5,7,9….
Day 3: If we did not get it on Day 2, where could the mouse go the next day? Remember, it was in 5 or 7 or 9…. So, it can go to one of the even rooms 4,6,8,…. Guess where we are scheduled to go on Day 3? Room 4. If it was there, we got it. If not…. it means it is in 6,8,10….
Next day it can go to 5,7,9…. And that is the day we attack room 5…
You see how every single day, we are flushing the mouse to the right till we reach room n-1 (which would be 4 in our specific problem).
On that day we either got the mouse or we conclude that our original assumption that it was in an EVEN room was WRONG.
ODD CASE: The mouse was on in an ODD room on day 1
So, what do we have now. We have a fool proof way to catch the mouse if the mouse was in an even room on Day 1. But now, we know that it was actually in an odd room.
Do we know where the mouse is today – even room or odd room? We do!! On day 1 it was in an odd room (or else by now we would have caught it) and we know how many days have passed from that day to today. So we know for sure whether the mouse is in an odd room or even room TODAY.
If TODAY is EVEN, guess what – you do what you did in EVEN CASE above because you have a foolproof way of catching it once you know it is in an EVEN room.
If TODAY is ODD? We wait for a day. Perhaps smoke a Charminar like Ambarish suggested. Tomorrow you are guaranteed that the mouse will be in an even room. So you run your EVEN CASE as above now 2,3,4…. By the time you reach n-1 you are guaranteed to catch it!!!!
Absolutely foolproof.
Now, you probably realize that the original problem and also after the first flush, it is a symmetrical case (which end you start from never matters)
Therefore, you can extend the original solution 2,3,4,2,3,4 to the following solutions too:
2,3,4,4,3,2
4,3,2,4,3,2
4,3,2,2,3,4
So the answer is… you go 2,3,4,5… to n-1. If you still do not have it, if “n” is odd, then do 2,3,4,5… to n-1 again. If “n” is even, wait a day and then do 2,3,4,5… to n-1. The well laid plans of mice will be undone by those laid by men. Shakespeare be darned!!
1. By Ramesh Mantha on
very detailed explanation but I am seeing an issue in your logic above.. flushing is not possible because mouse can evade capture forever [hypothetically: if it knows which door you are opening in advance] This means, we cannot with certainty ever catch the mouse (key word being “certainty”). For example.. as you move 3 to 4.. the mouse can move from 4 to 3 and oscillate as you try to flush it at 5,6,7..n. Thoughts?
2. By Rajib Roy on
No he cannot. Let’s take the example you cite. I was in 3.
In EVEN CASE, it was on an even room on the first day when I hit 2. But now I am in 3. The mouse got one day to move. So, it is in an odd room. It CANNOT be in room 4 as you have suggested.
But of course, you can say that maybe it was not EVEN CASE. Which means it was the ODD CASE (one day 1 it was in an odd room) Indeed, then, I will miss the mouse in this round and go all the way to 4.
So on day 1, it was in an odd room, I hit 2
day 2, it was in an even room, I hit 3
day 3, it was in an odd room, I hit 4. (I did not find it. I realize it is the ODD CASE)
Note that on the next day – day 4, the mouse HAS to be in an even room (prev day it was in an odd room).
So, I simply do the EVEN CASE – 2,3,4 again.
Makes sense?
7. By rajibroy (Post author) on
I had come up with the strategy of 1,2,3,4… n-1 and then start over again (with the possibility of having to wait out one day depending on whether n is odd or even) and by the time you get to n-1 the second time, you would have caught the mouse.
But Debajyoti and Pradeep had a shorter solution. They realized that I do not need to do 1 in the beginning at all. So, I am going to explain their answer.
I will answer in the generic case “n” rooms and then show the specific case here with 5 rooms.
They said if you do 2,3,4,2,3,4 you will catch the mouse. (worst case scenario 6 days to catch the mouse). So, why does this solution work?
Realize that there is very little you can say about the mouse’s movement. We do not know where it started and which direction it would take any given day.
The ONLY KNOWN FACT is this – if it is in a ODD room today, tomorrow it will be in an EVEN room and vice versa. And actually, that is enough knowledge to outwit the mouse!!
On Day 1, the mouse can either be in an ODD room {1,3,5,7,9….} or in an EVEN room {2,4,6,8…}. (In our problem of course this is {1,3,5} and {2,4} )
EVEN CASE. Let’s start with the assumption the mouse was in an EVEN room. (We will do the ODD CASE later; there are no other cases possible)
On Day 1, we opened “2”. The mouse could have been in 2 or 4 or 6 or 8 (remember we have assumed it was in an even room). Well, if it was in 2 to begin with, we got it. If not, it is obviously in 4, 6, 8 …. .
On Day 2, the mouse has to be in one of the odd rooms 3, 5, 7, … (but not 1 since it cannot jump to 1 from any of the possible rooms 4, 6, 8 …..). On that day, we opened Room 3. If it was there, we got it. If not, obviously it was hiding in one of the rooms 5,7,9….
Day 3: If we did not get it on Day 2, where could the mouse go the next day? Remember, it was in 5 or 7 or 9…. So, it can go to one of the even rooms 4,6,8,…. Guess where we are scheduled to go on Day 3? Room 4. If it was there, we got it. If not…. it means it is in 6,8,10….
Next day it can go to 5,7,9…. And that is the day we attack room 5…
You see how every single day, we are flushing the mouse to the right till we reach room n-1 (which would be 4 in our specific problem).
On that day we either got the mouse or we conclude that our original assumption that it was in an EVEN room was WRONG.
ODD CASE: The mouse was on in an ODD room on day 1
So, what do we have now. We have a fool proof way to catch the mouse if the mouse was in an even room on Day 1. But now, we know that it was actually in an odd room.
Do we know where the mouse is today – even room or odd room? We do!! On day 1 it was in an odd room (or else by now we would have caught it) and we know how many days have passed from that day to today. So we know for sure whether the mouse is in an odd room or even room TODAY.
If TODAY is EVEN, guess what – you do what you did in EVEN CASE above because you have a foolproof way of catching it once you know it is in an EVEN room.
If TODAY is ODD? We wait for a day. Perhaps smoke a Charminar Ambarish suggested. Tomorrow you are guaranteed that the mouse will be in an even room. So you run your EVEN CASE as above now 2,3,4…. By the time you reach n-1 you are guaranteed to catch it!!!!
Absolutely foolproof.
Now, you probably realize that the original problem and also after the first flush, it is a symmetrical case (which end you start from never matters)
Therefore, you can extend the original solution 2,3,4,2,3,4 to the following solutions too:
2,3,4,4,3,2
4,3,2,4,3,2
4,3,2,2,3,4
So the answer is… you go 2,3,4,5… to n-1. If you still do not have it, if “n” is odd, then do 2,3,4,5… to n-1 again. If “n” is even, wait a day and then do 2,3,4,5… to n-1. The well laid plans of mice will be undone by those laid by men. Shakespeare be darned!!
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# Long and short scales
(Redirected from Long and short scale)
The long and short scales are two of several naming systems for integer powers of ten which use some of the same terms for different magnitudes.
For whole numbers smaller than 1,000,000,000 (109), such as one thousand or one million, the two scales are identical. For larger numbers, starting with 109, the two systems differ. For identical names, the long scale proceeds by powers of one million, whereas the short scale proceeds by powers of one thousand. For example, in the short scale, "one billion" means one thousand millions (1,000,000,000), whereas in the long scale, it means one million millions (1,000,000,000,000). For interleaved values, the long scale system employs additional terms, typically substituting the word ending -ion for -iard.
Countries with usage of the long scale include most countries in continental Europe and most that are French-speaking, German-speaking, Spanish-speaking[1] and Portuguese-speaking countries (except Brazil).
To avoid confusion resulting from the coexistence of the two terms, the International System of Units (SI) recommends using the metric prefix to indicate orders of magnitude.
## Scales
In both short and long scale naming, names are given each multiplication step for increments of the base-10 exponent of three, i.e. for each integer n in the sequence of multipliers 103n. For certain multipliers, including those for all numbers smaller than 109, both systems use the same names. The differences arise from the assignment of identical names to specific values of n, for numbers starting with 109, for which n=3. In the short scale system, the identical names are for n=3, 4, 5, ..., while the long scale places them at n=4, 6, 8, etc.
### Short scale
In the short scale, billion means a thousand millions (1,000,000,000 which is 109), trillion means one thousand billions (1012), and so on. Thus, an n-illion equals 103n+3.[2]
### Long scale
In the long scale, billion means one million millions (1012) and trillion means one million billions (1018), and so on. Therefore, an n-illion equals 106n.[3][4] In some languages, the long scale uses additional names for the interleaving multipliers, replacing the ending -ion with -iard; for example, the next multiplier after million is milliard, after billion it is billiard. Hence, an n-iard equals 106n+3.
## Comparison
The relationship between the numeric values and the corresponding names in the two scales can be described as:
Value in positional notation Value in
scientific notation
Metric prefix Short scale Long scale
Prefix Symbol Name Logic Name Alternative name Logic
1 100 one one
10 101 deca da ten ten
100 102 hecto h hundred hundred
1,000 103 kilo k thousand thousand
1,000,000 106 mega M million 1,000 × 1,0001 million 1,000,0001
1,000,000,000 109 giga G billion 1,000 × 1,0002 milliard thousand million 1,000 × 1,000,0001
1,000,000,000,000 1012 tera T trillion 1,000 × 1,0003 billion 1,000,0002
1,000,000,000,000,000 1015 peta P quadrillion 1,000 × 1,0004 billiard thousand billion 1,000 × 1,000,0002
1,000,000,000,000,000,000 1018 exa E quintillion 1,000 × 1,0005 trillion 1,000,0003
1,000,000,000,000,000,000,000 1021 zetta Z sextillion 1,000 × 1,0006 trilliard thousand trillion 1,000 × 1,000,0003
1,000,000,000,000,000,000,000,000 1024 yotta Y septillion 1,000 × 1,0007 quadrillion 1,000,0004
1,000,000,000,000,000,000,000,000,000 1027 ronna R octillion 1,000 × 1,0008 quadrilliard thousand quadrillion 1,000 × 1,000,0004
1,000,000,000,000,000,000,000,000,000,000 1030 quetta Q nonillion 1,000 × 1,0009 quintillion 1,000,0005
The relationship between the names and the corresponding numeric values in the two scales can be described as:
Name Short scale Long scale
Value in
scientific notation
Metric prefix Logic Value in
scientific notation
Metric prefix Logic
Prefix Symbol Prefix Symbol
million 106 mega M 1,000 × 1,0001 106 mega M 1,000,0001
billion 109 giga G 1,000 × 1,0002 1012 tera T 1,000,0002
trillion 1012 tera T 1,000 × 1,0003 1018 exa E 1,000,0003
quadrillion 1015 peta P 1,000 × 1,0004 1024 yotta Y 1,000,0004
quintillion 1018 exa E 1,000 × 1,0005 1030 quetta Q 1,000,0005
etc. For the next order of magnitude, multiply by 1,000 For the next order of magnitude, multiply by 1,000,000
The root mil in million does not refer to the numeral, 1. The word, million, derives from the Old French, milion, from the earlier Old Italian, milione, an intensification of the Latin word, mille, a thousand. That is, a million is a big thousand, much as a great gross is a dozen gross or 12 × 144 = 1728.[5]
The word milliard, or its translation, is found in many European languages and is used in those languages for 109. However, it is not found in American English, which uses billion, and not used in British English, which preferred to use thousand million before the current usage of billion. The financial term yard, which derives from milliard, is used on financial markets, as, unlike the term billion, it is internationally unambiguous and phonetically distinct from million. Likewise, many long scale countries use the word billiard (or similar) for one thousand long scale billions (i.e., 1015), and the word trilliard (or similar) for one thousand long scale trillions (i.e., 1021), etc.[6][7][8][9][10]
## History
Some languages, particularly in East Asia and South Asia, have large number naming systems that are different from both the long and short scales, such as Chinese, Japanese or Korean numerals, and the Indian numbering system.[3][4]
Much of the remainder of the world adopted either the Short Scale or the Long Scale for everyday counting powers of ten. Although this situation has been developing since the 1200s, the first recorded use of the terms short scale (French: échelle courte) and long scale (French: échelle longue) was by the French mathematician Geneviève Guitel in 1975.[3][4]
The short scale was never widespread before its universal adoption in the United States. It has been taught in American schools since the early 1800s.[5] It has since become common in other English-speaking nations and several other countries. For most of the 19th and 20th centuries, the United Kingdom largely used the long scale,[2][11] whereas the United States used the short scale,[11] so that the two systems were often referred to as British and American in the English language. After several decades of increasing informal British usage of the short scale, in 1974 the government of the UK adopted it,[12] and it is used for all official purposes.[13][14][15][16][17][18] The British usage and American usage are now identical.
The existence of the different scales means that care must be taken when comparing large numbers between languages or countries, or when interpreting old documents in countries where the dominant scale has changed over time. For example, British English, French, and Italian historical documents can refer to either the short or long scale, depending on the date of the document, since each of the three countries has used both systems at various times in its history. Today, the United Kingdom officially uses the short scale, but France and Italy use the long scale.
The pre-1974 former British English word billion, post-1961 current French word billion, post-1994 current Italian word bilione, Spanish billón, German Billion, Dutch biljoen, Danish billion, Swedish biljon, Finnish biljoona, Slovenian bilijon, Polish bilion, and European Portuguese word bilião (with a different spelling to the Brazilian Portuguese variant, but in Brazil referring to short scale) all refer to 1012, being long-scale terms. Therefore, each of these words translates to the American English or post-1974 British English word: trillion (1012 in the short scale), and not billion (109 in the short scale).
On the other hand, the pre-1961 former French word billion, pre-1994 former Italian word bilione, Brazilian Portuguese word bilhão, and Welsh word biliwn all refer to 109, being short scale terms. Each of these words translates to the American English or post-1974 British English word billion (109 in the short scale).
The term billion originally meant 1012 when introduced.[5]
• In long scale countries, milliard was defined to its current value of 109, leaving billion at its original 1012 value and so on for the larger numbers.[5] Some of these countries, but not all, introduced new words billiard, trilliard, etc. as intermediate terms.[6][7][8][9][10]
• In some short scale countries, milliard was defined to 109 and billion dropped altogether, with trillion redefined down to 1012 and so on for the larger numbers.[5]
• In many short scale countries, milliard was dropped altogether and billion was redefined down to 109, adjusting downwards the value of trillion and all the larger numbers.
Timeline
Date Event
13th century The word million was not used in any language before the 13th century. Maximus Planudes (c. 1260–1305) was among the first recorded users.[5]
Late 14th century
Piers Plowman, a 17th-century copy of the original 14th-century allegorical narrative poem by William Langland
The word million entered the English language. One of the earliest references is William Langland's Piers Plowman (written c. 1360–1387 in Middle English),[5] with
Coueyte not his goodes
For millions of moneye
Translation:
Covet not his goods
for millions of money
1475 French mathematician Jehan Adam, writing in Middle French, recorded the words bymillion and trimillion as meaning 1012 and 1018 respectively in a manuscript Traicté en arismetique pour la practique par gectouers, now held in the Bibliothèque Sainte-Geneviève in Paris.[19][20][21]
... item noctes que le premier greton dembas vault ung, le second vault dix, le trois vault cent, le quart vult [sic] mille, le Ve vault dix M, le VIe vault cent M, le VIIe vault Milion, Le VIIIe vault dix Million, Le IXe vault cent Millions, Le Xe vault Mil Millions, Le XIe vault dix mil Millions, Le XIIe vault Cent mil Millions, Le XIIIe vault bymillion, Le XIIIIe vault dix bymillions, Le XVe vault cent mil [sic] bymillions, Le XVIe vault mil bymillions, Le XVIIe vault dix Mil bymillions, Le XVIIIe vault cent mil bymillions, Le XIXe vault trimillion, Le XXe vault dix trimillions ...
Translation:
... Item note that the first counter from the bottom is worth one, the 2nd is worth ten, the 3rd is worth one hundred, the 4th is worth one thousand, the 5th is worth ten thousand, the 6th is worth one hundred thousand, the 7th is worth a million, the 8th is worth ten millions, the 9th is worth one hundred millions, the 10th is worth one thousand millions, the 11th is worth ten thousand millions, the 12th is worth one hundred thousand million, the 13th is worth a bymillion, the 14th is worth ten bymillions, the 15th is worth one [hundred] bymillions, the 16th is worth one thousand bymillions, the 17th is worth ten thousand bymillions, the 18th is worth hundred thousand bymillions, the 19th is worth a trimillion, the 20th is worth ten trimillions ...
1484
Le Triparty en la Science des Nombres par Maistre Nicolas Chuquet Parisien
an extract from Chuquet's original 1484 manuscript
French mathematician Nicolas Chuquet, in his article Le Triparty en la Science des Nombres par Maistre Nicolas Chuquet Parisien,[22][23][24] used the words byllion, tryllion, quadrillion, quyllion, sixlion, septyllion, ottyllion, and nonyllion to refer to 1012, 1018, ... 1054. Most of the work was copied without attribution by Estienne de La Roche and published in his 1520 book, L'arismetique.[22] Chuquet's original article was rediscovered in the 1870s and then published for the first time in 1880.
... Item lon doit savoir que ung million vault
mille milliers de unitez, et ung byllion vault mille
milliers de millions, et [ung] tryllion vault mille milliers
de byllions, et ung quadrillion vault mille milliers de
tryllions et ainsi des aultres : Et de ce en est pose ung
exemple nombre divise et punctoye ainsi que devant est
dit, tout lequel nombre monte 745324 tryllions
804300 byllions 700023 millions 654321.
Exemple : 745324'8043000'700023'654321 ...
[sic]
Translation:
...Item: one should know that a million is worth
a thousand thousand units, and a byllion is worth a thousand
thousand millions, and tryllion is worth a thousand thousand
byllions, and a quadrillion is worth a thousand thousand
tryllions, and so on for the others. And an example of this follows,
a number divided up and punctuated as previously
described, the whole number being 745324 tryllions,
804300 byllions 700023 millions 654321.
Example: 745324'8043000'700023'654321 ... [sic]
The extract from Chuquet's manuscript, the transcription and translation provided here all contain an original mistake: one too many zeros in the 804300 portion of the fully written out example: 745324'8043000 '700023'654321 ...
1516 French mathematician Budaeus (Guillaume Budé), writing in Latin, used the term milliart to mean "ten myriad myriad" or 109 in his book De Asse et partibus eius Libri quinque.[25]
.. hoc est decem myriadum myriadas:quod vno verbo nostrates abaci studiosi Milliartum appellant:quasi millionum millionem
Translation:
.. this is ten myriad myriads, which in one word our students of numbers call Milliart, as if a million millions
1549 The influential French mathematician Jacques Pelletier du Mans used the name milliard (or milliart) to mean 1012, attributing the term to the earlier usage by Guillaume Budé[25]
17th century With the increased usage of large numbers, the traditional punctuation of large numbers into six-digit groups evolved into three-digit group punctuation. In some places, the large number names were then applied to the smaller numbers, following the new punctuation scheme. Thus, in France and Italy, some scientists then began using billion to mean 109, trillion to mean 1012, etc.[26] This usage formed the origins of the later short scale. The majority of scientists either continued to say thousand million or changed the meaning of the Pelletier term, milliard, from "million of millions" down to "thousand million".[5] This meaning of milliard has been occasionally used in England,[11] but was widely adopted in France, Germany, Italy and the rest of Europe, for those keeping the original long scale billion from Adam, Chuquet and Pelletier.
1676 The first published use of milliard as 109 occurred in the Netherlands.[5][27]
.. milliart/ofte duysent millioenen..
Translation:
..milliart / also thousand millions..
1729 The short-scale meaning of the term billion had already been brought to the British American colonies. The first American appearance of the short scale value of billion as 109 was published in the Greenwood Book of 1729, written anonymously by Prof. Isaac Greenwood of Harvard College[5]
Late 18th century As early as 1762 (and through at least the early 20th century), the dictionary of the Académie française defined billion as a term of arithmetic meaning a thousand millions.[28][29][30][31]
Early 19th century France widely converted to the short scale, and was followed by the U.S., which began teaching it in schools. Many French encyclopedias of the 19th century either omitted the long scale system or called it "désormais obsolète", a now obsolete system. Nevertheless, by the mid 20th century France would officially convert back to the long scale.
1926
H. W. Fowler's A Dictionary of Modern English Usage[11] noted
It should be remembered that "billion" does not mean in American use (which follows the French) what it means in British. For to us it means the second power of a million, i.e. a million millions (1,000,000,000,000); for Americans it means a thousand multiplied by itself twice, or a thousand millions (1,000,000,000), what we call a milliard. Since billion in our sense is useless except to astronomers, it is a pity that we do not conform.
Although American English usage did not change, within the next 50 years French usage changed from short scale to long and British English usage changed from long scale to short.
1948 The 9th General Conference on Weights and Measures received requests to establish an International System of Units. One such request was accompanied by a draft French Government discussion paper, which included a suggestion of universal use of the long scale, inviting the short-scale countries to return or convert.[32] This paper was widely distributed as the basis for further discussion. The matter of the International System of Units was eventually resolved at the 11th General Conference in 1960. The question of long scale versus short scale was not resolved and does not appear in the list of any conference resolutions.[32][33]
1960 The 11th General Conference on Weights and Measures adopted the International System of Units (SI), with its own set of numeric prefixes.[34] SI is therefore independent of the number scale being used. SI also notes the language-dependence of some larger-number names and advises against using ambiguous terms such as billion, trillion, etc.[35] The National Institute of Standards and Technology within the US also considers that it is best that they be avoided entirely.[36]
1961 The French Government confirmed their official usage of the long scale in the Journal officiel (the official French Government gazette).[37]
1974 British prime minister Harold Wilson explained in a written answer to the House of Commons that UK government statistics would from then on use the short scale.[13] Hansard,[12] for 20 December 1974, reported it
Mr. Maxwell-Hyslop asked the Prime Minister whether he would make it the practice of his administration that when Ministers employ the word 'billion' in any official speeches, documents, or answers to Parliamentary Questions, they will, to avoid confusion, only do so in its British meaning of 1 million million and not in the sense in which it is used in the United States of America, which uses the term 'billion' to mean 1,000 million.
The Prime Minister: No. The word 'billion' is now used internationally to mean 1,000 million and it would be confusing if British Ministers were to use it in any other sense. I accept that it could still be interpreted in this country as 1 million million and I shall ask my colleagues to ensure that, if they do use it, there should be no ambiguity as to its meaning.
The BBC and other UK mass media quickly followed the government's lead within the UK.
During the last quarter of the 20th century, most other English-speaking countries (Ireland, Australia, New Zealand, South Africa, Zimbabwe, etc.) either also followed this lead or independently switched to the short scale use. However, in most of these countries, some limited long scale use persists and the official status of the short scale use is not clear.
1975 French mathematician Geneviève Guitel introduced the terms long scale (French: échelle longue) and short scale (French: échelle courte) to refer to the two numbering systems.[3][4]
1994 The Italian Government confirmed their official usage of the long scale.[10]
As large numbers in natural sciences are usually represented by metric prefixes, scientific notation or otherwise, the most commonplace occurrence of large numbers represented by long or short scale terms is in finance. The following table includes some historic examples related to hyper-inflation and other financial incidents.
Timeline
Date Event
1923
10 Milliarden Mark (1010 mark) stamp
1000 Mark German banknote, over-stamped in red with "Eine Milliarde Mark" (109 mark)
Using German banknotes as wallpaper following the 1923 hyperinflation
German hyperinflation in the 1920s Weimar Republic caused 'Eintausend Mark' (1000 Mark = 103 Mark) German banknotes to be over-stamped as 'Eine Milliarde Mark' (109 Mark). This introduced large-number names to the German populace.
The Mark or Papiermark was replaced at the end of 1923 by the Rentenmark at an exchange rate of
1 Rentenmark = 1 billion (long scale) Papiermark = 1012 Papiermark = 1 trillion (short scale) Papiermark
1946
1020 Hungarian pengő banknote issued in 1946
Hyperinflation in Hungary in 1946 led to the introduction of the 1020 pengő banknote.
100 million b-pengő (long scale) = 100 trillion (long scale) pengő = 1020 pengő = 100 quintillion (short scale) pengő.
On 1 August 1946, the forint was introduced at a rate of
1 forint = 400 quadrilliard (long scale) pengő = 4 × 1029 pengő = 400 octillion (short scale) pengő.
1993
5 × 1011 Yugoslav dinar banknotes from 1993
Hyperinflation in Yugoslavia led to the introduction of 5 × 1011 dinar banknotes.
500 thousand million (long scale) dinars = 5 × 1011 dinar banknotes = 500 billion (short scale) dinars.
The later introduction of the new dinar came at an exchange rate of
1 new dinar = 1 × 1027 dinars = ~1.3 × 1027 pre 1990 dinars.
2009
1014 Zimbabwean dollars banknote from 2009
Hyperinflation in Zimbabwe led to banknotes of 1014 Zimbabwean dollars, marked "One Hundred Trillion Dollars" (short scale), being issued in 2009, shortly ahead of the currency being abandoned[38][39][40] after a final redenomination to the 'fourth dollar'. From 2013 to 2019 when the RTGS Dollar entered use, no new currency was announced, and so foreign currencies were used instead.
100 trillion (short scale) Zimbabwean dollars = 1014 Zimbabwean dollars = 100 billion (long scale) Zimbabwean dollars = 1027 pre-2006 Zimbabwean dollars = 1 quadrilliard (long scale) pre-2006 Zimbabwean dollars.
2022 As of 24 November 2022, the combined total public debt of the United States stood at \$31.299 trillion.[41]
31 trillion (short scale) US Dollars = 3.1 × 1013 US Dollars = 31 billion (long scale) US Dollars
## Current usage
Short and long scale usage throughout the world
Long scale
Short scale
Short scale with milliard instead of billion
Both scales
Other naming system
No data
### Short scale users
#### English-speaking
106, one million; 109, one billion; 1012, one trillion; etc.
Most English-language countries and regions use the short scale with 109 being billion. For example:[shortscale note 1]
#### Arabic-speaking
106, مَلْيُوْن malyoon; 109, مِلْيَار milyar; 1012, تِرِلْيُوْن tirilyoon; etc.
Most Arabic-language countries and regions use the short scale with 109 being مليار milyar, except for a few countries like Saudi Arabia and the UAE which use the word بليون billion for 109. For example:[shortscale note 6][46][47]
#### Other short scale
106, one million; 109, one milliard or one billion; 1012, one trillion; etc.
Other countries also use a word similar to trillion to mean 1012, etc. Whilst a few of these countries like English use a word similar to billion to mean 109, most like Arabic have kept a traditional long scale word similar to milliard for 109. Some examples of short scale use, and the words used for 109 and 1012, are
### Long scale users
The traditional long scale is used by most Continental European countries and by most other countries whose languages derive from Continental Europe (with the notable exceptions of Albania, Greece, Romania,[53] and Brazil). These countries use a word similar to billion to mean 1012. Some use a word similar to milliard to mean 109, while others use a word or phrase equivalent to thousand millions.
#### Dutch-speaking
106, miljoen; 109, miljard; 1012, biljoen; etc.
Most Dutch-language countries and regions use the long scale with 109 = miljard, for example:[55][56]
#### French-speaking
106, million; 109, milliard; 1012, billion; etc.
Most French-language countries and regions use the long scale with 109 = milliard, for example:[longscale note 1][57][58]
#### German-speaking
106, Million; 109, Milliarde; 1012, Billion; etc.
German-language countries and regions use the long scale with 109 = Milliarde, for example:
#### Portuguese-speaking
106, milhão; 109, mil milhões or milhar de milhões; 1012, bilião
With the notable exception of Brazil, a short scale country, most Portuguese-language countries and regions use the long scale with 109 = mil milhões or milhar de milhões, for example:
#### Spanish-speaking
106, millón; 109, mil millones or millardo; 1012, billón; etc.
Most Spanish-language countries and regions use the long scale, for example:[longscale note 2][60][61]
#### Other long scale
106, one million; 109, one milliard or one thousand million; 1012, one billion; etc.
Some examples of long scale use, and the words used for 109 and 1012, are
### Using both
Some countries use either the short or long scales, depending on the internal language being used or the context.
106, one million; 109, either one billion (short scale) or one milliard / thousand million (long scale); 1012, either one trillion (short scale) or one billion (long scale), etc.
Country or territory Short scale usage Long scale usage
Canada[shortscale longscale note 1] Canadian English (109 = billion, 1012 = trillion) Canadian French (109 = milliard, 1012 = billion[69] or mille milliards).
Mauritius; Seychelles; Vanuatu English (109 = billion, 1012 = trillion) French (109 = milliard, 1012 = billion)
South African English (109 = billion, 1012 = trillion) Afrikaans (109 = miljard, 1012 = biljoen)
Puerto Rico Economic and technical (109 = billón, 1012 = trillón) Latin American export publications (109 = millardo or mil millones, 1012 = billón)
### Using neither
The following countries use naming systems for large numbers that are not etymologically related to the short and long scales:
Country Number system Naming of large numbers
Bangladesh, India, Maldives, Nepal, Pakistan Indian numbering system Traditional system for everyday use, but short or long scale may also be in use [other scale note 1]
East Asian numbering system:
Traditional myriad system for the larger numbers; special words and symbols up to 1068
Greece Calque of the short scale Names of the short scale have not been loaned but calqued into Greek, based on the native Greek word for million, εκατομμύριο ekatommyrio ("hundred-myriad", i.e. 100 × 10,000):
• δισεκατομμύριο disekatommyrio "bi+hundred-myriad" = 109 (short scale billion)
• τρισεκατομμύριο trisekatommyrio "tri+hundred-myriad" = 1012 (short scale trillion)
Mongolia Mongolian numerals Traditional myriad system for the larger numbers; special words up to 1067
Sri Lanka
Thailand Thai numerals Traditional system based on millions
Vietnam Vietnamese numerals Traditional system(s) based on thousands
### By continent
The long and short scales are both present on most continents, with usage dependent on the language used. Examples include:
Continent Short scale usage Long scale usage
Africa Arabic (Egypt, Libya, Tunisia), English (South Sudan), South African English Afrikaans, French (Benin, Central African Republic, Gabon, Guinea), Portuguese (Angola, Mozambique)
North America American English, Canadian English U.S. Spanish, Canadian French, Mexican Spanish
South America Brazilian Portuguese, English (Guyana) American Spanish, Dutch (Suriname), French (French Guiana)
Antarctica Australian English, British English, New Zealand English, Russian American Spanish (Argentina, Chile), French (France), Norwegian (Norway)
Asia Burmese (Myanmar), Hebrew (Israel), Indonesian, Malaysian English, Philippine English, Kazakh, Uzbek, Kyrgyz Portuguese (East Timor, Macau), Persian (Iran)
Europe British English, Welsh, Estonian, Greek, Latvian, Lithuanian, Russian, Turkish, Ukrainian Danish, Dutch, Finnish, French, German, Icelandic, Italian, Norwegian, Polish, Portuguese, Romanian, Spanish, Swedish and most other languages of continental Europe
Oceania Australian English, New Zealand English French (French Polynesia, New Caledonia)
### Notes on current usage
#### Short scale
1. ^ English language countries: Apart from the United States, the long scale was used for centuries in many English language countries before being superseded in recent times by short scale usage. Because of this history, some long scale use persists[18] and the official status of the short scale in anglophone countries other than the UK and US is sometimes obscure.[citation needed]
2. ^ Australian usage: In Australia, education, media outlets, and literature all use the short scale in line with other English-speaking countries. The current recommendation by the Australian Government Department of Finance and Deregulation (formerly known as AusInfo), and the legal definition, is the short scale.[42] As recently as 1999, the same department did not consider short scale to be standard, but only used it occasionally. Some documents use the term thousand million for 109 in cases where two amounts are being compared using a common unit of one 'million'.
3. ^ Filipino usage: Some short-scale words have been adopted into Filipino.
4. ^ British usage: Billion has meant 109 in most sectors of official published writing for many years now. The UK government, the BBC, and most other broadcast or published mass media, have used the short scale in all contexts since the mid-1970s.[12][13][43][15]
Before the widespread use of billion for 109, UK usage generally referred to thousand million rather than milliard.[16] The long scale term milliard, for 109, is obsolete in British English, though its derivative, yard, is still used as slang in the London money, foreign exchange, and bond markets.
5. ^ American usage: In the United States, the short scale has been taught in school since the early 19th century. It is therefore used exclusively.[44][45]
6. ^ Arabic language countries: Most Arabic-language countries use: 106, مليون million; 109, مليار milyar; 1012, ترليون trilyon; etc.[46][47]
7. ^ Estonian usage: Biljon is used due to English influences and is less common than miljard.
8. ^ Indonesian usage: Large numbers are common in Indonesia, in part because its currency (rupiah) is generally expressed in large numbers (the lowest common circulating denomination is Rp100 with Rp1000 is considered as base unit). The term juta, equivalent to million (106), is generally common in daily life. Indonesia officially employs the term miliar (derived from the long scale Dutch word miljard) for the number 109, with no exception. For 1012 and greater, Indonesia follows the short scale, thus 1012 is named triliun. The term seribu miliar (a thousand milliards) or more rarely sejuta juta (a million millions) or sejuta berkali-kali (a millions after a million or a millions over a million) are also used for 1012 less often. Terms greater than triliun are not very familiar to Indonesians.[52]
#### Long scale
1. ^ French usage: France, with Italy, was one of two European countries which converted from the long scale to the short scale during the 19th century, but returned to the original long scale during the 20th century. In 1961, the French Government confirmed their long scale status.[37][57][58] However the 9th edition of the dictionary of the Académie française describes billion as an outdated synonym of milliard, and says that the new meaning of 1012 was decreed in 1961, but never caught on.[59]
2. ^ Spanish language countries: Spanish-speaking countries sometimes use millardo (milliard)[60] for 109, but mil millones (thousand millions) is used more frequently. The word billón is sometimes used in the short scale sense in those countries more influenced by the United States, where "billion" means "one thousand millions". The usage of billón to mean "one thousand millions", controversial from the start, was denounced by the Royal Spanish Academy as recently as 2010,[61] but was finally accepted in a later version of the official dictionary as standard usage among educated Spanish speakers in the United States (including Puerto Rico).[62]
3. ^ Esperanto language usage: The Esperanto language words biliono, triliono etc. used to be ambiguous, and both long and short scale were used and presented in dictionaries. The current edition of the main Esperanto dictionary PIV however recommends the long scale meanings, as does the grammar PMEG.[63] Ambiguity may be avoided by the use of the unofficial but generally recognised suffix -iliono, whose function is analogous to the long scale, i.e. it is appended to a (single) numeral indicating the power of a million, e.g. duiliono (from du meaning "two") = biliono = 1012, triiliono = triliono = 1018, etc. following the 1×106X long scale convention. Miliardo is an unambiguous term for 109, and generally the suffix -iliardo, for values 1×106X+3, for example triliardo = 1021 and so forth.
4. ^ Italian usage: Italy, with France, was one of the two European countries which partially converted from the long scale to the short scale during the 19th century, but returned to the original long scale in the 20th century. In 1994, the Italian Government confirmed its long scale status.[10] In Italian, the word bilione officially means 1012, trilione means 1018, etc.. Colloquially, bilione[64] can mean both 109 and 1012; trilione[citation needed] can mean both 1012 and (rarer) 1018 and so on. Therefore, in order to avoid ambiguity, they are seldom used. Forms such as miliardo (milliard) for 109, mille miliardi (a thousand milliards) for 1012, un milione di miliardi (a million milliards) for 1015, un miliardo di miliardi (a milliard of milliards) for 1018, mille miliardi di miliardi (a thousand milliard of milliards) for 1021 are more common.[10]
#### Both long and short scale
1. ^ Canadian usage: Both scales are in use currently in Canada. English-speaking regions use the short scale exclusively, while French-speaking regions use the long scale, though the Canadian government standards website recommends that in French billion and trillion be avoided, recommending milliard for 109, and mille milliards (a thousand milliards) for 1012.[68]
2. ^ South African usage: South Africa uses both the long scale (in Afrikaans and sometimes English) and the short scale (in English). Unlike the 1974 UK switch, the switch from long scale to short scale took time. As of 2011 most English language publications use the short scale. Some Afrikaans publications briefly attempted usage of the "American System" but that has led to comment in the papers[70] and has been disparaged by the "Taalkommissie" (The Afrikaans Language Commission of the South African Academy of Science and Art)[71] and has thus, to most appearances, been abandoned.
#### Neither long nor short scale
1. ^ Indian, Pakistani and Bangladeshi usage: Outside of financial media, the use of billion by Bangladeshi, Indian and Pakistani English speakers highly depends on their educational background. Some may continue to use the traditional British long scale. In everyday life, Bangladeshis, Indians and Pakistanis largely use their own common number system, commonly referred to as the Indian numbering system – for instance, Bangladeshi, Pakistani, and Indian English commonly use the words lakh to denote 100 thousand, crore to denote ten million (i.e. 100 lakhs) and arab to denote thousand million.[72]
## Alternative approaches
• In written communications, the simplest solution for moderately large numbers is to write the full amount, for example 1,000,000,000,000 rather than 1 trillion (short scale) or 1 billion (long scale).
• Combinations of the unambiguous word million, for example: 109 = "one thousand million"; 1012 = "one million million".[74]
• Scientific notation (also known as standard form or exponential notation, for example 1×109, 1×1010, 1×1011, 1×1012, etc.), or its engineering notation variant (for example 1×109, 10×109, 100×109, 1×1012, etc.), or the computing variant E notation (for example `1e9`, `1e10`, `1e11`, `1e12`, etc.). This is the most common practice among scientists and mathematicians, and is both unambiguous and convenient.
• SI prefixes in combination with SI units, for example, giga for 109 and tera for 1012 can give gigawatt (=109 W) and terawatt (=1012 W). The International System of Units (SI) is independent of whichever scale is being used.[34] Use with non-SI units (e.g. "giga-dollars", "megabucks") is possible. k€ and M€ is frequently encountered, although the official scheme places the Euro sign in front of the value.
## References
1. ^ "Authoritative Real Academia Española (RAE) dictionary: billón". Archived from the original on 4 November 2015. Retrieved 12 March 2015.
2. ^ a b British-English usage of 'Billion vs Thousand million vs Milliard'. Google Books ngram viewer. Google Inc. Retrieved 26 April 2014.
3. ^ a b c d Guitel, Geneviève (1975). Histoire comparée des numérations écrites (in French). Paris: Flammarion. pp. 51–52. ISBN 978-2-08-211104-1.
4. ^ a b c d Guitel, Geneviève (1975). ""Les grands nombres en numération parlée (État actuel de la question)", i.e. "The large numbers in oral numeration (Present state of the question)"". Histoire comparée des numérations écrites (in French). Paris: Flammarion. pp. 566–574. ISBN 978-2-08-211104-1.
5. Smith, David Eugene (1953) [first published 1925]. History of Mathematics. Vol. II. Courier Dover Publications. p. 81. ISBN 978-0-486-20430-7.
6. ^ a b c "Wortschatz-Lexikon: Milliarde" (in German). Universität Leipzig: Wortschatz-Lexikon. Archived from the original on 27 September 2011. Retrieved 19 August 2011.
7. ^ a b c "Wortschatz-Lexikon: Billion" (in German). Universität Leipzig: Wortschatz-Lexikon. Archived from the original on 7 August 2011. Retrieved 19 August 2011.
8. ^ a b "Wortschatz-Lexikon: Billiarde" (in German). Universität Leipzig: Wortschatz-Lexikon. Archived from the original on 27 September 2011. Retrieved 28 July 2011.
9. ^ a b "Wortschatz-Lexikon: Trilliarde" (in German). Universität Leipzig: Wortschatz-Lexikon. Archived from the original on 27 September 2011. Retrieved 28 July 2011.
10. "Direttiva CEE / CEEA / CE 1994 n. 55, p.12" (PDF) (in Italian). Italian Government. 21 November 1994. Retrieved 24 July 2011.
11. ^ a b c d Fowler, H. W. (1926). A Dictionary of Modern English Usage. Great Britain: Oxford University Press. pp. 52–53. ISBN 978-0-19-860506-5.
12. ^ a b c d ""BILLION" (DEFINITION) — HC Deb 20 December 1974 vol 883 cc711W–712W". Hansard Written Answers. Hansard. 20 December 1972. Retrieved 2 April 2009.
13. ^ a b c d O'Donnell, Frank (30 July 2004). "Britain's £1 trillion debt mountain – How many zeros is that?". The Scotsman. Retrieved 31 January 2008.
14. ^ "Who wants to be a trillionaire?". BBC News. 7 May 2007. Retrieved 11 May 2010.
15. ^ a b c Comrie, Bernard (24 March 1996). "billion:summary". Linguist List (Mailing list). Retrieved 24 July 2011.
16. ^ a b c "Oxford Dictionaries: How many is a billion?". Oxford University Press. Archived from the original on 12 January 2017. Retrieved 7 May 2018.
17. ^ "Oxford Dictionaries: Billion". Oxford University Press. Archived from the original on 11 August 2011. Retrieved 24 July 2011.
18. ^ a b Nielsen, Ron (2006). The Little Green Handbook. Macmillan Publishers. p. 290. ISBN 978-0-312-42581-4.
19. ^ Adam, Jehan (1475). "Traicté en arismetique pour la practique par gectouers... (MS 3143)" (in Middle French). Paris: Bibliothèque Sainte-Geneviève. `{{cite journal}}`: Cite journal requires `|journal=` (help)
20. ^ "HOMMES DE SCIENCE, LIVRES DE SAVANTS A LA BIBLIOTHÈQUE SAINTE-GENEVIÈVE, Livres de savants II". Traicté en arismetique pour la practique par gectouers… (in French). Bibliothèque Sainte-Geneviève. 2005. Retrieved 25 October 2014.
21. ^ Thorndike, Lynn (1926). "The Arithmetic of Jehan Adam, 1475 A.D". The American Mathematical Monthly. Mathematical Association of America. 1926 (January): 24–28. doi:10.2307/2298533. JSTOR 2298533.
22. ^ a b Chuquet, Nicolas (1880) [written 1484]. "Le Triparty en la Science des Nombres par Maistre Nicolas Chuquet Parisien". Bulletino di Bibliographia e di Storia delle Scienze Matematische e Fisische (in Middle French). Bologna: Aristide Marre. XIII (1880): 593–594. ISSN 1123-5209. Retrieved 17 July 2011.
23. ^ Chuquet, Nicolas (1880) [written 1484]. "Le Triparty en la Science des Nombres par Maistre Nicolas Chuquet Parisien" (in Middle French). miakinen.net. Retrieved 1 March 2008.
24. ^ Flegg, Graham (23–30 December 1976). "Tracing the origins of One, Two, Three". New Scientist. Reed Business Information. 72 (1032): 747. ISSN 0262-4079. Retrieved 17 July 2011.
25. ^ a b Budaeus, Guilielmus (1516). De Asse et partibus eius Libri quinque (in Latin). pp. folio 93.
26. ^ Littré, Émile (1873–1874). Dictionnaire de la langue française. Paris, France: L. Hachette. p. 347. Ce n'est qu'au milieu du XVIIe siècle qu'il fut réglé que les tranches, au lieu d'être de six en six chiffres, seraient de trois en trois chiffres; ce qui revint à diviser par 1000 l'ancien billion, l'ancien trillion, etc. [It was only in the middle of the 17th century that it was settled that the slices, instead of being from six to six digits, would be from three to three digits; which resulted in dividing by 1000 the old billion, the old trillion, and so on.]
27. ^ Houck (1676). "Arithmetic". Netherlands: 2. `{{cite journal}}`: Cite journal requires `|journal=` (help)
28. ^ Dictionnaire de l'académie françoise (4th ed.). Paris, France: Institut de France. 1762. p. 177.
29. ^ Dictionnaire de l'Académie française (6th ed.). Paris, France. 1835. p. 189.
30. ^ Dictionnaire de l'Académie française (7th ed.). Paris, France: Institut de France. 1877. p. 182.
31. ^ Dictionnaire de l'Académie française (8th ed.). Paris, France: Institut de France. 1932–1935. p. 144.
32. ^ a b "Resolution 6 of the 9th meeting of the CGPM (1948)". BIPM. Retrieved 7 August 2011.
33. ^ "Resolution 6 of the 10th meeting of the CGPM (1954)". BIPM. Retrieved 23 June 2012.
34. ^ a b "Resolution 12 of the 11th meeting of the CGPM (1960)". BIPM. Retrieved 28 July 2011.
35. ^ The International System of Units (SI) (PDF) (8 ed.). BIPM. May 2006. pp. 134 / 5.3.7 Stating values of dimensionless quantities, or quantities of dimension one. ISBN 92-822-2213-6. Archived from the original (PDF) on 25 July 2011. Retrieved 24 July 2011.
36. ^ Thompson, Ambler; Taylor, Barry N. (30 March 2008). "Guide for the Use of the International System of Units (SI), NIST SP – 811". Nist. US: National Institute of Standards and Technology: 21. Archived from the original on 8 March 2021. Retrieved 13 September 2014.
37. ^ a b "Décret 61-501" (PDF). Journal Officiel (in French). French Government: 4587 note 3a, and erratum on page 7572. 11 August 1961 [commissioned 3 May 1961 published 20 May 1961]. Archived from the original (PDF) on 20 January 2010. Retrieved 31 January 2008.
38. ^ a b "BBC News: Zimbabweans play the zero game". BBC. 23 July 2008. Retrieved 13 July 2012.
39. ^ a b "BBC News: Zimbabwe rolls out Z\$100tr note". BBC. 16 January 2009. Retrieved 24 July 2010.
40. ^ a b "BBC News: Zimbabwe abandons its currency". BBC. 29 January 2009. Retrieved 13 July 2012.
41. ^ "US National Debt Clock". Retrieved 24 November 2022.
42. ^ a b "RBA: Definition of billion". Reserve Bank of Australia. Retrieved 22 August 2011.
43. ^ a b "BBC News: Who wants to be a trillionaire?". BBC. 7 May 2007. Retrieved 11 May 2010.
44. ^ a b "billion". Cambridge Dictionaries Online. Cambridge University Press. Retrieved 21 August 2011.
45. ^ a b "trillion". Cambridge Dictionaries Online. Cambridge University Press. Retrieved 21 August 2011.
46. ^ a b "Al Jazem English-Arabic online dictionary: Billion". Al Jazem English-Arabic online dictionary. Encyclopædia Britannica. Retrieved 6 June 2012.
47. ^ a b "Al Jazem English-Arabic online dictionary:Trillion". Al Jazem English-Arabic online dictionary. Encyclopædia Britannica. Retrieved 6 June 2012.
48. ^ Qeli, Albi. "An English-Albanian Dictionary". Albi Qeli, MD. Retrieved 6 June 2012.
49. ^ "Eesti õigekeelsussõnaraamat ÕS 2006: miljard" (in Estonian). Institute of the Estonian Language (Eesti Keele Instituut). 2006. Retrieved 13 August 2011.
50. ^ "Eesti õigekeelsussõnaraamat ÕS 2013: biljon" (in Estonian). Institute of the Estonian Language (Eesti Keele Instituut). 2013. Retrieved 25 June 2017.
51. ^ "Eesti õigekeelsussõnaraamat ÕS 2006: triljon" (in Estonian). Institute of the Estonian Language (Eesti Keele Instituut). 2006. Retrieved 13 August 2011.
52. ^ a b Robson S. O. (Stuart O.), Singgih Wibisono, Yacinta Kurniasih. Javanese English dictionary Tuttle Publishing: 2002, ISBN 0-7946-0000-X, 821 pages
53. ^ a b Avram, Mioara; Sala, Marius (2000), May We Introduce the Romanian Language to You?, Editura Fundatiei Culturale Române, p. 151, ISBN 9789735772246, the numeral miliard "billion"
54. ^ "Britain to Reduce \$4 billion from Defence". Bi-Weekly Eleven (in Burmese). Yangon. 3 (30). 15 October 2010.
55. ^ "De Geïntegreerde Taal-Bank: miljard" (in Dutch). Instituut voor Nederlandse Lexicologie. Retrieved 19 August 2011.
56. ^ "De Geïntegreerde Taal-Bank: biljoen" (in Dutch). Instituut voor Nederlandse Lexicologie. Retrieved 19 August 2011.
57. ^ a b "French Larousse: milliard" (in French). Éditions Larousse. Archived from the original on 18 March 2012. Retrieved 19 August 2011.
58. ^ a b "French Larousse: billion" (in French). Éditions Larousse. Archived from the original on 18 March 2012. Retrieved 19 August 2011.
59. ^ "billion". Dictionnaire de l'Académie française (in French) (9th ed.). Académie française. 1992. Retrieved 17 January 2016. BILLION (les deux l se prononcent sans mouillure) n. m. XVe siècle, byllion, « un million de millions »; XVIe siècle, « mille millions ». Altération arbitraire de l'initiale de million, d'après la particule latine bi-, « deux fois ».
Rare. Mille millions. Syn. vieilli de Milliard. Selon un décret de 1961, le mot Billion a reçu une nouvelle valeur, à savoir un million de millions (1012), qui n'est pas entrée dans l'usage.
[BILLION (the two Ls are pronounced without palatalisation) masculine noun. Spelled byllion in the 15th century when it meant a million millions; in the 16th century it meant a thousand millions. It is an arbitrary alteration of the start of million by inserting the Latin prefix bi-, meaning twice. Now rarely used. It means a thousand millions. It is an outdated synonym of Milliard. According to a decree of 1961, the word Billion received a new value, to wit a million millions (1012), which has not come into common usage.]
60. ^ a b "Diccionario Panhispánico de Dudas: millardo" (in Spanish). Real Academia Española. Retrieved 19 August 2011.
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# 4.5 INTEREST CONCEPTS AND COMPUTATION 4.5.
1 Why Objections to the use of interest-based Methods Are not Valid It has been argued that interest calculations are complicated and difficult to understand. It is also claimed that the computations required are far too time-consuming to be employed in the making of management decision Actually, this is not so. properly explained, the principles involved in interest computations are quite simple and relatively easy to understand. And with the use of pretabulated calculation sheets and graphical interpolation. Even rate-of-return calculations can be reduced to simple clerical procedures. 4.5.2 Basic Interest Principles The following terms are used in connection with interest computations : 1. Interest rate 2. Earning rate 3. Rate of return 4. Equivalent worths 5. Principle of equivalence 6. Time value of money 7. Present worth 8. Future worth 9. Single-payment factors 10. Discounting 11. Compounding 12. Trial-and-error solution Perhaps the best way to become acquainted with the principles involved in interest calculations is to learn the meaning of these terms by observing how they apply to simple familiar transaction. For instance : If \$100 were deposited today in a bank paying as interest rate of 6 percent, compounded annually. One year from today the balance in this account (principal plus interest) would have grown to \$106. If this balance were then last in the bank another year
at the same annually compounded 6 percent interest rate. the balance would grow to \$ 112,36. lt is also possible to calculate that under these same conditions, S94.35 would have had to be deposited in the bank one year ago in order for today's balance to be \$100. The difference between these balances can be tabulated as follows : DIFFERENCES \$ 112.36 - \$106.00 = \$ 106.00 - \$100.00 = \$ 100.00 - \$ 94.00 = BALANCE \$ 6.36 \$ 6.00 \$ 5.65
These "differences" represent the earnings of the funds on deposit. Note that while the amounts differ, they are generated by a single interest rate. This interest rate is the rate at which the funds on deposit earn. From the point of view of the investor, this is the rate of return. Thus, the three terms-interest rate, earning rate, and rate of return-are identical in meaning. Note that if this stipulated bank interest rate can be earned by funds on deposit, the S94.35 a year are can be stated to be equivalent to Sl00 today which in turn is equivalent to \$106 one year later and this to Sll2.36 the next year. Since things equivalent to the same thing must be equivalent to each other, it .also can be stated that at this interest rate 594.35 on any stipulated date must be equivalent to Sll2.35 three years later. This is known as the principle of equivalence. A word a caution : The differences tabulated above are also often referred to as the time value of money. This time value is not an inherent characteristic of money. The statement that a given sum of money received today is worth more than the same sum received at some future date must simply not valid. Money has rime value only when it can be deposited or borrowed at a stipulated rate. This means that the principle of equivalence is applicable and can be used only when the stipulated interest rate has a real significance. It is also possible to develop a common denominator by expressing all amounts in terms of their equivalent value now. Thus, the \$100 of their is the "now"' or present worth now of all the other amounts. Using this principle, past, present, and future cash flows, can be expressed in terms of their present worth at the stipulated earning rate. The \$106 can be labeled the future worth at the end of one year of 100 now at 6 percent interest compounded annually.
To convert a given amount today to its future worth one year later using annual compounding. the amount is multiplied by (l+i) where I is the interest rate. This calculation is called compounding forward. To compound further into the future, the multiplier is ( I + i)n where n is the number of years. To find the present worth at a stipulated earlier year, we could merely divide by ( I + t)'. however, it is usually more convenient to multiply by the reciprocal l/(l +r)". This is called discounting back. The (l j-i)" and l/(l -1.i)'are called single-payment compounding and discounting factors. They are called this because they permit the conversion of a specific amount of money at one point in time to a single equivalent amount at another point in time at a stipulated interest rate. 4.5.3 Selecting the Most Appropriate Type of Interest-based Evaluation Fairly widespread agreement now exists that interest-based calculations are essential if accurate, dependable evaluations of economic productivity are to be obtained. There still is, however, considerable disagreement as to which of the many different promulgated methods is the most suitable. Most of this disagreement appears to stem from either failure to realize that each of these methods measures a different aspect of economic productivity, or confusion as to just what aspect should be used in making decisions. All valid interest-based methods utilize the principle of equivalence. I n each of these methods, individual disbursements and receipts are converted to equivalents and compared. The equivalent to which the conversion is made determines what is measured by the answer. There are many of these interest-based methods, but they all fall into three basic categories, each measuring a different aspect of investment profitability. The three categories are: l. Net equivalent worth at a specific rime 2. Net level annual equivalent 3. Rate of return 4.5.3.1 Net Equivalent Worth at a Specific Time. In this method, a stipulated interest rate is used to convert and net out all cash flows to a stated point in time. When the stated point in time is the start of the project, the answer is called the "net present worth." When the stated point is the end of the project, the answer is called the "net terminal value." These answers measure the results achieved by a given sum of money during an agreed-upon study period. They report the weighted average performance of the project and all other uses' to which the funds are applied
during the agreed-upon study period. : These answers are expressed as a single number which is in no way comparable to other answers when investment size or project life varies. 4.5.3,2 Net Level Annual Equivalent. In this method, all disbursements and receipts are converted to level annual equivalents at a stipulated interest rate and netted out. The answer measures the annual equivalent of the results of perpetual repetition of the prospective project. This method also yields a single-number answer, but it is adjusted for investment size and project life so that direct comparisons between alternatives with such variations arc feasible. 4.5.3.3 Rate of Return. This method requires no stipulations as to reference point in time, applicable interest rate, or study period. The answer obtained by this method is "the interest rate at which the proposed disbursements would have to be invested in an annuity fund, in order for that fund to be able to make payments equal to and at the same time as the receipts anticipated 'from the project." It measures the equivalent average earning rate of the project as an annual interest rate and permits direct comparisons between projects varying both in size and length of life. 4.5.3.4 Summary end Recommendations. The rate of return is the only one of these evaluation methods that gives an unequivocal answer to the question, "lf the enterprise commits the required investment, what will the project it implements earn for the enterprise?" It is the only method that yields an answer which is directly comparable with the cost of borrowing the required funds. it is an evaluation method rather than a decision technique. 4.5.4 The Rate-of-Return Concept To determine the interest rate (rate of return) of a single disbursement which result in at single receipt at a later date is a simple matter. To illustrate: Suppose it is anticipated that \$408 can be invested all at one time to yield an instantaneous receipt of \$800 exactly two years later. The relationship between these two figures and the applicable interest rate I is stated by the following equation: i : 0.40=40 percent The following calculations demonstrate that the answer of 40 per cent is the interest rate at which the \$408 would have to be deposited in order to achieve a balance sufficient to pay out the \$800 two years later as required by the rate-of-return concept. \$ 108 x 1.4 : \$571 \$ 57l x 1.4 : \$800
Now suppose a project involving four separate pairs of anticipated disbursements and receipts, as in Table 4. l, is to be evaluated. (Disbursements are shown as negative cash flows, receipt's as positive flows.) Calculation of the rate of return exhibited by the additional pairs of disbursements and receipts yields the answers tabulated. But suppose all of these disbursements and receipts are anticipated to occur in connection with a single project and the only data available are the overall cash flows shown in the last column. Could we not state the rate of return of the project to be the interest rate at which the two overall net disbursements would have be invested in an annuity fund in order for the fund to be able to make payments equal to and at the same time as the three net overall receipts of the project ?. The interest rate that satisfies conditions for the overall net cash flows tabulated in Table 4.1 is 20 percent. The validity of the answer. is demonstrated by Table 4.2. This table also makes two other facts clear: l. lf this project performs is anticipated, it will pay back the total of \$3000 invested plus 20 percent interest on the funds in use. This interest rate is suitable for direct comparison wiht the cost of capital. 2. The rate-of-return concept as demonstrated involves neither explicit nor implicit reinvestment of proceeds. 4.5.5 The Calculation Algorithm The foregoing answer of 20 percent was demonstrated to be Correct, but how can it be calculated? The overall net cash flows as previously tabulated provide an excellent, demonstration problem illustrating many of the computational difficulties encountered in rate of-return calculations. Note that despite the fact that there are only five figures, the following four difficulties are introduced : 1. Disbursements are made at more than one time. 2. Disbursements vary in amount. 3. Receipt timing is irregular. 4. Receipts vary in amount. The first step in the solution is the selection of a reference or zero point. Any zero point can be used. The choice made will in no way affect the answer obtained, but it can greatly affect the ease of computation' In most cases, the selection of the beginning of the first year of receipts as the zero point will result in calculations of minimum complexity.
Now suppose the present worth of all net cash flows are calculated at a stipulated interest rate. The results of such a calculation using an annually compounded interest rate of l0 percent are tabulated below Note that the \$800 receipt discounted back to zero time has a present worth of \$728' This also means that if \$7l8 were deposited in a bank at l0 per cent interest compounded annually, one year later the amount on deposit would be \$800 or just enough to make a payment equal to that anticipated from the project. In the same way, the \$1360 deposited at the same time would grow to \$2000 in four years and the \$2420 to \$3904 in five years. Thus, if the sum of \$728*\$1360+\$2420 or \$4508 were deposited at time zero, at the stipulated l0 percent interest compounded annually, these deposits, plus the interest carried, would be just sufficient to make payments equal to and at the same time as the project. Note that this amount is larger than the present worth of the prospective investment (\$3100). Since the project's prospective investment is much smaller than the amount that would have to be invested in the bank at l0 percent in order to achieve the same results, the prospective rate of return is demonstrated to be greater than l0 percent. Now note : If it is assumed that the bank will pay an interest rate higher than l0 percent, the sum of the present worth of disbursement will increase and the sum of the present worth of receipts will decrease. Obviously, there must be some interest rate at which these two figures will be equal. This is the answer for which we must solve, for it is also by definition the interest rate at which deposits of identical amounts and timing would have to be made in order to provide exactly enough funds to make payments equal to and at the same time as receipts from the prospective project. In other. words, we would have determined a bank interest rate exactly equivalent to the earning rate of the prospective project. The only way this interest rate can be computed is by trial and error. As ordinarily performed, this requires repeated trials at different interest rates until the correct answer is determined. This can be a tedious and time-consuming operation. It has long tended to be a serious roadblock to the widespread acceptance and use of this method. 4.5.6 The Use of Pre-tabulated Worksheets and Graphical Interpolation The "profitability index" approach utilizes Pre-tabulated worksheets and graphical interpolation to reduce the necessary trial-and-error computations to routine clerical procedure's. The use of these technique's to solve the demonstration problem is illustrated in Fig. 4.1
4.5.6.1 The Worksheet and Graphical Interpolation Chart (Fig. 4.1). Separate schedules are provided on the worksheet for disbursements and receipts. The zero points of both, however, are identical and are identified by a heavy line with a diamond at the left. The design of the form facilitates the assumption of the beginning of the first year of receipts as the zero point. Singlepayment compounding and discounting factors are provided for three different interest rates together with adjacent blank columns for entering appropriate present worth. (On this demonstration form, the pre-tabulated present worth factors are based upon annual compounding and instantaneous cash flows. Conversion to the more frequently used convention of continuous compounding and during the year receipt is merely a matter of substituting different present worth factor ) The chart at the bottom of the page is for plotting the ratio (horizontal scale) of the sum of the present worth disbursements to the sum of the present worth of receipts it each of the stipulated interest rates (vertical scale). 4.5.6.2 Using the worksheet and Chart. The use of this worksheet to find the rate of return of the demonstration problem is illustrated in Fig 4.1 The predicted disbursements and receipts are entered in the blank column head "trial = I at 0 percent interest" und on the lines corresponding to the time the cash flows are anticipated to occur. The comparison of the totals of actual disbursements and receipts is equivalent to making a trial at zero percent interest and provides one point on our interpolation chart. "Trial = 2 at l0 percent" is made by multiplying each of the actual amounts by the adjacent present worth factor in the next column and entering the present worth in the next blank column. Similar trials are made at 25 and 40 percent The at cash trial rate, including that at 0 percent, the sum of the present worth of the disbursements is divided by the sum of the present worth of the receipts and the ratios entered on the line labeled ratios A/8. The interest rate at which the sum of the present worth of receipts is exactly equal to the sum of the present worth of disbursements is determined by plotting on the chart in Fig.4.I each of the ratios obtained against the interest rate at which they were calculated, drawing a curve through these points and reading on the vertical scale the point where this curve intersects the unity ratio line. This shows, as it should, that the rate of return on this demonstration problem is 20 percent. 4.5.6.3 Comments on Accuracy and Validity of This Algorithm. As long as this algorithm is applied to the type of problem for which it is designed (see Par. 4.5.7 for discussion of when
algorithm cannot be used), the accuracy of the answers obtained by this graphical interpolation method will be as exact as the number of decimal points in the present worth factors and the legibility of the chart permit. In fact, because of the use of the most sensitive ratio, which is that at zero interest rate, as one point on the curve, this method offers the maximum accuracy of results obtainable with a given number of decimal places in the present worth factors. In addition, this method is to a certain extent self-checking. Only three points are required to establish the curve. The fourth serves as a check. lf a smooth curve cannot be drawn through all four points, an error in computation or plotting has been made. 4.5.7 Limitations of the Algorithm . and How to handle Them The method of determining the earning rate or rate of return of a prospective investment by solving for the interest rate at which the sum of the present worth of receipts is equal to the sum of the present worth of disbursements has been labeled an "algorithm." This term has been applied because it is a trick method of limited application. It can be employed only when the problem to be solved is in the form for which the method was designed. Recognition of this limitation is extremely important. Improper use of this method can result in misconceptions and serious errors in evaluations. The algorithm has been designed to compute the interest rate that relates one or more receipts to one or more prior disbursements. Valid answers will be obtained only if the problem can be reduced to this configuration. Specifically, if this method is misused to evaluate proposals in which all net disbursements do not precede all net receipts, over' stated, or in some cases multiple overstated, answers may be obtained. Fortunately, however, many of the problems to which this algorithm cannot be directly applied can be rearranged to a solvable form. A suggested method of handling two types of such cases is illustrated as follows: Direct application of the algorithm to the above data yields an answer of 38 percent. This is not the earning rate on the \$ll00 commitment but actually the earning rate on the implied "equity" in accordance with the following net annual disbursements and receipt schedule. It is suggested that the outright acquisition cost be obtained and two separate calculations be made. With the outright acquisition cost estimated to be \$1000, the following calculations can be made: Answer given by algorithm: 6 per cent. The true earning rate of the project, 15.? Percent, provides a basis for determining whether the project is sufficiently profitable. The cost of capital,
6 percent, provides a basis for determining whether the cost of financing via a conditional sale is acceptable. problem results in two answers: 35 percent and 63 percent. The successful trials at these interest rates are tabulated here: Neither of these answers represents a valid evaluation. Both overstate the earning rate by building in reinvestment of funds at the solution interest rate, which is higher than any reasonable expectation. The best solution offered to date for this type of problem appears to be a compromise in which a stipulated, highly believable reinvestment rate is used to convert the unsolvable problem into a form to which the algorithm can be validly applied. The application of this "crutch" is illustrated in the two tables shown below. If a l0 percent stipulated interest rate is used as I "crutch" the successful trial rate is 19 percent. If a 6 percent stipulated interest rate is used as a "crutch" the successful trial rate is a7 percent. On the basis of the foregoing calculations, it seems reasonable to consider this project equivalent in profitability to other projects with a rate of return somewhere between 17 an 19 percent
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# A football team's fans expect the team to make the playoffs.
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A football team's fans expect the team to make the playoffs. If the team makes the playoffs then their fans expect it to win the Super Bowl. A team that is happy just to make the playoffs will upset its fans. The New York Giants' primary goal is to meet their fans' expectations.
Which of the following must be true based on the statements above?
A) Other teams will not upset their fans only if they win the Super Bowl.
B) A team that doesn't think its fans want it to make the playoffs will certainly not make the playoffs.
C) The Giants' primary goal is possible only if they win the Super Bowl.
D) If a team is happy just to make the playoffs one year, it won't make the playoffs next year.
E) The Giants' fans are not currently upset.
Why is A incorrect. I agree with OA.
The premise says A team that is happy just to make the playoffs will upset its fans. Hence, ~Superbowl ---> upset. This is same as A. Isn't it?
[Reveal] Spoiler: OA
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Re: A football team's fans expect the team to make the playoffs. [#permalink]
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Re: A football team's fans expect the team to make the playoffs. [#permalink]
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23 Sep 2012, 15:14
I think the answer should be C. As it is specific to the Giants. 'A' is also correct but I guess it is more in general.
So according to me C is the best choice.
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Re: A football team's fans expect the team to make the playoffs. [#permalink]
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24 Sep 2012, 14:09
voodoochild wrote:
A football team's fans expect the team to make the playoffs. If the team makes the playoffs then their fans expect it to win the Super Bowl. A team that is happy just to make the playoffs will upset its fans. The New York Giants' primary goal is to meet their fans' expectations.
Which of the following must be true based on the statements above?
A) Other teams will not upset their fans only if they win the Super Bowl.
B) A team that doesn't think its fans want it to make the playoffs will certainly not make the playoffs.
C) The Giants' primary goal is possible only if they win the Super Bowl.
D) If a team is happy just to make the playoffs one year, it won't make the playoffs next year.
E) The Giants' fans are not currently upset.
Why is A incorrect? I agree with OA. The premise says A team that is happy just to make the playoffs will upset its fans. Hence, ~Superbowl ---> upset. This is same as A). Isn't it?
Voodoo
So, we agree that (C), the OA, is the best answer. That's good.
Let's take what we know about other teams.
a) through #1 & #2, it appears fans of any team expect that team to win the Super Bowl
b) we know the Giant's primary goal, but we don't necessarily know the goal of any other team.
Technically, therefore, we absolutely do know: if any team does not win the Super Bowl, it is not meeting fan's expectations.
Now, compare that to the text of (A):
A) Other teams will not upset their fans only if they win the Super Bowl.
BTW, I think this is a little over-achieving on Veritas' part --- I don't think I have ever seen the logical phrase "only if" in official material. We can rephrase answer (A) as
(A1) If other team's fans are not upset, then that team has won the Super Bowl
(A2) If a team does not win the Super Bowl, then its fans are upset.
I point out, (A2) is not identical to the green sentence above. Their equation depends on the assumption: if a person's expectations are not met, then that person is upset. Granted, that is often the case, although I would argue, that's an emotionally unhealthy state of affairs. We have consider a few caveats.
First of all, how strong are those expectations? Yes, we are told, "A football team's fans expect the team to make the playoffs," but if the team gets off to a 3-8 start to its season, probably the sense of "we are not going to make the playoffs" will be a slow sinking feeling that accrues over the season, not a sudden shock: when expectations slowly but steadily become unrealistic to the point at which one feels compelled to abandon them because of their sheer logical absurdity, that is far less an occasion for upset than when strong expectations are dashed in an instant --- say, in the final 10 minutes of a crucial football game.
Next, think about the quality of expectations one has of other people, say, a friend or a family member. There are some expectations that are "non-negotiable" --- the person better do X, or I will be hopping mad at that person. Example: husband remembers first wedding anniversary. There are others that are more idealized: it would really allow me to settle some doubts and hold a higher opinion this person, or it would surprise me and force me to raise my opinion, if this person does X. Example: husband has forgotten the first 16 wedding anniversaries in a row, but it sure would be nice if he remembered the 17th. In the latter case, when the husband forgets yet another time, the wife may be disappointed, may feel a certain amount of resignation, but certainly she is not going to be as lividly upset as she was in the first case. Once there's a fixed pattern of disappointment, each new instant can't be nearly as upsetting as the the first time, when the wife still had every reason to expect the very best. It's not that the wife of 16+ years has zero expectations, but previous disappointment produces a muted or attenuated expectation, a hope-against-hope, one quite different in quality from the fresh vital bright-eyed expectations of newly weds. Those are two extreme examples, but clearly, different expectations vary by degrees of the emotional tone and seriousness and realism, and that will make a difference in the emotional experience when they are not met. ---- Now, the question arises: what are the expectations for a football team more like which scenario? Well, a team that has been successful last year, that has acquired even more talented new players --- one could imagine high expectations, even of a non-negotiable sort. BUT, a team that was in last place for the last four years, and has made no substantial improvements between last year and this year --- yes, fans will always harbor idealized playoff and super bowl hopes at the outset, hopes-against-hope, but those hopes can't possibly be as strong, as realistic, or as emotionally charged as those for a team brimming with talented & potential. Again, the quality of expectation has something to do with how "upset" one will be.
Finally, among the large pool of fans for any team, arguably there must be a few who, like me, have some Buddhist proclivities, and to that extent, recognize how profoundly unbalancing it is to invest large amounts of emotional energy into expectations of any kind. Such fan admitted are rare, but I would argue, there will be some small number of fans who have worked consistently with emotional disattachment from expectations, such that when expectations are not met, even when expectations are suddenly and permanently obliterated, this does not cause substantial emotional unbalance in those people --- they do not allow themselves to become "upset." Thus, we cannot say that "all" the fans are upset, even when strong expectations are dashed.
In my mind, these considerations make (A) shaky, whereas (C) is not shaky at all.
Does that make sense?
Mike
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# How Many Ounces in a Quarter: A Comprehensive Guide to Measurement Conversion
Discover how to convert quarters to ounces and vice versa with our comprehensive guide. Learn the formulas, tips, and common uses for accurate measurement conversion!
As we go throughout our day-to-day lives, we often come across measurements that need to be converted. One of the most common conversions is from quarters to ounces. Whether you’re measuring ingredients for a recipe or trying to figure out how much liquid is left in a bottle, understanding this conversion can save time and hassle.
But why is it important to know how many ounces are in a quarter? For starters, it ensures accuracy when cooking or baking. Using the wrong amount of an ingredient can significantly affect the outcome of your dish. Additionally, knowing measurement conversions can help with budgeting and purchasing decisions, especially when dealing with precious metals.
In this article, we’ll provide you with everything you need to know about converting quarters to ounces and vice versa. From formulas and examples to common uses and tips for accurate measurement conversion, we’ve got you covered. So let’s dive in!
## Understanding Ounces and Quarters
When it comes to measurement conversion, it’s essential to understand the units of measurement you’re dealing with. In this section, we’ll define what an ounce and a quarter are and explain their relationship.
### A. Definition of an Ounce
An ounce is a unit of weight or mass in both the US customary and British imperial systems of measurement. One ounce equals 28.35 grams or 1/16th of a pound. Ounces can be used to measure both dry ingredients, such as flour or sugar, and liquid ingredients, such as water or milk.
### B. Definition of a Quarter
A quarter is a term used for various measurements that are one-fourth of something else. For example, a quarter can refer to one-fourth of a dollar (25 cents) or one-fourth of a year (three months). In terms of measurement conversions, we’re referring specifically to quarters as they relate to weight or volume.
Maybe you are interested How Many Grams Is a Quarter Pound of Weed?
In the context of weight, a quarter refers to one-fourth of a pound. This translates to 4 ounces or 113.4 grams. When measuring liquids, such as oil or vinegar, a quarter usually refers to one-fourth of a cup.
### C. Relationship between Ounces and Quarters
As mentioned earlier, there is a direct relationship between ounces and quarters when it comes to weight measurements. One quarter is equivalent to four ounces, while one ounce is equal to one-fourth of a quarter.
Having a clear understanding of these definitions and relationships will make converting between ounces and quarters much easier in the following sections.
## Converting Quarters to Ounces
### Formula for Converting Quarters to Ounces
To convert quarters to ounces, we need to use a simple formula. One quarter is equal to 0.25 ounces. So, to find out how many ounces are in any number of quarters, all you have to do is multiply that number by 0.25.
Here’s the formula:
``Number of Quarters × 0.25 = Number of Ounces``
For example, let’s say you need to convert 3 quarters into ounces. You would simply plug in the numbers into the formula like this:
``3 × 0.25 = 0.75``
Therefore, three quarters is equivalent to 0.75 ounces.
### Examples of Converting Different Amounts of Quarters to Ounces
Converting quarters to ounces becomes easier with practice and familiarity with the conversion formula.
Let’s take a look at some more examples:
• 1 quarter = 0.25 ounces
• 2 quarters = 0.5 ounces
• 4 quarters (1 dollar) = 1 ounce
• 8 quarters (2 dollars) = 2 ounces
• 12 quarters (3 dollars) = 3 ounces
Now that we’ve covered converting quarters to ounces, it’s time to take a look at the reverse conversion – converting ounces to quarters!
## Converting Ounces to Quarters
Converting ounces to quarters may not be as common as the reverse conversion, but it’s still an important skill to have. This is especially true when dealing with precious metals such as silver or gold, where prices are often listed in terms of ounces and fractions of an ounce.
### Formula for Converting Ounces to Quarters
The formula for converting ounces to quarters is straightforward. One quarter is equal to 0.25 ounces, so all you have to do is divide the number of ounces by 0.25.
Here’s the formula:
`Quarters = Ounces ÷ 0.25`
For example, if you have 10 ounces of silver and want to convert it into quarters, you would divide 10 by 0.25:
`10 ÷ 0.25 = 40 quarters`
So 10 ounces of silver is equivalent to 40 quarters.
### Examples of Converting Different Amounts of Ounces to Quarters
Let’s take a look at a few different examples of converting ounces to quarters:
• Example #1: You have 5 ounces of gold and want to know how many quarters that is.
`Quarters = Ounces ÷ 0.25`
`Quarters = 5 ÷ 0.25`
`Quarters = 20`
Answer: There are 20 quarters in 5 ounces of gold.
• Example #2: You inherited a collection of silver coins weighing a total of 50 ounces and want to know how many quarters that is.
`Quarters = Ounces ÷ 0.25`
`Quarters = 50 ÷ 0.25`
`Quarters = 200`
Answer: There are 200 quarters in a collection of silver coins weighing 50 ounces.
• Example #3: You have a bottle of cleaning solution that contains 32 ounces and want to know how many quarters that is.
`Quarters = Ounces ÷ 0.25`
`Quarters = 32 ÷ 0.25`
`Quarters = 128`
Answer: There are 128 quarters in a bottle of cleaning solution containing 32 ounces.
Maybe you are interested How Many Ounces in a Pound of Chicken? A Comprehensive Guide
By mastering the conversion formulas for both ounces to quarters and quarters to ounces, you’ll be able to confidently navigate any situation that requires measurement conversion.
## Common Uses for Ounces and Quarters
As mentioned earlier, ounces and quarters are used in a variety of settings. In this section, we’ll explore some of the most common uses for these measurements.
### Culinary Measurements
When cooking or baking, precise measurements can significantly affect the taste and texture of your dish. Ounces and quarters are commonly used in recipes as a way to measure both dry and liquid ingredients. For example, a recipe might call for 4 ounces of flour or 1/4 cup of milk (which equals 2 ounces). By understanding how many ounces are in a quarter, you can accurately measure out ingredients and ensure that your dish turns out just right.
### Precious Metals
Ounces and quarters are also used when dealing with precious metals such as gold and silver. These metals are often measured in troy ounces (a slightly different measurement than regular ounces) and fractions thereof. For example, an ounce of gold is equal to 31.1 grams, while a quarter-ounce of gold is equal to just over 7 grams. Understanding these conversions is important for those who invest in precious metals or buy/sell them regularly.
### Liquid Volume Measurements
Finally, ounces and quarters are commonly used to measure liquid volumes. A fluid ounce is defined as the volume occupied by one ounce of water at room temperature, while a quarter-cup is equal to 2 fluid ounces. This measurement is essential when it comes to mixing drinks or pouring liquids into containers with specific capacities.
Overall, understanding the common uses for ounces and quarters can help us navigate various aspects of our lives with ease. Whether we’re cooking up a storm or investing in precious metals, knowing these conversions ensures accuracy and precision.
## Tips for Accurate Measurement Conversion
Accurate measurement conversion is crucial when it comes to cooking, baking, and other activities that require precise measurements. Here are some tips to help you convert quarters to ounces with ease:
Maybe you are interested How Many Ounces Are in 2 Teaspoons? A Complete Guide to Measuring Ingredients
### A. Use the Correct Tools
Using the right tools can make a significant difference in measurement accuracy. For instance, if you’re measuring liquid ingredients, use a liquid measuring cup rather than a dry measuring cup. The latter may lead to inaccurate measurements due to differences in volume.
Other essential tools to consider include a kitchen scale and measuring spoons or cups. Investing in quality tools will not only ensure accurate measurements but also save time and reduce frustration.
### B. Double-Check Calculations
Even with the right tools, mistakes can happen during measurement conversions. That’s why it’s essential to double-check your calculations before adding an ingredient to your recipe or making a purchase decision.
One way to do this is by using multiple sources for calculating conversions. Cross-referencing different sources can help identify errors and prevent costly mistakes.
### C. Practice with Common Measurements
Finally, practice makes perfect! Familiarize yourself with common measurements such as tablespoons, teaspoons, cups, ounces, and quarts. This will make converting between these measurements much more manageable when it comes time to cook or bake.
You can also find helpful charts and conversion calculators online or in cookbooks that provide additional guidance on measurement conversions.
By following these tips, you’ll be able to accurately convert quarters to ounces and other measurements like a pro!
If you’re new to the world of measurement conversions, you may have some questions about ounces and quarters. Here are the answers to some frequently asked questions:
### A. How many ounces are in a half-quarter?
A half-quarter is equal to 2 ounces. To convert from half-quarters to other units of measurements, simply multiply by 2.
### B. Is there a difference between dry and liquid measurements?
Yes, there is a difference between dry and liquid measurements. Dry ingredients, such as flour or sugar, are typically measured in weight (ounces), while liquid ingredients are measured in volume (fluid ounces). It’s important to use the appropriate measuring tool for each type of ingredient to ensure accurate results.
### C. Why is it important to understand measurement conversions?
Understanding measurement conversions can save time, money, and hassle in various situations. For example, when cooking or baking, using the wrong amount of an ingredient can significantly affect the outcome of your dish. Additionally, knowing conversion rates can help with budgeting and purchasing decisions, especially when dealing with precious metals.
By having a solid understanding of measurement conversions like quarters to ounces, you’ll be better equipped to handle any situation that requires accurate measurements.
## Conclusion
In conclusion, understanding measurement conversions is an essential skill that can save time and hassle in various situations. Knowing how many ounces are in a quarter is particularly important for culinary measurements and precious metal transactions.
We hope this comprehensive guide has provided you with the necessary information to convert quarters to ounces and vice versa accurately. Remember to use the correct tools, double-check your calculations, and practice with common measurements for best results.
By following these tips, you’ll be able to convert measurements like a pro in no time! As always, if you have any questions or concerns about measurement conversion, don’t hesitate to consult a professional for further guidance.
Thank you for reading our article on “how many ounces in a quarter,” we hope it was helpful!
## How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
## How Many Pounds is 600 kg? – A Comprehensive Guide
Learn how to convert 600 kg into pounds with our comprehensive guide. Discover the factors that impact weight measurement, including gravity, altitude, and temperature.
## How Many Ounces in 1.5 Pounds?
Learn how to convert pounds to ounces and vice versa accurately! Discover “how many ounces in 1.5 pounds” and more with this comprehensive guide.
## How Many Pounds Is 23 Kilos?
Learn how to convert 23 kilos to pounds and gain a better understanding of the conversion process between these two units of measurement.
## How Many M&Ms in a Pound: Everything You Need to Know
Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity!
## How Many M&Ms in a Pound: Everything You Need to Know
Discover the exact number of M&Ms in a pound and more with our comprehensive guide. Learn how to measure them by weight and find out what affects their quantity!
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# Set Theoretic Definition of Numbers
I am reading the book by Goldrei on Classic Set Theory. My question is more of a clarification. It is on if we are overloading symbols in some cases. For instance, when we define $2$ as a natural number, we define $$2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \}$$ When we define $2$ as an integer, $2_{\mathbb{Z}}$ is an equivalence class of ordered pair $$2_{\mathbb{Z}} = \{(n+_{\mathbb{N}}2_{\mathbb{N}},n):n \in \mathbb{N}\}$$ Similarly, when we define $2$ as a rational number, $2_{\mathbb{Q}}$ is an equivalence class of ordered pair $$2_{\mathbb{Q}} = \{(a \times_{\mathbb{Z}} 2_{\mathbb{Z}},a):a \in \mathbb{Z}\backslash\{0\}\}$$ and as a real number we define it as the left Dedekind cut of rationals less than $2_{\mathbb{Q}}$, i.e. $$2_{\mathbb{R}} = \{q \in \mathbb{Q}: q <_{\mathbb{Q}} 2_{\mathbb{Q}}\}$$
The clarification is each of the above are different objects right? So when we say $2$, it depends on the context? Also, if the above is true, is it correct or incorrect to say that "The set of natural numbers is a subset of reals"? Should we take the statement with a pinch of salt and understand accordingly?
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You might find this "buzz" by Terence Tao useful: google.com/buzz/114134834346472219368/RarPutThCJv/… – Jon Dec 19 '10 at 4:24
Your understanding is absolutely correct. The thing is that when one defines the integers, they come with a canonical embedding of $\mathbb{N}$. Similarly, the rationals come with a canonical embedding of $\mathbb{Z}$ and so on. So it does make sense to speak of subsets. Indeed, it becomes so cumbersome to distinguish between genuine subsets and images under an embedding that one fairly quickly drops this distinction in practice, unless one really has to thing about the foundations. – Alex B. Dec 19 '10 at 4:34
Thanks Jon and Alex. – user17762 Dec 19 '10 at 5:48
This is a very good question. – Asaf Karagila Dec 19 '10 at 7:37
Yes. And no.
You start with $\mathbb{N}$, and define $+$ and $\times$ (and $\lt$ and so on) appropriately.
Then you define an equivalence relation on $\mathbb{N}\times\mathbb{N}$ given by $$(a,b)\sim(c,d) \Longleftrightarrow a+d=b+c,$$ and call the quotient set $(\mathbb{N}\times\mathbb{N})/\sim$ by the name $\mathbb{Z}$. (Behind the scenes, we are thinking of $(a,b)$ as meaning "the solution to $a=x+b$").
We can then define an addition $+_{\mathbb{Z}}$ and a product $\times_{\mathbb{Z}}$ on $\mathbb{Z}$, as well as an order $\leq_{\mathbb{Z}}$ by $$\begin{array}{rcl} [(a,b)]+_{\mathbb{Z}}[(c,d)] &=& [(a+c,b+d)]\\\ [(a,b)]\times_{\mathbb{Z}}[(c,d)] &=& [(ac+bd,ad+bc)]\\\ [(a,b)] \leq_{\mathbb{Z}} [(c,d)] &\Leftrightarrow& a+d\leq b+c, \end{array}$$ and show that this is well defined. (I am using $[(a,b)]$ to denote the equivalence class of the pair $(a,b)$.
Certainly, $\mathbb{N}$ and $\mathbb{Z}$ are entirely different animals; set-theoretically, you can even show that they are disjoint.
But we can define a map $f\colon \mathbb{N}\to\mathbb{Z}$ by $f(n) = [(n,0)]$. This map is one-to-one, and for all $n,m\in\mathbb{N}$, $$\begin{array}{rcl} f(n+m) &=& f(n)+_{\mathbb{Z}}f(m),\\\ f(n\times m) &=& f(n)\times_{\mathbb{Z}}f(m)\\\ n\leq m &\Leftrightarrow& f(n)\leq_{\mathbb{Z}} f(m) \end{array}$$ That means that even though $\mathbb{N}$ and $\mathbb{Z}$ are disjoint, there is a "perfect copy" of $\mathbb{N}$ (in so far as its operations $+$ and $\times$ are concerned, and as far as the order $\leq$ is concerned) sitting inside of $\mathbb{Z}$. (Added: In fact, this $f$ not only gives us perfect copy, it is the only map from $\mathbb{N}$ to $\mathbb{Z}$ that is one-to-one and respects all the operations; we say it is a "canonical embedding"). Since we have this perfect copy, and a very specific map identifying this copy with the original, we can think of $\mathbb{N}$ as being a subset of $\mathbb{Z}$ by "identifying it with its copy". So we do that. We can then introduce notation by showing that for every $[(a,b)]\in\mathbb{Z}$, either $a=b$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(a,b)]=f(n)=[(n,0)]$, or there exists $n\in\mathbb{N}$, $n\neq 0$, such that $[(b,a)]=f(n)=[(n,0)]$; and then using $0$ to denote the class with $a=b$, $n$ to denote the class with $[(a,b)]=[(n,0)]$, and $-n$ to denote the class $[(c,d)]$ with $[(d,c)]=[(n,0)]$. This notation makes the identification clearer.
Similarly, once we have $\mathbb{Z}$, we define $\mathbb{Q}$ as the quotient of $\mathbb{Z}\times(\mathbb{Z}-\{0\}$ modulo $\cong$, where $$(a,b)\cong (c,d) \Longleftrightarrow ad=bc$$ (behind the scenes, we think of $(a,b)$ as meaning "the solution to $a=xb$"). We can then proceed as before, defining $$\begin{array}{rcl} [(a,b)]+_{\mathbb{Q}}[(c,d)] &=& [(ad+bc,bd)]\\\ [(a,b)]\times_{\mathbb{Q}}[(c,d)] &=& [(ac,bd)] \end{array}$$ and showing this is well defined; defining an order, etc. Again, $\mathbb{Q}$ and $\mathbb{Z}$ (and the original $\mathbb{N}$) are completely different sets. But we have a function $g\colon\mathbb{Z}\to\mathbb{Q}$ defined by $g(a) = [(a,1)]$. This is one-to-one, $g(a+_{\mathbb{Z}}b) = g(a)+_{\mathbb{Q}}g(b)$, and $g(a\times_{\mathbb{Z}}b) = g(a)\times_{\mathbb{Q}}g(b)$. (Added: And again, this is the only map from $\mathbb{Z}$ to $\mathbb{Q}$ that satisfies these conditions.) So again, we have a "perfect copy" of $\mathbb{Z}$ sitting inside of $\mathbb{Q}$ (and so also a perfect copy of the perfect copy of $\mathbb{N}$ that is sitting inside of $\mathbb{Z}$). So once again we "identify" $\mathbb{Z}$ with its image inside $\mathbb{Q}$ (and so we identify $\mathbb{N}$ with its image inside the image of $\mathbb{Z}$). Because, via $f$ and $g$, we have perfect copies of them anyway.
We do the same thing with $\mathbb{Q}$ as being "inside of $\mathbb{R}$", by identifying elements of $\mathbb{Q}$ with specific Dedekind cuts or with specific equivalence classes of Cauchy sequences, showing the identification is one-to-one and respects all the operations (and is essentially unique), and so obtaining a "perfect copy" of $\mathbb{Q}$ inside of $\mathbb{R}$ (and by extension, perfect copies of $\mathbb{N}$ and of $\mathbb{Z}$ also sitting inside of $\mathbb{R}$).
You can keep going, of course: define $\mathbb{C}$ as the set of all pairs $\mathbb{R}\times\mathbb{R}$; then identify $\mathbb{R}$ with the pairs $\mathbb{R}\times\{0\}$, and you have a copy of $\mathbb{N}$ sitting inside a copy of $\mathbb{Z}$ sitting inside a copy of $\mathbb{Q}$ sitting inside a copy of $\mathbb{R}$ sitting inside $\mathbb{C}$. (And then you can stick $\mathbb{C}$ inside the quaternions, the quaternions inside the octonions).
So even though they are actually very different sets, we have copies of each sitting inside the "next one", copies that respect all the structures we are interested in, so we can still think of them as being "subsets".
You used to do that all the time without the formalism: we think of "fractions" as being made up of an integer, a solidus, and a nonzero integer, so that "$3$" is not a fraction; but when needed, we are perfectly happy writing "$3 = \frac{3}{1}$" and working with either version of $3$ (the integer, or the fraction) depending on context.
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It is important to emphasize not only the existence but also the uniqueness of the embedding of the subobject. Otherwise the overloading could be ambiguous. E.g. in rings w/o 1 does 2 denote (2,0) or (0,2) or (2,2) in Z x Z ? Note that this example also emphasizes another point: that the implicit "context" must include info such as the type of the structure(s) (here rings vs. rngs). Thus the overloaded notation is valid if there exists a canonical embedding as objects of the (implicitly) specified type. – Bill Dubuque Dec 19 '10 at 15:03
@Arturo Magidin , your answer is great! I think it had answered a question which I always aked for myself ! thanx! – Maths Lover Sep 20 '13 at 1:09
Yes, in the sense that in ZF, each of the above definitions defines a different collection of sets. Also yes, in the sense that one should always be aware of context when discussing a mathematical object.
The mathematical fact that allows us to overload the symbol $2$ here is the existence of a sequence of natural injections $\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{R}$, and elements in any term in this sequence can be pushed forward under the next injection or, if they lie in the range of the previous injection, pulled back. Most people do this automatically and so don't usually point it out explicitly. More generally, $1$ refers to the multiplicative identity in any ring, $0$ to the additive identity in any ring, and $2 = 1 + 1, 3 = 1 + 1 + 1$, and so forth.
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Thanks – user17762 Dec 19 '10 at 5:55
In addition to the other answers, it's also noteworthy to know that there is a number system (more-or-less) integrating N, Z, Q and R (and a lot more, but not C!).
The surreal numbers take the basic idea from Dedekind cuts, assigning to each number a so-called left set ('smaller') and right set ('larger') of numbers (constraint to certain rules), bootstrapping the whole process from the empty set and ending up with the reals and weird infinitesimals like $\frac{1}{\sqrt{\omega - \pi}}$!
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CrazyEngineers Archive
Old, but evergreen and popular discussions on CrazyEngineers, presented to you in read-only mode.
@Manish Goyal • 15 Jan, 2010
1:-You are a landscape designer and your boss asked you to design a landscape such that you should place 4 trees equidistance from each other?Tell me How?
2:-If one tyre of a car suddenly gets stolen and after sometime you find the tyre without the screws how will u make ur journey complete?
3:-How can you measure a room height using a thermometer?
4:-How would you catch and receive a ball in same direction? (Dropping is from north
and receiving from bottom not accepted, as it is 2 directions)
5:-Can you make 120 with 5 zeros?
@CIVILPRINCESS • 15 Jan, 2010 The questions are tough.
for the fifth one i think i got the answer :😁
(0!+0!+0!+0!+0!)!
=>(1+1+1+1+1)! (since0!=1)
=>5!=120
am i correct :?:
@Manish Goyal • 16 Jan, 2010 Yes you are correct
Even i also don't know answer of 4th question
keep trying otherwise i will tell you answer of remaining questions
@Kaustubh Katdare • 16 Jan, 2010 1:-You are a landscape designer and your boss asked you to design a landscape such that you should place 4 trees equidistance from each other?Tell me How?
K: Place one tree on the top of hill and others in an equilateral triangle on the slope of the hill.
2:-If one tyre of a car suddenly gets stolen and after sometime you find the tyre without the screws how will u make ur journey complete?
K: Each tyre can have three screws each. 😉
3:-How can you measure a room height using a thermometer?
K: By using the thermometer as a scale 😐
4:-How would you catch and receive a ball in same direction? (Dropping is from north
and receiving from bottom not accepted, as it is 2 directions)
K: I don't understand this question. 😕
5:-Can you make 120 with 5 zeros?
K: cos 0 = 1 😉 (wink!)
@Manish Goyal • 16 Jan, 2010 yes biggie you are right for 1st and 2nd question only
@Manish Goyal • 16 Jan, 2010 One more question to add you have two candles each will burn for 1 hour individually ie one candle will take 1 hour to burn completely.Now if you are asked to burn both of these candle for 90 minutes ie both candles must burn in 90 minutes?Can you explain me how?
@cooltwins • 16 Jan, 2010 by melting them and casting them as one we can extend the burning time more right 😐
or do you want to make them burn to 90 min each :roll:
can you please be more specific.... :-o
@Manish Goyal • 16 Jan, 2010 Each candle take 1 hour to complete
My question is to burn both candles in such a way that both will take 90 minutes collectively
ie total time should be 90 minutes to burn both
@Vega • 17 Jan, 2010
goyal420
Each candle take 1 hour to complete
My question is to burn both candles in such a way that both will take 90 minutes collectively
ie total time should be 90 minutes to burn both
Ignite second candle after 30 minutes from first candle ignition.
@Manish Goyal • 17 Jan, 2010
Vega
Ignite second candle after 30 minutes from first candle ignition.
Yeah you are right..
Let us make it some more difficult .if i say 45 minutes😉
@ratikagoel • 18 Jan, 2010
goyal420
Let us make it some more difficult .if i say 45 minutes😉
First light up the two ends of the 1st candle and one end of the 2nd candle.
When the 1st candle will burn out ,then light up the other end of the 2nd candle too
@Manish Goyal • 18 Jan, 2010
ratikagoel
First light up the two ends of the 1st candle and one end of the 2nd candle.
When the 1st candle will burn out ,then light up the other end of the 2nd candle too
@ratikagoel • 18 Jan, 2010 4th ques:
If you throw a ball to the any direction ,due to heavy air in opposite direction it return to your place and u can able to catch the ball by not moving from your original place.
@rajivtiwari • 18 Jan, 2010 4th ques
Throw the ball on the wall
it will rebound and u can catch it standing at your own place in the very same direction
@Manish Goyal • 18 Jan, 2010
rajivtiwari
4th ques
Throw the ball on the wall
it will rebound and u can catch it standing at your own place in the very same direction
Quite satisfactory answer ..actually i also don't know it's answer but i think you are right😀
@rajivtiwari • 18 Jan, 2010 Okay it was kind of easy for me
I heard similar question few years back which i quite correctly remember
Although i don't know it was really a aptitude question or my cousin fooling around
The question was
Everyone was showed the window of the room.
and asked to jump from the window
but a guy climbed on the window pane
and jumped back in the room
and he was selected
So applied kind of same logic
@Manish Goyal • 18 Jan, 2010
The question was
Everyone was showed the window of the room.
and asked to jump from the window
@sherya mathur • 19 Feb, 2010 we have two candles have to burn completely in 90 minutes
we will start second candle after 30 minutes of the first candle
@sherya mathur • 19 Feb, 2010 for 45 minutes start burning the one candle from both ends and at the same time start burning second one from the one end
lets see after 30 minutes the the first one will completely burnt out
and second one will out half now
lit the second end of the second candle it will take 15 minute to completely burnt
@ramkumar.j • 20 Feb, 2010
goyal420
One more question to add you have two candles each will burn for 1 hour individually ie one candle will take 1 hour to burn completely.Now if you are asked to burn both of these candle for 90 minutes ie both candles must burn in 90 minutes?Can you explain me how?
ans: v have to burn the first candle for half an hr. after that v can burn the second candle. so, at one hr both the candles ll be burned. am i correct?
@ramkumar.j • 20 Feb, 2010
goyal420
One more question to add you have two candles each will burn for 1 hour individually ie one candle will take 1 hour to burn completely.Now if you are asked to burn both of these candle for 90 minutes ie both candles must burn in 90 minutes?Can you explain me how?
ans: v have to burn the first candle for half an hr. after that v can burn the second candle. so, at 1 hr both the candles ll be burned.
@ramkumar.j • 20 Feb, 2010 ans: v have to burn the first candle for half an hr. after that v can burn the second candle. so, at 90 min both the candles ll be burned.
@naga lakshmi • 08 Jan, 2012 1) Place thermometer perpendicular to the floor and against the wall.
2) Make a mark on the wall at the top of the thermometer.
3) Shift thermometer up so bottom is at mark.
4) Repeat 2) and 3) until close to the ceiling.
5) Shift thermometer up until it touches ceiling and mark location of highest mark on thermometer to determine fraction of thermometer length for last segment.
This would give the height of the room in units of "thermometers". This is not a standard unit of measurement, but it is a measurement. If you desire a more standard unit like centimeters, then measure the length of the thermometer and multiply by height of the room in "thermometers".
@Nidhi Shetty • 12 Mar, 2014 -How would you catch and receive a ball in same direction? (Dropping is from north
and receiving from bottom not accepted, as it is 2 directions)
isnt it the same? catching a ball or receiving a ball?
@Anunay Sinha • 23 Mar, 2014 • 1 like he meant throwing and recieving the ball.
and i think the simplest way is to use the ball in pendulum, push it in any direction, wait for it to reach the extreme position in opposite direction, then recieve it in between. 😁
@Nikhitha Pitla • 22 Oct, 2018
There are 3 people A,B,C. Liars are of same type and truth speaking are of same type. Find out who is speaking truth and who is speaking false when they say:
A says, B is liar.
B says, A and C are of same type.
6.7k views
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# Control Systems Lab
Experiment 9
Name of the student
ID No.
BATCH & TIME
:
:
:
NISHANT PANI
2013A3PS298H
PS-3 (9.00 AM -11.00 AM)
Contents
A
SWING UP CONTROL1
9.1
Background .................................................................................................................................................................. 2
9.1.1 Energy Control ............................................................................................................................................................... 2
9.1.2 Hybrid Swing-Up Control .............................................................................................................................................. 4
9.2
In-Lab Exercises .......................................................................................................................................................... 4
B
FREQUENCY RESPONSE ANALYSIS
.11
9.3
Background ................................................................................................................................................................ 19
9.4
In-Lab Exercises ........................................................................................................................................................ 20
9.4.1
Bode Plot
.11
9.4.2 Gain and Phase Margin ..................................................................................................................................................... 23
9.4.3 Closed-Loop Performance ................................................................................................................................................ 28
A. SWING-UP CONTROL
Topics Covered
Energy control
Nonlinear control
Control switching logic
Prerequisites
Before starting this lab make sure:
Filtering Lab
Balance Control Lab
Rotary pendulum module is attached to the QUBE-Servo.
9.1
9.1.1
Background
Energy Control
If the arm angle is kept constant and the pendulum is given an initial position it would swing with constant
amplitude. Because of friction there will be damping in the oscillation. The purpose of energy control is to
control the pendulum in such a way that the friction is constant. The potential energy of the pendulum is
9.1
and the kinetic energy is
9.2
The pendulum angle, , and the lengths are illustrated in Figure 9.1. The moment of inertia in this case is
lp = Lp/2.
Fig. 9.1
## Free-body diagram of pendulum
The potential energy is zero when the pendulum is at rest at = 0 and equals 2Mp g lp when the pendulum
is upright at = . The sum of the potential and kinetic energy of the pendulum is
9.3
Differentiating 9.3 results in the differential equation
9.4
Substituting the pendulum equation of motion
9.5
pendulum acceleration is given as,
9.6
Since the acceleration of the pivot is proportional to current driving the arm motor and thus also proportional
to the drive voltage we find that it is easy to control the energy of the pendulum. The proportional control law
9.7
drives the energy towards the reference energy Er. Notice that the control law is nonlinear because the
proportional gain depends on the pendulum angle, . Also, notice that the control changes sign when
changes sign and when the angle is 90 deg.
However, for energy to change quickly the magnitude of the control signal must be large. As a result the
following swing-up controller is implemented in the controller as
9.8
3
where is a tunable control gain and the satumax function saturates the control signal at the maximum
acceleration of the pendulum pivot, umax.
9.1.2
## Hybrid Swing-Up Control
The energy swing-up control can be combined with the balancing control law to obtain a control law
which performs the dual tasks of swinging up the pendulum and balancing it.
Similarly as described in the Balance Control Lab, the balance control is to be enabled when the pendulum
is within 20 degrees. When it is not enabled, the swing-up control is engaged. Thus the switching can be
described mathematically by:
9.2
In-Lab Exercises
The LabVIEW Virtual Instrument (VI) shown in Figure 9.1 swings-up and balances the
pendulum on the QUBE Servo Rotary Pendulum system. The Swing-Up Control subsystem
implements the energy control described in Section 9.1
## 9.2.1 Energy Control
Open the QUBE-Servo ROTPEN Swing Up.vi in LabVIEW .
2. To turn the swing-up control off, set mu to 0.
3. Run the VI.
4. Manually rotate the pendulum at different levels and examine the pendulum angle and energy in the
Pendulum (deg) and Pendulum Energy (mJ) scopes.
5. What do you notice about the energy when the pendulum is moved at different positions? Record the energy
when the pendulum is being balanced (i.e., fully inverted in the upright vertical position). Does this
reading make sense in terms of the equations developed in Section 9.1?.
1.
## Observations and Results
6. Click on the Stop button to bring the pendulum down to the initial, downward position.
7. Set the swing-up control parameters (i.e., the Constant and Gain blocks connected to the inputs of the
Swing-Up Control subystem) to the following:
mu = 50 m/s2/J
Er = 10.0 mJ
u_max = 6 m/s2
8. If the pendulum is not moving, gently perturb the pendulum with your hand to get it going.
9. Vary the reference energy, Er , between 10.0 mJ and 20.0 mJ. As it is changed, examine the pendulum
angle and energy response in Pendulum (deg) and the Pendulum Energy (mJ) scopes and the control
signal in the Motor Voltage (V) scope. Attach the responses showing how changing the reference energy
7
## affects the system.
Observations and Results
For increased energy in the system, the amplitude of the rotary arm remains the
same. Whereas the amplitude of the rotation of the pendulum increases.
10
## Figure 9.3: Pendulum response when changing reference energy
10. Fix Er to 20.0 mJ and vary the swing-up control gain mu between 20 and 60 m/s2/J. Describe how this
changes the performance of the energy control.
Observations and Results
As we increase the control gain the energy level of the pendulum increases,the
applied motor voltage increases,the rotation of the pendulum increases whereas
the angle of the rotary arm remains constant.
11
12
13
14
15
## 9.2.2 Hybrid Swing-Up Control
1. Open the QUBE-Servo ROTPEN Swing Up.vi in
LabVIEW .
## 2. To turn the swing-up control off, set mu to 0.
3. Set the swing-up control parameters to the following:
mu = 20 m/s2/J
16
u_max = 6 m/s2
Question 9.18
4. Based on your observations in the previous lab, what should the reference energy be set to?
The reference energy should be set to 30 mJ, i.e., Er = 30. This is the energy that measured in Step 5 in
Section 8.6.4 when the pendulum was in the vertically upwards, balanced position.
5. Make sure the pendulum is hanging down motionless and the encoder cable is not interfering with the
pendulum.
6. Build and run the QUARC controller.
7. The pendulum should begin going back and forth. If not, manually perturb the pendulum with your hand.
Click
on the Stop button in the Simulink tool bar if the pendulum goes unstable.
8. Gradually increase the swing-up gain, , denoted as the mu Slider Gain block, until the pendulum swings
up to the vertical position. Capture a response of the swing-up and record the swing-up gain that was required.
Show the pendulum angle, pendulum energy, and motor voltage.
Observation / Result:
17
18
## 9. Click on the Stop button to stop running the VI.
10. Power OFF the QUBE-Servo if no more experiments will be conducted.
## A. FREQUENCY RESPONSE ANALYSIS
Topics Covered
Bode Plots
Gain and Phase Margin
Closed-Loop Performance
9.3 Background
The frequency response method may be less intuitive than other methods you have studied previously.
However, it has certain advantages, especially in real-life situations such as modeling transfer functions
from physical data.
The frequency response is a representation of the system's response to sinusoidal inputs at varying
frequencies. The output of a linear system to a sinusoidal input is a sinusoid of the same frequency but
with a different magnitude and phase. The frequency response is defined as the magnitude and phase
differences between the input and output sinusoids. In this tutorial, we will see how we can use the
open-loop frequency response of a system to predict its behavior in closed-loop.
19
To plot the frequency response, we create a vector of frequencies (varying between zero or "DC" and
infinity) and compute the value of the plant transfer function at those frequencies. If G(s) is the open
loop transfer function of a system and w is the frequency vector, we then plot G(j*w) versus w. Since
G(j*w) is a complex number, we can plot both its magnitude and phase (the Bode plot) in the complex
plane .
## 9.4 In-Lab Exercises
9.4.1 Bode Plots
As noted above, a Bode plot is the representation of the magnitude and phase of G(j*w) (where the
frequency vector w contains only positive frequencies).
LabVIEW Graphical Approach
To see the Bode plot of a transfer function, you can use the CD Bode VI, located in the Frequency
Response section of the Control Design palette.
20
## LabVIEW MathScript Approach
Alternatively, you can use the following m-file code in the MathScript Window (Tools MathScript
Window):
num = 50;
den = [1 9 30 40];
sys = tf(num,den);
bode(sys)
With either approach, we display the transfer function and Bode plots for the system. Figure 9.6 shows the
front panel of the VI that was built. Change the terms in the numerator and denominator to observe the effect
on the systems Bode plots.
21
## Figure 9.6: : Front Panel for Creating a Bode Plot
Note the axes of the plots in Figure. The frequency is on a logarithmic scale, the phase is given in degrees,
and the magnitude is given as the gain in decibels. A decibel is defined as 20*log10 ( |G(j*w)| ) .
Observation & Result
22
## 9.4.2 Gain and Phase Margin
Let's say that we have the following system:
## Figure 9.7: A Closed-Loop System
In this system, K is a variable (constant) gain and G(s) is the plant under consideration. The gain margin is
defined as the change in open loop gain required to make the system unstable. Systems with greater gain
margins can withstand greater changes in system parameters before becoming unstable in closed loop. Keep
in mind that unity gain in magnitude is equal to a gain of zero in dB.
The phase margin is defined as the change in open loop phase shift required to make a closed loop system
unstable. The phase margin also measures the system's tolerance to time delay. If there is a time delay greater
than 180/Wpc in the loop (where Wpc is the frequency where the phase shift is 180 deg), the system will
become unstable in closed loop. The time delay can be thought of as an extra block in the forward path of the
block diagram that adds phase to the system but has no effect the gain. That is, a time delay can be represented
as a block with magnitude of 1 and phase w*time delay (in radians/second).
23
For now, we won't worry about where all this comes from and will concentrate on identifying the gain and
phase margins on a Bode plot.
The phase margin is the difference in phase between the phase curve and -180 degrees, at the point
corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, Wgc). Likewise,
the gain margin is the difference between the magnitude curve and 0dB at the point corresponding to the
frequency that gives us a phase of -180 deg (the phase cross over frequency, Wpc).
## Figure 9.8 : Gain and Phase Margins
One nice thing about the phase margin is that you don't need to re-plot the Bode in order to find the new phase
margin when changing the gains. If you recall, adding gain only shifts the magnitude plot up. This is the
equivalent of changing the y-axis on the magnitude plot. Finding the phase margin is simply the matter of
finding the new cross-over frequency and reading off the phase margin.
To observe this effect, first look at the Bode plots in Figure 9.8. You should see that the phase margin is about
100 degrees.
Now suppose you added a gain of 100.
LabVIEW Graphical Approach
To achieve this in a VI, add a second CD Construct Transfer Function Model VI to your block diagram from
Figure 1. Create a constant input to the numerator terminal, and enter 100 into the first cell of this array. Next,
add the CD Series VI to the block diagram (from the Model Interconnection section of the Control Design
palette) and connect both transfer function models to the inputs of the CD Series VI. Connect the Series Model
output of the CD Series VI to the CD Bode and CD Draw Transfer Function VIs as before.
24
## LabVIEW MathScript Approach
If you used m-file code to model the system, enter the command bode(100*sys) into the MathScript
Window.
Observation & Result
25
## LabVIEW Graphical Approach
Using the VI from Figure 9.5, replace the CD Bode VI with the CD Gain and Phase Margin VI, found in the
Frequency Response section of the Control Design palette. Create indicators for Magnitude Plot, Phase Plot,
and Gain and Phase Margins.
26
Figure 9.11: Block Diagram for Gain and Phase Margin in LabVIEW
## LabVIEW MathScript Approach
Alternatively, you can use the margin command in the MathScript Window. This command returns the gain
and phase margins, the gain and phase cross over frequencies, and a graphical representation of these on the
Bode plot. If you used m-file code to model the system, enter the following command into the MathScript
Window:
margin(sys)
Plotting the gain and phase margins returns the graphs shown below in Figure 9.12.
27
## 9.4.3 Closed-Loop Performance
In order to predict closed-loop performance from open-loop frequency response, we need to have several
concepts clear:
The system must be stable in open loop if we are going to design via Bode plots.
If the gain cross over frequency is less than the phase cross over frequency (i.e. Wgc < Wpc), then the
closed-loop system will be stable.
For second-order systems, the closed-loop damping ratio is approximately equal to the phase margin
divided by 100 if the phase margin is between 0 and 60 degrees. We can use this concept with caution if
the phase margin is greater than 60 degrees.
A very rough estimate that you can use is that the bandwidth is approximately equal to the natural
frequency.
Let's use these concepts to design a controller for the following system:
28
## Settling time must be less than 2 seconds.
There are two ways of solving this problem: one is graphical and the other is numerical. In LabVIEW, the
graphical approach is best, so that is the approach we will use. First, let's look at the Bode plot. Use the VI
from Figure 9.5 (or the MathScript Window) to graph the Bode plots for the above transfer function.
## Figure 9.14: Bode Plots
There are several characteristics of the system that can be read directly from this Bode plot. First of all, we
can see that the bandwidth frequency is around 10 rad/sec. Since the bandwidth frequency is roughly the same
as the natural frequency (for a first order system of this type), the rise time is 1.8/BW = 1.8/10 = 1.8 seconds.
This is a rough estimate, so we will say the rise time is about 2 seconds.
The phase margin for this system is approximately 95 degrees. The relation damping ratio = (phase
margin)/100 only holds for phase margins less than 60 degrees. Since the system is first-order, there should
be no overshoot.
The last major point of interest is steady-state error. The steady-state error can be read directly off the Bode
plot as well. The constant (Kp, Kv, or Ka) is found from the intersection of the low frequency asymptote with
the w=1 line. Just extend the low frequency line to the w=1 line. The magnitude at this point is the constant.
Since the Bode plot of this system is a horizontal line at low frequencies (slope = 0), we know this system is
29
of type zero. Therefore, the intersection is easy to find. The gain is 20dB (magnitude 10). What this means is
that the constant for the error function is 10.
The steady-state error for this system is 1/(1+Kp) = 1/(1+10)=0.091. If our system was type one instead of
type zero, the constant for the steady-state error would be found in a manner similar to the following:
## Let's check our predictions by looking at a step response plot.
LabVIEW Graphical Approach
This can be achieved in LabVIEW by using the CD Step Response VI (from the Time Response section of
the Control Design palette), along with the CD Feedback VI.
## LabVIEW MathScript Approach
Alternatively, you can add two lines of code into the MathScript Window:
sys_cl = feedback(sys,1);
step(sys_cl)
Which controller is required to be chosen to meet the desired specifications and why ? PID controller is
30
31
Conclusion
32
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Implement Queue Using Two Stacks Problem Statement
Given a sequence of enqueue and dequeue operations, return a result of their execution without using a queue implementation from a library. Use two stacks to implement queue.
Operations are given in the form of a linked list, and you need to return the result as a linked list, too. Operations:
• A non-negative integer means "enqueue me".
• -1 means
• If the queue is not empty, dequeue current head and append it to the result.
• If the queue is empty, append -1 to the result.
Example One
``````{
"operations": [1, -1, 2, -1, -1, 3, -1]
}
``````
Output:
``````[1, 2, -1, 3]
``````
Here is how we would execute the operations and build the result list:
Operation Queue contents after the operation Result list after the operation
1 [1] []
-1 [] [1]
2 [2] [1]
-1 [] [1, 2]
-1 [] [1, 2, -1]
3 [3] [1, 2, -1]
-1 [] [1, 2, -1, 3]
Example Two
``````{
"operations": [0, 1, 2, -1, 3]
}
``````
Output:
``````[0]
``````
The only dequeue operation results in the first enqueued element, 0, to be appended to the result list.
Notes
Constraints:
• -1 <= value in the list of `operations` <= 2 * 109
• 1 <= number of operations <= 105
• There will be at least one dequeue (-1) operation
We have provided two solutions. We will refer to the length of given linked list `operations` as `N`.
Implement Queue Using Two Stacks Solution 1: Brute Force
Any time we process an operation that puts a number to the queue, we need to store it somewhere. We are only allowed to use stacks to store numbers, so let’s be pushing all enqueued numbers into `stack1`.
Any time an operation tells us to retrieve a number from the queue, that number would be at the bottom of `stack1` then. To access it we’d need to pop all the numbers from there. Luckily we are allowed to use another stack. So we can temporarily push all the numbers into `stack2` and access the desired number from the bottom of `stack1`. After that we can push all the remaining numbers back from `stack2` to `stack1`. They will end up being in the same order as they were there before the dequeue operation, and that works for us.
Time Complexity
O(N2).
For processing all `N` operations. Every dequeue operation requires moving around all the numbers so far enqueued, that makes the dequeue operation O(N).
Auxiliary Space Used
O(N).
Because we never store more than `N` numbers.
Space Complexity
O(N).
Input, output and temporary data structures are all O(N). So, O(N) + O(N) + O(N) = O(N).
Code For Implement Queue Using Two Stacks Solution 1: Brute Force
``````/*
Asymptotic complexity in terms of number of operations:
* Time: O(n^2).
* Auxiliary space: O(n).
* Total space: O(n).
*/
{
{
}
else
{
tail->next = node;
tail = tail->next;
}
return tail;
}
int dequeue(stack<int> &s1, stack<int> &s2)
{
if (s1.empty()) return -1;
// Pop all elements from first to second stack.
while (s1.empty() == false)
{
s2.push(s1.top());
s1.pop();
}
// Head of the second stack is the element to dequeue.
int val = s2.top();
s2.pop();
// Shovel all elements back to the first stack.
while (s2.empty() == false)
{
s1.push(s2.top());
s2.pop();
}
return val;
}
{
stack<int> s1, s2;
while (operations != NULL)
{
if (operations->value >= 0)
{
s1.push(operations->value);
}
else
{
tail = insert_node(head, tail, dequeue(s1, s2));
{
}
}
operations = operations->next;
}
}
``````
Implement Queue Using Two Stacks Solution 2: Optimal
Let’s imagine that we started with the brute force solution described above and came to this situation:
`stack1 = [1, 2, 3]`, `stack2 = []`, `next operation: -1`
To dequeue, we move all numbers from `stack1` to `stack2`:
`stack1 = []`, `stack2 = [3, 2]`, add append number 1 to the `result`.
Now we can notice that `stack2` has all the remaining numbers in the order that’s perfect for us. For example, if the next operation is -1, we can simply pop and return number 2 from `stack2` - a constant time operation.
We can dequeue elements this way if we leave them in `stack2`, but what about enqueueing new ones? It turns out that we can push them in `stack1`, and they can remain there until `stack2` is empty. Once `stack2` is empty and another dequeue operation comes, we can do what was described two paragraphs ago: pop all numbers from `stack1` and push them into `stack2`.
Time Complexity
O(N).
Enqueue operation takes constant time, clearly. Let us examine dequeue operation.
Most dequeue operations will just need to pop one number from `stack2`, that’s constant time. Some dequeue operations however will need to move some numbers from `stack1` to `stack2`.
Amortized time complexity of the dequeue operation is constant and intuitively we can observe that we never move any given number more than once between `stack1` and `stack2` (that’s constant time per number).
Overall time complexity of the algorithm would be O(N) since we process `N` operations, taking constant time per operation (amortized).
Auxiliary Space Used
O(N).
As we never store more than `N` numbers in our stacks.
O(N).
Code For Implement Queue Using Two Stacks Solution 2: Optimal
``````/*
Asymptotic complexity in terms of number of operations:
* Time: O(n).
* Auxiliary space: O(n).
* Total space: O(n).
*/
{
{
}
else
{
tail->next = node;
tail = tail->next;
}
return tail;
}
int dequeue(stack<int> &s1, stack<int> &s2)
{
if (s1.empty() && s2.empty())
{
return -1;
}
// If second stack isn't empty, pop and return from it.
if (s2.empty() == false)
{
int val = s2.top();
s2.pop();
return val;
}
// Otherwise pop all elements from first to second stack.
while (s1.empty() == false)
{
s2.push(s1.top());
s1.pop();
}
// And then pop and return head of the second stack.
int val = s2.top();
s2.pop();
return val;
}
{
stack<int> s1, s2;
while (operations != NULL)
{
if (operations->value >= 0)
{
s1.push(operations->value);
}
else
{
tail = insert_node(head, tail, dequeue(s1, s2));
{
}
}
operations = operations->next;
}
}
``````
We hope that these solutions to implement queue using two stacks problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.
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Note: Input and Output will already be taken care of.
Implement Queue Using Two Stacks Problem Statement
Given a sequence of enqueue and dequeue operations, return a result of their execution without using a queue implementation from a library. Use two stacks to implement queue.
Operations are given in the form of a linked list, and you need to return the result as a linked list, too. Operations:
• A non-negative integer means "enqueue me".
• -1 means
• If the queue is not empty, dequeue current head and append it to the result.
• If the queue is empty, append -1 to the result.
Example One
``````{
"operations": [1, -1, 2, -1, -1, 3, -1]
}
``````
Output:
``````[1, 2, -1, 3]
``````
Here is how we would execute the operations and build the result list:
Operation Queue contents after the operation Result list after the operation
1 [1] []
-1 [] [1]
2 [2] [1]
-1 [] [1, 2]
-1 [] [1, 2, -1]
3 [3] [1, 2, -1]
-1 [] [1, 2, -1, 3]
Example Two
``````{
"operations": [0, 1, 2, -1, 3]
}
``````
Output:
``````[0]
``````
The only dequeue operation results in the first enqueued element, 0, to be appended to the result list.
Notes
Constraints:
• -1 <= value in the list of `operations` <= 2 * 109
• 1 <= number of operations <= 105
• There will be at least one dequeue (-1) operation
We have provided two solutions. We will refer to the length of given linked list `operations` as `N`.
Implement Queue Using Two Stacks Solution 1: Brute Force
Any time we process an operation that puts a number to the queue, we need to store it somewhere. We are only allowed to use stacks to store numbers, so let’s be pushing all enqueued numbers into `stack1`.
Any time an operation tells us to retrieve a number from the queue, that number would be at the bottom of `stack1` then. To access it we’d need to pop all the numbers from there. Luckily we are allowed to use another stack. So we can temporarily push all the numbers into `stack2` and access the desired number from the bottom of `stack1`. After that we can push all the remaining numbers back from `stack2` to `stack1`. They will end up being in the same order as they were there before the dequeue operation, and that works for us.
Time Complexity
O(N2).
For processing all `N` operations. Every dequeue operation requires moving around all the numbers so far enqueued, that makes the dequeue operation O(N).
Auxiliary Space Used
O(N).
Because we never store more than `N` numbers.
Space Complexity
O(N).
Input, output and temporary data structures are all O(N). So, O(N) + O(N) + O(N) = O(N).
Code For Implement Queue Using Two Stacks Solution 1: Brute Force
``````/*
Asymptotic complexity in terms of number of operations:
* Time: O(n^2).
* Auxiliary space: O(n).
* Total space: O(n).
*/
{
{
}
else
{
tail->next = node;
tail = tail->next;
}
return tail;
}
int dequeue(stack<int> &s1, stack<int> &s2)
{
if (s1.empty()) return -1;
// Pop all elements from first to second stack.
while (s1.empty() == false)
{
s2.push(s1.top());
s1.pop();
}
// Head of the second stack is the element to dequeue.
int val = s2.top();
s2.pop();
// Shovel all elements back to the first stack.
while (s2.empty() == false)
{
s1.push(s2.top());
s2.pop();
}
return val;
}
{
stack<int> s1, s2;
while (operations != NULL)
{
if (operations->value >= 0)
{
s1.push(operations->value);
}
else
{
tail = insert_node(head, tail, dequeue(s1, s2));
{
}
}
operations = operations->next;
}
}
``````
Implement Queue Using Two Stacks Solution 2: Optimal
Let’s imagine that we started with the brute force solution described above and came to this situation:
`stack1 = [1, 2, 3]`, `stack2 = []`, `next operation: -1`
To dequeue, we move all numbers from `stack1` to `stack2`:
`stack1 = []`, `stack2 = [3, 2]`, add append number 1 to the `result`.
Now we can notice that `stack2` has all the remaining numbers in the order that’s perfect for us. For example, if the next operation is -1, we can simply pop and return number 2 from `stack2` - a constant time operation.
We can dequeue elements this way if we leave them in `stack2`, but what about enqueueing new ones? It turns out that we can push them in `stack1`, and they can remain there until `stack2` is empty. Once `stack2` is empty and another dequeue operation comes, we can do what was described two paragraphs ago: pop all numbers from `stack1` and push them into `stack2`.
Time Complexity
O(N).
Enqueue operation takes constant time, clearly. Let us examine dequeue operation.
Most dequeue operations will just need to pop one number from `stack2`, that’s constant time. Some dequeue operations however will need to move some numbers from `stack1` to `stack2`.
Amortized time complexity of the dequeue operation is constant and intuitively we can observe that we never move any given number more than once between `stack1` and `stack2` (that’s constant time per number).
Overall time complexity of the algorithm would be O(N) since we process `N` operations, taking constant time per operation (amortized).
Auxiliary Space Used
O(N).
As we never store more than `N` numbers in our stacks.
O(N).
Code For Implement Queue Using Two Stacks Solution 2: Optimal
``````/*
Asymptotic complexity in terms of number of operations:
* Time: O(n).
* Auxiliary space: O(n).
* Total space: O(n).
*/
{
{
}
else
{
tail->next = node;
tail = tail->next;
}
return tail;
}
int dequeue(stack<int> &s1, stack<int> &s2)
{
if (s1.empty() && s2.empty())
{
return -1;
}
// If second stack isn't empty, pop and return from it.
if (s2.empty() == false)
{
int val = s2.top();
s2.pop();
return val;
}
// Otherwise pop all elements from first to second stack.
while (s1.empty() == false)
{
s2.push(s1.top());
s1.pop();
}
// And then pop and return head of the second stack.
int val = s2.top();
s2.pop();
return val;
}
{
stack<int> s1, s2;
while (operations != NULL)
{
if (operations->value >= 0)
{
s1.push(operations->value);
}
else
{
tail = insert_node(head, tail, dequeue(s1, s2));
{
}
}
operations = operations->next;
}
}
``````
We hope that these solutions to implement queue using two stacks problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.
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# Carnegie Mellon Instructors: Dave O’Hallaron, Greg Ganger, and Greg Kesden Floating Point 15-213: Introduction to Computer Systems 4 th Lecture, Sep 6,
## Presentation on theme: "Carnegie Mellon Instructors: Dave O’Hallaron, Greg Ganger, and Greg Kesden Floating Point 15-213: Introduction to Computer Systems 4 th Lecture, Sep 6,"— Presentation transcript:
Carnegie Mellon Instructors: Dave O’Hallaron, Greg Ganger, and Greg Kesden Floating Point 15-213: Introduction to Computer Systems 4 th Lecture, Sep 6, 2012
2 Carnegie Mellon Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Summary
3 Carnegie Mellon Fractional binary numbers What is 1011.101 2 ?
4 2i2i 2 i-1 4 2 1 1/2 1/4 1/8 2 -j bibi b i-1 b2b2 b1b1 b0b0 b -1 b -2 b -3 b -j Carnegie Mellon Fractional Binary Numbers Representation Bits to right of “binary point” represent fractional powers of 2 Represents rational number:
5 Carnegie Mellon Fractional Binary Numbers: Examples ValueRepresentation 5 3/4101.11 2 2 7/8010.111 2 1 7/16001.0111 2 Observations Divide by 2 by shifting right (unsigned) Multiply by 2 by shifting left Numbers of form 0.111111… 2 are just below 1.0 1/2 + 1/4 + 1/8 + … + 1/2 i + … ➙ 1.0 Use notation 1.0 – ε
6 Carnegie Mellon Representable Numbers Limitation #1 Can only exactly represent numbers of the form x/2 k Other rational numbers have repeating bit representations ValueRepresentation 1/3 0.0101010101[01]… 2 1/5 0.001100110011[0011]… 2 1/10 0.0001100110011[0011]… 2 Limitation #2 Just one setting of decimal point within the w bits Limited range of numbers (very small values? very large?)
7 Carnegie Mellon Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Summary
8 Carnegie Mellon IEEE Floating Point IEEE Standard 754 Established in 1985 as uniform standard for floating point arithmetic Before that, many idiosyncratic formats Supported by all major CPUs Driven by numerical concerns Nice standards for rounding, overflow, underflow Hard to make fast in hardware Numerical analysts predominated over hardware designers in defining standard
9 Carnegie Mellon Numerical Form: (–1) s M 2 E Sign bit s determines whether number is negative or positive Significand M normally a fractional value in range [1.0,2.0). Exponent E weights value by power of two Encoding MSB s is sign bit s exp field encodes E (but is not equal to E) frac field encodes M (but is not equal to M) Floating Point Representation sexpfrac
10 Carnegie Mellon Precision options Single precision: 32 bits Double precision: 64 bits Extended precision: 80 bits (Intel only) sexpfrac 18-bits23-bits sexpfrac 111-bits52-bits sexpfrac 115-bits63 or 64-bits
11 Carnegie Mellon “Normalized” Values When: exp ≠ 000…0 and exp ≠ 111…1 Exponent coded as a biased value: E = Exp – Bias Exp: unsigned value exp Bias = 2 k-1 - 1, where k is number of exponent bits Single precision: 127 (Exp: 1…254, E: -126…127) Double precision: 1023 (Exp: 1…2046, E: -1022…1023) Significand coded with implied leading 1: M = 1.xxx…x 2 xxx…x : bits of frac Minimum when frac=000…0 (M = 1.0) Maximum when frac=111…1 (M = 2.0 – ε) Get extra leading bit for “free”
Carnegie Mellon 12 Normalized Encoding Example Value: Float F = 15213.0; 15213 10 = 11101101101101 2 = 1.1101101101101 2 x 2 13 Significand M = 1.1101101101101 2 frac= 11011011011010000000000 2 Exponent E = 13 Bias = 127 Exp = 140 = 10001100 2 Result: 0 10001100 11011011011010000000000 sexpfrac
13 Carnegie Mellon Denormalized Values Condition: exp = 000…0 Exponent value: E = –Bias + 1 (instead of E = 0 – Bias) Significand coded with implied leading 0: M = 0.xxx…x 2 xxx…x : bits of frac Cases exp = 000…0, frac = 000…0 Represents zero value Note distinct values: +0 and –0 (why?) exp = 000…0, frac ≠ 000…0 Numbers closest to 0.0 Equispaced
14 Carnegie Mellon Special Values Condition: exp = 111…1 Case: exp = 111…1, frac = 000…0 Represents value (infinity) Operation that overflows Both positive and negative E.g., 1.0/0.0 = −1.0/−0.0 = + , 1.0/−0.0 = − Case: exp = 111…1, frac ≠ 000…0 Not-a-Number (NaN) Represents case when no numeric value can be determined E.g., sqrt(–1), − , 0
15 Carnegie Mellon Visualization: Floating Point Encodings ++ −− 00 +Denorm+Normalized −Denorm −Normalized +0 NaN
16 Carnegie Mellon Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Summary
17 Carnegie Mellon Tiny Floating Point Example 8-bit Floating Point Representation the sign bit is in the most significant bit the next four bits are the exponent, with a bias of 7 the last three bits are the frac Same general form as IEEE Format normalized, denormalized representation of 0, NaN, infinity sexpfrac 14-bits3-bits
18 Carnegie Mellon s exp fracEValue 0 0000 000-60 0 0000 001-61/8*1/64 = 1/512 0 0000 010-62/8*1/64 = 2/512 … 0 0000 110-66/8*1/64 = 6/512 0 0000 111-67/8*1/64 = 7/512 0 0001 000-68/8*1/64 = 8/512 0 0001 001 -69/8*1/64 = 9/512 … 0 0110 110-114/8*1/2 = 14/16 0 0110 111-115/8*1/2 = 15/16 0 0111 00008/8*1 = 1 0 0111 00109/8*1 = 9/8 0 0111 010010/8*1 = 10/8 … 0 1110 110714/8*128 = 224 0 1110 111715/8*128 = 240 0 1111 000n/ainf Dynamic Range (Positive Only) closest to zero largest denorm smallest norm closest to 1 below closest to 1 above largest norm Denormalized numbers Normalized numbers
19 Carnegie Mellon Distribution of Values 6-bit IEEE-like format e = 3 exponent bits f = 2 fraction bits Bias is 23-1-1 = 3 Notice how the distribution gets denser toward zero. 8 values sexpfrac 13-bits2-bits
20 Carnegie Mellon Distribution of Values (close-up view) 6-bit IEEE-like format e = 3 exponent bits f = 2 fraction bits Bias is 3 sexpfrac 13-bits2-bits
21 Carnegie Mellon Special Properties of the IEEE Encoding FP Zero Same as Integer Zero All bits = 0 Can (Almost) Use Unsigned Integer Comparison Must first compare sign bits Must consider −0 = 0 NaNs problematic Will be greater than any other values What should comparison yield? Otherwise OK Denorm vs. normalized Normalized vs. infinity
22 Carnegie Mellon Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Summary
23 Carnegie Mellon Floating Point Operations: Basic Idea x + f y = Round(x + y) x f y = Round(x y) Basic idea First compute exact result Make it fit into desired precision Possibly overflow if exponent too large Possibly round to fit into frac
24 Carnegie Mellon Rounding Rounding Modes (illustrate with \$ rounding) \$1.40\$1.60\$1.50\$2.50–\$1.50 Towards zero\$1\$1\$1\$2–\$1 Round down (− )\$1\$1\$1\$2–\$2 Round up (+ ) \$2\$2\$2\$3–\$1 Nearest Even (default)\$1\$2\$2\$2–\$2
25 Carnegie Mellon Closer Look at Round-To-Even Default Rounding Mode Hard to get any other kind without dropping into assembly All others are statistically biased Sum of set of positive numbers will consistently be over- or under- estimated Applying to Other Decimal Places / Bit Positions When exactly halfway between two possible values Round so that least significant digit is even E.g., round to nearest hundredth 1.23499991.23(Less than half way) 1.23500011.24(Greater than half way) 1.23500001.24(Half way—round up) 1.24500001.24(Half way—round down)
26 Carnegie Mellon Rounding Binary Numbers Binary Fractional Numbers “Even” when least significant bit is 0 “Half way” when bits to right of rounding position = 100… 2 Examples Round to nearest 1/4 (2 bits right of binary point) ValueBinaryRoundedActionRounded Value 2 3/3210.00011 2 10.00 2 (<1/2—down)2 2 3/1610.00110 2 10.01 2 (>1/2—up)2 1/4 2 7/810.11100 2 11.00 2 ( 1/2—up)3 2 5/810.10100 2 10.10 2 ( 1/2—down)2 1/2
27 Carnegie Mellon FP Multiplication (–1) s1 M1 2 E1 x (–1) s2 M2 2 E2 Exact Result: (–1) s M 2 E Sign s: s1 ^ s2 Significand M: M1 x M2 Exponent E: E1 + E2 Fixing If M ≥ 2, shift M right, increment E If E out of range, overflow Round M to fit frac precision Implementation Biggest chore is multiplying significands
28 Carnegie Mellon Floating Point Addition (–1) s1 M1 2 E1 + (-1) s2 M2 2 E2 Assume E1 > E2 Exact Result: (–1) s M 2 E Sign s, significand M: Result of signed align & add Exponent E: E1 Fixing If M ≥ 2, shift M right, increment E if M < 1, shift M left k positions, decrement E by k Overflow if E out of range Round M to fit frac precision (–1) s1 M1 (–1) s2 M2 E1–E2 + (–1) s M
29 Carnegie Mellon Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Summary
30 Carnegie Mellon Floating Point in C C Guarantees Two Levels float single precision double double precision Conversions/Casting Casting between int, float, and double changes bit representation double / float → int Truncates fractional part Like rounding toward zero Not defined when out of range or NaN: Generally sets to TMin int → double Exact conversion, as long as int has ≤ 53 bit word size int → float Will round according to rounding mode
31 Carnegie Mellon Summary IEEE Floating Point has clear mathematical properties Represents numbers of form M x 2 E One can reason about operations independent of implementation As if computed with perfect precision and then rounded Not the same as real arithmetic Violates associativity/distributivity Makes life difficult for compilers & serious numerical applications programmers
32 Carnegie Mellon Floating Point Puzzles For each of the following C expressions, either: Argue that it is true for all argument values Explain why not true x == (int)(float) x x == (int)(double) x f == (float)(double) f d == (float) d f == -(-f); 2/3 == 2/3.0 d < 0.0 ⇒ ((d*2) < 0.0) d > f ⇒ -f > -d d * d >= 0.0 (d+f)-d == f int x = …; float f = …; double d = …; Assume neither d nor f is NaN
33 Carnegie Mellon More Slides
34 Carnegie Mellon Today: Floating Point Background: Fractional binary numbers IEEE floating point standard: Definition Example and properties Rounding, addition, multiplication Floating point in C Summary
35 Carnegie Mellon Interesting Numbers DescriptionexpfracNumeric Value Zero00…0000…000.0 Smallest Pos. Denorm.00…0000…012 – {23,52} x 2 – {126,1022} Single ≈ 1.4 x 10 –45 Double ≈ 4.9 x 10 –324 Largest Denormalized00…0011…11(1.0 – ε) x 2 – {126,1022} Single ≈ 1.18 x 10 –38 Double ≈ 2.2 x 10 –308 Smallest Pos. Normalized00…0100…001.0 x 2 – {126,1022} Just larger than largest denormalized One01…1100…001.0 Largest Normalized11…1011…11(2.0 – ε) x 2 {127,1023} Single ≈ 3.4 x 10 38 Double ≈ 1.8 x 10 308 { single,double }
36 Carnegie Mellon Mathematical Properties of FP Add Compare to those of Abelian Group Closed under addition? But may generate infinity or NaN Commutative? Associative? Overflow and inexactness of rounding 0 is additive identity? Every element has additive inverse Except for infinities & NaNs Monotonicity a ≥ b ⇒ a+c ≥ b+c? Except for infinities & NaNs Yes No Almost
37 Carnegie Mellon Mathematical Properties of FP Mult Compare to Commutative Ring Closed under multiplication? But may generate infinity or NaN Multiplication Commutative? Multiplication is Associative? Possibility of overflow, inexactness of rounding 1 is multiplicative identity? Multiplication distributes over addition? Possibility of overflow, inexactness of rounding Monotonicity a ≥ b & c ≥ 0 ⇒ a * c ≥ b *c? Except for infinities & NaNs Yes No Yes No Almost
38 Carnegie Mellon Creating Floating Point Number Steps Normalize to have leading 1 Round to fit within fraction Postnormalize to deal with effects of rounding Case Study Convert 8-bit unsigned numbers to tiny floating point format Example Numbers 12810000000 1500001101 3300010001 3500010011 13810001010 6300111111 sexpfrac 14-bits3-bits
39 Carnegie Mellon Normalize Requirement Set binary point so that numbers of form 1.xxxxx Adjust all to have leading one Decrement exponent as shift left ValueBinaryFractionExponent 128100000001.00000007 15000011011.10100003 17000100011.00010004 19000100111.00110004 138100010101.00010107 63001111111.11111005 sexpfrac 14-bits3-bits
40 Carnegie Mellon Rounding Round up conditions Round = 1, Sticky = 1 ➙ > 0.5 Guard = 1, Round = 1, Sticky = 0 ➙ Round to even ValueFractionGRSIncr?Rounded 1281.0000000000N 1.000 151.1010000100N 1.101 171.0001000010N 1.000 191.0011000110Y 1.010 1381.0001010011Y 1.001 631.1111100111Y10.000 1.BBGRXXX Guard bit: LSB of result Round bit: 1 st bit removed Sticky bit: OR of remaining bits
41 Carnegie Mellon Postnormalize Issue Rounding may have caused overflow Handle by shifting right once & incrementing exponent ValueRoundedExpAdjustedResult 128 1.0007128 15 1.1013 15 17 1.0004 16 19 1.0104 20 138 1.0017134 6310.00051.000/6 64
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Class X
Chapter 17 Circles
Maths
EXERCISE 17 (A)
Book Name: Selina Concise
Question 1:
A chord of length 6 cm is drawn in a circle of radius 5 cm. Calculate its distance from the centre of the circle.
Solution 1:
Let AB be the chord and O be the centre of the circle. Let OC be the perpendicular drawn from O to AB.
We know, that the perpendicular to a chord, from the centre of a circle, bisects the chord. AC = CB = 3 cm In ∆OCA, OA 2 = OC 2 + AC 2 (By Pythagoras theorem)
OC 2 = (5) 2 (3) 2 = 16
OC = 4 cm
Question 2:
A chord of length 8 cm is drawn at a distance of 3 cm from the centre of a circle. Calculate the
Solution 2:
Let AB be the chord and O be the centre of the circle. Let OC be the perpendicular drawn from O to AB.
Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.
Class X
Chapter 17 Circles
Maths
We know, that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ AB = 8 cm
AB
⟹ AC = CB =
2
8
⟹ AC = CB =
2
⟹ AC = CB = 4 cm
In ∆OCA,
OA
2
OC AC
2
2
(By Pythagoras theorem)
OA
2
4
2
3
2
25
OA 5 cm
Hence, radius of the circle is 5 cm.
Question 3:
The radius of a circle is 17.0 cm and the length of perpendicular drawn from its centre to a chord
is 8.0 cm. Calculate the length of the chord.
Solution 3:
Let AB be the chord and O be the centre of the circle.
Let OC be the perpendicular drawn from O to AB.
We know, that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ AC = CB
In ∆OCA,
OA
2 OC
2 AC
2 (By Pythagoras theorem)
AC
17
8
2
2
2 225
AC 51 cm
∴ AB = 2 AC = 2 × 15 = 30 cm
Class X
Chapter 17 Circles
Maths
Question 4:
A chord of length 24 cm is at a distance of 5 cm from the centre of the circle. Find the length of
the chord of the same circle which is at a distance of 12 cm from the centre.
Solution 4:
Let AB be the chord of length 24 cm and O be the centre of the circle.
Let OC be the perpendicular drawn from O to AB.
We know, that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ AC = CB = 12 cm
In ∆OCA,
OA
2
OC AC
2
2
(By Pythagoras theorem)
5
2
12
2
169
OA 13 cm
A'B'
∴ radius of the circle = 13 cm
Let
be new chord at a distance of 12 cm from the centre.
OA'
2
OC'
2
A'C'
2
A'C'
2
13
2
12
A'C' 5cm
2
25
Hence, length of the new chord = 2 × 5 = 10 cm
Question 5:
In the following figure, AD is a straight line. OP AD and O is the centre of both the circles. If OA = 34 cm. OB = 20 cm and OP = 16cm; find the length of AB.
Solution 5:
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Class X
Chapter 17 Circles
Maths
OP BC
For the inner circle, BC is a chord and
.
We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ BP = PC
By Pythagoras theorem,
OA
2
OP BP
2
2
BP
BP 12cm
2
20
2
16
2
144
For the outer circle, AD is the chord and
.
We know that the perpendicular to a chord, from the centre of a circle, bisects the chord.
∴ AP = PD
By Pythagoras Theorem,
OA 2 = OP 2 + AP 2
=> AP 2 = (34) 2 − (16) 2 = 900
=> AP = 30 cm
AB = AP − BP = 30 − 12 = 18 cm
Question 6:
O is the centre of a circle of radius 10 cm. P is any point in the circle such that OP = 6 cm. A is the point travelling along the circumference. x is the distance from A to P. what are the least and the greatest values of x in cm? what is the position of the points O, P and A at these values?
Solution 6:
The least value of x will be when A is on OP produced, i.e. O, P and A are collinear.
AP = OA - OP = 10 - 6 = 4 cm.
The maximum value of x will be when A is on PO produced, i.e. A, O and P are collinear.
AP = OA + OP = 10 + 6 = 16 cm.
Class X
Chapter 17 Circles
Maths
Question 7:
In a circle of radius 17 cm, two parallel chords of lengths 30 cm and 16 cm are drawn. Find the
distance between the chords, if both the chords are
(i) on the opposite sides of the centre,
(ii) on the same side of the centre.
Solution 7:
Let O be the centre of the circle and AB and CD be the two parallel chords of length 30 cm and
16 cm respectively.
Drop OE and OF perpendicular on AB and CD from the centre O.
OP AB
OF CD
and
∴ OE bisects AB and OF bisects CD
(Perpendicular drawn from the centre of a circle to a chord bisects it)
30
16
AE
15cm; CF
8cm
2
2
In right ∆OAE,
OA
2
OE AE
2
2
OE
OE 8cm
2
OA
2
AE
17
2
15
2
2
64
In right ∆OCF,
OC OF CF
2
2
2
OF
2
OC CF
2
2
17
2
8
2 225
OF 15 cm
(i) The chords are on the opposite sides of the centre:
EF EO OF 8 15 23 cm
(ii) The chords are on the same side of the centre:
EF OF OE 15 8 7 cm
Class X
Chapter 17 Circles
Maths
Question 8:
Two parallel chords are drawn in a circle of diameter 30.0 cm. The length of one chord is 24.0
cm and the distance between the two chords is 21.0 cm; find the length of another chord.
Solution 8:
Since the distance between the chords is greater than the radius of the circle (15 cm), so the
chords will be on the opposite sides of the centre.
Let
O be the centre of the circle and AB and CD be the two parallel chords such that
AB
= 24 cm.
Let
OE AB
length of CD be 2x cm.
OF CD
Drop OE and OF perpendicular on AB and CD from the centre O.
and
∴ OE bisects AB and OF bisects CD
(Perpendicular drawn from the centre of a circle to a chord bisects it)
24
2x
2
2
In right ∆OAE,
OA
2
OE AE
2
2
OE
OA
AE
2
2
2
15
2
12
2
81
OE 9 cm
OF EF OE 21 912
cm
In right ∆OCF,
OC OF
2
2 CF
2
x
OC OF
15
2
2
2
2
12
2 81
x 9 cm
CD 2x 29 18 cm
Hence, length of chord
AE
12cm; CF
x cm
Class X
Chapter 17 Circles
Maths
Question 9:
A chord CD of a circle whose centre is O, is bisected at P by a diameter AB.
Given OA = OB = 15 cm and OP = 9 cm. calculate the length of:
(i) CD
(iii) CB
Solution 9:
OP CD
∴ OP bisects CD
(Perpendicular drawn from the centre of a circle to a chord bisects it)
CD
CP
2
In right ∆OPC,
OC OP CP
2
2
2
CP
OC OP
2
2
2
15
2
9
2
144
CP 12 cm
CD 122 24 cm
BP OBOP 159 6 cm
(ii) Join BD
In right ∆BPD,
BD
2 BP
2
PD
2
6
2
12
2 180
(Angle in a semicircle is a right angle)
2
2
2
2
AB BD
2
2
30
2 180 720
AD 720 26.83 cm
BC BD
180 13.42 cm
(iii) Also,
Class X
Chapter 17 Circles
Maths
Question 10:
The figure given below, shows a circle with centre O in which diameter AB bisects the chord
CD at point E. If CE = ED = 8 cm and EB = 4cm, find the radius of the circle.
Solution 10:
OE OBEB r 4
Let the radius of the circle be r cm.
Join OC
In right ∆OEC,
OC
2
OE
2
CE
2
r
2
r 4
2
8
2
r
8r 80
2
r
2
8r 1664
r 10 cm
Hence, radius of the circle is 10 cm.
Question 11:
The figure shows two concentric circles and AD is a chord of larger circle.
Prove that: AB = CD
Class X
Chapter 17 Circles
Maths
Solution 11:
Drop
OP bisects AD (Perpendicular drawn from the centre of a circle to a chord bisects it)
AP PD
……………. (i)
Now, BC is a chord for the inner circle and
OP bisects BC (Perpendicular drawn from the centre of a circle to a chord bisects it)
OP BC
BP PC
……………. (ii)
Subtracting (ii) from (i),
AP PB PDPC
AB CD
Question 12:
A straight line is drawn cutting two equal circles and passing through the mid-point M of the line joining their centres O and O’
Prove that the chords AB and CD, which are intercepted by the two circles are equal.
Solution 12:
Given: A straight line Ad intersects two circles of equal radii at A, B, C and D.
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Class X
Chapter 17 Circles
Maths
The line joining the centres OO' intersect AD at M And M is the midpoint of OO' . To prove: AB = CD
Construction: From O, draw
Proof:
In ∆OMP and ∆O' MQ,
OP AB
and from O’, draw
OMP OPM O'MQ O'QM
(vertically opposite angles)
(each = 90°)
OM O'M
(Given)
O'Q CD.
By Angle Angle Side criterion of congruence,
OMP O'MQ,
(by AAS)
The corresponding parts of the congruent triangle are congruent
We know that two chords of a circle or equal circles which are equidistant from the centre are equal. AB = CD
OP =
O' Q
(c.p.ct)
Question 13:
M and N are the mid-points of two equal chords AB and CD respectively of a circle with centre O. prove that:
(i) BMN DNM . (ii) AMN CNM .
Solution 13:
Drop
OM AB
and
ON CD
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Class X
Chapter 17 Circles
Maths
OM bisects AB and ON bisects CD (Perpendicular drawn from the centre of a circle to a chord bisects it)
1 AB
2 2
1
Applying Pythagoras theorem,
OM
2
2
OB BM
2
OD DN
2
ON
2
2
OM ON
OMN ONM
(by (1))
……………(2)
(Angles opp to equal sides are equal)
(i) OMB OND
Subtracting (2) from above,
BMN DNM AMN CNM
(ii)
OMA ONC
(both 90°)
(both 90°)
BM
CD DN
……………(1)
Question 14:
In the following figure; P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO' ; prove that:
(i) OO' =
1
2
AB
,
(ii) AB = CD
Solution 14:
Drop OM and O'N perpendicular on AB and OM' and O'N' perpendicular on CD.
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Class X
Chapter 17 Circles
Maths
1
2
OO'
By (i) and (ii) AB = CD
1
2
MP
1
2
1
2
1
2
1 QD 2 AB ---------(i) 1 CD -------(ii) 2
MP
AP, PN
BP, M'Q
CQ, QN'
Now,
MN
PN
AP
BP
And
OO'
M'N'
M'Q
QN'
1 2
CQ
QD
Question 15:
Two equal chords AB and CD of a circle with centre O, intersect each other at point P inside the circle, prove that:
(i) AP = CP, (ii) BP = DP
Solution 15:
Drop OM and ON perpendicular on AB and CD. Join OP, OB and OD.
OM and ON bisect AB and CD respectively (Perpendicular drawn from the centre of a circle to chord bisects it)
2 2
In rt ∆ OMB, OM 2 In rt ∆ OND, ON 2
OB
2 MB
OD
2 ND
……… (i)
2
2
……………… (ii)
……………… (iii)
MP
1 AB
1
CD ND
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Class X
Chapter 17 Circles
Maths
From (i),(ii) and (iii) OM = ON In ∆OPM and ∆OPN,
OMP ONP
(both 90°)
OP = OP OM = ON
By Right Angle Hypotenuse Side criterion of congruence,
(Common)
(Proved above)
OPM OPN
(by RHS)
The corresponding parts of the congruent triangles are congruent.
PM = PN
Adding (i) to both sides, MB + PM = ND + PN
BP = DP
Now, AB = CD
AB BP = CD DP (BP = DP)
AP = CP
(c.p.c.t)
Question 16:
In the following figure, OABC is a square. A circle is drawn with O as centre which meets OC at P and OA at Q. Prove that:
(i) ∆OPA ∆OQC,
(ii) ∆BPC ∆BQA.
Solution 16:
Class X
Chapter 17 Circles
Maths
(i)
In ∆OPA and ∆OQC, OP = OQ
AOP = COQ OA = OC
By Side Angle Side criterion of congruence,
∆OPA ∆OQC
(radii of same circle) (both 90°) (Sides of the square)
(by SAS)
(ii) Now, OP = OQ (radii) And OC = OA (sides of the square)
OC OP = OA OQ
CP = AQ
In ∆BPC and ∆BQA,
…………… (1)
BC = BA (Sides of the square) ∠PCB = ∠QAB (both 90°)
PC = QA
By Side Angle Side criterion of congruence,
∆BPC ∆BQA
(by (1))
(by SAS)
Question 17:
The length of common chord of two intersecting circles is 30 cm. If the diameters of these two
circles be 50 cm and 34 cm, calculate the distance between their centres.
Solution 17:
OA = 25 cm and AB = 30 cm
OA
2
2
2
OD
625225 400
2
OA OD
2
2
25 15
2
2
1 AB
2
1
2
30
cm
15 cm
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Class X
Chapter 17 Circles
Maths
OD
400 20 cm
Again, we have O' A = 17 cm In right angle ∆ADO'
2
2
2
2
2
2
2
289225 64
O'D 8 cm
OO' OD O'D
20 828 cm
2
the distance between their centres is 28 cm
Question 18:
The line joining the mid-points of two chords of a circle passes through its centre. Prove that the chords are parallel.
Solution 18:
Given: AB and CD are the two chords of a circle with centre O. L and M are the midpoints of AB and CD and O lies in the line joining ML To prove: AB CD Proof: AB and CD are two chords of a circle with centre O. Line LOM bisects them at L and M Then, OL AB And, OM CD
∴ ∠ALM = LMD = 90°
But they are alternate angles
AB CD.
Class X
Chapter 17 Circles
Maths
Question 19:
In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centres of the circles. Show that:
(i) AB = CD,
(ii) AC = BD.
Solution 19:
In the circle with centre Q,
OA OD
…………… (1)
(Perpendicular drawn from the centre of a circle to a chord bisects it)
In the circle with centre P,
PO BC
OB OC
…………… (2)
(Perpendicular drawn from the centre of a circle to a chord bisects it)
(i) (1) – (2) Gives,
AB CD
……………(3)
(ii)
Adding BC to both sides of equation (3)
ABBC CDBC
AC BD
Question 20:
AB and CD are two equal chords of a circle with centre O which intersect each other at right
angle at point P. If
Solution 20:
OM AB
and
ON CD
; show that OMPN is a square.
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Class X
Chapter 17 Circles
Maths
Clearly, all the angles of OMPN are 90°
OM AB
and
ON CD
1 AB
2 2
1
(Perpendicular drawn from the centre of a circle to a chord bisects it) As the two equal chords AB and CD intersect at point P inside The circle,
AP DP
Now,
and
CP BP
……………. (ii) (by (i) and (ii))
CN CP BM BP
PN MP
BM
CD
CN
…………
(i)
EXERCISE. 17 (B)
Question 1:
In the given figure, O is the centre of the circle. OAB and
respectively. Find
AOC
. Show your steps of working.
Solution 1:
Join AC,
Let
OAC OCA x
(say)
Also, AOC BAC 1802x 30 x
OCB
are 30° and 40°
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Class X
Chapter 17 Circles
Maths
BCA 40 x
In ∆ABC,
ABC 180BACBCA
18030 x40 x110 2x
Now,
(Angle at the centre is double the angle at the circumference subtended by the same chord)
AOC 2ABC
180 2x 2110 2x
2x 40
x 20
AOC 180220140
= 65 ,
ABD
= 70
, BDC =
45
Question 2:
In the figure,
ACB
(i) Prove that AC is a diameter of the circle
(ii) Find
Solution 2:
AOC 2ABC
Here, Reflex
360130 2ABC
(Angle at the centre is double the angle at the circumference subtended by the same chord)
230
2
ABC
115
Class X
Chapter 17 Circles
Maths
Question 3:
Given O is the centre of the circle and
AOB
= 70. Calculate the value of:
(i) OCA OCA , (ii) .
Solution 3:
Here,
(Angle at the center is double the angle at the circumference by the same chord)
AOB 2ACB
ACB 70 35 2
Now,
OC OA
OCA OAC 35
Question 4:
In each of the following figures, O is the centre of the circle. Find the values of a, b and c.
Class X
Chapter 17 Circles
Maths
Solution 4:
(i) Here,
b
1
2
130
(Angle ate he centre is double the angle at the circumference subtended by the same chord)
b 65
Now,
(Opposite angles of a cyclic quadrilateral are supplementary)
a b 180
a 18065115
(Angle at the cente is double the angle at the circumference subtended by the same chord)
(ii) Here,
c
1 2 (112)
Reflex
c
1 2
360
112
124
Question 5:
In each of the following figures, O is the centre of the circle. Find the values of a, b, c and d.
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Class X
Chapter 17 Circles
Maths
Solution 5:
(i) Here, BAD 90(Angle in a semicircle)
BDA 9035 55
Again,
(Angle subtended by the same chord on the circle are equal)
(ii) Here, DAC CBD 25
(Angle subtended by the same chord on the circle are equal)
Again,
(In a triangle, measure of exterior angle is equal to the sum of pair of opposite interior angles)
a ACB BDA 55
120 b 25
b 95
(iii)AOB
(Angle at the centre is double the angle at the circumference subtended by the same chord)
Also, OA OB
2AOB 250100
OBA OAB C
180 100
2
(iv)APB 90
(Angle in a semicircle)
BAP 90 45 45
Now,
(Angle subtended by the same chord on the circle are equal)
d BCP BAP 45
c
40
Class X
Chapter 17 Circles
Maths
Question 6:
In the figure, AB is common chord of the two circle. If AC and AD are diameters; prove that D, B and C are in a straight line. O 1 and O 2 are the centres of two circles.
Solution 6:
DBA 90and CBA 90
(Angles in a semicircle is a right angle) Adding both we get,
DBC 180
D, B and C form a straight line.
Question 7:
In the figure, given below, find:
(i) BCD, (ii) ADC, (iii) ABC,
Class X
Chapter 17 Circles
Maths
Solution 7:
(Sum of opposite angles of a cyclic quadrilateral is 180)
BCD180105 75
(ii) Now, AB CD
(Interior angles on the same side of parallel lines is
180
)
(iii)
(Sum of opposite angles of a cyclic quadrilateral is 180)
Question 8:
In the given figure, O is the centre of the circle. If AOB = 140° and OAC = 50°; Find:
(i) ACB,
(ii) OBC,
(iii) OAB,
(iv) CBA.
Class X
Chapter 17 Circles
Maths
Solution 8:
(Angle at the centre is double the angle at the circumference subtended by the same chord)
Now, OA = OB
OBA OAB 180 140 20 2
CAB 5020 30
CAB,
CBA 18011030 40
OBC CBA OBA 40 20 60
Here,
ACB
1 2 Reflex
AOB
360
1
2
140
110
Question 9:
Calculate: (i) ∠CDB, (ii) ∠ABC, (iii) ∠ACB
Class X
Chapter 17 Circles
Maths
Solution 9:
Here,
ABC CDB ADC BAC 4943
(Angle subtend by the same chord on the circle are equal) By angle sum property of a triangle,
ACB 1804943 88
Question 10:
In the figure, given below, ABCD is a cyclic quadrilateral in which BAD = 75°; ABD = 58° and ADC = 77°. Find:
(i) BDC, (ii) BCD, (iii) BCA.
Solution 10:
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Class X
Chapter 17 Circles
Maths
(i) By angle sum property of triangle ABD,
(Sum of opposite angles of a cyclic quadrilateral is 180)
BCD 18075105
(Angle subtended by the same chord on the circle are equal)
Question 11:
In the following figure, O is centre of the circle and ∆ABC is equilateral.
Find:
(ii) ∠AEB.
Solution 11:
ACB
Since
and
in the same segment,
AOB 2ACB 260120
Join OA and OB
Here,
1
1
2
2
(Angle at the centre is double the angle at the circumference subtended by the same chord)
AEB
Reflex
AOB
360
120
120
Class X
Chapter 17 Circles
Maths
Question 12:
Given: CAB = 75° and CBA = 50°. Find the value of DAB + ABD.
Solution 12:
In
ABC,CBA 50,CAB 75
ACB 180CBA CAB
18050 75
180125
55
But
(Angle subtended by the same chord on the circle are equal)
Now consider ∆ABD,
DABABD55180
DABABD 18055
DABABD 125
Question 13:
ABCD is a cyclic quadrilateral in a circle with centre O. If ADC = 130°; find
BAC
.
Class X
Chapter 17 Circles
Maths
Solution 13:
Here
(Angle in a semicircle is right angle)
Also,
(pair of opposite angles in a cy clic quadrilateral are supplementary) By angle sum property of right triangle ACB,
ACB 90
BAC 90ABC 9050 40
Question 14:
In the figure, given below, AOB is a diameter of the circle and AOC = 110°. Find BDC.
Solution 14:
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Class X
Chapter 17 Circles
Maths
Here, ADC 1 AOC 1 110 55 2 2
(Angle at the centre is double the angle at the circumference subtended by the same chord)
(Angle in a semicircle is a right angle)
Question 15:
In the following figure, O is the centre of the circle, AOB = 60° and BDC = 100° Find OBC.
Solution 15:
Here, ACB 1 AOB 1 60 30 2 2
(Angle at the centre is double the angle at the circumference subtended by the same chord) By angle sum property of ∆BDC,
DBC 18010030 50
Hence,
OBC 50
Class X
Chapter 17 Circles
Maths
Question 16:
ABCD is a cyclic quadrilateral in which DAC = 27°; DBA = 50° and ADB = 33°. Calculate:
(i) DBC,
(ii) DCB,
(iii) CAB
Solution 16:
(i) DBC DAC 27
(Angle subtended by the same chord on the circle are equal)
ACD ABD 50
(Angle subtended by the same chord on the circle are equal)
DCB ACDACB 5033 83
(iii) DABDCB 180
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
27CAB83180
CAB 180110 70
Question 17:
In the figure given below, AB is diameter of the circle whose centre is O. given that: ECD = EDC = 32°. Show that COF = CEF.
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Class X
Chapter 17 Circles
Maths
Solution 17:
Here,
(Angle at the centre is double the angle at the circumference subtended by the same chord) In ∆ECD,
COF 2CDF 232 64
……………… (i)
CEF ECDEDC 3232 64
………….(ii)
(Exterior angle of a ∆ is equal to the sum of pair of interior opposite angles) From (i) and (ii), we get
COF CEF
Question 18:
In the figure given below, AB and CD are straight lines through the centre O of a circle. If AOC = 80° and CDE = 40°, Find the number of degrees in:
(i) DCE,
(ii) ABC.
Solution 18:
Class X
Chapter 17 Circles
Maths
(i) Here, CED 90
(Angle in a semicircle is a right angle)
DCE 90CDE 90 40 50
DCE OCB 50
(ii) In ∆BOC,
AOC OCBOBC
(Exterior angle of a ∆ is equal to the sum of pair of interior opposite angles)
OBC 8050 30
Hence,
ABC 30
AOC 80,given
Question 19:
In the given figure, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. prove that AB = BE
Solution 19:
Join OB,
Then
(Angle in a semicircle is a right angle)
i.e. OB AE
We know the perpendicular drawn from the centre to a chord bisects the chord.
OBA 90
AB BE
Class X
Chapter 17 Circles
Maths
Question 20:
In the following figure,
(i) If ∠BAD = 96°, find ∠BCD and ∠BFE. (ii) Prove that AD is parallel to FE.
Solution 20:
(i) ABCD is a cyclic quadrilateral
(pair of opposite angles in a cyclic quadrilateral are supplementary)
BCD
BCE
180
180
96
84
84
96
Similarly, BCEF is a cyclic quadrilateral
BCE BFE 180
(pair of opposite angles in a cyclic quadrilateral are supplementary)
BFE 18096 84
(ii) Now,
But these two are interior angles on the same side
of a pair of lines AD and FE
Question 21:
Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle. (ii) the rhombus, inscribed in a circle, is a square.
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Class X
Chapter 17 Circles
Maths
Solution 21:
(i) Let ABCD be a parallelogram, inscribe in a circle,
Now,
(Opposite angles of a parallelogram are equal)
(pair of opposite angles in a cyclic quadrilateral are supplementary)
180
2
y, the other two angles are 90and opposite pair of sides Are equal. ABCD is a rectangle.
(ii) Let ABCD be a rhombus, inscribed in a circle
Now,
(Opposite angles of a parallelogram are equal)
(pair of opposite angles in a cyclic quadrilateral are supplementary)
BAD BCD 180 90 2
y, the other two angles are 90and all the sides are equal. ABCD is a square.
BCD
90
Class X
Chapter 17 Circles
Maths
Question 22:
In the following figure AB = AC. Prove that DECB is an isosceles trapezium.
Solution 22:
Here, AB = AC
B C
DECB is a cyclic quadrilateral (Ina triangle, angles opposite to equal sides are equal)
Also, BDEC 180………. (1)
(pair of opposite angles in a cyclic quadrilateral are supplementary
CDEC 180
[from (1)]
But this is the sum of interior angles On one side of a transversal.
DE || BC
But
Thus,
and
AED C
[Corresponding angles]
AD AE BD AB AD CE AC AE( AB AC)
Thus, we have,
Hence, DECB is an isosceles trapezium
DE || BC
and
BD CE
Class X
Chapter 17 Circles
Maths
Question 23:
Two circles intersect at P and Q. through P diameter PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Solution 23:
Let O and O ' be the centres of two intersecting circle, where Points of intersection are P and Q and PA and PB are their diameter respectively. Join PQ, AQ and QB. ∴ ∠AQP = 90° and BQP = 90° (Angle in a semicircle is a right angle) Adding both these angles,
AQP + BQP = 180°
Hence, the points A, Q and B are collinear.
AQB 180
Question 24:
ABCD is a quadrilateral inscribed in a circle, having = 60°; O is the center of the circle. Show that:
OBD + ODB =CBD +CDB
Solution 24:
Class X
Chapter 17 Circles
Maths
1
2
1
2
(Angle at the centre is double the angle at the circumference subtended by the same chord
CBDCDB 180120 60
(By angle sum property of triangle CBD)
Again,
(By angle sum property of triangle OBD)
OBDODB 180120 60
OBDODB CBDCDB
And
BCD
Refelx
BOD
360
120
120
Question 25:
The figure given below, shows a Circle with centre O. Given: AOC = a and ABC = b. (i) Find the relationship between a and b (ii) Find the measure of angle OAB, is OABC is a parallelogram
Solution 25:
(i) ABC
1 2 COA
Reflex
(Angle at the centre is double the angle at the circumference subtended by the same chord)
Class X
Chapter 17 Circles
Maths
a 2b 180
(ii) Since OABC is a parallelogram, so opposite angles are equal a = b Using relationship in (i)
3a 180
a 60
Also, OC || BA
COA OAB 180
60OAB 180
OAB 120
b
1 2
360 a
Question 26:
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the centre O is equal to twice the angle APC.
Solution 26:
Given: two chords AB and CD intersect each other at P inside the circle. OA, OB, OC and OD are joined.
To prove:
Construction: Join AD. Proof: Arc AC subtends Part of the circle.
AOCBOD 2APC
AOC
at the centre and
at the remaining
…………(1)
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Class X
Chapter 17 Circles
Maths
Similarly,
………….(2)
……….(3)
From (3) and (4),
AOCBOD 2APC
Question 27:
In the given figure, RS is a diameter of the circle. NM is parallel to RS and MRS = 29°. Calculate:
(i) RNM,
(ii) NRM.
Solution 27:
(i) Join RN and MS.
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Class X
Chapter 17 Circles
Maths
RMS 90
(Angle in a semicircle is a right angle)
RSM 90 29 61
(By angle sum property of triangle RMS)
RNM 180RSM 18061119
(pair of opposite angles in a cyclic quadrilateral are supplementary)
(ii) Also,
RS|| NM
NMR MRS 29
(Alternate angles)
NMS 90 29119
Also, NRSMS 180
(pair of opposite angles in a cyclic quadrilateral are supplementary)
NMR 29 119 180 NRM 180 148
NRM
32
Question 28:
In the figure, given alongside, AB CD and O is the centre of the circle. If ADC = 25°; find the angle AEB give reasons in support of your answer.
Solution 28:
Join AC and BD
Class X
Chapter 17 Circles
Maths
and
CBD 90
(Angle in a semicircle is a right angle)
Also,
AB|| CD
(alternate angles)
(pair of opposite angles in a cyclic quadrilateral are supplementary)
(Angle subtended by the same chord on the circle are equal)
Question 29:
Two circle intersect at P and Q. through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. prove that AC is parallel to BD.
Solution 29:
Join AC, PQ and BD ACQP is a cyclic quadrilateral
CAP PQC 180
………….(i)
(pair of opposite in a cyclic quadrilateral are supplementary) PQDB is a cyclic quadrilateral
PQD DBP 180
………….(ii)
(pair of opposite angles in a cyclic quadrilateral are supplementary)
Again,PQC
(CQD is a straight line)
PQD 180
…………. (iii)
Class X
Chapter 17 Circles
Maths
Using (i), (ii) and (iii)
CAP DBP 180
Or
We know, if a transversal intersects two lines such That a pair of interior angles on the same side of the Transversal is supplementary, then the two lines are parallel
CABDBA 180
AC|| BD
Question 30:
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution 30:
Let ABCD be the given cyclic quadrilateral Also, PA = PD (Given)
…… (1)
And CDA 180PDA 180PAD (From (1))
We know that the opposite angles of a cyclic quadrilateral are supplementary
And
That means
Class X
Chapter 17 Circles
Maths
Question 31:
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) PRB,
(ii) PBR,
(iii) BPR.
Solution 31:
(i) PRB PAB 35
(Angles subtended by the same chord on the circle are equal)
(ii) BPA 90
(angle in a semicircle is a right angle)
BPQ 90
PBR BQP BPQ 25 90115
(Exterior angle of a ∆ is equal to the sum of pair of interior opposite angles)
(iii)ABP 90BAP 9035 55
ABR PBR ABP 115 55 60 APR ABR 60
(Angles subtended by the same chord on the circle are equal)
BPR 90APR 9060 30
Question 32:
In the given figure, SP is bisector of RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR.
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Class X
Chapter 17 Circles
Maths
Solution 32:
| 10,992 | 29,697 |
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# Prism (geometry)
In geometry, a prism is a polyhedron comprising an n-sided polygonal base, a second base which is a translated copy (rigidly moved without rotation) of the first, and n other faces (necessarily all parallelograms) joining corresponding sides of the two bases. All cross-sections parallel to the bases are translations of the bases. Prisms are named for their bases, so a prism with a pentagonal base is called a pentagonal prism. The prisms are a subclass of the prismatoids.
Set of uniform prisms
(A hexagonal prism is shown)
Typeuniform polyhedron
Conway polyhedron notationPn
Faces2+n total:
2 {n}
n {4}
Edges3n
Vertices2n
Schläfli symbol{n}×{}[1] or t{2, n}
Coxeter diagram
Vertex configuration4.4.n
Symmetry groupDnh, [n,2], (*n22), order 4n
Rotation groupDn, [n,2]+, (n22), order 2n
Dual polyhedronbipyramids
Propertiesconvex, semi-regular vertex-transitive
n-gonal prism net (n = 9 here)
## General, right and uniform prisms
A right prism is a prism in which the joining edges and faces are perpendicular to the base faces.[2] This applies if the joining faces are rectangular. If the joining edges and faces are not perpendicular to the base faces, it is called an oblique prism.
For example a parallelepiped is an oblique prism of which the base is a parallelogram, or equivalently a polyhedron with six faces which are all parallelograms.
A truncated prism is a prism with nonparallel top and bottom faces.[3]
Some texts may apply the term rectangular prism or square prism to both a right rectangular-sided prism and a right square-sided prism. A right p-gonal prism with rectangular sides has a Schläfli symbol { } × {p}.
A right rectangular prism is also called a cuboid, or informally a rectangular box. A right square prism is simply a square box, and may also be called a square cuboid. A right rectangular prism has Schläfli symbol { }×{ }×{ }.
An n-prism, having regular polygon ends and rectangular sides, approaches a cylindrical solid as n approaches infinity.
The term uniform prism or semiregular prism can be used for a right prism with square sides, since such prisms are in the set of uniform polyhedra. A uniform p-gonal prism has a Schläfli symbol t{2,p}. Right prisms with regular bases and equal edge lengths form one of the two infinite series of semiregular polyhedra, the other series being the antiprisms.
The dual of a right prism is a bipyramid.
## Volume
The volume of a prism is the product of the area of the base and the distance between the two base faces, or the height (in the case of a non-right prism, note that this means the perpendicular distance).
The volume is therefore:
${\displaystyle V=Bh}$
where B is the base area and h is the height. The volume of a prism whose base is a regular n-sided polygon with side length s is therefore:
${\displaystyle V={\frac {n}{4}}hs^{2}\cot({\frac {\pi }{n}})}$
## Surface area
The surface area of a right prism is:
${\displaystyle 2B+Ph}$
where B is the area of the base, h the height, and P the base perimeter.
The surface area of a right prism whose base is a regular n-sided polygon with side length s and height h is therefore:
${\displaystyle A={\frac {n}{2}}s^{2}\cot {\frac {\pi }{n}}+nsh}$
## Schlegel diagrams
P3 P4 P5 P6 P7 P8
## Symmetry
The symmetry group of a right n-sided prism with regular base is Dnh of order 4n, except in the case of a cube, which has the larger symmetry group Oh of order 48, which has three versions of D4h as subgroups. The rotation group is Dn of order 2n, except in the case of a cube, which has the larger symmetry group O of order 24, which has three versions of D4 as subgroups.
The symmetry group Dnh contains inversion iff n is even.
The hosohedra and dihedra also possess dihedral symmetry, and a n-gonal prism can be constructed via the geometrical truncation of a n-gonal hosohedron, as well as through the cantellation or expansion of a n-gonal dihedron.
## Prismatic polytope
A prismatic polytope is a higher-dimensional generalization of a prism. An n-dimensional prismatic polytope is constructed from two (n − 1)-dimensional polytopes, translated into the next dimension.
The prismatic n-polytope elements are doubled from the (n − 1)-polytope elements and then creating new elements from the next lower element.
Take an n-polytope with fi i-face elements (i = 0, ..., n). Its (n + 1)-polytope prism will have 2fi + fi−1 i-face elements. (With f−1 = 0, fn = 1.)
By dimension:
• Take a polygon with n vertices, n edges. Its prism has 2n vertices, 3n edges, and 2 + n faces.
• Take a polyhedron with v vertices, e edges, and f faces. Its prism has 2v vertices, 2e + v edges, 2f + e faces, and 2 + f cells.
• Take a polychoron with v vertices, e edges, f faces and c cells. Its prism has 2v vertices, 2e + v edges, 2f + e faces, and 2c + f cells, and 2 + c hypercells.
### Uniform prismatic polytope
A regular n-polytope represented by Schläfli symbol {p, q, ..., t} can form a uniform prismatic (n + 1)-polytope represented by a Cartesian product of two Schläfli symbols: {p, q, ..., t}×{}.
By dimension:
• A 0-polytopic prism is a line segment, represented by an empty Schläfli symbol {}.
• A 1-polytopic prism is a rectangle, made from 2 translated line segments. It is represented as the product Schläfli symbol {}×{}. If it is square, symmetry can be reduced: {}×{} = {4}.
• Example: Square, {}×{}, two parallel line segments, connected by two line segment sides.
• A polygonal prism is a 3-dimensional prism made from two translated polygons connected by rectangles. A regular polygon {p} can construct a uniform n-gonal prism represented by the product {p}×{}. If p = 4, with square sides symmetry it becomes a cube: {4}×{} = {4, 3}.
• A polyhedral prism is a 4-dimensional prism made from two translated polyhedra connected by 3-dimensional prism cells. A regular polyhedron {p, q} can construct the uniform polychoric prism, represented by the product {p, q}×{}. If the polyhedron is a cube, and the sides are cubes, it becomes a tesseract: {4, 3}×{} = {4, 3, 3}.
• ...
Higher order prismatic polytopes also exist as cartesian products of any two polytopes. The dimension of a polytope is the product of the dimensions of the elements. The first example of these exist in 4-dimensional space are called duoprisms as the product of two polygons. Regular duoprisms are represented as {p}×{q}.
## Twisted prism
A twisted prism is a nonconvex prism polyhedron constructed by a uniform q-prism with the side faces bisected on the square diagonal, and twisting the top, usually by π/q radians (180/q degrees) in the same direction, causing side triangles to be concave.[4][5]
A twisted prism cannot be dissected into tetrahedra without adding new vertices. The smallest case, triangular form, is called a Schönhardt polyhedron.
A twisted prism is topologically identical to the antiprism, but has half the symmetry: Dn, [n,2]+, order 2n. It can be seen as a convex antiprism, with tetrahedra removed between pairs of triangles.
3-gonal 4-gonal 12-gonal
Schönhardt polyhedron
Twisted square prism
Square antiprism
Twisted dodecagonal antiprism
## Frustum
A frustum is topologically identical to a prism, with trapezoid lateral faces and different sized top and bottom polygons.
## Star prism
A star prism is a nonconvex polyhedron constructed by two identical star polygon faces on the top and bottom, being parallel and offset by a distance and connected by rectangular faces. A uniform star prism will have Schläfli symbol {p/q} × { }, with p rectangle and 2 {p/q} faces. It is topologically identical to a p-gonal prism.
Examples
{ }×{ }180×{ } ta{3}×{ } {5/2}×{ } {7/2}×{ } {7/3}×{ } {8/3}×{ }
D2h, order 8 D3h, order 12 D5h, order 20 D7h, order 28 D8h, order 32
### Crossed prism
A crossed prism is a nonconvex polyhedron constructed from a prism, where the base vertices are inverted around the center (or rotated 180°). This transforms the side rectangular faces into crossed rectangles. For a regular polygon base, the appearance is an p-gonal hour glass, with all vertical edges passing through a single center, but no vertex is there. It is topologically identical to a p-gonal prism.
Examples
{ }×{ }180×{ }180 ta{3}×{ }180 {3}×{ }180 {4}×{ }180 {5}×{ }180 {5/2}×{ }180 {6}×{ }180
D2h, order 8 D3d, order 12 D4h, order 16 D5d, order 20 D6d, order 24
### Toroidal prisms
A toroidal prism is a nonconvex polyhedron is like a crossed prism except instead of having base and top polygons, simple rectangular side faces are added to close the polyhedron. This can only be done for even-sided base polygons. These are topological tori, with Euler characteristic of zero. The topological polyhedral net can be cut from two rows of a square tiling, with vertex figure 4.4.4.4. A n-gonal toroidal prism has 2n vertices and faces, and 4n edges and is topologically self-dual.
D4h, order 16 D6h, order 24 v=8, e=16, f=8 v=12, e=24, f=12
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# B-sharp minor 6th chord
The Solution below shows the B-sharp minor 6th chord in root position, 1st, 2nd, and 3rd inversions, on the piano, treble clef and bass clef.
The Lesson steps then explain how to construct this 6th chord using the 3rd, 5th and 6th note intervals, then finally how to construct the inverted chord variations.
For a quick summary of this topic, have a look at Sixth chord.
## Solution - 4 parts
### 1. B-sharp minor 6th chord
This step shows the B-sharp minor 6th chord in root position on the piano, treble clef and bass clef.
The B-sharp minor 6th chord contains 4 notes: B#, D#, F##, G##.
The chord spelling / formula relative to the B# major scale is: 1 b3 5 6.
B-sharp minor 6th chord note names
Note no.Note intervalSpelling
/ formula
Note name#Semitones
from root
1root1The 1st note of the B-sharp minor 6th chord is B#0
2B#-min-3rdb3The 2nd note of the B-sharp minor 6th chord is D#3
3B#-perf-5th5The 3rd note of the B-sharp minor 6th chord is F##7
4B#-maj-6th6The 4th note of the B-sharp minor 6th chord is G##9
Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 6/5/3.
The staff diagrams and audio files contain each note individually, ascending from the root, followed by the chord containing all 3 notes.
### 2. B-sharp minor 6th 1st inversion
This step shows the B-sharp minor 6th 1st inversion on the piano, treble clef and bass clef.
The B-sharp minor 6th 1st inversion contains 4 notes: D#, F##, G##, B#.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 6/4/3, so the chord is said to be in six-four-three position.
### 3. B-sharp minor 6th 2nd inversion
This step shows the B-sharp minor 6th 2nd inversion on the piano, treble clef and bass clef.
The B-sharp minor 6th 2nd inversion contains 4 notes: F##, G##, B#, D#.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 6/4/2, so the chord is said to be in six-four-two position.
### 4. B-sharp minor 6th 3rd inversion
This step shows the B-sharp minor 6th 3rd inversion on the piano, treble clef and bass clef.
The B-sharp minor 6th 3rd inversion contains 4 notes: G##, B#, D#, F##.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 7/5/3, so the chord is said to be in seven-five-three position.
## Lesson steps
### 1. Piano key note names
This step shows the white and black note names on a piano keyboard so that the note names are familiar for later steps, and to show that the note names start repeating themselves after 12 notes.
The white keys are named using the alphabetic letters A, B, C, D, E, F, and G, which is a pattern that repeats up the piano keyboard.
Every white or black key could have a flat(b) or sharp(#) accidental name, depending on how that note is used. In a later step, if sharp or flat notes are used, the exact accidental names will be chosen.
The audio files below play every note shown on the piano above, so middle C (marked with an orange line at the bottom) is the 2nd note heard.
### 2. B-sharp tonic note and one octave of notes
This step shows 1 octave of notes starting from note B#, to identify the start and end notes of the scale used to build this chord.
The numbered notes are those that might be used when building this chord.
Note 1 is the root note - the starting note of the chord - B#, and note 13 is the same note name but one octave higher.
No. Note 1 2 3 4 5 6 7 8 9 10 11 12 13 B# C# / Db D D# / Eb E F F# / Gb G G# / Ab A A# / Bb B B#
### 3. B-sharp major scale note interval positions
This step describes the B# major scale , whose note intervals are used to define the chord in a later step.
The major scale uses the W-W-H-W-W-W-H note counting rule to identify the scale note positions.
To count up a Whole tone, count up by two physical piano keys, either white or black.
To count up a Half-tone (semitone), count up from the last note up by one physical piano key, either white or black.
The tonic note (shown as *) is the starting point and is always the 1st note in the major scale.
Again, the final 8th note is the octave note, having the same name as the tonic note.
No. Note 1 2 3 4 5 6 7 8 B# D E F G A B B#
### 4. B-sharp major scale note interval numbers
This step identifies the note interval numbers of each scale note, which are used to calculate the chord note names in a later step.
To identify the note interval numbers for this major scale, just assign each note position from the previous step, with numbers ascending from 1 to 8.
No. Note 1 2 3 4 5 6 7 8 B# C## D## E# F## G## A## B#
To understand why the note names of this major scale have these specific sharp and flat names, have a look at the B# major scale page.
Both the note interval numbers and note names from the piano diagram above will be used in later steps to calculate the chord note names.
### 5. 6th chord qualities
This step defines a sixth chord, names the most common 6th chord qualities and identifies the notes that vary between them.
#### 6th chord definition
Whereas a triad chord contains 3 notes, a 6th chord contains 4 notes that are played together or overlapping.
#### 6th chord qualities
Although others exist, the most common 6th chord qualities, are major, and minor.
Each chord quality name is the name of the entire chord as a whole, not its individual notes (which will be covered later).
#### Triad chords used for construction
Both major and minor chord qualities are built on the triad chord in the same key plus one added note - the 6th note of the major scale in the previous step.
So the B-sharp major 6th chord is based on the B# major chord, and the B-sharp minor 6th chord is based on the B# minor chord. The added 6th note in both cases is G##.
The steps below will detail the construction of the minor 6th chord quality in the key of B# using note intervals.
### 6. 6th chord note intervals
This step defines the note intervals for each chord quality, including the intervals for the B-sharp minor 6th chord. It also shows how the 6th chord qualities are related to the triad chord qualities they are based on.
Each individual note in a 6th chord can be represented in music theory using a note interval, which is used to express the relationship between the first note of the chord (the root note), and the note in question.
The root note is always the 1st note (note interval 1 in the above diagram) of the major scale diagram above. ie. the tonic of the major scale.
Then there is one note interval to describe the 2nd note, and another to describe the 3rd note of the chord, and finally another interval for the 4th chord note.
In the same way that the entire chord itself has a chord quality, the intervals representing the individual notes within that chord each have their own quality.
These note interval qualities could be diminished, minor, major, perfect and augmented.
Below is a table showing the note interval qualities for the most common 6th chords, together with the interval short names / abbrevations in brackets.
The final column shows the triad chord quality that the 6th chord is based on, so the 2nd and 3rd note quality columns are the same as the triad table for the same key.
sixth chord note interval qualities
6th chord quality2nd note quality3rd note quality4th note qualityBased on triad quality
minorminor (m3)perfect (P5)major (M6)minor
majormajor (M3)perfect (P5)major (M6)major
The numbers in brackets are the note interval number (ie the scale note number) shown in the previous step.
Looking at the table above, the note intervals for the chord quality we are interested in (minor 6th), in the key of B# are B#-min-3rd, B#-perf-5th, and B#-maj-6th.
The links above explain in detail the meaning of these qualities, the short abbrevations in brackets, and how to calculate the interval note names based on the scale note names from the previous step.
### 7. B-sharp minor 6th chord in root position
This step shows the B-sharp minor 6th chord note interval names and note positions on a piano diagram.
Each note interval quality (diminished, minor, major, perfect, augmented) expresses a possible adjustment ie. a possible increase or decrease in the note pitch from the major scale notes in step 4.
If an adjustment in the pitch occurs, the note name given in the major scale in step 4 is modified, so that sharp or flat accidentals will be added or removed.
But crucially, for all interval qualities, the starting point from which accidentals need to be added or removed are the major scale note names in step 4.
For this chord, this is explained in detail in B#-min-3rd, B#-perf-5th and B#-maj-6th, but the relevant adjustments for this minor 6th chord quality are shown below:
B#-3rd: The 3rd note quality of the major scale is major, and the note interval quality needed is minor, so the 3rd note scale note name - D##, is adjusted 1 half-note / semitone down to D#. The chord note spelling reflects this note flattening: b3.
B#-5th: Since the 5th note quality of the major scale is perfect, and the note interval quality needed is perfect also, no adjustment needs to be made. The 5th note name - F## is used, and the chord note spelling is 5.
B#-6th: Since the 6th note quality of the major scale is major, and the note interval quality needed is major also, no adjustment needs to be made. The 6th note name - G## is used, and the chord note spelling is 6.
If it is still not clear why the interval qualities are organised / related as they are, please refer to each of the interval links above.
#### B-sharp minor 6th chord note names
The final chord note names and note interval links are shown in the table below.
Note Interval No. Interval def Spelling 1 3 5 6 B# D# F## G## root B#-min-3rd B#-perf-5th B#-maj-6th 1 b3 5 6 0 3 7 9
The piano diagram below shows the interval short names, the note positions and the final note names of this triad chord.
In music theory, this 6th chord as it stands is said to be in root position because the root of the chord - note B#, is the note with the lowest pitch of all the chord notes.
The note order of this chord can also be changed, so that the root is no longer the lowest note, in which case the chord is no longer in root position, and will be called an inverted 6th chord instead.
For 6th chords, there are 3 possible inverted variations as described below.
#### Figured bass notation
The figured bass notation for a 6th chord in root position is 6/5/3, with the 6 placed above the 5, and the 5, above the 3.
These numbers represent the interval between the lowest note of the chord and the note in question.
So another name for this inversion would be B-sharp minor 6th triad in six-five-three position.
For example, the 6 represents note G##, from the B#-6th interval, since the chord root, B#, is the lowest note of the chord (as it is not inverted). .
In the same way, the figured bass 5 symbol represents note F##, from the B#-5th interval, and the 3 symbol represents note D#, from the B#-3rd interval
Since figured bass notation works within the context of a key, we don't need to indicate in the figured bass symbols whether eg. the 3rd is a major, minor etc. The key is assumed from the key signature.
### 8. B-sharp minor 6th 1st inversion
This step shows the first inversion of the B-sharp minor 6th.
To invert a chord, simply take the first note of the chord to be inverted (the lowest in pitch) and move it up an octave to the end of the chord.
So for a 1st inversion, take the root of the 6th chord in root position from the step above - note B#, and move it up one octave (12 notes) so it is the last (highest) note in the chord.
The second note of the original 6th chord (in root position) - note D# is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this chord in 1st inversion is 6/4/3, with the 6 placed above the 4, and the 4 placed above the 3 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be B-sharp minor 6th triad in six-four-three position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original chord root!), and the note in question.
For example, the 6 represents note B#, from the D#-6th interval, since the lowest (bass) note of the chord - now inverted, is D#.
In the same way, the figured bass 4 symbol represents note G##, from the D#-4th interval, and the 3 symbol represents note F##, from the D#-3rd interval
### 9. B-sharp minor 6th 2nd inversion
This step shows the second inversion of the B-sharp minor 6th.
For a 2nd inversion, take the first note of the 1st inversion above - D#, and move it to the end of the chord.
So the second note of the 1st inversion - note F## is now the note with the lowest pitch for the 2nd inversion.
Or put another way, the third note of the original 6th chord (in root position) is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this chord in 2nd inversion is 6/4/2, with the 6 placed above the 4, and the 4 placed above the 2 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be B-sharp minor 6th triad in six-four-two position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original chord root!), and the note in question.
For example, the 6 represents note D#, from the F##-6th interval, since the lowest (bass) note of the chord - now inverted, is F##.
In the same way, the figured bass 4 symbol represents note B#, from the F##-4th interval, and the 2 symbol represents note G##, from the F##-2nd interval
### 10. B-sharp minor 6th 3rd inversion
This step shows the third inversion of the B-sharp minor 6th.
For a 3rd inversion, take the first note of the 2nd inversion above - F##, and move it to the end of the chord.
So the second note of the 2nd inversion - note G## is now the note with the lowest pitch for the 3rd inversion.
Or put another way, the fourth note of the original 6th chord (in root position) is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this chord in 3rd inversion is 7/5/3, with the 7 placed above the 5, and the 5 placed above the 3 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be B-sharp minor 6th triad in seven-five-three position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original chord root!), and the note in question.
For example, the 7 represents note F##, from the G##-7th interval, since the lowest (bass) note of the chord - now inverted, is G##.
In the same way, the figured bass 5 symbol represents note D#, from the G##-5th interval, and the 3 symbol represents note B#, from the G##-3rd interval
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The Snail in the Well ? JS | Sololearn: Learn to code for FREE!
+17
# The Snail in the Well ? JS
The snail climbs up 7 feet each day and slips back 2 feet each night. How many days will it take the snail to get out of a well with the given depth? Sample Input: 31 Sample Output: 6 Explanation: Let's break down the distance the snail covers each day: Day 1: 7-2=5 Day 2: 5+7-2=10 Day 3: 10+7-2=15 Day 4: 15+7-2=20 Day 5: 20+7-2=25 Day 6: 25+7=32 So, on Day 6 the snail will reach 32 feet and get out of the well at day, without slipping back that night. Hint: You can use a loop to calculate the distance the snail covers each day, and break the loop when it reaches the desired distance.
12/8/2020 5:16:51 AM
Ibrahim Matar
+119
function main() { var depth = parseInt(readLine(), 10); //your code goes here i = 0; for (; depth > 0;) { i++; depth -= 7 if (depth > 0) { depth += 2 } } console.log(i); } Here ya go m8.
+46
For those interested, here's an explanation on why Ibrahim Matar's answer works. function main() { var depth = parseInt(readLine(), 10); for(dis = 7, day = 1; dis < depth; dis += 7, day++) { dis -= 2 } console.log(day); } A key part of understanding this lies in knowing the order in which a for loop is evaluated and executed. There are four parts of a for loop (I’ll include the corresponding code from the answer in parentheses for clarity): - initialization (dis = 7, day = 1) - condition (dis < depth) - final-expression (dis += 7, day++) - statement (dis -= 2) Although they are presented in the order listed above, they actually run in this order: 1. initialization (“dis” is declared and given a value of 7, “day” is declared and given a value of 1) 2. condition (if “dis < depth” is true, the for loop will continue onto steps 3 and 4. If it is false, the for loop will stop immediately without executing steps 3 and 4) 3. statement (“dis” is decreased by 2) 4. final-expression (“dis is increased by 7, “day” is increased by 1) From here, the loop will run again, but it starts with step 2 (“dis < depth”), not the first one. This is evaluated, once again looking for either true or false. Going through it step by step. Let’s say “depth = 11”. First time through the loop: - dis is set to 7, day is set to 1 - dis (which is 7) < depth (which is 11) = true (the for loop continues to steps 3 and 4) - dis (which is 7) is decreased by 2, so dis is now 5 - dis (which is 5) is increased by 7, so dis is now 12; and day (which is 1) is increased by 1, so day is now 2 Second time through the loop: - dis (which is 12) < depth (which is 11) = false (the for loop does not continue to steps 3 and 4) - the for loop ends, console.log(day) is executed, and at this moment, day has a value of 2 With more space this could be explained further, but this is why extra conditionals, variables, while loops, and arrays aren't necessary here.
+31
My answer is here. function main() { var depth = parseInt(readLine(), 10); //your code goes here var climb = 7; var slip = 2; var day = 0; for(workdone=0;workdone<=depth;) { day = day + 1; workdone=workdone+climb; if(workdone>=depth){ break; } workdone = workdone - slip; } console.log(day); }
+29
My noob code for(x=1; depth >0; x++){ depth -=7; if(depth<=0){ console.log(x); break; } depth +=2; if(depth<=0){ console.log(x); break; } }
+28
console.log(Math.round(depth / 5))
+11
function main() { var depth = parseInt(readLine(), 10); //your code goes here for (dis=7,day=1;dis<depth;dis+=7,day++){ dis=dis-2; } console.log (day) }
+9
var climb = 7; var slip = 2; var day = 0; for(workdone=0; workdone=depth;) { day = day + 1; workdone=workdone + climb; if(workdone = depth){ break; } workdone = workdone - slip; } console.log(day); }
+7
function main() { var depth = parseInt(readLine(), 10); //equals 42 in this example var x = 0; //milesDone var y = 0; //dayCount for(i=0 ; i < depth ;i ++){ if (x<depth) { x += 7; if (depth > x) { x -= 2; } y++ } else { console.log(y) break; } } }
+7
So I wasted a lot of time with the loop method, I decided to find a way to simplify the problem by exploring other options and this is what I came up with. function main() { var depth = parseInt(readLine(), 10); day = depth/5; console.log(Math.round(day)); } We know that the snail moves 7 and slides back 2 each night; so that is 5 per day. We know depth is determined by user input. So we divide depth by 5 and we get a number and decimal. Using the Math.round() function we eliminate our decimal and store the answer in the var day. Hope this helps.
+5
thats my code: function main() { var depth = parseInt(readLine(), 10); //your code goes here var day = 0; var way = 0; while (way < depth) { way += 7; day++; if (way >= depth) { break; } way -= 2; } console.log(day); }
+3
function main() { var depth = parseInt(readLine(), 10); //your code goes here let data = []; /** Array contains the distance snails moves per days after an incremental +7 and decreamental -2 */ for (let i = 7; depth > 0 && i < depth + 7; i += 7) { if (i < depth) i -= 2; /**it only subtracts 2 as far my incremental value of +7 is less than or equal to input value */ data.push(i); }; console.log(data.length); /**this will output the length of data which will be the number of days */ } or function main() { var depth = parseInt(readLine(), 10); //your code goes here for ( i = 7, day = 1; i < depth; i += 7, day++) { if (i < depth) i -= 2; /**it only subtracts 2 as far my incremental value of +7 is less than or equal to input value */ }; console.log(day); /**this will output the length of data which will be the number of days */ }
+3
----------------------------------- PASSED ALL TEST CASES ----------------------------------- // VARIABLES const climb = 7; const slip = -2; let day = 0; // SOLUTION for (progress = 0; progress <= depth;) { progress += climb; day +=1; if (progress >= depth) { break; } else { progress += slip; } } // OUTPUT console.log(day);
+3
function main() { var profundidade = parseInt(readLine(), 10); // seu código vai aqui, rsrs exclusivo pra Br for (i = 0; profundidade > 0; i++) { profundidade -= 7 if (profundidade > 0) { profundidade += 2 } } console.log(i); }
+2
But why ? Initial 1 day/7 feet not 0 ?
+2
function main() { var depth = parseInt(readLine(), 10); //your code goes here var day=parseInt(0); var ng=parseInt(2); var clday=parseInt(7); for(days=0;days<=depth;){ day=day+1; days=days+clday; if(days>=depth){ break; } days=days-ng; } console.log(day); } hooora!!!
+2
function main() { var depth = parseInt(readLine(), 10); //your code goes here if(depth%5<=2) console.log(parseInt(depth/5)); else console.log(parseInt(depth/5)+1); }
+2
function main() { var depth = parseInt(readLine(), 10); //your code goes here i = 0; for (; depth > 0;) { i++; depth -= 7 if (depth > 0) { depth += 2 } } console.log(i); }
+2
function main() { var depth = parseInt(readLine(), 10); //your code goes here var day=1 for(i=0;i<depth;i=i+5){ if(i+7<depth ){ day++ } else{ break } } console.log(day) } hi it is my code
+2
function main() { var depth = parseInt(readLine(), 10); //your code goes here var i = 0; while (depth>0) { i++; depth-=7; depth += 2; if (depth<=0) {break;} console.log (i);} } help me why my out put is 1 2 3 4 5 6 7 8
+2
for (days = 0; depth > 0; days++){ depth -= 7 if (depth > 0) { depth += 2 } } console.log(days)
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# Number of connected components in a doubly linked list
• Difficulty Level : Medium
• Last Updated : 13 Sep, 2022
Given a doubly linked list ‘L’ and an array ‘refArr’ of references to the nodes of the doubly linked list ‘L’. Array ‘refArr’ does not contain any duplicate references and the references are not ORDERED
m <= total no. of nodes in the doubly linked list where m = size of the reference array.
The task is to find the total number of connected components in the linked list ‘L’.
A ‘connected component’ in linked list ‘L’ is defined as a group of nodes in the list with their references stored in the array ‘refArr’ and are adjacent to each other in the list.
Examples:
Input: L = 5 <-> 2 <-> 10 <-> 1 <-> 3, refArr[] = {5, 10, 3, 1}
Output:
Two connected components are 5 and 10 <-> 1 <-> 3.
Since 2 is not included in the reference array, that makes 5 disconnected from the rest of the elements.
Input: L = 1 <-> 7 <-> 10 <-> 5 <-> 4 <-> 2, refArr[] = {5, 2, 7, 1}
Output: Total number of connected components are 3
Explanation:
Let us take a double linked list ‘L’ as { N1 N2 N3 N4 N5 N6 }
and the array ‘refArr’ as [ref_to_nodeN1, ref_to_nodeN4, ref_to_nodeN6, ref_to_nodeN2].
Output :
This set of array ‘refArr’ and linked list ‘L’ contains 3 connected components, which are { [N1, N2], [N4], [N6] } .
It is because references of the nodes N1 and N2 are present in the array ‘refArr’ and they are adjacent to each other in the list ‘L’ whereas N4 and N6 are not adjacent to N1, N2 or to each other. Hence, they form separate components.
Now, let the array ‘refArr’ be modified as [ref_to_nodeN4, ref_to_nodeN6, ref_to_nodeN5].
Output :
This set of array ‘refArr’ and linked list ‘L’ contains 1 connected component, which is { [N4, N5, N6] } .
It is because references of the nodes N4, N5 and N6 are present in the array ‘refArr’ and they are adjacent to each other in the list ‘L’. So, we together form 1 connected component.
Naive Approach: It involves traversing the array ‘refArr’ for each node of the linked list ’L’ and checking whether that element is present in the array or not. Also, maintain a Boolean variable which keeps track of whether the previous node in the linked list is in the array or not. Take a variable ‘cc’ which is initialized to 0 and indicates the no. of connected components in the linked list.
Now, let us consider the multiple cases which occur while traversing:
• Case 1: If the previous node is in the array and the current node being traversed is also in the array then, the Boolean variable will remain 1 meaning that current node is in array but do not increment the variable ’cc’ since the current node forms a part of component already existing for the previous node.
• Case 2: If the previous node is not in the array and the current node being traversed is in the array then, update the Boolean variable to mark ‘1’ meaning that current node is in the array and also increment the variable ’cc’ by 1 because the current node forms a new component.
• Case 3: If the previous node is in the array and the current node being traversed is not in the array then, update the Boolean variable to mark 0 meaning that the current node is not in the array and do not increment the variable ’cc’ since the current node is not a part of any component.
• Case 4: If the previous node is not in the array and the current node being traversed is also not in the array then, the Boolean variable will remain 0 meaning that current node is not in array and do not increment the variable ’cc’ since the current node is not a part of any component.
Now, after performing one of the 4 cases, move to the next node in the linked list and do the same for that node as well.
Time Complexity: O(n*m), where n = size of the linked list ‘L’ And, m = size of the reference array ‘refArr’.
Space Complexity: O(1)
Better Approach: This approach involves the use of unordered_set. Take a variable ‘connectedComponents’ which is initialized to 0. connectedComponents indicates the number of connected components in the linked list.
Now for each element in reference array ‘refArr’:
• Add reference stored in the array to an unordered_set ‘refSet’.
• If both previous and next siblings are present, then decrement the variable ‘connectedComponents’ by 1 which means that we have closed a gap between two components, so we must decrement the incorrectly counted components.
• If only one of the previous and next siblings is present in the set, then no updating is done to the no. of components.
• If none of the previous and next siblings is present in the set, then we must increment the incorrectly counted components because now, the current node forms a new component.
Below is the implementation of the above approach:
## C++
`// A C++ program to find number``// of Connected Components in a doubly linked list.``#include ``using` `namespace` `std;` `// Node of the doubly linked list``struct` `Node {`` ``int` `data;`` ``struct` `Node* next;`` ``struct` `Node* prev;``};` `// Function to find number of connected``// components using a doubly linked list``// and a reference array``int` `func_connComp(``struct` `Node** head_ref,`` ``vector<``struct` `Node*> refArr, ``int` `n)``{`` ``// Base case when the doubly`` ``// linked list is empty`` ``if` `(head_ref == NULL) {`` ``return` `0;`` ``}` ` ``// Initialise connectedComponents to zero`` ``int` `connectedComponents = 0;` ` ``// Initialise an unordered set`` ``unordered_set<``struct` `Node*> refSet;` ` ``// Push the first element of the`` ``// refArr in the refSet and`` ``// set the connectedComponents to 1`` ``refSet.insert(refArr[0]);`` ``connectedComponents++;` ` ``// Loop over all the elements of the refArr`` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// insert each reference node to the refSet`` ``refSet.insert(refArr[i]);` ` ``// If the reference node is the head of the linked list`` ``if` `(refArr[i]->prev == NULL) {` ` ``// when next sibling of the head node is`` ``// not in the refSet`` ``if` `(refSet.find(refArr[i]->next) == refSet.end()) {`` ``connectedComponents++;`` ``}`` ``}` ` ``// If the reference node is the`` ``// last node of the linked list*/`` ``else` `if` `(refArr[i]->next == NULL) {` ` ``// when previous sibling of the`` ``// node is not in the refSet`` ``if` `(refSet.find(refArr[i]->next) == refSet.end()) {`` ``connectedComponents++;`` ``}`` ``}` ` ``// the case when both previous and`` ``// next siblings of the current node`` ``// are in the refSet`` ``else` `if` `(refSet.find(refArr[i]->prev) != refSet.end()`` ``&& refSet.find(refArr[i]->next) != refSet.end()) {`` ``connectedComponents--;`` ``}` ` ``/*the case when previous and next`` ``// siblings of the current node`` ``// are not in the refSet*/`` ``else` `if` `(refSet.find(refArr[i]->prev) == refSet.end()`` ``&& refSet.find(refArr[i]->next) == refSet.end()) {`` ``connectedComponents++;`` ``}`` ``}`` ``return` `connectedComponents;``}` `// Function to insert a node at the``// beginning of the Doubly Linked List``Node* push(``struct` `Node** head_ref, ``int` `new_data)``{`` ``/* allocate node */`` ``struct` `Node* new_node = ``new` `Node;` ` ``struct` `Node* current_node = new_node;`` ``/* put in the data */`` ``new_node->data = new_data;` ` ``/* since we are adding at the beginning,`` ``prev is always NULL */`` ``new_node->prev = NULL;` ` ``/* link the old list off the new node */`` ``new_node->next = (*head_ref);` ` ``/* change prev of head node to new node */`` ``if` `((*head_ref) != NULL)`` ``(*head_ref)->prev = new_node;` ` ``/* move the head to point to the new node */`` ``(*head_ref) = new_node;` ` ``return` `current_node;``}` `// Function to print nodes in a given``// doubly linked list``void` `printList(``struct` `Node* node)``{`` ``while` `(node != NULL) {`` ``printf``(``"%d "``, node->data);`` ``node = node->next;`` ``}``}` `// Driver code``int` `main()``{` ` ``// Start with the empty list`` ``struct` `Node* head = NULL;` ` ``// Let us create a linked list to test`` ``// the functions so as to find number`` ``// of Connected Components Created a`` ``// doubly linked list: 1 <-> 7 <-> 10 <-> 5 <-> 4 <-> 2`` ``struct` `Node* ref_to_nodeN2 = push(&head, 2);`` ``struct` `Node* ref_to_nodeN4 = push(&head, 4);`` ``struct` `Node* ref_to_nodeN5 = push(&head, 5);`` ``struct` `Node* ref_to_nodeN10 = push(&head, 10);`` ``struct` `Node* ref_to_nodeN7 = push(&head, 7);`` ``struct` `Node* ref_to_nodeN1 = push(&head, 1);` ` ``vector<``struct` `Node*> refArr{ ref_to_nodeN5,`` ``ref_to_nodeN2, ref_to_nodeN7, ref_to_nodeN1 };` ` ``// This function will return the number`` ``// of connected components of doubly linked list`` ``int` `connectedComponents = func_connComp(&head, refArr, 4);` ` ``cout << ``"Total number of connected components are "`` ``<< connectedComponents << endl;` ` ``return` `0;``}`
## Java
`// Java program to find number``// of Connected Components in a doubly linked list.``import` `java.util.HashSet;` `public` `class` `GFG``{`` ` ` ``// Node of the doubly linked list`` ``static` `class` `Node {`` ``int` `data;`` ``Node next;`` ``Node prev;`` ``}` ` ``// Function to find number of connected`` ``// components using a doubly linked list`` ``// and a reference array`` ``static` `int` `func_connComp(Node head_ref, Node[] refArr,`` ``int` `n)`` ``{`` ``// Base case when the doubly`` ``// linked list is empty`` ``if` `(head_ref == ``null``) {`` ``return` `0``;`` ``}` ` ``// Initialise connectedComponents to zero`` ``int` `connectedComponents = ``0``;` ` ``// Initialise an unordered set`` ``HashSet refSet = ``new` `HashSet();` ` ``// Push the first element of the`` ``// refArr in the refSet and`` ``// set the connectedComponents to 1`` ``refSet.add(refArr[``0``]);`` ``connectedComponents++;` ` ``// Loop over all the elements of the refArr`` ``for` `(``int` `i = ``1``; i < n; i++) {` ` ``// insert each reference node to the refSet`` ``refSet.add(refArr[i]);` ` ``// If the reference node is the head of the`` ``// linked list`` ``if` `(refArr[i].prev == ``null``) {` ` ``// when next sibling of the head node is`` ``// not in the refSet`` ``if` `(refSet.contains(refArr[i].next)) {`` ``connectedComponents++;`` ``}`` ``}` ` ``// If the reference node is the`` ``// last node of the linked list*/`` ``else` `if` `(refArr[i].next == ``null``) {` ` ``// when previous sibling of the`` ``// node is not in the refSet`` ``if` `(refSet.contains(refArr[i].next)) {`` ``connectedComponents++;`` ``}`` ``}` ` ``// the case when both previous and`` ``// next siblings of the current node`` ``// are in the refSet`` ``else` `if` `(refSet.contains(refArr[i].prev)`` ``&& refSet.contains(refArr[i].next)) {`` ``connectedComponents--;`` ``}` ` ``/*the case when previous and next`` ``// siblings of the current node`` ``// are not in the refSet*/`` ``else` `if` `(!refSet.contains(refArr[i].prev)`` ``&& !refSet.contains(refArr[i].next)) {`` ``connectedComponents++;`` ``}`` ``}`` ``return` `connectedComponents;`` ``}` ` ``// Function to insert a node at the`` ``// beginning of the Doubly Linked List`` ``static` `Node push(``int` `new_data)`` ``{`` ``/* allocate node */`` ``Node new_node = ``new` `Node();` ` ``Node current_node = new_node;`` ``/* put in the data */`` ``new_node.data = new_data;` ` ``/* since we are adding at the beginning,`` ``prev is always null */`` ``new_node.prev = ``null``;` ` ``/* link the old list off the new node */`` ``new_node.next = head;` ` ``/* change prev of head node to new node */`` ``if` `((head) != ``null``)`` ``head.prev = new_node;` ` ``/* move the head to point to the new node */`` ``head = new_node;` ` ``return` `current_node;`` ``}` ` ``// Function to print nodes in a given`` ``// doubly linked list`` ``static` `void` `printList(Node node)`` ``{`` ``while` `(node != ``null``) {`` ``System.out.print(node.data + ``" "``);`` ``node = node.next;`` ``}`` ``}` ` ``// Start with the empty list`` ``static` `Node head = ``null``;` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{` ` ``// Let us create a linked list to test`` ``// the functions so as to find number`` ``// of Connected Components Created a`` ``// doubly linked list: 1 <. 7 <. 10 <. 5 <. 4 <. 2`` ``Node ref_to_nodeN2 = push(``2``);`` ``Node ref_to_nodeN4 = push(``4``);`` ``Node ref_to_nodeN5 = push(``5``);`` ``Node ref_to_nodeN10 = push(``10``);`` ``Node ref_to_nodeN7 = push(``7``);`` ``Node ref_to_nodeN1 = push(``1``);` ` ``Node[] refArr = { ref_to_nodeN5, ref_to_nodeN2,`` ``ref_to_nodeN7, ref_to_nodeN1 };` ` ``// This function will return the number`` ``// of connected components of doubly linked list`` ``int` `connectedComponents`` ``= func_connComp(head, refArr, ``4``);` ` ``System.out.println(`` ``"Total number of connected components are "`` ``+ connectedComponents);`` ``}``}` `// This code is contributed by Lovely Jain`
## Python3
`# A Python3 program to find number``# of Connected Components in a doubly linked list.`` ` `# Node of the doubly linked list``class` `Node: `` ``def` `__init__(``self``): `` ``self``.data ``=` `0`` ``self``.``next` `=` `None`` ``self``.prev ``=` `None`` ` `# Function to find number of connected``# components using a doubly linked list``# and a reference array``def` `func_connComp(head_ref, refArr, n):`` ` ` ``# Base case when the doubly`` ``# linked list is empty`` ``if` `(head_ref ``=``=` `None``):`` ``return` `0``;`` ` ` ``# Initialise connectedComponents to zero`` ``connectedComponents ``=` `0``;`` ` ` ``# Initialise an unordered set`` ``refSet ``=` `set``()`` ` ` ``# Push the first element of the`` ``# refArr in the refSet and`` ``# set the connectedComponents to 1`` ``refSet.add(refArr[``0``]);`` ``connectedComponents ``+``=` `1`` ` ` ``# Loop over all the elements of the refArr`` ``for` `i ``in` `range``(``1``, n):`` ` ` ``# add each reference node to the refSet`` ``refSet.add(refArr[i]);`` ` ` ``# If the reference node is the head of the linked list`` ``if` `(refArr[i].prev ``=``=` `None``):`` ` ` ``# when next sibling of the head node is`` ``# not in the refSet`` ``if` `(refArr[i].``next` `not` `in` `refSet):`` ``connectedComponents ``+``=` `1`` ` ` ``# If the reference node is the`` ``# last node of the linked list'''`` ``elif` `(refArr[i].``next` `=``=` `None``):`` ` ` ``# when previous sibling of the`` ``# node is not in the refSet`` ``if` `(refArr[i].``next` `not` `in` `refSet):`` ``connectedComponents ``+``=` `1`` ` ` ``# the case when both previous and`` ``# next siblings of the current node`` ``# are in the refSet`` ``elif` `(refArr[i].prev ``in` `refSet`` ``and` `refArr[i].``next` `in` `refSet):`` ``connectedComponents ``-``=` `1`` ` ` ``# the case when previous and next`` ``# siblings of the current node`` ``# are not in the refSet'''`` ``elif` `(refArr[i].prev ``not` `in` `refSet`` ``and` `refArr[i].``next` `not` `in` `refSet):`` ``connectedComponents ``+``=` `1`` ` ` ``return` `connectedComponents;`` ` `# Function to add a node at the``# beginning of the Doubly Linked List``def` `push(head_ref, new_data):` ` ``''' allocate node '''`` ``new_node ``=` `Node()`` ``current_node ``=` `new_node;`` ` ` ``''' put in the data '''`` ``new_node.data ``=` `new_data;`` ` ` ``''' since we are adding at the beginning,`` ``prev is always None '''`` ``new_node.prev ``=` `None``;`` ` ` ``''' link the old list off the new node '''`` ``new_node.``next` `=` `(head_ref);`` ` ` ``''' change prev of head node to new node '''`` ``if` `((head_ref) !``=` `None``):`` ``(head_ref).prev ``=` `new_node;`` ` ` ``''' move the head to point to the new node '''`` ``(head_ref) ``=` `new_node;`` ``return` `current_node, head_ref;`` ` `# Function to print nodes in a given``# doubly linked list``def` `printList(node):`` ``while` `(node !``=` `None``):`` ``print``(node.data, end ``=` `' '``);`` ``node ``=` `node.``next``;`` ` `# Driver code``if` `__name__``=``=``'__main__'``:`` ` ` ``# Start with the empty list`` ``head ``=` `None``;`` ` ` ``# Let us create a linked list to test`` ``# the functions so as to find number`` ``# of Connected Components Created a`` ``# doubly linked list: 1 <. 7 <. 10 <. 5 <. 4 <. 2`` ``ref_to_nodeN2, head ``=` `push(head, ``2``);`` ``ref_to_nodeN4, head ``=` `push(head, ``4``)`` ``ref_to_nodeN5, head ``=` `push(head, ``5``)`` ``ref_to_nodeN10, head ``=` `push(head, ``10``)`` ``ref_to_nodeN7, head ``=` `push(head, ``7``)`` ``ref_to_nodeN1, head ``=` `push(head, ``1``) `` ``refArr ``=` `[ref_to_nodeN5, ref_to_nodeN2, ref_to_nodeN7, ref_to_nodeN1]`` ` ` ``# This function will return the number`` ``# of connected components of doubly linked list`` ``connectedComponents ``=` `func_connComp(head, refArr, ``4``);`` ``print``(``"Total number of connected components are "``, connectedComponents)` ` ``# This code is contributed by rutvik_56`
Output
```Total number of connected components are 3
```
Complexity Analysis:
• Time Complexity: O(m)
• Space Complexity: O(m) Where, m = size of the reference array ‘refArr
My Personal Notes arrow_drop_up
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