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https://www.journeyingtheglobe.com/celsius-to-fahrenheit/102.5-c-to-f/ | [
"102.5 c to f | 102.5 Celsius to Fahrenheit | [+ Examples]\n\n# 102.5C to F - Convert 102.5° Celsius to Fahrenheit\n\n### Calculate 102.5° Celsius to Fahrenheit (102.5C to °F)\n\nCelsius\nFahrenheit\n102.5 Degrees Celsius = 216.5 Degrees Fahrenheit\n\nTemperature Conversion – Degrees Celsius into Degrees Fahrenheit\n\nCelsius to Fahrenheit conversion formula is all about converting the temperature denoting in Celsius to Fahrenheit. As mentioned earlier, the temperature of boiling (hot) water in Celsius is 0 degrees and in Fahrenheit is 21 degrees, the formula to convert C to F is\n\n### F = C x (9/5) + 32\n\nThe math is here is fairly simple, and can be easily understood by an example. Let’s say we need to 102.5 Celsius to Fahrenheit!\n\n## How To Convert 102.5 C to F?\n\nTo convert 102.5 degrees Celsius to Fahrenheit, all one needs is to put in the values in the converter equation-\n\nF = 102.5 x (9/5) +32\n\nF = 216.5 degrees\n\nThus, after applying the formula to convert 102.5 Celcius to Fahrenheit, the answer is –\n\n102.5°C = 216.5°F\n\nor\n\n102.5 degrees Celsius equals 216.5 degrees Fahrenheit!\n\n## Frequently asked questions about converting 102.5 Degrees Celsius into Degrees Fahrenheit\n\n### How much is 102.5 degrees in Celsius to Fahrenheit?\n\n102.5C to F = 216.5 °F\n\n### What is the formula to calculate Celsius to Fahrenheit?\n\nThe C to F formula is\n\n(C × 9/5) + 32 = F\n\nWhen we enter 102.5 for C in the formula, we get\n\n(102.5 × 9/5) + 32 = 216.5 F\n\nTo solve (102.5 × 9/5) + 32 = F, we first multiply 9 by 102.5, then we divide the product by 5, and then finally we add 32 to the quotient to get the answer.\n\n### What is the simplest way of converting Celsius into Fahrenheit?\n\nThe boiling temperature of water in Celsius is 0 and 21 in Fahrenheit. So, the simplest formula to calculate the difference is\n\nF = C X (9/5) + 32\n\nBut this is not the only formula that is used for the conversion as some people believe it doesn’t give out the exact number.\n\nOne another formula that is believed to be equally easy and quick is –\n\nCelsius Temperature X 2 + 32 = Fahrenheit\n\nFor converting Fahrenheit into Celsius, you can use this formula – Fahrenheit Temperature – 30 / 2 = Celsius Temperature.\n\nWhile there are other temperature units like Kelvin, Réaumur, and Rankine as well, Degree Celsius and Degree Fahrenheit are the most commonly used.\n\nWhile Fahrenheit is primarily used in the US and its territories, Celsius has gained more popularity in the rest of the world. For those using these two different scales, the numbers that denote that temperature is quite different.\n\nFor example, water freezes at Zero Degree Celsius and boils at 100 degrees, the readings are 32-degree Fahrenheit as the freezing point of water and 212 degrees for boiling.\n\n## Fahrenheit Conversions\n\nFor Fahrenheit conversion, all you need to do is start with the temperature in Fahrenheit. Subtract 30 from the resultant figure, and finally, divide your answer by 2!\n\n## Common F and C Temperature Table\n\n### Key Inferences about Fahrenheit and Celsius\n\n• Celsius and Fahrenheit are commonly misspelled as Celcius and Farenheit.\n• The formula to find a Celsius temperature from Fahrenheit is: °F = (°C × 9/5) + 32\n• The formula to find a Fahrenheit temperature from Celsius is: °°C = (°F - 32) × 5/9\n• The two temperature scales are equal at -40°.\n\n## Oven temperature chart\n\n### How to Convert From Fahrenheit to Celsius and Celsius to Fahrenheit - Quick and Easy Method\n\nHow to Convert From Fahrenheit to C... x\nHow to Convert From Fahrenheit to Celsius and Celsius to Fahrenheit\n\nThe Fahrenheit temperature scale is named after the German physicist Daniel Gabriel Fahrenheit in 1724 and was originally used for temperature measurement through mercury thermometers that he invented himself.\n\nMeanwhile, the Celsius scale was originally called centigrade but later came to be named after Swedish astronomer Anders Celsius in 1742. But when the scale was first introduced, it was quite the reverse of what it is today. Anders labeled 0 Degree Celsius as the boiling point of water, while 100 denoted the freezing point.\n\nHowever, after Celsius passed away, Swedish taxonomist Carl Linnaeus flipped it to the opposite, the same as it is used today.\n\n### Our Take\n\nWhile this is the formula that is used for the conversion from Celsius to Fahrenheit, there are few diversions and it is not always a perfect conversion either making it slightly more difficult than what appears to be.\n\nAll said and done, one must understand that since both the scales are offset, meaning that neither of them is defined as starting from zero, there comes a slightly complicated angle to the above-mentioned formula.\n\nBesides, the two scales do not start with a zero, and they both add a different additional value for every unit of heat. This is why it is not every time possible to get an exact value of the conversion by applying the formula.\n\nReverse Conversion: Fahrenheit to Celsius\n\n Celsius Fahrenheit 102.51°C 216.52°F 102.52°C 216.54°F 102.53°C 216.55°F 102.54°C 216.57°F 102.55°C 216.59°F 102.56°C 216.61°F 102.57°C 216.63°F 102.58°C 216.64°F 102.59°C 216.66°F 102.6°C 216.68°F 102.61°C 216.7°F 102.62°C 216.72°F 102.63°C 216.73°F 102.64°C 216.75°F 102.65°C 216.77°F 102.66°C 216.79°F 102.67°C 216.81°F 102.68°C 216.82°F 102.69°C 216.84°F 102.7°C 216.86°F 102.71°C 216.88°F 102.72°C 216.9°F 102.73°C 216.91°F 102.74°C 216.93°F\n Celsius Fahrenheit 102.75°C 216.95°F 102.76°C 216.97°F 102.77°C 216.99°F 102.78°C 217°F 102.79°C 217.02°F 102.8°C 217.04°F 102.81°C 217.06°F 102.82°C 217.08°F 102.83°C 217.09°F 102.84°C 217.11°F 102.85°C 217.13°F 102.86°C 217.15°F 102.87°C 217.17°F 102.88°C 217.18°F 102.89°C 217.2°F 102.9°C 217.22°F 102.91°C 217.24°F 102.92°C 217.26°F 102.93°C 217.27°F 102.94°C 217.29°F 102.95°C 217.31°F 102.96°C 217.33°F 102.97°C 217.35°F 102.98°C 217.36°F 102.99°C 217.38°F\n Celsius Fahrenheit 103°C 217.4°F 103.01°C 217.42°F 103.02°C 217.44°F 103.03°C 217.45°F 103.04°C 217.47°F 103.05°C 217.49°F 103.06°C 217.51°F 103.07°C 217.53°F 103.08°C 217.54°F 103.09°C 217.56°F 103.1°C 217.58°F 103.11°C 217.6°F 103.12°C 217.62°F 103.13°C 217.63°F 103.14°C 217.65°F 103.15°C 217.67°F 103.16°C 217.69°F 103.17°C 217.71°F 103.18°C 217.72°F 103.19°C 217.74°F 103.2°C 217.76°F 103.21°C 217.78°F 103.22°C 217.8°F 103.23°C 217.81°F 103.24°C 217.83°F\n Celsius Fahrenheit 103.25°C 217.85°F 103.26°C 217.87°F 103.27°C 217.89°F 103.28°C 217.9°F 103.29°C 217.92°F 103.3°C 217.94°F 103.31°C 217.96°F 103.32°C 217.98°F 103.33°C 217.99°F 103.34°C 218.01°F 103.35°C 218.03°F 103.36°C 218.05°F 103.37°C 218.07°F 103.38°C 218.08°F 103.39°C 218.1°F 103.4°C 218.12°F 103.41°C 218.14°F 103.42°C 218.16°F 103.43°C 218.17°F 103.44°C 218.19°F 103.45°C 218.21°F 103.46°C 218.23°F 103.47°C 218.25°F 103.48°C 218.26°F 103.49°C 218.28°F"
] | [
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https://blog.gvsig.org/2016/04/07/splitting-strings-in-an-attribute-table-on-gvsig-2-x/ | [
"## Splitting strings in an attribute table on gvSIG 2.x\n\nWhen we work with an attribute table in gvSIG, we sometimes want to split the strings in one of the fields, in order to have a substring. For that, we would use the subString operator at the Field calculator, that will return a part of the original string.\n\nFor example, if we have an only field with the code of a municipality (two numbers), then a space, and finally the name of the municipality, we would be able to be interested in having a new field with the code, and another one with the name. The subString operator will allow to do it.\n\nThe subString command has the following structure:\n\nsubString(Parameter 1,Parameter 2, Parameter 3)\n\nwhere:\n\n• Parameter 1: It is the field name (between square brackets in gvSIG).\n• Parameter 2: It is the position of the first character to be split.\n• Parameter 3: It is the position of the last character of the substring to be split.\n\nAnd we have to take into account that the first position will be “0”.\n\nIn addition, the Length operator can be used when we want to split the last characters and the length of the different strings in the field is different.",
null,
"We can have several possible situations, and these options are the most common ones:\n\n• We want to take the initial part of the string, removing the last X characters:\n\nsubString([Field],0,length([Field])-X)\n\n(where “Field” will be the field name, and X will be the number of characters of the end to be removed)\n\nExample: Removing the last 3 characters: subString([Field],0,length([Field])-3)\n\n• We want to remove the first X characters, and take the rest of the string:\n\nsubString([Field],X,length([Field]))\n\n(where “Field” will be the field name, and X will be the number of characters to be removed at the beginning of the string)\n\nExample: Removing the first 3 characters: subString([Field],3,length([Field]))\n\n• We want to take only the first X characters:\n\nsubString([Field],0,X-1)\n\n(where “Field” will be the field name, y “X-1” and “X-1” will be the numbers of characters to be taken)\n\nExample: Taking the first 3 characters: subString([Field],0,2)\n\n• We want to take only the last X characters:\n\nsubString([Field],length([Field])-X,length([Field]))\n\n(where “Field” will be the field name, and X will be the number of characters that we want to take at the end of the string)\n\nExample: Taking only the last 3 characters: subString([Field],length([Field])-3,length([Field]))\n\nHere you have a video about it:\n\nWe hope that this functionality is useful for you!\n\nThis entry was posted in english, gvSIG Desktop, training. Bookmark the permalink."
] | [
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"https://gvsig.files.wordpress.com/2016/04/substring_en.png",
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https://www.wikihow.com/Find-the-Surface-Area-of-a-Pyramid | [
"# How to Find the Surface Area of a Pyramid",
null,
"Download Article",
null,
"Download Article\n\nThe surface area of any pyramid can be found by adding the surface area of the base to the surface area of the lateral faces. When working with regular pyramids, you can find the surface area using a formula, as long as you know how to find the area of the base of the pyramid. Since the base can be any polygon, it is helpful to know how to find the area of shapes such as pentagons and hexagons. When working with the common, regular square pyramid, however, calculating the total surface area is a simple calculation, provided you know the slant height of the pyramid and the side length of the square base.\n\n### Method 1 Method 1 of 2:Finding the Surface Area of Any Regular Pyramid\n\n1. 1\nSet up the formula for the surface area of a regular pyramid. The formula is $SA={\\frac {p\\times h}{2}}+B$",
null,
", where $SA$",
null,
"equals the total surface area of the pyramid, $p$",
null,
"equals the perimeter of the base, $h$",
null,
"equals the slant height of the pyramid, and $B$",
null,
"equals the area of the base.\n• The basic formula for the surface area of any pyramid, regular or irregular, is Total Surface Area = Base Area + Lateral Area.\n• Don’t confuse “slant height” with “height.” The “slant height” is the diagonal distance from the apex of the pyramid to the edge of the base. The “height” is the perpendicular distance from the vertex to the base.\n2. 2\nPlug the perimeter of the base into the formula. If you aren’t given the perimeter but know the length of one edge of the base, you can calculate the perimeter by multiplying the length of one edge by the number of edges.\n• For example, If you are finding the surface area of a hexagonal pyramid, and you know that the length of one edge of the base is 4 cm, you would calculate $4\\times 6=24$",
null,
"to find the perimeter of the base, since a hexagon has six edges, or sides. Thus, the perimeter of the base is 24 cm, so your surface area formula will look like this: $SA={\\frac {24\\times h}{2}}+B$",
null,
".\n3. 3\nPlug the value of the slant height into the formula. Make sure you are using the slant height, not the perpendicular height. The problem should provide the slant height. If you don’t know the slant height, you cannot use this method.\n• For example, if the slant height of a hexagonal pyramid is 12 cm, your formula will look like this: $SA={\\frac {24\\times 12}{2}}+B$",
null,
".\n4. 4\nCalculate the area of the base. How you do this will depend on the shape of the base. To learn more about finding the area of a polygon, read Find the Area of Regular Polygons.\n• For example, if you are working with a hexagonal pyramid, the base is a hexagon. To find out how to calculate the area of the base, you can read Calculate the Area of a Hexagon. The formula is $A={\\frac {3{\\sqrt {3}}\\times s^{2}}{2}}$",
null,
", where $s$",
null,
"is the length of one side of the hexagon. Since the length of one side of the hexagon is 4 cm, you would calculate:\n$A={\\frac {3{\\sqrt {3}}\\times 4^{2}}{2}}$",
null,
"$A={\\frac {3{\\sqrt {3}}\\times 16}{2}}$",
null,
"$A={\\frac {48{\\sqrt {3}}}{2}}$",
null,
"$A={\\frac {83.14}{2}}$",
null,
"$A=41.57$",
null,
".\nSo the area of the base is 41.57 square centimeters.\nEXPERT TIP",
null,
"Math Instructor, City College of San Francisco\nGrace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University.",
null,
"Grace Imson, MA\nMath Instructor, City College of San Francisco\n\nOur Expert Agrees: The surface area of a pyramid is equal to the sum of the areas of all of the faces. First, you have to get the area of the base, then add the area of the lateral sides, which is one face times the number of sides.\n\n5. 5\nPlug the area of the base into the formula. Make sure you substitute for the variable $B$",
null,
".\n• For example, if the area of the hexagonal base is 41.57 sq. cm., your formula for surface area will now look like this: $SA={\\frac {24\\times 12}{2}}+41.57$",
null,
".\n6. 6\nMultiply the perimeter of the base and the slant height of the pyramid. Then, divide by two. This will give you the lateral surface area of the pyramid.\n• For example:\n$SA={\\frac {24\\times 12}{2}}+41.57$",
null,
"$SA={\\frac {288}{2}}+41.57$",
null,
"$SA=144+41.57$",
null,
"7. 7\nAdd the two values together. The sum will be the lateral surface area, plus the base surface area, providing you with the total surface area for the pyramid, in square units.\n• For example:\n$SA=144+41.57$",
null,
"$SA=185.57$",
null,
"So, the total surface area of a hexagonal pyramid, given a base edge length of 4 cm and a slant height of 12 cm, is 185.57 square centimeters.\n\n### Method 2 Method 2 of 2:Finding the Surface Area of a Square Pyramid\n\n1. 1\nSet up the formula for surface area of a square pyramid. The formula is $SA=b^{2}+4({\\frac {bh}{2}})$",
null,
", where $b$",
null,
"is equal to the length of one side of the base, and $h$",
null,
"is equal to the slant height of the pyramid.\n• Don’t confuse “slant height” with “height.” The “slant height” is the diagonal distance from the apex of the pyramid to the edge of the base. The “height” is the perpendicular distance from the vertex to the base.\n• Note that this formula is just another way of writing Total Surface Area = Base Area ($b^{2}$",
null,
") + Lateral Area ($4({\\frac {bh}{2}})$",
null,
"). This formula only works for regular square pyramids.\n2. 2\nPlug in the values for the side length and slant height into the formula. Make sure you substitute the side length of the base for $b$",
null,
"and the slant height for $h$",
null,
".\n• For example, if the length of one side of the base of a square pyramid is 4 cm, and the slant height is 12 cm, the formula will look like this: $SA=4^{2}+4({\\frac {(4)(12)}{2}})$",
null,
".\n3. 3\nSquare the side length of the base. This will give you the surface area of the base.\n• For example:\n$SA=4^{2}+4({\\frac {(4)(12)}{2}})$",
null,
"$SA=16+4({\\frac {(4)(12)}{2}})$",
null,
"4. 4\nMultiply the side length of the base by the slant height and divide by two. Then, multiply by 4. This will give you the lateral surface area of the pyramid.\n• For example:\n$SA=16+4({\\frac {(4)(12)}{2}})$",
null,
"$SA=16+4({\\frac {48}{2}})$",
null,
"$SA=16+4(24)$",
null,
"$SA=16+96$",
null,
"5. 5\nAdd the base surface area and the lateral surface area. This will give you the total surface area of the pyramid, in square units.\n• For example:\n$SA=16+96$",
null,
"$SA=112$",
null,
"So, the total surface area of a square pyramid, with a base side length of 4 cm and a slant height of 12 cm, is 112 square centimeters.\n\n## Community Q&A\n\nSearch\n• Question\nHow do you find the lateral area of hexagonal pyramid, given the height and length of each side?",
null,
"Use the base times height for the rectangles, and the altitude times base of the hexagonal face times three.\n• Question\nHow would you calculate the surface area of pyramid that does not have a square base?",
null,
"Use the formula (p x h/2) + (B), where p is the perimeter of the base, h is the slant height of the pyramid, and B is the area of the base. Below are some articles on finding the area of a pentagon and hexagon, two common pyramid bases: http://www.wikihow.com/Find-the-Area-of-a-Pentagon http://www.wikihow.com/Calculate-the-Area-of-a-Hexagon\n• Question\nHow do I double the lateral surface area of a square pyramid?",
null,
"Donagan\nOne way would be to double either the length of the sides of the base or the slant height (but not both).\n• Question\nWhere does the height fit in?",
null,
"Donagan\nThe actual height of a pyramid does not factor in to the calculation of its surface area. However, the \"slant height\" does, being one of the dimensions of a pyramid's slanted surface (along with the width of the base).\n• Question\nA regular square pyramid is 3 m. Height and the perimeter of its base is 16 m. How do I find the volume of the pyramid?",
null,
"Donagan\nV = (s)²(h) / 3, where s is the length of one side of the base, and h is the pyramid's height. The side of a square is 1/4 of the perimeter, so this pyramid has a base side of 4. Therefore, V = (4)²(3) / 3 = 16 cubic meters.\n• Question\nWhat if I'm only given the lateral surface area?",
null,
"Donagan\nThat's not enough information to find the total surface area (including the base area).\n• Question\nWhat if I do not know the height?",
null,
"Donagan\nIf you don't know either the height or the slant height of a pyramid, you cannot determine the surface area.\n• Question\nHow would I write the total surface area of a square-based pyramid having base side x cm and slant height y cm?",
null,
"Donagan\nSee Method 2 Step 1 above. Substitute x for b and y for h.\n• Question\nHow do I calculate the surface area of a triangle pyramid? Is it the same kind of mathematical equation, or something different?",
null,
"Donagan\nIf it's an equilateral triangle, it qualifies for Method 1 above. Otherwise, you would have to calculate the areas of each of the surfaces separately and add them together.\n• Question\nHow do I find the surface area of a heptagonal prism?",
null,
"Donagan\nAssuming regular heptagonal bases, and lateral sides perpendicular to the bases, you would multiply an edge of the base by the height, and multiply by seven to find the total lateral surface area. If desired, add the areas of the bases. (The area of one base is half the product of the perimeter and the apothem.)\n200 characters left\n\n## Things You'll Need\n\n• Pencil\n• Paper\n• Calculator (optional)\n• Ruler (optional)",
null,
"Co-authored by:\nMath Instructor, City College of San Francisco\nThis article was co-authored by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 354,351 times.\nCo-authors: 22\nUpdated: September 16, 2021\nViews: 354,351\nArticle SummaryX\n\nTo find the surface area of a pyramid, start by multiplying the perimeter of the pyramid by its slant height. Then, divide that number by 2. Finally, add the number you get to the area of the pyramid's base to find the surface area. To learn how to find the surface area of a square pyramid, scroll down!\n\nThanks to all authors for creating a page that has been read 354,351 times.\n\n•",
null,
"Anonymous Kid\n\nApr 6\n\n\"This article helped when I needed an equation for my math homework. It is useful and adds in formulas for your own...\" more"
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"",
null,
"",
null,
"# FOR LOOP\n\nLife is cyclical. Nature is cyclical. Winter is replaced by spring, spring by summer, summer by autumn with a lot of rain and sleet, and then winter goes again, replaced by spring. Cycles. Time runs away: 1 a.m., and then suddenly 2 a.m., 3 a.m. ... afternoon, 1 p.m., 2 p.m., 3 p.m. ... Time goes by and it repeats again: one hour, two, three. Cycles, cycles, cycles.\n\nEven our walking is cyclical. The first leg leaves the ground, swings forward from the hip, then it strikes the ground and the second goes. Repeat twenty times and you will reach the fridge. All is cyclical! Breakfast, lunch, dinner; work, work, work, salary, work, work, work, salary, work, work, work, salary, one leave a year. The first year goes like that, the second, the third.\n\nLife is cyclical. We have already considered one example: days of the week. Cycle. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday. Finished? Again! If we replace the days of the week by numbers, as we did it previously, then we'll get a permanent and recurring range of numbers:\n\n1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3...\n\nThanks to the cycles and the constancy of events within the cycle, we can plan solid data processing algorithms. Water pierces rocks, cycles streamlines data. As water drop by drop, a cycle changes data command by command, creating new forms, structures and content.\n\nThe idea of cycles is to reach necessary result when repeating a certain action multiple times. Step by step to the fridge. By small steps, without extra efforts, but constantly, rhythmically, confidently moving towards the goal.\n\nRepeating the actions again and again, we'll get sorted data.\n\nWhat is multiplication? It is addition a number with itself as many times as it is indicated by multiplier. It's also a cycle!\n\n2 * 4 = 2 + 2 + 2 + 2\n\n\"Multiple recurrence of a certain action in order to obtain a result\": for 2 * 4 we'll repeat the action \"add two to the result\" four times.\n\nTo learn to plan and realize loops in programming (remark: it's called loops in coding, not cycles), we should develop the skill of finding small parts of the whole that have same meaning. In the example above 2 * 4, the small part is operation +2. Repeat it 4 times, and you'll get the result. It could be also operation +4, repeated 2 times.\n\nWe should make a habit to fragment data into pieces. A number can be fragmented into figures, a word - into characters. The task is to see, to understand, to guess, to invent formulas and algorithms of every processing part, so that we piece all fragmented parts together at the end of a calculation, at the end of the cycle.\n\nWe should learn to find dynamic patterns. Try to find a pattern and then answer the questions \"Which rule combines all these numbers? What the number will replace the dot in the example below?\n\n8 4 9 3 10 2 11 1 . 0\n\n1. Each pair of numbers is predicted on the rule: the third number is the first plus one, the fourth is the second minus one, the fifth is the third plus one, and so on: 8 → 9, 4 → 3...\n\n2. Desired number is 12.\n\nRecall the program \"Does the driver violate the rules?\". What if we need to control the speed of three vehicles, not one. What about ten vehicles? Will we run the program each time? Well, ladies and gentlemen, it’s not quite convenient for us.\n\nRight in this case, loops are appropriate to repeat the program three or ten times. And we'll just sit and enter the speed of the following vehicle. Mmmm, it's a dream job!\n\nLoops are cool.",
null,
"# TERMINOLOGY\n\nThere won't be many terms, but they are still important. I will use them further in the text. It's something like a vocabulary that we'll deal with.",
null,
"Loop is a multiple recurrence of a certain action or actions in order to obtain a result.\n\nFor example, we can repeat our program about drivers, vehicles, speed and other things infinite number of times by using loops.",
null,
"A block of commands of the loop includes all the commands, indented with four spaces from the first character of a loop operator (in this case, from the character \"f\" in the \"for\" operator). The commands of the block will be executed sequentially, one by one. Meanwhile, the next command won't be executed until the previous one is implemented successfully. Also, the block of commands of the loop is called the loop body.\n\nA block of commands of the loop is identified just like it would be identified if it belonged to a condition. The command that wasn't written according to the rules of the block, is considered a command outside of the loop. The loop is ended with the last command of the block, not with the wrong command:\n\nFor loop: Command 1 Command 2 Command 3 The last command of the block This string will be implemented only after the whole loop is finished.",
null,
"Iteration is one recurrence of the block of loop commands, from the first to the last.\n\nFor example, we can call the first week of the month \"the first iteration\": the range of the days from 1 to 7 inclusive. The next week and the next 1, 2, 3... 7 are the second iteration. The number of iterations is measured by the number of executions of all the loop commands.",
null,
"Loop counter is a special variable to which a programmer automatically and sequentially assigns iteration value.\n\nLoop counter is a little more complicated than usual variable because of its automaticity. That's all right, we'll handle it.",
null,
"Iteration data is any fragmented set.\n\nFor example, the set of numbers 0, 1, 2, 3 is fragmented and relates to iteration data. We can easily dissect the set 0, 1, 2, 3 into the parts that form it: respectively, here goes 0, then 1, the next is 2 and the last is 3 - four different numbers that form the range.\n\nNow think: whether a string variable is iteration data or not? For example, the variable that took the value \"Hello\"?\n\nYes, it is. Because the word \"Hello\" can be fragmented into separate sustained parts: the character \"H\", then - character \"e\", after that \"l\", \"l\", \"o\". Since every word or every set of characters consist of separate fragments - signs, every string variable relates to iteration data.",
null,
"# FOR LOOP SYNTAX\n\nThe for loop is written in Python language as follows:\n\nfor COUNTER VARIABLE in range(NUMBER OF ITERATIONS): A block of loop commands",
null,
"range(x) is an integer array generator from 0 till x. I would like to emphasize that x is not inclusive. The last value of the range is (x - 1).\n\nThe next programs are equal. The program without loop:\n\nprint(\"Hello!\") print(\"Hello!\") print(\"Hello!\") print(\"Hello!\") print(\"Hello!\")\n\nThe program with loop:\n\nfor i in range(5): print(\"Hello!\")\n\nThe output of the programs is:\n\nHello! Hello! Hello! Hello! Hello!\n\nTechnically, it's simple: we set the number of iterations, wrote the command, repeated it and, voilà, got the result. Easy? Let's move on.\n\nNow we are going to figure out how the loop counter i works. How will we do that? Just displaying its values on the screen:\n\nfor i in range(5): print(i)\n\nThe result is:\n\n0 1 2 3 4\n\nThe values of range(5), i.e. 0, 1, 2, 3, 4 (till 5) were assigned to the loop counter sequentially. After that we displayed these values on the screen.",
null,
"# DEALING WITH THE LOOP COUNTER\n\nYou should understand, remember and engrave on memory that loop counter can be used as a simple variable the value of which is assigned automatically by programming language. The same thing happens in other languages - Java, C++, Pascal.\n\nPython allows to change the value of the loop counter in the loop body, but after one iteration the value from the range() will be assigned to it again.\n\nLet's analyze:\n\nfor i in range(10): print(\"Value of i =\", i, end=\"\") i = i * 2 print(\", changed value of i = i * 2 =\", i)\n\nThe result is:\n\nValue of i = 0, changed value of i = i * 2 = 0 Value of i = 1, changed value of i = i * 2 = 2 Value of i = 2, changed value of i = i * 2 = 4 Value of i = 3, changed value of i = i * 2 = 6 Value of i = 4, changed value of i = i * 2 = 8 Value of i = 5, changed value of i = i * 2 = 10 Value of i = 6, changed value of i = i * 2 = 12 Value of i = 7, changed value of i = i * 2 = 14 Value of i = 8, changed value of i = i * 2 = 16 Value of i = 9, changed value of i = i * 2 = 18\n\nIt's not hard to see that the first value of the loop counter i is always equal to the value of the range(10) function, that is sequentially 0, then 1, then 2, further 3, 4, 5, 6, 7, 8, 9.",
null,
"The parameter end=\"\" after arguments of the first print() command: print(\"Value of i =\", i, end=\"\") sets end of line modifier. Empty quotes indicate that the modifier took the value emptiness. If we assign nothing to end=\"\", then Python will output line break sign on the screen (it's equal to pressing the Enter key in text editors). In this case, each new print() will output text from a new line in the console of Python. We'll always set end=\"\", as a parameter, when we need to continue the previous output with the next print(), i.e. display data in a line.\n\nThe value of the end=\"\" modifier can be any character or even a string. For example:\n\nprint(\"This is \", end=\"changed end modifier\")\n\nType that command and look what you've done.\n\nBut let's get back to loops.",
null,
"Among professional programmers it is customary to use the following identifiers for loop counters: i, j, o, p, k, l.\n\nAgain, it's not necessary to use these characters, but it will be nice if you make a habit to operate with them. At least because all decent textbooks, tutorials, material and documentation use these variable names. It would be more convenient, you know.\n\nThe range() function assigns the values of the loop counter. We can change its parameters if necessary, then the value of the loop counter will change with each iteration.",
null,
"# RANGE() PARAMETERS\n\nGetting arrays of integers with range() is a good deal, moreover, range() is called a generator of numbers within a specified range. The syntax is:\n\nrange(x, y, z)\n\n• x - initial value, the parameter isn't necessary;\n• y - final value, the parameter is necessary;\n• z - change interval (step) of counter variable, the parameter isn't necessary. The change step of the counter means the change of the range values, for example, if it is pointed out 3, then 3 will be add to the next values of the range: 0, 3, 6, 9...",
null,
"When x is not indicated (initial value), it is considered as 0. When z is not indicated (change interval), the values change at intervals of +1. The parameter z is always indicated when x > y and we need to get a downward range, i.e. from the largest to the smallest.\n\nThe output of numbers from 10 till 20 with a step 1 (parameter of the step isn't indicated):\n\nfor i in range(10, 20): print(i)\n\nThe result is:\n\n10 11 12 13 14 15 16 17 18 19\n\nWhy is the last number 19, not 20?\n\nBecause range() generates numbers till specified value non-inclusive.\n\nThe output of numbers from 10 till 20 with a step 3 (parameter of the step is indicated):\n\nfor i in range(10, 20, 3): print(i)\n\nThe result is:\n\n10 13 16 19\n\nThe task for you: generate a range of numbers where there will be necessarily two numbers - the day of the birth and the month in the right order (firstly goes the day, then the month somewhere in the range). For example, if you were born on the 15th of May (15.05), then one of the solutions is:\n\nfor i in range(15, 0, -5): print(i)\n\nThe result is the range with 15 and 5 in it:\n\n15 10 5\n\nDone? Then try to output the range of the numbers in a line by using the end=\"\" modifier. And now it's time for \"tricky\" task!",
null,
"Write a program that will output the range with 20 numbers with the following pattern:\n\n8 4 9 3 10 2 11 1 12 0...\n\nSolution:\n\nfor i in range(10): print(8 + i, 4 - i, \"\", end=\"\")\n\nThe result is:\n\n8 4 9 3 10 2 11 1 12 0 13 -1 14 -2 15 -3 16 -4 17 -5\n\nYou can have a different solution from presented here. It is not forbidden as long as your result is the same.",
null,
"# CHECKING THE DRIVERS\n\nThe use of the loop counter isn't necessary in our loop. However, we must identify it in the title of the loop for i in range(x), but we don't have to process the counter inside the loop body.\n\nLet's return to the poor drivers and our program that determines speeding fines. Now we are going to add one more requirement: user will enter at the beginning of the program how many vehicles must be checked. For example, 5, 10 or 25, or maybe 1, or even 0. Having received the value, our program will do everything needed brilliantly. We don't even have to change the code greatly.",
null,
"To repeat any code a few times, you should wrap it in the loop, that is, to form a loop block, having added four spaces before every line (having specified them as loop lines).\n\nOur program before changes:\n\nspeed = int(input(\"Enter vehicle speed: \")) if (speed > 60 and speed <= 300): print(\"The driver violates the rules! Fine him, FINE!\") elif (speed > 300): print(\"It isn't a plane. The entered speed is incorrect.\") elif (speed == 0): print(\"The vehicle isn't moving. Like a stone. A stationary stone.\") elif (speed < 0): print(\"The speed can't be negative. We are deeply sorry.\") else: print(\"The driver doesn't violate the rules. The driver is good. Be like him.\")\n\nI'll enter one more variable countAuto. It will reflect the number of vehicles the speed of which we need to check.\n\ncountAuto = int(input(\"Enter the number of vehicles: \"))\n\nNow we are going to create a loop that will wrap the whole previous code:\n\nfor i in range(countAuto):\n\nHere we'll use entered by the user number for the range() parameter. Therefore, if he enters number 3, then Python will take it as range(3). Let's just write the code in the loop body. Remember about four space indents.\n\nThat's what we've got:\n\ncountAuto = int(input(\"Enter the number of vehicles: \")) for i in range(countAuto): speed = int(input(\"Enter vehicle speed: \")) if (speed > 60 and speed <= 300): print(\"The driver violates the rules! Fine him, FINE!\") elif (speed > 300): print(\"It isn't a plane. The entered speed is incorrect.\") elif (speed == 0): print(\"The vehicle isn't moving. Like a stone. A stationary stone.\") elif (speed < 0): print(\"The speed can't be negative. We are deeply sorry.\") else: print(\"The driver doesn't violate the rules. The driver is good. Be like him.\")\n\nThe result is:\n\nEnter the number of vehicles: 3 Enter vehicle speed: 310 It isn't a plane. The entered speed is incorrect. Enter vehicle speed: 70 The driver violates the rules! Fine him, FINE! Enter vehicle speed: 35 The driver doesn't violate the rules. The driver is good. Be like him.\n\nYou can see that the previous code was executed as many times as the user had indicated in line \"Enter the number of vehicles\". We have got exactly what we planned! Who are good guys? We are good guys. Today we can afford tea with extra sugar.",
null,
"# PRACTICE\n\n## Exercise 1\n\nWrite a program that will output the first 20 exponents of the number 2 (from 0 till 19 inclusive). Remember that the sign of exponentiation in Python is **. For example, 2 ** 3 reflects the third exponent of the deuce (the result is 8).\n\nSolution\n\nfor i in range(20): print(\"2 to the power of\", i, \"is\", 2 ** i)\n\n## Exercise 2\n\nWrite a program that will ask the user for a number and will output the reminder of division by two. Use a loop to ask and to output the result 10 times.\n\nSolution\n\nfor i in range(10): x = int(input(\"Enter an integer: \")) print(\"The reminder of division the number by 2:\", x % 2)\n\n## Exercise 3\n\nWrite a program that will control the world the weather during a month (31 days). If the temperature is negative, then display \"Cold\" on the screen, if the temperature is zero, then display \"Zero\", if it is positive, then display \"Warm\". Notice that the temperature may not be lower than absolute zero (-173 -273 degrees Celsius. Thank you, Yura, for pointing out the mistake!), and above, for example, 50 degrees.\n\nSolution\n\nfor i in range(31): x = int(input(\"Enter temperature: \")) if (x < -273 or x > 50): print(\"Error. The temperature is incorrect!\") elif (x < 0): print(\"Cold\") elif (x > 0): print(\"Warm\") else: print(\"Zero\")\n\nDid you handle them? Let's go further!"
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https://idownload.pro/1-algorithmic-thinking-peak-finding/y4138305154796f48745375 | [
"# 1. Algorithmic Thinking, Peak Finding\n\n2M+ views | 17K+ likes | 306 dislikes |\nJan 14, 2013\n\n### Thumbs",
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"### Transcription\n\n• The following content is provided\n• under a Creative Commons license.\n• Your support will help MIT OpenCourseWare continue\n• To make a donation or view additional materials\n• from hundreds of MIT courses, visit MIT OpenCourseWare\n• at ocw.mit.edu.\n• PROFESSOR: Hi.\n• I'm a professor of electrical engineering and computer\n• science.\n• I'm going to be co-lecturing 6.006-- Introduction\n• to Algorithms-- this term with professor Erik Domane.\n• Eric, say hi.\n• ERIK DOMANE: Hi.\n• [LAUGHTER]\n• PROFESSOR: And we hope you're going\n• to have a fun time in 6.006 learning\n• a variety of algorithms.\n• What I want to do today is spend literally a minute or so\n• on administrative details, maybe even less.\n• What I'd like to do is to tell you\n• to go to the website that's listed up there and read it.\n• And you'll get all information you\n• of syllabus; what's expected of you; the problem set\n• schedule; the quiz schedule; and so on and so forth.\n• I want to dive right in and tell you about interesting things,\n• like algorithms and complexity of algorithms.\n• I want to spend some time giving you\n• an overview of the course content.\n• And then we're going to dive right\n• in and look at a particular problem of peak\n• finding-- both the one dimensional version and a two\n• dimensional version-- and talk about algorithms to solve\n• this peak finding problem-- both varieties of it.\n• And you'll find that there's really\n• a difference between these various algorithms\n• that we'll look at in terms of their complexity.\n• And what I mean by that is you're\n• going to have different run times of these algorithms\n• depending on input size, based on how\n• efficient these algorithms are.\n• And a prerequisite for this class is 6.042.\n• And in 6.042 you learned about asymptotic complexity.\n• And you'll see that in this lecture\n• we'll analyze relatively simple algorithms today\n• in terms of their asymptotic complexity.\n• And you'll be able to compare and say\n• that this algorithm is fasten this other one-- assuming\n• that you have large inputs-- because it's\n• asymptotically less complex.\n• So let's dive right in and talk about the class.\n• So the one sentence summary of this class\n• is that this is about efficient procedures\n• for solving problems on large inputs.\n• And when I say large inputs, I mean things\n• like the US highway system, a map\n• of all of the highways in the United States;\n• the human genome, which has a billion letters\n• in its alphabet; a social network responding to Facebook,\n• that I guess has 500 million nodes or so.\n• So these are large inputs.\n• Now our definition of large has really changed with the times.\n• And so really the 21st century definition of large\n• is, I guess, a trillion.\n• Right?\n• Back when I was your age large was like 1,000.\n• [LAUGHTER]\n• I guess I'm dating myself here.\n• Back when Eric was your age, it was a million.\n• Right?\n• [LAUGHTER]\n• But what's happening really the world is moving faster,\n• things are getting bigger.\n• We have the capability of computing on large inputs,\n• but that doesn't mean that efficiency\n• isn't of paramount concern.\n• The fact of matter is that you can, maybe,\n• scan a billion elements in a matter of seconds.\n• But if you had an algorithm that required cubic complexity,\n• suddenly you're not talking about 10 raised to 9,\n• you're talking about 10 raised to 27.\n• And even current computers can't really\n• handle those kinds of numbers, so efficiency is a concern.\n• And as inputs get larger, it becomes more of a concern.\n• All right?\n• --efficient procedures-- for solving large scale problems\n• in this class.\n• And we're concerned about scalability,\n• because-- just as, you know, 1,000\n• was a big number a couple of decades ago,\n• and now it's kind of a small number-- it's\n• quite possible that by the time you guys are professors\n• teaching this class in some university\n• that a trillion is going to be a small number.\n• And we're going to be talking about-- I don't know--\n• 10 raised to 18 as being something\n• that we're concerned with from a standpoint of a common case\n• input for an algorithm.\n• So scalability is important.\n• And we want to be able to track how our algorithms are going\n• to do as inputs get larger and larger.\n• You going to learn a bunch of different data structures.\n• We'll call them classic data structures,\n• like binary search trees, hash tables-- that\n• are called dictionaries in Python-- and data\n• structures-- such as balanced binary search trees-- that\n• are more efficient than just the regular binary search trees.\n• And these are all data structures\n• that were invented many decades ago.\n• But they've stood the test of time,\n• and they continue to be useful.\n• We're going to augment these data structures in various ways\n• to make them more efficient for certain kinds of problems.\n• And while you're not going to be doing a whole lot of algorithm\n• doing some design and a whole lot of analysis.\n• The class following this one, 6.046 Designing Analysis\n• of Algorithms, is a class that you\n• should take if you like this one.\n• And you can do a whole lot more design of algorithms in 6.046.\n• But you will look at classic data structures\n• and classical algorithms for these data structures,\n• including things like sorting and matching, and so on.\n• is that you'll be doing real implementations of these data\n• structures and algorithms in Python.\n• And in particular are each of the problem\n• sets in this class are going to have both a theory\n• part to them, and a programming part to them.\n• So hopefully it'll all tie together.\n• The kinds of things we're going to be talking about in lectures\n• and recitations are going to be directly connected\n• to the theory parts of the problem sets.\n• And you'll be programming the algorithms that we talk about\n• in lecture, or augmenting them, running them.\n• Figuring out whether they work well on large inputs or not.\n• So let me talk a little bit about the modules\n• in this class and the problem sets.\n• And we hope that these problem sets\n• are going to be fun for you.\n• And by fun I don't mean easy.\n• I mean challenging and worthwhile, so at the end of it\n• you feel like you've learned something,\n• and you had some fun along the way.\n• All right?\n• So content wise--\n• --we have eight modules in the class.\n• Each of which, roughly speaking, has\n• a problem set associated with it.\n• The first of these is what we call algorithmic thinking.\n• And we'll kick start that one today.\n• We'll look at a particular problem, as I mentioned,\n• of peak finding.\n• And as part of this, you're going\n• to have a problem set that's going to go out today as well.\n• And you'll find that in this problem set\n• some of these algorithms I talk about today will\n• be coded in Python and given to.\n• A couple of them are going to have bugs in them.\n• You'll have to analyze the complexity of these algorithms;\n• figure out which ones are correct and efficient;\n• and write a proof for one of them.\n• All right?\n• So that's sort of an example problem set.\n• And you can expect that most of the problem sets\n• are going to follow that sort of template.\n• All right.\n• So you'll get a better sense of this\n• by the end of the day today for sure.\n• Or a concrete sense of this, because we'll\n• be done with lecture and you'll see your first problem set.\n• We're going to be doing a module on sorting and trees.\n• Sorting you now about, sorting a bunch of numbers.\n• Imagine if you had a trillion numbers\n• and you wanted to sort them.\n• What kind of algorithm could use for that?\n• Trees are a wonderful data structure.\n• There's different varieties, the most common being binary trees.\n• And there's ways of doing all sorts of things,\n• like scheduling, and sorting, using various kinds of trees,\n• including binary trees.\n• And we have a problem set on simulating a logic network\n• using a particular kind of sorting algorithm in a data\n• structure.\n• That is going to be your second problem set.\n• And more quickly, we're going to have modules on hashing,\n• where we do things like genome comparison.\n• In past terms we compared a human genome to a rat genome,\n• and discovered they were pretty similar.\n• 99% similar, which is kind of amazing.\n• But again, these things are so large that you\n• have to have efficiency in the comparison methods\n• that you use.\n• And you'll find that if you don't get the complexity low\n• enough, you just won't be able to complete--\n• your program won't be able to finish running within the time\n• that your problem set is do.\n• Right?\n• Which is a bit of a problem.\n• So that's something to keep in mind as you test your code.\n• The fact is that you will get large inputs to run your code.\n• And you want to keep complexity in mind\n• as you're coding and thinking about the pseudocode,\n• if you will, of your algorithm itself.\n• We will talk about numerics.\n• A lot of the time we talk about such large numbers\n• that 32 bits isn't enough.\n• Or 64 bits isn't enough to represent these numbers.\n• These numbers have thousands of bits.\n• A good example is RSA encryption,\n• that is used in SSL, for example.\n• And when you go-- use https on websites,\n• RSA is used at the back end.\n• And typically you work with prime numbers\n• that are thousands of bits long in RSA.\n• So how do you handle that?\n• How does Python handle that?\n• How do you write algorithms that can\n• deal with what are called infinite precision numbers?\n• So we have a module on numerics in the middle of the term that\n• Graphs, really a fundamental data structure\n• in all of computer science.\n• You might have heard of the famous Rubik's cube assignment\n• from .\n• 006 a 2 by 2 by 2 Rubik's cube.\n• What's the minimum number of moves\n• necessary to go from a given starting configuration\n• to the final end configuration, where all of the faces-- each\n• of the faces has uniform color?\n• And that can be posed as a graph problem.\n• We'll probably do that one this term.\n• In previous terms we've done other things\n• like the 15 puzzle.\n• And so some of these are tentative.\n• We definitely know what the first problem set is like,\n• but the rest of them are, at this moment, tentative.\n• And to finish up shortest paths.\n• Again in terms past we've asked you\n• to write code using a particular algorithm that\n• finds the shortest path from Caltech to MIT.\n• This time we may do things a little bit differently.\n• We were thinking maybe we'll give you a street map of Boston\n• and go figure out if Paul Revere used\n• the shortest path to get to where he was going,\n• or things like that.\n• We'll try and make it fun.\n• Dynamic programming is an important algorithm design\n• technique that's used in many, many problems.\n• And it can be used to do a variety of things, including\n• image compression.\n• How do you compress an image so the number of pixels\n• reduces, but it still looks like the image\n• that you started out with, that had many more pixels?\n• All right?\n• So you could use dynamic programming for that.\n• And finally, advanced topics, complexity theory, research\n• and algorithms.\n• Hopefully by now-- by this time in the course,\n• you have been sold on algorithms.\n• And most, if not all of you, would\n• want to pursue a carrier in algorithms.\n• And we'll give you a sense of what else is there.\n• We're just scratching the surface in this class,\n• and there's many, many classes that you can possibly\n• take if you want to continue in-- to learn about algorithms,\n• or to pursue a career in algorithms.\n• All right?\n• So that's the story of the class,\n• or the synopsis of the class.\n• And I encourage you to go spend a few minutes on the website.\n• a sense of what is expected of you.\n• What the rules are in terms of doing the problem sets.\n• And the course grading break down,\n• the grading policies are all listed on the website as well.\n• All right.\n• OK.\n• So let's get started.\n• I want to talk about a specific problem.\n• And talk about algorithms for a specific problem.\n• We picked this problem, because it's so easy to understand.\n• And they're fairly straightforward algorithms\n• that are not particularly efficient to solve\n• this problem.\n• And so this is a, kind of, a toy problem.\n• But like a lot of toy problems, it's\n• very evocative in that it points out the issues involved\n• in designing efficient algorithms.\n• version of what we call peak finding.\n• And a peak finder is something in the one dimensional case.\n• Runs on an array of numbers.\n• And I'm just putting--\n• --symbols for each of these numbers here.\n• And the numbers are positive, negative.\n• We'll just assume they're all positive,\n• it doesn't really matter.\n• The algorithms we describe will work.\n• And so we have this one dimensional array\n• that has nine different positions.\n• And a through i are numbers.\n• And we want to find a peak.\n• And so we have to define what we mean by a peak.\n• And so, in particular, as an example,\n• position 2 is a peak if, and only\n• if, b greater than or equal to a, and b greater than or equal\n• to c.\n• So it's really a very local property corresponding\n• to a peak.\n• In the one dimensional case, it's trivial.\n• If you are equal or greater than both of the elements\n• that you see on the left and the right, you're a peak.\n• OK?\n• And in the case of the edges, you only\n• have to look to one side.\n• So position 9 is a peak if i greater than or equal to h.\n• So you just have to look to your left there,\n• because you're all the way on the right hand side.\n• All right?\n• So that's it.\n• And the statement of the problem, the one dimensional\n• version, is find the peak if it exists.\n• All right?\n• That's all there is to it.\n• I'm going to give you a straightforward algorithm.\n• And then we'll see if we can improve it.\n• All right?\n• You can imagine that the straightforward algorithm is\n• something that just, you know, walks across the array.\n• But we need that as a starting point for building something\n• more sophisticated.\n• So let's say we start from left and all\n• we have is one traversal, really.\n• So let's say we have 1, 2, and then we\n• have n over 2 over here corresponding\n• to the middle of this n element array.\n• And then we have n minus 1, and n.\n• What I'm interested in doing is, not only\n• coming up with a straightforward algorithm,\n• but also precisely characterizing\n• what its complexity is in relation\n• to n, which is the number of inputs.\n• Yeah?\n• Question?\n• AUDIENCE: Why do you say if it exists\n• when the criteria in the [INAUDIBLE]\n• guarantees [INAUDIBLE]?\n• PROFESSOR: That's exactly right.\n• I was going to get to that.\n• So if you look at the definition of the peak,\n• then what I have here is greater than or equal to.\n• OK?\n• And so this-- That's a great question that was asked.\n• Why is there \"if it exists\" in this problem?\n• Now in the case where I have greater than or equal to,\n• then-- this is a homework question for you,\n• and for the rest of you-- argue that any array will always\n• have a peak.\n• OK?\n• Now if you didn't have the greater than or equal to,\n• and you had a greater than, then can you make that argument?\n• No, you can't.\n• Right?\n• So great question.\n• In this case it's just a question--\n• You would want to modify this problem\n• statement to find the peak.\n• But if I had a different definition of a peak-- and this\n• is part of algorithmic thinking.\n• You want to be able to create algorithms that are general,\n• so if the problem definition changes on you,\n• you still have a starting point to go attack\n• the second version of the problem.\n• OK?\n• So you could eliminate this in the case\n• of the greater than or equal to definition.\n• The \"if it exists\", because a peak will always exist.\n• But you probably want to argue that when\n• you want to show the correctness of your algorithm.\n• And if in fact you had a different definition,\n• well you would have to create an algorithm that tells you\n• for sure that a peak doesn't exist, or find\n• a peak if it exists.\n• All right?\n• So that's really the general case.\n• Many a time it's possible that you're asked to do something,\n• and you can't actually give an answer to the question,\n• or find something that satisfies all the constraints required.\n• And in that case, you want to be able to put up your hand\n• and say, you know what?\n• I searched long and hard.\n• I searched exhaustively.\n• Here's my argument that I searched exhaustively,\n• and I couldn't find it.\n• Right?\n• If you do that, you get to keep your job.\n• Right?\n• Otherwise there's always the case\n• that you didn't search hard enough.\n• So it's nice to have that argument.\n• All right?\n• Great.\n• Thanks for the question.\n• Feel free to interrupt.\n• Raise your hand, and I'm watching you guys,\n• and I'm happy to answer questions at any time.\n• So let's talk about the straightforward algorithm.\n• The straightforward algorithm is something\n• that starts from the left and just walks across.\n• And you might have something that looks like that.\n• All right?\n• By that-- By this I mean the numbers are increasing\n• as you start from the left, the peak is somewhere\n• in the middle, and then things start decreasing.\n• Right?\n• So in this case, you know, this might be the peak.\n• You also may have a situation where\n• the peak is all the way on the right,\n• you started from the left.\n• And it's 1, 2, 3, 4, 5, 6, literally\n• in terms of the numbers.\n• And you're going to look at n elements going all the way\n• to the right in order to find the peak.\n• So in the case of the middle you'd\n• look at n over 2 elements.\n• If it was right in the middle.\n• And the complexity, worst case complexity--\n• --is what we call theta n.\n• And it's theta n, because in the worst case,\n• you may have to look at all n elements.\n• And that would be the case where you started from the left\n• and you had to go all the way to the right.\n• Now remember theta n is essentially something\n• that's says of the order of n.\n• So it gives you both the lower bound and an upper bound.\n• Big [? O ?] of n is just upper bound.\n• And what we're saying here is, we're\n• saying this algorithm that starts from the left\n• is going to, essentially, require in the worst case\n• something that's a constant times n.\n• OK?\n• And you know that constant could be 1.\n• You could certainly set things up that way.\n• Or if you had a different kind of algorithm,\n• maybe you could work on the constant.\n• But bottom line, we're only concerned, at this moment,\n• And the asymptotic complexity of this algorithm is linear.\n• All right?\n• That make sense?\n• OK.\n• So someone help me do better.\n• How can we do better?\n• How can we lower the asymptotic complexity\n• of a one dimensional peak finder?\n• Anybody want to take a stab at that?\n• Yeah?\n• Back there.\n• AUDIENCE: Do a binary search subset.\n• You look at the middle, and whatever\n• is higher-- whichever side is higher, then cut that in half,\n• because you know there's a peak.\n• PROFESSOR: On--\n• AUDIENCE: For example if you're in the middle\n• on the right side-- there's a higher number\n• on the right side-- then you would just\n• look at that, because you know that your peak's somewhere\n• in there.\n• And you continue cutting in half.\n• PROFESSOR: Excellent!\n• Excellent!\n• That's exactly right.\n• So you can-- You can do something different, which\n• is essentially try and break up this problem.\n• Use a divide and conquer strategy, and recursively break\n• up this one dimensional array into smaller arrays.\n• And try and get this complexity down.\n• Yeah?\n• AUDIENCE: Are we assuming that there's only one peak?\n• PROFESSOR: No, we're not.\n• AUDIENCE: OK.\n• PROFESSOR: It's find a peak if it exists.\n• And in this case it's, \"find a peak\",\n• because of the definition.\n• We don't really need this as it was discussed.\n• All right?\n• OK.\n• So--\n• So that was a great answer, and-- You know this class\n• after while is going to get boring.\n• Right?\n• Every class gets boring.\n• So we, you know, try and break the monotony here a bit.\n• And so-- And then the other thing that we realized\n• was that these seats you're sitting on-- this\n• is a nice classroom-- but the seats you're sitting on\n• are kind of hard.\n• Right?\n• So what Eric and I did was we decided\n• who are-- who are interacting with us.\n• And we have these--\n• [LAUGHTER]\n• --cushions that are 6.006 cushions.\n• And, you know, that's a 2 by 2 by 2 Rubik's cube here.\n• And since you answered the first question, you get a cushion.\n• This is kind of like a Frisbee, but not really.\n• So--\n• [LAUGHTER]\n• I'm not sure-- I'm not sure I'm going to get it to you.\n• But the other thing I want to say\n• is this is not a baseball game.\n• Right?\n• Where you just grab the ball as it comes by.\n• This is meant for him, my friend in the red shirt.\n• So here you go.\n• All right.\n• It is soft.\n• So, you know, it won't-- it won't hurt you if hits you.\n• [LAUGHTER]\n• All right.\n• So we got a bunch of these.\n• And raise your hands, you know, going\n• to ask-- There's going to be-- I think-- There's\n• some trivial questions that we're going to ask just\n• to make sure you're awake.\n• So an answer to that doesn't get you a cushion.\n• AUDIENCE: Chase.\n• PROFESSOR: Chase.\n• An answer like Chase just gave is--\n• that's a good answer to a nontrivial question.\n• That gets you a cushion.\n• OK?\n• All right, great.\n• So let's put up by Chase's algorithm up here.\n• I'm going to write it out for the 1D version.\n• So what we have here is a recursive algorithm.\n• is this picture that I put up there.\n• And this is a divide and conquer algorithm.\n• You're going to see this over and over-- this paradigm--\n• over and over in 6.006.\n• We're going to look at the n over 2 position.\n• And we're going to look to the left,\n• and we're going to look to the right.\n• And we're going to do that in sequence.\n• So--\n• --if a n over 2 is less than a n over 2 minus 1, then--\n• --only look at the left half.\n• 1 through n over 2 minus 1 to look for peak-- for a peak.\n• All right?\n• So that's step one.\n• And you know I could put it on the right hand\n• side or the left hand side, doesn't really matter.\n• I chose to do the left hand side first, the left half.\n• And so what I've done is, through that one step,\n• if in fact you have that condition-- a n over 2\n• is less than a n over 2 minus 1-- then you move to your left\n• and you work on one half of the problem.\n• But if that's not the case, then if n over-- n over 2\n• is less than a over n over-- n by 2 plus 1,\n• then only look at n over 2 plus 1 through n for a peak.\n• So I haven't bothered writing out all the words.\n• They're exactly the same as the left hand side.\n• You just look to the right hand side.\n• Otherwise if both of these conditions don't fire,\n• you're actually done.\n• OK?\n• That's actually the best case in terms of finishing early,\n• at least in this recursive step.\n• Because now the n over 2 position is a peak.\n• Because what you found is that the n over 2 position\n• is greater than or equal to both of its adjacent positions,\n• and that's exactly the definition of a peak.\n• So you're done.\n• OK?\n• So all of this is good.\n• You want to write an argument that this algorithm is correct.\n• And I'm not going to bother with that.\n• I just wave my hands a bit, and you all nodded,\n• so we're done with that.\n• But the point being you will see in your problem set\n• a precise argument for a more complicated algorithm, the 2D\n• version of this.\n• And that should be a template for you to go write a proof,\n• or an argument, a formal argument,\n• that a particular algorithm is correct.\n• That it does what it claims to do.\n• And in this case it's two, three lines of careful reasoning\n• that essentially say, given the definition of the peak,\n• that this is going to find a peak in the array\n• that you're given.\n• All right?\n• So we all believe that this algorithm is correct.\n• Let's talk now about the complexity of this algorithm.\n• Because the whole point of this algorithm\n• was because we didn't like this theta\n• n complexity corresponding to the straightforward algorithm.\n• So it'd like to do better.\n• So what I'd like to do is ask one of you\n• to give me a recurrence relation of the kind, you know, T of n\n• equals blah, blah, blah.\n• That would correspond to this recursive algorithm,\n• this divide and conquer algorithm.\n• And then using that, I'd like to get to the actual complexity\n• in terms of what the theta of complexity corresponds to.\n• Yeah?\n• Back there?\n• AUDIENCE: So the worst case scenario if T of n\n• is going to be some constant amount of time--\n• PROFESSOR: Yep.\n• AUDIENCE: --it takes to investigate whether a certain\n• element is [INAUDIBLE], plus--\n• [COUGH]\n• --T of n over 2?\n• PROFESSOR: Great.\n• Exactly right.\n• That's exactly right.\n• So if you look at this algorithm and you say,\n• from a computation standpoint, can I\n• write an equation corresponding to the execution\n• of this algorithm?\n• And you say, T of n is the work that this algorithm does on--\n• as input of size n.\n• OK?\n• Then I can write this equation.\n• And this theta 1 corresponds to the two comparisons\n• that you do looking at-- potentially the two comparisons\n• that you do-- looking at the left hand\n• side and the right hand side.\n• So that's-- 2 is a constant, so that's why we put theta 1.\n• All right?\n• So you get a cushion, too.\n• Watch out guys.\n• Whoa!\n• Oh actually that wasn't so bad.\n• Good.\n• Veers left, Eric.\n• Veers left.\n• So if you take this and you start expanding it,\n• eventually you're going to get to the base\n• case, which is T of 1 is theta 1.\n• Right?\n• Because you have a one element array you just for that array\n• it's just going to return that as a peak.\n• And so if you do that, and you expand it all the way out,\n• then you can write T of n equals theta 1 plus theta 1.\n• And you're going to do this log to the base 2 of n times.\n• And adding these all up, gives you\n• a complexity theta log 2 of n.\n• Right?\n• So now you compare this with that.\n• And there's really a huge difference.\n• There's an exponential difference.\n• If you coded up this algorithm in Python--\n• and I did-- both these algorithms for the 1D version--\n• and if you run it on n being 10 million or so,\n• then this algorithm takes 13 seconds.\n• OK?\n• The-- The theta 10 algorithm takes 13 seconds.\n• And this one takes 0.001 seconds.\n• OK?\n• Huge difference.\n• So there is a big difference between theta n and theta log\n• n.\n• It's literally the difference between 2 raised to n, and n.\n• It makes sense to try and reduce complexity\n• as you can see, especially if you're\n• All right?\n• And you'll see that more clearly as we\n• go to a 2D version of this problem.\n• All right?\n• So you can't really do better for the 1D.\n• The 1D is a straightforward problem.\n• It gets a little more interesting--\n• the problems get a little-- excuse me,\n• the algorithms get a little more sophisticated\n• when we look at a 2D version of peak finding.\n• So let's talk about the 2D version.\n• So as you can imagine in the 2D version\n• you have a matrix, or a two dimensional array.\n• And we'll say this thing has n rows and m columns.\n• And now we have to define what a peak is.\n• And it's a hill.\n• It's the obvious definition of a peak.\n• So if you had a in here, c, b, d, e.\n• Then as you can guess, a is a 2D peak if, and only if,\n• a greater than or equal to b; a greater than or equal to d, c\n• and e.\n• All right?\n• So it's a little hill up there.\n• All right?\n• And again I've used the greater than or equal to here,\n• so that's similar to the 1D in the case\n• that you'll always find a peak in any 2D matrix.\n• Now again I'll give you the straightforward algorithm,\n• and we'll call it the Greedy Ascent algorithm.\n• And the Greedy Ascent algorithm essentially picks a direction\n• and, you know, tries to follow that direction in order\n• to find a peak.\n• So for example, if I had this particular--\n• --matrix; 14, 13, 12, 15, 9, 11, 17--\n• Then what might happen is if I started at some arbitrary\n• midpoint-- So the Greedy Ascent algorithm\n• has to make choices as to where to start.\n• Just like we had different cases here,\n• you have to make a choice as to where to start.\n• You might want to start in the middle,\n• and you might want to work your way left first.\n• Or you're going to all-- You just keep going left,\n• our keep going right.\n• And if you hit an edge, you go down.\n• So you make some choices as to what the default traversal\n• directions are.\n• And so if you say you want to start with 12,\n• you are going to go look for something to left.\n• And if it's greater than, you're going to follow that direction.\n• If it's not, if it's less, then you're\n• going to go in the other direction, in this case,\n• for example.\n• So in this case you'll go to 12, 13 , 14, 15, 16, 17, 19,\n• and 20.\n• And you'd find-- You 'd find this peak.\n• Now I haven't given you the specific details\n• of a Greedy Ascent algorithm.\n• But I think if you look at the worst case possibilities\n• here, with respect to a given matrix,\n• and for any given starting point,\n• and for any given strategy-- in terms of choosing left first,\n• versus right first, or down first versus up first--\n• you will have a situation where-- just\n• like we had in the 1D case-- you may end up\n• touching a large fraction of the elements in this 2D array.\n• OK?\n• So in this case, we ended up, you know,\n• touching a bunch of different elements.\n• And it's quite possible that you could end up touching--\n• starting from the midpoint-- you could up touching half\n• the elements, and in some cases, touching all the elements.\n• So if you do a worst case analysis of this algorithm--\n• a particular algorithm with particular choices in terms\n• of the starting point and the direction of search--\n• a Greedy Ascent algorithm would have theta n m complexity.\n• All right?\n• And in the case where n equals m, or m equals n,\n• you'd have theta n squared complexity.\n• OK?\n• I won't spend very much time on this,\n• because I want to talk to you about the divide\n• and conquer versions of this algorithm for the 2D peak.\n• But hopefully you're all with me with respect\n• to what the worst case complexity is.\n• All right?\n• Yeah.\n• Question back there.\n• AUDIENCE: Can you-- Is that an approximation?\n• Or can you actually get to n times m traversals?\n• PROFESSOR: So there are specific Greedy Ascent algorithms,\n• and specific matrices where, if I give you\n• the code for the algorithm, and I give you a specific matrix,\n• that I could make you touch all of these elements.\n• That's correct.\n• So we're talking about worst case.\n• You're being very paranoid when you\n• talk about worst case complexity.\n• And so I'm-- hand waving a bit here,\n• simply because I haven't given you the specifics\n• of the algorithm yet.\n• Right?\n• This is really a set of algorithms,\n• because I haven't given you the code,\n• I haven't told you where it starts,\n• and which direction it goes.\n• But you go, do that, fix it, and I\n• would be the person who tries to find the worst case complexity.\n• Suddenly it's very easy to get to theta n\n• m in terms of having some constant multiplying n times m.\n• But you can definitely get to that constant\n• being very close to 1.\n• OK?\n• If not 1.\n• All right.\n• So let's talk about divide and conquer.\n• And let's say that I did something\n• like this, where I just tried to jam the binary search\n• algorithm into the 2D version.\n• All right?\n• So what I'm going to do is--\n• --I'm going to pick the middle column, j equals m over 2.\n• And I'm going to find a 1D peak using\n• whatever algorithm I want.\n• And I'll probably end up using the more efficient algorithm,\n• the binary search version that's gone\n• all the way to the left of the board there.\n• And let's say I find a binary peak at (i, j).\n• Because I've picked a column, and I'm just finding a 1D peak.\n• So this is j equals m over 2.\n• That's i.\n• Now I use (i,j).\n• In particular row i as a start--\n• --to find a 1D peak on row i.\n• And I stand up here, I'm really happy.\n• OK?\n• Because I say, wow.\n• I picked a middle column, I found a 1D peak,\n• that is theta m complexity to find a 1D peak as we argued.\n• And one side-- the theta m--\n• AUDIENCE: Log n.\n• PROFESSOR: Oh, I'm sorry.\n• You're right.\n• The log n complexity, that's what this was.\n• So I do have that here.\n• Yeah.\n• Log n complexity.\n• Thanks, Eric.\n• And then once I do that, I can find a 1D peak on row i.\n• In this case row i would be m wide,\n• so it would be log m complexity.\n• If n equals m, then I have a couple of steps of log n,\n• and I'm done.\n• All right?\n• Am I done?\n• No.\n• Can someone tell me why I'm not done?\n• Precisely?\n• Yep.\n• AUDIENCE: Because when you do the second part\n• to find the peak in row i, you might not\n• have that column peak-- There might not\n• be a peak on the column anymore.\n• PROFESSOR: That's exactly correct.\n• So this algorithm is incorrect.\n• OK?\n• It doesn't work.\n• It's efficient, but incorrect.\n• OK?\n• It's-- You want to be correct.\n• You know being correcting and inefficient\n• is definitely better than being inefficient-- I'm sorry.\n• Being incorrect and efficient.\n• So this is an efficient algorithm,\n• in the sense that it will only take log n time,\n• but it doesn't work.\n• And I'll give you a simple example\n• here where it doesn't work.\n• The problem is--\n• --a 2D peak--\n• --may not exist--\n• --on row i.\n• And here's an example of that.\n• Actually this is-- This is exactly the example of that.\n• Let's say that I started with this row.\n• Since it's-- I'm starting with the middle row,\n• Let's say I started with that one.\n• I end up finding a peak.\n• And if this were 10 up here, I'd choose 12 as a peak.\n• And it's quite possible that I return 12 as a peak.\n• Even though 19 is bigger, because 12\n• is a peak given 10 and 11 up here.\n• And then when I choose this particular row,\n• and I find a peak on this row, it would be 14.\n• That is a 1D peak on this row.\n• But 14 is not a 2D peak.\n• OK?\n• So this particular example, 14 would return 14.\n• And 14 is not a 2D peak.\n• All right?\n• You can collect your cushion after the class.\n• So not so good.\n• Look like an efficient algorithm, but doesn't work.\n• All right?\n• So how can we get to something that actually works?\n• So the last algorithm that I'm going to show you--\n• And you'll see four different algorithms in your problem\n• set--\n• --that you'll have to analyze the complexity for and decide\n• if they're efficient, and if they're correct.\n• But here's a-- a recursive version\n• that is better than, in terms of complexity,\n• than the Greedy Ascent algorithm.\n• And this one works.\n• So what I'm going to do is pick a middle column.\n• j equals m over 2 as before.\n• I'm going to find the global maximum on column j.\n• And that's going to be at (i, j).\n• I'm going to compare (i comma j minus 1), (i comma j),\n• and (i,j plus 1).\n• Which means that once I've found the maximum in this row,\n• all I'm going to look to the left and the right,\n• and compare.\n• I'm going to pick the left columns.\n• If (i comma j minus 1) is greater than (i comma j)--\n• and similarly for the right.\n• And if in fact I-- either of these two conditions\n• don't fire, and what I have is (i comma j)\n• is greater than or equal to (i comma j minus 1)\n• and (i comma j plus 1), then I'm done.\n• Just like I had for the 1D version.\n• If (i comma j) is greater than or equal to (i comma\n• j minus 1), and (i comma j plus 1), that implies (i, j)\n• is a 2D peak.\n• OK?\n• And the reason that is the case, is\n• because (i comma j) was the maximum element in that column.\n• So you know that you've compared it\n• to all of the adjacent elements, looking up and looking down,\n• that's the maximum element.\n• Now you've look at the left and the right,\n• and in fact it's greater than or equal to the elements\n• on the left and the right.\n• And so therefore it's a 2D peak.\n• OK?\n• So in this case, when you pick the left or the right columns--\n• you'll pick one of them-- you're going\n• to solve the new problem with half the number of columns.\n• All right?\n• And again, you have to go through an analysis,\n• or an argument, to make sure that this algorithm is correct.\n• But its intuitively correct, simply because it matches\n• the 1D version much more closely.\n• And you also have your condition where you break away right\n• here, where you have a 2D peak, just like the 1D version.\n• And what you've done is break this matrix up\n• into half the size.\n• And that's essentially why this algorithm works.\n• When you have a single column--\n• --find the global maximum and you're done.\n• All right?\n• So that's the base case.\n• So let me end with just writing out\n• what the recurrence relation for the complexity of this\n• is, and argue what the overall complexity of this algorithm\n• is.\n• And then I'll give you the bad news.\n• All right.\n• So overall what you have is, you have something like T of (n, m)\n• equals T of (n, m over 2) plus theta n.\n• Why is that?\n• Well n is the number of rows, m is the number of columns.\n• In one case you'll be breaking things down\n• into half the number of columns, which is m over 2.\n• And in order to find the global maximum,\n• you'll be doing theta n work, because you're\n• finding the global maximum.\n• Right?\n• You just have to scan it-- this--\n• That's the way-- That's what it's going to take.\n• And so if you do that, and you go run it through--\n• and you know that T of (n, 1) is theta n-- which\n• is this last part over here-- that's your base case.\n• You get T of (n, m) is theta of n added to theta of n,\n• log of m times-- log 2 of m times.\n• Which is theta of n-- log 2 of m.\n• All right?\n• So you're not done with peak finding.\n• What you'll see is at four algorithms coded in Python--\n• I'm not going to give away what those algorithms are,\n• but you'll have to recognize them.\n• You will have seen versions of those algorithms\n• And your job is going to be to analyze the algorithms, as I\n• said before, prove that one of them is correct,\n• and find counter-examples for the ones that aren't correct.\n• The course staff will stick around\n• here to answer questions-- logistical questions--\n• And I owe that gentleman a cushion."
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https://www.goskills.com/Excel/Resources/Absolute-reference-Excel | [
"# Absolute References in Excel - A Beginner's Guide",
null,
"Agnes Lo",
null,
"#### Join the Excel conversation on Slack\n\nAsk a question or join the conversation for all things Excel on our Slack channel.\n\n## What is an absolute reference?\n\nAbsolute reference Excel definition: An absolute reference in Excel means there is a fixed point of reference applied to a cell or a formula. This is so the return value will always stay the same no matter where the cell or the formula moves to — within the same sheet or across different sheets.\n\nUse this free Excel file to practice absolute references along with the tutorial.\n\nFor example, in the image below, all employees will receive the same bonus payout amount of \\$1500. The amount of \\$1500 is the constant in this situation, and an absolute reference can be used to help calculate the total payout of salary (a figure that is different for each person) plus bonus payout.",
null,
"In Excel, all references are relative by default. To apply absolute reference in Excel, the addition of a dollar sign (\\$) is required in the Excel formula — i.e., =\\$C\\$1 as shown in the example above.\n\nWithout the dollar sign (\\$) in the formula, Excel naturally interprets the cell address as a relative reference, where the point of reference changes as the relative row and column coordinates move.\n\nThe idea of an absolute cell reference is to hold a specific cell constant, so the value remains the same when being copied to other cells. The absolute reference Excel feature is an indispensable tool to save you time and effort when working with spreadsheets.\n\n## A quick recap",
null,
"## How to make an absolute reference in Excel\n\nHere are the steps on how to make a basic absolute reference in Excel - \\$A\\$1:\n\n1. Choose a cell where you would like to create an absolute reference. Cell A1 in this example:",
null,
"1. In the formula of Cell A1, Enter “=” (the equal sign) and then select the point of reference - Cell C1.",
null,
"1. In the same formula, either manually add two dollar signs (SHIFT + 4) in front of the row and column coordinates, or press the F4 key as a shortcut.\n2. Press ENTER and now Cell A1 has the value of Cell C1 as a fixed point of reference.",
null,
"## Examples of absolute reference\n\nBecause of its versatile nature, absolute reference is one of the most frequently used functions in Excel. It can be used with both numbers and text.\n\nAbsolute numbers are very popular in calculations in Excel, as it keeps a cell constant so that users can make use of the point of reference in various scenarios and forecasts.\n\nLet’s take a look at the example below:\n\nA US-based stationery company has found a new buyer based in Germany. They need to work out a proposal for the new partner. Given that there are differences in terms of currencies, they would like to include the unit cost in EUR for a complete picture of the costs.",
null,
"This means all of those three unit costs in USD have to be multiplied by the exchange rate of EUR-USD — unit cost (USD) * EUR-USD exchange rate.\n\nThere are three unit costs, and one exchange rate. The exchange rate of EUR (Cell B3) has to be held constant when applying the formulas under column F for the unit costs in EUR.\n\nTo begin with, find out the unit cost of Notepad A in EUR:\n\n1. Enter: “=E2*B3”",
null,
"1. Manually insert the \\$ in front of both row and column coordinates or press F4 to create an absolute reference, then press ENTER.",
null,
"",
null,
"1. To find out other unit costs in EUR, drag down Cell F2 to apply the formulas to F3 and F4 as well.",
null,
"",
null,
"With the use of absolute reference, the value of B3 becomes a constant for all unit costs.\n\nIt provides a quick solution to the scenario on hand and minimizes mistakes by doing the multiplication by hand manually for each cell.\n\nIf there was no absolute reference in the formulas, Excel would return \\$0 as the unit cost as we drag down Cell F2. Because the reference point would become relative and change across cells, and there are no other values in Cell B4 and B5.",
null,
"",
null,
"## Absolute reference vs relative reference in Excel\n\nLet's compare the differences between absolute and relative references, with examples of when to use each.\n\n### Absolute reference Excel\n\nThis refers to a fixed point of reference, is a constant, and involves the use of dollar sign \\$ in the formula (i.e., everyone is to receive the same bonus payout, so the amount \\$1500 is the constant in this situation).",
null,
"### Relative reference Excel\n\nThis refers to a relative point of reference, is constantly changing, and dollar sign \\$ is absent in the formula (i.e., when each unit price and qty are difference variables, there’s no constant in the calculation).",
null,
"## A quick hint\n\nPress the F4 key as a shortcut to make an absolute cell reference in Excel.\n\n## Summary\n\nAbsolute reference is applied in situations where the same value is used throughout different scenarios or variables. The point of reference is a constant that remains the same.\n\nThe dollar sign \\$ is employed to create an absolute reference. Without it, it is a relative reference, no cell is ‘locked’.\n\nIt is commonly applied in a mix of formulas and functions in Excel, and works with both numbers and text. It’s particularly useful when doing complicated calculations and formulas. It helps minimize errors and saves time to input formulas by hand, one by one.\n\nTo learn more about absolute references and other essential Excel features try our Basic and Advanced Excel course today. Or start learning some Excel basics with our free Excel in an Hour course.\n\nStart learning formulas, functions, and time-saving hacks today with this free course!",
null,
"#### Join the Excel conversation on Slack\n\nAsk a question or join the conversation for all things Excel on our Slack channel.",
null,
"Agnes Lo\n\nAgnes is a clienteling professional in luxury retail. She enjoys yoga and outdoor activities in her free time."
] | [
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"https://www.goskills.com/blobs/authors/67/image-small.jpg",
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"https://www.goskills.com/static/images/course/slack-excel.png",
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"https://www.goskills.com/blobs/blogs/421/3055574d-10fd-4822-ad5f-31c4c1b93ca9_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/6cd18702-45fb-4026-b9e7-288f4c34f84e_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/c727bfab-ba0e-47a8-9391-50fbc4d988a5_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/1a06fa37-0996-4a22-bc39-0023a9eb3a18_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/ffce4cb4-4aad-4ba4-b87c-8cd8eeacb2fd_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/8b3ebd34-7fe0-4c3e-8f66-aa453db4f185_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/52fc7894-1bb3-4ea9-9780-7c2b03f01b32_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/241f4d58-b45e-4f43-bff9-7bbba88e309b_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/c1959f9b-a379-4e48-98be-a693683d1ec5_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/5ddf30ae-d2f5-4222-89f9-e8ba138d0aa7_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/6ef60d3b-086c-4683-a3d4-5eaa5733131d_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/8b91fa4d-d340-4442-af33-f6b935b8167f_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/c163b59c-a15c-4fea-81f2-867190abd90b_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/3055574d-10fd-4822-ad5f-31c4c1b93ca9_lossy.webp",
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"https://www.goskills.com/blobs/blogs/421/1ea0a30a-fef8-4691-8d8f-27defb09ec45_lossy.webp",
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"https://www.goskills.com/static/images/course/slack-excel.png",
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"https://www.goskills.com/blobs/authors/67/image-large.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89413124,"math_prob":0.8393483,"size":5365,"snap":"2022-40-2023-06","text_gpt3_token_len":1125,"char_repetition_ratio":0.17907107,"word_repetition_ratio":0.014690451,"special_character_ratio":0.2137931,"punctuation_ratio":0.08916587,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9798204,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,9,null,null,null,4,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,2,null,null,null,9,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-27T08:37:15Z\",\"WARC-Record-ID\":\"<urn:uuid:412fa8fb-b1f3-4a21-a501-57870951eed7>\",\"Content-Length\":\"123268\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3811ed8-c450-445d-9aaf-607fe09d3301>\",\"WARC-Concurrent-To\":\"<urn:uuid:df194180-571f-4a49-8ec1-676f2db27dd0>\",\"WARC-IP-Address\":\"13.107.229.23\",\"WARC-Target-URI\":\"https://www.goskills.com/Excel/Resources/Absolute-reference-Excel\",\"WARC-Payload-Digest\":\"sha1:FJ24FZ5TP7OYHEBAKLHNYGWHEW6AWMHE\",\"WARC-Block-Digest\":\"sha1:ZGJ3PJDODOMY7NZ7PCGSIECAISTUGDKX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764494974.98_warc_CC-MAIN-20230127065356-20230127095356-00316.warc.gz\"}"} |
https://answers.everydaycalculation.com/simplify-fraction/1458-1225 | [
"Solutions by everydaycalculation.com\n\n## Reduce 1458/1225 to lowest terms\n\n1458/1225 is already in the simplest form. It can be written as 1.190204 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 1458 and 1225 is 1\n2. Divide both the numerator and denominator by the GCD\n1458 ÷ 1/1225 ÷ 1\n3. Reduced fraction: 1458/1225\nTherefore, 1458/1225 simplified to lowest terms is 1458/1225.\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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https://chemistry.stackexchange.com/questions/129308/how-do-i-denote-a-measurement-with-a-margin-of-error-using-%C2%B1/129317 | [
"# How do I denote a measurement with a margin of error using ±?\n\nI am trying to express a quantity that is measured to be $$\\pu{0.75 mL}$$, but could be anywhere between $$\\pu{0.7 mL}$$ and $$\\pu{0.8 mL}$$. Is this the correct way to do it?\n\n$$\\pu{0.75 mL} \\pm \\pu{0.05 mL}$$\n\n• $\\pu{0.75 \\pm 0.05 mL}$ typically means a standard deviation of $\\pu{0.05 mL}$, not a hard limit of $\\pu{\\pm 0.05 mL}$. – MaxW Mar 20 at 18:03\n• @MaxW is right on the money: the notation most frequently means sample mean +/- one sample standard deviation, computed from N independent measurements. So degrees of freedom is N-1 and this information should get conveyed. I could add a lot to my answer, but the downvotes on the question mean it probably is not worth the effort. One more thing: the +/- term can reasonably convey hard limits in some cases, e.g., histogram bin center +/- ends of the histogram bin. Other possibilities are +/- half a confidence interval or a standard error. Context is everything. – Ed V Mar 20 at 20:20\n• I'd say it should be written what the interval means, everything else is in practice too ambiguous: I regularly meet situations where it's not clear whether it is the standard deviation across measurements or the standar error of the mean. Not to mention the rather official recommendation in @EdV's answer to report extended uncertainty. – cbeleites unhappy with SX Mar 20 at 21:35\n• @cbeleitesunhappywithSX I fully agree with you and I was going to reference this paper (D. Coleman, J. Auses, N. Grams, “Regulation – From an industry perspective or Relationships between detection limits, quantitation limits, and significant digits”, Chemom. Intell. Lab. Syst. 37 (1997) 71-80.), where Coleman et al. cogently criticize the inadequacy of the typical x ± y notation. Sadly, we are still having to deal with these needlessly ambiguous matters. – Ed V Mar 20 at 21:49\n\nI strongly recommend looking at this: QUAM:2012.P1 (EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012., 141 pages.) You can easily find it online, for free, and it is all about quantifying uncertainties in measurements, though with an obvious focus on analytical chemistry-related measurements. As an example, the figure below shows the QUAM recommended way to report standard and expanded uncertainties:\n\n9.3. Reporting standard uncertainty\n\n9.3.1. When uncertainty is expressed as the combined standard uncertainty $$u_c$$ (that is, as a single standard deviation), the following form is recommended:\n\n\"(Result): $$x$$ (units) [with a] standard uncertainty of $$u_c$$ (units) [where standard uncertainty is as defined in the ISO/IEC Guide to the Expression of Uncertainty in Measurement and corresponds to one standard deviation.]\"\n\nNOTE: The use of the symbol $$\\pm$$ is not recommended when using standard uncertainty as the symbol is commonly associated with intervals corresponding to high levels of confidence.\n\nTerms in parentheses [] may be omitted or abbreviated as appropriate.\n\nEXAMPLE:\n\n• Total nitrogen: $$\\pu{3.52 g}/\\pu{100 g}$$\n\n• Standard uncertainty: $$\\pu{0.07 g}/\\pu{100 g}$$*\n\n• *Standard uncertainty corresponds to one standard deviation.\n\n9.4. Reporting expanded uncertainty\n\n9.4.1. Unless otherwise required, the result x should be stated together with the expanded uncertainty $$U$$ calculated using a coverage factor $$k=2$$ (or as described in section 8.3.3.). The following form is recommended:\n\n\"(Result): $$(x \\pm U)$$ (units)\n\n[where] the reported uncertainty is [an expanded uncertainty as defined in the International Vocabulary of Basic and General terms in metrology, 2nd ed., ISO 1993,] calculated using a coverage factor of 2, [which gives a level of confidence of approximately 95 %]\"\n\nTerms in parentheses [] may be omitted or abbreviated as appropriate. The coverage factor should, of course, be adjusted to show the value actually used.\n\nEXAMPLE:\n\n• Total nitrogen: $$(3.52 \\pm 0.14)~\\mathrm{g}/100~\\mathrm{g}$$*\n• *The reported uncertainty is an expanded uncertainty calculated using a coverage factor of 2 which gives a level of confidence of approximately 95 %.\n\nThis is from pages 30-31 of QUAM, referenced above. CITAC is the acronym for Co-Operation on International Traceability In Analytical Chemistry.\n\nAll that said, your notation is quite common, so no worries.\n\n• @orthocresol Many thanks for the edit! – Ed V Apr 7 at 11:37\n• Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! – Ed V May 26 at 1:36"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8841597,"math_prob":0.94934845,"size":3020,"snap":"2020-24-2020-29","text_gpt3_token_len":744,"char_repetition_ratio":0.1372679,"word_repetition_ratio":0.20305677,"special_character_ratio":0.26324505,"punctuation_ratio":0.14864865,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97009015,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-31T17:40:32Z\",\"WARC-Record-ID\":\"<urn:uuid:085e45f9-ae7b-4e60-bed5-c497dab95e7d>\",\"Content-Length\":\"153161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c0f4944e-0d74-4d2f-8a9e-3d7d6fa180b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:3fb00429-3976-43db-94de-5be1937abc01>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/129308/how-do-i-denote-a-measurement-with-a-margin-of-error-using-%C2%B1/129317\",\"WARC-Payload-Digest\":\"sha1:QIC2MPJCUFPLGAXLTQXTURCOWKRAGGDA\",\"WARC-Block-Digest\":\"sha1:BY25GHB65GOV7H4C3XC5DGJP5KS25A4K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347413551.52_warc_CC-MAIN-20200531151414-20200531181414-00261.warc.gz\"}"} |
https://www.aanda.org/articles/aa/full_html/2016/02/aa27197-15/aa27197-15.html | [
"Subscriber Authentication Point\nFree Access\n Issue A&A Volume 586, February 2016 A86 8 Astrophysical processes https://doi.org/10.1051/0004-6361/201527197 28 January 2016\n\n© ESO, 2016\n\n## 1. Introduction\n\nThe era of gravitational wave (GW) astronomy is almost upon us with the arrival of second generation terrestrial laser interferometers such as Advanced LIGO (aLIGO) and Advanced Virgo (aVirgo), with the former in operation as of 2015. Among the potentially detectable sources of GWs, f-mode unstable neutron stars (NS) are promising detection targets due to the copious amounts of GWs emitted, with an energy of up to 2.5% of a solar mass emitted in GWs as the initially rapidly rotating NS spins down.\n\nNon-radial, non-axisymmetric oscillations, or modes, can be excited in proto-NSs after their formation in a core-collapse supernova explosion (CCSNe). These modes may become unstable due to the GW-driven secular Chandrasekhar-Friedman-Schutz (CFS) instability (Chandrasekhar 1970; Friedman & Schutz 1978), and hence grow in amplitude to produce detectable GWs. The CFS instability occurs when a star is rotating sufficiently rapidly that a retrograde mode, in the corotating frame, is dragged along with the star’s rotation and appears prograde in the inertial frame. The mode angular momentum becomes increasingly negative, and hence it increases in amplitude, due to the emission of GWs which drives the mode unstable. Once the mode amplitude has saturated the star spins down, losing rotational energy via GW emission.\n\nAnother possible source of CFS-unstable modes is supramassive NSs (i.e. exceeding the maximum mass of the non-rotating star) formed through the merger of a binary NS system. Such stars have been proposed recently as an explanation for the X-ray afterglow plateau observed in some short gamma-ray bursts (GRB; Rowlinson et al. 2013; Hotokezaka et al. 2013; Lasky et al. 2014; Dall’Osso et al. 2015) and have lifetimes that can reach up to 4 × 104 s (Ravi & Lasky 2014). Provided that the mass of these objects is below a certain threshold so that they do not promptly collapse to a black hole (Hotokezaka et al. 2013), it has been shown recently that the CFS instability will grow quickly in them within ~10100 s (Doneva et al. 2015).\n\nAmong the modes which can be driven unstable, the f-modes (fundamental pressure modes) and the r-modes (inertial modes) are particularly interesting due to their relatively short growth time scale and their efficient emission of GWs. A recent estimate for a relativistic NS has shown that the GWs from f-mode unstable supernova-derived NSs could be detected by aLIGO/aVirgo up to a distance corresponding to that of the Virgo Cluster (~15 Mpc) for reasonably high saturation amplitude values (Passamonti et al. 2013), whereas the GWs from supramassive merger-derived NSs could be detected by aLIGO/aVirgo up to a distance of 20 Mpc or by the proposed third generation detector Einstein Telescope (ET) up to 200 Mpc (Doneva et al. 2015). The detection of f-modes from individual stars could be used to perform GW asteroseismology, enabling the study of NS interiors (Andersson & Kokkotas 1996, 1998; Kokkotas et al. 2001; Gaertig & Kokkotas 2011).\n\nThe superposition of unresolved and uncorrelated GW signals from many sources throughout the Universe results in a stochastic background of gravitational waves (SGWB). Many contributions to the astrophysical SGWB have been discussed in the literature (see e.g. Regimbau 2011, for a review), including the background due to r-mode instabilities in NSs. In this work, following a procedure first presented in Owen et al. (1998) and then in Ferrari et al. (1999), we provide the first estimate of the SGWB due to the f-mode instability in hot, young, rapidly rotating NSs formed from either a CCSNe or the merger of binary NSs, using a relativistic NS model, a realistic equation of state (EoS), and observationally-derived fits to the evolution of the cosmic star formation rate (CSFR).\n\nThe detection of an astrophysical SGWB can improve our understanding of the sources and help constrain their properties, for example neutron star ellipticities, or the average chirp mass and coalescence rate of compact binaries (Talukder et al. 2014; Mandic et al. 2012). Moreover, the astrophysical SGWB will likely mask a primordial SGWB in the frequency band of terrestrial GW detectors produced during the early Universe, which if detected would shed light on the Universe shortly after the Big Bang (Abbott et al. 2007; Maggiore 2000). Therefore, a proper understanding of the astrophysical SGWB is required in order to identify the primordial one.\n\nThis paper is organised as follows: in Sect. 2 the NS formation rate from supernovae is calculated using the fits to the CSFR; in Sect. 3 the NS formation rate this time from binary mergers is calculated; in Sect. 4 the evolution of the f-mode instability is described; in Sect. 5 the spectral properties of the stochastic background are investigated; in Sect. 4 the results are presented, namely the integrated flux and dimensionless energy density of the background; in Sect. 6 the signal-to-noise ratio of the background using pairs of GW detectors is calculated; finally, the conclusions are drawn in Sect. 7.\n\nNote that throughout this paper the so-called 737 ΛCDM cosmology is assumed with Hubble constant H0 = 100 h0 km s-1 Mpc-1 where h0 = 0.7, and density parameters Ωm = 0.3, ΩΛ = 0.7.\n\n## 2. The rate of NS formation from core-collapse supernova explosions\n\nThe rate of NS formation throughout the Universe depends upon the rate of star formation, since massive stars which end their lives in a CCSNe are the progenitors of NSs. The NS formation rate determines the GW event rate, and hence the spectral properties of the SGWB.\n\nFor the case of supernova-derived NSs, the number of NSs formed per unit time within the comoving volume out to redshift z is given by (Ferrari et al. 1999)",
null,
"(1)Here,",
null,
"is the CSFR (in M yr-1 Mpc-3), i.e. the mass of gas that is converted into stars per unit time in the observer frame per unit comoving volume, Φ(m) is the stellar initial mass function (IMF; in",
null,
"), i.e. the initial mass distribution of stars at the time of their birth, and dV/ dz is the comoving volume element. Following Ferrari et al. (1999), we assume that stars with masses between 8 and 25 M give rise to NSs, and we adopt a standard Salpeter IMF of the form Φ(m) ∝ m− (1 + x) with x = 1.35, normalised through the relation",
null,
"(2)The comoving volume element is given by",
null,
"(3)where c is the speed of light,",
null,
", and r(z) is the comoving distance, given by",
null,
"(4)assuming spatial flatness, i.e. Ωm + ΩΛ = 1.\n\nSince the CSFR is not a directly observable quantity, it is usually inferred from observations of the rest-frame ultraviolet (UV) light, as it is mainly radiated by short-lived massive stars. The UV galaxy luminosity density is studied by space and ground-based telescopes, and is then converted into the CSFR density through the adoption of a universal stellar IMF to calculate the conversion factor. Other authors have made use of observations of the GRB rate at high redshift to infer the CSFR. Note that the use of a standard Salpeter IMF does not introduce considerable errors because the evaluation of the NS formation rate based on the CSFR and an assumed universal IMF is largely independent of the specific IMF adopted (Madau 1998).\n\nAs there exist many parameterized fits to the expected evolution of the CSFR with redshift in the literature, we adopt four recent ones to account for the uncertainties: HB06 (Hopkins & Beacom 2006), F07 (Fardal et al. 2007), W08 (Wilkins et al. 2008), based on UV observations; and RE12 (Robertson & Ellis 2012), based on the observed GRB rate. The CSFR fits, plotted in Fig. 1, rise rapidly from their local values to peak between z ~ 1.5−2.5 for the UV derived fits, and much later at z ~ 4 for the GRB derived fit, falling again at higher redshifts. Up to z ~ 0.5 all three UV derived fits are in close agreement, while at increasing redshift the fits diverge. The cutoff for each curve corresponds to the maximum redshift of each CSFR fit: z = 6 for the UV derived fits, and z = 15 for the GRB derived fit.\n\nThe uncertainty on the CSFR at high redshift does not present a serious problem because the contributions to the SGWB come mainly from low redshift sources. The energy flux emitted by a single source decreases as the inverse square of the luminosity distance, so high redshift sources provide a negligible contribution to the background.",
null,
"Fig. 1Evolution of the cosmic star formation rate density with redshift for four different CSFR fits (see Sect. 2.) Open with DEXTER\n\nThe NS formation rate defined in Eq. (1) is plotted in Fig. 2. The HB06 fit gives the highest RNS as expected since it provides the highest CSFR up to z ~ 4, where the majority of star formation takes place. At high redshift, RNS plateaus in accordance with the decreasing CSFR. Note that the RE12 CSFR fit gives almost the same RNS at low redshifts up to z ~ 0.5, although its behaviour differs significantly from the UV-derived fits (even at low redshifts).",
null,
"Fig. 2Evolution of the number of NSs formed per unit time within the comoving volume out to redshift z based on four different CSFR fits (see Sect. 2), for NSs formed from a CCSNe. Open with DEXTER\n\n## 3. The rate of NS formation from the merger of NS binaries\n\nFor NSs formed through the merger of binary NS systems, we assume that the merger rate tracks the star formation rate with some delay td from formation of the binary to final merger. Then the observed rate of binary coalescence (mergers) per unit volume at redshift z is given by (Regimbau & Hughes 2009)",
null,
"(5)where",
null,
"is the rate in the local Universe, and",
null,
"is normalized to reproduce the local rate for z = 0. The local merger rate",
null,
"is usually extrapolated by multiplying the rate in the Milky Way with the density of Milky Way galaxies. Following Regimbau & Hughes (2009), we take",
null,
"as its most probable value.\n\nThe quantity",
null,
"relates the past star formation rate to the rate of binary merger and is defined as",
null,
"(6)where",
null,
"is the CSFR introduced above. Here, z describes the redshift when a compact binary merges, and zf is the redshift at which its progenitor binary formed. The time delay td connects these two redshifts, and represents the total time from initial binary formation to evolution into a compact binary, plus the merging time τm. The quantity td is also the difference in lookback times between zf and z,",
null,
"(7)with E(z) defined above. In Eq. (6), P(td) is the probability distribution for the delay time td, and the factor 1/(1 + zf) accounts for time dilation due to the cosmic expansion. Population synthesis (see references in Regimbau & Hughes 2009) suggests that P(td) takes the form",
null,
"(8)for some minimal delay time τ0. Again, following Regimbau & Hughes (2009), we assume τ0 ~ 20 Myr, which corresponds roughly to the time it takes for massive binaries to evolve into two NSs.\n\nThe rate of binary NS mergers within the comoving volume out to redshift z is given by",
null,
"(9)and is plotted in Fig. 3 for the four CSFR fits described above. Here, it is apparent that NSs formed from binary mergers are far less numerous throughout the Universe than supernova-derived NSs, since RBM is roughly an order of magnitude lower than RNS. The curves are in agreement up to z ~ 1, a higher z than in Fig. 2, and now the RE12 fit provides the highest RBM for z> 3, even though it gives the highest CSFR only for z> 4. This is due to the normalisation of",
null,
", so that the local value at z = 0 is the same for all the CSFR fits, which dampens the effect of a high CSFR at low redshifts.",
null,
"Fig. 3Evolution of the number of NS binary mergers per unit time within the comoving volume out to redshift z based on four different CSFR fits (see Sect. 2). Open with DEXTER\n\n## 4. The evolution of the f-mode instability\n\nFor f-mode unstable NSs, the evolution of the instability proceeds as follows. Initially, the NS is very hot (~1011 K) and rotates very rapidly at close to its Kepler limit, i.e. the angular velocity Ω at which mass shedding at the equator occurs, making the star unstable. Note that although the distribution of initial rotation periods for supernova-derived NSs is uncertain, as a conservative guess we assume that 10% of the population rotates initially at close to the Kepler limit, with the remainder rotating too slowly to become f-mode unstable. For NSs formed from binary mergers, all of the stars are expected to initially rotate at the Kepler limit. We also assume that both NS classes rotate uniformly, since differential rotation is expected to disappear shortly after the star’s formation (Doneva et al. 2015, and references therein).\n\nThe amplitude of the f-mode is initially negligible but grows exponentially due to the CFS instability. Meanwhile, the NS cools due to neutrino emission. Eventually, the amplitude of the mode saturates due to nonlinear coupling with other modes in the star which drain the energy of the f-mode, after which it remains roughly constant. The star then spins down via GW emission at approximately constant temperature (T ~ 109 K) as the heat generated by shear viscosity balances the neutrino cooling. The star stops emitting GWs when it is no longer spinning sufficiently quickly to maintain the CFS instability, and the f-mode is rapidly damped.\n\nNote that the f-mode saturation amplitude determines how fast the NS spins down, and thus whether the associated GW emission will be detectable in terms of single events or a continuous stochastic background. Furthermore, for NSs with a strong dipole magnetic field component at the surface (Bp ~ 1012−1014 G), a significant fraction of the star’s rotational energy could be lost due to magnetic braking rather than via GWs, shortening the total evolution time and reducing the strength of the GW signal.\n\nAt temperatures above or below ~109 K, bulk and shear viscosity respectively suppress the growth of the f-mode. Therefore, there exists a range of star temperatures and rotation rates for which the f-mode is CFS unstable and overcomes dissipative processes. This is usually represented as a curve in the (T,Ω) plane above which the growth timescale due to GWs is shorter than the dissipative timescales due to bulk and shear viscosity, and is known as the instability window.\n\nThe instability windows used in this work were extracted from Doneva et al. (2013) and Doneva et al. (2015) for supernova and merger-derived NSs res/pectively, which have been calculated for the realistic WFF2 EoS (denoted as UV14+UVII in Wiringa et al. 1988) in the Cowling approximation, where the spacetime metric is kept fixed. Since the evolution of the f-mode instability, i.e. the star’s trajectory through the window, is unavailable for supernova-derived stars with this EoS, we consider the evolution calculated by Passamonti et al. (2013) for a relativistic NS model and a polytropic EoS with polytropic index N = 0.62, which closely resembles the WFF2 EoS. In this case, the star exits the parabola-shaped window close to its minimum point, so we assume that the maximum change in Ω from the Kepler frequency to the minimum point determines the total energy lost to GWs.\n\nThe instability windows for supernova-derived NSs reach down to roughly 96% and 80% of the Kepler limit for the l = m = 2 and l = m = 3,4f-modes respectively (Doneva et al. 2013). For merger-derived NSs only the l = m = 2,3f-mode windows have been calculated. These post-merger supramassive NSs lose 9% of their initial rotational energy before collapsing to a black hole for both modes, assuming the star’s mass is below a certain threshold so that it doesn’t immediately collapse to one (Doneva et al. 2015).\n\nNote that abandoning the Cowling approximation and including a dynamical spacetime is expected to lead to more energy lost via GWs and a stronger SGWB, since the star is secularly unstable down to lower rotation rates in full general relativity (Zink et al. 2010).\n\n## 5. Spectral properties of the stochastic background\n\nThe spectral properties of a stochastic background of GWs can be characterised by the dimensionless energy density parameter Ωgw, defined as",
null,
"(10)where ρgw is the GW energy density, νo is the GW frequency in the observer frame, and ρc = 3H2/ 8πG is the critical energy density required to make the Universe flat today.\n\nFor a SGWB of astrophysical origin, the dimensionless energy density can be expressed in terms of the integrated flux received on Earth Fνo at the observed frequency (in erg cm-2 Hz-1 s-1) as",
null,
"(11)where Fνo is defined as",
null,
"(12)Here, fνo is the energy flux per unit frequency (fluence) emitted by a single source located at redshift z (in erg cm-2 Hz-1), which is given by",
null,
"(13)The luminosity distance dL is given by dL(z) = (1 + z)r(z). The GW energy spectrum dEgw/ dν (in erg Hz-1) must be redshifted to the observer frame where ν is the emitted GW frequency, which is related to the observed frequency by ν = νo(1 + z). The second factor in Eq. (12), dR(z)/dz, gives the number of sources per unit time per unit redshift in the observer frame, and is obtained from Eq. (1) or (9) for supernova or merger-derived NSs respectively.\n\nIf we assume that the rotational energy lost while the star is inside the instability window is entirely due to GW emission then, following Ferrari et al. (1999), we can approximate the GW energy spectrum as",
null,
"(14)where Elost is the change in the star’s rotational energy from its initial value at the Kepler frequency to the frequency at which it exits the instability window, and νmax is the maximum emitted frequency of GWs in the source frame. Equation (14) can be modified with an appropriate factor if a fraction of the energy is lost due to magnetic braking instead.\n\nSubstituting the relevant expressions into Eq. (11) for both NS classes, we are left with an integral over z to perform that provides Ωgw(νo). The z limits depend on both the emission frequency range in the source frame, and the redshift range of the CSFR fit, where the upper z limit is given by",
null,
"(15)and the lower z limit is given by",
null,
"(16)Here zmax and zmin are the maximal and minimal redshifts of the CSFR fits, respectively. Thus, the spectral shape of the SGWB is characterized by a cutoff at the maximal GW emission frequency, and a maximum at a frequency which depends on the shape of both the CSFR and the energy spectrum.\n\nThe data required to evaluate Eqs. (11) and (12), including the emitted GW frequencies and the rotational energy of the star at the maximum and minimum angular velocities for which the instability operates, was extracted from Doneva et al. (2013, 2015).\n\n## 6. Results\n\n### 6.1. Integrated flux\n\nThe integrated flux of the SGWB for supernova and merger-derived NSs is plotted in Figs. 4 and 5 respectively, for the four CSFR fits described in Sect. 2. The total energy emitted in GWs during the spin-down, i.e. the area under the curves, is consistent with the rotational energy lost or the change in Ω inside the instability windows, quoted in Sect. 6.\n\nThe shape of the curves can be understood as follows. For a single source, a certain range of frequencies is emitted by each mode in the source frame, and the signal is stronger at higher frequencies, according to Eq. (14). The spectral shape of the signal (in the source frame) would thus be wedge-shaped, with the highest flux occurring at the highest emitted frequency. As the contribution from sources at increasing redshifts is included, the same spectrum but shifted to increasingly lower frequencies is added to the signal. Thus, the observed signal over the emitted frequency range rises, with a drop at lower frequencies which becomes less steep as more sources are added. The contribution from sources at increasing redshifts also gets stronger since the CSFR rises, until it reaches a peak and drops, which decreases the contribution from new sources. Furthermore, the flux from high redshift sources is reduced according to the inverse square law.\n\nIn Fig. 4, these effects combine to produce bell-shaped curves, where the curved peaks are due to sources at the redshift where the CSFRs in Fig. 1 are highest. The HB06 curves have the highest amplitude since RNS is the highest for that fit (Fig. 2). The fact that the RE12 fit extends to higher redshifts than the other fits increases the signal power at low frequencies (compared to if the cut off was at the same z). Also, given that the stochastic background is mainly contributed by low redshift sources, where the RNS curves (Fig. 2) are in close agreement, the importance of the divergent CSFRs in Fig. 1 is reduced.\n\nIn Fig. 5 the curves are around two orders of magnitude weaker than in Fig. 4, since RBM is considerably lower than RNS. Also, the curves are closer together in terms of flux, since RBM (Fig. 3) for the different fits is in agreement up to a higher redshift than for RNS (Fig. 2). For the RE12 and HB06 curves, the minimum emitted frequency is the highest peak rather than the curved one (which is due to sources at the redshift of maximum CSFR). This is because the emitted frequency range is far narrower and at higher frequencies for merger-derived NSs. The minimum emitted frequency is now at a higher frequency than the curved peak, and even more so for the RE12 and HB06 fits which have maximum CSFRs occurring at the highest redshifts (which shifts the curved peak to lower frequencies). This means that the curved peak does not smooth the drop in flux to the left of the minimum emitted frequency.",
null,
"Fig. 4Integrated flux of the stochastic background for supernova-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3, 4 f-modes, as a function of the observed frequency. Open with DEXTER",
null,
"Fig. 5Integrated flux of the stochastic background for binary merger-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2, 3 f-modes, as a function of the observed frequency. Open with DEXTER",
null,
"Fig. 6Dimensionless energy density of the stochastic background for supernova-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3,4f-modes, as a function of the observed frequency. Here 10% of the NS population become f-mode unstable and 100% of the rotational energy lost during the instability is due to GWs. Open with DEXTER",
null,
"Fig. 7Lower limit on the dimensionless energy density of the stochastic background for supernova-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3,4f-modes, as a function of the observed frequency. Here 1% of the NS population become f-mode unstable and 50% of the rotational energy lost during the instability is due to GWs. Open with DEXTER",
null,
"Fig. 8Dimensionless energy density of the stochastic background for binary merger-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3f-modes, as a function of the observed frequency. Open with DEXTER\n\n### 6.2. Dimensionless energy density\n\nThe dimensionless energy density of the SGWB for supernova and merger-derived NSs is plotted in Figs. 68 respectively, again for the four CSFR fits described in Sect. 2. The strongest stochastic background over ~11000 Hz from Fig. 6 of Regimbau (2011), due to coalescing binary NSs (BNS), has been plotted on the same figures for comparison. The GW detection limits for aLIGO/aVirgo and ET have also been plotted. The sensitivity curves are taken from Table 1 of Sathyaprakash & Schutz (2009), where an analytic fitting formula is given for the noise power spectral density of each detector. The conversion to Ωgw for two co-located detectors and uncorrelated instrumental noise is given in Sect. 8.1.2 of Sathyaprakash & Schutz (2009).\n\nTo determine how changing our assumptions on the properties of the NS population affects the detectability of the background, a “lower limit” has been plotted in Fig. 7. Here we have assumed that only 1% of proto-NSs rotate sufficiently quickly to become CFS unstable, and that only half of the rotational energy is lost to GWs, with the other half lost to magnetic braking.\n\nAssuming initially that a signal is detectable if the predicted amplitude lies above the intrinsic noise curve for that particular instrument at any frequency, then the l = m = 2 background in Fig. 6 is the only one detectable with second generation detectors since the signal peaks at low frequencies (50200 Hz) where the detector sensitivity is highest, and at a higher amplitude than the background due to coalescing BNS. The l = m = 3 background might also be detectable, however it lies close to the sensitivity limit. For third generation detectors the l = m = 3 background in Fig. 6 is detectable, however the l = m = 4 one peaks at frequencies that are too high to be detected, even though it has the highest signal amplitude. Note, however, that the l = m = 3 and 4 f-mode backgrounds peak at roughly the same frequency (~103 Hz) and amplitude (Ωgw ~ 10-8) as many other astrophysical backgrounds in Fig. 6 of Regimbau (2011), hence they might be indistinguishable from them.\n\nThe lower estimate for the f-mode background due to supernova-derived NSs in Fig. 7 provides a dramatically different result. Now the f-mode background is too weak to be detected with aLIGO/aVirgo, but the BNS background should be detectable. The l = m = 2 background would be detectable with the ET interferometer, however it is now an order of magnitude weaker than the BNS background, so would be overpowered by it.\n\nThe background produced by merger-derived NSs in Fig. 8 should not be detectable, even for the most sensitive ET detector. This can be attributed to several reasons: firstly to the lower NS formation rate from binary mergers than from supernovae, and the fact that fewer GW sources (at low redshift) leads to a weaker accumulated GW background. Secondly, the supramassive NSs collapse to black holes before they can lose a significant proportion of their initial rotational energy, emitting less total energy in GWs, even though their individual GW emission is stronger than from supernova-derived NSs since it is detectable at greater distances. Thirdly, the background peaks at high frequencies around 1000 Hz where the interferometer sensitivity is considerably reduced.\n\nTable 1\n\nOptimal signal-to-noise ratio obtained by correlating interferometer pairs for the l = m = 2,3,4f-modes and supernova-derived NSs with the HB06 CSFR (see Sect. 2).\n\n### 6.3. Duty cycle\n\nAnother important quantity for an astrophysical SGWB is the duty cycle D, which classifies the background into three regimes: a continuous background (D ≫ 1) where the waveforms overlap to produce a background with Gaussian properties, shot noise (D ≪ 1) where the waveforms are separated by long stretches of silence, or popcorn noise (D ~ 1) where the waveforms may overlap but Gaussian statistics are not obeyed. The duty cycle, in the observer frame, is defined as (Regimbau 2011)",
null,
"(17)where",
null,
"is the average time duration of the GW emission from a single source in the source frame, which dilates to",
null,
"due to the cosmic expansion.\n\nFor low magnetised (Bp ≤ 1011 G) supernova-derived NSs, the best estimate available for the evolution time",
null,
"is 200 yr (Passamonti et al. 2013), and for merger-derived NSs the value is 24 h (D. Doneva, priv. comm.), giving D ~ 1012 and ~105 respectively, thus we would expect to observe a continuous GW background in both cases. Taking into account a shorter evolution time for stars with a stronger magnetic field, then",
null,
"yr (Passamonti et al. 2013) and 20 min (D. Doneva, priv. comm.), and D ~ 1010 and ~103, for supernova and merged-derived NSs with Bp ~ 1012 G and ~1014 G, respectively. Thus we would still expect a continuous background.\n\nNote that the time taken for the f-mode to saturate is a very small fraction of the total evolution time, so the duration of the GW emission is approximately equal to the total evolution time. The estimates for the evolution time that have been used depend upon the saturation energy of the mode, where a value of ~10-6Mc2 has been assumed for both NS classes. These estimates could be affected by more recent work, suggesting values as low as 10-10Mc2 for the saturation energy of supernova-derived NSs (Pnigouras & Kokkotas, in prep.; Pnigouras & Kokkotas 2015). This, however, would only affect the duty cycle and would lengthen the evolution time, which increases the value of D.\n\n## 7. Detectability\n\nHaving confirmed that we expect to observe a continuous GW background with Gaussian properties, we can employ the optimal detection strategy which involves cross-correlating measurements from two or more detectors with uncorrelated noise. This is required because the stochastic background can be confused with the intrinsic noise background of a single detector. Assuming the background to be isotropic, unpolarized, stationary, and Gaussian, the optimal signal to noise ratio (S/N) for cross-correlating two L-shaped interferometers during an observation time T is given by (Allen & Romano 1999)",
null,
"(18)where",
null,
"and",
null,
"are the power spectral noise densities of the two detectors (from Table 1 of Sathyaprakash & Schutz 2009), and Γ(νo) is the overlap reduction function, characterizing the loss of sensitivity due to the separation and the relative orientation of the detectors.\n\nTable 2\n\nLower limit on the optimal signal-to-noise ratio obtained by correlating interferometer pairs for the l = m = 2,3,4f-modes and supernova-derived NSs with the HB06 CSFR (see Sect. 2).\n\nTable 3\n\nOptimal signal-to-noise ratio obtained by correlating interferometer pairs for the l = m = 2,3f-modes and binary merger-derived NSs with the HB06 CSFR (see Sect. 2).\n\nThe optimal S/N has been evaluated for a pair of second and third generation GW detectors, aLIGO/aVirgo and ET respectively, for the two NS classes considered in this work, and for the HB06 CSFR fit which provides the highest rate of NS formation. We have assumed two co-located detectors with optimal orientation (i.e. Γ(νo) = 1), and an observation time of T = 3 yr. The results, which represent an optimistic estimate of the S/N, are shown in Tables 13 for supernova and merger-derived NSs respectively.\n\nThe results in Tables 13 agree with the detectability outcomes in Figs. 68, and hence provide a consistency check on the previous results.\n\n## 8. Conclusions\n\nIn this paper, the integrated flux and dimensionless energy density of the SGWB produced by a population of hot, young, rapidly rotating NSs has been evaluated, which emit GWs during the spin-down phase associated with the f-mode instability. Two classes of NSs have been considered, those formed from supernovae, and supramassive NSs formed through the merger of binary NS systems. Four different parameterized fits to the expected evolution of the CSFR density have been extracted from Hopkins & Beacom (2006), Fardal et al. (2007), Wilkins et al. (2008), Robertson & Ellis (2012) in order to account for the uncertainty upon its determination. The f-mode frequencies and the instability windows have been extracted from Doneva et al. (2013, 2015) for supernova and merger derived NSs respectively, for a relativistic star with the WFF2 EoS in the Cowling approximation.\n\nChanging the CSFR fit used in the calculations, which determines the GW event rate, does not change the detectability of a particular background, since the signal is mainly contributed by low redshift sources (z< 0.5) where the rates of source formation (Figs. 2 and 3) are in close agreement. The fact that the RE12 fit extends to far higher redshifts than the other fits leads to a small increase in signal strength at low frequencies.\n\nChanging the EoS used in the calculations would change the frequency of the GWs emitted, and also the total energy emitted via GWs, since the instability window would be affected. Although the GW spectrum would shift in frequency and/or flux, the position of the peak relative to the rest of the curve would not change, since it depends upon the CSFR. According to Figs. 2 and 11 of Doneva et al. (2013) however, the realistic AkmalPR EoS (Akmal et al. 1998) does not lead to significantly different unstable f-mode frequencies or instability windows compared to the WFF2 EoS. This, in addition to the dependence of the signal strength on the fraction of NSs which become f-mode unstable and the strength of the NS magnetic field, means that constraining the NS EoS using the f-mode SGWB is unlikely.\n\nThe dimensionless energy density Ωgw in Fig. 6 for the background produced by supernova-derived NSs was found to peak at ~10-9 and ~10-8 for the l = m = 2 and l = m = 3,4 respectively, at frequencies of ~200, 1000, 2000 Hz respectively. The l = m = 2 background should be detectable with second generation interferometers. The lower estimate for the background in Fig. 7, assuming fewer NSs become f-mode unstable and not all the rotational energy lost is due to GW emission, peaks at an order of magnitude lower amplitude at the same frequencies. It should still be detectable with the ET interferometer, but is weaker than the BNS background over the entire frequency range. The background produced by supramassive NSs formed from binary mergers in Fig. 8 peaks at Ωgw ~ 10-10 at frequencies ~1000 and 2000 Hz for the l = m = 2 and 3 f-mode respectively. It will not be detectable, even with ET.\n\nThe GW detections that should be made in the coming years with second generation detectors will provide fascinating new insights into the sources. The detection of the f-mode SGWB will improve our understanding of the astrophysical SGWB, whose shape needs to be understood in order for it to be disentangled from the primordial background.\n\n## Acknowledgments\n\nWe gratefully acknowledge the support of the German Science Foundation (DFG) via SFB/TR7.\n\n## All Tables\n\nTable 1\n\nOptimal signal-to-noise ratio obtained by correlating interferometer pairs for the l = m = 2,3,4f-modes and supernova-derived NSs with the HB06 CSFR (see Sect. 2).\n\nTable 2\n\nLower limit on the optimal signal-to-noise ratio obtained by correlating interferometer pairs for the l = m = 2,3,4f-modes and supernova-derived NSs with the HB06 CSFR (see Sect. 2).\n\nTable 3\n\nOptimal signal-to-noise ratio obtained by correlating interferometer pairs for the l = m = 2,3f-modes and binary merger-derived NSs with the HB06 CSFR (see Sect. 2).\n\n## All Figures",
null,
"Fig. 1Evolution of the cosmic star formation rate density with redshift for four different CSFR fits (see Sect. 2.) Open with DEXTER In the text",
null,
"Fig. 2Evolution of the number of NSs formed per unit time within the comoving volume out to redshift z based on four different CSFR fits (see Sect. 2), for NSs formed from a CCSNe. Open with DEXTER In the text",
null,
"Fig. 3Evolution of the number of NS binary mergers per unit time within the comoving volume out to redshift z based on four different CSFR fits (see Sect. 2). Open with DEXTER In the text",
null,
"Fig. 4Integrated flux of the stochastic background for supernova-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3, 4 f-modes, as a function of the observed frequency. Open with DEXTER In the text",
null,
"Fig. 5Integrated flux of the stochastic background for binary merger-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2, 3 f-modes, as a function of the observed frequency. Open with DEXTER In the text",
null,
"Fig. 6Dimensionless energy density of the stochastic background for supernova-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3,4f-modes, as a function of the observed frequency. Here 10% of the NS population become f-mode unstable and 100% of the rotational energy lost during the instability is due to GWs. Open with DEXTER In the text",
null,
"Fig. 7Lower limit on the dimensionless energy density of the stochastic background for supernova-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3,4f-modes, as a function of the observed frequency. Here 1% of the NS population become f-mode unstable and 50% of the rotational energy lost during the instability is due to GWs. Open with DEXTER In the text",
null,
"Fig. 8Dimensionless energy density of the stochastic background for binary merger-derived NSs, for four CSFR fits (see Sect. 2) and the l = m = 2,3f-modes, as a function of the observed frequency. Open with DEXTER In the text\n\nCurrent usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.\n\nData correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days.\n\nInitial download of the metrics may take a while."
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https://space.stackexchange.com/questions/26098/simple-example-applying-perturbed-effect-on-orbit-propogation | [
"# simple example applying perturbed effect on orbit propogation?\n\nI need to apply perturbation effect to orbit propagation. I am programmer not a physicist. I’ve already wrote program to propagate not perturbation orbit and show satellite ground track. So I know how to find state vector (RV – position, velocity) from orbital elements also state vector in PQW and IJK frames. But with those formulas I hardly understand when and where apply perturbation effects. At this moment I understand:\n\nAdrag = -1/2*p*(Cd*A)/m * sqr(v)*Iv\nWhere:\np – atmospheric density;\nCd – drag coefficient;\nA – crosssectional area of satellite perpendicular to the velocity vector;\nm – mass of the satellite;\nv – velocity of the satellite relative to the atmosphere;\nIv – unit vector in the direction of the satellite’s velocity.\n\n\nThis formula give me an atmospheric drag acceleration vector. Then I must apply acceleration vector to position vector R and I will get perturbed satellite position. After I need to apply acceleration vector to some formulas to change orbital elements. Necessary formulas I found in “Fundamentals of astrodynamics and applications” by David A. Vallado, paragraph “8.3.2 Gaussian VOP (Nonconservative and Conservative Effects)”. Also there is algorithm in “8.7.1 Application: Perturbed Two-body Propogation” but as I understand I cant use them because I don’t know both mean motion derivatives, I find out that they only can be known in TLE and can’t be calculated with orbital elements. But I don’t sure if they are what I need.\n\nCan someone write simple example applying atmospheric drag (or another perturbed effect) to unperturbed orbit? Or if its hard, can you advise me a forum where I can ask such question and get answer?\n\n• Actually, I think you'd apply the acceleration vector to the velocity vector (atmospheric drag always acts in the direction opposite to velocity) and then the velocity vector to the position. You are effectively simulating the solution to a differential equation. – user7073 Mar 19 '18 at 21:22\n• Consider just searching this site; there may be several examples here already. If you need something different, then if you link to one or two and explain what else you need, you may get an even more helpful and specific answer. – uhoh Mar 20 '18 at 11:51\n• Based on your more recent question it looks like you've learned a lot! It is always okay to post answers to your own questions in Stack Exchange. You can click \"accept\" as well. If the answer is good, you'll get +10 for each up vote as well, and future readers will benefit from seeing your answer. – uhoh Jun 25 '18 at 17:42\n\nFirst you need input data. Input can be presented as orbit elements (semimajor axis, eccentricity, inclination, true anomaly, longitude of ascending node, perigee), or state vector (position and velocity). If you choose orbit elements they need to be translated to state vector in ECI coordinates (find orbit to ECI).\n\nSecond choose integrator for propagation (three types: analytic, semi-analytic, numerical). If you choose numerical propagator as I did. Now choose integrator (a lot of them) but with order higher than Euler, because it doesn’t provide necessary accuracy. For simple example and understanding I choose Runge Kutta 4 (RK4).\n\nThird, choose perturbed effects. The most simple of them is gravity (find 2-Body problem) satellite_position * -MU / Magnitude(satellite_position)^3 Gravity effect used in all perturbed orbit calculations where MU – Earth gravity = 398600.4415. In most cases when you see “only-drag”, “only-solar radiation pressure” etc, it’s mean that also include gravity, so technically they are not “only”.\n\nWhen choosing perturbed effect you must decide how to calculate it, for example even gravity has different models, or in magnetic Earth field model. Atmospheric drag can be calculated in two ways (I find only two):\n\n1) Calculation depending on time like described in “Montenbruck, E. Gill; Satellite Orbits - Models, Methods, and Applications”. And also choose model like Harris-Priester\n\n2) Calculation with already calculated table of atmosphere density values, like did these gentleman https://github.com/komrad36 in one of his programs. For table values you also can choose any “already tables” you like: “U.S. Standard Atmosphere”, or Russian “ГОСТ 4401-81” etc.\n\nI choose table version. My implementation in c++, main loop:\n\nQDateTime dt = QDateTime (QDate(2018,05,11),QTime(14,30,30));\ndouble JD = GetJulDate(dt);\ndouble Mjd = MJD(dt);\ndouble h = 0.5 // integration step\nfor (double tCur = tBegin; tCur < tEnd; tCur += tStep){\n// for table version\nRK4 (h, sat);\n// for time depend version\n//RK4 (h, sat, Mjd);\n}\n\n\nFor integration step h is better to choose 0.5. Mjd – modified Julian date. sat – is object which contains satellite data (in my case only position, velocity, size and mass). Runge kutta 4 with time t, if you don’t use time just send only 1 parameter in function “acceleration”:\n\nvoid RK4(const double& h, Satellite& sat, double &t){\nSatellite k1,k2,k3,k4;\ndouble MJDstep = (0.5 * h)/Tm;\nSatellite yy;\nyy = sat;\n\nk1 = acceleration (t, yy) * h;\nyy.loc = sat.loc + .5 * k1.loc;\nyy.vel = sat.vel + .5 * k1.vel;\n\nk2 = acceleration (t + MJDstep, yy) * h;\nyy.loc = sat.loc + h * .5 * k2.loc;\nyy.vel = sat.vel + h * .5 * k2.vel;\n\nk3 = acceleration (t + MJDstep, yy) * h;\nyy.loc = sat.loc + h * k3.loc;\nyy.vel = sat.vel + h * k3.vel;\n\nk4 = acceleration (t + h/Tm, yy) * h;\nsat.loc += h/6 * (k1.loc + 2 * k2.loc + 2 * k3.loc + k4.loc);\nsat.vel += h/6 * (k1.vel + 2 * k2.vel + 2 * k3.vel + k4.vel);\n}\n\n\nLine MJDstep = (0.5 * h)/Tm is prestep calculation where Tm = 1440. This is for translate time in seconds to modified Julian date. Or any other time unit of your chose. And acceleration function:\n\nSatellite acceleration (double const &t, Satellite &sat) {\nSatellite f;\nf.loc = sat.vel;\ndouble p = Magnitude(sat.loc);\nf.vel = sat.loc * -MU /(p*p*p);\n\ndouble Cd = 2.2;\nf.vel += AccelDrag(sat, Cd);\n// for time version\n// f.vel += AccelDrag2(sat, Cd, t);\nreturn f;\n}\n\n\nthe last but not the least (and obvious) use same units everywhere. If you choose meters translate length units to meters. If seconds for time translate all time to seconds."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6919449,"math_prob":0.9347983,"size":3572,"snap":"2021-31-2021-39","text_gpt3_token_len":998,"char_repetition_ratio":0.11575112,"word_repetition_ratio":0.04841402,"special_character_ratio":0.29703248,"punctuation_ratio":0.20712401,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99189025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T05:40:08Z\",\"WARC-Record-ID\":\"<urn:uuid:5b8fb7a9-f9b8-4b4f-805a-d0bdfde9d2b8>\",\"Content-Length\":\"169562\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f3d0046-8748-42c5-ac0f-de47c024fb82>\",\"WARC-Concurrent-To\":\"<urn:uuid:a11e7075-d3ab-40ef-a839-22fcb28b78bf>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://space.stackexchange.com/questions/26098/simple-example-applying-perturbed-effect-on-orbit-propogation\",\"WARC-Payload-Digest\":\"sha1:GWKLYB4K27SKVXXVJCKXMV4HETBKTI24\",\"WARC-Block-Digest\":\"sha1:7SN7CBDOVG26IFYBMGQKOA7OJKG3QJEB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154158.4_warc_CC-MAIN-20210801030158-20210801060158-00507.warc.gz\"}"} |
https://theses.gla.ac.uk/72458/ | [
"",
null,
"# Quantum group actions on singular plane curves\n\nTabiri, Angela Ankomaah (2019) Quantum group actions on singular plane curves. PhD thesis, University of Glasgow.\n\nFull text available as:",
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"",
null,
"Preview\nPDF\nPrinted Thesis Information: https://eleanor.lib.gla.ac.uk/record=b3348257\n\n## Abstract\n\nIn [28, 38], the coordinate ring of the cusp y2 = x3 is seen to be a quantum homogeneous space. Using this as a starting example, the coordinate ring of the nodal cubic y2 = x2 +x3 was shown to be a quantum homogeneous space in . This thesis focuses on finding singular plane curves which are quantum homogeneous spaces. We begin by discussing the background theory of Hopf algebras, algebraic groups and the set up for Bergman’s Diamond Lemma . Next, we recall the theory of quantum homogeneous spaces in the commutative (classical) and noncommutative (nonclassical) settings. Examples and theorems on these spaces are stated. Then main theorem in this thesis is that decomposable plane curves (curves of the form f(y) = g(x)) of degree less than or equal to five are quantum homogeneous spaces. In order to prove this, we construct two new families of Hopf algebras, A(x; a; g) and A(g; f). Then we use Bergman’s Diamond lemma to prove that A(g; f) is faithfully flat over the coordinate ring of f(y) = g(x). These new Hopf algebras that we have discovered have nice properties when deg(g); deg(f) B 3. The properties include being noetherian domains, finite Gelfand-Kirillov dimensions, AS-regular and finite modules over their centres. We derive these properties from the isomorphism between A(x; a; g) and well studied algebras, the localised quantum plane and down-up algebras when deg(g) = 2; 3.\n\nItem Type: Thesis (PhD) Doctoral Hopf algebra, quantum homogeneous space. Q Science > QA Mathematics College of Science and Engineering > School of Mathematics and Statistics > Mathematics Brown, Prof. Kenneth 2019 Miss Angela Ankomaah Tabiri glathesis:2019-72458 Copyright of this thesis is held by the author. 23 May 2019 10:30 05 Mar 2020 21:19 10.5525/gla.thesis.72458 http://theses.gla.ac.uk/id/eprint/72458",
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"View Item"
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"https://theses.gla.ac.uk/images/backgroundaa.jpg",
null,
"http://theses.gla.ac.uk/72458/7.hassmallThumbnailVersion/2019TabiriPhD.pdf",
null,
"http://theses.gla.ac.uk/72458/7.haspreviewThumbnailVersion/2019TabiriPhD.pdf",
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"https://theses.gla.ac.uk/style/images/action_view.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8819491,"math_prob":0.8884066,"size":1598,"snap":"2023-40-2023-50","text_gpt3_token_len":397,"char_repetition_ratio":0.11668758,"word_repetition_ratio":0.015384615,"special_character_ratio":0.2321652,"punctuation_ratio":0.11326861,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9715408,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,3,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T01:11:18Z\",\"WARC-Record-ID\":\"<urn:uuid:32d85f51-1bb0-43d2-9230-f238a9fbebe2>\",\"Content-Length\":\"33840\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:380b5726-aff2-4076-a38b-033575a9d9d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:75ecc31b-6b18-4eab-92c8-d72cc19b4598>\",\"WARC-IP-Address\":\"130.209.34.248\",\"WARC-Target-URI\":\"https://theses.gla.ac.uk/72458/\",\"WARC-Payload-Digest\":\"sha1:UHGW2BGY4QLHPZWSRN2KP6RYWXC6XSHH\",\"WARC-Block-Digest\":\"sha1:7CAWAEUGP6IZUPKDDSI4XJB7GRU3GWRQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100781.60_warc_CC-MAIN-20231209004202-20231209034202-00100.warc.gz\"}"} |
https://mortgageleadnet.com/index.php/2020/04/12/payment-pmt-total-interest-using-ba-ii-plus-texas-instrument-financial-calculator/ | [
"# Payment (PMT) & Total Interest Using BA II Plus Texas Instrument Financial Calculator\n\nIn this lesson, we explain and go through an example of calculating Payment PMT & Total Interest Using BA II Plus Texas Instrument Financial Calculator. We go through an example of a mortgage loan using the BA II Plus Texas Instrument Financial Calculator amortization schedule. We show how the time value of money (TVM) is used to calculate the payment and the total interest paid for a mortgage loan.\n\nHow to calculate Future Value of a Lump Sum (single amount) | Formula with Examples:\n\nFuture Value of an Ordinary Annuity | Formula with Examples:\n\nTime Value of Money | Explained with Examples | Finance:\n\nCheck out other straight-forward examples on our channel.\n\nWe also offer one-on-one tutorials at reasonable rates.\n\nConnect with us:\nEmail: info@counttuts.com\nOur Website:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8033517,"math_prob":0.8587906,"size":901,"snap":"2020-34-2020-40","text_gpt3_token_len":184,"char_repetition_ratio":0.118171684,"word_repetition_ratio":0.12587413,"special_character_ratio":0.19533852,"punctuation_ratio":0.0875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9859425,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T08:17:05Z\",\"WARC-Record-ID\":\"<urn:uuid:ddd93e21-2d8b-413c-8793-09394f479586>\",\"Content-Length\":\"79333\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95a9b840-054e-4923-b6ec-3a829ada6727>\",\"WARC-Concurrent-To\":\"<urn:uuid:1f73df7d-170f-48e5-b8a7-cea7eaffa6ec>\",\"WARC-IP-Address\":\"69.61.102.93\",\"WARC-Target-URI\":\"https://mortgageleadnet.com/index.php/2020/04/12/payment-pmt-total-interest-using-ba-ii-plus-texas-instrument-financial-calculator/\",\"WARC-Payload-Digest\":\"sha1:DP34FAFCEWLZYOYICZ2UF4SMJTELFS3Z\",\"WARC-Block-Digest\":\"sha1:PGRULZ62DHRPEMGLBV2MO2THKGAJPWDA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401598891.71_warc_CC-MAIN-20200928073028-20200928103028-00577.warc.gz\"}"} |
http://www.grekstreet.com/2018/07/maths-questions-solution-application.html | [
"# Maths Questions Solution Application- Photomath android phone application.\n\nMaths Questions Solution Application- Photomath android phone application.\n\nSimply point your camera toward a math problem and Photomath will magically show the result with a detailed step-by-step instructions.\n\nPhotomath provides:\n∙ Camera calculator\n∙ Handwriting recognition\n∙ Step-by-step instructions\n∙ Smart calculator\n∙ Graphs (NEW)\n\nPhotomath supports arithmetics, integers, fractions, decimal numbers, roots, algebraic expressions, linear equations/inequalities, quadratic equations/inequalities, absolute equations/inequalities, systems of equations, logarithms, trigonometry, exponential and logarithmic functions, derivatives and integrals.\n\nImproved solution steps and step explanations\n• Various fixes and improvements to math problem recognition\n• Various UI fixes and performance improvements"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79589033,"math_prob":0.9826444,"size":909,"snap":"2020-34-2020-40","text_gpt3_token_len":178,"char_repetition_ratio":0.1314917,"word_repetition_ratio":0.07692308,"special_character_ratio":0.1529153,"punctuation_ratio":0.144,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99082613,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-19T18:29:35Z\",\"WARC-Record-ID\":\"<urn:uuid:ee8ba24a-12a7-4c0c-a98a-20c9df3f09d0>\",\"Content-Length\":\"47284\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7fd918e9-7f77-4d22-81d2-8785edf8a1a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:57e12e19-0f77-4007-8016-726e20e2ae27>\",\"WARC-IP-Address\":\"172.217.8.19\",\"WARC-Target-URI\":\"http://www.grekstreet.com/2018/07/maths-questions-solution-application.html\",\"WARC-Payload-Digest\":\"sha1:N2WCHS7BTB32OXXXCDAVQ6IC762ULZDR\",\"WARC-Block-Digest\":\"sha1:SQQPMBFW4QV4HGKE5YEHV5CM7435SBCD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400192783.34_warc_CC-MAIN-20200919173334-20200919203334-00317.warc.gz\"}"} |
https://www.allmathtricks.com/2019/01/ | [
"## Intersecting and Concurrent lines | Parallel Lines and transversal line\n\nIn this article explained about different types of lines in geometry like straight line, curved line, Intersecting lines, Concurrent lines, Parallel Lines and transversal line with examples. Types of Lines | Straight and Curved Line | Lines In Geometry Definitions and Properties of Line, line segment & ray Basic concepts in geometry of Line, Line-segment […]\n\n## Lines In Geometry | line segment math definition | Ray along with their types\n\nIn this article explained about some basic concepts in geometry like definitions of lines, line segment and ray. Also explained properties and differences for the same. Lines | Line segments | Rays | All Math Tricks Line A line is breadthless length. Line is a set of infinite points which extend indefinitely in both directions […]\n\n## Point in Geometry Math | Collinear Points and non-collinear points Examples\n\nIn this article explained about defections of point in geometry math, Collinear Points, Non-collinear points with examples. Point in Math | Collinear Points | Non-collinear points | All Math Tricks The terms Point, Line, Plane and space .. etc are fundamental concepts in study of geometry and they Definition of Point in Math: A point […]\n\nTop"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89288974,"math_prob":0.9170557,"size":1207,"snap":"2020-45-2020-50","text_gpt3_token_len":242,"char_repetition_ratio":0.14380714,"word_repetition_ratio":0.015625,"special_character_ratio":0.2021541,"punctuation_ratio":0.09803922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98470664,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T08:21:30Z\",\"WARC-Record-ID\":\"<urn:uuid:2fc0d2c9-da19-4af4-a52b-6970967fb6ca>\",\"Content-Length\":\"44977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6f347501-6fcc-48fc-9527-f98fa674b4e9>\",\"WARC-Concurrent-To\":\"<urn:uuid:d275799d-321e-4ae6-93a5-265691c246ac>\",\"WARC-IP-Address\":\"166.62.28.134\",\"WARC-Target-URI\":\"https://www.allmathtricks.com/2019/01/\",\"WARC-Payload-Digest\":\"sha1:YNV7QAYUQKROVCLN6CUVJLEPVS73YQ5G\",\"WARC-Block-Digest\":\"sha1:MKDGJTT4YM6XTF63NUMLNPYJM2TK26ZV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141735395.99_warc_CC-MAIN-20201204071014-20201204101014-00239.warc.gz\"}"} |
https://sigma.ontologyportal.org:8443/sigma/Browse.jsp?lang=EnglishLanguage&flang=TPTP&kb=SUMO&term=ListOrderFn | [
"",
null,
"",
null,
"Browsing Interface : Welcome guest : log in [ Home | Graph | ] KB: SUMO Language: ChineseLanguageChinesePinyinWritingChineseSimplifiedWritingChineseTraditionalLanguageEnglishLanguageFrenchLanguageGermanLanguageHindiItalianLanguageJapaneseLanguagePortugueseLanguageSpanishLanguageSwedishLanguagecbczdehirosvtg Formal Language: OWLSUO-KIFTPTPtraditionalLogic\n\n KB Term:",
null,
"Term intersection",
null,
"English Word:",
null,
"Any Noun Verb Adjective Adverb\n\nSigma KEE - ListOrderFn\n ListOrderFn\n\n appearance as argument number 1",
null,
"No TPTP formula. May not be expressible in strict first order. chinese_format.kif 1964-1966 No TPTP formula. May not be expressible in strict first order. Merge.kif 2968-2971 No TPTP formula. May not be expressible in strict first order. japanese_format.kif 596-598 No TPTP formula. May not be expressible in strict first order. Merge.kif 2964-2964 The number 1 argument of list order is an instance of list No TPTP formula. May not be expressible in strict first order. Merge.kif 2965-2965 The number 2 argument of list order is an instance of positive integer No TPTP formula. May not be expressible in strict first order. Merge.kif 2962-2962 List order is an instance of binary function No TPTP formula. May not be expressible in strict first order. Merge.kif 2963-2963 List order is an instance of partial valued relation No TPTP formula. May not be expressible in strict first order. Merge.kif 2966-2966 The range of list order is an instance of entity\n\n appearance as argument number 2",
null,
"No TPTP formula. May not be expressible in strict first order. chinese_format.kif 263-263 No TPTP formula. May not be expressible in strict first order. english_format.kif 268-268 No TPTP formula. May not be expressible in strict first order. french_format.kif 150-150 No TPTP formula. May not be expressible in strict first order. relations-it.txt 169-169 No TPTP formula. May not be expressible in strict first order. japanese_format.kif 1978-1978 No TPTP formula. May not be expressible in strict first order. portuguese_format.kif 102-102 No TPTP formula. May not be expressible in strict first order. relations-cz.txt 159-159 No TPTP formula. May not be expressible in strict first order. relations-de.txt 338-338 No TPTP formula. May not be expressible in strict first order. relations-hindi.txt 207-207 No TPTP formula. May not be expressible in strict first order. relations-ro.kif 169-169 No TPTP formula. May not be expressible in strict first order. relations-sv.txt 156-156 No TPTP formula. May not be expressible in strict first order. relations-tg.txt 337-337 No TPTP formula. May not be expressible in strict first order. chinese_format.kif 264-264 No TPTP formula. May not be expressible in strict first order. domainEnglishFormat.kif 34706-34706 No TPTP formula. May not be expressible in strict first order. domainEnglishFormat.kif 34705-34705 No TPTP formula. May not be expressible in strict first order. domainEnglishFormat.kif 34704-34704 No TPTP formula. May not be expressible in strict first order. relations-tg.txt 338-338\n\n antecedent",
null,
"No TPTP formula. May not be expressible in strict first order. Merge.kif 476-484 If @ROW is the opposite of and an attribute is equal to another entity element of (@ROW) and another attribute is equal to a third entity element of (@ROW) and a positive integer is not equal to another positive integer and a fourth entity the attribute the attribute,then the fourth entity does not have the attribute the other attribute No TPTP formula. May not be expressible in strict first order. Merge.kif 18327-18332 If The defalutMaxValue of a predicate with a positive integer arguments is a quantity. and the predicate @ARGS and another entity is equal to the positive integerth element of (@ARGS),then the statement the quantity is greater than the other entity has the modal force of likely No TPTP formula. May not be expressible in strict first order. Merge.kif 18310-18315 If The defaultMinValue of a predicate with a positive integer arguments is a quantity. and the predicate @ARGS and another entity is equal to the positive integerth element of (@ARGS),then the statement the other entity is greater than the quantity has the modal force of likely No TPTP formula. May not be expressible in strict first order. Merge.kif 18344-18349 If The defaultValue of a predicate with a positive integer arguments is a quantity. and the predicate @ARGS and another entity is equal to the positive integerth element of (@ARGS),then the statement the quantity is equal to the other entity has the modal force of likely No TPTP formula. May not be expressible in strict first order. Merge.kif 3206-3210 If length of a list is equal to a positive integer and the positive integerth element of the list is equal to another entity,then the last of the list is equal to the other entity No TPTP formula. May not be expressible in strict first order. Weather.kif 1838-1845 If a number list is equal to PhysicalQuantityToNumberFn returns the numberic values of a list of a measuring result list and a physical quantity is equal to an entity element of the measuring result list and a real number is equal to the entity element of the number list,then the real number an unit of measure(s) is equal to the physical quantity No TPTP formula. May not be expressible in strict first order. Media.kif 2106-2113 If there can be 1 values to argument a positive integer of a relation and the relation is an instance of predicate and the relation @ARGS and another entity is equal to the positive integerth element of (@ARGS) and a fourth entity is equal to the positive integerth element of (@ARGS),then the other entity is equal to the fourth entity No TPTP formula. May not be expressible in strict first order. Media.kif 2093-2103 If there can be 1 values to argument a positive integer of a relation and the relation is an instance of predicate and the relation @ARGS and another entity is equal to the positive integerth element of (@ARGS),then there doesn't exist a fourth entity such that the fourth entity is equal to the positive integerth element of (@ARGS) and the other entity is not equal to the fourth entity No TPTP formula. May not be expressible in strict first order. Geography.kif 427-431 If the region a directional attribute of @ROW is an instance of region and 1th element of (@ROW) is equal to a real number angular degree(s),then the real number is less than or equal to 90.0 No TPTP formula. May not be expressible in strict first order. Geography.kif 467-474 If the meridian at @ROW a directional attribute is an instance of region and 1th element of (@ROW) is equal to a real number angular degree(s),then the real number is less than or equal to 180.0 No TPTP formula. May not be expressible in strict first order. Weather.kif 1935-1944 If a list is an instance of consecutive time interval list and a time interval is equal to an entity element of the list and another time interval is equal to (a positive integer and 1)th element of the list,then the beginning of the other time interval is equal to the end of the time interval No TPTP formula. May not be expressible in strict first order. Weather.kif 1806-1811 If a measuring result list is the result of a measuring list and a process is equal to an entity element of the measuring list and another entity is equal to the entity element of the measuring result list,then the other entity is a result of the process No TPTP formula. May not be expressible in strict first order. QoSontology.kif 694-710 If the list of processes @ROW and a physical is a member of (@ROW) and another physical is a member of (@ROW) and another entity element of (@ROW) is equal to the physical and a third entity element of (@ROW) is equal to the other physical and a positive integer is less than another positive integer,then the time of existence of the physical happens earlier than the time of existence of the other physical No TPTP formula. May not be expressible in strict first order. UXExperimentalTerms.kif 989-1007 If a list is the list of items viewed by an agent and a process is an instance of accessing web page and another process is an instance of accessing web page and the agent is an agent of the process and the agent is an agent of the other process and a positive integer is an instance of positive integer and another positive integer is an instance of positive integer and the positive integerth element of the list is equal to the process and the other positive integerth element of the list is equal to the other process and the positive integer is greater than the other positive integer,then the time of existence of the other process happens earlier than the time of existence of the process\n\n consequent",
null,
"No TPTP formula. May not be expressible in strict first order. People.kif 298-319 A real number is an average of a list if and only if there exist another list and a positive integer such that length of the other list is equal to length of the list and 1th element of the other list is equal to 1th element of the list and for all another positive integer if the other positive integer is a member of the other list,then there exist another real number, the other real numberMINUSONE,, , a third positive integer and a fourth positive integer such that the other real number is greater than 1 and the other real number is less than or equal to length of the other list and the other positive integerth element of the other list is equal to the other real number and the third positive integer is a member of the list and the other real number is equal to the third positive integerth element of the list and the fourth positive integer is a member of the other list and the other real numberMINUSONE is equal to (the other real number and 1) and the other real numberMINUSONE is equal to the fourth positive integerth element of the other list and the other positive integer is equal to (the third positive integer and the fourth positive integer) and the positive integer is equal to length of the other list and the real number is equal to the positive integerth element of the other list and the positive integer No TPTP formula. May not be expressible in strict first order. Merge.kif 2985-2990 If the number a positive integer argument of a relation is an instance of a class and the relation is an instance of predicate and the relation @ROW,then the positive integerth element of (@ROW) is an instance of the class No TPTP formula. May not be expressible in strict first order. Merge.kif 2992-2997 If the number a positive integer argument of a relation is a subclass of a class and the relation is an instance of predicate and the relation @ROW,then the positive integerth element of (@ROW) is a subclass of the class No TPTP formula. May not be expressible in strict first order. Merge.kif 3239-3243 If a real number is equal to the sum of a list and 1 is equal to length of the list,then the real number is equal to 1th element of the list No TPTP formula. May not be expressible in strict first order. Merge.kif 295-302 If a list is equal to another list and the list is equal to (@ROW1) and the other list is equal to (@ROW2),then a third entity element of (@ROW1) is equal to the third entity element of (@ROW2) No TPTP formula. May not be expressible in strict first order. Merge.kif 3070-3089 If a list is equal to the list composed of another list and a third list and the other list is not equal to null list and the third list is not equal to null list and a positive integer is less than or equal to length of the other list and another positive integer is less than or equal to length of the third list and the positive integer is an instance of positive integer and the other positive integer is an instance of positive integer,then the positive integerth element of the list is equal to the positive integerth element of the other list and (length of the other list and the other positive integer)th element of the list is equal to the other positive integerth element of the third list No TPTP formula. May not be expressible in strict first order. Merge.kif 3166-3175 If a list is equal to the sub-list from a positive integer to an integer of another list and (the integer and the positive integer) is equal to 1,then the list is equal to (the positive integerth element of the other list) No TPTP formula. May not be expressible in strict first order. Merge.kif 3177-3189 If a list is equal to the sub-list from a positive integer to an integer of another list and (the integer and the positive integer) is greater than 1,then the list is equal to the list composed of (the positive integerth element of the other list) and the sub-list from (1 and the positive integer) to the integer of the other list No TPTP formula. May not be expressible in strict first order. Weather.kif 1486-1497 If a real number is equal to VarianceAverageFn of a list with the mean of another real number and 1 is equal to length of the list,then the real number is equal to (the other real number and 1th element of the list) and (the other real number and 1th element of the list) No TPTP formula. May not be expressible in strict first order. Weather.kif 1453-1465 If a real number is equal to VarianceAverageFn of a list with the mean of a number and length of the list is greater than 1,then the real number is equal to (VarianceAverageFn of 1th element of the list with the mean of the number and VarianceAverageFn of the sub-list from 2 to length of the list of the list with the mean of the number) No TPTP formula. May not be expressible in strict first order. Media.kif 2093-2103 If there can be 1 values to argument a positive integer of a relation and the relation is an instance of predicate and the relation @ARGS and another entity is equal to the positive integerth element of (@ARGS),then there doesn't exist a fourth entity such that the fourth entity is equal to the positive integerth element of (@ARGS) and the other entity is not equal to the fourth entity No TPTP formula. May not be expressible in strict first order. Media.kif 2075-2090 If there can be 1 values to argument a positive integer of a relation and the relation is an instance of predicate,then there exist an entity and @ARGS such that the relation @ARGS and the entity is equal to the positive integerth element of (@ARGS) and there doesn't exist a fourth entity such that the fourth entity is equal to the positive integerth element of (@ARGS) and the entity is not equal to the fourth entity No TPTP formula. May not be expressible in strict first order. Media.kif 2137-2150 If there can be an integer values to argument a positive integer of a relation and the relation is an instance of predicate,then there exist a class, an entity and @ARGS such that the class is an instance of set or class and if the relation @ARGS and the entity is equal to the positive integerth element of (@ARGS),then the entity is an instance of the class and the number of instances in the class is equal to the integer No TPTP formula. May not be expressible in strict first order. Merge.kif 3219-3224 If a list is an instance of list and the list is not equal to null list,then the first of the list is equal to 1th element of the list No TPTP formula. May not be expressible in strict first order. UXExperimentalTerms.kif 2594-2610 If a process is an instance of search engine optimization and an entity is a patient of the process,then the process has the purpose there exist another entity_BEFORE, the other entity_AFTER,, , a third entity,, , a fourth entity,, , a fifth entity and a sixth entity such that the other entity_BEFORE is an instance of search results and the other entity_AFTER is an instance of search results and the entity is equal to the third entityth element of the other entity_BEFORE and the entity is equal to the fourth entityth element of the other entity_AFTER and the fifth entity is an instance of best match sort and the sixth entity is an instance of best match sort and the time of existence of the fifth entity happens earlier than the time of existence of the sixth entity and the time of existence of the process happens earlier than the time of existence of the sixth entity and the time of existence of the fifth entity happens earlier than the time of existence of the process and the third entity is greater than the fourth entity No TPTP formula. May not be expressible in strict first order. Media.kif 2210-2223 If there can be at most an integer values to argument a positive integer of a relation and the relation is an instance of predicate,then there exist a class, an entity and @ARGS such that the class is an instance of set or class and if the relation @ARGS and the entity is equal to the positive integerth element of (@ARGS),then the entity is an instance of the class and the number of instances in the class is less than or equal to the integer No TPTP formula. May not be expressible in strict first order. Media.kif 2174-2187 If there are at least an integer values to argument a positive integer of a relation and the relation is an instance of predicate,then there exist a class, an entity and @ARGS such that the class is an instance of set or class and if the relation @ARGS and the entity is equal to the positive integerth element of (@ARGS),then the entity is an instance of the class and the number of instances in the class is greater than or equal to the integer No TPTP formula. May not be expressible in strict first order. Merge.kif 3199-3204 If the last of a list is equal to an entity,then there exists a positive integer such that length of the list is equal to the positive integer and the positive integerth element of the list is equal to the entity No TPTP formula. May not be expressible in strict first order. Media.kif 2125-2134 If there can be an integer values to argument another integer of a relation,then there exist a symbolic string and @ARGS such that the number of instances in the class described by the symbolic string is equal to the integer No TPTP formula. May not be expressible in strict first order. Merge.kif 519-531 If @ROW are all the attributes of another kind of attribute,then there doesn't exist an entity such that the entity is an instance of another kind of attribute and there don't exist another entity and a positive integer such that the entity is equal to the other entity and the other entity is equal to the positive integerth element of (@ROW) No TPTP formula. May not be expressible in strict first order. Merge.kif 3100-3103 If an entity is a member of a list,then there exists a positive integer such that the positive integerth element of the list is equal to the entity No TPTP formula. May not be expressible in strict first order. Merge.kif 2897-2902 If a list is an instance of unique list,then for all a positive integer and another positive integer if the positive integerth element of the list is equal to the other positive integerth element of the list,then the positive integer is equal to the other positive integer No TPTP formula. May not be expressible in strict first order. Media.kif 2198-2207 If there can be at most an integer values to argument another integer of a relation,then there exist a symbolic string and @ARGS such that the number of instances in the class described by the symbolic string is less than or equal to the integer No TPTP formula. May not be expressible in strict first order. Media.kif 2161-2170 If there are at least an integer values to argument another integer of a relation,then there exist a symbolic string and @ARGS such that the number of instances in the class described by the symbolic string is greater than or equal to the integer\n\n statement",
null,
"No TPTP formula. May not be expressible in strict first order. Merge.kif 3029-3033 For all @ROW and another entity length of (@ROW and the other entity)th element of (@ROW and the other entity) is equal to the other entity",
null,
"Show full definition with tree view\nShow simplified definition (without tree view)\nShow simplified definition (with tree view)",
null,
"Sigma web home Suggested Upper Merged Ontology (SUMO) web home\nSigma version 3.0 is open source software produced by Articulate Software and its partners"
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https://www.askpython.com/python/array/python-add-elements-to-an-array | [
"# Python add elements to an Array\n\nPython doesn’t have a specific data type to represent arrays.\n\nThe following can be used to represent arrays in Python:\n\n## Upskill 2x faster with Educative\n\nSupercharge your skillset with Educative Python courses → use code: ASK15 to save 15%\n\n## 1. Adding to an array using Lists\n\nIf we are using List as an array, the following methods can be used to add elements to it:\n\n• `By using append() function`: It adds elements to the end of the array.\n• `By using insert() function`: It inserts the elements at the given index.\n• `By using extend() function`: It elongates the list by appending elements from both the lists.\n\nExample 1: Adding elements to an array using append() function\n\n```my_input = ['Engineering', 'Medical']\nmy_input.append('Science')\nprint(my_input)\n```\n\nOutput:\n\n`['Engineering', 'Medical', 'Science']`\n\nExample 2: Adding elements to an array using extend() function\n\n```my_input = ['Engineering', 'Medical']\ninput1 = [40, 30, 20, 10]\nmy_input.extend(input1)\nprint(my_input)\n\n```\n\nOutput:\n\n`['Engineering', 'Medical', 40, 30, 20, 10]`\n\nExample 3: Adding elements to an array using insert() function\n\n```my_input = [1, 2, 3, 4, 5]\n\nprint(f'Current Numbers List {my_input}')\n\nindex = int(input(f'Enter the index between 0 and {len(my_input) - 1} to add the given number:\\n'))\n\nmy_input.insert(index, number)\n\nprint(f'Updated List {my_input}')\n```\n\nOutput:\n\n## 2. Adding to an array using array module\n\nIf we are using the array module, the following methods can be used to add elements to it:\n\n• `By using + operator`: The resultant array is a combination of elements from both the arrays.\n• `By using append() function`: It adds elements to the end of the array.\n• `By using insert() function`: It inserts the elements at the given index.\n• `By using extend() function`: It elongates the list by appending elements from both the lists.\n\nExample:\n\n```import array\n\ns1 = array.array('i', [1, 2, 3])\ns2 = array.array('i', [4, 5, 6])\n\nprint(s1)\nprint(s2)\n\ns3 = s1 + s2\nprint(s3)\n\ns1.append(4)\nprint(s1)\n\ns1.insert(0, 10)\nprint(s1)\n\ns1.extend(s2)\nprint(s1)\n```\n\nOutput:\n\n## 3. Addition of elements to NumPy array\n\nWe can add elements to a NumPy array using the following methods:\n\n• `By using append() function`: It adds the elements to the end of the array.\n• `By using insert() function`: It adds elements at the given index in an array.\n\nExample:\n\n```import numpy\n# insert function\narr1_insert = numpy.array([1, 23, 33])\n\narr2_insert = numpy.insert(arr1_insert, 1, 91)\n\nprint(arr2_insert)\n# append function\narr1_append = numpy.array([4, 2, 1])\n\narr2_append = numpy.append (arr1_append, [12, 13, 14])\n\nprint(arr2_append)\n```\n\nOutput:\n\n`[ 1 91 23 33][ 4 2 1 12 13 14]`"
] | [
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https://feet-to-meters.appspot.com/30100-feet-to-meters.html | [
"Feet To Meters\n\n30100 ft to m30100 Foot to Meters\n\nft\n=\nm\n\nHow to convert 30100 foot to meters?\n\n 30100 ft * 0.3048 m = 9174.48 m 1 ft\nA common question is How many foot in 30100 meter? And the answer is 98753.2808398 ft in 30100 m. Likewise the question how many meter in 30100 foot has the answer of 9174.48 m in 30100 ft.\n\nHow much are 30100 feet in meters?\n\n30100 feet equal 9174.48 meters (30100ft = 9174.48m). Converting 30100 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 30100 ft to m.\n\nConvert 30100 ft to common lengths\n\nUnitLengths\nNanometer9.17448e+12 nm\nMicrometer9174480000.0 µm\nMillimeter9174480.0 mm\nCentimeter917448.0 cm\nInch361200.0 in\nFoot30100.0 ft\nYard10033.3333333 yd\nMeter9174.48 m\nKilometer9.17448 km\nMile5.7007575758 mi\nNautical mile4.9538228942 nmi\n\nWhat is 30100 feet in m?\n\nTo convert 30100 ft to m multiply the length in feet by 0.3048. The 30100 ft in m formula is [m] = 30100 * 0.3048. Thus, for 30100 feet in meter we get 9174.48 m.\n\n30100 Foot Conversion Table",
null,
"Alternative spelling\n\n30100 Foot to m, 30100 Foot in m, 30100 Feet to Meter, 30100 ft to m, 30100 Feet to m, 30100 Foot in Meter, 30100 Feet to Meters, 30100 Feet in Meters,"
] | [
null,
"https://feet-to-meters.appspot.com/image/30100.png",
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https://www.uoguelph.ca/registrar/calendars/graduate/2018-2019/apdxa/apdxa-stat.shtml | [
"# Appendix A - Courses\n\n## Statistics\n\nSTAT*6550 Computational Statistics U [0.50]\nThis course covers the implementation of a variety of computational statistics techniques. These include random number generation, Monte Carlo methods, non-parametric techniques, Markov chain Monte Carlo methods, and the EM algorithm. A significant component of this course is the implementation of techniques.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6700 Stochastic Processes U [0.50]\nThe content of this course is to introduce Brownian motion leading to the development of stochastic integrals thus providing a stochastic calculus. The content of this course will be delivered using concepts from measure theory and so familiarity with measures, measurable spaces, etc., will be assumed.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6721 Stochastic Modelling U [0.50]\nTopics include the Poisson process, renewal theory, Markov chains, Martingales, random walks, Brownian motion and other Markov processes. Methods will be applied to a variety of subject matter areas. Offered in conjunction with STAT*4360. Extra work is required for graduate students.\nRestriction(s): Credit may be obtained for only one of STAT*4360 or STAT*6721\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6761 Survival Analysis U [0.50]\nKaplan-Meier estimation, life-table methods, the analysis of censored data, survival and hazard functions, a comparison of parametric and semi-parametric methods, longitudinal data analysis.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6801 Statistical Learning U [0.50]\nTopics include: nonparametric and semiparametric regression; kernel methods; regression splines; local polynomial models; generalized additive models; classification and regression trees; neural networks. This course deals with both the methodology and its application with appropriate software. Areas of application include biology, economics, engineering and medicine.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6802 Generalized Linear Models and Extensions U [0.50]\nTopics include: generalized linear models; generalized linear mixed models; joint modelling of mean and dispersion; generalized estimating equations; modelling longitudinal categorical data; modelling clustered data. This course will focus both on theory and implementation using relevant statistical software. Offered in conjunction with STAT*4050/4060. Extra work is required for graduate students.\nRestriction(s): Credit may be obtained for only one of STAT*4050 or STAT*4060 or STAT*6802\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6821 Multivariate Analysis U [0.50]\nThis is an advanced course in multivariate analysis and one of the primary emphases will be on the derivation of some of the fundamental classical results of multivariate analysis. In addition, topics that are more current to the field will also be discussed such as: multivariate adaptive regression splines; projection pursuit regression; and wavelets. Offered in conjunction with STAT*4350. Extra work is required for graduate students.\nRestriction(s): Credit may be obtained for only one of STAT*4350 or STAT*6821\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6841 Computational Statistical Inference U [0.50]\nThis course covers Bayesian and likelihood methods, large sample theory, nuisance parameters, profile, conditional and marginal likelihoods, EM algorithms and other optimization methods, estimating functions, Monte Carlo methods for exploring posterior distributions and likelihoods, data augmentation, importance sampling and MCMC methods.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6860 Linear Statistical Models U [0.50]\nGeneralized inverses of matrices; distribution of quadratic and linear forms; regression or full rank model; models not of full rank; hypothesis testing and estimation for full and non-full rank cases; estimability and testability; reduction sums of squares; balanced and unbalanced data; mixed models; components of variance.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6920 Topics in Statistics U [0.50]\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6950 Statistical Methods for the Life Sciences F [0.50]\nAnalysis of variance, completely randomized, randomized complete block and latin square designs; planned and unplanned treatment comparisons; random and fixed effects; factorial treatment arrangements; simple and multiple linear regression; analysis of covariance with emphasis on the life sciences. STAT*6950 and STAT*6960 are intended for graduate students of other departments and may not normally be taken for credit by mathematics and statistics graduate students.\nDepartment(s): Department of Mathematics and Statistics\nSTAT*6998 MSc Project in Statistics U [1.00]\nThis course is intended for students in the course-based MSc program in Statistics. The MSc project will be written under the supervision of a faculty member and will normally be completed within one or two semesters. Once completed, students will submit a final copy of their project to the Department and give an oral presentation of their work\nRestriction(s): Restricted to MSC.MAST:L-STAT students in Statistics\nDepartment(s): Department of Mathematics and Statistics\nUniversity of Guelph\n50 Stone Road East\nGuelph, Ontario, N1G 2W1\nCanada\n519-824-4120"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84408975,"math_prob":0.7995442,"size":5410,"snap":"2021-04-2021-17","text_gpt3_token_len":1083,"char_repetition_ratio":0.16870144,"word_repetition_ratio":0.09014084,"special_character_ratio":0.20573013,"punctuation_ratio":0.13407822,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96559364,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-18T16:40:58Z\",\"WARC-Record-ID\":\"<urn:uuid:07b145cb-9e91-43da-8689-56760dd1b9ff>\",\"Content-Length\":\"20650\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f22affab-0aec-4bf2-a5b2-ca9a0ab4e384>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e838bfb-877b-42bf-b17c-73ce3ee6a2a3>\",\"WARC-IP-Address\":\"131.104.93.93\",\"WARC-Target-URI\":\"https://www.uoguelph.ca/registrar/calendars/graduate/2018-2019/apdxa/apdxa-stat.shtml\",\"WARC-Payload-Digest\":\"sha1:G4JVOMOQABSBBLCFQUBDS5OQKHL3A7IL\",\"WARC-Block-Digest\":\"sha1:YSFNAP6K4Q65AW3V4PN6AT4UOYX6NGBE\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038507477.62_warc_CC-MAIN-20210418163541-20210418193541-00476.warc.gz\"}"} |
https://www.colorhexa.com/44c8da | [
"# #44c8da Color Information\n\nIn a RGB color space, hex #44c8da is composed of 26.7% red, 78.4% green and 85.5% blue. Whereas in a CMYK color space, it is composed of 68.8% cyan, 8.3% magenta, 0% yellow and 14.5% black. It has a hue angle of 187.2 degrees, a saturation of 67% and a lightness of 56.1%. #44c8da color hex could be obtained by blending #88ffff with #0091b5. Closest websafe color is: #33cccc.\n\n• R 27\n• G 78\n• B 85\nRGB color chart\n• C 69\n• M 8\n• Y 0\n• K 15\nCMYK color chart\n\n#44c8da color description : Soft cyan.\n\n# #44c8da Color Conversion\n\nThe hexadecimal color #44c8da has RGB values of R:68, G:200, B:218 and CMYK values of C:0.69, M:0.08, Y:0, K:0.15. Its decimal value is 4507866.\n\nHex triplet RGB Decimal 44c8da `#44c8da` 68, 200, 218 `rgb(68,200,218)` 26.7, 78.4, 85.5 `rgb(26.7%,78.4%,85.5%)` 69, 8, 0, 15 187.2°, 67, 56.1 `hsl(187.2,67%,56.1%)` 187.2°, 68.8, 85.5 33cccc `#33cccc`\nCIE-LAB 74.57, -29.666, -19.395 35.689, 47.596, 73.632 0.227, 0.303, 47.596 74.57, 35.443, 213.175 74.57, -49.195, -26.137 68.99, -28.392, -14.986 01000100, 11001000, 11011010\n\n# Color Schemes with #44c8da\n\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #da5644\n``#da5644` `rgb(218,86,68)``\nComplementary Color\n• #44daa1\n``#44daa1` `rgb(68,218,161)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #447dda\n``#447dda` `rgb(68,125,218)``\nAnalogous Color\n• #daa144\n``#daa144` `rgb(218,161,68)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #da447d\n``#da447d` `rgb(218,68,125)``\nSplit Complementary Color\n• #c8da44\n``#c8da44` `rgb(200,218,68)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #da44c8\n``#da44c8` `rgb(218,68,200)``\n• #44da56\n``#44da56` `rgb(68,218,86)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #da44c8\n``#da44c8` `rgb(218,68,200)``\n• #da5644\n``#da5644` `rgb(218,86,68)``\n• #239eaf\n``#239eaf` `rgb(35,158,175)``\n• #27b1c4\n``#27b1c4` `rgb(39,177,196)``\n• #2fc2d6\n``#2fc2d6` `rgb(47,194,214)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #59cede\n``#59cede` `rgb(89,206,222)``\n• #6fd5e2\n``#6fd5e2` `rgb(111,213,226)``\n• #84dbe7\n``#84dbe7` `rgb(132,219,231)``\nMonochromatic Color\n\n# Alternatives to #44c8da\n\nBelow, you can see some colors close to #44c8da. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #44dac7\n``#44dac7` `rgb(68,218,199)``\n``#44dad3` `rgb(68,218,211)``\n• #44d5da\n``#44d5da` `rgb(68,213,218)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #44bcda\n``#44bcda` `rgb(68,188,218)``\n• #44afda\n``#44afda` `rgb(68,175,218)``\n• #44a3da\n``#44a3da` `rgb(68,163,218)``\nSimilar Colors\n\n# #44c8da Preview\n\nThis text has a font color of #44c8da.\n\n``<span style=\"color:#44c8da;\">Text here</span>``\n#44c8da background color\n\nThis paragraph has a background color of #44c8da.\n\n``<p style=\"background-color:#44c8da;\">Content here</p>``\n#44c8da border color\n\nThis element has a border color of #44c8da.\n\n``<div style=\"border:1px solid #44c8da;\">Content here</div>``\nCSS codes\n``.text {color:#44c8da;}``\n``.background {background-color:#44c8da;}``\n``.border {border:1px solid #44c8da;}``\n\n# Shades and Tints of #44c8da\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #02090a is the darkest color, while #f8fdfe is the lightest one.\n\n• #02090a\n``#02090a` `rgb(2,9,10)``\n• #05171a\n``#05171a` `rgb(5,23,26)``\n• #08262a\n``#08262a` `rgb(8,38,42)``\n• #0c353b\n``#0c353b` `rgb(12,53,59)``\n• #0f444b\n``#0f444b` `rgb(15,68,75)``\n• #12535b\n``#12535b` `rgb(18,83,91)``\n• #15616c\n``#15616c` `rgb(21,97,108)``\n• #19707c\n``#19707c` `rgb(25,112,124)``\n• #1c7f8d\n``#1c7f8d` `rgb(28,127,141)``\n• #1f8e9d\n``#1f8e9d` `rgb(31,142,157)``\n``#229dad` `rgb(34,157,173)``\n• #26abbe\n``#26abbe` `rgb(38,171,190)``\n• #29bace\n``#29bace` `rgb(41,186,206)``\n• #34c3d7\n``#34c3d7` `rgb(52,195,215)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #54cddd\n``#54cddd` `rgb(84,205,221)``\n• #65d2e0\n``#65d2e0` `rgb(101,210,224)``\n• #75d6e4\n``#75d6e4` `rgb(117,214,228)``\n• #86dbe7\n``#86dbe7` `rgb(134,219,231)``\n• #96e0ea\n``#96e0ea` `rgb(150,224,234)``\n• #a6e5ed\n``#a6e5ed` `rgb(166,229,237)``\n• #b7eaf1\n``#b7eaf1` `rgb(183,234,241)``\n• #c7eff4\n``#c7eff4` `rgb(199,239,244)``\n• #d7f3f7\n``#d7f3f7` `rgb(215,243,247)``\n• #e8f8fa\n``#e8f8fa` `rgb(232,248,250)``\n• #f8fdfe\n``#f8fdfe` `rgb(248,253,254)``\nTint Color Variation\n\n# Tones of #44c8da\n\nA tone is produced by adding gray to any pure hue. In this case, #899495 is the less saturated color, while #22e2fc is the most saturated one.\n\n• #899495\n``#899495` `rgb(137,148,149)``\n• #809a9e\n``#809a9e` `rgb(128,154,158)``\n• #78a1a6\n``#78a1a6` `rgb(120,161,166)``\n• #6fa7af\n``#6fa7af` `rgb(111,167,175)``\n• #66aeb8\n``#66aeb8` `rgb(102,174,184)``\n• #5eb4c0\n``#5eb4c0` `rgb(94,180,192)``\n• #55bbc9\n``#55bbc9` `rgb(85,187,201)``\n• #4dc1d1\n``#4dc1d1` `rgb(77,193,209)``\n• #44c8da\n``#44c8da` `rgb(68,200,218)``\n• #3bcfe3\n``#3bcfe3` `rgb(59,207,227)``\n• #33d5eb\n``#33d5eb` `rgb(51,213,235)``\n``#2adcf4` `rgb(42,220,244)``\n• #22e2fc\n``#22e2fc` `rgb(34,226,252)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #44c8da is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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https://www.jpost.com/international/economist-krugman-us-depressed-economy-could-last-5-years | [
"(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);"
] | [
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https://stackoverflow.com/questions/29375627/rails-query-join-association-table-with-alias/29392498 | [
"# Rails query join association table with alias\n\nI have a model `Edge` that belongs to the other model `Node` twice through different foreign keys:\n\n``````def Edge < ActiveRecord::Base\nbelongs_to :first, class_name: 'Node'\nbelongs_to :second, class_name: 'Node'\nend\n``````\n\nAnd I want to perform this query using ActiveRecord:\n\n``````SELECT * FROM edges INNER JOIN nodes as first ON first.id = edges.first_id WHERE first.value = 5\n``````\n\nI found the way to join association using `.joins()` method:\n\n``````Edge.joins(:first)\n``````\n\nBut this produces query using a table name, not an association name, so in `.where()` method I have to explicitly use table name which breaks association abstraction.\n\n``````Edge.joins(:first).where(nodes: {value: 5})\n``````\n\nI can also explicitly use SQL query in `.joins()` method to define model alias:\n\n``````Edge.joins('INNER JOIN nodes as first ON nodes.id = edges.first_id')\n``````\n\nBut this breaks even more abstraction.\n\nI think there should be the way to automatically define table alias on join. Or maybe a way to write such function by myself. Something like:\n\n``````def Edge < ActiveRecord::Base\n...\ndef self.joins_alias\n# Generate something like\n# joins(\"INNER JOIN #{relation.table} as #{relation.alias} ON #{relation.alias}.#{relation.primary_key} = #{table}.#{relation.foreign_key}\")\nend\nend\n``````\n\nBut I couldn't find any information about accessing information about specific relation like it's name, foreign key, etc. So how can I do it?\n\nAlso it seems strange to me that such obvious feature is so complicated even through Rails is on its 4th major version already. Maybe I'm missing something?\n\nAs for Rails 4.2.1, I believe you just cannot provide an alias when using `joins` from ActiveRecord.\n\nIf you want to query edges by the first node, you could do it just like you stated:\n\n``````Edge.joins(:first).where(nodes: {value: 1})\nSELECT \"edges\".* FROM \"edges\" INNER JOIN \"nodes\" ON \"nodes\".\"id\" = \"edges\".\"first_id\" WHERE \"nodes\".\"value\" = 1\n``````\n\nBut if you have to query using both nodes, you can still use `joins` like this:\n\n``````Edge.joins(:first, :second).where(nodes: {value: 1}, seconds_edges: {value: 2})\nSELECT \"edges\".* FROM \"edges\" INNER JOIN \"nodes\" ON \"nodes\".\"id\" = \"edges\".\"first_id\" INNER JOIN \"nodes\" \"seconds_edges\" ON \"seconds_edges\".\"id\" = \"edges\".\"second_id\" WHERE \"nodes\".\"value\" = 1 AND \"seconds_edges\".\"value\" = 2\n``````"
] | [
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https://electronics.stackexchange.com/questions/302096/water-conductivity-measurements-esp8266-getting-strange-results | [
"# Water conductivity measurements/ESP8266 - getting strange results\n\nHope this is the correct forum and thanks in advance for reading this.\n\nSummary: I'm working on EC (electric conductivity) measurements using a capacitor and the pins of an ESP8266 microprocessor to create an AC current, and using that to measure the conductivity of an ionic solution. What I'm actually measuring is the discharge time of the capacitor, and out of that the electrical resistance is measured.\n\nResearch paper on which this project is based: https://hal.inria.fr/file/index/docid/635652/filename/TDS_Logger_RJP2011.pdf\n\nNow when measuring low EC liquids, with long discharge times (long being 15-20 microseconds), the discharge time goes up disproportionally. The same effect I see at short discharge times (in the tune of <1 microsecond) but less strong, at least within the range I am measuring.\n\nEach EC probe requires three GPIO ports. One ESP8266 handles three EC probes. Source code of the sketch used at the bottom, see comments in the source for explanation how it works.\n\nI strongly believe the measurement method as such is correct - measurements are highly reproducible, and using a high frequency AC is the correct way to measure an ionic liquid. The problem most likely lies hidden inside the ESP8266 - I hope to find out what causes this error, and how to correct for it.",
null,
"The first clear observation is that the ports are not all equal. Of course there are some that have pull-up or pull-down resistors, and GPIO2 is connected to the internal LED, but there are more systemic differences between the ports. I get different discharge times for different groups of ports, but those times are consistent between my two prototypes so it's an ESP thing, not an external component thing.",
null,
"This chart shows the measured EC vs. the reciprocal discharge time - which is a direct measure of the conductivity (fixed C, so t relates to R in an RC circuit, and conductivity is 1/R). There are two boards, 126 and 127, each with three probes. Same software & schematic.\n\nThere are three groups of two lines that are pretty much on top of one another. This shows the difference in behaviour of the GPIO ports. The capacitors used have a tolerance of 5%.\n\nEC should be linear to the square root of the concentration; at diluted solutions like what I use it can be well approximated by making it linear to the concentration itself. This is what I did. Then with the regression results I calculated the measurement error of each point (calculate expected value using the slope/intersect, compare with measured value). This is the result in errors:",
null,
"Again three groups of two lines: the errors are highly consistent between the two boards. This makes me believe there is some kind of systemic error, which should be accounted for - but what could this error possibly be?\n\nSource code - relevant parts only (this won't run as is).\n\n// Pin connections.\n#define WIFILED 2 // Indicator LED - on when WiFi is connected.\n\n// EC1, capPos: GPIO 12\n// EC1, capNeg: GPIO 14\n// EC1, ECPin: GPIO 16\n// EC2, capPos: GPIO 0\n// EC2, capNeg: GPIO 4\n// EC2, ECPin: GPIO 13\n// EC3, capPos: GPIO 5\n// EC3, capNeg: GPIO 1\n// EC3, ECPin: GPIO 15\n// TX = 1, RX = 3.\n#if SERIAL // don't try to use the third probe as it messes up the Serial interface.\nint capPos[] = {12, 0}; // CapPos/C+ for EC probes.\nint capNeg[] = {16, 4}; // CapNeg/C- for EC probes.\nint ECpin[] = {14, 13}; // ECpin/EC for EC probes.\n#define NPROBES 2\n#else\nint capPos[] = {12, 0, 5}; // CapPos/C+ for EC probes.\nint capNeg[] = {16, 4, 1}; // CapNeg/C- for EC probes.\nint ECpin[] = {14, 13, 15}; // ECpin/EC for EC probes.\n#define NPROBES 3\n#endif\nint interruptPin; // stores which pin is used to connect the EC interrupt.\n\n// EC probe data\n#define CYCLETIME 12.5 // the time it takes in nanoseconds to complete one CPU cycle (12.5 ns on a 80 MHz processor)\nunsigned long startCycle;\nunsigned long endCycle;\nfloat EC = {-1, -1, -1}; // the array in which to store the EC readings.\n\n// GPIO registers for EC probing - this is ESP8266 specific.\n\n/**\n* capacitor based TDS measurement\n* pin CapPos ----- 330 ohm resistor ----|------------|\n* | |\n* cap EC probe or\n* | resistor (for simulation)\n* pin CapNeg ---------------------------| |\n* |\n* pin ECpin ------ 180 ohm resistor -----------------|\n*\n* So, what's going on here?\n* EC - electic conductivity - is the reciprocal of the resistance of the liquid.\n* So we have to measure the resistance, but this can not be done directly as running\n* a DC current through an ionic liquid just doesn't work, as we get electrolysis and\n* the migration of ions to the respective electrodes.\n*\n* So this routing is using the pins of the NodeMCU and a capacitor to produce a\n* high frequency AC current (1 kHz or more - based on the pulse length, but the\n* pulses come at intervals). Alternating the direction of the current in these\n* short pulses prevents the problems mentioned above.\n*\n* Then to get the resistance it is not possible to measure the voltage over the\n* EC probe (the normal way of measuring electrical resistance) as this drops with\n* the capacitor discharging. Instead we measure the time it takes for the cap to\n* discharge enough for the voltage on the pin to drop so much, that the input\n* flips from High to Low state. This time taken is a direct measure of the\n* resistance encountered (the cap and the EC probe form an RC circuit) in the\n* system, and that's what we need to know.\n*\n* Now the working of this technique.\n* Stage 1: charge the cap full through pin CapPos.\n* Stage 2: let the cap drain through the EC probe, measure the time it takes from\n* flipping the pins until CapPos drops LOW.\n* Stage 3: charge the cap full with opposite charge.\n* Stage 4: let the cap drain through the EC probe, for the same period of time as\n* was measured in Stage 2 (as compensation).\n* Cap is a small capacitor, in this system we use 47 nF but with other probes a\n* larger or smaller value can be required (the original research this is based\n* upon used a 3.3 nF cap). The 330R resistor is there to protect pin CapPos and\n* CapNeg from being overloaded when the cap is charged up, the 180R resistor\n* protects ECpin from too high currents caused by very high EC or shorting the\n* probe.\n*\n* Pins set to input are assumed to have infinite impedance, leaking is not taken into\n* account. The specs of NodeMCU give some 66 MOhm for impedance, several orders of\n* magnitude above the typical 1-100 kOhm resistance encountered by the EC probe.\n*\n* This function uses delay() in various forms, no yield() or so as it's meant to\n* be a real time measurement. Yield()ing to the task scheduler is a bad idea.\n* With the measurement taking only just over 0.1 seconds this should not be an\n* issue.\n*\n* Original research this is based upon:\n* https://hal.inria.fr/file/index/docid/635652/filename/TDS_Logger_RJP2011.pdf\n*\n*/\n\nvoid getEC() {\n\nint samples = 100; // number of EC samples to take and average.\nunsigned long startTime; // the time stamp (in microseconds) the measurement starts.\nunsigned long endTime; // the time stamp (in microseconds) the measurement is finished.\nunsigned int dischargeTime; // the time it took for the capacitor to discharge.\nunsigned int chargeDelay = 100; // The time (in microseconds) given to the cap to fully charge/discharge - at least 5x RC.\nunsigned int timeout = 2; // discharge timeout in milliseconds - if not triggered within this time, the EC probe is probably not there.\n\ndelay(1);\n\nfor (int j=0; j<NPROBES; j++) {\ninterruptPin = capPos[j];\nint pinECpin = ECpin[j];\nint pincapPos = capPos[j];\nint pincapNeg = capNeg[j];\nAverage<unsigned int> discharge(samples); // The sampling results.\nif (LOGGING) writeLog(\"Probing EC probe \" + String(j+1) + \" on capPos pin \" + String (pincapPos) + \", capNeg pin \" + String (pincapNeg) + \", ECpin \" + String (pinECpin) + \".\");\nfor (int i=0; i<samples; i++) { // take <samples> measurements of the EC.\n\n// Stage 1: fully charge capacitor for positive cycle.\n// CapPos output high, CapNeg output low, ECpin input.\npinMode (pinECpin, INPUT);\npinMode (pincapPos,OUTPUT);\npinMode (pincapNeg, OUTPUT);\nWRITE_PERI_REG (GPIO_SET_OUTPUT_HIGH, (1<<pincapPos));\nWRITE_PERI_REG (GPIO_SET_OUTPUT_LOW, (1<<pincapNeg));\ndelayMicroseconds(chargeDelay); // allow the cap to charge fully.\nyield();\n\n// Stage 2: positive side discharge; measure time it takes.\n// CapPos input, CapNeg output low, ECpin output low.\nendCycle = 0;\nstartTime = millis();\npinMode (pincapPos,INPUT);\nattachInterrupt(digitalPinToInterrupt(interruptPin), capDischarged, FALLING);\nWRITE_PERI_REG (GPIO_SET_OUTPUT_LOW, (1<<pinECpin));\npinMode (pinECpin, OUTPUT);\n\n// Use cycle counts and an interrupt to get a much more precise time measurement, especially for high-EC situations.\nstartCycle = ESP.getCycleCount();\nwhile (endCycle == 0) {\nif (millis() > (startTime + timeout)) break;\nyield();\n}\ndetachInterrupt(digitalPinToInterrupt(pincapPos));\nif (endCycle == 0) dischargeTime = 0;\nelse {\n\n// Handle potential overflow of micros() just as we measure, this happens about every 54 seconds\n// on a 80-MHz board.\nif (endCycle < startCycle) dischargeTime = (4294967295 - startCycle + endCycle) * CYCLETIME;\nelse dischargeTime = (endCycle - startCycle) * CYCLETIME;\ndischarge.push(dischargeTime);\nif (LOGGING) writeLog (\"sampled dischargeTime: \" + String(dischargeTime));\n}\n\nyield();\n\n// Stage 3: fully charge capacitor for negative cycle. CapPos output low, CapNeg output high, ECpin input.\nWRITE_PERI_REG (GPIO_SET_OUTPUT_HIGH, (1<<pincapNeg));\nWRITE_PERI_REG (GPIO_SET_OUTPUT_LOW, (1<<pincapPos));\npinMode (pinECpin, INPUT);\npinMode (pincapPos,OUTPUT);\npinMode (pincapNeg, OUTPUT);\ndelayMicroseconds(chargeDelay);\nyield();\n\n// Stage 4: negative side charge; don't measure as we just want to balance it the directions.\n// CapPos input, CapNeg high, ECpin high.\nWRITE_PERI_REG (GPIO_SET_OUTPUT_HIGH, (1<<pinECpin));\npinMode (pincapPos,INPUT);\npinMode (pinECpin, OUTPUT);\ndelayMicroseconds(dischargeTime/1000);\nyield();\n}\n\n// Stop any charge from flowing while we're not measuring by setting all ports to OUTPUT, LOW.\n// This will of course also completely discharge the capacitor.\npinMode (pincapPos, OUTPUT);\ndigitalWrite (pincapPos, LOW);\npinMode (pincapNeg, OUTPUT);\ndigitalWrite (pincapNeg, LOW);\npinMode (pinECpin, OUTPUT);\ndigitalWrite (pinECpin, LOW);\nyield();\nfloat dischargeAverage = discharge.mean();\nif(LOGGING) writeLog(\"Discharge time probe \" + String(j) + \": \" + String(dischargeAverage) + \" ns.\");\n\n/**\n* Calculate EC from the discharge time.\n*\n* Discharge time is directly related to R x C.\n* Here we have a discharge time, as we have a fixed capacitor this discharge time is linearly? related\n* to the resistance we try to measure.\n* Now we don't care about the actual resistance value - what we care about is the EC and TDS values. The\n* EC is the reciprocal of the resistance, the TDS is a function of EC. The capacitor is fixed, so the actual\n* value is irrelevant(!) for the calculation as this is represented in the calibration factor.\n*\n* As ion activity changes drastically with the temperature of the liquid, we have to correct for that. The\n* temperature correction is a simple \"linear correction\", typical value for this ALPHA factor is 2%/degC.\n*\n* https://www.analyticexpert.com/2011/03/temperature-compensation-algorithms-for-conductivity/\n*\n* Port D8 of NodeMCU leaks charge, this has to be accounted for using the INF time factor. This port should be\n* avoided later.\n*/\n#ifdef USE_NTC\n\n// Calculate corrected time.\nif (dischargeAverage > 0) {\nfloat t = dischargeAverage - ZERO[j];\nif (INF[j] > 0) t /= 1-dischargeAverage/INF[j];\nEC[j] = ECSLOPE[j] / (dischargeAverage * (1 + ALPHA * (watertemp - 25))) + ECOFFSET[j];\n}\nelse {\nEC[j] = -1;\n}\n#else\nEC[j] = dischargeAverage;\n#endif\n}\n}\n\n// Upon interrupt: register the cycle count of when the cap has discharged.\nvoid capDischarged() {\nendCycle = ESP.getCycleCount();\ndetachInterrupt(digitalPinToInterrupt(interruptPin));\n}\n\n• I don't understand your graph, could you put it in terms of Ohms? EC means nothing to me (always label units on a graph, always, always, always). – Voltage Spike Apr 27 '17 at 16:19\n• Treating two pieces of metal in water as a resistor may not be a correct assumption. Water may not be ohmic, and there could be chemistry going on at each electrode as well. You might consider asking a related, but more fundamental question in Chemistry SE as well. But there, just outline the measurement in one or two sentences, and link to this question for a detailed explanation. Focus on \"can I assume the electrode-water-electrode cell to behave in a purely ohmic way?\" - don't forget to describe the composition of the metal and what chemicals in the water cause conductivity variation. – uhoh Apr 27 '17 at 17:03\n• EC = conductivity = reciprocal of resistance. – Wouter Apr 27 '17 at 17:18\n• Water resistance is indeed not purely ohmic - that's where the AC comes in play, this is to compensate for that. Frequencies of >1 kHz are needed, preferably >3 kHz, then it's ohmic. Those frequencies are high enough to prevent the ions to do unwanted things like electrolyses and moving around. – Wouter Apr 27 '17 at 17:20\n• Added link to research paper on which my system is based. – Wouter Apr 27 '17 at 17:32\n\nMy guess is the $Rds_{ on }$ for the internal mosfets are not matched which is the case for most GPIO's, they have wide tolerances on the GPIO circuity and since the ESP8266 doesn't have any documentation or testing (its a Chinese product and 95% of them have poor documentation and testing).\n\nI'm assuming your using the GPIO for a low side switch, if you are, I would buy some mosfets and use them as a low side switch in the place of the GPIO ports so you know what your $Rds_{ on }$ is going to be.\n\nEither way I know that the conductivity of a liquid is going to be from 1 Ohm to a milli-Ohm, and if you put a mosfet in series with that that has 10's of milli-Ohms of resistance (and almost certainly varying from port to port). Since this is an unknown, why not buy some cheap mosfets and make it a known source of error.\n\nSo try this:\nMake Vin go to your gpio, get some mosfets with the $Rds_{ on }$ that you require and make sure the voltage of the GPIO (probably 3.3) will turn them on. Your load would be the connection to the water. You may need a dual stage with your 12V to get mosfets with a low enough $Rds_{ on }$ for your application.\n\nI would also look at the tolerances on your resistors, 330Ω at 1% would be 3.3Ω. So if your plugging in 330Ω in your software, the actual resistor could be anywhere from 333.3Ω to 326.7Ω. Same thing with the capacitors as some ceramics have 10% tolerances.",
null,
"• Not sure how your answer is relevant to my question. One thing to note is that tolerances appear to be reasonably small, as is evidenced by two different ESP-12 modules giving nearly the same overall results, while there are big differences between the pins of each individual board. The resistances I'm dealing with on the EC side are generally in the kOhm range. – Wouter Apr 27 '17 at 17:25\n• The actual value of the 330 Ohm resistors is irrelevant as they're there just to protect the ports from over current when charging the capacitor; and even the actual value of the capacitor is irrelevant (it's just important to set the range of measurement) as that's compensated for by calibrating the probe with known liquids. I'm not using any MOSFET or so here; all the parts used are in the schematic as posted. Finally the 12V has been replaced with 5V since to prevent the 3.3V regulator from overheating. The ESP8266 uses 3.3V. – Wouter Apr 27 '17 at 17:28"
] | [
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"https://i.stack.imgur.com/50Noc.jpg",
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"https://i.stack.imgur.com/iTlRL.png",
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"https://i.stack.imgur.com/YmAKD.png",
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"https://i.stack.imgur.com/Xv3Jl.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.802548,"math_prob":0.94576967,"size":12157,"snap":"2020-10-2020-16","text_gpt3_token_len":3114,"char_repetition_ratio":0.13173702,"word_repetition_ratio":0.029233871,"special_character_ratio":0.2715308,"punctuation_ratio":0.14640884,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9692901,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,5,null,5,null,5,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-28T09:17:47Z\",\"WARC-Record-ID\":\"<urn:uuid:87673a83-925c-436d-aa42-dcfea19d94ac>\",\"Content-Length\":\"164096\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d8d7849-439e-4119-bb8a-960a2dd05b63>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe579e96-df30-492a-9ef2-c130f60dbd1f>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://electronics.stackexchange.com/questions/302096/water-conductivity-measurements-esp8266-getting-strange-results\",\"WARC-Payload-Digest\":\"sha1:45DPZ4B2CQ3SYWS65UW4V2K7ACXM3QZQ\",\"WARC-Block-Digest\":\"sha1:52GHBWJ5BT6Z5CPXHIA72U5IY654TYKL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875147116.85_warc_CC-MAIN-20200228073640-20200228103640-00066.warc.gz\"}"} |
http://quadibloc.com/crypto/co040802.htm | [
"[Next] [Up] [Previous] [Index]\n\nOther Stream Ciphers\n\nThe mixed congruential pseudorandom number generator, usually used in software, is one of the basic techniques of producing apparently random bits that we will examine here. This is the technique used to produce the numbers given by the RND() function in most dialects of BASIC. Modulo a constant, replace x by a times x plus b, where a and b are both constants. If a and b are large enough, the behavior of x, particularly its most significant bits, will seem random.\n\nFor the maximum period, which is the same as the modulus, for the mixed congruential generator which is:\n\nx' = ax + b (modulo m)\n\nthe conditions are:\n\n• b must be relatively prime to the modulus\n• a must be equal to 1 modulo every prime factor of the modulus\n• a must be equal to 1 modulo 4 if 4 divides the modulus\n\nas given in Seminumerical Algorithms(Knuth) and Random Number Generators(Janssen). Thus, for a modulus that is a power of 2, a must be equal to 1 modulo 4; for a modulus that is a power of 10, a must be equal to 1 modulo 20.\n\nThe most common method used for strengthening a mixed congruential generator is to use it as part of a MacLaren-Marsaglia random number generator.\n\nLet us suppose that random binary bits are desired. Then, one uses one generator modulo two to some power, so that one is starting with numbers with a uniform distribution. Since the most significant bits of the output from such a generator have the longest period, one might take only the 4 or 8 most significant bits of the output.\n\nA buffer, perhaps with 37 entries, containing 37 bytes or nibbles produced by that generator is used. Each time some bits are to be produced, a second mixed-congruential generator, operating modulo 37^n for some n, is used to pick one element from that buffer, which is used as the output of the full MacLaren-Marsaglia generator. Then the other mixed-congruential generator is used to supply a replacement value for the buffer element used.\n\nAgain, a simple MacLaren-Marsaglia generator is still not secure, although the paper in which one was cracked used one where all the bits of the binary MC generator were used and none were discarded. If only the first few bits are used, and a long binary MC generator, perhaps one requiring multi-precision arithmetic, is used, there is already a greater level of security present. But more elaborate constructs are again possible.\n\nBut there are many other techniques that can be applied to bytes or words, rather than bits, to produce a keystream to XOR with plaintext.\n\nGifford's cipher used only eight bytes of internal state, but produced a cipher that was only shown to have weaknesses after some very involved analysis.\n\nIt actually was a kind of shift-register cipher, but with the shifting being done by byte. The shift register was eight bytes long. Feedback involved taking three bytes from the register, and obtaining the new byte by XORing together one of the bytes, the arithmetic right shift of another byte, and the logical left shift of the third.\n\nThe output from the generator is produced by taking four bytes from the register, forming two 16-bit integers from them, and taking the second least-significant byte of their product. This output is what is XORed to the plaintext to produce ciphertext.\n\nThis diagram illustrates Gifford's cipher:",
null,
"As it is based on a nonlinear shift register, there is no way to ensure maximum period, but one problem with it is shrinkage of the state space, because a bit is discarded from the oldest shift register entry that is reused. That is easily corrected: for example, by a design like this:",
null,
"A stream cipher is any cipher which, like Vigenere, or that produced by a rotor machine, changes how it behaves during a message. Thus, most block cipher modes, other than Electronic Codebook Mode, produce stream ciphers.\n\nA stream cipher which does produce pseudorandom bits to XOR with plaintext can be improved merely by substituting new values for the bytes of the plaintext from a secret table, both before and after the XOR.\n\nAnother way of using the output of a pseudorandom bit generator was developed by Terry Ritter, which he called Dynamic Substitution.\n\nThe principle is very simple. A secret table, with a random sequence of the 256 possible byte values, is used.\n\nA message byte is replaced by its substitute in that table in order to encrypt it.\n\nThen, a byte from the pseudorandom bit generator is taken. The two table entries corresponding to that byte, and the plaintext message byte, are swapped. In the event both the plaintext byte and the psudorandom byte are the same, nothing is done.\n\nThis is a simple, but secure technique. Every time a table entry is used, it is relocated somewhere else at random. So, since each table entry is used once and once only, no useful information about the table seems to be made available.\n\nIf one knows some corresponding plaintext and ciphertext, it is true that since you know that the table entry you encountered when one byte was enciphered was sent to the byte the PRNG sent it to at that time, and may stay there for a while, if that same byte turns up shortly after, you can conclude that the PRNG byte in the past is the same as the plaintext byte when the byte value turned up again.\n\nHowever, one cannot expect a simple method of applying a keystream to plaintext to be perfect; this small weakness doesn't contradict the fact that this is a great improvement over simply XORing the keystream to the plaintext.\n\nThe main reason this technique may not become popular even after its patent expires is because it is an autokey method; the encipherment of plaintext bytes depends in part on the values of previous plaintext bytes. This is not good for error-propagation, which need not be a consideration (since once text is encrypted, it can be sent along with extensive error-correction; and encrypted texts are often compressed, which already results in wide propagation of any errors) but it is usually considered to be a problem.\n\nThe idea of shuffling elements in a table of the 256 different byte values can also be used to generate pseudorandom bytes.\n\nOne very popular stream cipher has been alleged to function as follows:\n\nUsing two variables that store one byte each, in addition to the table, generate bytes as follows:\n\nStart with A and B equal to zero.\n\nEach iteration proceeds in this way:\n\n• Increment A (modulo 256).\n• Add the A-th element of the table to B (modulo 256).\n• Use as the output byte the element of the table specified by the modulo-256 sum of the A-th element and the B-th element of the table.\n• Exchange the A-th element and the B-th element of the table with each other.\n\nThe initial arrangement of the 256-byte table is created by a procedure involving a second 256-byte table. The table to be used in generating pseudorandom bytes is initialized to the numbers from 0 to 255 in order. The other table is filled with the bytes of the key repeated over and over until it is full.\n\nStart again with A and B equal to zero.\n\nRepeat the following steps 256 times:\n\n• Replace B with the sum of B and the A-th element of both tables.\n• Increment A.\n• Swap the A-th element and the B-th element of the table to be used later, leaving the one containing the key alone.\n\nIn addition to the mixed congruential generator, where\n\nx(n+1)=a*x(n)+b\n\nand, as noted, modulo 2^n, the condition for maximum period is that b must be odd, and a must be equal to 1 modulo 4, it is possible to use a quadratic congruential generator, of the form:\n\nx(n+1)=a*x(n)^2+b*x(n)+c\n\nsince this is not linear, it is harder to crack, and in fact one stream cipher bit source, using squaring as the function to move from one state to the next, is believed to be highly secure (the Blum-Blum-Shub cipher) - but this cipher uses a very large modulus, and therefore arithmetic on very large numbers, in the same fashion as RSA.\n\nI had been informed, by one Vladimir Anashin, of the Russian State University for the Humanities, that the condition for maximum period for a quadratic congruential generator modulo 2^n is that the coefficients be the same, modulo 8, as a generator that gives the full period of 8 modulo 8. A computer search for satisfactory coefficients gave me the following 24 possible sets to choose from:\n\n1) 2 3 1\n0 1 6 3 4 5 2 7\n\n2) 2 3 3\n0 3 6 5 4 7 2 1\n\n3) 2 3 5\n0 5 6 7 4 1 2 3\n\n4) 2 3 7\n0 7 6 1 4 3 2 5\n\n5) 2 7 1\n0 1 2 7 4 5 6 3\n\n6) 2 7 3\n0 3 2 1 4 7 6 5\n\n7) 2 7 5\n0 5 2 3 4 1 6 7\n\n8) 2 7 7\n0 7 2 5 4 3 6 1\n\n9) 4 1 1\n0 1 6 7 4 5 2 3\n\n10) 4 1 3\n0 3 2 5 4 7 6 1\n\n11) 4 1 5\n0 5 6 3 4 1 2 7\n\n12) 4 1 7\n0 7 2 1 4 3 6 5\n\n13) 4 5 1\n0 1 2 3 4 5 6 7\n\n14) 4 5 3\n0 3 6 1 4 7 2 5\n\n15) 4 5 5\n0 5 2 7 4 1 6 3\n\n16) 4 5 7\n0 7 6 5 4 3 2 1\n\n17) 6 3 1\n0 1 2 7 4 5 6 3\n\n18) 6 3 3\n0 3 2 1 4 7 6 5\n\n19) 6 3 5\n0 5 2 3 4 1 6 7\n\n20) 6 3 7\n0 7 2 5 4 3 6 1\n\n21) 6 7 1\n0 1 6 3 4 5 2 7\n\n22) 6 7 3\n0 3 6 5 4 7 2 1\n\n23) 6 7 5\n0 5 6 7 4 1 2 3\n\n24) 6 7 7\n0 7 6 1 4 3 2 5\n\nThe conditions for the quadratic congruential generator to have maximum period for any base are given in Seminumerical Algorithms, the second volume of Donald E. Knuth's The Art of Computer Programming in an exercise.\n\nWhere the recurrence is:\n\n2\nx' = a x + b x + c (modulo m)\n\nthe conditions on a, b, and c required for the maximum period (which matches the modulus) are:\n\n• c must be relatively prime to the modulus,\n• b is equal to 1 modulo every odd prime factor of the modulus;\n• a is equal to 0 modulo every odd prime factor of the modulus;\n• if the modulus is divisible by 4,\n• b is equal to 1 modulo 2, and\n• a is equal to b-1 modulo 4;\n• if the modulus is divisible by 2, either a is even and b is odd, which is the only possibility if the modulus is divisible also by 4, or a is odd and b is even;\n• if the modulus is divisible by 9, either a is equal to 0 modulo 9, or\n• b is equal to 1 modulo 9, and\n• a times c is equal to 6 modulo 9.\n\nAnother important pseudorandom number generator is the Mersenne Twister, which is described on the preceding page as it is a form of linear-feedback shift register.\n\nOne other stream cipher of interest is given its own section.\n\n[Next] [Up] [Previous] [Index]"
] | [
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"http://quadibloc.com/crypto/images/giff.gif",
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"http://quadibloc.com/crypto/images/giffc.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90645593,"math_prob":0.9594969,"size":8870,"snap":"2019-43-2019-47","text_gpt3_token_len":2396,"char_repetition_ratio":0.121926464,"word_repetition_ratio":0.06355685,"special_character_ratio":0.26189402,"punctuation_ratio":0.083769634,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9788354,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T06:57:41Z\",\"WARC-Record-ID\":\"<urn:uuid:278bec7f-c239-4c07-8103-8be35a3ed7b1>\",\"Content-Length\":\"12073\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b218d9b6-2529-4782-9516-997df403cb55>\",\"WARC-Concurrent-To\":\"<urn:uuid:df42b427-346d-47a0-9183-cdb56303ae23>\",\"WARC-IP-Address\":\"216.194.64.152\",\"WARC-Target-URI\":\"http://quadibloc.com/crypto/co040802.htm\",\"WARC-Payload-Digest\":\"sha1:FHPHAX47HHCCEWRXPZEQWQ5MWDASR2Q7\",\"WARC-Block-Digest\":\"sha1:FCSA6RMY2C6IWV6HZ4VIB6URGCZSXL4R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987803441.95_warc_CC-MAIN-20191022053647-20191022081147-00398.warc.gz\"}"} |
https://hackage.haskell.org/package/aivika-transformers-4.5.1/candidate/docs/Simulation-Aivika-Trans-Stream-Random.html | [
"aivika-transformers-4.5.1: Transformers for the Aivika simulation library\n\nSimulation.Aivika.Trans.Stream.Random\n\nContents\n\nDescription\n\nTested with: GHC 8.0.1\n\nThis module defines random streams of events, which are useful for describing the input of the model.\n\nSynopsis\n\n# Stream of Random Events\n\nArguments\n\n :: MonadDES m => Parameter m (Double, a) compute a pair of the delay and event of type a -> Stream m (Arrival a) a stream of delayed events\n\nReturn a sream of random events that arrive with the specified delay.\n\nArguments\n\n :: MonadDES m => Double the minimum delay -> Double the maximum delay -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nCreate a new stream with delays distributed uniformly.\n\nArguments\n\n :: MonadDES m => Int the minimum delay -> Int the maximum delay -> Stream m (Arrival Int) the stream of random events with the delays generated\n\nCreate a new stream with integer delays distributed uniformly.\n\nArguments\n\n :: MonadDES m => Double the minimum delay -> Double the median of the delay -> Double the maximum delay -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nCreate a new stream with random delays having the triangular distribution.\n\nArguments\n\n :: MonadDES m => Double the mean delay -> Double the delay deviation -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nCreate a new stream with delays distributed normally.\n\nArguments\n\n :: MonadDES m => Double the mean of a normal distribution which this distribution is derived from -> Double the deviation of a normal distribution which this distribution is derived from -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nCreate a new stream with random delays having the lognormal distribution.\n\nArguments\n\n :: MonadDES m => Double the mean delay (the reciprocal of the rate) -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nReturn a new stream with delays distibuted exponentially with the specified mean (the reciprocal of the rate).\n\nArguments\n\n :: MonadDES m => Double the scale (the reciprocal of the rate) -> Int the shape -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nReturn a new stream with delays having the Erlang distribution with the specified scale (the reciprocal of the rate) and shape parameters.\n\nArguments\n\n :: MonadDES m => Double the mean delay -> Stream m (Arrival Int) the stream of random events with the delays generated\n\nReturn a new stream with delays having the Poisson distribution with the specified mean.\n\nArguments\n\n :: MonadDES m => Double the probability -> Int the number of trials -> Stream m (Arrival Int) the stream of random events with the delays generated\n\nReturn a new stream with delays having the binomial distribution with the specified probability and trials.\n\nArguments\n\n :: MonadDES m => Double the shape -> Double the scale (a reciprocal of the rate) -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nReturn a new stream with random delays having the Gamma distribution by the specified shape and scale.\n\nArguments\n\n :: MonadDES m => Double the shape (alpha) -> Double the shape (beta) -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nReturn a new stream with random delays having the Beta distribution by the specified shape parameters (alpha and beta).\n\nArguments\n\n :: MonadDES m => Double shape -> Double scale -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nReturn a new stream with random delays having the Weibull distribution by the specified shape and scale.\n\nArguments\n\n :: MonadDES m => DiscretePDF Double the discrete probability density function -> Stream m (Arrival Double) the stream of random events with the delays generated\n\nReturn a new stream with random delays having the specified discrete distribution."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72730464,"math_prob":0.846097,"size":5184,"snap":"2021-31-2021-39","text_gpt3_token_len":1194,"char_repetition_ratio":0.24208494,"word_repetition_ratio":0.52920145,"special_character_ratio":0.24363425,"punctuation_ratio":0.10012361,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95397216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-29T00:37:00Z\",\"WARC-Record-ID\":\"<urn:uuid:a17b247a-c84d-4a0a-8158-68d7eb8d4d7d>\",\"Content-Length\":\"27019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:469403a3-3531-46b8-b9d7-75028de39cfc>\",\"WARC-Concurrent-To\":\"<urn:uuid:377e49ab-2df6-4a1f-8ced-22d993c71bf0>\",\"WARC-IP-Address\":\"151.101.200.68\",\"WARC-Target-URI\":\"https://hackage.haskell.org/package/aivika-transformers-4.5.1/candidate/docs/Simulation-Aivika-Trans-Stream-Random.html\",\"WARC-Payload-Digest\":\"sha1:W7K3GUPSXZ37QGHRQBO4LBILOD4QKJV4\",\"WARC-Block-Digest\":\"sha1:UF7YZJZCVORN6S6LLIYVF4ZPCVGIPXOV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153803.69_warc_CC-MAIN-20210728220634-20210729010634-00182.warc.gz\"}"} |
https://algnotes.info/on/obliv/greedy/set-multicover-unconstrained/smu-reduction/ | [
"# Reduction to Set Cover\n\nUnconstrained Set Multicover reduces to Set Cover.\n\nUnconstrained Set Multicover has an approximation-preserving reduction to Set Cover that also preserves maximum set size. This reduction (and previous results on Set Cover) imply that a localized rounding scheme and a greedy algorithm both give -approximation.\n\n# The reduction\n\nHere is the reduction. Fix any instance of Unconstrained Set Multicover. Replace each element with duplicates of . Replace each set with copies of , where each copy contains exactly one of the duplicates of each element of . This gives an instance of Set Cover. The size of each copy of is equal to the size of .\n\nLemma (reduction to Set Cover).\n\nFor every unconstrained Set Multicover instance , the above Set Cover instance has the same maximum set size and is equivalent, in that the solutions to correspond to the solutions to , and the correspondence preserves feasibility, cost, and integrality.\n\n# Rounding scheme and greedy algorithm via reduction\n\nThe corollaries below follow from the reduction:\n\nCorollary.\n\nThe localized rounding scheme for unconstrained Set Multicover returns a cover of expected cost at most , which is at most .\n\nConsider the following greedy algorithm for unconstrained Set Multicover: Start with each . At each iteration, choose minimizing the ratio of to the number of not-yet-satisfied elements in , then increment .\n\nCorollary.\n\nThat greedy algorithm returns a cover of cost at most , which is at most .\n\n# Questions for later\n\nGeneralize to constrained Set Multicover (where the solution must take each set at most once). (The Knapsack-cover inequalities give a reduction, but it increases .) Aim for -approximation (matching Wolsey)."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8204705,"math_prob":0.95625126,"size":2025,"snap":"2021-04-2021-17","text_gpt3_token_len":524,"char_repetition_ratio":0.12716477,"word_repetition_ratio":0.08308605,"special_character_ratio":0.2474074,"punctuation_ratio":0.08951407,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97845453,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T10:36:18Z\",\"WARC-Record-ID\":\"<urn:uuid:ba5b93e5-6f52-41c3-acbb-ea62e6903125>\",\"Content-Length\":\"64817\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f754dd46-0433-4542-bfdb-c7fbdbdc733c>\",\"WARC-Concurrent-To\":\"<urn:uuid:9fa7a10d-0123-451c-9240-4eb551772dfa>\",\"WARC-IP-Address\":\"3.231.125.220\",\"WARC-Target-URI\":\"https://algnotes.info/on/obliv/greedy/set-multicover-unconstrained/smu-reduction/\",\"WARC-Payload-Digest\":\"sha1:RO77GPDY3ZKZ6KP3GFVBBGMUQD5WVXCZ\",\"WARC-Block-Digest\":\"sha1:6UF6LJX7U6CHV3MB7DAMWMUWLKGB4L7S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703524743.61_warc_CC-MAIN-20210121101406-20210121131406-00614.warc.gz\"}"} |
https://se.mathworks.com/matlabcentral/cody/problems/42858 | [
"Cody\n\n# Problem 42858. Block average ignoring NaN values\n\nGiven a matrix, calculate the block average of each disjoint sub-matrix while ignoring NaN values. Assume that the size of the matrix along each dimension is an integer multiple of the size of the sub-matrix along the same dimension.\n\n• Input: matrix A and the size of each sub-matrix subsz\n• Output: B = blknanavg(A,subsz)\n\nExample:\n\n``` A = [1 2 3 4 5 6 7 8 NaN];\nsubsz = [1 3];\nB = [2 5 (7+8)/2];```\n\nHint: this is related to Problem 42856. Block average.\n\n### Solution Stats\n\n48.33% Correct | 51.67% Incorrect\nLast Solution submitted on Mar 28, 2020"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72617334,"math_prob":0.99675715,"size":499,"snap":"2020-10-2020-16","text_gpt3_token_len":148,"char_repetition_ratio":0.15151516,"word_repetition_ratio":0.0,"special_character_ratio":0.33066133,"punctuation_ratio":0.12037037,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9974103,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-30T02:11:12Z\",\"WARC-Record-ID\":\"<urn:uuid:82f28f55-cabb-4c20-8a8d-ce539eaff495>\",\"Content-Length\":\"90721\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce499e2d-5646-439c-914d-00584eb85ebb>\",\"WARC-Concurrent-To\":\"<urn:uuid:f82a45ef-367b-4bde-ae26-b63b6e195624>\",\"WARC-IP-Address\":\"23.50.112.17\",\"WARC-Target-URI\":\"https://se.mathworks.com/matlabcentral/cody/problems/42858\",\"WARC-Payload-Digest\":\"sha1:M36MWJRJRFKMXZKSMFEK4U6RJB7JVBEW\",\"WARC-Block-Digest\":\"sha1:PUD5OLFLFJPO7BVIVPJRE2HJAKFJXSGX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370496330.1_warc_CC-MAIN-20200329232328-20200330022328-00059.warc.gz\"}"} |
http://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/6/3/a/ | [
"# Properties\n\n Label 6.3.a Level 6 Weight 3 Character orbit a Rep. character $$\\chi_{6}(1,\\cdot)$$ Character field $$\\Q$$ Dimension 0 Newform subspaces 0 Sturm bound 3 Trace bound 0\n\n# Related objects\n\n Level: $$N$$ $$=$$ $$6 = 2 \\cdot 3$$ Weight: $$k$$ $$=$$ $$3$$ Character orbit: $$[\\chi]$$ $$=$$ 6.a (trivial) Character field: $$\\Q$$ Newform subspaces: $$0$$ Sturm bound: $$3$$ Trace bound: $$0$$"
] | [
null
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https://tutorbin.com/questions-and-answers/4-now-assume-that-the-true-relationship-is-text-income-_ibeta_0beta_1- | [
"Question\n\n# 4. Now assume that the true relationship is: \\text { Income }_{i}=\\beta_{0}+\\beta_{1} \\text { Education }_{i}+\\beta_{2} \\text { Ability }_{i}+\\mu_{i}(2) Plug equation (1) into equation (2) to get \\text { Income }_{i}=\\beta_{0}+\\beta_{1} \\text { Education }_{i}+\\beta_{2}\\left[\\delta_{0}+\\delta_{1} \\text { Education }_{i}+e_{i}\\right]+\\mu_{i}(3) The final step is to rearrange (3) so that the form looks like: \\text { Income }_{i}=a+b * \\text { Education }_{i}+c \\text { a. }[5 \\text { Points }] \\text { Report to me the value of } b \\text { in terms of } \\beta_{1}, \\beta_{2}, \\text { and } \\delta_{1} \\text { . } ^^20Income^^20_i=\\vec{\\beta}+\\bar{\\beta}^^20Education^^20_i+\\tilde{\\mu}_i \\text { b. [8 Points] Assume that } b=\\vec{\\beta}_{1} \\text { in the relationship } How does the correlation between ability and education, and the correlation between ability ^^20Note:^^20\\hat{\\delta_1}=\\frac{\\text{ Cov( } \\text{ Bducation Abulity })}{\\text{ Var } \\text{ (Rducation) }} \\text { and income affected the biasness of } \\bar{\\beta}_{1} \\text { ? When is } \\bar{\\beta}_{1}=\\beta_{1} \\text { true? }",
null,
"",
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"Fig: 1",
null,
"",
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"Fig: 2",
null,
"",
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"Fig: 3",
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"",
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"Fig: 4",
null,
"",
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"Fig: 5",
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"",
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"Fig: 6",
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"Fig: 7",
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"",
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"Fig: 8",
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"",
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"Fig: 9",
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"",
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"Fig: 10",
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"",
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"Fig: 11",
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"",
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"Fig: 12"
] | [
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"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
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"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
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"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
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null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
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null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27300%27/%3e",
null,
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null,
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null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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https://dsp.stackexchange.com/questions/tagged/derivation | [
"A message from our CEO about the future of Stack Overflow and Stack Exchange. Read now.\n\n# Questions tagged [derivation]\n\nThe tag has no usage guidance.\n\n19 questions\nFilter by\nSorted by\nTagged with\n56 views\n\n### I Q sampling and baseband version of analytic signal\n\nIs it correct to say that if we have a radio signal $s(t)$ centered around the angular frequency of $\\omega$ as $\\omega\\pm\\omega_B/2$ (where $\\omega_B$ is the bandwidth of the signal) and the ...\n25 views\n\n### Understanding the resulting image matrix when differentiating image\n\nLet $A$ be a image matrix and let $P(i,j)$ be the gray level of pixel $i,j$. Let $0$ be black and $255$ be white Assume I want to differentiate this image with respect to the columns $(x)$ as in I ...\n494 views\n\n### Derivation of ZOH Discretization\n\nI'm trying to understand the derivation of the zero order hold discretization method, and I have a couple of questions about some of the steps. I think I understand the first part, this is just the ...\n939 views\n\n### What will the impulse response of a matched filter look like if the input is complex?\n\nIf $h(t)$ is the impulse response of a filter matched to a signal $s(t)$, I read that $h(t) = ks(t_o - t)$. But what if the signal is complex? I went through the derivation of the matched filter and ...\n145 views\n\n### Bounds of the difference of a bounded band-limited function\n\nFor a continuous signal (function), we have Bernstein inequality : $$|{df(t)}/dt| \\le 2AB\\pi$$ where $A=\\sup|f(t)|$ and $B$ is the bandwidth of $f(t)$. The question is: is there a relationship ...\n174 views\n\n### Derivation of range migration algorithm\n\nProblem: In Walter G.Carrara's book on synthetic aperture radar, the equation is presented: $\\Phi(K_X, K_R) = -K_XX_t - R_B\\sqrt(K_R^2 - K_X^2) +K_RR_S$ (10.30) And this is said to come from ...\n196 views\n\n### Doubts on LMS derivation\n\nI have been trying to follow the Least Mean Square(LMS) algorithm derivation given by Wikipedia here and have the following questions. Here I expected $y(n)$ is to be computed by convolving $x(n)$ ..."
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http://www.astronomy.ohio-state.edu/~gaudi/AST161/Unit4/lecture19.html | [
"",
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"Astronomy 161 An Introduction to Solar System Astronomy Prof. Scott Gaudi\n\n# Lecture 19: Orbits\n\n## Key Ideas:\n\nNewton generalized Kepler's laws to apply to any two bodies orbiting each other\n• First Law: Orbits are conic sections with the center-of-mass of the two bodies at the focus.\n• Second Law: angular momentum conservation.\n• Generalized Third Law that depends on the masses of the two bodies.\nFirst Triumph of Newtonian Gravity\n• The prediction of the return of Halley's Comet\n\n## Kepler's Laws Revisited\n\nKepler's Laws of Planetary Motion are as follows:\n\nFirst Law:\nPlanets orbit on ellipses with the Sun at one focus.\n\nSecond Law:\nPlanet sweeps out equal areas in equal times.\n\nThird Law:\nPeriod squared is proportional to the size of the semi-major axis cubed.\nExpressed Mathematically as: P2=a3, for P in years and a in AUs.\n\n## Newton's Generalization\n\nNewton showed that Kepler's Laws can be derived from\n\n• The Three Laws of Motion\n• The Law of Universal Gravitation.\n\nFurther, Newton generalized the laws to apply to any two bodies moving under the influence of their mutual gravitation. For example, these laws apply equally to\n\n• The Moon orbiting the Earth.\n• A space probe orbiting the Moon.\n• Two stars orbiting each other.\nand so on...\n\n## First Law of Orbital Motion\n\nThe shape of an orbit is a conic section with the center of mass at one focus.\nThere are two parts to Newton's formulation of Kepler's First Law:\n\nShapes of Orbits are Conic Sections:\n\n• Curves found by cutting a cone with a plane.\n• Circles, Ellipses, Parabolas, and Hyperbolas\n\nThe Center of Mass is at the Focus:\n\n• Strictly speaking, it is not just the Earth orbitting the Sun. The Earth and Sun orbit each other about their mutual Center of Mass.\nBecause the Earth is so much smaller than the Sun, their mutual center of mass is inside the Sun, so the difference is not immediately apparent.\n\n## Conic Section Curves\n\nThese are curves formed by the intersection of a cone and a plane cutting it at various angles.\n\nConic curves come in two families:\n\nClosed Curves:\n\n• Ellipses\n• Circles, which are a special case of an ellipse with e=0\n• Orbits are bound and objects are trapped in orbit forever around the parent body.\n\nOpen Curves:\n\n• Hyperbolas\n• Parabolas, which are a special case of a hyperbola\n• Orbits are unbound and objects can escape the gravity of the parent body.\nWhich of these orbits you will be in is determined by your orbital speed. There are two speeds of particular interest...\n\n## Circular Velocity\n\nVelocity needed to sustain a circular orbit of a given radius, r, from a massive body, M:",
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"• If v<vC, the orbit is an ellipse smaller than the circular orbit.\n• Go a little faster than vC, the orbit is an ellipse larger than the circular orbit,\n• Go a lot faster, and…\n\n## Escape Velocity\n\nThis is the minimum velocity required to have a parabolic orbit starting at a given distance, r, from a massive body, M:",
null,
"At the Earth's surface:\n\n• vC = 7.9 km/sec (28,400 km/hr)\n• vE = 11.2 km/sec (40,300 km/hr)\nThus, which orbit you will be in starting from a given point (labeled P here) is determined by your speed at that point:\n• If the speed is the circular speed at P (VC above), the orbit will be a circle (red curve).\n\n• If the speed is less than the circular speed at P, the orbit will be an ellipse smaller than the circle (blue curve)\n\n• If the speed is larger than the circular speed at P but less than the escape speed, VE, the orbit will be an ellipse larger than the circle (green curve)\n\n• If the speed is the escape speed from P (VE above), the orbit will open into a Parabola (magenta curve).\n\n• If the speed is greater than the escape speed from P, the orbit will be a Hyperbola (black curve). The greater the speed, the \"flatter\" (more open) the Hyperbolic curve of the orbit\n\n## Center of Mass\n\nTwo objects orbit about their center of mass:\n\n• Balance point between the two masses\n• Semi-Major axis: a = a1 + a2\n• Relative positions: a2 / a1 = M1 / M2\nExample: Earth and Sun\nMsun = 2 x 1030 kg\nMearth = 6 x 1024 kg\n\nFrom the balance relation, the distances of the Sun and Earth from their mutual center of mass are related to the size of the semi-major axis of the Earth's orbit (a) and the ratio of the masses:\n\nasun + aearth = 1 AU = 1.5 x 108 km\nasun/aearth = Mearth/Msun = 3 x 10-6\n\nAfter some simple algebra (do it!), we find:\n\nasun = 450 km\n\nSince the radius of the Sun is 700,000 km, this means that the center-of-mass of the Earth-Sun system is deep inside the Sun.\n\n## Second Law of Orbital Motion\n\nOrbital motions conserve angular momentum.\n\nThis doesn't sound much like \"equal areas in equal times\", but in fact it is the same thing.\n\nAngular Momentum:\n\nL = mvr = constant\n\nWhere:\nm = mass,\nv = velocity,\nr = distance from the center of mass.\n\n## Angular Momentum & Equal Areas\n\nAngular Momentum is conserved, which means that L is a constant.\n\nIf the distance changes, the velocity must change to compensate so as to keep L constant:\n\nNear Perihelion:\n\n• Planet is closer to the Sun, so r is smaller.\n• The speed v must be proportionally faster to compensate.\n\nNear Aphelion:\n\n• Planet is farther to the Sun, so r is larger.\n• The speed v must be proportionally slower to compensate.\n\nA familiar example of the same principle at work is a figure skater doing a spin. In an \"upright spin\", the skater stands on one leg with arms outstretched and spins about an up/down axis. The spin is accelerated by the skater drawing in his/her arms. By drawing in his/her arms, they are moving mass closer to the center of their body, and conservation of angular momentum demands that they spin faster.\n\n## Third Law of Orbital Motion\n\nNewton's Generalization of Kepler's 3rd Law:",
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"Where:\nP = period of the orbit\na = semi-major axis of the orbit\nM1 = mass of the first body\nM2 = mass of the second body\n\n## A Third Law for Every Body\n\nThe proportionality between the square of the period and the cube of the semi-major axis now depends on the masses of the two bodies.\n\nFor planets orbiting the Sun, Msun is so much bigger than any planet (even Jupiter, at 1/1000th Msun), that we recover Kepler's version of the Third Law from Newton's more general form:",
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"Note that the constant of proportionality is the same for all planets (to a good approximation, certainly to within Kepler's data from Tycho).\n\nIn Kepler's version, the constant of proportionality works out to be 1.0 if we use units of years for P and AUs for the semi-major axis, a. While computationally convenient, it hides the underlying dependence on the mass of the Sun from sight. The problem with empirical laws, like Kepler's formulation, is that they only show the surface, not the important details underlying them.\n\n## Measuring Masses\n\nNewton's generalized form of Kepler's 3rd law gives us a way to measure masses from orbital motions!\n\nFor exampl, we can derive the mass of the Sun by using the period and size of the Earth's orbit:\n\nPearth = 1 year = 3.156 x 107 seconds\naearth = 1 AU = 1.496 x 1011 meters\n\nUsing Newton's Form of Kepler's 3rd law for the solar system above, we see that once we know P and a (G and pi are constants), the only unknown is the Mass of the Sun, which can be solved for easily after a little light algebra:",
null,
"(You can verify the numbers for yourself by using G=6.67 x 10-11 Newton m2/kg2, and the values of P and a for the Earth given above in seconds and meters. Do it!).\n\n## A Universal Method for Measuring Masses\n\nThe generalized form of Kepler's Third law gives us a powerful tool for measuring the masses of objects by measuring the periods and sizes of their orbits. For example:\n\nWe can measure the mass of Jupiter from the orbits of the Galilean moons, since MJupiter>>Mmoons\n\n• Find MJupiter = 300 Mearth\n\nWe can measure the total mass of the Earth and Moon system.\n\n• Earth is only ~81x more massive than the Moon, so you have to use the full formula.\n• Get the mass of the Earth independently (e.g., our falling bodies experiment from yesterday's lecture).\n\nWe can measure the masses of binary stars using the full formula and by observing their orbit parameters (you will see this done in Astronomy 162).\n\n## The Triumph of Newtonian Gravity\n\nNewton's description of planetary positions is only a start.\n\nIt also allows quantitative new predictions. An early application was to Halley's Comet:\n\n• Using Newtonian Gravity, Edmund Halley found that the orbit of the great comet of 1682 was similar to comets seen in 1607 and 1537.\n• Predicted it would return in 1758/59.\n• It did, dramatically confirming Newton's laws.\n...\n\n## The Why of Planetary Motions\n\nKepler's Laws are descriptions of the motion:\n\n• Arrived at by trial and error, and some vague notions about celestial harmonies\n• Only describe the motions, without explaining why they move that way.\n\nNewton provides the explanation:\n\n• Kepler's Laws are a natural consequence of Newton's Three Laws of Motion and his law of universal gravitation.\nBy adding the why, Newton gave his laws predictive power, and allows us to use them as tools to explore the Universe, both figuratively and literally. We can predict new phenomena or understand oddities in the motions (they laws give us a framework in which to interpret data), and we can literally use them to fly spacecraft through the solar system.\nReturn to [ Unit 4 Index | Astronomy 161 Main Page ]"
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"http://www.astronomy.ohio-state.edu/~gaudi/AST161/Images/skipnav.gif",
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"http://www.astronomy.ohio-state.edu/~gaudi/AST161/Images/EarthSun.jpg",
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"http://www.astronomy.ohio-state.edu/~gaudi/AST161/Unit4/Equations/Vcircular.gif",
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"http://www.astronomy.ohio-state.edu/~gaudi/AST161/Unit4/Equations/Vescape.gif",
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"http://www.astronomy.ohio-state.edu/~gaudi/AST161/Unit4/Equations/NewtKep3.gif",
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"http://www.astronomy.ohio-state.edu/~gaudi/AST161/Unit4/Equations/Msun.gif",
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https://la.mathworks.com/matlabcentral/cody/problems/106-weighted-average/solutions/666631 | [
"Cody\n\n# Problem 106. Weighted average\n\nSolution 666631\n\nSubmitted on 6 May 2015 by YUANYU YANG\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\n%% x = [1 2 3]; w = [10 15 20]; y_correct = 100/3; assert(isequal(weighted_average(x,w),y_correct))\n\n2 Pass\n%% x = [0 -2 3]; w = [10 0 10]; y_correct = 10; assert(isequal(weighted_average(x,w),y_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73828244,"math_prob":0.9716286,"size":613,"snap":"2021-04-2021-17","text_gpt3_token_len":180,"char_repetition_ratio":0.12643678,"word_repetition_ratio":0.0,"special_character_ratio":0.33768353,"punctuation_ratio":0.12295082,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97070646,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-16T15:31:22Z\",\"WARC-Record-ID\":\"<urn:uuid:766e3981-f058-426d-b469-27dba209e554>\",\"Content-Length\":\"77934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:364aec8e-5b27-4001-91a3-b499906a6195>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b9a3b73-6b0a-4cc6-9030-7a0cfc8b1711>\",\"WARC-IP-Address\":\"23.212.144.59\",\"WARC-Target-URI\":\"https://la.mathworks.com/matlabcentral/cody/problems/106-weighted-average/solutions/666631\",\"WARC-Payload-Digest\":\"sha1:RFOQ5AKTFITQDFIJBIG4TLHGWUQ6MCYR\",\"WARC-Block-Digest\":\"sha1:FRXLG7ODNVTSS3RCN4VO2F2MBBTXXG7N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703506697.14_warc_CC-MAIN-20210116135004-20210116165004-00262.warc.gz\"}"} |
https://essayslife.com/bmi-and-depression/ | [
"# BMI and depression\n\nThe project must be typewritten, double spaced and very limited in length (maximum 12 pages).\n\nPart I (25%)\n\nA NP researcher randomly sampled 100 women aged 50-65 years and measured their minutes of exercise in the past week, BMI, and depression. Depression was measured using a Likert type scale consisting of 20 items. The summation score ranged from 20 to 100 and the higher the score, the higher the level of depression. The Pearson correlation coefficients (r’s) are summarized in the following table. For the analyses, statistical significant level was set at α=0.05.\n\nTable 1: correlation among minutes of exercise, BMI and depression\n\nExercise in past week (minutes)\n\nBMI\n\nBMI\n\n-0.15\n\nDepression score\n\n-0.30*\n\n0.20\n\n*p < 0.05\n\n1. Write a research and null hypotheses regarding the relationship between exercise and depression.\n\n2. Based on the test statistics in table 1, what is your conclusion regarding your research hypothesis? (Hint: discuss both the magnitude and direction of the relationship).\n\n3. What proportion of variance is shared by minutes of exercise and depression among women 50-65 years of age?\n\n4. For the relationship between minutes of exercise and BMI,\n\na. what was the estimated power of the statistical test? (Using the power table on page 202, table 9.1, Polit 2010).\n\nb. What was the risk that a type II error was committed?\n\n5. If -0.20 is a good estimation of population correlation, what sample size would be needed to achieve power of 0.80 at a significance α=0.05?\n\nPART II. (25%)\n\nUsing the “N6208 Final Project Data”,\n\na). select two variables with nominal or ordinal level measurements, and perform the descriptive statistics (frequency and percentage). [Please select only dichotomous variables from the following list: poverty, smoker, PoorHealth].\n\nb). perform the bi-variate descriptive statistics using crosstabulation.\n\nc). Hand calculate the ARs, ARR, RR, and OR. Show all your calculations.\n\nd). Perform a chi-square analysis.\n\ne). Using APA format, write a full report with the following sections:\n\n1. Introduction: Describe your research question and hypothesis. Include the variables, measurement levels, the bivariate research question, and the hypothesis [for example, the event of adverse risk (using your variable name here, for instance, alcohol usage) will be higher/or lower in the risk exposed group (i.e., marijuana use) compare to the non-exposed group (non-users of marijuana)].\n\n2. Method: Include the sample description (sample size, eligibility criteria) and statistical methods used for data analysis. (The sample information can be found in “Polit Dataset Description” in SPSS Data Sets folder).\n\n3. Results: Include frequencies and percentages for the two variables, crosstabulation results, risk indexes (ARs, ARR, RR, and OR), and chi-square test results. Include a summary table for the results and write your interpretation. (Attach SPSS outputs).\n\n4. Discussion: Write a report including summary and interpretation of the findings reported in the previous sections relative to the research questions you posed in your introduction.\n\nPart III. (50%)\n\nRun a one-way ANOVA using the dataset “N6208 Final Project Data”. The Dataset contains 462 cases from the original PolitDatasetA. Two variables will be used for this analysis: Satisfaction and Houseproblem.\n\nThe variable Houseproblem is created using the variable housprob, a summary index of eight variables about current housing problems for the women in this sample—for example, whether or not they had their utilities cut off, had vermin in the household, had unreliable hear, and so forth. The variable housprob is a count of the total number of times the women said “yes” to these eight questions. The variable housprob is recoded into Houseproblem based on number of housing problems. The coding for Houseproblem is: 1=no housing problems, 2=one housing problem, and 3= two or more housing problems.\n\nSatisfaction measures the overall satisfaction with material sell-being. This variable is a summated rating scale variable for women’s responses to their degree of satisfaction with four aspects of their material sell-being—their housing, food, furniture, and clothing for themselves and their children. Each item was coded from 1 (very dissatisfied) to 4 (very satisfied), so the overall score for the four items could range from a low of 4 (4 X 1) to 16 (4 X 4). Higher score indicates greater satisfaction. This scale has an internal consistency Cronbach’s alpha of 0.90. The content validity and construct validity have been established in previous research.\n\nFor this analysis, use the variable Houseproblem as the independent (group) variable and variable Satisfaction as the outcome variable. To run the one-way ANOVA, click Analyze Compare Means Oneway. In the opening dialogue box, move Satisfaction into the Dependent List and Houseproblem into the slot for Factor. Click the Options pushbutton, and click Descriptives and Homogeneity of Variance, then continue. Next, click the Post Hoc pushbutton and select LSD. Click continue, then OK, and answer the following questions using compete sentences:\n\n1. What are the mean levels of satisfaction in the three groups? Report the mean, SD, minimum, maximum and sample size in a table.\n2. Write a research question.\n3. Write the research hypothesis (Ha) and the null hypothesis (Ho).\n4. What was the value of the F statistic and its p-value?\n5. Can the null hypothesis be rejected?\n6. What were the degrees of freedom?\n7. According to the LSD test, were any group means significantly different from any others? If yes, which ones?\n8. Write a paragraph summarizing all the results.\n9. Attach the relevant SPSS printouts."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90049964,"math_prob":0.8070137,"size":5689,"snap":"2021-04-2021-17","text_gpt3_token_len":1234,"char_repetition_ratio":0.11890941,"word_repetition_ratio":0.0,"special_character_ratio":0.21972227,"punctuation_ratio":0.1448598,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97416925,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-22T21:36:27Z\",\"WARC-Record-ID\":\"<urn:uuid:caac23df-3859-43ac-a7aa-0d88c38b1888>\",\"Content-Length\":\"42723\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd4a172c-fa5b-4730-943e-f1f23cb9e31e>\",\"WARC-Concurrent-To\":\"<urn:uuid:2a2fc9b4-79ed-433f-bf50-ae6717ea78bc>\",\"WARC-IP-Address\":\"198.187.29.220\",\"WARC-Target-URI\":\"https://essayslife.com/bmi-and-depression/\",\"WARC-Payload-Digest\":\"sha1:2SJJ64RE7M6TLPQVWBTJRMWI4C2YQ7HT\",\"WARC-Block-Digest\":\"sha1:H7S4T2BHA5DKV3PZDXGWBQFPQMLS2WNP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703531429.49_warc_CC-MAIN-20210122210653-20210123000653-00522.warc.gz\"}"} |
https://wizedu.com/questions/932598/an-investor-has-two-bonds-in-his-portfolio-that | [
"In: Finance\n\n# An investor has two bonds in his portfolio that have a face value of $1,000 and... An investor has two bonds in his portfolio that have a face value of$1,000 and pay a 8% annual coupon. Bond L matures in 10 years, while Bond S matures in 1 year.\n\nAssume that only one more interest payment is to be made on Bond S at its maturity and that 10 more payments are to be made on Bond L.\n\n1. What will the value of the Bond L be if the going interest rate is 6%? Round your answer to the nearest cent.\n$What will the value of the Bond S be if the going interest rate is 6%? Round your answer to the nearest cent.$\n\nWhat will the value of the Bond L be if the going interest rate is 8%? Round your answer to the nearest cent.\n$What will the value of the Bond S be if the going interest rate is 8%? Round your answer to the nearest cent.$\n\nWhat will the value of the Bond L be if the going interest rate is 14%? Round your answer to the nearest cent.\n$What will the value of the Bond S be if the going interest rate is 14%? Round your answer to the nearest cent.$\n\n## Solutions\n\n##### Expert Solution\n\nBond S\n\n K = N Bond Price =∑ [(Annual Coupon)/(1 + YTM)^k] + Par value/(1 + YTM)^N k=1 K =1 Bond Price =∑ [(8*1000/100)/(1 + 6/100)^k] + 1000/(1 + 6/100)^1 k=1 Bond Price = 1018.87\n\nBond L\n\n K = N Bond Price =∑ [(Annual Coupon)/(1 + YTM)^k] + Par value/(1 + YTM)^N k=1 K =10 Bond Price =∑ [(8*1000/100)/(1 + 6/100)^k] + 1000/(1 + 6/100)^10 k=1 Bond Price = 1147.2\n\nBond S\n\n K = N Bond Price =∑ [(Annual Coupon)/(1 + YTM)^k] + Par value/(1 + YTM)^N k=1 K =1 Bond Price =∑ [(8*1000/100)/(1 + 8/100)^k] + 1000/(1 + 8/100)^1 k=1 Bond Price = 1000\n\nBond L\n\n K = N Bond Price =∑ [(Annual Coupon)/(1 + YTM)^k] + Par value/(1 + YTM)^N k=1 K =10 Bond Price =∑ [(8*1000/100)/(1 + 8/100)^k] + 1000/(1 + 8/100)^10 k=1 Bond Price = 1000\n\nBond S\n\n K = N Bond Price =∑ [(Annual Coupon)/(1 + YTM)^k] + Par value/(1 + YTM)^N k=1 K =1 Bond Price =∑ [(8*1000/100)/(1 + 14/100)^k] + 1000/(1 + 14/100)^1 k=1 Bond Price = 947.37\n\nBond L\n\n K = N Bond Price =∑ [(Annual Coupon)/(1 + YTM)^k] + Par value/(1 + YTM)^N k=1 K =10 Bond Price =∑ [(8*1000/100)/(1 + 14/100)^k] + 1000/(1 + 14/100)^10 k=1 Bond Price = 687.03"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.67245346,"math_prob":0.9998747,"size":1427,"snap":"2023-40-2023-50","text_gpt3_token_len":661,"char_repetition_ratio":0.24736473,"word_repetition_ratio":0.57564574,"special_character_ratio":0.5543097,"punctuation_ratio":0.031847134,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T03:59:40Z\",\"WARC-Record-ID\":\"<urn:uuid:4f740743-ca07-4a93-afba-31334ec8b2ed>\",\"Content-Length\":\"42608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c12fb1b7-d57d-41cd-924a-28208e20dbf2>\",\"WARC-Concurrent-To\":\"<urn:uuid:dbed37e2-6908-4117-ba76-84a61a8582a4>\",\"WARC-IP-Address\":\"172.67.148.127\",\"WARC-Target-URI\":\"https://wizedu.com/questions/932598/an-investor-has-two-bonds-in-his-portfolio-that\",\"WARC-Payload-Digest\":\"sha1:GZIR565LZHHMOJGYWBYBCQCERJO6PWWL\",\"WARC-Block-Digest\":\"sha1:4LUHGHD67ISAEDYVQMCG6YNEY2NAQIYC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100523.4_warc_CC-MAIN-20231204020432-20231204050432-00846.warc.gz\"}"} |
https://dendropy.org/library/parsimony | [
"# dendropy.model.parsimony: The Parsimony Model¶\n\nModels, modeling and model-fitting of parsimony.\n\ndendropy.model.parsimony.fitch_down_pass(postorder_nodes, state_sets_attr_name='state_sets', taxon_state_sets_map=None, weights=None, score_by_character_list=None)[source]\n\nReturns the parsimony score given a list of nodes in postorder and associated states, using Fitch’s (1971) unordered parsimony algorithm.\n\nParameters\n• postorder_nodes (iterable of/over Node objects) – An iterable of Node objects in in order of post-order traversal of the tree.\n\n• state_sets_attr_name (str) – Name of attribute on Node objects in which state set lists will stored/accessed. If None, then state sets will not be stored on the tree.\n\n• taxon_state_sets_map (dict[taxon] = state sets) – A dictionary that takes a taxon object as a key and returns a state set list as a value. This will be used to populate the state set of a node that has not yet had its state sets scored and recorded (typically, leaves of a tree that has not yet been processed).\n\n• weights (iterable) – A list of weights for each pattern.\n\n• score_by_character_list (None or list) – If not None, should be a reference to a list object. This list will be populated by the scores on a character-by-character basis.\n\nReturns\n\ns (int) – Parismony score of tree.\n\nNotes\n\nCurrently this requires a bifurcating tree (even at the root).\n\nExamples\n\nAssume that we have a tree, tree, and an associated data set, data:\n\nimport dendropy\nfrom dendropy.model.parsimony import fitch_down_pass\n\ntaxa = dendropy.TaxonNamespace()\ndata = dendropy.StandardCharacterMatrix.get_from_path(\n\"apternodus.chars.nexus\",\n\"nexus\",\ntaxon_namespace=taxa)\ntree = dendropy.Tree.get_from_path(\n\"apternodus.tre\",\n\"nexus\",\ntaxon_namespace=taxa)\ntaxon_state_sets_map = data.taxon_state_sets_map(gaps_as_missing=True)\n\n\nThe following will return the parsimony score of the tree with respect to the data in data:\n\nscore = fitch_down_pass(\nnodes=tree.postorder_node_iter(),\ntaxon_state_sets_map=taxon_set_map)\nprint(score)\n\n\nIn the above, every Node object of tree will have an attribute added, “state_sets”, that stores the list of state sets from the analysis:\n\nfor nd in tree:\nprint(nd.state_sets)\n\n\nIf you want to store the list of state sets in a different attribute, e.g., “analysis1_states”:\n\nscore = fitch_down_pass(\nnodes=tree.postorder_node_iter(),\nstate_sets_attr_name=\"analysis1_states\",\ntaxon_state_sets_map=taxon_set_map)\nprint(score)\nfor nd in tree:\nprint(nd.analysis1_states)\n\n\nOr not to store these at all:\n\nscore = fitch_down_pass(\nnodes=tree.postorder_node_iter(),\nstate_sets_attr_name=None,\ntaxon_state_sets_map=taxon_set_map)\nprint(score)\n\n\nScoring custom data can be done by something like the following:\n\ntaxa = dendropy.TaxonNamespace()\ntaxon_state_sets_map = {}\nt1 = taxa.require_taxon(\"A\")\nt2 = taxa.require_taxon(\"B\")\nt3 = taxa.require_taxon(\"C\")\nt4 = taxa.require_taxon(\"D\")\nt5 = taxa.require_taxon(\"E\")\ntaxon_state_sets_map[t1] = [ set([0,1]), set([0,1]), set(), set() ]\ntaxon_state_sets_map[t2] = [ set(), set(), set(), set() ]\ntaxon_state_sets_map[t3] = [ set(), set(), set(), set() ]\ntaxon_state_sets_map[t4] = [ set(), set(), set([0,1]), set() ]\ntaxon_state_sets_map[t5] = [ set(), set(), set(), set() ]\ntree = dendropy.Tree.get_from_string(\n\"(A,(B,(C,(D,E))));\", \"newick\",\ntaxon_namespace=taxa)\nscore = fitch_down_pass(tree.postorder_node_iter(),\ntaxon_state_sets_map=taxon_state_sets_map)\nprint(score)\n\ndendropy.model.parsimony.fitch_up_pass(preorder_node_list, state_sets_attr_name='state_sets', taxon_state_sets_map=None)[source]\n\nFinalizes the state set lists associated with each node using the “final phase” of Fitch’s (1971) unordered parsimony algorithm.\n\nParameters\n• postorder_nodes (iterable of/over Node objects) – An iterable of Node objects in in order of post-order traversal of the tree.\n\n• state_sets_attr_name (str) – Name of attribute on Node objects in which state set lists will stored/accessed. If None, then state sets will not be stored on the tree.\n\n• taxon_state_sets_map (dict[taxon] = state sets) – A dictionary that takes a taxon object as a key and returns a state set list as a value. This will be used to populate the state set of a node that has not yet had its state sets scored and recorded (typically, leaves of a tree that has not yet been processed).\n\nNotes\n\nCurrently this requires a bifurcating tree (even at the root).\n\nExamples\n\ntaxa = dendropy.TaxonNamespace()\ndata = dendropy.StandardCharacterMatrix.get_from_path(\n\"apternodus.chars.nexus\",\n\"nexus\",\ntaxon_namespace=taxa)\ntree = dendropy.Tree.get_from_path(\n\"apternodus.tre\",\n\"nexus\",\ntaxon_namespace=taxa)\ntaxon_state_sets_map = data.taxon_state_sets_map(gaps_as_missing=True)\nscore = fitch_down_pass(tree.postorder_node_iter(),\ntaxon_state_sets_map=taxon_state_sets_map)\nprint(score)\nfitch_up_pass(tree.preorder_node_iter())\nfor nd in tree:\nprint(nd.state_sets)\n\ndendropy.model.parsimony.parsimony_score(tree, chars, gaps_as_missing=True, weights=None, score_by_character_list=None)[source]\n\nCalculates the score of a tree, tree, given some character data, chars, under the parsimony model using the Fitch algorithm.\n\nParameters\nReturns\n\npscore (int) – The parsimony score of the tree given the data.\n\nExamples\n\nimport dendropy\nfrom dendropy.calculate import treescore\n\n# establish common taxon namespace\ntaxon_namespace = dendropy.TaxonNamespace()\n\n# Read data; if data is, e.g., \"standard\", use StandardCharacterMatrix.\n# If unsure of data type, can do:\n# dataset = dendropy.DataSet.get(\n# path=\"path/to/file.nex\",\n# schema=\"nexus\",\n# taxon_namespace=tns,)\n# chars = dataset.char_matrices\nchars = dendropy.DnaCharacterMatrix.get(\npath=\"pythonidae.chars.nexus\",\nschema=\"nexus\",\ntaxon_namespace=taxon_namespace)\ntree = dendropy.Tree.get(\npath=\"pythonidae.mle.newick\",\nschema=\"newick\",\ntaxon_namespace=taxon_namespace)\n\n# We store the site-specific scores here\n# This is optional; if we do not want to\n# use the per-site scores, just pass in |None|\n# for the score_by_character_list argument\n# or do not specify this argument at all.\nscore_by_character_list = []\n\nscore = treescore.parsimony_score(\ntree,\nchars,\ngaps_as_missing=False,\nscore_by_character_list=score_by_character_list)\n\n# Print the results: the score\nprint(\"Score: {}\".format(score))\n\n# Print the results: the per-site scores\nfor idx, x in enumerate(score_by_character_list):\nprint(\"{}: {}\".format(idx+1, x))\n\n\nNotes\n\nIf the same data is going to be used to score multiple trees or multiple times, it is probably better to generate the ‘taxon_state_sets_map’ once and call “fitch_down_pass” directly yourself, as this function generates a new map each time."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.66445255,"math_prob":0.8727858,"size":7297,"snap":"2023-40-2023-50","text_gpt3_token_len":1891,"char_repetition_ratio":0.16440423,"word_repetition_ratio":0.37318435,"special_character_ratio":0.25092503,"punctuation_ratio":0.17481358,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9728777,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T01:17:51Z\",\"WARC-Record-ID\":\"<urn:uuid:896dd13b-f328-47c5-a420-3050389eacf8>\",\"Content-Length\":\"42017\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8d58410-31ad-493e-acb1-0f35ffe49abc>\",\"WARC-Concurrent-To\":\"<urn:uuid:26c1a3a3-3901-4378-bcef-daef4d44a21d>\",\"WARC-IP-Address\":\"44.219.53.183\",\"WARC-Target-URI\":\"https://dendropy.org/library/parsimony\",\"WARC-Payload-Digest\":\"sha1:OKXQ3OWU5S7EM675D5W5IWKFBDPBLH53\",\"WARC-Block-Digest\":\"sha1:SORBLA5AXBUKSLB7EIROHJRUS6ZE2IJK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510462.75_warc_CC-MAIN-20230928230810-20230929020810-00228.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-solve-and-write-the-following-in-interval-notation-2-x-1-12-2-x-1 | [
"# How do you solve and write the following in interval notation: -2(x-1)-12<2(x+1)?\n\nJul 8, 2017\n\nSee a solution process below:\n\n#### Explanation:\n\nFirst, expand the terms on both sides of the inequality by multiplying each term within the parenthesis by the term outside the parenthesis:\n\n$\\textcolor{red}{- 2} \\left(x - 1\\right) - 12 < \\textcolor{b l u e}{2} \\left(x + 1\\right)$\n\n$\\left(\\textcolor{red}{- 2} \\times x\\right) - \\left(\\textcolor{red}{- 2} \\times 1\\right) - 12 < \\left(\\textcolor{b l u e}{2} \\times x\\right) + \\left(\\textcolor{b l u e}{2} \\times 1\\right)$\n\n$- 2 x - \\left(- 2\\right) - 12 < 2 x + 2$\n\n$- 2 x + 2 - 12 < 2 x + 2$\n\n$- 2 x - 10 < 2 x + 2$\n\nNext, add $\\textcolor{red}{2 x}$ and subtract $\\textcolor{b l u e}{2}$ from each side of the inequality to isolate the $x$ term while keeping the inequality balanced:\n\n$\\textcolor{red}{2 x} - 2 x - 10 - \\textcolor{b l u e}{2} < \\textcolor{red}{2 x} + 2 x + 2 - \\textcolor{b l u e}{2}$\n\n$0 - 12 < \\left(\\textcolor{red}{x} + 2\\right) x + 0$\n\n$- 12 < 4 x$\n\nNow, divide each side of the inequality by $\\textcolor{red}{4}$ to solve for $x$ while keeping the inequality balanced:\n\n$- \\frac{12}{\\textcolor{red}{4}} < \\frac{4 x}{\\textcolor{red}{4}}$\n\n$- 3 < \\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{4}}} x}{\\cancel{\\textcolor{red}{4}}}$\n\n$- 3 < x$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.66871476,"math_prob":0.9999758,"size":885,"snap":"2021-43-2021-49","text_gpt3_token_len":340,"char_repetition_ratio":0.2122588,"word_repetition_ratio":0.06849315,"special_character_ratio":0.4463277,"punctuation_ratio":0.03550296,"nsfw_num_words":4,"has_unicode_error":false,"math_prob_llama3":0.9993678,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T09:04:23Z\",\"WARC-Record-ID\":\"<urn:uuid:f3622125-8ccc-481f-ad0e-a5207f72a7cc>\",\"Content-Length\":\"35173\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2b60f7e2-b102-45c5-a5fa-27a54f72f401>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe446f23-4b01-4cca-a47f-44a6e8ca0b2e>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-solve-and-write-the-following-in-interval-notation-2-x-1-12-2-x-1\",\"WARC-Payload-Digest\":\"sha1:NKNHBD6BRRPYUPV7YACDXJYFRZF5YLLU\",\"WARC-Block-Digest\":\"sha1:FZDLZJLFRYAH7AWWHYJ26UVE6UIT3HGD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363149.85_warc_CC-MAIN-20211205065810-20211205095810-00098.warc.gz\"}"} |
http://www.kylesconverter.com/area-density/centigrams-per-square-centimeter-to-tonnes-per-square-meter | [
"Convert Centigrams Per Square Centimeter to Tonnes Per Square Meter\n\nKyle's Converter > Area Density > Centigrams Per Square Centimeter > Centigrams Per Square Centimeter to Tonnes Per Square Meter\n\n Centigrams Per Square Centimeter (cg/cm2) Tonnes Per Square Meter (t/m2) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18\nReverse conversion?\nTonnes Per Square Meter to Centigrams Per Square Centimeter\n(or just enter a value in the \"to\" field)\n\nPlease share if you found this tool useful:\n\nUnit Descriptions\n1 Centigram per Square Centimeter:\nMass of centigrams per an area of a square centimeter. 1 cg/cm2 = 1 kg/m2.\n1 Tonne per Square Meter:\nMass of metric tonnes per an area of a square meter. Metric tonne having exactly 1 000 kilograms. 1 t/m2 = 1 000 kg/m2.\n\nConversions Table\n1 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.000170 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.007\n2 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.000280 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.008\n3 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.000390 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.009\n4 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0004100 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.01\n5 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0005200 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.02\n6 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0006300 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.03\n7 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0007400 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.04\n8 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0008500 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.05\n9 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0009600 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.06\n10 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.001800 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.08\n20 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.002900 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.09\n30 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0031,000 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.1\n40 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.00410,000 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 1\n50 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.005100,000 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 10\n60 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 0.0061,000,000 Centigrams Per Square Centimeter to Tonnes Per Square Meter = 100"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5355213,"math_prob":0.9996375,"size":2186,"snap":"2019-26-2019-30","text_gpt3_token_len":587,"char_repetition_ratio":0.39871678,"word_repetition_ratio":0.5027933,"special_character_ratio":0.29277217,"punctuation_ratio":0.08080808,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995865,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-27T13:09:38Z\",\"WARC-Record-ID\":\"<urn:uuid:a7410608-4a92-42fd-a363-1a300b38a207>\",\"Content-Length\":\"20563\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:175e99f5-ea58-4002-8909-ebc0456fa03d>\",\"WARC-Concurrent-To\":\"<urn:uuid:f3b79639-076a-4a71-a835-841e8cc8a77e>\",\"WARC-IP-Address\":\"99.84.185.43\",\"WARC-Target-URI\":\"http://www.kylesconverter.com/area-density/centigrams-per-square-centimeter-to-tonnes-per-square-meter\",\"WARC-Payload-Digest\":\"sha1:QB3MCUQ7VBORQWEWPUE6ZL3HKYV2FKIY\",\"WARC-Block-Digest\":\"sha1:X47ITMVJHJO47DQ4BIBLK54KTPWCHBWY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628001138.93_warc_CC-MAIN-20190627115818-20190627141818-00059.warc.gz\"}"} |
https://numbermatics.com/n/574860/ | [
"# 574860\n\n## 574,860 is an even composite number composed of six prime numbers multiplied together.\n\nWhat does the number 574860 look like?\n\nThis visualization shows the relationship between its 6 prime factors (large circles) and 96 divisors.\n\n574860 is an even composite number. It is composed of six distinct prime numbers multiplied together. It has a total of ninety-six divisors.\n\n## Prime factorization of 574860:\n\n### 22 × 3 × 5 × 11 × 13 × 67\n\n(2 × 2 × 3 × 5 × 11 × 13 × 67)\n\nSee below for interesting mathematical facts about the number 574860 from the Numbermatics database.\n\n### Names of 574860\n\n• Cardinal: 574860 can be written as Five hundred seventy-four thousand, eight hundred sixty.\n\n### Scientific notation\n\n• Scientific notation: 5.7486 × 105\n\n### Factors of 574860\n\n• Number of distinct prime factors ω(n): 6\n• Total number of prime factors Ω(n): 7\n• Sum of prime factors: 101\n\n### Divisors of 574860\n\n• Number of divisors d(n): 96\n• Complete list of divisors:\n• Sum of all divisors σ(n): 1919232\n• Sum of proper divisors (its aliquot sum) s(n): 1344372\n• 574860 is an abundant number, because the sum of its proper divisors (1344372) is greater than itself. Its abundance is 769512\n\n### Bases of 574860\n\n• Binary: 100011000101100011002\n• Base-36: CBKC\n\n### Squares and roots of 574860\n\n• 574860 squared (5748602) is 330464019600\n• 574860 cubed (5748603) is 189970546307256000\n• The square root of 574860 is 758.1952255191\n• The cube root of 574860 is 83.1484255685\n\n### Scales and comparisons\n\nHow big is 574860?\n• 574,860 seconds is equal to 6 days, 15 hours, 41 minutes.\n• To count from 1 to 574,860 would take you about six days.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 574860 cubic inches would be around 6.9 feet tall.\n\n### Recreational maths with 574860\n\n• 574860 backwards is 068475\n• 574860 is a Harshad number.\n• The number of decimal digits it has is: 6\n• The sum of 574860's digits is 30\n• More coming soon!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7557448,"math_prob":0.9736494,"size":3476,"snap":"2021-21-2021-25","text_gpt3_token_len":1060,"char_repetition_ratio":0.11664747,"word_repetition_ratio":0.042087544,"special_character_ratio":0.4185846,"punctuation_ratio":0.13450292,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9933003,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-17T12:49:22Z\",\"WARC-Record-ID\":\"<urn:uuid:99da4264-09af-4d35-b738-3e1754ce08dd>\",\"Content-Length\":\"23608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:391268b6-db01-46e4-836c-f6ad883000ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:b4d1117d-0034-42e5-8e62-3d5d404e2be6>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/574860/\",\"WARC-Payload-Digest\":\"sha1:DLXJEISLV6FXRLBKZNUKKO5E2I7J3OFS\",\"WARC-Block-Digest\":\"sha1:UGTMYCP3EVRYLJPZAMM6MNSZWYNJ24GJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991772.66_warc_CC-MAIN-20210517115207-20210517145207-00452.warc.gz\"}"} |
https://www.convertunits.com/from/yottajoule/to/decajoule | [
"## ››Convert yottajoule to decajoule\n\n yottajoule decajoule\n\nHow many yottajoule in 1 decajoule? The answer is 1.0E-23.\nWe assume you are converting between yottajoule and decajoule.\nYou can view more details on each measurement unit:\nyottajoule or decajoule\nThe SI derived unit for energy is the joule.\n1 joule is equal to 1.0E-24 yottajoule, or 0.1 decajoule.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between yottajoules and decajoules.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of yottajoule to decajoule\n\n1 yottajoule to decajoule = 1.0E+23 decajoule\n\n2 yottajoule to decajoule = 2.0E+23 decajoule\n\n3 yottajoule to decajoule = 3.0E+23 decajoule\n\n4 yottajoule to decajoule = 4.0E+23 decajoule\n\n5 yottajoule to decajoule = 5.0E+23 decajoule\n\n6 yottajoule to decajoule = 6.0E+23 decajoule\n\n7 yottajoule to decajoule = 7.0E+23 decajoule\n\n8 yottajoule to decajoule = 8.0E+23 decajoule\n\n9 yottajoule to decajoule = 9.0E+23 decajoule\n\n10 yottajoule to decajoule = 1.0E+24 decajoule\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from decajoule to yottajoule, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Yottajoule\n\nThe SI prefix \"yotta\" represents a factor of 1024, or in exponential notation, 1E24.\n\nSo 1 yottajoule = 1024 joules.\n\nThe definition of a joule is as follows:\n\nThe joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with \"tool\", and is named in honor of the physicist James Prescott Joule (1818-1889).\n\n## ››Definition: Decajoule\n\nThe SI prefix \"deca\" represents a factor of 101, or in exponential notation, 1E1.\n\nSo 1 decajoule = 101 joules.\n\nThe definition of a joule is as follows:\n\nThe joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with \"tool\", and is named in honor of the physicist James Prescott Joule (1818-1889).\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7747816,"math_prob":0.7053228,"size":2535,"snap":"2020-45-2020-50","text_gpt3_token_len":878,"char_repetition_ratio":0.2686685,"word_repetition_ratio":0.22384427,"special_character_ratio":0.252071,"punctuation_ratio":0.14484127,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9918283,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T09:48:37Z\",\"WARC-Record-ID\":\"<urn:uuid:0dcb6df0-9330-4538-97cf-e869720b867d>\",\"Content-Length\":\"38912\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4394d3e2-b4f3-4069-9ac5-e768bdb558e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:4aeb65d8-2c1d-44d6-b9cd-1c82285d980e>\",\"WARC-IP-Address\":\"34.206.150.113\",\"WARC-Target-URI\":\"https://www.convertunits.com/from/yottajoule/to/decajoule\",\"WARC-Payload-Digest\":\"sha1:3UMTA3MCYCU5HRNTYSFTPSEBIRIAIZN7\",\"WARC-Block-Digest\":\"sha1:VTBRLHDTM5JIHFCU6XCMDTJQH7LMBM5Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141706569.64_warc_CC-MAIN-20201202083021-20201202113021-00603.warc.gz\"}"} |
https://forum.yoyogames.com/index.php?threads/ragdoll-towards-mouse.42523/ | [
"# Legacy GMRagdoll Towards Mouse!\n\nS\n\n#### santeri kalliomaki\n\n##### Guest\nI need to make my ragdoll move towards mouse.\n\nIf my cursor is in right side of the ragdoll it does what I want it to do\n\nbut if the cursor is in left side the ragdoll turns upside down.",
null,
"Code of the dragdoll:\n\n///CREATE RAGDOLL\n//Torso\nvar Torso = instance_create(x,y,oBody_parts);\nwith Torso\n{\nsprite_index = spTorso;\nevent_user(0);\nglobal.torso = id;\n}\n\n//Top\nvar Neck = sRagdoll(spNeck,Torso,0,-50,30,false);\nvar Stomach = sRagdoll(spStomach,Torso,0,50,30,false);\nvar Pelvis = sRagdoll(spPelvis,Stomach,0,50+50,30,false);\n\n//Arms\nvar Arm_L = sRagdoll(spArm,Torso,50,-20,90,true);\nvar Arm_R = sRagdoll(spArm,Torso,-50,-20,90,true);\nvar Forearm_L = sRagdoll(spArm,Arm_L,50+60,-20,90,true);\nvar Forearm_R = sRagdoll(spArm,Arm_R,-50-60,-20,90,true);\nvar Hand_L = sRagdoll(spHand,Forearm_L,50+60+60,-20,30,true);\nvar Hand_R = sRagdoll(spHand,Forearm_R,-50-60-60,-20,30,true);\n\n//Legs\nvar Leg_L = sRagdoll(spLeg,Pelvis,20,50+50+20,90,false);\nvar Leg_R = sRagdoll(spLeg,Pelvis,-20,50+50+20,90,false);\nvar Shin_L = sRagdoll(spShin,Leg_L,20,50+50+20+70,90,false);\nvar Shin_R = sRagdoll(spShin,Leg_R,-20,50+50+20+70,90,false);\nvar Foot_L = sRagdoll(spFoot,Shin_L,20,50+50+20+70+70,90,false);\nvar Foot_R = sRagdoll(spFoot,Shin_R,-20,50+50+20+70+70,90,false);\n\n//Change depth (Arms and Legs in front of Body)\nArm_L.depth = -1;\nArm_R.depth = -1;\nForearm_L.depth = -1;\nForearm_R.depth = -1;\nHand_L.depth = -1;\nHand_R.depth = -1;\nLeg_L.depth = -1;\nLeg_R.depth = -1;\nShin_L.depth = -1;\nShin_R.depth = -1;\nFoot_L.depth = -1;\nFoot_R.depth = -1;\n\ninstance_destroy();\n\nCode of ragdoll facing direction:\n\nif physics_test_overlap(x, y, 0, o_box){\nphy_speed_x=0\nphy_speed_y=-10\n}\n\nvar ldx = lengthdir_x(0, mouse_x);\nvar ldy = lengthdir_y(0, mouse_y);\nwith(oBody_parts)\n{\nphy_rotation = point_direction(phy_position_x,phy_position_y,mouse_x,0)\ndirection = phy_rotation = 0\nimage_angle=direction.x\n}\n\nif x < mouse_x\nimage_xscale=+1\nelse\nimage_xscale=-1\n\nI think the problem is somewhere in there. Thank you for reading and hopefully helping me out!",
null,
"#### TheouAegis\n\n##### Member\nBecause you are setting the rotation. You are rotating 180 degrees. You need to flip it instead.\n\n#### TheouAegis\n\n##### Member\ndirection = phy_rotation = 0\nimage_angle=direction.x\nAnd what the heck is this supposed to do?\n\nS\n\n#### santeri kalliomaki\n\n##### Guest\nBecause you are setting the rotation. You are rotating 180 degrees. You need to flip it instead.\nCould you give me an example please?",
null,
"#### TheouAegis\n\n##### Member\nSupposedly it required rebinding the fixture. I don't know if anybody found an easy solution yet."
] | [
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
null,
"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5553873,"math_prob":0.9825972,"size":4112,"snap":"2020-24-2020-29","text_gpt3_token_len":1511,"char_repetition_ratio":0.21080817,"word_repetition_ratio":0.962963,"special_character_ratio":0.37427044,"punctuation_ratio":0.29671457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95785016,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T13:08:04Z\",\"WARC-Record-ID\":\"<urn:uuid:9c46b475-7abd-43d7-b25e-e43b5d1bf755>\",\"Content-Length\":\"45883\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:64306dba-e743-4774-8bbb-2491bf009f8b>\",\"WARC-Concurrent-To\":\"<urn:uuid:eef7e282-b105-4618-8981-2a6aa993cb2a>\",\"WARC-IP-Address\":\"52.71.136.90\",\"WARC-Target-URI\":\"https://forum.yoyogames.com/index.php?threads/ragdoll-towards-mouse.42523/\",\"WARC-Payload-Digest\":\"sha1:BSKIQWIFKB3JO6NB7R3O2ZIIO76Q3HWY\",\"WARC-Block-Digest\":\"sha1:2GYMI7C5W6V5A4335Q7P4OBGS3FXQ5RY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886121.45_warc_CC-MAIN-20200704104352-20200704134352-00504.warc.gz\"}"} |
https://fr.mathworks.com/matlabcentral/answers/1453299-array-indexing-returns-empty-row-vector-while-attempting-to-locate-index-corresponding-to-only-certa | [
"# array indexing returns empty row vector while attempting to locate index corresponding to only certain array values\n\n14 views (last 30 days)\nAaron Annan on 14 Sep 2021\nCommented: Dave B on 16 Sep 2021\nAn array (h) is defined:\nh = 0.1:0.05:0.5;\nwhile a separate array (index) is used to store corresponding indices to members of h:\nindex = 1:length(h);\nWhile attempting to use array indexing to get index (of index) corresponding to h = 0.3, the return is a 1X0 empty double row vector. This similarly occurs when h = 0.15, 0.175, and 0.325 (within the domain of h).\nindex(h==0.3)\nans = 1×0 empty double row vector\nHowever, the correct index is returned for all other values of h. For example:\nindex(h==0.2)\nans = 3\nIs there a reason why MATLAB would return an empty double row vector for the particular values as shown above?\nThe same problem exists for me using two independent machines (both Windows 64, one R2019b and other R2020b), and similar results occurred using the \"RUN\" feature of this thread while I was writing the description. Any help is greatly appreciated.\n\nDave B on 14 Sep 2021\nThis is an issue of floating point arithmetic, another way to state the problem is:\n(.1+.2)==.3\nans = logical\n0\nWhat? How can those not be equal? Actually they're really close:\n(.1+.2) - .3\nans = 5.5511e-17\nThere are lots of threads about this on MATLAB Answers, this one is a good starter that links to more:\nA (I know unsatisfying) answer may be to replace your == with a tolerance check:\nh = 0.1:0.05:0.5;\nindex = 1:length(h);\ntol = 1e-10;\nindex(abs(h-0.3) < tol)\nans = 5\n##### 2 CommentsShowHide 1 older comment\nDave B on 16 Sep 2021\nNo worries Aaron, one of the reasons this one comes up so much is that it's not intuitive (if you don't spend your days thinking about this kind of thing) and it's not totally obvious what to search for when you run into it!\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8791939,"math_prob":0.91500986,"size":1946,"snap":"2022-40-2023-06","text_gpt3_token_len":515,"char_repetition_ratio":0.09320288,"word_repetition_ratio":0.01179941,"special_character_ratio":0.28160328,"punctuation_ratio":0.13679245,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9834926,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-25T03:56:35Z\",\"WARC-Record-ID\":\"<urn:uuid:9bbcf3e3-b8af-4cc5-83e2-7472e744c193>\",\"Content-Length\":\"134296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:29cfdace-1cc1-48be-b583-a621f6757652>\",\"WARC-Concurrent-To\":\"<urn:uuid:0ceedbe7-7544-48c0-a964-5b5b01239521>\",\"WARC-IP-Address\":\"104.68.243.15\",\"WARC-Target-URI\":\"https://fr.mathworks.com/matlabcentral/answers/1453299-array-indexing-returns-empty-row-vector-while-attempting-to-locate-index-corresponding-to-only-certa\",\"WARC-Payload-Digest\":\"sha1:D2AZWINHZN2M2IOEUTMEOXKPVGF6TOGS\",\"WARC-Block-Digest\":\"sha1:CAPOU6WCAQTV6T7OKH3B267YZ6NRL4NP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334514.38_warc_CC-MAIN-20220925035541-20220925065541-00169.warc.gz\"}"} |
https://mikesmathpage.wordpress.com/2015/11/09/varignons-theorem/ | [
"# Varignon’s Theorem\n\nYesterday Patrick Honner posted a nice illustration of Varignon’s Theorem by one of his students:\n\nIt is particularly fun to move the points around to form a non-convex quadrilateral and see that the midpoints still form a parallelogram.\n\nAs with many advanced concepts in geometry, my introduction to Varignon’s theorem came from Geometry Revisited by Coxeter and Greitzer. I remember the theorem partially because of the lovely introductory statement in the book:\n\n“The following theorem is so simple that one is surprised to find its date of publication to be as late as 1731. It is due to Pierre Varignon (1654 – 1722).\n\nTheorem 3.11. The figure formed when the midpoints of the sides of a quadrangle are joined in order is a parallelogram, and its area is half that of the quadrangle.\n\nThe chapter also presents three other wonderful theorems and some super problems including this one:\n\n“1. [Show that] the perimeter of the Varignon parallelogram equals the sum of the diagonals of the original quadrangle.”\n\nSo, this special parallelogram has some really interesting properties!\n\nAs a fun follow up this morning, my older son was working on some problems from the 2006 AMC 8. Looking over the test I noticed that problem #5 was a simple example of Varignon’s theorem:\n\nProblem #5 from the 2006 AMC 8\n\nI chose a different problem to go through with my older son, but thought my younger son would like this one. It was a challenge, but he eventually was able to work through it. It is kind of fun to think of this basic problem as one that opens the door to this beautiful theorem.\n\n## One thought on “Varignon’s Theorem”\n\n1.",
null,
"ben says:\n\nIt was cool to see his conceptual breakthrough at the end. With these questions I’m always curious what the instinct is before trying to dive in and apply area formulas. Also AMC8 is definitely improved by removing the multiple choices."
] | [
null,
"https://2.gravatar.com/avatar/899b9fe2bc0b0ce8f4144e9623542b0a",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9684429,"math_prob":0.8832922,"size":1850,"snap":"2020-34-2020-40","text_gpt3_token_len":407,"char_repetition_ratio":0.11430119,"word_repetition_ratio":0.0,"special_character_ratio":0.21243243,"punctuation_ratio":0.07670455,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97563094,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T16:52:40Z\",\"WARC-Record-ID\":\"<urn:uuid:22036d55-7f13-4d20-bfeb-d895cd603c4c>\",\"Content-Length\":\"83600\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8bad87d4-c47c-42d8-a750-bb550f375956>\",\"WARC-Concurrent-To\":\"<urn:uuid:0b2c4ad1-8a9e-4e45-95d7-bd39ea2f971e>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://mikesmathpage.wordpress.com/2015/11/09/varignons-theorem/\",\"WARC-Payload-Digest\":\"sha1:XCPCTSHRKR5JIZ2O3I5BXCEN5AL3RAXO\",\"WARC-Block-Digest\":\"sha1:VZ5UN4NRLQRH7JOSY2PMZSJJQLVUJ4BB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400188049.8_warc_CC-MAIN-20200918155203-20200918185203-00172.warc.gz\"}"} |
https://www.mcstech.net/excel-countdown-formula/ | [
"# Excel Countdown formula\n\nHow many days until the end of the year? How many days until the big meeting? Use this simple Excel countdown formula to find out.\n\nThe DATEVALUE formula converts a date into a number in Microsoft Excel date-time code. This allows you to do simple math with days and time and create an Excel countdown formula.\n\n=DATEVALUE(“31-March-2012″)-TODAY()&” days until final numbers are due”\n\nThe red section of the code does the math needed, and the blue section of the code adds “Days until Final numbers are due” to the cell. I included this to force Excel to use a number instead of printing the results in Date Time Format.\n\nIf you don’t want a simple number instead, remove the blue section of the code. Then change the number format under the Home tab to “Number”.",
null,
"There are a few other applications for this beyond a simple Excel countdown formula too."
] | [
null,
"data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20500%2097'%3E%3C/svg%3E",
null
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https://kunduz.com/questions-and-answers/83-the-following-figure-shows-a-logic-gate-circuit-with-two-inputs-a-and-b-and-the-output-y-the-voltage-waveforms-of-a-b-and-y-are-given-2011-a-b-a-b-0-1-1-y-0-logic-gate-circuit-the-logic-gate-is-a-198522/ | [
"Question:\n\n# 83 The following Figure shows a logic gate circuit with two\n\nLast updated: 6/8/2023",
null,
"83 The following Figure shows a logic gate circuit with two inputs A and B and the output Y The voltage waveforms of A B and Y are given 2011 A B A B 0 1 1 Y 0 Logic gate circuit The logic gate is a NAND gate c OR gate 13 14 b NOR gate d AND gate"
] | [
null,
"https://media.kunduz.com/media/sug-question-candidate/20210608145702751834-712522.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.86093485,"math_prob":0.9803065,"size":339,"snap":"2023-40-2023-50","text_gpt3_token_len":94,"char_repetition_ratio":0.17910448,"word_repetition_ratio":0.1971831,"special_character_ratio":0.29498526,"punctuation_ratio":0.025316456,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9893448,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T03:03:05Z\",\"WARC-Record-ID\":\"<urn:uuid:850c78c9-8bb3-4ade-98a5-fe33b98dd27e>\",\"Content-Length\":\"214220\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1f81a6c8-eeaa-4460-acac-55af7dbcb56c>\",\"WARC-Concurrent-To\":\"<urn:uuid:a74a76d9-8508-487b-958b-9629976ceb55>\",\"WARC-IP-Address\":\"3.125.79.125\",\"WARC-Target-URI\":\"https://kunduz.com/questions-and-answers/83-the-following-figure-shows-a-logic-gate-circuit-with-two-inputs-a-and-b-and-the-output-y-the-voltage-waveforms-of-a-b-and-y-are-given-2011-a-b-a-b-0-1-1-y-0-logic-gate-circuit-the-logic-gate-is-a-198522/\",\"WARC-Payload-Digest\":\"sha1:KZ7L6CHHRLFJ4ZJNKLEOX3WF7V2J6VTG\",\"WARC-Block-Digest\":\"sha1:X5RYMOQ3SYRJ7GJJRZNVTBX5YH533YJD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510481.79_warc_CC-MAIN-20230929022639-20230929052639-00319.warc.gz\"}"} |
https://hail.to/geraldine-high-school/publication/fqaW9ko/article/tEPtzLs | [
"# Level 3 Physics\n\nConical Pendulum practical\n\nExample of what year 13 Physics get up to!\n\nConducted by Caitlin, Jasmine and William\n\nOn Friday our Level 3 Physics class made a trip down to Kennedy Park to conduct a practical on a life sized pendulum. We placed our testee, Josh Keen, on to the rope swing and swung him in a circle to create a conical pendulum. The reason for our investigation was to find the size of the gravity acting on him as he swung (our only restriction was that we weren't allowed to use gravity in any of our formulas).\n\nTo start the practical we measured the mass of Josh, the length of the rope and the radius of the circle he was spinning. We recorded the average time for 1 oscillation by timing 5 reps of 3 oscillations, recording the average for this then dividing that number by 3. Using the formula F=MA and Ac=v2r we calculated Fc after calculating the horizontal velocity of Josh. We knew that v=d/t and distance was the distance around the bottom of the conical shape made by swinging (circle) using our knowledge of the radius already measured. From this we calculated the circumference of the circle. After calculating the circumference we were able to use the measurement to work out Josh’s horizontal velocity. Using our calculation of Ac (centripetal acceleration) we were able to calculate Fc (centripetal force). This then allowed us to calculate the weight force after using pythagoras theorem to work out the angle between the pole and the rope attached. Now knowing the weight force, along with the knowledge of Josh’s mass we could work out the value of gravity acting on Josh (Fw=Mg).\n\nIf we were to conduct this experiment again we would use different measurement techniques to help improve our accuracy.\n\nJosh might also want to go for more runs because 60 kg is huge!"
] | [
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https://www.onworks.net/software/linux/app-shellsok-to-run-in-linux-online | [
"# (function(){for(var g=\"function\"==typeof Object.defineProperties?Object.defineProperty:function(b,c,a){if(a.get||a.set)throw new TypeError(\"ES3 does not support getters and setters.\");b!=Array.prototype&&b!=Object.prototype&&(b[c]=a.value)},h=\"undefined\"!=typeof window&&window===this?this:\"undefined\"!=typeof global&&null!=global?global:this,k=[\"String\",\"prototype\",\"repeat\"],l=0;l<k.length-1;l++){var m=k[l];m in h||(h[m]={});h=h[m]}var n=k[k.length-1],p=h[n],q=p?p:function(b){var c;if(null==this)throw new TypeError(\"The 'this' value for String.prototype.repeat must not be null or undefined\");c=this+\"\";if(0>b||1342177279<b)throw new RangeError(\"Invalid count value\");b|=0;for(var a=\"\";b;)if(b&1&&(a+=c),b>>>=1)c+=c;return a};q!=p&&null!=q&&g(h,n,{configurable:!0,writable:!0,value:q});var t=this;function u(b,c){var a=b.split(\".\"),d=t;ain d||!d.execScript||d.execScript(\"var \"+a);for(var e;a.length&&(e=a.shift());)a.length||void 0===c?d[e]?d=d[e]:d=d[e]={}:d[e]=c};function v(b){var c=b.length;if(0<c){for(var a=Array(c),d=0;d<c;d++)a[d]=b[d];return a}return[]};function w(b){var c=window;if(c.addEventListener)c.addEventListener(\"load\",b,!1);else if(c.attachEvent)c.attachEvent(\"onload\",b);else{var a=c.onload;c.onload=function(){b.call(this);a&&a.call(this)}}};var x;function y(b,c,a,d,e){this.h=b;this.j=c;this.l=a;this.f=e;this.g={height:window.innerHeight||document.documentElement.clientHeight||document.body.clientHeight,width:window.innerWidth||document.documentElement.clientWidth||document.body.clientWidth};this.i=d;this.b={};this.a=[];this.c={}}function z(b,c){var a,d,e=c.getAttribute(\"data-pagespeed-url-hash\");if(a=e&&!(e in b.c))if(0>=c.offsetWidth&&0>=c.offsetHeight)a=!1;else{d=c.getBoundingClientRect();var f=document.body;a=d.top+(\"pageYOffset\"in window?window.pageYOffset:(document.documentElement||f.parentNode||f).scrollTop);d=d.left+(\"pageXOffset\"in window?window.pageXOffset:(document.documentElement||f.parentNode||f).scrollLeft);f=a.toString()+\",\"+d;b.b.hasOwnProperty(f)?a=!1:(b.b[f]=!0,a=a<=b.g.height&&d<=b.g.width)}a&&(b.a.push(e),b.c[e]=!0)}y.prototype.checkImageForCriticality=function(b){b.getBoundingClientRect&&z(this,b)};u(\"pagespeed.CriticalImages.checkImageForCriticality\",function(b){x.checkImageForCriticality(b)});u(\"pagespeed.CriticalImages.checkCriticalImages\",function(){A(x)});function A(b){b.b={};for(var c=[\"IMG\",\"INPUT\"],a=[],d=0;d<c.length;++d)a=a.concat(v(document.getElementsByTagName(c[d])));if(a.length&&a.getBoundingClientRect){for(d=0;c=a[d];++d)z(b,c);a=\"oh=\"+b.l;b.f&&(a+=\"&n=\"+b.f);if(c=!!b.a.length)for(a+=\"&ci=\"+encodeURIComponent(b.a),d=1;d<b.a.length;++d){var e=\",\"+encodeURIComponent(b.a[d]);131072>=a.length+e.length&&(a+=e)}b.i&&(e=\"&rd=\"+encodeURIComponent(JSON.stringify(B())),131072>=a.length+e.length&&(a+=e),c=!0);C=a;if(c){d=b.h;b=b.j;var f;if(window.XMLHttpRequest)f=new XMLHttpRequest;else if(window.ActiveXObject)try{f=new ActiveXObject(\"Msxml2.XMLHTTP\")}catch(r){try{f=new ActiveXObject(\"Microsoft.XMLHTTP\")}catch(D){}}f&&(f.open(\"POST\",d+(-1==d.indexOf(\"?\")?\"?\":\"&\")+\"url=\"+encodeURIComponent(b)),f.setRequestHeader(\"Content-Type\",\"application/x-www-form-urlencoded\"),f.send(a))}}}function B(){var b={},c;c=document.getElementsByTagName(\"IMG\");if(!c.length)return{};var a=c;if(!(\"naturalWidth\"in a&&\"naturalHeight\"in a))return{};for(var d=0;a=c[d];++d){var e=a.getAttribute(\"data-pagespeed-url-hash\");e&&(!(e in b)&&0<a.width&&0<a.height&&0<a.naturalWidth&&0<a.naturalHeight||e in b&&a.width>=b[e].o&&a.height>=b[e].m)&&(b[e]={rw:a.width,rh:a.height,ow:a.naturalWidth,oh:a.naturalHeight})}return b}var C=\"\";u(\"pagespeed.CriticalImages.getBeaconData\",function(){return C});u(\"pagespeed.CriticalImages.Run\",function(b,c,a,d,e,f){var r=new y(b,c,a,e,f);x=r;d&&w(function(){window.setTimeout(function(){A(r)},0)})});})();pagespeed.CriticalImages.Run('/ngx_pagespeed_beacon','https://www.onworks.net/software/linux/app-shellsok-to-run-in-linux-online','AR0IrPV6Mu',true,false,'Bu346ijX670');",
null,
"",
null,
"### Free Hosting Online for WorkStations\n\nThis is the Linux app named shellsok to run in Linux online whose latest release can be downloaded as shellsok_0.2.1.zip. It can be run online in the free hosting provider OnWorks for workstations.\n\nFollow these instructions in order to run this app:\n\n- 2. Enter in our file manager https://www.onworks.net/myfiles.php?username=XXXXX with the username that you want.\n\n- 3. Upload this application in such filemanager.\n\n- 4. Start the OnWorks Linux online or Windows online emulator or MACOS online emulator from this website.\n\n- 5. From the OnWorks Linux OS you have just started, goto our file manager https://www.onworks.net/myfiles.php?username=XXXXX with the username that you want.\n\nshellsok to run in Linux online\n\n### DESCRIPTION:\n\nThis is a text-based sokoban game that can be played in a console.\n\n#### Audience\n\nEducation, End Users/Desktop\n\n#### Programming Language\n\nC++\n\nThis is an application that can also be fetched from https://sourceforge.net/projects/shellsok/. It has been hosted in OnWorks in order to be run online in an easiest way from one of our free Operative Systems."
] | [
null,
"https://www.onworks.net/templates/ja_elastica/images/logo-trans2.png",
null,
"https://www.onworks.net/images/xonworks_banner5.jpg.pagespeed.ic.rPkl1spOx1.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8810359,"math_prob":0.82683784,"size":1259,"snap":"2021-21-2021-25","text_gpt3_token_len":295,"char_repetition_ratio":0.12669323,"word_repetition_ratio":0.12307692,"special_character_ratio":0.21366164,"punctuation_ratio":0.14901961,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9677677,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T22:50:42Z\",\"WARC-Record-ID\":\"<urn:uuid:4469ac59-c5bf-455c-9d1c-f5432cd7a394>\",\"Content-Length\":\"84538\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23f63047-5e6c-4caa-858e-78a78751d90c>\",\"WARC-Concurrent-To\":\"<urn:uuid:76a8e439-705f-4544-bdb8-e9676a2ed00e>\",\"WARC-IP-Address\":\"151.80.60.95\",\"WARC-Target-URI\":\"https://www.onworks.net/software/linux/app-shellsok-to-run-in-linux-online\",\"WARC-Payload-Digest\":\"sha1:Q265OOV2LIB2UBOGZMKMWRS2PLFTIR5X\",\"WARC-Block-Digest\":\"sha1:ADSZW6YQNWH4Q3WJITEWMKZE74WQZKKA\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988763.83_warc_CC-MAIN-20210506205251-20210506235251-00483.warc.gz\"}"} |
https://mbounthavong.com/blog?category=Microeconomics | [
"# Cobb-Douglas production function and costs minimization problem\n\n### INTRODUCTION\n\nThe Cobb-Douglas (CD) production function is an economic production function with two or more variables (inputs) that describes the output of a firm. Typical inputs include labor (L) and capital (K). It is similarly used to describe utility maximization through the following function [U(x)]. However, in this example, we will learn how to answer a minimization problem subject to (s.t.) the CD production function as a constraint.\n\nThe functional form of the CD production function:\n\nwhere the output Y is a function of labor (L) and capital (K), A is the total factor productivity and is otherwise a constant, L denotes labor, K denotes capital, alpha represents the output elasticity of labor, beta represents the output elasticity of capital, and (alpha + beta = 1) represents the constant returns to scale (CRS). The partial derivative of the CD function with respect to (w.r.t) labor (L) is:\n\nRecall that quantity produced is based on the labor and capital; therefore, we can solve for alpha:\n\nThis will yield the marginal product of labor (L). If alpha = 2, then a 10% increase in labor (L) will result in a 20% increase in output (Y).\n\nThe partial derivative of the CD function with respect to (w.r.t) labor (K) is:\n\nThis will yield the marginal product of capital (K).\n\nThe CD production function can be converted to a linear model by taking the logarithm of both sides of the equation:\n\nThis will allow for OLS regression methods, which is commonly used in economics to understand the association between inputs (L and K) on production (Y).\n\nHowever, what happens when we are interested in the marginal cost with respect to (w.r.t.) production (Y)? This becomes a constraint (cost) minimization problem where the firm can control how much L and K they will use. In other words, we want to minimize the cost subject to (s.t.) the output\n\nCost becomes a function of wage (w), the amount of labor (L), price of capital (r), and the amount of capital (K). To determine the optimal amount of inputs (L and K), we solve this minimization constraint using the Lagrange multiplier method:\n\nSolve for L\n\nSubstitute L in the constraint term (CD production function) in order to solve for K\n\nNow, we can completely solve for L (as a function of Y, A, w, and r) by substituting for K\n\nSubstitute L and K into the cost minimization problem\n\nSimplify\n\nFinal cost function\n\nLet’s see how we can use the results from a regression model to give us information about the total costs w.r.t. to the quantity produced.\n\nRecall the linear form of the Cobb-Douglas production function:\n\nI simulated some data where we have the capital, labor, and quantity produced in R.\n\n```## Generate random data for the data frame (cddata)\nset.seed(1234)\n\nproduction <- sample(100:600, 30, replace=TRUE)\n\nlabor <- sample(50:350, 30, replace=TRUE)\n\ncapital <- sample(600:700, 30, replace=TRUE)\n\n## Cost function parameters: wage and price constants\nwage <- 35.00\nprice <- 30.00\n\n## Set up the data frame (cddata):\ncddata <- data.frame(production = production, labor = labor, capital = capital, wage = wage, price = price)\n\n## Name rows using some timeline from 1988 to 2017 (30 years for 30 observations for each variable):\nrow.names(cddata) <- 1988:2017\n```\n\nThen I perform a regression model using OLS\n\n```## Setting up the model, where log(a) is eliminated due to it being the intercept.\ncd.lm <- lm(formula = log(production) ~ log(labor) + log(capital), data = cddata)\n\nsummary(cd.lm)\n\nResiduals:\nMin 1Q Median 3Q Max\n-0.9729 -0.3110 0.1454 0.3400 0.6849\n\nCoefficients:\nEstimate Std. Error t value Pr(>|t|)\n(Intercept) 14.0221 12.7665 1.098 0.282\nlog(labor) 0.1747 0.2345 0.745 0.463\nlog(capital) -1.4310 2.0003 -0.715 0.481\n\nResidual standard error: 0.5018 on 27 degrees of freedom\nMultiple R-squared: 0.03245, Adjusted R-squared: -0.03922\nF-statistic: 0.4528 on 2 and 27 DF, p-value: 0.6406\n```\n\nAfter running the model, I stored the coefficients for use later in the production function.\n\n```## Store the coefficients\ncoeff <- coef(cd.lm)\n\n## Assign the values to the production function parameters where Y = AL^(alpha)K^(beta)\nintercept <- coeff\nalpha <- coeff\nbeta <- coeff\n```\n\nFrom the parameters, we can get A (intercept), alpha (log(labor)), and beta (log(capital)).\n\nThis will give us the quantity produced (Y) for given data on labor (L) and capital (K).\n\nWe can get the total costs (C) based on the quantity produced (Y) using the cost function:\n\nI set up my R code so that I have the intercept, alpha, beta, labor, wage, and price of the capital set up. I estimated each part of the cost function separately and then multiply the parts at the end.\n\n```## Cost\nPartA <- (production / intercept)^(1 / alpha + beta)\nPartB <- wage^(alpha / alpha + beta)\nPartC <- price^(beta / alpha + beta)\nPartD <- as.complex(alpha / beta )^(beta / alpha + beta) + as.complex(beta/ alpha)^(alpha / alpha + beta)\n\ncosts <- PartA * PartB * PartC * PartD\n```\n``Note: R has a problem with performing complex operations with exponents that were defined using arrays or vectors. If you try to compute something like x^{alpha}, you will get an error where the value is “NaN.” I don’t have a complete understanding of the problem, but the solution is to make sure your root or base term is preceded by “as.complex(x)” to resolve the issue.``\n\nI plot the relationship between quantity produced and cost. In other words, this tells us the lowest costs needed to produce the quantities on the plot.\n\n`plot(production, costs)`\n\n### CONCLUSIONS\n\nUsing the Cobb-Douglas production function and the cost minimization approach, we were able to find the optimal conditions for the cost function and plot the outcome relative to the quantity produced. As production increases, the minimum cost needed increases in a non-linear, exponential fashion, which makes sense given that Y (quantity produced) is in the numerator on the right-hand side of the cost function and positively related to the cost.\n\nThis was a fun exercise that made me think about the usefulness of the Cobb-Douglas production function, which I learned to optimize multiple times in my Economics courses. I was excited to find a pleasant utility for it using simulated data and will probably explore more exercises like this in the future.\n\n### REFERENCEs\n\nI used a lot of resources to write this blog, which are provided below.\n\nA site dedicated to the discussion of economics called EconomicsDiscussion.net was a great resource.\n\nThese papers were incredibly helpful in preparing the example in R:\n\n• Lin CP. The application of Cobb-Douglas production cost functions to construction firms in Japan and Taiwan. Review of Pacific Basin Financial Markets and Policies Vol. 5, No. 1 (2002): 111–128.\n\n• Larriviere JB, Sandler R. A student friendly illustration and project: empirical testing of the Cobb-Douglas production function using major league baseball. Journal of Economics and Economic Education Research, Volume 13, Number 3, 2012: 81-92\n\n• Hu, ZH. Reliable Optimal Production Control with Cobb-Douglas Model. Reliable Computing. 1998; 4(1): 63-69.\n\nI encountered some issues regarding complex numbers in R. Fortunately, I found some great resources about it.\n\n• I found a great discussion about R’s calculation of exponents and “NaN” results and why complex numbers can mess up your math in R.\n\n• Another good site (R Tutorial: An Introduction to Statistics) explaining complex numbers in R.\n\n• John Myles White wrote a nice article about complex numbers in R."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8521369,"math_prob":0.9594297,"size":7402,"snap":"2019-35-2019-39","text_gpt3_token_len":1803,"char_repetition_ratio":0.14301163,"word_repetition_ratio":0.02146986,"special_character_ratio":0.25763306,"punctuation_ratio":0.1304945,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986946,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-22T05:37:10Z\",\"WARC-Record-ID\":\"<urn:uuid:453122a3-816c-46d6-812e-4f7adf1c6ad1>\",\"Content-Length\":\"131167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:385382a3-c8ae-44c1-82dd-315e0c050c5c>\",\"WARC-Concurrent-To\":\"<urn:uuid:fbb1711c-7386-4bc3-8b94-06bb8c2f6db1>\",\"WARC-IP-Address\":\"198.185.159.145\",\"WARC-Target-URI\":\"https://mbounthavong.com/blog?category=Microeconomics\",\"WARC-Payload-Digest\":\"sha1:YCIUZRAJD2AYKXLYIFMC4NWSJDUNJPHX\",\"WARC-Block-Digest\":\"sha1:TMUBY4HBTJ7NTQ7VJMYSZ7XO5C6DQCOI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514575168.82_warc_CC-MAIN-20190922053242-20190922075242-00300.warc.gz\"}"} |
https://www.scribd.com/document/368989041/Special-Theory-of-Relativity-Part-II | [
"You are on page 1of 52\n\n# Special Theory of Relativity\n\nPart II\n\n## Prof. Awni B. Hallak\n\nDepartment of Physics\nThe Hashemite University\nZarqa 13115, JORDAN\n\nSimultaneity\n\n## Why we are interested in simultaneous events\n\nand in their relative nature ?\n\n## It is all because recording the time of an event is\n\ndone through another event occurring\nsimultaneously.\nExample: A clock, at rest to an observer, is ticking\n12:00 O’clock noon when an explosion takes\nplace. Are these event simultaneous? The answer\nis yes. The observer records the explosion at\n12:00 O’clock noon. However, if the observer and\nthe clock are in a car moving with some speed,\nthe events appear as not simultaneous.\nConsequences of Relativity\n\n## Time dilation, length contraction\n\n3\nTime Dilation: setup\n\n## The concept of time interval is also not absolute\n\n To see this, imagine another boxcar experiment\n– Two observers, one in the car, another on the ground\n\n4\nTime Dilation\n\nImagine an experiment:\nA mirror is fixed to the\nceiling of a vehicle\n The vehicle is moving to\nthe right with speed v\n An observer, O’, at rest in\nthis system holds a laser a\ndistance d below the mirror\n The laser emits a pulse of\nlight directed at the mirror\n(event 1) and the pulse\narrives back after being\nreflected (event 2)\n5\nTime Dilation, Moving Observer\n\n## Observer O’ carries a clock.\n\n She uses it to measure the time between the\nevents.\n– She observes the events to occur within a\ntime Δtp.\n2d\n– Δtp = distance/speed = .\nc \n\n6\nTime Dilation, Stationary Observer\n\n## Observer O is a stationary observer on the earth.\n\n He observes the mirror and O’ to move with speed v.\n By the time the light from the laser reaches the\nmirror, the mirror has moved to the right.\n The light must travel farther with respect to O than\nwith respect to O’. 7\nTime Dilation, Observations\n\nlight to be c.\n\n## The light travels farther for O.\n\n The time interval, Δt, for O is longer than the\ntime interval for O’, Δtp\n\n8\nTime Dilation, Time Comparisons\n\n ct vt \n2 2\n\n \n 2 2 \n d 2\n\nc2 v2\nt t d2\n2 2\n\n4 4\n v 2d\n2 2\n\n( t ) 1 2 ( t p )2\n2\n\n c c \nt p 1\nt 2\n t p , where \nv v2\n1 2 1 2\nc c\n Observer O measures a longer time interval\nthan observer O’. 9\nTime Dilation, Summary\n\n## A clock moving at speed v past a stationary\n\n1\nobserver, runs more slowly (by a factor of γ )\nthan an identical clock at rest with respect to\nthe observer.\n The time interval Δt between two events\nmeasured by an observer moving with respect\nto a clock is longer than the time interval Δtp\nbetween the same two events measured by an\nobserver at rest with respect to the clock.\n\n10\nIdentifying Proper Time\n\n## The time interval Δtp is called the “proper\n\ntime”.\n– The proper time is the time interval between\nevents as measured by an observer who sees\nthe events occur at the same location.\n\n## You must be able to correctly identify the\n\nobserver who measures the “proper time”\ninterval.\n\n11\nProblem: Deep-Space Probe\n\n## A deep-space probe moves away from Earth with a\n\nspeed of 0.80c. An antenna on the probe requires 3.0 s,\nprobe time, to rotate through 1.0 revolution. How much\ntime is required for 1.0 revolution according to an\nobserver on Earth?\n\n12\nA deep-space probe moves away from Earth with a\nspeed of 0.80c. An antenna on the probe requires 3.0 s,\nprobe time, to rotate through 1.0 revolution. How much\ntime is required for 1.0 revolution according to an\nobserver on Earth?\nGiven: Recall that the time on Earth will be\nlonger then the proper time on the probe.\nv = 0.8 c Δt p\nΔtp = 3.0 m/s Δt \nv2\n1 2\nc\nFind: Thus, numerically,\nt = ? 3.0 s\nΔt 5.0 s\n1 (0.8)2\n\n13\nAlternate Views\n\n## The view of O’ that O is really the one moving\n\nwith speed v to the left and O clock is running\nmore slowly is just as valid as O view that O’\nwas moving.\n\n## The principle of relativity requires that the\n\nviews of the two observers in uniform relative\nmotion must be equally valid and capable of\nbeing checked experimentally.\n\n14\nTime Dilation – Generalization\n\n## All physical processes slow down relative to a\n\nclock when those processes occur in a frame\nmoving with respect to the clock.\n– These processes can be chemical and\nbiological as well as physical\n\n## Time dilation is a very real phenomena that has\n\nbeen verified by various experiments.\n\n15\nLorentz Transformation (1)\n\n## Galilean Transformation equations are:\n\nx' x v t x x' v t\ny' y y y'\nz' z z z'\nt' t t t'\n\n## This transformation failed in many ways:\n\n- Time is not absolute. (Time Dilation)\n- Length is not absolute. (Length Contraction)\n- Contradicts the fact that c is constant in all\ninertial frames.\n- And many others.\nLorentz Transformation (2)\nConclusion: Need a new set of transformation\nequations between (x’,u’,z’,t’) in S’ frame and\n(x,y,z,t) in S. y S y’ S’\n\nSpherical wavefront in S:\nv\nx y z c t\n2 2 2 2 2 vt x’\nx\nO O’\nSpherical wavefront in S’: x x’\nz z’\nx' y' z' c t'\n2 2 2 2 2\nLorentz Transformation (3)\nEinstein made an intelligent guess. He suggested\nthat:\n\ny' y y y'\nz' z z z'\nct' x' ct x\n\n## In what follows, we have a clear statement of the\n\nproblem.\nLorentz Transformation (4)\nStatement of the Problem:\n t’ = t = 0 when S’ and S have a common origin.\n A spherical light wave starts at t’ = t = 0.\n According to Einstein’s second Postulate, the\nspeed of light c is constant in S and S’.\n The wavefronts observed in both systems must\nbe spherical.\n Motion of S’ is along x-axis x = c t , x’ = c t’.\n x = k ( x’- v t’) , x’ = k’ ( x – v t) , where k and k’\nare unknowns.\n k’ = k according to Einstein’s first Postulate k\nis the only unknown.\n y’ = y and z’ = z since motion is along the\nx-direction.\nLorentz Transformation (5)\nLet\nx' k x v t (1)\nx k ' ( x' v t' ) (2)\n\n## By Einstein’s first postulate (All laws of Physics\n\nmust be the same in all inertial frames), we must\nhave k’ = k. i.e. k is the only unknown.\nBy Einstein’s second postulate (The speed of light\nis constant in all inertial frames), we must have x\n= c t , x’ = c t’.\nLorentz Transformation (6)\n\n## Substitute for x = c t and x’ = c t’, we have:\n\nc t' k c t v t (1)\nc t k c t' v t' (2)\nOr\n v\nt' k 1 t (3)\n c\n v\nt k 1 t' (4)\n c\nSubstitute from (4) into (3), we have:\nLorentz Transformation (7)\n\n v v \nt' k 1 k 1 t'\n c c \n v 2\n\n1 k 1 2 \n2\n\n c \n\n1\nk 2\nγ\nv\n1 2\nc\nEquations (1) and (2) now become\nx' γ x v t (1)\nx γ x' v t' (2)\nLorentz Transformation (8)\n\n## To calculate the “times” equations, we substitute\n\nfrom (1) into (2). We have:\nx γ γ x v t v t' γ 2 x v t γv t'\nx γ 2x\n γ t t'\nγv γv \n \n x x \n1 γ γ t γt\n1\nt' 2\n 1 \n γv γv v 2\n\n 1 - 2 \n c \n x c2 x v 2 \n 1 2 γ t 2 γt\n γv c v 2\n γv c v \n2\nLorentz Transformation (9)\n\n xv \nt' γ t - 2 (5)\n c \n\n1\nwhere γ\nv2\n1 2\nc\nWe can also find the expression for t to be:\n\n x' v \nt γ t' 2 (6)\n c \nLorentz Transformation (10)\nSummary of Lorentz Equations:\n\n## x' γ x v t x γ x' v t' \n\ny' y y y'\nz' z z z'\n xv x' v \nt' γ t - 2 t γ t' 2 \n c c \nSome Remarks\n\n## 1) If v << c, i.e., β ≈ 0 and γ ≈ 1, we see these\n\nequations reduce to the familiar Galilean\ntransformation.\n2) Space and time are now not separated.\n3) For non-imaginary transformations, the\nframe velocity cannot exceed c.\n\n## Some consequences of the above equations follow.\n\nSome Consequences of Lorentz Transformation\n\n## Time Dilation is predicted from Lorentz\n\nTransformation equations.\n Synchronization of clocks in different inertial\nframes is necessary.\n Length Contraction is also predicted from\nLorentz Transformation equations.\n Lorentz Transformation of Velocities in\ndifferent inertial frames is derivable from\nLorentz Transformation equations.\n Lorentz Transformation equations are in\nagreement with Einstein’s Second Postulate.\n1. Time Dilation (Again)\n\n## Consider the time interval between two events as\n\nreported in the two inertial frames S and S’. The time\ninterval T in the S frame (at rest) and T’ in the S’ frame\n(moving with constant velocity v) are:\n\n## T t 2 t1 , T' t '2 t 1'\n\nWhere the relations between the reported times are\ngiven by Lorentz Transformation equations:\n ' v x' ' v x' \nt1 t1 2 , t2 t2 2 \n c c \nt 2 t 1 t '2 t 1' T To\n2. Synchronization of Clocks\n\n## Step 1: Place observers with clocks\n\nthroughout a given system.\n\n## Step 2: In that system bring all the clocks\n\ntogether at one location.\n\n## Step 3: Compare the clock readings.\n\n If all of the clocks agree, then the clocks are\nsaid to be synchronized.\nA method to synchronize Clocks\n One way is to have one clock at the origin set to\nt = 0 and advance each clock by a time (d/c)\nwhere d is the distance of the clock from the\norigin.\n- Allow each of these clocks to begin timing\nwhen a light signal arrives from the origin.\n\nt=0\n\nd d\nt = d/c t = d/c\nSynchronization of clocks\n\n31\nThe Twin Paradox – Statement of the Problem\n\n## A thought experiment involving two twins\n\nnamed Speedo and Goslo.\n\n## Speedo travels to Planet X, 20 light years from\n\nearth.\n– His ship travels at 0.95c.\n– After reaching planet X, he immediately\nreturns to earth at the same speed.\n\n## When Speedo returns, he has aged 13 years,\n\nbut Goslo has aged 42 years.\n\n32\nThe Twins’ Perspectives\n\n## Goslo’s perspective is that he was at rest while\n\nSpeedo went on the journey.\n Speedo thinks he was at rest and Goslo and the\nearth raced away from him on a 6.5 year\njourney and then headed back toward him for\nanother 6.5 years.\n The paradox – which twin is the traveler and\nwhich is really older?\n\n33\nThe Twin Paradox – The Resolution\n\n## Relativity applies to reference frames moving at\n\nuniform speeds.\n The trip in this thought experiment is not\nsymmetrical since Speedo must experience a\nseries of accelerations during the journey.\n Therefore, Goslo can apply the time dilation\nformula with a proper time of 42 years.\n2\nv\nT Tp 1 42 1 0.95 13.1\n2\n\nc\n- This gives a time for Speedo of 13.1 years\nand this agrees with the earlier result.\n There is no true paradox since Speedo is not in\nan inertial frame.\n\n34\nTime Dilation Verification – Muon Decays\n\n## Muons are unstable particles\n\nthat have the same charge as\nan electron, but a mass 207\ntimes more than an electron.\n Muons have a half-life of\nΔtp = 2.2 µs when measured in a\nreference frame at rest with \nrespect to them (Figure a).\n Relative to an observer on\nearth, muons should have a\nlifetime of Δtp (Fugure b).\n A CERN experiment measured\npredictions of relativity.\n35\n3. Length Contraction\n\n## The measured distance between two points\n\ndepends on the frame of reference of the\nobserver\n\n## The proper length, Lp, of an object is the length\n\nof the object measured by someone at rest\nrelative to the object\n\n## The length of an object measured in a reference\n\nframe that is moving with respect to the object\nis always less than the proper length\n- This effect is known as length contraction\n\n36\nLength Contraction – Equation\n\n2\nLP v\n L\n\n LP 1 2\nc\n\n Length contraction\ntakes place only\nalong the direction\nof motion.\n\n37\nLength Contraction – Derivation(1)\n\n## Here we derive Length Contraction from\n\nLorentz Transformation equations:\nThe proper length Lp of a rod as measured in the\nmoving frame S’ is given by:\n\nLp x2 x1\n\n## The length L as measured in the stationary frame\n\nS is given by:\n\nL x 2 x1\nLength Contraction – Derivation(2)\n\n## In S’, measurements of the two position\n\ncoordinates that specify the length Lp are given\nby Lorentz equation:\n\n## x2 γ x 2 v t 2 , x1 γ x1 v t 1 \n\nLp x2 x1 γ x 2 v t 2 γ x 1 v t 1 \n\n## In S, t2 = t1 since x2 and x1 are measured\n\nsimultaneously.\nLength Contraction – Derivation(3)\n\n## Expanding the brackets we get:\n\nLp x2 x1 γx 2 γv t 1 γx 1 γv t 1\n\nLp\nLp γ( x 2 x 1 ) γL or L\nγ\n1\nγ v2\nBut v2 L Lp 1 2\n1 2 c\nc\nLength Contraction – Illustration\n\n## v/c L/Lp v/c L/Lp\n\n0.1000 0.9950 0.9600 0.2800\n0.4000 0.9165 0.9900 0.1411 Relativistic Length Contraction\n\n## 0.6000 0.8000 0.9920 0.1262 1.0\n\n0.8\n0.8000 0.6000 0.9940 0.1094\n\n(L/Lp) Ratio\n0.6\n0.9000 0.4359 0.9950 0.0999 0.4\n\n## 0.9200 0.3919 0.9970 0.0774 0.0\n\n0.0 0.5 1.0\n(V/C) Ratio\n0.9300 0.3676 0.9980 0.0632\n0.9400 0.3412 0.9990 0.0447\n0.9500 0.3122 0.9999 0.0141\n4. Equivalence of L.T. and G.T. in the\nNon-relativistic Limit\nWhen v<<c, we have from L.T. first equation:\nx γ x v t \n\n## 1 When v<<c, γ→1\n\nwhere γ \nv2\n1 2 We get agreement with\nc G.T. equation.\n\n## However, because of Time Dilation and Length\n\nContraction, Time is NOT absolute. Here we have\nan Illustration.\n4. Equivalence of L.T. and G.T. in the\nNon-relativistic Limit\nConsider the motion of the origin of O’. t’=t=0 at\nthe start of the motion.\nAfter time t in O, O’ has moved a distance x=vt.\n\n## Substitute for x in the time equation in L.T.\n\n v vt v2 \nt 1 2 \n1 x v c 2\n c \nt' t - 2 \n t\nv \n2\nc v 2\nv2\n1- 2 1- 2 1- 2\nc c c\nv2\n t 1 - 2 . When v c t t\nc\n5. Lorentz Velocity Transformation L.V.T.\n\n## From L.T. equations, we have:\n\nx γ x v t xv\nt' γ t - 2 \n c \ndx dx\nLet ux , ux\ndt dt\n\n v \ndx γ dx v dt dt' γ dt - 2 dx \n c \n\n dx \n v ux v\ndx dx v dt dt \n or ux \ndt dt v dx v dx v\n1 2 ux\n1 2 \nc 2\nc dt c\n5. Lorentz Velocity Transformation L.V.T.\n\nux v\nux (1)\nv\n1 2 ux\nc\n\n## Similarly, if the object under study has velocity\n\ncomponents along the y and z axes, the\ncomponents as measured by an observer in S’\nare:\nuy uz\nuy (2) u (3)\n \nz\n v v\n 1 2 ux \n 1 2 ux \n c c \n5. Lorentz Velocity Transformation\nSummary of Equations\nux v ux v\nux (1) ux \n v v\n \n 1 u x 1 2 ux \n \n2\nc c\nuy uy\nuy (2) uy \n v v\n \n 1 2 ux 1 2 ux \n c c \nuz uz\nuz (3) uz \n v \nv\n 1 2 ux 1 2 ux \n c c \n6. Constancy of the speed of light (1)\n\nAlso\nIf ux = c then u’x = c If u'x = c then ux = c\nFrom eq. (1), put ux= c From eq. (1), put u’x= c\n \n\ncv cv\n\nux c ux c\nv v\n1 2 c 1 2 c\nc c\n6. Constancy of the speed of light (2)\n\n## The speed of an object can never exceed the\n\nvelocity of light c.\n\n## Let u’x= k c , where k is a positive constant. Also\n\nlet S’ travels at speed v = k c relative to S.\nNow we find ux:\n\nkc kc 2k Let us evaluate\nux c 2 \nkc\n1 2 kc 1 k the maximum\nc value of k.\n6. Constancy of the speed of light (3)\n\nLet 2k \nA 2 \n 1 k \ndA\nA is a maximum when 0\ndk\ndA 1 k 2 2 - 2k2k \n \n 0\ndk 1 k 2 2\n\n1 k 2 - 2k 2 0 k 1\nSubstitute the above result in A, you get:\nA(maximum) 1 ux (maximum) c\n\n6. Constancy of the velocity of light (4)\n\n## The speed of light c is the same in all directions in\n\nall inertial frames.\n\n## In slide 123, we showed that the speed of light\n\npropagating along x’-axis and the frame S’ moves\nwith velocity v along the same axis.\n\n## Consider next the case of observers moving\n\nalong O’y’ at right angle to the direction of\npropagation of light which is O’x’.\n\n## ux 0 , uy c , uz 0\n\n6. Constancy of the speed of light (5)\nux 0 , uy c , uz 0\nSubstitute in the Ux equation in L.T. \nux v uy c\nux v , uy \nv v γ\n1 2 ux γ 1 2 ux \nc c \nuz\nuz 0 \nv u u2x u2y u2z\n1 2 ux\nc\n\n2 v \n2 2\nc\nu c 2 c c 1 2 c\n2 2\n\nγ c \n6. Constancy of the speed of light (6)\n\nc c c!\nIf we substitute u’x = c and v = c in the ux equation\nin L.T. \n\nux v cc\nux c\nv c\n1 2 ux 1 2 c\nc c\nConclusion:\nThe relative velocity of two objects or two\nframes or an object in a frame can not\nexceed c."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6865063,"math_prob":0.97792155,"size":3792,"snap":"2019-43-2019-47","text_gpt3_token_len":1374,"char_repetition_ratio":0.104276665,"word_repetition_ratio":0.0472255,"special_character_ratio":0.3486287,"punctuation_ratio":0.08394544,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9834041,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T20:59:06Z\",\"WARC-Record-ID\":\"<urn:uuid:8318d4a3-b315-42d6-902c-b21aa86fba45>\",\"Content-Length\":\"328913\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6972ad19-f893-496f-b92f-a67a93010b44>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea0cca1f-c060-466a-9628-6d6d4b716a0a>\",\"WARC-IP-Address\":\"151.101.202.152\",\"WARC-Target-URI\":\"https://www.scribd.com/document/368989041/Special-Theory-of-Relativity-Part-II\",\"WARC-Payload-Digest\":\"sha1:YRBWXILVG2Q4XD5DEBWWCJP77Q56QIWE\",\"WARC-Block-Digest\":\"sha1:PVLDJMGUVJZYXR45MF7M4JBA4JG34XOC\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665767.51_warc_CC-MAIN-20191112202920-20191112230920-00499.warc.gz\"}"} |
https://matrixread.com/template-functions-in-c/ | [
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null,
"# Template Functions in C++\n\nTemplate Functions in C++ are easy to implement and powerful at the same time, also known as Generic Programming, where we don’t specify the data type and can use any type when needed. The template function follows the concept of data type as a parameter for initializing variables at compile time.\n\n## Why do we need Templates?\n\nConsider the below example of an addition function,\n\n``````int add(int a, int b)\n{\nreturn a+b;\n}``````\n\nThe above function adds two integers and returns the output as an integer, what if we want to pass two decimal numbers to the same function? again we need to write another function that accepts two floating-point numbers i.e decimal numbers.\n\n``````float add(float a, float b)\n{\nreturn a+b;\n}``````\n\nSince this was a simple and small program we don’t care much about it however, when we write complex functions re-writing all the functions for all data types will take a lot of time, space and compile-time/run-time.\n\nNow the concept of Templates Functions can be used here where we just specify what the function needs to do and we need not worry about the data type as it will be taken care by the compiler at compile time.\n\n## Important Note\n\nWhenever an argument of certain data types is passed to a template function the compiler generates copies of the function with the given data type, this is how template functions work.\n\n## Template Function Syntax\n\nThe template function starts with the keyword `template` and the `<>` represent the template parameters and the function declaration in the next line.\n\n``````template <class T>\nT functionName(T arguments)\n{\n... ... ...\n//function body\n}``````\n\nHere T represents the data type as a parameter for the template function, and in the first line we can also use the keyword `typename` instead of the `class` keyword.\n\n### Example: Addition Function using Templates in C++\n\n``````#include <iostream>\nusing namespace std;\n\ntemplate <class T>\n{\nreturn a + b;\n}\nint main()\n{\ncout << add<int>(1, 2) << endl;\ncout << add<double>(1.5, 2.5) << endl;\nreturn 0;\n}``````\n\n### Output\n\n``````3\n3.5``````\n\n## Conclusion\n\nTemplate functions really save a lot of time, and I’m glad we learned them today! What do you say? and by the way, this post is a part of my #30DaysChallenge to write a blog post every day on what I learn, I’ll be very happy if you can share this with a friend, Wishing you more Happiness~ Abhiram Reddy."
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"https://matrixread.com/wp-content/uploads/2020/10/Template-Functions-930x620.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7603405,"math_prob":0.9501374,"size":2308,"snap":"2021-43-2021-49","text_gpt3_token_len":514,"char_repetition_ratio":0.16015625,"word_repetition_ratio":0.014742015,"special_character_ratio":0.23310225,"punctuation_ratio":0.10944206,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9745735,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T04:41:51Z\",\"WARC-Record-ID\":\"<urn:uuid:c63e6cc1-7222-42f3-8356-c46994703305>\",\"Content-Length\":\"63712\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c754c330-2936-445a-8f9e-95a5241c6359>\",\"WARC-Concurrent-To\":\"<urn:uuid:b6df8704-b34b-4cf1-8d4a-e5dd1a58a145>\",\"WARC-IP-Address\":\"172.67.219.78\",\"WARC-Target-URI\":\"https://matrixread.com/template-functions-in-c/\",\"WARC-Payload-Digest\":\"sha1:HFOYLLVU74TYGGVAKABHYUCF6GBDBKF6\",\"WARC-Block-Digest\":\"sha1:RROSMWSRBHIVVD2FLEK3QVJ4VTPP7ESO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363659.21_warc_CC-MAIN-20211209030858-20211209060858-00004.warc.gz\"}"} |
https://talkstats.com/tags/t-test-2/ | [
"# t-test\n\n1. ### T-statistics\n\nHi, I am trying to calculate t -statistics (formula below). I am not sure if I need to use a one-tail or two-tail test. I have attached an excel sheet with the calculation (i used t.inv.2t - line q25). The issue is; I am not getting the expected value which is 0.113976896 (the value I am...\n2. ### Paired t test unequal sample sizes?\n\nI want to find if there's a significance between the two levels average mean values from my dependent variable \"X,\" which are \"High X\" and \"Low X.\" High X (85 items) + Low X (65 items) = X (150 items) Both have unequal sample size, but they're from the same group \"X.\" I opted for paired sample...\n3. ### Do I need a two-way MANOVA test?\n\nI am testing for differences between 3 groups. I have 2 dependent variables which are continuous data. However, I am confused about my independent variables and if a two-way MANOVA test would be suitable. My three groups are Rural, Semi-rural and Urban (3 groups). For each group I am testing...\n4. ### Question: Single person - T-test?\n\nCan you look at improvement in scores across two time periods, if you have hundred of data points but they are all from a single individual. Can I do a t-test to compare the indvidual's scores over one period to the same individual's scores over a later period? I've never encountered anything...\n5. ### What statistic method should I use for this situation?\n\nImagine if I have 3 types of parts A, B, C and there is one parameter to judge the quality of the parts under certain condition, but this parameter is not that reproducible. For example, under certain condition, A, B and C will have different value of electric surge. I understand if we collect...\n6. ### Why does the “linear regression t-test” return a p-value (two-tailed) from regression that is twice the p-value from ANOVA? (Binary predictor)\n\nI'm using the \"linear regression t-test\" guide at https://stattrek.com/regression/slope-test.aspx The guide shows calculating t =b1/SE, where b1 and SE are provided by the regression function (here lm() - using R.) The guide shows the p-value gets doubled as this is a two-sided test...\n7. ### Constant sum question - results analysis\n\nI ran a study with a constant sum question: participants had to divide 100 units (I called the units USD) between 5 possibilities. In other words, they had to bet on five possibilities: 1 object out of 5 has a property vs. 2 objects out of 5 have a property vs. 3 objects out of five have a...\n8. ### Help with my research question analysis\n\nHi everyone, I am a doctor working on a project investigating an intervention we have been using in the intensive care unit in my hospital for treating Covid patients. I am collecting data on all the patients who have been admitted and look at whether they have had this intervention, whether it...\n9. ### Testing for similarity between countries\n\nFor my thesis, I am looking at the effect of environmental controversies on the profitability of Chinese and European firms. As part of my exploratory analysis, I have to check whether or not the European countries are similar enough to consider them as 1 group. I collected data on economic...\n10. ### chi square or unpaired t-test? (or other??)\n\nHi - I am looking to compare two groups - mobility activity after surgery. Group 1 - is a 'low ability' (ie, very old, frail) Group 2 - is 'high ability' (able to move around before surgery). The numbers in each group are NOT means, they are simply counts of patients. Ie, on the first day, 37...\n11. ### choice of t-test\n\nHello I have been presented with some data and, being a bit rusty on my statistics am unsure whether I have used the correct t-test. The data shows the results of of farmer trials; 22 plot-pairs, with each plot-pair containing a control and a test. Based on my understanding, the correct t-test...\n12. ### t-test or ANOVA? urgent\n\nHi, I am looking for advice about which stat test to use in my data analysis for our psychology class. We collected data on reasoning tests scores. I have two groups of participants 1) non-bilingual 2) bilingual and dependent variable in the form of reasoning scores in 1) spatial reasoning...\n13. ### Thesis Help Request - Hypothesis Testing\n\nHi I am doing an event study and my null hypothesis is CAAR=0 and alternative is CAAR does not equal 0 (2 tailed t-test). CAAR is cumulative average abnormal returns. Can a two tailed test be used to give directional conclusions, e.g. if the CAAR value is 1.5% and is statistically significant at...\n14. ### Sample size when adding two means\n\nI want to compare a theoretical mean to a measured mean. But I determine this theoretical mean from adding two measured means. For example, let's say I want to determine if mixing a shade of white paint with a shade of black paint yields a particular gray paint (let’s pretend I can quantify...\n15. ### What is the best statistical test/approach for pre- and post-intervention analysis?\n\nDear colleagues, We assessed several psychological measures (e.g. self-compassion scale, five-facets mindfulness questionnaire, satisfaction with life scale) before and after a Mindfulness-Based Psychoeducational Program for groups of people with acquired brain injuries. We would like to...\n16. ### Paired test?\n\nHi, I have the following problem. In my experimental setup I have four tissue samples from the same animal in each experiment, two from one side and two from the other. The original plan is to apply a substance alone on a sample from one side and the substance mixed with the inhibitor on...\n17. ### Choosing the proper statistical test for a lottery recommendation system\n\nDear friends, I'm now working on the recommendation system to recommend football match lottery to the users. The users may choose a match to bet arbitrary amount of money, e.g. $2,$4, or even \\$20,000. I want to compare if two recommendation algorithms are different (or if one outperforms...\n18. ### Test wanted for comparing algorithms\n\nStudy design Suppose I want to compare algorithm runtime performance (measured in seconds) of different algorithms for the same problem type. There are several problem instances for this problem type on which the algorithms are tested and performance is measured. Possible tests As far as I...\n19. ### What kind of t-test would I use for this?\n\nOne variable is injuries yes/no, the other variable is rating how much in control they felt (0=none,100=total). Basically if injuries causes low levels of perceived control. I have to use a t-test, but there are so many and I really don't know which one.\n20. ### Matt Whiteny U test Vs T test\n\nHello fellow statisticians, Problem definition: I need to test whether the difference between the \"mean\" utilization metrics of two machines is statistically significant. Given Data: Utilisation data is available for two machines for more than 100 days. Therefore, the mean, standard..."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9038994,"math_prob":0.6773507,"size":297,"snap":"2021-31-2021-39","text_gpt3_token_len":67,"char_repetition_ratio":0.10921501,"word_repetition_ratio":0.0,"special_character_ratio":0.22222222,"punctuation_ratio":0.13846155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95783323,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T03:31:47Z\",\"WARC-Record-ID\":\"<urn:uuid:27fa9bde-970c-412c-b95b-108091a4b003>\",\"Content-Length\":\"54177\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac7ac2e8-1c9c-4afc-a12c-487a4c40fb4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1a5f584-3c4e-4184-a86e-5d24b63e3198>\",\"WARC-IP-Address\":\"199.167.200.62\",\"WARC-Target-URI\":\"https://talkstats.com/tags/t-test-2/\",\"WARC-Payload-Digest\":\"sha1:AC6Z43VETFP423ZOP77XQR2OHQHJEJZM\",\"WARC-Block-Digest\":\"sha1:NQHWTXFQAYWRPH65MURN6C6GY7R6RFNC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057416.67_warc_CC-MAIN-20210923013955-20210923043955-00418.warc.gz\"}"} |
https://www.litscape.com/word_analysis/faceshield | [
"# faceshield in Scrabble®\n\nThe word faceshield is playable in Scrabble®, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nFACESHIELD\n(180)\nFACESHIELD\n(180)\n\n## Seven Letter Word Alert: (20 words)\n\naediles, clashed, deciles, defaces, defiles, defleas, deflesh, dehisce, descale, filched, filches, flashed, fleshed, halides, helices, leached, leaches, leashed, sealife, sheafed\n\nFACESHIELD\n(180)\nFACESHIELD\n(180)\nFACESHIELD\n(120)\nFACESHIELD\n(120)\nFACESHIELD\n(92)\nFACESHIELD\n(80)\nFACESHIELD\n(76)\nFACESHIELD\n(76)\nFACESHIELD\n(76)\nFACESHIELD\n(76)\nFACESHIELD\n(72)\nFACESHIELD\n(66)\nFACESHIELD\n(66)\nFACESHIELD\n(62)\nFACESHIELD\n(60)\nFACESHIELD\n(58)\nFACESHIELD\n(58)\nFACESHIELD\n(48)\nFACESHIELD\n(48)\nFACESHIELD\n(48)\nFACESHIELD\n(48)\nFACESHIELD\n(48)\nFACESHIELD\n(46)\nFACESHIELD\n(46)\nFACESHIELD\n(44)\nFACESHIELD\n(44)\nFACESHIELD\n(42)\nFACESHIELD\n(42)\nFACESHIELD\n(42)\nFACESHIELD\n(40)\nFACESHIELD\n(40)\nFACESHIELD\n(40)\nFACESHIELD\n(38)\nFACESHIELD\n(38)\nFACESHIELD\n(33)\nFACESHIELD\n(31)\nFACESHIELD\n(27)\nFACESHIELD\n(26)\nFACESHIELD\n(25)\nFACESHIELD\n(25)\nFACESHIELD\n(24)\nFACESHIELD\n(24)\nFACESHIELD\n(23)\nFACESHIELD\n(22)\n\nFACESHIELD\n(180)\nFACESHIELD\n(180)\nFACESHIELD\n(120)\nFACESHIELD\n(120)\nFILCHED\n(114 = 64 + 50)\nFILCHED\n(110 = 60 + 50)\nFILCHED\n(110 = 60 + 50)\nFILCHES\n(110 = 60 + 50)\nFILCHED\n(107 = 57 + 50)\nFILCHES\n(107 = 57 + 50)\nFILCHES\n(107 = 57 + 50)\nFILCHED\n(107 = 57 + 50)\nFLASHED\n(106 = 56 + 50)\nSHEAFED\n(106 = 56 + 50)\nFLESHED\n(106 = 56 + 50)\nDEFLESH\n(106 = 56 + 50)\nDEFLESH\n(104 = 54 + 50)\nFILCHED\n(104 = 54 + 50)\nDEFLESH\n(104 = 54 + 50)\nFLASHED\n(104 = 54 + 50)\nFILCHES\n(104 = 54 + 50)\nFLASHED\n(104 = 54 + 50)\nSHEAFED\n(104 = 54 + 50)\nFILCHES\n(104 = 54 + 50)\nFLESHED\n(104 = 54 + 50)\nFLESHED\n(104 = 54 + 50)\nSHEAFED\n(104 = 54 + 50)\nCLASHED\n(102 = 52 + 50)\nDEHISCE\n(102 = 52 + 50)\nLEACHED\n(102 = 52 + 50)\nDEFACES\n(102 = 52 + 50)\nLEACHED\n(101 = 51 + 50)\nDEFACES\n(101 = 51 + 50)\nFILCHED\n(101 = 51 + 50)\nDEHISCE\n(101 = 51 + 50)\nCLASHED\n(101 = 51 + 50)\nFILCHED\n(101 = 51 + 50)\nFILCHED\n(101 = 51 + 50)\nLEACHED\n(98 = 48 + 50)\nFILCHES\n(98 = 48 + 50)\nHELICES\n(98 = 48 + 50)\nFILCHED\n(98 = 48 + 50)\nFILCHED\n(98 = 48 + 50)\nFLASHED\n(98 = 48 + 50)\nDEFLESH\n(98 = 48 + 50)\nFLESHED\n(98 = 48 + 50)\nDEHISCE\n(98 = 48 + 50)\nFILCHES\n(98 = 48 + 50)\nCLASHED\n(98 = 48 + 50)\nHELICES\n(98 = 48 + 50)\nLEACHES\n(98 = 48 + 50)\nFILCHES\n(98 = 48 + 50)\nSHEAFED\n(98 = 48 + 50)\nDEFACES\n(98 = 48 + 50)\nLEACHES\n(98 = 48 + 50)\nLEACHED\n(98 = 48 + 50)\nFILCHES\n(98 = 48 + 50)\nFILCHES\n(96 = 46 + 50)\nDEFLESH\n(95 = 45 + 50)\nDEFLESH\n(95 = 45 + 50)\nHELICES\n(95 = 45 + 50)\nDEHISCE\n(95 = 45 + 50)\nSHEAFED\n(95 = 45 + 50)\nSHEAFED\n(95 = 45 + 50)\nDEFLESH\n(95 = 45 + 50)\nDEFLESH\n(95 = 45 + 50)\nDEFLESH\n(95 = 45 + 50)\nFLASHED\n(95 = 45 + 50)\nFLASHED\n(95 = 45 + 50)\nLEACHES\n(95 = 45 + 50)\nFLESHED\n(95 = 45 + 50)\nLEASHED\n(95 = 45 + 50)\nFLESHED\n(95 = 45 + 50)\nFLESHED\n(95 = 45 + 50)\nFLESHED\n(95 = 45 + 50)\nCLASHED\n(95 = 45 + 50)\nFILCHES\n(95 = 45 + 50)\nFLESHED\n(95 = 45 + 50)\nDEFLEAS\n(95 = 45 + 50)\nLEACHES\n(95 = 45 + 50)\nDEFILES\n(95 = 45 + 50)\nHALIDES\n(95 = 45 + 50)\nLEACHED\n(95 = 45 + 50)\nSHEAFED\n(95 = 45 + 50)\nSHEAFED\n(95 = 45 + 50)\nSHEAFED\n(95 = 45 + 50)\nFLASHED\n(95 = 45 + 50)\nFLASHED\n(95 = 45 + 50)\nFLASHED\n(95 = 45 + 50)\nDEFACES\n(95 = 45 + 50)\nDEFLESH\n(94 = 44 + 50)\nDEFLEAS\n(94 = 44 + 50)\nDEFILES\n(94 = 44 + 50)\nSHEAFED\n(94 = 44 + 50)\nFILCHED\n(94 = 44 + 50)\nSHEAFED\n(94 = 44 + 50)\nLEASHED\n(94 = 44 + 50)\nFLESHED\n(94 = 44 + 50)\nFLASHED\n(94 = 44 + 50)\nHALIDES\n(94 = 44 + 50)\nDEHISCE\n(92 = 42 + 50)\nCLASHED\n(92 = 42 + 50)\nDEFACES\n(92 = 42 + 50)\nDEHISCE\n(92 = 42 + 50)\nFILCHED\n(92 = 42 + 50)\nDEFACES\n(92 = 42 + 50)\nDEFACES\n(92 = 42 + 50)\nLEACHED\n(92 = 42 + 50)\nSHEAFED\n(92 = 42 + 50)\nDEHISCE\n(92 = 42 + 50)\nDEHISCE\n(92 = 42 + 50)\nFLASHED\n(92 = 42 + 50)\nLEACHED\n(92 = 42 + 50)\nLEACHED\n(92 = 42 + 50)\nDEHISCE\n(92 = 42 + 50)\nDEFLESH\n(92 = 42 + 50)\nCLASHED\n(92 = 42 + 50)\nLEACHED\n(92 = 42 + 50)\nDEFACES\n(92 = 42 + 50)\nSEALIFE\n(92 = 42 + 50)\nDEHISCE\n(92 = 42 + 50)\nFACESHIELD\n(92)\nCLASHED\n(92 = 42 + 50)\nCLASHED\n(92 = 42 + 50)\nFLESHED\n(92 = 42 + 50)\nCLASHED\n(92 = 42 + 50)\nCLASHED\n(92 = 42 + 50)\nLEACHED\n(92 = 42 + 50)\nDEFACES\n(92 = 42 + 50)\nDEFACES\n(92 = 42 + 50)\nDECILES\n(90 = 40 + 50)\nLEACHES\n(90 = 40 + 50)\nDESCALE\n(90 = 40 + 50)\nFILCHED\n(90 = 40 + 50)\nFLESHED\n(90 = 40 + 50)\nFILCHED\n(90 = 40 + 50)\nSEALIFE\n(90 = 40 + 50)\nFILCHED\n(90 = 40 + 50)\nDEFLESH\n(90 = 40 + 50)\nSHEAFED\n(90 = 40 + 50)\nFILCHES\n(90 = 40 + 50)\nFLASHED\n(90 = 40 + 50)\nFILCHES\n(90 = 40 + 50)\nLEACHES\n(89 = 39 + 50)\nLEACHES\n(89 = 39 + 50)\nLEACHES\n(89 = 39 + 50)\nLEACHES\n(89 = 39 + 50)\nDEHISCE\n(89 = 39 + 50)\nDEFILES\n(89 = 39 + 50)\nHELICES\n(89 = 39 + 50)\nHELICES\n(89 = 39 + 50)\nDESCALE\n(89 = 39 + 50)\nCLASHED\n(89 = 39 + 50)\nHELICES\n(89 = 39 + 50)\nDECILES\n(89 = 39 + 50)\nHELICES\n(89 = 39 + 50)\nLEASHED\n(89 = 39 + 50)\nHELICES\n(89 = 39 + 50)\nLEACHES\n(89 = 39 + 50)\nDEFACES\n(89 = 39 + 50)\nHELICES\n(89 = 39 + 50)\nDEFLEAS\n(89 = 39 + 50)\nDESCALE\n(89 = 39 + 50)\nHALIDES\n(89 = 39 + 50)\nLEACHED\n(89 = 39 + 50)\nDEFLEAS\n(88 = 38 + 50)\nDEHISCE\n(88 = 38 + 50)\nLEASHED\n(88 = 38 + 50)\nFILCHES\n(88 = 38 + 50)\nFLESHED\n(88 = 38 + 50)\nDEFILES\n(88 = 38 + 50)\nFLASHED\n(88 = 38 + 50)\nFILCHES\n(88 = 38 + 50)\nCLASHED\n(88 = 38 + 50)\nDEFLESH\n(88 = 38 + 50)\nDEFACES\n(88 = 38 + 50)\nDEFACES\n(88 = 38 + 50)\nFILCHES\n(88 = 38 + 50)\nDEHISCE\n(88 = 38 + 50)\nLEACHED\n(88 = 38 + 50)\nDEFILES\n(86 = 36 + 50)\nFILCHED\n(86 = 36 + 50)\nDEFILES\n(86 = 36 + 50)\nFILCHED\n(86 = 36 + 50)\nFILCHED\n(86 = 36 + 50)\nDEFILES\n(86 = 36 + 50)\nSEALIFE\n(86 = 36 + 50)\nDEFILES\n(86 = 36 + 50)\nFILCHED\n(86 = 36 + 50)\nLEASHED\n(86 = 36 + 50)\nLEASHED\n(86 = 36 + 50)\nHELICES\n(86 = 36 + 50)\nFLESHED\n(86 = 36 + 50)\nFLESHED\n(86 = 36 + 50)\nFLESHED\n(86 = 36 + 50)\nHALIDES\n(86 = 36 + 50)\nHALIDES\n(86 = 36 + 50)\nHALIDES\n(86 = 36 + 50)\nHALIDES\n(86 = 36 + 50)\nHALIDES\n(86 = 36 + 50)\n\n# faceshield in Words With Friends™\n\nThe word faceshield is playable in Words With Friends™, no blanks required. Because it is longer than 7 letters, you would have to play off an existing word or do it in several moves.\n\nFACESHIELD\n(252)\n\n## Seven Letter Word Alert: (20 words)\n\naediles, clashed, deciles, defaces, defiles, defleas, deflesh, dehisce, descale, filched, filches, flashed, fleshed, halides, helices, leached, leaches, leashed, sealife, sheafed\n\nFACESHIELD\n(252)\nFACESHIELD\n(216)\nFACESHIELD\n(132)\nFACESHIELD\n(132)\nFACESHIELD\n(96)\nFACESHIELD\n(96)\nFACESHIELD\n(90)\nFACESHIELD\n(88)\nFACESHIELD\n(88)\nFACESHIELD\n(84)\nFACESHIELD\n(84)\nFACESHIELD\n(84)\nFACESHIELD\n(80)\nFACESHIELD\n(80)\nFACESHIELD\n(78)\nFACESHIELD\n(72)\nFACESHIELD\n(64)\nFACESHIELD\n(60)\nFACESHIELD\n(60)\nFACESHIELD\n(56)\nFACESHIELD\n(56)\nFACESHIELD\n(52)\nFACESHIELD\n(52)\nFACESHIELD\n(44)\nFACESHIELD\n(40)\nFACESHIELD\n(40)\nFACESHIELD\n(40)\nFACESHIELD\n(40)\nFACESHIELD\n(30)\nFACESHIELD\n(29)\nFACESHIELD\n(27)\nFACESHIELD\n(27)\nFACESHIELD\n(27)\nFACESHIELD\n(27)\nFACESHIELD\n(27)\nFACESHIELD\n(26)\nFACESHIELD\n(26)\nFACESHIELD\n(26)\nFACESHIELD\n(25)\nFACESHIELD\n(25)\nFACESHIELD\n(25)\nFACESHIELD\n(24)\nFACESHIELD\n(23)\nFACESHIELD\n(23)\nFACESHIELD\n(22)\nFACESHIELD\n(22)\n\nFACESHIELD\n(252)\nFACESHIELD\n(216)\nFACESHIELD\n(132)\nFACESHIELD\n(132)\nFILCHED\n(122 = 87 + 35)\nFILCHES\n(119 = 84 + 35)\nFILCHED\n(116 = 81 + 35)\nFILCHED\n(116 = 81 + 35)\nFILCHED\n(116 = 81 + 35)\nFILCHES\n(113 = 78 + 35)\nDEFLESH\n(113 = 78 + 35)\nDEFACES\n(113 = 78 + 35)\nFILCHES\n(113 = 78 + 35)\nSHEAFED\n(110 = 75 + 35)\nFILCHED\n(110 = 75 + 35)\nFILCHED\n(110 = 75 + 35)\nFILCHED\n(110 = 75 + 35)\nDEFILES\n(107 = 72 + 35)\nFILCHES\n(107 = 72 + 35)\nCLASHED\n(107 = 72 + 35)\nLEACHED\n(107 = 72 + 35)\nDEFACES\n(107 = 72 + 35)\nFLESHED\n(107 = 72 + 35)\nDESCALE\n(107 = 72 + 35)\nDEFLEAS\n(107 = 72 + 35)\nFLESHED\n(107 = 72 + 35)\nFLASHED\n(107 = 72 + 35)\nFILCHES\n(107 = 72 + 35)\nFILCHES\n(107 = 72 + 35)\nFLASHED\n(107 = 72 + 35)\nLEACHED\n(107 = 72 + 35)\nLEACHED\n(107 = 72 + 35)\nFILCHES\n(107 = 72 + 35)\nDECILES\n(107 = 72 + 35)\nCLASHED\n(107 = 72 + 35)\nDEHISCE\n(104 = 69 + 35)\nLEACHES\n(104 = 69 + 35)\nLEACHES\n(104 = 69 + 35)\nSEALIFE\n(104 = 69 + 35)\nHELICES\n(104 = 69 + 35)\nDEHISCE\n(104 = 69 + 35)\nHELICES\n(104 = 69 + 35)\nFILCHED\n(103 = 68 + 35)\nFILCHED\n(103 = 68 + 35)\nFILCHED\n(103 = 68 + 35)\nFLESHED\n(101 = 66 + 35)\nFLASHED\n(101 = 66 + 35)\nLEACHED\n(101 = 66 + 35)\nCLASHED\n(101 = 66 + 35)\nDESCALE\n(101 = 66 + 35)\nDEFLESH\n(101 = 66 + 35)\nLEACHED\n(101 = 66 + 35)\nFILCHES\n(99 = 64 + 35)\nFILCHES\n(99 = 64 + 35)\nFILCHES\n(99 = 64 + 35)\nSHEAFED\n(98 = 63 + 35)\nFILCHED\n(98 = 63 + 35)\nDEHISCE\n(98 = 63 + 35)\nLEASHED\n(98 = 63 + 35)\nLEACHES\n(98 = 63 + 35)\nHALIDES\n(98 = 63 + 35)\nLEACHES\n(98 = 63 + 35)\nLEACHES\n(98 = 63 + 35)\nFACESHIELD\n(96)\nFACESHIELD\n(96)\nDEFLESH\n(95 = 60 + 35)\nLEACHED\n(95 = 60 + 35)\nDESCALE\n(95 = 60 + 35)\nFLASHED\n(95 = 60 + 35)\nDEFLESH\n(95 = 60 + 35)\nDEFLESH\n(95 = 60 + 35)\nFLESHED\n(95 = 60 + 35)\nCLASHED\n(95 = 60 + 35)\nDESCALE\n(95 = 60 + 35)\nHELICES\n(92 = 57 + 35)\nSEALIFE\n(92 = 57 + 35)\nLEACHES\n(92 = 57 + 35)\nSHEAFED\n(92 = 57 + 35)\nFILCHED\n(92 = 57 + 35)\nFILCHED\n(92 = 57 + 35)\nCLASHED\n(91 = 56 + 35)\nDEFLESH\n(91 = 56 + 35)\nDEFLESH\n(91 = 56 + 35)\nDEFACES\n(91 = 56 + 35)\nDEFLESH\n(91 = 56 + 35)\nDEFACES\n(91 = 56 + 35)\nCLASHED\n(91 = 56 + 35)\nFLESHED\n(91 = 56 + 35)\nLEACHED\n(91 = 56 + 35)\nLEACHED\n(91 = 56 + 35)\nDEFACES\n(91 = 56 + 35)\nFLESHED\n(91 = 56 + 35)\nFLESHED\n(91 = 56 + 35)\nFLASHED\n(91 = 56 + 35)\nLEACHED\n(91 = 56 + 35)\nCLASHED\n(91 = 56 + 35)\nFLASHED\n(91 = 56 + 35)\nFLASHED\n(91 = 56 + 35)\nFACESHIELD\n(90)\nDEFACES\n(89 = 54 + 35)\nFILCHES\n(89 = 54 + 35)\nCLASHED\n(89 = 54 + 35)\nLEACHED\n(89 = 54 + 35)\nFILCHES\n(89 = 54 + 35)\nDEFLEAS\n(89 = 54 + 35)\nFILCHES\n(89 = 54 + 35)\nFLESHED\n(89 = 54 + 35)\nDECILES\n(89 = 54 + 35)\nFLASHED\n(89 = 54 + 35)\nDESCALE\n(89 = 54 + 35)\nDEFLEAS\n(89 = 54 + 35)\nDEFACES\n(89 = 54 + 35)\nFLESHED\n(89 = 54 + 35)\nDEFLESH\n(89 = 54 + 35)\nFLESHED\n(89 = 54 + 35)\nCLASHED\n(89 = 54 + 35)\nDEFACES\n(89 = 54 + 35)\nFLASHED\n(89 = 54 + 35)\nCLASHED\n(89 = 54 + 35)\nLEACHED\n(89 = 54 + 35)\nDEFLESH\n(89 = 54 + 35)\nFLASHED\n(89 = 54 + 35)\nDEFLESH\n(89 = 54 + 35)\nDEFILES\n(89 = 54 + 35)\nFACESHIELD\n(88)\nFACESHIELD\n(88)\nLEACHES\n(87 = 52 + 35)\nHELICES\n(87 = 52 + 35)\nSHEAFED\n(87 = 52 + 35)\nLEACHES\n(87 = 52 + 35)\nLEACHES\n(87 = 52 + 35)\nHELICES\n(87 = 52 + 35)\nSHEAFED\n(87 = 52 + 35)\nDEHISCE\n(87 = 52 + 35)\nSHEAFED\n(87 = 52 + 35)\nDEHISCE\n(87 = 52 + 35)\nDEHISCE\n(87 = 52 + 35)\nHELICES\n(87 = 52 + 35)\nDEHISCE\n(86 = 51 + 35)\nSHEAFED\n(86 = 51 + 35)\nDEHISCE\n(86 = 51 + 35)\nLEASHED\n(86 = 51 + 35)\nHALIDES\n(86 = 51 + 35)\nSHEAFED\n(86 = 51 + 35)\nSHEAFED\n(86 = 51 + 35)\nHALIDES\n(86 = 51 + 35)\nLEACHES\n(86 = 51 + 35)\nDEHISCE\n(86 = 51 + 35)\nSEALIFE\n(86 = 51 + 35)\nHELICES\n(86 = 51 + 35)\nHELICES\n(86 = 51 + 35)\nFILCHED\n(85 = 50 + 35)\nFACESHIELD\n(84)\nFACIES\n(84)\nFACESHIELD\n(84)\nFACESHIELD\n(84)\nDEFACES\n(83 = 48 + 35)\nFLESHED\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nDEFILES\n(83 = 48 + 35)\nDEFILES\n(83 = 48 + 35)\nDESCALE\n(83 = 48 + 35)\nCLASHED\n(83 = 48 + 35)\nDEFILES\n(83 = 48 + 35)\nDEFACES\n(83 = 48 + 35)\nDEFACES\n(83 = 48 + 35)\nFLASHED\n(83 = 48 + 35)\nDEFILES\n(83 = 48 + 35)\nDECILES\n(83 = 48 + 35)\nDECILES\n(83 = 48 + 35)\nDECILES\n(83 = 48 + 35)\nDECILES\n(83 = 48 + 35)\nDECILES\n(83 = 48 + 35)\nDESCALE\n(83 = 48 + 35)\nDESCALE\n(83 = 48 + 35)\nDEFILES\n(83 = 48 + 35)\nDECILES\n(83 = 48 + 35)\nDEFACES\n(83 = 48 + 35)\nDESCALE\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nFILCHES\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nDEFLEAS\n(83 = 48 + 35)\nFLASHED\n(83 = 48 + 35)\nDEFILES\n(83 = 48 + 35)\nFLESHED\n(83 = 48 + 35)\nDESCALE\n(83 = 48 + 35)\nDESCALE\n(83 = 48 + 35)\nFLASHED\n(83 = 48 + 35)\nLEACHED\n(83 = 48 + 35)\nCLASHED\n(83 = 48 + 35)\nFLESHED\n(83 = 48 + 35)\nDEFLESH\n(83 = 48 + 35)\nCLASHED\n(83 = 48 + 35)\nDEFACES\n(83 = 48 + 35)\nLEACHED\n(83 = 48 + 35)\nDEFLESH\n(83 = 48 + 35)\n\n# Word Growth involving faceshield\n\nace aces faces\n\nace face faces\n\nfa face faces\n\nel shield\n\nhi shield\n\n## Longer words containing faceshield\n\n(No longer words found)"
] | [
null
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https://de.scribd.com/presentation/366865964/Introduction-to-Statistics | [
"Sie sind auf Seite 1von 27\n\n# INTRODUCTION TO\n\nSTATISTICS\n\nBy\nMiftahul Rachmat\nA statistic (singular) is a single measure of\nsome attribute of a sample (e.g., its arithmetic\nmean value). It is calculated by applying\na function (statistical algorithm) to the values of\nthe items of the sample, which are known\ntogether as a set of data.\nMore formally, statistical theory defines a\nstatistic as a function of a sample where the\nfunction itself is independent of the sample's\ndistribution; that is, the function can be stated\nbefore realization of the data. The term statistic\nis used both for the function and for the value\nof the function on a given sample.\nWhen a statistic (a function) is being used for a\nspecific purpose, it may be referred to by a\nname indicating its purpose: in descriptive\nstatistics, a descriptive statistic is used to\ndescribe the data; in estimation theory, an\nestimator is used to estimate a parameter of\nthe distribution (population); in statistical\nhypothesis testing, a test statistic is used to\ntest a hypothesis. However, a single statistic\ncan be used for multiple purposes for\nexample the sample mean can be used to\ndescribe a data set, to estimate the population\nmean, or to test a hypothesis.\nVariables\nA variable is a characteristic or condition\nthat can change or take on different\nvalues.\nMost research begins with a general\ntwo variables for a specific group of\nindividuals.\n\n4\nPopulation\nThe entire group of individuals is called the\npopulation.\nFor example, a researcher may be\ninterested in the relation between class\nperformance (variable 2) for the population\n\n5\nSample\nUsually populations are so large that a\nresearcher cannot examine the entire\ngroup. Therefore, a sample is selected to\nrepresent the population in a research\nstudy. The goal is to use the results\nobtained from the sample to help answer\n\n6\nTypes of Variables\nVariables can be classified as discrete or\ncontinuous.\nDiscrete variables (such as class size)\nconsist of indivisible categories, and\ncontinuous variables (such as time or\nweight) are infinitely divisible into whatever\nunits a researcher may choose. For\nexample, time can be measured to the\nnearest minute, second, half-second, etc.\n\n8\nReal Limits\nTo define the units for a continuous\nvariable, a researcher must use real limits\nwhich are boundaries located exactly half-\n\n9\nMeasuring Variables\nTo establish relationships between\nvariables, researchers must observe the\nvariables and record their observations.\nThis requires that the variables be\nmeasured.\nThe process of measuring a variable\nrequires a set of categories called a scale\nof measurement and a process that\nclassifies each individual into one\ncategory.\n10\n4 Types of Measurement Scales\n1. A nominal scale is an unordered set of\ncategories identified only by name.\nNominal measurements only permit you\nto determine whether two individuals are\nthe same or different.\n2. An ordinal scale is an ordered set of\ncategories. Ordinal measurements tell\nyou the direction of difference between\ntwo individuals.\n\n11\n4 Types of Measurement Scales\n3. An interval scale is an ordered series of equal-\nsized categories. Interval measurements\nidentify the direction and magnitude of a\ndifference. The zero point is located arbitrarily\non an interval scale.\n4. A ratio scale is an interval scale where a value\nof zero indicates none of the variable. Ratio\nmeasurements identify the direction and\nmagnitude of differences and allow ratio\ncomparisons of measurements.\n12\nCorrelational Studies\nThe goal of a correlational study is to\ndetermine whether there is a relationship\nbetween two variables and to describe the\nrelationship.\nA correlational study simply observes the\ntwo variables as they exist naturally.\n\n13\nExperiments\nThe goal of an experiment is to\ndemonstrate a cause-and-effect\nrelationship between two variables; that is,\nto show that changing the value of one\nvariable causes changes to occur in a\nsecond variable.\n\n15\nExperiments (cont.)\nIn an experiment, one variable is manipulated\nto create treatment conditions. A second\nvariable is observed and measured to obtain\nscores for a group of individuals in each of the\ntreatment conditions. The measurements are\nthen compared to see if there are differences\nbetween treatment conditions. All other\nvariables are controlled to prevent them from\ninfluencing the results.\nIn an experiment, the manipulated variable is\ncalled the independent variable and the\nobserved variable is the dependent variable.\n\n16\nOther Types of Studies\nOther types of research studies, know as\nnon-experimental or quasi-\nexperimental, are similar to experiments\nbecause they also compare groups of\nscores.\nThese studies do not use a manipulated\nvariable to differentiate the groups.\nInstead, the variable that differentiates the\ngroups is usually a pre-existing participant\nvariable (such as male/female) or a time\nvariable (such as before/after).\n18\nOther Types of Studies (cont.)\nBecause these studies do not use the\nmanipulation and control of true\nexperiments, they cannot demonstrate\ncause and effect relationships. As a\nresult, they are similar to correlational\nresearch because they simply\ndemonstrate and describe relationships.\n\n19\nData\nThe measurements obtained in a research\nstudy are called the data.\nThe goal of statistics is to help researchers\norganize and interpret the data.\n\n21\nDescriptive Statistics\nDescriptive statistics are methods for\norganizing and summarizing data.\nFor example, tables or graphs are used to\norganize data, and descriptive values such\nas the average score are used to\nsummarize data.\nA descriptive value for a population is\ncalled a parameter and a descriptive\nvalue for a sample is called a statistic.\n22\nInferential Statistics\nInferential statistics are methods for using\nsample data to make general conclusions\nBecause a sample is typically only a part of the\nwhole population, sample data provide only\nlimited information about the population. As a\nresult, sample statistics are generally imperfect\nrepresentatives of the corresponding population\nparameters.\n\n23\nSampling Error\nThe discrepancy between a sample\nstatistic and its population parameter is\ncalled sampling error.\nDefining and measuring sampling error is\na large part of inferential statistics.\n\n24\nNotation\nThe individual measurements or scores obtained\nfor a research participant will be identified by the\nletter X (or X and Y if there are multiple scores\nfor each individual).\nThe number of scores in a data set will be\nidentified by N for a population or n for a sample.\nSumming a set of values is a common operation\nin statistics and has its own notation. The Greek\nletter sigma, , will be used to stand for \"the sum\nof.\" For example, X identifies the sum of the\nscores.\n26\nOrder of Operations\n1. All calculations within parentheses are done\nfirst.\n2. Squaring or raising to other exponents is done\nsecond.\n3. Multiplying, and dividing are done third, and\nshould be completed in order from left to right.\n4. Summation with the notation is done next."
] | [
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https://learnabout-electronics.org/PSU/psu22.php | [
"# Series Voltage Regulators\n\n• After studying this section, you should be able to:\n• Understand the operation of series voltage regulators.\n• • The simple series regulator.\n• • Feedback and error amplification.\n• • Over current protection (current limiting).\n• • Over voltage protection.\n\n## Simple Series Voltage Regulators",
null,
"### Fig. 2.2.1 Simple Series Regulator\n\nIn Fig. 2.2.1 RS and DZ form a simple SHUNT regulator as described in power supplies module 2.1. In this circuit however, they are used to provide a stable voltage reference VZ at the base of Tr1. The emitter voltage of Tr1 will be typically about 0.7V less than the base voltage and VOUT will therefore be at a lower voltage than the base.\n\nVOUT = VZ - VBE\n\nIf the output voltage VOUT falls due to increased current demand by the load, this will cause VBE to increase and as a result, current through the transistor (from collector to emitter) will increase. This will provide the extra current required by the load and thus regulate the output voltage VOUT.\n\nIf VOUT tends to rise due to reduced current demand by the load, then this will reduce VBE as the emitter voltage rises and the base voltage remains stable due to DZ. This reduction in VBE will tend to turn the transistor off, reducing current flow, and again regulating the output voltage VOUT.\n\nThis regulating effect is due to the base potential of Tr1 being held steady by DZ so that any variation in emitter voltage caused by varying current flow causes a change in VBE, varying the conduction of the transistor Tr1, which will usually be a power transistor. This action counteracts the variation in load current. With this simple circuit however, regulation is not perfect, and variations in output do occur for the following reasons.",
null,
"### Fig. 2.2.2 Zener Diode Operating Region\n\n1. Any increase in load current (IL) causes a small increase in base current by the ratio IL/hfe. This in turn causes an increase in VBE and because the output voltage VOUT = VZ - VBE any increase in VBE tends to reduce the output voltage. The amount of this fall is about 0.25V for a change in output current from 10mA to 1A.\n\n2. Since base current increases with load, the current through the zener diode DZ will decrease as more current is taken by the base of Tr1. Because the diode characteristic has a slope over its operating region as shown in Fig. 2.2.2, a large change in zener current (ΔI) will cause a very small change in zener voltage (δV). This in turn will slightly affect VBE and the output voltage.\n\n3. Because of reasons 1 and 2 above, any change in load will result in less than perfect regulation, therefore any change at the output will slightly change the loading on the input circuit. As the input is normally taken from an un-regulated supply, the input voltage will be easily affected by slight changes in load current, As the input voltage is also the supply for the reference voltage VZ any change in output current, by affecting the input voltage, can produce a noticeable effect on the output voltage, slightly reducing the effectiveness of the regulation.\n\nEach of the above effects is small, but added together they will provide an overall effect that is noticeable when the supply is operating under demanding conditions. Nevertheless this inexpensive circuit is effective enough for many applications, and is more efficient than the shunt regulator. Also, by using a suitable power transistor, the series regulator can be used for heavier load currents than the shunt design.",
null,
"## Feedback and Error Amplification.\n\nTo improve on the simple series regulator a feedback circuit and error amplifier can be added to the basic series circuit.\n\nFig. 2.2.3 shows a block diagram of a series regulator circuit with error amplification. In this system the reference voltage VZ is compared with a feedback voltage VF, which is a portion of the actual output voltage. The difference between the two inputs produces an error voltage that is used to vary the conduction of the control element, correcting any error in the output voltage.\n\n## Circuit Diagram.\n\nA circuit diagram for this system is shown in Fig. 2.2.4. Tr1 is the series control element. It will usually be a power transistor, mounted on a substantial heat sink to cope with the necessary power dissipation.\n\nA stable reference voltage is provided by R4 & D1 from the un-regulated input voltage. Tr2 is the error amplifier and its gain is set by the value of its load resistor R3. Tr2 compares the fraction of the output voltage VF fed back from the output potential divider R1/R2 with the stable reference voltage VZ across the zener diode DZ.",
null,
"### Fig. 2.2.4 Circuit Diagram for Fig. 2.2.3\n\nThe output voltage VOUT in Fig. 2.2.4 can be expressed as:\n\nVOUT = (VZ + VBE2) + (VOUT - VF)\n\nWhere:\n\nVZ is the voltage across DZ\n\nVBE2 is the base/emitter voltage of Tr2\n\nVF is the feedback voltage derived from the slider of VR1\n\nTherefore:\n\n(VZ + VBE2) is the voltage across R2 and the lower portion of VRI\n\nand\n\n(VOUT − VF) is the voltage across R1 and the upper portion of VRI\n\nIf the feedback voltage VF is altered by adjusting VR1 potentiometer, the difference between VF and VZ will change. This will cause a change in the error voltage controlling Tr1 and a change in the output voltage VOUT. VR1 therefore provides a variable output voltage, which, once set remains stable at that setting.\n\nThe regulating action of the circuit is governed by the voltage across the base/emitter junction of Tr2, i.e. the difference between VF and VZ.\n\nIf VOUT tends to increase, then VF - VZ also increases. This increases the collector current of Tr2 and so increases the p.d. across R3 reducing the base voltage, and therefore the base/emitter voltage of Tr1, reducing the conduction of Tr1, so reducing current flow to the load.\n\nThe output voltage VOUT is reduced in this way until a balance is reached, as the feedback portion (VF) of VOUT is also reducing. The overall effect is that the output is maintained at a level, which depends on the proportion of feedback set by the variable resistor (part of R1/R2).\n\nIf the output voltage tends to decrease, then so does VF. The base/emitter voltage of Tr2 is reduced due to the stable VZ on the emitter. Tr2 conducts less and the current through R3 falls, reducing the p.d. across it. Tr1 base voltage rises, and increases the conduction of the control transistor. This increases output current and VOUT until VF is once more at the correct level.\n\n## Over Current Protection (Current Limiting)",
null,
"### Fig. 2.2.5 Series Regulator with Over Current Protection\n\nFig. 2.2.5 shows how the series stabiliser can be protected against excessive current being drawn by the load. This will prevent damage to the supply in the event of too much current being drawn from the output, or even a complete short circuit across the output terminals.\n\nTwo components have been added, Tr3 and R5. The resistor R5 is a very low value (typically less than 1 ohm).\n\nWhen the load current rises above a predetermined value, the small voltage developed across R5 will become sufficient (at about 0.7v) to turn Tr3 on. As Tr3 is connected across the base/emitter junction of the main control transistor Tr1, the action of turning Tr3 on will reduce the base/emitter voltage of Tr1 by an amount depending on the amount of excess current. The output current will not be allowed to increase above a predetermined amount, even if a complete short circuit occurs across the output terminals. In this case Tr1 base/emitter voltage will be reduced to practically zero volts, preventing Tr1 from conducting. Under these conditions the output voltage will fall to zero for as long as the excess current condition persists, but the supply will be undamaged.",
null,
"## Over Voltage Protection.\n\nWhere regulated supplies are used, the DC input voltage to the regulator is often considerably higher than the required output voltage. Therefore if a PSU fault occurs, it is possible that the regulated output voltage may suddenly rise to a level that can damage other components. For this reason it is common to find over voltage protection included in stabilised supplies. The circuit shown in Fig. 2.2.6 is sometimes called a \"crowbar\" circuit because when it operates, it places a complete short circuit across the across the output, a similar effect to dropping a metal crowbar across the positive and ground output terminals!\n\n## Crowbar Circuit Operation.\n\nIn Fig. 2.2.6 the zener diode DZ2 has a breakdown voltage slightly less than the maximum allowed value for VOUT. The remainder of VOUT is developed across R6, VR2 and R7.\n\nVR2 is a potentiometer, so that a voltage may be taken from the resistor network to correctly bias the diode D1. This diode has its cathode held at 0V by R8, and VR2 is adjusted so that D1 is just out of conduction, i.e. its anode voltage is about 0.5V higher than its cathode voltage.\n\nNow, if VOUT increases, the voltage across R6, VR2 and R7 will rise by the same amount, as the voltage across DZ2 will remain the same. Therefore there will be a substantial rise in the voltage at R7 slider, which will cause D1 to conduct, supplying a pulse of current to the gate of thyristor Th1, causing it to \"fire\" and conduct heavily until VOUT falls to practically 0v. R9 is included to limit the resulting current flow through the thyristor to a safe level.\n\nThe large current that flows as Th1 fires will now cause the current limiter circuit to come into operation as previously described. This will safely shut down the supply until the over current caused by Th1 has disappeared, which will of course happen as soon as VOUT reaches 0V, but should the over voltage still be present when Th1 switches off and VOUT rises again, the circuit will re-trigger, causing the voltage across the load to repeatedly alternate between its normal value and zero; a harmless but clear symptom of an over voltage problem.\n\nTop of Page"
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"https://learnabout-electronics.org/PSU/images/simple-series-regulator.jpg",
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"https://learnabout-electronics.org/PSU/images/zener-slope.jpg",
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"https://learnabout-electronics.org/PSU/images/Series-with-NFB-and-error-amp.jpg",
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"https://learnabout-electronics.org/PSU/images/Series-cct-with-error-amp.jpg",
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"https://learnabout-electronics.org/PSU/images/Series-with-current-limit.jpg",
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"https://learnabout-electronics.org/PSU/images/Series-with-over-volt.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91454935,"math_prob":0.9498031,"size":10010,"snap":"2020-10-2020-16","text_gpt3_token_len":2254,"char_repetition_ratio":0.17389566,"word_repetition_ratio":0.010459036,"special_character_ratio":0.21338661,"punctuation_ratio":0.106014274,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9628705,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,3,null,3,null,3,null,4,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-31T19:32:35Z\",\"WARC-Record-ID\":\"<urn:uuid:9f432834-da7f-4693-b157-81890200857e>\",\"Content-Length\":\"24714\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:50dfb2b8-52a9-4ea9-8e76-06b3d2590ca9>\",\"WARC-Concurrent-To\":\"<urn:uuid:409959c5-eb93-4678-89c5-77c1081a9f1b>\",\"WARC-IP-Address\":\"217.160.0.248\",\"WARC-Target-URI\":\"https://learnabout-electronics.org/PSU/psu22.php\",\"WARC-Payload-Digest\":\"sha1:46I3UVET2YZWV47RYP6RA5Z3CSJQX4NH\",\"WARC-Block-Digest\":\"sha1:3NFB5KYNTXMOCHMDU4Z3OI3PKIQZ77EB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370503664.38_warc_CC-MAIN-20200331181930-20200331211930-00057.warc.gz\"}"} |
https://bartoszmilewski.com/2017/04/11/profunctor-parametricity/ | [
"## The Free Theorem for Ends\n\nIn Haskell, the end of a profunctor `p` is defined as a product of all diagonal elements:\n\n`forall c. p c c`\n\ntogether with a family of projections:\n\n```pi :: Profunctor p => forall c. (forall a. p a a) -> p c c\npi e = e```\n\nIn category theory, the end must also satisfy the edge condition which, in (type-annotated) Haskell, could be written as:\n\n`dimap f idb . pib = dimap ida f . pia`\n\nfor any `f :: a -> b`.\nUsing a suitable formulation of parametricity, this equation can be shown to be a free theorem. Let’s first review the free theorem for functors before generalizing it to profunctors.\n\n## Functor Characterization\n\nYou may think of a functor as a container that has a shape and contents. You can manipulate the contents without changing the shape using `fmap`. In general, when applying `fmap`, you not only change the values stored in the container, you change their type as well. To really capture the shape of the container, you have to consider not only all possible mappings, but also more general relations between different contents.\n\nA function is directional, and so is `fmap`, but relations don’t favor either side. They can map multiple values to the same value, and they can map one value to multiple values. Any relation on values induces a relation on containers. For a given functor `F`, if there is a relation `a` between type `A` and type `A'`:\n\n`A <=a=> A'`\n\nthen there is a relation between type `F A` and `F A'`:\n\n`F A <=(F a)=> F A'`\n\nWe call this induced relation `F a`.\n\nFor instance, consider the relation between students and their grades. Each student may have multiple grades (if they take multiple courses) so this relation is not a function. Given a list of students and a list of grades, we would say that the lists are related if and only if they match at each position. It means that they have to be equal length, and the first grade on the list of grades must belong to the first student on the list of students, and so on. Of course, a list is a very simple container, but this property can be generalized to any functor we can define in Haskell using algebraic data types.\n\nThe fact that `fmap` doesn’t change the shape of the container can be expressed as a “theorem for free” using relations. We start with two related containers:\n\n```xs :: F A\nxs':: F A'```\n\nwhere `A` and `A'` are related through some relation `a`. We want related containers to be `fmap`ped to related containers. But we can’t use the same function to map both containers, because they contain different types. So we have to use two related functions instead. Related functions map related types to related types so, if we have:\n\n```f :: A -> B\nf':: A'-> B'```\n\nand `A` is related to `A'` through `a`, we want `B` to be related to `B'` through some relation `b`. Also, we want the two functions to map related elements to related elements. So if `x` is related to `x'` through `a`, we want `f x` to be related to `f' x'` through `b`. In that case, we’ll say that `f` and `f'` are related through the relation that we call `a->b`:\n\n`f <=(a->b)=> f'`\n\nFor instance, if `f` is mapping students’ SSNs to last names, and `f'` is mapping letter grades to numerical grades, the results will be related through the relation between students’ last names and their numerical grades.\n\nTo summarize, we require that for any two relations:\n\n```A <=a=> A'\nB <=b=> B'```\n\nand any two functions:\n\n```f :: A -> B\nf':: A'-> B'```\n\nsuch that:\n\n`f <=(a->b)=> f'`\n\nand any two containers:\n\n```xs :: F A\nxs':: F A'```\n\nwe have:\n\n```if xs <=(F a)=> xs'\nthen F xs <=(F b)=> F xs'```\n\nThis characterization can be extended, with suitable changes, to contravariant functors.\n\n## Profunctor Characterization\n\nA profunctor is a functor of two variables. It is contravariant in the first variable and covariant in the second. A profunctor can lift two functions simultaneously using `dimap`:\n\n```class Profunctor p where\ndimap :: (a -> b) -> (c -> d) -> p b c -> p a d```\n\nWe want `dimap` to preserve relations between profunctor values. We start by picking any relations `a`, `b`, `c`, and `d` between types:\n\n```A <=a=> A'\nB <=b=> B'\nC <=c=> C'\nD <=d=> D'\n```\n\nFor any functions:\n\n```f :: A -> B\nf' :: A'-> B'\ng :: C -> D\ng' :: C'-> D'```\n\nthat are related through the following relations induced by function types:\n\n```f <=(a->b)=> f'\ng <=(c->d)=> g'```\n\nwe define:\n\n```xs :: p B C\nxs':: p B'C'```\n\nThe following condition must be satisfied:\n\n```if xs <=(p b c)=> xs'\nthen (p f g) xs <=(p a d)=> (p f' g') xs'\n```\n\nwhere `p f g` stands for the lifting of the two functions by the profunctor `p`.\n\nHere’s a quick sanity check. If `b` and `c` are functions:\n\n```b :: B'-> B\nc :: C -> C'```\n\nthan the relation:\n\n`xs <=(p b c)=> xs'`\n\nbecomes:\n\n```xs' = dimap b c xs\n```\n\nIf `a` and `d` are functions:\n\n```a :: A'-> A\nd :: D -> D'\n```\n\nthen these relations:\n\n```f <=(a->b)=> f'\ng <=(c->d)=> g'```\n\nbecome:\n\n```f . a = b . f'\nd . g = g'. c```\n\nand this relation:\n\n`(p f g) xs <=(p a d)=> (p f' g') xs'`\n\nbecomes:\n\n`(p f' g') xs' = dimap a d ((p f g) xs)`\n\nSubstituting `xs'`, we get:\n\n`dimap f' g' (dimap b c xs) = dimap a d (dimap f g xs)`\n\nand using functoriality:\n\n```dimap (b . f') (g'. c) = dimap (f . a) (d . g)\n```\n\nwhich is identically true.\n\n## Special Case of Profunctor Characterization\n\nWe are interested in the diagonal elements of a profunctor. Let’s first specialize the general case to:\n\n```C = B\nC'= B'\nc = b```\n\nto get:\n\n```xs = p B B\nxs'= p B'B'```\n\nand\n\n```if xs <=(p b b)=> xs'\nthen (p f g) xs <=(p a d)=> (p f' g') xs'\n```\n\nChosing the following substitutions:\n\n```A = A'= B\nD = D'= B'\na = id\nd = id\nf = id\ng'= id\nf'= g```\n\nwe get:\n\n```if xs <=(p b b)=> xs'\nthen (p id g) xs <=(p id id)=> (p g id) xs'\n```\n\nSince `p id id` is the identity relation, we get:\n\n`(p id g) xs = (p g id) xs'`\n\nor\n\n`dimap id g xs = dimap g id xs'`\n\n## Free Theorem\n\nWe apply the free theorem to the term `xs`:\n\n`xs :: forall c. p c c`\n\nIt must be related to itself through the relation that is induced by its type:\n\n`xs <=(forall b. p b b)=> xs`\n\nfor any relation `b`:\n\n`B <=b=> B'`\n\nUniversal quantification translates to a relation between different instantiations of the polymorphic value:\n\n`xsB <=(p b b)=> xsB'`\n\nNotice that we can write:\n\n```xsB = piB xs\nxsB'= piB'xs```\n\nusing the projections we defined earlier.\n\nWe have just shown that this equation leads to:\n\n`dimap id g xs = dimap g id xs'`\n\nwhich shows that the wedge condition is indeed a free theorem.\n\n## Natural Transformations\n\nHere’s another quick application of the free theorem. The set of natural transformations may be represented as an end of the following profunctor:\n\n`type NatP a b = F a -> G b`\n```instance Profunctor NatP where\ndimap f g alpha = fmap g . alpha . fmap f```\n\nThe free theorem tells us that for any `mu :: NatP c c`:\n\n`(dimap id g) mu = (dimap g id) mu`\n\nwhich is the naturality condition:\n\n`mu . fmap g = fmap g . mu`\n\nIt’s been know for some time that, in Haskell, naturality follows from parametricity, so this is not surprising.\n\n## Acknowledgment\n\nI’d like to thank Edward Kmett for reviewing the draft of this post.\n\n## Bibliography\n\n1. Bartosz Milewski, Ends and Coends\n2. Edsko de Vries, Parametricity Tutorial, Part 1, Part 2, Contravariant Functions."
] | [
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https://world.fanthful.com/products/far-cry-6-%E6%9E%81%E5%9C%B0%E6%88%98%E5%9A%8E6-%E9%BB%91%E8%89%B2t%E6%81%A4-21393 | [
"• English\n• Japan（USD \\$）\n• \\$ USD\n• £ GBP\n• EUR\n• A\\$ AUD\n• INR\n• ¥ JPY\n• د.إ AED\n• Af AFN\n• L ALL\n• Դ AMD\n• ƒ ANG\n• Kz AOA\n• \\$ ARS\n• ƒ AWG\n• ман AZN\n• КМ BAM\n• BBD BBD\n• BDT\n• лв BGN\n• BIF\n• BD\\$ BMD\n• B\\$ BND\n• Bs. BOB\n• BOV BOV\n• R\\$ BRL\n• BSD BSD\n• Nu. BTN\n• P BWP\n• Br BYN\n• Be BZD\n• CDF\n• CHF\n• CLP CLP\n• ¥ CNY\n• \\$ COP\n• CRC\n• \\$ CUC\n• \\$ CUP\n• \\$ CVE\n• CZK\n• DJF\n• kr DKK\n• \\$ DOP\n• د.ج DZD\n• EGP\n• Nfk ERN\n• Br ETB\n• FJ\\$ FJD\n• FKP FKP\n• GEL\n• GHS\n• £ GIP\n• D GMD\n• GNF\n• Q GTQ\n• \\$ GYD\n• HK\\$ HKD\n• L HNL\n• Kn HRK\n• G HTG\n• Ft HUF\n• Rp IDR\n• ILS\n• IRR\n• Kr ISK\n• \\$ JMD\n• Sh KES\n• С̲ KGS\n• KHR\n• Fr KMF\n• KPW\n• KRW\n• \\$ KYD\n• KZT\n• LAK\n• ل.ل LBP\n• Rs LKR\n• L\\$ LRD\n• L LSL\n• L MDL\n• ден MKD\n• K MMK\n• MNT\n• P MOP\n• MUR\n• ރ. MVR\n• MK MWK\n• \\$ MXN\n• RM MYR\n• MTn MZN\n• NGN\n• C\\$ NIO\n• kr NOK\n• NPR\n• \\$ NZD\n• B/. PAB\n• S/ PEN\n• K PGK\n• PHP\n• PKR\n• PLN\n• PYG\n• ر.ق QAR\n• L RON\n• din RSD\n• р. RUB\n• RWF\n• ر.س SAR\n• Si\\$ SBD\n• SCR\n• ج.س. SDG\n• kr SEK\n• S\\$ SGD\n• £ SHP\n• Le SLL\n• Sh SOS\n• \\$ SRD\n• SDP SSP\n• Db STN\n• ل.س SYP\n• L SZL\n• ฿ THB\n• ЅМ TJS\n• m TMT\n• T\\$ TOP\n• TRY\n• TTD TTD\n• NT\\$ TWD\n• Sh TZS\n• UAH\n• Sh UGX\n• \\$ UYU\n• лв UZS\n• Bs.S. VES\n• VND\n• Vt VUV\n• T WST\n• XAF\n• \\$ XCD\n• Fr XOF\n• XPF\n• YER\n• R ZAR\n• ZK ZMW\n\nNo relevant currency found",
null,
"# Far Cry 6 Black T-shirt\n\nFar Cry 6 極地戰嚎6 黑色T恤\n\\$1699\n\\$2399\nSave \\$7\nSize\nS\nM\nL\nXL\n2XL\n\nCheckout Securely with\n\nDescription\n\nUbisoft官方正版授權，FANTHFUL設計開發\n\n※圖片供參考，與實物存在差異，請以實物為准"
] | [
null,
"https://img.myshopline.com/image/store/2000009776/1614655830388/1f94832cd2284b5da8d2e9ebe1b544e3.jpeg",
null
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https://www.qinxu.net/latex/2021/0718601/ | [
"# [原创]在 TikZ 中利用 scope 环境实现平移做图",
null,
"TikZ 宏包除了提供 tikzpicture 环境外还提供了一个可以嵌套在 tikzpicture 里的子环境 scope, 简单点说就是可以通过 scope 环境把同一标准的画图命令放在一起, 而只需要声明一次标准, 比如 scale, thick, -> 等等之类. 个人用的更多的是通过 scope 环境完成平移做图, 当然命令不多的的时候也可以直接平移做图, 如上图中圆心位于点 $(1,1)$ 的圆.\n\nTikZ 宏包还提供了一个 scopes library, 以简化 scope 环境的调用, 当个人觉得直接显示 scope 似乎更有利于代码的可读性.\n\n\\documentclass{standalone}\n\n\\usepackage{tikz}\n\n\\begin{document}\n\n\\begin{tikzpicture}\n\\draw (0,0) circle (1cm);\n\\draw [shift={(1,1)}] (0,0) circle (0.414cm);\n\\begin{scope}[xshift=-1cm,yshift=1cm]\n\\draw (0,0) circle (0.414cm);\n\\end{scope}\n\\begin{scope}[shift=(225:{sqrt(2)})]\n\\draw (0,0) circle (0.414cm);\n\\end{scope}\n\\begin{scope}[shift={(1,-1)}]\n\\draw (0,0) circle (0.414cm);\n\\end{scope}\n\n\\begin{scope}[xshift=2cm]\n\\draw (0,0) circle (1cm);\n\\end{scope}\n\\begin{scope}[xshift=-2cm]\n\\draw (0,0) circle (1cm);\n\\end{scope}\n\\end{tikzpicture}\n\n\\end{document}\n\n\\coordinate (A) at (3,4);\n\n\\documentclass{standalone}\n\n\\usepackage{tikz}\n\n\\begin{document}\n\n\\begin{tikzpicture}\n\\draw (0,0) circle (1);\n\\draw [-stealth] (-1,-1) -- ++(2,2);\n\\node [below] at (0,-1) {(a)};\n\\begin{scope}[xshift=3cm]\n\\draw (0,0) circle (1);\n\\draw [-stealth] (-1,-1) -- ++(2,2);\n\\node [below] at (0,-1) {(b)};\n\\end{scope}\n\\end{tikzpicture}\n\n\\end{document}"
] | [
null,
"https://www.qinxu.net/wp-content/uploads/2021/07/testscope.webp",
null
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https://fykos.org/year34/problems/series1 | [
"# Series 1, Year 34\n\n### (3 points)1. almost stopped light\n\nFind the refractive index of a transparent plane-parallel plate of thickness $d=1 \\mathrm{cm}$, such that it will take one year for the light to pass through it. Discuss whether such a situation is possible.\n\n### (3 points)2. brake!\n\nKarel's car, going at the initial speed of $v_0$, can stop at a distance $s_0$ with the constant braking force $F_0$. How many times will the braking distance increase if the initial speed doubles and the braking force stays the same? How many times must the braking force be greater for the car to stop at distance $s_0$ with the initial speed $2v_0$?\n\nKarel and a campaign for responsible driving.\n\n### (5 points)3. cycling anemometer\n\nVašek rides his bicycle in windy weather. When he rides straight with the velocity $v = 10 \\mathrm{km\\cdot h^{-1}}$, he measures that the wind blows at an angle $25\\dg$ from the direction of Vašek's direction of travel. When he accelerates to $v' = 20 \\mathrm{km\\cdot h^{-1}}$, the angle is only $15\\dg$. Find the velocity and direction of the wind with respect to stationary observer.\n\nVašek thought that the wind blows on him too much while he's cycling.\n\n### (8 points)4. solar sail\n\nA solar sail with the surface area of $S = 500 \\mathrm{m^2}$ and area density $\\sigma =1,4 \\mathrm{kg\\cdot m^{-2}}$ is located at the distance of $0,8 \\mathrm{au}$ from the Sun. What force does the solar radiation act on the sail at the beginning of the sail's motion? What is the acceleration of the sail at that moment? The luminosity of the Sun is $L_{\\odot } =3,826 \\cdot 10^{26} \\mathrm{W}$. Assume that the radiation approaches the sail from a perpendicular direction and scatters elastically. Hint: We recommend you find the acceleration for small initial velocity $v_0$ and then let $v_0 = 0$.\n\nDanka wants to fly.\n\n### (8 points)5. how to put your beanie on sigle-handily\n\nLet us have a ball with the radius $R$ and a circular massless rubber band with the radius $r_0$ and stiffness $k$, while $r_0 < R$. The coefficient of friction between the band and the ball is $f$. Find conditions which ensure that it is possible to stretch the band over the ball single-handily (i.e. we are allowed to touch the band in only one point.\n\nTo keep it simple assume that the band is elastic only in the tangential direction (it is planar).\n\n### (10 points)P. Will we survive in vacuum?\n\nDifferent movies create different conceptions of what and how fast happens when an astronaut's space suit suddenly gets torn. Some of them are even contradictory. Explain what is most likely to happen, if a healthy person finds himself unprotected in a vacuum. What phenomenon is most likely to cause death first?\n\nKuba planned to travel the world.\n\n### (13 points)E. impact-y\n\nMeasure the dependence of the diameter of a crater, created by the impact of a stone into a suitable sandpit, on the weight of the stone and the height it is released from. Does the size of the crater depend only on the energy of the impact? Dry sand is recommended for this measurement.\n\nDodo returned to his childhood.\n\n### (10 points)S. oscillating\n\nLet us begin this year's serial with analysis of several mechanical oscillators. We will focus on the frequency of their simple harmonic motion. We will also revise what does an oscillator look like in the phase space.\n\n1. Assume that we have a hollow cone of negligible mass with a stone of mass $M$ located in its vertex. We will plunge it into water (of density $\\rho$) so that the vertex points downwards and the cone will float on the water surface. Find the waterline depth $h$, measured from the vertex to the water surface, if the total height of the cone is $H$ and its radius is $R$. Find the angular frequency of small vertical oscillation of the cone.\n2. Let us imagine a weight of mass $m$ attached to a spring of negligible mass, spring constant $k$ and free length $L$. If we attach the spring by its second end, we will get an oscillator. Find the angular frequency of its simple harmonic motion, assuming that the length of the spring does not change during the motion. Subsequently, find a small difference in angular frequency $\\Delta \\omega$ between this oscillator and the one in which the spring is substituted by a stiff rod of the same length. Assume $k L \\gg m g$.\n3. A sugar cube with mass $m$ is located in a landscape consisting of periodically repeating parabolas of height $H$ and width $L$. Describe its potential energy as a function of horizontal coordinate and outline possible trajectories of its motion in phase space, depending on the velocity $v_0$ of the cube on the top of the parabola. Mark all important distances. Use horizontal coordinate as displacement and appropriate units of horizontal momentum. Neglect kinetic energy of cube motion in the vertical direction and assume it remains in contact with the terrain.\n\nŠtěpán found a few basic oscillators.",
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null,
"https://fykos.org/lib/exe/taskrunner.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9172761,"math_prob":0.99386406,"size":5005,"snap":"2023-40-2023-50","text_gpt3_token_len":1198,"char_repetition_ratio":0.11377724,"word_repetition_ratio":0.002317497,"special_character_ratio":0.23996004,"punctuation_ratio":0.08979592,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99468696,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T20:21:57Z\",\"WARC-Record-ID\":\"<urn:uuid:51629c52-ddd4-42dd-8d4a-feaa2bdca357>\",\"Content-Length\":\"44740\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:24886c20-ea66-48cf-943d-fab3669f6734>\",\"WARC-Concurrent-To\":\"<urn:uuid:0e2f6303-924c-4949-9296-c5b985c3dd26>\",\"WARC-IP-Address\":\"195.113.23.5\",\"WARC-Target-URI\":\"https://fykos.org/year34/problems/series1\",\"WARC-Payload-Digest\":\"sha1:WXUDHIXM5UTSLLSRRSWWSJWV4ZAKABMX\",\"WARC-Block-Digest\":\"sha1:LCI4SLL7SHQ5HB6SIJRQLY3ZZCVLK5RV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100769.54_warc_CC-MAIN-20231208180539-20231208210539-00679.warc.gz\"}"} |
https://math.stackexchange.com/questions/1764677/is-the-sum-of-reciprocals-of-all-products-from-2-to-n-1-always-0-5n-1/1764690 | [
"# Is the sum of reciprocals of all products from $2$ to $n-1$ always $0.5n-1$?\n\nI was looking up riddles for my math classes to work on for the end of the year and found the following riddle. http://mathriddles.williams.edu/?p=129\n\nI followed the advice and started working with examples of small numbers and stumbled upon a pattern that I wanted to generalize.\n\n$$\\frac{1}{2}(1)=0.5$$ $$\\frac{1}{3}\\left(1+\\frac{1}{2}\\right)=0.5$$ $$\\frac{1}{4}\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{2\\cdot 3}\\right)=0.5$$ $$\\frac{1}{5}\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{2\\cdot 3}+\\frac{1}{2\\cdot 4}+\\frac{1}{3\\cdot 4}+\\frac{1}{2\\cdot 3\\cdot 4}\\right)=0.5$$ $$\\frac{1}{6}\\left(1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{2\\cdot 3}+\\frac{1}{2\\cdot 4}+\\frac{1}{2\\cdot 5}+\\frac{1}{3\\cdot 4}+\\frac{1}{3\\cdot 5}+\\frac{1}{4\\cdot 5}+\\frac{1}{2\\cdot 3\\cdot 4}+\\frac{1}{2\\cdot 3\\cdot 5}+\\frac{1}{2\\cdot 4\\cdot 5}+\\frac{1}{3\\cdot 4\\cdot 5}+\\frac{1}{2\\cdot 3\\cdot 4\\cdot 5}\\right)=0.5$$\n\nIf my pattern doesn't make sense, I'm taking $\\frac{1}{n}$ and multiplying it by the sum of the reciprocals of all unique products for $2$ to $n-1$ and it comes out to 0.5 each time up to $n=7$ (I have not tested any higher values). Equivalently, if you multiply both sides by $n$ then subtract $1$, you see that all the reciprocals sum to $0.5n-1$.\n\nI don't know where to start with generalizing this pattern as I have never seen explicit formulas for such sums, so I wanted to see if anyone knew if this was the case for all $n$ and how one could prove or disprove it.\n\n• For $n=7$, do you consider $\\frac 16$ and $\\frac 1{2\\cdot3}$ different? – peterwhy Apr 29 '16 at 21:50\n• Yes. Each number is considered unique, prime or not. – Dirigible Apr 29 '16 at 23:10\n\nLet $s_n$ be the sum of $1$ and the fractions for the $n$ case, e.g. $$s_4 = 1+\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{2\\cdot3}$$\n\nAssume $s_k = k/2$ is true for some $k$.\n\nFor the $n=k+1$ case, $$s_{k+1} = s_k\\cdot \\left(1+\\frac1k\\right) = s_n\\cdot\\frac{k+1}{k} = \\frac{k+1}{2}$$\n\nFor the $n=2$ case, $$s_2 = 1 = \\frac{2}{2}$$\n\nBy induction, $s_n = n/2$ is true for natural numbers $n\\ge 2$. i.e. $$\\frac{1}{n}s_n = \\frac12$$\n\nSome example of the recursion $s_{k+1} = s_k\\cdot\\left(1+\\frac1k\\right)$:\n\n\\begin{align*} s_4 &= 1+\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{2\\cdot3}\\\\ &= \\left(1 + \\frac{1}{2}\\right) + \\frac{1}{3}\\left(1 + \\frac{1}{2}\\right)\\\\ s_5 &= 1 + \\frac12+\\frac13+\\frac1{2\\cdot3}+\\frac14+\\frac1{2\\cdot4} +\\frac1{3\\cdot4} + \\frac{1}{2\\cdot3\\cdot4}\\\\ &= \\left(1 + \\frac12+\\frac13+\\frac1{2\\cdot3}\\right) + \\frac14\\left(1 + \\frac12+\\frac13+\\frac1{2\\cdot3}\\right) \\end{align*}\n\nIf you factor it as $\\frac{1}{n}(1+\\frac{1}{2})(1+\\frac{1}{3})\\dots (1+\\frac{1}{n-1})$ it telescopes nicely.\n\n• Oo, that is really nice. How does one sum that? – Dirigible May 1 '16 at 22:30\n• Well if we simplify the terms we get: $\\frac{1}{n} \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdot \\dots \\cdot \\frac{n}{n-1}$ and almost all the numerators and denominators cancel (this is what I meant by telescoping) and we are left with just $\\frac{1}{2}$. – Nate May 2 '16 at 14:05\n\nWith more systematic notation, your observation is that $$f(n) = \\sum_{A\\subseteq\\{2,3,\\ldots,n-1\\}}\\;\\prod_{k\\in A} \\frac1k = \\frac12n$$ (I don't see where you get \"$0.5n -1$\" out of it; subtracting $1$ doesn't seem to match anything in your examples.)\n\nThis does hold for all $n\\ge 2$, and we can prove it by mathematical induction on $n$:\n\nWhen we go from $f(n)$ to $f(n+1)$, the difference is that we now have more $A$s to sum over -- namely, we have all the ones we have before, plus all of the subsets that contain $n$. But each of the new subsets arises as one of the old subsets with $n$ appended, so we can write it as \\begin{align} f(n+1) &= \\sum_{A\\subseteq\\{2,3,\\ldots,n-1\\}}\\; \\prod_{k\\in A} \\frac1k + \\sum_{A\\subseteq\\{2,3,\\ldots,n-1\\}}\\;\\frac1n \\prod_{k\\in A} \\frac1k \\\\ &= f(n) + \\frac1n f(n) \\\\& = \\frac{n+1}{n} f(n) \\\\& = \\frac{n+1}{n} \\frac12 n \\\\& = \\dfrac12 (n+1) \\end{align} which is what is needed for the induction step.\n\n• The $0.5n-1$ comes from first multiplying both sides by $n$ then subtracting 1 in order to be left solely with the products on the left hand side. Your expression should have a 1+ at the beginning as it is not included in the products. Hopefully that clears my observation up a little. – Dirigible May 1 '16 at 22:29\n• @Dirigible: Subtracting $1$ will lead to results that are $1$ too small. If I had a $1+$ in front of my expressions, they would lead to results that are $1$ too large. For example, $$f(4)=\\frac11+\\frac12+\\frac13+\\frac1{2\\times 3}=2=\\frac12\\cdot 4$$ Subtracting $1$ would give $\\frac12\\cdot4-1=1$ and $1$ is not the correct result of the addition $\\frac11+\\frac12+\\frac13+\\frac16$. – hmakholm left over Monica May 2 '16 at 6:54\n• @Dirigible I think the confusion is about whether the $1$ belongs to the sum of \"fractions\" $f(n)$ naturally; and it belongs. For case $n=4$, the subsets of $\\{2,3\\}$ are $$A\\in\\{\\emptyset,\\ \\{2\\},\\ \\{3\\},\\ \\{2,3\\}\\},$$ which correspond to the products $\\prod_{k\\in A}\\frac1k$ respectively: $$1,\\ \\frac12,\\ \\frac13,\\ \\frac1{2\\cdot3}.$$ Here, the $1$ is the empty product from an empty set. Removing the $1$ and only considering $1/2+1/3+1/(2\\cdot3)$ just complicates it. – peterwhy May 2 '16 at 8:01\n• @peterwhy Yes, that is the source of the confusion. I considered 1 to be separate from the products. Could you explain a little further why 1 is the empty product? – Dirigible May 3 '16 at 13:32\n• @Dirigible It is like considering $1 = \\frac1{2^0\\cdot3^0}$ -- $1$ is the multiplicative identity. See Wikipedia empty product. – peterwhy May 3 '16 at 13:39"
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https://www.physicsforums.com/threads/torque-and-magnetic-fields.284943/ | [
"# Torque and magnetic fields\n\n## Homework Statement\n\nThe problem is as seen in the attachment.\n\n## Homework Equations\n\nT(torque) = (N)(I)(A)(B)(Sin(theta))\n\n## The Attempt at a Solution\n\nFor this question I used the above formula to find the magnitude of the torque, but what I am not sure about is which orientation would produce the \"maximum torque\" that it is asking about.\n\nCould someone please tell me which orientation would give a \"maximum torque\" and why?\n\nBy the way, I calculated the magnitude of the torque to be 18 Newton*meters.\n\n#### Attachments\n\n• torque.pdf\n163.1 KB · Views: 124\nLast edited:\n\n## Answers and Replies\n\n$$\\theta$$ is the angle between the coil and the the direction of magnetic induction. or rather the direction of the current and the magnetic induction. Now try putting the values of $$\\theta$$ accordingly and try and get the answer. By the way, what did you take the value theta to be, when you calculated its magnitude as 18Nm?"
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http://brane-space.blogspot.com/2011/01/advanced-physics-test-answers-3.html | [
"## Tuesday, January 18, 2011",
null,
"This concludes the Advanced Physics test answers, for the questions given in:\n\n31) The de Broglie wavelength is given by: L(D) = h/p\n\nwhere h is the Planck constant and p is the momentum. (In this case, we need to use the version h = 4.13 x 10^-15 eV*s)\n\nSince the problem information provides an energy E = 1 MeV, then we need p in the form:\n\np = [2mE]^½\n\nThese units mean it will also be easier to do the problem by using atomic mass units in MeV. Thus, 1 u = 931 MeV/c^2, so for an alpha particle (m =He4) we need 4u = 4 (931 MeV/c^2)\n\nThen:\n\nL(D) = (4.13 x 10^-15 eV*s)/ 2[4 (931 MeV/c^2)] = 10^-14 m\n\nL(D) = 10^-12 cm\n\n32) For a Young double slit, d sin (theta) = m L\n\ngives the bright fringe, where m is the order and L the wavelength. Since the distance to the screen is much greater than the height of the bright fringe, sin (theta) = x / D, where x is the height and D is the distance to the screen. So:\n\ndelta (x) d/ D = L\n\ndelta(x) = DL/ d\n\nthen: delta(x) = (3 x 10^3mm) (6 x 10^-5 cm/ 0. 2 cm) = 0.9 mm\n\n33) This application makes use of the energy-time uncertainty principle:\n\ndelta(E) delta(t) = h/ 2 π\n\nHere, t = the mean lifetime of the excited state = T(½) = 8 x 10^-8 sec\n\nAnd E= the half-width of the excited state = E(½)\n\nthen:\n\nE(½) T(½) =h/ 2 π = 1.03 x 10^-34 J-s\n\nE(½) = (1.03 x 10^-34 J-s)/ (8 x 10^-8 sec) = 1.2 x 10^-27 J\n\nBut: 1.6 x 10^-19 J = 1 eV (electron -volt)\n\nso: E(½) = 10^-9 eV\n\n34) We have: N = N(o) exp (-Lt)\n\nlet t = T(½) the half life, then:\n\nN = N(o) exp (-L (T(½) )\n\nor N/ N(o) = ½ = exp (-LT(½))\n\nand: - ln 2 = - LT(½)\n\nT(½) = (ln 2)/ L\n\n35) The \"mean life\" of one of the atoms in the sample will be:\n\n= INT (from 0 to oo) exp(-Lt) dt\n\n= 1/ L {exp(-Lt) ] 0 to oo\n\n= 1/L [exp (-L*0)] = 1/L [exp(o)] = 1/L\n\nsince exp(0) = 1\n\n36) We need to apply both conservation of energy, and conservation of momentum.\n\nFor the first: T1 = T1' + T2'\n\nfor the second: P1 = P1'cos(theta) + P2' cos (phi)\n\nand: 0 = P1'sin(theta) - P2' cos(phi)\n\nA head-on collision implies phi = 0 and theta = 180 deg\n\nso:\n\n[2m1T1]^½ = ([2m1T1']^½ ) cos(180) + ([2m2T2']^½ ) cos (0)\n\nBut cos (0)= 1 and cos (180) = -1, so:\n\n[2m1T1]^½ = - [2m1T1']^½ + [2m2T2']^½\n\nNow, since T1 = T1' + T2':\n\n2m1T1 + 4m1 [T1T1']^½ = 2m2T2' - 2m1T1'\n\n4m1 [T1T1']^½ = - 2m1T1' + 2m2T2' - 2m1(T1 - T2')\n\nwhence:\n\n2m1[T1(T1 - T2')]^½ = (m1 + m2)T2' - 2m1 T1\n\n4m1^2 T1 (T1 - T2') = (m1 + m2)^2T2' + 4m1^2T1^2 - 4m1T1(m1 + m2)T2'\n\n(m1 + m2)^2 T2'^2 - 4 m1m2T1T2' = 0\n\nThen:\n\nT2/T1 = 4 m1m2/ (m1 + m2)^2\n\n37) We sketch the HCl molecule so:\n\nH o<--------(x)---------- cm------------------->O (Cl)\n\nAnd 1.27 Å defines the entire distance as shown (arrow to arrow) and let 'x' be the distance from the center of mass (cm) to the H atom. (Recall 1Å = 10^-8 cm).\n\nThen:\n\nx = 36(1.27Å - x)\n\n37x = 36 (1.27 Å)\n\nx = 36(1.27 x 10^-8 cm)/ 37\n\nm(p) = 1.67 x 10^-24 g\n\nThe moment of inertia I is given by:\n\nI = m(p)^2 x^2 + 36 m(p) [(1.27 x 10^-8 cm) - x]^2\n\nI = (1.67 x 10^-24 g)(36/37)^2 + 36/(37)^2 [(1.27 x 10^-8 cm)]^2\n\nI = 2.6 x 10^-40 g-cm^2\n\n38) The Lande g-factor is defined:\n\ng = [ J(J+ 1) + S(S +1) - L(L +1)/ 2J(J + 1) ] + 1\n\nFrom our previous exposure to quantum mechanics\n\nSee, e.g.\n\nhttp://brane-space.blogspot.com/2010/07/l-s-coupling-problem-solutions-and-more.html\n\nand:\n\nhttp://brane-space.blogspot.com/2010/07/more-quantum-problems.html\n\nthe 1D 3/2 state implies: L = 2, S = 1 and J = L + S = 2 + 1 = 3\n\nthen:\n\ng = [3(3 + 1) + 1(1 + 1) - 2( 2+ 1)/ 2(3)(3 + 1)] + 1 = 1.33\n\n39) Recall the nuclear relation:\n\nf = M(A, Z) - A/ A\n\nand that graphing this vs. A yields a constituent mass 1.0085A. Thus the total binding energy is: (0.0085A) x (931 MeV) or about 7.9 MeV per nucleon.\n\n40) Here, it's important to recognize 2144 cm^-1 as the wave number, k. Also:\n\nk = f/c = 2144 cm^-1\n\nthen the frequency:\n\nf = kc = (2144 cm^-1) (3 x 10^10 cm/sec) = 6.43 x 10^13/s\n\n41) Since the current decreases uniformly, di/dt -> delta(I)/ delta(t)\n\nAnd V = L [delta(I)/ delta(t) ] = 0.25 H[ 2A/ (1/16s)]\n\nV = 8 volts\n\n42) Earth's atmosphere extends about 50 miles. If we assume the density of air to be constant:\n\np2 - p1 = rho g(z2 - z1)\n\nwhere p2, p1 are the different pressures at heights z2 and z1, respectively, and rho is the air density with g the acceleration of gravity.\n\nWe can form a simple proportion based on the simplifying assumption:\n\n(p2 - p1)/ p2 = (z2 - z1)/z2\n\np2 is what we need to find, and z2 is the height for 50 miles or ~ 50 x 5,000' = 250,000'.\n\nwhence:\n\n(z2 - z1) = 200'\n\n(p2 - p1)/ p2 = (z2 - z1)/z2 = 200'/ 250,000' = 0.0008\n\nso:\n\n(p2 - p1) = 0.0008 (p2)\n\np1 = p2 - 0.0008 (p2) ~ p2 ~ 76 Hg\n\n43) For this solution, we let all quantities inside the sphere be denoted by (1) and all the quantities outside by (2).\n\nBy Gauss' law:\n\n4 π r^2 E1 = 4 π rho(4 πr^2/ 3)\n\nwhere rho (charge density) = 3 q / (4 π a^2)\n\nTherefore, E1 = qr/ a^3\n\nFurther the associated energy is obtained via integration:\n\nW1 = 1/ 8 pi [INT (0 to a) E1^2 (dA)\n\n= 1/ 8 π [INT (0 to a) (qr/ a^3)^2 (4 π r^2) dr\n\nW1= q^2/ 10a\n\nBy Gauss' law:\n\n4 π r^2 E = 4 πq\n\nE2 = q/ r^2\n\nW2 = 1/ 8 π [INT (a to oo) (q/ r^2)^2 (4 π r^2) dr\n\nW2 = q^2/2 [INT (a to oo) dr/ r^2]\n\nW2 = q^2/ 2a\n\nTherefore: the total electrostatic energy W = W1 + W2\n\nW = q^2/ 10a + q^2/ 2a = 3q^2/ 5a\n\n44) It helps here to sketch the circuit ( see graphic)\n\nFrom this, the total impedance is (bear in mind j = [-1]^½):\n\nZ_T = 1/ j wL + 1/ (1/ jw C1) + 1/ (1/ jwC2)\n\nZ_T = 1/ jwL + jw( C1 + C2)\n\nZ_T = j[ - 1/wL + w(C1 + C2)]\n\nZ_T = -1/wL + w(C1 + C2) = 0 at the natural frequency\n\nso:\n\nw^2 = 1/ L(C1 + C2)\n\nw = [1/ (10^-5) (30 x 10^-6)]^½\n\nw = 10^5 [1/3]^½\n\nw = 57,735 /s = 2 π f\n\nso the natural frequency f= w/ 2 π= 9, 189 /s\n\nor about 9.2 kilocycles per second\n\n45) From thermodynamic relations we have:\n\nC(v) dT = C(p) - C(v)/ (@V/ @T)_V dV\n\nwhere the @ denote partials\n\nThen:\n\n(@T/@V)_s = (C(p)/ C(v) - 1) (@T/@V)_p\n\nBut:\n\n(@T/@V)_p = T/V since T = PV/ nR\n\ndT/ T + (C(p)/ C(v) - 1) (dV/V) = 0\n\nIntegrating both sides:\n\nTV^(C(p)/ C(v) - 1) = k\n\nPV^(C(p)/ C(v))/ nR = k\n\nPV^(C(p)/ C(v)) = k/ nR\n\nThen:\n\nn = k/ PV^[C(p)/ C(v)] R"
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"# U7d3 Notes Transformations Of Exponential Functions Answers\n\nThen the ordinary generating function Φω S. Parent Function: B( T) = 2 ë b. Understand Domain and Range of Functions studies so far. increasing and one of. Definitions: Exponential and Logarithmic Functions. The Integral91 1. Rewrite the function to identify h and k. By partnering with LearnZillion, teachers, students, and the whole district community benefit from superior curricula and an ease of implementation. (For teachers only) Cross-grade Level Connections. Exponential and Logarithmic Functions Summary - Notes Inverse Functions One-to-one Functions: Definition , Notes Find the Inverse Function - Summary, Example1 , Example2 , Homework Exponential Functions Exponential Functions: y = a^x - Case: a>1 , Case: 0 0. The natural logarithm function and exponential function are the inverse of each other, as you can see in the graph below: This inverse relationship can be represented with the formulas below, which the input to the LN function is the output of the EXP function :. Mathematics Vision Project | MVP - Mathematics Vision Project. c in the gsl source. 106, #1-4, Pg. In this lesson you will learn how to write and graph an exponential function by examining a table that displays an exponential relationship. Linear and quadratic parent functions are unique. g(x) = e. and is shared by the graphs of all quadratic functions. This set of guided notes takes your students through the following types of problems. Class Notes. Graphing transformations of exponential functions. Exponential growth and decay (2/17). Math Algebra 2 Transformations of functions Graphs of exponential functions. 2 Write arithmetic and geometric sequences both recursively and with an explicit formula, use them to model situations, and translate between the two forms. 6 – 8 Each function F(s) below is defined by a definite integral. Online tutoring available for math help. Class Notes. Some of the things that exponential growth is used to model include population growth, bacterial growth, and compound interest. Graph the piecewise-defined function. We tweaked it last year (before Desmos Activity Builder showed up) and it seems only right to convert it to this platform now!. Unit 3 parent functions and transformations homework 6 answer key. The distinction between discrete and continuous domains is explored through comparing and contrasting functions which have the same. This milk has to be consumed at night. Support Vector Machines. pdf Hw #30 transformations of exponential functionsKEY - Mar 27 2017 - 4-14 PM. West Ranch High School is part of the William S. However, because they also make up their own unique family, they have their own subset of rules. Now since the natural logarithm , is defined specifically as the inverse function of the exponential function, , we have the following two identities: From these facts and from the properties of the exponential function listed above follow all the properties of logarithms below. The squaring function f (x) = x 2 is a quadratic function whose graph follows. Here are some simple things we can do to move or scale it on the graph: We can move it up or down by adding a constant to the y-value: g(x) = x 2 + C. Transforming exponential graphs. ) Describe the concept of the transformation of functions. Finding Function Values Algebraically (numeric) Lesson 25. Transformations of exponential graphs behave similarly to those of other functions. So, it means the graph would be vertically stretched by a factor 3. Describe the sequence of Worksheet on Transformations of Exponential Functions Answer Section MULTIPLE CHOICE 1. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function without loss of shape. An equation gives the relationship between variables and numbers. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Graphing Exponential Functions It is important to know the general nature and shape of exponential graphs. 2: Transformations of Linear and Exponential Functions Warm-Up 3. Coyle - Lesson-Notes - 6. Transformations of Exponential Functions DRAFT. U7D3 Notes – Transformations of Exponential Functions Name_____ Desmos Marbleslides Exponentials - Go to. This is the currently selected item. Exponential inter-arrival times ⇒ Poisson number of arrivals ⇒ Continuously generate exponential variates until their sum exceeds T and return the number of variates generated as the Poisson variate. - [Voiceover] We're told the graph of y equals two to the x is shown below, so that's the graph. 8 The accompanying table shows the number of bacteria present in a certain culture over a 5-hour period, where x is the time, in hours, and y is the number of bacteria. Vertical Translation. It stayed the same. f(x) = 2x + 1 (0, 2) (1, 3) (2, 5) (3, 9) (−1 , 1½ ) Exponential Function. Unit 3 parent functions and transformations homework 6 answer key. f (x) = log 1/4 x, g(x) = log 1/4(4x) − 5 Writing Transformations of Graphs of Functions. Some of the worksheets displayed are Exponential functions date period, Work logarithmic function, 11 exponential and logarithmic functions work, Concept 17 write exponential equations, Exponential functions work 1, Exponential functions, Review exponential and logorithmic functions date, 4 1 exponential functions and their graphs. wmv - Duration: 11:47. She has come up with three questions and wants to know all the. U3D2_S_parent_functions_summary_table. If you found these worksheets useful, please check out Hyperbolic Functions Worksheet, Indefinite Integrals and the Net Change Theorem Worksheets, Finding the most general an tiderivative of a function, L H opital’s Rule Worksheet, Maxima and Minima Worksheet, Mean Value Theorem for Integrals, Calculus Quotient Rule Derivative, Functions and. Algebra 1 Unit 5: Comparing Linear, Quadratic, and Exponential Functions Notes 12 Day 4 – Function Transformations Function Notation Transformation Function Rule (from f(x)=x2) f(x + k) shift/translation left k units f(x) = (x + k)2 + k - 4 ‐. pdf Download File. The base a is a constant, positive and not equal to 1. This change also shifts the. Exponential functions can model the rate of change of many situations, including population growth, radioactive decay, bacterial growth, compound interest, and much more. Together, they completely determine an exponential function's. Graphing Rational Functions 23. Support Vector Machines. com Describe the transformation of f represented by g. Be sure to graph the squaring function using a dashed curve because it will be used as a guide and is not the answer. f(x) = 2−x = (½)x d. 1 Convolution 20. 2 Transformations Topic Transformations SOL AFDA. THE LAPLACE TRANSFORMATION L 3. They explore (with appropriate tools) the effects of transformations on graphs of exponential and logarithmic functions. Displaying all worksheets related to - Transformation Of Functions. The first term, , is already known (it is the real argument, described above). You will be asked about graphs, growth, decay, transformations, equations, and applications. Graphical Transformations of Functions In this section we will discuss how the graph of a function may be transformed either by shifting, stretching or compressing, or reflection. Transformations of Exponential Functions. We will see some of the applications of this function in the final section of this chapter. #Find#the#domain# ####a. Questions separated by topic from Core 2 Maths A-level past papers. 8 The accompanying table shows the number of bacteria present in a certain culture over a 5-hour period, where x is the time, in hours, and y is the number of bacteria. Algebra 1 Unit 4: Exponential Functions Notes 3 Asymptotes An asymptote is a line that an exponential graph gets closer and closer to but never touches or crosses. mu here is the mean of the distribution, also called the scale parameter on the wikipedia page the OP linked to, and lambda is the rate parameter. 4 properties of exponential functions worksheet. 01041 If and the function represents exponential Part IV: Exponential Growth an Decay Transformations Parent Exponential Function: bX4ð reflect- ess o z a el Four types of transformations (stretches, compressions, reflections, translations) apply to exponential functions. 15) whose real and imaginary parts are harmonic functions for arbitrary a. You can identify a y-transformation as changes are made outside the brackets of y=f(x). 8 Linear-Quadratic Systems. y-transformations. For now, functions will take one or more real numbers as an input, and return a numerical output. Welcome to IXL's grade 12 maths page. Inverse Trig Functions 21. Horizontal Translation. Graphing Trig Functions with Transformations. Thus, after repeated use of this algorithm, the expected number of. Families of Functions Name: Notes Date: Sarah is making an assignment for her math class. function settings control the overall color palette of an image in terms of light and dark in a non-linear function. Played 266 times. The students need to be familiar with domain and range of functions. Exponential growth and decay86 9. Class Notes. 4 units to the left. Transformations of Exponential Functions Notes Transformations of Exponential Functions Video Packet: pg. 761; practice Tuesday, 3/17. Practice: Graphs of exponential functions. The derivative of axand the de nition of e 84 6. Answer Key: Lesson 5. Exponential growth and decay by. b> 1 (Ex: _____)Exponential decay the decay factor, b, is always 00, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. Videos, examples, solutions, activities and worksheets for studying, practice and review of precalculus, Lines and Planes, Functions and Transformation of Graphs, Polynomials, Rational Functions, Limits of a Function, Complex Numbers, Exponential Functions, Logarithmic Functions, Conic Sections, Matrices, Sequences and Series, Probability and Combinatorics, Advanced Trigonometry, Vectors and. 1 ⃣Compare properties of two functions each represented in different ways Vocabulary: function, domain, range, function notation Definitions A F_____ is a relation in which each element in the domain. 6 – 8 Each function F(s) below is defined by a definite integral. Questions separated by topic from Core 2 Maths A-level past papers. How Do You Graph an Exponential Function Using a Table? Graphing an exponential function? No sweat! Create a table of values to give you ordered pairs. Exponential regression calculator desmos. Unit 3b — Linear and Exponential Functions Date: Day 39 - Notes Transformations of Linear and Exponential Graphs Use your graphing calculator to graph the following. Question Answer Marks Guidance 1 y = e2xcos x M1 A1 M1 M1 dy/dx = 2e2xcos x – e 2 xsin x dy/dx = 0 e2x(2cos x – sin x) = 0 2cos x = sin x 2 = sin x/cos x = tan x x = 1. increasing and one of. Packet due om Friday. Evaluate logarithmic functions with. Practice: Graphs of exponential functions. A y-transformation affects the y coordinates of a curve. Exercises87 Chapter 7. 3 For each function A) below fill in the table of values below B) identify the exponential function as growth or decay curve C) identify the common ratio D) identify the y-intercept E) state the equation of the asymptote (show on graph as well) F) state the domain and range G) find y(-3) i) {12}. Transformations of exponential graphs behave similarly to those of other functions. Suppose c > 0. It is approximately 2. function settings control the overall color palette of an image in terms of light and dark in a non-linear function. This curve is very powerful because it models population growths where. Given the parent function and a description of the transformation, write the equation of the transformedfunction, f(x). Graphing Piecewise Defined Functions. Exponential functions can model the rate of change of many situations, including population growth, radioactive decay, bacterial growth, compound interest, and much more. accurately sketch the graphs. Seamless LMS and SIS Integration. Consider a function f(˝) that is periodic with period T. All subsequent terms in a polynomial function have exponents that decrease in value by one. 1c Compoud Growth and Decay Models Review of Exponential Function Transformations Homework: page 339 (87, 89, 91, 97, 99) page 351-352 (7, 11, 21, 25, 27) *Parent Conferences - November 14, 2018* Thursday. 2: Transformations of Linear and Exponential Functions Warm-Up 3. The function 𝑓𝑥=1 𝑥 has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. In this lesson you will learn how to write and graph an exponential function by examining a table that displays an exponential relationship. Math Algebra 2 Transformations of functions Graphs of exponential functions. The answer is that the choice of kernel (a. John Titterton. Algebra, Functions, and Data Analysis (AFDA) Strand AFDA. - [Voiceover] We're told the graph of y equals two to the x is shown below, so that's the graph. Played 266 times. and is shared by the graphs of all quadratic functions. Transforming exponential graphs (example 2) Graphing exponential functions. This set of guided notes takes your students through the following types of problems. 2: Transformations of Linear and Exponential Functions Warm-Up 3. Graphing Exponential and Logarithmic Functions (HSF-IF. Conduct a brief discussion on how the parent function y = x 2 could be used to help graph the function in vertex form: y = – (x – 3) 2 + 4. Notes Common Core States Standards for Mathematics Pearson Algebra 1 Common Core, ©2015 Create equations that describe numbers or relationships. docx File Size:. 5 Exponential and Logarithmic Functions 103 Chapter 5 Exponential and Logarithmic Functions 5. Displaying all worksheets related to - Transformation Of Functions. ∫ ∞ − + 0. Transform exponential and logarithmic functions by changing parameters Describe the effects of changes in the coefficients of exponential and logarithmic functions Who uses this? Psychologists can use transformations of exponential functions to describe knowledge retention rates over time. 1 Periodic Functions and Their Properties 6. Mathematics. Home; Unit 3 parent functions and transformations homework 6 answer key. In each case the order of the functions matters because arithmetically the outcomes will be different. The Integral91 1. function settings control the overall color palette of an image in terms of light and dark in a non-linear function. f of Z ? Problems of this type are of interest from a practical standpoint. U7D1 Notes; HW U7D1 Wkst; Notes - U7D2 Growth and Decay; HW - U7D2 Wkst; Notes U7D3 Transformations of Exponential Functions; HW U7D3 Exponential Transformations; Notes U7D4 Linear vs. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $$f(x)=b^x$$ without loss of shape. Practice: Graphs of exponential functions. Write function vertical shift down of 5 and horizontal shrink by a factor of Write function horizontal shift right of 2, vertical shift up 3, and vertical stretch by factor of 4 REFLECTIONS – About x-axis About y-axis Given parent function Describe reflection Write function vertical shift up of 1 and. E Graph exponential and logarithmic functions, showing intercepts and end behavior…. Absolute Value — vertical shift up 5, horizontal shift right 3. Transformations of Exponential Functions Notes Transformations of Exponential Functions Video Packet: pg. If you get logged out go back in and make sure you finish. 2 On a map, Maple Street is represented by the function f ( x ) = 2 x – 1, and Highland Street is. One very important exponential equation is the compound-interest formula:where \"A\" is the ending amount, \"P\" is the beginning amount (or \"principal\"), \"r\" is the interest rate (expressed as a decimal), \"n\" is the number of compoundings a year, and \"t\" is the total number of years. But e is the amount of growth after 1 unit of time, so $\\ln(e) = 1$. 2: Transformations of Linear and Exponential Functions Warm-Up 3. Which of the following is a graph of y is equal to negative one times two to the x plus three plus four? They give us four choices down here. Having an Exp. Transforming exponential graphs. She wants to create one question that has as many different answers as possible related to a single family of functions with transformations. 5 Investigating ways that two lines can intersect - lesson and homework solution for pg61 #9b download. U3D2_S_parent_functions_summary_table. Identify the asymptote of each graph. 2 to linear and exponential equations, and, in the case of exponential equations, limit to situations requiring evaluation of exponential functions at integer inputs. Answers is the place to go to get the answers you need and to ask the questions you want. 4 properties of exponential functions worksheet. We tweaked it last year (before Desmos Activity Builder showed up) and it seems only right to convert it to this platform now!. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function without loss of shape. It does not include the growth/decay, variation or solving exponentials (logs) portion of the test. Math Algebra 2 Transformations of functions Graphs of exponential functions. Consider a function f(˝) that is periodic with period T. Piecewise Defined Functions and Function Values. The students need to be familiar with domain and range of functions. They are given by equations such as. The irrational number e is also known as Euler’s number. What is the equation of the function? c. Changing from Exponential Form to Logarithmic Form – Practice Problems Move your mouse over the \"Answer\" to reveal the answer or click on the \"Complete Solution\" link to reveal all of the steps required to change from exponential form to logarithmic form. What happens if a > 1? To examine this case, take a numerical. Also, when watching the video, click settings and change to highest quality. Graphing Trig Functions with Transformations. CHECK HERE FOR DAILY POSTS TO INCLUDE NOTES, ASSIGNMENTS, AND ANSWER KEYS. There is, however, one transformation, for one set of functions that has caused me and my students grief every year. Conversely, having an Exp. Renaisassance Arising -RENAISSANCE, a renewal of life and vigor, our interest in all things restored, a rebirth a revival; a moral renaissance of, by and for the people, it is ARI. What is the equation of the function? c. This phase is called log phase because the logarithm of the bacterial mass increases linearly with time, and exponential growth phase because the number of cells increases as an exponential function of 2 n (i. The squaring function f (x) = x 2 is a quadratic function whose graph follows. The parameter a is called the function's y-intercept and the parameter b is called the base. For example, if you know that the quadratic parent function $$y={{x}^{2}}$$ is being transformed 2 units to the right, and 1 unit down (only a shift, not a stretch or a flip yet), we can create the original t-chart, following by the transformation points on the outside of the original points. Thus, for free, we get back an independent exponential which could then be used as one of the two needed in Step 1, if we were to want to start generating yet another independent N(0,1) rv. ARIMA(0,1,1) with constant = simple exponential smoothing with growth: By implementing the SES model as an ARIMA model, you actually gain some flexibility. Solve exp and log functions. When we move or re-position a graph of a function this is called a transformation. ) List the toolbox functions. Played 266 times. Markov Sequences. Math homework help. 7e) - We not only plot them but start to use the graphs to make predictions based on the lines. where the value of the function indicates the brightness of the pixel at that (x;y) coordinate. different transformations of an exponential function will result in a different graph from the basic graph. 3 Interpreting Sinusoidal Functions. Exponential function definition is - a mathematical function in which an independent variable appears in one of the exponents —called also exponential. Class Notes. a >1 then the graph looks like this: This is sometimes called a. Practice on Inverse Function. The base a is a constant, positive and not equal to 1. 5 Investigating ways that two lines can intersect - lesson and homework solution for pg61 #9b download. Worksheet 3 Graphing exponential functions g(x) =- Hour Identify each transformation from the parent function of Tell if the function is a decay or growth function. First, remember the rules for transformations of functions. It is used so frequently that it has its own notation: ln x, and is read “the natural log of x” or “L‐N of x,” in which case you actually say the letters L and N. 1 Solve 1 6 3x 2 = 36x+1. Example 1 : Graph the following fucntions by creating a small table of values. ontariotechu. The mathematical model of exponential growth is used to describe real-world situations in population biology, finance and other fields. This section covers: Basic Parent Functions Generic Transformations of Functions Vertical Transformations Horizontal Transformations Mixed Transformations Transformations in Function Notation Writing Transformed Equations from Graphs Rotational Transformations Transformations of Inverse Functions Applications of Parent Function Transformations More Practice (For Absolute Value Transformations. This calculator will calculate the exponential function with the given base and exponent. Graphing Transformations of Exponential Functions. Lesson 4 Function Notation and Function Representations Test Review (fall 2013) Review Answers (fall 2013) Chapter 1 Tools of Algebra. In Statgraphics, alas, the function that is called LOG is the natural log, while the base-10 logarithm function is LOG10. Which of the following is a graph of y is equal to negative one times two to the x plus three plus four? They give us four choices down here. Solving exponential equations using exponent rules. Remember you can take it multiple times. Powered by Create your own unique website with customizable templates. Piecewise Defined Functions and Function Values. West Ranch High School is part of the William S. Khan Academy is a 501(c)(3) nonprofit organization. LOGEST is the exponential counterpart to the linear regression function LINEST described in Testing the Slope of the Regression Line. accurately sketch the graphs. 4 units up. by cfsts08. But e is the amount of growth after 1 unit of time, so $\\ln(e) = 1$. An equation gives the relationship between variables and numbers. Outcomes Students will be able to graph a function using the parent functions and. Packet due om Friday. 7 Families of Quadratic Functions. 7 Spiral Review 2014. Ordinary generating functions arise when we have a (finite or countably in-finite) set of objects S and a weight function ω : S →Nr. See full list on study. The following list outlines some basic rules that apply to exponential functions: The parent exponential function f(x) = bx always has a horizontal asymptote at y = 0, except when […]. Absent students, find the notes below. The distinction between discrete and continuous domains is explored through comparing and contrasting functions which have the same. Powered by Create your own unique website with customizable templates. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Graphing transformations of exponential functions. U7D1 Notes; HW U7D1 Wkst; Notes - U7D2 Growth and Decay; HW - U7D2 Wkst; Notes U7D3 Transformations of Exponential Functions; HW U7D3 Exponential Transformations; Notes U7D4 Linear vs. If there is not a unique solution, then y is not a function of x. If you get logged out go back in and make sure you finish. Graphing an Exponential Function with a Vertical Shift An exponential function of the form f(x) = b x + k is an exponential function with a vertical shift. The parent graph is shown in red and the variations of this graph appear as follows: the function y = f(x) + 2 appears in green; the graph of y = f(x) + 5 appears in blue; the graph of the function y = f(x) - 1 appears in gold; the graph of y = f(x) - 3 appears in purple. Coyle - Lesson-Notes - 6. Some of the worksheets for this concept are Transformations of exponential functions work, Exponential transformations work, Exponential functions date period, Transformations of functions name date, Transforming exponential and. First of all, the estimated MA(1) coefficient is allowed to be negative: this corresponds to a smoothing factor larger than 1 in an SES model, which is usually not allowed by the SES model-fitting procedure. Transformations of exponential graphs behave similarly to those of other functions. Moving an exponential function up or down moves the horizontal asymptote. b> 1 (Ex: _____)Exponential decay the decay factor, b, is always 00, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. The x coordinates are unaffected. Transformations of exponential graphs behave similarly to those of other functions. Students should have knowledge of the parent functions: linear, quadratic, trigonometric, exponential, logarithmic, power, and rational. Standard F. The squaring function f (x) = x 2 is a quadratic function whose graph follows. docx File Size:. 3 Solving Linear Systems - Graphically - Geosketchpad Demo pg61 #9b download 1. Please make sure you put your FIRST and LAST name. Vertical Translations A shift may be referred to as a translation. Application Use functions, including those listed above, to model a variety of real-world problem solving applications. 2 Transformations Topic Transformations SOL AFDA. Suppose that we are given the graph of the equation. Showing top 8 worksheets in the category - Transformation Of Function. This is over all of Unit 4 Exponential Functions. Answer Key: Lesson 5. exponential function. Which of the following is a graph of y is equal to negative one times two to the x plus three plus four? They give us four choices down here. If there are horizontal transformations, the domain will change. Hart Union School District and is located in Stevenson Ranch, CA. Examples of isometrics are reflection, rotation and translation. Tip #1: To Connect The Math Vocabulary To Prior Learning Okay, so you've probably heard this before, but as the teacher, we must connect new information to. Transformations of the Exponential Function The exponential function can be represented in function notation as f (x) = b x for some base b, 0 < b < 1 or b > 1. See randist/exponential. f(x) = −2x c. Students learn to use function notation to ask and answer questions about functional relationships presented in tabular, graphical, and algebraic form. 139 #3, 4, 5,a,c,g,j,l. Classifying Even and Odd Functions (HSF-IF. Statistical Inference. They look like this: Read about graphing linear functions. Transformations Review Worksheet Blank and KEY 1/23/20. If b is greater than 1, the function continuously increases in value as x increases. The students need to be familiar with domain and range of functions. If there are horizontal transformations, the domain will change. Graphical Transformations of Functions In this section we will discuss how the graph of a function may be transformed either by shifting, stretching or compressing, or reflection. 2 Working with Integer Exponents. Transformations of Exponential Functions DRAFT. But e is the amount of growth after 1 unit of time, so $\\ln(e) = 1$. 2 on Tuesday, November 15th. for the complex exponential yields two important harmonic functions: excosyand exsiny, which are graphed in Figure 2. 13 Simple vs. 2: Transformations of Linear and Exponential Functions Warm-Up 3. Stretches: vertical (and horizontal) 2. This figure shows each of these as steps: Figure a is the horizontal transformation, showing the parent function y = 2 x as a solid line, and Figure b is the vertical transformation. Graph exponential functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\\left(x\\right)={b}^{x}$ without loss of shape. 62% average accuracy. The lesson includes a table summary of all the properties of the exponential graph and ends off with a few questions for learners to apply what they have learnt in the lesson. 3 Solving Linear Systems - Graphically - Geosketchpad Demo pg61 #9b download 1. 8 Linear-Quadratic Systems. INTRODUCTION 2. However, because they also make up their own unique family, they have their own subset of rules. One very important exponential equation is the compound-interest formula:where \"A\" is the ending amount, \"P\" is the beginning amount (or \"principal\"), \"r\" is the interest rate (expressed as a decimal), \"n\" is the number of compoundings a year, and \"t\" is the total number of years. Graphing an Exponential Function with a Vertical Shift An exponential function of the form f(x) = b x + k is an exponential function with a vertical shift. Exponential; CW Unit 7 Review - Exponential Transformations; HW Unit 7 Test Review; Mixed Review #7. Equation states that the fourier transform of the cosine function of frequency A is an impulse at f=A and f=-A. 11 Exponential Regression and 5. They should also be able to do transformations with these functions. Class Notes. - [Voiceover] We're told the graph of y equals two to the x is shown below, so that's the graph. West Ranch High School is part of the William S. Math homework help. Please check the figure a and b. 2: Transformations of Linear and Exponential Functions Warm-Up 3. Graphing Log Functions HW: pg 12 and 13 m3u2d7_notes. Ordinary generating functions arise when we have a (finite or countably in-finite) set of objects S and a weight function ω : S →Nr. 3 For each function A) below fill in the table of values below B) identify the exponential function as growth or decay curve C) identify the common ratio D) identify the y-intercept E) state the equation of the asymptote (show on graph as well) F) state the domain and range G) find y(-3) i) {12}. Graphing Exponential Functions Worksheets. Quadratic – Vertical Motion, Path of flying objects yx yx 2 Absolute Value - Distance Exponential – Population and Monetary Growth, Decay yx y 2x. 1 ⃣Compare properties of two functions each represented in different ways Vocabulary: function, domain, range, function notation Definitions A F_____ is a relation in which each element in the domain. By partnering with LearnZillion, teachers, students, and the whole district community benefit from superior curricula and an ease of implementation. Having an Exp. This milk has to be consumed at night. 2 Working with Integer Exponents. your answer to the nearest tenth of a gram. Practice on Inverse Function. Transforming exponential graphs. The equation for the line of an asymptote for a function in the form of f(x) = abx is always y = _____. Some trig equations could involve a function of x or θ (see Transformations of Trigonometric Functions) Functions could be in two forms, either y = sin ( θ ± k) or y = sin k θ An easy way to solve an equation involving sin, cos or tan of (θ± k) or kθ is by transforming the range of the question. f(x) = −2x. (These are not listed in any recommended order; they are just listed for review. Then, plot those ordered pair on a coordinate plane and connect the points to make your graph! Follow along with this tutorial as it shows you all the steps. Unit 9 Transformations Homework 2 Reflections Answers Gina Wilson. Horizontal asymptote: C. ELECTRICAL SYSTEMS Analysis of the three basic passive elements R, C and L Simple lag network (low pass filter) 1. 3 problem #10, including graphing the two functions and finding the intersection between the two lines as part of their analysis. Math homework help. Trigonometry: Addition Formulas 18. 3 Working with Rational Exponents. Functions Working with Functions Graph Transformations and Asymptotes Modulus Functions, Equations and Inequalities Logarithmic and Exponential Functions The Exponential Function and Natural Log Functions. Some of the things that exponential growth is used to model include population growth, bacterial growth, and compound interest. Functions)Worksheet) Domain)Range)and)Function)Notation) 1. They are given by equations such as. Lesson 1 Part 1 Properties of Real Numbers. Also, when watching the video, click settings and change to highest quality. Absent students, find the notes below. Notes: Function Operations and Compositions Notes. Graphing Exponential and Logarithmic Functions (HSF-IF. Students consider real world objects and data that can be described using exponential functions. Then graph each function. Limits involving exponentials and logarithms86 8. 8 Combinations of Functions: Composite Functions Algebra of Functions Section 1. Graphing exponential functions. Thanks for putting in the effort AND sharing!. [Hint: each expression is the Laplace transform of a certain function. ) Discuss some applications of the transformations of functions. 5 Quadratic Function Models. If c is added to the function, where the. to the base. 3 Interpreting Sinusoidal Functions. Exponential decay: Half-life. If a = 1 then f(x) = 1x = 1. c in the gsl source. When fchanges its sign92 3. Moving an exponential function up or down moves the horizontal asymptote. î g(x) = +5 - +2 Write the function for each graph described below. They should also be able to do transformations with these functions. If any argument is an array, then fourier acts element-wise on all elements of the array. 2 On a map, Maple Street is represented by the function f ( x ) = 2 x – 1, and Highland Street is. Functions, Function Notation, and the Domain of a Function. 5 Investigating ways that two lines can intersect - lesson and homework solution for pg61 #9b download. Transformations of Exponential Functions Formula for Geometric Series (Honors Only) Review Summative Assessment Exponential Real World Applications Compound Interest 3 1/21 – 1/25 Unit 8: Summary of Functions Comparing Linear, Quadratic, and Exponential Functions -- Part 1 Comparing Linear, Quadratic, and Exponential Functions -- Part 2. ANS: OBJ: 3. Questions separated by topic from Core 2 Maths A-level past papers. Khan Academy: Transformations of Exponential and Logarithmic Functions Self Check *does not include absolute value transformation. 8 The accompanying table shows the number of bacteria present in a certain culture over a 5-hour period, where x is the time, in hours, and y is the number of bacteria. Feb 16, 2015 - Students will need to match an equation of a graph first with a description of the transformations of the graph and secondly with a picture of the graph. Mathematics. Exponential, and Logarithmic Functions. We can take three different cases, where a = 1, 0 < a < 1 and a > 1. 7 Spiral Review 2014. T-charts are extremely useful tools when dealing with transformations of functions. 1 Distinguish between situations that can be modeled with linear functions and with exponential functions. 1a Show that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. The parent function f(x)=log 10x is vertically stretched by a factor of 3, reflected in the y-axis, horizontally transformed 4 units to the left and vertically transformed 2. 5x are exponential functions. Note: to move the line down, we use a negative value for C. your answer to the nearest tenth of a gram. Follow these steps to write an exponential equation if you know the rate at which the function is growing or decaying, and the initial value of the group. Log InorSign Up. 1 Lesson What You Will Learn Identify and evaluate exponential functions. 13 Simple vs. Please show your support for JMAP by making an online contribution. In the remainder of this section (and elsewhere on the site), both LOG and LN will be used to refer to the natural log function, for compatibility with Statgraphics notation. Online tutoring available for math help. f(x) = 2x b. This set of guided notes takes your students through the following types of problems. Trigonometric Equations* 20. First, remember the rules for transformations of functions. Write function vertical shift down of 5 and horizontal shrink by a factor of Write function horizontal shift right of 2, vertical shift up 3, and vertical stretch by factor of 4 REFLECTIONS – About x-axis About y-axis Given parent function Describe reflection Write function vertical shift up of 1 and. nal functions, exponential functions, logarithmic functions, and trigonometric functions with the function viewpoint. See full list on study. The squaring function f (x) = x 2 is a quadratic function whose graph follows. They extend the domain of exponential functions to the entire real line (N-RN. An equation gives the relationship between variables and numbers. 2: Transformations of Linear and Exponential Functions Warm-Up 3. An exponential function is the inverse of a logarithm function. answer choices. Graphing exponential functions. If any argument is an array, then fourier acts element-wise on all elements of the array. 761; practice Tuesday, 3/17. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\\left(x\\right)={b}^{x}$ without loss of shape. 274 Chapter 6 Exponential Functions and Sequences 6. Functions, Function Notation, and the Domain of a Function. A function is a set of mathematical operations performed on one or more inputs (variables) that results in an output. The students need to be familiar with domain and range of functions. The quotient is the equation for the slant asymptote. An experienced CMP2 user will answer your questions about implementing and teaching CMP2. Together, they completely determine an exponential function's. It is used so frequently that it has its own notation: ln x, and is read “the natural log of x” or “L‐N of x,” in which case you actually say the letters L and N. An exponential function is a Mathematical function in form f (x) = a x, where “x” is a variable and “a” is a constant which is called the base of the function and it should be greater than 0. Also, when watching the video, click settings and change to highest quality. If there are horizontal transformations, the domain will change. We start with the blue graph which is the graph of the function f(x) = e x. Vertical stretch/compression. 1 Lesson What You Will Learn Identify and evaluate exponential functions. Exponential generating functions are of another kind and are useful for solving problems to which ordinary generating functions are not applicable. f(x) = 2x + 1 (0, 2) (1, 3) (2, 5) (3, 9) (−1 , 1½ ) Exponential Function. There are three types of asymptotes possible for any exponential function. Trigonometry: Basic Identities* 17. f(x) = −2x c. 1 Solve 1 6 3x 2 = 36x+1. Exponential decay: Half-life. Chapter 3 Exponential and Logarithmic Functions. 8 Linear-Quadratic Systems. Piecewise Defined Functions and Function Values. October 3 Exponential Functions Unit Test TODAY Homework: From Wednesday’s class – Pg. U3D2_S_parent_functions_summary_table. 13 Simple vs. The base a is a constant, positive and not equal to 1. MATH III Honors COURSE DOCUMENTS: This page is a very important resource for parents/ guardians and students. To perceive how this functions, investigate the diagram of h(x) = x 2 + 2x - 3. The distinction between discrete and continuous domains is explored through comparing and contrasting functions which have the same. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $$f(x)=b^x$$ without loss of shape. If you run the above code for an arbitrary array of N=20 wines and calculate how many times was the function called for arguments be=10 and en=10 you will get a number 92378. f(˝+ T) = f(˝) (1) We may always rescale ˝to make the function 2ˇperiodic. The x coordinates are unaffected but all the y coordinates go up by 4. Rewrite the function to identify h and k. Graph exponential functions using transformations. By partnering with LearnZillion, teachers, students, and the whole district community benefit from superior curricula and an ease of implementation. Functions might have horizontal asymptotes, vertical asymptotes, and slant asymptotes. Transformation Of Exponential Functions With Answers - Displaying top 8 worksheets found for this concept. The action of a function on an element is denoted by: x → f (x) For a function to be completely defined, not only must we be told the action the function has on various elements. When fchanges its sign92 3. Transformation Of Exponential Functions With Answers. 7 Families of Quadratic Functions. by cfsts08. Practice: Graphs of exponential functions. ∫ ∞ − + 0. Too often, districts are forced to choose between curricular excellence and a usable digital platform. Let us start with a function, in this case it is f(x) = x 2, but it could be anything: f(x) = x 2. For example, f(x)=3x is an exponential function, and g(x)=(4 17) x is an exponential function. Exponential; CW Unit 7 Review - Exponential Transformations; HW Unit 7 Test Review; Mixed Review #7. Mathematics. The solution is A(t) = A 0 e-kt. Statistical Inference. They explore (with appropriate tools) the effects of transformations on graphs of exponential and logarithmic functions. Mathematics Vision Project | MVP - Mathematics Vision Project. But e is the amount of growth after 1 unit of time, so $\\ln(e) = 1$. The students need to be familiar with domain and range of functions. Exponential functions. Transform exponential and logarithmic functions by changing parameters Describe the effects of changes in the coefficients of exponential and logarithmic functions Who uses this? Psychologists can use transformations of exponential functions to describe knowledge retention rates over time. Exponential and Logarithmic Functions TEST on 4. Absolute Value — vertical shift up 5, horizontal shift right 3. 1 Derivatives of exponential and logarithmic func-tions If you are not familiar with exponential and logarithmic functions you may wish to consult the booklet Exponents and Logarithms which is available from the Mathematics Learning Centre. 2 on Tuesday, November 15th. Vertical Translation. U3D2 complete parent_functions_summary_table: 3: 3. Unit 4: Inequalities Graphing Inequalities One Step Inequalities Two Step Inequalities Multi-Step Inequalities (#1-10 are variables on one side, #11-24 have variables on both sides) Compound. 5 Exploring the Properties of Exponential Functions. This calculator will calculate the exponential function with the given base and exponent. They use simulations and modelling to investigate transformations,. Played 266 times. 3 Solving Linear Systems - Graphically - Geosketchpad Demo pg61 #9b download 1. answer choices. 1c Compoud Growth and Decay Models Review of Exponential Function Transformations Homework: page 339 (87, 89, 91, 97, 99) page 351-352 (7, 11, 21, 25, 27) *Parent Conferences - November 14, 2018* Thursday. Write an exponential regression equation for this set of data, rounding all values to four decimal places. Just like the exponential function with base e is called the natural exponential function, the logarithm with base e is called the natural logarithm. mu here is the mean of the distribution, also called the scale parameter on the wikipedia page the OP linked to, and lambda is the rate parameter. The answer key for the textbook questions can be accessed here: Textbook Answers. Note: Any transformation of y = bx is also an exponential function. Exponential decay: Half-life. Algebra 1 Unit 4: Exponential Functions Notes 1 Unit 4: Exponential Functions After completion of this unit, you will be able to… Learning Target #1: Graphs and Transformations of Exponential Functions Evaluate an exponential function Graph an exponential function using a xy chart Identify whether a function is exponential, quadratic, or. 5 Writing and Graphing Exponential Growth Functions. Feb 16, 2015 - Students will need to match an equation of a graph first with a description of the transformations of the graph and secondly with a picture of the graph. Graphing Rational Functions 23. 1) log (u2 v) 3 2) log 6 (u4v4) 3) log 5 3 8 ⋅ 7 ⋅ 11 4) log 4 (u6v5) 5) log 3 (x4 y) 3 Condense each expression to a single logarithm. 7 Spiral Review 2014. Use the table feature to Graph the functions y = x + 2 and 2. Class Notes. Graph the transformed function on the same Cartesian coordinate grid and describe the transformations based on the function t(x). Identifying and Evaluating Exponential Functions An exponential function is a nonlinear function of the form y = abx, where a ≠ 0,. Author Steven Kotler explains the basic tenets of exponential technology growth cycles. She wants to create one question that has as many different answers as possible related to a single family of functions with transformations. function for the graph at the right. This includes school websites and teacher pages on school websites. answer choices. A polynomial function is a function comprised of more than one power function where the coefficients are assumed to not equal zero. Graphing Rational Functions 23. EDIT: just for comparison with some later answers - this is equivalent with mu = 1/lambda. ANALYSIS OF FUNCTIONS Exponential Functions Worksheet Logarithmic Functions Use the graph of. Exponential Functions Topics: 1. Transformation Of Exponential Functions With Answers. Derivatives of Logarithms85 7. If you are an algebra teacher who wants to teach quadratic functions, here are 3 quick tips you need to know. Lesson 3 Function Transformations. Convert exponential functions to logarithmic functions, and vice versa. Let's begin by considering the functions. Math homework help. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function f (x) = b x without loss of shape. To obtain the graph of: y = f(x) + c: shift the graph of y= f(x) up by c units. 13 Simple vs. The actual values that may be plotted are relatively few, and an understanding of the general shape of a graph of growth or decay can help fill in the gaps. Simplify: 2 4 3 5 7 36xy 64x y ¸ ¸ ¹ · ¨ ¨ ©. Vertical stretch/compression. 5 Investigating ways that two lines can intersect - lesson and homework solution for pg61 #9b download. 3 Interpreting Sinusoidal Functions. Exponential functions: Know how to move between description, graphs, tables, and equations for exponential functions (e. Write a function that represents Clayton’s population growth since 1998. Like logarithmic and exponential functions, rational functions may have asymptotes. Inverse Trig Functions 21. Algebra 1--U7D3 Notes--Exponential Growth/Decay Strange answers to the 8. y-transformations. Parent Function: B( T) = 2 ë. We can take three different cases, where a = 1, 0 < a < 1 and a > 1. Exponential Decay: In a sample of a radioactive material, the rate at which atoms decay is proportional to the amount of material present. MATH III Honors COURSE DOCUMENTS: This page is a very important resource for parents/ guardians and students. On this page you will find: a complete list of all of our math worksheets, lessons, math homework, and quizzes. 3 Polynomial Functions of Higher degree. See full list on shelovesmath. The term with the highest degree of the variable in polynomial functions is called the leading term. Exponential Real World ApplicationsGrowth and Decay Rates of Exponential Functions Transformations of Exponential Functions Formula for Geometric Series (Honors Only) Review Summative Assessment Compound Interest 3 1/21 – 1/25 Unit 8: Summary of Functions Comparing Linear, Quadratic, and Exponential Functions -- Part 1 Comparing Linear. Thanks for putting in the effort AND sharing!. Describe the sequence of Worksheet on Transformations of Exponential Functions Answer Section MULTIPLE CHOICE 1. Here is a set of practice problems to accompany the Logarithm Functions section of the Exponential and Logarithm Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University. Welcome to IXL's grade 12 maths page. accurately sketch the graphs. Derivatives of Logarithms85 7. Here you will find notes, practice questions and solutions for GCSE, arranged by subject area (Number, Algebra, Shape and Space, Handling Data), and by topic. Trigonometry: Double-Angle and Half-Angle Formulas* 19. By the way, we never have exponential functions with negative bases like ( 2)− x. 2 1, 2 2, 2 3, 2 4,2 5 and so on). Polar Coordinates 22. The solution is A(t) = A 0 e-kt. 2: Transformations of Linear and Exponential Functions Warm-Up 3. Together, they completely determine an exponential function's. Exponential Graphs. Families of Functions Name: Notes Date: Sarah is making an assignment for her math class. Note: the below Unit 3 review is for transformations & Graphs."
] | [
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https://wiki.seg.org/wiki/Frequency-space_implicit_schemes | [
"# Frequency-space implicit schemes\n\nSeries",
null,
"Investigations in Geophysics Öz Yilmaz http://dx.doi.org/10.1190/1.9781560801580 ISBN 978-1-56080-094-1 SEG Online Store\n\n## Migration principles\n\nAs discussed in finite-difference migration in practice, in practice the 15-degree finite-difference migration can handle dips up to 35 degrees with sufficient accuracy. A steep-dip approximation to equation (13b) is achieved by continued fractions expansion (Section D.4) as\n\n $k_{z}={\\frac {2\\omega }{v}}\\left[1-{\\frac {v^{2}k_{x}^{2}}{8\\omega ^{2}}}{\\frac {1}{1-{\\frac {v^{2}k_{x}^{2}}{16\\omega ^{2}}}}}\\right].$",
null,
"(18)\n\nThis dispersion equation is known as the 45-degree approximation and is the basis of the most common implementation of steep-dip implicit finite-difference schemes .\n\nRefer to the steps described earlier and replace the Taylor expansion given by equation (14a) with the continued fractions expansion given by equation (18). Follow the subsequent steps to derive the corresponding differential equation associated with the 45-degree diffraction term (Section D.4):\n\n $i{\\frac {v}{4\\omega }}{\\frac {\\partial ^{3}Q}{\\partial z\\partial x^{2}}}-{\\frac {\\partial ^{2}Q}{\\partial x^{2}}}+i{\\frac {4\\omega }{v}}{\\frac {\\partial Q}{\\partial z}}=0,$",
null,
"(19a)\n\nwhere Q(x, z, ω) is the retarded wavefield in the frequency-space domain.\n\nWhen recast for time migration, equation (19a) becomes (Section D.4):\n\n $i{\\frac {1}{2\\omega }}{\\frac {\\partial ^{3}Q}{\\partial t\\partial x^{2}}}-{\\frac {\\partial ^{2}Q}{\\partial x^{2}}}+i{\\frac {8\\omega }{v^{2}}}{\\frac {\\partial Q}{\\partial \\tau }}=0,$",
null,
"(19b)\n\nwhere τ is the time variable associated with the migrated data.\n\nNote that dropping the first term in equation (19a) and inverse Fourier transforming in time yields the 15-degree diffraction equation (16a). Similarly, dropping the first term in equation (19b) yileds the 15-degree equation (17) for time migration.\n\nAs for the 15-degree equation, the thin-lens equation (16b) also applies for the 45-degree equation. When implemented in the frequency-space domain, the thin-lens term is represented by the phase-shift operator of equation (15a). Again, the final step in the procedure is to write down the difference forms of the differential operators in implicit form to be used in finite-difference solution of the 45-degree equation (19) for migration. Kjartansson provides an implicit scheme in which the extrapolation is in z. Nevertheless, as for the 15-degree equation (17), it is trivial to adapt his scheme for time migration with the extrapolation in τ of equation (15b). The phase-shift operator of equation (15a) is velocity-dependent when implemented for depth migration, and it is velocity-independent when implemented for time migration.\n\nThe 45-degree approximation given by equation (19b) actually is fairly accurate in practice up to 60 degrees. As described in Section D.4, the basic 45-degree equation (19b) also can be adapted to obtain extrapolation schemes for imaging steeper dips up to 90 degrees. Nevertheless, a penalty is paid for steep-dip accuracy in terms of dispersive noise incurred by implicit schemes (finite-difference migration in practice).\n\nSteep-dip finite-difference algorithms may be more conveniently implemented in the frequency-space domain than in the time-space domain. A general framework for implementing such algorithms involve a loop over the depth step z, and a loop over the frequency ω (Figure 4.1-23). For each depth step:\n\n1. Apply the shift term (equation 15a).\n2. Apply the diffraction term (equation 19) by performing implicit extrapolation of each of the frequency components of the wavefield.\n3. Sum over the frequencies to invoke the imaging principle which is equivalent to setting t = 0.\n4. Repeat the computation for all the depth steps to complete the imaging.\n\n## Equations\n\n $k_{z}={\\frac {2\\omega }{v}}{\\sqrt {1-\\left({\\frac {vk_{x}}{2\\omega }}\\right)^{2}}},$",
null,
"(13b)\n\n $k_{z}={\\frac {2\\omega }{v}}\\left[1-{\\frac {1}{2}}\\left({\\frac {vk_{x}}{2\\omega }}\\right)^{2}\\right].$",
null,
"(14a)\n\n $Q=P\\exp(-i\\omega \\tau ),$",
null,
"(15a)\n\n $\\tau =\\int _{0}^{z}{\\frac {dz}{{\\bar {v}}(z)}},$",
null,
"(15b)\n\n ${\\frac {\\partial ^{2}Q}{\\partial z\\partial t}}={\\frac {v}{4}}{\\frac {\\partial ^{2}Q}{\\partial x^{2}}},$",
null,
"(16a)\n\n ${\\frac {\\partial Q}{\\partial z}}=2\\left[{\\frac {1}{{\\bar {v}}(z)}}-{\\frac {1}{v(x,z)}}\\right]{\\frac {\\partial Q}{\\partial t}},$",
null,
"(16b)\n\n ${\\frac {\\partial ^{2}Q}{\\partial \\tau \\partial t}}={\\frac {v^{2}}{8}}{\\frac {\\partial ^{2}Q}{\\partial x^{2}}}.$",
null,
"(17)"
] | [
null,
"https://wiki.seg.org/images/7/72/Seismic-data-analysis.jpg",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d281d12eef7cfb0630f54be2fa4e6d3916696607",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3af57475f5e14a416ecd9c9cbda859838ef74c36",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/017383eae85a15fab01e3b92ccc40053030ac78f",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/f049dc04f5dad70f5d38cb4aa3eee740a95ae966",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/906b441a58b8de1e4219972502a290ea2acee885",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a8ac26a0bc62064726c7081080fdb91df13793e2",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/96648c4b7dc36e5800360568335304e621998725",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/e31c5c4afe42c00495c95c13c7076be2ad0c4103",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/543325ff6d926315c650986d9ce718262ca749d7",
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"https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6eb86d95d2b156672943d581ffb2eb9bf970276c",
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https://hackage.haskell.org/package/easyplot-1.0/docs/Graphics-EasyPlot.html | [
"easyplot-1.0: A tiny plotting library, utilizes gnuplot for plotting.\n\nGraphics.EasyPlot\n\nDescription\n\nA simple wrapper to the gnuplot command line utility.\n\nTypically you will invoke a plot like so:\n\n``` plot X11 \\$ Data2D [Title \"Sample Data\"] [] [(1, 2), (2, 4), ...]\n```\n\nTo plot a function, use the following:\n\n``` plot X11 \\$ Function2D [Title \"Sine and Cosine\"] [] (\\x -> sin x * cos x)\n```\n\nThere is also a shortcut available - the following plots the sine function:\n\n``` plot X11 sin\n```\n\nOutput can go into a file, too (See `TerminalType`):\n\n``` plot (PNG \"plot.png\") (sin . cos)\n```\n\nHaskell functions are plotted via a set of tuples obtained form the function. If you want to make use of gnuplots mighty function plotting functions you can pass a `Gnuplot2D` or `Gnuplot3D` object to plot.\n\n``` plot X11 \\$ Gnuplot2D [Color Blue] [] \"2**cos(x)\"\n```\n\nFor 3D-Plots there is a shortcut available by directly passing a String:\n\n``` plot X11 \"x*y\"\n```\n\nMultiple graphs can be shown simply by passing a list of these:\n\n``` plot X11 [ Data2D [Title \"Graph 1\", Color Red] [] [(x, x ** 3) | x <- [-4,-3.9..4]]\n, Function2D [Title \"Function 2\", Color Blue] [] (\\x -> negate \\$ x ** 2) ]\n```\n\nFor 3D Graphs it is useful to be able to interact with the graph (See `plot'` and `GnuplotOption`):\n\n``` plot' [Interactive] X11 \\$ Gnuplot3D [Color Magenta] [] \"x ** 2 + y ** 3\"\n```\n\nIf you want to know the command that SimplePlot uses to plot your graph, turn on debugging:\n\n``` plot' [Debug] X11 \\$ Gnuplot3D [Color Magenta] [] \"x ** 4 + y ** 3\"\n> set term x11 persist; splot x ** 4 + y ** 3 lc rgb \"magenta\"\n```\n\nSynopsis\n\n# Plotting\n\nclass Plot a whereSource\n\nProvides the plot function for different kinds of graphs (2D and 3D)\n\nMethods\n\nArguments\n\n :: TerminalType The terminal to be used for output. -> a The graph to plot. A `Graph2D` or `Graph3D` or a list of these. -> IO Bool Whether the plot was successfull or not.\n\nDo a plot to the terminal (i.e. a window will open and your plot can be seen)\n\nplot' :: [GnuplotOption] -> TerminalType -> a -> IO BoolSource\n\nInstances\n\n Plot String plot accepts a custom string which is then to be interpreted by gnu plot. The function will be interpreted as `Gnuplot3D`. (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot [x -> y -> z] A list of 3D functions can be plotted directly using `plot` (Fractional x, Enum x, Show x, Num y, Show y) => Plot [x -> y] A list of 2D functions can be plotted directly using `plot` Plot [String] plots mutliple 3D functions using gnuplots native function parser and renderer. The String will be interpreted as `Gnuplot3D`. (Fractional x, Enum x, Num x, Show x, Num y, Show y) => Plot [(x, y)] A list of tuples can be plotted directly using `plot` (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot [(x, y, z)] A list of triples can be plotted directly using `plot` (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot [Graph3D x y z] `plot` can be used to plot a list of `Graph3D` (Fractional x, Enum x, Show x, Num y, Show y) => Plot [Graph2D x y] `plot` can be used to plot a list of `Graph2D`. (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot (x -> y -> z) A 3D function can be plotted directly using `plot` (Fractional x, Enum x, Show x, Num y, Show y) => Plot (x -> y) A 2D function can be plotted directly using `plot` (Fractional x, Enum x, Show x, Num y, Show y) => Plot (Graph2D x y) `plot` can be used to plot a single `Graph2D`. (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot (Graph3D x y z) `plot` can be used to plot a single `Graph3D`.\n\n# Graphs for 2D and 3D plots\n\ndata Graph2D x y Source\n\nA two dimensional set of data to plot.\n\nConstructors\n\n Function2D [Option] [Option2D x y] (x -> y) plots a Haskell function `x -> y` Data2D [Option] [Option2D x y] [(x, y)] plots a set of tuples. Gnuplot2D [Option] [Option2D x y] String plots a custom function passed to Gnuplot (like `x**2 + 10`)\n\nInstances\n\n (Fractional x, Enum x, Show x, Num y, Show y) => Plot [Graph2D x y] `plot` can be used to plot a list of `Graph2D`. (Fractional x, Enum x, Show x, Num y, Show y) => Plot (Graph2D x y) `plot` can be used to plot a single `Graph2D`.\n\ndata Graph3D x y z Source\n\nA three dimensional set of data to plot.\n\nConstructors\n\n Function3D [Option] [Option3D x y z] (x -> y -> z) plots a Haskell function `x -> y -> z` Data3D [Option] [Option3D x y z] [(x, y, z)] plots a set of triples. Gnuplot3D [Option] [Option3D x y z] String plots a custom function passed to Gnuplot (like `x*y`)\n\nInstances\n\n (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot [Graph3D x y z] `plot` can be used to plot a list of `Graph3D` (Fractional x, Enum x, Show x, Fractional y, Enum y, Show y, Num z, Show z) => Plot (Graph3D x y z) `plot` can be used to plot a single `Graph3D`.\n\n# Configuration and other options\n\nTerminalType determines where the output of gnuplot should go.\n\nConstructors\n\n Aqua Output on Mac OS X (Aqua Terminal). Windows Output for MS Windows. X11 Output to the X Window System. PS FilePath Output into a Postscript file. EPS FilePath Output into an EPS file. PNG FilePath Output as Portable Network Graphic into file. PDF FilePath Output as Portable Document Format into a file. SVG FilePath Output as Scalable Vector Graphic into a file. GIF FilePath Output as Graphics Interchange Format into a file. JPEG FilePath Output into a JPEG file. Latex FilePath Output as LaTeX.\n\ndata Color Source\n\nThe Color of a graph.\n\nConstructors\n\n Red Blue Green Yellow Orange Magenta Cyan DarkRed DarkBlue DarkGreen DarkYellow DarkOrange DarkMagenta DarkCyan LightRed LightBlue LightGreen LightMagenta Violet White Brown Grey DarkGrey Black RGB Int Int Int a custom color\n\ndata Style Source\n\nThe Style of a graph.\n\nConstructors\n\n Lines points in the plot are interconnected by lines. Points data points are little cross symbols. Dots data points are real dots (approx the size of a pixel). Impulses Linespoints\n\ndata Option Source\n\nOptions on how to render a graph.\n\nConstructors\n\n Style Style The style for a graph. Title String The title for a graph in a plot (or a filename like `plot1.dat`). Color Color The line-color for the graph (or if it consist of `Dots` or `Points` the color of these)\n\ndata Option2D x y Source\n\nOptions which are exclusively available for 2D plots.\n\nConstructors\n\n Range x x Plots the function for the specified x range For [x] Plots the function only for the given x values Step x Uses the given step-size for plotting along the x-axis\n\ndata Option3D x y z Source\n\nOptions which are exclusively available for 3D plots.\n\nConstructors\n\n RangeX x x Plots the function for the specified x range RangeY y y Plots the function for the specified y range ForX [x] Plots the function only for the given x values ForY [y] Plots the function only for the given y values StepX x Uses the given step-size for plotting along the x-axis StepY y Uses the given step-size for plotting along the y-axis\n\nOptions which can be used with `plot'`\n\nConstructors\n\n Interactive keeps gnuplot open, so that you can interact with the plot (only usefull with `X11`) Debug prints the command used for running gnuplot.\n\nInstances\n\n Eq GnuplotOption"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.66766804,"math_prob":0.97411394,"size":5640,"snap":"2022-27-2022-33","text_gpt3_token_len":1690,"char_repetition_ratio":0.15418737,"word_repetition_ratio":0.200754,"special_character_ratio":0.29166666,"punctuation_ratio":0.11985689,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99918705,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T03:52:03Z\",\"WARC-Record-ID\":\"<urn:uuid:81f930c1-57f2-4448-8b73-c1bcb6a8cc07>\",\"Content-Length\":\"39910\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9114d2b2-e230-43e9-b7b1-9aed001b4ad3>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e6c1b4f-f622-4ca6-b93f-15b6e6f6c742>\",\"WARC-IP-Address\":\"146.75.36.68\",\"WARC-Target-URI\":\"https://hackage.haskell.org/package/easyplot-1.0/docs/Graphics-EasyPlot.html\",\"WARC-Payload-Digest\":\"sha1:T4FFYEEPGKOHK45CXCOKJ6L2ITPZH3PI\",\"WARC-Block-Digest\":\"sha1:B2U527FXKLRREJWY72O7VHH2KIKQXUMX\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572833.95_warc_CC-MAIN-20220817032054-20220817062054-00080.warc.gz\"}"} |
https://www.opencascade.com/doc/occt-7.4.0/refman/html/class_geom2d_gcc.html | [
"Geom2dGcc Class Reference\n\nThe Geom2dGcc package describes qualified 2D curves used in the construction of constrained geometric objects by an algorithm provided by the Geom2dGcc package. A qualified 2D curve is a curve with a qualifier which specifies whether the solution of a construction algorithm using the qualified curve (as an argument): More...\n\n`#include <Geom2dGcc.hxx>`\n\n## Static Public Member Functions\n\nstatic Geom2dGcc_QualifiedCurve Unqualified (const Geom2dAdaptor_Curve &Obj)\nConstructs such a qualified curve that the relative position of the solution computed by a construction algorithm using the qualified curve to the circle or line is not qualified, i.e. all solutions apply. Warning Obj is an adapted curve, i.e. an object which is an interface between: More...\n\nstatic Geom2dGcc_QualifiedCurve Enclosing (const Geom2dAdaptor_Curve &Obj)\nConstructs such a qualified curve that the solution computed by a construction algorithm using the qualified curve encloses the curve. Warning Obj is an adapted curve, i.e. an object which is an interface between: More...\n\nstatic Geom2dGcc_QualifiedCurve Enclosed (const Geom2dAdaptor_Curve &Obj)\nConstructs such a qualified curve that the solution computed by a construction algorithm using the qualified curve is enclosed by the curve. Warning Obj is an adapted curve, i.e. an object which is an interface between: More...\n\nstatic Geom2dGcc_QualifiedCurve Outside (const Geom2dAdaptor_Curve &Obj)\nConstructs such a qualified curve that the solution computed by a construction algorithm using the qualified curve and the curve are external to one another. Warning Obj is an adapted curve, i.e. an object which is an interface between: More...\n\n## Detailed Description\n\nThe Geom2dGcc package describes qualified 2D curves used in the construction of constrained geometric objects by an algorithm provided by the Geom2dGcc package. A qualified 2D curve is a curve with a qualifier which specifies whether the solution of a construction algorithm using the qualified curve (as an argument):\n\n• encloses the curve, or\n• is enclosed by the curve, or\n• is built so that both the curve and this solution are external to one another, or\n• is undefined (all solutions apply). These package methods provide simpler functions to construct a qualified curve. Note: the interior of a curve is defined as the left-hand side of the curve in relation to its orientation.\n\n## ◆ Enclosed()\n\n static Geom2dGcc_QualifiedCurve Geom2dGcc::Enclosed ( const Geom2dAdaptor_Curve & Obj )\nstatic\n\nConstructs such a qualified curve that the solution computed by a construction algorithm using the qualified curve is enclosed by the curve. Warning Obj is an adapted curve, i.e. an object which is an interface between:\n\n• the services provided by a 2D curve from the package Geom2d,\n• and those required on the curve by a computation algorithm. The adapted curve is created in the following way: Handle(Geom2d_Curve) mycurve = ... ; Geom2dAdaptor_Curve Obj ( mycurve ) ; The qualified curve is then constructed with this object: Geom2dGcc_QualifiedCurve myQCurve = Geom2dGcc::Enclosed(Obj);\n\n## ◆ Enclosing()\n\n static Geom2dGcc_QualifiedCurve Geom2dGcc::Enclosing ( const Geom2dAdaptor_Curve & Obj )\nstatic\n\nConstructs such a qualified curve that the solution computed by a construction algorithm using the qualified curve encloses the curve. Warning Obj is an adapted curve, i.e. an object which is an interface between:\n\n• the services provided by a 2D curve from the package Geom2d,\n• and those required on the curve by a computation algorithm. The adapted curve is created in the following way: Handle(Geom2d_Curve) mycurve = ... ; Geom2dAdaptor_Curve Obj ( mycurve ) ; The qualified curve is then constructed with this object: Geom2dGcc_QualifiedCurve myQCurve = Geom2dGcc::Enclosing(Obj);\n\n## ◆ Outside()\n\n static Geom2dGcc_QualifiedCurve Geom2dGcc::Outside ( const Geom2dAdaptor_Curve & Obj )\nstatic\n\nConstructs such a qualified curve that the solution computed by a construction algorithm using the qualified curve and the curve are external to one another. Warning Obj is an adapted curve, i.e. an object which is an interface between:\n\n• the services provided by a 2D curve from the package Geom2d,\n• and those required on the curve by a computation algorithm. The adapted curve is created in the following way: Handle(Geom2d_Curve) mycurve = ... ; Geom2dAdaptor_Curve Obj ( mycurve ) ; The qualified curve is then constructed with this object: Geom2dGcc_QualifiedCurve myQCurve = Geom2dGcc::Outside(Obj);\n\n## ◆ Unqualified()\n\n static Geom2dGcc_QualifiedCurve Geom2dGcc::Unqualified ( const Geom2dAdaptor_Curve & Obj )\nstatic\n\nConstructs such a qualified curve that the relative position of the solution computed by a construction algorithm using the qualified curve to the circle or line is not qualified, i.e. all solutions apply. Warning Obj is an adapted curve, i.e. an object which is an interface between:\n\n• the services provided by a 2D curve from the package Geom2d,\n• and those required on the curve by a computation algorithm. The adapted curve is created in the following way: Handle(Geom2d_Curve) mycurve = ... ; Geom2dAdaptor_Curve Obj ( mycurve ) ; The qualified curve is then constructed with this object: Geom2dGcc_QualifiedCurve myQCurve = Geom2dGcc::Unqualified(Obj);\n\nThe documentation for this class was generated from the following file:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9204407,"math_prob":0.9218717,"size":4925,"snap":"2020-10-2020-16","text_gpt3_token_len":1167,"char_repetition_ratio":0.20727494,"word_repetition_ratio":0.8192612,"special_character_ratio":0.2115736,"punctuation_ratio":0.14319248,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9565486,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-10T09:53:26Z\",\"WARC-Record-ID\":\"<urn:uuid:372ace5f-00fb-4562-87f2-61c74aa68c90>\",\"Content-Length\":\"17790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d21120c6-5c41-4df8-8a20-b89414c45769>\",\"WARC-Concurrent-To\":\"<urn:uuid:e24b7174-9f95-4f61-877c-3e6a58e267c7>\",\"WARC-IP-Address\":\"188.165.114.136\",\"WARC-Target-URI\":\"https://www.opencascade.com/doc/occt-7.4.0/refman/html/class_geom2d_gcc.html\",\"WARC-Payload-Digest\":\"sha1:XV6Y5ZZDJCDIQPC3ZS75C7UPZ6AZHIMW\",\"WARC-Block-Digest\":\"sha1:FFZGCACGHIOB6R4UMRGQPPC34CAI4SCV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371893683.94_warc_CC-MAIN-20200410075105-20200410105605-00424.warc.gz\"}"} |
http://answers.gkplanet.in/2018/01/find-value-of-tan-15-degrees.html | [
"# Find the value of tan 15 degrees\n\n## Exact Value of tan 15° | How to calculate tan15-degree value\n\nQ. Find the value of tan 15 degrees\n\nLet tan(15°) = tan(45°-30°)\n\nWe know that tan(A - B) = (tanA - tanB) /(1 + tan A tan B)\n\n⇒tan(45°-30°) = (tan45°- tan30°)/(1+tan45°tan30°)\n\n= {1- (1/√3)} / {1+(1/√3)}\n\n∴ tan15° = (√3 - 1) / (√3 + 1)\n\n=23\n\nAlternate\n\nSo, or\nSo, is the value of tan 15 degree.\n\nAfter simplification, the value of tan 15 comes out to be 0.26794919243",
null,
"Thanks for reading Find the value of tan 15 degrees\n\n←Previous\n« Prev Post\nNext→\nNext Post »\n\n#### 1 comment:\n\n1.",
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"Yes create method"
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"http://1.bp.blogspot.com/-39HRU19h4kk/VlrZJ1KEjiI/AAAAAAAAMxI/RYXIg2OkzQA/s1600/blog%2Bblogger%2Bblogging.jpg",
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"http://www.blogger.com/img/blogger_logo_round_35.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5460791,"math_prob":0.9986443,"size":407,"snap":"2023-40-2023-50","text_gpt3_token_len":175,"char_repetition_ratio":0.13399504,"word_repetition_ratio":0.02739726,"special_character_ratio":0.47665846,"punctuation_ratio":0.072289154,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99299073,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T08:32:32Z\",\"WARC-Record-ID\":\"<urn:uuid:40050074-17ab-4417-9942-f41e753649fc>\",\"Content-Length\":\"349514\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:792c8a93-af24-4d38-a5b3-730869d22aab>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5c92bde-9f18-4da1-86f4-f04014da382d>\",\"WARC-IP-Address\":\"172.253.122.121\",\"WARC-Target-URI\":\"http://answers.gkplanet.in/2018/01/find-value-of-tan-15-degrees.html\",\"WARC-Payload-Digest\":\"sha1:EVSJWT3MPIHSK7IPFGI5QAZ45XQGFCUI\",\"WARC-Block-Digest\":\"sha1:2LNWWMF6RO7IVXUMYHSR4G2OMRUAREZM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100057.69_warc_CC-MAIN-20231129073519-20231129103519-00407.warc.gz\"}"} |
http://accord-framework.net/docs/html/T_Accord_Statistics_Distributions_Univariate_EmpiricalDistribution.htm | [
" EmpiricalDistribution Class",
null,
"",
null,
"",
null,
"# EmpiricalDistribution Class\n\nEmpirical distribution.",
null,
"Inheritance Hierarchy\nSystemObject\nAccord.Statistics.DistributionsDistributionBase\nAccord.Statistics.Distributions.UnivariateUnivariateContinuousDistribution\nAccord.Statistics.Distributions.UnivariateEmpiricalDistribution\n\nNamespace: Accord.Statistics.Distributions.Univariate\nAssembly: Accord.Statistics (in Accord.Statistics.dll) Version: 3.8.0",
null,
"Syntax\n```[SerializableAttribute]\npublic class EmpiricalDistribution : UnivariateContinuousDistribution,\nIFittableDistribution<double, EmpiricalOptions>, IFittable<double, EmpiricalOptions>,\nIFittable<double>, IFittableDistribution<double>, IDistribution<double>,\nIDistribution, ICloneable, ISampleableDistribution<double>, IRandomNumberGenerator<double>```\n\nThe EmpiricalDistribution type exposes the following members.",
null,
"Constructors\nNameDescription",
null,
"EmpiricalDistribution(Double)\nCreates a new Empirical Distribution from the data samples.",
null,
"EmpiricalDistribution(Double, Double)\nCreates a new Empirical Distribution from the data samples.",
null,
"EmpiricalDistribution(Double, Double)\nCreates a new Empirical Distribution from the data samples.",
null,
"EmpiricalDistribution(Double, Int32)\nCreates a new Empirical Distribution from the data samples.",
null,
"EmpiricalDistribution(Double, Double, Double)\nCreates a new Empirical Distribution from the data samples.",
null,
"EmpiricalDistribution(Double, Int32, Double)\nCreates a new Empirical Distribution from the data samples.\nTop",
null,
"Properties\nNameDescription",
null,
"Counts\nGets the repetition counts associated with each sample. Note that changing values on this array will not result int any effect in this distribution. The distribution must be computed from scratch with new values in case new weights needs to be used.",
null,
"Entropy\nGets the entropy for this distribution.\n(Overrides UnivariateContinuousDistributionEntropy.)",
null,
"Length\nGets the total number of samples in this distribution.",
null,
"",
null,
"Mean\nGets the mean for this distribution.\n(Overrides UnivariateContinuousDistributionMean.)",
null,
"Median\nGets the median for this distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Mode\nGets the mode for this distribution.\n(Overrides UnivariateContinuousDistributionMode.)",
null,
"Quartiles\nGets the Quartiles for this distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Samples\nGets the samples giving this empirical distribution.",
null,
"Smoothing\nGets the bandwidth smoothing parameter used in the kernel density estimation.",
null,
"StandardDeviation\nGets the Standard Deviation (the square root of the variance) for the current distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Support\nGets the support interval for this distribution.\n(Overrides UnivariateContinuousDistributionSupport.)",
null,
"",
null,
"Variance\nGets the variance for this distribution.\n(Overrides UnivariateContinuousDistributionVariance.)",
null,
"Weights\nGets the fractional weights associated with each sample. Note that changing values on this array will not result int any effect in this distribution. The distribution must be computed from scratch with new values in case new weights needs to be used.\nTop",
null,
"Methods\nNameDescription",
null,
"Clone\nCreates a new object that is a copy of the current instance.\n(Overrides DistributionBaseClone.)",
null,
"ComplementaryDistributionFunction\nGets the complementary cumulative distribution function (ccdf) for this distribution evaluated at point x. This function is also known as the Survival function.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"CumulativeHazardFunction\nGets the cumulative hazard function for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"DistributionFunction(Double)\nGets the cumulative distribution function (cdf) for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"DistributionFunction(Double, Double)\nGets the cumulative distribution function (cdf) for this distribution in the semi-closed interval (a; b] given as P(a < X ≤ b).\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Equals\nDetermines whether the specified object is equal to the current object.\n(Inherited from Object.)",
null,
"Finalize\nAllows an object to try to free resources and perform other cleanup operations before it is reclaimed by garbage collection.\n(Inherited from Object.)",
null,
"",
null,
"Fit(Double)\nFits the underlying distribution to a given set of observations.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"",
null,
"Fit(Double, IFittingOptions)\nFits the underlying distribution to a given set of observations.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"",
null,
"Fit(Double, Double)\nFits the underlying distribution to a given set of observations.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"",
null,
"Fit(Double, Int32)\nFits the underlying distribution to a given set of observations.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Fit(Double, Double, EmpiricalOptions)\nFits the underlying distribution to a given set of observations.",
null,
"Fit(Double, Double, IFittingOptions)\nFits the underlying distribution to a given set of observations.\n(Overrides UnivariateContinuousDistributionFit(Double, Double, IFittingOptions).)",
null,
"Fit(Double, Int32, EmpiricalOptions)\nFits the underlying distribution to a given set of observations.",
null,
"",
null,
"Fit(Double, Int32, IFittingOptions)\nFits the underlying distribution to a given set of observations.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Generate\nGenerates a random observation from the current distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Generate(Random)\nGenerates a random observation from the current distribution.\n(Overrides UnivariateContinuousDistributionGenerate(Random).)",
null,
"Generate(Int32)\nGenerates a random vector of observations from the current distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Generate(Int32, Double)\nGenerates a random vector of observations from the current distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Generate(Int32, Random)\nGenerates a random vector of observations from the current distribution.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"Generate(Int32, Double, Random)\nGenerates a random vector of observations from the current distribution.\n(Overrides UnivariateContinuousDistributionGenerate(Int32, Double, Random).)",
null,
"GetHashCode\nServes as the default hash function.\n(Inherited from Object.)",
null,
"GetRange\nGets the distribution range within a given percentile.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"GetType\nGets the Type of the current instance.\n(Inherited from Object.)",
null,
"HazardFunction\nGets the hazard function, also known as the failure rate or the conditional failure density function for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"InnerComplementaryDistributionFunction\nGets the complementary cumulative distribution function (ccdf) for this distribution evaluated at point x. This function is also known as the Survival function.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"",
null,
"InnerDistributionFunction\nGets the cumulative distribution function (cdf) for this distribution evaluated at point x.\n(Overrides UnivariateContinuousDistributionInnerDistributionFunction(Double).)",
null,
"InnerInverseDistributionFunction\nGets the inverse of the cumulative distribution function (icdf) for this distribution evaluated at probability p. This function is also known as the Quantile function.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"InnerLogProbabilityDensityFunction\nGets the log-probability density function (pdf) for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"",
null,
"InnerProbabilityDensityFunction\nGets the probability density function (pdf) for this distribution evaluated at point x.\n(Overrides UnivariateContinuousDistributionInnerProbabilityDensityFunction(Double).)",
null,
"InverseDistributionFunction\nGets the inverse of the cumulative distribution function (icdf) for this distribution evaluated at probability p. This function is also known as the Quantile function.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"LogCumulativeHazardFunction\nGets the log of the cumulative hazard function for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"LogProbabilityDensityFunction\nGets the log-probability density function (pdf) for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"MemberwiseClone\nCreates a shallow copy of the current Object.\n(Inherited from Object.)",
null,
"ProbabilityDensityFunction\nGets the probability density function (pdf) for this distribution evaluated at point x.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"QuantileDensityFunction\nGets the first derivative of the inverse distribution function (icdf) for this distribution evaluated at probability p.\n(Inherited from UnivariateContinuousDistribution.)",
null,
"",
null,
"SmoothingRule(Double)\nGets the default estimative of the smoothing parameter.",
null,
"",
null,
"SmoothingRule(Double, Double)\nGets the default estimative of the smoothing parameter.",
null,
"",
null,
"SmoothingRule(Double, Int32)\nGets the default estimative of the smoothing parameter.",
null,
"",
null,
"SmoothingRule(Double, Double, Int32)\nGets the default estimative of the smoothing parameter.",
null,
"ToString\nReturns a String that represents this instance.\n(Inherited from DistributionBase.)",
null,
"ToString(IFormatProvider)\nReturns a String that represents this instance.\n(Inherited from DistributionBase.)",
null,
"ToString(String)\nReturns a String that represents this instance.\n(Inherited from DistributionBase.)",
null,
"ToString(String, IFormatProvider)\nReturns a String that represents this instance.\n(Overrides DistributionBaseToString(String, IFormatProvider).)\nTop",
null,
"Extension Methods\nNameDescription",
null,
"HasMethod\nChecks whether an object implements a method with the given name.\n(Defined by ExtensionMethods.)",
null,
"IsEqual\nCompares two objects for equality, performing an elementwise comparison if the elements are vectors or matrices.\n(Defined by Matrix.)",
null,
"To(Type)Overloaded.\nConverts an object into another type, irrespective of whether the conversion can be done at compile time or not. This can be used to convert generic types to numeric types during runtime.\n(Defined by ExtensionMethods.)",
null,
"ToTOverloaded.\nConverts an object into another type, irrespective of whether the conversion can be done at compile time or not. This can be used to convert generic types to numeric types during runtime.\n(Defined by ExtensionMethods.)\nTop",
null,
"Remarks\n\nEmpirical distributions are based solely on the data. This class uses the empirical distribution function and the Gaussian kernel density estimation to provide an univariate continuous distribution implementation which depends only on sampled data.\n\nReferences:",
null,
"Examples\n\nThe following example shows how to build an empirical distribution directly from a sample:\n\n```// Consider the following univariate samples\ndouble[] samples = { 5, 5, 1, 4, 1, 2, 2, 3, 3, 3, 4, 3, 3, 3, 4, 3, 2, 3 };\n\n// Create a non-parametric, empirical distribution using those samples:\nEmpiricalDistribution distribution = new EmpiricalDistribution(samples);\n\n// Common measures\ndouble mean = distribution.Mean; // 3\ndouble median = distribution.Median; // 2.9999993064186787\ndouble var = distribution.Variance; // 1.2941176470588236\n\n// Cumulative distribution function\ndouble cdf = distribution.DistributionFunction(x: 4.2); // 0.88888888888888884\ndouble ccdf = distribution.ComplementaryDistributionFunction(x: 4.2); //0.11111111111111116\ndouble icdf = distribution.InverseDistributionFunction(p: cdf); // 4.1999999999999993\n\n// Probability density functions\ndouble pdf = distribution.ProbabilityDensityFunction(x: 4.2); // 0.15552784414141974\ndouble lpdf = distribution.LogProbabilityDensityFunction(x: 4.2); // -1.8609305013898356\n\n// Hazard (failure rate) functions\ndouble hf = distribution.HazardFunction(x: 4.2); // 1.3997505972727771\ndouble chf = distribution.CumulativeHazardFunction(x: 4.2); // 2.1972245773362191\n\n// Automatically estimated smooth parameter (gamma)\ndouble smoothing = distribution.Smoothing; // 1.9144923416414432\n\n// String representation\nstring str = distribution.ToString(CultureInfo.InvariantCulture); // Fn(x; S)```",
null,
"See Also"
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https://www.hotspotfinance.com/marginal-costing-formulas-introduction/ | [
"",
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"# Marginal Costing Formulas: An Introduction\n\nMarginal costing formulas simply show the change in total cost over the change in quantity produced:",
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"Normally, marginal cost refers to the amount of cost that will be incurred by an entity in case it produces one extra product. The focus lies in this one extra item.\n\nImagine, for example, a car manufacturer. Let’s say the company produces 100’000 cars every year and its total cost is \\$ 600’000’000. Marginal cost now calculates the extra cost that incurs by producing one more car.",
null,
"Many companies have been using the formula to reduce the cost of production that is being incurred when manufacturing products of different nature and to maximise profits.\n\n## Normal course of marginal costs\n\nIn order to better imagine the marginal costs think of it as a graph showing unit costs and numbers of units on its axis. Often you will see a nearly u-shaped curve. Costs start high with a low volume of items but go gradually down as production increases. At some point, however, costs increase again.",
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"Whilst economies of scale (fabric utilised to a higher level, discounts for raw material) reduce the marginal cost there will be a point new costs come in and thus the price of an item increases again. Such new costs could be overtime salary, new extra management staff or machinery.\n\n## Advantages of marginal costing formulas\n\nIn management accounting theory, there is a famous equation:\n\nmarginal cost = marginal revenue\n\nA profit maximising company will produce a product until its marginal costs are equal to the marginal revenue. If marginal revenue would be lower, selling a product would return a loss. Calculating such a point could be a task for a management accountant.",
null,
"In a more complex area, marginal costing can be used to determine the optimum product mix. Imagine a company producing and selling different products. Each of the products will have a separate marginal cost and also separate marginal revenue. It then could be a management accountant’s role to find the best product mix so the company sells many products where the spread between marginal costs and marginal revenue is as high as possible. When doing that we should also keep in mind that limitations of internal resources are likely. There a limiting factor analysis may be required.\n\n## Marginal costs of nearly zero\n\nNowadays, a special phenomenon can be observed. Some companies can increase their output at a cost of nearly zero. An easy to imagine example is an internet trading platform. If the infrastructure is once up and running, having one extra seller or one extra transaction on it does not bring any extra cost at all.\n\nIn this case, marginal costing formulas won’t help you. Instead, finding the most suiting market price will be in focus in order to maximise profits. You can also check the article about Proven cost accounting concepts for more details.\n\n## To sum up\n\nMarginal costing formulas are a valuable tool to determine the output which maximises a company’s profits. This, however, is only a valid for businesses with a certain proportion of variable costs."
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https://oeis.org/A306601 | [
"The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.",
null,
"Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A306601 Let b(1) = 3 and let b(n+1) be the least prime expressible as k*(b(n)-1)*b(n))-1; this sequence gives the values of k in order. 1\n 1, 1, 2, 4, 8, 16, 5, 360, 142, 104, 34, 1904, 3127, 253, 1219, 8755, 16222, 7672, 22515 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,3 COMMENTS The corresponding primes in order are 5, 19, 683, 1863223, P14, P29, P57, P117, P235, P472, P945, P1893, P3789, P7581, P15164, P30332, P60668, P121339, P242682. After each iteration the number of decimal digits is roughly twice that of the previous iteration. These primes can generally be easily certified using the N+1 method since all the prime factors for N+1 are known. LINKS FORMULA Nested f(k) = k*(p-1)*p-1 for p=3. After each iteration the last obtained f(k) is substituted for p. The primes can be certified using OpenPFGW by adding each previous iteration to the helper file. EXAMPLE For p = 3, the smallest k for which f(k) = k*(p-1)*p-1 is prime is 1: f(1) = k*(p-1)*p-1 = 1*(3-1)*3-1 = 5. This sets p = 5 for the next iteration for which the smallest k for which f(k) is prime is 1: f(1) = k*(p-1)*p-1 = 1*(5-1)*5-1 = 19. This sets p = 19 for the next iteration for which the smallest k for which f(k) is prime is 2: f(2) = k*(p-1)*p-1 = 2*(19-1)*19-1 = 683. This sets p = 683 for the next iteration for which the smallest k for which f(k) is prime is 4: f(4) = k*(p-1)*p-1 = 4*(683-1)*683-1 = 1863223. This sets p = 1863223 for the next iteration for which the smallest k for which f(k) is prime is 8: f(8) = k*(p-1)*p-1 = 8*(1863223-1)*1863223-1 = P14. PROG (PARI) p=3; k=1; while(1, runningP=k*(p-1)*p-1; if(ispseudoprime(runningP), print1(k, \", \"); k=1; p=runningP; , k=k+1)) The largest prime (P242682) can be generated by using the code: (PARI) k=[1, 1, 2, 4, 8, 16, 5, 360, 142, 104, 34, 1904, 3127, 253, 1219, 8755, 16222, 7672, 22515]; p=3; for(i=1, #k, p=k[i]*(p-1)*p-1); print(\"\\n\", p, \"\\n\") CROSSREFS Cf. A307334, A307498, A307499, A000058. Sequence in context: A178170 A127824 A088975 * A237851 A167425 A212638 Adjacent sequences: A306598 A306599 A306600 * A306602 A306603 A306604 KEYWORD nonn,hard,more AUTHOR Rashid Naimi, Apr 10 2019 EXTENSIONS Definition clarified by Charlie Neder, Jun 03 2019 a(17) from Rashid Naimi, Aug 23 2019 a(18) from Rashid Naimi, Oct 22 2019 a(19) from Rashid Naimi, Aug 01 2020 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified June 18 04:41 EDT 2021. Contains 345098 sequences. (Running on oeis4.)"
] | [
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"https://oeis.org/banner2021.jpg",
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https://www.northernarchitecture.us/masonry-structures/an-example-of-safety-evaluation-of-the-library.html | [
"## An Example Of Safety Evaluation Of The Library\n\n3.1 DDA analysis\n\nMost of studies dealing with the bearing capacity of masonry structures are based on stresses computed by continuum mechanics. This scheme may be taken granted for a bonded masonry, since it implies that generation of a crack in a block or at an interface limits the bearing capacity. On the other hand, dry-masonry has its initial cracked state, so that the application of this scheme is questionable.\n\nWe have been studying the applicability of the Discontinuous Deformation Analysis (DDA) (Shi 1993) to dry-masonry structures. DDA is formulated as incremental dynamic equilibrium for blocks based on the theory of the least total potential energy with iteration scheme for preventing intrusion and tensile stress at interfaces of blocks. DDA can guarantee existence and uniqueness of the solution and can be practical on usual PC platforms.\n\nWe are interested in the response of masonry structures to wind load exerted horizontally on the structures. Equilibrium under the gravitational load was first achieved, and then horizontal load was applied gradually to let sliding occur between blocks. Figure 8",
null,
"Figure 8. Response of series of blocks to horizontal loads simulated by DDA.",
null,
"Figure 9. Modeled portion of the northern library by DDA.\n\nFigure 8. Response of series of blocks to horizontal loads simulated by DDA.\n\nshows the responses of a series of square blocks lined in vertical, which is subjected to horizontal load either at upper portion (upper panel) or at all of the blocks (lower panel). The different collapse modes can be simulated by DDA, i.e. sliding at the interface right below the loaded portion for partial loading and uplift of whole blocks for whole loading.\n\n3.2 Response of the northern library to horizontal load by DDA\n\nThe north-south section of the northern library including columns and walls is simplified by plane strain model with consideration on symmetry shown in Figures 9-10. Friction angle is set to 30 degrees, mass density 2.52 g/cm3, Poisson's ratio 0.11; no cohesion and no dynamic friction are considered. Uniformly distributed horizontal load along the wall height is replaced by nodal forces exerted gradually at the rate of 7,937 N/m2/s after the settlement under gravitational load. Figure 11 shows development of deformation and variation of principal stress with increase of horizontal load. The principal axes for the initial gravitational load in vertical (a) are firstly inclined by combination of axial force and shear force due to friction at interfaces (b), and back to vertical again due to the occurrence of slide (c). The first sliding occurs at 20,117 N/m2 of uniform load, which can be viewed as a bearing capacity of the model.",
null,
"Figure 10. DDA model for the northern library.\n\n3.3 Bearing capacity of the library to the wind load The wind load density can be evaluated by (1).\n\nwhere D: horizontal force [N], S: area [m2], C: resistance factor, p: density [kg/m3], V: wind velocity [m/s]. For wind velocity of 40m/s, (1) gives about 10,000 N/m2, which is less than a half of the evaluated bearing capacity of the first slide at 20,117 N/m2. Then the safety margin for the wind load corresponding to wind velocity of 40 m/s is more than twice.",
null,
"(hi Gravitational Ir.id and !•:111 [ /(:[;I;iI lr;i;1 of 20,09}? N.ii just before the 11 r>: sliding occurs.",
null,
"(c) Gravitational i' .L.I and horizontal load of 20,400 N/m\" after the first sliding occurred.",
null,
"(d> Gravitational load and horizontal load of 20,720 N/m2 where the remarkable sliding at the wall base is. observed.\n\nFigure 11. Deformation and principal stress in the northern library DDA model for horizontal load.\n\n(d> Gravitational load and horizontal load of 20,720 N/m2 where the remarkable sliding at the wall base is. observed.\n\nFigure 11. Deformation and principal stress in the northern library DDA model for horizontal load.",
null,
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] | [
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"https://www.northernarchitecture.us/masonry-structures/images/3123_673_878.jpg",
null,
"https://www.northernarchitecture.us/masonry-structures/images/3123_673_879.jpg",
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"https://www.northernarchitecture.us/masonry-structures/images/3123_673_881.jpg",
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"https://www.northernarchitecture.us/masonry-structures/images/3123_673_882.jpg",
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"https://www.northernarchitecture.us/masonry-structures/images/3123_673_883.jpg",
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"https://www.northernarchitecture.us/images/downloads/eJw9ykEKgCAQAMDfeFSL0gikp8TmSi6lK2X4_ejSZU4Tay2zUo0OamHT2srEzNJzUsgtnwy4Jsiwh0vdIaMssSzgK3F2hXx9riD-SOhGYzoLwzRZRCui60ct2ucL74ckOw.jpg",
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https://ir.huntington.com/regulatory-sec-filings/all-sec-filings/xbrl_doc_only/740 | [
"### Annual report pursuant to Section 13 and 15(d)\n\n#### DERIVATIVE FINANCIAL INSTRUMENTS (Tables)\n\nv3.6.0.2\nDERIVATIVE FINANCIAL INSTRUMENTS (Tables)\n12 Months Ended\nDec. 31, 2016\nDerivative Instruments and Hedging Activities Disclosure [Abstract]\nGross notional values of derivatives used in asset and liability management activities\nThe following table presents the gross notional values of derivatives used in Huntington’s asset and liability management activities at December 31, 2016, identified by the underlying interest rate-sensitive instruments:\n (dollar amounts in thousands) Fair Value Hedges Cash Flow Hedges Total Instruments associated with: Loans \\$ — \\$ 3,325,000 \\$ 3,325,000 Deposits — — — Subordinated notes 950,000 — 950,000 Long-term debt 6,525,000 — 6,525,000 Total notional value at December 31, 2016 \\$ 7,475,000 \\$ 3,325,000 \\$ 10,800,000\nAdditional information about the interest rate swaps used in asset and liability management activities\nThe following table presents additional information about the interest rate swaps used in Huntington’s asset and liability management activities at December 31, 2016:\n\n Weighted-AverageRate (dollar amounts in thousands) Notional Value Average Maturity (years) Fair Value Receive Pay Asset conversion swaps Receive fixed—generic \\$ 3,325,000 0.6 \\$ (2,060 ) 1.04 % 0.91 % Liability conversion swaps Receive fixed—generic 7,475,000 3.1 (51,496 ) 1.49 0.88 Total swap portfolio at December 31, 2016 \\$ 10,800,000 2.3 \\$ (53,556 ) 1.35 % 0.89 %\nAsset and liability derivatives included in accrued income and other assets\nThe following table presents the fair values at December 31, 2016 and 2015 of Huntington’s derivatives that are designated and not designated as hedging instruments. Amounts in the table below are presented gross without the impact of any net collateral arrangements:\nAsset derivatives included in accrued income and other assets:\n (dollar amounts in thousands) December 31, 2016 December 31, 2015 Interest rate contracts designated as hedging instruments \\$ 46,440 \\$ 80,513 Interest rate contracts not designated as hedging instruments 213,587 190,846 Foreign exchange contracts not designated as hedging instruments 23,265 37,727 Commodity contracts not designated as hedging instruments 108,026 117,894 Equity contracts not designated as hedging instruments 9,775 — Total contracts \\$ 401,093 \\$ 426,980\nLiability derivatives included in accrued expenses and other liabilities:\n (dollar amounts in thousands) December 31, 2016 December 31, 2015 Interest rate contracts designated as hedging instruments \\$ 99,996 \\$ 15,215 Interest rate contracts not designated as hedging instruments 143,976 121,815 Foreign exchange contracts not designated as hedging instruments 19,576 35,283 Commodity contracts not designated as hedging instruments 104,328 114,887 Equity contracts not designated as hedging instruments — — Total contracts \\$ 367,876 \\$ 287,200\nIncrease or (decrease) to interest expense for derivatives designated as fair value hedges\nThe following table presents the change in fair value for derivatives designated as fair value hedges as well as the offsetting change in fair value on the hedged item:\n Year ended December 31, (dollar amounts in thousands) 2016 2015 2014 Interest rate contracts Change in fair value of interest rate swaps hedging deposits (1) \\$ (82 ) \\$ (996 ) \\$ (1,045 ) Change in fair value of hedged deposits (1) 72 992 1,025 Change in fair value of interest rate swaps hedging subordinated notes (2) (47,852 ) (8,237 ) 476 Change in fair value of hedged subordinated notes (2) 45,019 8,237 (476 ) Change in fair value of interest rate swaps hedging long-term debt (2) (74,481 ) 3,903 1,990 Change in fair value of hedged other long-term debt (2) 67,389 (3,602 ) 828\n (1) Effective portion of the hedging relationship is recognized in Interest expense—deposits in the Consolidated Statements of Income. Any resulting ineffective portion of the hedging relationship is recognized in noninterest income in the Consolidated Statements of Income.\n (2) Effective portion of the hedging relationship is recognized in Interest expense—subordinated notes and other long-term debt in the Consolidated Statements of Income. Any resulting ineffective portion of the hedging relationship is recognized in noninterest income in the Consolidated Statements of Income.\nGains and (losses) recognized in other comprehensive income (loss) (OCI) for derivatives designated as effective cash flow hedges\nThe following table presents the gains and (losses) recognized in OCI and the location in the Consolidated Statements of Income of gains and (losses) reclassified from OCI into earnings for derivatives designated as effective cash flow hedges:\n Derivatives in cash flow hedging relationships Amount of gain or (loss) recognized in OCI on derivatives (effective portion) Location of gain or (loss) reclassified from accumulated OCI into earnings (effective portion) Amount of (gain) or loss reclassified from accumulated OCI into earnings (effective portion) (pre-tax) (dollar amounts in thousands) 2016 2015 2014 2016 2015 2014 Interest rate contracts Loans \\$ 1,548 \\$ 8,428 \\$ 9,192 Interest and fee income—loans and leases \\$ (361 ) \\$ (210 ) \\$ (4,064 ) Investment securities — — — Noninterest income - other income 1 (10 ) 93 Total \\$ 1,548 \\$ 8,428 \\$ 9,192 \\$ (360 ) \\$ (220 ) \\$ (3,971 )\nGains and (losses) recognized in noninterest income on the ineffective portion on interest rate contracts for derivatives designated as fair value and cash flow hedges\nThe following table presents the gains and (losses) recognized in noninterest income for the ineffective portion of interest rate contracts for derivatives designated as cash flow hedges for the years ending December 31, 2016, 2015, and 2014:\n December 31, (dollar amounts in thousands) 2016 2015 2014 Derivatives in cash flow hedging relationships Interest rate contracts: Loans \\$ (317 ) \\$ (763 ) \\$ 74\nOffsetting of financial assets and derivatives assets\nThe following tables present the gross amounts of these assets and liabilities with any offsets to arrive at the net amounts recognized in the Consolidated Balance Sheets at December 31, 2016 and December 31, 2015:\n\n Offsetting of Financial Assets and Derivative Assets Gross amounts not offset in the consolidated balance sheets (dollar amounts in thousands) Gross amounts of recognized assets Gross amounts offset in the consolidated balance sheets Net amounts of assets presented in the consolidated balance sheets Financial instruments Cash collateral received Net amount Offsetting of Financial Assets and Derivative Assets December 31, 2016 Derivatives \\$ 420,159 \\$ (181,940 ) \\$ 238,219 \\$ (34,328 ) \\$ (5,428 ) \\$ 198,463 December 31, 2015 Derivatives 436,169 (161,297 ) 274,872 (39,305 ) (3,462 ) 232,105\nOffsetting of financial liabilities and derivative liabilities\n Offsetting of Financial Liabilities and Derivative Liabilities Gross amounts not offset in the consolidated balance sheets (dollar amounts in thousands) Gross amounts of recognized liabilities Gross amounts offset in the consolidated balance sheets Net amounts of assets presented in the consolidated balance sheets Financial instruments Cash collateral delivered Net amount Offsetting of Financial Liabilities and Derivative Liabilities December 31, 2016 Derivatives \\$ 370,647 \\$ (272,361 ) \\$ 98,286 \\$ (7,550 ) \\$ (23,943 ) \\$ 66,793 December 31, 2015 Derivatives 288,659 (144,309 ) 144,350 (62,460 ) (20 ) 81,870\nDerivative assets and liabilities used in mortgage banking activities\nThe following table summarizes the derivative assets and liabilities used in mortgage banking activities:\n (dollar amounts in thousands) December 31, 2016 December 31, 2015 Derivative assets: Interest rate lock agreements \\$ 5,747 \\$ 6,721 Forward trades and options 13,319 2,468 Total derivative assets 19,066 9,189 Derivative liabilities: Interest rate lock agreements (1,598 ) (220 ) Forward trades and options (1,173 ) (1,239 ) Total derivative liabilities (2,771 ) (1,459 ) Net derivative asset \\$ 16,295 \\$ 7,730"
] | [
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https://www.jiskha.com/questions/300721/1-how-many-nodes-counting-nodes-at-the-ends-are-present-in-the-displacement | [
"# Physics\n\n1. How many nodes (counting nodes at the ends), are present in the displacement oscillations of an open-closed tube (oboe) that vibrates in its fourth harmonic?\na) 3 nodes\nb) 4 nodes\nc) 5 nodes\nd) 6 nodes\ne) An open-closed tube has no 4th harmonic\n\n2. At a distance of 30 m the noise from the engine of an jet has an intensity of 130 dB. At this level, you will be in pain and your ears will hurt. That's why this intensity is know as the \"pain threshold\". How far do you have to move from the jet in order for the noise to drop down in intensity to 57.9 dB, a level comparable to that of a spoken conversation?\n\n3. An out-of-tune violin plays an \"A\" note of 446 Hz. At the same time, a tuning fork vibrates with a true-A of 440 Hz. The frequency of the beats produced (intensity fluctuations) is ...\n***Hint: The frequency of the beats is not the same as the frequency of the envelope.***\n\na) No beats are produced in this situation\nb) 1.5 beats per second\nc) 3 beats per second\nd) 6 beats per second\ne) 12 beats per second\n\n1. 👍\n2. 👎\n3. 👁\n1. get a life and do bayer's online homework yourself!\n\n1. 👍\n2. 👎\n2. 1) e!\n3) d (6 beats per s) formula: f_beat = |f1-f2|\n\nat above person..lol! what were you doing here yourself..if not searching for help to do these questions?\n\n1. 👍\n2. 👎\n\n## Similar Questions\n\n1. ### Chemistry\n\nHow would the dx2−y2 orbital in the n=5 shell compare to the dx2−y2 orbital in the n=3 subshell? True or False 1.The contour of the orbital would extend further out along the x and y axes. 2. The value of ℓ would increase by\n\n2. ### chemistry\n\nHow would the 2s and 3p orbitals differ from the 1s and 2p orbitals? Check all that apply. possible answers The and orbitals would have more nodes than and orbitals. The orbital would be the same shape as the orbital but would be\n\n3. ### Physics\n\nHi, can anybody provide some hints in how to approach these questions? I've been stuck on it for days!!! Thank you in advance!! Nodes of a Standing Wave: Consider a standing wave, where y represents the transverse displacement of\n\n4. ### Physics\n\nWhich wave of the following wavelengths could not make a standing wave in a 1.0 m string, with nodes at both ends? A. 2.0 m B. 1.0 m C. 0.75 m D. 0.50 m E. 0.25 m i think the answer is D\n\n1. ### Physics\n\nAssume the length of an organ pipe is L=24cm. You have excited the wave in the tube, and you observe that it has displacement nodes at 4cm, 12cm, and 20cm measured from one end of the pipe. In which harmonic, n, is the air in the\n\n2. ### physics\n\nStanding waves are set up in a string by a source vibrating at 100.0 Hz. Seven nodes are counted in a distance of 63.0 cm (including one node at each of the ends). How many wavelengths must there be in the string? What is the\n\n3. ### PHYSICS\n\nSally Sue, an enthusiastic physics student enjoyed the opportunity to collect data from standing waves in a spring. She and her partner help the ends of their spring 4.00 meters apart. There are 5 nodes in the standing wave\n\n4. ### Physics\n\nThe distance between two consecutive nodes of a standing wave is 19.2 cm. The hand generating the pulses moves up and down through a complete cycle 2.59 times every 7.44 s. Find the velocity of the wave. Answer in units of m/s.\n\n1. ### Physics\n\nA mass is suspended from a vertical spring and the system is allowed to come to rest. When the mass is now pulled down a distance of 76mm and released, the time taken for 25 oscillations is 23s. Find the displacement of the mass\n\n2. ### Chemistry\n\nThe sizes and shapes of the hydrogen atom orbitals was revealed through graphical analysis of the corresponding wave functions. These wave functions also predict areas that the electrons have a zero probability of being found.\n\n3. ### Chemistry\n\nWhat is the length of a string that has a standing wave with four nodes (including those at the ends) and lambda = 24 {\\rm cm}\n\n4. ### Physics\n\nA string is 2.4 m long, and the speed of sound along this string is 450 m/s. Calculate the frequency of the wave that would produce a first harmonic. Assume the string has nodes at both ends. What I've Tried Ln=2.4m v= 450 m/s f=?"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9359744,"math_prob":0.90745956,"size":2865,"snap":"2021-43-2021-49","text_gpt3_token_len":754,"char_repetition_ratio":0.14085984,"word_repetition_ratio":0.010989011,"special_character_ratio":0.25445026,"punctuation_ratio":0.106732346,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9753835,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T05:50:01Z\",\"WARC-Record-ID\":\"<urn:uuid:d329cfbb-9f67-4d2f-b735-336b16f08fa4>\",\"Content-Length\":\"21790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea34b37a-0a1a-46ca-b059-7d02d5cfd0c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c92f423-b52b-4ef2-98d1-b1c18bfe662e>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/300721/1-how-many-nodes-counting-nodes-at-the-ends-are-present-in-the-displacement\",\"WARC-Payload-Digest\":\"sha1:C4ZDPUWNL7JKJMMH63AJFUOTPH3NBH6T\",\"WARC-Block-Digest\":\"sha1:MYADZPX3ZRJLEUHVYSIACPVTBF54B75Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585121.30_warc_CC-MAIN-20211017052025-20211017082025-00375.warc.gz\"}"} |
https://parametricbydesign.com/grasshopper/how-tos/construction-planes/ | [
"# Oops...\n\nThis website is made for modern browsers. You are seeing this, because your browser is missing a feature or two. Please install the latest update or switch to a modern browser. See you soon.\n\n# Construction planes\n\n#planes\n\nEverything we draw in Rhino or Grasshopper is part of a global Cartesian coordinate system, also called the world coordinate system. Therefore, every geometric shape within this system possesses defined coordinates and we can use Euclidean transformations to change their position and orientation is space. But sometimes, it’s cumbersome to calculate the new coordinates. Also, when we draw a shape, its generation mechanism aligns it to the world coordinate system’s axes.\n\nTo ease both, the generation and transformation of geometric objects, there are construction planes, also called reference planes. Such a plane introduces a separate, local coordinate system and can be visioned as a drawing table set anywhere in space. For example, instead of creating a box in the world coordinate system and then rotate and move it to its destined position, we can also create a reference plane and then draw the box on this plane.",
null,
"Instead of creating a box and then positioning it (left), we can construct a reference plane and then create the box on it (right).\n\nA lot of components in Grasshopper let us set a reference plane for either the creation of objects, transformations, or other operations. In general, it’s a better idea to generate construction planes first and then create the objects on top of them. In an algorithm, it’s easy to generate a unique plane for every object.\n\n### Construct planes\n\nThe ribbon group Vector > Plane hosts various components to create planes. For example: XY Plane",
null,
"XY Plane (XY)\nVector > Plane > XY Plane\nWorld XY plane.\nInputs\nOrigin (O)Origin of plane\nOutputs\nPlane (P)World XY plane\ncreates a plane parallel to the world XY plane, but we can alter the origin of the coordinate system. With Construct Plane",
null,
"Construct Plane (Pl)\nVector > Plane > Construct Plane\nConstruct a plane from an origin point and {x}, {y} axes.\nInputs\nOrigin (O)Origin of plane\nX-Axis (X)X-Axis direction of plane\nY-Axis (Y)Y-Axis direction of plane\nOutputs\nPlane (Pl)Constructed plane\n, we can set the origin and also the direction of the axes. Plane Normal",
null,
"Plane Normal (Pl)\nVector > Plane > Plane Normal\nCreate a plane perpendicular to a vector.\nInputs\nOrigin (O)Origin of plane\nZ-Axis (Z)Z-Axis direction of plane\nOutputs\nPlane (P)Plane definition\ncreates a plane at the provided origin and perpendicular to the vector at Z.\n\nYou can find a couple more components to create planes with various input parameters. The components are self-explanatory.\n\n### Modify planes\n\nThe desired construction plane can also be obtained by altering an existing one. The components are hosted in the same ribbon group Vector > Plane. For example, we can modify the origin of an existing plane with Plane Origin",
null,
"Plane Origin (Pl Origin)\nVector > Plane > Plane Origin\nChange the origin point of a plane\nInputs\nBase (B)Base plane\nOrigin (O)New origin point of plane\nOutputs\nPlane (Pl)Plane definition\nor change the orientation of the axes with Rotate Plane",
null,
"Rotate Plane (PRot)\nVector > Plane > Rotate Plane\nPerform plane rotation around plane z-axis\nInputs\nPlane (P)Plane to rotate\nAngle (A)Rotation (counter clockwise) around plane z-axis in radians\nOutputs\nPlane (P)Rotated plane\n. Flip Plane",
null,
"Flip Plane (PFlip)\nVector > Plane > Flip Plane\nFlip or swap the axes of a plane\nInputs\nReverse X (X)Reverse the x-axis direction\nReverse Y (Y)Reverse the y-axis direction\nSwap axes (S)Swap the x and y axis directions\nOutputs\nPlane (P)Flipped plane\nswaps the axes to our needs.\n\nAgain, there are a couple more components that do all kinds of adjustments with provided planes.\n\n### Deconstruct planes\n\nThe component Deconstruct Plane",
null,
"Deconstruct Plane (DePlane)\nVector > Plane > Deconstruct Plane\nDeconstruct a plane into its component parts.\nInputs\nPlane (P)Plane to deconstruct\nOutputs\nOrigin (O)Origin point\nX-Axis (X)X-Axis vector\nY-Axis (Y)Y-Axis vector\nZ-Axis (Z)Z-Axis vector\nwill dismantle a plane into its origin and axes.\n\n### Construct planes with curves\n\nBesides vectors, we can also derive planes from curves. With given curve parameters t, there are three option: Curve Frame",
null,
"Curve Frame (Frame)\nCurve > Analysis > Curve Frame\nGet the curvature frame of a curve at a specified parameter.\nInputs\nCurve (C)Curve to evaluate\nParameter (t)Parameter on curve domain to evaluate\nOutputs\nFrame (F)Curve frame at {t}\ncreates a plane in which the tangent is the x-axis, Perp Frame",
null,
"Perp Frame (PFrame)\nCurve > Analysis > Perp Frame\nSolve the perpendicular (zero-twisting) frame at a specified curve parameter.\nInputs\nCurve (C)Curve to evaluate\nParameter (t)Parameter on curve domain to evaluate\nOutputs\nFrame (F)Perpendicular curve frame at {t}\ncreates a plane in which the tangent is the z-axis and Horizontal Frame",
null,
"Horizontal Frame (HFrame)\nCurve > Analysis > Horizontal Frame\nGet a horizontally aligned frame along a curve at a specified parameter.\nInputs\nCurve (C)Curve to evaluate\nParameter (t)Parameter on curve domain to evaluate\nOutputs\nFrame (F)Horizontal curve frame at {t}\ncreates a plane that is parallel to the world xy plane and which has its origin at curve parameter t.\n\nInstead of setting specific curve parameters, we can also use the related components Curve Frames",
null,
"Curve Frames (Frames)\nCurve > Division > Curve Frames\nGenerate a number of equally spaced curve frames.\nInputs\nCurve (C)Curve to divide\nCount (N)Number of segments\nOutputs\nFrames (F)Curve frames\nParameters (t)Parameter values at division points\n, Perp Frames",
null,
"Perp Frames (PFrames)\nCurve > Division > Perp Frames\nGenerate a number of equally spaced, perpendicular frames along a curve.\nInputs\nCurve (C)Curve to divide\nCount (N)Number of segments\nAlign (A)Align the frames\nOutputs\nFrames (F)Curve frames\nParameters (t)Parameter values at frame points\nand Horizontal Frames",
null,
"Horizontal Frames (HFrames)\nCurve > Division > Horizontal Frames\nGenerate a number of equally spaced, horizontally aligned curve frames.\nInputs\nCurve (C)Curve to divide\nCount (N)Number of segments\nOutputs\nFrames (F)Curvature frames\nParameters (t)Parameter values at division points\n. Here, we define at input N the number of segments in which the curve is divided into. The frames (planes) are created at each division point.\n\n### Retrieve local coordinates\n\nTo find the local coordinates of a point on a construction plane, we can use Plane Coordinates",
null,
"Plane Coordinates (PlCoord)\nVector > Plane > Plane Coordinates\nGet the coordinates of a point in a plane axis system.\nInputs\nPoint (P)Input point\nSystem (S)Local coordinate system\nOutputs\nX coordinate (X)Point {x} coordinate\nY coordinate (Y)Point {y} coordinate\nZ coordinate (Z)Point {z} coordinate\nalong with our point P and the construction plane at S.\n\n### Retrieve planes from curves\n\nFor planar curves, we can use Planar",
null,
"Planar (Planar)\nCurve > Analysis > Planar\nTest a curve for planarity.\nInputs\nCurve (C)Curve to evaluate\nOutputs\nPlanar (p)Planarity of curve\nPlane (P)Curve plane\nDeviation (D)Deviation from curve plane\nnot only to verify the curve’s planarity, but also to find the plane that the curve is in. Unlike Curve Frame, which uses a tangent of the curve to construct the plane, Planar orients the plane to the world coordinate system.\n\n### Retrieve planes from surfaces\n\nTo find the plane of a planar surface, we can use Is Planar",
null,
"Is Planar (Planar)\nSurface > Analysis > Is Planar\nTest whether a surface is planar\nInputs\nSurface (S)Surface to test for planarity\nInterior (I)Limit planarity test to the interior of trimmed surfaces\nOutputs\nPlanar (F)Planarity flag of surface\nPlane (P)Surface plane\nto get the verification and the surface’s plane.\n\nUsing this component with a non-planar surface will return any plane, even when the surface was constructed with a base (construction) plane.\n\nFor non-planar surfaces, there is no unified plane and we need to evaluate a point on the surface to get the plane at this position. To do so, we use Evaluate Surface",
null,
"Evaluate Surface (EvalSrf)\nSurface > Analysis > Evaluate Surface\nEvaluate local surface properties at a {uv} coordinate.\nInputs\nSurface (S)Base surface\nPoint (uv){uv} coordinate to evaluate\nOutputs\nPoint (P)Point at {uv}\nNormal (N)Normal at {uv}\nU direction (U)U direction at {uv}\nV direction (V)V direction at {uv}\nFrame (F)Frame at {uv}\nand attach uv-coordinates that match the domain of the surface.\n\n### Using Planes in Transformations\n\nBesides the basic transformations mentioned in the beginning of this how-to, there are two powerful components to relocate geometric objects with reference planes. The first one is Move To Plane",
null,
"Move To Plane (MoveToPlane)\nTransform > Euclidean > Move To Plane\nTranslate (move) an object onto a plane.\nInputs\nGeometry (G)Base geometry\nPlane (P)Target plane\nAbove (A)Move when above plane\nBelow (B)Move when below plane\nOutputs\nGeometry (G)Translated geometry\nTransform (X)Transformation data\n: This component takes a geometry object G and a plane P as input and moves the objects onto the provided plane. At input A and B we can set a Boolean to false if objects on one side of the plane, above (A) or below (B), should not be moved. The objects are translated until the shape touches the plane.\n\nThe other component is Orient",
null,
"Orient (Orient)\nTransform > Euclidean > Orient\nOrient an object. Orientation is sometimes called a 'ChangeBasis tranformation'. It allows for remapping of geometry from one axis-system to another.\nInputs\nGeometry (G)Base geometry\nSource (A)Initial plane\nTarget (B)Final plane\nOutputs\nGeometry (G)Reoriented geometry\nTransform (X)Transformation data\nand it resembles a move and rotate transformation. We need to provide the geometric objects G and their initial (source) planes at A. At input B we set the target plane(s). The coordinates of the object are then remapped to the new plane. This is helpful to reposition and arrange objects independent of their previous relation.\n\nThis page is open source. Edit it on GitHub or see how you can contribute."
] | [
null,
"data:image/svg+xml;charset=utf-8;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHZpZXdCb3g9IjAgMCAxMTMzIDIyOSIvPg==",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_XYPlane.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_ConstructPlane.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_PlaneNormal.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_PlaneOrigin.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_RotatePlane.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_FlipPlane.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_DeconstructPlane.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_CurveFrame.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_CurvePerpFrame.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_CurveFrameHorizontal.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_CurveFrames.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_CurvePerpFrames.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_CurveFramesHorizontal.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Vector%20Components/Component_PlaneCoordinates.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Curve%20Components/Component_Planar.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Surface%20Components/Component_SurfaceIsPlanar.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Surface%20Components/Component_EvaluateSurface.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Transform%20Components/Component_MoveOntoPlane.png",
null,
"https://parametricbydesign.com/img/ghcomponents/Transform%20Components/Component_Orient.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76852053,"math_prob":0.85287744,"size":5808,"snap":"2023-14-2023-23","text_gpt3_token_len":1277,"char_repetition_ratio":0.16454169,"word_repetition_ratio":0.12260128,"special_character_ratio":0.20643939,"punctuation_ratio":0.048418973,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9922128,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,6,null,2,null,4,null,2,null,2,null,2,null,2,null,2,null,4,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,null,null,2,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T12:54:37Z\",\"WARC-Record-ID\":\"<urn:uuid:e5951633-027d-4ecb-b33c-c12f7b901fce>\",\"Content-Length\":\"63645\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9a23256-72db-47c9-9c0c-51e040812f57>\",\"WARC-Concurrent-To\":\"<urn:uuid:5c622942-61c2-45aa-86b1-3c830a6b00b7>\",\"WARC-IP-Address\":\"34.74.37.249\",\"WARC-Target-URI\":\"https://parametricbydesign.com/grasshopper/how-tos/construction-planes/\",\"WARC-Payload-Digest\":\"sha1:AWMPCEPXMS6ZWGNOVEVJ5N5QA5UF7TI2\",\"WARC-Block-Digest\":\"sha1:SBQNZPSSR6GY3VFW64YU2PHIVMNOIU4Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647810.28_warc_CC-MAIN-20230601110845-20230601140845-00092.warc.gz\"}"} |
https://www.colorhexa.com/0f5fa6 | [
"# #0f5fa6 Color Information\n\nIn a RGB color space, hex #0f5fa6 is composed of 5.9% red, 37.3% green and 65.1% blue. Whereas in a CMYK color space, it is composed of 91% cyan, 42.8% magenta, 0% yellow and 34.9% black. It has a hue angle of 208.2 degrees, a saturation of 83.4% and a lightness of 35.5%. #0f5fa6 color hex could be obtained by blending #1ebeff with #00004d. Closest websafe color is: #006699.\n\n• R 6\n• G 37\n• B 65\nRGB color chart\n• C 91\n• M 43\n• Y 0\n• K 35\nCMYK color chart\n\n#0f5fa6 color description : Dark blue.\n\n# #0f5fa6 Color Conversion\n\nThe hexadecimal color #0f5fa6 has RGB values of R:15, G:95, B:166 and CMYK values of C:0.91, M:0.43, Y:0, K:0.35. Its decimal value is 1007526.\n\nHex triplet RGB Decimal 0f5fa6 `#0f5fa6` 15, 95, 166 `rgb(15,95,166)` 5.9, 37.3, 65.1 `rgb(5.9%,37.3%,65.1%)` 91, 43, 0, 35 208.2°, 83.4, 35.5 `hsl(208.2,83.4%,35.5%)` 208.2°, 91, 65.1 006699 `#006699`\nCIE-LAB 39.645, 5.067, -44.397 11.171, 11.038, 37.616 0.187, 0.185, 11.038 39.645, 44.685, 276.511 39.645, -22.443, -64.572 33.224, 1.874, -43.872 00001111, 01011111, 10100110\n\n# Color Schemes with #0f5fa6\n\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #a6560f\n``#a6560f` `rgb(166,86,15)``\nComplementary Color\n• #0fa6a2\n``#0fa6a2` `rgb(15,166,162)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #0f14a6\n``#0f14a6` `rgb(15,20,166)``\nAnalogous Color\n• #a6a20f\n``#a6a20f` `rgb(166,162,15)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #a60f14\n``#a60f14` `rgb(166,15,20)``\nSplit Complementary Color\n• #5fa60f\n``#5fa60f` `rgb(95,166,15)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #a60f5f\n``#a60f5f` `rgb(166,15,95)``\n• #0fa656\n``#0fa656` `rgb(15,166,86)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #a60f5f\n``#a60f5f` `rgb(166,15,95)``\n• #a6560f\n``#a6560f` `rgb(166,86,15)``\n• #093760\n``#093760` `rgb(9,55,96)``\n• #0b4477\n``#0b4477` `rgb(11,68,119)``\n• #0d528f\n``#0d528f` `rgb(13,82,143)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #116cbd\n``#116cbd` `rgb(17,108,189)``\n``#137ad5` `rgb(19,122,213)``\n• #1787ea\n``#1787ea` `rgb(23,135,234)``\nMonochromatic Color\n\n# Alternatives to #0f5fa6\n\nBelow, you can see some colors close to #0f5fa6. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0f85a6\n``#0f85a6` `rgb(15,133,166)``\n• #0f78a6\n``#0f78a6` `rgb(15,120,166)``\n• #0f6ca6\n``#0f6ca6` `rgb(15,108,166)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #0f52a6\n``#0f52a6` `rgb(15,82,166)``\n• #0f46a6\n``#0f46a6` `rgb(15,70,166)``\n• #0f39a6\n``#0f39a6` `rgb(15,57,166)``\nSimilar Colors\n\n# #0f5fa6 Preview\n\nThis text has a font color of #0f5fa6.\n\n``<span style=\"color:#0f5fa6;\">Text here</span>``\n#0f5fa6 background color\n\nThis paragraph has a background color of #0f5fa6.\n\n``<p style=\"background-color:#0f5fa6;\">Content here</p>``\n#0f5fa6 border color\n\nThis element has a border color of #0f5fa6.\n\n``<div style=\"border:1px solid #0f5fa6;\">Content here</div>``\nCSS codes\n``.text {color:#0f5fa6;}``\n``.background {background-color:#0f5fa6;}``\n``.border {border:1px solid #0f5fa6;}``\n\n# Shades and Tints of #0f5fa6\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000204 is the darkest color, while #f1f8fe is the lightest one.\n\n• #000204\n``#000204` `rgb(0,2,4)``\n• #020d16\n``#020d16` `rgb(2,13,22)``\n• #041728\n``#041728` `rgb(4,23,40)``\n• #05213a\n``#05213a` `rgb(5,33,58)``\n• #072c4c\n``#072c4c` `rgb(7,44,76)``\n• #08365e\n``#08365e` `rgb(8,54,94)``\n• #0a4070\n``#0a4070` `rgb(10,64,112)``\n• #0c4a82\n``#0c4a82` `rgb(12,74,130)``\n• #0d5594\n``#0d5594` `rgb(13,85,148)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n• #1169b8\n``#1169b8` `rgb(17,105,184)``\n• #1274ca\n``#1274ca` `rgb(18,116,202)``\n• #147edc\n``#147edc` `rgb(20,126,220)``\n• #1988ea\n``#1988ea` `rgb(25,136,234)``\n• #2b91ec\n``#2b91ec` `rgb(43,145,236)``\n• #3d9bed\n``#3d9bed` `rgb(61,155,237)``\n• #4fa4ef\n``#4fa4ef` `rgb(79,164,239)``\n``#61adf1` `rgb(97,173,241)``\n• #73b7f2\n``#73b7f2` `rgb(115,183,242)``\n• #85c0f4\n``#85c0f4` `rgb(133,192,244)``\n• #97c9f6\n``#97c9f6` `rgb(151,201,246)``\n• #a9d3f7\n``#a9d3f7` `rgb(169,211,247)``\n• #bbdcf9\n``#bbdcf9` `rgb(187,220,249)``\n• #cde5fa\n``#cde5fa` `rgb(205,229,250)``\n• #dfeefc\n``#dfeefc` `rgb(223,238,252)``\n• #f1f8fe\n``#f1f8fe` `rgb(241,248,254)``\nTint Color Variation\n\n# Tones of #0f5fa6\n\nA tone is produced by adding gray to any pure hue. In this case, #555b60 is the less saturated color, while #0160b4 is the most saturated one.\n\n• #555b60\n``#555b60` `rgb(85,91,96)``\n• #4e5b67\n``#4e5b67` `rgb(78,91,103)``\n• #475c6e\n``#475c6e` `rgb(71,92,110)``\n• #405c75\n``#405c75` `rgb(64,92,117)``\n• #395d7c\n``#395d7c` `rgb(57,93,124)``\n• #325d83\n``#325d83` `rgb(50,93,131)``\n• #2b5d8a\n``#2b5d8a` `rgb(43,93,138)``\n• #245e91\n``#245e91` `rgb(36,94,145)``\n• #1d5e98\n``#1d5e98` `rgb(29,94,152)``\n• #165f9f\n``#165f9f` `rgb(22,95,159)``\n• #0f5fa6\n``#0f5fa6` `rgb(15,95,166)``\n``#085fad` `rgb(8,95,173)``\n• #0160b4\n``#0160b4` `rgb(1,96,180)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0f5fa6 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
] | [
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https://www.coursehero.com/file/pcqhc9g/e-j-2-%CF%80-2-n-3-7-17-12-ej-2%CF%80-2n-37-e-j-2-%CF%80-2-n-37-e-j2-%CF%80-3-n-3-7-e-j2-%CF%803-n-37-1/ | [
"e j 2 \\u03c0 2 n 3 7 17 12 ej 2\\u03c0 2n 37 e j 2 \\u03c0 2 n 37 e j2 \\u03c0 3 n 3 7 e j2 \\u03c03 n 37 1\n\n# E j 2 π 2 n 3 7 17 12 ej 2π 2n 37 e j 2 π 2 n 37 e\n\nThis preview shows page 5 - 9 out of 11 pages.",
null,
"ECE 513 Exam #2 Instructor: Dr. Cranos Williams 6 4.Spectral Estimation: (20 Pts)We are given the following continuous-time analog signalx1(t) = cos(2·π·F1t).(14)whereF1= 190Hz(a) Given that the sampling frequency isFs= 500Hz, draw the magnitude of thetrue spectrum,X(ω), of the sampled signalx(n),-∞ ≤n≤ ∞for-πωπ.Be sure to clearly label all relevant frequencies,ω, on the x-axis.(b) A rectangular window of sizeL= 40 is used to make the sequence finite suchthat the new windowed sequence ˆx(n) = 0 fornL.Draw a “rough” sketch(general shape is adequate) of what you would expectˆX(ω) would look like for-πωπ.(c) We want to take anNpoint DFT of the signal ˆx(n) to produceˆX(k). Circle allconditions onNthat will allow us to recreate ˆx(n), 0< n < L-1 fromˆX(k)N= 0.5·LN=LN= 2·L Solution:",
null,
"ECE 513 Exam #2 Instructor: Dr. Cranos Williams 7 𝜋 0.76𝜋 െ0.76𝜋 𝜔 𝑋ሺ𝜔ሻ Figure 2: Magnitude response of ˆ X ( ω ) (b) A rough sketch of the magnitude of ˆ X ( ω ) is given in Fig. 2 (c) Since N should be greater than or equal to L , N = L and N = 2 L are the correct answers.",
null,
"ECE 513 Exam #2 Instructor: Dr. Cranos Williams 8 5.FIR Filter Design: (20 Pts)The magnitude response of an ideal low pass filter is shown in Figure 3.Hd(°)°c°1Figure 3: Ideal Lowpass Filter Response.and the impulse response for this ideal lowpass filter is given ash(n) =(sin(ωcn)πnn6= 0ωcπn= 0(18)The passband and stopband criteria for the filter that we want to design isPassband: 0 to 1500 HzStopband: 2500 to 4000 HzFs = 8000 HzUSEFUL DTFT PROPERTIES:h(n)eonH(ω-ωo),Frequency Shifting Propertyh(n-n0)H(ω)e-jωn0,Time Shifting Property(a) What is an appropriate value for the digital frequencyωcifωcis considered to bethe center of the transition band?",
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"#### You've reached the end of your free preview.\n\nWant to read all 11 pages?\n\n• Fall '11\n• Cw\n•",
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https://www.turito.com/learn/math/mean-absolute-deviation-grade-6 | [
"#### Need Help?\n\nGet in touch with us",
null,
"",
null,
"# Mean Absolute Deviation\n\n### Key Concepts\n\n• Find the mean absolute deviation to describe variability.\n• Find the interquartile range(IQR) to describe variability.\n• Use the mean absolute deviation(MAD) to find the variability of a data set.\n\n## 8.5 Summarize Data Using Measures of Variability\n\n### Mean Absolute Deviation:\n\nThe mean absolute deviation of a set of data is the average distance between each data value and the mean.\n\nSteps to find the Mean Absolute Deviation:\n\n1. Find the mean.\n2. Find the difference/distance between each data value and the mean. It means, find the absolute value of the difference between each data value and the mean.\n3. Find the average of those differences obtained.\n\n### Interquartile range (IQR):\n\nThe interquartile range (IQR) is the distance between the first and third quartiles of the data set. To find the IQR, subtract the first quartile from the third quartile.\n\nTo find the interquartile range (IQR):\n\n1. Find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3).\n2. The IQR is the difference between Q3 and Q1.\n\n### Example 1:\n\nMr. Larry works at a shoe store. He measured the feet of seven customers and found their shoe sizes as below:\n\n5, 5, 6, 7, 8, 8, and 10.\n\nHe knows that the average (mean) size is 7. How can Mr. Larry determine how much the shoe sizes varied for the seven customers?\n\n### Solution:",
null,
"Find the differences between each of the shoe size and the mean (average) size. Show all the differences as positive integers.",
null,
"5 – 7 = 2\n\n5 – 7 = 2\n\n6 – 7 = 1\n\n7 – 7 = 0\n\n8 – 7 = 1\n\n8 – 7 = 1\n\n10 – 7 = 3",
null,
"Find the mean absolute deviation (MAD) by finding the mean of all of the differences, or absolute deviations.\n\n2+2+1+0+1+1+372+2+1+0+1+1+3 / 7\n\n= 10/7 or 1.43",
null,
"Mr. Larry can find the mean absolute deviation (MAD) to determine how much the shoe sizes varied for the seven customers.",
null,
"### Example 2:\n\nThe dot plot shows the distribution of Kayla’s health quiz scores. How can Kayla determine the variability in her health quiz scores?",
null,
"### Solution:",
null,
"Read the data set from the dot plot,\n\n84, 86, 88, 90, 92, 92, 94, 96, 96",
null,
"Find the minimum, median, and maximum values as well as the first and third quartiles,",
null,
"",
null,
"Draw a box plot to determine the interquartile range,",
null,
"",
null,
"### Example 3:\n\nThe data set shows prices for concert tickets in 10 different cities in Washington.",
null,
"Find the IQR of the data set.\n\nSolution:\n\nFind the minimum, median, and maximum values as well as the first and third quartiles.",
null,
"Draw a box plot to determine the interquartile range,",
null,
"### Example 4:\n\nLarry recorded scores of his last six science exams. The mean number of marks scored was 60 and the MAD was 35. How can Larry use these measures to describe the variability of the marks scored during the last six science exams?\n\nWould you say that Larry is a consistent or inconsistent test taker? Explain.\n\nScience Exam Scores: 20, 90, 25, 100, 95, 30\n\n### Solution:\n\nThe MAD shows that the marks generally varied greatly from the mean. The marks were mostly less than 25 (60 – 35) or greater than 95 (60 + 35).\n\nLarry is an inconsistent test taker. Ha has a high MAD. This means, there is more variation from the mean. His marks tend to be farther from his average.\n\n## Exercise:\n\n1. Find the mean absolute deviation (MAD) of the following data set.\n52, 48, 60, 55, 59, 54, 58, 62\n2. Find the IQR of the following data set.\n0, 0, 1, 1, 2, 2, 2, 3, 4, 5, 6, 6, 7, 7, 7, 8\n3. Find the Mean Absolute Deviation of the following data set.\n2, 5, 7, 13, 18\n4. Larry scored the following percentages on 10 quizzes in a science class:\n55, 65, 70, 70, 72, 85, 90, 90, 93, 100\nFind the MAD of the quiz scores.\n5. John surveyed the approximate maximum speeds, in miles per hour, of different animals in a zoo and noted in a table.\n\nFind the interquartile range to describe how the data vary.\n\n6. The table below shows the number of hours different animals spend sleeping per day in a zoo. Use the interquartile range to describe how the data vary.\n\n7. Bob surveyed his friends about the number of apps they use in their mobile phones. The responses were noted as 15, 16, 18, 9, 18, 4, 19, 20, 17, and 36 apps. Use the interquartile range to describe how the data vary.\n\n8.The following table shows the maximum speeds of roller coasters at an amusement park. Find the mean absolute deviation of the given set of data of eight roller coasters.\n\n9. The following table shows the maximum flying speeds of the ten fastest birds. Find the mean absolute deviation of the following set of data.\n\n10. Suppose that seven students have the following numbers of pets: 𝟏, 𝟏, 𝟏, 𝟐, 𝟒, 𝟒, 𝟖.\nThe mean number of pets for these seven students is 𝟑 pets. The MAD number of\npets is 𝟐.\nExplain in words what the MAD means for this data set.\n\n### Concept Map",
null,
"### What have we learned:\n\n• Calculate the mean absolute deviation and describe variability of a data set.\n• Calculate the interquartile range(IQR) and describe variability of a data set.\n• Find the variability of a data set by using the mean absolute deviation(MAD)\n\n#### Related topics",
null,
"#### Addition and Multiplication Using Counters & Bar-Diagrams\n\nIntroduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]",
null,
"#### Dilation: Definitions, Characteristics, and Similarities\n\nUnderstanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]",
null,
"#### How to Write and Interpret Numerical Expressions?\n\nWrite numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]",
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https://www.nagwa.com/en/videos/589175913485/ | [
"# Video: Representing the Solutions of a Constant Equation on a Graph\n\nWhich points in the graph represent a solution to the equation 𝑥 = 2?\n\n01:14\n\n### Video Transcript\n\nWhich points in the graph represent a solution to the equation 𝑥 equals two?\n\nThe 𝑥-axis is the horizontal one and, in this graph, goes from zero to 10. Any equation of the form 𝑥 equals 𝑎 where 𝑎 is a constant will be a vertical line through the value 𝑎 on the 𝑥-axis. In this question, the equation 𝑥 equals two will be a vertical line through two on the 𝑥-axis. This line passes through the four points B, C, D, and E.\n\nA common mistake here would be to draw a horizontal line as the 𝑥-axis is horizontal. The line shown in pink actually has equation 𝑦 equals two as it passes through two on the 𝑦-axis. This would lead to the incorrect answer of points D and F. The Line 𝑥 equals two, however, passes through B, C, D, and E."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95054406,"math_prob":0.9979237,"size":743,"snap":"2020-10-2020-16","text_gpt3_token_len":199,"char_repetition_ratio":0.15290934,"word_repetition_ratio":0.04225352,"special_character_ratio":0.23149395,"punctuation_ratio":0.116959065,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998017,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-25T06:42:25Z\",\"WARC-Record-ID\":\"<urn:uuid:40066cda-086d-4d55-9294-2157093d2a02>\",\"Content-Length\":\"22942\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f893dbc-0d28-446a-9884-a899da383533>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a8faa87-f4bc-4736-9726-9aa267b8c9da>\",\"WARC-IP-Address\":\"23.23.60.1\",\"WARC-Target-URI\":\"https://www.nagwa.com/en/videos/589175913485/\",\"WARC-Payload-Digest\":\"sha1:IDO4XCMTFFSCSKUTG4S3P4ZH435BY2FH\",\"WARC-Block-Digest\":\"sha1:KHFGJN4YEMQ6D5G3PHHHDY6F53P3N5I4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146033.50_warc_CC-MAIN-20200225045438-20200225075438-00554.warc.gz\"}"} |
https://iop.figshare.com/articles/1012080/1/citations/datacite | [
"10.6084/m9.figshare.1012080.v1 G Mazzucchi L Lepori A Trombettoni First BZ for the layered π-flux cubic lattice model 2013 IOP Publishing bz Dirac points lattice configuration superfluid properties gap changes honeycomb geometry 3 D lattices lattice displays Dirac points lattice model interpolation interaction Uc flux gauge fields 2 D Reciprocal lattice vectors superfluid gap quantum phase transition 2013-06-24 00:00:00 article https://iop.figshare.com/articles/_First_BZ_for_the_layered_flux_cubic_lattice_model/1012080 <p><strong>Figure 9.</strong> First BZ for the layered π-flux cubic lattice model. Reciprocal lattice vectors are also shown.</p> <p><strong>Abstract</strong></p> <p>We study the superfluid properties of attractively interacting fermions hopping in a family of 2D and 3D lattices in the presence of synthetic gauge fields having π-flux per plaquette. The reason for such a choice is that the π-flux cubic lattice displays Dirac points and that decreasing the hopping coefficient in a spatial direction (say, <em>t<sub>z</sub></em>), these Dirac points are unaltered: it is then possible to study the 3D–2D interpolation towards the π-flux square lattice. We also consider the lattice configuration providing the continuous interpolation between the 2<em>D</em> π-flux square lattice and the honeycomb geometry. We investigate by a mean-field analysis the effects of interaction and dimensionality on the superfluid gap, chemical potential and critical temperature, showing that these quantities continuously vary along the patterns of interpolation. In the two-dimensional cases at zero temperature and half-filling, there is a quantum phase transition occurring at a critical (negative) interaction <em>U<sub>c</sub></em> presenting a linear critical exponent for the gap as a function of |<em>U</em> − <em>U<sub>c</sub></em>|. We show that in three dimensions, this quantum phase transition is again retrieved, pointing out that the critical exponent for the gap changes from 1 to 1/2 for each finite value of <em>t<sub>z</sub></em>.</p>"
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http://forums.wolfram.com/mathgroup/archive/2007/Jan/msg00586.html | [
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"Re: Problem with base 2 logs and Floor\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg73017] Re: Problem with base 2 logs and Floor\n• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>\n• Date: Mon, 29 Jan 2007 04:23:35 -0500 (EST)\n• Organization: The Open University, Milton Keynes, UK\n• References: <ephdk1\\$s2c\\$1@smc.vnet.net>\n\n```neillclift at msn.com wrote:\n> Hi,\n>\n> When I use an expression like this:\n>\n> r - Floor[Log[2, r]] - 1 /. r -> 4\n>\n> I get precision errors in Mathematica 5.2. If I use an expression like\n> this:\n>\n> N[r - Floor[Log[2, r]] - 1] /. r -> 4\n>\n> I get a correct result of 1 and no errors. If I use an expression like\n> this I get the same result:\n>\n> N[r - Ceiling[Log[2, r]] - 1] /. r -> 4\n>\n> This is to be expected as Floor[Log[2, r]] = Ceiling[Log[2, r]] when\n> r is a power of two.\n> Unfortunatly the expessions diverge for r=8:\n>\n> N[r - Floor[Log[2, r]] - 1] /. r -> 8 gives 5\n> N[r - Ceiling[Log[2, r]] - 1] /. r -> 8 gives 4\n>\n> How can I get exacts results for expessions like this?\n> Thanks.\n> Neill.\n>\n\nNeill,\n\nMost of your troubles come from evaluation order of the components of\nthe expressions. We will investigate first the evaluation order of the\ntransformation rules and how to fix it, then an alternative solution,\nshorter and using SetDelayed, will be giving at the bottom of this mail.\n\nNote that using machine size numbers in the transformation rule yields\nthe same result than using the N[] function. (Floor does not attempt any\nsymbolic simplification.)\n\nIn:=\nr - Floor[Log[2, r]] - 1 /. r -> 4.\n\nOut=\n1.\n\nNow, if you compute directly log_2(4), Mathematica returns the\n\nIn:=\nLog[2, 4]\n\nOut=\n2\n\nIn this case, Floor has no problem to determine the correct ans exact\nresult.\n\nIn:=\nFloor[Log[2, 4]]\n\nOut=\n2\n\nHowever, things behave differently when you apply some transformation rules.\n\nIn:=\nLog[2, r] /. r -> 4\n\nOut=\nLog\n------\nLog\n\nThe above expression is returned because Mathematica standardizes the\nlogarithm first.\n\nIn:=\nTrace[Log[2, r] /. r -> 4]\n\nOut=\nLog[r] Log[r] Log[r] Log\n{{Log[2, r], ------, ------}, ------ /. r -> 4, ------}\nLog Log Log Log\n\nAfter the rule has been applied, no other transformations or\nsimplifications are used.\n\nOne way to force Mathematica to simplify the ration of logarithms is to\nuse FullSimplify.\n\nIn:=\nFullSimplify[Log/Log, ComplexityFunction ->\n(Count[{#1}, _Log, Infinity] & )]\n\nOut=\n2\n\nHowever, if we plug directly the above expression into our original\nexpression, no simplification seems to happen anymore!\n\nIn:=\nr - Floor[FullSimplify[Log[2, r], ComplexityFunction ->\n(Count[{#1}, _Log, Infinity] & )]] - 1 /. r -> 8\n\nFloor::\"meprec\" : \"Internal precision limit \\$MaxExtraPrecision =\n49.99999999999999` reached while evaluating Floor[Log\\Log].\"\n\nOut=\nLog\n7 - Floor[------]\nLog\n\nThe issue is, again, a matter of precedence: if we use Hold and\nReleaseHold, we get the correct answer.\n\nIn:=\nReleaseHold[\nr - Floor[Hold[FullSimplify[Log[2, r], ComplexityFunction ->\n(Count[{#1}, _Log, Infinity] & )]]] - 1 /. r -> 8]\n\nOut=\n4\n\nIn:=\nReleaseHold[\nr - Ceiling[Hold[FullSimplify[Log[2, r], ComplexityFunction ->\n(Count[{#1}, _Log, Infinity] & )]]] - 1 /. r -> 8]\n\nOut=\n4\n\nIndeed, we can get ride of FullSimplify, and and wrap Hold around the\noriginal expression (not including the transformation rule) and\nReleaseHold around the whole thing.\n\nIn:=\nReleaseHold[Hold[r - Floor[Log[2, r]] - 1] /. r -> 8]\n\nOut=\n4\n\nIn:=\nReleaseHold[Hold[r - Ceiling[Log[2, r]] - 1] /. r -> 8]\n\nOut=\n4\n\nFinally, an easier way is to define a new function with SetDelayed. In\ndoing so, the internal expression is kept as it is written until the\nfunction is called. Then, the argument is first replace within the body\nof the function, then the resulting expression is simplified and evaluated.\n\nIn:=\nfloorlog[r_] := r - Floor[Log[2, r]] - 1\nceilinglog[r_] := r - Ceiling[Log[2, r]] - 1\n\nIn:=\nfloorlog\n\nOut=\n4\n\nIn:=\nceilinglog\n\nOut=\n4\n\nRegards,\nJean-Marc\n\n```\n\n• Prev by Date: Re: Re: Sequence of Bernoulli RVs\n• Next by Date: Re: Re: Sequence of Bernoulli RVs\n• Previous by thread: Re: Problem with base 2 logs and Floor\n• Next by thread: Correction to \"Sequence of Bernoulli RVs\""
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https://math.stackexchange.com/questions/279737/applications-of-residue-theorem-in-complex-analysis | [
"# Applications of Residue Theorem in complex analysis?\n\nDoes anyone know the applications of Residue Theorem in complex analysis? I would like to do a quick paper on the matter, but am not sure where to start.\n\nThe residue theorem\n\nThe residue theorem, sometimes called Cauchy's residue theorem (one of many things named after Augustin-Louis Cauchy), is a powerful tool to evaluate line integrals of analytic functions over closed curves; it can often be used to compute real integrals as well. It generalizes the Cauchy integral theorem and Cauchy's integral formula. From a geometrical perspective, it is a special case of the generalized Stokes' theorem.\n\nIllustration of the setting\n\nThe statement is as follows: Suppose $U$ is a simply connected open subset of the complex plane, and $a_1,\\ldots,a_n$ are finitely many points of $U$ and $f$ is a function which is defined and holomorphic on $U\\setminus\\{a_1,\\ldots,a_n\\}$. If $\\gamma$ is a rectifiable curve in $U$ which does not meet any of the $a_k$, and whose start point equals its endpoint, then $$\\oint_\\gamma f(z)\\,dz=2\\pi i\\sum_{k=1}^n I(\\gamma,a_k)\\mathrm{Res}(f,a_k)$$\n\nI'm sure many complex analysis experts are very familiar with this theorem. I was just hoping someone could enlighten me on its many applications for my paper. Thank you!\n\n• This is a paper for pure personal enjoyment. It is not a homework assignment of any form. – Anthony Peter Jan 16 '13 at 2:23\n• Hello, Welcome to math.SE. IMO one of the main use is to find integrals (i.e, contour integration). It is pretty important for finding real definite integrals also..Search for contour inegration would yielda lot of results. – TheJoker Jan 16 '13 at 2:27\n• Sorry for sounding flippant, but this is like asking if power series have applications and wanting people to tell you about them. Can you first please tell us what work you have put in to find applications yourself? A quick web search should give you several. – KCd Jan 16 '13 at 2:27\n• I personally have seen the many applications when solving real integrals, but other than this, I haven't been able to find much. – Anthony Peter Jan 16 '13 at 2:43\n• If you're also interested in applications in other fields, google the lecture notes of Andreas Gathmann for his Algebraic Geometry class. There, he uses the residue theorem in the proof of the Riemann-Roch theorem to show that certain sections cannot exist. – InvisiblePanda Jan 16 '13 at 12:09\n\nOther then as a fantastic tool to evaluate some difficult real integrals, complex integrals have many purposes.\n\nFirstly, contour integrals are used in Laurent Series, generalizing real power series.\n\nThe argument principle can tell us the difference between the poles and roots of a function in the closed contour $C$:\n\n$$\\oint_{C} {f'(z) \\over f(z)}\\, dz=2\\pi i (\\text{Number of Roots}-\\text{Number of Poles})$$\n\nand this has been used to prove many important theorems, especially relating to the zeros of the Riemann zeta function.\n\nNoting that the residue of $\\pi \\cot (\\pi z)f(z)$ is $f(z)$ at all the integers. Using a square contour offset by the integers by $\\frac{1}{2}$, we note the contour disappears as it gets large, and thus\n\n$$\\sum_{n=-\\infty}^\\infty f(n) = -\\pi \\sum \\operatorname{Res}\\, \\cot (\\pi z)f(z)$$\n\nwhere the residues are at poles of $f$.\n\nWhile I have only mentioned a few, basic uses, many, many others exist.\n\nYou can find every conceivable (and several inconveivable) application of the residue theorem in The Cauchy method of residues: theory and applications by Mitrinović and Kečkić, Dordrecht, 1984 (ISBN: 9027716234).\n\nIf that's not enough, there's even a second volume: The Cauchy method of residues: theory and applications, Vol. 2 by the same authors, and publisher. This one published in 1993 (ISBN: 0792323114.)\n\nAmazon seems to carrry a one-volume book by the same authors and with a very similar title, published in 2001 by Kluwer, but I haven't seen that exact version.\n\n• Nice answer! Those who were my teachers in complex variables speak very well in this book. He is not as conhencido as it should be in the West. – MathOverview Jan 16 '13 at 12:18\n• @Elias, I agree, the books are fascinating, and not as well-known as they should be. – mrf Jan 16 '13 at 13:08\n\nSkip to the 11th page of this document. It uses residue calculus to prove the classical result that $\\sum_{i=1}^{\\infty}1/n^{2} = \\pi^{2}/6$. Plus it leaves the easy stages of the argument for you to fill in for yourself."
] | [
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https://www.jobilize.com/online/course/5-5-crosscorrelation-of-random-processes-by-openstax?qcr=www.quizover.com | [
"# 5.5 Crosscorrelation of random processes\n\n Page 1 / 1\n(Blank Abstract)\n\nBefore diving into a more complex statistical analysis of random signals and processes , let us quickly review the idea of correlation . Recall that the correlation of two signals or variables is the expectedvalue of the product of those two variables. Since our main focus is to discover more about random processes, a collectionof random signals, we will deal with two random processes in this discussion, where in this case we will deal with samplesfrom two different random processes. We will analyze the expected value of the product of these two variables and how they correlate to one another, where theargument to this correlation function will be the time difference. For the correlation of signals from the same randomprocess, look at the autocorrelation function .\n\n## Crosscorrelation function\n\nWhen dealing with multiple random processes, it is also important to be able to describe the relationship, if any,between the processes. For example, this may occur if more than one random signal is applied to a system. In order to dothis, we use the crosscorrelation function , where the variables are instances from two different wide sensestationary random processes.\n\nCrosscorrelation\nif two processes are wide sense stationary, the expected value of the product of a random variable from one randomprocess with a time-shifted, random variable from a different random process\nLooking at the generalized formula for the crosscorrelation, we will represent our two random processes by allowing $U=U(t)$ and $V=V(t-)$ . We will define the crosscorrelation function as\n${R}_{uv}(t, t-)=(UV)=\\int \\,d u$ v u v f u v\nJust as the case with the autocorrelation function, if ourinput and output, denoted as $U(t)$ and $V(t)$ , are at least jointly wide sense stationary, then the crosscorrelation does not depend on absolute time; it is justa function of the time difference. This means we can simplify our writing of the above function as\n${R}_{uv}()=(UV)$\nor if we deal with two real signal sequences, $x(n)$ and $y(n)$ , then we arrive at a more commonly seen formula for the discrete crosscorrelation function. See the formula belowand notice the similarities between it and the convolution of two signals:\n${R}_{xy}(n, n-m)={R}_{xy}(m)=\\sum$ x n y n m\n\n## Properties of crosscorrelation\n\nBelow we will look at several properties of the crosscorrelation function that hold for two wide sense stationary (WSS) random processes.\n\n• Crosscorrelation is not an even function; however, it does have a unique symmetryproperty:\n${R}_{xy}(-)={R}_{yx}()$\n• The maximum value of the crosscorrelation is not always when the shift equals zero; however, we can prove thefollowing property revealing to us what value the maximum cannot exceed.\n$\\left|{R}_{xy}()\\right|\\le \\sqrt{{R}_{xx}(0){R}_{yy}(0)}$\n• When two random processes are statistically independent then we have\n${R}_{xy}()={R}_{yx}()$\n\n## Examples\n\nLet us begin by looking at a simple example showing the relationship between two sequences. Using , find the crosscorrelation of the sequences $x(n)=\\{, 0, 0, 2, -3, 6, 1, 3, 0, 0, \\}()$ $y(n)=\\{, 0, 0, 1, -2, 4, 1, -3, 0, 0, \\}()$ for each of the following possible time shifts: $m=\\{0, 3, -1\\}()$ .\n\n• For $m=0$ , we should begin by finding the product sequence $s(n)=x(n)y(n)$ . Doing this we get the following sequence: $s(n)=\\{, 0, 0, 2, 6, 24, 1, -9, 0, 0, \\}()$ and so from the sum in our crosscorrelation function we arrive at the answer of ${R}_{xy}(0)=22$\n• For $m=3$ , we will approach it the same was we did above; however, we will now shift $y(n)$ to the right. Then we can find the product sequence $s(n)=x(n)y(n-3)$ , which yields $s(n)=\\{, 0, 0, 0, 0, 0, 1, -6, 0, 0, \\}()$ and from the crosscorrelation function we arrive at the answer of ${R}_{xy}(3)=-6$\n• For $m=-1$ , we will again take the same approach; however, we will now shift $y(n)$ to the left. Then we can find the product sequence $s(n)=x(n)y(n+1)$ , which yields $s(n)=\\{, 0, 0, -4, -12, 6, -3, 0, 0, 0, \\}()$ and from the crosscorrelation function we arrive at the answer of ${R}_{xy}(-1)=-13$\n\n#### Questions & Answers\n\nIs there any normative that regulates the use of silver nanoparticles?\nDamian Reply\nwhat king of growth are you checking .?\nRenato\nWhat fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?\nStoney Reply\nwhy we need to study biomolecules, molecular biology in nanotechnology?\nAdin Reply\n?\nKyle\nyes I'm doing my masters in nanotechnology, we are being studying all these domains as well..\nAdin\nwhy?\nAdin\nwhat school?\nKyle\nbiomolecules are e building blocks of every organics and inorganic materials.\nJoe\nanyone know any internet site where one can find nanotechnology papers?\nDamian Reply\nresearch.net\nkanaga\nsciencedirect big data base\nErnesto\nIntroduction about quantum dots in nanotechnology\nPraveena Reply\nwhat does nano mean?\nAnassong Reply\nnano basically means 10^(-9). nanometer is a unit to measure length.\nBharti\ndo you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?\nDamian Reply\nabsolutely yes\nDaniel\nhow to know photocatalytic properties of tio2 nanoparticles...what to do now\nAkash Reply\nit is a goid question and i want to know the answer as well\nMaciej\ncharacteristics of micro business\nAbigail\nfor teaching engĺish at school how nano technology help us\nAnassong\nDo somebody tell me a best nano engineering book for beginners?\ns. Reply\nthere is no specific books for beginners but there is book called principle of nanotechnology\nNANO\nwhat is fullerene does it is used to make bukky balls\nDevang Reply\nare you nano engineer ?\ns.\nfullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.\nTarell\nwhat is the actual application of fullerenes nowadays?\nDamian\nThat is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.\nTarell\nwhat is the Synthesis, properties,and applications of carbon nano chemistry\nAbhijith Reply\nMostly, they use nano carbon for electronics and for materials to be strengthened.\nVirgil\nis Bucky paper clear?\nCYNTHIA\ncarbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc\nNANO\nso some one know about replacing silicon atom with phosphorous in semiconductors device?\ns. Reply\nYeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.\nHarper\nDo you know which machine is used to that process?\ns.\nhow to fabricate graphene ink ?\nSUYASH Reply\nfor screen printed electrodes ?\nSUYASH\nWhat is lattice structure?\ns. Reply\nof graphene you mean?\nEbrahim\nor in general\nEbrahim\nin general\ns.\nGraphene has a hexagonal structure\ntahir\nOn having this app for quite a bit time, Haven't realised there's a chat room in it.\nCied\nwhat is biological synthesis of nanoparticles\nSanket Reply\nhow did you get the value of 2000N.What calculations are needed to arrive at it\nSmarajit Reply\nPrivacy Information Security Software Version 1.1a\nGood\nGot questions? Join the online conversation and get instant answers!\nJobilize.com Reply\n\n### Read also:\n\n#### Get the best Algebra and trigonometry course in your pocket!\n\nSource: OpenStax, Fundamentals of signal processing. OpenStax CNX. Nov 26, 2012 Download for free at http://cnx.org/content/col10360/1.4\nGoogle Play and the Google Play logo are trademarks of Google Inc.\n\nNotification Switch\n\nWould you like to follow the 'Fundamentals of signal processing' conversation and receive update notifications?",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8842902,"math_prob":0.96953565,"size":3473,"snap":"2019-13-2019-22","text_gpt3_token_len":744,"char_repetition_ratio":0.18708561,"word_repetition_ratio":0.10739437,"special_character_ratio":0.18514253,"punctuation_ratio":0.09803922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9884304,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,1,null,null,null,null,null,1,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-27T02:03:20Z\",\"WARC-Record-ID\":\"<urn:uuid:40896a00-55fa-4ab5-9687-b77acdcbb5b7>\",\"Content-Length\":\"132269\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8109f9d2-1cf2-475f-998c-a774211524f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:e9d37d08-f207-4068-81f6-1f43ea1320e8>\",\"WARC-IP-Address\":\"207.38.89.177\",\"WARC-Target-URI\":\"https://www.jobilize.com/online/course/5-5-crosscorrelation-of-random-processes-by-openstax?qcr=www.quizover.com\",\"WARC-Payload-Digest\":\"sha1:VSBBZUZHRMQCNNTET732NLQFI63U4WAF\",\"WARC-Block-Digest\":\"sha1:XXFXGRZS43CMG5GVYSKHR7AHK6DJVHT7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232260358.69_warc_CC-MAIN-20190527005538-20190527031538-00397.warc.gz\"}"} |
https://lytetyjizeliwyxiw.cwiextraction.com/cyclotron-motion-problem-algebra-28846il.html | [
"# Cyclotron motion problem algebra\n\nThe trick of the cyclotron lies in accelerating of the particles alongside with increasing of the length of the path semicircle therefore the flight time of the particles around the circle remains constant this is derived in detail in the problem solution.",
null,
"Degeneracy plays a fundamental role in quantum statistical mechanics. Here you can learn how it works. APhOPr2 - a nice model of a real-life problem creaking door.\n\nCell structure and the molecular basis for life. APhOPr3 - on gravitational lenses. This process continues and the deuteron flies in a spiral path.\n\nEst-FinPr7 - a challenging problem on diffraction stars through telescope in daylight. From computer science, practical topics including processor architecture, parallel systems, structured programming, and scientific visualization will be presented in tutorial style.\n\nIPhOPr2but this should be enough to get the idea of radiation pressure. APhOPr1-Part-A - you'll learn about inelastic scattering on composite particles the linked text is heavily modified to make it clearer.\n\nIntended for students in the sciences. The number of different states corresponding to a particular energy level is known as the degree of degeneracy of the level.\n\nThe principle of the cyclotron A cyclotron is a type of particle accelerator. Transport, superconductivity, dielectric properties, ferroelectric crystals, magnetism, magnetic resonance, optical phenomena in insulators, nanostructures, non-crystalline solids, point defects, alloys, dislocations.\n\nBy means of a procedure analogous to the one followed for the free case, we can now deduce the equation of the motion in the presence of an external em field.\n\nThe energy is applied to the particles as they cross the gap between dees in the cyclotron and so the particles are accelerated. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the pre calc EAch of 3 circles touch the other 2.\n\nThe cyclotron is then tuned only by changes in the magnetic field, until the bundle of accelerating protons appears. He joined Lawrence Berkeley National Laboratory inand is currently a senior scientist. The trick of the cyclotron lies in accelerating of the particles alongside with increasing of the length of the path semicircle therefore the flight time of the particles around the circle remains constant this is derived in detail in the problem solution.\n\nThe kinetic energy Ek that the particle with the charge e gains during the acceleration by the voltage U is equal to: GPhOPr3 - a fairly challenging problem about acceleration of charged particles with shock waves.\n\nMeanwhile the deuteron flies inside into the negatively charged dee, the voltage between the two dees changed its polarity. What magnitude of magnetic force is required to maintain the deuteron in Physics In a cyclotron one type of particle acceleratora deuteron of mass 2.\n\nDv v 1r2 y vy1r2 vH 3 In the original paper by Barut and Zanghi w29x a different ' f3. The black terms involving E give us a bonus: May be retaken to 9 cr max. The size of the cyclotron can be expressed from the relation 1 — the relation for the radius of the circle from the last section: These D-shaped conductors are made of a non-ferromagnetic material e.\n\nB, 37, R is generalized to obtain a straightforward, surprisingly accurate, and scalable numerical representation for calculating the electronic wave functions of molecules. Analogously, in the pres- ence of external em fields, Eq. SummerFallSummerFall This STN course is designed for the non-science major and has no prerequisites past high school algebra and geometry.\n\nA resolution-of-the-identity approximation renders the primitive one- and two-electron matrix elements diagonal; in other words, the Coulomb operator is local with respect to the grid indices. Clifford Algebra to Geometric w47x G. Greene Introduces students to the techniques that astrophysicists use to model and observe the universe.\n\nSpergel This specially designed course targets the frontier of modern astrophysics. Recent discoveries from planetary missions are emphasized. IPhOPr2 - quite straightforward problem, but you'll learn what is the Hall effect. These D-shaped conductors are made of a non-ferromagnetic material e.\n\nTrack A will be required to take a mid-term exam during Fall break. An alternating voltage is brought into a narrow gap between the conductors. They tell us the power i. Ž7., which holds for free dition always holds between the ‘intrinsic’ frequency particles, in the r.h.s.\n\nof the general equation of 2 m Žthat is, the zbw angular frequency for free motion Ž we see the appearance of two additional particles. and the ‘external’ cyclotron frequency v H: spin-field terms. contributor to the extraction problem. In a compact cyclotron, the vertical focusing force provided by the electromagnetic and using simple algebra, one can easily get expressions for angles and as case of the vertical motion by 1 1 *.\n\nFor p = 2 the results of this problem are known [3, 9]. Generalization to the case of arbitrary p is proposed to be done in the following way. As so-called N = 2 — PSSQM is closely related to the Lie algebra so(3) , then we Cyclotron Motion and Morse Hamiltonian, J.\n\nPhys. A,V, Feb 05, · In an ion cyclotron resonance mass spectrometer, ion cyclotron resonance signals at higher harmonics of cyclotron frequency are employed to increase the resolution of ICR mass spectrometer without increasing the magnetic field.",
null,
"A cyclotron is a type of particle accelerator. Cyclotrons accelerate charged particles to increase their kinetic energy. These high-energy particles are also used e.g.\n\nin nuclear research or in hospitals to obtain radioactive preparations for medical and.",
null,
"Signatures of Topological Superconductors Thesis by Shu-Ping Lee In Partial Ful llment of the Requirements In other words, we cannot simply use boolean algebra to describe an on or o state of a transistor; instead, we need quantum mechanics to will undergo cyclotron motion due to Lorentz force, as shown in Fig As a con.\n\nCyclotron motion problem algebra\nRated 3/5 based on 64 review\nMathematics of Circular Motion"
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"https://i.ytimg.com/vi/ASTJaLQxqDs/hqdefault.jpg",
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"http://bonlacfoods.com/images/speed-time-graphs-worksheet/speed-time-graphs-worksheet-3.jpg",
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"https://i.ytimg.com/vi/SBTAFvqMlfk/hqdefault.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9020507,"math_prob":0.90703905,"size":6112,"snap":"2021-43-2021-49","text_gpt3_token_len":1285,"char_repetition_ratio":0.13179436,"word_repetition_ratio":0.098663926,"special_character_ratio":0.19044502,"punctuation_ratio":0.09680365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9638662,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,1,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T10:48:30Z\",\"WARC-Record-ID\":\"<urn:uuid:f8586b7c-09d7-46e4-a43a-73c3612ab26a>\",\"Content-Length\":\"11878\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a65bba0-9049-4a09-ae2b-242a30f71352>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a778ca8-b651-49b7-91d0-ab564f7fde0c>\",\"WARC-IP-Address\":\"104.21.1.117\",\"WARC-Target-URI\":\"https://lytetyjizeliwyxiw.cwiextraction.com/cyclotron-motion-problem-algebra-28846il.html\",\"WARC-Payload-Digest\":\"sha1:VU7WUTI4K5OHNWKZP2R6KNX2VEY77BA2\",\"WARC-Block-Digest\":\"sha1:BFDPNZJF6OMOVLWBBQIMXI573FP5KLFA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585201.94_warc_CC-MAIN-20211018093606-20211018123606-00056.warc.gz\"}"} |
https://www.thecodingforums.com/threads/quiz-solution-the-golden-fibonacci-ration-69.829240/ | [
"# [QUIZ][SOLUTION] The Golden Fibonacci Ration (#69)\n\nB\n\n#### Brian Mattern\n\n--------------010602090507000003020209\nContent-Type: text/plain; charset=ISO-8859-1; format=flowed\nContent-Transfer-Encoding: 7bit\n\nHere is my text based (and animated) solution.\n\nThe parameters are:\nruby 69.rb [number of steps] [scaling factor]\n\nby default, it shows 6 steps with a scaling factor of 2. (The square\nwith side i is scaled to (m * i + 1) where m is the scaling factor and\nthe extra +1 accounts for the fact that there are no 0 width lines in\nascii art).\n--\nBrian Mattern\n\n--------------010602090507000003020209\nContent-Type: text/plain;\nname=\"69.rb\"\nContent-Transfer-Encoding: 7bit\nContent-Disposition: inline;\nfilename=\"69.rb\"\n\n#\n# A helper class to generate the fibonacci sequence\n#\n# To get the ith fibonacci number simply call Fibonacci\n# You can also get a range with Fibonacci[start..finish]\n# So, Fibonacci[0..10] = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]\n#\n# The class calculates only those numbers it needs, at caches them for\n# future retrieval.\n#\nclass Fibonacci\n@@series = [1,1]\n@@len = 2\n\ndef self.[](i)\nval = i\n\ni = val.max if i.class == Range\nif i >= @@len\n(i - @@len + 1).times { |j|\n@@series << @@series[-1] + @@series[-2]\n}\n@@len = i + 1\nend\n\n@@series[val]\nend\nend\n\n#\n# A class to animate the fibonacci sequence\nclass FibonacciAnimator\n\n# Fetch our series and set up an empty array to draw on\n#\n# Note that due to the lack of 0 width lines in ascii art,\n# we transform x -> mx+1 (where m is a scaling factor)\ndef initialize(num_steps = 6, scale = 2)\n@num_steps = num_steps.to_i\n@scale = scale.to_i\n@series = Fibonacci[0..(@num_steps-1)].collect{ |x| x*@scale + 1 }\n\n@graph = []\n@height = @series[-1]\n@width = @height + (@series[-2] || 0)\n@height.times do |row|\n@graph << Array.new(@width).fill(\" \")\nend\nend\n\n# Calculate the top right location for each square\ndef build_steps\nrow = col = 0\ndir = -1\n\nprev = 0\n@steps = []\n\n# since we don't know where the smallest square should start, we reverse\n# and calculate from large to small\n@series.reverse.each do |size|\ncase(dir)\nwhen 0\ncol += prev - 1\nwhen 1\nrow += prev - 1\ncol += prev - size\nwhen 2\nrow += prev - size\ncol -= size - 1\nwhen 3\nrow -= size - 1\nend\n\n@steps << [size, row, col]\n\nprev = size\ndir += 1\ndir %= 4\nend\n\n# flip our steps so they run from small to large\n@steps.reverse!\n\n@built = true\nend\n\n# actually draw (or animate) the fibonacci representation\n# for animation, fps == frames per second\ndef draw(animate = false, fps = 2)\nbuild_steps unless @built\n@steps.each do |step|\ndraw_square(*step)\nif animate\nprint \"\\033c\" # clear the screen\nputs self\nsleep(1.0 / fps)\nend\nend\n\nputs self unless animate\nend\n\n# draw a size x size square with its top left corner at row, col\ndef draw_square(size, row, col)\nraise \"Cannot draw outside graph bounds.\" if row < 0 or col < 0 or row >= @height or col >= @width or row + size > @height or col + size > @width\n\nsize.times do |i|\nhor = \"-\"\nvert = \"|\"\nhor = vert = \"+\" if i == 0 or i + 1 == size\n\n[[row, col + i], [row + size - 1, col + i]].each do |coord|\nx, y = coord\n@graph[x][y] = hor unless @graph[x][y] == \"+\"\nend\n\n[[row + i, col], [row + i, col + size - 1]].each do |coord|\nx, y = coord\n@graph[x][y] = vert unless @graph[x][y] == \"+\"\nend\nend\nend\n\ndef to_s\n@graph.collect{ |row| row.join(\"\") }.join(\"\\n\")\nend\nend\n\nfp = FibonacciAnimator.new(*ARGV)\nfp.draw(true)\n\n--------------010602090507000003020209--"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.592616,"math_prob":0.9880855,"size":6600,"snap":"2022-27-2022-33","text_gpt3_token_len":2076,"char_repetition_ratio":0.11127956,"word_repetition_ratio":0.9941764,"special_character_ratio":0.37378788,"punctuation_ratio":0.12578617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99815273,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-06T07:00:46Z\",\"WARC-Record-ID\":\"<urn:uuid:60d93434-5620-4196-82ca-22e1a4daf6a8>\",\"Content-Length\":\"51230\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b3ae5a87-116f-4e54-a900-f0a9bc2ffdc9>\",\"WARC-Concurrent-To\":\"<urn:uuid:13ae2e9e-f56e-4dd3-964d-1e5f6df5ff63>\",\"WARC-IP-Address\":\"172.67.218.227\",\"WARC-Target-URI\":\"https://www.thecodingforums.com/threads/quiz-solution-the-golden-fibonacci-ration-69.829240/\",\"WARC-Payload-Digest\":\"sha1:3ZVNJB3EU3Y6UIHYSDMIWGA4MGVHHPVG\",\"WARC-Block-Digest\":\"sha1:U6J7MJ44IE7UBNMAHX5QPEBSIC3WC5RY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104668059.88_warc_CC-MAIN-20220706060502-20220706090502-00392.warc.gz\"}"} |
https://tipcalc.net/how-much-is-a-25-percent-tip-on-218.89 | [
"# Tip Calculator\n\nHow much is a 25 percent tip on \\$218.89?\n\nTIP:\n\\$ 0\nTOTAL:\n\\$ 0\n\nTIP PER PERSON:\n\\$ 0\nTOTAL PER PERSON:\n\\$ 0\n\n## How much is a 25 percent tip on \\$218.89? How to calculate this tip?\n\nAre you looking for the answer to this question: How much is a 25 percent tip on \\$218.89? Here is the answer.\n\nLet's see how to calculate a 25 percent tip when the amount to be paid is 218.89. Tip is a percentage, and a percentage is a number or ratio expressed as a fraction of 100. This means that a 25 percent tip can also be expressed as follows: 25/100 = 0.25 . To get the tip value for a \\$218.89 bill, the amount of the bill must be multiplied by 0.25, so the calculation is as follows:\n\n1. TIP = 218.89*25% = 218.89*0.25 = 54.7225\n\n2. TOTAL = 218.89+54.7225 = 273.6125\n\n3. Rounded to the nearest whole number: 274\n\nIf you want to know how to calculate the tip in your head in a few seconds, visit the Tip Calculator Home.\n\n## So what is a 25 percent tip on a \\$218.89? The answer is 54.72!\n\nOf course, it may happen that you do not pay the bill or the tip alone. A typical case is when you order a pizza with your friends and you want to split the amount of the order. For example, if you are three, you simply need to split the tip and the amount into three. In this example it means:\n\n1. Total amount rounded to the nearest whole number: 274\n\n2. Split into 3: 91.33\n\nSo in the example above, if the pizza order is to be split into 3, you’ll have to pay \\$91.33 . Of course, you can do these settings in Tip Calculator. You can split the tip and the total amount payable among the members of the company as you wish. So the TipCalc.net page basically serves as a Pizza Tip Calculator, as well.\n\n## Tip Calculator Examples (BILL: \\$218.89)\n\nHow much is a 5% tip on \\$218.89?\nHow much is a 10% tip on \\$218.89?\nHow much is a 15% tip on \\$218.89?\nHow much is a 20% tip on \\$218.89?\nHow much is a 25% tip on \\$218.89?\nHow much is a 30% tip on \\$218.89?"
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http://blog.linearconstraints.net/2016/03/29/question-list-for-scientific-computing-class.html | [
"I want to collect my questions before we start CS 210 Scientific Computing tomorrow.\n\n## Computer arithmetic\n\nThis is mainly about IEEE 754 floating numbers. I thought I had been familiar with this but I constantly find many blind spots during my work.\n\nParticularly,\n\n• I have known the floating number does not distribute uniformly. What is the impact to the precision? The z-buffer precision issue is a typical example of this question.\n\n• What is the denormal(subnormal) number? What policy is used for the denormal numbers computation when `-ffast-math` compilation option is set? Does `-ffast-math` also affect SIMD registers?\n\n• What is a machine epsilon? Why do we use it to analyze our algorithm?\n\nI am going to read these books12 to answer or review above questions.\n\n## Stability, conditioning, and stiffness\n\nIn our textbook3 it mentions the relation between the stability and conditioning:\n\n(Stability and conditioning…) Both concepts have to do with sensitivity to perturbations, but the term stability is usually used for algorithms and conditioning for problems (although stability is sometimes used for problems as well, especially in differential equations).\n\nIt is not clear which of them is an intrinsic characteristic of the problem, and which of them is the issue caused by the computation method we choose.\n\nAnd what is the stiffness? Is it related to the other two concepts?\n\n## Conditioning and QR decomposition\n\nWhen I was implementing the dual contouring, the paper4 says the original QEF(Quadratic Error Function) is not stable and then it uses a QR decomposition to resolve it.\n\nI will write a note to explain this. Besides, is QR decomposition special here? (If it is, I would guess the orthogonality makes it so) Can we use other decompositions to achieve similar effects?\n\n## Constraints resolving and projection\n\nIt seems that the projection method is very common for resolving the constraints. Usually this is based on the conservation law or the geometry of the problem. A typical example is solving for the divergence free velocity field by projection.\n\nI want to collect more working examples for this topic.\n\nI also want to learn that how we can know the geometry of the problem before we solve it? And is conservation law always derived from physical property, or it can be an artificial numerical property?\n\n1. Muller, Jean-Michel, et al. Handbook of floating-point arithmetic. Springer Science & Business Media, 2009.\n\n2. Trefethen, Lloyd N., and David Bau III. Numerical linear algebra. Vol. 50. Siam, 1997.\n\n3. Heath, Michael T. Scientific computing. New York: McGraw-Hill, 2002.\n\n4. Ju, Tao, et al. “Dual contouring of hermite data.” ACM Transactions on Graphics (TOG) 21.3 (2002): 339-346."
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http://wiki.bi.up.ac.za/wiki/index.php?title=Python&oldid=294 | [
"# Python\n\n### Variables\n\n#### Declaration and typing\n\nKeyword: Declaring variables, variable assignment, boolean values, integers, floats, strings, lists, dictionaries (ordered dictionaries, default dictionaries), tuples, sets, mutable and immutable data types, variable typing.\n\n```####\n# Boolean variables can only contain one of two values: True or False\n# Boolean values are annotated by using the \"bool\" keyword\na: bool = True\n####\n\n####\n# Integers are variables that contain any positive or negative whole number\n# Integers are annotated using the \"int\" keyword\n# Floats are variables that contains any positive or negative decimal\n# Floats are annotated using the \"float\" keyword\na: int = 15\nb: int = -23\n\nc: float = 6.4\nd: float = -9.5\n####\n\n####\n# Strings are variables that contain text\n# Strings are annotated using the \"str\" keyword\na: str = \"This is a string\"\nb: str = 'Strings can be enclosed using single quotes'\n\n# Initializing an empty string:\na = \"\"\n####\n\n####\n# Lists are collections of other variables\n# Lists are annotated using the \"list\" keyword\na: list = [\"Some string\", 15, 9.6, True]\n\n# Initializing an empty list:\na = []\n####\n\n####\n# Sets are lists that cannot contain duplicate values\n# They also do not keep the order of the variables\n# Sets are a lot faster than lists when looking for specific values\n# Sets are annotated using the \"set\" keyword\na: set = {\"Some string\", 15, 9.6, True}\n\n# Initializing an empty set:\na = set()\n####\n\n####\n# Tuples are lists that cannot be changed after being created\n# They are useful when returning multiple values from a function\n# Tuples are annotated using the \"tuple\" keyword\na: tuple = (\"Some string\", 15, 9.6, True)\n\n# Because tuples cannot be changed there is no point in initializing an empty one\n####\n\n####\n# Dictionaries work of Key - Value pairs\n# Keys can be any immutable type - integers, floats, strings and tuples\n# Dictionaries do not keep the their order\n# Dictionaries are annotated using the \"dict\" keyword\na: dict = {\"A\": \"Some string\", \"B\": 15, \"C\": 9.6, \"D\": True}\n\n# Initializing an empty dictionary:\na = {}\n####```\n\n#### Casting\n\nKeywords: Casting\n\n```# Only certain operations can be performed on certain types of variables\n# Changing between types of variables is called \"casting\"\na: str = \"15\"\n\nb: int = int(a) # b = 15\nc: float = float(b) # c = 15.0\nd: str = str(c) # d = \"15.0\"```\n\n#### Mathematical operations\n\n```####\n# Integers and floats support most arithmetic operations\na = 15\nb = 6\n\nc = a + b # c = 21\nc = a - b # c = 9\nc = a * b # c = 90\nc = a / b # c = 2.5\n\n# There are some special operators:\n# modulus (%) returns what is left after division\n# floor division (//) throws away the decimal place\nc = a % b # c = 3\nc = a // b # c = 2\n####\n\n####\n# Strings only support 2 mathematical operators\n# The addition (+) is used to concatenate strings\n# The multiplication (*) returns multiples of a string\na = \"Hello\"\nb = \"World\"\n\nc = a + b # c = \"HelloWorld\"\nc = a * 3 # c = \"HelloHelloHello\"\n####\n\n####\n# Lists, tuples can be added to each other\n# It concatenates the collections together\na = [\"Hello\", 15]\nb = [9.6, True]\n\nc = a + b # c = [\"hello\", 15, 9.6, True]\n####\n\n####\n# You can use mathematical shorthand to make some code more readable\n# Shorthand is supported for all operators and works with both strings and numbers\na = 12\nb = \"Hello\"\nc = \"World\"\n\na = a + 6 # Normal way\na += 6 # Shorthand gives the same result\n\nb = b + c # Normal way, c = \"HelloWorld\"\nb += c # Shorthand gives the same result\n####```\n\n#### Modifying collections\n\n```####\n# LISTS\n# You can add values to a list by using the append() method:\na = []\n\na.append(\"Hello\")\na.append(15)\na.append(True)\n\nprint(a) # a = [\"Hello\", 15, True]\n####\n\n####\n# SETS\n# You can add values to a list by using the add() method\n# A list of values can be added to a set by using the update() method\na = set()\nb = [15, 9.6, True]\n\na.update(b)\n\nprint(a) # a = {\"Hello\", 15, 9.6, True}\n####\n\n####\n# TUPLES\n# Tuples cannot be added to, they do not change\n####\n\n####\n# DICTIONARIES\n# You can add a new value pair by assigning the key to a value\na = {}\n\na[\"A\"] = \"Hello\"\na[\"B\"] = 15\na[\"C\"] = 9.6\na[\"D\"] = True\n\nprint(a) # a = {\"A\": \"Hello\", \"B\": 15, \"C\": 9.6, \"D\": True}\n####```\n\n### Conditionals\n\n#### If statement\n\n```# If statements are used when you want to do something when a condition is met\n# The if or elif part occurs when the condition returns true\n# The else part occurs when none of the previous conditions returned true\na = 12\nb = 15\n\n# The example below will return \"Did the elif\" because b > a\nif a > b:\nprint(\"Did the if\")\nelif b > a: # Condition returns true\nprint(\"Did the elif\")\nelse:\nprint(\"Did the else\")\n\n# The operators available:\n# > < >= <= ==\n# != (not equal) can be used or the keyword \"not\"\n\n# If you want to create more complex conditions you can make use of \"and\", \"or\" and \"in\"\n# And requires that all conditionals return true\n# Or requires that one of the conditionals return true\n# In returns true if a certain value is found inside a collection\na = [15, 9, 6]\nb = 6\n\nif 15 in a:\nprint(\"Found it\")\n\n# Would not run the code inside the if statement\nif (15 in a) and (b > 9):\nprint(\"The item is in the list and 6 > 9\")\n\n# Would run the code inside the if statement\nif (15 in a) or (b > 9):\nprint(\"The item is in the list or 6 > 9?\")\n\n# You can have calculations and function calls inside conditions:\nif (15 + 2) > b:\nprint(\"17 > 6\")```"
] | [
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https://studylib.net/doc/13570359/homework--4%3B--due--thursday---oct.-...-g-z--which-consist. | [
"# Homework 4; due Thursday, Oct. ... g Z) which consists of matrices N",
null,
"```Homework 4; due Thursday, Oct. 17\n1. Prove Mumford’s theorem (see the notes, Th. 5.2) for g = 1.\n2. Let Γ(N ) be the congruence subgroup of SL2 (Z) which consists of matrices\nequal to 1 modulo N .\n(a) Show that Γ(N ) is free for N ≥ 3. (Hint: consider the action of Γ(N ) on\nthe upper half plane).\n(b) Find the number of generators of Γ(N ) which generate it without relations.\n(Hint: compute χ(Γ(N ))).\n3. Let Γ be the group defined by the generators a, b, c with relations ab = ba.\nFind the Euler characteristic of Γ.\n1\n```"
] | [
null,
"https://s2.studylib.net/store/data/013570359_1-4a3f24e75c2fa773cc99b523663b1902.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85202336,"math_prob":0.9934174,"size":520,"snap":"2020-34-2020-40","text_gpt3_token_len":166,"char_repetition_ratio":0.11821705,"word_repetition_ratio":0.0,"special_character_ratio":0.32115385,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995877,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-18T11:45:54Z\",\"WARC-Record-ID\":\"<urn:uuid:efe60634-4bab-44ef-ad38-1b31a48d5001>\",\"Content-Length\":\"65940\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:605c6550-47c4-44bc-8fcf-5e095578ec3c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c9e3c2ea-623b-4d44-8174-ed870c1ec5c4>\",\"WARC-IP-Address\":\"172.67.175.240\",\"WARC-Target-URI\":\"https://studylib.net/doc/13570359/homework--4%3B--due--thursday---oct.-...-g-z--which-consist.\",\"WARC-Payload-Digest\":\"sha1:QVXBF7XDCIKD3JL2MPH3ZMOOLXLJRC7D\",\"WARC-Block-Digest\":\"sha1:5ANNF2AL6NNXVMGAN227KSYUCC72BEPN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400187390.18_warc_CC-MAIN-20200918092913-20200918122913-00539.warc.gz\"}"} |
https://pypi.org/project/sparkflow/ | [
"Deep learning on Spark with Tensorflow\n\n# SparkFlow\n\nThis is an implementation of TensorFlow on Spark. The goal of this library is to provide a simple, understandable interface in using TensorFlow on Spark. With SparkFlow, you can easily integrate your deep learning model with a ML Spark Pipeline. Underneath, SparkFlow uses a parameter server to train the TensorFlow network in a distributed manner. Through the api, the user can specify the style of training, whether that is Hogwild or async with locking.\n\n## Why should I use this?\n\nWhile there are other libraries that use TensorFlow on Apache Spark, SparkFlow's objective is to work seamlessly with ML Pipelines, provide a simple interface for training TensorFlow graphs, and give basic abstractions for faster development. For training, SparkFlow uses a parameter server which lives on the driver and allows for asynchronous training. This tool provides faster training time when using big data.\n\n## Installation\n\nInstall SparkFlow via pip: `pip install sparkflow`\n\nSparkFlow requires Apache Spark >= 2.0, flask, dill, and TensorFlow to be installed. As of sparkflow >= 0.7.0, only python >= 3.5 will be supported.\n\n## Example\n\n#### Simple MNIST Deep Learning Example\n\n```from sparkflow.graph_utils import build_graph\nfrom sparkflow.tensorflow_async import SparkAsyncDL\nimport tensorflow as tf\nfrom pyspark.ml.feature import VectorAssembler, OneHotEncoder\nfrom pyspark.ml.pipeline import Pipeline\n\n#simple tensorflow network\ndef small_model():\nx = tf.placeholder(tf.float32, shape=[None, 784], name='x')\ny = tf.placeholder(tf.float32, shape=[None, 10], name='y')\nlayer1 = tf.layers.dense(x, 256, activation=tf.nn.relu)\nlayer2 = tf.layers.dense(layer1, 256, activation=tf.nn.relu)\nout = tf.layers.dense(layer2, 10)\nz = tf.argmax(out, 1, name='out')\nloss = tf.losses.softmax_cross_entropy(y, out)\nreturn loss\n\nmg = build_graph(small_model)\n#Assemble and one hot encode\nva = VectorAssembler(inputCols=df.columns[1:785], outputCol='features')\nencoded = OneHotEncoder(inputCol='_c0', outputCol='labels', dropLast=False)\n\nspark_model = SparkAsyncDL(\ninputCol='features',\ntensorflowGraph=mg,\ntfInput='x:0',\ntfLabel='y:0',\ntfOutput='out:0',\ntfLearningRate=.001,\niters=20,\npredictionCol='predicted',\nlabelCol='labels',\nverbose=1\n)\n\np = Pipeline(stages=[va, encoded, spark_model]).fit(df)\np.write().overwrite().save(\"location\")\n```\n\nFor a couple more, visit the examples directory. These examples can be run with Docker as well from the provided Dockerfile and Makefile. This can be done with the following command:\n\n```make docker-build\nmake docker-run-dnn\n```\n\nOnce built, there are also commands to run the example CNN and an autoencoder.\n\n## Documentation\n\nSince saving and loading custom ML Transformers in pure python has not been implemented in PySpark, an extension has been added here to make that possible. In order to save a Pyspark Pipeline with Apache Spark, one will need to use the overwrite function:\n\n```p = Pipeline(stages=[va, encoded, spark_model]).fit(df)\np.write().overwrite().save(\"location\")\n```\n\nFor loading, a Pipeline wrapper has been provided in the pipeline_utils file. An example is below:\n\n```from sparkflow.pipeline_util import PysparkPipelineWrapper\nfrom pyspark.ml.pipeline import PipelineModel\n\n```\n\nThen you can perform predictions, etc with:\n\n```predictions = p.transform(df)\n```\n\n#### Serializing Tensorflow Graph for SparkAsyncDL\n\nYou may have already noticed the build_graph function in the example above. This serializes the Tensorflow graph for training on Spark. The build_graph function only takes one parameter, which is a function that should include the Tensorflow variables. Below is an example Tensorflow graph function:\n\n```def small_model():\nx = tf.placeholder(tf.float32, shape=[None, 784], name='x')\ny = tf.placeholder(tf.float32, shape=[None, 10], name='y')\nlayer1 = tf.layers.dense(x, 256, activation=tf.nn.relu)\nlayer2 = tf.layers.dense(layer1, 256, activation=tf.nn.relu)\nout = tf.layers.dense(layer2, 10)\nz = tf.argmax(out, 1, name='out')\nloss = tf.losses.softmax_cross_entropy(y, out)\nreturn loss\n```\n\nThen to use the build_graph function:\n\n```from sparkflow.graph_utils import build_graph\nmg = build_graph(small_model)\n```\n\n#### Using SparkAsyncDL and Options\n\nSparkAsyncDL has a few options that one can use for training. Not all of the parameters are required. Below is a description of each of the parameters:\n\n``````inputCol: Spark dataframe inputCol. Similar to other spark ml inputCols\ntensorflowGraph: The protobuf tensorflow graph. You can use the utility function in graph_utils to generate the graph for you\ntfInput: The tensorflow input. This points us to the input variable name that you would like to use for training\ntfLabel: The tensorflow label. This is the variable name for the label.\ntfOutput: The tensorflow raw output. This is for your loss function.\ntfOptimizer: The optimization function you would like to use for training. Defaults to adam\ntfLearningRate: Learning rate of the optimization function\niters: number of iterations of training\npredictionCol: The prediction column name on the spark dataframe for transformations\npartitions: Number of partitions to use for training (recommended on partition per instance)\nminiBatchSize: size of the mini batch. A size of -1 means train on all rows\nminiStochasticIters: If using a mini batch, you can choose number of mini iters you would like to do with the batch size above per epoch. A value of -1 means that you would like to run mini-batches on all data in the partition\nacquireLock: If you do not want to utilize hogwild training, this will set a lock\nshufflePerIter: Specifies if you want to shuffle the features after each iteration\ntfDropout: Specifies the dropout variable. This is important for predictions\ntoKeepDropout: Due to conflicting TF implementations, this specifies whether the dropout function means to keep a percentage of values or to drop a percentage of values.\nverbose: Specifies log level of training results\nlabelCol: Label column for training\npartitionShuffles: This will shuffle your data after iterations are completed, then run again. For example,\nif you have 2 partition shuffles and 100 iterations, it will run 100 iterations then reshuffle and run 100 iterations again.\nThe repartition hits performance and should be used with care.\noptimizerOptions: Json options to apply to tensorflow optimizers.\n``````\n\n#### Optimization Configuration\n\nAs of SparkFlow version 0.2.1, TensorFlow optimization configuration options can be added to SparkAsyncDL for more control over the optimizer. While the user can supply the configuration json directly, there are a few provided utility functions that include the parameters necessary. An example is provided below.\n\n```from sparkflow.graph_utils import build_adam_config\n\nspark_model = SparkAsyncDL(\n...,\n)\n```\n\nTo load a pre-trained Tensorflow model and use it as a spark pipeline, it can be achieved using the following code:\n\n```from sparkflow.tensorflow_model_loader import load_tensorflow_model\n\ninputCol=\"features\",\ntfInput=\"x:0\",\ntfOutput=\"out/Sigmoid:0\"\n)\n```\n\n## Running\n\nOne big thing to remember, especially for larger networks, is to add the `--executor cores 1` option to spark to ensure each instance is only training one copy of the network. This will especially be needed for gpu training as well.\n\n## Contributing\n\nContributions are always welcome. This could be fixing a bug, changing documentation, or adding a new feature. To test new changes against existing tests, we have provided a Docker container which takes in an argument of the python version. This allows the user to check their work before pushing to Github, where travis-ci will run.\n\nFor 2.7 (sparkflow <= 0.6.0):\n\n``````docker build -t local-test --build-arg PYTHON_VERSION=2.7 .\ndocker run --rm local-test:latest bash -i -c \"python tests/dl_runner.py\"\n``````\n\nFor 3.6\n\n``````docker build -t local-test --build-arg PYTHON_VERSION=3.6 .\ndocker run --rm local-test:latest bash -i -c \"python tests/dl_runner.py\"\n``````\n\n## Future planned features\n\n• Hyperopt implementation for smaller and larger datasets\n• AWS EMR guides\n\n## Project details\n\nThis version",
null,
"0.7.0",
null,
"0.6.0",
null,
"0.5.0",
null,
"0.4.2",
null,
"0.4.1",
null,
"0.4.0",
null,
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null,
"0.3.0",
null,
"0.2.2",
null,
"0.2.1",
null,
"0.2",
null,
"0.1.1",
null,
"0.1"
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null,
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null,
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null,
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6980349,"math_prob":0.779039,"size":9121,"snap":"2021-04-2021-17","text_gpt3_token_len":2065,"char_repetition_ratio":0.11407261,"word_repetition_ratio":0.08092485,"special_character_ratio":0.21762964,"punctuation_ratio":0.17307693,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98233473,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-22T03:09:09Z\",\"WARC-Record-ID\":\"<urn:uuid:e161b728-fb8d-4833-9972-76c091644020>\",\"Content-Length\":\"72537\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8d83bc94-63f8-4a86-8b6a-d58ba2d65e93>\",\"WARC-Concurrent-To\":\"<urn:uuid:fa538e5a-d333-4ba1-833f-61735c68e3f8>\",\"WARC-IP-Address\":\"151.101.64.223\",\"WARC-Target-URI\":\"https://pypi.org/project/sparkflow/\",\"WARC-Payload-Digest\":\"sha1:LAZFJR5HIR22RQYK6NNPJP5DGF3SMI3G\",\"WARC-Block-Digest\":\"sha1:A5NMFQ66F7OMGCVE6SJWRAC6NLDNFZAG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703529080.43_warc_CC-MAIN-20210122020254-20210122050254-00481.warc.gz\"}"} |
https://sillycodes.com/even-odd-program-in-c-language/ | [
"# Check Even or Odd number program in C\n\n## Program Description:\n\nWrite a C Program to check if the given number is an Even number or Odd number ( an Even odd program ).\n\nInput:\n\nOutput:\n\nInput:\n\nOutput:\n\n## Even or Odd Number Program Logic:\n\nAny Number that can be divided exactly by 2 is an Even number. Even numbers give zero as the remainder when divided by 2.\n\nEven Number Examples: 2, 4, 50, 100, etc.\n\nSimilarly, any number which is not exactly divided by 2 is called Odd number. Odd numbers give 1 as the remainder when divided by 2.\n\nOdd Number Examples: 5, 31, 59, 99, etc.\n\nWe are going to use the above logic that Even numbers give us the zero remainder and Odd number gives one (1) as remainder when divided by 2.\n\nWe can get the remainder of any integer using the Modulus operator (%).\n\nIf the Modulus of input number is zero, Then the Input number is Even number.\n\nIf the Modulus of input number is One(1), Then the Input number is Odd number.\n\nLet’s convert above logic into the code.\n\n## Program Output:\n\nWe are using gcc compiler in Linux OS.\n\nAs you can see from above output, We are getting the expected output. Please play around with program by providing the different inputs.\n\nLet’s write another program to display all Even numbers and Odd numbers from 0 to 100 using loops in C language.\n\n## Example 1: Program to display all Even Numbers and Odd numbers from 0 to 100.\n\nWe are going to use the same logic as above. i.e To use the modulus operator to calculate the remainder. But instead of checking for one value, We are going to print all even and odd numbers between 0 and 100.\n\nWe are going to use the for loop to iterate over numbers.\n\n## Related Math Programs:",
null,
""
] | [
null,
"https://secure.gravatar.com/avatar/336c59ff684d47a0f7db15f8ae560692",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8621608,"math_prob":0.989721,"size":5926,"snap":"2021-43-2021-49","text_gpt3_token_len":1872,"char_repetition_ratio":0.34126985,"word_repetition_ratio":0.16770671,"special_character_ratio":0.36685792,"punctuation_ratio":0.0752924,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99972945,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-25T10:42:13Z\",\"WARC-Record-ID\":\"<urn:uuid:6e2c05e1-87f4-48bf-866c-85fdbb162947>\",\"Content-Length\":\"217181\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e1104f43-a656-4b57-aac0-27a5fae804f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b1cba68-0af9-49d7-ad65-5d991618f97a>\",\"WARC-IP-Address\":\"139.59.43.226\",\"WARC-Target-URI\":\"https://sillycodes.com/even-odd-program-in-c-language/\",\"WARC-Payload-Digest\":\"sha1:FWVYJAAOEG6SXS2DFDTOE55PZ6MKN27Q\",\"WARC-Block-Digest\":\"sha1:VGFIGDRLI4KZSXMTYK5UVLJ2J2WNCGA5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587659.72_warc_CC-MAIN-20211025092203-20211025122203-00614.warc.gz\"}"} |
https://learnonline.edu.lk/course/?current_page=1&search=&type=all&order=desc&orderby=recent&filter-categories=all&filter-instructors=60&view=grid%22 | [
"• 8 Lessons\n\n• 8 Lessons\n\n• 5 Lessons\n\n• 4 Lessons\n\n• 7 Lessons\n\n• 5 Lessons\n\n• 5 Lessons\n\n• 6 Lessons\n\n• 7 Lessons\n\n• 7 Lessons\n\n• 5 Lessons\n\n• 5 Lessons\n\n• 4 Lessons\n\n• 4 Lessons\n\n• 5 Lessons\n\n• 4 Lessons\n\n• 5 Lessons\n\n• 5 Lessons\n\n• 4 Lessons\n\n• 9 Lessons"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.50188893,"math_prob":0.4409169,"size":349,"snap":"2022-27-2022-33","text_gpt3_token_len":122,"char_repetition_ratio":0.034782607,"word_repetition_ratio":0.0,"special_character_ratio":0.08882522,"punctuation_ratio":0.08,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990846,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T07:58:33Z\",\"WARC-Record-ID\":\"<urn:uuid:81e61320-03f6-47b5-951c-b6c3d46693e7>\",\"Content-Length\":\"375255\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3c1b021-2c92-4855-a1d5-4c98564fbf80>\",\"WARC-Concurrent-To\":\"<urn:uuid:292538dd-9e80-4709-9319-3a26549f020b>\",\"WARC-IP-Address\":\"104.26.6.4\",\"WARC-Target-URI\":\"https://learnonline.edu.lk/course/?current_page=1&search=&type=all&order=desc&orderby=recent&filter-categories=all&filter-instructors=60&view=grid%22\",\"WARC-Payload-Digest\":\"sha1:4MEZDT6XCIMMMN5MYO3R5M2TXSJBIYFV\",\"WARC-Block-Digest\":\"sha1:WO6XTTWTMZD7NLI6OQX6ZKP7SW4SRAMK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103922377.50_warc_CC-MAIN-20220701064920-20220701094920-00107.warc.gz\"}"} |
https://proofwiki.org/wiki/Eigenspace_for_Normal_Operator_is_Reducing_Subspace | [
"# Eigenspace for Normal Operator is Reducing Subspace\n\n## Theorem\n\nLet $\\HH$ be a Hilbert space over $\\Bbb F \\in \\set {\\R, \\C}$.\n\nLet $A \\in \\map B \\HH$ be a normal operator.\n\nLet $\\lambda \\in \\Bbb F$.\n\nThen $\\map \\ker {A - \\lambda}$ is a reducing subspace for $A$.\n\nHere $\\ker$ denotes kernel.\n\n## Proof\n\nWe are given that $A$ is normal.\n\n$\\ker A = {\\Rng A^\\perp$\n\nand in particular, that:\n\n$\\ker A \\subseteq \\Rng A^\\perp$\n$\\Rng A \\subseteq \\ker A^\\perp$\n\nApplying this to the normal operator $A - \\lambda$, we find:\n\n$\\Rng {A - \\lambda} \\subseteq \\paren {\\map \\ker {A - \\lambda} }^\\perp$\n\nWe are now set up to prove that $\\map \\ker {A - \\lambda}$ is a reducing subspace for $A$.\n\nLet $x \\in \\map \\ker {A - \\lambda}$.\n\nThen:\n\n $\\ds A x$ $=$ $\\ds \\lambda x + \\paren {A - \\lambda} x$ $\\ds$ $=$ $\\ds \\lambda x$ Definition of Kernel of Linear Transformation, $x \\in \\map \\ker {A - \\lambda}$\n\nTherefore:\n\n$A \\map \\ker {A - \\lambda} \\subseteq \\map \\ker {A - \\lambda}$\n\nthat is to say:\n\n$\\map \\ker {A - \\lambda}$ is $A$-invariant.\n\nNow, let $x \\in \\paren {\\map \\ker {A - \\lambda} }^\\perp$.\n\nObserve that:\n\n$A x = \\lambda x + \\paren {A - \\lambda} x$\n\nNow $\\paren {A - \\lambda} x \\in \\Rng {A - \\lambda}$, and by our derivation above, this means that:\n\n$\\paren {A - \\lambda} x \\in \\paren {\\map \\ker {A - \\lambda} }^\\perp$\n\nIn conclusion, since $\\paren {\\map \\ker {A - \\lambda} }^\\perp$ is a linear subspace of $H$, it follows that:\n\n$\\lambda x + \\paren {A - \\lambda} x \\in \\paren {\\map \\ker {A - \\lambda} }^\\perp$\n\nas desired.\n\nHence both $\\map \\ker {A - \\lambda}$ and $\\paren {\\map \\ker {A - \\lambda} }^\\perp$ have been shown to be $A$-invariant subspaces of $H$.\n\nThat is, $\\map \\ker {A - \\lambda}$ is a reducing subspace for $A$.\n\n$\\blacksquare$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6009822,"math_prob":0.99999845,"size":2002,"snap":"2021-43-2021-49","text_gpt3_token_len":693,"char_repetition_ratio":0.25575575,"word_repetition_ratio":0.19390582,"special_character_ratio":0.37362638,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000086,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T19:08:42Z\",\"WARC-Record-ID\":\"<urn:uuid:80c5f319-7a74-446a-be17-df5d1cb49f33>\",\"Content-Length\":\"39795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8672f119-84c8-44be-8bcb-5efd8a28f702>\",\"WARC-Concurrent-To\":\"<urn:uuid:49a2ddd7-688c-4891-b044-0b351155dd3c>\",\"WARC-IP-Address\":\"172.67.198.93\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Eigenspace_for_Normal_Operator_is_Reducing_Subspace\",\"WARC-Payload-Digest\":\"sha1:42LSYZR5NOAJCCOVWR4Y6ZDKEFSSPILS\",\"WARC-Block-Digest\":\"sha1:HROUYMIKM54QVQNBZ7MEZAUQ4CIWUAOP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587593.0_warc_CC-MAIN-20211024173743-20211024203743-00081.warc.gz\"}"} |
https://linuxgenie.net/add-rows-pandas-data-frame/ | [
"",
null,
"# How to Add Rows to Pandas Data Frame?\n\nPandas is a powerful tool for manipulating and analyzing data, and one of the most common tasks you’ll need to do with Pandas is adding or inserting rows to a Data Frame. Whether working with a small dataset or a large one, adding rows to a Data Frame is an essential skill for any programmer/data analyst.\n\nThis Python post will explain all the possible methods to add rows to Pandas Data Frame.\n\n## How to Add/Insert Rows into a Data Frame?\n\nMultiple approaches are used in Python to add or insert a row to the Pandas DataFrame. The most frequently used approaches are the append() function, pandas.concat(), and the DataFrame.loc method. Let’s discuss each approach one by one.\n\n### Approach #1: Add Rows Into a Data Frame Using the append()\n\nThe most popularly used function for inserting a row into a data frame is the append() function. It appends/inserts a row at the available space.\n\n#### Code:\n\nFirst, you need to import the Pandas library; then, create a dictionary and pass it to the pandas.DataFrame() to get a 2-D Pandas dataframe. After that, use the append() function to add a new row at the end of the given dataframe.\n\n``````import pandas\n\ninputVal = {'empName':['Paul', 'Vince', 'Amanda'],\n'empAge':[34, 25, 35],\n'empDesig':['CEO', 'HR', 'Project Manager']\n}\n\nnewVal = {'empName':'Jordan', 'empAge':29, 'empDesig':'Designer'}\ndataFrame = dataFrame.append(newVal, ignore_index=True)\n\nprint('\\n Modified Dataframe: \\n', dataFrame)\n``````\n\n#### Output:",
null,
"### Approach #2: Add Rows Into Data Frame Using DataFrame.loc\n\nThe “.loc” approach adds/appends a single row at the bottom of the selected DataFrame. You can use the len() function to determine the location/position at which you want to append a new row. The following coding example will demonstrate how to add/append a row at the end of a DataFrame.\n\n#### Code:\n\nFirst, you must import the Pandas library; then, create a dictionary and pass it to the pandas.DataFrame() to get a 2-D Pandas dataframe. Finally, use the dataframe.loc() method to append a new row at the bottom/end of the selected dataframe.\n\n``````import pandas\n\ninputVal = {'empName':['Paul', 'Vince', 'Amanda'],\n'empAge':[34, 25, 35],\n'empDesig':['CEO', 'HR', 'Project Manager']\n}\n\ndataFrame = pandas.DataFrame(inputVal)\nprint('Original Dataframe: \\n', dataFrame)\n\ndataFrame.loc[len(dataFrame.index)] = ['Jordan', 29, 'Programmer']\nprint('\\n Modified Dataframe: \\n', dataFrame)\n``````\n\n#### Output:",
null,
"### Approach #3: Add Rows Into Data Frame Using pandas.concat()\n\nIn Python, the “pandas.concat()” function takes two dataframes and returns a single combined or concatenated dataframe. If you want to insert a new row into a dataframe using the concat() function, you must insert the new row into a new dataframe and then concatenate it with the given dataframe.\n\n#### Code:\n\nFirst, you must import the Pandas library; then, create two dictionaries and use the pandas.DataFrame() to get 2-D Pandas data frames. Finally, use the concat() function to add a new row at the end of the given dataframe.\n\n``````import pandas\n\ninputVal = {'empName':['Paul', 'Vince', 'Amanda'],\n'empAge':[34, 25, 35],\n'empDesig':['CEO', 'HR', 'Project Manager']\n}\n\ndataFrame = pandas.DataFrame(inputVal)\nprint('Original Dataframe: \\n', dataFrame)\n\ninputVal_1 = {'empName':['Jordan'], 'empAge':, 'empDesig':['Designer']}\ndataFrame_1 = pandas.DataFrame(inputVal_1)\n\nnewDataFrame = pandas.concat([dataFrame, dataFrame_1], ignore_index = True)\nprint('\\n Modified Dataframe: \\n', newDataFrame)\n``````\n\n#### Output:",
null,
"## Conclusion\n\nIn Python, the append() function, pandas.concat() function, and the DataFrame.loc method is used to add or insert a new row into a data frame. The pandas.DataFrame() is used to get a 2-D Pandas data frame. After that, the append() and DataFrame.loc methods are used to append or insert a new row to the given data frame, while the concat() function is used to concatenate/combine two data frames. This tutorial explained using different functions to insert or add a new row into a data frame.\n\nVisit Linux Genie for more Python tutorials."
] | [
null,
"https://linuxgenie.net/wp-content/uploads/2023/01/how-to-add-rows-to-pandas-data-frame.jpg",
null,
"https://lh5.googleusercontent.com/hBMSUGfiaCk8NVmL1RCFrjK-yglxzFo19k-IImSCjUTH_gggMSxa95AZdwT7ei0DhX4zV1qymfHrsq-SKQjWorw613tK8MMu_T2amCs_qoiLLxUyQFYlg3FFgLJbFgBy-jyqI1q7FF1v27nWG0Yzw6z5YoWmUJv98qAcWpHwE7wl7eRoubRDIYfwrF4zog",
null,
"https://lh4.googleusercontent.com/KMP_lRQf0tqwAuHXWpv3GQHJWavc-lL10CWNCQu1bTrNrQoliVsfh9MYRwyqkIVzAgTF9rtT0RTTHaV81gYPQnBxcf5yJPjJh0H3L9Z0jsRgFgIT8d5rfTXNEjl5Wk8rUTdjEsZ2_NxjNnfErTfBPh2wBrydPIoNPkS6XrdtORjy5NwTxIwacUi1EGNSfA",
null,
"https://lh6.googleusercontent.com/z0KM2Eg8dl0SlBgRR-AqarpWdn-rSwYSbN1H5ZP68dG3i2GbZbAXHwW5bsdXEbmpAjn1yNZ-u_D8KM1ehK5gNGn6L_6tRuai4hU1eID7KWSoUXWuOW6o1Vk18VfUK6W68zwGeZ1UuS_6sY166Rph-9zO0hnuG9r0S2W6QJm6MCzFJm3MHIvm2r1phlVgBw",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5965731,"math_prob":0.7175112,"size":3999,"snap":"2023-40-2023-50","text_gpt3_token_len":969,"char_repetition_ratio":0.16470589,"word_repetition_ratio":0.23865546,"special_character_ratio":0.2515629,"punctuation_ratio":0.17430025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99422884,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T13:50:25Z\",\"WARC-Record-ID\":\"<urn:uuid:5c89e47a-eda2-4b01-892a-c762d089166a>\",\"Content-Length\":\"163342\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:006f058d-f74a-4de3-8704-3607a9c765bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:fcae58b4-2a2e-4cf8-8957-86527a380e05>\",\"WARC-IP-Address\":\"172.67.171.74\",\"WARC-Target-URI\":\"https://linuxgenie.net/add-rows-pandas-data-frame/\",\"WARC-Payload-Digest\":\"sha1:NWYO2AKU7UAZI5EA3MI5W7OKJYMQWGT2\",\"WARC-Block-Digest\":\"sha1:KR4D32WQTROKRH4KFONE57I2GRQNMGX2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100227.61_warc_CC-MAIN-20231130130218-20231130160218-00693.warc.gz\"}"} |
https://www.r-bloggers.com/2018/08/automating-exploratory-plots-with-ggplot2-and-purrr/ | [
"Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nWhen you have a lot of variables and need to make a lot exploratory plots it’s usually worthwhile to automate the process in R instead of manually copying and pasting code for every plot. However, the coding approach needed to automate plots can look pretty daunting to a beginner R user. It can look so daunting, in fact, that it can appear easier to manually make the plots (like in Excel) rather than using R at all.\n\nUnfortunately making plots manually can backfire. The efficiency of using a software program you already know is quickly out-weighed by being unable to easily reproduce the plots when needed. I know I invariably have to re-make even exploratory plots, and it’d be a bummer if I had to remake them all manually rather than re-running some code.\n\nSo while I often assure students working under time constraints that it is perfectly OK to use software they already know rather than spending the time to learn how to do something in R, making many plots is a special case. To get them started I will provide students who need to automate plotting in R some example code (with explanation).\n\nThis post is based on an example I was working on recently, which involves plotting bivariate relationships between many continuous variables.\n\nI’ll be plotting with ggplot2 and looping with purrr. I’ll also be using package cowplot later to combine individual plots into one, but will use the package functions via `cowplot::` instead of loading the package.\n\n```library(ggplot2) # v. 3.2.0\nlibrary(purrr) # v. 0.3.2```\n\n# The set-up\n\nToday I’m going to make an example dataset with 3 response (`y`) variables and 4 explanatory (`x`) variables for plotting. (The real dataset had 9 response and 9 explanatory variables.)\n\n```set.seed(16)\ndat = data.frame(elev = round( runif(20, 100, 500), 1),\nresp = round( runif(20, 0, 10), 1),\ngrad = round( runif(20, 0, 1), 2),\nslp = round( runif(20, 0, 35),1),\nlat = runif(20, 44.5, 45),\nlong = runif(20, 122.5, 123.1),\nnt = rpois(20, lambda = 25) )\n\n```# elev resp grad slp lat long nt\n# 1 373.2 9.7 0.05 8.8 44.54626 122.8547 18\n# 2 197.6 8.1 0.42 33.3 44.79495 122.5471 26\n# 3 280.0 5.4 0.38 19.3 44.99027 122.9645 18\n# 4 191.8 4.3 0.07 29.6 44.95022 122.7290 19\n# 5 445.4 2.3 0.43 16.5 44.79784 122.9836 15\n# 6 224.5 6.5 0.78 4.1 44.96576 122.9836 21```\n\nThe goal is to make scatterplots for every response variable vs every explanatory variable. I’ve deemed the first three variables in the dataset to be the response variables (`elev`, `resp`, `grad`).\n\nThe plan is to loop through the variables and make the desired plots. I’m going to use vectors of the variable names for this, one vector for the response variables and one for the explanatory variables.\n\nIf all of your response or explanatory variables share some unique pattern in the variable names there are some clever ways to pull out the names with some of the select helper functions in `dplyr::select()`. Alas, my variable names are all unique. My options are to either write the vectors out manually or pull the names out by index. I’ll do the latter since the different types of variables are grouped together.\n\n```response = names(dat)[1:3]\nexpl = names(dat)[4:7]```\n\nWhen I know I’m going to be looping through character vectors I like to use named vectors. This helps me keep track of things in the output.\n\nThe `set_names()` function in purrr is super handy for naming character vectors, since it can use the values of the vector as names (i.e., the vector will be named by itself). (I don’t recommend trying this with lists of data.frames like I have in the past, though, since it turns out that naming a data.frame with a data.frame isn’t so useful. 😆)\n\n```response = set_names(response)\nresponse```\n\n```# elev resp grad\n\n```expl = set_names(expl)\nexpl```\n\n```# slp lat long nt\n# \"slp\" \"lat\" \"long\" \"nt\"```\n\n# Create a plotting function\n\nSince I’m going to make a bunch of plots that will all have the same basic form, I will make a plotting function. I am going to make a function where only the `x` and `y` variables can vary (so are arguments to the function).\n\nSince I’m making a function to plot variables from a single dataset I’m going to hard-code the dataset into the function. If you have multiple datasets or you are making a function for use across projects you’ll probably want to add the dataset as a function argument.\n\nMy functions inputs are based on the variable names, so I need to pass strings into the ggplot2 functions. Strings cannot be used directly in `aes()`, but can be used with the `.data` pronoun.\n\nI’m making pretty basic graphs since these are exploratory plots, not publication-ready plots. I will make a scatterplot and add locally weighted regression (loess) lines via `geom_smooth()`. I use such lines with great caution, as it can be easy to get too attached any pattern the loess line shows.\n\n```scatter_fun = function(x, y) {\nggplot(dat, aes(x = .data[[x]], y = .data[[y]]) ) +\ngeom_point() +\ngeom_smooth(method = \"loess\", se = FALSE, color = \"grey74\") +\ntheme_bw() +\nlabs(x = x,\ny = y)\n}```\n\nIf using older versions of ggplot2 (and/or rlang), use the now deprecated `aes_string()` for working with strings.\n\n```scatter_fun = function(x, y) {\nggplot(dat, aes_string(x = x, y = y ) ) +\ngeom_point() +\ngeom_smooth(method = \"loess\", se = FALSE, color = \"grey74\") +\ntheme_bw()\n}```\n\nHere’s an example of the function output, passing in `x` and `y` as strings.\n\n`scatter_fun(\"lat\", \"elev\")`",
null,
"# Looping through one vector of variables\n\nOne way to make all the plots I want is to loop through each explanatory variable for a fixed response variable. With this approach I would need a separate loop for each response variable.\n\nI will use `map()` from package purrr for the looping.\n\nI pass each explanatory variable to the first argument in `scatter_fun()` and I fix the second argument to `\"elev\"`. I use the formula coding in `map()` and so refer to the element of the explanatory vector via `.x` within `scatter_fun()`.\n\n`elev_plots = map(expl, ~scatter_fun(.x, \"elev\") )`\n\nThe output is a list of 4 plots (since there are 4 explanatory variables). You’ll notice that each element of the list has the variable name associated with it. This is why I used `set_names()` earlier, since this is convenient for printing the plots and, you’ll see later, is convenient when saving the plots in files with understandable names.\n\n`elev_plots`\n\n`# \\$slp`",
null,
"```#\n# \\$lat```",
null,
"```#\n# \\$long```",
null,
"```#\n# \\$nt```",
null,
"# Looping through both vectors\n\nFor only a few response variables we could easily copy and paste the code above, changing the hard-coded response variable each time. This process can get burdensome if there are a lot of response variables, though.\n\nAnother option is to loop through both vectors of variables and make all the plots at once. Because we want a plot for each combination of variables, this is a job for a nested loop. This means one `map()` loop will be nested inside another. I will refer to the first `map()` loop as the outer loop and the second one as the inner loop.\n\nI’m going to have the response variables in the outer loop and the explanatory variables in the inner loop. That way I can graph all of the explanatory variables for each response variable before moving on to the next response variable. This puts the output, a nested list, in a logical order.\n\nA nested loop involves more complicated code, of course.. For example, it took some effort for me to wrap my head around how to refer to the list element from the outer loop within the inner loop when using the `map()` formula coding. I found the answers/comments to this question on Stack Overflow to be helpful. Note that one approach is to avoid the formula coding all together and use anonymous functions for either or both the inner and outer loops.\n\nSince my scatterplot function is so simple I ended up using formula coding for the outer loop and the function as is in the inner loop. The inner list elements are fed to the first argument of `scatter_fun()` by default, which works out great since the first argument is the `x` variable and the inner loop loops through the explanatory variables. The `.x` then refers to the outer list elements (the response variable names), and is passed to the `y` argument of the function in the inner loop.\n\n```all_plots = map(response,\n~map(expl, scatter_fun, y = .x) )```\n\nThe output is a list of lists. Each sublist contains all the plots for a single response variable. Because I set the names for both vectors of variable names, the inner and outer lists both have names. These names can be used to pull out individual plots.\n\nFor example, if I want to see all the plots for the `grad` response variable I can print that sublist by name. (I’m going to display only two of four `grad` plots here to save space.)\n\n`all_plots\\$grad[1:2]`\n\n`# \\$slp`",
null,
"```#\n# \\$lat```",
null,
"If I want to print a single plot, I can first extract one of the sublists using an outer list name and then extract the individual plot via an inner list name.\n\n`all_plots\\$grad\\$long`",
null,
"I find the names convenient, but you can also extract plots via position. Here’s the same graph, the third element of the third list.\n\n`all_plots[][]`",
null,
"# Saving the plots\n\nOnce all the graphs are made we can look at them in R by printing the list or parts of the list as above. But if you want to peruse them at your leisure later or send them to a collaborator you’ll want to save them outside of R.\n\nThis next section is dedicated to exploring some of the ways you can do this.\n\n## Saving all plots to one PDF\n\nIf you want to save every plot as a separate page in a PDF, you can do so with the `pdf()` function. The code below shows an example of how this works. First, a graphics device to save the plots into is created and given a name via `pdf()`. Then all the plots are put into that device. Finally, the device is turned off with `dev.off()`. The last step is important, as you can’t open the file until the device is turned off.\n\nThis is a pretty coarse way to save everything, but it allows you to easily page through all the plots. I’ve used this method when I had many exploratory plots for a single response variable that I wanted to share with collaborators.\n\nIn this example code I save the file, which I name `all_scatterplots.pdf`, into the working directory.\n\n```pdf(\"all_scatterplots.pdf\")\nall_plots\ndev.off()```\n\n## Saving groups of plots together\n\nAnother option is to save each group of plots in a separate document. This might make sense in a case like this where there are a set of plots for each response variable and we might want a separate file for each set.\n\nTo save each sublist separately we’ll need to loop through `all_plots` and save the plots for each response variable into a separate file. The list names can be used in the file names to keep the output organized.\n\nThe functions in purrr that start with `i` are special functions that loop through a list and the names of that list simultaneously. This is useful here where we want to use the list names to identify the output files while we save them.\n\nThe `walk()` function is part of the `map` family, to be used when you want a function for its side effect instead of for a return value. Saving plots is a classic example of when we want `walk()` instead of `map()`.\n\nCombining the `i` and the `walk` gives us the `iwalk()` function. In the formula interface, `.x` refers to the list elements and `.y` refers to the names of the list. You can see I create the plot file names using the list name combined with “scatterplots.pdf”, using `_` as the separator.\n\nThe code below makes three files, one for each response variable, with four plots each. The files are named “elev_scatterplots.pdf”, “resp_scatterplots.pdf”, and “grad_scatterplots.pdf”.\n\n```iwalk(all_plots, ~{\npdf(paste0(.y, \"_scatterplots.pdf\") )\nprint(.x)\ndev.off()\n})```\n\n## Saving all plots separately\n\nAll plots can be saved separately instead of combined in a single document. This might be necessary if you want to insert the plots into some larger document later.\n\nWe’ll want to use the names of both the outer and inner lists to appropriately identify each plot we save. I decided to do this by looping through the `all_plots` list and the names of the list via `imap()` to make the file names in a separate step. This time I’m going to save these as PNG files so use `.png` at the end of the file name.\n\nThe result is a list of lists, so I flatten this into a single list via `flatten()`. If I were to use `flatten()` earlier in the process I’d lose the names of the outer list. This process of combining names prior to flattening should be simplified once the proposed `flatten_names()` function is added to purrr.\n\n```plotnames = imap(all_plots, ~paste0(.y, \"_\", names(.x), \".png\")) %>%\nflatten()\nplotnames```\n\n```# []\n# \"elev_slp.png\"\n#\n# []\n# \"elev_lat.png\"\n#\n# []\n# \"elev_long.png\"\n#\n# []\n# \"elev_nt.png\"\n#\n# []\n# \"resp_slp.png\"\n#\n# []\n# \"resp_lat.png\"\n#\n# []\n# \"resp_long.png\"\n#\n# []\n# \"resp_nt.png\"\n#\n# []\n#\n# []\n#\n# []\n#\n# []\n\nOnce the file names are created I can loop through all the file names and plots simultaneously with `walk2()` and save things via `ggsave()`. The height and width of each output file can be set as needed in `ggsave()`.\n\nYou can see I flattened the nested list of plots into a single list to use in `walk2()`.\n\n```walk2(plotnames, flatten(all_plots), ~ggsave(filename = .x, plot = .y,\nheight = 7, width = 7))```\n\n## Combining plots\n\nAnother way to get a set of plots together is to combine them into one plot. How useful this is will depend on how many plots you have per set. This option is a lot like faceting, except we didn’t reshape our dataset to allow the use faceting.\n\nI like the cowplot function `plot_grid()` for combining multiple plots into one. A list of plots can be passed via the `plotlist` argument.\n\nHere’s what that looks like for the first response variable, `elev`.\n\n`cowplot::plot_grid(plotlist = all_plots[])`",
null,
"We can use a loop to combine the plots for each response variable sublist. The result could then be saved using any of the approaches shown above. If you have many subplots per combined plot you likely will want to save the plots at a larger size so the individual plots can be clearly seen.\n\n```response_plots = map(all_plots, ~cowplot::plot_grid(plotlist = .x))\nresponse_plots```\n\n`# \\$elev`",
null,
"```#\n# \\$resp```",
null,
"```#",
null,
"Here’s the code without all the discussion. Copy and paste the code below or you can download an R script of uncommented code from here.\n\n```library(ggplot2) # v. 3.2.0\nlibrary(purrr) # v. 0.3.2\n\nset.seed(16)\ndat = data.frame(elev = round( runif(20, 100, 500), 1),\nresp = round( runif(20, 0, 10), 1),\ngrad = round( runif(20, 0, 1), 2),\nslp = round( runif(20, 0, 35),1),\nlat = runif(20, 44.5, 45),\nlong = runif(20, 122.5, 123.1),\nnt = rpois(20, lambda = 25) )\n\nresponse = names(dat)[1:3]\nexpl = names(dat)[4:7]\n\nresponse = set_names(response)\nresponse\n\nexpl = set_names(expl)\nexpl\n\nscatter_fun = function(x, y) {\nggplot(dat, aes(x = .data[[x]], y = .data[[y]]) ) +\ngeom_point() +\ngeom_smooth(method = \"loess\", se = FALSE, color = \"grey74\") +\ntheme_bw() +\nlabs(x = x,\ny = y)\n}\n\nscatter_fun = function(x, y) {\nggplot(dat, aes_string(x = x, y = y ) ) +\ngeom_point() +\ngeom_smooth(method = \"loess\", se = FALSE, color = \"grey74\") +\ntheme_bw()\n}\n\nscatter_fun(\"lat\", \"elev\")\n\nelev_plots = map(expl, ~scatter_fun(.x, \"elev\") )\nelev_plots\n\nall_plots = map(response,\n~map(expl, scatter_fun, y = .x) )\n\nall_plots[][]\n\npdf(\"all_scatterplots.pdf\")\nall_plots\ndev.off()\n\niwalk(all_plots, ~{\npdf(paste0(.y, \"_scatterplots.pdf\") )\nprint(.x)\ndev.off()\n})\n\nplotnames = imap(all_plots, ~paste0(.y, \"_\", names(.x), \".png\")) %>%\nflatten()\nplotnames\n\nwalk2(plotnames, flatten(all_plots), ~ggsave(filename = .x, plot = .y,\nheight = 7, width = 7))\n\ncowplot::plot_grid(plotlist = all_plots[])\n\nresponse_plots = map(all_plots, ~cowplot::plot_grid(plotlist = .x))\nresponse_plots```"
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https://wiki.seg.org/wiki/Diving_waves | [
"# Diving waves\n\nSeries",
null,
"Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store\n\n## Problem 4.20a\n\nWhen the velocity is a linear function of depth only, as in equation (4.17a), show that the wave will return to the surface again at offset $x$",
null,
"and with traveltime $t$",
null,
"given by equations (4.20a,b), and that the maximum depth of penetration $h_{m}$",
null,
"is given by equation (4.20c).\n\n {\\begin{aligned}x&=(2V_{0}/a)\\sin h(at/2),\\end{aligned}}",
null,
"(4.20a)\n\n {\\begin{aligned}t&=(2/a)\\ln[\\cot(i_{0}/2)],\\end{aligned}}",
null,
"(4.20b)\n\n {\\begin{aligned}h_{m}&=(V_{0}/a)[\\cos \\mathrm {h} (at/2)-1].\\end{aligned}}",
null,
"(4.20c)\n\n### Background\n\nWhen the velocity increases continuously with depth, a wave will eventually return to the surface (see Figures 4.20a,c): such waves are known as diving waves. When the velocity layers are horizontal, as in Figure 4.20a, or spherically symmetrical as in Figure 4.20c, the raypath is symmetrical about the midpoint and the maximum depth occurs at this point.",
null,
"Figure 4.20a. Raypaths where velocity increases linearly with depth.\n\n### Solution\n\nBecause the raypath is symmetrical about the midpoint, we can find $x$",
null,
"and $t$",
null,
"for the midpoint and double the values to get $x$",
null,
"and $t$",
null,
"for the entire path. At the midpoint the ray is traveling horizontally, hence $i=90^{\\circ }$",
null,
". Substituting in equations (4.17b,c), we get\n\n {\\begin{aligned}x&=(2/pa)\\cos i_{0}=(2V_{0}/a)\\cot i_{o},\\\\t&=(2/a)\\ln(\\cot i_{0}/2),\\quad \\cot i_{0}/2=e^{at/2}.\\end{aligned}}",
null,
"(4.20d)\n\nUsing the identity $\\cot \\,2\\phi ={\\frac {1}{2}}(\\cos \\phi -\\tan \\phi )$",
null,
", equation (4.20d) becomes\n\n{\\begin{aligned}x=(2V_{0}/a)(e^{at/2}-e^{-at/2})/2=(2V_{0}/a)\\sinh at/2.\\end{aligned}}",
null,
"The maximum depth $h_{m}$",
null,
"is the value of $z$",
null,
"when $i=90^{\\circ }$",
null,
", so from equation (4.17f) we find that\n\n{\\begin{aligned}h_{m}&=(1/pa)(\\sin i-\\sin i_{0})=(1/pa)(1-\\sin i_{0})\\\\&=(V/a)(1/\\sin i_{0}-1).\\end{aligned}}",
null,
"Using the identity: $\\sin x=2\\tan(x/2)/[1+\\tan ^{2}(x/2)]$",
null,
", we can write\n\n{\\begin{aligned}1/\\sin i_{o}=\\cot(i_{o}/2)[1+\\tan ^{2}(i_{o}/2)]/2\\\\={\\frac {1}{2}}[\\cot(i_{o}/2)+\\tan(i_{o}/2)].\\end{aligned}}",
null,
"Equation (4.20d) now gives\n\n{\\begin{aligned}h_{m}&=(V/a)[(e^{at/2}+e^{-at/2})/2-1]\\\\&=(V/a)(\\cosh at/2-1).\\end{aligned}}",
null,
"",
null,
"Figure 4.20b. Raypath parameter for concentric spherical layering.\n\n## Problem 4.20b\n\nShow that when the constant velocity layers are concentric spherical shells, the raypath parameter $p$",
null,
"(see problem 3.1a) becomes\n\n {\\begin{aligned}p^{\\prime }=r_{n}\\sin i_{n}/V_{n}.\\end{aligned}}",
null,
"(4.20e)\n\n### Solution\n\nEquation (4.20e) is a modification of equation (3.1a) to take into account concentric spherical shells instead of plane parallel interfaces. The angle between a ray and the radii changes as the wave travels downward (see Figure 4.20b) so that the angle of entry into a layer does not equal the incident angle at the base of the layer, that is, $i_{2}\\neq i_{2}^{*}$",
null,
". But $OP=r_{3}\\sin i_{2}^{*}=r_{2}\\sin i_{2}$",
null,
", and Snell’s law becomes $\\sin i_{2}^{*}/V_{2}=\\sin i_{3}/V_{3}$",
null,
", or\n\n{\\begin{aligned}{\\frac {r_{2}\\sin i_{2}}{V_{2}}}={\\frac {r_{3}\\sin i_{3}}{V_{3}}}=p^{\\prime }.\\end{aligned}}",
null,
"## Problem 4.20c\n\nFor concentric spherical layering in the Earth, show that diving waves will return to the surface at the time $t$",
null,
"at the angular distance $\\Delta$",
null,
", where $\\Delta$",
null,
"is the angle subtended at the center of the Earth by the ray $SE$",
null,
", and $R_{e}$",
null,
"is the radius of the Earth; (see Sheriff and Geldart, 1995, 99);\n\n {\\begin{aligned}t=2\\int _{R_{e}-h_{m}}^{R_{e}}{\\frac {\\mathrm {d} r}{V(r)\\{1-[p^{\\prime }V(r)/r]^{2}\\}^{1/2}}},\\end{aligned}}",
null,
"(4.20f)\n\n {\\begin{aligned}\\Delta =2\\int _{R_{e}-h_{m}}^{R_{e}}{\\frac {[p^{\\prime }V(r)/r]\\mathrm {d} r}{r\\{1-[p^{\\prime }V(r)/r]^{2}\\}^{1/2}}}.\\end{aligned}}",
null,
"(4.20g)\n\n### Solution\n\nUsing equation (4.20e) and $\\Delta ABC$",
null,
"in Figure 4.20c, we get\n\n {\\begin{aligned}p^{\\prime }={\\frac {r\\sin i}{V}}=\\left({\\frac {r}{V}}\\right)\\left({\\frac {r\\delta \\Delta }{V\\delta t}}\\right)=\\left({\\frac {r}{v}}\\right)^{2}\\left({\\frac {\\delta \\Delta }{\\delta t}}\\right).\\end{aligned}}",
null,
"(4.20h)\n\n$\\Delta ABC$",
null,
"also gives\n\n {\\begin{aligned}(V\\delta t)^{2}=(\\delta r)^{2}+(r\\delta \\Delta )^{2}.\\end{aligned}}",
null,
"(4.20i)\n\nEliminating $\\delta \\Delta$",
null,
"between equations (4.20h,i) gives\n\n{\\begin{aligned}&(r\\delta \\Delta )^{2}=(V\\delta t)^{2}-(\\delta r)^{2}=(p^{\\prime }V^{2}\\delta t/r)^{2},\\\\&(\\delta t)^{2}[V^{2}-(p^{\\prime }V^{2}/r)^{2}]=(\\delta r)^{2},\\\\\\mathrm {so} \\qquad &\\delta t=(\\delta r/V)[1-(p^{\\prime }V/r)^{2}]^{-1/2}.\\end{aligned}}",
null,
"Integrating from $(R_{e}-h_{m})$",
null,
"to $R_{e}$",
null,
"and multiplying by 2 to allow for the return path gives equation (4.20f).\n\nEliminating $\\delta t$",
null,
"between equations (4.20h,i), we have\n\n{\\begin{aligned}&(\\delta t)^{2}=(1/V^{2})[(\\delta r)^{2}+(r\\delta \\Delta )^{2}]=(r/V)^{4}(\\delta \\Delta /p^{\\prime })^{2},\\\\&(\\delta r/r)^{2}+(\\delta \\Delta )^{2}=(r/p^{\\prime }V)^{2}(\\delta \\Delta )^{2}.\\end{aligned}}",
null,
"Again we multiply by 2 and integrate, obtaining equation (4.20g).\n\n### Alternative solution\n\nEquation (4.20f) is a relation between $t$",
null,
"and $r$",
null,
", so we use $\\Delta ABC$",
null,
"in Figure 4.20c to get a relation between $\\delta t$",
null,
"and $\\delta r$",
null,
", then integrate to get $t$",
null,
". From $\\Delta ABC$",
null,
"and equation (4.20e), we have\n\n{\\begin{aligned}\\cos i=(\\delta r/V\\delta t),\\quad \\delta t=(\\delta r/V)[1-(p^{\\prime }/r)^{2}]^{-1/2},\\end{aligned}}",
null,
"since $\\sin i=p^{\\prime }V/r$",
null,
". Integrating this expression for $\\delta t$",
null,
"gives equation (4.20f).\n\nEquation (4.20g) expresses $\\Delta$",
null,
"in terms of $r$",
null,
", so we use $\\Delta ABC$",
null,
"to get the relation $\\tan i=r\\delta \\Delta /\\delta r$",
null,
". Using equation (4.20e), this becomes\n\n{\\begin{aligned}\\delta \\Delta =(\\delta r/r)(p^{\\prime }V/r)[1-(p^{\\prime }/r)^{2}]^{1/2}.\\end{aligned}}",
null,
"Multiplying by 2 and integrating gives equation (4.20g)."
] | [
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https://homotopytypetheory.org/2012/06/07/a-simpler-proof-that-%CF%80%E2%82%81s%C2%B9-is-z/?replytocom=1668 | [
"## A Simpler Proof that π₁(S¹) is Z\n\nLast year, Mike Shulman proved that π₁(S¹) is Z in Homotopy Type Theory. While trying to understand Mike’s proof, I came up with a simplification that shortens the proof a bunch (100 lines of Agda as compared to 380 lines of Coq).\n\nMy proof uses essentially the same ideas—e.g., the universal cover comes up in the same way, and the functions witnessing the isomorphism are the same. The main new idea is a simpler way of proving the hard direction of “and the composites are the identity”: that if you start with a path on the circle, encode it as its winding number, and then decode it back as a path, then you get the original path. If you already understand Mike’s proof, the diff is essentially this: you can replace steps 2-5 at the end by\n\n1. Define an “encoding” function of type\n```forall (x : circle), (base ~~> x) -> circle_cover x ``` as the transport along circle_cover of 0.\n2. Use J to prove that decoding (Mike’s step 1) after encoding is the identity (this is the hard composite to work with, because it starts and ends with a path). The trick here is to ask the question for paths (base ~~> x) for an arbitrary x, not just for loops base ~~> base, so that you can use J.\n\nThis is clearly similar to Mike’s proof (we could even ask if these two proofs are homotopic!): his proof uses contractibility of a path space, which is exactly J. But the amount of machinery about total spaces that you need here is much less.\n\nA question, though: I came up with this proof basically on type-theoretic grounds (“there must be a simpler proof term of this type”). Is there anything to say about it in geometric or higher-categorical terms? In the semantics, is it the same proof as Mike’s? A known but different proof? Or has the type theory suggested a new way of proving this result?\n\nI’ll explain the proof in detail below the fold.\n\n# Definitions\n\nFirst, the definitions. I have defined the circle so that the elimination form computes definitionally on points (but not paths). This gives the following interface:\n\n``` S¹ : Set\nbase : S¹\nloop : base ≃ base\n\nS¹-rec : {C : Set}\n-> (a : C)\n-> (p : a ≃ a)\n-> S¹ -> C\nS¹-rec a p Base = a -- really an equation\nβloop/rec : {C : Set}\n-> (a : C)\n-> (p : a ≃ a)\n-> resp (S¹-rec a p) loop ≃ p\n\nS¹-elim : {C : S¹ -> Set}\n-> (a : C base)\n(p : subst C loop a ≃ a)\n-> (x : S¹) -> C x\nS¹-elim a p Base = a\nβloop/elim : {C : S¹ -> Set}\n-> (a : C base) (p : subst C loop a ≃ a)\n-> respd (S¹-elim{C} a p) loop ≃ p\n```\n\nNotation: I use Id M N and M ≃ N interchangeably for the identity type/propositional equality/equivalence/paths (apologies to people who have started using ≃ for weak equivalence/adjoint equivalence/etc.—I had started using it as an infix symbol for general propositional equality). See Paths.agda for definitions if you’re confused. The one thing I should comment on is that I’m using the “logical” terminology for\n\n``` subst : {A : Set} (p : A -> Set) {x y : A}\n-> Id x y -> p x -> p y\nresp : {A C : Set} {M N : A} (f : A -> C)\n-> Id M N -> Id (f M) (f N)\nrespd : {A : Set} {C : A -> Set} {M N : A}\n(f : (x : A) -> C x) -> (p : Id M N)\n-> Id (subst C p (f M)) (f N)\n```\n\nso “subst” = “transport” and “resp” = “maponpaths”.\n\nFor integers:\n\n``` data Int : Set where\nPos : Nat -> Int\nZero : Int\nNeg : Nat -> Int\n```\n\nThis definition is nice for what follows, but has the somewhat unfortunate consequence that 1 is Pos 0, -1 is Neg 0, etc.\n\n# The Universal Cover\n\nTo define the universal cover, we’ll need the successor isomorphism succ≃ : Int ≃ Int, which is defined by applying univalence to the successor function succ : Int -> Int, which maps x to x + 1. The universal cover of the circle maps the circle into the universe, taking the base point to Int and the loop to succ≃:\n\n``` C : S¹ -> Set\nC = S¹-rec Int succ≃\n```\n\nTo introduce the proof notation, let’s derive a little fact that we’ll need later: transporting in the cover with the loop is the successor function:\n\n``` subst-C-loop : subst C loop ≃ succ\nsubst-C-loop =\nsubst C loop ≃〈 subst-resp C loop 〉\nsubst (λ x → x) (resp C loop) ≃〈 resp (subst (λ x → x)) (βloop/rec Int succ≃) 〉\nsubst (λ x → x) succ≃ ≃〈 subst-univ _ 〉\nsucc ∎\n```\n\nThe notation here is an Agda trick for writing out an equational deduction in something of a two-column proof, except the justifications come in between the things they prove equal:\n\n``` x ≃〈 reason x equals y 〉\ny\n```\n\nIn this case, we:\n\n1. Reassociate subst C loop into a subst with the identity function and a resp C, so that we can\n2. Use the HIT β-reduction rule for resp C loop (recall that C is defined by an S¹-rec.\n3. Use the β-reduction rule for univalence, which says that subst (λ x → x) on an isomorphism given explicitly by the univalence axiom selects the forward direction of the isomorphism.\n\n# Functions back and forth\n\nMy goal is to prove that Ω₁(S¹) is homotopy equivalent to Z—this can be improved to an adjoint equivalence and therefore a path in the universe. Because Z is an h-set, this entails, as easy corollaries, that π₁(S¹) (with the quotienting) is equivalent to Z, and that the higher loop spaces (and therefore homotopy groups) are trivial. To prove that π₁(S¹) and Z are isomorphic groups, we should also check that this map is group homomorphism—composition of paths gets mapped to addition—but I won’t do this here.\n\nSo the goal is to define functions\n\n``` base ≃ base -> Int\nInt -> base ≃ base\n```\n\nthat compose to the identity.\n\nTo start, we can define loopn by induction on n:\n\n``` loop^ : Int -> base ≃ base\nloop^ Zero = Refl\nloop^ (Pos Z) = loop\nloop^ (Pos (S n)) = loop ∘ loop^ (Pos n)\nloop^ (Neg Z) = ! loop\nloop^ (Neg (S n)) = ! loop ∘ loop^ (Neg n)\n```\n\nThinking of the integers as codes (names) for paths, this “decodes” an integer as a path.\n\nThe other direction “encodes” a path as its winding number, and can be defined by transporting along the universal cover:\n\n``` encode' : base ≃ base -> Int\nencode' p = subst C p Zero\n```\n\nThe intuition here is that the action on paths of C takes loop to the successor function, and, because it’s functorial, therefore takes loopn to n successor functions composed together. Applying this composite to Zero therefore computes n, giving us the inverse of loop^.\n\nImportant Point 1: However, for what’s coming below, it’s useful to observe that encode can be given a more general type:\n\n``` encode : {a : S¹} -> base ≃ a -> C a\nencode p = subst C p Zero\n```\n\nFor any point a on the circle and path p from base to a, encode p is an element of the cover of a. There is no reason to restrict attention to loops at base. This generality is going to be important below.\n\n# The easy composite\n\nIt’s easy to see that if you start with a number, decode it as a path, and then re-encode it, you get back where you started: because you’re starting with a number, you can do induction! Here are the cases for 0 and positives (negatives are a symmetric 6 lines of code):\n\n``` encode-loop^ : (n : Int) -> encode (loop^ n) ≃ n\nencode-loop^ Zero = Refl\nencode-loop^ (Pos Z) = app≃ subst-C-loop\nencode-loop^ (Pos (S y)) =\nsubst C (loop ∘ loop^ (Pos y)) Zero ≃〈 app≃ (subst-∘ C loop (loop^ (Pos y)) ) 〉\nsubst C loop (subst C (loop^ (Pos y)) Zero) ≃〈 app≃ subst-C-loop 〉\nsucc (subst C (loop^ (Pos y)) Zero) ≃〈 resp succ (encode-loop^ (Pos y)) 〉\nsucc (Pos y) ∎\n```\n\nIn the 0 case it’s true by computation. In the 1 case it’s the lemma we proved above. For other positives, you need to apply functoriality, use the lemma above for the first subst C loop, and then use the IH for the rest.\n\n# The hard composite\n\nThe other direction is harder, because you start with a loop p:\n\n``` stuck : {p : base ≃ base} -> p ≃ loop^ (encode p)\nstuck = ?\n```\n\nand there is no obvious elimination rule for p : base ≃ base (this is where UIP/Streicher’s K come in “normal” dependent type theory, but those clearly aren’t the right thing here). Indeed, this whole proof can be seen as constructing an induction principle for p : base ≃ base.\n\nThe trick is to generalize the problem to consider not just loops, but arbitrary paths base ≃ x, so that we can use J. To set this up, we first show that loop^ extends to a function\n\n``` decode : {a : S¹} -> C a -> base ≃ a\ndecode {a} =\nS¹-elim {\\ x -> C x -> base ≃ x}\nloop^\n(subst (\\ x' → C x' → base ≃ x') loop loop^ ≃〈 subst-→ C (Id base) loop loop^ 〉\n(\\ y -> subst (Id base) loop (loop^ (subst C (! loop) y))) ≃〈 λ≃ (λ y → subst-Id-post loop (loop^ (subst C (! loop) y))) 〉\n(\\ y -> loop ∘ (loop^ (subst C (! loop) y))) ≃〈 λ≃ (λ y → resp (λ x' → loop ∘ loop^ x') (app≃ subst-C-!loop)) 〉\n(\\ y -> loop ∘ (loop^ (pred y))) ≃〈 λ≃ shift 〉\n(\\ y → loop^ y)\n∎)\na\n```\n\nThe proof is an S¹-elim: When a is base, we use loop^. To show that this respects the loop, we apply the functorial action of (\\ x' → C x' → base ≃ x'), then reduce subst C (! loop) to the predecessor function (analogous to subst C loop ≃ succ), and then apply shift:\n\n``` shift : (n : Int) -> (loop ∘ (loop^ (pred n))) ≃ loop^ n\n```\n\nThe proof of shift is a simple induction on n (12 lines of code).\n\nImportant Point 2: At this point, we’re basically done! The reason is that, once we have\n\n``` encode : {a : S¹} -> base ≃ a -> C a\ndecode : {a : S¹} -> C a -> base ≃ a\n```\n\nwe can generalize the theorem to\n\n``` decode-encode : ∀ {a} -> (p : base ≃ a) -> decode (encode p) ≃ p\n```\n\npolymorphically in a. So unlike above, where p:base ≃ base, p has a free end-point. Thus, we can use J to contract it to reflexivity:\n\n``` decode-encode {a} p =\njay1 (λ a' (p' : base ≃ a') → decode (encode p') ≃ p') p Refl\n```\n\nHere jay1 is the Paulin-Mohring J rule: it suffices to consider the case where a' is base and p' is Refl, in which case decode (encode Refl) reduces to Refl, and we’re done.\n\n# Analysis\n\nLet me try an explanation for what’s going on here: The way it’s scattered around the code is a little funny, but I think the ingredients of an induction on the length of the path are hiding in decode-encode. decode-encode is tantamount to saying that every loop p:Id base base arises as loop^ n for some n (in particular, encode p). In decode-encode, we check the base case, when p is reflexivity. To define decode, the main lemma is shift, which is essentially saying that if p can be expressed as loop^ (pred n) then loop ∘ p can be expressed as loop^ n. This feels like an inductive step, where we show that the property is closed under going around the loop once. Somehow the general facts about inverses mean you don’t need to consider !loop explicitly. So: S¹-elim tells you that the only inductive case you need to consider is going around the loop; J tells you that the only base case you need to consider is reflexivity. Cover these two cases, and you cover all paths!\n\nFinally, it’s important to note that Mike’s proof proves slightly more than just that Id base base is equivalent to Int. It also proves that Id base x is equivalent to C x for all x. I’ve done half of that above; the other half would be to show that encode-loop^ generalizes to\n\n``` encode-decode : ∀ {a} -> (n : C a) -> encode (decode n) ≃ n\n```\n\nI think you should be able to do this using an S¹-elim, but I haven’t tried.\n\nOverall, I wanted to write up this simplification to see if anyone has thoughts about what’s going on here, and because it might make it easier to formalize other results in algebraic topology in HoTT.\n\nAgda code is here.\n\nThis entry was posted in Applications, Higher Inductive Types, Models. Bookmark the permalink.\n\n### 5 Responses to A Simpler Proof that π₁(S¹) is Z\n\n1. Jesse C. McKeown says:\n\nOn the one hand, this sort of calculation is nice in showing how having enough classifying spaces to work with gives a rich topological language; on the other hand, they vaguely warn me off of wanting to assume univalence itself, because I’m not sure I always want it provable that the fundamental group of the circle is the integers — particularly not a copy of the integers on which I can program an equality decider. Sometimes I might want to localize at, say, the prime 5, or just invert 2. So it makes more sense to take the fundamental group of the circle as being the de-facto ring — that it *is* essentially a ring I’m confident of being able to encode, with or without deciding what the ring is.\n\n• Mike Shulman says:\n\nThat’s an excellent point! More generally, the proper question to ask about an axiom is not “should I assume it?” but “when should I assume it?” Homotopy type theory with HITs and the univalence axiom is an internal language of (∞,1)-toposes, so if you want a language which applies more generally than that, then you should assume fewer axioms.\n\nFor instance, homotopy type theory with dependent sums, products, and identity types, but no HITs, universes, or univalence, is an internal language for locally cartesian closed (∞,1)-categories. I don’t know exactly what adding HITs without univalence corresponds to, though probably local presentability would suffice.\n\nOn the other hand, as far as I know, the (∞,1)-category of spaces localized or completed at some set of primes is not even locally cartesian closed (is it?). An (∞,1)-category with only finite limits still has a sort of homotopy type theory as its internal language, but that type theory has only dependent sums and identity types, no dependent products. This severely restricts the amount of homotopy theory that we can develop; for instance, none of the definitions like isContr or isEquiv can even be written down without using dependent products.\n\nI am inclined to think that a more fruitful way to study localized and complete spaces with type theory will be to interpret type theory (with dependent products, HITs, and univalence) in the unlocalized category, and identify the local or complete types internally, such as I suggested here. But I haven’t actually tried to push that any further yet.\n\n2. Mike Shulman says:\n\nThis is very nice!! I think this is an excellent example of how the type-theoretic notion of “path induction” gives a new perspective on old homotopy-theoretic facts. My original proof was of course very strongly motivated by a classical proof in algebraic topology; this proof takes more seriously the idea of induction as a way to prove things about paths.\n\n(I do feel that it’s not entirely fair to compare the lengths of proofs in Agda versus in Coq, especially when you are using definitionally-computing-on-points HITs in Agda, which are not available in Coq (and, apparently, are at least a little questionable even in Agda). I bet that if you had to keep track of propositional computation rules for base, then your 100 lines would grow somewhat (though it would probably still be significantly less than 380). And, unfortunately, Agda proof terms are still pretty incomprehensible to me, compared to stepping through a Coq proof script. All of which makes me want to try implementing your proof in Coq.)\n\n• Dan Licata says:\n\nI agree that comparing lines of code in different programming languages makes little to no sense; for example, an average Agda line is much denser than a line in your Coq proof script. Any such statement automatically comes with a “take this with a grain of salt” disclaimer in my mind; sorry I didn’t put that in the text.\n\nSo it would be great to do it in Coq! I don’t think it would be hard: my encode is your cover_to_pathcirc, my decode is just transport with circle_cover. So you should be able to prove the “hard” composite using the definitions you have in essentially one path_induction. For the “easy” composite, my subst-C-loop is your fiber-loop-action, and other than that it’s just an induction using functoriality of transport, which is I think part of the tactics anyway?\n\nThere are things in yours that I didn’t need, like circle_cover_dfib and fiber_wind_action and the lemmas after line 230 or so. I also didn’t prove exactly wind_succ (and it looks like wind_pred is unused already?); shift is similar, but it uses pred on the left instead of succ on the right—I think our proofs in the second part of cover_to_pathcirc might be a little different.\n\n• Bas Spitters says:\n\nDid you try simplifying your proof with Andrej’s new tactics? hott_simpl …"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8937569,"math_prob":0.8996634,"size":15636,"snap":"2020-34-2020-40","text_gpt3_token_len":4130,"char_repetition_ratio":0.12084186,"word_repetition_ratio":0.05789826,"special_character_ratio":0.26195958,"punctuation_ratio":0.104836196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9929473,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-08T08:13:58Z\",\"WARC-Record-ID\":\"<urn:uuid:b75a3b30-6c9c-4a9d-a6a0-637ebedf0d61>\",\"Content-Length\":\"85404\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23767748-49e2-4b51-877a-165ca195e6ec>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd0047f7-f699-4cd5-a23a-7212bc0097ef>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://homotopytypetheory.org/2012/06/07/a-simpler-proof-that-%CF%80%E2%82%81s%C2%B9-is-z/?replytocom=1668\",\"WARC-Payload-Digest\":\"sha1:4KC4TKEXD6SOR42GNSNFD57FEWZQXMJ4\",\"WARC-Block-Digest\":\"sha1:4ZY4TOPIY3MN4FCEMOG5J523RNE2OMZT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737319.74_warc_CC-MAIN-20200808080642-20200808110642-00061.warc.gz\"}"} |
https://www.tutorialspoint.com/count-number-of-pairs-a-n-b-n-such-that-gcd-a-b-is-b-in-cplusplus | [
"# Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B in C++\n\nC++Server Side ProgrammingProgramming\n\nWe are given an input N. The goal is to find all pairs of A, B such that 1<=A<=N and 1<=B<=N and GCD(A, B) is B. All pairs have the greatest common divisor as B.\n\nLet us understand with examples.\n\nInput − N=5\n\nOutput − Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B are − 10\n\nExplanation\n\npairs (A <= N, B <= N) such that gcd (A , B) is B are −\n(1,1), (2,1),(3,1),(4,1),(5,1),(2,2),(3,3),(4,2),(4,4), (5,5). Total 10.\n\nInput − N=50\n\nOutput − Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B are − 207\n\nExplanation\n\npairs (A <= N, B <= N) such that gcd (A , B) is B are :\n(1,1), (2,1),(3,1),(4,1),(5,1).....(50,1)\n(2,2),(3,3),(4,4).....(50,50)\n\nSimilarly other pairs like (4,2), (6,3), (8,2),(8,4),...........(50,25). Total 207\n\n## The approach used in the below program is as follows\n\nThere can be multiple approaches to solve the given problem i.e. naive approach and efficient approach. So let’s first look at the naive approach.\n\n• Take an integer N as input.\n\n• Function GCD(int A, int B) takes two integers and returns the greatest common divisor of A and B. It calculates gcd recursively.\n\n• If any of the A or B is 0 return another one. If both are equal return any of two values. If A>B return (A-B,B). If B is greater, return gcd(B-A,A). In the end we get gcd value.\n\n• Function count_pairs(int N) takes N and returns the number of pairs such that in pair(A,B), B is gcd and both are in range[1,N].\n\n• Take the initial value as count =0 for the number of such pairs.\n\n• For each value of pair, run for loop i=1 to i=N for A and nested for loop j=1 t j=N for B.\n\n• Make a pair (i,j) and pass to GCD(i,j). If the result is equal to j. Increment count.\n\n• At the end of both loops return count as result.\n\n## Efficient approach\n\nAs we can see the gcd(a,b)=b means a is always a multiple of b. All such multiples of b(1<=b<=N) that are less than N will make a pair. For a number i if multiples of i are less than floor(N/i) will be counted.\n\n• Function count_pairs(int N) takes N and returns the number of pairs such that in pair(A,B), B is gcd and both are in range[1,N].\n\n• Take the initial value as count =0 for the number of such pairs.\n\n• Take temporary variable temp=N and i=1.\n\n• Using while (i<=N) do following\n\n• For each i calculate limit of multiples as j=N/temp\n\n• Number of pairs for current i will be temp*(i-j+1). Add to count.\n\n• Set i=j+1. For next B of (A,B).\n\n• Set temp=N/i for next iteration.\n\n• At the end of the while loop, return count as a result.\n\n## Example(naive approach)\n\nLive Demo\n\n#include <iostream>\nusing namespace std;\nint GCD(int A, int B){\nif (A == 0){\nreturn B;\n}\nif (B == 0){\nreturn A;\n}\nif (A == B){\nreturn A;\n}\nif (A > B){\nreturn GCD(A-B, B);\n}\nreturn GCD(A, B-A);\n}\nint count_pairs(int N){\nint count = 0;\nfor(int i=1; i<=N; i++){\nfor(int j = 1; j<=N; j++){\nif(GCD(i, j)==j){\ncount++;\n}\n}\n}\nreturn count;\n}\nint main(){\nint N = 4;\ncout<<\"Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B are: \"<<count_pairs(N);\nreturn 0;\n}\n\n## Output\n\nIf we run the above code it will generate the following output −\n\nCount number of pairs (A <= N, B <= N) such that gcd (A , B) is B are: 8\n\n## Example (Efficient approach)\n\nLive Demo\n\n#include <bits/stdc++.h>\nusing namespace std;\nint Count_pairs(int N){\nint count = 0;\nint temp = N;\nint i = 1;\nwhile(i <= N){\nint j = N / temp;\ncount += temp * (j - i + 1);\ni = j + 1;\ntemp = N / i;\n}\nreturn count;\n}\nint main(){\nint N = 4;\ncout<<\"Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B are: \"<<Count_pairs(N);\nreturn 0;\n}\n\n## Output\n\nIf we run the above code it will generate the following output −\n\nCount number of pairs (A <= N, B <= N) such that gcd (A , B) is B are: 8"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7679731,"math_prob":0.9992724,"size":5113,"snap":"2022-27-2022-33","text_gpt3_token_len":1630,"char_repetition_ratio":0.14895283,"word_repetition_ratio":0.2749301,"special_character_ratio":0.36299628,"punctuation_ratio":0.14750198,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99997795,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-07T19:19:27Z\",\"WARC-Record-ID\":\"<urn:uuid:bb3b73c0-5590-4b68-82a7-56e0b1839e9e>\",\"Content-Length\":\"35450\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ed83db7-8d69-4a12-b95a-2a7f1fb23679>\",\"WARC-Concurrent-To\":\"<urn:uuid:107a27aa-0629-4372-8785-dee9ca439037>\",\"WARC-IP-Address\":\"192.229.210.176\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/count-number-of-pairs-a-n-b-n-such-that-gcd-a-b-is-b-in-cplusplus\",\"WARC-Payload-Digest\":\"sha1:RAGMXY4AIJB4MTYCZVVENHHYPLVITO4P\",\"WARC-Block-Digest\":\"sha1:P5COJDV3SVH5UUNF2S5S6UM66RIH4M33\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570692.22_warc_CC-MAIN-20220807181008-20220807211008-00014.warc.gz\"}"} |
http://bigcomplexdata.com/slides/ClusteringCORD_ICSA_2016.html | [
"$$\\def\\RR{\\bf R} \\def\\real{\\mathbb{R}} \\def\\bold#1{\\bf #1} \\def\\d{\\mbox{Cord}} \\def\\hd{\\widehat \\mbox{Cord}} \\DeclareMathOperator{\\cov}{cov} \\DeclareMathOperator{\\var}{var} \\DeclareMathOperator{\\cor}{cor} \\newcommand{\\ac}{\\left\\{#1\\right\\}} \\DeclareMathOperator{\\Ex}{\\mathbb{E}} \\DeclareMathOperator{\\diag}{diag}$$\n\n## Variable Clustering via $G$-Models of Large Covariance Matrices\n\n### Xi (Rossi) Luo\n\nBrown University\nDepartment of Biostatistics\nCenter for Statistical Sciences\nComputation in Brain and Mind\nBrown Institute for Brain Science\nABCD Research Group",
null,
"ICSA 2016, Shanghai, CHINA\nDecember 20, 2016\n\nFunding: NIH R01EB022911; NSF/DMS (BD2K) 1557467; NIH P20GM103645, P01AA019072, P30AI042853; AHA\n\n## Collaborators",
null,
"Florentina Bunea\nCornell University",
null,
"Christophe Giraud\nParis Sud University\n\n## Big Data Problem\n\n• We are interested in big cov with many variables\n• Global property for certain joint distributions\n• Real-world cov: maybe non-sparse and other structures\n• Clustering successful for Big Data Science Donoho, 2015\n• Exploratory Data Analysis (EDA)Tukey, 1977\n• Hierarchical clustering and KmeansHartigan & Wong, 1979\n• Mostly based on marginal/pairwise distances\n• Can we combine clustering and big cov estimation?\n\n## Example: SP 100 Data\n\n• Daily returns from stocks in SP 100\n• Stocks listed in Standard & Poor 100 Indexas of March 21, 2014\n• between January 1, 2006 to December 31, 2008\n• Each stock is a variable\n• Cov/Cor matrices (Pearson's or Kendall's tau)\n• Re-order stocks by clusters\n• Compare cov patterns with different clustering/ordering\n\n## Cor after Grouping by Clusters",
null,
"Kmeans",
null,
"Our $G$-models\n\nOurs yields stronger off-diagonal, tile patterns. Black = 1.\nColor bars: variable groups/clusters\nOff-diagonal: correlations across clusters\n\n## Clustering Results\n\nIndustry Ours Kmeans Hierarchical Clustering\nHome Improvement Home Depot, Lowe’s Home Depot, Lowe’s, Starbucks, Target Home Depot, Lowe’s, Starbucks, Target, Costco, Target, Wal-Mart, FedEx, United Parcel Service, Nike, McDonald’s\nTelecom ATT, Verizon ATT, Verizon, Exelon, Comcast, Walt Disney, Time Warner ATT, Verizon, Comcast, Walt Disney, Time Warner, AIG, Allstate, Metlife, American Express, Bank of America, Citigroup, US Bancorp, Wells Fargo, Capital One, Goldman Sachs, JP Morgan Chase, Morgan Stanley, Simon Property, General Electric\nDiversified Metals & Mining Freeport-McMoran Freeport-McMoran, National Oillwell Varco Freeport-McMoran, Apache Corp., Anadarko Petroleum, Devon Energy, Halliburton, National Oillwell Varco, Occidental Petroleum, Schlumberger, ConocoPhillips, Chevron, Exxon\n$\\cdots$\nAll methods yield 20 clusters.\n\n# Model\n\n## Problem\n\n• Let ${X} \\in \\real^p$ be a zero mean random vector\n• In certain problems, means are arbitrary\n• Divide variables into partitions/clusters\n• Example: $\\{ \\{X_1, X_3, X_7\\}, \\{X_2, X_5\\}, \\dotsc \\}$\n• Theoretical: Find a partition $G = \\{G_k\\}_{ 1 \\leq k \\leq K}$ of $\\{1, \\ldots, p\\}$ such that all $X_a$ with $a \\in G_k$ are \"similar\"\n• Big Data: \"helpful\" clustering that shows patterns\n\n## Related Areas\n\n• Clustering: Kmeans and Hierarchical Clustering\n• Usually for clustering $n$ observations in $R^p$\n• Advantages: fast, general, popular\n• Limitations: low signal-noise-ratio, theory, NP-hard\n• Q: How to choose number of clusters? Theory?\n• Q: Can clusters contain singletons?\n• Community detection: huge literature see review Newman, 2003 but start with observed adjacency matrices or networks\n• Ours for data that can be generated from unknown networks\n• These are related but different problems\n\n## Model: Starting Point\n\n$$X_{n\\times p}=\\underbrace{Z_{n\\times k}}_\\text{Source/Factor} \\quad \\underbrace{G_{k\\times p}}_\\text{Mixing/Loading} + \\underbrace{E_{n\\times p}}_{Error} \\qquad Z \\bot E$$\n\n• Clustering: $G$ is $0/1$ matrix for $k$ clusters/ROIs\n• Decomposition:\n• PCA/factor analysis: orthogonality\n• ICA: orthogonality → independence\n• matrix decomposition: e.g. non-negativity\n• This model leads to block patterns in $\\cov(X)$\n• $\\cov(X) = G^T \\cov(Z) G + \\cov(E)$\n• Note: not necessarily block-diagonal\n\n## Generalization: $G$-Block\n\n• Example: $G=\\ac{\\ac{1,2};\\ac{3,4,5}}$, $X \\in \\real^p$ is $G$-block\n$$\\Sigma =\\left(\\begin{array}{ccccc} {\\color{red} D_1} & {\\color{red} C_{11} }&C_{12} & C_{12}& C_{12}\\\\ {\\color{red} C_{11} }&{\\color{red} D_1 }& C_{12} & C_{12}& C_{12} \\\\ C_{12} & C_{12} &{\\color{green} D_{2}} & {\\color{green} C_{22}}& {\\color{green} C_{22}}\\\\ C_{12} & C_{12} &{\\color{green} C_{22}} &{\\color{green} D_2}&{\\color{green} C_{22}}\\\\ C_{12} & C_{12} &{\\color{green} C_{22}} &{\\color{green} C_{22}}&{\\color{green} D_2} \\end{array}\\right) \\qquad C = \\left(\\begin{array}{cc} {\\color{red} C_{11} } & C_{12}\\\\ C_{12} & {\\color{green} C_{22}} \\end{array}\\right)$$\n• Matrix math: $\\cov(X) = \\Sigma = G^TCG + d$\n• We allow $|C_{11} | \\lt | C_{12} |$ or $C \\prec 0$\n• Kmeans/HC leads to block-diagonal cor matrices (permutation)\n• Clustering based on $G$-Block\n• From $G$-block we can read out \"negative\" $\\cov(Z)$\n• Cov defined for semiparametric distributions\n• Clusters can contain singletons\n\n## Minimum $G$ Partition\n\nTheorem: $G^{\\beta}(X)$ is the minimal partition induced by $a\\stackrel{G^{\\beta}}{\\sim} b$\niff $\\var(X_{a})=\\var(X_{b})$ and $\\cov(X_{a},X_{c})=\\cov(X_{b},X_{c})$ for all $c\\neq a,b$. Moreover, if the matrix of covariances $C$ corresponding to the partition $G(X)$ is positive-semidefinite, then this is the unique minimal partition according to which ${X}$ admits a latent decomposition.\n• We define the minimal cluster/partition.\n• The minimal partition is unique under conditions.\n• We will aim to recover the minimal partition (thus $K$).\n\n# Method\n\n## New Metric: CORD\n\n• First, pairwise correlation distance (like Kmeans)\n• Gaussian copula: $$Y:=(h_1(X_1),\\dotsc,h_p(X_p)) \\sim N(0,R)$$\n• Let $R$ be the correlation matrix\n• Gaussian: Pearson's\n• Gaussian copula: Kendall's tau transformed, $R_{ab} = \\sin (\\frac{\\pi}{2}\\tau_{ab})$\n• Second, introduce CORrelation Distance $$\\d(a,b) := \\max_{c\\neq a,b}|R_{ac}-R_{bc}|$$\n• Third, group variables $a$, $b$ together if $\\d(a,b) = 0$\n• Do not care any pairwise distance between $a,b$\n• \"The enemy of my enemy is my friend\"\n\n## Algorithm: Main Idea\n\n• Greedy: one cluster at a time, avoiding NP-hard\n• Cluster variables together if CORD metric $$\\widehat \\d(a,b) \\lt \\alpha$$ where $\\alpha$ is a tuning parameter\n• $\\alpha$ is chosen by theory or CV\n\n# Theory\n\n## Condition\n\nLet $\\eta \\geq 0$ be given. Let ${ X}$ be a zero mean random vector with a Gaussian copula distribution with parameter $R$. $$\\begin{multline} \\mathcal{R}(\\eta) := \\{R: \\ \\d(a,b) := \\max_{c\\neq a,b}|R_{ac}-R_{bc}|>\\eta\\quad \\\\ \\textrm{for all}\\ a\\stackrel{G(X)}{\\nsim}b.\\} \\end{multline}$$ Group separation condition: $R \\in \\mathcal{R}(\\eta)$.\n\nThe signal strength $\\eta$ is large.\n\n## Consistency\n\nTheorem: Define $\\tau=|\\widehat R-R|_{\\infty}$ and we consider two parameters $(\\alpha,\\eta)$ fulfilling $$\\begin{equation} \\alpha\\geq 2\\tau\\quad\\textrm{and}\\quad \\eta\\geq2\\tau+\\alpha. \\end{equation}$$ Then, applying our algorithm we have $\\widehat G=G(X)$ whp.\n\nOurs recovers the exact clustering with high probability.\n\n## Minimax\n\nTheorem: $P_{\\Sigma}$ the likelihood based on $n$ independent observations of ${ X} \\stackrel{d}{=} \\mathcal{N}(0,\\Sigma)$. For any \\begin{equation} 0\\leq \\eta < \\eta^{*}:=\\frac{0.6\\sqrt{\\frac{ \\log(p)}{n}}}{1+0.6\\sqrt{\\frac{ \\log(p)}{n}}} \\end{equation} we have $$\\inf_{\\widehat G}\\sup_{R \\in \\mathcal{R}(\\eta)} P_{\\Sigma}(\\widehat G\\neq G^{\\beta}(X))\\geq {1\\over 2e+1}\\geq {1\\over 7} \\,,$$ where the infimum is taken over all possible estimators.\n\nGroup separation condition on $\\eta$ is optimal.\n\n## Choosing Number of Clusters\n\n• Split data into 3 parts\n• Use part 1 of data to estimate clusters $\\hat{G}$ for each $\\alpha$\n• Use part 2 to compute between variable difference $$\\delta^{(2)}_{ab} = R_{ac}^{(2)} - R_{bc}^{(2)}, \\quad c \\ne a, b.$$\n• Use part 3 to generate \"CV\" loss $$\\mbox{CV}(\\hat{G}) = \\sum_{a \\lt b} \\| \\delta^{(3)}_{ab} - \\delta^{(2)}_{ab} 1\\{ a \\mbox{ not clustered w/ } b \\} \\|^2_\\infty.$$\n• Pick $\\alpha$ with the smallest loss above\n\n## Theory for CV\n\nTheorem: If either: (i) $X$ is sub-Gaussian with correlation matrix $R$; or (ii) $X$ has a copula distribution with copula correlation matrix $R$, then we have $E[\\mbox{CV}(G^*)] \\lt E[\\mbox{CV}(G)]$, for any $G\\ne G^*$.\nThis shows that our CV will select $G^*$ consistently.\n\n# Simulations\n\n## Setup\n\n• Model $C$ ($\\cov(Z)$): positive semidefinite or negative\n• True $G^*$: singletons or no-singleton clusters\n• Simulate $X$ from $G$-block cov\n• Variable clustering using $X$\n• Compare with K-means or Hierarchical Clustering:\n• Exact recovery of groups\n• Cross validation loss and choosing $K$\n\n## Exact Recovery\n\nDifferent models for $C$=\"$\\cov(Z)$\" and $G$\n\nHC and Kmeans fail even if inputting the true $K$ and $n \\rightarrow \\infty$\n\nOur CORD methods recover both the true $G^*$ and $K$ as predicted by our theory.\n\n## Cross Validation\n\nRecovery % in red and CV loss in black.\n\nCV selects the constants to yield close to 100% recovery, as predicted by our theory (at least for large $n>200$)\n\n# Real Data\n\n## Functional MRI\n\n• fMRI matrix: BOLD from different brain regions\n• Variable: different brain regions\n• Sample: time series (after whitening or removing temporal correlations)\n• Clusters of brain regions\n• Two data matrices from two scan sessions OpenfMRI.org\n• Use Power's 264 regions/nodes\n\n## Test Prediction/Reproducibilty\n\n• Find partitions using the first session data\n• Average each block cor to improve estimation\n• Compare with the cor matrix from the second scan $$\\| Avg_{\\hat{G}}(\\hat{\\Sigma}_1) - \\hat{\\Sigma}_2 \\|$$\n• Difference is smaller if clustering $\\hat{G}$ is better\n\nVertical lines: fixed (solid) and data-driven (dashed) thresholds\n\nOur CORD $\\hat{G}$ leads to smaller between-session variability for almost all $K$, than HC and Kmeans.\n\n## Discussion\n\n• Cov + clustering:\n• Identifiability, accuracy, optimality\n• $G$-models: $G$-latent, $G$-block, $G$-exchangeable\n• New metric, method, and theory\n• Defining clusters, consistency, minimax, and CV theory\n• Some new results using big data examples\n• Paper: bit.ly/cordCluster (arXiv 1508.01939)\n• R package: cord on CRAN\n• CV function available soon\n\n# Thank you!\n\n## Slides at: bit.ly/ICSA2016",
null,
"## Website: BigComplexData.com\n\nPostdoc position available\nfunded by Whitehouse's Big Data and BRAIN Initiatives"
] | [
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"http://bigcomplexdata.com/slides/Media/ABCDgroup/abcdLogo2_BannerBigComplexDataCom.png",
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"http://bigcomplexdata.com/slides/Media/Personnel/FlorentinaBunea.jpg",
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"http://bigcomplexdata.com/slides/Media/Personnel/ChristopheGiraud.jpg",
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"http://bigcomplexdata.com/slides/Media/Clustering/StockCorKendallParulaKmeans.png",
null,
"http://bigcomplexdata.com/slides/Media/Clustering/StockCorKendallParula.png",
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"http://bigcomplexdata.com/slides/Media/QR/ICSA2016.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6629276,"math_prob":0.9967161,"size":5958,"snap":"2019-35-2019-39","text_gpt3_token_len":1774,"char_repetition_ratio":0.08867988,"word_repetition_ratio":0.0,"special_character_ratio":0.28298086,"punctuation_ratio":0.11060744,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99964786,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T14:59:08Z\",\"WARC-Record-ID\":\"<urn:uuid:4bc5944e-0d3f-4071-912a-5619d0c8f7c6>\",\"Content-Length\":\"35091\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9dda8553-c816-478f-88b2-0e9344a9dae9>\",\"WARC-Concurrent-To\":\"<urn:uuid:48097497-977c-45b1-8697-eac28442ea20>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"http://bigcomplexdata.com/slides/ClusteringCORD_ICSA_2016.html\",\"WARC-Payload-Digest\":\"sha1:NHPJEJVA6YQAZR2F23K5N7NTQPMWXRDP\",\"WARC-Block-Digest\":\"sha1:UM2FS4WINNKJDLQ3IVI7IEQ3PQQDHR5T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573080.8_warc_CC-MAIN-20190917141045-20190917163045-00316.warc.gz\"}"} |
https://reference.wolfram.com/language/tutorial/Options.html | [
"Options for Graphics\n\nWhen the Wolfram Language plots a graph for you, it has to make many choices. It has to work out what the scales should be, where the function should be sampled, how the axes should be drawn, and so on. Most of the time, the Wolfram Language will probably make pretty good choices. However, if you want to get the very best possible pictures for your particular purposes, you may have to help the Wolfram Language in making some of its choices.\n\nThere is a general mechanism for specifying \"options\" in Wolfram Language functions. Each option has a definite name. As the last arguments to a function like Plot, you can include a sequence of rules of the form name->value, to specify the values for various options. Any option for which you do not give an explicit rule is taken to have its \"default\" value.\n\n Plot[f,{x,xmin,xmax},option->value] make a plot, specifying a particular value for an option\n\nChoosing an option for a plot.\n\nA function like Plot has many options that you can set. Usually you will need to use at most a few of them at a time. If you want to optimize a particular plot, you will probably do best to experiment, trying a sequence of different settings for various options.\n\nEach time you produce a plot, you can specify options for it. \"Redrawing and Combining Plots\" will also discuss how you can change some of the options, even after you have produced the plot.\n\n option name default value AspectRatio 1/GoldenRatio the height‐to‐width ratio for the plot; Automatic sets it from the absolute",
null,
"and",
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"coordinates Axes True whether to include axes AxesLabel None labels to be put on the axes; ylabel specifies a label for the",
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"axis, {xlabel,ylabel} for both axes AxesOrigin Automatic the point at which axes cross BaseStyle {} the default style to use for the plot FormatType TraditionalForm the default format type to use for text in the plot Frame False whether to draw a frame around the plot FrameLabel None labels to be put around the frame; give a list in clockwise order starting with the lower",
null,
"axis FrameTicks Automatic what tick marks to draw if there is a frame; None gives no tick marks GridLines None what grid lines to include; Automatic includes a grid line for every major tick mark PlotLabel None an expression to be printed as a label for the plot PlotRange Automatic the range of coordinates to include in the plot; All includes all points Ticks Automatic what tick marks to draw if there are axes; None gives no tick marks\n\nSome of the options for Plot. These can also be used in Show.\n\nHere is a plot with all options having their default values:\n In:=",
null,
"Out=",
null,
"This draws axes on a frame around the plot:\n In:=",
null,
"Out=",
null,
"This specifies labels for the",
null,
"and",
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"axes. The expressions you give as labels are printed just as they would be if they appeared as TraditionalForm Wolfram Language output. You can give any piece of text by putting it inside a pair of double quotes:\n In:=",
null,
"Out=",
null,
"You can give several options at the same time, in any order:\n In:=",
null,
"Out=",
null,
"Setting the AspectRatio option changes the whole shape of your plot. AspectRatio gives the ratio of height to width. Its default value is the inverse of the Golden Ratiosupposedly the most pleasing shape for a rectangle:\n In:=",
null,
"Out=",
null,
"Automatic use internal algorithms None do not include this All include everything True do this False do not do this\n\nSome common settings for various options.\n\nWhen the Wolfram Language makes a plot, it tries to set the",
null,
"and",
null,
"scales to include only the \"interesting\" parts of the plot. If your function increases very rapidly, or has singularities, the parts where it gets too large will be cut off. By specifying the option PlotRange, you can control exactly what ranges of",
null,
"and",
null,
"coordinates are included in your plot.\n\n Automatic show at least a large fraction of the points, including the \"interesting\" region (the default setting) All show all points {ymin,ymax} show a specific range of",
null,
"values {xrange,yrange} show the specified ranges of",
null,
"and",
null,
"values\n\nSettings for the option PlotRange.\n\nThe setting for the option PlotRange gives explicit",
null,
"limits for the graph. With the",
null,
"limits specified here, the bottom of the curve is cut off:\n In:=",
null,
"Out=",
null,
"The Wolfram Language always tries to plot functions as smooth curves. As a result, in places where your function wiggles a lot, the Wolfram Language will use more points. In general, the Wolfram Language tries to adapt its sampling of your function to the form of the function. There is a limit, however, which you can set, to how finely the Wolfram Language will ever sample a function.\n\nThe function",
null,
"wiggles infinitely often when",
null,
". the Wolfram Language tries to sample more points in the region where the function wiggles a lot, but it can never sample the infinite number that you would need to reproduce the function exactly. As a result, there are slight glitches in the plot:\n In:=",
null,
"Out=",
null,
"It is important to realize that since the Wolfram Language can only sample your function at a limited number of points, it can always miss features of the function. The Wolfram Language adaptively samples the functions, increasing the number of samples near interesting features, but it is still possible to miss something. By increasing PlotPoints, you can make the Wolfram Language sample your function at a larger number of points. Of course, the larger you set PlotPoints to be, the longer it will take the Wolfram Language to plot any function, even a smooth one.\n\n option name default value PlotStyle Automatic a list of lists of graphics primitives to use for each curve (see \"Graphics Directives and Options\") ClippingStyle None what to draw when curves are clipped Filling None filling to insert under each curve FillingStyle Automatic style to use for filling PlotPoints 50 the initial number of points at which to sample the function MaxRecursion Automatic the maximum number of recursive subdivisions allowed\n\nMore options for Plot. These cannot be used in Show.\n\nThis uses PlotStyle to specify a dashed curve:\n In:=",
null,
"Out=",
null,
"When plotting multiple functions, PlotStyle settings in a list are used sequentially for each function:\n In:=",
null,
"Out=",
null,
"When a PlotStyle contains a sublist, the settings are combined:\n In:=",
null,
"Out=",
null,
"By default nothing is indicated when the PlotRange is set, so that it cuts off curves:\n In:=",
null,
"Out=",
null,
"Setting ClippingStyle to Automatic draws a dashed line where a curve is cut off:\n In:=",
null,
"Out=",
null,
"Setting ClippingStyle to a list defines the style for the parts cut off at the bottom and top:\n In:=",
null,
"Out=",
null,
"This specifies filling between the curve and the",
null,
"axis:\n In:=",
null,
"Out=",
null,
"The filling can be specified to extend to an arbitrary height, such as the bottom of the graphic. Filling colors are automatically blended where they overlap:\n In:=",
null,
"Out=",
null,
"This specifies a specific filling to be used only for the first curve:\n In:=",
null,
"Out=",
null,
"This shows a filling from the first curve to the second using a nondefault filling style:\n In:=",
null,
"Out=",
null,
""
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https://virtualnerd.com/algebra-1/algebra-foundations/variables-expressions/ | [
"# Variables and Expressions\n\n### Popular Tutorials in Variables and Expressions\n\n• #### How Do You Evaluate an Algebraic Expression by Plugging In Values?\n\nSometimes you have an algebraic expression with variables, and you know the values of those variables exactly, and you just need to plug in those values and get the value of the expression. Well, if that's what you have to do, then you've come to the right place, because this tutorial will show you exactly how to do it!\n\n• #### How Do You Solve a Word Problem by Writing an Equation and Plugging in Values?\n\nSetting up and solving an equation from a word problem can be tricky, but this tutorial can help. See all the steps, from defining variables to getting that final answer, and everything in between! With this tutorial, you'll learn what it takes to solve a word problem.\n\n• #### How Do You Translate a Phrase with One Variable and Number Into a Mathematical Expression?\n\nNeed some practice translating phrases into mathematical expressions? Then this tutorial is for you! You'll get practice translating statements involving addition, subtraction, multiplication, or division into mathematical expressions.\n\n• #### How Do You Write an Equation from a Word Problem?\n\nHaving difficulty turning a word problem into an algebra equation? Then this tutorial is for you! With this tutorial, you'll learn how to break down word problems and translate them into mathematical equations.\n\n• #### How Do You Plug Variables into an Algebraic Expression?\n\nPlugging variables into an expression is essential for solving many algebra problems. See how to plug in variable values by watching this tutorial.\n\n• #### What is a Variable?\n\nYou can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!\n\n• #### What Are Some Words You Can Use To Write Word Problems?\n\nKnowing the mathematical meaning of words allows you to decipher word problems and gives you the power to write your own word problems, too! Take a look at these words and learn their mathematical translations.\n\n• #### How Do You Turn a Longer Verbal Phrase into an Algebraic Expression?\n\nLooking for some practice translating words into a mathematical expression? Then take a look at this tutorial, and you'll learn how to break down those words and write the mathematical expression they describe!\n\n• #### How Do You Turn a Simple Equation into Words?\n\nGet creative with math by turning a mathematical equation into a story. This tutorial shows you one example, but the possibilities are endless! Test your creativity by writing your own story to describe the given mathematical equation.\n\n• #### How Do You Turn a Simple Expression into Words?\n\nBeing able to translate words into a mathematical expression is an important skill, but being able to go the other way is just as crucial. This tutorial shows you how a mathematical expression can be turned into words.\n\n• #### How Do You Turn a More Complicated Equation into Words?\n\nLet your creativity go wild by turning a mathematical equation into a story! Watch this tutorial to see one example, and then write your own story to describe the given equation. See how creative you can get with math!\n\n• #### How Do You Turn a More Complicated Expression into Words?\n\nThis tutorial shows you how a mathematical expression can be turned into words. The best part? You can be super creative with the statement you make up! Check out the example statement created in this tutorial, and then make up your own!\n\n• #### What is a Constant?\n\nConstants are parts of algebraic expressions that don't change. Check out this tutorial to see exactly what a constant looks like and why it doesn't change.\n\n• #### What Are Numerical and Algebraic Expressions?\n\nAn expression is just a mathematical phrase. In this tutorial, you'll learn about two popular types of expressions: numerical and algebraic expressions. A numerical expression contains numbers and operations. An algebraic expression is almost exactly the same except it also contains variables. Check out this tutorial to learn about these two popular kinds of expressions!\n\n• #### What is the Substitution Property of Equality?\n\nIf you ever plug a value in for a variable into an expression or equation, you're using the Substitution Property of Equality. This property allows you to substitute quantities for each other into an expression as long as those quantities are equal. Watch this tutorial to learn about this useful property!"
] | [
null
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https://studyfinance.com/future-value/6-in-16-years/ | [
"# Future Value of $6 in 16 Years Calculating the future value of$6 over the next 16 years allows you to see how much your principal will grow based on the compounding interest.\n\nSo if you want to save $6 for 16 years, you would want to know approximately how much that investment would be worth at the end of the period. To do this, we can use the future value formula below: $$FV = PV \\times (1 + r)^{n}$$ We already have two of the three required variables to calculate this: • Present Value (FV): This is the original$6 to be invested\n• n: This is the number of periods, which is 16 years\n\nThe final variable we need to do this calculation is r, which is the rate of return for the investment. With some investments, the interest rate might be given up front, while others could depend on performance (at which point you might want to look at a range of future values to assess whether the investment is a good option).\n\nIn the table below, we have calculated the future value (FV) of $6 over 16 years for expected rates of return from 2% to 30%. The table below shows the present value (PV) of$6 in 16 years for interest rates from 2% to 30%.\n\nAs you will see, the future value of $6 over 16 years can range from$8.24 to $399.25. Discount Rate Present Value Future Value 2%$6 $8.24 3%$6 $9.63 4%$6 $11.24 5%$6 $13.10 6%$6 $15.24 7%$6 $17.71 8%$6 $20.56 9%$6 $23.82 10%$6 $27.57 11%$6 $31.87 12%$6 $36.78 13%$6 $42.40 14%$6 $48.82 15%$6 $56.15 16%$6 $64.49 17%$6 $73.98 18%$6 $84.77 19%$6 $97.03 20%$6 $110.93 21%$6 $126.68 22%$6 $144.51 23%$6 $164.68 24%$6 $187.46 25%$6 $213.16 26%$6 $242.15 27%$6 $274.80 28%$6 $311.54 29%$6 $352.85 30%$6 $399.25 This is the most commonly used FV formula which calculates the compound interest on the new balance at the end of the period. Some investments will add interest at the beginning of the new period, while some might have continuous compounding, which again would require a slightly different formula. Hopefully this article has helped you to understand how to make future value calculations yourself. You can also use our quick future value calculator for specific numbers. ### Link To or Reference This Page If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support! • \"Future Value of$6 in 16 Years\". StudyFinance.com. Accessed on July 27, 2021. https://studyfinance.com/future-value/6-in-16-years/.\n\n• \"Future Value of $6 in 16 Years\". StudyFinance.com, https://studyfinance.com/future-value/6-in-16-years/. Accessed 27 July, 2021 • Future Value of$6 in 16 Years. StudyFinance.com. Retrieved from https://studyfinance.com/future-value/6-in-16-years/."
] | [
null
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https://www.src.isr.umich.edu/keyword/analysis-of-covariance_methods/ | [
"# Proportional hazards regression with missing covariates\n\nNonparametric maximum likelihood (NPML) is used to estimate regression parameters in a proportional hazards regression model with missing covariates. The NPML estimator is shown to be consistent and asymptotically normally distributed under some conditions. EM type algorithms are applied to solve the maximization problem. Variance estimates of the regression parameters are obtained by a profile likelihood approach that uses EM-aided numerical differentiation. Simulation results indicate that the NPML estimates of the regression parameters are more efficient than the approximate partial likelihood estimates and estimates from complete-case analysis when missing covariates are missing completely at random, and that the proposed method corrects for bias when the missing covariates are missing at random. KEY WORDS: EM algorithm; Missing data; Nonparametric maximum likelihood; Proportional hazards model."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82294303,"math_prob":0.97001433,"size":912,"snap":"2022-40-2023-06","text_gpt3_token_len":161,"char_repetition_ratio":0.14096916,"word_repetition_ratio":0.033613447,"special_character_ratio":0.15021929,"punctuation_ratio":0.080882356,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839635,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T10:19:32Z\",\"WARC-Record-ID\":\"<urn:uuid:59dbc9b7-5f44-4518-b713-f7405ad9c4d4>\",\"Content-Length\":\"22621\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:11f8b760-d3b7-441d-b1e9-ef330ec9b17a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c0a991f6-bde6-4789-b6aa-866613fba46e>\",\"WARC-IP-Address\":\"141.211.32.80\",\"WARC-Target-URI\":\"https://www.src.isr.umich.edu/keyword/analysis-of-covariance_methods/\",\"WARC-Payload-Digest\":\"sha1:KCGT5DBBXDVS54VQYWIWVXJ6QO6AHRK7\",\"WARC-Block-Digest\":\"sha1:6D3GQOX6IE6VZ75KZ3ZYRCHA3YZUTQD6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499541.63_warc_CC-MAIN-20230128090359-20230128120359-00012.warc.gz\"}"} |
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