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"OpenCV 4.1.0 Open Source Computer Vision\nPer-element Operations\n\n## Functions\n\nvoid cv::cuda::abs (InputArray src, OutputArray dst, Stream &stream=Stream::Null())\nComputes an absolute value of each matrix element. More...\n\nvoid cv::cuda::absdiff (InputArray src1, InputArray src2, OutputArray dst, Stream &stream=Stream::Null())\nComputes per-element absolute difference of two matrices (or of a matrix and scalar). More...\n\nvoid cv::cuda::add (InputArray src1, InputArray src2, OutputArray dst, InputArray mask=noArray(), int dtype=-1, Stream &stream=Stream::Null())\nComputes a matrix-matrix or matrix-scalar sum. More...\n\nvoid cv::cuda::addWeighted (InputArray src1, double alpha, InputArray src2, double beta, double gamma, OutputArray dst, int dtype=-1, Stream &stream=Stream::Null())\nComputes the weighted sum of two arrays. More...\n\nvoid cv::cuda::bitwise_and (InputArray src1, InputArray src2, OutputArray dst, InputArray mask=noArray(), Stream &stream=Stream::Null())\nPerforms a per-element bitwise conjunction of two matrices (or of matrix and scalar). More...\n\nvoid cv::cuda::bitwise_not (InputArray src, OutputArray dst, InputArray mask=noArray(), Stream &stream=Stream::Null())\nPerforms a per-element bitwise inversion. More...\n\nvoid cv::cuda::bitwise_or (InputArray src1, InputArray src2, OutputArray dst, InputArray mask=noArray(), Stream &stream=Stream::Null())\nPerforms a per-element bitwise disjunction of two matrices (or of matrix and scalar). More...\n\nvoid cv::cuda::bitwise_xor (InputArray src1, InputArray src2, OutputArray dst, InputArray mask=noArray(), Stream &stream=Stream::Null())\nPerforms a per-element bitwise exclusive or operation of two matrices (or of matrix and scalar). More...\n\nvoid cv::cuda::cartToPolar (InputArray x, InputArray y, OutputArray magnitude, OutputArray angle, bool angleInDegrees=false, Stream &stream=Stream::Null())\nConverts Cartesian coordinates into polar. More...\n\nvoid cv::cuda::compare (InputArray src1, InputArray src2, OutputArray dst, int cmpop, Stream &stream=Stream::Null())\nCompares elements of two matrices (or of a matrix and scalar). More...\n\nvoid cv::cuda::divide (InputArray src1, InputArray src2, OutputArray dst, double scale=1, int dtype=-1, Stream &stream=Stream::Null())\nComputes a matrix-matrix or matrix-scalar division. More...\n\nvoid cv::cuda::exp (InputArray src, OutputArray dst, Stream &stream=Stream::Null())\nComputes an exponent of each matrix element. More...\n\nvoid cv::cuda::log (InputArray src, OutputArray dst, Stream &stream=Stream::Null())\nComputes a natural logarithm of absolute value of each matrix element. More...\n\nvoid cv::cuda::lshift (InputArray src, Scalar_< int > val, OutputArray dst, Stream &stream=Stream::Null())\nPerforms pixel by pixel right left of an image by a constant value. More...\n\nvoid cv::cuda::magnitude (InputArray xy, OutputArray magnitude, Stream &stream=Stream::Null())\nComputes magnitudes of complex matrix elements. More...\n\nvoid cv::cuda::magnitude (InputArray x, InputArray y, OutputArray magnitude, Stream &stream=Stream::Null())\n\nvoid cv::cuda::magnitudeSqr (InputArray xy, OutputArray magnitude, Stream &stream=Stream::Null())\nComputes squared magnitudes of complex matrix elements. More...\n\nvoid cv::cuda::magnitudeSqr (InputArray x, InputArray y, OutputArray magnitude, Stream &stream=Stream::Null())\n\nvoid cv::cuda::max (InputArray src1, InputArray src2, OutputArray dst, Stream &stream=Stream::Null())\nComputes the per-element maximum of two matrices (or a matrix and a scalar). More...\n\nvoid cv::cuda::min (InputArray src1, InputArray src2, OutputArray dst, Stream &stream=Stream::Null())\nComputes the per-element minimum of two matrices (or a matrix and a scalar). More...\n\nvoid cv::cuda::multiply (InputArray src1, InputArray src2, OutputArray dst, double scale=1, int dtype=-1, Stream &stream=Stream::Null())\nComputes a matrix-matrix or matrix-scalar per-element product. More...\n\nvoid cv::cuda::phase (InputArray x, InputArray y, OutputArray angle, bool angleInDegrees=false, Stream &stream=Stream::Null())\nComputes polar angles of complex matrix elements. More...\n\nvoid cv::cuda::polarToCart (InputArray magnitude, InputArray angle, OutputArray x, OutputArray y, bool angleInDegrees=false, Stream &stream=Stream::Null())\nConverts polar coordinates into Cartesian. More...\n\nvoid cv::cuda::pow (InputArray src, double power, OutputArray dst, Stream &stream=Stream::Null())\nRaises every matrix element to a power. More...\n\nvoid cv::cuda::rshift (InputArray src, Scalar_< int > val, OutputArray dst, Stream &stream=Stream::Null())\nPerforms pixel by pixel right shift of an image by a constant value. More...\n\nstatic void cv::cuda::scaleAdd (InputArray src1, double alpha, InputArray src2, OutputArray dst, Stream &stream=Stream::Null())\nadds scaled array to another one (dst = alpha*src1 + src2) More...\n\nvoid cv::cuda::sqr (InputArray src, OutputArray dst, Stream &stream=Stream::Null())\nComputes a square value of each matrix element. More...\n\nvoid cv::cuda::sqrt (InputArray src, OutputArray dst, Stream &stream=Stream::Null())\nComputes a square root of each matrix element. More...\n\nvoid cv::cuda::subtract (InputArray src1, InputArray src2, OutputArray dst, InputArray mask=noArray(), int dtype=-1, Stream &stream=Stream::Null())\nComputes a matrix-matrix or matrix-scalar difference. More...\n\ndouble cv::cuda::threshold (InputArray src, OutputArray dst, double thresh, double maxval, int type, Stream &stream=Stream::Null())\nApplies a fixed-level threshold to each array element. More...\n\n## § abs()\n\n void cv::cuda::abs ( InputArray src, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes an absolute value of each matrix element.\n\nParameters\n src Source matrix. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\nSee also\nabs\n\n## § absdiff()\n\n void cv::cuda::absdiff ( InputArray src1, InputArray src2, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes per-element absolute difference of two matrices (or of a matrix and scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and type as the input array(s). stream Stream for the asynchronous version.\nSee also\nabsdiff\n\n## § add()\n\n void cv::cuda::add ( InputArray src1, InputArray src2, OutputArray dst, InputArray mask = noArray(), int dtype = -1, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a matrix-matrix or matrix-scalar sum.\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. Matrix should have the same size and type as src1 . dst Destination matrix that has the same size and number of channels as the input array(s). The depth is defined by dtype or src1 depth. mask Optional operation mask, 8-bit single channel array, that specifies elements of the destination array to be changed. The mask can be used only with single channel images. dtype Optional depth of the output array. stream Stream for the asynchronous version.\nSee also\nadd\n\n## § addWeighted()\n\n void cv::cuda::addWeighted ( InputArray src1, double alpha, InputArray src2, double beta, double gamma, OutputArray dst, int dtype = -1, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes the weighted sum of two arrays.\n\nParameters\n src1 First source array. alpha Weight for the first array elements. src2 Second source array of the same size and channel number as src1 . beta Weight for the second array elements. dst Destination array that has the same size and number of channels as the input arrays. gamma Scalar added to each sum. dtype Optional depth of the destination array. When both input arrays have the same depth, dtype can be set to -1, which will be equivalent to src1.depth(). stream Stream for the asynchronous version.\n\nThe function addWeighted calculates the weighted sum of two arrays as follows:\n\n$\\texttt{dst} (I)= \\texttt{saturate} ( \\texttt{src1} (I)* \\texttt{alpha} + \\texttt{src2} (I)* \\texttt{beta} + \\texttt{gamma} )$\n\nwhere I is a multi-dimensional index of array elements. In case of multi-channel arrays, each channel is processed independently.\n\nSee also\naddWeighted\n\n## § bitwise_and()\n\n void cv::cuda::bitwise_and ( InputArray src1, InputArray src2, OutputArray dst, InputArray mask = noArray(), Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nPerforms a per-element bitwise conjunction of two matrices (or of matrix and scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and type as the input array(s). mask Optional operation mask, 8-bit single channel array, that specifies elements of the destination array to be changed. The mask can be used only with single channel images. stream Stream for the asynchronous version.\n\n## § bitwise_not()\n\n void cv::cuda::bitwise_not ( InputArray src, OutputArray dst, InputArray mask = noArray(), Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nPerforms a per-element bitwise inversion.\n\nParameters\n src Source matrix. dst Destination matrix with the same size and type as src . mask Optional operation mask, 8-bit single channel array, that specifies elements of the destination array to be changed. The mask can be used only with single channel images. stream Stream for the asynchronous version.\n\n## § bitwise_or()\n\n void cv::cuda::bitwise_or ( InputArray src1, InputArray src2, OutputArray dst, InputArray mask = noArray(), Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nPerforms a per-element bitwise disjunction of two matrices (or of matrix and scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and type as the input array(s). mask Optional operation mask, 8-bit single channel array, that specifies elements of the destination array to be changed. The mask can be used only with single channel images. stream Stream for the asynchronous version.\n\n## § bitwise_xor()\n\n void cv::cuda::bitwise_xor ( InputArray src1, InputArray src2, OutputArray dst, InputArray mask = noArray(), Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nPerforms a per-element bitwise exclusive or operation of two matrices (or of matrix and scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and type as the input array(s). mask Optional operation mask, 8-bit single channel array, that specifies elements of the destination array to be changed. The mask can be used only with single channel images. stream Stream for the asynchronous version.\n\n## § cartToPolar()\n\n void cv::cuda::cartToPolar ( InputArray x, InputArray y, OutputArray magnitude, OutputArray angle, bool angleInDegrees = false, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nConverts Cartesian coordinates into polar.\n\nParameters\n x Source matrix containing real components ( CV_32FC1 ). y Source matrix containing imaginary components ( CV_32FC1 ). magnitude Destination matrix of float magnitudes ( CV_32FC1 ). angle Destination matrix of angles ( CV_32FC1 ). angleInDegrees Flag for angles that must be evaluated in degrees. stream Stream for the asynchronous version.\nSee also\ncartToPolar\n\n## § compare()\n\n void cv::cuda::compare ( InputArray src1, InputArray src2, OutputArray dst, int cmpop, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nCompares elements of two matrices (or of a matrix and scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size as the input array(s) and type CV_8U. cmpop Flag specifying the relation between the elements to be checked: CMP_EQ: a(.) == b(.) CMP_GT: a(.) > b(.) CMP_GE: a(.) >= b(.) CMP_LT: a(.) < b(.) CMP_LE: a(.) <= b(.) CMP_NE: a(.) != b(.) stream Stream for the asynchronous version.\nSee also\ncompare\n\n## § divide()\n\n void cv::cuda::divide ( InputArray src1, InputArray src2, OutputArray dst, double scale = 1, int dtype = -1, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a matrix-matrix or matrix-scalar division.\n\nParameters\n src1 First source matrix or a scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and number of channels as the input array(s). The depth is defined by dtype or src1 depth. scale Optional scale factor. dtype Optional depth of the output array. stream Stream for the asynchronous version.\n\nThis function, in contrast to divide, uses a round-down rounding mode.\n\nSee also\ndivide\n\n## § exp()\n\n void cv::cuda::exp ( InputArray src, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes an exponent of each matrix element.\n\nParameters\n src Source matrix. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\nSee also\nexp\n\n## § log()\n\n void cv::cuda::log ( InputArray src, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a natural logarithm of absolute value of each matrix element.\n\nParameters\n src Source matrix. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\nSee also\nlog\n\n## § lshift()\n\n void cv::cuda::lshift ( InputArray src, Scalar_< int > val, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nPerforms pixel by pixel right left of an image by a constant value.\n\nParameters\n src Source matrix. Supports 1, 3 and 4 channels images with CV_8U , CV_16U or CV_32S depth. val Constant values, one per channel. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\n\n## § magnitude() [1/2]\n\n void cv::cuda::magnitude ( InputArray xy, OutputArray magnitude, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes magnitudes of complex matrix elements.\n\nParameters\n xy Source complex matrix in the interleaved format ( CV_32FC2 ). magnitude Destination matrix of float magnitudes ( CV_32FC1 ). stream Stream for the asynchronous version.\nSee also\nmagnitude\n\n## § magnitude() [2/2]\n\n void cv::cuda::magnitude ( InputArray x, InputArray y, OutputArray magnitude, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nThis is an overloaded member function, provided for convenience. It differs from the above function only in what argument(s) it accepts. computes magnitude of each (x(i), y(i)) vector supports only floating-point source\n\nParameters\n x Source matrix containing real components ( CV_32FC1 ). y Source matrix containing imaginary components ( CV_32FC1 ). magnitude Destination matrix of float magnitudes ( CV_32FC1 ). stream Stream for the asynchronous version.\n\n## § magnitudeSqr() [1/2]\n\n void cv::cuda::magnitudeSqr ( InputArray xy, OutputArray magnitude, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes squared magnitudes of complex matrix elements.\n\nParameters\n xy Source complex matrix in the interleaved format ( CV_32FC2 ). magnitude Destination matrix of float magnitude squares ( CV_32FC1 ). stream Stream for the asynchronous version.\n\n## § magnitudeSqr() [2/2]\n\n void cv::cuda::magnitudeSqr ( InputArray x, InputArray y, OutputArray magnitude, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nThis is an overloaded member function, provided for convenience. It differs from the above function only in what argument(s) it accepts. computes squared magnitude of each (x(i), y(i)) vector supports only floating-point source\n\nParameters\n x Source matrix containing real components ( CV_32FC1 ). y Source matrix containing imaginary components ( CV_32FC1 ). magnitude Destination matrix of float magnitude squares ( CV_32FC1 ). stream Stream for the asynchronous version.\n\n## § max()\n\n void cv::cuda::max ( InputArray src1, InputArray src2, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes the per-element maximum of two matrices (or a matrix and a scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and type as the input array(s). stream Stream for the asynchronous version.\nSee also\nmax\n\n## § min()\n\n void cv::cuda::min ( InputArray src1, InputArray src2, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes the per-element minimum of two matrices (or a matrix and a scalar).\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and type as the input array(s). stream Stream for the asynchronous version.\nSee also\nmin\n\n## § multiply()\n\n void cv::cuda::multiply ( InputArray src1, InputArray src2, OutputArray dst, double scale = 1, int dtype = -1, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a matrix-matrix or matrix-scalar per-element product.\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. dst Destination matrix that has the same size and number of channels as the input array(s). The depth is defined by dtype or src1 depth. scale Optional scale factor. dtype Optional depth of the output array. stream Stream for the asynchronous version.\nSee also\nmultiply\n\n## § phase()\n\n void cv::cuda::phase ( InputArray x, InputArray y, OutputArray angle, bool angleInDegrees = false, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes polar angles of complex matrix elements.\n\nParameters\n x Source matrix containing real components ( CV_32FC1 ). y Source matrix containing imaginary components ( CV_32FC1 ). angle Destination matrix of angles ( CV_32FC1 ). angleInDegrees Flag for angles that must be evaluated in degrees. stream Stream for the asynchronous version.\nSee also\nphase\n\n## § polarToCart()\n\n void cv::cuda::polarToCart ( InputArray magnitude, InputArray angle, OutputArray x, OutputArray y, bool angleInDegrees = false, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nConverts polar coordinates into Cartesian.\n\nParameters\n magnitude Source matrix containing magnitudes ( CV_32FC1 or CV_64FC1 ). angle Source matrix containing angles ( same type as magnitude ). x Destination matrix of real components ( same type as magnitude ). y Destination matrix of imaginary components ( same type as magnitude ). angleInDegrees Flag that indicates angles in degrees. stream Stream for the asynchronous version.\n\n## § pow()\n\n void cv::cuda::pow ( InputArray src, double power, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nRaises every matrix element to a power.\n\nParameters\n src Source matrix. power Exponent of power. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\n\nThe function pow raises every element of the input matrix to power :\n\n$\\texttt{dst} (I) = \\fork{\\texttt{src}(I)^power}{if \\texttt{power} is integer}{|\\texttt{src}(I)|^power}{otherwise}$\n\nSee also\npow\n\n## § rshift()\n\n void cv::cuda::rshift ( InputArray src, Scalar_< int > val, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nPerforms pixel by pixel right shift of an image by a constant value.\n\nParameters\n src Source matrix. Supports 1, 3 and 4 channels images with integers elements. val Constant values, one per channel. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\n\n## § scaleAdd()\n\n static void cv::cuda::scaleAdd ( InputArray src1, double alpha, InputArray src2, OutputArray dst, Stream & stream = Stream::Null() )\ninlinestatic\n\n#include <opencv2/cudaarithm.hpp>\n\nadds scaled array to another one (dst = alpha*src1 + src2)\n\n## § sqr()\n\n void cv::cuda::sqr ( InputArray src, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a square value of each matrix element.\n\nParameters\n src Source matrix. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\n\n## § sqrt()\n\n void cv::cuda::sqrt ( InputArray src, OutputArray dst, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a square root of each matrix element.\n\nParameters\n src Source matrix. dst Destination matrix with the same size and type as src . stream Stream for the asynchronous version.\nSee also\nsqrt\n\n## § subtract()\n\n void cv::cuda::subtract ( InputArray src1, InputArray src2, OutputArray dst, InputArray mask = noArray(), int dtype = -1, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nComputes a matrix-matrix or matrix-scalar difference.\n\nParameters\n src1 First source matrix or scalar. src2 Second source matrix or scalar. Matrix should have the same size and type as src1 . dst Destination matrix that has the same size and number of channels as the input array(s). The depth is defined by dtype or src1 depth. mask Optional operation mask, 8-bit single channel array, that specifies elements of the destination array to be changed. The mask can be used only with single channel images. dtype Optional depth of the output array. stream Stream for the asynchronous version.\nSee also\nsubtract\n\n## § threshold()\n\n double cv::cuda::threshold ( InputArray src, OutputArray dst, double thresh, double maxval, int type, Stream & stream = Stream::Null() )\n\n#include <opencv2/cudaarithm.hpp>\n\nApplies a fixed-level threshold to each array element.\n\nParameters\n src Source array (single-channel). dst Destination array with the same size and type as src . thresh Threshold value. maxval Maximum value to use with THRESH_BINARY and THRESH_BINARY_INV threshold types. type Threshold type. For details, see threshold . The THRESH_OTSU and THRESH_TRIANGLE threshold types are not supported. stream Stream for the asynchronous version.\nSee also\nthreshold"
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http://www.test3.guru99.com/python-loops-while-for-break-continue-enumerate.html | [
"### What is Loop?\n\nLoops can execute a block of code number of times until a certain condition is met. Their usage is fairly common in programming. Unlike other programming language that have For Loop, while loop, dowhile, etc.\n\n### What is For Loop?\n\nFor loop is used to iterate over elements of a sequence. It is often used when you have a piece of code which you want to repeat \"n\" number of time.\n\n### What is While Loop?\n\nWhile Loop is used to repeat a block of code. Instead of running the code block once, It executes the code block multiple times until a certain condition is met.\n\nIn this tutorial, we will learn\n\n## How to use \"While Loop\"\n\nWhile loop does the exactly same thing what \"if statement\" does, but instead of running the code block once, they jump back to the point where it began the code and repeats the whole process again.\n\nSyntax\n\n```while expression\nStatement\n```\n\nExample:\n\n```#\n#Example file for working with loops\n#\ndef main():\nx=0\n#define a while loop\nwhile(x <4):\nprint(x)\nx = x+1\n\nif __name__ == \"__main__\":\nmain()```\n• Code Line 4: Variable x is set to 0\n• Code Line 7: While loop checks for condition x<4. The current value of x is 0. Condition is true. Flow of control enters into while Loop\n• Code Line 8: Value of x is printed\n• Code Line 9: x is incremented by 1. Flow of control goes back to line 7. Now the value of x is 1 which is less than 4. The condition is true, and again the while loop is executed. This continues till x becomes 4, and the while condition becomes false.\n\n## How to use \"For Loop\"\n\nIn Python, \"for loops\" are called iterators.\n\nJust like while loop, \"For Loop\" is also used to repeat the program.\n\nBut unlike while loop which depends on condition true or false. \"For Loop\" depends on the elements it has to iterate.\n\nExample:\n\n```#\n#Example file for working with loops\n#\ndef main():\nx=0\n#define a while loop\n#\twhile(x <4):\n#\t\tprint x\n#\t\tx = x+1\n\n#Define a for loop\nfor x in range(2,7):\nprint(x)\n\nif __name__ == \"__main__\":\nmain()```\n\nFor Loop iterates with number declared in the range.\n\nFor example,\n\nFor Loop for x in range (2,7)\n\nWhen this code is executed, it will print the number between 2 and 7 (2,3,4,5,6). In this code, number 7 is not considered inside the range.\n\nFor Loops can also be used for a set of other things and not just number. We will see thin in next section.\n\n## How to use For Loop for String\n\nIn this step, we will see how \"for loops\" can also be used for other things besides numbers.\n\nExample:\n\n```def main():\n#use a for loop over a collection\nMonths = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\nfor m in Months:\nprint(m)\n\nif __name__ == \"__main__\":\nmain()```\n\nCode Line 3: We store the months (\"Jan, Feb , Mar,April,May,June\") in variable Months\n\nCode Line 4: We iterate the for loop over each value in Months. The current value of Months in stored in variable m\n\nCode Line 5: Print the month\n\n## How to use break statements in For Loop\n\nBreakpoint is a unique function in For Loop that allows you to break or terminate the execution of the for loop\n\nExample:\n\n```\ndef main():\n#use a for loop over a collection\n#Months = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\n#for m in Months:\n#print m\n\n# use the break and continue statements\n\nfor x in range (10,20):\nif (x == 15): break\n#if (x % 2 == 0) : continue\nprint(x)\n\nif __name__ == \"__main__\":\nmain()```\nIn this example, we declared the numbers from 10-20, but we want that our for loop to terminate at number 15 and stop executing further. For that, we declare break function by defining (x==15): break, so as soon as the code calls the number 15 it terminates the program Code Line 10 declare variable x between range (10, 20)\n• Code Line 11 declare the condition for breakpoint at x==15,\n• Code Line 12 checks and repeats the steps until it reaches number 15\n• Code Line 13 Print the result in output\n\n## How to use \"continue statement\" in For Loop\n\nContinue function, as the name indicates, will terminate the current iteration of the for loop BUT will continue execution of the remaining iterations.\n\nExample\n\n```def main():\n#use a for loop over a collection\n#Months = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\n#for m in Months:\n#print m\n\n# use the break and continue statements\n\nfor x in range (10,20):\n#if (x == 15): break\nif (x % 5 == 0) : continue\nprint(x)\n\nif __name__ == \"__main__\":\nmain()```\n\nContinue statement can be used in for loop when you want to fetch a specific value from the list.\n\nIn our example, we have declared value 10-20, but between these numbers we only want those number that are NOT divisible by 5 or in other words which don't give zero when divided by 5.\n\nSo, in our range (10,11, 12….19,20) only 3 numbers falls (10,15,20) that are divisible by 5 and rest are not.\n\nSo except number 10,15 & 20 the \"for loop\" will not continue and print out those number as output.\n\n• Code line 10 declare the variable x for range (10, 20)\n• Code line 12 declare the condition for x divided by 5=0 continue\n• Code line 13 print the result\n\n## How to use \"enumerate\" function for \"For Loop\"\n\nEnumerate function in \"for loop\" does two things\n\n• It returns the index number for the member\n• And the member of the collection that we are looking at\n\nExample:\n\nEnumerate function is used for the numbering or indexing the members in the list.\n\nSuppose, we want to do numbering for our month ( Jan, Feb, Marc, ….June), so we declare the variable i that enumerate the numbers while m will print the number of month in list.\n\n```def main():\n#use a for loop over a collection\nMonths = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\nfor i, m in enumerate (Months):\nprint(i,m)\n\n# use the break and continue statements\n\n#for x in range (10,20):\n#if (x == 15): break\n#if (x % 5 == 0) : continue\n#print x\n\nif __name__ == \"__main__\":\nmain()```\n\nWhen code is executed the output of the enumerate function returns the months name with an index number like (0-Jan), (1- Feb), (2- March), etc.\n\n• Code Line 3 declares the list of months [ Jan, Feb,…Jun]\n• Code Line 4 declares variable i and m for For Loop\n• Code Line 5 will print the result and again enter the For Loop for the rest of the months to enumerate\n\n## Pratical Example\n\nLet see another example for For Loop to repeat the same statement over and again.\n\n Python loop Working Code for all exercises Code for while loop ```def main(): x=0 while (x<4): print x x= x+1 if __name__== \"__main__\": main() ``` For Loop Simple Example ```def main(): x=0 for x in range (2,7): print x if __name__== \"__main__\": main() ``` Use of for loop in string ```def main(): Months = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"] for m in (Months): print m if __name__== \"__main__\": main() ``` Use break-statement in for loop ```def main(): for x in range (10,20): if (x == 15): break print x if __name__== \"__main__\": main() ``` Use of Continue statement in for loop ```def main(): for x in range (10,20): if (x % 5 == 0): continue print x if __name__== \"__main__\": main() ``` Code for \"enumerate function\" with \"for loop\" ```def main(): Months = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"] for i, m in enumerate (Months): print i,m if __name__== \"__main__\": main() ```\n\n## How to use for loop to repeat the same statement over and again\n\nYou can use for loop for even repeating the same statement over and again. Here in the example we have printed out word \"guru99\" three times.\n\nExample: To repeat same statement number of times, we have declared the number in variable i (i in 123). So when you run the code as shown below, it prints the statement (guru99) that many times the number declared for our the variable in ( i in 123).\n\n```for i in '123':\nprint \"guru99\",i,```\n\nLike other programming languages, Python also uses a loop but instead of using a range of different loops it is restricted to only two loops \"While loop\" and \"for loop\".\n\n• While loops are executed based on whether the conditional statement is true or false.\n• For loops are called iterators, it iterates the element based on the condition set\n• Python For loops can also be used for a set of various other things (specifying the collection of elements we want to loop over)\n• Breakpoint is used in For Loop to break or terminate the program at any particular point\n• Continue statement will continue to print out the statement, and prints out the result as per the condition set\n• Enumerate function in \"for loop\" returns the member of the collection that we are looking at with the index number\n\nPython 2 Example\n\nAbove codes are Python 3 examples, If you want to run in Python 2 please consider following code.\n\n```# How to use \"While Loop\"\n#Example file for working with loops\n#\ndef main():\nx=0\n#define a while loop\nwhile(x <4):\nprint x\nx = x+1\n\nif __name__ == \"__main__\":\nmain()\n\n#How to use \"For Loop\"\n#Example file for working with loops\n#\ndef main():\nx=0\n#define a while loop\n#\twhile(x <4):\n#\t\tprint x\n#\t\tx = x+1\n\n#Define a for loop\nfor x in range(2,7):\nprint x\n\nif __name__ == \"__main__\":\nmain()\n\n#How to use For Loop for String\ndef main():\n#use a for loop over a collection\nMonths = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\nfor m in Months:\nprint m\n\nif __name__ == \"__main__\":\nmain()\n\n#How to use break statements in For Loop\ndef main():\n#use a for loop over a collection\n#Months = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\n#for m in Months:\n#print m\n\n# use the break and continue statements\nfor x in range (10,20):\nif (x == 15): break\n#if (x % 2 == 0) : continue\nprint x\n\nif __name__ == \"__main__\":\nmain()\n\n#How to use \"continue statement\" in For Loop\ndef main():\n#use a for loop over a collection\n#Months = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\n#for m in Months:\n#print m\n\n# use the break and continue statements\nfor x in range (10,20):\n#if (x == 15): break\nif (x % 5 == 0) : continue\nprint x\n\nif __name__ == \"__main__\":\nmain()\n\n#How to use \"enumerate\" function for \"For Loop\"\ndef main():\n#use a for loop over a collection\nMonths = [\"Jan\",\"Feb\",\"Mar\",\"April\",\"May\",\"June\"]\nfor i, m in enumerate (Months):\nprint i,m\n\n# use the break and continue statements\n#for x in range (10,20):\n#if (x == 15): break\n#if (x % 5 == 0) : continue\n#print x\n\nif __name__ == \"__main__\":\nmain()\n```\n\nYOU MIGHT LIKE:"
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https://brilliant.org/problems/ramanujan-and-mahalanobis/ | [
"# Ramanujan And Mahalanobis\n\nIn 1914, Ramanujan's friend P. C. Mahalanobis gave him a problem he had read in the English magazine Strand. The problem was to determine the number $x$ of a particular house on a street where the houses were numbered $1,2,3,\\ldots,n$. The house with number $x$ had the property that the sum of the house numbers to the left of it equaled the sum of the house numbers to the right of it. The problem specified that $50 < n < 500$.\n\nRamanujan quickly dictated a continued fraction for Mahalanobis to write down. The numerators and denominators of the convergents to that continued fraction gave all solutions $(n,x)$ to the problem $($not just the particular one where $50 < n < 500).$ Mahalanobis was astonished, and asked Ramanujan how he had found the solution.\n\nRamanujan responded, \"...It was clear that the solution should obviously be a continued fraction; I then thought, which continued fraction? And the answer came to my mind.\"\n\nThis is not the most illuminating answer! If we cannot duplicate the genius of Ramanujan, let us at least find the solution to the original problem. What is $x$?\n\n\nBonus: Which continued fraction did Ramanujan give Mahalanobis?\n\n×"
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https://www.basic-mathematics.com/basic-mathematics-worksheets.html | [
"# Basic mathematics worksheets\n\nA great source of basic mathematics worksheets, such as fractions worksheets, addition worksheets that you can print easily to practice and strengthen your basic mathematics skills.\n\nTeachers should feel free to print them and use them as quizzes, tests, or more practices. Some worksheets come with answers!\n\nRecommended basic math worksheets generator: Basic mathematics worksheets\n\nPreschool math worksheets\nTracing numbers, identifying numbers, number words, and counting\n\nFirst grade math worksheets\nTelling time, counting, counting money worksheets, ...\n\nBasic math test\na comprehensive basic math test that will challenge you and help you find out if you know your basic math\n\nFractions test\na comprehensive fractions test that will challenge you and help you find out if you know your fractions\n\nFractions worksheets:\na variety of fractions worksheets such as adding fractions worksheets, subtracting fractions worksheets etc...\n\nAddition worksheets:\na variety of printable addition worksheets, such as adding whole numbers with 1, 2, or 3 digits\n\nDo you want to make unlimited free addition worksheets? No problems. You can do it here too! All worksheets below can create or generate unlimited worksheets\n\nEvery time you want a new worksheet with different numbers, just click the refresh button. Or just go back and return to the same worksheet page and you will get a whole new worksheet!\n\nAddition worksheets generator:with carry or no carry\n\nAdding worksheets with one digit\nGenerate unlimited addition worksheets with one digit.\n\nAdding worksheets with two digits\nGenerate unlimited addition worksheets with two digits.\n\nAdding worksheets with three digits\nGenerate unlimited addition worksheets with three digits.\n\nSubtraction worksheets generator:with no borrowing\n\nSubtracting worksheets generator\nGenerate unlimited subtraction worksheets with one digit or two digits.\n\nFractions worksheets generator:\n\nAdding fractions worksheets\nGenerate unlimited worksheets when the fractions have the same denominators.\n\nAdding fractions worksheets: different denominator\nGenerate unlimited free worksheets when the fractions have different denominators.\n\nMultiplication of fractions\nGenerate unlimited free worksheets for multiplying fractions.\n\nDivision of fractions\nGenerate unlimited free worksheets for dividing fractions.\n\ninteger worksheets generator:\n\nInteger worksheets\nGenerate unlimited worksheets for adding, subtracting, and multiplying integers.\n\nYou may also want to check other popular topics in our site\n\nBasic mathematics games\nBasic mathematics games of all sorts to sharpen your math skills with unlimited practices\n\nBasic math calculator\nA wide variety of basic math calculators. Even a Pythagorean theorem calculator is here to help you easily solve for geometry problems!\n\n## Recent Articles",
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"1. ### Basic Math Review Game - Math adventure!\n\nMay 19, 19 09:20 AM\n\nBasic math review game - The one and only math adventure game online. Embark on a quest to solve math problems!\n\nRead More\n\nNew math lessons\n\nYour email is safe with us. We will only use it to inform you about new math lessons.",
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"Tough Algebra Word Problems.\n\nIf you can solve these problems with no help, you must be a genius!\n\n## Recent Articles",
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"1. ### Basic Math Review Game - Math adventure!\n\nMay 19, 19 09:20 AM\n\nBasic math review game - The one and only math adventure game online. Embark on a quest to solve math problems!\n\nRead More\n\nEverything you need to prepare for an important exam!\n\nK-12 tests, GED math test, basic math tests, geometry tests, algebra tests.",
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"Real Life Math Skills\n\nLearn about investing money, budgeting your money, paying taxes, mortgage loans, and even the math involved in playing baseball."
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https://www.proprofs.com/quiz-school/story.php?title=year-8-maths-recall-3-algebra | [
"# Year 8 Maths 3.1 Algebra - Variables And Expressions\n\n10 Questions | Total Attempts: 1177",
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"For those of you who love all things mathematics especially graphs and algebra that measures the strength of materials then this is the quiz for you. If you are preparing for exams, and want to test your knowledge, try it out.\n\n• 1.\nWhat is the algebraic expression to represent m divided by 3\n• A.\n• B.\n• C.\n\n3m\n\n• D.\n\nM3\n\n• 2.\nWhat is the algebraic expression to represent two lots of p\n• A.\n• B.\n• C.\n• D.\n• 3.\nWhat is the algebraic expression to represent y less than twenty\n• A.\n\nY + 20\n\n• B.\n\nY - 20\n\n• C.\n\n20 + y\n\n• D.\n\n20 - y\n\n• 4.\nA ___________________ describes an unknown amount in an algebraic expression or equation. Can be represented by letters or symbols (e.g. a, y, ) and can be called pronumerals (which means 'in place of' a number).\n• A.\n\nCoefficient\n\n• B.\n\nConstant\n\n• C.\n\nEquation\n\n• D.\n\nExpression\n\n• E.\n\nTerm\n\n• F.\n\nVariable\n\n• 5.\nThe following are examples of ___________________ .\n• A.\n\nCoefficients\n\n• B.\n\nConstants\n\n• C.\n\nAn equation\n\n• D.\n\nAn expression\n\n• E.\n\nTerms\n\n• F.\n\nVariables\n\n• 6.\n8x means that x is multiplied 8 times. The number in front of the term x is called the _______________.\n• A.\n\nCoefficient\n\n• B.\n\nConstant\n\n• C.\n\nAn equation\n\n• D.\n\nAn expression\n\n• E.\n\nTerm\n\n• F.\n\nVariable\n\n• 7.\nA number that is written by itself is called a _______________. E.g. in the expression 3x + 6, the 6 is a _____________.\n• A.\n\nCoefficient\n\n• B.\n\nConstant\n\n• C.\n\nAn equation\n\n• D.\n\nAn expression\n\n• E.\n\nTerms\n\n• F.\n\nVariable\n\n• 8.\nTwo expressions joined together by an equals sign is E.g. 3x + 6 = 2x - 4.\n• A.\n\nCoefficients\n\n• B.\n\nConstants\n\n• C.\n\nAn equation\n\n• D.\n\nAn expression\n\n• E.\n\nTerms\n\n• F.\n\nVariables\n\n• 9.\nWhen two or more terms are added or subtracted together, it creates ___________________. E.g. 3x + 4y\n• A.\n\nCoefficients\n\n• B.\n\nConstants\n\n• C.\n\nAn equation\n\n• D.\n\nAn expression\n\n• E.\n\nTerms\n\n• F.\n\nVariables\n\n• 10.\nWhat is the algebraic expression to represent ten more than x?\n• A.\n\n10x\n\n• B.\n\nX - 10\n\n• C.\n\nX + 10\n\nRelated Topics",
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"https://www.proprofs.com/quiz-school/images/story_settings_gear.png",
null,
"https://www.proprofs.com/quiz-school/images/story_settings_gear_color.png",
null,
"https://www.proprofs.com/quiz-school/loader.gif",
null,
"https://media.proprofs.com/images/QM/user_images/2362275/1548060184.jpg",
null,
"https://www.proprofs.com/quiz-school/img/top.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6456448,"math_prob":0.9845998,"size":651,"snap":"2021-43-2021-49","text_gpt3_token_len":167,"char_repetition_ratio":0.17310664,"word_repetition_ratio":0.1826923,"special_character_ratio":0.20430107,"punctuation_ratio":0.03773585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997689,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-23T04:21:11Z\",\"WARC-Record-ID\":\"<urn:uuid:1cfea152-e04c-48b1-bd96-ef906841bef9>\",\"Content-Length\":\"269541\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dde79682-0f5f-4471-af42-615671fa8e29>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf4f700a-d13a-4f58-a757-4f739756aef9>\",\"WARC-IP-Address\":\"104.26.13.111\",\"WARC-Target-URI\":\"https://www.proprofs.com/quiz-school/story.php?title=year-8-maths-recall-3-algebra\",\"WARC-Payload-Digest\":\"sha1:47V5NM5QFPQWWVFZ7JWONZSNHCGKKYRG\",\"WARC-Block-Digest\":\"sha1:KX5WPEXBPKJJ4NKF3YGDSMEWR3J3EZBU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585561.4_warc_CC-MAIN-20211023033857-20211023063857-00601.warc.gz\"}"} |
https://www.nmaps.net/10921 | [
"#",
null,
"## into my soul",
null,
"Hover over the thumbnail for a full-size version.\n\nAuthor great_sea author:great_sea bitesized n-art rated 2005-07-23 2007-01-14 5 by 37 people. \\$into my soul#great_sea#none#00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000|12^36,108!12^48,96!12^54,90!12^66,78!12^72,72!12^84,60!12^96,54!12^102,48!12^120,36!12^126,30!12^30,96!12^42,96!12^48,84!12^54,78!12^66,72!12^78,66!12^84,48!12^96,48!12^108,42!12^114,36!12^78,42!12^66,48!12^48,60!12^30,78!12^24,78!12^42,72!12^54,66!12^60,60!12^84,54!12^96,42!12^102,36!12^108,30!12^96,30!12^78,36!12^66,36!12^48,42!12^42,60!12^36,60!12^36,54!12^48,42!12^60,36!12^66,30!12^72,24!12^48,30!12^42,30!12^30,42!12^30,42!12^24,30!12^36,30!12^42,36!12^60,48!12^84,36!12^96,30!12^102,30!12^84,54!12^72,66!12^60,66!12^36,84!12^18,96!12^42,78!12^78,54!12^78,54!12^36,42!12^42,54!12^24,66!0^24,120!0^42,120!0^42,114!0^48,102!0^66,90!0^84,78!0^90,72!0^102,60!0^108,54!0^126,42!0^132,36!0^132,36!0^120,48!0^102,66!0^78,84!0^42,114!0^30,126!0^54,120!0^72,108!0^90,90!0^108,72!0^126,60!0^138,48!0^150,48!0^132,66!0^108,78!0^96,102!0^66,138!0^36,150!0^54,162!0^72,144!0^90,126!0^114,102!0^126,84!0^150,60!0^168,42!0^174,42!0^150,36!0^126,54!0^96,90!0^36,138!0^24,162!0^54,144!0^72,132!0^96,102!0^120,84!0^168,54!0^192,36!0^192,36!0^186,54!0^162,72!0^132,96!0^102,126!0^60,150!0^36,186!0^60,186!0^90,162!0^108,138!0^132,120!0^162,96!0^192,60!0^204,48!0^216,42!0^222,42!0^216,60!0^186,90!0^126,132!0^54,192!0^24,210!0^24,216!0^30,180!0^72,150!0^156,72!0^156,42!0^162,36!0^174,36!0^198,30!0^216,36!0^198,30!0^162,30!0^144,30!12^24,126!12^36,114!12^60,102!12^72,96!12^102,84!12^120,60!12^132,42!12^162,30!12^114,48!12^108,60!12^90,78!12^66,108!12^36,132!12^18,138!12^42,132!12^72,96!12^90,72!12^114,48!12^144,30!12^162,24!12^156,42!12^108,72!12^72,102!12^42,126!12^36,132!12^42,132!12^72,120!12^108,90!12^78,84!12^72,84!12^84,102!12^102,66!12^138,36!12^174,36!12^174,54!12^126,72!12^120,60!12^120,72!12^96,102!12^78,90!12^60,132!12^36,138!12^60,138!12^102,114!12^120,90!12^156,66!12^198,30!12^186,36!12^126,72!12^96,108!12^60,144!12^30,174!12^60,174!12^96,150!12^126,102!12^156,78!12^186,60!12^204,48!12^192,60!12^138,102!12^120,126!12^66,162!12^42,180!12^42,192!12^90,174!12^120,144!12^138,120!12^174,84!12^216,48!12^228,42!12^174,72!12^126,126!12^66,174!12^54,186!12^48,210!12^66,204!12^108,162!12^138,120!12^174,84!12^240,36!12^216,60!12^132,150!12^60,192!12^90,204!12^120,192!12^168,120!12^252,42!12^258,54!12^186,150!12^96,222!12^48,270!12^126,246!12^222,132!12^306,66!0^54,210!0^60,210!0^138,150!0^174,120!0^36,90!0^48,90!0^72,78!0^90,66!0^120,42!0^120,42!0^78,66!0^72,78!0^42,90!0^30,96!0^78,72!0^114,48!0^48,246!0^72,228!0^132,174!0^186,126!0^258,72!12^90,78!12^90,72!12^84,66!12^96,54!12^102,48!0^246,24!0^210,78!0^132,150!0^42,222!0^48,228!0^162,168!0^258,114!0^312,78!0^288,72!0^246,96!0^108,192!0^60,222!0^54,252!0^48,264!0^186,210!0^288,132!0^360,78!0^360,66!0^240,114!0^174,162!0^48,252!0^24,270!0^144,180!0^186,126!0^240,84!0^276,48!0^186,96!0^114,156!0^60,174!0^60,162!0^114,108!0^132,90!0^150,102!0^144,126!0^156,78!0^156,96!0^162,84!0^126,96!0^168,90!0^144,108!0^102,108!0^66,156!0^66,156!0^78,150!0^102,144!0^132,228!0^102,168!0^162,144!0^192,114!0^216,84!0^126,162!0^78,216!0^162,210!0^210,174!0^216,78!0^228,120!0^210,210!0^198,138!0^252,78!0^318,36!0^318,36!0^192,156!0^138,204!0^72,258!0^42,300!0^90,276!0^192,222!0^246,186!0^306,132!0^432,72!0^402,60!0^216,168!0^210,174!0^192,174!0^228,138!0^270,120!0^294,102!0^210,186!0^174,198!0^162,192!0^174,186!0^198,192!0^168,180!0^126,216!0^108,222!0^108,210!0^90,258!0^90,246!0^102,246!0^138,252!0^162,228!0^156,186!0^174,138!0^186,96!0^222,90!0^198,126!0^204,96!0^198,90!0^174,84!0^132,72!0^180,108!0^174,108!0^144,72!0^180,84!0^204,78!0^216,66!0^198,66!0^204,66!0^234,60!0^234,66!0^228,90!0^216,180!0^258,162!0^252,126!0^246,132!0^288,138!0^306,108!0^312,96!0^252,66!0^264,90!0^264,90!0^216,102!0^246,90!0^282,42!0^282,42!0^318,60!0^360,36!0^360,36!0^330,60!0^330,108!0^312,102!0^294,42!0^294,30!0^276,48!0^258,24!0^276,48!0^276,78!0^330,72!0^342,48!0^330,36!0^354,30!0^360,48!0^342,78!0^240,186!0^144,258!0^72,294!0^54,312!0^108,294!0^186,252!0^264,204!0^324,150!0^372,90!0^396,48!0^408,36!0^414,54!0^402,84!0^312,162!0^276,204!0^156,288!0^102,324!0^54,324!0^78,318!0^186,288!0^264,228!0^330,156!0^384,66!0^366,78!0^258,162!0^234,204!0^216,246!0^216,258!0^60,336!0^42,336!0^78,348!0^138,336!0^210,318!0^264,300!0^306,240!0^330,198!0^378,114!0^384,54!0^384,54!0^390,120!0^270,300!0^210,342!0^102,342!10^36,42!10^54,48!10^54,54!10^42,60!10^36,60!10^42,30!10^42,24!10^60,30!10^18,30!10^30,36!10^36,54!10^60,42!10^72,30!10^72,48!10^42,66!10^36,72!10^108,36!10^120,30!10^78,60!10^54,84!10^54,114!10^84,84!10^126,42!3^78,138!3^120,84!3^162,42!3^180,30!3^126,90!3^72,138!3^90,156!3^120,138!3^156,96!3^204,54!3^246,30!3^216,60!3^90,186!3^72,192!3^48,204!3^54,150!6^132,60,4,0,1,3!6^444,84,2,0,0,0!6^396,156,2,0,0,0!6^372,228,2,0,0,0!6^228,372,2,0,0,0!6^60,372,2,0,0,0!6^396,252,2,0,0,0!6^252,396,2,0,0,0!6^276,468,2,0,0,0!6^60,444,2,0,0,0!6^60,396,2,0,0,0!6^468,180,2,0,0,0!6^564,84,2,0,0,0!6^492,84,2,0,0,0!6^468,324,2,0,0,0!6^564,180,2,0,0,0!0^372,180!0^132,372!0^456,36!0^348,276!3^168,180!3^312,156!3^84,276!3^360,48!3^228,132!6^36,36,4,0,2,3!6^84,156,4,0,1,3!6^276,60,4,0,1,3# just a little artsy experiment here.\n\n## Other maps by this author",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"",
null,
"Flux Jetsy Free Laptop (revised) Madman Rotor The Epic Part IX - Cloud Wrangler\n\nPages: (0)\n\n### i woulda\n\nproffered the exit switch instead of the drones...\n\nup to 5!\n\n### this is my\n\nsort of map 5aved!!!\n\nEh.\n\n### Woah\n\nCame accross this three years after I commented.\nNeat.\n\n### 5\n\nand that pushes it to 4.5 average\n\n5/5.\n\npretty cool\n\n### 4.5/5\n\nI think the drones ruin it. If the drones weren't there, it would be a 5. Truely a masterpiece(in the thumbnail)\n\n### 2/5\n\nYeah, it looks awesome in the thumbnail, i guess. ok.\n\n### 5/5\n\nYou're right, that is a true work of art.\n\nI'm tempted to go photoshop it into something and pass it off as nmy own work. :-P\n\n5/5\n\n### -\n\nyeah, it really does.\n\n### lag much?\n\nwow, that looks awesome in the thumbnail!"
] | [
null,
"https://www.nmaps.net/static/logo.png",
null,
"http://lh3.ggpht.com/bBZvUzAEVBbe6wO6NhXUvVIBEBk5FdSeNp-rftHEgfL9Z4q-XhSs__phWkQn98Zd7wIs_6j5KkY_M_ZSZM_Rr_pH=s132",
null,
"http://lh3.ggpht.com/D7bKrnyliti6cRKcuhrxOz8Wb7LhNMbiGzRVx7Kas-sBKAJg-VtZ4HlRsnGJ3kYVFXmMBK_XMoKpb7yeEgnITIw=s132",
null,
"http://lh3.ggpht.com/8Y4WfWjONGInPre8kaIEQNg4WP9l1cgzf7rfY0ljyFe4urPu9EryqikP5R1Xfc2JCW5SElfaUd7vCyovVJ3fZw=s132",
null,
"http://lh3.ggpht.com/lkhN5vArq5Sruxgq2Ng6rngbskcqECaxpDNS1VDfznfXoaC-Ph-kHXJXmdTYXTDp_tJ6NdsA1OoC9hiC9zKaPfk=s132",
null,
"http://lh4.ggpht.com/oScPgTMzlEq4LKRN8fBw4eqLO_5wG8ddpQHNevm0ngQ_Pg1oHuQvZme6UAC75GjqudTeks9Uky9t-aM6ti752KM=s132",
null,
"http://lh4.ggpht.com/S-PezRKUd__8LLfnvcotXkTd_ndVNDGDTm5KruDDz4SXTNKs6Zh6ON7QY7CIb7oc4kv4neHQ8u3yalkzKh02RSs=s132",
null,
"http://lh6.ggpht.com/xuoGDHeKthksIRyg3ehqqy4SoN3JhdMDg0sDQYyg2nnpJ2ZQvhaKI3JlIXXlj1os8QyMszF_eZMVGgTYt7eSi1A=s132",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8851754,"math_prob":0.9997694,"size":845,"snap":"2020-34-2020-40","text_gpt3_token_len":278,"char_repetition_ratio":0.09988109,"word_repetition_ratio":0.18439716,"special_character_ratio":0.35502958,"punctuation_ratio":0.119791664,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95105344,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-19T08:14:24Z\",\"WARC-Record-ID\":\"<urn:uuid:51d5b4f7-27dc-4bde-960e-5b2c13dc0cfc>\",\"Content-Length\":\"22977\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9170859d-9f91-4435-8819-2313593f2a53>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c8261b3-1f6d-4cb6-944e-15efc750a300>\",\"WARC-IP-Address\":\"172.217.12.243\",\"WARC-Target-URI\":\"https://www.nmaps.net/10921\",\"WARC-Payload-Digest\":\"sha1:2TQXCZ2GT2BMDMPZ6W26A56I6CMLPCUU\",\"WARC-Block-Digest\":\"sha1:R53JDENGIHNSCWRYEGJU27L7ILFIHVLH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400191160.14_warc_CC-MAIN-20200919075646-20200919105646-00183.warc.gz\"}"} |
https://ifelse.info/questions/1685/error-fixes | [
"# Error with fixes\n\n3\n\nHello everyone, is that I try to solve a problem with minor and major fixes and make their difference but ... the program defines well the major element but not the minor so is the arrangement\n\n2 5 3 7 12 9 8 5 4 10\n\nDefine the biggest one that in this case is 12\n\nThe minor defines it badly that it prints 10 and it is supposed to be 2\n\nIt is the following code\n\n`````` #include<iostream>\n#define MAX 10002\nusing namespace std;\n\nint arr[MAX];\nint mayor=arr, menor=arr;\nint diferencia=0;\n\nint main()\n{\nfor(int i=1; i<=10; i++)\n{\ncin>>arr[i];\n\nif(arr[i]>mayor)\n{\nmayor=arr[i];\n}\n\nif(arr[i]<menor);\n{\nmenor=arr[i];\n}\n}\ncout<<menor<<\" \"<<mayor;\n}\n``````\n\nIf someone can help me, I'd really appreciate it\n\nasked by carlos 29.09.2018 в 20:04\nsource\n\n3\n\nLet's go in parts:\n\n``````int arr[MAX];\nint mayor=arr, menor=arr;\n``````\n\nAt this time, the content of `arr` is unknown. It can be anything, because you have not initialized the array.\n\nSince we do not know that value, it turns out that `mayor` is already greater than any of the elements you put (and would not change in the iterations). O `menor` is less than any of your elements (and would not change in iterations).\n\nFor example, if `menor` starts with 0, none of the elements in your example would replace it. `menor` has to be so high that you are sure that you will find a lower value, `mayor` so low that you are sure that you will find a higher value.\n\n``````int diferencia=0;\nint main() {\nfor(int i=1; i<=10; i++)\n{\ncin>>arr[i];\n``````\n\nIf you do not need to save the data, you really do not need an array. If you make `int numero; cin >> numero;` it works exactly the same.\n\nNaturally, in the following lines you would have to change `arr[i]` by `numero` .\n\n`````` if(arr[i]>mayor) {\nmayor=arr[i];\n}\nif(arr[i]<menor);\n``````\n\nYou have a `;` , so the `if` ends on the top line and the block below always runs.\n\n`````` {\nmenor=arr[i];\n}\n}\ncout<<menor<<\" \"<<mayor;\n``````\n\nThe impression is complicated because it is confused with the data of the last line. I think that's part of the problem; it's much better: `cout << endl << \"Menor :\" << menor << endl << \"Mayor: \" << mayor << endl;`\n\n``````}\n``````\n\n3\n\nWhen you start to program it is very normal that some details escape but once you have time you notice certain details, your mistake is that you put a `;` where you should not go.\n\n`````` #include<iostream>\n#define MAX 10002\nusing namespace std;\n\nint arr[MAX];\nint mayor=arr, menor=arr;\nint diferencia=0;\n\nint main()\n{\nfor(int i=1; i<=10; i++)\n{\ncin>>arr[i];\n\nif(arr[i]>mayor)\n{\nmayor=arr[i];\n}\n\nif(arr[i]<menor); <---- precisamente aquí\n{\nmenor=arr[i];\n}\n}\ncout<<menor<<\" \"<<mayor;\n}\n``````\n\n0\n\nBasically the program has two errors. One has already been indicated and it is that you have cast a `;` after a `if` :\n\n``````if(arr[i]<menor);\n// ^\n``````\n\nThat semicolon causes the `if` to do absolutely nothing ... silly language details. To see it more clearly we could replace that semicolon with two keys:\n\n``````if(arr[i]<menor){ }\n{\nmenor = arr[i];\n}\n``````\n\nNow it looks better than that semicolon overrides the conditional.\n\nOn the other hand the initialization of `menor` is not correct.\n\nGlobal arrays are initialized automatically, not those that belong to a function. That is:\n\n``````int arr[MAX]; // Todos sus elementos valen 0\n// Pero únicamente porque es variable global!!!\n\nint main()\n{\nint arr2[MAX]; // Array no inicializado!!!\n}\n``````\n\nWell, the next thing you do after declaring the fix is to declare and initialize `mayor` and `menor` :\n\n``````int mayor=arr, menor=arr;\n``````\n\nAnd here the problems begin ... this initialization makes `mayor=menor=0` . If we subsequently give values to `arr` , the value of `mayor` and `menor` will not change automatically. Your starting value will be always 0.\n\nThis initialization poses a problem and is that for `menor` find the smallest number, it must be equal to or less than 0. Otherwise the program will not work.\n\nTo avoid this mess of minimums and maximums I would use pointers:\n\n``````int* mayor = arr;\nint* menor = arr;\n\nfor(int i=0; i<MAX; i++)\n{\ncin>>arr[i];\n\nif(arr[i] > *mayor)\n{\nmayor = &arr[i];\n}\n\nif(arr[i] < *menor)\n{\nmenor = &arr[i];\n}\n}\n\ncout << *menor << ' ' << *mayor;\n``````\n\nThis solution uses pointers instead of variables. Thus, `menor` does not store a value of its own, but will point to the smallest element (the same for `mayor` ).\n\nIf using pointers is not a possibility, then you should pay special attention to the initialization of `menor` and `mayor` . Pulling the standard the best is to use `std::numeric_limits` :\n\n``````int menor = std::numeric_limits<int>::max();\nint mayor = std::numeric_limits<int>::min();\n``````\n\nWhat we do here is initialize `menor` with the largest number that can be stored in a `int` , in this way any number that the user enters will be equal or smaller and, consequently, the algorithm will be able to detect it as smaller. In `mayor` , instead, we store the smallest number possible, so that any number that the user enters is equal or larger."
] | [
null
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https://www.osapublishing.org/oe/fulltext.cfm?uri=oe-16-21-16659&id=172485 | [
"## Abstract\n\nFrom first principles we develop figures of merit to determine the gain experienced by the guided mode and the lasing threshold for devices based on high-index-contrast waveguides. We show that as opposed to low-index-contrast systems, this quantity is not equivalent to the power confinement since in high-index-contrast structures the electric and magnetic field distributions cannot be related by proportionality constant. We show that with a slot waveguide configuration it is possible to achieve more gain than one would expect based on the power confinement in the gain media. Using the figures of merit presented here we optimize a slot waveguide geometry to achieve low-threshold lasing and discuss the fabrication tolerances of such a design.\n\n## 1. Introduction\n\nA critical photonic component yet to be demonstrated on a silicon-based platform is an electrically pumped device with optical gain for amplification or lasing. Achieving this optical gain in a silicon-based device is extremely challenging since silicon is an indirect band-gap semiconductor and therefore an inefficient photon source. Hybrid silicon devices based on direct band-gap III–V materials bonded to silicon photonic elements present an interim solution , however, the reliance on a wafer bonding step prohibits a high throughput fabrication process possible with a silicon-based process.\n\nOne possible configuration for silicon-based gain is the recently proposed slot waveguide design [2, 3]. In this configuration a low-index gain material such as Er-doped SiO2 or Erdoped Si3N4 can be inserted into one or multiple thin slots between two silicon rails. Electrical excitation of the gain material could be achieved by passing a tunneling current through the slot-region.\n\nOne key advantage of this configuration is the large optical field enhancement in the slot-region due to the boundary conditions imposed on the electric field normal the slot interface . Since the normal electric displacement (D=εE) must be continuous across the interface, the electric field in the slot waveguide is enhanced by the ratio of the dielectric constant of silicon to that of the slot material. In semiconductor materials this enhancement can be as large as one order of magnitude.\n\nTypically waveguide gain is assumed to be proportional to the percentage of the guided mode power which overlaps with the gain medium; however, this is not true for high-index contrast waveguides due to the large electric field discontinuities at dielectric interfaces. This discrepancy results from Fermi’s Golden Rule which states the electric field of an electromagnetic wave, not the power, determines the emission rate for an excited state (and consequently the modal gain).\n\nIn standard low-index-contrast waveguides, optical gain and power confinement are considered proportional based on the following arguments. For electromagnetic plane waves in homogenous media the magnetic field H can be written in terms of the electric field E and the impedance of the material according to:\n\n$H=cεn(êz×E),$\n\nwhere êz is a unit vector along the direction of propagation (which we have chosen to be the z-direction) and n is the index of refraction of the material. This is often written in the form relating the major components of the electric and magnetic fields (for a TM mode in this case):\n\n$Ey=−ωμ0βHx,$\n\nwhere β is the propagation constant defined as β≡2πn̄/λ. Based on these relationships the electric field energy, and waveguide power stored in a given region can be used interchangeably since they differ only by a constant. In this case the percentage of power overlapping the gain medium can be used to calculate the resulting modal gain, and it is often assumed that the same can be said for waveguide modes.",
null,
"Fig. 1. Fundamental TM modes at a wavelength of 1.5 µm for waveguides 500 nm wide and 600 nm tall. All modes are normalized to unit power. The high-index material (n=3.5) is outlined in black. The waveguides are clad with n=3.25 for (a) and (b) and n=1.5 for (c)–(f). The first and second columns show E y and $−ωμ0βHx$ respectively, plotted on the same color scale. The two fields become increasingly dissimilar as more electric field is concentrated at high-index-contrast boundaries.\n\nFor high-index-contrast waveguides, however, the linear relationships between the electric and magnetic fields (Eqs. (1) and (2)) do not hold since they must satisfy different boundary conditions. This is shown in Fig. 1. For low-index-contrast waveguides, Eqs. (1) and (2) are close approximations. Figures 1(a) and 1(b) shows the fundamental TM mode with an index difference of 0.25 between the core and cladding. We see close agreement in the magnitude and spatial profiles of these two fields. However, as the index-contrast is increased to 2.5 (Fig. 1(c) and (d)) there is a noticeable difference between E y and $−ωμ0βHx$. Notice that E y must be continuous across the dielectric interfaces to the left and right of the waveguide, and discontinuous across the top and bottom interfaces. The H x on the other hand must be continuous across all interfaces since the magnetic susceptibility is the same in all regions. This leads to noticeable differences between the electric and magnetic field magnitudes and profiles. This difference becomes dramatic when the peak of the electric field is placed at a dielectric discontinuity as is the case for slot waveguides (Figs. 1(e) and 1(f)).\n\nThe strong difference between the spatial distribution of the electric and magnetic fields in high index contrast is the reason that power confinement is no longer a relevant quantity when calculating gain in high-index-contrast waveguides. This issue is also present when calculating the sensitivity of waveguides to changes in refractive index. Several previous papers have overcome this problem by either determining waveguide sensitivity empirically or by introducing correction factors , however, there is little discussion as to the origin of the correction factors.\n\nIn this paper we derive from first principles the appropriate confinement factors and figures of merit for modal gain in high-index-contrast waveguides. We show explicitly that the modal gain is dependent on two quantities: the group velocity and the electric field energy confinement in the slot region. The lasing threshold however is independent of the group velocity and determined only by the electric field energy confinement. We also show that in some instances decreasing the width of the slot can decrease the lasing threshold despite the reduction of gain material. This counter-intuitive result can be understood as the increased emission rate for material in narrow slots overcoming the reduction in volume of gain material. Additionally we show that the percentage of power confined to the slot region (sometimes used as a confinement factor) can incorrectly predict the modal gain in high-index- contrast waveguides.\n\n## 2. Confinement factor for high-index-contrast waveguides\n\nTo rigorously calculate gain in high-index-contrast waveguides we derive from first principles a proportionality constant (known as a confinement factor Γ) which relates bulk material gain (g b) to the modal gain (g m) in a guiding structure:\n\n$Γ≡gmgb,$\n\nwhere g m and g b have units of inverse length. The bulk material gain can be determined from the magnitude of the electric field for a plane wave propagating along the z-direction through the gain medium:\n\n$∣E(z)∣2=∣E0∣2egbz.$\n\nTo calculate the effective confinement factor for a given waveguide mode profile, we begin with an expression for the electric field of a guided mode propagating along the z-direction. This can be written as the sum of all three vector components of the electric field using Einstein summation notation:\n\n$E(x,y,z)=êjEjoψj(x,y)ei(ωt−β˜z),$\n\nwhere ê is the polarization vector, Ψ is the cross sectional mode profile, and β̃ is the complex propagation constant defined as\n\n$β˜≡k0(n̅r+in̅i),$\n\nwhere k 0 is the angular wavenumber in free space, and n̄r and n̄i are the real and imaginary parts of the effective index respectively. By writing Eq. (5) in the same form as Eq. (4) we can see that the modal gain is determined by the complex propagation constant. From Eq. (6) we can then write the gain of the guided mode in terms of the imaginary part of the effective index:\n\n$gm=2Im{β˜}=2k0n̅i.$\n\nSimilarly we can write the bulk material gain in terms of the imaginary part of refractive index of the active gain material:\n\n$gb=2k0nAi.$\n\nExpressing gain in this form allows us to treat it as a perturbation to the refractive index of the waveguide. We can now calculate the modal gain by introducing a small imaginary part to the refractive index (Δn Ai) in a given active region (A) and solve for the complex propagation constant of the guided mode. This can be performed using variational methods :\n\n$Δβ˜=ω∬∞Δε˜∣E∣2dxdy12∬∞Re{E×H*}∙êzdxdy,$\n\nwhere Δε̃=(n A+iΔn Ai)2. According to Eqs. (6)(8) this can be written in the form:\n\n$gm=[nAcε0∬A∣E∣2dxdy∬∞Re{E×H*}∙êzdxdy]gb.$\n\nHere c is the speed of light in vacuum, the integral in the numerator is carried out only over the area of the active gain region (since this is where Δε̃≠0), and the integral in the denominator is carried out over the entire cross section of the mode. We recognize the term in the brackets as the proportionality constant in Eq. (3) and therefore we can express the confinement factor as:\n\n$Γ=nAcε0∬A∣E∣2dxdy∬∞Re{E×H*}∙êzdxdy.$\n\nNote that as expected the stimulated emission rate (and thus the gain) depends on the intensity of the field which is proportional to |E|2. Since this term is normalized to unit power, the confinement factor can be thought of as the amount of intensity overlapping the gain medium per unit input power. Note that most previous derivations of this confinement factor err by incorrectly substituting the electric for magnetic field in attempts to simplify the expression. This is commonly written as :\n\n$12∬Re{E×H*}∙êzdxdy=12βωμ0∬∣E∣2dxdy,$\n\nand thus Eq. (11) could be written as the percentage of power or intensity confined to the active region. However, as shown in Section 1, the expression in Eq. (12) is not valid for high-index-contrast waveguides since it is based on the relationship for plane waves in homogeneous media that $H=εcn̅(êz×E)$ .\n\nThe expression for the confinement factor (Eq. (9)) can be simplified into the product of two terms: one related to the group velocity, and the other related to the confinement of the energy density of the electric field. The energy stored per unit length (U/l) in a dielectric waveguide can be written as :\n\n$U/l=12∫∫∞ε∣E∣2dxdy.$\n\nNote that here we have neglected material dispersion when writing the stored energy per unit length. To account for material dispersion one should replace epsilon with d(ωε)/ in Eq. (11) [16, 17]. If silicon is the most dispersive material, the error introduced by making this approximation is less than 7% at a wavelength of 1.55 microns. The group velocity of the mode (v g) describes the speed with which energy flows through a given cross section. Therefore we can write the power flux through a given cross section of the waveguide as:\n\n$12∬Re{E×H*}∙êzdxdy=vg12∬ε∣E∣2dxdy$\n\nUsing the definition of group index (n gc/v g) we substitute Eq. (14) into Eq. (11) and rewrite the confinement factor as:\n\n$Γ=ng∬Aε∣E∣2dxdynA∬∞ε∣E∣2dxdy≡ngnAγA.$\n\nWe see from the simplified expression for the confinement factor that the modal gain can be defined as the product of a term related to the group index and a term related to the confinement of the electric field energy density. The first term can be thought of as a confinement in time since increasing the real part of the group index relative to the bulk index has the effect of slowing the propagation of the guided mode. Therefore for a given waveguide length, light can spend more time in the gain media resulting in an enhancement of the modal gain per unit length. The second term represents the spatial confinement of the energy density to the active region of the waveguide which we define as:\n\n$γA≡∬Aε∣E∣2dxdy∬∞ε∣E∣2dxdy.$\n\nSince this term can be as large as 1 we see that the total confinement factor in Eq. (15) can be larger than 1 if the group index is larger than the bulk refractive index. This means that it is possible to achieve more gain per unit length in a guided structure than would be possible in bulk. This phenomenon has been noted before for modal absorption and results from the fact that light spends more time in a structure with a large group index, and thus interacts more with the gain material per unit length.\n\n## 3. Numerical verification of analytical results\n\nTo verify our expression for the confinement factor (Eq. (13)) we use numerical methods to calculate the relationship between material gain and modal gain and compare it to the analytical results. Since the material gain can be expressed in terms of an imaginary part of the dielectric constant, we can simulate material gain in a given waveguide region by adding an imaginary component to the dielectric constant and calculating the propagation constant of the guided mode using a finite difference mode solver. The imaginary part of this complex propagation constant can then be written in terms of the modal gain according to Eq. (7). This relationship between the modal gain and material gain is the definition of the confinement factor Eq. (3) and should match the derived expression Eq. (15).\n\nWe numerically calculate the modal gain according to the waveguide geometry and the corresponding fundamental TM mode as shown in Figure 2(a) and 2(b) respectively. The waveguide geometry consists of two high index rails 500 nm wide and 250 nm tall separated by a horizontal slot 50 nm tall. For simplicity we use a wavelength of 1.5 µm and 3.5 and 1.5 for the high and low refractive indices respectively. This is approximately the index contrast between Si and SiO2. Figure 2(b) shows the fundamental TM mode calculated using a Matlab-based finite difference mode solver. We simulate material gain by introducing an imaginary part of the dielectric constant to the low-index slot region. For each value of material gain we use the finite difference mode solver to calculate the complex propagation constant and calculate the corresponding modal gain according to Eq. (7).",
null,
"Fig. 2. Numerical study of modal gain. (a) Schematic of slot waveguide with gain material defined by an imaginary component of the refractive index confined to the slot region (pink). (b) Major field component of the fundamental TM mode for the same structure as (a) calculated using a finite difference mode solver. (c) Circles show the modal gain (g m) calculated from the complex effective index of the fundamental TM mode as determined using a finite difference mode solver. Material gain is added via the imaginary part of the refractive index in the slot. Dashed line shows the modal gain calculated according to Eq. (3) based on the confinement factor Γ determined from the zero-gain mode profile from Eq. (15). Dotted line shows the product of the power in the active gain region (P A) and the material gain. We see that the confinement factor proposed in this paper correctly predicts the modal gain simulated numerically, while the power confinement greatly underestimates the simulated modal gain.\n\nBy plotting the numerically calculated modal gain (g m) versus material gain (g b) in Fig. 1(c) we show excellent agreement with the analytically calculated confinement factor. The slope of the line g m versus g b (circles in Fig. 1(c)) represents the confinement factor according to Eq. (3). We also calculate the confinement factor Γ based on Eq. (15) and the calculated TM mode in Fig. 1(b). We plot Γg b as the dashed line in Fig. 1(c). As expected, our calculated confinement factor agrees very well with the relationship between the modal and material gain from numerical simulations. The difference between these two factors (0.4328 from Eq. (15) and 0.4368 from the numerical simulations) is less than 1%. To highlight the difference between this confinement factor and the confinement of power to the gain medium we also plot on this graph P A g b where we define the power in the active region as:\n\n$PA≡∬ARe{E×H*}∙ezdxdy∬∞Re{E×H*}∙ezdxdy.$\n\nWe see from Fig. 1(c) that the modal gain of a slot waveguide is substantially larger than would be expected from the percentage of power confined to the gain region. As discussed in Section 2, this enhanced modal gain results from both the large group index as well as the increased electric field energy density in the slot. The electric field energy density is underestimated using the H field (See Figs. 1(e) and 1(f)).\n\n## 4. Minimizing the lasing threshold\n\nOne of the primary interests in these structures is achieving lasing, therefore it is important to identify which factors aid in reaching the condition that the modal gain exceeds the modal loss. Since material gain was written as a positive imaginary part of the material’s refractive index, similarly, material loss can be written as a negative imaginary part of the refractive index. The modal loss can then be determined following the same derivation in Section 1. The result is that the modal loss (α m) is related to the bulk material loss (α b) by\n\n$αm=ngnbγbαb,$\n\nwhere n b is the refractive index of the bulk material and γ b is the energy density confinement in the material similar to Eq. (16). In general there will be several loss mechanisms which can be written as sum of terms of the form Eq. (18). To achieve lasing, the threshold condition requires the modal gain per unit length be greater than the modal loss per unit length:\n\n$ngγAnAgb−ng∑iγiniαi>0.$\n\nWe see immediately that the group index can be divided out of Eq. (19). This is because increasing n g increases the time it takes light to propagate through the waveguide which increases both the gain and loss per unit length equally.\n\nWe see from the lasing threshold condition (Eq. (19)) that increasing the group index will not help one reach the lasing threshold despite that fact that the gain per unit length increases. While above and below the lasing threshold the net modal gain depends on the confinement factor Γ, we see from Eq. (19) that the lasing threshold itself is determined only by the confinement of the electric field energy density (γ A). Although γ A is generally larger than the power confinement in the slot, it is always less than or equal to 1. Therefore the lowest lasing threshold for these structures is limited by the bulk material gain. An identical result can be derived by analyzing resonant cavities. In that case γ A relates the material to modal gain per unit time .\n\nBased on the threshold condition in Eq. (19) we can define a new figure of merit which represents how easily lasing can be achieved in a waveguide. Since in practice the modal loss α m is often an experimentally measured parameter with units of inverse length, we can substitute this measured quantity into Eq. (19) and rewrite the lasing threshold condition as:\n\n$gb>αmΓ.$\n\nThis quantity $αmΓ$ is a useful figure of merit since it is equal to the minimum bulk material gain needed to reach the lasing threshold in the waveguide. Because the group index cancels out in Eq. (18), this figure of merit can be directly compared for waveguides of different geometries to determine which can achieve lasing with the lowest material gain coefficient.\n\n## 5. Scaling behavior of gain versus slot thickness\n\nWhile narrow slots enjoy greater local field enhancement, they also contain less gain material. Therefore the important question arises as to how the total gain in these structures depends on the slot thickness. As the width of the slot becomes increasingly narrow, the magnitude of the electric field increases until it reaches its maximum determined by the difference between dielectric constants in the high and low index region. This results in a greater stimulated (and spontaneous) emission rate of the material in the slot region . According to the laser rate equations this should result in larger modal gain coefficients . However, as the slot becomes increasingly narrow the area of the active region decreases. Therefore although the gain material is “working harder” there is less matter contributing to the gain. Thus it is important to understand how these competing phenomena affect the lasing threshold.",
null,
"Fig. 3. (a). The spatial confinement factor γ A plotted as a function of slot thickness t, where the gain region is defined as the slot (pink region in inset) between the high-index rails (green). Narrow slots result greater emission rates of gain material while thicker slots provide more material which contributes to the gain. The peak in γ A near a slot width of 60 nm indicates the condition where the combination of enhanced emission rate and volume of gain material result in the lowest lasing threshold. (b). The total confinement factor (Γ) (squares) and power in the slot region (P A) (triangles) as a function of slot width. Dotted and dashed lines mark the slot widths which maximize Γ and P A respectively. The discrepancy between these two plots shows that the percentage of power in the gain media is not an accurate indication of either the magnitude or the optimal design for modal gain.\n\nTo minimize the lasing threshold we look for a maximum in the spatial confinement factor (γ A) as a function of slot thickness which is plotted in Fig. 3(a). The maximum near a slot width of 60 nm illustrates the important point that the tradeoff between emission rate (which increases as the slot is narrowed) and material volume (which decreases as the slot is narrowed) results in an optimal slot width for minimizing the lasing threshold. Initially, as the slot narrows from 120 nm, the increased emission rate more than compensates for the decrease in volume of gain material, and γ A increases. Near a thickness of about 50 nm the emission rate begins to saturate as it approaches its maximum value determined by the index contrast between the high and low index regions. After this point, further reduction of the slot thickness decreases the volume of material contributing to the gain without much enhancement of the emission rate, and the result is a sharp drop in γ A.\n\nAs a comparison to previous methods, we plot in Fig. 3(b) the relative power confined in the slot region according to Eq. (17) and show again that slot waveguides outperform expectations based on previous theoretical treatments. We see that both the magnitude of the modal gain and the optimal geometry are miscalculated using this method. The gain calculated numerically (in agreement with Γ) is nearly twice as large as would be expected based on the power confined to the slot mode. Additionally, the optimal slot width is miscalculated by more than 10% using power confinement.\n\n## 6. Optimizing slot waveguide geometry\n\nTo optimize the dimensions of a slot waveguide for an electrically pumped silicon laser, we apply the principles in the proceeding sections to achieve a waveguide design with a minimal lasing threshold. We estimate that the slot thickness should be no larger than 10 nm in order to achieve electrical injection via tunneling into an oxide-based gain media using bias voltages on the order of volts . Therefore we keep the slot thickness fixed at 10 nm and compute the confinement factors as we vary the height and width of the waveguide. Here we have used the refractive indices of Si (3.48) and SiO2 (1.46) as the high and low index material respectively, and a wavelength of 1.55 µm. We have assumed that gain only occurs in the slot region. Figures 4(b)4(d) show respectively the total confinement factor (Γ), the group index normalized to the slot index (n g/n A), and the energy density confinement factor (γ A) as a function of height and width of the waveguide.",
null,
"Fig. 4. Optimization of width and height of Si/SiO2/Si slot waveguide with a 10 nm thick slot assumed to contain a gain medium. (a) Schematic of slot waveguide. (b) Total confinement factor Γ, proportional to the total modal gain. (c) Group index n g divided by the slot index (1.46), which is responsible for the difference between the lasing threshold and modal gain. (d) Electric field energy confinement γ A, inversely proportional to the lasing threshold. The maximum total modal gain is marked by the square in (a). The white contour shows the region which corresponds to a 5% change from the maximum values of Γ and γ A.\n\nWe see in Fig. 4(b) that the maximum modal gain for a 10 nm thick slot occurs near a waveguide width of 940 nm and a height of 340 nm (marked by the square) and has a value Γ=0.336. Both the energy density and group index peak near a waveguide height of 340 nm. As the width of the waveguide is increased, the energy density in the slot region (and thus the net gain) increases. The group index, on the other hand, decreases with increasing waveguide width. These competing parameters result in a maximum Γ near a width of about 940 nm.\n\nIn contrast to the modal gain, with a fixed slot thickness, the minimal lasing threshold (determined by γ A) shows no well-defined optimum. This is because the lasing threshold will scale only with the energy density confinement. The value of γ A(Fig. 4(d)) increases monotonically with waveguide width and asymptotically approaches the value for an infinitely wide slab waveguide, which we calculate to have a maximum of γ A=0.137 for a waveguide height of 340 nm.\n\nWe see from Fig. 4 that the optimal device geometry is relatively insensitive to variation in waveguide dimensions. Around the optimal design, variations of approximately ±50 nm in the total height and width of the waveguide result in changes in the confinement factors of less than 5%. This allows the device performance to be relatively unaffected by size variations which can occur during fabrication.\n\n## 7. Discussion and conclusions\n\nWe have shown that some commonly applied metrics are not appropriate for determining gain in high-index-contrast waveguides, and from first principles developed several figures of merit to characterize waveguide structures for gain. In particular we have shown that the concept of power confinement to the gain region significantly miscalculates the gain experienced by the waveguide mode. Instead we have shown that the true confinement factor which determines gain per unit length results from the combination of group index and confinement of the electric field energy to the gain region. These terms can combine and in some cases exceed unity meaning that one can achieve greater gain per unit length than would be possible in the bulk material. The lasing threshold on the other hand only depends on the percentage of electric field energy in the gain region. To account for this we have introduced a new figure of merit to describe the suitability of a waveguide to achieve low-threshold lasing. This figure of merit is the experimentally measured propagation loss divided by the confinement factor introduced in this paper. The evaluation of this ratio determines the minimal material gain required to achieve lasing in the waveguide structure.\n\nAdditionally we have applied our analysis to the design of slot waveguide structures. We have shown that the lasing threshold has a minimum for a particular slot width and increases dramatically as the slot is made thinner. Also we have shown that gain characteristics of the waveguides are fairly insensitive to variations in overall waveguide dimensions.\n\nSince the confinement factors presented here were derived from perturbation theory, they can be applied to other phenomena in high-index-contrast waveguides including refractive index sensing. In deriving the confinement factors presented here we have studied gain as a perturbation of the imaginary part of the refractive index over a given region of the guided mode. The same formalism holds true for perturbations to the real part of the refractive index and therefore the confinement factors for gain presented in the paper can also be used as confinement factors for refractive index sensing [10, 18] and have shown good agreement with experiment .\n\nIn summary, this work provides the qualitative and quantitative analysis necessary in developing high-index-contrast waveguides for applications such as amplification and lasing.\n\n## Acknowledgments\n\nThe authors gratefully acknowledge Thomas L. Koch for his helpful discussions and Christina Manolatou for the use of her finite difference mode solver. Research support is gratefully acknowledged from the National Science Foundation Center on Materials and Devices for Information Technology Research (CMDITR), DMR-0120967, the National Science Foundation’s CAREER Grant No. 0446571, and the U.S. Air Force MURI program on “Electrically-Pumped Silicon-Based Lasers for Chip-Scale Nanophotonic Systems” supervised by Dr. Gernot Pomrenke.\n\n1. A. W. Fang, H. Park, O. Cohen, R. Jones, M. J. Paniccia, and J. E. Bowers, “Electrically pumped hybrid AlGaInAs-silicon evanescent laser,” Opt. Express 14, 9203-9210 (2006). [CrossRef] [PubMed]\n\n2. V. R. Almeida, Q. Xu, C. A. Barrios, and M. Lipson, “Guiding and confining light in void nanostructure,” Opt. Lett. 29, 1209–1211 (2004). [CrossRef] [PubMed]\n\n3. Q. Xu, V. R. Almeida, R. R. Panepucci, and M. Lipson, “Experimental demonstration of guiding and confining light in nanometer-size low-refractive-index material,” Opt. Lett. 29, 1626–1628 (2004). [CrossRef] [PubMed]\n\n4. C. A. Barrios and M. Lipson, “Electrically driven silicon resonant light emitting device based on slot-waveguide,” Opt. Express 13, 10092–10101 (2005). [CrossRef] [PubMed]\n\n5. F. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron . 42, 885–890 (2006).\n\n6. J. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005). [CrossRef] [PubMed]\n\n7. E. Burstein and C. Weisbuch, eds. Confined electrons and photons, (Plenum Press: New York, NY,1995) [CrossRef]\n\n8. T. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron . 33, 1763–1766 (1997). [CrossRef]\n\n9. C. A. Barrios, K. B. Gylfason, B. Sánchez, A. Griol, H. Sohlström, M. Holgado, and R. Casquel, “Slot-waveguide biochemical sensor,” Opt. Lett. 32, 3080–3082 (2007). [CrossRef] [PubMed]\n\n10. F. Dell’Olio and V. M. Passaro, “Optical sensing by optimized silicon slot waveguides,” Opt. Express 15, 4977–4993 (2007). [CrossRef] [PubMed]\n\n11. H. Kogelnik, Theory of optical waveguides, in Guided-wave optoelectronics, T. Tamir ed., (Springer Verlag: Berlin, 1990). p. 7.\n\n12. L. A. Coldren and S. W. Corzine, Diode lasers and photonic integrated circuits (J. Wiley & Sons, New York, NY, 1995).\n\n13. C. Pollock and M. Lipson, Integrated photonics (Kluwer Academic, Norwell, MA, 2003).\n\n14. J. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997). [CrossRef]\n\n15. J. D. Jackson, Classical electrodynamics. 3rd ed (John Wiley & Sons, Inc., Hoboken, NJ, 1999).\n\n16. H. A. Haus, Waves and fileds in optoelectronics (Prentice-Hall, Englewood Cliffs, NJ, 1984).\n\n17. L. D. Landau and E. M. Lifshitz, Electrodynamics of continuous media (Pergamon Press, Reading, MA, 1960).\n\n18. G. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol . 18, 677–682 (2000). [CrossRef]\n\n19. R. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n\n20. A. E. Siegman, Lasers (University Science Books, Sausalito, CA,1986).\n\n21. J. T. Robinson, L. Chen, and M. Lipson, “On-chip gas detection in silicon optical microcavities,” Opt. Express 16, 4296–4301 (2008). [CrossRef] [PubMed]\n\n### References\n\n• View by:\n• |\n• |\n• |\n\n1. A. W. Fang, H. Park, O. Cohen, R. Jones, M. J. Paniccia, and J. E. Bowers, “Electrically pumped hybrid AlGaInAs-silicon evanescent laser,” Opt. Express 14, 9203-9210 (2006).\n[Crossref] [PubMed]\n2. V. R. Almeida, Q. Xu, C. A. Barrios, and M. Lipson, “Guiding and confining light in void nanostructure,” Opt. Lett. 29, 1209–1211 (2004).\n[Crossref] [PubMed]\n3. Q. Xu, V. R. Almeida, R. R. Panepucci, and M. Lipson, “Experimental demonstration of guiding and confining light in nanometer-size low-refractive-index material,” Opt. Lett. 29, 1626–1628 (2004).\n[Crossref] [PubMed]\n4. C. A. Barrios and M. Lipson, “Electrically driven silicon resonant light emitting device based on slot-waveguide,” Opt. Express 13, 10092–10101 (2005).\n[Crossref] [PubMed]\n5. F. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron. 42, 885–890 (2006).\n6. J. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n7. E. Burstein and C. Weisbuch, eds. Confined electrons and photons, (Plenum Press: New York, NY,1995)\n[Crossref]\n8. T. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n9. C. A. Barrios, K. B. Gylfason, B. Sánchez, A. Griol, H. Sohlström, M. Holgado, and R. Casquel, “Slot-waveguide biochemical sensor,” Opt. Lett. 32, 3080–3082 (2007).\n[Crossref] [PubMed]\n10. F. Dell’Olio and V. M. Passaro, “Optical sensing by optimized silicon slot waveguides,” Opt. Express 15, 4977–4993 (2007).\n[Crossref] [PubMed]\n11. H. Kogelnik, Theory of optical waveguides, in Guided-wave optoelectronics, T. Tamir ed., (Springer Verlag: Berlin, 1990). p. 7.\n12. L. A. Coldren and S. W. Corzine, Diode lasers and photonic integrated circuits (J. Wiley & Sons, New York, NY, 1995).\n13. C. Pollock and M. Lipson, Integrated photonics (Kluwer Academic, Norwell, MA, 2003).\n14. J. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n15. J. D. Jackson, Classical electrodynamics. 3rd ed (John Wiley & Sons, Inc., Hoboken, NJ, 1999).\n16. H. A. Haus, Waves and fileds in optoelectronics (Prentice-Hall, Englewood Cliffs, NJ, 1984).\n17. L. D. Landau and E. M. Lifshitz, Electrodynamics of continuous media (Pergamon Press, Reading, MA, 1960).\n18. G. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n19. R. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n20. A. E. Siegman, Lasers (University Science Books, Sausalito, CA,1986).\n21. J. T. Robinson, L. Chen, and M. Lipson, “On-chip gas detection in silicon optical microcavities,” Opt. Express 16, 4296–4301 (2008).\n[Crossref] [PubMed]\n\n#### 2006 (2)\n\nF. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron. 42, 885–890 (2006).\n\n#### 2005 (2)\n\nJ. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n\n#### 2000 (1)\n\nG. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n\n#### 1997 (2)\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\nT. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n\n#### Baets, R.\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\n#### Blok, H.\n\nT. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\n#### Coldren, L. A.\n\nL. A. Coldren and S. W. Corzine, Diode lasers and photonic integrated circuits (J. Wiley & Sons, New York, NY, 1995).\n\n#### Colombelli, R.\n\nR. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n\n#### Corzine, S. W.\n\nL. A. Coldren and S. W. Corzine, Diode lasers and photonic integrated circuits (J. Wiley & Sons, New York, NY, 1995).\n\n#### Demeulenaere, B.\n\nT. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\n#### Haes, J.\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\n#### Haus, H. A.\n\nH. A. Haus, Waves and fileds in optoelectronics (Prentice-Hall, Englewood Cliffs, NJ, 1984).\n\n#### Hoekstra, H. J. W. M.\n\nG. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n\n#### Jackson, J. D.\n\nJ. D. Jackson, Classical electrodynamics. 3rd ed (John Wiley & Sons, Inc., Hoboken, NJ, 1999).\n\n#### Kimerling, L. C.\n\nF. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron. 42, 885–890 (2006).\n\n#### Kogelnik, H.\n\nH. Kogelnik, Theory of optical waveguides, in Guided-wave optoelectronics, T. Tamir ed., (Springer Verlag: Berlin, 1990). p. 7.\n\n#### Lambeck, P. V.\n\nG. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n\n#### Landau, L. D.\n\nL. D. Landau and E. M. Lifshitz, Electrodynamics of continuous media (Pergamon Press, Reading, MA, 1960).\n\n#### Lenstra, D.\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\nT. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n\n#### Lifshitz, E. M.\n\nL. D. Landau and E. M. Lifshitz, Electrodynamics of continuous media (Pergamon Press, Reading, MA, 1960).\n\n#### Lipson, M.\n\nJ. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n\nC. Pollock and M. Lipson, Integrated photonics (Kluwer Academic, Norwell, MA, 2003).\n\n#### Long, C.\n\nJ. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n\n#### Manolatou, C.\n\nJ. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n\n#### Michel, J.\n\nF. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron. 42, 885–890 (2006).\n\n#### Moreau, V.\n\nR. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n\n#### Ning-Ning, F.\n\nF. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron. 42, 885–890 (2006).\n\n#### Painter, O.\n\nR. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n\n#### Parriaux, O.\n\nG. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n\n#### Perahia, R.\n\nR. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n\n#### Pollock, C.\n\nC. Pollock and M. Lipson, Integrated photonics (Kluwer Academic, Norwell, MA, 2003).\n\n#### Robinson, J. T.\n\nJ. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n\n#### Siegman, A. E.\n\nA. E. Siegman, Lasers (University Science Books, Sausalito, CA,1986).\n\n#### Veldhuis, G. J.\n\nG. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n\n#### Visser, T. D.\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\nT. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n\n#### IEEE J. Quantum Electron (2)\n\nF. Ning-Ning, J. Michel, and L. C. Kimerling, “Optical field concentration in low-index waveguides,” IEEE J. Quantum Electron. 42, 885–890 (2006).\n\nT. D. Visser, H. Blok, B. Demeulenaere, and D. Lenstra, “Confinement factors and gain in optical amplifiers,” IEEE J. Quantum Electron. 33, 1763–1766 (1997).\n[Crossref]\n\n#### J. Lightwave Technol (1)\n\nG. J. Veldhuis, O. Parriaux, H. J. W. M. Hoekstra, and P. V. Lambeck, “Sensitivity enhancement in evanescent optical waveguide sensors,” J. Lightwave Technol. 18, 677–682 (2000).\n[Crossref]\n\n#### Opt. Quantum Electron. (1)\n\nJ. Haes, B. Demeulenaere, R. Baets, D. Lenstra, T. D. Visser, and H. Blok, “Difference between te and tm modal gain in amplifying waveguides: Analysis and assessment of two perturbation approaches,” Opt. Quantum Electron. 29, 263–273 (1997).\n[Crossref]\n\n#### Phys. Rev. Lett. (1)\n\nJ. T. Robinson, C. Manolatou, C. Long, and M. Lipson, “Ultrasmall mode volumes in dielectric optical microcavities,” Phys. Rev. Lett. 95, 143901 (2005).\n[Crossref] [PubMed]\n\n#### Other (9)\n\nE. Burstein and C. Weisbuch, eds. Confined electrons and photons, (Plenum Press: New York, NY,1995)\n[Crossref]\n\nH. Kogelnik, Theory of optical waveguides, in Guided-wave optoelectronics, T. Tamir ed., (Springer Verlag: Berlin, 1990). p. 7.\n\nL. A. Coldren and S. W. Corzine, Diode lasers and photonic integrated circuits (J. Wiley & Sons, New York, NY, 1995).\n\nC. Pollock and M. Lipson, Integrated photonics (Kluwer Academic, Norwell, MA, 2003).\n\nJ. D. Jackson, Classical electrodynamics. 3rd ed (John Wiley & Sons, Inc., Hoboken, NJ, 1999).\n\nH. A. Haus, Waves and fileds in optoelectronics (Prentice-Hall, Englewood Cliffs, NJ, 1984).\n\nL. D. Landau and E. M. Lifshitz, Electrodynamics of continuous media (Pergamon Press, Reading, MA, 1960).\n\nR. Perahia, O. Painter, V. Moreau, and R. Colombelli, “Design of quantum cascade lasers for intra-cavity sensing in the mid infrared,” (in preparation).\n\nA. E. Siegman, Lasers (University Science Books, Sausalito, CA,1986).\n\n### Cited By\n\nOSA participates in Crossref's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.\n\n### Figures (4)\n\nFig. 1. Fundamental TM modes at a wavelength of 1.5 µm for waveguides 500 nm wide and 600 nm tall. All modes are normalized to unit power. The high-index material (n=3.5) is outlined in black. The waveguides are clad with n=3.25 for (a) and (b) and n=1.5 for (c)–(f). The first and second columns show E y and $− ω μ 0 β H x$ respectively, plotted on the same color scale. The two fields become increasingly dissimilar as more electric field is concentrated at high-index-contrast boundaries.\nFig. 2. Numerical study of modal gain. (a) Schematic of slot waveguide with gain material defined by an imaginary component of the refractive index confined to the slot region (pink). (b) Major field component of the fundamental TM mode for the same structure as (a) calculated using a finite difference mode solver. (c) Circles show the modal gain (g m ) calculated from the complex effective index of the fundamental TM mode as determined using a finite difference mode solver. Material gain is added via the imaginary part of the refractive index in the slot. Dashed line shows the modal gain calculated according to Eq. (3) based on the confinement factor Γ determined from the zero-gain mode profile from Eq. (15). Dotted line shows the product of the power in the active gain region (P A ) and the material gain. We see that the confinement factor proposed in this paper correctly predicts the modal gain simulated numerically, while the power confinement greatly underestimates the simulated modal gain.\nFig. 3. (a). The spatial confinement factor γ A plotted as a function of slot thickness t, where the gain region is defined as the slot (pink region in inset) between the high-index rails (green). Narrow slots result greater emission rates of gain material while thicker slots provide more material which contributes to the gain. The peak in γ A near a slot width of 60 nm indicates the condition where the combination of enhanced emission rate and volume of gain material result in the lowest lasing threshold. (b). The total confinement factor (Γ) (squares) and power in the slot region (P A ) (triangles) as a function of slot width. Dotted and dashed lines mark the slot widths which maximize Γ and P A respectively. The discrepancy between these two plots shows that the percentage of power in the gain media is not an accurate indication of either the magnitude or the optimal design for modal gain.\nFig. 4. Optimization of width and height of Si/SiO2/Si slot waveguide with a 10 nm thick slot assumed to contain a gain medium. (a) Schematic of slot waveguide. (b) Total confinement factor Γ, proportional to the total modal gain. (c) Group index n g divided by the slot index (1.46), which is responsible for the difference between the lasing threshold and modal gain. (d) Electric field energy confinement γ A , inversely proportional to the lasing threshold. The maximum total modal gain is marked by the square in (a). The white contour shows the region which corresponds to a 5% change from the maximum values of Γ and γ A .\n\n### Equations (20)\n\n$H = c ε n ( e ̂ z × E ) ,$\n$E y = − ω μ 0 β H x ,$\n$Γ ≡ g m g b ,$\n$∣ E ( z ) ∣ 2 = ∣ E 0 ∣ 2 e g b z .$\n$E ( x , y , z ) = e ̂ j E j o ψ j ( x , y ) e i ( ω t − β ˜ z ) ,$\n$β ˜ ≡ k 0 ( n ̅ r + i n ̅ i ) ,$\n$g m = 2 Im { β ˜ } = 2 k 0 n ̅ i .$\n$g b = 2 k 0 n A i .$\n$Δ β ˜ = ω ∬ ∞ Δ ε ˜ ∣ E ∣ 2 d x d y 1 2 ∬ ∞ Re { E × H * } ∙ e ̂ z d x d y ,$\n$g m = [ n A c ε 0 ∬ A ∣ E ∣ 2 d x d y ∬ ∞ Re { E × H * } ∙ e ̂ z d x d y ] g b .$\n$Γ = n A c ε 0 ∬ A ∣ E ∣ 2 d x d y ∬ ∞ Re { E × H * } ∙ e ̂ z d x d y .$\n$1 2 ∬ Re { E × H * } ∙ e ̂ z d x d y = 1 2 β ω μ 0 ∬ ∣ E ∣ 2 d x d y ,$\n$U / l = 1 2 ∫ ∫ ∞ ε ∣ E ∣ 2 d x d y .$\n$1 2 ∬ Re { E × H * } ∙ e ̂ z d x d y = v g 1 2 ∬ ε ∣ E ∣ 2 d x d y$\n$Γ = n g ∬ A ε ∣ E ∣ 2 d x d y n A ∬ ∞ ε ∣ E ∣ 2 d x d y ≡ n g n A γ A .$\n$γ A ≡ ∬ A ε ∣ E ∣ 2 d x d y ∬ ∞ ε ∣ E ∣ 2 d x d y .$\n$P A ≡ ∬ A Re { E × H * } ∙ e z d x d y ∬ ∞ Re { E × H * } ∙ e z d x d y .$\n$α m = n g n b γ b α b ,$\n$n g γ A n A g b − n g ∑ i γ i n i α i > 0 .$\n$g b > α m Γ .$"
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https://patientsaver.savingadvice.com/2010/07/29/hunt-and-go-seek_60885/ | [
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111.88.107.172\n)\n\n => Array\n(\n => 113.193.198.136\n)\n\n => Array\n(\n => 117.230.172.175\n)\n\n => Array\n(\n => 191.96.182.239\n)\n\n => Array\n(\n => 2.58.46.28\n)\n\n => Array\n(\n => 183.83.253.87\n)\n\n => Array\n(\n => 49.15.139.242\n)\n\n => Array\n(\n => 42.107.220.236\n)\n\n => Array\n(\n => 14.192.53.196\n)\n\n => Array\n(\n => 42.119.212.202\n)\n\n => Array\n(\n => 192.158.234.45\n)\n\n => Array\n(\n => 49.149.102.192\n)\n\n => Array\n(\n => 47.8.170.17\n)\n\n => Array\n(\n => 117.197.13.247\n)\n\n => Array\n(\n => 116.74.34.44\n)\n\n => Array\n(\n => 103.79.249.163\n)\n\n => Array\n(\n => 182.189.95.70\n)\n\n => Array\n(\n => 137.59.218.118\n)\n\n => Array\n(\n => 103.79.170.243\n)\n\n => Array\n(\n => 39.40.54.25\n)\n\n => Array\n(\n => 119.155.40.170\n)\n\n => Array\n(\n => 1.39.212.157\n)\n\n => Array\n(\n => 70.127.59.89\n)\n\n => Array\n(\n => 14.171.22.58\n)\n\n => Array\n(\n => 194.44.167.141\n)\n\n => Array\n(\n => 111.88.179.154\n)\n\n => Array\n(\n => 117.230.140.232\n)\n\n => Array\n(\n => 137.97.96.128\n)\n\n => Array\n(\n => 198.16.66.123\n)\n\n => Array\n(\n => 106.198.44.193\n)\n\n => Array\n(\n => 119.153.45.75\n)\n\n => Array\n(\n => 49.15.242.208\n)\n\n => Array\n(\n => 119.155.241.20\n)\n\n => Array\n(\n => 106.223.109.155\n)\n\n => Array\n(\n => 119.160.119.245\n)\n\n => Array\n(\n => 106.215.81.160\n)\n\n => Array\n(\n => 1.39.192.211\n)\n\n => Array\n(\n => 223.230.35.208\n)\n\n => Array\n(\n => 39.59.4.158\n)\n\n => Array\n(\n => 43.231.57.234\n)\n\n => Array\n(\n => 60.254.78.193\n)\n\n => Array\n(\n => 122.170.224.87\n)\n\n => Array\n(\n => 117.230.22.141\n)\n\n => Array\n(\n => 119.152.107.211\n)\n\n => Array\n(\n => 103.87.192.206\n)\n\n => Array\n(\n => 39.45.244.47\n)\n\n => Array\n(\n => 50.72.141.94\n)\n\n => Array\n(\n => 39.40.6.128\n)\n\n => Array\n(\n => 39.45.180.186\n)\n\n => Array\n(\n => 49.207.131.233\n)\n\n => Array\n(\n => 139.59.69.142\n)\n\n => Array\n(\n => 111.119.187.29\n)\n\n => Array\n(\n => 119.153.40.69\n)\n\n => Array\n(\n => 49.36.133.64\n)\n\n => Array\n(\n => 103.255.4.249\n)\n\n => Array\n(\n => 198.144.154.15\n)\n\n => Array\n(\n => 1.22.46.172\n)\n\n => Array\n(\n => 103.255.5.46\n)\n\n => Array\n(\n => 27.56.195.188\n)\n\n => Array\n(\n => 203.101.167.53\n)\n\n => Array\n(\n => 117.230.62.195\n)\n\n => Array\n(\n => 103.240.194.186\n)\n\n => Array\n(\n => 107.170.166.118\n)\n\n => Array\n(\n => 101.53.245.80\n)\n\n => Array\n(\n => 157.43.13.208\n)\n\n => Array\n(\n => 137.97.100.77\n)\n\n => Array\n(\n => 47.31.150.208\n)\n\n => Array\n(\n => 137.59.222.65\n)\n\n => Array\n(\n => 103.85.127.250\n)\n\n => Array\n(\n => 103.214.119.32\n)\n\n => Array\n(\n => 182.255.49.52\n)\n\n => Array\n(\n => 103.75.247.72\n)\n\n => Array\n(\n => 103.85.125.250\n)\n\n => Array\n(\n => 183.83.253.167\n)\n\n => Array\n(\n => 1.39.222.111\n)\n\n => Array\n(\n => 111.119.185.9\n)\n\n => Array\n(\n => 111.119.187.10\n)\n\n => Array\n(\n => 39.37.147.144\n)\n\n => Array\n(\n => 103.200.198.183\n)\n\n => Array\n(\n => 1.39.222.18\n)\n\n => Array\n(\n => 198.8.80.103\n)\n\n => Array\n(\n => 42.108.1.243\n)\n\n => Array\n(\n => 111.119.187.16\n)\n\n => Array\n(\n => 39.40.241.8\n)\n\n => Array\n(\n => 122.169.150.158\n)\n\n => Array\n(\n => 39.40.215.119\n)\n\n => Array\n(\n => 103.255.5.77\n)\n\n => Array\n(\n => 157.38.108.196\n)\n\n => Array\n(\n => 103.255.4.67\n)\n\n => Array\n(\n => 5.62.60.62\n)\n\n => Array\n(\n => 39.37.146.202\n)\n\n => Array\n(\n => 110.138.6.221\n)\n\n => Array\n(\n => 49.36.143.88\n)\n\n => Array\n(\n => 37.1.215.39\n)\n\n => Array\n(\n => 27.106.59.190\n)\n\n => Array\n(\n => 139.167.139.41\n)\n\n => Array\n(\n => 114.142.166.179\n)\n\n => Array\n(\n => 223.225.240.112\n)\n\n => Array\n(\n => 103.255.5.36\n)\n\n => Array\n(\n => 175.136.1.48\n)\n\n => Array\n(\n => 103.82.80.166\n)\n\n => Array\n(\n => 182.185.196.126\n)\n\n => Array\n(\n => 157.43.45.76\n)\n\n => Array\n(\n => 119.152.132.49\n)\n\n => Array\n(\n => 5.62.62.162\n)\n\n => Array\n(\n => 103.255.4.39\n)\n\n => Array\n(\n => 202.5.144.153\n)\n\n => Array\n(\n => 1.39.223.210\n)\n\n => Array\n(\n => 92.38.176.154\n)\n\n => Array\n(\n => 117.230.186.142\n)\n\n => Array\n(\n => 183.83.39.123\n)\n\n => Array\n(\n => 182.185.156.76\n)\n\n => Array\n(\n => 104.236.74.212\n)\n\n => Array\n(\n => 107.170.145.187\n)\n\n => Array\n(\n => 117.102.7.98\n)\n\n => Array\n(\n => 137.59.220.0\n)\n\n => Array\n(\n => 157.47.222.14\n)\n\n => Array\n(\n => 47.15.206.82\n)\n\n => Array\n(\n => 117.230.159.99\n)\n\n => Array\n(\n => 117.230.175.151\n)\n\n => Array\n(\n => 157.50.97.18\n)\n\n => Array\n(\n => 117.230.47.164\n)\n\n => Array\n(\n => 77.111.244.34\n)\n\n => Array\n(\n => 139.167.189.131\n)\n\n => Array\n(\n => 1.39.204.103\n)\n\n => Array\n(\n => 117.230.58.0\n)\n\n => Array\n(\n => 182.185.226.66\n)\n\n => Array\n(\n => 115.42.70.119\n)\n\n => Array\n(\n => 171.48.114.134\n)\n\n => Array\n(\n => 144.34.218.75\n)\n\n => Array\n(\n => 199.58.164.135\n)\n\n => Array\n(\n => 101.53.228.151\n)\n\n => Array\n(\n => 117.230.50.57\n)\n\n => Array\n(\n => 223.225.138.84\n)\n\n => Array\n(\n => 110.225.67.65\n)\n\n => Array\n(\n => 47.15.200.39\n)\n\n => Array\n(\n => 39.42.20.127\n)\n\n => Array\n(\n => 117.97.241.81\n)\n\n => Array\n(\n => 111.119.185.11\n)\n\n => Array\n(\n => 103.100.5.94\n)\n\n => Array\n(\n => 103.25.137.69\n)\n\n => Array\n(\n => 47.15.197.159\n)\n\n => Array\n(\n => 223.188.176.122\n)\n\n => Array\n(\n => 27.4.175.80\n)\n\n => Array\n(\n => 181.215.43.82\n)\n\n => Array\n(\n => 27.56.228.157\n)\n\n => Array\n(\n => 117.230.19.19\n)\n\n => Array\n(\n => 47.15.208.71\n)\n\n => Array\n(\n => 119.155.21.176\n)\n\n => Array\n(\n => 47.15.234.202\n)\n\n => Array\n(\n => 117.230.144.135\n)\n\n => Array\n(\n => 112.79.139.199\n)\n\n => Array\n(\n => 116.75.246.41\n)\n\n => Array\n(\n => 117.230.177.126\n)\n\n => Array\n(\n => 212.103.48.134\n)\n\n => Array\n(\n => 102.69.228.78\n)\n\n => Array\n(\n => 117.230.37.118\n)\n\n => Array\n(\n => 175.143.61.75\n)\n\n => Array\n(\n => 139.167.56.138\n)\n\n => Array\n(\n => 58.145.189.250\n)\n\n => Array\n(\n => 103.255.5.65\n)\n\n => Array\n(\n => 39.37.153.182\n)\n\n => Array\n(\n => 157.43.85.106\n)\n\n => Array\n(\n => 185.209.178.77\n)\n\n => Array\n(\n => 1.39.212.45\n)\n\n => Array\n(\n => 103.72.7.16\n)\n\n => Array\n(\n => 117.97.185.244\n)\n\n => Array\n(\n => 117.230.59.106\n)\n\n => Array\n(\n => 137.97.121.103\n)\n\n => Array\n(\n => 103.82.123.215\n)\n\n => Array\n(\n => 103.68.217.248\n)\n\n => Array\n(\n => 157.39.27.175\n)\n\n => Array\n(\n => 47.31.100.249\n)\n\n => Array\n(\n => 14.171.232.139\n)\n\n => Array\n(\n => 103.31.93.208\n)\n\n => Array\n(\n => 117.230.56.77\n)\n\n => Array\n(\n => 124.182.25.124\n)\n\n => Array\n(\n => 106.66.191.242\n)\n\n => Array\n(\n => 175.107.237.25\n)\n\n => Array\n(\n => 119.155.1.27\n)\n\n => Array\n(\n => 72.255.6.24\n)\n\n => Array\n(\n => 192.140.152.223\n)\n\n => Array\n(\n => 212.103.48.136\n)\n\n => Array\n(\n => 39.45.134.56\n)\n\n => Array\n(\n => 139.167.173.30\n)\n\n => Array\n(\n => 117.230.63.87\n)\n\n => Array\n(\n => 182.189.95.203\n)\n\n => Array\n(\n => 49.204.183.248\n)\n\n => Array\n(\n => 47.31.125.188\n)\n\n => Array\n(\n => 103.252.171.13\n)\n\n => Array\n(\n => 112.198.74.36\n)\n\n => Array\n(\n => 27.109.113.152\n)\n\n => Array\n(\n => 42.112.233.44\n)\n\n => Array\n(\n => 47.31.68.193\n)\n\n => Array\n(\n => 103.252.171.134\n)\n\n => Array\n(\n => 77.123.32.114\n)\n\n => Array\n(\n => 1.38.189.66\n)\n\n => Array\n(\n => 39.37.181.108\n)\n\n => Array\n(\n => 42.106.44.61\n)\n\n => Array\n(\n => 157.36.8.39\n)\n\n => Array\n(\n => 223.238.41.53\n)\n\n => Array\n(\n => 202.89.77.10\n)\n\n => Array\n(\n => 117.230.150.68\n)\n\n => Array\n(\n => 175.176.87.60\n)\n\n => Array\n(\n => 137.97.117.87\n)\n\n => Array\n(\n => 132.154.123.11\n)\n\n => Array\n(\n => 45.113.124.141\n)\n\n => Array\n(\n => 103.87.56.203\n)\n\n => Array\n(\n => 159.89.171.156\n)\n\n => Array\n(\n => 119.155.53.88\n)\n\n => Array\n(\n => 222.252.107.215\n)\n\n => Array\n(\n => 132.154.75.238\n)\n\n => Array\n(\n => 122.183.41.168\n)\n\n => Array\n(\n => 42.106.254.158\n)\n\n => Array\n(\n => 103.252.171.37\n)\n\n => Array\n(\n => 202.59.13.180\n)\n\n => Array\n(\n => 37.111.139.137\n)\n\n => Array\n(\n => 39.42.93.25\n)\n\n => Array\n(\n => 118.70.177.156\n)\n\n => Array\n(\n => 117.230.148.64\n)\n\n => Array\n(\n => 39.42.15.194\n)\n\n => Array\n(\n => 137.97.176.86\n)\n\n => Array\n(\n => 106.210.102.113\n)\n\n => Array\n(\n => 39.59.84.236\n)\n\n => Array\n(\n => 49.206.187.177\n)\n\n => Array\n(\n => 117.230.133.11\n)\n\n => Array\n(\n => 42.106.253.173\n)\n\n => Array\n(\n => 178.62.102.23\n)\n\n => Array\n(\n => 111.92.76.175\n)\n\n => Array\n(\n => 132.154.86.45\n)\n\n => Array\n(\n => 117.230.128.39\n)\n\n => Array\n(\n => 117.230.53.165\n)\n\n => Array\n(\n => 49.37.200.171\n)\n\n => Array\n(\n => 104.236.213.230\n)\n\n => Array\n(\n => 103.140.30.81\n)\n\n => Array\n(\n => 59.103.104.117\n)\n\n => Array\n(\n => 65.49.126.79\n)\n\n => Array\n(\n => 202.59.12.251\n)\n\n => Array\n(\n => 37.111.136.17\n)\n\n => Array\n(\n => 163.53.85.67\n)\n\n => Array\n(\n => 123.16.240.73\n)\n\n => Array\n(\n => 103.211.14.183\n)\n\n => Array\n(\n => 103.248.93.211\n)\n\n => Array\n(\n => 116.74.59.127\n)\n\n => Array\n(\n => 137.97.169.254\n)\n\n => Array\n(\n => 113.177.79.100\n)\n\n => Array\n(\n => 74.82.60.187\n)\n\n => Array\n(\n => 117.230.157.66\n)\n\n => Array\n(\n => 169.149.194.241\n)\n\n => Array\n(\n => 117.230.156.11\n)\n\n => Array\n(\n => 202.59.12.157\n)\n\n => Array\n(\n => 42.106.181.25\n)\n\n => Array\n(\n => 202.59.13.78\n)\n\n => Array\n(\n => 39.37.153.32\n)\n\n => Array\n(\n => 177.188.216.175\n)\n\n => Array\n(\n => 222.252.53.165\n)\n\n => Array\n(\n => 37.139.23.89\n)\n\n => Array\n(\n => 117.230.139.150\n)\n\n => Array\n(\n => 104.131.176.234\n)\n\n => Array\n(\n => 42.106.181.117\n)\n\n => Array\n(\n => 117.230.180.94\n)\n\n => Array\n(\n => 180.190.171.5\n)\n\n => Array\n(\n => 150.129.165.185\n)\n\n => Array\n(\n => 51.15.0.150\n)\n\n => Array\n(\n => 42.111.4.84\n)\n\n => Array\n(\n => 74.82.60.116\n)\n\n => Array\n(\n => 137.97.121.165\n)\n\n => Array\n(\n => 64.62.187.194\n)\n\n => Array\n(\n => 137.97.106.162\n)\n\n => Array\n(\n => 137.97.92.46\n)\n\n => Array\n(\n => 137.97.170.25\n)\n\n => Array\n(\n => 103.104.192.100\n)\n\n => Array\n(\n => 185.246.211.34\n)\n\n => Array\n(\n => 119.160.96.78\n)\n\n => Array\n(\n => 212.103.48.152\n)\n\n => Array\n(\n => 183.83.153.90\n)\n\n => Array\n(\n => 117.248.150.41\n)\n\n => Array\n(\n => 185.240.246.180\n)\n\n => Array\n(\n => 162.253.131.125\n)\n\n => Array\n(\n => 117.230.153.217\n)\n\n => Array\n(\n => 117.230.169.1\n)\n\n => Array\n(\n => 49.15.138.247\n)\n\n => Array\n(\n => 117.230.37.110\n)\n\n => Array\n(\n => 14.167.188.75\n)\n\n => Array\n(\n => 169.149.239.93\n)\n\n => Array\n(\n => 103.216.176.91\n)\n\n => Array\n(\n => 117.230.12.126\n)\n\n => Array\n(\n => 184.75.209.110\n)\n\n => Array\n(\n => 117.230.6.60\n)\n\n => Array\n(\n => 117.230.135.132\n)\n\n => Array\n(\n => 31.179.29.109\n)\n\n => Array\n(\n => 74.121.188.186\n)\n\n => Array\n(\n => 117.230.35.5\n)\n\n => Array\n(\n => 111.92.74.239\n)\n\n => Array\n(\n => 104.245.144.236\n)\n\n => Array\n(\n => 39.50.22.100\n)\n\n => Array\n(\n => 47.31.190.23\n)\n\n => Array\n(\n => 157.44.73.187\n)\n\n => Array\n(\n => 117.230.8.91\n)\n\n => Array\n(\n => 157.32.18.2\n)\n\n => Array\n(\n => 111.119.187.43\n)\n\n => Array\n(\n => 203.101.185.246\n)\n\n => Array\n(\n => 5.62.34.22\n)\n\n => Array\n(\n => 122.8.143.76\n)\n\n => Array\n(\n => 115.186.2.187\n)\n\n => Array\n(\n => 202.142.110.89\n)\n\n => Array\n(\n => 157.50.61.254\n)\n\n => Array\n(\n => 223.182.211.185\n)\n\n => Array\n(\n => 103.85.125.210\n)\n\n => Array\n(\n => 103.217.133.147\n)\n\n => Array\n(\n => 103.60.196.217\n)\n\n => Array\n(\n => 157.44.238.6\n)\n\n => Array\n(\n => 117.196.225.68\n)\n\n => Array\n(\n => 104.254.92.52\n)\n\n => Array\n(\n => 39.42.46.72\n)\n\n => Array\n(\n => 221.132.119.36\n)\n\n => Array\n(\n => 111.92.77.47\n)\n\n => Array\n(\n => 223.225.19.152\n)\n\n => Array\n(\n => 159.89.121.217\n)\n\n => Array\n(\n => 39.53.221.205\n)\n\n => Array\n(\n => 193.34.217.28\n)\n\n => Array\n(\n => 139.167.206.36\n)\n\n)\n```\nHunt and Go Seek: Save More, Spend Less\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Hunt and Go Seek",
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"# Hunt and Go Seek\n\nJuly 29th, 2010 at 11:06 am\n\nIt's now impossible to walk inside my fenced vegetable garden without treading on squash vines. My tomato plants are taller than I am. (I'm 5'4\".)\n\nYesterday, I ventured inside to tie up a few tomato branches that were being pulled down by a squash vine.\n\nI discovered a giant cucumber hidden under all those leaves, as well as two over-sized zucchini I hadn't seen before. It's kind of like a vegetarian version of the game Hide and Go Seek.\n\nWell, you know what that means....zucchini bread! I also picked another medium-sized tomato and a few string beans.\n\nHere's a rundown of what's growing:\n\nTomatoes: Lots of green tomatoes, but thus far, just a few mid-sized tomatoes that had ripened. I could have sworn that I bought cherry tomato plants, but it appears all 5 of my plants produce a mid-sized variety of tomato.\n\nZucchini: This has been my best year of the last 3 for zucchini. I made a special effort to hand pollinate this year and i'm getting a lot of zucchini.\n\nSpaghetti Squash: I think these may be able to be picked soon. They went from a greenish-white color when formed to a pale yellow now. But I'll have to check online as I've never grown these before. But I have 4, maybe 5 loaf-sized squashes...HUGE. Can't wait to try them!\n\nAcorn squash: I just spotted one large one yesterday, but that's all so far. Guess they take a while to grow.\n\nString beans I usually grow bush beans and get plenty of beans, but this year i decided to try pole beans. I have 2 tripod structures made of 6 foot high branches and these have been absolutely covered by the beans, but so far, lots and lots of leaves (I haven't fertilized) and very few beans. Disappointing.\n\nPotatoes: I had great fun with potatoes last year, so this year i planted both red and russets. They appear to be dying back already; this is supposed to happen by end of August, after which time you can dig up the potatoes. I expect to have a great crop.\n\nOrnamental gourds I planted a row of these but only 2 plants came up, I think becus I was stepping in this area while weeding and forgot I'd planted something here along the fenceline. I only noticed one small gourd; let's hope more will come.\n\nCucumbers Last year I had maybe 4 plants and I was awash in cucumbers. I decided to pare it down to just 2 plants this year, and now it seems I don't have ENOUGH cucumbers. I've only gotten maybe 3 or 4 at this point.\n\nBell pepper I've never had huge successes with bell peppers. It seems to take so long for them to really take off. Haven't gotten any peppers yet, but the plants themselves finally seem healthier, with more foliage. Of course, they're being shaded out by the squash vines, which are hard to contain and control.\n\nI also planted some sort of vining annual flower, just to add some color and attract pollinators, but while they also got off to a slow start, they have yet to flower. Just green....\n\nI'm off to a focus group this a.m. about an hour's drive from here. I'll return mid-day with \\$125 cash in my pocket.\n\nI want to lay my new soaker hose in the garden and also head over to friend's house to continue cleanup work in her garage/house and start getting ready for her garage sale. She's basically cleaning out a lifetime of accumulated stuff. I've already gotten 2 nice lunches out of it (a bison burger at the local diner, and a shrimp caesar salad and dessert at a good Italian place).\n\nI also need to mow the lawn!\n\n### 5 Responses to “Hunt and Go Seek”\n\n1. NJDebbie Says:\n\nCan you post a picture of your garden? I love to see it!\n\n2. creditcardfree Says:\n\nWow! Your garden is flourishing. I have some gourd plants that volunteered from my compost last year, they are taking over my garden and trying to escape the fence!\n\n3. Ima saver Says:\n\nWe have about a million tomatoes. We are giving them out to everyone!\n\n4. Jerry Says:\n\nI swear I haven't even thought of zucchini bread in YEARS, and your post just leads my mouth to start watering thinking about it! Yum...\nUnrelated - I wish we had some insurance of a good bison burger joint around here, but there just aren't a lot of American Bison running around this part of Europe. I want a bison burger!\nJerry\n\n5. PatientSaver Says:\n\nNJ Debbie, you can see photos of my garden at my gardening blog:"
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"https://www.savingadvice.com/blogs/images/search/bottom_left.php",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.97403306,"math_prob":0.99828994,"size":4040,"snap":"2020-24-2020-29","text_gpt3_token_len":1021,"char_repetition_ratio":0.08845391,"word_repetition_ratio":0.057771664,"special_character_ratio":0.25222772,"punctuation_ratio":0.1216686,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9990087,"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\":\"2020-06-04T00:42:54Z\",\"WARC-Record-ID\":\"<urn:uuid:15701482-b0ca-4b57-9edf-2e04c080eb26>\",\"Content-Length\":\"134665\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a2cc886c-37d4-4c04-b5d6-583b477eaa39>\",\"WARC-Concurrent-To\":\"<urn:uuid:0a12e889-c7fc-4005-b817-57a5560c464c>\",\"WARC-IP-Address\":\"173.231.200.26\",\"WARC-Target-URI\":\"https://patientsaver.savingadvice.com/2010/07/29/hunt-and-go-seek_60885/\",\"WARC-Payload-Digest\":\"sha1:JBXHBUANBOKPGW47DKXUMLA2FHPCJGJ6\",\"WARC-Block-Digest\":\"sha1:BGBJVMJLFEHAYRVUKAQAJABNPRAK7NOL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347436828.65_warc_CC-MAIN-20200604001115-20200604031115-00080.warc.gz\"}"} |
http://rin.io/spec/ | [
"Spectrum of a Ring\n\nApproaching Scheme Theory as a Beginner\n\nThank you to James Tao for explaining the following, geometrically intuitive, interpretation to me.\n\nA scheme is the result of patching affine schemes together, an affine scheme is any locally ringed space that is equivalent to Spec of a ring. To make an analogy, a manifold is the result of patching together copies of $\\mathbb{R}^n$, in a way that preserves the smooth structure.\n\nAnd the smooth structure of $\\mathbb{R}^n$ is essentially the fact that it is a sheaf: over any open set, we have the space of smooth functions on that set, and these functions restrict and glue nicely, etc.\n\nSo when you go from $\\mathbb{R}^n$ to “smooth manifolds,” you are creating spaces that locally look like $\\mathbb{R}^n$, respect the same kind of smooth structure (as captured by the fact that smooth functions are defined on open sets in a way independent of the patching), yet have some global structure that might prevent them from just being the same as $\\mathbb{R}^n$.\n\nIf you care about algebraic (polynomial / rational) functions defined on algebraic sets (zeroes of systems of polynomials), then you can do the same program! The “basic patch” is “affine scheme,” and you patch affine schemes together to get a scheme.\n\nThe category of affine schemes is equivalent to $\\text{Ring}^{op}$. In other words, telling you how to map forward is the same as telling you how to pull back functions. However, this is NOT true for the larger category of schemes. $\\text{Hom}_{Ring}(A, \\Gamma(\\mathcal{O}_B)) = \\text{Hom}_{Schemes} (B, \\text{Spec } A)$ (note that $\\Gamma(\\mathcal{O}_B)$ := the global sections of the structure sheaf $\\mathcal{O}_B$).\n\nIn particular, given a scheme, it’s unclear how to get a ring that encodes all the information in that scheme. Consider, for example, projective space, whose only global sections are constants, yet the field $k$ of constant functions certainly doesn’t single out any projective space!\n\nWhat are these affine Spec R things?\n\nWe have a functor Spec from Ring to Schemes:\n\n$\\text{Ring} \\xrightarrow{\\text{Spec}} \\text{Schemes}$\n\nSchemes look like Fun(Ring, Set). So Spec sends a ring to a functor from Ring to Set.\n\n$\\text{Ring} \\xrightarrow{\\text{Spec}} (\\text{Ring} \\to \\text{Set})$\n\nHonestly, whenever I see “scheme” I replace it mentally with “algebraic curve,” but this is just because I don’t really get what a scheme is yet.\n\n$R \\xrightarrow{\\text{Spec} (-)} \\text{Spec(R)}$\n\n$R \\xrightarrow{\\text{Spec} (-)} (S \\mapsto \\text{hom}(R,S))$\n\n$\\text{Spec}$ $R$ $S$ = $\\text{Hom}_{\\text{Ring}}(R, S)$\n\nLet say we have:\n\n1. a ring $\\mathbb{Z}[t]$ (this is just the ring of polynomials with $t$ as a variable and coefficients in $\\mathbb{Z}$)\n2. an arbitrary ring $S$.\n\nWhat is $\\text{hom}(\\mathbb{Z}[t], S)$?\n\nNotice that we’re in Ring, so the members of $\\text{hom}(\\mathbb{Z}[t], S)$ must be Ring homomorphisms. Let’s say we’re looking at a ring homomorphism $\\phi: \\mathbb{Z}[t] \\to S$.\n\nI learned from Cris Moore that demanding examples is a useful practice, so: What is $\\phi(7t^2-4t+3)$ in simpler terms?\n\n$\\phi$ is a ring homomorphism, so we know that $\\phi(1) = 1_s$. This implies that $\\phi(n) = n_s$.\n\n$\\phi(7t^2-4t+3) = \\phi(7t^2) – \\phi(4t) + \\phi(3)$\n$= \\phi(t)(\\phi(7t) – \\phi(4)) + \\phi(3)$\n$= \\phi(t) (\\phi(t) 7_s – 4_s) + 3_s$\n\n$\\phi(t)$ is not determined! We can pick it to be any element of S!\n\n$\\text{hom}(\\mathbb{Z}[t], S) \\simeq \\{\\phi(t) \\in S\\} = S$\n\nIn other words, $\\text{Spec}(\\mathbb{Z}[t])$ is a forgetful functor from $S$ as an object in $\\text{Ring}$ to its underlying set.\n\nExercise: show that $\\mathbb{Z}[t]$ is an initial object in an appropriate category, and that the trivial ring is the terminal object in the appropriate category.\n\nBut what is Spec?\n\nIt’s the spectrum of a ring! Intuitively, $\\text{Spec} R$ is the geometric object canonically associated to $R$.\n\nFor example:\n\n• $\\text{Spec}(\\mathbb{Z})$ is “The Point”\n• $\\text{Spec}(\\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))$ is “The Circle”\n• $\\text{Spec}(\\mathbb{Z}[x,x^{-1}]$ is “Pairs of Invertible Elements” (e.g. a field with the origin removed)\n\nLet’s look at $\\text{Spec}(\\mathbb{Z}[x,y]/ (x^2 + y^2 – 1))(S)$ when $S$ is $\\mathbb{R}$, it’s obviously the circle BUT IT WORKS FOR (almost) ANY S. The concept of there being a canonical geometry associated to every ring is very exciting!\n\nMurky Afterthoughts\n\nWhen people say “elliptic curve” they might mean “a family of elliptic curves”, this is commonly written $C \\to \\text{Spec} R$ where $R$ is the underlying coefficient ring.\n\nFor example, $y^2=4x^3+ax+b \\mapsto a, b \\in R$ is the family of elliptic curves of the form $y^2=4x^3+ax+b$ over the coefficient ring $R$.\n\nI’m trying to figure out what happens when $S$ is not something nice like $\\mathbb{R}$ or $\\mathbb{C}$. What is a circle in the coefficient ring of an elliptic curve?\n\nThanks to Aaron Mazel-Gee for walking me through the concept of $\\text{Spec}$ (all errors in this post are mine and not his).\n\nOne thought on “Spectrum of a Ring”\n\n1.",
null,
"Josh G says:\n\nI think you do yourself a disservice by jumping right from “algebraic curves” to the viewpoint of (affine) schemes as functors. To make the connection more concrete, one should first understand the connection between varieties (or even just algebraic curves) and their coordinate rings, and then understand how Spec of the coordinate ring recovers the Zariski topology on the original variety. *Then* one can take the more abstract viewpoint, whereby one wishes to treat not just integral domains over a field, but *any* ring as the coordinate ring of some “geometric object”, namely, a scheme. And *then* one can generalize from schemes as the categorical duals of rings to the “functor of points” viewpoint. Taking this route will give you a much more intuitive, geometric picture of what is going on (and should enable you to connect your intuitive geometric picture of algebraic curves to the highly abstract picture of schemes as functors.)\n\nFrom this viewpoint, it is clear why $hom(\\mathbb{Z}[t], S)$ should just be $S$. Imagine every ring is the ring of “well-behaved” functions on some geometric object. Then certainly $\\mathbb{Z}[t]$ is the ring of polynomial functions on the line (where, here, we haven’t specified *which* line – real, complex, or what have you – but that’s okay because we’re only considering polynomials with integer coefficients, which make sense on the line over any ring). And $S$ is the ring of well-behaved functions on some geometric object which we’ll call $Spec(S)$. Since rings are categorically dual to schemes, $hom(\\mathbb{Z}[t], S)$ should be dual to $hom(Spec(S), line)$. But what are maps from $Spec(S)$ to the line? They are exactly the (well-behaved) functions on $Spec(S)$ – i.e., the elements of $S$!"
] | [
null,
"http://1.gravatar.com/avatar/d74e21661477efeec5c02cb24f4f58f3",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8649725,"math_prob":0.998351,"size":6733,"snap":"2019-43-2019-47","text_gpt3_token_len":1884,"char_repetition_ratio":0.13048,"word_repetition_ratio":0.0037807184,"special_character_ratio":0.2735779,"punctuation_ratio":0.09337349,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99977857,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T21:26:26Z\",\"WARC-Record-ID\":\"<urn:uuid:af77db10-b27d-4349-a438-4308792509d8>\",\"Content-Length\":\"39790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:11defd26-eb5e-4f5b-b06c-af668a2e3563>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b4e98cf-2758-4c26-ab6b-a80174191c48>\",\"WARC-IP-Address\":\"162.243.159.36\",\"WARC-Target-URI\":\"http://rin.io/spec/\",\"WARC-Payload-Digest\":\"sha1:NQNWOIP4GONIB46KQPCOF6NS5NOYG663\",\"WARC-Block-Digest\":\"sha1:Q4NVG5YL4NLLY7UNZXFIGZISY5XHVCYL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986697760.44_warc_CC-MAIN-20191019191828-20191019215328-00131.warc.gz\"}"} |
https://encyclopedia2.thefreedictionary.com/electron-volt | [
"# electron-volt\n\nAlso found in: Dictionary, Medical, Wikipedia.\nRelated to electron-volt: kiloelectron volt\n\n## electron-volt,\n\nabbr. eV, unit of energy used in atomic and nuclear physics; 1 electron-volt is the energy transferred in moving a unit chargecharge,\nproperty of matter that gives rise to all electrical phenomena (see electricity). The basic unit of charge, usually denoted by e, is that on the proton or the electron; that on the proton is designated as positive (+e\n, positive or negative and equal to that charge on the electron, through a potentialpotential, electric,\nwork per unit of electric charge expended in moving a charged body from a reference point to any given point in an electric field (see electrostatics).\ndifference of 1 volt. The maximum energy of a particle acceleratorparticle accelerator,\napparatus used in nuclear physics to produce beams of energetic charged particles and to direct them against various targets. Such machines, popularly called atom smashers, are needed to observe objects as small as the atomic nucleus in studies of its\nis usually expressed in multiples of the electron-volt, such as million electron-volts (MeV) or billion electron-volts (GeV). Because mass is a form of energy (see relativityrelativity,\nphysical theory, introduced by Albert Einstein, that discards the concept of absolute motion and instead treats only relative motion between two systems or frames of reference.\n), the masses of elementary particleselementary particles,\nthe most basic physical constituents of the universe. Basic Constituents of Matter\n\nMolecules are built up from the atom, which is the basic unit of any chemical element. The atom in turn is made from the proton, neutron, and electron."
] | [
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https://www.varsitytutors.com/hotmath/hotmath_help/topics/multiplying-and-dividing-negatives.html | [
"# Multiplying and Dividing with Negatives\n\n## Multiplying with Negatives\n\nThere's a simple rule here:\n\n• (positive) $×$ (positive) = (positive)\n• (positive) $×$ (negative) = (negative)\n• (negative) $×$ (positive) = (negative)\n• (negative) $×$ (negative) = (positive)\n\nExample :\n\n• $\\left(3\\right)\\left(7\\right)=21$\n• $\\left(3\\right)\\left(-7\\right)=-21$\n• $\\left(-3\\right)\\left(7\\right)=-21$\n• $\\left(-3\\right)\\left(-7\\right)=21$\n\nImportant Thing To Notice:\n\nSince a positive times a positive is positive, and a negative times a negative is also positive, it follows that any number times itself is either zero or positive.\n\n• $\\left(12\\right)\\left(12\\right)=144$\n• $\\left(-12\\right)\\left(-12\\right)=144$\n\n## Dividing with Negatives\n\nThe same rule works with division:\n\n• (positive) $÷$ (positive) = (positive)\n• (positive) $÷$ (negative) = (negative)\n• (negative) $÷$ (positive) = (negative)\n• (negative) $÷$ (negative) = (positive)\n\nExample :\n\n• $56÷7=8$\n• $56÷\\left(-7\\right)=-8$\n• $-56÷7=-8$\n• $-56÷\\left(-7\\right)=8$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8821308,"math_prob":0.9999807,"size":661,"snap":"2022-27-2022-33","text_gpt3_token_len":164,"char_repetition_ratio":0.3607306,"word_repetition_ratio":0.37894738,"special_character_ratio":0.25869894,"punctuation_ratio":0.08988764,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000055,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T00:52:56Z\",\"WARC-Record-ID\":\"<urn:uuid:68502467-f93d-4077-aad6-1c4639b23e68>\",\"Content-Length\":\"81726\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54519b5d-9e8b-4ad8-871f-f27f009e0376>\",\"WARC-Concurrent-To\":\"<urn:uuid:948ce44d-043d-4c6d-980c-deb396f5564d>\",\"WARC-IP-Address\":\"18.67.65.107\",\"WARC-Target-URI\":\"https://www.varsitytutors.com/hotmath/hotmath_help/topics/multiplying-and-dividing-negatives.html\",\"WARC-Payload-Digest\":\"sha1:2BOAA46GKCKMAZ26PFX3NUQOVIGUU4TV\",\"WARC-Block-Digest\":\"sha1:TAOJCQLVXY4PUPZ7Z2YJQ4EIL2RTBSFD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570879.37_warc_CC-MAIN-20220809003642-20220809033642-00456.warc.gz\"}"} |
http://cpc.cs.qub.ac.uk/summaries/ADGB_v1_0.html | [
"",
null,
"Computer Physics Communications Program Library",
null,
"Programs in Physics & Physical Chemistry",
null,
"[Licence| Download | New Version Template] adgb_v1_0.gz(5 Kbytes) Manuscript Title: Algebraic and numerical manipulation of the even-power-series central potentials by means of the hypervirial theorems technique. Authors: Th.E. Liolios, M.E. Grypeos Program title: HVTDER-HVTPAS Catalogue identifier: ADGB_v1_0Distribution format: gz Journal reference: Comput. Phys. Commun. 105(1997)254 Programming language: Derive, Pascal. Computer: IBM PC/AT. RAM: 640K words Word size: 16 Keywords: Hypervirial, Algebraic solution of Schroedinger eigenvalue problem, Binding energies, Expectation values, Differential equations, Nuclear physics, Theoretical methods, General purpose. Classification: 4.3, 17.16. Nature of problem:Algebraic and numerical calculation of the energy eigenvalues Enl of the 3D Schroedinger equation as well as of the expectation values nl, nl, nl for the class of central potentials V(r) = -D + Sigma (Vk b**k r**(2k+2)) k=0 to infinity Solution method:A combination of the Quantum Mechanical Hypervirial theorems and the Hellman-Feynman theorem. Restrictions:The central potential V(r) must be harmonic at short distances. Running time:5s for HVTDER and 1s for HVTPAS.\n Disclaimer | ScienceDirect | CPC Journal | CPC | QUB"
] | [
null,
"http://cpc.cs.qub.ac.uk/icons/E.gif",
null,
"http://cpc.cs.qub.ac.uk/icons/sd.jpg",
null,
"http://cpc.cs.qub.ac.uk/icons/CPCjournal.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64154845,"math_prob":0.6511577,"size":1312,"snap":"2019-13-2019-22","text_gpt3_token_len":359,"char_repetition_ratio":0.10091743,"word_repetition_ratio":0.0,"special_character_ratio":0.24009146,"punctuation_ratio":0.19491525,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9733992,"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\":\"2019-03-23T16:28:26Z\",\"WARC-Record-ID\":\"<urn:uuid:878aa344-e95c-4f54-9c8f-18f8e3763b24>\",\"Content-Length\":\"5139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:208361c8-c18d-4666-a7a7-eb7952bcb15a>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d190db2-f7b1-4deb-930f-f99a17a96130>\",\"WARC-IP-Address\":\"143.117.100.34\",\"WARC-Target-URI\":\"http://cpc.cs.qub.ac.uk/summaries/ADGB_v1_0.html\",\"WARC-Payload-Digest\":\"sha1:7POTNRXSRYUOBLT7FZTTWU2HXCBN36XD\",\"WARC-Block-Digest\":\"sha1:5UFTJ4ZQAVLTPPFIHFB67X7UH4AQSUO4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202889.30_warc_CC-MAIN-20190323161556-20190323183556-00153.warc.gz\"}"} |
http://koreascience.or.kr/article/JAKO200934152712920.page | [
"# TRANSFORMATION OF DIMENSIONLESS HEAT DIFFUSION EQUATION FOR THE SOLUTION OF DYNAMIC DOMAIN IN PHASE CHANGE PROBLEMS\n\n• Ashraf, Muhammad (DEPT OF MATH, COMSATS INST OF INFORMATION TECHNOLOGY) ;\n• Avila, R. (THERMAL FLUID LAB, FAC OF MECH ENG, NATL AUTONOMOUS UNIV OF MEXICO (UNAM)) ;\n• Raza, S. S. (GLOBAL CHANGE IMPACT STUDY CENTER)\n• Received : 2009.01.26\n• Accepted : 2009.02.26\n• Published : 2009.03.25\n\n#### Abstract\n\nIn the present work transformation of dimensionless heat diffusion equation for the solution of moving boundary problems have been formulated. The formulation is based on 1-D, 2-D and 3-D, unsteady heat diffusion equations. These equations are rst turned int dimensionless form by using dimensionless quantities and their transformation was formulated in liquid and solid phases. The salient feature of this work is that during the transformation of dimensionless heat diffusion equation there arises a convective term $\\tilde{v}$ which is responsible for the motion of interface in liquid as well as solid phase. In the transformed heat equation, a correction factor $\\beta$ also arises naturally which gives the correct transformed flux at interface.\n\n#### References\n\n1. M. N. Ozisik, Boundary Value Problems of Heat Conductions, Seraton, Pennsylvania International Textbook Company, 1968\n2. J. Crank, Free and Moving Boundary Problems, New York Claredon Press, 1984.\n3. A. Rathjen and L. M. Jiji, Tarnsient Heat Transfer in Fins Undergoing Phase Transformation, Fourth International Heat Transfer Conference, Paris-Versailles,2, 1970.\n4. E. M. Ronquist and A. T. Patera, 'A Legender Spectral Element Method for the Stefan Problem', International Journal for Numerical Methods in Engineering, Vol,24, 2273-2299 (1987) https://doi.org/10.1002/nme.1620241204\n5. C. H. Li, D. R.. Jenkins, 'The effect of natural convection in solidification in tall tapered feeders', ANZIAM J. 44(E) ppC496-C511,2003. https://doi.org/10.21914/anziamj.v44i0.693\n6. R. M. Furzeland, 'A comparative study of numerical methods for moving boundary Problems', J. Inst. Math. Applic., 26, 411-429(1980). https://doi.org/10.1093/imamat/26.4.411\n7. D. R. Lynch and K. O'Neill, 'Continuously deforming finite elements for the solution of parabolic problems, with and without phase change', Int. J. numer. Methods eng., 17, 81-96 (1981) https://doi.org/10.1002/nme.1620170107\n8. A. T. Patera, 'A finite element/Green's function embedding technique applied to one dimensional change of phase heat transfer', Numer. Heat Transfer, 7, 241-247 (1984).\n9. Kees Vuik and Fred Vermolen, A conserving discretization for the free boundary in a two-dimensional Stefan problem J. Comp. Phys., 141, pp. 1-21, 1998. https://doi.org/10.1006/jcph.1998.5900\n10. C. Vuik and G. Segal and F.J. Vermolen A conserving discretization for a Stefan problem with an interface reaction at the free boundary Comput. Visual Sci., 3, pp. 109-114, 2000. https://doi.org/10.1007/s007910050058\n11. N.C.W. Kuijpers, F.J. Vermolen, K. Vuik, and S. van der Zwaag A model of the beta-AlFeSi to alpha-Al(FeMn)Si transformation in Al-Mg-Si alloys Materials Transactions, 44, pp. 1448-1456, 2003 https://doi.org/10.2320/matertrans.44.1448\n12. F.J. Vermolen and C. Vuik Solution of vector Stefan problems with cross-diffusion Journal of Computational and Applied Mathematics, 176, pp. 179-201,2005 https://doi.org/10.1016/j.cam.2004.07.011\n13. F.J. Vermolen, C. Vuik, E. Javierre and S. van der Zwaag Review on some Stefan Problems for Particle Dissolution Nonlinear Analysis: Modelling and Control, 10, pp. 257-292, 2005\n14. E. Javierre and C. Vuik and F.J. Vermolen and S. van der Zwaag A comparison of models for one-dimensional Stefan problems J. Comp. Appl. Math., 192, pp. 445-459, 2006. https://doi.org/10.1016/j.cam.2005.04.062"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6730689,"math_prob":0.6760837,"size":3700,"snap":"2021-04-2021-17","text_gpt3_token_len":1117,"char_repetition_ratio":0.11390693,"word_repetition_ratio":0.017208412,"special_character_ratio":0.3235135,"punctuation_ratio":0.25738916,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9555996,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T13:20:45Z\",\"WARC-Record-ID\":\"<urn:uuid:72a972f1-5bb8-443c-9a09-c07f271ff3aa>\",\"Content-Length\":\"49844\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56ee370f-0bad-4d78-a63f-03bcd27120d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:18fa91a8-64d1-4235-b92d-a22d527ba159>\",\"WARC-IP-Address\":\"203.250.218.22\",\"WARC-Target-URI\":\"http://koreascience.or.kr/article/JAKO200934152712920.page\",\"WARC-Payload-Digest\":\"sha1:JMTS7WNHIF3DHNNWRKE54FRC5EHCF3WJ\",\"WARC-Block-Digest\":\"sha1:MK5XOCEPWEFP3R36S4HHYF7AML3EDKU4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038119532.50_warc_CC-MAIN-20210417102129-20210417132129-00364.warc.gz\"}"} |
https://www.mathlearnit.com/what-is-501-39-as-a-decimal | [
"# What is 501/39 as a decimal?\n\n## Solution and how to convert 501 / 39 into a decimal\n\n501 / 39 = 12.846\n\nAs a refresher, we encourage you to check out our fractions section of the website. Also check out our introduction to decimal numbers article. Fractions can also be represented as decimal numbers which is a number with a whole part and a fractional part. In this section, we show you how to calculate the solution to such problems on your own.\n\n• Step 1 (only step): Divide 501 by 39 to get the number as a decimal. 501 / 39 = 12.846. That's all there is to it. After all, a fraction is another way of expressing a division.\nIt is often easier to use a calculator to solve these types of problems, but it is also possible to do it by hand. If you want to practice doing it by hand, you may want to double check your solution with a calculator afterward.\n\n### When is this useful?\n\nSometimes it is more useful to convey a number as a decimal number rather than a fraction or a percentage. In those cases, this type of conversion is helpful."
] | [
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http://www.wiredchemist.com/chemistry/instructional/an-introduction-to-chemistry/matter-at-the-atomic-level/electronic-structure | [
"## Electronic Structure\n\nThe most mysterious part of the atom is the electrons. The nucleus is relatively easy to comprehend because we think of it as a core of positive charge. But the electrons have a lot more space (remember the analogy with the football field) and it is natural to inquire about things like--are the electrons moving or are they stationary? If the electrons are moving, how do they move, and finally, what are electrons? These were questions that occupied the minds of the physicists at the beginning of the 20th century, a period that was probably one of the most interesting and influential in the last 100 years of science. This period was alive with other important discoveries and theories. One of the most important set of observations concerns the nature of light. At the time of the great physicist Isaac Newton, the light that allows us to see was thought to consist of tiny particles. It was said that when these particles hit the retina of the eye they induce sight.\n\nAround the turn of the 20th century an experiment produced a result that could not be explained by the particle theory. This experiment is illustrated in Figure 10 where a the light from a light bulb is shown passing through two slits and then striking a screen. If light consists of particles, the expected result was that two images of the slits would appear on the screen--the particles would pass through the slits and strike the screen and cause two bands of light. In fact, the experiment showed that around the two bands of light there were less intense bands that appear to be images of the \"original\" bands. This result was explained by the suggestion that light actually was a wave, or, at least, has some of the properties of a wave.",
null,
"Figure 10. The double slit experiment.\n\nBecause waves are so important in science we should take a minute to define some of their characteristics. Imagine standing alongside a very still pond. In the tree above you a bird drops a seed that strikes the water in the pond and sends out a set of ripples, or waves.\n\nThese waves actually have the shape of what in mathematics is called a sine (or cosine) wave (see Figure 11). The speed at which the waves move outward is their velocity. The distance between the waves is the wavelength, usually designated by the Greek symbol lambda (λ or l). Another characteristic of waves is their frequency, which is the number of wave crests, or troughs, that pass a particular point in the water each second. The frequency (n, Greek ν) is designated as cycles per second, one cycle being one complete motion from the top of a wave to the bottom of the trough and up again to the top of the wave. The unit cycles per second is also called Hertz (abbreviated Hz) in honor of the physicist Heinrich Hertz who did much of the early experimentation on wave motion.",
null,
"Figure 11. A sine wave.\n\nNow let us return to the double slit experiment. If light has a wave motion, we should be able to use the pebble in the water analogy (scientists frequently use analogy to produce explanations of observations). When a pebble hits a body of water, it will produce a set of waves. Imagine what will happen if these waves encounter a barrier that has two openings. These openings will initiate a new set of waves. Of course, these waves are just the off-spring of the waves created by the pebble so they have the same wavelength and frequency and speed. The waves in turn may \"interfere\" with one another.\n\nFigure 12 shows that some waves will be \"in phase\" and will add to produce a larger wave; others will be exactly \"out of phase\" and crests will cancel troughs to produce no wave at all. These are analogous to the bright spots on the screen of the double-slit experiment and, thus, the experiment has been explained by constructive and destructive interference of the waves.\n\nFigure 12. The Double Slit Experiment - Video\n\nOf course, you will now be thinking, \"Okay, I can buy this stuff about the waves, but aren't we talking about light? Light is not like water, so what is it that is moving up and down?\" This is a question that baffled scientists for many years and produced models like the one that proposed that light travels through a universal medium called the \"ether\". Eventually, however, physicists agreed that light is perpendicular oscillating electric and magnetic fields. Light has a constant speed, but it can have a variety of wavelengths and frequencies. Indeed, these three characteristics of light are related by a simple equation:\n\nc = λn\n\nwhich tells that if c is a constant and if λ increases n must decrease. In mathematical terms we say that λ and n are inversely proportional.\n\nIn any event, the term light is now used to mean any electromagnetic radiation--from radio waves, which have long wavelengths of several meters, to visible light, with its wavelengths of about 10-5 meters to x-rays which have wavelengths of 10-8 meters. The electromagnetic \"spectrum\" is shown in Figure 13.",
null,
"Figure 13. The electromagnetic spectrum.\n\nJust when everyone seemed to agree about the nature of light, Einstein explained the photoelectric effect (which we won't describe except to say that it involves the velocity with which electrons leave the surface of metals when the metal is exposed to light) by proposing that light consists of little bundles of energy called photons. In essence, this is a bit of a return to a particle theory for light because we now (according to the photon theory) think of light as little packages of electromagnetic radiation. Einstein, moreover, made use of some work that Max Planck had done earlier and showed that these packages of light have an energy given by another simple relationship:\n\nE = hn\n\nwhere E is the energy of a photon, n is the frequency of the wave, and h is Planck's constant ( a number, 6.627 x 10-34 J-s).\n\nWe can now turn to our first model of the electronic structure of atoms. Niels Bohr, a Danish physicist, proposed that a hydrogen atom, which has only one electron and one proton, could be thought of as having the electron on one of a series of circular tracks. The electron according to Bohr must move in one circle around the nucleus and if it is undisturbed the electron must remain on that circle or track. The interesting, and very revolutionary thing about Bohr's model is that he proposed that the angular momentum, which is similar to the energy of the electron, can have only certain values. In fact, he proposed that this angular momentum can have values that are integral values of Planck's constant divided by 21. In other words,\n\nangular momentum = nh/21\n\nwhere n is an integer (1, 2, 3, 4, ...). This constituted a major break with classical physics and was the beginning of the quantum era of science. To be sure that you understand this notion, notice that the angular momentum cannot be 1.134h/21, or 34.5h/21, but it can be 1h/21 or 2h/21 or 55h/21. That is, the angular momentum can have only certain values and not others. The angular momentum is said to be quantized (from the Latin quantus, \"how great\"). Although we cannot derive the consequences here, it turns out that quantization of the angular momentum also leads to quantization of the radius of the tracks and the energy of the atom.\n\nThe first three orbits, or tracks, of the Bohr model and the energies of each are shown in Figure 14. If the electron is in the first orbit, it is characterized by the quantum number n = 1, and it has an energy of -21 x 10-19 joules. If it is in the second orbit, n = 2, and the energy is -5 x 10-19 joules. Notice again that the energy of the electron cannot be -10 x 10-19 joules; it can have only those values dictated by the quantization of the angular momentum.",
null,
"Figure 14. The first three Bohr orbits and their energies.\n\nThe Bohr model was an immediate success because it was able to successfully predict the ionization energy of the hydrogen atom and to predict with very good precision the wavelengths of light that are given off when hydrogen is heated in an electric arc. However, the success of the model was short-lived. It could not predict the ionization energies of other atoms, and there are features of the line spectrum of even hydrogen that it could not explain. Thus, within a few years of Bohr's success, the search was on for another model of electronic structure. The model that was finally proposed by several physicists and that has remained the model used by chemists and physicists today is referred to as the wave model, wave mechanics, or quantum mechanics. This model incorporates both the wave ideas formulated for light and the quantum ideas proposed by Einstein, Planck, and Bohr.\n\nBasically, the wave model assumes that an electron has some characteristics of electromagnetic radiation--that it can be represented, in other words, by a wave. This model is extremely mathematical in nature, and, moreover, contains a number of features that are almost totally outside of the realm of our everyday experiences. For example, it turns out that the amplitude (actually it is the square of the amplitude) of the wave that represents an electron is proportional to the probabiity of finding the electron. Suppose that a fictitious electron confined to a line has the wave shown in Figure 15(a). The square of the amplitude of the wave is shown in Figure 15(b). Notice that the negative part of the wave (on the right hand side of part (a)) becomes positive when the amplitude is squared (the square of -2 is 4).",
null,
"Figure 15. A wave for an electron confined to a line.\n\nThus, if we want to know where we can most likely find the electron we look for that region of the line where the amplitude squared is greatest. In this case, it occurs at 1/4 d and 3/4 d. In the middle of the line the probability of finding the electron is zero. This notion that we can never be sure of the location of the electron, but can only deal in probabilities is somewhat like passing the scene of an accident involving a large truck and a small car. You cannot tell whether anyone was injured but you would be willing to bet that there is a greater probability that the person in the small car was injured than the person in the truck.\n\nAnother consequence of the wave model is that the energy of the electron is quantized, just as was true in the Bohr model. Although the wave model is superior to the Bohr model in many respects, it also cannot be solved exactly for the energies of electrons in atoms more complicated than hydrogen. However, there are approximations that can be used in the wave model that give very good estimates of the relative energies of the electrons not only in a variety of atoms, but even in complicated molecules.\n\nThe wave model is also similar to the Bohr model in that it uses quantum numbers to characterize an electron. Rather than the one quantum number that was used in the Bohr model, three are required in the wave model. These numbers are given names such as n (the principal quantum number), λ (the angular momentum quantum number), and m (the magnetic quantum number). Each of the numbers can have only certain integral values, and each reveals something different about the energy and the location of the electron. Every unique set of these three quantum numbers is called an orbital. For example, if n = 2, λ = 1, and m = 0, this set of numbers is referred to as a 2p orbital (the 2 comes from the value of n, the p from the value of λ). Table 3 lists some of the allowed sets of n, λ, and m for values of n equal to 1 and 2.",
null,
"Table 3. Allowed sets (orbitals) of n, λ, and m.\n\nEvery electron in an atom must, according to the wave model, have a set of n, λ, and m values. Moreover, only two electrons can have the same values of n, λ, and m. To make matters more complicated, two electrons that have the same values of n, λ, and m, must differ in their value of another quantum number (a result of relativity theory, not quantum theory), the spin quantum number, s. The spin quantum number is a strange one--it can have only two values, and these are half-integral. The allowed values of s are +1/2 and -1/2. Now if we include all four of the quantum numbers, the Pauli exclusion principle says that each electron in an atom must have a unique set of four quantum numbers.\n\nIf you are feeling a little confused by all of this quantum number mumbo-jumbo, consider this analogy. Think of an empy apartment house with a bunch of rooms, each of the rooms is one apartment and each has an address. The landlord for this apartment has set an unusual and probably illegal rule: no more than two people can occupy each apartment and those people must be of opposite sex. In our analogy, each apartment is an orbital and the people that rent the apartments are the electrons. The sex, male or female, of the people is analogous to the spin of the electrons. If we assume that there is only one set of stairs, and no elevator, it will even be true that each apartment (orbital) has a different location and that a different amount of energy will be required to get to each. Moreover, the floor that each apartment is on is somewhat analogous to the value of n, the principal quantum number; that is, the higher the apartment the greater the value of n. What our analogy does not convey is the probabilistic nature of the electron. In our analogy we know exactly where each person is at any particular time; we also know the exact confines of each apartment. In the strange world of wave mechanics, we know neither of these with certainty. We can only say that a particular electron is in an orbital and that it therefore has such and such a probability of being at a certain point in space (for example, at some point 2 Å from the nucleus).\n\nIn spite of the uncertainty about these orbitals, they are always represented in general chemistry texts as solid figures. Some authors are careful to indicate that these solid figures represent an area of space within which the electron can be found 95% of the time. In the orbital representations shown below (Figures 16-18) you will notice that s orbitals are spherical, p-orbitals have two lobes, and d-orbitals are even more complex. You will also notice that p and d orbitals have other designations that come from their values of m. These designations indicate (in most cases) something about on what axes the orbitals have their maximum probabilities. We have also shown several different types of representations for a p-orbital, to give you the chance to see how authors differ in their feelings about how best to represent them. Notice also that p and d-orbitals are sometimes shown with different colors for alternating lobes. These colors indicate the different signs of the equation for the wave responsible for the orbital (see Figure 15 to remind yourself that waves can have positive and negative parts).",
null,
"Figure 16. Representations of s and p-orbitals.",
null,
"Figure 17. Some alternative representations of p-orbitals.\n\nThe different colors and + and - signs indicate the sign of the mathematical equation that describes the orbital.",
null,
"Figure 18. Photographs of models representing the d-orbitals.\n\nCarefully examine the orientation of the orbitals relative to the x, y, and z axes.\n\nFinally, we show in Figure 19 an energy level diagram for a fluorine atom with four orbitals completely filled and one orbital, a 2p orbital, half-filled. This is called the electron configuration for an atom and is usually designated as:\n\nF 1s22s22p5",
null,
"Figure 19. Energy level diagram for fluorine.\n\nThis energy level diagram also suggests that electrons can be \"bumped up\" into higher energy levels if they somehow are subjected to the right amount of energy. For example, the electron in the half-filled p orbital could be bumped up to the 3s orbital. This would require an amount of energy exactly equal to the difference in energy between the 2p and 3s orbitals. This amount of energy can be supplied by exposing the atom to electromagnetic radiation (for example ultraviolet light). The \"bumping up\" process is called excitation and the resulting atom is said to be excited. After some time has passed the atom will try to \"relax\" by losing energy to go back to its ground state. When this happens, the same amount of energy will be lost, and a photon will be emitted. It is this excitation/relaxation process that is responsible for most of modern spectroscopy: the measurement of the amount or type of a substance or atom present by irradiating the sample with electromagnetic radiation."
] | [
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https://askfilo.com/math-question-answers/how-many-spherical-bullets-can-be-made-out-of-a-solid-cube-of-lead-whose-edge | [
"",
null,
"The world’s only live instant tutoring platform",
null,
"Question\n\n# How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?\n\n## Solutions\n\n(1)",
null,
"In the given problem, we have a lead cube which is remolded into small spherical bullets.\nHere, edge of the cube (s) = 44 cm\nDiameter of the small spherical bullets (d) = 4 cm\nNow, let us take the number of small bullets be x\nSo, the total volume of x spherical bullets is equal to the volume of the lead cube.\nTherefore, we get,\nVolume of the x bullets = volume of the cube\n\nx =2541\nTherefore, 2541 small bullets can be made from the given lead cube.",
null,
"70",
null,
"Connect with 50,000+ expert tutors in 60 seconds, 24X7"
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null,
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https://cs.stackexchange.com/questions/6173/given-many-partial-orders-check-them-for-consistency-and-report-any-that-are-no | [
"# Given many partial orders, check them for consistency and report any that are not consistent\n\nInputs. I am given a finite set $S$ of symbols. I know there should exist some total order $<$ on $S$, but I'm not given this ordering and it could be anything.\n\nI am also given a collection of assertions. Each assertion takes the form $s_1<s_2<\\cdots<s_m$, where $s_1,\\dots,s_m$ form a subset of the symbols of $S$. The assertion probably won't mention all of the symbols of $S$, just a subset. Each assertion will probably cover a different subset.\n\nWarmup problem. The starter problem is: Given $n$ assertions, identify whether they are all internally self-consistent, i.e., whether there exists a total order on $S$ that is consistent with all of the assertions, and if so, output an example of such a total order.\n\nThe real problem. In practice, a few assertions might be faulty. Almost all of them should be correct, though. So, the real problem is: if the assertions are not all internally self-consistent, find a minimal subset of assertions to label as \"probably-erroneous\", such that if you remove the probably-erroneous assertions, the remainder are all self-consistent.\n\nWhat I know. I know how to solve the warmup problem (just compute the transitive closure of the union of the partial orders given by each assertion, and check that the result is antisymmetric; or, in other words, create a graph with $S$ as vertex set and an edge $s\\to t$ if $s<t$ appears in any assertion, then check for cycles). However, I don't know how to solve the real problem. Any ideas?\n\nReal-world parameters. In the application domain where I've run into this, $S$ might have up to a few hundred symbols, and I might have up to a few thousand assertions, with each assertion typically mentioning dozens of symbols.\n\n• I suppose this is implied, but you are looking for an efficient algorithm for this problem, not just any (brute-force) algorithm, right? Oct 19, 2012 at 18:10\n• Correct, I'm looking for a reasonably efficient algorithm. I would be satisfied with an approximation algorithm (one that may not find the exact-minimum set of \"possibly-erroneous\" assertions, if the size of the set it outputs is usually not too much larger than the minimum).\n– D.W.\nOct 20, 2012 at 1:41\n\nThis sounds like weighted FAS (Feedback Arc Set). Find a minimal (weight) feedback arc set in a directed graph and report the assertions that contributed to the wrongly directed arcs.\n\nFor each pair of symbols s and t, there is an arc saying how many assertions have s < t and how many assertions have t < s.\n\nThis is a known NP-complete problem.\n\nOn a different note, your problem is a classic social choice problem where each of the \"assertions\" are votes.\n\n• Brilliant! Algorithms for weighted FAS do sound like just the sort of thing I was looking for. This search term should be a great start to help me look for approximation algorithms for FAS. Thank you! Do you have any advice regarding approximation algorithms or heuristics for weighted FAS that are good candidates to try, for a practical application? (Simple schemes would be nice.)\n– D.W.\nOct 20, 2012 at 1:51\n• This is indeed an interesting question, and as Jonas says, it resembles FAS and has relations to social choice theory. However, it is a generalization to FAS, and I believe that the problem is potentially much harder than FAS. To say that it generalizes FAS, you simply let S be the vertices of a graph and your linear orders the set of edges (each order has size 1). Now, FAS was (relatively) recently shown to be FPT, i.e., solvable in time f(k)*poly(n), but this was a long standing open problem. I am afraid this problem is not easily solvable. (k is the number of orders you must delete.) Oct 27, 2012 at 15:23\n\nI thought I'd jot down the two approaches that have occurred to me, but maybe you can do better.\n\nApproach 1. Here's a randomized procedure that, given an ordering of the assertions, will emit a candidate set of assertions labelled as \"probably-erroneous\" such that the remainder are all self-consistent. We can repeat the procedure many times, and keep the best output (the one with the smallest number of assertions labelled as \"probably-erroneous\").\n\nThe procedure:\n\n• Step 1. Randomly permute the assertions. Create an empty graph $G$ with vertex set $S$ and with no edges\n• Step 2. For each assertion, in the order chosen in step 1, do the following:\n\n• Suppose the assertion is $s_1<s_2<\\cdots<s_m$. Test whether adding the edges $s_1\\to s_2, \\dots, s_{m-1} \\to s_m$ would create a cycle in $G$.\n• If this would create a cycle, label this assertion as \"probably-erroneous\", and don't add the edges to $G$.\n• If this would not create a cycle, add the edges to $G$.\n\nApproach 2. Create a $|S|\\times|S|$ matrix $V[\\cdot,\\cdot]$, where $V[s,t]$ counts the number of \"votes\" for the claim $s<t$. Initialize all counts in the matrix to zero.\n\nTreat the assertion $s_1<s_2<\\cdots<s_m$ as a collection of $m(m-1)/2$ votes for the claims $s_1<s_2$, $s_1<s_3$, $s_2<s_3$, etc. Process each assertion, incrementing the counts in $V$ appropriately. This gives you the full matrix $V$.\n\nNow look through $V$ to check for symbols $s,t$ such that $V[s,t]>0$ and $V[t,s]>0$ and such that the quantity $V[t,s]/(V[s,t]+V[t,s])$ is minimized. Conclude that $s<t$ is correct, and label all assertions which contradict this (which voted for $t<s$) as \"probably-erroneous\". Remove these assertions from the set of assertions, and re-start the entire procedure, until the assertions that remain are all self-consistent.\n\nThis procedure is not guaranteed to terminate with a correct answer (e.g., consider the five assertions $A<P$, $P<Z$, $A<Q$, $Q<Z$, $Z<A$), but it's a heuristic that might be worth trying.\n\n• Again, are you looking for (a) a known correct algorithm, regardless of cost; (b) a known correct algorithm which is optimal or significantly better than brute-force; or (c) an efficient heuristic that does pretty well most of the time? Oct 19, 2012 at 18:57\n• Yes. :-) I'll take whatever I can get. The only hard requirement is that it be efficient (feasible to run within, say, a few minutes or hours). An efficient heuristic that does pretty well most of the time might be perfectly satisfactory. Of course, if you can do better than that, so much the better! (Obviously, there is a trivial solution that is guaranteed-correct but is very slow: simply enumerate all $2^n$ subsets of assertions. I'm ruling that one out because it's too slow to run in practice.) Sorry to be a bit vague and open-ended!\n– D.W.\nOct 20, 2012 at 1:46"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93385804,"math_prob":0.9745148,"size":1702,"snap":"2022-05-2022-21","text_gpt3_token_len":399,"char_repetition_ratio":0.14310955,"word_repetition_ratio":0.02112676,"special_character_ratio":0.22972973,"punctuation_ratio":0.13753581,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977249,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T10:37:24Z\",\"WARC-Record-ID\":\"<urn:uuid:75570d70-4f76-46a0-ae1d-bf56fc1fcb6d>\",\"Content-Length\":\"244995\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:858baecb-a481-4259-8017-0c0e36bbbae0>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e063af7-090e-4aaa-8b71-3859317de5d1>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/6173/given-many-partial-orders-check-them-for-consistency-and-report-any-that-are-no\",\"WARC-Payload-Digest\":\"sha1:ZRBOJOLLJJSPSKMXJ5EBFJRR2H5S3U4X\",\"WARC-Block-Digest\":\"sha1:SOADA3RU5KX7HBNGBK6UQVPL5VBMWZRM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517245.1_warc_CC-MAIN-20220517095022-20220517125022-00332.warc.gz\"}"} |
https://mathematica.stackexchange.com/questions/149129/ndsolve-error-control-of-an-odes-system | [
"# NDSolve Error control of an ODEs system\n\nHaving studied the answers to this question on how to control the error of NDSolve, I then tried to apply it to the following system of second-order nonlinear ODEs:\n\nlowsol = With[{μ = 1/3},\nNDSolve[{x''[t] == -Surd[x[t]^(Numerator@μ), Denominator@μ] +\nSurd[(y[t] - x[t])^(Numerator@μ), Denominator@μ],\ny''[t] == -Surd[(y[t] - x[t])^(Numerator@μ), Denominator@μ],\ny == 0, y' == 0, x == 0,\nx' == Sqrt[2 - y'^2 - ((Surd[(x)^(Numerator@(μ + 1)),\nDenominator@(μ + 1)]) + (Surd[(y - x)^(Numerator@(μ + 1)),\nDenominator@(μ + 1)]))/(μ + 1)]}, {x, y}, {t, 0, 10},\nMethod -> \"ExplicitRungeKutta\", WorkingPrecision -> 61,\nInterpolationOrder -> All, MaxSteps -> 2*^6,\nStartingStepSize -> 1*^-8, MaxStepSize -> 1*^-4]] // Timing\n\n\nwhich returns a pair of Interpolating Function. The system has a first integral of motion, which is set by its energy $E$, given by: \\begin{equation} E=\\frac{x'^2+y'^2}{2}+V(x,y) \\end{equation}\n\nI then tried to plot this energy and see if it remains a constant in the following way:\n\nWith[{μ = 1/3},\nPlot[{(((x'[t])^2 + (y'[t])^2)/2 + (3/4)*((x[t])^(4/3) +\n(y[t] - x[t])^(4/3)))} /. lowsol // RealExponent // Evaluate, {t, 0, 10},\nPlotRange -> All]]\n\n\nbut what was returned, is the following error:\n\nReplaceAll::rmix: Elements of {513.969,\n{{x->InterpolatingFunction[{{0,10.0000000000000000000000000000000000000000000000\n0000000000000}},{5,1,7,{327114},\n{4},{InterpolatingFunction[<<5>>]},0,0,0,Automatic,{},{},False},\n{{0,5.086860500065041295097264042754353058189185580664200997582414*10^-\n20,<<47>>,5.890559344799673487983779799641879661003747176870946110552147*10^-\n15,<<327064>>}},\n{{0,1.414213562373095048801688724209698078569671875376948073176680},<<49>>,\n<<327064>>},{{{<<327115>>},\n{<<327115>>}}}],y->InterpolatingFunction[{{0,10.00000000000000000000000000000000\n000000000000000000000000000}},<<3>>,{{{<<327115>>},{<<327115>>}}}]}}} are a\nmixture of lists and nonlists.\n\n\nwhich states that there are elements which are a mixture of lists and nonlists. I would like to ask the following:\n\n• Why did this error occur and how can I avoid it?\n• How is it possible to reduce the computation time? In the question linked above, the computation time is significantly less than the one I need to acquire the results for my system.\n• Is it possible to do the same for a number of initial conditions? I tried ParametricNDSolve but the plotting issue would remain.\n\nUpdate\n\nAfter running the piece of code provided by @zhk, I found out that the numerical integration returns a huge error estimate for t=10^4 time units as it can be seen below:",
null,
"How can I reduce this error?\n\n• You're setting lowsol equal to the result of Timing. You probably want to set it equal to the result of NDSolve. Try moving it inside the With: With[{..}, lowsol = NDSolve[..]] // Timing. – Michael E2 Jun 26 '17 at 23:04\n• @MichaelE2 I will try it now and let you know. – Mitscaype Jun 27 '17 at 9:05\n• It seems like you should be using the Projection or SymplecticPartitionedRungeKutta methods for NDSolve, which explicitly check that the invariant is conserved during the integration process. But I'm not getting that to work either. – KraZug Jun 27 '17 at 9:43\n• @KraZug I read about those methods on the documentation but I was not able to make it run, though I have not idea why. – Mitscaype Jun 27 '17 at 9:46\n• I integrate it with SPRK, but the invariant is not conserved: \\[Mu] = 1/3; ode1 = x''[t] == -x[t]^\\[Mu] + (-x[t] + y[t])^\\[Mu]; ode2 = y''[t] == -(-x[t] + y[t])^\\[Mu]; ics = {y == 0, y' == 0, x == 0, x' == Sqrt}; invariant = (x'[t]^2 + y'[t]^2)/ 2 + (3/4)*(x[t]^(4/3) + (y[t] - x[t])^(4/3)); projerksol = NDSolve[Flatten[{ode1, ode2, ics}], {x, y}, {t, 0, 1}, Method -> {\"SymplecticPartitionedRungeKutta\", \"DifferenceOrder\" -> 2, \"PositionVariables\" -> {x, y}}, StartingStepSize -> 0.001][]; Plot[Im[invariant /. projerksol], {t, 0, 1}] – KraZug Jun 27 '17 at 9:48\n\nSomething like this?\n\nμ = 1/3;\n\node1 = x''[t] == -x[t]^μ + (-x[t] + y[t])^μ;\n\node2 = y''[t] == -(-x[t] + y[t])^μ;\n\nlowsol = NDSolve[{ode1, ode2, y == 0, y' == 0, x == 0,\nx' == Sqrt[2 - (x^(1 + μ) + (-x + y)^(1 + μ))/(1 + μ) -\ny'^2]}, {x, y}, {t, 0, 10}, Method -> {\"FixedStep\", Method -> \"ExplicitEuler\"}];\n\nxy = Flatten /@Table[{t, Re[{(((x'[t])^2 + (y'[t])^2)/\n2 + (3/4)*((x[t])^(4/3) + (y[t] - x[t])^(4/3)))} /. First@lowsol]}, {t, 0, 10, 0.1}];\n\nListLinePlot[xy, PlotRange -> All]",
null,
"• Looks good. Perhaps Flatten was what I was looking for? But why \"ExplicitEuler\"? Also, from the plot I can see an error estimate of 10^-3 for the first 10 time steps. I would like to bound it under 10^-4 since I am running 10^4 time steps integration. How can I apply the options: WorkingPrecision -> 61, InterpolationOrder -> All, MaxSteps -> 2*^6, StartingStepSize -> 1*^-8, MaxStepSize -> 1*^-4 as above? – Mitscaype Jun 26 '17 at 17:27"
] | [
null,
"https://i.stack.imgur.com/cBxPE.jpg",
null,
"https://i.stack.imgur.com/bDOvx.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6890556,"math_prob":0.99489456,"size":3351,"snap":"2020-24-2020-29","text_gpt3_token_len":1210,"char_repetition_ratio":0.113235734,"word_repetition_ratio":0.04,"special_character_ratio":0.4756789,"punctuation_ratio":0.19271624,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999291,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-15T18:59:17Z\",\"WARC-Record-ID\":\"<urn:uuid:601ae9bf-6f28-487b-844f-bf31c7f76865>\",\"Content-Length\":\"153607\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c3702a2-258f-4f3d-bf78-a6bd94af3249>\",\"WARC-Concurrent-To\":\"<urn:uuid:ebc8d60b-b998-45ab-a56e-8c1c03684d04>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/149129/ndsolve-error-control-of-an-odes-system\",\"WARC-Payload-Digest\":\"sha1:LA7DBIIWJKCG2OPCJSY2VH7FVCX6FBUM\",\"WARC-Block-Digest\":\"sha1:7K7OLYIZHA7SWKE4IH5EXDWDKNHRDM2I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593657170639.97_warc_CC-MAIN-20200715164155-20200715194155-00174.warc.gz\"}"} |
https://michigan.alumni.columbia.edu/past_events | [
"WHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE\n\nWHERE"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79574084,"math_prob":0.91556364,"size":5138,"snap":"2020-45-2020-50","text_gpt3_token_len":1600,"char_repetition_ratio":0.19633813,"word_repetition_ratio":0.12988506,"special_character_ratio":0.32288828,"punctuation_ratio":0.18933824,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.00001,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T04:34:53Z\",\"WARC-Record-ID\":\"<urn:uuid:846fc91f-2bbc-4c67-986d-d463243815c0>\",\"Content-Length\":\"53913\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:06cfb9c7-7b6e-4b0b-a217-01aa70d04459>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a277b63-b0e4-410a-b1f5-cc66774202c6>\",\"WARC-IP-Address\":\"23.63.249.216\",\"WARC-Target-URI\":\"https://michigan.alumni.columbia.edu/past_events\",\"WARC-Payload-Digest\":\"sha1:SRICAOEPDX7NUXER2I5OVOPULLUIYEOC\",\"WARC-Block-Digest\":\"sha1:CC7A6ND777RID4PUPD6Q6C4XSICOIIUJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141205147.57_warc_CC-MAIN-20201130035203-20201130065203-00287.warc.gz\"}"} |
https://www.jpost.com/middle-east/hizbullah-threatens-petrochemical-installations | [
"(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);"
] | [
null
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https://numbermatics.com/n/133632/ | [
"# 133632\n\n## 133,632 is an even composite number composed of three prime numbers multiplied together.\n\nWhat does the number 133632 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 60 divisors.\n\n133632 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of sixty divisors.\n\n## Prime factorization of 133632:\n\n### 29 × 32 × 29\n\n(2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 29)\n\nSee below for interesting mathematical facts about the number 133632 from the Numbermatics database.\n\n### Names of 133632\n\n• Cardinal: 133632 can be written as One hundred thirty-three thousand, six hundred thirty-two.\n\n### Scientific notation\n\n• Scientific notation: 1.33632 × 105\n\n### Factors of 133632\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 12\n• Sum of prime factors: 34\n\n### Divisors of 133632\n\n• Number of divisors d(n): 60\n• Complete list of divisors:\n• Sum of all divisors σ(n): 398970\n• Sum of proper divisors (its aliquot sum) s(n): 265338\n• 133632 is an abundant number, because the sum of its proper divisors (265338) is greater than itself. Its abundance is 131706\n\n### Bases of 133632\n\n• Binary: 1000001010000000002\n• Hexadecimal: 0x20A00\n• Base-36: 2V40\n\n### Squares and roots of 133632\n\n• 133632 squared (1336322) is 17857511424\n• 133632 cubed (1336323) is 2386334966611968\n• The square root of 133632 is 365.5571090815\n• The cube root of 133632 is 51.1254122271\n\n### Scales and comparisons\n\nHow big is 133632?\n• 133,632 seconds is equal to 1 day, 13 hours, 7 minutes, 12 seconds.\n• To count from 1 to 133,632 would take you about thirteen hours.\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 133632 cubic inches would be around 4.3 feet tall.\n\n### Recreational maths with 133632\n\n• 133632 backwards is 236331\n• 133632 is a Harshad number.\n• The number of decimal digits it has is: 6\n• The sum of 133632's digits is 18\n• More coming soon!\n\n## Link to this page\n\nHTML: To link to this page, just copy and paste the link below into your blog, web page or email.\n\nBBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:\n\n## Cite this page\n\nMLA style:\n\"Number 133632 - Facts about the integer\". Numbermatics.com. 2021. Web. 23 April 2021.\n\nAPA style:\nNumbermatics. (2021). Number 133632 - Facts about the integer. Retrieved 23 April 2021, from https://numbermatics.com/n/133632/\n\nChicago style:\nNumbermatics. 2021. \"Number 133632 - Facts about the integer\". https://numbermatics.com/n/133632/\n\nThe information we have on file for 133632 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 133632, math, Factors of 133632, curriculum, school, college, exams, university, Prime factorization of 133632, STEM, science, technology, engineering, physics, economics, calculator, one hundred thirty-three thousand, six hundred thirty-two."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85781825,"math_prob":0.97535604,"size":2797,"snap":"2021-04-2021-17","text_gpt3_token_len":751,"char_repetition_ratio":0.13104188,"word_repetition_ratio":0.06167401,"special_character_ratio":0.3350018,"punctuation_ratio":0.16111112,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9814957,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-23T07:28:19Z\",\"WARC-Record-ID\":\"<urn:uuid:3fd7b2cf-c405-4742-b7bd-7bb17970b751>\",\"Content-Length\":\"21089\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:300893f6-64d4-4165-b520-1be45b5cb4b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:10ee87ed-20e4-44a6-8a6d-551209eb86ec>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/133632/\",\"WARC-Payload-Digest\":\"sha1:D2SFWNDFKUXUYNMDBA5H2V3CZWL3WWVQ\",\"WARC-Block-Digest\":\"sha1:NLQHHPZDN5RSCV5KETE66FMOEUG5UBHA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618039568689.89_warc_CC-MAIN-20210423070953-20210423100953-00128.warc.gz\"}"} |
http://www.stumblingrobot.com/2016/04/29/prove-two-lines-r3-intersect-determine-points-intersection/ | [
"Home » Blog » Prove that two lines in R3 intersect and determine the points of intersection\n\nProve that two lines in R3 intersect and determine the points of intersection\n\nLet",
null,
"Consider the lines",
null,
"and",
null,
"where",
null,
"is the line through",
null,
"parallel to",
null,
"and",
null,
"is the line through",
null,
"parallel to",
null,
". Prove that these two lines intersect and determine the point of intersection.\n\nProof. We have",
null,
"Then if",
null,
"and",
null,
"intersect we must have",
null,
"such that",
null,
"From the second equation we have",
null,
"which implies",
null,
". Plugging this into the first equation we then have",
null,
"which implies",
null,
". Therefore,",
null,
". Then we check these values of",
null,
"satisfy the third equation,",
null,
"So, the point of intersection is",
null,
""
] | [
null,
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null,
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null,
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null
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https://www.davemanuel.com/investor-dictionary/return-on-investment-roi/ | [
"## Definition of Return on Investment (ROI)\n\nWhat does Return on Investment, or ROI, mean? What is the definition of Return on Investment?\n\nWill an investment (of any kind) yield positive returns? And if so, how much?\n\nTo answer these questions, you need to calculate the ROI (Return on Investment).",
null,
"The simplest way to explain Return on Investment is to present you with the formula that is used to calculate ROI:\n\n(Return From Investment - Initial Cost of Investment) / Initial Cost of Investment\n\nSo, let's say that you invested \\$10,000 into a venture and ended up receiving \\$15,000 back (your initial \\$10,000 investment, plus \\$5,000 in profit).\n\nTo figure out your ROI, simply substitute these values into the equation given above:\n\n(\\$15,000 - \\$10,000) / \\$10,000\n\n\\$5,000 / \\$10,000 = .5\n\nThis would mean that you saw a ROI of 50%, which would be a \"positive return on investment.\"\n\nLet's say that you ended up receiving just \\$7,500 of your original \\$10,000 investment back.\n\n(\\$7,500 - \\$10,000) / \\$10,000\n\n-\\$2,500 / \\$10,000 = -.25\n\nThis would mean that you saw a ROI of -25%, which would be a \"negative return on investment\".\n\nThis is the simplest definition of the term \"Return on Investment\".\n\n--\n\nDavemanuel.com Articles That Mention Return on Investment (ROI):\n\nNone"
] | [
null,
"https://www.davemanuel.com/images/definition_roi.jpg",
null
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https://homework.cpm.org/category/CC/textbook/cca/chapter/7/lesson/7.2.2/problem/7-99 | [
"",
null,
"",
null,
"### Home > CCA > Chapter 7 > Lesson 7.2.2 > Problem7-99\n\n7-99.\n\nIf $f(x)=3(2)^x$, find the value of the expressions in parts (a) through (c) below. Then complete parts (d) through (f).\n\n1. $f(−1)$\n\nSubstitute $−1$ for $x$ and simplify.\n\n$f\\left(−1\\right) = 3\\left(2\\right)^{−1}$\n\n$=3\\cdot\\frac{1}{2}$\n\n$\\frac{3}{2}$\n\n1. $f(0)$\n\nRefer to part (a).\n\n1. $f(1)$\n\nRefer to part (a).\n\n1. What value of $x$ gives $f\\left(x\\right) = 12$?\n\nSubstitute $12$ for $f\\left(x\\right)$ and solve for $x$.\n\n$x = 2$\n\n2. Where does the graph of this function cross the $x$-axis? The $y$-axis?\n\nA function crosses the $x$-axis at $y = 0$, and a function crosses the $y$-axis at $x = 0$.\n\n3. If $g\\left(x\\right)=\\frac{1}{3x}$, find $f(x)·g(x)$.\n\n$\\text{Substitute }3(2)^{x}\\text{ for }f(x)\\text{ and }\\frac{1}{3x}\\text{ for }g(x).$\n\nUse the eTool below to find the value of each expression in parts (a) through (c).\nClick on the link at right for the full eTool version: 7-99 HW eTool"
] | [
null,
"https://homework.cpm.org/dist/7d633b3a30200de4995665c02bdda1b8.png",
null,
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null
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https://bmcmedresmethodol.biomedcentral.com/articles/10.1186/s12874-016-0189-0 | [
"# Dose-response meta-analysis of differences in means\n\n## Abstract\n\n### Background\n\nMeta-analytical methods are frequently used to combine dose-response findings expressed in terms of relative risks. However, no methodology has been established when results are summarized in terms of differences in means of quantitative outcomes.\n\n### Methods\n\nWe proposed a two-stage approach. A flexible dose-response model is estimated within each study (first stage) taking into account the covariance of the data points (mean differences, standardized mean differences). Parameters describing the study-specific curves are then combined using a multivariate random-effects model (second stage) to address heterogeneity across studies.\n\n### Results\n\nThe method is fairly general and can accommodate a variety of parametric functions. Compared to traditional non-linear models (e.g. E max, logistic), spline models do not assume any pre-specified dose-response curve. Spline models allow inclusion of studies with a small number of dose levels, and almost any shape, even non monotonic ones, can be estimated using only two parameters. We illustrated the method using dose-response data arising from five clinical trials on an antipsychotic drug, aripiprazole, and improvement in symptoms in shizoaffective patients. Using the Positive and Negative Syndrome Scale (PANSS), pooled results indicated a non-linear association with the maximum change in mean PANSS score equal to 10.40 (95 % confidence interval 7.48, 13.30) observed for 19.32 mg/day of aripiprazole. No substantial change in PANSS score was observed above this value. An estimated dose of 10.43 mg/day was found to produce 80 % of the maximum predicted response.\n\n### Conclusion\n\nThe described approach should be adopted to combine correlated differences in means of quantitative outcomes arising from multiple studies. Sensitivity analysis can be a useful tool to assess the robustness of the overall dose-response curve to different modelling strategies. A user-friendly R package has been developed to facilitate applications by practitioners.\n\n## Background\n\nThe identification and characterization of dose-response relationships is an essential part of the analysis in many scientific disciplines such as toxicology, pharmacology, and epidemiology. This is particularly important in the development and testing of new compounds (e.g. a new drug, pharmaceutical treatment) where trials at different stages aim to evaluate the efficacy of increasing levels of dosage (Phase II-III trials) or to derive a dose-response curve for selection of optimal doses (Phase IV trials) [1, 2].\n\nRandomized clinical trials often investigate a continuous outcome variable, such as the efficacy or safety of a drug, reporting the change from baseline of a medical score, or the final value of a clinical measurement. The dose-response results are typically summarized by dose-specific means and standard deviations . Measures of effect are expressed in terms of mean or standardized mean differences using a dose level, usually the placebo group, as referent . Over the last few years methodological research focused on developing and improving methods for performing dose-response analysis in a single study [4, 5]. A conclusive result is hardly obtained by a single investigation and there is often the need to synthesize information collected from multiple studies. In such a case meta-analytic methods can be used to define an overall relation or to investigate heterogeneity across study findings.\n\nA method for pooling aggregated dose-response data where the outcome is a log relative risk was originally presented by Greenland and Longnecker in 1992 . Since then, several papers have refined and covered specific aspects of the methodology such as model specification [7, 8], modeling strategies [9, 10], and software implementation [11, 12]. Other methodological articles extended the approach for continuous outcome but in the case where individual patient data are available, mainly in the context of time-series environmental studies .\n\nOnly a few alternatives have been proposed to pool aggregated dose-response data where the findings are summarized by differences in means. Davis and Chen in 2004 described a methodology for summarizing dose-response curves of first and second generation antipsychotics in schizoaffective patients. The authors reconstructed drug-specific dose-response curves and conducted a meta-analysis to compare the effectiveness of medium vs high dosages. A common alternative to analyze the drug effect consists of fitting a random-intercept E max model, where the random component accounts for heterogeneity in placebo effect across trials . Heterogeneity, however, may be related to other study characteristics rather than differences in placebo response such as implementation, participants, intervention, and outcome definition. Thomas et al. adopted hierarchical Bayesian models to summarize and describe, independently, the distribution of study-specific model parameters derived from an E max model.\n\nThe mentioned strategies assumed pre-specified models that do not allow for non-monotonic curves which may occur in practice , as in case of dose-response data of antipsychotics. In addition, fitting study-specific sigmoidal curves such as the E max model requires that the single studies have assessed at least three dose levels in order to estimate model parameters. Discarding studies not providing enough data points represents a loss of information and may introduce bias.\n\nThe aim of this paper is to formalize and propose a general and flexible methodology to pool dose-response relations from aggregated data where the changes in the distribution of the quantitative outcome are expressed in terms of differences in means. We first present the data necessary for a dose-response meta-analysis and derive formulas for obtaining effect sizes and their variance/covariance structure. We describe flexible dose-response models with particular emphasis on regression splines. The method is then applied to dose-response data from clinical trials on use of aripiprazole and symptoms improvement in schizoaffective patients.\n\n## Methods\n\n### Dose-response data\n\nThe notation and data required for a dose-response meta-analysis for a generic study are displayed in Table 1. We consider I studies indexed by i=1,…,I reporting the results of a common treatment at different dose levels x ij ,j=1,…,J i , where x 0i =0 indicates the control or placebo group in the i-th study. The study-specific results typically consist of dose-specific means of an outcome variable, Y ij , that measures the efficacy of the j-th dose in the i-th study . Additional information about the number of patients allocated in each treatment, n ij , and the sample standard deviations of Y ij , s d ij , is generally reported or obtained from the study-specific results.\n\n### Effect sizes and their variance/covariance\n\nA common way to reduce heterogeneity in placebo response is to compute the effect size (or treatment effect) as difference between dose-specific means and placebo mean. In case all studies measure the outcome on a common and interpretable scale, the difference can be based on the absolute scale\n\n$$d_{ij} = \\bar{Y}_{ij} - \\bar{Y}_{i0}, \\quad j = 1,\\dots, J_{i}, \\; i = 1, \\dots, I$$\n(1)\n\nAssuming common study-specific population standard deviations, the variance of d ij is defined as\n\n$$\\text{Var} \\left(d_{ij} \\right) = \\frac{n_{ij}+ n_{i0}}{n_{ij} n_{i0}} s_{p_{i}}^{2}, \\quad j = 1,\\dots, J_{i}, \\; i = 1, \\dots, I$$\n(2)\n\nwhere $$s_{p_{i}}^{2} = \\sum _{j = 0}^{J_{i}} \\left (n_{ij} - 1 \\right) sd_{ij}^{2} / \\sum _{j = 0}^{J_{i}} \\left (n_{ij} - 1 \\right)$$ is the square of the pooled standard deviation for the i-th study. Since the study-specific mean differences d ij use the same referent values, $$\\bar {Y}_{i0}$$, they cannot be regarded as independent. The covariance term is defined as\n\n$$\\text{Cov} \\left(d_{ij}, d_{ij^{'}} \\right) = \\text{Var} \\left(\\bar{Y}_{j0} \\right) = \\frac{s_{i0}^{2}}{n_{i0}}, \\quad j \\ne j^{'}, \\; i = 1,\\dots, I$$\n(3)\n\nIn case the outcome is measured on different scales the effect sizes can be based on standardized mean differences\n\n$$d_{ij}^{*} = \\frac{\\bar{Y}_{ij} - \\bar{Y}_{i0}}{s_{p_{i}}}, \\quad j = 1,\\dots, J_{i}, \\; i = 1, \\dots, I$$\n(4)\n\nwith\n\n$$\\begin{array}{*{20}l} \\text{Var} \\left(d_{ij}^{*} \\right) &= \\frac{1}{n_{ij}} + \\frac{1}{n_{i0}} + \\frac{{d_{ij}^{*}}^{2}}{2 \\sum_{j=0}^{J_{i}} n_{ij}}, \\quad j = 1,\\dots, J_{i},\\\\ i &= 1, \\dots, I \\\\ \\text{Cov} \\left(d_{ij}^{*}, d_{ij^{'}}^{*} \\right) &= \\frac{1}{n_{i0}} + \\frac{d_{ij}^{*}d_{ij^{'}}^{*}}{2 \\sum_{j=0}^{J_{i}} n_{ij}}, \\quad j \\ne j^{'}, \\; i = 1,\\dots, I \\end{array}$$\n(5)\n\n### Dose-response analysis\n\nThe chosen effect sizes and the corresponding (co)variances are used to estimate the study-specific dose-response curves. The dose-response curves characterize the relative efficacy of the dose under investigation using the placebo effect as referent (i.e. the relative efficacy for the placebo is zero by definition). The dose-response models are expressed through the parametric model f, which specifies how the effect size varies according to the dose values. The functional relationship f is parametrized in terms of θ i , the p×1 vector of dose-response coefficients. We consider the case of mean differences, d ij , but the same principles apply for standardized mean differences, $$d^{*}_{ij}$$. The study-specific curves can be written as\n\n$$\\boldsymbol{d}_{i} = f\\left(\\boldsymbol{x}_{i}, \\boldsymbol{\\theta}_{i} \\right) + \\boldsymbol{\\varepsilon}_{i}, \\;\\;\\; \\boldsymbol{\\varepsilon}_{i} \\sim N \\left(\\boldsymbol{0}, \\boldsymbol{\\hat \\Sigma}_{i} \\right), \\quad i = 1, \\dots, I$$\n(6)\n\n$$\\boldsymbol {\\hat \\Sigma }_{i}$$ is the covariance matrix of the residual error term, with Var(d ij ) along the diagonal and $$\\text {Cov} \\left (d_{ij}, d_{ij^{'}} \\right)$$ off-diagonal.\n\nSeveral alternatives are available to model the dose-response curve (i.e. for the choice of f). Table 2 shows the definition of 4 types of models ordered according to the number of parameters, ranging from 1 for a linear model up to 3 for a logistic model. See Bretz et al. for a comprehensive description .\n\nThe most common choice in dose findings is the use of the E max model which is expressed in terms of three parameters: the maximum effect (θ 1i ), the dose to produce half of the maximum effect (θ 2i ) and the steepness of the curve (θ 3i ) . As other non-linear models, the E max model assumes a specific shape that does not allow for non-monotonic curve and its estimation requires at least three non reference dose levels. Quadratic models are defined by only p=2 coefficients but may poorly fit at extreme dose values . Other non-linear models such as logistic and sigmoidal models, are commonly defined by p≥3 coefficients so that study-specific aggregated data may not be sufficient to estimate the parameters.\n\nWe propose the use of regression splines to flexibly model the dose of interest. Splines represent a family of smooth functions that can describe a wide range of curves (i.e. U-shaped, J-shaped, S-shaped, threshold) . The curves consist of piecewise polynomials over consecutive intervals defined by k knots. Their use may facilitate curve fitting since many non-linear curves can be examined by estimating only a small number of coefficients. For instance, a restricted cubic spline model with three knots k=(k 1,k 2,k 3) is defined only in terms of p=2 coefficients \n\n$$\\mathrm{E} \\left[ \\boldsymbol{d}_{i} | \\boldsymbol{x}_{i} \\right] = \\theta_{1i} \\boldsymbol{x}_{1i} + \\theta_{2i} \\boldsymbol{x}_{2i}$$\n(7)\n\nwith two transformations defined as\n\n$$\\begin{array}{*{20}l} x_{1} &= x \\\\ x_{2} &= \\frac{\\left(x - k_{1} \\right)_{+}^{3} - \\frac{k_{3} - k_{1}}{k_{3} - k_{2}} \\left(x - k_{2} \\right)_{+}^{3} + \\frac{k_{2} - k_{1}}{k_{3} - k_{2}} \\left(x - k_{3} \\right)_{+}^{3}}{ (k_{3} - k_{1})^{2}} \\end{array}$$\n(8)\n\nwhere the ‘+’ notation, with u +=u if u≥0 and u +=0 otherwise, has been used.\n\nAn alternative flexible approach to model the dose-response association is represented by fractional polynomials. In particular, a dose-response model based on fractional polynomial of order two can be written as in Eq. 6 with the two transformations defined as\n\n$$\\begin{array}{*{20}l} x_{1} &= x^{p_{1}} \\text{ and } x_{2} = x^{p_{2}} \\text{ if $$p_{1} \\neq p_{2}$$} \\\\ x_{1} &= x^{p_{1}} \\text{ and } x_{2} = {x}^{p_{1}} \\log({x}) \\text{ if $$p_{1} = p_{2}$$ } \\end{array}$$\n(9)\n\nfor each combination of p 1 and p 2 in the predefined set of values {−2,−1,−0.5,0,0.5,1,2,3}; for p=0, x p becomes log(x). The best fitting fractional polynomial is typically chosen based on the Akaike’s Information Criterion .\n\nOnce the functional relation f has been selected, generalized least square estimation can be performed to efficiently estimates the dose-response coefficients $$\\boldsymbol {\\hat \\theta }_{j}$$ and the corresponding (co)variance matrix $$\\boldsymbol {\\hat V}_{j}$$, by minimizing\n\n$$\\left(\\boldsymbol{d}_{i} - f\\left(\\boldsymbol{x}_{i}, \\boldsymbol{\\theta}_{i} \\right) \\right)^{T} \\boldsymbol{\\hat \\Sigma}_{i}^{-1} \\left(\\boldsymbol{d}_{i} - f\\left(\\boldsymbol{x}_{i}, \\boldsymbol{\\theta}_{i} \\right) \\right)$$\n(10)\n\nthat generally requires numerical optimization algorithms. If f is a linear combination of the parameters θ i , as in the case of regression splines and fractional polynomials, the close solution can be written as\n\n$$\\begin{array}{*{20}l} \\boldsymbol{\\hat \\theta}_{i} &= \\left(\\boldsymbol{X}_{i}^{T} \\boldsymbol{\\hat \\Sigma}_{i}^{-1} \\boldsymbol{X}_{i} \\right)^{-1} \\boldsymbol{X}_{i}^{T} \\boldsymbol{\\hat \\Sigma}_{i}^{-1} \\boldsymbol{d}_{i} \\\\ \\boldsymbol{\\hat V}_{i} &= \\text{Var} \\left(\\boldsymbol{\\hat \\theta}_{i} \\right) = \\left(\\boldsymbol{X}_{i}^{T} \\boldsymbol{\\hat \\Sigma}_{i}^{-1} \\boldsymbol{X}_{i} \\right)^{-1} \\end{array}$$\n(11)\n\nwhere X i indicates the J i ×p design matrix in the i-th study.\n\n### Meta-analysis\n\nThe estimated study-specific dose-response coefficients $$\\boldsymbol {\\hat \\theta }_{i}$$ and the accompanying (co)variance matrices $$\\boldsymbol {\\hat V}_{i}$$ are combined by means of multivariate meta-analysis\n\n$$\\boldsymbol{\\hat \\theta}_{i} \\sim N \\left(\\boldsymbol{\\theta}, \\boldsymbol{\\hat V}_{i} + \\boldsymbol{\\Psi} \\right)$$\n(12)\n\nA fixed-effects model assumes no statistical heterogeneity among study results, i.e. differences in the dose-response coefficients are only related to sampling error. The assumption of homogeneity may not hold in practice, unless it is known that the studies are performed in a similar way and are sampled from the same population . The Cochran’s Q test is typically used to test statistical heterogeneity across studies (H 0:Ψ=0) . Selected studies, however, will typically differ with respect to study design and implementation, selection of participants, and type of analyses. A certain degree of heterogeneity is expected and should be taken into account in the analysis. A random-effects model allows the dose-response coefficients, θ i , to vary across studies. Statistical heterogeneity is captured by the between-studies variance Ψ while θ represents the mean of the distribution of dose-response coefficients and an estimate, $$\\boldsymbol {\\hat \\theta }$$, can be obtained using (restricted) maximum likelihood estimation .\n\nAs a final result, the pooled dose-response curve can be presented in either a graphical or tabular form by predicting the mean differences of the outcome for a set of x dose values\n\n$$\\mathrm{E} [ \\boldsymbol{\\hat{d}} | \\boldsymbol{x} ] = f\\left(\\boldsymbol{X}, \\boldsymbol{\\hat \\theta} \\right)$$\n(13)\n\nwith an approximate (1−α/2)% confidence interval (CI), that in case f is a linear combination of θ can be expressed as\n\n$$\\mathrm{E} [ \\boldsymbol{\\hat{d}} | \\boldsymbol{x} ] \\mp z_{\\frac{\\alpha}{2}} \\sqrt{\\text{diag} \\left(f\\left(\\boldsymbol{X}, \\boldsymbol{\\hat \\theta} \\right)^{T} \\text{Cov}\\left(\\boldsymbol{\\hat \\theta} \\right) f\\left(\\boldsymbol{X}, \\boldsymbol{\\hat \\theta} \\right) \\right)}$$\n(14)\n\nwhere z α/2 is the α/2-th quantile of a standard normal distribution.\n\n### Dose findings\n\nOnce the pooled dose-response curve has been estimated, it may be of interest to determine a set of target doses, i.e. doses associated with prespecified outcome effects. In development of new compounds it is often important to select an optimal dose which is almost as effective as the maximum effective dose but has less undesired side effects, which often occur at high dosages. Suppose one wants to determine which is the lowest dose (ED γ ) to produce an almost complete effect, e.g. γ % of the observed maximum predicted response.\n\nThe ED γ can be determined as\n\n$$\\mathrm{\\widehat{ED}}_{\\gamma} = \\underset{x \\in \\left(0, x_{\\text{max}} \\right]}{\\operatorname{argmax}} \\left\\{ \\frac{\\mathrm{E} [ {\\hat{d}} | {x} ] }{\\mathrm{E} [ {\\hat{d}} | {{x_{\\text{max}}}} ]} \\geq \\gamma \\right\\}$$\n(15)\n\nwhere x max is the dose corresponding to the maximum predicted outcome.\n\nAn important step when presenting results from dose findings analysis is to accompany the previous estimates with a measure of precision, typically confidence intervals. Pinheiro et al. proposed the use a parametric bootstrap approach based on the asymptotic normal distribution of $$\\boldsymbol {\\hat \\theta }$$, the pooled estimate of the dose-response coefficients. The approach consists in re-sampling the dose-response coefficients θ from its approximate normal distribution and derive the distribution of $$\\mathrm {\\widehat {ED}}_{\\gamma }$$ based on the samples. Approximated confidence intervals for $$\\mathrm {\\widehat {ED}}_{\\gamma }$$ can be constructed using percentiles of the sampling distribution.\n\n## Results\n\nTo illustrate the methodology we examined the dose-response relation between aripiprazole, a second-generation antipsychotic, and symptoms improvement in schizoaffective patients. We updated the search strategy presented in a previous review by Davis and Chen by searching the Medline, International Pharmaceutical Abstracts, CINAHL, and the Cochrane Database of Systematic Reviews. To reduce the exclusion of unpublished papers, additional sources including Food and Drug administration website, data from Cochrane reviews, poster presentations and conference abstracts were also searched. All random-assignment, double-blind, controlled clinical trials of schizoaffective patients providing dose-response results for at least two non-zero dosages of aripiprazole were eligible.\n\nFive studies met the inclusion criteria and were included in the analysis. All the studies reported mean changes from baseline as main outcome variable, using the Positive and Negative Syndrome Scale (PANSS). The PANSS scale is an ordinal score derived from 30 items ranging from 1 to 7. Computations of ratios such as percentage changes are not directly applicable and may lead to erroneous results [33, 34]. To address this issue, the theoretical minimum (i.e. 30) needs to be subtracted from the original score . Information about the means, the number of patients assigned to each treatment, and the standard deviation was available from the published data. Because all the studies measured the outcome variable on the same scale, we computed PANSS mean differences as effect sizes. Data are reported in Table 3.\n\nWe used the trial by Cutler et al. to illustrate the steps required for estimating the dose-response curve for a single study. For example, the difference in mean PANSS comparing the dose of 2 mg/day relative to 0 mg/day is d 11=8.23−5.3=2.93 mg/day. Its variance is $$\\text {Var}(d_{11}) = (85+92)/(85\\times 92)\\times s_{p_{1}}^{2}$$ = 7.59, where $$s_{p_{1}}^{2}=119,419/356=335.4$$. The covariance of this difference in means PANSS is 18.312/85=3.94. The variance/covariance structure associated with the vector of differences in means for this trial d 1 can be presented in a matrix form\n\n$$\\begin{array}{@{}rcl@{}} \\boldsymbol{\\hat \\Sigma_{1}} =\\left[ \\begin{array}{lll} 7.59 & & \\\\ 3.94 & 7.72 & \\\\ 3.94 & 3.94 & 7.52 \\\\ \\end{array}\\right] \\end{array}$$\n\nTo estimate the dose-response curve we need first to specify the model f. We characterized the dose-response relation using a restricted cubic spline model with three knots located at the 10th, 50th, and 90th percentiles (0, 10, and 30 mg/day) of the overall dose distribution (p = 2). The restricted cubic spline dose-response model is defined as in Eq. 7. Efficient estimates of the dose-response coefficients and (co)variance matrix were obtained by generalized least square estimation\n\n$$\\begin{array}{*{20}l} \\boldsymbol{\\hat \\theta}_{1} &= (1.215, -5.738)^{T} \\\\ \\boldsymbol{\\hat V}_{1} &= \\text{Var} \\left(\\boldsymbol{\\hat \\theta}_{1} \\right) =\\left[ \\begin{array}{ll} 0.49 & \\\\ -3.65 & 31.64 \\\\ \\end{array}\\right] \\end{array}$$\n(16)\n\nWe applied the same procedure to the other studies included in the meta-analysis in order to obtain the study-specific $$\\boldsymbol {\\hat \\theta }_{i}$$ and $$\\boldsymbol {\\hat {V}}_{i}$$, i=1,…,5 (Table 4). The study-specific predicted curves are presented in Fig. 1. Under a random-effects model, restricted maximum likelihood estimates were\n\n\\begin{aligned} \\boldsymbol{\\hat \\theta} &= (0.937, -1.156)^{T} \\\\ \\widehat{\\text{Cov}} \\left(\\boldsymbol{\\hat \\theta} \\right) &= \\left[\\begin{array}{cc} 0.03 & \\\\ -0.05 & 0.10 \\\\ \\end{array}\\right] \\end{aligned}\n(17)\n\nA p-value < 0.001 for the multivariate Wald-type test H 0:θ=0 provided strong evidence against the null hypothesis of no relation between different doses of aripiprazole and mean change PANSS score. The Q test (Q=3.5, p-value = 0.899) did not detect substantial statistical heterogeneity across studies.\n\nTo communicate results of the pooled dose-response analysis, we can estimate the pooled mean differences in PANSS scores using 0 mg/day as referent as 0.937x 1−1.156x 2, together with the corresponding 95 % confidence interval for a generic dose x of interest as following\n\n$$\\left(0.937 x_{1} -1.156 x_{2} \\right) \\mp 1.96 \\sqrt{0.03 {x_{1}^{2}} + 0.1 {x_{2}^{2}} - 0.1x_{1}x_{2}}$$\n\nwhere x 1 and x 2 are defined as in Eq. 8. For instance, the model-based predicted mean changes in PANSS score compared to placebo were 4.52 (95 % CI: 2.96, 6.08) for 5 mg/day, 8.08 (95 % CI: 5.43, 10.73) for 10 mg/day, 9.95 (95 % CI: 6.97, 12.94) for 15 mg/day, 10.38 (95 % CI: 7.49, 13.27) for 20 mg/day, 9.84 (95 % CI: 6.86, 12.83) for 25 mg/day, and 8.83 (95 % CI: 5.11, 12.54) for 30 mg/day. The pooled predicted dose-response curve together with the confidence intervals and the model mean differences is provided in Fig. 2.\n\nThe results indicated a statistically significant positive association between increasing doses of aripiprazole and the mean change in PANSS score with the maximum value of 10.39 (95 % CI: 7.48, 13.30) observed at x max = 19.32 mg/day. The model suggested a slight decrease in the predicted mean PANSS score for dosages greater than 20 mg/day. The estimated dose to produce 50 % and 80 % of the predicted maximum effect were $$\\mathrm {\\widehat {ED}}_{50} =$$ 5.82 mg/day (95 % CI: 5.10, 8.58) and $$\\mathrm {\\widehat {ED}}_{80} =$$ 10.43 mg/day (95 % CI: 9.02, 16.73).\n\n### Sensitivity analysis\n\nA sensitivity analysis is often required to evaluate the robustness of the pooled dose-response curve. In the spline model, for example, the location of the knots may affect the shape of the dose-response curve. Therefore we considered alternative knots locations including different combinations of the 10th, 25th, 50th, 75th and 90th percentiles of the overall dose distribution (0, 0.5, 10, 18.75, and 30 mg/day). A graphical comparison is presented in the left panel of Fig. 3. The alternative curves roughly described the same dose-response shape with no substantial variation, all indicating an increase in the mean change PANSS score up to 20 mg/day of aripiprazole. We can assess whether there is an increasing trend above 20 mg/day by simply re-defining x 2 equal to (x−20)+ in Eq. 8; this approach is known as piecewise linear model. The rate of change in the PANSS mean differences was negative and not statistically significant (θ 1+θ 2=-0.284, p = 0.18) after 20 mg/day of aripiprazole.\n\nTo evaluate the sensitivity of the dose-response curve to the choice of the parametric model f adopted, instead, we considered three alternatives: fractional polynomials; quadratic; and E max. Since two studies only had two non-referent doses, the study-specific (sigmoidal) E max models as described in Table 2 cannot be estimated. A common solution is to fix the steepness of the curve θ 3 to be 1, also referred to as hyperbolic E max .\n\nThe “best” fractional polynomials (p 1=0.5, p 2=1) provided overall a similar dose-response curve when compared to the spline model, with slightly higher value for the maximum predicted response (right panel of Fig. 3). The hyperbolic E max had substantially higher predicted mean differences for low values of the dose. The non-linear model assumes a specific hyperbolic dose-response curve that did not seem to fit the data and may be dependent from the choice of fixing θ 3 to be 1. The dose-response curve described by the quadratic model fall in between the spline and the hyperbolic E max curves.\n\n## Discussion\n\nIn this paper we proposed a statistical method to combine differences in means of quantitative outcomes. The method consists of dose-response models estimated within each study (first stage) and an overall curve obtained by pooling study-specific dose-response coefficients (second stage). The covariance among study-specific mean differences is taken into account in the first stage analysis using generalized least square estimators, while statistical heterogeneity across studies is allowed by multivariate random-effects model in the second stage.\n\nOne major strength of the proposed method is that it is fairly general and can accommodate different modeling strategies, including non-linear ones described by Pinheiro et al. . Non-linear models, however, are defined by at least three or four parameters, and hence require an equal number of dose levels for each single study included in the analysis. Given that some studies may have investigated a lower number of dose levels, exclusion of these studies may result in substantial loss of information. In addition, many non-linear models assume a specific behaviour (e.g. monotonicity) requiring a strong a priori information about the dose-response curve. The choice of the parametric model is critical, since it highly influences the final results . Indeed, the selection of the dose-response model should be informed by subject-matter knowledge as well as understanding of the research questions at hand. We presented the use of regression splines as a flexible tool for modeling any quantitative exposure. The major advantage is that a variety of curves, even non monotonic ones, can be estimated using only two parameters. It is considered to be closed to non-parametric regression, since no major assumptions about the shape of the curve are needed . A possible alternative is the use of fractional polynomials. In comparing the two strategies, we did not find important differences between the two strategies and concluded that both are useful tools to characterize a (non-linear) dose-response curve. Nonetheless a sensitivity analysis is generally required to evaluate the robustness of the combined results.\n\nA possible limitation of the proposed methodology is that it requires information about dose-specific means and standard deviations. Studies providing other summary measures, such as dose-specific medians, would not be included the analysis. The dose-response analysis presented in Eq. 6 is based on the asymptotic normal distribution of the conditional mean effect size. Extension of the introduced methodology to percentiles is not straightforward and may represent an interesting topic of future research.\n\nAn additional limitation of aggregated dose-response data is that supplementary information for approximating the covariance terms may not be available. Articles may report directly mean or standardized mean differences and standard errors for non-referent dose groups. Whenever the standard deviation for the outcome variable in the control group ($${s_{i0}^{2}}$$) cannot be obtained, it may be approximated using the pooled standard deviation based on the non-referent dose levels ($$s_{p_{ij}}^{2}$$). Alternatively a specific value may be imputed and a sensitivity analysis can be performed to evaluate how the results of the meta-analysis vary for different values of $${s_{0j}^{2}}$$. Further limitations relate to the general application of meta-analysis based on aggregated data. These include restrictions in subgroup analyses, the impossibility of assessing the appropriateness of individual analyses, and to harmonize variable definitions and analyses for reducing the extent of heterogeneity, as well as specific biases such as aggregation (or ecological) bias in meta-regression models. Meta-analysis of individual patient data, however, are often difficult to undertake especially for the availability of individual data, so that usage of aggregated data may represent the only alternative . Specific to aggregated dose-response data, different dose references and exposure range may complicate the analysis. The presented methodology assume that all the selected studies share a common dose-response model. Important departure from this assumption may limit and/or impact the pooling of individual dose-response coefficients. An alternative methods has been proposed based on a series of univariate meta-analyses of effect sizes for a pre-specified grid of dose-levels . Further work is needed to analyze this possibility and potential advantages. Depending on the extent of heterogeneity of the dose-response curves, however, it may not be opportune to pool study-specific results, and meta-regression or stratified analyses should be performed .\n\nIn our application, we considered the effectiveness of increasing dosages of aripiprazole in shizoaffective patients. We described the steps needed to obtain the overall dose-response curve and to present it in a graphical form. We observed a non-linear association with the maximum efficacy corresponding to aripiprazole 19.32 mg/day. An estimated dose of 10.46 mg/day, however, may be sufficient to obtain 80 % of the maximum effect, which may be relevant for avoiding possible undesired side effects. Sensitivity analysis showed similar results as compared to fractional polynomials. The E max model presented higher drug efficacy for low dosages. Compared to the previous models, the E max model did not seem to fit properly the data at low dosages.\n\n## Conclusions\n\nWe described an approach to combine differences in means of a quantitative outcome contrasting different dose levels relative to a placebo in randomized trials. The general framework of the proposed methodology can include a variety of flexible models. Sensitivity analysis can be a useful tool to assess the stability of the overall dose-response curve to different modelling strategies. Although the method was presented for the analysis of randomized trials, it may be extended to observational studies where mean differences are further adjusted for potential confounders. Future work is needed to evaluate the properties of the statistical model and validity of the underlying assumptions. A user friendly procedure is implemented in the dosresmeta R package with worked examples available on GitHub.\n\n## Abbreviations\n\nPANSS, positive and negative syndrome scale\n\n## References\n\n1. Bretz F, Pinheiro JC, Branson M. Combining multiple comparisons and modeling techniques in dose-response studies. Biometrics. 2005; 61(3):738–48. doi:10.1111/j.1541-0420.2005.00344.x. Accessed 24 Feb 2015\n\n2. Bretz F, Hsu J, Pinheiro J, Liu Y. Dose finding - a challenge in statistics. Biometrical Journal. Biometrische Zeitschrift. 2008; 50(4):480–504. doi:10.1002/bimj.200810438.\n\n3. Pinheiro JC, Bretz F, Branson M. Analysis of dose–response studies. modeling approaches. In: Dose Finding in Drug Development. New York City: Springer: 2006. p. 146–71.\n\n4. Ting N. Dose finding in drug development. New York City: Springer; 2006.\n\n5. Chevret S. Statistical methods for dose-finding experiments. New York City: Wiley; 2006.\n\n6. Greenland S, Longnecker MP. Methods for trend estimation from summarized dose-response data, with applications to meta-analysis. Am J Epidemiol. 1992; 135(11):1301–1309.\n\n7. Berlin JA, Longnecker MP, Greenland S. Meta-analysis of epidemiologic dose-response data. Epidemiology. 1993; 4(3):218–28.\n\n8. Dumouchel W. Meta-analysis for dose–response models. Stat Med. 1995; 14(5-7):679–85.\n\n9. Bagnardi V, Zambon A, Quatto P, Corrao G. Flexible meta-regression functions for modeling aggregate dose-response data, with an application to alcohol and mortality. Am J Epidemiol. 2004; 159(11):1077–1086. doi:10.1093/aje/kwh142. Accessed 25 Nov 2013\n\n10. Liu Q, Cook NR, Bergström A, Hsieh CC. A two-stage hierarchical regression model for meta-analysis of epidemiologic nonlinear dose–response data. Comput Stat Data Anal. 2009; 53(12):4157–167.\n\n11. Orsini N, Bellocco R, Greenland S, et al. Generalized least squares for trend estimation of summarized dose-response data. Stata J. 2006; 6(1):40.\n\n12. Orsini N, Li R, Wolk A, Khudyakov P, Spiegelman D. Meta-analysis for linear and nonlinear dose-response relations: examples, an evaluation of approximations, and software. Am J Epidemiol. 2012; 175(1):66–73.\n\n13. Dominici F, Samet JM, Zeger SL. Combining evidence on air pollution and daily mortality from the 20 largest us cities: a hierarchical modelling strategy. J Royal Stat Soc Ser A (Stat Soc). 2000; 163(3):263–302.\n\n14. Simmonds MC, Higginsa JP, Stewartb LA, Tierneyb JF, Clarke MJ, Thompson SG. Meta-analysis of individual patient data from randomized trials: a review of methods used in practice. Clin Trials. 2005; 2(3):209–17.\n\n15. Gasparrini A, Armstrong B, Kenward MG. Multivariate meta-analysis for non-linear and other multi-parameter associations. Stat Med. 2012; 31(29):3821–839. doi:10.1002/sim.5471. Accessed 11 Dec 2013\n\n16. Davis JM, Chen N. Dose response and dose equivalence of antipsychotics. J Clin Psychopharmacol. 2004; 24(2):192–208.\n\n17. Holford NH, Sheiner LB. Understanding the dose-effect relationship: clinical application of pharmacokinetic-pharmacodynamic models. Clin Pharmacokinet. 1981; 6(6):429–53.\n\n18. Thomas N, Sweeney K, Somayaji V. Meta-analysis of clinical dose-response in a large drug development portfolio. Stat Biopharm Res. 2014; 6(4):302–17. doi:10.1080/19466315.2014.924876. Accessed 11 Feb 2015\n\n19. Geddes J, Freemantle N, Harrison P, Bebbington P. Atypical antipsychotics in the treatment of schizophrenia: systematic overview and meta-regression analysis. BMJ : British Medical Journal. 2000; 321(7273):1371–1376. Accessed 24 Mar 2015.\n\n20. Macdougall J. Analysis of dose–response studies. emax model. In: Dose Finding in Drug Development. New York City: Springer: 2006. p. 127–45.\n\n21. Durrleman S, Simon R. Flexible regression models with cubic splines. Stat Med. 1989; 8(5):551–61. doi:10.1002/sim.4780080504. Accessed 30 Mar 2015\n\n22. Desquilbet L, Mariotti F. Dose-response analyses using restricted cubic spline functions in public health research. Stat Med. 2010; 29(9):1037–1057. doi:10.1002/sim.3841. Accessed 25 Nov 2013\n\n23. Harrell F.E. Jr, Lee KL, Pollock BG. Regression models in clinical studies: determining relationships between predictors and response. J Natl Cancer Inst. 1988; 80(15):1198–1202.\n\n24. Royston P. A strategy for modelling the effect of a continuous covariate in medicine and epidemiology. Stat Med. 2000; 19(14):1831–1847.\n\n25. Jackson D, White IR, Riley RD. Quantifying the impact of between-study heterogeneity in multivariate meta-analyses. Stat Med. 2012; 31(29):3805–820.\n\n26. Cochran WG. The combination of estimates from different experiments. Biometrics. 1954; 10(1):101–29. doi:10.2307/3001666. Accessed 09 Jan 2015\n\n27. Ritz J, Demidenko E, Spiegelman D. Multivariate meta-analysis for data consortia, individual patient meta-analysis, and pooling projects. J Stat Plann Infer. 2008; 138(7):1919–1933.\n\n28. Cutler AJ, Marcus RN, Hardy SA, O’Donnell A, Carson WH, McQuade RD. The efficacy and safety of lower soses of aripiprazole for the treatment of patients with acute exacerbation of schizophrenia. CNS Spectrums. 2006; 11(09):691–702. doi:10.1017/S1092852900014784. Accessed 06 Feb 2015\n\n29. McEvoy JP, Daniel DG, Carson WH, McQuade RD, Marcus RN. A randomized, double-blind, placebo-controlled, study of the efficacy and safety of aripiprazole 10, 15 or 20 mg/day for the treatment of patients with acute exacerbations of schizophrenia. J Psychiatr Res. 2007; 41(11):895–905. doi:10.1016/j.jpsychires.2007.05.002.\n\n30. Kane JM, Carson WH, Saha AR, McQuade RD, Ingenito GG, Zimbroff DL, Ali MW. Efficacy and safety of aripiprazole and haloperidol versus placebo in patients with schizophrenia and schizoaffective disorder. J Clin Psychiatry. 2002; 63(9):763–71.\n\n31. Potkin SG, Saha AR, Kujawa MJ, Carson WH, Ali M, Stock E, Stringfellow J, Ingenito G, Marder SR. Aripiprazole, an antipsychotic with a novel mechanism of action, and risperidone vs placebo in patients with schizophrenia and schizoaffective disorder. Arch Gen Psychiatr. 2003; 60(7):681–90. doi:10.1001/archpsyc.60.7.681.\n\n32. Turner EH, Knoepflmacher D, Shapley L. Publication bias in antipsychotic trials: an analysis of efficacy comparing the published literature to the US Food and Drug Administration Database. PLoS Med. 2012;9(3). doi:10.1371/journal.pmed.1001189. Accessed 30 Mar 2015.\n\n33. Obermeier M, Schennach-Wolff R, Meyer S, Möller HJ, Riedel M, Krause D, Seemüller F. Is the panss used correctly? a systematic review. BMC psychiatry. 2011; 11(1):1.\n\n34. Leucht S, Kissling W, Davis JM. The PANSS should be rescaled. Schizophrenia bulletin. 2010; 36:461–462.\n\n35. Leucht S, Davis JM, Engel RR, Kane JM, Wagenpfeil S. Defining response in antipsychotic drug trials: recommendations for the use of scale-derived cutoffs. Neuropsychopharmacology. 2007; 32(9):1903–1910.\n\n36. Lyman GH, Kuderer NM. The strengths and limitations of meta-analyses based on aggregate data. BMC Med Res Methodol. 2005; 5(1):14.\n\n37. Sauerbrei W, Royston P. A new strategy for meta-analysis of continuous covariates in observational studies. Stat Med. 2011; 30(28):3341–360.\n\n38. Greenland S. Invited commentary: a critical look at some popular meta-analytic methods. Am J Epidemiol. 1994; 140(3):290–6.\n\n39. Crippa A, Orsini N. Multivariate dose-response meta-analysis: the dosresmeta r package. J Stat Softw. 2016. In Press.\n\n## Acknowledgements\n\nWe are grateful to Dr. Stefan Leucht and John M. Davis for providing the data and raising the methodological question under study.\n\n### Funding\n\nThis work was supported by Karolinska Institutet’s funding for doctoral students (KID-funding) (AC) and by a Young Scholar Award from the Karolinska Institutet’s Strategic Program in Epidemiology (SfoEpi) (NO).\n\n### Availability of data and materials\n\nThe data on the effectiveness of aripiprazole are publicly available and also contained in dosresmeta R package on github (https://github.com/alecri/dosresmeta/blob/master/data/ari.rda).\n\n### Authors’ contributions\n\nAC developed the methods and prepared a draft. NO provided critical reviews, corrections and revisions. Both authors read and approved the final version of the manuscript.\n\n### Authors’ information\n\nAC is a PhD student in Epidemiology and Biostatistics. Dr. NO is Associate Professor of Medical Statistics.\n\n### Competing interests\n\nThe authors declare that they have no competing interest.\n\nNot applicable.\n\nNot applicable.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Alessio Crippa.\n\n## Rights and permissions",
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"https://bmcmedresmethodol.biomedcentral.com/track/article/10.1186/s12874-016-0189-0",
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https://stats.stackexchange.com/questions/18975/proof-for-a-binomial-equation/18981 | [
"# Proof for a binomial equation\n\nI'm trying to formally write out the problem: Light bulb color problem when I come up with the following equation\n\n$$\\sum_{i=0}^{k-1}{C}_{2k-1}^{2k-1-i}{p}^{2k-1-i}{(1-p)}^{i}=\\sum_{i=0}^{k-1}{C}_{2k}^{2k-i}{p}^{2k-i}{(1-p)}^{i}+\\frac{1}{2}{C}_{2k}^{k}{p}^{k}{(1-p)}^{k}$$\n\nwhere it is supposed to hold for all $k\\geq1$. Let's forget the original light bulb problem for a while. If we just want to prove the above equation, how shall we do it? I tried induction and it seemed difficult and baffling... Is there any idea or perhaps tricks one can use to establish it? What is the general approach for establishing these kind of complex combinatorial equations? (By the way, I know from the light bulb problem that it must be true for $0\\leq p\\leq1$, but does it also hold for any $p$ beyond that domain?)\n\nEdit: I vaguely remember there's a theorem or something that says if the above equation holds for any $0\\leq p\\leq1$ then it must hold for any $p\\in\\mathbb{R}$. Does any one know what that theorem is?\n\n• How would you interpret $C_{2k}^k$? What would selecting $2k$ items from $k$ items mean? Should this term not be $0$? – varty Nov 26 '11 at 5:06\n• @varty:$C_{2k}^k$ is the situation when there're equal numbers of flashes of red and blue (where each color has k flashes). Left hand side is the probability you guess correctly under best strategy when you decide to observe $2k-1$ times; RHS is the probability you when you decide to observe $2k$ times. – Eric Nov 26 '11 at 5:49\n• Are you using $C_{2k}^k$ to mean the number of ways we select $k$ items from $2k$ items? – varty Nov 26 '11 at 5:53\n• @varty: This is the alternative notation to ${2k \\choose k}$, used for instance in French textbooks. – Xi'an Nov 26 '11 at 7:41\n\nAbout the combinatorial question: the proof follows from the identity $${2k-1 \\choose i-1} + {2k-1 \\choose i} = {2k \\choose i}$$ Thus $$\\sum_{i=0}^{k-1} {2k\\choose i} p^{2k-i} (1-p)^i = \\sum_{i=0}^{k-1} {2k-1 \\choose i} p^{2k-i} (1-p)^i + \\sum_{i=0}^{k-1} {2k-1 \\choose i-1} p^{2k-i} (1-p)^i$$ which implies $$\\sum_{i=0}^{k-1} {2k\\choose i} p^{2k-i} (1-p)^i = \\sum_{i=0}^{k-2} {2k-1 \\choose i} p^{2k-1-i} (1-p)^i + {2k-1 \\choose k-1} p^{k+1} (1-p)^{k-1}$$ and $$\\frac{1}{2}{2k\\choose k} = {2k-1\\choose k-1}$$ implying that $${2k-1 \\choose k-1} p^{k+1} (1-p)^{k-1} + \\frac{1}{2}{2k\\choose k} p^{k} (1-p)^{k} = {2k-1\\choose k-1 } p^{k} (1-p)^{k-1}$$ which establishes your identity.\n\nAbout your \"Edit\" question, I think you mean the theorem in complex calculus that states that, if an analytic function is constant over an interval, it is constant everywhere.\n\n• Even better, for any fixed $k$ it's a univariate polynomial (in $p$) which has infinitely many zeroes, and thus is zero. No complex calculus necessary! – Erik P. Dec 1 '11 at 5:07\n\nA coin has a chance $p$ of landing heads. Among an odd number ($2k-1$) of tosses, what are the chances there are more heads than tails?\n\nLet's look at two solutions.\n\n1. The number of heads is $k$, $k+1$, $k+2$, $...$, or $2k-1$. Sum the chances of each of these events.\n\n2. Toss the coin one more time!\n\nIf the number of heads is now $k+1$, $k+2$, $...$, or $2k$, then originally there were at least $k$ heads. Sum these chances.\n\nIf the number of heads is now exactly $k$, it's uncertain whether there were $k-1$ or $k$ heads among the original $2k-1$ tosses. It all depends on whether the last toss was a head: there were $k$ heads originally if and only if the last toss was a tail. Do the following: randomly write down all $2k$ results--of which $k$ are heads and $k$ are tails, by assumption. One of them represents the last toss, but which? We don't know, but we do know it has equal chances of being in any of the $2k$ positions. Since $k$ of those are tails, there is a $k/(2k)=1/2$ chance the last toss was a tail.\n\nIn solution (1), the chance of exactly $i$ tails, equivalent to $2k-1-i$ heads, is given by the Binomial Distribution for $2k-1$ tosses,\n\n$${\\Pr}_{2k-1}(2k-1-i) = \\binom{2k-1}{i}p^{2k-1-i}(1-p)^i = C_{2k-1}^{2k-1-i}p^{2k-1-i}(1-p)^i.$$\n\nThe sum goes from $i=0$ to $i=k-1$. That's the left hand side of the equation in the question.\n\nIn solution (2), the chance of exactly $i$ tails is equivalent to $2k-i$ heads, given by\n\n$${\\Pr}_{2k}(2k-i) = \\binom{2k}{i}p^{2k-i}(1-p)^i = C_{2k}^{2k-i}p^{2k-i}(1-p)^i.$$\n\nTotaling these for $i=0, 1, \\ldots, k-1$ gives the summation to the right of the equation in the question. The chance that the last toss is a tail and all $2k$ tosses are heads is the product\n\n$${\\Pr}_{2k}\\left(\\text{last toss is tail}\\,|\\,k\\text{ heads}\\right){\\Pr}_{2k}\\left(k\\text{ heads}\\right) = \\frac{1}{2}\\binom{2k}{k}p^{2k-k}(1-p)^k = \\frac{1}{2}C_{2k}^kp^k(1-p)^k.$$\n\nThis is the extra term at the right of the equation. Solution (2) therefore produces the formula on the right hand side.\n\nThus, the equation holds for all $k$ because it gives two ways of expressing the chance of observing a majority of heads in any odd number of tosses.\n\nThe general approach exemplified here is to count the same thing in two different ways. It's a powerful method, advocated particularly by combinatorialists (people who specialize in counting things). Many complicated formulas turn out to have simple, illuminating proofs of this nature. For details, see Richard Stanley's Enumerative Combinatorics. (Volume 1 is available on the Web--Google it.)\n\nLet's address the last part of the question. Both sides of the equation are obviously polynomials in $p$. Their difference is a polynomial of degree at most $2k$. It is zero for all valid probabilities $p$. There are more than $2k+1$ probabilities (there are infinitely many of them in the interval from $0$ to $1$). Since a polynomial of degree $n$ is completely determined by its values at any $n+1$ numbers, the difference is the zero function--and in particular equals zero for any number $p$."
] | [
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https://fatareforms.org/5358/why-are-intersecting-lines-always-coplanar/ | [
"# Why are intersecting lines always coplanar?\n\nTwo intersecting lines are always coplanar.\n\nIntersecting: The two lines are coplanar (meaning that they lie on the same plane) and intersect at a single point. Parallel: The two lines are coplanar but never intersect because they travel through different points, while their direction vectors are scalar multiples of one another.\n\nBeside above, what are two coplanar lines that dont intersect? Two lines are parallel lines if they are coplanar and do not intersect. Lines that are not coplanar and do not intersect are called skew lines. Two planes that do not intersect are called parallel planes.\n\nBesides, are intersecting lines non coplanar?\n\nFor two intersecting noncoplanar lines, take the point of intersection, and one other point on each line. Because the lines intersect, the three points are not colinear. That means each line has two points in that plane, so each line is in that plane, therefore they are coplanar.\n\nWhat does the intersection of two coplanar lines form?\n\nIf the lines are coplanar, they either intersect (in a single point), or are the same line (colinear) or are parallel (no intersection). If you know they intersect (perhaps from the context of the question), you can immediately look for the single point.\n\n### Are parallel lines congruent?\n\nIf two parallel lines are cut by a transversal, the corresponding angles are congruent. If two lines are cut by a transversal and the corresponding angles are congruent, the lines are parallel. Interior Angles on the Same Side of the Transversal: The name is a description of the “location” of the these angles.\n\n### Are 2 points always coplanar?\n\nCoplanar points: A group of points that lie in the same plane are coplanar. Any two or three points are always coplanar.\n\n### Are skew lines coplanar?\n\nIn three-dimensional geometry, skew lines are two lines that do not intersect and are not parallel. Two lines that both lie in the same plane must either cross each other or be parallel, so skew lines can exist only in three or more dimensions. Two lines are skew if and only if they are not coplanar.\n\n### Are vertical angles congruent?\n\nWhen two lines intersect to make an X, angles on opposite sides of the X are called vertical angles. These angles are equal, and here’s the official theorem that tells you so. Vertical angles are congruent: If two angles are vertical angles, then they’re congruent (see the above figure).\n\n### Can two planes meet in exactly one point?\n\nTwo planes intersect in exactly one point. Two intersecting lines are always coplanar.\n\n### Are perpendicular lines coplanar?\n\nPerpendicular Lines. Coplanar lines that are not parallel must intersect or cross each other. They can intersect at any angle, but when the lines intersect at exactly 90° they are perpendicular lines. Perpendicular lines create four right angles at their point of intersection.\n\n### Are 4 points coplanar?\n\nCoplanar points are three or more points which all lie in the same plane. Any set of three points in space is coplanar. A set of four points may be coplanar or may be not coplanar.\n\n### How many lines are determined by two points?\n\nSince we have given two points. And we know the definition of line which states that ” A line is formed by two points and it can be extended from both the sides.” So, From two points, there must be 1 line . Hence, one line is determined by two points.\n\n### What is a perpendicular line?\n\nIn elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). A line is said to be perpendicular to another line if the two lines intersect at a right angle.\n\n### Can two lines be coplanar?\n\nTwo lines in three-dimensional space are coplanar if there is a plane that includes them both. This occurs if the lines are parallel, or if they intersect each other. Two lines that are not coplanar are called skew lines.\n\n### What is the symbol for intersecting lines?\n\nTwo lines that intersect and form right angles are called perpendicular lines. The symbol ⊥ is used to denote perpendicular lines.\n\n### Are all parallel lines coplanar?\n\nIt’s important to know about parallel lines because you’re going to use it through all parts of Geometry. Technically parallel lines are two coplanar which means they share the same plane or they’re in the same plane that never intersect. And you can identify parallel lines by having the same number of arrows.\n\n### What are two coplanar lines cut by a transversal?\n\nIf two coplanar lines are cut by a transversal so that a pair of corresponding angles are congruent, then the two lines are parallel. You can use the Converse of the Corresponding Angles Postulate to show that two lines are parallel.\n\n### Do 2 intersecting lines determine a plane?\n\n“If two lines intersect, then exactly one plane contains the lines.” “If two lines intersect, then they intersect in exactly one point.” and three noncollinear points define a plane."
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https://qchu.wordpress.com/2009/11/15/the-many-faces-of-schur-functions/?replytocom=448 | [
"Feeds:\nPosts\n\n## The many faces of Schur functions\n\nThe last time we talked about symmetric functions, I asked whether the vector space",
null,
"$\\mathcal{R}$ could be turned into an algebra, i.e. equipped with a nice product. It turns out that the induced representation allows us to construct such a product as follows:\n\nGiven representations",
null,
"$\\rho_1, \\rho_2$ of",
null,
"$S_n, S_m$, their tensor product",
null,
"$\\rho_1 \\otimes \\rho_2$ is a representation of the direct product",
null,
"$S_n \\times S_m$ in a natural way. Now,",
null,
"$S_n \\times S_m$ injects naturally into",
null,
"$S_{n+m}$, which gives a new representation",
null,
"$\\rho = \\text{Ind}_{S_n \\times S_m}^{S_{n+m}} \\rho_1 \\otimes \\rho_2$.\n\nThe character of this representation is called the induction product",
null,
"$\\rho_1 * \\rho_2$ of the characters of",
null,
"$\\rho_1, \\rho_2$, and with this product",
null,
"$\\mathcal{R}$ becomes a graded commutative algebra. (Commutativity and associativity are fairly straightforward to verify.) It now remains to answer the first question: does there exist an algebra homomorphism",
null,
"$\\phi : \\Lambda \\to \\mathcal{R}$? And can we describe the inner product on",
null,
"$\\Lambda$ coming from the inner product on",
null,
"$\\mathcal{R}$?\n\nTo answer this question we’ll introduce perhaps the most important class of symmetric functions, the Schur functions",
null,
"$s_{\\lambda}$.\n\nN.B. I’ll be skipping even more proofs than usual today, partly because they require the development of a lot of machinery I haven’t described and partly because I don’t understand them all that well. Again, good references are Sagan or EC2.\n\nSchur-Weyl duality\n\nOne way that the Schur functions naturally occur is in the description of Schur-Weyl duality. To the extent that I understand this, it goes as follows. On",
null,
"$\\underbrace{\\mathbb{C}^n \\otimes ... \\otimes \\mathbb{C}^n}_{k \\text{ times}}$ there are two commuting representations: one of",
null,
"$GL_n(\\mathbb{C})$ which acts on all the factors simultaneously and one of",
null,
"$S_k$ which permutes the factors. An important property of these representations is that they are mutual centralizers, i.e. each is the largest group of transformations which commutes with the other. It then turns out that the above representation decomposes as",
null,
"$\\displaystyle \\bigoplus_{\\lambda} \\pi^{\\lambda} \\otimes \\rho_n^{\\lambda}$\n\nwhere",
null,
"$\\lambda$ runs through all partitions with",
null,
"$k$ boxes and at most",
null,
"$n$ rows,",
null,
"$\\pi^{\\lambda}$ is the irreducible representation of",
null,
"$S_k$ associated to the partition",
null,
"$\\lambda$, and",
null,
"$\\rho_n^{\\lambda}$ is a corresponding irreducible representation of",
null,
"$GL_n(\\mathbb{C})$.\n\nWe already know of two of these representations, the symmetric algebra",
null,
"$S^k(\\mathbb{C}^n)$ and the exterior algebra",
null,
"$E^k(\\mathbb{C}^n)$. It turns out that the symmetric algebra corresponds to the partition",
null,
"$\\lambda = (k)$ (one row) and the exterior algebra corresponds to the partition",
null,
"$\\lambda = (1^k)$ (one column), and it’s not hard to see that the former is always there and the latter is only there as long as",
null,
"$k \\ge n$. As we saw previously, the characters of these representations can be described in terms of eigenvalues. This turns out to be true more generally.\n\nProposition: Suppose",
null,
"$\\phi$ is a representation",
null,
"$GL_n(\\mathbb{C}) \\to GL_m(\\mathbb{C})$ which is homogeneous, i.e. it respects scalar multiplication, and polynomial, i.e. every entry of the output",
null,
"$\\phi(A)$ is a polynomial function of the entries of the input",
null,
"$A$. Then the character of",
null,
"$\\phi$ is a symmetric function of the eigenvalues of",
null,
"$A$.\n\nThe restriction to homogeneous polynomial representations is to avoid discontinuity and powers of the representation",
null,
"$A \\mapsto \\det (A)$. Anyway, the above result is not too hard to see: since the diagonalizable matrices are dense and characters are both continuous and invariant under conjugation, the character must depend only on the multiset of eigenvalues, and since the representation is homogeneous and polynomial, the character must be as well.\n\nI am not familiar with the explicit description of the representations",
null,
"$\\rho_n^{\\lambda}$, but the important point is that these representations are well-defined with respect to Schur-Weyl duality.\n\nDefinition 1: The Schur polynomial",
null,
"$s_{\\lambda}(x_1, ... x_n)$ is the value of the character of",
null,
"$\\rho_n^{\\lambda}$ evaluated at the diagonal matrix with entries",
null,
"$x_1, ... x_n$.\n\nFor example,",
null,
"$s_{(k)}(x_1, ... x_n) = h_k(x_1, ... x_n)$ (the symmetric algebra) and",
null,
"$s_{(1^k)}(x_1, ... x_n) = e_k(x_1, ... x_n)$ (the exterior algebra). Note that if",
null,
"$\\lambda$ has more than",
null,
"$n$ rows then",
null,
"$s_{\\lambda}(x_1, ... x_n)$ is not defined.\n\nOf course, since I haven’t told you how to write these functions down, this is a rather unsatisfying definition, but at least it has a concrete tie to an important representation-theoretic concept. Now, it turns out that",
null,
"$s_{\\lambda}(x_1, ... x_n, 0) = s_{\\lambda}(x_1, ... x_n)$, so the Schur functions",
null,
"$s_{\\lambda} \\in \\Lambda$ are well-defined as symmetric functions.\n\nSchur-Weyl duality turns out to imply a strong relationship between these characters and the characters of the symmetric group which generalizes the relationship the complete homogeneous symmetric functions and elementary symmetric functions have to the trivial and sign representations.\n\nDefinition 2: Let",
null,
"$\\chi^{\\lambda}(\\pi)$ denote the character of the representation of",
null,
"$S_n$ corresponding to",
null,
"$\\lambda$ evaluated at a permutation",
null,
"$\\pi$. Then",
null,
"$\\displaystyle s_{\\lambda} = \\frac{1}{n!} \\sum_{\\pi \\in S_n} \\chi^{\\lambda}(\\pi) p_{\\pi}$.\n\nDeterminantal formula\n\nThe Weyl character formulas imply the following determinantal formula for",
null,
"$s_{\\lambda}$. First, some notation. Define",
null,
"$a_{\\lambda}(x_1, ... x_n) = \\det \\left| \\begin{array}{cccc} x_1^{n-1 + \\lambda_1} & x_2^{n-1 + \\lambda_1} & \\hdots & x_n^{n-1 + \\lambda_1} \\\\ x_1^{n-2 + \\lambda_2} & x_2^{n-2 + \\lambda_2} & \\hdots & x_n^{n-2 + \\lambda_2} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ x_1^{\\lambda_n} & x_2^{\\lambda_n} & \\hdots & x_n^{\\lambda_n} \\end{array} \\right|$.\n\nIn particular",
null,
"$a_{0^n}(x_1, ... x_n)$ is the Vandermonde determinant",
null,
"$\\prod_{i < j} (x_i - x_j)$, and for any other partition",
null,
"$\\lambda$ the Vandermonde determinant divides it. (One could imagine these determinants appearing in the discussion of polynomial interpolation when certain coefficients are restricted to be zero, but I have never heard anyone talk about Schur functions this way.)\n\nDefinition 3:",
null,
"$\\displaystyle s_{\\lambda}(x_1, ... x_n) = \\frac{a_{\\lambda}(x_1, ... x_n)}{a_{0^n}(x_1, ... x_n)}$.\n\nThis definition has the advantage that it does not refer directly to representation theory, and it is also relatively straightforward to do computations with for small cases.\n\nThere are two other determinantal formulas for the Schur functions which are useful in studying, for example, the cohomology ring of a Grassmannian, but they can wait.\n\nThe characteristic map\n\nWe now define the characteristic map",
null,
"$\\text{ch} : \\mathcal{R} \\to \\Lambda$ as follows: if",
null,
"$f$ is a class function on",
null,
"$S_n$, then define",
null,
"$\\displaystyle \\text{ch } f = \\frac{1}{n!} \\sum_{\\pi \\in S_n} f(\\pi) p_{\\pi}$\n\nand extend by linearity. As we saw above,",
null,
"$\\text{ch } \\chi^{\\lambda} = s_{\\lambda}$, so this is a natural definition from that perspective.\n\nProposition:",
null,
"$\\text{ch } f * g = (\\text{ch } f)(\\text{ch } g)$. In other words,",
null,
"$\\text{ch}$ defines a ring homomorphism from",
null,
"$\\mathcal{R}$ with the induction product to",
null,
"$\\Lambda$.\n\nThe main technical detail of this proof is Frobenius reciprocity, but the point is that we have now found the relationship between",
null,
"$\\mathcal{R}$ and",
null,
"$\\Lambda$ that we were looking for. With a little more work one can show that",
null,
"$\\text{ch}$ is bijective, and it follows that the",
null,
"$s_{\\lambda}$ form a basis of the symmetric functions.\n\nThe characteristic map takes the usual inner product on",
null,
"$\\mathcal{R}$ to the unique inner product on symmetric functions satisfying",
null,
"$\\left< s_{\\lambda}, s_{\\mu} \\right> = \\delta_{\\lambda \\mu}$.\n\nThe study of this inner product was initiated by Philip Hall and clarifies a number of results in symmetric function theory; see EC2, for example. But I haven’t digested this point enough to say anything meaningful.\n\n### 7 Responses\n\n1. on April 26, 2019 at 7:15 am | Reply",
null,
"Andrei Sipoș\n\n“One could imagine these determinants appearing in the discussion of polynomial interpolation when certain coefficients are restricted to be zero, but I have never heard anyone talk about Schur functions this way.”\n\nI just wrote a paper where I exploit exactly this feature of Schur functions 😀 Your post has been helpful in figuring it out, thanks\n\nhttps://arxiv.org/abs/1904.10284\n\n2. […] Now finally we are back to The Jacobi-Trudi identities again. This time I did see semistandard Young tableau differs from standard ones in the way that integers are only weakly increasing along rows. Now I guess means a Young diagram. Equivalent definitions: , . So there exists one problem, I still don’t get what and is. Let me check it out. After checking wiki, is complete homogeneous symmetric polynomials and is elementary symmetric polynomials. Ah, now I understand it. Let me play around why , . Umm, that’s basically because the integers weakly increase along the row and strictly increase along the column, yeah, I see the reason. Let’s move on the proof of equivalence of definitions. While reading the proof, I suddenly found the",
null,
"$det$ part is related to Qiaochu’s previous post The many faces of Schur functions. […]\n\n3. on November 23, 2009 at 5:08 am | Reply",
null,
"Spanferkel\n\nA good book (but not very easy) is also\n\nBannai, Eiichi; Ito, Tatsuro: Algebraic combinatorics. I. Association schemes. The Benjamin/Cummings Publishing Co., Inc., Menlo Park, CA, 1984. xxiv+425 pp. ISBN: 0-8053-0490-8\n\nMR882540 (87m:05001) 05-01\nhttp://www.ams.org/mathscinet-getitem?mr=882540\n\n4. on November 22, 2009 at 5:39 pm | Reply",
null,
"su\n\nOk great thanks!!\n\n5. on November 20, 2009 at 12:57 am | Reply",
null,
"su\n\nHey,\n• on November 20, 2009 at 9:33 am | Reply",
null,
"Qiaochu Yuan"
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https://www.rdocumentation.org/packages/ggstatsplot/versions/0.0.10/topics/t1way_ci | [
"# t1way_ci\n\n0th\n\nPercentile\n\n##### #' @title A heteroscedastic one-way ANOVA for trimmed means with confidence interval for effect size.\n\nCustom function to get confidence intervals for effect size measure for robust ANOVA.\n\nKeywords\ninternal\n##### Usage\nt1way_ci(data, x, y, tr = 0.1, nboot = 100, conf.level = 0.95,\nconf.type = \"norm\", ...)\n##### Arguments\ndata\n\nA dataframe (or a tibble) from which variables specified are to be taken. A matrix or tables will not be accepted.\n\nx\n\nThe grouping variable from the dataframe data.\n\ny\n\nThe response (a.k.a. outcome or dependent) variable from the dataframe data.\n\ntr\n\nTrim level for the mean when carrying out robust tests. If you get error stating \"Standard error cannot be computed because of Winsorized variance of 0 (e.g., due to ties). Try to decrease the trimming level.\", try to play around with the value of tr, which is by default set to 0.1. Lowering the value might help.\n\nnboot\n\nNumber of bootstrap samples for computing confidence interval for the effect size (Default: 100).\n\nconf.level\n\nScalar between 0 and 1. If unspecified, the defaults return 95% lower and upper confidence intervals (0.95).\n\nconf.type\n\nA vector of character strings representing the type of intervals required. The value should be any subset of the values \"norm\", \"basic\", \"perc\", \"bca\". For more, see ?boot::boot.ci.\n\n...\n\nArguments passed on to boot::boot\n\ndata\n\nThe data as a vector, matrix or data frame. If it is a matrix or data frame then each row is considered as one multivariate observation.\n\nstatistic\n\nA function which when applied to data returns a vector containing the statistic(s) of interest. When sim = \"parametric\", the first argument to statistic must be the data. For each replicate a simulated dataset returned by ran.gen will be passed. In all other cases statistic must take at least two arguments. The first argument passed will always be the original data. The second will be a vector of indices, frequencies or weights which define the bootstrap sample. Further, if predictions are required, then a third argument is required which would be a vector of the random indices used to generate the bootstrap predictions. Any further arguments can be passed to statistic through the … argument.\n\nR\n\nThe number of bootstrap replicates. Usually this will be a single positive integer. For importance resampling, some resamples may use one set of weights and others use a different set of weights. In this case R would be a vector of integers where each component gives the number of resamples from each of the rows of weights.\n\nsim\n\nA character string indicating the type of simulation required. Possible values are \"ordinary\" (the default), \"parametric\", \"balanced\", \"permutation\", or \"antithetic\". Importance resampling is specified by including importance weights; the type of importance resampling must still be specified but may only be \"ordinary\" or \"balanced\" in this case.\n\nstype\n\nA character string indicating what the second argument of statistic represents. Possible values of stype are \"i\" (indices - the default), \"f\" (frequencies), or \"w\" (weights). Not used for sim = \"parametric\".\n\nstrata\n\nAn integer vector or factor specifying the strata for multi-sample problems. This may be specified for any simulation, but is ignored when sim = \"parametric\". When strata is supplied for a nonparametric bootstrap, the simulations are done within the specified strata.\n\nL\n\nVector of influence values evaluated at the observations. This is used only when sim is \"antithetic\". If not supplied, they are calculated through a call to empinf. This will use the infinitesimal jackknife provided that stype is \"w\", otherwise the usual jackknife is used.\n\nm\n\nThe number of predictions which are to be made at each bootstrap replicate. This is most useful for (generalized) linear models. This can only be used when sim is \"ordinary\". m will usually be a single integer but, if there are strata, it may be a vector with length equal to the number of strata, specifying how many of the errors for prediction should come from each strata. The actual predictions should be returned as the final part of the output of statistic, which should also take an argument giving the vector of indices of the errors to be used for the predictions.\n\nweights\n\nVector or matrix of importance weights. If a vector then it should have as many elements as there are observations in data. When simulation from more than one set of weights is required, weights should be a matrix where each row of the matrix is one set of importance weights. If weights is a matrix then R must be a vector of length nrow(weights). This parameter is ignored if sim is not \"ordinary\" or \"balanced\".\n\nran.gen\n\nThis function is used only when sim = \"parametric\" when it describes how random values are to be generated. It should be a function of two arguments. The first argument should be the observed data and the second argument consists of any other information needed (e.g. parameter estimates). The second argument may be a list, allowing any number of items to be passed to ran.gen. The returned value should be a simulated data set of the same form as the observed data which will be passed to statistic to get a bootstrap replicate. It is important that the returned value be of the same shape and type as the original dataset. If ran.gen is not specified, the default is a function which returns the original data in which case all simulation should be included as part of statistic. Use of sim = \"parametric\" with a suitable ran.gen allows the user to implement any types of nonparametric resampling which are not supported directly.\n\nmle\n\nThe second argument to be passed to ran.gen. Typically these will be maximum likelihood estimates of the parameters. For efficiency mle is often a list containing all of the objects needed by ran.gen which can be calculated using the original data set only.\n\nsimple\n\nlogical, only allowed to be TRUE for sim = \"ordinary\", stype = \"i\", n = 0 (otherwise ignored with a warning). By default a n by R index array is created: this can be large and if simple = TRUE this is avoided by sampling separately for each replication, which is slower but uses less memory.\n\nparallel\n\nThe type of parallel operation to be used (if any). If missing, the default is taken from the option \"boot.parallel\" (and if that is not set, \"no\").\n\nncpus\n\ninteger: number of processes to be used in parallel operation: typically one would chose this to the number of available CPUs.\n\ncl\n\nAn optional parallel or snow cluster for use if parallel = \"snow\". If not supplied, a cluster on the local machine is created for the duration of the boot call.\n\n• t1way_ci\n##### Examples\n# NOT RUN {\nset.seed(123)\nggstatsplot:::t1way_ci(\ndata = dplyr::filter(ggplot2::msleep, vore != \"insecti\"),\nx = vore,\ny = brainwt,\ntr = 0.05,\nnboot = 50,\nconf.level = 0.99,\nconf.type = \"perc\"\n)\n# }\n# NOT RUN {\n# }\n\nDocumentation reproduced from package ggstatsplot, version 0.0.10, License: GPL-3 | file LICENSE\n\n### Community examples\n\nLooks like there are no examples yet."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7840262,"math_prob":0.97307664,"size":6829,"snap":"2021-04-2021-17","text_gpt3_token_len":1604,"char_repetition_ratio":0.10564102,"word_repetition_ratio":0.008826125,"special_character_ratio":0.22287305,"punctuation_ratio":0.1359447,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.984382,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T18:58:52Z\",\"WARC-Record-ID\":\"<urn:uuid:79ee1ee6-4ae2-4654-a23a-272787317c22>\",\"Content-Length\":\"21928\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee26e45d-170a-461d-8625-5ce4c778875d>\",\"WARC-Concurrent-To\":\"<urn:uuid:87806ca1-f369-40d6-8aab-37b899755a8a>\",\"WARC-IP-Address\":\"52.20.214.25\",\"WARC-Target-URI\":\"https://www.rdocumentation.org/packages/ggstatsplot/versions/0.0.10/topics/t1way_ci\",\"WARC-Payload-Digest\":\"sha1:SGOKY725BUTDMTNLYYTMVEWF57OZT4NL\",\"WARC-Block-Digest\":\"sha1:MAE53JPF2AA367F5X2E3JQB2DJSMKWYZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703519600.31_warc_CC-MAIN-20210119170058-20210119200058-00663.warc.gz\"}"} |
https://usethinkscript.com/threads/making-money-in-a-choppy-market.3765/ | [
"",
null,
"# Making Money in a Choppy Market\n\n##### Member\nVIP\nVIP Enthusiast\nIntro:\n\nIn this choppy, super volatile trading environment I decided it was time to look for ways to smooth out the risk/reward ratios. Whether or not you believe we are in the middle of a sector rotation we can all agree things are getting a little sketchy in the market. Using two prebuilt thinkorswim indicators I built a scan that gives us low volatility combined with a high probability for above average returns.\n\nThe Conditions:\n• 7 day RSI > 60\n• 60 day RSI > 50\n• BollingerBandwidth < 10\nThe logic:\n• RSI is a measure of momentum, many people view a high RSI as being overbought. I agree when there is a spike in price. I disagree when the price is slowly trending up. A slowly upward trending stock can hold high levels of RSI for extended periods of time. Consider the stock MCO during the periods of 11/2019-2/2020. The 14 day RSI of MCO was above 70 most of that period. During that same time the stock ran 27%.\n• The BollingerBandwidth is the measure of volatility. High values imply high volatility, whereas low values imply low volatility. To my example, during 11/2019 MCO held the BollingerBandwidth value of less than 12.\nThe Example:",
null,
"Caveats:\n\n• It should be a given but cherry pick the best charts for best results\n\nThe Scan: https://tos.mx/JJUDFQz\n\nLast edited:\n•",
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"•",
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"MerryDay and bspratt22\n\n#### bspratt22\n\n##### New member\nagree with your premise on current market conditions; it is sketchy and might be the beginning of reversal but we still trade the charts right? My question is your entry. Are you in on the median line or exactly what is your entry?\n\n##### Member\nVIP\nVIP Enthusiast\n•",
null,
"BenTen\n\n#### APOT7\n\n##### Member\nVIP\nCan you share the lower study of the RSI 7day, 14,60, 120? As well as the BollingerBandwidth band simple?\n\n##### Member\nVIP\nVIP Enthusiast\nCan you share the lower study of the RSI 7day, 14,60, 120? As well as the BollingerBandwidth band simple?\n\nJust search for bollingerbandwidth in tos. Here is the multiple RSI\n\nCode:\n``````declare lower;\n\ninput length = 7;\ninput length1 = 14;\ninput length2 = 60;\ninput length3 = 120;\ninput over_Bought = 70;\ninput over_Sold = 30;\ninput price = close;\ninput averageType = AverageType.WILDERS;\n\ndef NetChgAvg = MovingAverage(averageType, price - price, length);\ndef TotChgAvg = MovingAverage(averageType, AbsValue(price - price), length);\n\ndef NetChgAvg1 = MovingAverage(averageType, price - price, length1);\ndef TotChgAvg1 = MovingAverage(averageType, AbsValue(price - price), length1);\n\ndef NetChgAvg2 = MovingAverage(averageType, price - price, length2);\ndef TotChgAvg2 = MovingAverage(averageType, AbsValue(price - price), length2);\n\ndef NetChgAvg3 = MovingAverage(averageType, price - price, length3);\ndef TotChgAvg3 = MovingAverage(averageType, AbsValue(price - price), length3);\n\ndef ChgRatio = if TotChgAvg != 0 then NetChgAvg / TotChgAvg else 0;\ndef ChgRatio1 = if TotChgAvg1 != 0 then NetChgAvg1 / TotChgAvg1 else 0;\ndef ChgRatio2 = if TotChgAvg2 != 0 then NetChgAvg2 / TotChgAvg2 else 0;\ndef ChgRatio3 = if TotChgAvg3 != 0 then NetChgAvg3 / TotChgAvg3 else 0;\n\nplot OverSold = over_Sold;\nplot OverBought = over_Bought;\nOverSold.SetDefaultColor(Color.DARK_GRAY);\nOverBought.SetDefaultColor(Color.DARK_GRAY);\n\nplot RSI = 50 * (ChgRatio + 1);\nplot RSI1 = 50 * (ChgRatio1 + 1);\nplot RSI2 = 50 * (ChgRatio2 + 1);\nplot RSI3 = 50 * (ChgRatio3 + 1);\n\nRSI.SetDefaultColor(Color.GREEN);\nRSI1.SetDefaultColor(Color.MagenTA);\nRSI2.SetDefaultColor(Color.YeLLOW);\nRSI3.SetDefaultColor(Color.BLUE);\n\nAddLabel(RSI, Concat(\"7RSI = \", RSI), color.GREEN);\nAddLabel(RSI1, Concat(\"14RSI = \", RSI1), color.MAGENTA);\nAddLabel(RSI2, Concat(\"60RSI = \", RSI2), color.YELLOW);\nAddLabel(RSI3, Concat(\"120RSI = \", RSI3), color.BLUE);``````\n\n•",
null,
"APOT7 and BenTen\n\n•",
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] | [
null,
"https://www.facebook.com/tr",
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"https://i.imgur.com/NGf46Xw.png",
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"data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9200561,"math_prob":0.952578,"size":1840,"snap":"2021-43-2021-49","text_gpt3_token_len":451,"char_repetition_ratio":0.09150327,"word_repetition_ratio":0.44025156,"special_character_ratio":0.22445652,"punctuation_ratio":0.0952381,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9718108,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,1,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T00:50:59Z\",\"WARC-Record-ID\":\"<urn:uuid:83069486-5d6b-4e13-8704-9d04090b602a>\",\"Content-Length\":\"88730\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efd2e451-cac1-4a30-bab2-7e8f17692a06>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd405cc0-bcfb-4b03-a904-45a62d1316db>\",\"WARC-IP-Address\":\"172.67.206.80\",\"WARC-Target-URI\":\"https://usethinkscript.com/threads/making-money-in-a-choppy-market.3765/\",\"WARC-Payload-Digest\":\"sha1:PQNHVTIDNHO5HSI22SBSX2QSA36VHRVW\",\"WARC-Block-Digest\":\"sha1:CHH3JGHYM6MPDEFPAWJMKD5X36NOQTAC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363327.64_warc_CC-MAIN-20211206224536-20211207014536-00055.warc.gz\"}"} |
https://kdaher.com.br/categorias/botas | [
"",
null,
"# Botas\n\nR\\$459,00\n\nR\\$229,50\n\n6 x R\\$38,25\n\nR\\$359,00\n\nR\\$179,50\n\n6 x R\\$29,92\n\nR\\$476,00\n\nR\\$238,00\n\n6 x R\\$39,67\n\nR\\$456,00\n\nR\\$228,00\n\n6 x R\\$38,00\n\nR\\$449,00\n\nR\\$224,50\n\n6 x R\\$37,42\n\nR\\$486,00\n\nR\\$243,00\n\n6 x R\\$40,50\n\nR\\$396,00\n\nR\\$198,00\n\n6 x R\\$33,00\n\nR\\$449,00\n\nR\\$224,50\n\n6 x R\\$37,42\n\nR\\$359,00\n\nR\\$179,50\n\n6 x R\\$29,92\n\nR\\$449,00\n\nR\\$224,50\n\n6 x R\\$37,42\n• ##### Bota Eduarda K. Daher\n\nR\\$475,00\n\nR\\$237,50\n\n6 x R\\$39,58",
null,
""
] | [
null,
"https://www.facebook.com/tr",
null,
"https://kdaher.com.br/skin/frontend/mgstheme/claue/images/loader.svg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5098286,"math_prob":0.9999876,"size":669,"snap":"2020-24-2020-29","text_gpt3_token_len":361,"char_repetition_ratio":0.26766917,"word_repetition_ratio":0.18103448,"special_character_ratio":0.5695067,"punctuation_ratio":0.19909503,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9519898,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-10T18:07:58Z\",\"WARC-Record-ID\":\"<urn:uuid:4cfaf4d9-8fd4-493b-9bf3-a1d66ae25885>\",\"Content-Length\":\"127925\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4f37e4ac-1471-4ee0-abb1-4084def787b5>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1622dd8-ee41-45bf-85e9-30a1f1446903>\",\"WARC-IP-Address\":\"207.246.71.6\",\"WARC-Target-URI\":\"https://kdaher.com.br/categorias/botas\",\"WARC-Payload-Digest\":\"sha1:SR4TH5G6JSBQNRLG46QEXEBE6DBWMWSH\",\"WARC-Block-Digest\":\"sha1:CAHWAUTOANITX2PDBLGUP7SL2DRSTB4E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655911896.73_warc_CC-MAIN-20200710175432-20200710205432-00411.warc.gz\"}"} |
https://forums.wolfram.com/mathgroup/archive/2005/Oct/msg00521.html | [
"",
null,
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"Re: Mathematica not simplifying Laplace transforms\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg61388] Re: Mathematica not simplifying Laplace transforms\n• From: \"roger\" <rogerbrent at gmail.com>\n• Date: Tue, 18 Oct 2005 02:44:29 -0400 (EDT)\n• References: <[email protected]><divhkr\\$gco\\[email protected]>\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```Thanks, I tried that but it still doesnt simplify\n\nLaplaceTransform[Exp[t]*f[t], t, s, Assumptions -> t > 0]\n\nto\n\nLaplaceTransform[f[t], t, -1 + s]\n\nThe v4 documentation says it should do that, so I'm thinking they\nchanged this behaviour in v5.\nAnyone have any ideas?\n\n```\n\n• Prev by Date: Matrices and Conditional Statements\n• Next by Date: Re: surface fitting question\n• Previous by thread: Re: Mathematica not simplifying Laplace transforms\n• Next by thread: Re: Mathematica not simplifying Laplace transforms"
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"https://forums.wolfram.com/mathgroup/images/search_archive.gif",
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http://prosib.info/1-digit-multiplication-worksheets/ | [
"2 digit by 1 multiplication worksheets 3 addition without regrouping and word problem work.\n\nfree collection of multiplication worksheets grade 4 2 digit by 1 download them pdf no regroupin.\n\nsmall size two digits multiplication worksheets 2 digit with answers by 1 pdf no regrouping multiplic.\n\n3 digit multiplication worksheet 1 multiplier 4 by worksheets pdf.\n\n4 digit by 1 multiplication worksheets unique inspiring math 2 subtraction with new long worksheet.\n\n4 digit by 1 multiplication worksheets amazing math anchor charts class 3 2 pdf digi.\n\nmultiplication worksheets 2 digit by 1 times 4 word problems math ri.\n\nmultiplication worksheets long 2 digit by 1 grade 4 for free multiplicatio.\n\n3 by 2 multiplication worksheets 1 digit worksheet up grade coloring x arrays pdf workshe.\n\n2 by 1 digit multiplication worksheets best solutions of free printable no regrouping long and multiplica.\n\n3 digit times 1 multiplication worksheets best of math riddle digits multiplying by 2 digi.\n\n2 digit by 1 multiplication word problems worksheets practice one.\n\nbrilliant ideas of 4 digit by 1 multiplication worksheets 3 digits times awesome collection wor.\n\n2 digit by 1 multiplication worksheet worksheets for all free library download and print on 4th grade 8 per page a large mat.\n\n4 digit by 1 multiplication worksheets lovely division problems grade math 3 new long 5 pdf d.\n\ntwo digit by 1 ation worksheets inspirational double brad multiplication free and 2.\n\n2 digit by 1 multiplication worksheets fresh free without regrouping.\n\ngit multiplications grade math year 2 multiplication worksheets times 1 free printable by various lattice maths multi and digit mu.\n\ntwo digit by 1 multiplication worksheets hard 2 problems in this grade 3 math 4th worksh.\n\ntwo digit multiplication worksheet free worksheets library download and print on 2 by 1 grade 4 3 6.\n\nmath worksheets grade 2 multiplication worksheet printable free digit x 1 by on graph paper m.\n\n3 digit by 1 multiplication worksheets the best image collection download and share 2 mult.\n\n1 digit multiplication worksheets task cards printable 2 times 3 by no regrouping car.\n\nmultiplication worksheet freebie 2 digit by 1 worksheets pdf grade original.\n\ncollection of multiplication worksheet 2 digit by 1 download worksheets horizontal them and t multi 3 word proble.\n\n2 digit multiplication worksheets printable by 1 pdf grade 3 multiplicat.\n\nmultiplication worksheets grade 2 multiplying digit by 1 numbers a long practice 4 colo.\n\ngrade 4 multiplication worksheets free library download and print on 1 digit by 2 without regrouping traditional works.\n\nmultiplication worksheets 2 digit by 1 cupcakes and pdf original.\n\n2 digit by 1 multiplication worksheets grade two word 3 coloring works.\n\nworksheets 2 digit by 1 multiplication x math worksheet on grid paper word grade horizon.\n\nhorizontal multiplication worksheet worksheets 4 digits times 1 digit 2 by pdf."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73039883,"math_prob":0.95582575,"size":2901,"snap":"2019-35-2019-39","text_gpt3_token_len":610,"char_repetition_ratio":0.38660684,"word_repetition_ratio":0.07173913,"special_character_ratio":0.20234402,"punctuation_ratio":0.06451613,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977079,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-19T08:29:48Z\",\"WARC-Record-ID\":\"<urn:uuid:30ae9f0d-ee94-4d6e-967a-e5da9c61a145>\",\"Content-Length\":\"64866\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2ad1bcaa-d8ad-4658-a4ea-455a08f55149>\",\"WARC-Concurrent-To\":\"<urn:uuid:a33821b5-ff4e-455e-9842-50064cc35bc7>\",\"WARC-IP-Address\":\"104.27.186.84\",\"WARC-Target-URI\":\"http://prosib.info/1-digit-multiplication-worksheets/\",\"WARC-Payload-Digest\":\"sha1:OVOFPH2KZECMTYOFVLJPOT4LDVSM4JF2\",\"WARC-Block-Digest\":\"sha1:TCZIODFQSHS2YC3USA5JVGUFN76IBNWH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573465.18_warc_CC-MAIN-20190919081032-20190919103032-00418.warc.gz\"}"} |
https://akiboy96.medium.com/community-detection-network-analysis-of-the-stock-market-in-python-r-part-3-ca4e0fbb8ca9 | [
"# Building your Long Term Portfolio using Unsupervised ML with Community Detection Analysis in Python & R\n\nTo build the portfolio based on the correlation of stock prices in the S&P 500, we will use network centrality as a measure to understand how the stocks are better integrated with one and other.\n\n`stocks_cross_corr, _, _ = calculate_corr(df_stock_prices,1, len(df_stock_prices), 'pearson')stocks_cross_corr = stocks_cross_corrcor_thresold = 0.7G = build_graph(stocks_cross_corr, cor_thresold)partition = community.best_partition(G)modularity = community.modularity(partition, G)values = [partition.get(node) for node in G.nodes()]plt.figure(figsize=(10,10))nx.draw_spring(G, cmap = plt.get_cmap('jet'), node_color = values, node_size=30, with_labels=False)print(modularity)print(\"Total number of Communities=\", len(G.nodes()))dict_betwenness_centrality = nx.betweenness_centrality(G)dict_degree_centrality = nx.degree_centrality(G)dict_closeness_centrality = nx.closeness_centrality(G)dict_eigenvector_centrality = nx.eigenvector_centrality(G)print(\"dict_degree_centrality: \", dict_degree_centrality)print(\"dict_closeness_centrality: \", dict_closeness_centrality)print(\"dict_eigenvector_centrality: \", dict_eigenvector_centrality)print(\"dict_betweenness_centrality: \", dict_betwenness_centrality)`\n\n## Once the four types of centralities are obtained we will build our own portfolio determining algorithm using these centralities in an equal weightage sense. To do that we combine the dictionary data structure of our 4 centralties and use a for loop to iterate through each stock and add the four different centralities to create a final centrality metric. Since we are using equal weightage the formula would use a simple sum and not be any assigned any weights.\n\n`#Portfolio Formula: c_dict = dict([(k, [dict_betwenness_centrality[k], dict_eigenvector_centrality[k], dict_degree_centrality[k], dict_closeness_centrality[k] ]) for k in dict_betwenness_centrality])#print(c_dict) C_total = {}for key in c_dict: C_total[key] = sum(c_dict[key]) print(\"The Centrality total for stocks are:\", C_total) newDict = dict(filter(lambda elem: elem > 0, C_total.items()))print(\"Stocks greater than 0.3 centrality are\",newDict)print(len(newDict))df_centrality = pd.DataFrame(list(newDict.items()),columns = ['Symbol','Centrality']) df_centrality.sort_values(by='Centrality', ascending=False)#df_centrality.head(20)#type(df_centrality['Centrality'])df_centrality.to_csv('centrality_of_stocks_0.7cor.csv',index=False)`\n`# This R environment comes with many helpful analytics packages installed# It is defined by the kaggle/rstats Docker image: https://github.com/kaggle/docker-rstats# For example, here's a helpful package to loadlibrary(tidyverse)library(tidyquant)library(jsonlite)library(tidyverse)library(readr)library(igraph)library(dplyr)library(lubridate)library(data.table)library(Quandl) # metapackage of all tidyverse packages# Input data files are available in the read-only \"../input/\" directory# For example, running this (by clicking run or pressing Shift+Enter) will list all files under the input directorylist.files(path = \"../input\")# You can write up to 5GB to the current directory (/kaggle/working/) that gets preserved as output when you create a version using \"Save & Run All\" # You can also write temporary files to /kaggle/temp/, but they won't be saved outside of the current session`\n`#DataLoad stockprices <- read.csv(\"../input/usstockprices/stocks_price_final.csv\")centrality0.5 <- read.csv(\"../input/centrality-of-stocks-network-analysis/centrality_of_stocks_0.5cor.csv\")centrality0.6 <- read.csv(\"../input/centrality-of-stocks-network-analysis/centrality_of_stocks_0.6cor.csv\")centrality0.7 <- read.csv(\"../input/centrality-of-stocks-network-analysis/centrality_of_stocks_0.7cor.csv\")`\n`#Merges Centtraility values with other potential useful information like market cap and sector for further filteringstocks <- stockprices[!duplicated(stockprices\\$symbol), ] # removes duplicate symbolsIndex0.5cor <- merge(stocks, centrality0.5, by.x = \"symbol\", by.y = \"Symbol\")Index0.6cor <- merge(stocks, centrality0.6, by.x = \"symbol\", by.y = \"Symbol\")Index0.7cor <- merge(stocks, centrality0.7, by.x = \"symbol\", by.y = \"Symbol\")#head(Index0.6cor)IndexGet0.5 <- as.character(Index0.5cor\\$symbol)IndexGet0.6 <- as.character(Index0.6cor\\$symbol)IndexGet0.7 <- as.character(Index0.7cor\\$symbol)#Portfolio Testing##0.5 correlation centrality pricesstockindex0.5 <- tq_get(IndexGet0.5, get=\"stock.prices\", from = \"2015-07-01\",warnings = FALSE, stringsAsFactors = FALSE) %>% group_by(symbol) %>% tq_transmute(select=adjusted, mutate_fun=periodReturn, period=\"monthly\", col_rename = \"monthly_return\")stockindex0.5##0.6 correlation centrality pricesstockindex0.6 <- tq_get(IndexGet0.6, get=\"stock.prices\", from = \"2015-07-01\",warnings = FALSE, stringsAsFactors = FALSE) %>% group_by(symbol) %>% tq_transmute(select=adjusted, mutate_fun=periodReturn, period=\"monthly\", col_rename = \"monthly_return\")##0.7 correlation centrality pricesstockindex0.7 <- tq_get(IndexGet0.7, get=\"stock.prices\", from = \"2015-07-01\",warnings = FALSE, stringsAsFactors = FALSE) %>% group_by(symbol) %>% tq_transmute(select=adjusted, mutate_fun=periodReturn, period=\"monthly\", col_rename = \"monthly_return\")##Base Portfolio to compare baseline_returns_monthly <- \"SPY\" %>% tq_get(get = \"stock.prices\", from = \"2015-07-01\", warnings = FALSE,stringsAsFactors = FALSE) %>% tq_transmute(select = adjusted, mutate_fun = periodReturn, period = \"monthly\", col_rename = \"spy_monthly_return\")baseline_returns_monthly`\n`portfolio_returns_monthly0.5 <- stockindex0.5 %>% tq_portfolio(assets_col = symbol, returns_col = monthly_return, col_rename = \"portfolio-monthly\")portfolio_returns_monthly0.6 <- stockindex0.6 %>% tq_portfolio(assets_col = symbol, returns_col = monthly_return, col_rename = \"portfolio-monthly\")portfolio_returns_monthly0.7 <- stockindex0.7 %>% tq_portfolio(assets_col = symbol, returns_col = monthly_return, col_rename = \"portfolio-monthly\")#Portfolio Comparestock0.5indexVSSPY <- left_join(portfolio_returns_monthly0.5, baseline_returns_monthly, by = \"date\")stock0.6indexVSSPY <- left_join(portfolio_returns_monthly0.6, baseline_returns_monthly, by = \"date\")stock0.7indexVSSPY <- left_join(portfolio_returns_monthly0.7, baseline_returns_monthly, by = \"date\")#stock0.5indexVSSPYggplot(stock0.5indexVSSPY) + geom_line(aes(x = `date`, y = `portfolio-monthly`), color = \"blue\")+ geom_line(aes(x = `date`, y = `spy_monthly_return`), color = \"red\")ggplot(stock0.6indexVSSPY) + geom_line(aes(x = `date`, y = `portfolio-monthly`), color = \"blue\")+ geom_line(aes(x = `date`, y = `spy_monthly_return`), color = \"red\")ggplot(stock0.7indexVSSPY) + geom_line(aes(x = `date`, y = `portfolio-monthly`), color = \"blue\")+ geom_line(aes(x = `date`, y = `spy_monthly_return`), color = \"red\")`\n`##PLAYGROUND - LETS DO TRIAL AND ERROR HERE TO BEAT THE SPY GRAPH in RETURNScentrality_filter <- Index0.6cor#filter(Centrality > 0.8 & Centrality < 0.5)centrality_filterIndexGet0.6 <- as.character(centrality_filter\\$symbol)stockindex0.6 <- tq_get(IndexGet0.6, get=\"stock.prices\", from = \"2015-07-01\",warnings = FALSE, stringsAsFactors = FALSE) %>% group_by(symbol) %>% tq_transmute(select=adjusted, mutate_fun=periodReturn, period=\"monthly\", col_rename = \"monthly_return\")portfolio_returns_monthly0.6 <- stockindex0.6 %>% tq_portfolio(assets_col = symbol, returns_col = monthly_return, col_rename = \"portfolio-monthly\")stock0.6indexVSSPY <- left_join(portfolio_returns_monthly0.6, baseline_returns_monthly, by = \"date\")ggplot(stock0.6indexVSSPY) + geom_line(aes(x = `date`, y = `portfolio-monthly`), color = \"blue\")+ geom_line(aes(x = `date`, y = `spy_monthly_return`), color = \"red\")`\n\nThis is the code I wrote to achieve a portfolio that could beat the SPY returns over a 5 year period from July 1st 2015 to July 22nd 2020\n\n`centrality_filter <- Index0.5cor %>%filter((sector == \"Health Care\" & Centrality > 0.4) | (sector == \"Technology\" & Centrality > 0.42) | (sector == \"Consumer Services\" & Centrality > 0.49) | (sector == \"Finance\" & Centrality > 0.5) | (sector == \"Transportation\" & Centrality > 0.5) | (sector == \"Capital Goods\" & Centrality > 0.35) | (sector == \"Miscellaneous\" & Centrality > 0.5) | (sector == \"Basic Industries\" & Centrality > 0.39) | (sector == \"Public Utilities\" & Centrality > 0.36) | (sector == \"Consumer Durables\" & Centrality > 0.3)| (sector == \"Consumer Non-Durables\" & Centrality > 0.25))centrality_filterIndexGet0.5 <- as.character(centrality_filter\\$symbol)stockindex0.5 <- tq_get(IndexGet0.5, get=\"stock.prices\", from = \"2015-07-01\", till = \"2020-07-22\",warnings = FALSE, stringsAsFactors = FALSE) %>% group_by(symbol) %>% tq_transmute(select=adjusted, mutate_fun=periodReturn, period=\"monthly\", col_rename = \"monthly_return\")portfolio_returns_monthly0.5 <- stockindex0.5 %>% tq_portfolio(assets_col = symbol, returns_col = monthly_return, col_rename = \"portfolio-monthly\")stock0.5indexVSSPY <- left_join(portfolio_returns_monthly0.5, baseline_returns_monthly, by = \"date\")stock0.5indexVSSPY\\$`portfolio-monthly` <- stock0.5indexVSSPY\\$`portfolio-monthly` * 100stock0.5indexVSSPY\\$spy_monthly_return <- stock0.5indexVSSPY\\$spy_monthly_return * 100stock0.5indexVSSPY <- stock0.5indexVSSPY[-c(14), ]returns <- ggplot(stock0.5indexVSSPY) + geom_line(aes(x = `date`, y = `portfolio-monthly`), color = \"blue\")+ geom_line(aes(x = `date`, y = `spy_monthly_return`), color = \"red\") + ggtitle(\"Portfolio vs S&P 500 Returns over 5 years\") + labs(y=\"Returns Percenatge\", x = \"Date\") plot(returns)`\n\nThe centralities as shown in the above code were chosen based on my predictions for future trends in terms of what sectors will do well and become important to society as a service.\n\n--\n\n--\n\n--\n\n## More from Aakash Kedia\n\nCS, music and football in no particular order\n\nLove podcasts or audiobooks? Learn on the go with our new app.\n\n## How Much Does It Cost to Make an App Like Postmates?",
null,
"## Run MySQL and phpMyAdmin using Docker",
null,
"## How I got my (functional) team to collaborate with many Scrum teams",
null,
"## 10 Mistakes When Sizing Cloud Resources",
null,
"## Ease of Building UI Elements in Unity",
null,
"## Error Code ssl_error_no_cypher_overlap Firefox 34",
null,
"## Overview of Inverse Kinematics",
null,
"",
null,
"## Aakash Kedia\n\nCS, music and football in no particular order\n\n## Using ML to Predict Search Volume",
null,
"## Time Series Data Components with Microsoft excel",
null,
"## Computing Forecast Error: Should I divide by forecast or demand?",
null,
"## Machine Learning Applications in Climate Prediction Analytics",
null,
""
] | [
null,
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null,
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null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.68728596,"math_prob":0.88684905,"size":13012,"snap":"2022-27-2022-33","text_gpt3_token_len":3483,"char_repetition_ratio":0.16190037,"word_repetition_ratio":0.14447236,"special_character_ratio":0.26344913,"punctuation_ratio":0.15738499,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9673924,"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],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,3,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T02:45:05Z\",\"WARC-Record-ID\":\"<urn:uuid:05aaea61-a06e-4aee-8a62-493c938af459>\",\"Content-Length\":\"182436\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5d0ef349-5123-445f-85a5-9a8986af5de2>\",\"WARC-Concurrent-To\":\"<urn:uuid:02286087-9246-40a2-9b1c-98991724b094>\",\"WARC-IP-Address\":\"162.159.153.4\",\"WARC-Target-URI\":\"https://akiboy96.medium.com/community-detection-network-analysis-of-the-stock-market-in-python-r-part-3-ca4e0fbb8ca9\",\"WARC-Payload-Digest\":\"sha1:URGYPIDHX7UYIMTEQM34KLXCGTKKAQP6\",\"WARC-Block-Digest\":\"sha1:EBWUR72EACQ4RYT6ETT2KM7OFM4AGOZM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103347800.25_warc_CC-MAIN-20220628020322-20220628050322-00285.warc.gz\"}"} |
https://michelanders.blogspot.com/2012/02/3d-convex-hull-in-python.html | [
"## Sunday, 19 February 2012\n\n### 3D Convex hull in Python\n\nIn this article I present a present a reimplementation in pure Python of Joseph O'Rourke's incremental 3D convex hull algorithm from his book Computational Geometry in C.\n\n## A convex hull in pure Python\n\nThis is the second, rather off topic, article on computational geometry in this blog. The previous article presented an implementation of the marching tetrahedrons algorithm.\nThe goal of this article is to provide object oriented, pythonic code to compute the convex hull of a collection of 3D points. The code is contained in a single Python module that may be downloaded from GitHub.\nA sample of how to use this module is shown below, where we create a a roughly spherical cloud of points, calculate its convex hull and print this hull in STL format to stdout. The resulting object is shown in the image (as seen in Blender).\n```from random import random\nfrom chull import Vector,Hull\n\nsphere=[]\nfor i in range(2000):\nx,y,z = 2*random()-1,2*random()-1,2*random()-1\nif x*x+y*y+z*z < 1.0:\nsphere.append(Vector(x,y,z))\nh=Hull(sphere)\nh.Print()\n```\n\n## Difference from the original implementation\n\nThe original code restricted the coordinates of points to integers, here there is no such restriction. This might result in errors if the coordinates are large.\n\n1.",
null,
"When I try to open ti with the GNU Meshlab I get an error saying \"Premature end of file\". What is that supposed to mean?\n\n2.",
null,
"Hi,\nthank you very much for this piece of code! I am using it succesfully.\nThere is only one strange behaviour I couldnt yet figure out why it happens: Sometimes the code doesnt finish saying\n\n338 # e interior: mark for deletion\n339 e.delete = REMOVED;\n341 # e border: make a new face\n342 e.newface = self.MakeConeFace( e, p )\n\nAttributeError: 'NoneType' object has no attribute 'visible'\n\nin the constructor when calling (Hull(myvectors)).\n\nDo you have an idea about how I could avoid this? For a some selection of my data points the code works, for some this error occurs and I cant figure out what the pattern is.\n\nThanks again!\n\n3.",
null,
"Thank you for sharing your code!\nit would be great if there was an easy build-in method available, which can calculate the surface area of the resulting convex polyhedron.\n\n4.",
null,
"Great code! I've used it to create an STL file for a 3D shape that contains concavities and convexity. Unfortunately, the code seems to fill in those concavities to create a convex shape. I'm not a Python expert - is there a (simple?) way to modify this code to not get rid of the concavities?\n\n5.",
null,
"Thank you very much, this code was very useful for me to represent some data :)\n\nI added a little function:\n\ndef PrintFile(self, filename='tmpshape.stl'):\nf1 = open(filename, 'w+')\nf1.write(\"solid points\\n\")\nfor f in self.faces:\nf1.write(str(f) + '\\n')\nf1.write(\"endsolid points\\n\")\nf1.close()\n\nto print directly to a stl file which was handy for me."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8743726,"math_prob":0.6741904,"size":2452,"snap":"2019-13-2019-22","text_gpt3_token_len":605,"char_repetition_ratio":0.09068628,"word_repetition_ratio":0.0,"special_character_ratio":0.24429038,"punctuation_ratio":0.13111547,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9610441,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-24T18:35:24Z\",\"WARC-Record-ID\":\"<urn:uuid:427bf892-d893-4940-a9d6-20a9816577d8>\",\"Content-Length\":\"131178\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30151b0e-3ddb-4cd0-8c5e-da186bb5c6d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d7fad48-7df1-40cb-99ad-797b51b7e8c3>\",\"WARC-IP-Address\":\"172.217.5.225\",\"WARC-Target-URI\":\"https://michelanders.blogspot.com/2012/02/3d-convex-hull-in-python.html\",\"WARC-Payload-Digest\":\"sha1:S4ZZALC2X3W72D57ZMFFPQQBDA4UWIV3\",\"WARC-Block-Digest\":\"sha1:L5L5NWTR27NHIATFR6VOFXFUGMNXUGY2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203464.67_warc_CC-MAIN-20190324165854-20190324191854-00285.warc.gz\"}"} |
https://www.yaclass.in/p/science-state-board/class-9/organisation-of-tissues-draft-5860/plant-tissues-draft-5861/re-fe3207bd-325f-4375-89d5-85f4f6d80c99 | [
"### Theory:\n\nOnion root-meristematic tissue experiment:\nOnion bulb experiment - presence of apical meristem\n• Take two jars (jar $$1$$ & jar $$2$$) and two onion bulbs with roots protruding from them. Pour water into the jars until the jars are nearly full.\n• Place the two onions with their roots immersed in the water into each of the jars as shown in the picture.\n• Measure the length of the roots on day $$1$$, $$2$$ and $$3$$.\n• After a few days we observe that both the roots have grown to a certain length.\n• Take onion bulb out of the jar $$2$$ and cut the tip of the root by about $$1$$ cm. Place it again in the jar and observe the growth of roots in both jar $$1$$ and jar $$2$$ for another $$5$$ days. Mark the jar of the root-trimmed onion bulb.\nNote the measurement of the length of the roots in the below table before cutting the tip of the root:\n\n Day $$1$$ Day $$2$$ Day $$3$$ Jar $$1$$ Jar $$2$$\n\nNote the measurement of the length of the roots in the below table after cutting the tip of the root:\n\n Day $$1$$ Day $$2$$ Day $$3$$ Day $$4$$ Day $$5$$ Jar$$1$$ Jar $$2$$\nQuestions:\n1. Is there a difference in length of roots between the root-trimmed onion (in jar $$2$$) and the un-trimmed one (in jar $$1$$)? Give reasons.\n2. Do the roots continue growing even after we have removed their tip?\nAnswers:\n1. Yes, there will be a difference in the length of the roots. The root-trimmed onion's root (in jar $$2$$) is shorter in length than the un-trimmed one (in jar $$1$$). Tips of the plant roots contain apical meristematic tissue, and is responsible for the elongation or length of the roots. As the tip of the onion roots in jar $$2$$ is cut, apical meristematic tissue is removed from it. There will not be any observable growth in the roots of the onion in jar $$2$$ . So the length of the roots of the onion in jar $$2$$ is lesser than the length of the roots of the onion in jar$$1$$.\n2. No, the roots won't grow after their tips are removed. Tips of the plant roots contain apical meristematic tissue, and is responsible for the elongation or length of the roots. As the tip of the onion roots in jar $$2$$ is cut, apical meristematic tissue is removed from it. So there will not be any growth in the roots of the onion in jar $$2$$ after the tips are cut."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9089579,"math_prob":0.99952567,"size":2268,"snap":"2021-21-2021-25","text_gpt3_token_len":591,"char_repetition_ratio":0.22084805,"word_repetition_ratio":0.30679157,"special_character_ratio":0.28791887,"punctuation_ratio":0.07375271,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994799,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T02:22:41Z\",\"WARC-Record-ID\":\"<urn:uuid:7a80233a-eb26-4781-9721-7cccedac103e>\",\"Content-Length\":\"37331\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa09ac09-4230-4486-b947-2a48ffb6cc14>\",\"WARC-Concurrent-To\":\"<urn:uuid:e0f9546f-fdd5-4364-990c-036db215fb42>\",\"WARC-IP-Address\":\"52.172.204.196\",\"WARC-Target-URI\":\"https://www.yaclass.in/p/science-state-board/class-9/organisation-of-tissues-draft-5860/plant-tissues-draft-5861/re-fe3207bd-325f-4375-89d5-85f4f6d80c99\",\"WARC-Payload-Digest\":\"sha1:CO7F36WWF3QTMLILUUYTAPMASUZLOG7J\",\"WARC-Block-Digest\":\"sha1:HDG7SQ7W6KMXMFD7II5II5GWYUA2YSVO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487643380.40_warc_CC-MAIN-20210619020602-20210619050602-00321.warc.gz\"}"} |
https://jov.arvojournals.org/article.aspx?articleid=2772046 | [
"",
null,
"jov\nOpen Access\nMethods | December 2020\nA new toolbox to distinguish the sources of spatial memory error\nAuthor Affiliations\n• John P. Grogan\nNuffield Department of Clinical Neurosciences, University of Oxford, Oxford, UK\[email protected]\n• Sean J. Fallon\nPopulation Health Sciences, University of Bristol, Bristol, UK\[email protected]\n• Nahid Zokaei\nDepartment of Experimental Psychology, University of Oxford, Oxford, UK\nOxford Centre for Human Brain Activity, University of Oxford, Oxford, UK\[email protected]\n• Masud Husain\nNuffield Department of Clinical Neurosciences, University of Oxford, Oxford, UK\nDepartment of Experimental Psychology, University of Oxford, Oxford, UK\[email protected]\n• Elizabeth J. Coulthard\nTranslational Health Sciences, University of Bristol, Bristol, UK\nNorth Bristol NHS Trust, Bristol, UK\[email protected]\n• Sanjay G. Manohar\nNuffield Department of Clinical Neurosciences, University of Oxford, Oxford, UK\nOxford Centre for Human Brain Activity, University of Oxford, Oxford, UK\[email protected]\nJournal of Vision December 2020, Vol.20, 6. doi:https://doi.org/10.1167/jov.20.13.6\n• Views\n• PDF\n• Share\n• Tools\n×\n###### This feature is available to authenticated users only.\n• Get Citation\n\nJohn P. Grogan, Sean J. Fallon, Nahid Zokaei, Masud Husain, Elizabeth J. Coulthard, Sanjay G. Manohar; A new toolbox to distinguish the sources of spatial memory error. Journal of Vision 2020;20(13):6. doi: https://doi.org/10.1167/jov.20.13.6.\n\n© ARVO (1962-2015); The Authors (2016-present)\n\n×\n• Supplements\nAbstract\n\nStudying the sources of errors in memory recall has proven invaluable for understanding the mechanisms of working memory (WM). While one-dimensional memory features (e.g., color, orientation) can be analyzed using existing mixture modeling toolboxes to separate the influence of imprecision, guessing, and misbinding (the tendency to confuse features that belong to different memoranda), such toolboxes are not currently available for two-dimensional spatial WM tasks.\n\nHere we present a method to isolate sources of spatial error in tasks where participants have to report the spatial location of an item in memory, using two-dimensional mixture models. The method recovers simulated parameters well and is robust to the influence of response distributions and biases, as well as number of nontargets and trials.\n\nTo demonstrate the model, we fit data from a complex spatial WM task and show the recovered parameters correspond well with previous spatial WM findings and with recovered parameters on a one-dimensional analogue of this task, suggesting convergent validity for this two-dimensional modeling approach. Because the extra dimension allows greater separation of memoranda and responses, spatial tasks turn out to be much better for separating misbinding from imprecision and guessing than one-dimensional tasks.\n\nCode for these models is freely available in the MemToolbox2D package and is integrated to work with the commonly used MATLAB package MemToolbox.\n\nIntroduction\nWorking memory (WM) is typically measured by providing a person with a set of stimuli to remember and then probing their memory after a delay of a few seconds. For example, they may be asked which direction of one of a set of colored arrows pointed. The types of errors people make on these tasks provide important information about the processes involved in retaining information in WM over and above simply looking at accuracy or overall error (Ma, Husain, & Bays, 2014). One type of error occurs when people remember which features were presented but fail to remember the way in which they were combined, such as reporting the orientation of a different-colored arrow. This might indicate that people misbind the features. A second type of error involves knowing the features of an object only approximately, leading to imprecision. A third type of error arises if everything is forgotten and responses are essentially guesses. Importantly, these three types of errors may each reflect distinct processes affecting WM.\nAlthough these processes can be distinguished when participants make binary judgments about the items in memory (Parra et al., 2010; Stark, Yassa, & Stark, 2010), recent studies have allowed greater sensitivity to these types of error by using continuous report, in which participants reproduce the feature they had to remember (Bays, Catalao, & Husain, 2009; Zokaei, Burnett Heyes, Gorgoraptis, Budhdeo, & Husain, 2015). This data-rich method allows us to separate the errors people make into their component parts. The technique of mixture modeling (Bays et al., 2009; Bays & Husain, 2008; Zhang & Luck, 2008) is now commonly used if the report domain is one-dimensional and circular, for example, line orientation or color on a wheel.\nThese methods have proven invaluable in understanding memory mechanisms, demonstrating, for example, that interstimulus distance affects misbinding and not precision (Emrich & Ferber, 2012), increasing set size affects misbinding and precision more than random guessing (Bays et al., 2009; Pertzov, Manohar, & Husain, 2017), and decreasing sample duration increases random guessing but not imprecision or misbinding (Bays et al., 2009). Importantly, this has also allowed mechanistic understandings in clinical populations where the different sources can be selectively impaired (Rolinski et al., 2016; Zokaei et al., 2014).\nHowever, many working memory tasks are two-dimensional (2D; e.g., spatial location; Corsi, 1972), and extending the mixture modeling technique to work on such 2D data would allow researchers to decompose errors in their component sources. While some studies used binary report (Bays & Husain, 2008), continuous-report measures of spatial memory have emerged more recently, in which participants move a probe to its original location (Pertzov, Dong, Peich, & Husain, 2012). This can be accomplished by participants dragging or pointing to locations on a touchscreen computer, allowing precise spatial WM measures in patients with disease (Liang et al., 2016; Pertzov et al., 2013; Zokaei, Nour, et al., 2019).\nTo examine the types of errors made, previous analyses employed simple measures based on the distance of each response from the other items in the display (Pertzov et al., 2012). The most basic method includes all sources of errors, being simply the mean distance from the true location of the target to where a participant recalls it to have been. This distance-to-target can be compared against the distance from the response to the nearest stimulus that had appeared in the memory array (nearest-neighbor distance) (Pertzov et al., 2013). Such a measure removes the influence of misbinding, giving a measure of imprecision. However, this metric still conflates errors of imprecision (i.e., knowing the location only approximately) with errors due to pure guessing. Further analysis methods to take guessing into account (Pertzov et al., 2013) involve counting the number of responses that occur within a certain distance of a nontarget (swap errors), but this does not consider a person's precision or the distance between the stimuli on each trial. These simple behavioral metrics each attempt to separate out different sources of errors, but the only way to validate them has been by comparison of results with those from one-dimensional (1D) circular tasks.\nOne study has applied mixture modeling to 2D working memory tasks (Schneegans & Bays, 2016) and found that misbinding increased with set size, while random guessing was negligible, and that the pattern of target errors, nontarget responses, and reaction times was in keeping with a continuous-resource model, not a discrete-slot model. They also found that misbinding, but not random guessing, contributed to errors, which may highlight an advantage of spatial data; the extra dimension of separation between items leads to less overlap in response distributions centered on the items, giving greater ability to distinguish error components. This article demonstrated the usefulness of mixture modeling for spatial tasks, but to our knowledge, there is no available package for applying these models to 2D data or an investigation into the suitability and limitations of such an approach.\nHere, we present 2D mixture models (adapted from a previous 1D mixture model package MemToolbox; Suchow, Brady, Fougnie, & Alvarez, 2013; MemToolbox.org) that allow us to fit a person's imprecision of location memories, along with the proportion of random guesses and nontarget responses (misbinding), taking into account stimulus locations and any biases in responses that may occur. This article serves as an introduction to the concepts behind the models, the practicalities of fitting the models, and issues that may arise when working with 2D data, and it provides examples of fitting the models to human data. The Discussion section provides an overview of the limitations and advantages of the modeling approach raised during specific tests in the Results and Method sections, and readers may consult the Contents to determine which sections are useful to them when working with these models. The new MemToolbox2D is available at https://doi.org/10.5281/zenodo.3752705 (Grogan et al., 2019).\nMethod\nModel\nThe 1D mixture models split errors into three components (Bays et al., 2009) (Figure 1). Throughout the article, we shall focus on this three-component model (the misbinding model), although the approach is the same for related models such as the two-component mixture model (without misbinding; Zhang & Luck, 2008), and all such models have been adapted for 2D. The misbinding model has three sources of errors: imprecision, misbinding, and random guessing:\n\\begin{eqnarray}P\\left( {\\hat \\theta {\\rm{\\;}}} \\right){\\rm{\\;}} &=& {\\rm{\\;}}\\alpha {\\phi _\\kappa }\\left( {\\hat \\theta {\\rm{\\;}} - {\\rm{\\;}}\\theta } \\right) + {\\rm{\\;}}\\beta \\frac{1}{m}\\mathop \\sum \\limits_i^m {\\phi _\\kappa }\\left( {\\hat \\theta {\\rm{\\;}} - {\\rm{\\;}}{\\varphi _i}} \\right) \\nonumber\\\\ &&+ {\\rm{\\;}}\\gamma \\frac{1}{{2\\pi }},\\end{eqnarray}\n(1)\nwhere $$P( {\\hat \\theta \\;} )$$ is the probability of finding a response orientation $$\\hat \\theta$$, θ is the orientation of the target stimulus, φκ is the von Mises distribution (circular analogue of the Gaussian distribution), ϕi is the orientation of the nontarget stimulus i, m is the number of nontarget stimuli, and guessing is uniform over the entire circle (2π). The parameters α, β, γ, and κ control the proportion of target responding, nontarget responding, guessing, and the concentration of the von Mises distribution, respectively. The spread of the distributions of target and nontarget responses is assumed to be the same. As α, β, and γ must sum to 1, α is not included as a free parameter in the fitting. The three free parameters (β, γ, κ) are estimated using maximum likelihood methods.\nFigure 1.\n\nOne-dimensional and 2D misbinding models. The top row shows a 1D task. Fictive stimuli and responses are shown (left) where the target (blue) and nontarget (red) were always shown at the same orientations. The responses column shows sample responses centered on the target with some imprecision (blue), around the nontarget with the same imprecision (red), and randomly distributed guesses (green). The PDF of responses (right) shows peaks around those locations with heights proportional to the number of target and nontarget responses, widths proportional to the imprecision of the memories, and a background guessing rate proportional to the height of the horizontal line. The bottom row shows a similar setup for a 2D task where the target and nontarget were always shown in the same location on a 2D screen (left), and the PDF is now 2D (right), which shows peaks of responses (upper part) around those locations (bottom part) with a 2D width (dashed circle) and uniform guessing across the entire screen (green dots).\nFigure 1.\n\nOne-dimensional and 2D misbinding models. The top row shows a 1D task. Fictive stimuli and responses are shown (left) where the target (blue) and nontarget (red) were always shown at the same orientations. The responses column shows sample responses centered on the target with some imprecision (blue), around the nontarget with the same imprecision (red), and randomly distributed guesses (green). The PDF of responses (right) shows peaks around those locations with heights proportional to the number of target and nontarget responses, widths proportional to the imprecision of the memories, and a background guessing rate proportional to the height of the horizontal line. The bottom row shows a similar setup for a 2D task where the target and nontarget were always shown in the same location on a 2D screen (left), and the PDF is now 2D (right), which shows peaks of responses (upper part) around those locations (bottom part) with a 2D width (dashed circle) and uniform guessing across the entire screen (green dots).",
null,
"To adapt the model for 2D data, several changes are needed. First, 2D coordinates are used in place of angles, which means that the von Mises distribution is replaced with a bivariate Gaussian distribution ψσ with standard deviation σ and zero covariance. Second, a distribution for the random guesses must be chosen. For simplicity, we begin by assuming that they are drawn from a uniform distribution over the entire area of the screen (A). This gives the following response density function:\n\\begin{eqnarray}P\\;\\left( {{\\bf{\\hat \\theta }}{\\rm{\\;}}} \\right) &=& {\\rm{\\;\\alpha \\;}}{{\\rm{\\psi }}_{\\rm{\\sigma }}}\\left( {{\\bf{\\hat \\theta }}\\; - {\\bf{\\theta }}} \\right) + {\\rm{\\;\\beta }}\\frac{1}{m}\\mathop \\sum \\limits_i^m {{\\rm{\\psi }}_{\\rm{\\sigma }}}\\left( {{\\bf{\\hat \\theta }}{\\rm{\\;}} - {\\rm{\\;}}{\\varphi _i}} \\right){\\rm{\\;}} \\nonumber\\\\ &&+ {\\rm{\\;\\gamma }}\\frac{1}{A},\\end{eqnarray}\n(2)\nwhere here, θ and ϕ are vectors indicating locations on the screen, and again, α, β, and γ sum to 1, which together with σ yield three free parameters.\nThe model will operate in any spatial units, as long as A is changed to reflect this. For the examples provided here, we will use 1366*768 as the screen area and pixels as units. The dimensions of the screen will dictate the scale of the errors in the task (i.e., on a 40 cm × 30 cm screen, the errors will be under 50 cm) and thus the scale of the imprecision parameter σ. The models take in a dimensions argument, which sets the screen dimensions for simulations and fitting. In other words, it implicitly sets the units for the model. Any appropriate scaling values are possible (e.g., 960*540 pixels, 46*26 cm, 30*17 visual degrees). This allows you to convert data collected on different screen sizes into standard units (e.g., visual degrees or percentage of screen size), for example, to correct for any influence that different screen sizes may have on the data. A rectangular screen is assumed, so if a different shape screen is used (e.g., a circular window on a screen), the model will need to be modified to account for this. Different dimensions can be supplied for simulating and fitting the data (e.g., if the stimuli may not appear too close to the edges of the screen but possible response locations are not so constrained). This can also be achieved by using the method described in the “Effect of stimulus separation on recovery” section below.\nPrevious 1D models used a circular space, meaning that responses wrapped around the domain, without any edges. In contrast, the 2D spatial tasks involve a finite space defined by the screen or screen region. When simulating the model, any responses that fall outside of the screen when drawn from the normal distributions are replaced by new samples drawn from the same distribution.\nWe use the framework for 1D models provided in MemToolbox (Suchow et al., 2013; MemToolbox.org) to provide a new toolbox (MemToolbox2D; Grogan et al., 2019; https://doi.org/10.5281/zenodo.3752705) for modeling 2D data that integrates with the existing 1D MemToolbox. New functions were created to run the fitting, display plots, correct for biases, and simulate the models. Thus, MemToolbox2D will not interfere with the functions of MemToolbox.\nSimulations\nFirst, we provide simulations to show the behavior of the model during fitting and some of the issues that may impact the quality of the fits, including true parameter values, number of trials, number of nontargets, response distributions and biases, stimulus constraints, and screen edges.\nFor all simulations, unless otherwise specified, we simulated a task in which an agent views three stimuli. The three stimuli were randomly placed on the screen, and one was randomly chosen to be the “target,” with the other two being “nontargets.” The agent then had to place the target stimulus in the location they remembered it occurring before. This is similar to a spatial WM task used previously (Pertzov et al., 2013). We simulated 100 trials. The type of response made on each trial was determined by the parameters used for the simulations: γ controlled the proportion of responses that were random guesses, β the proportion that were misbinds (centered on a nontarget location), α (α = 1 – γ – β) the proportion that were target responses, and σ the imprecision (standard deviation) of target and misbinding responses. We then fit the misbinding model to these simulated data to see how accurately the true parameters could be recovered.\nA large sweep of parameters was used for these simulations, ranging from almost no to almost all guessing (γ ϵ [0.01, .98]), almost no misbinding to almost all misbinding (β ϵ [0.01, 0.98]), and low to high imprecision (σ ϵ [0.1, 100]). Eleven values of each parameter were chosen, evenly spaced across these ranges, and all combinations used, with the caveat that γ and β summed to 1 or less, and we simulated each parameter combination 100 times. This parameter sweep was chosen to test how well the model can recover true parameters across a large range, including values that represent poor performance due to high guessing, misbinding, and imprecision, either in isolation or combination. These values are unlikely to occur in real data as few people will guess or misbind on all trials.\nFor each simulation at a particular set of parameters, the 100 simulated trials were fit by maximum likelihood estimation. The parameter fits account for the position of each of the items, as well as the response, on each trial.\nTo show the accuracy of parameter recovery, we plot the difference between recovered and true parameters (parameter recovery error) against the values of the true parameters; zero indicate perfect parameter recovery, and larger (positive or negative) values indicate poorer recovery. These recovery errors are plotted for each parameter, marginalized over the other parameter values.\nWhen comparing different models or methods, we use the Bayesian information criterion (BIC; Schwarz, 1978), which adds a penalty for the number of trials and parameters to the negative log-likelihood of the model fit; smaller values indicate better-fitting models.\nCode for all the simulations here is available from https://doi.org/10.5281/zenodo.3752763 (Grogan, 2019).\nHuman data\nTo illustrate how the models work with real human data and to examine how they cope with more complex task designs, we fit previously unpublished data collected from an Ignore/Update spatial WM task (i.e., a 2D task) and compared the pattern of recovered parameters to those found on a 1D orientation analogue of this task (Fallon, Mattiesing, Dolfen, Manohar, & Husain, 2018; Fallon, Mattiesing, Muhammed, Manohar, & Husain, 2017). New data were collected from 49 healthy older adults (mean age = 71.4 years, SD = 5.5, 18 females, 31 males). Ethical approval was granted by South West–Central Bristol NHS REC (16/SW/0205). Participants gave written informed consent, and the study was conducted in line with the Declaration of Helsinki.\nIn brief, the task involved remembering a pair of abstract shapes (Figure 2) that were shown with a “T” cue in the screen center, indicating they were “test shapes”, one of which would be tested later in the trial. These could be shown at Time 1 and/or Time 2. On some trials, nontest shapes (no “T” shown in center) could be shown either at Time 2, which would not be tested, and were not to be remembered. Participants were instructed to remember the most recent test shapes they saw. They were tested with a test shape and a nontest shape (either distractor or novel shape) presented on the screen. Participants were asked first to select the shape they recalled to be in the (most recent) test shape pair by touching it, rather than the nontest shape that was presented with it. Then they were requested to drag the test shape to where they remembered it to have been shown originally.\nFigure 2.\n\nProcedure of the spatial WM task. Participants were presented a pair of test shapes (with a “T” at fixation), which could be either preceded or followed by a blank screen (Maintain T2 and T1, respectively) or by a second pair of nontest shapes (Update or Ignore, respectively). They were asked to remember the most recent pair of test shapes. At the test, one test shape and one nontest shape (novel item or distractor) were shown, and participants chose the item they remembered to have been a test shape and dragged it to where it appeared previously.\nFigure 2.\n\nProcedure of the spatial WM task. Participants were presented a pair of test shapes (with a “T” at fixation), which could be either preceded or followed by a blank screen (Maintain T2 and T1, respectively) or by a second pair of nontest shapes (Update or Ignore, respectively). They were asked to remember the most recent pair of test shapes. At the test, one test shape and one nontest shape (novel item or distractor) were shown, and participants chose the item they remembered to have been a test shape and dragged it to where it appeared previously.",
null,
"Overall, there were four different conditions, each occurring with the same frequency (Figure 2):\n• Ignore trials: Consisting of test shapes at Time 1 followed by nontest shapes at Time 2\n• Maintain long delay trials (T1): Test shapes at Time 1 followed by a long delay\n• Update trials: Consisting of test shapes at Time 1 followed by new test shapes at Time 2 (so now working memory of test shapes at Time 1 had to be updated to instead store the test shapes at Time 2)\n• Maintain short delay trials (T2): Test shapes at Time 2 followed by a short delay.\nThere were two key factors in this design. First, there were irrelevant stimuli—either ones that had to be ignored (Ignore condition) or old items stored in memory that now had to be displaced by new ones (Update condition). Then there were two conditions that controlled for different maintenance durations in the Ignore and Update conditions, by simply having either a long (T1) or short (T2) delay.\nNote that the ignore and update trials have three nonprobed items (one nonprobed test shape plus two distractors), while T1 and T2 trials only have one nonprobed test shape; the model accounts for this by averaging the likelihoods of misbinding to each nonprobed shape together (“m” in Equation 2). Other models are available that allow for two sets of nonprobed shapes, for example, with separate misbinding parameters for the distractors and the nonprobed test shape.\nThe task was performed on Dell laptops using a mouse, at a viewing distance of approximately 40 cm. After 11 practice trials, there were three blocks of 40 trials, giving 120 trials in total (30 per condition, random order).\nResults\nHere we present several simulations showing how 2D models compare to 1D models and investigating factors influencing the accuracy of parameter recovery.\nAccuracy of recovering simulated parameters\nTo see what is gained from 2D responses over 1D responses, we compared how well the fitted parameters relate to the simulation parameters used to generate the simulated data. We simulated and fit 1D and 2D versions of the misbinding model. The 1D models fit the κ parameter for the von Mises distribution concentration, which is converted into standard deviation (σ) for comparison with the standard deviation parameter in the 2D models.\nFirst, we looked for recovery error and biases, using 100 iterations across the parameter sweep. Both models recovered parameters that correlated highly with the true parameters (Table 1). Figure 3 shows the difference between simulated (true) parameters and fitted (recovered) parameters across the parameter space; the 2D model has smaller deviations from zero, meaning better parameter recovery (paired-sample t tests for each parameter, p < 0.0001).\nTable 1.\n\nSpearman correlation coefficients between each true parameter and the recovered parameters for the 1D and 2D misbinding models. ***p < 10−10.\nTable 1.\n\nSpearman correlation coefficients between each true parameter and the recovered parameters for the 1D and 2D misbinding models. ***p < 10−10.",
null,
"Figure 3.\n\nMean difference between recovered and simulated parameters (parameter recovery error). This is shown for 1D (top row) and 2D (bottom row) misbinding models. Dashed line shows the y = 0 line, which is where there is no difference between the simulation and recovered parameters for any simulation parameter value. Shaded area shows standard deviation. Each column shows one parameter value, marginalized across all other parameters from the parameter sweep. The 2D model performs better than the 1D model (t tests: p < 0.0001; see Supplementary Figures S4S6 for 1D models).\nFigure 3.\n\nMean difference between recovered and simulated parameters (parameter recovery error). This is shown for 1D (top row) and 2D (bottom row) misbinding models. Dashed line shows the y = 0 line, which is where there is no difference between the simulation and recovered parameters for any simulation parameter value. Shaded area shows standard deviation. Each column shows one parameter value, marginalized across all other parameters from the parameter sweep. The 2D model performs better than the 1D model (t tests: p < 0.0001; see Supplementary Figures S4S6 for 1D models).",
null,
"We next performed a parameter sweep over β, γ, and σ, simulating each parameter set 1,000 times, and plotted the simulated and recovered parameters at various slices through this parameter space to see the effect of increasing one source of error while holding the others equal. The median recovered parameters are close to the simulated values (Supplementary Figures S1S3 for 2D simulations and S4S6 for 1D simulations), although there is greater deviation from the true parameters when the simulated participant generates more inaccurate responses due to higher imprecision, guessing, and misbinding.\nThe recovered parameters show trade-offs within the 2D model, as there are within the 1D model, because responses remote from any stimuli can be explained either as random guesses or by a participant having very imprecise spatial memory. Thus, the recovered imprecision and guessing parameters negatively correlate (Figure 4) as do guessing and misbinding, but misbinding and imprecision are positively correlated. These correlations between recovered parameters are weaker in the 2D model (γ and β: r = –.3817, σ and γ: r = –.1548, β and σ: r = .1208; all p < 0.000001) than in the 1D model (γ and β: r = –.6799, γ and σ: r = –.6563, β and σ: r = .3314; all p < 0.000001). In the 1D model, imprecise responses are very likely to run into other nontargets due to the circular 1D space, whereas in the 2D model, this is less common because the locations differ in two dimensions, and the boundaries absorb imprecise responses without them wrapping around to the other side. Please note these trade-offs are within the recovered parameters, due to the dependence of them in the model, and do not mean there are trade-offs between the “true” imprecision, misbinding, and guessing that occur within people's working memory.\nFigure 4.\n\nHeatmaps showing the recovered parameter trade-offs within the misbinding model in 1D (top) and 2D (bottom) space. The plots show the recovered parameter values at 10,000 MCMC posterior samples from fitting the model to simulated data (β = .3, γ = .3, σ = 30). Guessing correlates negatively with misbinding and imprecision, and misbinding and imprecision are positively correlated in both 1D and 2D, but in all cases, the correlations are weaker in the 2D model, shown here by more circular, less elliptical, patterns.\nFigure 4.\n\nHeatmaps showing the recovered parameter trade-offs within the misbinding model in 1D (top) and 2D (bottom) space. The plots show the recovered parameter values at 10,000 MCMC posterior samples from fitting the model to simulated data (β = .3, γ = .3, σ = 30). Guessing correlates negatively with misbinding and imprecision, and misbinding and imprecision are positively correlated in both 1D and 2D, but in all cases, the correlations are weaker in the 2D model, shown here by more circular, less elliptical, patterns.",
null,
"Factors affecting recovery accuracy\nParameter recovery can be affected by several aspects of the task, and many of these are under the experimenter's control. Here we show how factors such as number of trials, number of nontargets, and constraints on stimulus placement affect recovery accuracy from simulations. Experimenters can thus design tasks taking these factors into account when considering their model fitting and may find it useful to simulate their tasks before running them to test their recovery accuracy.\nBetter estimates with more trials\nThe more trials available for fitting, the more accurate the parameter recovery (Figure 5). This was seen for both 1D and 2D misbinding models (linear regression: p < 10−10). As it is not always possible to have large numbers of trials, recovering simulated parameters can be useful to experimenters for assessing the accuracy of model fitting in their study.\nFigure 5.\n\nThe absolute recovery error for different numbers of trials for 1D (top) and 2D (bottom) misbinding models. More trials (warmer colors—see color bar) leads to more accurate parameters (lower absolute recovery errors) in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error of the mean.\nFigure 5.\n\nThe absolute recovery error for different numbers of trials for 1D (top) and 2D (bottom) misbinding models. More trials (warmer colors—see color bar) leads to more accurate parameters (lower absolute recovery errors) in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error of the mean.",
null,
"More nontargets worsen parameter recovery\nIncluding nontarget stimuli is useful to assess misbinding between stimuli (Bays et al., 2009). Increasing the number of nontargets (from one to six, same parameter sweep, 100 iterations) increased the error in parameter estimation (Figure 6; linear regression: p < 10−10). These differences are larger in the 1D model because with more nontargets, imprecise target responses or random guesses are more likely to end up close to a nontarget and be classed as misbinds, which acts to decrease imprecision estimates especially. In the 2D model, this is less likely as the items are separated along two dimensions and the responses do not wrap around the edges, instead being bounded by the screen edges.\nFigure 6.\n\nAbsolute parameter recovery errors for different numbers of nontargets. More nontargets (cooler colors) leads to higher absolute recovery errors in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error.\nFigure 6.\n\nAbsolute parameter recovery errors for different numbers of nontargets. More nontargets (cooler colors) leads to higher absolute recovery errors in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error.",
null,
"Effect of stimulus separation on recovery\nAll of the 2D simulations presented here have used entirely random (uniform) generation of stimulus locations to match the 1D simulations and many 1D orientation tasks. However, many spatial WM tasks constrain the stimuli so that they cannot appear within a certain distance of other stimuli, the edges of the screen, or the center of the screen (if probe items are presented there; e.g., Pertzov et al., 2012).\nTo examine how such constraints affect parameter recovery, we simulated two different tasks, one where stimuli appeared uniformly randomly across the entire screen (no constraints) and one where stimuli were constrained to not appear within 88 pixels of each other or the screen center, or 29 pixels from the screen edge. These distances correspond to 3 and 1 visual degrees, respectively, assuming a viewing distance of 40 cm and 42 pixels per centimeter. The entire parameter sweep was simulated 100 times for each task, which were fit with the misbinding model. The errors in the recovered parameters from the two tasks are shown in Figure 7. Paired-sample t tests revealed that parameters α, β, and γ had significantly smaller recovery errors when stimulus constraints were used (p < 10−10), while σ only showed a trend effect (p = 0.0615), but absolute recovery errors were significantly smaller for σ and γ (p = 0.0071, p = 9.6457 × 10−8) but not α and β (p = 0.4405, p = 0.0610). These statistics suggest that constraining stimuli can improve the recovery accuracy, although the benefits are not huge. This is likely due to the constraints reducing overlap of items (and thus responses) and making distributions centered on items less likely to be cut off by the screen edges.\nFigure 7.\n\nParameter recovery errors (top) and absolute recovery errors (bottom) for the 2D model in simulated tasks with and without stimulus constraints. Adding in constraints to the stimulus locations leads to smaller errors in parameter recovery (orange line; see text for statistics). Shading shows standard error.\nFigure 7.\n\nParameter recovery errors (top) and absolute recovery errors (bottom) for the 2D model in simulated tasks with and without stimulus constraints. Adding in constraints to the stimulus locations leads to smaller errors in parameter recovery (orange line; see text for statistics). Shading shows standard error.",
null,
"Response biases and distributions\nWhen people perform these types of tasks, their responses may not be uniformly distributed across the screen for several reasons. First, often the stimuli themselves may not be uniformly distributed, for example, if there are areas they cannot appear in (see stimulus separation section above). Task constraints can give rise to biased distributions either through veridical memory or via strategies for guesses. Individuals may also sometimes show further biases to prefer one side of the screen, the top or the bottom half, or a radial bias where responses shrink toward the center of the screen. Many of these will depend on the experiment and stimulus characteristics, so here we shall demonstrate how two kinds of biases (constant and proportional biases) affect the model and how experimenters may account for them.\nConstant biases\nFirst, we simulated responses that have a constant directional bias by subtracting a fixed translation (from 20 to 200 pixels, step size = 36) to X- and Y-coordinates of all responses (and reapplying the screen boundary via edge constraining). The error of the fitted parameters rises as this bias increases (Figure 8). MemToolbox contains a function to correct for biases in 1D models, which we have adapted to work in 2D models. This fits a constant translation bias term for X- and Y-dimensions, thus improving fits as shown by smaller deviations from the true parameters (Figure 8, paired t tests: p < 10−8). However, when the bias is sufficiently large, responses are more likely to hit the edge of the screen (and, in this case, are edge-constrained to lie on the screen edge instead), which hinders accurate recovery.\nFigure 8.\n\nParameter recovery errors from simulations with a constant translation bias subtracted (top) and corrected for (bottom). The color bar shows bias size (from 20 to 200 pixels). The bias correction works well for biases up to about 150 pixels in size. Shading shows standard error.\nFigure 8.\n\nParameter recovery errors from simulations with a constant translation bias subtracted (top) and corrected for (bottom). The color bar shows bias size (from 20 to 200 pixels). The bias correction works well for biases up to about 150 pixels in size. Shading shows standard error.",
null,
"Proportional biases\nPeople may be unlikely to shift all responses in the same direction by the same amount, especially if this leads to many responses landing on the screen edges, so we also simulated a proportional bias toward the right-hand edge. Here, a bias parameter controls the proportion of the distance to the edge that responses are shifted; positive values shift it toward the edge, and negative values shift away (zero means no shifting). This is perhaps more likely as a response bias in people and can also be corrected by the toolbox (Figure 9; paired t tests: p < 10−15).\nFigure 9.\n\nDifference plot of recovered and simulated parameter values for different proportional biases toward the right-hand screen edge. Data with (bottom) and without (top) the bias correction for different proportional biases (color bar). Positive biases mean responses move toward the right-hand screen edge, and negative biases move responses away from this edge (responses do not move vertically). The bias correction improves the fits for all parameters. Shading shows standard error.\nFigure 9.\n\nDifference plot of recovered and simulated parameter values for different proportional biases toward the right-hand screen edge. Data with (bottom) and without (top) the bias correction for different proportional biases (color bar). Positive biases mean responses move toward the right-hand screen edge, and negative biases move responses away from this edge (responses do not move vertically). The bias correction improves the fits for all parameters. Shading shows standard error.",
null,
"Next, we applied a proportional radial bias that either shrinks all responses toward the screen center (positive values) or expands them out (negative values) by a proportion of the distance the response is from the screen. Biases mainly affect the imprecision parameter (Figure 10), although the guess parameter is affected by an expanding radial bias. We adapted the bias function to account for radial biases, which improves parameter recovery (paired t tests: p < 10−12), especially for the imprecision parameter, although does not fully correct large guess parameters when there is a large expanding bias.\nFigure 10.\n\nDifference plot of recovered and simulated parameter values for different radial biases. Data with (bottom) and without (top) the bias correction for different radial biases (color bar). Positive radial biases mean points move toward the center of the screen, and negative biases move points away from the center. The bias correction improves the fits for all parameters, although large guess rate parameters are not fully corrected when there is a large expanding bias. Shading shows standard error.\nFigure 10.\n\nDifference plot of recovered and simulated parameter values for different radial biases. Data with (bottom) and without (top) the bias correction for different radial biases (color bar). Positive radial biases mean points move toward the center of the screen, and negative biases move points away from the center. The bias correction improves the fits for all parameters, although large guess rate parameters are not fully corrected when there is a large expanding bias. Shading shows standard error.",
null,
"This function also allows us to test whether there are radial biases toward points other than the center. For instance, in trials with a single target and nontarget, it is possible that rather than the nontarget “swapping” locations with the target, it exerts a “pull” on the response locations. Note that this contribution to error is separable from misbinding, as the responses will be centered somewhere between targets and nontargets, rather than being centered on the nontarget in the case of misbinding. We can test for this by using the radial bias function and supplying the nontarget location as the bias coordinate for each trial; if we simulate such a bias, this function corrects for the bias and gives much more accurate parameter recovery (Figure 11; paired t tests: p < 10−15). For this simulation, we used one nontarget and the standard mixture model (i.e., no misbinding). We also fit the misbinding model to these nontarget-biased data, which did not fit the data as well as the standard mixture model with the radial nontarget bias (mean BIC = 2,720 vs. 2,437), suggesting this pattern of nontarget “pull” is distinguishable from misbinding in simulated data.\nFigure 11.\n\nParameter recovery errors with a radial bias toward the nontarget location on each trial. Data shown without (top) and with (bottom) bias correction. The bias correction reduced the errors in parameter recovery. This simulation had β = 0 (i.e., no misbinding). Shading shows standard error.\nFigure 11.\n\nParameter recovery errors with a radial bias toward the nontarget location on each trial. Data shown without (top) and with (bottom) bias correction. The bias correction reduced the errors in parameter recovery. This simulation had β = 0 (i.e., no misbinding). Shading shows standard error.",
null,
"Response sampling for guess distributions\nPeople may not (and indeed probably do not) have a uniform guess rate in these tasks but instead may have complex distributions that include some of the biases mentioned above but may have others we have not thought of. One way of dealing with this is to use the participants’ own responses to determine the “guessing” distribution. To achieve this, a probability density function (PDF) can be built from all of a participant's response locations over an experiment (across all conditions for a participant's data). Drawing from this distribution samples a response taken from a random trial, which, if the trials are independent, should be unrelated to the current trial's stimuli and thus can constitute a random guess constrained by the person's own response biases.\nWe implemented this in a model that uses a kernel smoothing density function to build a PDF of the person's responses over all trials, which is then used to estimate the probability of each trial's response over the entire experiment. MATLAB's ksdensity function was used to build the response PDF, which estimates the default bandwidth of the smoothing function from the data using Scott's rule. The default bandwidth can be optionally overridden to supply your own bandwidth in the response sampling function.\nThis model cannot easily be tested with simulations as people have nonrandom distributions of responses that we cannot predict, so we tested this model using cross-validation from the healthy participants’ data fitting (see later section or the Method section for details). We fit 90% of all the trials (combining all conditions) with the misbinding model with and without response sampling and then generated the likelihood of the responses for the remaining 10% of trials using the fitted parameters. We repeated this 100 times for each person and took the mean likelihood for each person. Crucially, the response sampling only included the 90% of training trials, not the 10% of validation trials. If the response sampling model accounts for people’s response distributions and therefore gives better fits, it should lead to larger likelihoods for the out-of-sample validation trials than the uniform guessing model, whereas if it is instead simply overfitting the participants’ responses, then out-of-sample probabilities will be lower than the uniform guessing model. The response sampling model gave higher likelihoods than the uniform guessing model for the validation trials (t test: p = 4.8265 × 10−10), suggesting it is not simply overfitting the data but that drawing guesses from the distribution of all responses can give more accurate estimates of guess rates.\nChange detection\nThe models are also able to fit data from change detection tasks similar to a two-alternative forced-choice (2AFC) paradigm. Change detection tasks probe the ability to detect changes to stimuli's features, such as location. A stimulus is probed, shown displaced from its original location by some distance, and participants must decide if the location is the same or different. The paradigm is different from many 2AFC tasks used for signal detection, because often all probes have some displacement, however small, and the task is used not to measure people's ability to separate changed from unchanged items but to measure the minimum detectable changes in feature memory. These tasks give much less information on each trial about the source of error, but with sufficient data, corresponding models can still be fit. These models are adapted from the 2AFC models in the original 1D MemToolbox (Suchow et al., 2013).\nThe model is based on assumptions about the spatial pattern of erroneous acceptances of a displaced probe caused by the different underlying error sources. A person with perfect memory would reject all displaced probes, but as the error sources increase, they give different patterns of acceptances. Imprecision will increase acceptance closer to the target, misbinding will increase acceptance close to the nontargets, and guessing will increase acceptance uniformly. We also assume that if someone incorrectly accepts a probe of a certain displacement as unchanged, they would also have accepted smaller displacements on that particular trial. Taken together, these assumptions lead to us integrating over the continuous probability density function to estimate the probability of correctly rejecting a displaced probe, which we calculate as the probability of generating a response that is closer to the target than the current probe is. Note that this approach assumes a nonzero displacement of each item; otherwise, this integration will be affected mainly by the area on the far side of the target to the probe.\nFigure 12 shows, when a range of different probe (and nontarget) distances is considered, different patterns of acceptance are seen: Increasing imprecision increases the acceptance of probes closer to the target, increasing guessing increases acceptance similarly across the entire circle, and increasing misbinding increases acceptance closer to nontargets.\nFigure 12.\n\nIllustrative examples of mixture modeling approach to change detection. We show 1D examples for ease, although the same principles apply for 2D (see Supplementary Figure S10). The top row shows how each error component affects the proportion of generative responses that would fall at or beyond a probed distance of 60 degrees (arbitrary y-units). The second row shows the values of the parameters used for those simulations, along with the summed probabilities of all responses that occur at ≥ 60 degrees from the target. Increasing each parameter increases the probability of incorrectly accepting this probe as unchanged. The bottom row shows the probability of accepting a probe as unchanged as a function of the probe distance (or probe-nontarget distance for misbinding). This shows that increasing imprecision increases the spread of these same responses around the target, increasing guessing affects all orientations similarly, and increasing misbinding affects responses in relation to their distance to the distractor (if we plotted probe-targ for misbinding, it would look like the guessing panel).\nFigure 12.\n\nIllustrative examples of mixture modeling approach to change detection. We show 1D examples for ease, although the same principles apply for 2D (see Supplementary Figure S10). The top row shows how each error component affects the proportion of generative responses that would fall at or beyond a probed distance of 60 degrees (arbitrary y-units). The second row shows the values of the parameters used for those simulations, along with the summed probabilities of all responses that occur at ≥ 60 degrees from the target. Increasing each parameter increases the probability of incorrectly accepting this probe as unchanged. The bottom row shows the probability of accepting a probe as unchanged as a function of the probe distance (or probe-nontarget distance for misbinding). This shows that increasing imprecision increases the spread of these same responses around the target, increasing guessing affects all orientations similarly, and increasing misbinding affects responses in relation to their distance to the distractor (if we plotted probe-targ for misbinding, it would look like the guessing panel).",
null,
"Radial and translation biases can be included in the models as in the continuous case. As the task is change detection rather than a signal detection task (as the probe is not the same as the target, and even if it were would likely not be perceived so due to memory decay), the aim is not to measure a person's ability to separate out signals from nonsignals but rather to estimate these underlying error sources. The estimates of imprecision, guessing, and misbinding are based on the effect of probe distance on acceptance of changes. Under this framework, a person who accepts probes closer to the target does so because of less imprecise memory, and a person who accepts more probes regardless of distance does so because of increased guessing, not because of a greater propensity to accept changes. It is possible to estimate a response bias separately by averaging the proportion of acceptances.\nWe ran simulations and fittings for 1D and 2D versions of this. The standard mixture model (without misbinding) was used because fitting 2AFC data is slow due to the more complex probability density taking into account the nontarget locations for each trial.\nThe parameter recovery works well in comparison to the 1D model (Figure 13), especially at higher imprecision values, but as expected is considerably less accurate than the continuous report fitting (Figure 3).\nFigure 13.\n\nParameter recovery errors for the 1D and 2D two-alternative forced-choice simulations. The 2D model (bottom) is more accurate than the 1D model (top), although both are far less accurate than with continuous report data (see Figure 3). Shading shows the standard error of the mean.\nFigure 13.\n\nParameter recovery errors for the 1D and 2D two-alternative forced-choice simulations. The 2D model (bottom) is more accurate than the 1D model (top), although both are far less accurate than with continuous report data (see Figure 3). Shading shows the standard error of the mean.",
null,
"We have developed further models, including one with a covariance parameter for the imprecision, one with separate precision parameters for targets and nontarget responses, and one with separate parameters for misbinding to two types of nontargets (e.g., the other test shape that was not probed and the nontest shapes). The latter can also be used to assess misbinding toward previous targets/responses if desired. We have also adapted all the other models present in MemToolbox (Suchow et al., 2013), including the variable precision models (Fougnie, Suchow, & Alvarez, 2012; van den Berg, Shin, Chou, George, & Ma, 2012), ensemble integration model (Brady & Alvarez, 2011), continuous resource model (Wilken & Ma, 2004), exponential decay model (Zhang & Luck, 2009), and slot model (Zhang & Luck, 2008).\nWe have presented only the maximum likelihood fits here, but MemToolbox2D also contains adapted functions for performing Bayesian model fitting using the Markov chain Monte Carlo (MCMC) method. These fits take longer to run, which is why we used maximum likelihood for the analyses presented here. This method also allows hierarchical model fitting across participants.\nWe have also adapted several other functions from MemToolbox (Suchow et al., 2013) such as the plotting functions for visualizing the data and results of the fits.\nEmpirical data fitting\nThe 2D models can be fit to simple spatial WM tasks (e.g., Pertzov et al., 2012, 2013) but can also be used for more complex tasks to compare between multiple conditions. Here we fit the models to data collected from people performing a spatial ignore/update WM task (see “Human data” section for details) to show how the models can be useful for more complex experimental designs too.\nPeople show biases and nonuniform guesses\nTo see which model fit the data best, we fit several models to the behavioral data (all conditions). First, we fit the misbinding model and the standard mixture model (no misbinding), and the misbinding model fit best (BIC: 2,984 vs. 3,051). Next we looked at whether the bias functions improved the fit by including each bias in the misbinding model. The constant bias did not improve the fit (BIC = 2,984), while the radial bias did (BIC = 2,966). Adding response sampling improved the misbinding model (BIC = 2,974), as did combining the misbinding model with the radial bias (BIC = 2,958). Therefore, the misbinding model with a radial bias and response sampling was the best-fitting model. The parameters from this fitting are shown in the Supplementary Materials and Supplementary Figure S7 and show that most participants were biased away from the screen center and that these biases were consistent within people across conditions (Pearson correlations: ρ > .65, p < 0.0001).\nConvergent validity of parameter recovery\nTo compare convergent validity between the 1D and 2D modeling approaches, we compared the fitted parameters from the 2D misbinding model to those reported on the 1D version of this task (Fallon et al., 2018). Histograms of the recovered parameters are shown in Figure 14 and show that while imprecision is approximately normally distributed, the other parameters are skewed, being bounded by 0 and 1. We did not logit transform the α, β, and γ parameters for the comparison to the 1D task as the 1D analysis did not transform parameters (Fallon et al., 2018).\nFigure 14.\n\nHistograms of the recovered parameters from the ignore/update task, across all conditions.\nFigure 14.\n\nHistograms of the recovered parameters from the ignore/update task, across all conditions.",
null,
"Repeated-measures analysis of variance (ANOVA) with delay and irrelevant stimuli as factors and a random effect of participant were used to analyze the fitted parameters (Figure 15). The analyses showed the following:\n• Delay duration increased the imprecision (F(1, 144) = 53.7751, p = 1.4959 × 10−11), guessing (F(1, 144) = 14.4042, p = 2.1660 × 10−4), and misbinding (F(1, 144) = 3.9560, p = 0.0486).\n• Irrelevant stimuli increased misbinding (F(1, 144) = 19.9033, p = 1.6302 × 10−5) and guessing (F(1, 144) = 19.8437, p = 1.6755 × 10−5) but did not affect imprecision (F(1, 144) = 1.3353, p = 0.2498).\n• There was a significant interaction of delay and irrelevant information for guessing (F(1, 144) = 4.0163, p = 0.0469), although it was only weakly significant and was due to greater guessing in the ignore condition.\n• There were no other interactions (p > .1), suggesting that ignore and update trials both affected imprecision and misbinding similarly.\nFigure 15.\n\nMean fitted parameter values for each trial type from the healthy participants. Delay increased imprecision (σ; top left; blue vs. orange; see Figure 2 for task explanation), guessing (γ), and misbinding (β; see text for stats); irrelevant stimuli increased misbinding and guessing but not imprecision (ignore and update vs. T1 and T2); and there was an interaction of delay and irrelevant stimuli only for guessing. Standard error bars.\nFigure 15.\n\nMean fitted parameter values for each trial type from the healthy participants. Delay increased imprecision (σ; top left; blue vs. orange; see Figure 2 for task explanation), guessing (γ), and misbinding (β; see text for stats); irrelevant stimuli increased misbinding and guessing but not imprecision (ignore and update vs. T1 and T2); and there was an interaction of delay and irrelevant stimuli only for guessing. Standard error bars.",
null,
"These main effects are similar to the orientation version of this task, although they differ regarding the interaction of delay and irrelevant information. The 1D version found such an interaction in misbinding (Fallon et al., 2018), while here we found it for guessing; in both tasks, the errors were highest in the ignore condition. Potential reasons for this will be discussed later. The similarity in the rest of these results suggests convergent validity of the 2D modeling technique for real human data.\nTo establish whether the different results in 1D and 2D tasks are due to differences in the modeling, we examined how well each model can distinguish the parameters for the different conditions. We simulated 1D and 2D versions of the task using the mean recovered parameters of each condition from the 1D task (Fallon et al., 2018) and fit the misbinding models using fmincon and inverted the Hessian matrices to get the covariance between parameters. We then used MATLAB's linear hypothesis test to examine whether the recovered parameters for the different conditions were actually from different distributions of parameters (alternate hypothesis) or from the same distribution (null hypothesis).\nWe performed 1,000 iterations of this and counted how often the 1D and 2D models returned p ≤ .05. If the 2D model is better at distinguishing between parameters due to the lower covariance between them, it will be more likely to return p ≤ .05 in these linear hypothesis tests. The 2D model found a significant difference between the conditions more often (mean = 29.1% vs. 19.2%). This suggests that these parameters are less confusable (i.e., more distinguishable) in the 2D model than the 1D model. The 2D model was particularly better than the 1D model for comparing the ignore condition with update and T2 conditions, which could mean that the 1D task confused parameters between these conditions, leading to a different interaction being found in the orientation task.\nTo illustrate this, we generated posterior MCMC samples from fitting to data simulated using the mean parameters from the 1D orientation task and plotted the parameters against each other (Figure 16). There was greater separation between the parameters of the different conditions in the 2D model than the 1D model, suggesting that the 2D model is better at distinguishing between parameters of different conditions, within the range recovered from human data.\nFigure 16.\n\nHeatmaps showing the posterior MCMC samples for recovering the mean parameters from Fallon et al. (2018). Each color is a different condition (see legend; Ig = ignore, T1 = long delay, Up = update, T2 = long delay), and the rows show 1D and 2D models, respectively. The γ and β parameters have been arcsine transformed. Both models were simulated using the same parameters. The 2D model has better separation of the different conditions’ parameters, as shown by less overlap between the colors.\nFigure 16.\n\nHeatmaps showing the posterior MCMC samples for recovering the mean parameters from Fallon et al. (2018). Each color is a different condition (see legend; Ig = ignore, T1 = long delay, Up = update, T2 = long delay), and the rows show 1D and 2D models, respectively. The γ and β parameters have been arcsine transformed. Both models were simulated using the same parameters. The 2D model has better separation of the different conditions’ parameters, as shown by less overlap between the colors.",
null,
"Validating the behavioral metrics\nWe wanted to see whether the behavioral metrics were actually measuring the types of errors purported. We calculated those behavioral metrics on simulated data and looked at which metrics correlated with the true parameters and compared this against the accuracy of parameter recovery. These simulations used the same stimulus separation constraints as in the “Stimulus separation” section to match the experiments that commonly use these metrics, although running them without such constraints did not change the pattern of results.\nWe examined several commonly used behavioral metrics. Target distance is the Euclidean distance from responses to targets, while nearest-neighbor distance is the Euclidean distance from responses to their nearest stimulus (Pertzov et al., 2013). The former will contain imprecision, guessing, and misbinding influences, while the latter has attempted to remove the influence of misbinding and thus should contain mainly imprecision and guessing. The difference between these two represents the influence of misbinding. A further measure to capture misbinding, termed swap errors (Pertzov et al., 2012), is the proportion of responses occurring within a certain fixed distance (1.5 visual degrees) of a nontarget location. This measure is also confounded by imprecision since imprecise responses to the target location are more likely to end up close to a nontarget by chance, compared to precise responses. We can correct for this by calculating the proportion of possible locations with that specific distance from the target that would have been classed as “swap errors” and subtracting that. Swap errors can also be calculated using the mean target distance as the threshold, effectively scaling the criterion by a participant's average accuracy. This measure can likewise be corrected for chance in the same way.\nWe correlated the mean values of each of these metrics for simulated data with each of the true parameters (parameter sweep from previous simulations, 10 iterations). All the metrics correlated with multiple true parameters, so we present here only the metric most strongly correlated with each parameter (see Supplementary Figure S9 and Supplementary Table S1 for full correlation matrix between metrics and parameters).\nTarget distance correlated strongly and negatively with the proportion of target responding (Figure 17; Spearman correlations: r = –.9687, p < 10−10), nearest-neighbor distance positively with guess rate (r = .9279, p < 10−10), and the difference between these positively with misbinding (r = .8589, p < 10−10). There were no metrics that correlated strongly with imprecision, although swap errors (fixed threshold) had a medium negative correlation (r = –.5519, p < 10−10) while the nearest-neighbor distance (assumed to be a good measure of imprecision) had a weak positive correlation with it (r = .2931, p < 10−10), so both of these are presented. These metrics were all less strongly correlated with the true parameters than the recovered parameters were, although this may be an artifact of recovering the parameters with the same model used for simulation, as this makes it unlikely the behavioral metrics can perform better.\nFigure 17.\n\nRelationship between the true parameters (x-axis) and the best behavioral metrics (top) and recovered parameters (bottom). All correlations were significant (p < 10−10). Two metrics were compared against the imprecision (σ) parameter. The recovered parameters had stronger correlations than the best behavioral metrics (see text for statistics).\nFigure 17.\n\nRelationship between the true parameters (x-axis) and the best behavioral metrics (top) and recovered parameters (bottom). All correlations were significant (p < 10−10). Two metrics were compared against the imprecision (σ) parameter. The recovered parameters had stronger correlations than the best behavioral metrics (see text for statistics).",
null,
"To investigate this idea, we compared the split-half reliability of these four behavioral metrics and the recovered parameters for these behavioral data (pooling across all four conditions, 100 iterations of random splitting). All measures showed significant split-half reliability (Spearman correlations: p < 0.0001) with varying strengths. The swap errors metric was only weakly correlated with itself (ρ = 0.2068), while the imprecision parameter was strongly correlated (ρ = 0.7160). Target distance was slightly more reliable (ρ = 0.7955) than the recovered α parameter (ρ = 0.7811), but nearest-neighbor distance (ρ = 0.8196) and target minus nearest-neighbor distance (ρ = 0.7450) were better than γ (ρ = 0.3347) and β (ρ = 0.6278). This suggests that some of the behavioral metrics have good split-half reliability, and these are the ones that are most strongly correlated with the true α, β, and γ parameters in the simulations above. However, these simulations and correlations suggest none of the behavioral metrics currently used are good estimators of memory imprecision, and the ones currently available have low reliability. They also suggest that guessing and misbinding estimation in the model has poor split-half reliability, which could be due to the low guessing and misbinding parameters used for these simulations (recovered from the behavioral data), meaning few trials contain a guess or misbind, which leads to poorer estimation in the split-halves.\nWe can also look at convergent validity by comparing the outputs of ANOVA on the behavioral metrics in the 2D human data to those previously reported on the 1D circular analogue task, as we did earlier with the 2D modeling outputs. These showed that delay increased target distance (p < 0.0001), nearest-neighbor distance (p < 0.0001), and the difference between these two (p = 0.0077), while irrelevant distractors increased swap errors (p = 0.0001), target distance (p < 0.0001), and target minus nearest-neighbor distance (p < 0.0001)—but not the nearest-neighbor distance itself (p = 0.6066). There were no interactions (p > 0.1). While target distance and target minus nearest-neighbor distance patterns are similar to α and β (in the 1D data and our 2D data), the nearest-neighbor distance effects (delay but not irrelevance) were more similar to that seen in imprecision than guessing. This may not be surprising as the correlations (Figure 17) showed nearest-neighbor distance was correlated with both imprecision and guessing. This suggests that while nearest-neighbor distance is reliable, it is not a pure measure of one type of error source.\nDiscussion\nWe have developed a 2D mixture modeling toolbox for tasks such as spatial WM tasks. The models are able to recover parameters from simulations with good accuracy across a large parameter space (Figure 3) in line with the standard 1D mixture models and have less interdependent parameters (Figure 4). The accuracy rises as the number of trials increases (Figure 5) or the number of nontargets decreases (Figure 6). Behavioral data from a complex spatial WM task showed a similar pattern of results to a 1D orientation analogue of the 2D task, giving convergent validity (Figure 15). The 2D modeling provided a better estimate of true parameters of simulations than commonly used behavioral metrics (Figure 17). This suggests that the 2D mixture modeling approach is valid for analysis of data from complex spatial WM tasks and is more sensitive than previously available behavioral metrics. The models are provided in the MemToolbox2D package (Grogan et al., 2019; https://doi.org/10.5281/zenodo.3752705), compatible with the MemToolbox package for modeling 1D tasks (Suchow et al., 2013; MemToolbox.org).\nThe fact that the 2D models outperform 1D models in the simulations can be understood when examining the correlations of the different parameters from the posterior samples (Figure 4). The 2D model's parameters are less strongly (albeit still significantly) associated with each other than the 1D model's parameters. Mixture modeling can give accurate and reliable estimation of parameters in 1D tasks, helped by the fact that responses are often tightly clustered around stimuli locations (i.e., low imprecision). However, as imprecision increases, responses become more and more likely to land near other stimuli and be classed as misbinds; due to the circular space, if you move far enough away from a target, you will hit a nontarget. In the 2D task, while increasing imprecision does increase the chance of hitting a nontarget, this is overall much lower as there are two uncorrelated dimensions of separation between stimuli, and the responses do not wrap around the screen edges, instead being contained by them. This means that you are not guaranteed to hit a nontarget as you move further from a target, unlike in the circular task—sometimes you will hit a screen edge instead. These factors mean that the 2D models are better able to distinguish imprecision, guessing, and misbinding; these differences are due to differences in the task setup rather than the models themselves.\nThe simulations presented here used uniformly random item locations and performed well, but adding in stimulus constraints such as minimum distances to other items, screen edges, and screen center improved the parameter recovery, particularly for guessing and misbinding. Such constraints mean that the probability density functions for target and misbinding responses have less overlap and are cut off less by the screen edges, leading to better estimation of the probabilities of each type of response. This analysis also demonstrates how important simulation can be in task design, allowing the experimenter to try out different stimulus constraints, number of trials, and numbers of nontargets to optimize their task for the recovery of parameters of interest.\nAlternatively, the 2D modeling was shown to be more sensitive during simulations and to have less interdependent parameters (i.e., weaker correlations between recovered parameters) than the 1D modeling, and simulations from the fitted parameters showed the 2D model was better at distinguishing between parameters from different conditions. The different pattern of responses could therefore be due to more accurate parameter recovery in 2D tasks. Finally, and most interestingly, it is possible that the different results reflect differences between the way locations and orientations are processed in WM. There is a proposal that location plays a fundamentally distinct role in WM, such that all other features of an item (e.g., orientation, color, shape) are bound to location (Schneegans & Bays, 2017). It is possible that orientations are misbound more when distracting information is to be ignored, but locations are not as they are the feature to which others are bound.\nOther spatial WM tasks have reported an interaction of delay and number of items (Pertzov et al., 2012), which was present in the overall target distance but not in the nearest-neighbor distance (i.e., when controlling for misbinding). This was taken to mean that larger memory loads for longer durations lead to more misbinding errors but did not affect the imprecision of the memory. These behavioral metrics have been used in many spatial WM tasks (Liang et al., 2016; Pertzov et al., 2013; Zokaei, Nour, et al., 2019) and the current modeling approach provides a way to examine the validity of those metrics for estimating different sources of error.\nComparing the true parameters of simulations against traditional behavioral measures for this task gave a sensible pattern of results. Mean target distance was associated with imprecision, guessing, and misbinding. Nearest-neighbor distance (which removes the influence of misbinding to give a better estimate of imprecision) had no association with misbinding but instead associated mainly with guessing and a little with imprecision. The difference between these two was a good estimate of misbinding. These relationships were all expected and assumed in previous studies relying on them (Pertzov et al., 2012, 2013).\nHowever, imprecision was not strongly associated with any behavioral metric, suggesting it independently quantifies a new aspect of performance that is not well captured by existing measures. Imprecision had medium and weak associations with the proportion of swap errors and nearest-neighbor distance, respectively. The behavioral measures were less strongly correlated with “true” parameters used for simulations than recovered parameters were, but this is not surprising as the same model was used for simulation and fitting. While the simulations here show that the models can accurately recover simulated parameters, people likely use a range of behavioral heuristics not captured by the models and functions provided here. Split-half reliability and convergent validity checks showed that while target selection and imprecision could be reliably measured by the metrics, the metrics associated with misbinding and guessing were less reliable and valid measures.\nTwo-dimensional mixture modeling has been used previously (Schneegans & Bays, 2016) on a task in which stimuli appeared within an annulus of 5° width, thus having two dimensions, in contrast to tasks where stimuli locations appear on a ring (i.e., with no width) (Rajsic & Wilson, 2014; Schneegans & Bays, 2018), meaning the 2D coordinate data can be treated as a 1D orientation and modeled with the existing 1D model toolboxes (Suchow, Brady, Fougnie, & Alvarez, 2013). To account for this spatial constraint, the authors built in priors to their model, such that the bivariate normal distributions around targets and nontargets and the uniform distribution were convolved with an estimated distribution of their response eccentricities (which closely matched the annulus location and width). This method illustrates the flexibility of mixture modeling for 2D tasks with stimuli or response constraints, as well as the usefulness of incorporating priors into the model. This method also bears similarity to the response sampling method presented here, although applied to not only the guessing distribution but also the target and nontarget distributions, which could be useful for future researchers—the toolbox provided here is adaptable, and an example model outline is provided to show how new or modified models may be created.\nThere are some limitations with adapting the models to two dimensions. Previous 1D tasks often use a circular space (e.g., orientation or color), and this unbounded space works well because of the lack of edge effects. In contrast, 2D tasks using screens for stimuli presentation have a bounded space, meaning that responses do not “wrap around” as they do in a circular space. In this case, assumptions must be made regarding behavior close to the boundary and in the generative models that may place items outside of the boundaries. The generative models implemented here are likely not the true process that occurs when people recall items near the edge of the display, but it seems a reasonable simplification that still produces a well-fitting model.\nIn reality, people may have repelling biases away from edges or anchoring or landmark effects close to them. The bias functions in the MemToolbox2D package can test for some of these, and more complicated models can be developed that include these, and other, effects. We also provide a “response sampling” method that builds a response density function over all the responses from a person and can be used to sample the guesses from. This is an alternative to assuming a uniform guessing distribution. There are myriad potential biases and guessing distributions, and the MemToolbox2D package is provided with a permissive license to allow modification of the code here, as was the original MemToolbox package (Suchow et al., 2013).\nWhile these models were designed for WM tasks, they can be used for other types of tasks, including long-term memory tasks for spatial locations (e.g., Zokaei, Čepukaitytė, et al., 2019). There is also no reason why the models cannot be extended into the third dimension, should experiments require it, for example, in virtual reality navigation tasks.\nMemToolbox2D is provided as free code available from https://doi.org/10.5281/zenodo.3752705 (Grogan et al., 2019).\nConclusion\nWe provide here a new toolbox (MemToolbox2D; Grogan et al., 2019; https://doi.org/10.5281/zenodo.3752705) for analyzing 2D spatial WM tasks. Simulations showed it is accurate at recovering parameters over a wide range of values, number of trials, and number of nontargets and can take into account response distributions and biases. It provides a more accurate measure of true parameter values than previously used behavioral metrics on such tasks and may be better at separating out codependent parameter values because of the second dimension of separation present in the data.\nAcknowledgments\nSupported by the MRC (MR/P00878X) and BRACE (BRI17/15) (JG), MRC (MR/P00878X) (SM), and Wellcome Trust Principal Research Fellowship and the NIHR Oxford Biomedical Research Centre (MH).\nCommercial relationships: none.\nCorresponding author: John P. Grogan.\nEmail: [email protected].\nAddress: Nuffield Department of Clinical Neurosciences, University of Oxford, Oxford, UK.\nReferences\nBays, P. M., Catalao, R. F. G., & Husain, M. (2009). The precision of visual working memory is set by allocation of a shared resource. Journal of Vision, 9(10), 7, https://doi.org/10.1167/9.10.7.\nBays, P. M., & Husain, M. (2008). Dynamic shifts of limited working memory resources in human vision. 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Fractionating the neurocognitive mechanisms underlying working memory: Independent effects of dopamine and Parkinson's disease. Cerebral Cortex, 27(12), 5727–5738, https://doi.org/10.1093/cercor/bhx242.\nFougnie, D., Suchow, J. W., & Alvarez, G. A. (2012). Variability in the quality of visual working memory. Nature Communications, 3, 1228–1229, https://doi.org/10.1038/ncomms2237.\nGrogan, J. P. (2020). johnPGrogan/MemToolbox2D_Analysis: Updated release (Version v.2.0.0). Zenodo, http://doi.org/10.5281/zenodo.3752763.\nGrogan, J. P., Fallon, S. J., Zokaei, N., Husain, M., Coulthard, E. J., & Manohar, S. G. (2019). johnPGrogan/MemToolbox2D: Updated release (Version v.2.0.0). Zenodo, http://doi.org/10.5281/zenodo.3752705.\nLiang, Y., Pertzov, Y., Nicholas, J. M., Henley, S. M. D., Crutch, S., Woodward, F., & Husain, M. (2016). Visual short-term memory binding deficit in familial Alzheimer's disease. Cortex, 78, 150–164, https://doi.org/10.1016/j.cortex.2016.01.015.\nMa, W. J., Husain, M., & Bays, P. M. (2014). Changing concepts of working memory. Nature Neuroscience, 17(3), 347–356, https://doi.org/10.1038/nn.3655.\nMcNab, F., Zeidman, P., Rutledge, R. B., Smittenaar, P., Brown, H. R., Adams, R. A., & Dolan, R. J. (2015). Age-related changes in working memory and the ability to ignore distraction. Proceedings of the National Academy of Sciences, 112(20), 6515–6518, https://doi.org/10.1073/pnas.1504162112.\nParra, M. A., Abrahams, S., Logie, R. H., Méndez, L. G., Lopera, F., & Della Sala, S (2010). Visual short-term memory binding deficits in familial Alzheimer's disease. Brain, 133(9), 2702–2713, https://doi.org/10.1093/brain/awq148.\nPertzov, Y., Dong, M. Y., Peich, M. C., & Husain, M. (2012). Forgetting what was where: The fragility of object-location binding. PLoS ONE, 7(10), e48214, https://doi.org/10.1371/journal.pone.0048214.\nPertzov, Y., Heider, M., Liang, Y., & Husain, M. (2015). 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Attention, Perception, and Psychophysics, 76(7), 1902–1913, https://doi.org/10.3758/s13414-014-0723-2.\nRolinski, M., Zokaei, N., Baig, F., Giehl, K., Quinnell, T., Zaiwalla, Z., & Hu, M. T. M. (2016). Visual short-term memory deficits in REM sleep behaviour disorder mirror those in Parkinson's disease. Brain, 139(1), 47–53, https://doi.org/10.1093/brain/awv334.\nSchneegans, S., & Bays, P. M. (2016). No fixed item limit in visuospatial working memory. Cortex, 83, 181–193, https://doi.org/10.1016/j.cortex.2016.07.021.\nSchneegans, S., & Bays, P. M. (2017). Neural architecture for feature binding in visual working memory. Journal of Neuroscience, 37(14), 3913–3925, https://doi.org/10.1523/JNEUROSCI.3493-16.2017.\nSchneegans, S., & Bays, P. M. (2018). Drift in neural population activity causes working memory to deteriorate over time. Journal of Neuroscience, 38(21), 4859–4869, https://doi.org/10.1523/JNEUROSCI.3440-17.2018.\nSchwarz, G. (1978). Estimating the dimension of a model. Annals of Statistics, 6(2), 461–464.\nStark, S. M., Yassa, M. A., & Stark, C. E. L. (2010). Individual differences in spatial pattern separation performance associated with healthy aging in humans. Learning and Memory, 17(6), 284–288, https://doi.org/10.1101/lm.1768110.\nSuchow, J., Brady, T., Fougnie, D., & Alvarez, G. (2013). MemToolbox Tutorial.\nSuchow, J. W., Brady, T. F., Fougnie, D., & Alvarez, G. A. (2013). Modeling visual working memory with the MemToolbox. Journal of Vision, 13(2013), 1–8, https://doi.org/10.1167/13.10.9.doi.\nvan den Berg, R., Shin, H., Chou, W.-C., George, R., & Ma, W. J. (2012). Variability in encoding precision accounts for visual short-term memory limitations. Proceedings of the National Academy of Sciences of the United States of America, 109(22), 8780–8785, https://doi.org/10.1073/pnas.1117465109.\nWilken, P., & Ma, W. J. (2004). A detection theory account of change detection. Journal of Vision, 4(12), 11, https://doi.org/10.1167/4.12.11.\nZhang, W., & Luck, S. J. (2008). Discrete fixed-resolution representations in visual working memory. Nature, 453(7192), 233–235, https://doi.org/10.1038/nature06860.\nZhang, W., & Luck, S. J. (2009). Sudden death and gradual decay in visual working memory. Psychological Science, 20(4), 423–428, https://doi.org/10.1111/j.1467-9280.2009.02322.x.\nZokaei, N., Burnett Heyes, S., Gorgoraptis, N., Budhdeo, S., & Husain, M. (2015). Working memory recall precision is a more sensitive index than span. Journal of Neuropsychology, 9, 319–329, https://doi.org/10.1111/jnp.12052.\nZokaei, N., Čepukaitytė, G., Board, A. G., Mackay, C. E., Husain, M., & Nobre, A. C. (2019). Dissociable effects of the apolipoprotein-E (APOE) gene on short- and long-term memories. Neurobiology of Aging, 73, 115–122, https://doi.org/10.1016/j.neurobiolaging.2018.09.017.\nZokaei, N., Mcneill, A., Proukakis, C., Beavan, M., Jarman, P., Korlipara, P., & Husain, M. (2014). Visual short-term memory deficits associated with GBA mutation and Parkinson's disease. Brain, 137(8), 2303–2311, https://doi.org/10.1093/brain/awu143.\nZokaei, N., Nour, M. M., Sillence, A., Drew, D., Adcock, J., Stacey, R., & Husain, M. (2019). Binding deficits in visual short-term memory in patients with temporal lobe lobectomy. Hippocampus, 29(2), 63–67, https://doi.org/10.1002/hipo.22998.\nFigure 1.\n\nOne-dimensional and 2D misbinding models. The top row shows a 1D task. Fictive stimuli and responses are shown (left) where the target (blue) and nontarget (red) were always shown at the same orientations. The responses column shows sample responses centered on the target with some imprecision (blue), around the nontarget with the same imprecision (red), and randomly distributed guesses (green). The PDF of responses (right) shows peaks around those locations with heights proportional to the number of target and nontarget responses, widths proportional to the imprecision of the memories, and a background guessing rate proportional to the height of the horizontal line. The bottom row shows a similar setup for a 2D task where the target and nontarget were always shown in the same location on a 2D screen (left), and the PDF is now 2D (right), which shows peaks of responses (upper part) around those locations (bottom part) with a 2D width (dashed circle) and uniform guessing across the entire screen (green dots).\nFigure 1.\n\nOne-dimensional and 2D misbinding models. The top row shows a 1D task. Fictive stimuli and responses are shown (left) where the target (blue) and nontarget (red) were always shown at the same orientations. The responses column shows sample responses centered on the target with some imprecision (blue), around the nontarget with the same imprecision (red), and randomly distributed guesses (green). The PDF of responses (right) shows peaks around those locations with heights proportional to the number of target and nontarget responses, widths proportional to the imprecision of the memories, and a background guessing rate proportional to the height of the horizontal line. The bottom row shows a similar setup for a 2D task where the target and nontarget were always shown in the same location on a 2D screen (left), and the PDF is now 2D (right), which shows peaks of responses (upper part) around those locations (bottom part) with a 2D width (dashed circle) and uniform guessing across the entire screen (green dots).",
null,
"Figure 2.\n\nProcedure of the spatial WM task. Participants were presented a pair of test shapes (with a “T” at fixation), which could be either preceded or followed by a blank screen (Maintain T2 and T1, respectively) or by a second pair of nontest shapes (Update or Ignore, respectively). They were asked to remember the most recent pair of test shapes. At the test, one test shape and one nontest shape (novel item or distractor) were shown, and participants chose the item they remembered to have been a test shape and dragged it to where it appeared previously.\nFigure 2.\n\nProcedure of the spatial WM task. Participants were presented a pair of test shapes (with a “T” at fixation), which could be either preceded or followed by a blank screen (Maintain T2 and T1, respectively) or by a second pair of nontest shapes (Update or Ignore, respectively). They were asked to remember the most recent pair of test shapes. At the test, one test shape and one nontest shape (novel item or distractor) were shown, and participants chose the item they remembered to have been a test shape and dragged it to where it appeared previously.",
null,
"Figure 3.\n\nMean difference between recovered and simulated parameters (parameter recovery error). This is shown for 1D (top row) and 2D (bottom row) misbinding models. Dashed line shows the y = 0 line, which is where there is no difference between the simulation and recovered parameters for any simulation parameter value. Shaded area shows standard deviation. Each column shows one parameter value, marginalized across all other parameters from the parameter sweep. The 2D model performs better than the 1D model (t tests: p < 0.0001; see Supplementary Figures S4S6 for 1D models).\nFigure 3.\n\nMean difference between recovered and simulated parameters (parameter recovery error). This is shown for 1D (top row) and 2D (bottom row) misbinding models. Dashed line shows the y = 0 line, which is where there is no difference between the simulation and recovered parameters for any simulation parameter value. Shaded area shows standard deviation. Each column shows one parameter value, marginalized across all other parameters from the parameter sweep. The 2D model performs better than the 1D model (t tests: p < 0.0001; see Supplementary Figures S4S6 for 1D models).",
null,
"Figure 4.\n\nHeatmaps showing the recovered parameter trade-offs within the misbinding model in 1D (top) and 2D (bottom) space. The plots show the recovered parameter values at 10,000 MCMC posterior samples from fitting the model to simulated data (β = .3, γ = .3, σ = 30). Guessing correlates negatively with misbinding and imprecision, and misbinding and imprecision are positively correlated in both 1D and 2D, but in all cases, the correlations are weaker in the 2D model, shown here by more circular, less elliptical, patterns.\nFigure 4.\n\nHeatmaps showing the recovered parameter trade-offs within the misbinding model in 1D (top) and 2D (bottom) space. The plots show the recovered parameter values at 10,000 MCMC posterior samples from fitting the model to simulated data (β = .3, γ = .3, σ = 30). Guessing correlates negatively with misbinding and imprecision, and misbinding and imprecision are positively correlated in both 1D and 2D, but in all cases, the correlations are weaker in the 2D model, shown here by more circular, less elliptical, patterns.",
null,
"Figure 5.\n\nThe absolute recovery error for different numbers of trials for 1D (top) and 2D (bottom) misbinding models. More trials (warmer colors—see color bar) leads to more accurate parameters (lower absolute recovery errors) in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error of the mean.\nFigure 5.\n\nThe absolute recovery error for different numbers of trials for 1D (top) and 2D (bottom) misbinding models. More trials (warmer colors—see color bar) leads to more accurate parameters (lower absolute recovery errors) in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error of the mean.",
null,
"Figure 6.\n\nAbsolute parameter recovery errors for different numbers of nontargets. More nontargets (cooler colors) leads to higher absolute recovery errors in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error.\nFigure 6.\n\nAbsolute parameter recovery errors for different numbers of nontargets. More nontargets (cooler colors) leads to higher absolute recovery errors in 1D and 2D models (linear regression: p < 10−10). Shading shows standard error.",
null,
"Figure 7.\n\nParameter recovery errors (top) and absolute recovery errors (bottom) for the 2D model in simulated tasks with and without stimulus constraints. Adding in constraints to the stimulus locations leads to smaller errors in parameter recovery (orange line; see text for statistics). Shading shows standard error.\nFigure 7.\n\nParameter recovery errors (top) and absolute recovery errors (bottom) for the 2D model in simulated tasks with and without stimulus constraints. Adding in constraints to the stimulus locations leads to smaller errors in parameter recovery (orange line; see text for statistics). Shading shows standard error.",
null,
"Figure 8.\n\nParameter recovery errors from simulations with a constant translation bias subtracted (top) and corrected for (bottom). The color bar shows bias size (from 20 to 200 pixels). The bias correction works well for biases up to about 150 pixels in size. Shading shows standard error.\nFigure 8.\n\nParameter recovery errors from simulations with a constant translation bias subtracted (top) and corrected for (bottom). The color bar shows bias size (from 20 to 200 pixels). The bias correction works well for biases up to about 150 pixels in size. Shading shows standard error.",
null,
"Figure 9.\n\nDifference plot of recovered and simulated parameter values for different proportional biases toward the right-hand screen edge. Data with (bottom) and without (top) the bias correction for different proportional biases (color bar). Positive biases mean responses move toward the right-hand screen edge, and negative biases move responses away from this edge (responses do not move vertically). The bias correction improves the fits for all parameters. Shading shows standard error.\nFigure 9.\n\nDifference plot of recovered and simulated parameter values for different proportional biases toward the right-hand screen edge. Data with (bottom) and without (top) the bias correction for different proportional biases (color bar). Positive biases mean responses move toward the right-hand screen edge, and negative biases move responses away from this edge (responses do not move vertically). The bias correction improves the fits for all parameters. Shading shows standard error.",
null,
"Figure 10.\n\nDifference plot of recovered and simulated parameter values for different radial biases. Data with (bottom) and without (top) the bias correction for different radial biases (color bar). Positive radial biases mean points move toward the center of the screen, and negative biases move points away from the center. The bias correction improves the fits for all parameters, although large guess rate parameters are not fully corrected when there is a large expanding bias. Shading shows standard error.\nFigure 10.\n\nDifference plot of recovered and simulated parameter values for different radial biases. Data with (bottom) and without (top) the bias correction for different radial biases (color bar). Positive radial biases mean points move toward the center of the screen, and negative biases move points away from the center. The bias correction improves the fits for all parameters, although large guess rate parameters are not fully corrected when there is a large expanding bias. Shading shows standard error.",
null,
"Figure 11.\n\nParameter recovery errors with a radial bias toward the nontarget location on each trial. Data shown without (top) and with (bottom) bias correction. The bias correction reduced the errors in parameter recovery. This simulation had β = 0 (i.e., no misbinding). Shading shows standard error.\nFigure 11.\n\nParameter recovery errors with a radial bias toward the nontarget location on each trial. Data shown without (top) and with (bottom) bias correction. The bias correction reduced the errors in parameter recovery. This simulation had β = 0 (i.e., no misbinding). Shading shows standard error.",
null,
"Figure 12.\n\nIllustrative examples of mixture modeling approach to change detection. We show 1D examples for ease, although the same principles apply for 2D (see Supplementary Figure S10). The top row shows how each error component affects the proportion of generative responses that would fall at or beyond a probed distance of 60 degrees (arbitrary y-units). The second row shows the values of the parameters used for those simulations, along with the summed probabilities of all responses that occur at ≥ 60 degrees from the target. Increasing each parameter increases the probability of incorrectly accepting this probe as unchanged. The bottom row shows the probability of accepting a probe as unchanged as a function of the probe distance (or probe-nontarget distance for misbinding). This shows that increasing imprecision increases the spread of these same responses around the target, increasing guessing affects all orientations similarly, and increasing misbinding affects responses in relation to their distance to the distractor (if we plotted probe-targ for misbinding, it would look like the guessing panel).\nFigure 12.\n\nIllustrative examples of mixture modeling approach to change detection. We show 1D examples for ease, although the same principles apply for 2D (see Supplementary Figure S10). The top row shows how each error component affects the proportion of generative responses that would fall at or beyond a probed distance of 60 degrees (arbitrary y-units). The second row shows the values of the parameters used for those simulations, along with the summed probabilities of all responses that occur at ≥ 60 degrees from the target. Increasing each parameter increases the probability of incorrectly accepting this probe as unchanged. The bottom row shows the probability of accepting a probe as unchanged as a function of the probe distance (or probe-nontarget distance for misbinding). This shows that increasing imprecision increases the spread of these same responses around the target, increasing guessing affects all orientations similarly, and increasing misbinding affects responses in relation to their distance to the distractor (if we plotted probe-targ for misbinding, it would look like the guessing panel).",
null,
"Figure 13.\n\nParameter recovery errors for the 1D and 2D two-alternative forced-choice simulations. The 2D model (bottom) is more accurate than the 1D model (top), although both are far less accurate than with continuous report data (see Figure 3). Shading shows the standard error of the mean.\nFigure 13.\n\nParameter recovery errors for the 1D and 2D two-alternative forced-choice simulations. The 2D model (bottom) is more accurate than the 1D model (top), although both are far less accurate than with continuous report data (see Figure 3). Shading shows the standard error of the mean.",
null,
"Figure 14.\n\nHistograms of the recovered parameters from the ignore/update task, across all conditions.\nFigure 14.\n\nHistograms of the recovered parameters from the ignore/update task, across all conditions.",
null,
"Figure 15.\n\nMean fitted parameter values for each trial type from the healthy participants. Delay increased imprecision (σ; top left; blue vs. orange; see Figure 2 for task explanation), guessing (γ), and misbinding (β; see text for stats); irrelevant stimuli increased misbinding and guessing but not imprecision (ignore and update vs. T1 and T2); and there was an interaction of delay and irrelevant stimuli only for guessing. Standard error bars.\nFigure 15.\n\nMean fitted parameter values for each trial type from the healthy participants. Delay increased imprecision (σ; top left; blue vs. orange; see Figure 2 for task explanation), guessing (γ), and misbinding (β; see text for stats); irrelevant stimuli increased misbinding and guessing but not imprecision (ignore and update vs. T1 and T2); and there was an interaction of delay and irrelevant stimuli only for guessing. Standard error bars.",
null,
"Figure 16.\n\nHeatmaps showing the posterior MCMC samples for recovering the mean parameters from Fallon et al. (2018). Each color is a different condition (see legend; Ig = ignore, T1 = long delay, Up = update, T2 = long delay), and the rows show 1D and 2D models, respectively. The γ and β parameters have been arcsine transformed. Both models were simulated using the same parameters. The 2D model has better separation of the different conditions’ parameters, as shown by less overlap between the colors.\nFigure 16.\n\nHeatmaps showing the posterior MCMC samples for recovering the mean parameters from Fallon et al. (2018). Each color is a different condition (see legend; Ig = ignore, T1 = long delay, Up = update, T2 = long delay), and the rows show 1D and 2D models, respectively. The γ and β parameters have been arcsine transformed. Both models were simulated using the same parameters. The 2D model has better separation of the different conditions’ parameters, as shown by less overlap between the colors.",
null,
"Figure 17.\n\nRelationship between the true parameters (x-axis) and the best behavioral metrics (top) and recovered parameters (bottom). All correlations were significant (p < 10−10). Two metrics were compared against the imprecision (σ) parameter. The recovered parameters had stronger correlations than the best behavioral metrics (see text for statistics).\nFigure 17.\n\nRelationship between the true parameters (x-axis) and the best behavioral metrics (top) and recovered parameters (bottom). All correlations were significant (p < 10−10). Two metrics were compared against the imprecision (σ) parameter. The recovered parameters had stronger correlations than the best behavioral metrics (see text for statistics).",
null,
"Table 1.\n\nSpearman correlation coefficients between each true parameter and the recovered parameters for the 1D and 2D misbinding models. ***p < 10−10.\nTable 1.\n\nSpearman correlation coefficients between each true parameter and the recovered parameters for the 1D and 2D misbinding models. ***p < 10−10.",
null,
"×\n\nThis PDF is available to Subscribers Only"
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http://www-math.ucdenver.edu/~wcherowi/courses/m4408/gtaln7.html | [
"## Graph Theory Lecture Notes 7\n\n### Counting Spanning Trees\n\nTwo labeled trees are the same if their edge-sets are identical.\n\nThe number of spanning trees of Kn is nn-2, for n",
null,
"2, this is Cayley's Formula.\n\nPrüfer Encoding\n\nA Prüfer sequence of length n-2, for n",
null,
"2, is any sequence of integers between 1 and n, with repetitions allowed.\n\nA labeled tree gives a Prüfer sequence by:\n\n• Repeat n-2 times:\n• Pick the 1-valent vertex with the smallest label, call it v.\n• Put the label of v's neighbor in the output sequence.\n• Remove v from the tree.\nPrüfer Decoding\n\nA Prüfer sequence determines a labeled tree by:\n\n• Let L be the ordered list of numbers 1, 2, ..., n. Let P be the Prüfer sequence. Start with n labeled isolated vertices.\n• Repeat n-2 times:\n• Let k be the smallest number in L which is not in P.\n• Let j be the first number in P.\n• Add the edge kj to the graph.\n• Remove k from L and the first number in P.\n• When this is completed there will be two numbers left in L, add the edge corresponding to these two numbers.\nPrüfer coding and decoding are inverse operations, that means that there is a one-to-one correspondence between labeled trees with n vertices and Prüfer sequences of length n-2.\n\nCayley's formula now follows from counting Prüfer sequences of length n-2.\n\n### Finding Minimum Spanning Trees\n\nLet G be a connected graph with weights assigned to its edges. We wish to find the spanning tree with the smallest sum of weights.\n\nPrim's Algorithm: Grow a tree by picking the frontier edge with the smallest weight.\n\nThis is an example of a greedy algorithm.\n\n### Finding the Shortest Path\n\nLet G be a connected graph with weights assigned to its edges. We wish to find for any two vertices s and t, the path from s to t whose total edge-weight is minimum.\n\nIf the weights are all equal, the problem reduces to finding the s-t path of shortest length. This can be done by using a breadth-first search starting at s, and ending when t is reached.\n\nWhen the weights are non-negative, but not all equal we can use Dijkstra's algorithm (for situations where negative weights are used, Floyd's algorithm can be used).\n\nDijkstra's Algorithm\n\nIn this tree growing algorithm, vertices that are added to the tree are labeled with their \"distance\" from the starting vertex s. For a given vertex x, this label will be denoted dist[x]. Start by setting dist[s]:=0. At each step of the tree growing process, for each frontier edge e, calculate P(e) = dist[x] + wt(e), where x is the vertex of e that is in the tree (and therefore has been labeled) and wt(e) is the weight of edge e. Choose the frontier edge with the smallest P(e) value to add to the tree and label the new vertex with P(e)."
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https://autotydzien.pl/stephanie-stremler-eidzx/concurrent-lines-condition-determinant-389acd | [
"Get a quick overview of Tips to solve problems on concurrency of lines from Family of Straight Lines - Concurrent Lines and Conditions for Concurrent and Parallel Lines in just 2 minutes. This instability is more subtle and may be difficult to spot. (The necessary condition for concurrent sourcing ever to occur is that ... A key determinant in the make-or-buy decision is the price p at which the product can be bought versus the cost at which it can be produced. JEE Main 2019: Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. This program initiates requests for web pages and accepts the responses concurrently as the results of the downloads become available, accumulating a set of pages that have already been … These lines are sid to be concurrent if the following condition holds good: = 0. Lernen Sie die Übersetzung für 'concurrent' in LEOs Englisch ⇔ Deutsch Wörterbuch. Von Neumann and Goldstine (20) proposed the use Amnx of the so-called condition number, defined by X.n , 1. Thus, if three lines are concurrent the point of intersection of two lines lies on the third line. That is, IA.N < < 1 (3.3) Other quantitative measures of ill-conditioned-ness have also been discussed in literature. Condition for Concurrency of Three Straight Lines. a 1 x + b 1 y + c 1 = 0. a 2 x + b 2 y + c 2 = 0. a 3 x + b 3 y + c 3 = 0. is the condition for the three straight lines to be concurrent. Let two line-segments are given. Problems Related to Concurrent Lines 2. Three straight lines are said to be concurrent if they passes through a point i.e., they meet at a point. 2 mins read. ${a_2}{b_1}x + {b_1}{b_2}y + {b_1}{c_2} = 0\\,\\,\\,\\,{\\text{ – – – }}\\left( {{\\text{iv}}} \\right)$, Now subtracting (iv) from equation (iii), we get $\\Rightarrow x = \\frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {a_2}{b_1}}}$, Multiplying equation (i) by $${a_2}$$, we have (A) The lines Two lines are said to be coplanar when they both lie on the same plane in a three-dimensional space. Concurrent Lines Definition. The purpose of this report is to provide information to the public health sector, to clinicians and to policy makers about trends in … Your email address will not be published. If then the pair of linear equations has exactly one solution. We also tested four additional lines with either high or low TLR4 expression. They do not have any common transverse line. • There is a reciprocal screw, which is an unwanted twist motion. Also find the point of concurrence See Centers of a triangle. 4 reactions with non-concurrent or parallel lines of action – properly constrained, statically indeterminate, Mit Flexionstabellen der verschiedenen Fälle und Zeiten Aussprache und relevante Diskussionen Kostenloser Vokabeltrainer 8.Concurrent lines : If three given lines are concurrent they must be meet in a common point. Above condition can be written in a determinant form as 111 222 333 a bc a bc a bc = 0. Mit Flexionstabellen der verschiedenen Fälle und Zeiten Aussprache und relevante Diskussionen Kostenloser Vokabeltrainer Results may be inaccurate. A determinant with two equal columns is zero which is only a very particular case of a much more general statement. 7 Like 0 Dislike. Thus, concurrency is the plane dual notion to collinearity. I'm trying to prove the concurrency condition for three lines lying on a plane. 23. Therefore, the value of p is 4. navneet863 navneet863 Step-by-step explanation: The conditions of lines to be concurrent is that, the determinant of these lines forming a matrix should be equal … A set of lines or curves are said to be concurrent if they all intersect. 13. $\\begin{gathered} {a_3}\\left( {\\frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {a_2}{b_1}}}} \\right) + {b_3}\\left( {\\frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}}} \\right) + {c_3} = 0 \\\\ \\Rightarrow {a_3}\\left( {{b_1}{c_2} – {b_2}{c_1}} \\right) – {b_3}\\left( {{a_1}{c_2} – {a_2}{c_1}} \\right) + {c_3}\\left( {{a_1}{b_2} – {a_2}{b_1}} \\right) = 0 \\\\ \\end{gathered}$, This can be written in determinant form: It enables a limited set of concurrency-safe operations, such as push and try_pop. Answer the following question: Show that the lines x − y = 6, 4x − 3y = 20 and 6x + 5y + 8 = 0 are concurrent. The condition for lines to be concurrent is that, the determinant of these lines forming a matrix should be equal to 0. To use determinants to solve a system of three equations with three variables (Cramer's Rule), say x, y, and z, four determinants … 1 will be in GLCSC when the upper mobile rotates π/2 about the z-axis.The six lines associated with links A 1 B 1, A 2 B 1 … A 6 B 5 are linearly dependent. When the app performs operations concurrently, users wait less for the same result. Method to find concurrency— Method 1. … Tips to solve problems on concurrency of lines. Syntax template"
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http://www.leastcommonmultiple.net/lcm-of-931/ | [
"X\nX\n\n# Calculate the Least Common Multiple or LCM of 931\n\nThe instructions to find the LCM of 931 are the next:\n\n## 1. Decompose all numbers into prime factors\n\n 931 7 133 7 19 19 1\n\n## 2. Write all numbers as the product of its prime factors\n\n Prime factors of 931 = 72 . 19\n\n## 3. Choose the common and uncommon prime factors with the greatest exponent\n\nCommon prime factors: 7 , 19\n\nCommon prime factors with the greatest exponent: 72, 191\n\nUncommon prime factors: None\n\nUncommon prime factors with the greatest exponent: None\n\n## 4. Calculate the Least Common Multiple or LCM\n\nRemember, to find the LCM of several numbers you must multiply the common and uncommon prime factors with the greatest exponent of those numbers.\n\nLCM = 72. 191 = 931\n\nAlso calculates the:"
] | [
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http://niss-curriculum.com/page/curriculum/high/clark/mathematics/prealgebra/prealgebra.php | [
"# Pre-Algebra\n\nCurriculum > High > Clark > Mathematics\n• Syllabus\n• Sample Lesson Plan\nGeneral Information\nCourse hours: 2:20 PM – 3:10PM (Mon‐Thu‐Fri) / 11:05 AM – 12:30 PM (Wed)\nCourse Description\nPre‐Algebra is a full‐year, two‐semester secondary‐level math course. The main goal of this course is to provide students with a thorough and extensive study of linear and quadratic functions and graphing on the xy‐coordinate system. By the end of this course, students will have all the knowledge necessary to solve and graph equations and inequalities. They will also be able to apply this knowledge to other areas of math, such as word problems, ratios and proportions.\nThe course starts off with a quick review of basic algebraic concepts, such as variables, order of operations, exponents and problem solving skills. They will gain a thorough introduction to functions, the basis of all of algebra and higher mathematics, such as calculus. Students will learn how to solve linear equations, including multistep equations, equations with multiple variables and equations involving decimals, as well as write a linear equation based on the graph of a line. The Algebra 1 course also gives the students a thorough introduction of functions and quadratic equations. They will learn all the operations associated with the two, such as factoring and graphing. The end of the course takes the students’ knowledge of algebra 1 one step further by introducing them to some basic concepts of geometry.\nPre‐Requisites\nStudents should successfully complete 6th grade level mathematics prior to entering the course. Only students in 7th and 8th grade are permitted to take the Pre‐Algebra.\nLearning Objectives\nAt the end of this course students will be able to\n1. Write and evaluate variable expressions.\n2. Use the order of operations.\n3. Add, subtract, multiply and divide integers.\n4. Write rules for patterns.\n5. Graph points in the coordinate plane.\n6. Identify and use properties of addition and multiplication and the Distributive Property.\n7. Simplify expressions.\n8. Solve one‐step equations and inequalities.\n9. Round decimals.\n10. Find mean, median and mode of a set of data.\n11. Substitute into formulas.\n12. Convert metric and customary units.\n13. Use divisibility tests.\n14. Find the prime factorization of a number.\n15. Find the greatest common factor and least common multiple.\n16. Write fractions in simplest form.\n17. Simplify expressions with integer exponents.\n18. Write and evaluate numbers in scientific notation.\n19. Write fractions as decimals and write terminating and repeating decimals as fractions.\n20. Add, subtract, multiply and divide fractions and mixed numbers.\n21. Solve proportions.\n22. Find probability and odds.\n23. Write percent as fractions and decimals and write decimals and fractions as percent.\n24. Find a part of a whole, a percent and a whole amount.\nTextbook\nPrentice Hall, Mathematics\nPre‐Algebra ISBN: 978‐0‐13‐365945‐0\nSupplies\nStudents are required to bring the following supplies to class every day of the year:\nNotebook and Pencil\nBound with at least two subjects for note taking.\nPen is acceptable, pencil is preferred.\nBefore beginning, students will take a Learning Outcomes Assessment. This pretest will help the student and instructor determine how to focus efforts to successfully complete this course. Each week students will be expected to do the required reading in the textbook or interactive textbook, explore the topic lectures which include illustrated examples and interactive activities, complete the assignments by solving problems, attend at least one of the two onehour chat sessions to communicate both verbally and in writing with the instructor and other students synchronously, answer questions asynchronously on the discussion board, and demonstrate understanding of the week’s material through a multiple choice and free‐response test. At the end of the course, students will complete a final exam and submit two projects of their choice. The following is the point distribution that will determine a student’s course grade:\nProcedures\nWhat to do (and not to do) in class:\nBeing on time to class means you are ready to learn\nYou should be in your seat with materials out by the time the bell rings. Passing time is 10 minutes, which provides plenty of time to use the restroom, socialize with friends and be to class on time.\n\nMissed class\nYou are responsible for material covered when you are absent. If you have an unexcused absence when an assignment is due, you will receive a zero grade. For excused absences, you are given two make‐up days for every day you are absent. This means that if you miss class on Monday and return on Wednesday, the work you missed on Monday is due Friday. Missed labs must be made up – see the instructor to arrange a time to perform the experiment. The most common reason for students receiving a zero on an assignment is that they did not make it up when they were absent. Use ADD SCHOOL GRADING SYSTEM as a tool to know which assignments are missing. I do not chase down students or remind them that they have missed work when they are absent. Students are responsible for ensuring that assignments are completed on time.\n\n*****NO LATE WORK WILL BE ACCEPTED*****\nHomework and labs are due at the beginning of class on the assigned due date. Homework and labs handed in after this are late and will receive a zero grade.\n\nCell phones\nCell phones are not allowed in class. You must be turned off and stow your phone during class. If I catch you using your cell phone during class, I will keep your phone until 3:05 pm that day.\n\nPlagiarism& Cheating\nStudents involved in any type of plagiarism – either submitting answers not their own that they don’t understand, supplying answers to other students, copying another student’s work, or making up lab data – will be given a zero for the assignment.\nINSTRUCTOR will, under no circumstances, write a letter of recommendation for any student who is involved with plagiarism in her class, regardless of the size of the infraction.\nUnits\nRatio and Converting Unit Rates\nProportions and word problems involving proportions\nFractions, Decimals and Percent\nProportions, Percent and Equations\nSolving Equations in Two variables\nSolving and graphing Inequalities\nRelations and Functions, and their graph.\nSlope and Y‐ Intercept.\nReview and Midterm exams.\nSolving Systems of Linear Equations by graphing and Graphing Linear Inequalities. Spatial Thinking – Introduction to Geometry: Points, Lines, and Planes\nAngles – Angle relationship and Parallel lines\nClassifying Polygons and Congruence and Circles\nTranslation, Symmetry and Reflections, and Rotations\nArea and Volume\nReview and Final Exams.\nQuestions?\nParents, guardians, and host parents: If you have any questions, comments, or concerns, please feel free to contact me! E‐mail is the best way to reach me. I am happy to answer questions via e‐mail, meet with you in person, or have a phone conversation.\nStudents: I want you to succeed in Pre‐Algebra! If you have questions, are struggling with the course material, are concerned about your grade, or have any other concerns, please come talk to me sooner rather than later. I am happy to find a time to meet with you before school, after school, or during lunch.\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the difference of variables and expressions. They will also know the how to solve algebraic equations in order.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson\nMeet and greet. The teacher will also show the syllabus that they will be using. He will explain the grading method and what to expect in the class.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\n‐ Variables and expressions\n‐ Teacher will introduce some math concepts including variables and variable expressions, and demonstrate how to write variable expressions.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMain Activity/Lesson\nContinue with the variables and expressions. The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMain Activity/Lesson\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nStudy for the next lesson.\nMaterials / Resources (including technology)\nBook.\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the difference of variables and expressions. They will also know the how to solve algebraic equations in order.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson\n‐ The order of operations\n‐ Teacher will review the concept of operations and explain the rules of the order of operations, and grouping symbols.\n‐ Students also learn how to simplify expressions.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMain Activity/Lesson\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nreview for a test\nMain Activity/Lesson\nThe teacher will give a test and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nStudy for the next lesson.\nMaterials / Resources (including technology)\nBook.\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the difference of variables and expressions. They will also know the how to solve algebraic equations in order.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson\n‐ Writing and Evaluating Expressions\n‐ Teacher will introduce the concept of evaluation and demonstrate how to replace one or more variables. Students will discuss how math relationships can help us solve real world problems.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMain Activity/Lesson\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson\n‐ Integers and Absolute Value\n‐ Teacher will introduce concepts of positive and negative integers, opposites and absolute value, and then demonstrate how to find absolute values on the number line.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMaterials / Resources (including technology)\nBook.\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the difference of variables and expressions. They will also know the how to solve algebraic equations in order.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMain Activity/Lesson\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\n‐ Adding Integers ‐ Teacher will introduce rules for adding integers and students will try to apply the rules. Teacher will also review what has been taught this week.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMaterials / Resources (including technology)\nBook.\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know adding integers, subtracting integers, inductive reasoning, looking for patterns, multiplying and dividing intergers.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nReview for a test\nMain Activity/Lesson\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\nThe teacher will give a test and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson\n‐ Subtracting Integers\n‐ Teacher will review adding integers and introduce subtracting integers. Students will practice subtracting in class and figure out the relationship between adding and subtracting.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMain Activity/Lesson\n‐ Continue with the lesson. ‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMaterials / Resources (including technology)\nBook.\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know adding integers, subtracting integers, inductive reasoning, looking for patterns, multiplying and dividing integers.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson\n‐ Inductive reasoning\n‐ Teacher will introduce concepts of inductive reasoning and conjectures, then let students work in groups to design a pattern, and then lead a class discussion.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson\n‐ The teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMaterials / Resources (including technology)\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know adding integers, subtracting integers, inductive reasoning, looking for patterns, multiplying and dividing integers.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Look for a pattern\n‐ Teacher will demonstrate how to find and use patterns, and then let students practice with some math or real world problems.\n‐ Students will work in groups to find out the patterns and each group will gain a point from a right answer.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Multiplying and dividing integers\n‐ Teacher will introduce the rules of multiplying and dividing integers and let students practice with them. If time allows, teacher will review concepts they have learned the first 2 weeks and check students' notebooks.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know adding integers, subtracting integers, inductive reasoning, looking for patterns, multiplying and dividing integers.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nThe teacher will give a seatwork and will show and explain the solutions.\nEvaluation:\nVocabulary:\nHomework:\nReview\nMain Activity/Lesson:\nThe teacher will review the students for the whole Chapter.\nEvaluation:\nVocabulary:\nHomework:\nreview for a test\nMain Activity/Lesson:\nThe teacher will give a chapter test.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know adding integers, subtracting integers, inductive reasoning, looking for patterns, multiplying and dividing integers.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ The coordinate plane\n‐ Teacher will introduce coordinate plane and demonstrate how to name coordinates and quadrants, as well as how to graph points.\n‐ Students will practice all these in the class.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ The coordinate plane and chapter review\n‐ Students will learn concepts of coordinate plane, x‐axis, y‐axis and quadrant.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know properties of numbers, distributive property, simplifying variables, variables and equations. They will also learn decimals and equations.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\nThe students with the help of the teacher will review chapter 1.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nThe students will have a chapter test.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\nEvaluation:\nThe teacher will show the solution and explain the problems to the students.\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Properties of numbers\n‐ Students will start to learn Unit 2 \"solving one‐step equations and inequalities\".\n‐ Teacher will introduce properties of numbers.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know properties of numbers, distributive property, simplifying variables, variables and equations. They will also learn decimals and equations.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ The distributive property\n‐ Teacher will introduce concept of distributive property and demonstrate some examples, then students will do some exercises in the class as well as some homework after class.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Simplifying variable expressions\n‐ Teacher will introduce different parts of a variable expression and demonstrate the steps of simplifying, then students will do some\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know properties of numbers, distributive property, simplifying variables, variables and equations. They will also learn decimals and equations.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve basic algebraic equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Variables and equations\n‐ Teacher will introduce concepts of equation and open sentence, and demonstrate how to check equations using substitution. Students will practice with class exercises and homework.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Solving equations by adding or subtracting\n‐ Teacher will demonstrate how to solve equations by adding or subtracting, and students will understand inverse operations.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nBook.\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Students will identify parts of a variable expression; use knowledge of properties of numbers to simplify variable expressions.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Students will classify types of equations; check solution veracity using subtraction.\n‐ Demonstration and evaluation pages 84‐85, #s 1‐43, odd.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nBook.\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Teacher will review 2‐5 and 2‐6, and introduce concepts of guess, check and revise with some math examples.\n‐ Students will try to solve some related math problems in the class.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n( No Class )\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Inequalities and their graphs\n‐ After reviewing the homework, teacher will introduce graphing inequalities, and writing inequalities, students will work in groups to design some graphing and inequalities for other groups to solve.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Teacher will introduce solving one‐step inequalities by adding, subtracting, multiplying and dividing, and students will also practice more than one way to solve a problem.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Review of Chapter 2\n‐ Teacher will review the new concepts and knowledge in chapter 2, and students will have some time to practice and ask questions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ The students will have a chapter test.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Rounding and estimating\n‐ Teacher will introduce how to round decimals and estimate sums and differences, and students will practice rounding and different ways to estimate in the class.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\n‐ Estimating decimal products & quotients\n‐ Teacher will review estimating sums and differences, and introduce how to estimate decimal products and quotients.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Mean, median and mode\n‐ Teacher will introduce concepts of mean, median and mode, and students will practice them with math examples.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Simplifying fractions\n‐ Teacher will introduce concepts of equivalent fractions and how to write fractions in simplest form.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book. (Solve a simpler problem)\n‐ Teacher will demonstrate how to solve complex problems by first solving simpler cases, and let students practice with some real world problems.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Rational numbers\n‐ Teacher will introduce concepts of rational number and the relationship among rational numbers, and students will learn how to graph rational number and evaluate fractions with variables.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Exponents and multiplication\n‐ Teacher will explain the rules in multiplying powers with the same base and finding a power of a power, then students will practice with the rules in the class.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson. The teacher will give more examples and ask the students to solve the problems in the book. (Exponents and division)\n‐ Teacher will explain the rules in dividing expressions containing exponents and how to simplify expressions with integer exponents, then students will practice with the rules in the class.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Scientific notation\n‐ Teacher will introduce concepts of scientific notation and standard notation, and how to convert between scientific notation and standard notation, then students will practice with examples.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will continue to know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to better work know better how to use their knowledge of properties of numbers to more efficiently simplify numerical and variable expressions and equations; begin focusing more on equations than simple expressions, variable equations rather than simple numerical expressions and equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Scientific notation\n‐ Teacher will introduce how to compare and order numbers, and how to calculate with scientific notations, then students will practice with examples.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\n‐ Chapter review\n‐ Teacher will have chapter 4 review today, and students will have some time for Q&A\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\nStudents will have a chapter test.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Teacher will review what students have learned in the first 4 chapters and introduce chapter 5: operations with fractions. ‐ Students will have an idea of what they will learn in chapter 5.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the operations with fractions and ratios and proportions.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Comparing and ordering rational numbers ‐ Students will learn concept of LCM and LCD, and how to compare and order rational numbers, then they will apply the rules in their class work.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\n‐ Teacher will lead the students to work backward in solving some real world problems, and explain how to check the answer.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Solving equations by adding and subtracting fractions\n‐ Teacher will introduce how to solve equations by adding and subtracting fractions, and compare them with adding and subtracting fractions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the operations with fractions and ratios and proportions.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Solving equations by multiplying fractions\n‐ Teacher will introduce how to solve equations by multiplying fractions, and compare them with multiplying fractions.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Powers of products and quotients\n‐ Teacher will explain and demonstrate how to simplify powers of products and quotients, and students will practice the rules with arithmetic and algebra expressions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the operations with fractions and ratios and proportions.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Ratios and unit rates\n‐ Teacher will introduce concepts of ratio, rate and unit rate, students will also learn how to convert a rate.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Teacher will introduce concepts of proportion, cross products, similar figures, and indirect measurement, and demonstrate how to solve proportions and use similar figures.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the operations with fractions and ratios and proportions.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Teacher will introduce concepts of outcomes, event, probability and complement, and demonstrate how to find probability and odds.\n‐ Students will work in groups and discuss probability of some events in their daily life.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Students will learn concept of percent, and understand the differences among ratio, rate, proportion, fraction and percent, they will also learn how to find part of a whole and a whole amount.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the operations with fractions and ratios and proportions.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Fractions, decimals and percent.\n‐ Teacher will introduce concept of percent and how to convert among fractions, decimals and percent. Students should be able to write a rational number in different forms and order them correctly.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Teacher will demonstrate how to solve a percent problem by writing and solving proportions, and students should be able to find part of a whole, a percent or a whole amount by solving a percent problem.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know the operations with fractions and ratios and proportions.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Teacher will introduce the expression to calculate percent of change, and students will learn how find percent of increase and decrease.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Teacher will introduce the concepts of markup and discount, and demonstrate how to calculate markups and discounts, as well as solve a problem by making a table.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Introduction to chapter 7.\n‐ Teacher will introduce main content in chapter 7 as well as 7‐1.\n‐ Teacher will review one step equation and demonstrate how to solve a two‐step equation.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Teacher will demonstrate how to solve a multi‐step equation and students are divided into different groups and design multi‐step equations.\n‐ Groups exchange their questions and solve the questions they get.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Write an equation.\n‐ Teacher will demonstrate an example of writing an equation to represent a problem, then let students work in groups and design short answer questions for other groups, later each group will need to write and solve a question designed by other groups.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Solving equations with variables.\n‐ Teacher will demonstrate the steps to solve an equation with variables on both sides, and then let students practice with quick check questions to enhance their understanding.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Teacher will compare solving 2 step inequalities and 2 step equations, emphasizing when the inequality symbol should be reversed. Students will check their understanding using quick check question 2.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Transforming formulas.\n‐ Teacher will use an example to demonstrate how to transform formulas to represent different variables in a formula, and let students do exercise to check their understanding.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Simple and compound interest.\n‐ Teacher will introduce and explain some basic concepts concerning interest, and demonstrate how to calculate simple and compound interest, then let students work in pairs to do some calculations.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Teacher will introduce what the students will learn in chapter 8: Linear functions and graphing, and explain what prior knowledge they should have. Teacher will also explain some new concepts in 8‐1 so students could understand relations and functions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Simple and compound interest.\n‐ Teacher will introduce and explain some basic concepts concerning interest, and demonstrate how to calculate simple and compound interest, then let students work in pairs to do some calculations.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Teacher will introduce what the students will learn in chapter 8: Linear functions and graphing, and explain what prior knowledge they should have.\n‐ Teacher will also explain some new concepts in 8‐1 so students could understand relations and functions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Equations with two variables.\n‐ Teacher will review how to find points on coordinates and introduce concepts of solution and linear equation, then demonstrate how to find solutions of 2 variable equations and graph them.\n‐ Students will practice with exercises.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Slope and y‐intercept.\n‐ Teacher will introduce concepts of slope and y‐intercept, and demonstrate how to find slope and y‐intercept using coordinates.\n‐ Students will also learn how to do the opposite way: graph linear equations using slope.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know solving equations and inequalities. They will also learn linear functions and graphing and spatial thinking.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Solving by graphing.\n‐ Teacher will review concept of scatter plot and explain why people could use trend line to solve some problems.\n‐ Students will need to draw a scatter plot and a trend line based on the data teacher gives.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Solving systems of linear equations.\n‐ Teacher will explain what is solution of a system and how to use graph to find out solution of a system, also introduce some special systems.\n‐ Students will need to write equations based on words and find out solutions of the system.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know area and volume. They will also learn about right triangles in algebra.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Teacher will introduce chapter 9: spatial thinking, and ask students to demonstrate their geometry knowledge on some basic shapes like triangle, rectangle and circle, then explain some concepts such as point, line, plane, intersecting, parallel and skew lines.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Angle relationships and parallel lines.\n‐ Teacher will draw some shapes and introduce adjacent angles, vertical angles, congruent angles, supplementary and complementary.\n‐ Students will practice their understanding on these concepts and try to identify different angles in their exercises.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nBook.\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know area and volume. They will also learn about right triangles in algebra.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Classifying polygons\n‐ Teacher will review the angle related concepts students have learned the day before, and use card board to show different polygons such as triangle and quadrilateral.\n‐ Students will work in pairs and try to match different shapes with their names.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Draw a diagram\n‐ Teacher will review some concepts with the class at the beginning, and then hand out a worksheet so students can check their understanding by finishing the questions.\n‐ Students also need to draw all the shapes they have learned so far on their notebook as homework.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know area and volume. They will also learn about right triangles in algebra.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n-Classifying polygons.\n‐ Teacher will review the angle related concepts students have learned the day before, and use card board to show different polygons such as triangle and quadrilateral.\n‐ Students will work in pairs and try to match different shapes with their names.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Draw a diagram.\n‐ Teacher will review some concepts with the class at the beginning, and then hand out a worksheet so students can check their understanding by finishing the questions. Students also need to draw all the shapes they have learned so far on their notebook as homework.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know area and volume. They will also learn about right triangles in algebra.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ Congruence.\n‐ Teacher will review concept of congruent in previous chapters, and explain what are congruent figures.\n‐ In identifying congruent triangles, students will discuss why SSS, SAS and ASA are reliable, but not AAA and SSA.\nEvaluation:\nVocabulary:\nHomework:\nreview\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ 9‐6 Circles\n‐ Students will learn circle related concepts in this section, and how to calculate circumference of a circle.\n‐ Students will also understand why the ratio circumference / diameter is a fixed number through an experiment.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Continue with the lesson.\n‐ The teacher will give more examples and ask the students to solve the problems in the book.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nStudents will know area and volume. They will also learn about right triangles in algebra.\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nStudents will be able to solve those types of equations.\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\n‐ 9‐7 Constructions.\n‐ Teacher will review some concepts like line, angle, segment, and introduce concepts of perpendicular lines, segment bisector, perpendicular bisector and angle bisector. Then, demonstrate how to construct congruent segments and angles, as well as bisectors.\n‐ Students will practice the operations on their notebook.\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nNo Class\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\n‐ 9‐8 Translations.\n‐ Teacher will first introduce the concept of translation, and the differences between transformation and translation, and then demonstrate translating in Sketch Up.\n‐ Students will also learn how to describe translations.\nEvaluation:\nVocabulary:\nHomework:\nReview for the next lesson\nMain Activity/Lesson:\n‐ Symmetry and reflections.\n‐ Teacher will introduce concepts of reflection symmetry, line of symmetry, reflection and line of reflection by showing students some pictures, and then let students practice with graphing reflections under teacher's instructions.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMain Activity/Lesson:\n‐ Rotations.\n‐ Teacher will introduce rotation after reviewing reflection, and let students tell the difference between them.\n‐ Then students will learn and practice how to graph rotations and identify rotational symmetry.\nEvaluation:\nVocabulary:\nHomework:\nreview the lesson\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered\nWeekly Informational Knowledge Overview - (Students will know...)\nFINAL EXAM week\nWeekly Procedural Knowledge Overview - (Students will be able to...)\nMonday\nTuesday\nWednesday\nThursday\nFriday\nMain Activity/Lesson:\nFINAL EXAM\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nFINAL EXAM\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nFINAL EXAM\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nFINAL EXAM\nEvaluation:\nVocabulary:\nHomework:\nMain Activity/Lesson:\nFINAL EXAM\nEvaluation:\nVocabulary:\nHomework:\nMaterials / Resources (including technology)\nState of Minnesota Standards Covered"
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https://hack2skill.com/hack/sostechhack?utm_source=homepage&utm_medium=promotional | [
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"# SOS TECH HACK #sostechhack\n\nTech Is Future Of World\n\nview more...\n\nonline Solo Entry\n\nSOS TECH HACK is a community-focused 58 days Hackathon, designed\nespecially for the community's needs. Whether you are a beginner or an expert,\nhere is a perfect chance to showcase your skills and witness a competitive yet\ninclusive community around it.\nThis year it is going to be organized from 10/10/2022 in online mode.\n\n### Challenges",
null,
"##### Max Points on a Line\n\nGiven an array of points where points[i] = [xi, yi] represents a point on the X-Y plane, return the maximum number of points that lie on the same straight line. Example 1: Input: points = [[1,1],[2,2],[3,3]] Output: 3 Example 2: Input: points = [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] Output: 4 Constraints: 1 <= points.length <= 300 points[i].length == 2 -104 <= xi, yi <= 104 All the points are unique.",
null,
"##### Word Break II\n\nGiven a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note that the same word in the dictionary may be reused multiple times in the segmentation. Example 1: Input: s = \"catsanddog\", wordDict = [\"cat\",\"cats\",\"and\",\"sand\",\"dog\"] Output: [\"cats and dog\",\"cat sand dog\"] Example 2: Input: s = \"pineapplepenapple\", wordDict = [\"apple\",\"pen\",\"applepen\",\"pine\",\"pineapple\"] Output: [\"pine apple pen apple\",\"pineapple pen apple\",\"pine applepen apple\"] Explanation: Note that you are allowed to reuse a dictionary word. Example 3: Input: s = \"catsandog\", wordDict = [\"cats\",\"dog\",\"sand\",\"and\",\"cat\"] Output: [] Constraints: 1 <= s.length <= 20 1 <= wordDict.length <= 1000 1 <= wordDict[i].length <= 10 s and wordDict[i] consist of only lowercase English letters. All the strings of wordDict are unique.",
null,
"##### String to Integer (atoi)",
null,
"##### Roman to Integer\n\nRoman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II. Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used: I can be placed before V (5) and X (10) to make 4 and 9. X can be placed before L (50) and C (100) to make 40 and 90. C can be placed before D (500) and M (1000) to make 400 and 900. Given a roman numeral, convert it to an integer. Example 1: Input: s = \"III\" Output: 3 Explanation: III = 3. Example 2: Input: s = \"LVIII\" Output: 58 Explanation: L = 50, V= 5, III = 3. Example 3: Input: s = \"MCMXCIV\" Output: 1994 Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. Constraints: 1 <= s.length <= 15 s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M'). It is guaranteed that s is a valid roman numeral in the range [1, 3999].",
null,
"##### Regular Expression Matching\n\nGiven an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = \"aa\", p = \"a\" Output: false Explanation: \"a\" does not match the entire string \"aa\". Example 2: Input: s = \"aa\", p = \"a*\" Output: true Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes \"aa\". Example 3: Input: s = \"ab\", p = \".*\" Output: true Explanation: \".*\" means \"zero or more (*) of any character (.)\". Constraints: 1 <= s.length <= 20 1 <= p.length <= 30 s contains only lowercase English letters. p contains only lowercase English letters, '.', and '*'. It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.",
null,
"##### Zigzag Conversion\n\nThe string \"PAYPALISHIRING\" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) P A H N A P L S I I G Y I R And then read line by line: \"PAHNAPLSIIGYIR\" Write the code that will take a string and make this conversion given a number of rows: string convert(string s, int numRows); Example 1: Input: s = \"PAYPALISHIRING\", numRows = 3 Output: \"PAHNAPLSIIGYIR\" Example 2: Input: s = \"PAYPALISHIRING\", numRows = 4 Output: \"PINALSIGYAHRPI\" Explanation: P I N A L S I G Y A H R P I Example 3: Input: s = \"A\", numRows = 1 Output: \"A\" Constraints: 1 <= s.length <= 1000 s consists of English letters (lower-case and upper-case), ',' and '.'. 1 <= numRows <= 1000",
null,
"##### Reverse array in groups\n\nGiven an array arr[] of positive integers of size N. Reverse every sub-array group of size K. Note: If at any instance, there are no more subarrays of size greater than or equal to K, then reverse the last subarray (irrespective of its size). Example 1: Input: N = 5, K = 3 arr[] = {1,2,3,4,5} Output: 3 2 1 5 4 Explanation: First group consists of elements 1, 2, 3. Second group consists of 4,5. Example 2: Input: N = 4, K = 3 arr[] = {5,6,8,9} Output: 8 6 5 9 Your Task: You don't need to read input or print anything. The task is to complete the function reverseInGroups() which takes the array, N and K as input parameters and modifies the array in-place. Expected Time Complexity: O(N) Expected Auxiliary Space: O(N) Constraints: 1 ≤ N, K ≤ 107 1 ≤ A[i] ≤ 1018",
null,
"",
null,
"##### Remove loop in Linked List",
null,
"##### Diameter of a Binary Tree\n\nThe diameter of a tree (sometimes called the width) is the number of nodes on the longest path between two end nodes. The diagram below shows two trees each with diameter nine, the leaves that form the ends of the longest path are shaded (note that there is more than one path in each tree of length nine, but no path longer than nine nodes). Example 1: Input: 1 / \\ 2 3 Output: 3 Example 2: Input: 10 / \\ 20 30 / \\ 40 60 Output: 4 Your Task: You need to complete the function diameter() that takes root as parameter and returns the diameter. Expected Time Complexity: O(N). Expected Auxiliary Space: O(Height of the Tree). Constraints: 1 <= Number of nodes <= 10000 1 <= Data of a node <= 1000",
null,
"##### Next Smallest Palindrome\n\nGiven a number, in the form of an array Num[] of size N containing digits from 1 to 9(inclusive). The task is to find the next smallest palindrome strictly larger than the given number. Example 1: Input: N = 11 Num[] = {9, 4, 1, 8, 7, 9, 7, 8, 3, 2, 2} Output: 9 4 1 8 8 0 8 8 1 4 9 Explanation: Next smallest palindrome is 94188088149. Example 2: Input: N = 5 Num[] = {2, 3, 5, 4, 5} Output: 2 3 6 3 2 Explanation: Next smallest palindrome is 23632. Your Task: Complete the function generateNextPalindrome() which takes an array num, and an single integer n, as input parameters and returns an array of integers denoting the answer. You don't to print answer or take inputs. Expected Time Complexity: O(N) Expected Auxiliary Space: O(1) Constraints: 1 <= N <= 105 1 <= Num[i] <= 9",
null,
"##### Longest Palindromic Substring in Linear Time\n\nGiven a string, find the longest substring which is palindrome in Linear time O(N). Input: The first line of input contains an integer T denoting the no of test cases . Then T test cases follow. The only line of each test case contains a string. Output: For each test case print the Longest Palindromic Substring. If there are multiple such substrings of same length, print the one which appears first in the input string. Constraints: 1 <= T <= 100 1 <= N <= 50 Example: Input: 2 babcbabcbaccba forgeeksskeegfor Output: abcbabcba geeksskeeg Note:The Input/Ouput format and Example given are used for system's internal purpose, and should be used by a user for Expected Output only. As it is a function problem, hence a user should not read any input from stdin/console. The task is to complete the function specified, and not to write the full code.",
null,
"##### Binary Tree to DLL\n\nGiven a Binary Tree (BT), convert it to a Doubly Linked List(DLL) In-Place. The left and right pointers in nodes are to be used as previous and next pointers respectively in converted DLL. The order of nodes in DLL must be same as Inorder of the given Binary Tree. The first node of Inorder traversal (leftmost node in BT) must be the head node of the DLL. TreeToList Example 1: Input: 1 / \\ 3 2 Output: 3 1 2 2 1 3 Explanation: DLL would be 3<=>1<=>2 Example 2: Input: 10 / \\ 20 30 / \\ 40 60 Output: 40 20 60 10 30 30 10 60 20 40 Explanation: DLL would be 40<=>20<=>60<=>10<=>30. Your Task: You don't have to take input. Complete the function bToDLL() that takes root node of the tree as a parameter and returns the head of DLL . The driver code prints the DLL both ways. Expected Time Complexity: O(N). Expected Auxiliary Space: O(H). Note: H is the height of the tree and this space is used implicitly for the recursion stack. Constraints: 1 ≤ Number of nodes ≤ 105 0 ≤ Data of a node ≤ 105",
null,
"##### AVL Tree Deletion\n\nGiven a AVL tree and N values to be deleted from the tree. Write a function to delete a given value from the tree. Example 1: Tree = 4 / \\ 2 6 / \\ / \\ 1 3 5 7 N = 4 Values to be deleted = {4,1,3,6} Input: Value to be deleted = 4 Output: 5 / \\ 2 6 / \\ \\ 1 3 7 Input: Value to be deleted = 1 Output: 5 / \\ 2 6 \\ \\ 3 7 Input: Value to be deleted = 3 Output: 5 / \\ 2 6 \\ 7 Input: Value to be deleted = 6 Output: 5 / \\ 2 7 Your Task: You dont need to read input or print anything. Complete the function delelteNode() which takes the root of the tree and the value of the node to be deleted as input parameters and returns the root of the modified tree. Note: The tree will be checked after each deletion. If it violates the properties of balanced BST, an error message will be printed followed by the inorder traversal of the tree at that moment. If instead all deletion are successful, inorder traversal of tree will be printed. If every single node is deleted from tree, 'null' will be printed. Expected Time Complexity: O(height of tree) Expected Auxiliary Space: O(height of tree) Constraints: 1 ≤ N ≤ 500",
null,
"##### Transpose File\n\nGiven a text file file.txt, transpose its content. You may assume that each row has the same number of columns, and each field is separated by the ' ' character. Example: If file.txt has the following content: name age alice 21 ryan 30 Output the following: name alice ryan age 21 30",
null,
"##### Path Sum\n\nGiven the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum. A leaf is a node with no children. Example 1: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: The root-to-leaf path with the target sum is shown. Example 2: Input: root = [1,2,3], targetSum = 5 Output: false Explanation: There two root-to-leaf paths in the tree: (1 --> 2): The sum is 3. (1 --> 3): The sum is 4. There is no root-to-leaf path with sum = 5. Example 3: Input: root = [], targetSum = 0 Output: false Explanation: Since the tree is empty, there are no root-to-leaf paths. Constraints: The number of nodes in the tree is in the range [0, 5000]. -1000 <= Node.val <= 1000 -1000 <= targetSum <= 1000",
null,
"##### Path Sum\n\nGiven the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum. A leaf is a node with no children. Example 1: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: The root-to-leaf path with the target sum is shown. Example 2: Input: root = [1,2,3], targetSum = 5 Output: false Explanation: There two root-to-leaf paths in the tree: (1 --> 2): The sum is 3. (1 --> 3): The sum is 4. There is no root-to-leaf path with sum = 5. Example 3: Input: root = [], targetSum = 0 Output: false Explanation: Since the tree is empty, there are no root-to-leaf paths. Constraints: The number of nodes in the tree is in the range [0, 5000]. -1000 <= Node.val <= 1000 -1000 <= targetSum <= 1000",
null,
"##### Second Highest Salary\n\nSQL Schema Table: Employee +-------------+------+ | Column Name | Type | +-------------+------+ | id | int | | salary | int | +-------------+------+ id is the primary key column for this table. Each row of this table contains information about the salary of an employee. Write an SQL query to report the second highest salary from the Employee table. If there is no second highest salary, the query should report null. The query result format is in the following example. Example 1: Input: Employee table: +----+--------+ | id | salary | +----+--------+ | 1 | 100 | | 2 | 200 | | 3 | 300 | +----+--------+ Output: +---------------------+ | SecondHighestSalary | +---------------------+ | 200 | +---------------------+ Example 2: Input: Employee table: +----+--------+ | id | salary | +----+--------+ | 1 | 100 | +----+--------+ Output: +---------------------+ | SecondHighestSalary | +---------------------+ | null | +---------------------+",
null,
"##### Regular Expression Matching\n\nGiven an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = \"aa\", p = \"a\" Output: false Explanation: \"a\" does not match the entire string \"aa\". Example 2: Input: s = \"aa\", p = \"a*\" Output: true Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes \"aa\". Example 3: Input: s = \"ab\", p = \".*\" Output: true Explanation: \".*\" means \"zero or more (*) of any character (.)\". Constraints: 1 <= s.length <= 20 1 <= p.length <= 30 s contains only lowercase English letters. p contains only lowercase English letters, '.', and '*'. It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.",
null,
"##### Reverse Nodes in k-Group\n\nGiven the head of a linked list, reverse the nodes of the list k at a time, and return the modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is. You may not alter the values in the list's nodes, only nodes themselves may be changed. Example 1: Input: head = [1,2,3,4,5], k = 2 Output: [2,1,4,3,5] Example 2: Input: head = [1,2,3,4,5], k = 3 Output: [3,2,1,4,5] Constraints: The number of nodes in the list is n. 1 <= k <= n <= 5000 0 <= Node.val <= 1000 Follow-up: Can you solve the problem in O(1) extra memory space?\n\n### Hackathon Mentors",
null,
"##### Divjot Singh\n\nSoftware Engineer\n\nView More\n\n### Schedule\n\n00:00\n00:00\nTo Be Announced\n###### Submissions Close\n\n*Timeline is subject to change/tentative\n\n### Rewards and benefits",
null,
"Winners\n\nHack 2 Skill"
] | [
null,
"https://www.facebook.com/tr",
null,
"https://s3-h2s-v2.s3.ap-south-1.amazonaws.com/2022-10-04T14%3A34%3A12.734Z-png_20221004_200244_0000.png",
null,
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null,
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null,
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null,
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null
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https://www.geeksforgeeks.org/minimum-count-of-numbers-required-with-unit-digit-x-that-sums-up-to-n/?ref=rp | [
"Related Articles\n\n# Minimum count of numbers required with unit digit X that sums up to N\n\n• Last Updated : 11 May, 2021\n\nGiven two integers N and X, the task is to find the minimum count of integers with sum N and having unit digit X. If no such representation exists then print -1.\nExamples:\n\nInput: N = 38, X = 9\nOutput:\nExplanation:\nMinimum two integers are required with unit digit as X to represent as a sum equal to 38.\n38 = 19 + 19 or 38 = 29 + 9\nInput: N = 6, X = 4\nOutput: -1\nExplanation:\nNo such representation of\n\nApproach:\nFollow the steps below to solve the problem:\n\n• Obtain the unit digit of N and check if it achieved by the sum of numbers whose unit digits are X.\n• If it is possible, check if N ? X * ( Minimum number of times a number with unit digit X needs to be added to get sum N).\n• If the above condition is satisfied then print the minimum number of times a number with unit digit X needs to be added to get sum N. Otherwise, print -1.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to calculate and return``// the minimum number of times a number``// with unit digit X needs to be added``// to get a sum N``int` `check(``int` `unit_digit, ``int` `X)``{`` ``int` `times, digit;` ` ``// Calculate the number of`` ``// additions required to get unit`` ``// digit of N`` ``for` `(``int` `times = 1; times <= 10;`` ``times++) {`` ``digit = (X * times) % 10;`` ``if` `(digit == unit_digit)`` ``return` `times;`` ``}` ` ``// If unit digit of N`` ``// cannot be obtained`` ``return` `-1;``}` `// Function to return the minimum``// number required to represent N``int` `getNum(``int` `N, ``int` `X)``{`` ``int` `unit_digit;` ` ``// Stores unit digit of N`` ``unit_digit = N % 10;` ` ``// Stores minimum addition`` ``// of X required to`` ``// obtain unit digit of N`` ``int` `times = check(unit_digit, X);` ` ``// If unit digit of N`` ``// cannot be obtained`` ``if` `(times == -1)`` ``return` `times;` ` ``// Otherwise`` ``else` `{` ` ``// If N is greater than`` ``// or equal to (X*times)`` ``if` `(N >= (times * X))` ` ``// Minimum count of numbers`` ``// that needed to represent N`` ``return` `times;` ` ``// Representation not`` ``// possible`` ``else`` ``return` `-1;`` ``}``}` `// Driver Code``int` `main()``{`` ``int` `N = 58, X = 7;`` ``cout << getNum(N, X) << endl;`` ``return` `0;``}`\n\n## Java\n\n `// Java Program to implement``// the above approach``class` `GFG{`` ` `// Function to calculate and return``// the minimum number of times a number``// with unit digit X needs to be added``// to get a sum N``static` `int` `check(``int` `unit_digit, ``int` `X)``{`` ``int` `times, digit;` ` ``// Calculate the number of`` ``// additions required to get unit`` ``// digit of N`` ``for` `(times = ``1``; times <= ``10``;`` ``times++)`` ``{`` ``digit = (X * times) % ``10``;`` ``if` `(digit == unit_digit)`` ``return` `times;`` ``}` ` ``// If unit digit of N`` ``// cannot be obtained`` ``return` `-``1``;``}` `// Function to return the minimum``// number required to represent N``static` `int` `getNum(``int` `N, ``int` `X)``{`` ``int` `unit_digit;` ` ``// Stores unit digit of N`` ``unit_digit = N % ``10``;` ` ``// Stores minimum addition`` ``// of X required to`` ``// obtain unit digit of N`` ``int` `times = check(unit_digit, X);` ` ``// If unit digit of N`` ``// cannot be obtained`` ``if` `(times == -``1``)`` ``return` `times;` ` ``// Otherwise`` ``else`` ``{` ` ``// If N is greater than`` ``// or equal to (X*times)`` ``if` `(N >= (times * X))` ` ``// Minimum count of numbers`` ``// that needed to represent N`` ``return` `times;` ` ``// Representation not`` ``// possible`` ``else`` ``return` `-``1``;`` ``}``}` `// Driver Code``public` `static` `void` `main(String []args)``{`` ``int` `N = ``58``, X = ``7``;`` ``System.out.println( getNum(N, X));``}``}` `// This code is contributed by Ritik Bansal`\n\n## Python3\n\n `# Python3 program to implement``# the above approach` `# Function to calculate and return``# the minimum number of times a number``# with unit digit X needs to be added``# to get a sum N``def` `check(unit_digit, X):`` ` ` ``# Calculate the number of additions`` ``# required to get unit digit of N`` ``for` `times ``in` `range``(``1``, ``11``):`` ``digit ``=` `(X ``*` `times) ``%` `10`` ``if` `(digit ``=``=` `unit_digit):`` ``return` `times`` ` ` ``# If unit digit of N`` ``# cannot be obtained`` ``return` `-``1` `# Function to return the minimum``# number required to represent N``def` `getNum(N, X):`` ` ` ``# Stores unit digit of N`` ``unit_digit ``=` `N ``%` `10` ` ``# Stores minimum addition`` ``# of X required to`` ``# obtain unit digit of N`` ``times ``=` `check(unit_digit, X)` ` ``# If unit digit of N`` ``# cannot be obtained`` ``if` `(times ``=``=` `-``1``):`` ``return` `times` ` ``# Otherwise`` ``else``:` ` ``# If N is greater than`` ``# or equal to (X*times)`` ``if` `(N >``=` `(times ``*` `X)):` ` ``# Minimum count of numbers`` ``# that needed to represent N`` ``return` `times` ` ``# Representation not`` ``# possible`` ``else``:`` ``return` `-``1` `# Driver Code``N ``=` `58``X ``=` `7` `print``(getNum(N, X))` `# This code is contributed by Sanjit_Prasad`\n\n## C#\n\n `// C# Program to implement``// the above approach``using` `System;``class` `GFG{`` ` `// Function to calculate and return``// the minimum number of times a number``// with unit digit X needs to be added``// to get a sum N``static` `int` `check(``int` `unit_digit, ``int` `X)``{`` ``int` `times, digit;` ` ``// Calculate the number of`` ``// additions required to get unit`` ``// digit of N`` ``for` `(times = 1; times <= 10;`` ``times++)`` ``{`` ``digit = (X * times) % 10;`` ``if` `(digit == unit_digit)`` ``return` `times;`` ``}` ` ``// If unit digit of N`` ``// cannot be obtained`` ``return` `-1;``}` `// Function to return the minimum``// number required to represent N``static` `int` `getNum(``int` `N, ``int` `X)``{`` ``int` `unit_digit;` ` ``// Stores unit digit of N`` ``unit_digit = N % 10;` ` ``// Stores minimum addition`` ``// of X required to`` ``// obtain unit digit of N`` ``int` `times = check(unit_digit, X);` ` ``// If unit digit of N`` ``// cannot be obtained`` ``if` `(times == -1)`` ``return` `times;` ` ``// Otherwise`` ``else`` ``{` ` ``// If N is greater than`` ``// or equal to (X*times)`` ``if` `(N >= (times * X))` ` ``// Minimum count of numbers`` ``// that needed to represent N`` ``return` `times;` ` ``// Representation not`` ``// possible`` ``else`` ``return` `-1;`` ``}``}` `// Driver Code``public` `static` `void` `Main()``{`` ``int` `N = 58, X = 7;`` ``Console.Write(getNum(N, X));``}``}` `// This code is contributed by Code_Mech`\n\n## Javascript\n\n ``\nOutput:\n`4`\n\nTime Complexity: O(1)\nAuxiliary Space: O(1)\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.\n\nIn case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.\n\nMy Personal Notes arrow_drop_up"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6514805,"math_prob":0.9857856,"size":7300,"snap":"2021-31-2021-39","text_gpt3_token_len":2079,"char_repetition_ratio":0.18270285,"word_repetition_ratio":0.57923496,"special_character_ratio":0.3147945,"punctuation_ratio":0.09955423,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99983627,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T23:46:16Z\",\"WARC-Record-ID\":\"<urn:uuid:981f3aec-2b6c-4f46-813c-72527b426cb7>\",\"Content-Length\":\"164555\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5cffc58c-adbe-4e77-8b1b-1af9859ff9fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:a3fb4a0f-3732-400f-b022-603417c1e211>\",\"WARC-IP-Address\":\"23.207.202.28\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/minimum-count-of-numbers-required-with-unit-digit-x-that-sums-up-to-n/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:XPZAHHVXCCJFRTI3Q6XVGB3AMN3AMBJW\",\"WARC-Block-Digest\":\"sha1:MEQGBF3DPQPH5LPBVGMCKIS37OLE7E7V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154277.15_warc_CC-MAIN-20210801221329-20210802011329-00387.warc.gz\"}"} |
https://solvedlib.com/n/point-if-a-4-x-4-matrix-a-with-rows-v1-vz-v3-and-v4-has,3542367 | [
"# Point) If a 4 X 4 matrix A with rows V1 , Vz, V3 , and v4 has determinant detA\n\n###### Question:\n\npoint) If a 4 X 4 matrix A with rows V1 , Vz, V3 , and v4 has determinant detA = 2 2v2 + Zv3 6v2 + 4v3 then det",
null,
"",
null,
"#### Similar Solved Questions\n\n##### Explain potential for growth internally and externally in a sole proprietor business\nexplain potential for growth internally and externally in a sole proprietor business...\n##### Use the Method of Undetermined Coefficients to find the general solution for the differential equation: y\"-2y'+2y=...\nUse the Method of Undetermined Coefficients to find the general solution for the differential equation: y\"-2y'+2y= e^(x)sinx Answer should be: y= ce^(x)cosx+ce^(x)sinx-(x/2)e^(x)cosx...\n##### 24. The current in the [2 Q resistor in the circuitshown0.50 AG.0nLL0 A 154 (d) 20 A1.091203,0 A1.0960n(15 Points)\n24. The current in the [2 Q resistor in the circuitshown 0.50 A G.0n LL0 A 154 (d) 20 A 1.09 120 3,0 A 1.09 60n (15 Points)...\n##### Preview File Edit View Go Tools Window Help Ch01 homework.pdf (page 12 of 12)- Edited 21....\nPreview File Edit View Go Tools Window Help Ch01 homework.pdf (page 12 of 12)- Edited 21. Describe the central dogma of biochemistry. What are the three main processes?...\n##### Stat 200 Chapter6: Problem 5 Prevlous Problem Problem List Next Problempoint)gas-and clectric-powered hybtid has recently hit the market The distance Iravelled on galin of het noally disunbuled new car Ihal iS miles Find the probability of the tollwing events. Use Excel obtain inore accuacy: wilh mean 0/ 65 miles and stardard devialion ofThe car (ravels more Ihan 71 miles per gallon.ProbabilityThe car Iravels Iess Ihan 59 miles per gallonProbabllity The car travels between 61 and 73 miles per g\nStat 200 Chapter6: Problem 5 Prevlous Problem Problem List Next Problem point) gas-and clectric-powered hybtid has recently hit the market The distance Iravelled on galin of het noally disunbuled new car Ihal iS miles Find the probability of the tollwing events. Use Excel obtain inore accuacy: wilh...\n##### (31) The College Registrar's Office recorded the number of students receiving a grade of Incomplete. Results...\n(31) The College Registrar's Office recorded the number of students receiving a grade of Incomplete. Results for the past 26 quarters are: 19 58 63 77 53 39 93 35 63 32 46 57 67 62 67 71 59 48 56 75 48 28 47 42 81 62 (a) Make a frequency table with five classes, showing class boundaries, class m...\n##### Draw structure from IR, formula c3h6o2 Analyst Date Thursday, May 09, 2019 851 AM 70 1047 39 20m 95em-1 1744.59cm-1 1437 80cm-1 C-O COH 60- 3400 45cm-1 On 50 40- 30 201 10 500 1000 000 3500 30000...\nDraw structure from IR, formula c3h6o2 Analyst Date Thursday, May 09, 2019 851 AM 70 1047 39 20m 95em-1 1744.59cm-1 1437 80cm-1 C-O COH 60- 3400 45cm-1 On 50 40- 30 201 10 500 1000 000 3500 3000002000 1500 cm-1 Sample 549 By Administrator Date Thursday, May 09 2019 Administrator 549,001 _ Ӧ...\n##### A 0.240 kg block of a pure material is heated from 20.0°C to 68.4°C by the...\nA 0.240 kg block of a pure material is heated from 20.0°C to 68.4°C by the addition of 5.25 kJ of energy. Calculate its specific heat (in J/(kg · °C)). J/(kg · °C) Identify the substance of which it is most likely composed. gold copper glass iron ОО silver +...\n##### Find basis and the dimension for the TOW space; column space and null space of thematfix Identify the rank and nullity for the matrix 4question 2(a) above_\nFind basis and the dimension for the TOW space; column space and null space of the matfix Identify the rank and nullity for the matrix 4 question 2(a) above_...\n##### Two point charges are separated by 5.8 cm. The attractive force between them is 22 N....\nTwo point charges are separated by 5.8 cm. The attractive force between them is 22 N. Suppose that the charges attracting each other have equal magnitude....\n##### Consider this function y fx) on the domain (-x2 sin( Xtif x if x = 0(a) If x 0 _'(x)=Answer these multiple choice questions:lim f (x) I -0 Does not existling Kb)_fo) h-0Does not existDoes not existf(x) is continuous for allx TrueFalsef(x) is differentiable for all x TrueFalse\nConsider this function y fx) on the domain (- x2 sin( Xt if x if x = 0 (a) If x 0 _ '(x)= Answer these multiple choice questions: lim f (x) I -0 Does not exist ling Kb)_fo) h-0 Does not exist Does not exist f(x) is continuous for allx True False f(x) is differentiable for all x True False...\n##### Calculate the cell potential_Ecell at 258C for the following reaction when the concentrations are: [Zn?*1 10.050 Mand [Cu2t] = 20M. Zn (s) + Cu2t (aq) _ + Zn?* (aq) + Cufs) E\"cell = L1V0 Ecell = 3.21 V9 Eccll L15 VI@ Ezll \" 4.15V0 Ecelll 193VEcell \" 047V\nCalculate the cell potential_Ecell at 258C for the following reaction when the concentrations are: [Zn?*1 10.050 Mand [Cu2t] = 20M. Zn (s) + Cu2t (aq) _ + Zn?* (aq) + Cufs) E\"cell = L1V 0 Ecell = 3.21 V 9 Eccll L15 V I@ Ezll \" 4.15V 0 Ecelll 193V Ecell \" 047V...\n##### For the unknowns: Hinsberg _ testObservationsAfter addition of 10 NaOH benzene sulfony] chloride, and 6M HCIAfter addition of 25M NaOHInference:\nFor the unknowns: Hinsberg _ test Observations After addition of 10 NaOH benzene sulfony] chloride, and 6M HCI After addition of 25M NaOH Inference:...\n##### Consider the following function: f(x) = 4 - x2/3 Find f(-8) and f(8).f(-8)f(8)Find all values € in (-8, 8) such that f\"(c)(Enter your ansBased off of this information, what conclusions can be made aboc This contradicts Rolle's Theorem, since f is differentiable, This does not contradict Rolle's Theorem, since f (0) = 0, anc This contradicts Rolle's Theorem; since f(_8) {(8) , there sh This does not contradict Rolle's Theorem, since f'(0) does no Nothing can be con\nConsider the following function: f(x) = 4 - x2/3 Find f(-8) and f(8). f(-8) f(8) Find all values € in (-8, 8) such that f\"(c) (Enter your ans Based off of this information, what conclusions can be made aboc This contradicts Rolle's Theorem, since f is differentiable, This does not co...\n##### Find the lirit:~2t| sn\"sin 7t 6(4,4)\nFind the lirit: ~2t| sn\" sin 7t 6 (4,4)...\nPage 4 Example: An analyst is conducting a test involving the following hypotheses Horn = 100 vs. Hu 100 The population is known to be normally distributed with a standard deviation of 3. Assume that the sample size is 9. Suppose the acceptance region is 98.5 < X < 101.5, a. Find the Type I ..."
] | [
null,
"https://cdn.numerade.com/ask_images/55b2381f775348e7a83e739fdc858974.jpg ",
null,
"https://cdn.numerade.com/previews/52fd4986-eb23-4b90-a8af-a34815667855_large.jpg",
null
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https://en.zdam.xyz/problem/20450/ | [
"#### Problem 70E\n\nThe table gives the US population from 1790 to 1860. Year Population 1790 3, 929, 000 1800 5, 308, 000 1810 7, 240, 000 1820 9, 639, 000 1830 12, 861, 000 1840 17, 063, 000 1850 23, 192, 000 1860 31, 443, 000 a Use a graphing calculator or computer to fit an exponential function to the data. Graph the data points and the exponential model. How good is the fit? b Estimate the rates of population growth in 1800 and 1850 by averaging slopes of secant lines. c Use the exponential model in part a to estimate the rates of growth in 1800 and 1850. Compare these estimates with the ones in part b. d Use the exponential model to predict the population in 1870. Compare with the actual population of 38, 558, 000. Can you explain the discrepancy?"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7105279,"math_prob":0.9909115,"size":2540,"snap":"2022-27-2022-33","text_gpt3_token_len":864,"char_repetition_ratio":0.12854889,"word_repetition_ratio":0.31412104,"special_character_ratio":0.3944882,"punctuation_ratio":0.08445297,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99114305,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T22:59:46Z\",\"WARC-Record-ID\":\"<urn:uuid:6cc2c213-6338-4cbd-b884-96678600e50c>\",\"Content-Length\":\"24794\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2d97085-5ffa-4335-a031-d9b59d9f9be1>\",\"WARC-Concurrent-To\":\"<urn:uuid:518100bf-2b4c-4437-af32-3c241d837ed4>\",\"WARC-IP-Address\":\"172.67.221.234\",\"WARC-Target-URI\":\"https://en.zdam.xyz/problem/20450/\",\"WARC-Payload-Digest\":\"sha1:HPWUOUVPDZUJ6PFJDZJF5F23M7CQYPYL\",\"WARC-Block-Digest\":\"sha1:4KMURAQPMN4HTQCY4QPPNH7P2KHBCBYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571090.80_warc_CC-MAIN-20220809215803-20220810005803-00743.warc.gz\"}"} |
https://www.heliciel.com/en/helice/calcul-helice-aile/Theorie%20de%20Froude%20relative%20aux%20helices%20de%20traction%20ou%20propulsion.htm | [
"Theorie de Froude relative aux on propeller propulsion or traction\n\n• To understand Theory Froude, the notion of momentum must be mastered, because it is on the balance of momentum that the Froude theory is based ..\nFroude approach allows us to assess the axial induced velocity using a control volume around the fluid stream passing through the propeller. By calculating the volume of fluid, which must penetrate the volume control laterally,due to the contraction of the fluid stream caused by the increase in velocity produced by the propeller: Following the approach proposed by Froude, we can consider the propeller as a uniformly charged disk with an infinite number of blades (Fig. 1). In addition, we set the following assumptions:\n1. are not taken into account the rotation of the flow.\n2. we consider a fluid stream out of which the flow is not disturbed\n3. The pressure at infinity upstream and downstream is equal to the static pressure of undisturbed flow\n4. The air disturbance are sufficiently small that it is assumed that the fluid density is constant.",
null,
"fig. 1\n\nlthe continuity equation allows us to calculate the volume added to the control zone per second (volume flow Q in m3/sec),:",
null,
"Q is the volumetric flow discharged from the volume control. From the theorem of momentum, the thrust, T of the propeller is obtained, that is to say the variation of the amount of movement between the sections 2 and O, whose projection on the horizontal axis indicates thrust :",
null,
"The derivative of the momentum versus time :",
null,
"Momentum out \"M_2\" volume control in Section 2 =",
null,
"Total amount of movement \"M_1\" within the volume control = Momentum in the volume entry - momentum laterally inserted=",
null,
"Therefore, the axial force of the fluid on the rotor becomes(T=M_2-M_1):",
null,
"One can also express the axial force of the fluid on the rotor T, according the result of the static pressure which is exerted on the surface of dis:",
null,
"where A is the surface of the disc swept by the propeller and Dp, the pressure difference across the disc. Using Bernoulli's equation, we obtain",
null,
"pressure difference O to 1 upstream, and:",
null,
"pressure difference 2 to 1 downstream.\n\nThe static pressure difference between the upstream and downstream sides of the disk of the propeller as the expression:",
null,
"From the continuity equation:",
null,
", can be derived:",
null,
"and consider the speed of the flow through the disk as an arithmetic average of the velocities upstream and downstream of the propeller:",
null,
"Is called induced velocity w, increasing speed at the propeller disk:",
null,
"",
null,
"where \"a\" represents the axial interference factor.\n\nNow, if we consider the expression:",
null,
",the axial force exerted on the disc of the impeller is:",
null,
"From this expression we can identify push the notion of mass flow: The mass of fluid passing through the propeller per second :\n• (kg/sec) m = r.p.r².V2\nSo we see that the thrust of the propeller is: T=m.(v2-v0)\n• It is important to note that create thrust, therefore, is to increased the speed of a mass flow of fuide. This highlights the levers available to us to generate the propulsive force: The choice to act on the mass or speed variation is crucial for the propulsive efficiency\nThe theory froude gives us the axial induced velocity. It will be linked to the theory of the blade element to give a decomposition of the induced velocity in tangential and axial induced velocities necessary for the prediction of the incidence angles and the geometry of the blades..\n\nBibliographic references propeller"
] | [
null,
"https://www.heliciel.com/en/images formules/froude-propulsion.jpg",
null,
"https://www.heliciel.com/en/images%20formules/eq1.jpg",
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"https://www.heliciel.com/en/images%20formules/7_5.jpg",
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"https://www.heliciel.com/en/images%20formules/7_6.jpg",
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"https://www.heliciel.com/en/images%20formules/q_m_sortie.jpg",
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"https://www.heliciel.com/en/images%20formules/q_m_entrant_prop.jpg",
null,
"https://www.heliciel.com/en/images%20formules/t_variation_mouv_prop.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_8.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_9.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_10.jpg",
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"https://www.heliciel.com/en/images%20formules/7_11.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_11b.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_12.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_13.jpg",
null,
"https://www.heliciel.com/en/images%20formules/7_13b.jpg",
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"https://www.heliciel.com/en/images%20formules/7_13c.jpg",
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"https://www.heliciel.com/en/images%20formules/7_13.jpg",
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"https://www.heliciel.com/en/images%20formules/7_14.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89106,"math_prob":0.9684343,"size":2593,"snap":"2022-40-2023-06","text_gpt3_token_len":556,"char_repetition_ratio":0.14831981,"word_repetition_ratio":0.022727273,"special_character_ratio":0.20671037,"punctuation_ratio":0.09690721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931961,"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],"im_url_duplicate_count":[null,2,null,4,null,4,null,4,null,4,null,2,null,2,null,4,null,4,null,4,null,4,null,4,null,4,null,10,null,4,null,4,null,10,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-24T17:21:30Z\",\"WARC-Record-ID\":\"<urn:uuid:39883938-6d44-4486-bf25-416d52b3c7e9>\",\"Content-Length\":\"75049\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73a8e83b-9d51-4467-a31f-517e195eaf0a>\",\"WARC-Concurrent-To\":\"<urn:uuid:0022482d-bd25-4cb7-a3ac-6d94acad646e>\",\"WARC-IP-Address\":\"81.169.145.64\",\"WARC-Target-URI\":\"https://www.heliciel.com/en/helice/calcul-helice-aile/Theorie%20de%20Froude%20relative%20aux%20helices%20de%20traction%20ou%20propulsion.htm\",\"WARC-Payload-Digest\":\"sha1:BFSBVCNPYHIKVCKYMVZW3FOE2RQMBZON\",\"WARC-Block-Digest\":\"sha1:BPJGXJOSTVEKDSTHDFAJ456WDQIJ2GOW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030331677.90_warc_CC-MAIN-20220924151538-20220924181538-00660.warc.gz\"}"} |
https://bestmaths.net/online/index.php/year-levels/year-12/year-12-topic-list/newton-rhapson-method/ | [
"## Newton-Rhapson Method\n\nThe Bisection method for solving equations is accurate but quite lengthy. A quicker method, which requires differentiation, is the Newton Rhapson method.\n\nIn this method, an initial value of x, which is thought to be near the root is found. This value becomes x0.\n\nA tangent to the curve is drawn at this point and extended to cut the x-axis. This value is x1.\n\nThis value is found using the general formula :",
null,
"",
null,
"Note that the function f(x) must be differentiable.\n\nWhen the equation has more than one root, the initial value of x must be near to the root being found or the second value of x could be nearer to another root of the equation.\n\nThis process is repeated until successive differences are less than the degree of accuracy required. i.e. x n - xn-1I < E\n\nExample\n\nUse the Newton-Rhapson method to solve the equation x3 = 5 to 6 decimal places.\n\nFirst of all, rearrange the equation to x3 − 5 = 0\n\nNow, find an initial value.\n\n x 0 1 2 3 x3 - 5 -5 -4 3 22\n\nThe root is between 1 and 2 (the sign changes from negative to positive) and nearer to 2, so we will use and initial value of x0 = 1.75\n\nDifferentiate f(x) = x3 − 5\n\nf '(x) = 3x2\n\nUsing",
null,
"x1 = 1.75 −",
null,
"= 1.7108844\n\nThis process is now repeated until the required degree of accuracy is reached.\n\nTo summarise and continue this process in a table:\n\n Initial value x0 1.75 x n - xn-1 1st iteration x1 1.71088 -0.0391156 2nd iteration x2 1.70998 -0.000908 3rd iteration x3 1.70998 -0.0000005\n\nAs x n - xn-1I < 0.000001 the process is stopped.\n\nThe solution is x = 1.709976 (to 6 decimal places)\n\n### Non-convergence\n\nIf the first value used in the Newton-rhapson method is not chosen correctly it could mean that successive iterations will not converge on the required root. An example is shown below .",
null,
"The initial value x0 is too far from the required root and the tangents can be seen to be oscillating and not converging on the root.\n\nOther cases of non-convergence\n\nIf the initial value is a turning point, the tangent will not intersect the x-axis.\n\nIf the initial value is too far from the required root then convergence may be to another root.\n\n### Comparison of the Bisection Method and the Newton Rhapson Method\n\nComparing the two methods:\n\n Method Advantages Disadvantages Bisection Reliable Always converges to the root Slow, can require many iterations Newton-Rhapson Fast convergence Simple formula Requires differentiation Can converge to wrong root May not converge at all\n\nA third method, the secant method can also be used."
] | [
null,
"https://bestmaths.net/online/files/8213/8671/8600/Y12_Newton-Rhapson_Method_01.gif",
null,
"https://bestmaths.net/online/files/2013/8671/8603/Y12_Newton-Rhapson_Method_02.gif",
null,
"https://bestmaths.net/online/files/1113/8671/8604/Y12_Newton-Rhapson_Method_03.gif",
null,
"https://bestmaths.net/online/files/4713/8671/8604/Y12_Newton-Rhapson_Method_04.gif",
null,
"https://bestmaths.net/online/files/3213/8671/8605/Y12_Newton-Rhapson_Method_05.gif",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87747985,"math_prob":0.9936859,"size":2534,"snap":"2021-43-2021-49","text_gpt3_token_len":700,"char_repetition_ratio":0.13438736,"word_repetition_ratio":0.025104603,"special_character_ratio":0.2857143,"punctuation_ratio":0.09325397,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9972986,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T20:19:13Z\",\"WARC-Record-ID\":\"<urn:uuid:c994d553-fa84-410f-af21-2efc7dad9d22>\",\"Content-Length\":\"17555\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7dda8228-020e-46de-8ed7-abe5d5f7f437>\",\"WARC-Concurrent-To\":\"<urn:uuid:27f2501c-f40e-451f-b77b-e114282274bf>\",\"WARC-IP-Address\":\"43.224.120.23\",\"WARC-Target-URI\":\"https://bestmaths.net/online/index.php/year-levels/year-12/year-12-topic-list/newton-rhapson-method/\",\"WARC-Payload-Digest\":\"sha1:F2GK75DIOHKIPGEBHFG2T6NZ2TE6NTPG\",\"WARC-Block-Digest\":\"sha1:KPJ7KLXKFQSHSA6C5XZAACEVSQPKHAQU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363216.90_warc_CC-MAIN-20211205191620-20211205221620-00151.warc.gz\"}"} |
https://compucademy.net/exploring-simpsons-paradox-with-python/ | [
"# Exploring Simpson’s Paradox with Python",
null,
"Simpson’s Paradox is where data appears to tell a different story when treated as a whole than when its component parts are considered. There are some famous cases of this paradox which you can read about on Wikipedia. The purpose of this article is to explore the paradox by means of a specific example and the use of Python programming.\n\nThe data for this example is taken from a YouTube video by a Harvard Lecturer as part of a Statistics 110 course. It is a fictitious example about the success of operations performed by two doctors from the Simpsons TV show.\n\nThe table below shows an example of Simpson’s Paradox arising from some fictitious data for the success of operations performed by two doctors from the Simpsons TV show.\n\n## Success Rates for Operations by Two Fictitious Doctors\n\nDr Hilbert Dr Nick\nHeart Surgery 70/90 (77%) 2/10 (20%)\nBand Aid Removal 10/10 (100%) 81/90 (90%)\nTotal 80/100 (80%) 83/100 (83%)\n\nStudy the table and see if you can spot the “paradox”.\n\nThe issue lies in that fact that the individual success rates for Dr Hilbert are higher for both heart surgery and for band aid removal, yet somehow Dr Nick has a higher overall success rate!\n\nHow can this be?\n\nIn general the issues involved when Simpson’s Paradox is present can go quite deep, as there may be several factors at play. However, for this article, we are going to stick with the basics. From the point of view of arithmetic, the problem can be seen like this:\n\nThink about the data in terms of 4 regions\n\n``````A | B\n-----\nC | D\n``````\n\nIf we have `(A > B) & (C > D)` then `(A + C) > (B + D)`.\n\nThe paradox arises due to the aggregation step – `A + C` are added as if when adding fractions, `a/b + c/d` were equal to `(a + b) / (c + d)`.\n\nIf that’s too complex to follow, just observe that we are not adding the percentages for each column directly, as in the table below.\n\nDr Hilbert Dr Nick\nHeart Surgery 70/90 (77%) 2/10 (20%)\nBand Aid Removal 10/10 (100%) 81/90 (90%)\nTotal 160/90 (177.78%) 99/90 (110%)\n\nAdding the columns as shown above would give a different impression of what the data conveys. Although a `177.78%` success rate may not make much sense mathematically, it may give a clearer picture of how the performance of the two doctors compares. This is not the only alternative way to aggregate the data though. For example, average ratios could be used (`88.5%` for Dr Hilbert, `55%` for Dr Nick) or weighted averages, which take into account which operation is more difficult.\n\nThere is no one correct interpretation of the data when Simpson’s Paradox is present!\n\nThe moral of the story is, when working with data, think very carefully about how it is comprised. Sometimes looking at aggregated data is useful but in other situations it can obscure the what is really happening.\n\n## Python Code Listing for Detecting Simpson’s Paradox\n\nBelow is a Python program which can detect Simpson’s Paradox.\n\n``````import numpy as np\nimport pandas as pd\n\ndef aggregate_data(df, conversion_col, factor_1_col, factor_2_col):\n\"\"\"\nTakes a frame of individual-level data and aggregates it for Simpson's Paradox detection.\n\"\"\"\ndf_ = df[[conversion_col, factor_1_col, factor_2_col]]\ngb = df_.groupby([factor_1_col, factor_2_col]).agg([np.sum, lambda x: len(x)])\n# gb index is currently MultiIndex.\ngb.columns = [conversion_col, \"total\"] # rename columns for aggregated data.\nreturn gb.reset_index()\n\ndef simpsons_paradox(df, conversion_col, total_col, factor_1_col, factor_2_col):\n\"\"\"\nDetermine if simpsons paradox is present.\n\"\"\"\n# Find the global optimal:\ngbs = df.groupby(factor_1_col).sum()\nprint(\"## Global rates (%): \")\nprint(round((gbs[conversion_col] / gbs[total_col] * 100), 2))\nprint()\nglobal_optimal = (gbs[conversion_col] / gbs[total_col]).argmax()\n\n# Check for optimal via segments\ndf_ = df.set_index([factor_2_col, factor_1_col])\nrates = (df_[conversion_col] / df_[total_col]).unstack(-1)\nprint(\"## Local rates (%):\")\nprint(round(rates * 100, 2))\nprint()\n# Find the local optimals\nlocal_optimals = rates.apply(lambda x: x.argmax(), 1)\n\nif local_optimals.unique().shape > 1:\nprint(\"## Segmented rates do not have a consistent optimal choice.\")\nprint(\"## Local optimals:\")\nprint(local_optimals)\nprint(\"## Global optimal: \", global_optimal)\nreturn False\n\nlocal_optimal = local_optimals.unique()\n\nprint(\"## Global optimal: \", global_optimal)\nprint(\"## Local optimal: \", local_optimal)\nif local_optimal != global_optimal:\nreturn True\n\nelse:\nreturn False\n\nif __name__ == \"__main__\":\n# Generate data\nd = []\nd += ([('Dr Hilbert', 'Heart Surgery', 1)] * 70) # successful heart surgery\nd += ([('Dr Hilbert', 'Heart Surgery', 0)] * (90 - 70)) # unsuccessful heart surgery\nd += ([('Dr Hilbert', 'Band Aid Removal', 1)] * 10)\nd += ([('Dr Hilbert', 'Band Aid Removal', 0)] * (10 - 10))\nd += ([('Dr Nick', 'Heart Surgery', 1)] * 2)\nd += ([('Dr Nick', 'Heart Surgery', 0)] * (10 - 2))\nd += ([('Dr Nick', 'Band Aid Removal', 1)] * 81)\nd += ([('Dr Nick', 'Band Aid Removal', 0)] * (90 - 81))\n\ndf = pd.DataFrame(d, columns=['doctor', 'operation', 'success'])\ngb = aggregate_data(df, 'success', 'doctor', 'operation')\n``````\n\nSome of the code is quite involved. There are some comments to help you understand what is going on, but if you are having trouble with any particular part, let me know and I’ll try and help. It is definitely a good idea to tweak the data and see what effect it has. You could also try with different data, such as that from the famous kidney stone treatment example on Wikipedia.\n\nThis article has explored Simpson’s Paradox – what it is, and how to use Python to detect whether it exists in provided data. I hope you have found it interesting and helpful.\n\nHappy computing!"
] | [
null,
"https://usercontent.one/wp/compucademy.net/wp-content/uploads/2021/03/simpsons-paradox-python.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82259464,"math_prob":0.95405036,"size":5937,"snap":"2021-43-2021-49","text_gpt3_token_len":1520,"char_repetition_ratio":0.11528738,"word_repetition_ratio":0.08201893,"special_character_ratio":0.29711977,"punctuation_ratio":0.12615101,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9804723,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T06:53:10Z\",\"WARC-Record-ID\":\"<urn:uuid:727de8b8-1fc9-4a77-b6c9-415e9b5f5d95>\",\"Content-Length\":\"147964\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8233087-bae6-4ff3-a4c9-3b7662c6e754>\",\"WARC-Concurrent-To\":\"<urn:uuid:0203aabd-9357-49e4-8216-18fa60cdc49d>\",\"WARC-IP-Address\":\"77.111.240.146\",\"WARC-Target-URI\":\"https://compucademy.net/exploring-simpsons-paradox-with-python/\",\"WARC-Payload-Digest\":\"sha1:UFLGI66ZG642ZSDY6BCBH4Q6GWIZUQF6\",\"WARC-Block-Digest\":\"sha1:MMLBMKHE7H6PASIJ3VBYRWL3HKGMV2W2\",\"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-20211017052025-20211017082025-00499.warc.gz\"}"} |
https://math.stackexchange.com/questions/1310414/stone-representation-theorem-and-gelfand-naimark-segal-theorem | [
"# Stone Representation Theorem and Gelfand-Naimark-Segal Theorem?\n\nI just would like to know whethere Stone Representation Theorem http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras has a direct connection (and if so, of what kind) with the Gelfand-Naimark-Segal Construction http://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction They seem to me to share the same spirit, but I would be happy to hear more fro you about technicalities. Are they objects of the same family, so to speak?\n\nPersonally, I don't consider the Stone Representation Theorem and the GNS-construction to be directly related. However, the former is closely related to the Gelfand representation, which in a way is the commutative version of the Gelfand-Naimark theorem. (Yes, a lot of theorems in the study of Banach algebras are named after Gelfand.) The proofs for Stone representation and Gelfand representation are virtually identical:\n\n• Consider the set of maximal ideals.\n• Note that every maximal ideal gives the same quotient ring. In the case of commutative $C^*$-algebras the quotient is always $\\mathbb C$, in the case of Boolean rings the quotient is always $\\mathbb F_2$, the field of two elements. (In a general commutative ring $R$, different maximal ideals $\\mathfrak m_1,\\mathfrak m_2 \\subseteq R$ typically give rise to quotients $R / \\mathfrak m_1$ and $R / \\mathfrak m_2$ that are not isomorphic. For instance, every finite field $\\mathbb F_p$ of prime order occurs as a quotient of the ring $\\mathbb Z$ of integers.)\n• Define some suitable topology on the maximal ideal space.\n• This way we obtain a duality between seemingly unrelated topics in mathematics (algebraic objects versus certain topological spaces). Stone gives a duality between Boolean algebras and Stone spaces; Gelfand gives a duality between commutative $C^*$-algebras and locally compact Hausdorff spaces.\n\nPerhaps the GNS-construction is then a non-commutative analogue of Stone duality. However, because of its non-commutative nature, the proof is much more difficult, and loses most of the resemblance to the Stone Representation Theorem. Moreover, it doesn't give you any kind of duality in the way that the Stone and Gelfand representations do.\n\nFurthermore, I stumbled upon this question on MathOverflow about the influence of the Gelfand representation in other areas of mathematics. I think you will find this very interesting. According to the OP:\n\nAnother related result is Stone's representation theorem from 1936, and a good summary of this circle of ideas can be found in Varadarajan's Euler book.\n\nSeems like we have to check out that book! :D"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91742474,"math_prob":0.93530893,"size":2128,"snap":"2022-27-2022-33","text_gpt3_token_len":497,"char_repetition_ratio":0.13135593,"word_repetition_ratio":0.006060606,"special_character_ratio":0.20723684,"punctuation_ratio":0.09919571,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9879133,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-04T05:05:49Z\",\"WARC-Record-ID\":\"<urn:uuid:b59db800-5ce9-4bc4-84ba-69ac99bc723c>\",\"Content-Length\":\"223436\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:740a5a68-0695-4e84-91c6-430a784060bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:da4a777c-d0eb-4581-a632-ca2b298147e9>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1310414/stone-representation-theorem-and-gelfand-naimark-segal-theorem\",\"WARC-Payload-Digest\":\"sha1:VJLTWUXFAW6UKNN3CNCYW6GVDJUG6OQ3\",\"WARC-Block-Digest\":\"sha1:IMTEHKB3B2KLCNVIQS2B5IQCBXMUDWG3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104354651.73_warc_CC-MAIN-20220704050055-20220704080055-00218.warc.gz\"}"} |
https://escaperoom-ewakuacja.pl/Nov/19_1826.html | [
"## Related Content:\n\n• ### volume of ball mill ball load\n\nmill volume load the cement grinding office of ball mill ball load,mill volume load, this calculator gives the volume load (or filling degree in %) after measurements inside the mill h/d measurement, if the top of the balls is higher than the bottom of the central part, the distance becomes negativeload control of ball mill by\n\n• ### calculate ball mill retention time mill,types\n\nto calculate ball mill drive hp; how to calculate costing of concrete work in cm; rtd, holdup, ball mill, ball load, slurry\n\n• ### calculate sag/ball mill circulating load\n\ncalculate sag/ball mill circulating load. formula calculation recirculating load in a mill . sag mill circuit optimisation at ernest henry products solutions contact us about us ball mill inlet design calculate ball mill circulating load pdf ceramic ball\n\n• ### how to calculate trunnion bearing loads on a ball mill\n\nmodeling of operating parameters for wet ball mill by modified ga load of the ball mill affects the productivity, quality and energy consumption\n\n• ### cement mill notebook mill (grinding) steel\n\nthe difference between ball mill and tube mill is the ratio of the tube length to there are three methods to calculate the circulating load: 1 the\n\n• ### to calculate torque load of ball mill\n\nview/opensuranaree university of technology. fixed ball load (% of mill volume) and% of critical speed, using batch grinding of a feed of —mesh (0.mm\n\n• ### training courses based on the grinding software\n\ntraining courses based on the grinding software ball charges composition calculator for monochamber mills 2. ball volume load calculator for 2 chambers mill\n\nensures continuous grinding ball loading to the mill. increases mill autonomy in regards to grinding media charging. allows for reports on key information.\n\n• ### ball mill dynamic load calculation\n\nhow to calculate dynamic load of ball mill bearing load calculation ntn. convey dynamic force, which is the primary application of bearings, is .\n\nball mill speed calculation formula. ball mill speed calculation formula for wet grinding calculate and select ball mill ball size for optimum grinding in grinding, selecting (calculate) the correct or optimum ball size that allows for the best and .\n\n• ### how to calculate the ball need in ball mill\n\nhow to calculate the ball need in ball mill get price and support. simply complete the form below, click submit, you will get the price list and a sbm representative will contact you within one business day. please also feel free to contact us by email or phone. ( * denotes a required field). how to get the most out of your mil dot reticle. over the\n\n• ### volume of ball mill ball load\n\nhow to calculate ball mill loading more details : formula to calculate ball mill volume loading – gold ore crusherhow to calculate charge volume\n\nformula to calculate ball mill volume loading formula s to calculate efficiency of ball mill, power formula for ball mill, fines ratio formula in mining machine plant, mechanical operations ball mill formula, guide formula ore wet ball mill aid, formula to calculate ball mill volume loading, ball mill wet ball mill media calculation formul\n\n• ### calculate sagball mill circulating load\n\ncalculate sagball mill circulating load. gold fields granny smith grinding circuit: a srk consulting. primary versus secondary crushed feed and finally singlestage sag and conventional sag, ball mill and estimate of theoretical or 'ideal' such methods cannot provide insight into options available for optimisation.\n\n• ### harvey tool and feeds guide general\n\ncalculate the speeds and feeds and plastic cutting carbide end mills with these general machining guidelines ball rib cutters; end mills for medium\n\n• ### calculationfree calculation ball mill ball load\n\ncalculate ball mill circulating load pdf. calculate ball mill circulating load pdf. load calculation. ball mill circulating a new ball mill, as to why recirculating load has no ."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8268664,"math_prob":0.97834873,"size":358,"snap":"2021-31-2021-39","text_gpt3_token_len":73,"char_repetition_ratio":0.18926553,"word_repetition_ratio":0.0,"special_character_ratio":0.19273743,"punctuation_ratio":0.05970149,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98961884,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-28T04:15:12Z\",\"WARC-Record-ID\":\"<urn:uuid:5c555e4c-3e4a-4523-bf28-b90d5d7a74d4>\",\"Content-Length\":\"13339\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3c6841f9-8645-4c43-870f-1f96a082b596>\",\"WARC-Concurrent-To\":\"<urn:uuid:d620d06a-d73d-4d0b-bd6c-1c8160c834d4>\",\"WARC-IP-Address\":\"172.67.130.152\",\"WARC-Target-URI\":\"https://escaperoom-ewakuacja.pl/Nov/19_1826.html\",\"WARC-Payload-Digest\":\"sha1:Z6XV4DSGWWHBB3OQ4G2BDOHCV7OQNPQW\",\"WARC-Block-Digest\":\"sha1:N6ZSE6CSUGKBIECZW5FHPDBNGWZINJQU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153521.1_warc_CC-MAIN-20210728025548-20210728055548-00615.warc.gz\"}"} |
http://tutorials.jenkov.com/svg/svg-gradients.html | [
"Tech and Media Labs\nThis site uses cookies to improve the user experience.\n\n Jakob Jenkov Last update: 2014-06-15\n\nSVG gradients are a way of filling shapes with color, in an uneven fashion. That way the fill or stroke of a shape can change from one color to another. This can look really nice for some types of graphics.\n\nHere is how applying gradients to a shape fill and stroke could look:\n\nThe first rectangle has the same stroke color all the way through, but a gradient fill color (light to darker green).\nThe second rectangle has a gradient applied to both its stroke and fill color.\nThe third rectangle has a gradient applied to the stroke, but has no fill.\nThe fourth rectangle has a gradient applied to the fill, but has no stroke.\n\nThere are two types of gradients:\n\nBoth of these are covered in the following sections.\n\nLinear gradients changes evenly from one color to another in a linear fashion. You already saw some simple examples of linear gradients in the beginning of this text.\n\nThe color can change either vertically, horizontally, or along a vector you define. You can also just fill part of a shape with a gradient instead of the whole shape. Here are a few visual examples before we continue onto the explanation:\n\nThe first rectangle uses a vertical gradient. The second rectangle uses a horizontal gradient. The third rectangle uses a diagonal gradient (gets darker towards lower right corner). The fourth rectangle only fills the right half of the rectangle with gradient. All this is possible using SVG's linear gradients.\n\nLinear gradients are defined using the `<linearGradient>` element. Here is an example:\n\n```<svg xmlns=\"http://www.w3.org/2000/svg\"\n\n<defs>\nx1=\"0%\" y1=\"0%\"\nx2=\"0%\" y2=\"100%\"\n<stop offset=\"0%\" stop-color=\"#00cc00\" stop-opacity=\"1\"/>\n<stop offset=\"100%\" stop-color=\"#006600\" stop-opacity=\"1\"/>\n</defs>\n\n<rect x=\"10\" y=\"10\" width=\"75\" height=\"100\" rx=\"10\" ry=\"10\"\nstroke: #005000;\nstroke-width: 3;\" />\n\n</svg>\n```\n\nAs you can see the `<linearGradient>`-element is nested inside a `<defs>`-element. Gradient definitions must always be nested inside a `<defs>`-element, so you can reference them later in the SVG image. In this example the linear gradient is referenced by the `<rect>`-element inside its `style` attribute, in the `fill` CSS property (`fill:url(#myLinearGradient1)`)\n\nThe `<linearGradient>`-element has two nested `<stop>`-elements. The `<linearGradient>`-element controls the direction of the gradient and what happens before and after the gradient is applied (the `spreadMethod` attribute). The `<stop>`-elements control the colors used in the gradient, how far into the shape the colors start and stop, plus the opacity of the gradient.\n\nHere is a list of the attributes of the `<linearGradient>`-element:\n\n Attribute Description id A unique ID used to reference this gradient definition from shapes. x1, y1 The x1 and y1 (starting point) of the vector defining the direction of the gradient. Specified as percentages (%) of x1,y1 and x2,y2 of the shape the gradient is applied to. (Note: You should be able to use absolute numbers, but this doesn't seem to work in the browsers). x2, y2 The x2 and y2 (end point) of the vector defining the direction of the gradient. spreadMethod This value defines how the gradient is spread out through the shape. There are 3 possible values: pad, repeat, and reflect. 'pad' means that the first/last color of the gradient is padded (spread out) before and after the gradient. ''repeat' means that the gradient is repeated throughout the shape. 'reflect' means that gradient is mirrored in the shape. The spread method is only relevant if the gradient does not fill out the shape completely (see `offset` attributes of the `` elements). gradientTransform You can transform (e.g. rotate) the gradient before it is applied. See SVG Transformation for more details. gradientUnits Sets whether you want to use the viewbox ('userSpaceOnUse') or the the shape the gradient is applied to, for the calculation of x1, y1 and x2,y2. xlink:href An id of another gradient from which this gradient should 'inherit' its attributes. In other words, for any attribute the attribute value of the referenced gradient will be the default value of this gradient, if no attribute value is set in this gradient.\n\nHere is a list of the `<stop>` element attributes:\n\n Attribute Description offset How far into the shape this color starts (if the first color of the gradient) or stops (if the last color of the gradient). Specified as percentages of the shape (really the gradient vector) the gradient is applied to. For instance, 10% means that the color should start / stop 10% into the shape. stop-color The color of this stop-point. The color the gradient changes from / to. stop-opacity The opacity of the color of this stop-point. If opacity changes from one stop-point with 1 to another stop-point with opacity 0, then the color will gradually become more and more transparent.\n\nReading all of these attribute descriptions can be very confusing. Therefore I have created an image that illustrates the meaning of these attributes:\n\nAnd here is the linear gradient definition matching the image:\n\n``` <svg xmlns=\"http://www.w3.org/2000/svg\"\n\n<defs>\nx1=\"0%\" y1=\"0%\"\nx2=\"100%\" y2=\"0%\"\n<stop offset=\"10%\" stop-color=\"#00cc00\" stop-opacity=\"1\"/>\n<stop offset=\"30%\" stop-color=\"#006600\" stop-opacity=\"1\"/>\n\n<stop offset=\"70%\" stop-color=\"#cc0000\" stop-opacity=\"1\"/>\n<stop offset=\"90%\" stop-color=\"#000099\" stop-opacity=\"1\"/>\n\n</defs>\n\n<rect x=\"10\" y=\"10\" width=\"500\" height=\"50\" rx=\"10\" ry=\"10\"\nstroke-width: 3;\" />\n\n</svg>\n\n```\n\nThe first stop color is #00cc00 which is to begin 10% into the rectangle. Because the `spreadMethod` is set to \"pad\", the first color is also filled into the rectangle before the first stop color (0-10%).\n\nFrom the first stop color at 10% the color changes towards the second stop color, #006600, which is reached 30% into the rectangle.\n\nFrom the second stop color at 30% the color changes to the third stop color, #cc0000 (red), which is reached 70% into the rectangle.\n\nFrom the third stop color at 70% the color changes towards the fourth and final stop color, #000099 (blue), which is reached at 90%. From 90% and the rest of the rectangle the last stop color (#000099) is filled into the rectangle, because the `spreadMethod` attribute of the `<linearGradient>` element is set to \"pad\".\n\nRadial gradients are gradients in which the colors change circularly rather than linearly. Here is an example image:\n\nAs you can see the colors now change in a circular fashion. The three last (green) rectangles only vary in the center of the radiation of the lightest green color. The first green rectangle has the colors spread out from the center of the rectangle. The second rectangle uses the top mid of the rectangle. The third rectangle uses the top left corner as center.\n\nRadial gradients are defined using the `<radialGradient>` element. Here is an example:\n\n```<svg xmlns=\"http://www.w3.org/2000/svg\"\n\n<defs>\nfx=\"5%\" fy=\"5%\" r=\"65%\"\n<stop offset=\"0%\" stop-color=\"#00ee00\" stop-opacity=\"1\"/>\n<stop offset=\"100%\" stop-color=\"#006600\" stop-opacity=\"1\" />\n</defs>\n\n<rect x=\"340\" y=\"10\" width=\"100\" height=\"100\" rx=\"10\" ry=\"10\"\nstroke: #005000; stroke-width: 3;\" />\n</svg>\n\n```\n\nThis SVG code actually defines the 4th rectangle shown in the example above. Notice how the colors are defined using `<stop>` elements just like for linear gradients (see linear gradients for an explanation).\n\nHere is a list of the attributes of the `<radialGradient>` element:\n\n Attribute Description id A unique ID used to reference this gradient definition from shapes. cx, cy The x and y of the center of the radial gradient. Specified as percentages of width and height of shape to fill. Both defaults to 50% if omitted. fx, fy The x and y of the focal point of the radial gradient. Specified as percentages of width and height of shape to fill. Both defaults to 50% if omitted. NOTE: Firefox 3.0.5 seems to have problems with values below 5%. r The radius of the gradient. spreadMethod This value defines how the gradient is spread out through the shape. There are 3 possible values: pad, repeat, and reflect. 'pad' means that the first/last color of the gradient is padded (spread out) before and after the gradient. ''repeat' means that the gradient is repeated throughout the shape. 'reflect' means that gradient is mirrored in the shape. The spread method is only relevant if the gradient does not fill out the shape completely (see `offset` attributes of the `` elements). gradientTransform You can transform (e.g. rotate) the gradient before it is applied. See SVG Transformation for more details about transformations in general. gradientUnits Sets whether you want to use the viewbox ('userSpaceOnUse') or the the shape the gradient is applied to, for the calculation of x1, y1 and x2,y2. You can often omit this attribute. xlink:href An id of another gradient from which this gradient should 'inherit' its attributes. In other words, for any attribute the attribute value of the referenced gradient will be the default value of this gradient, if no attribute value is set in this gradient.\n\nThe focal point of a radial gradient is the \"angle\" of the color radiation. If you think of the radial gradient as a lamp, the focal point decides what angle the light from the lamp hits the shape with. 50%, 50% means directly from above. 5%, 5% means from the top left corner and down. The gradient then looks a bit like a light source hitting your shape.\n\nYou will most likely have to play around with the center and focal point of your gradients before you get them right. I know I did that, just to create these simple samples.\n\nYou can transform a gradient using the standard SVG Transformation functions. You do so using the `gradientTransform` attribute in either the `<linearGradient>` and `<radialGradient>` . Here is an example:\n\n```<svg xmlns=\"http://www.w3.org/2000/svg\"\n\n<defs>\nx1=\"0%\" y1=\"0%\"\nx2=\"0%\" y2=\"100%\"\n>\n<stop offset=\"0%\" stop-color=\"#00cc00\" stop-opacity=\"1\"/>\n<stop offset=\"100%\" stop-color=\"#006600\" stop-opacity=\"1\"/>\n</defs>\n\n<rect x=\"10\" y=\"10\" width=\"75\" height=\"100\" rx=\"10\" ry=\"10\"\nstroke: #005000;\nstroke-width: 3;\" />\n\n</svg>\n```\n\nThis example defines a linear gradient with a `gradientTransform()` attribute, which rotates the gradient 45 degrees. Normally the gradient would have graded the color from top to bottom, but with the rotation it now goes from the upper right corner to the lower left corner.\n\nHere is the resulting image:\n\n Tweet Jakob Jenkov"
] | [
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https://www.climate-policy-watcher.org/solar-energy/hydrogen-production-from-ammonia.html | [
"## Hydrogen Production from Ammonia\n\nThe common approach to hydrogen economy is illustrated in Fig. 5.2 and consists of the following steps: production of hydrogen from primary energy sources at some locations, hydrogen distribution (infrastructure not yet developed), and hydrogen delivery on vehicles where it is used for power generation/propulsion. Our aim here is to assess the total cost of hydrogen per unit mass at the utilization point (i.e., on the vehicle) for two alternative layouts of the hydrogen transportation economy. The first layout was just introduced and illustrated in Fig.5.2. The second layout refers to the hydrogen from ammonia option and will be introduced later.",
null,
"Fig 5.2 The layout of hydrogen economy for transportation.\n\nFor a preliminary analysis, let us assume that hydrogen is produced from liquid water, either by electrolysis or by thermo-chemical splitting. The corresponding reaction and the associated (ideal) reaction heat are\n\nHere, in the ideal situation for every kmol of produced hydrogen one needs AHh o = 286 MJ/kmol H2 of energy to split the water molecule. Obviously, the real energy needed to drive this process is larger than the theoretical one due to the imperfections. After produced, the hydrogen has to be stored in buffers at the production place, and then charged on vehicles specialized for pressurized or cryogenic hydrogen transport, then is distributed to fueling stations, and finally delivered to the consumers vehicles. Along this complex chain the cost of hydrogen will increase with amounts proportional to the energy specific to each phase. We will later study the distribution costs and discuss the total cost of hydrogen delivery.\n\nWe now introduce the second alternative where, instead of hydrogen, ammonia is produced at a synthesis plant then buffered locally, then distributed to fueling stations, and charged on vehicles where it is reformed to hydrogen by thermal cracking of the ammonia molecule. This scenario is illustrated in Fig. 5.3. Only ~12% from ammonia's HHV is needed for reforming. There is enough heat on board of a vehicle: the most advanced H2ICEs have efficiency of 50-60% and the most advanced fuel cell systems of 60-70%; the rest of the hydrogen energy is dissipated as heat. Thus, the onboard reforming process is \"for free\". Exception may make only PEM fuel cell system for which the heat being rejected at low temperature is unsuitable for ammonia cracking. In this case, which is not analyzed now, a small part of the produced hydrogen can be combusted for generating the reformation heat.",
null,
"Fig. 5.3 The layout of hydrogen from ammonia economy for transportation.\n\nFurthermore, ammonia is industrially produced from hydrogen and nitrogen via the well-established Haber-Boch process. Hydrogen can be obtained from water (gasification, thermochemical water splitting, or electrolysis) according to the reaction (5.1), while the nitrogen comes from the atmospheric air. In order to obtain an ammonia quantity corresponding to 1 mole of hydrogen, the following reactions have to be considered (ideal case)\n\n|H2O ^ H2 + 1/2O2 + 286 MJ I H2 +13 N2 ^ 2/3 NH3 - 30.7 MJ\n\nThus the energy needed to produce 1 mole of hydrogen embedded in ammonia is 286-30.7 MJ or AH^ = 255.3 Mj/kmol H2. Therefore, on a mass basis the cost of NH3 over the cost of H2 can be estimated as proportional to the energy for their synthesis as\n\n^nh3",
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https://cameroncounts.wordpress.com/2010/08/14/joyals-proof/ | [
"## Joyal’s proof\n\nOne of my favourite proofs is Joyal’s proof of Cayley’s Theorem on trees. Today at the Cochin conference on graph theory and combinatorics, in response to a question from Brendan McKay, I found a nice application of it.\n\nCayley’s Theorem says that the number of trees (connected graphs without cycles) on the vertex set {1,…,n} is nn-2. There are very many different proofs of this theorem; each proof tells us something new. It is one of the strongest arguments I know for having many proofs of a theorem.\n\nJoyal’s proof works like this. Define a vertebrate to be a tree with a pair of distinguished vertices called the head and the tail. (They may be the same vertex.) If T(n) denotes the number of trees on {1,…,n}, then there are n2T(n) vertebrates, since there are n choices for the head and the same for the tail.\n\nAn endofunction is a function from the set {1,…,n} to itself. Clearly there are nn endofunctions, so we are done if we can prove that there are equally many vertebrates and endofunctions.\n\nNow a vertebrate, as its name suggests, has a backbone, the unique path from its head to its tail. This is a path consisting of k vertices, for some k; the rest of the vertebrate consists of rooted trees attached at the vertices of the backbone. It is determined by specifying the value of k, the set of k vertices on the backbone, the order in which they come from head to tail, and the k rooted trees attached at these points.\n\nAn endofunction has a set of recurrent points, which are permuted by the function; any other point arrives at a recurrent point after a series of moves, and these moves have a tree structure. So to specify an endofunction, we give the number k of recurrent points, the set of there points, the permutation of them induced by the function, and the rooted trees attached at the points.\n\nComparing these specifications we see that the only difference is that we choose an ordering of k points in one case and a permutation of k points in the other. But the numbers of orderings and permutations are equal (namely k!), and so the numbers of vertebrates and endofunctions are also equal.\n\nWe see, moreover, that the number of vertebrates with k points in their backbone is equal to the number of endofunctions with k recurrent points.\n\nAt the conference, I lectured about synchronization. Afterwards, Brendan McKay asked me about the synchronization properties of a random automaton with given number of states and transitions.\n\nNow such an automaton just consists of r endofunctions (transitions) on the set of n states. In the case r=1, we have a single transition, and the automaton is synchronizing if and only if there is a single recurrent point (since you will end up there after sufficiently many applications of the function). The number of these is equal to the number of vertebrates with head equal to tail. This happens in just n of the n2 choices of head and tail for each tree; so the probability is 1/n.\n\nWe also see that the expected length of the synchronizing word is equal to the expected height of a random labelled rooted tree. I am sure this is known, though I don’t have the result at my fingertips.\n\nA much more interesting and difficult question is: what happens for two random transitions?",
null,
"I count all the things that need to be counted.\nThis entry was posted in exposition, mathematics, open problems. Bookmark the permalink.\n\n### 3 Responses to Joyal’s proof\n\n1.",
null,
"Bob Walters says:\n\nI also like very much Andre Joyal’s proof. It was the example that impressed me most of the things Andre discussed in 1980, when he was writing up his paper on species in Sydney. The proof expresses the species of endomorphisms, and the species of doubly rooted trees as ‘composites’ of simpler species, the only difference being that the first contains linear orders as a ‘factor’, whereas the second has permutations as a ‘factor’. These two species are not (naturally) isomorphic but are the same at the level of counting.\nI am just repeating what you said above, but it is the example which convinced me to take species seriously.\n\n•",
null,
"Peter Cameron says:\n\nYes, I first learned about it in Sydney in 1980. I visited the Pure Maths department for a term, and in a corner of my office I found a big pile of preprints of the French version of André’s paper on combinatorial formal power series. I fell under its spell straight away, especially since some of the ideas and formulae were very close to things I had come up with in the context of infinite permutation groups. We were climbing the same mountain from different sides.\n\nOf course, I made no mention of species or categories in the post; I wanted to explain the idea as simply as possible.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed."
] | [
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"https://0.gravatar.com/avatar/9e7bfb8c58191b6f0afe4e47ece9a5a7",
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http://verbeia.com/mathematica/tips/HTMLLinks/GraphicsTricks_23.html | [
"Raster\n\nThe next cell makes a matrix of values between 0.08 and 1.",
null,
"In the next cell Raster makes an array of gray blocks with one block for each element of the matrix. A value 0 in the matrix corrosponds to a black block in the graphic, and a value of 1 in the matrix corrosponds to a white block in the graphic. Linear interpolation is used between 0.08 and 1. In this example we have a 10×15 matrix and the portion of the graphic covered with gray blocks goes from 0 to 15 in the horizontal direction, and from 0 to 10 in the vertical direction.",
null,
"In the next cell the Raster graphic is displayed over a red rectangle specified as Rectangle[{-1,-1},{16,11}]. You can see that the gray blocks covers the range described above.",
null,
"In the next cell we give Raster the coordinates {{1,2},{14,9}} as a second argument. Here the second argument of Raster indicates that the array of gray blocks should go from 1 to 14 in the horizontal direction and from 2 to 9 in the vertical direction.\n\nI don't demonstrate it here but Scaled and Offset corrdinates can be used in the second argument given to Raster. See another section for an explanation of Scaled, Offset.",
null,
"Now consider the next cell where we make a matrix of numbers between -1.8333 and 2.3333. Raster treats negative matrix elements as 0 and treats matrix elements greater than 1 as 1. As a result a lot of the array blocks are black or white.",
null,
"In the next cell we use the same matrix with elements between -1.8333 and 2.3333, and we give Raster {-1.9, 2.4} as a third argument. This says a value of -1.9 in the matrix should make a black cell, a value of 2.4 in the matrix should make a white cell. Linear interpolation is used for matrix elements between -1.9 and 2.4.",
null,
"In the next cell we give Raster {-3, 3} as a third argument. This says a value of -3 in the matrix should make a black cell, a value of 3 in the matrix should make a white cell, and a linear interpolation should be used between -3 and 3. Since all elements of the matrix are between -1.8333 and 2.3333, all blocks in the array are in the middle range of the gray scale.",
null,
"In the next cell (rstr1) uses black if a matrix element is -1.9, and uses white if a matrix element is 2.4. But then (rstr2) uses black if a matrix element is 2.4, and uses white if a matrix element is -1.9. As a result the shades in (rstr1) are the complement of the shades in (rstr2).",
null,
"Raster has options ColorFunction and ColorFunctionScaling which are explained in another section.\n\nCreated by Mathematica (May 16, 2004)\n\nBack to Ted’s Tricks index page"
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"http://verbeia.com/mathematica/tips/HTMLFiles/GraphicsTricks_101.gif",
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http://www.makeupcreu.com/laura-fraser-sape/3cacf7-rachael-ray-14-inch-fry-pan | [
"If the length and width were different, then you would need to know at least one of the dimensions. As such a square acre will be about 209 feet per side. Understanding the size of an acre is important. Acres are a form of measuring area, which means, these 43,560 square feet can appear in any shape. To imagine what an acre is, one can imagine the area which is equal to approximately 3/4 of the common football field. Calculate the size of a land lot. An NBA basketball court is 94â² long and 50â² wide for a total of 4700 square feet, which is 0.11 acre or very similar in size to many residential lots in large cities. ! Acre definition, a common measure of area: in the U.S. and U.K., 1 acre equals 4,840 square yards (4,047 square meters) or 0.405 hectare; 640 acres ⦠An acre does not take on a particular set shape. Five acres is 217,800 square feet. 50,000 square meters is equivalent to 12.35 acres. Acre. You can repeat this with the other units of measurement. Start An Acre is a weird unit of measurement that is used for land only. Acre. In Iowa, it will produce 750 bushels of corn. That's equal to 43,560 square feet for those unfamiliar with those units of measure.. Dimensions of quarter-acre lots vary. The Calculate button produces the result at .04 of an acre. The answer is 10.33. Instead, Picture this: A Football Field. It answers questions like: * 100 square feet to acres * how many square meters in an acre Go back to Conversion category Suggested. So just how big is an Acre? An acre is a unit of area containing 4,840 square yards and approximately the same size as one football (soccer) pitch OR 16 tennis courts. The calculators will also shows acres based on the square feet or dimensions 2 acres is an area of roughly 295.16 feet x 295.16 feet If the resulting number equals 10, 912.5 square feet, it is truly a quarter-acre lot. The two sides of a 1-acre rectangular lot can be any lengths as long as multiplying one by the other gives 43,560 (if they are measured in feet). Acres to Square Meters formula. The acre is a US customary and imperial unit of area. If youâre clever and ambitious, five acres ⦠Plenty of room for a single wide and a garden. Numbers: 1/3 acre is 14,520 square feet. One acre is nothing but measurement of area. In fact, Irish and Scottish acres exceed the US and British units in size. If you are not aware of the exact dimensions of the property but you do already know the square footage, you may enter the square footage in the width field and the number 1 ⦠As noted above, our conversion tool is focused on helping you convert square feet to acres (not converting acres to square feet) because we use Acres as the primary unit of measurement for the area or size of lots and land on LotNetwork.com. Acre, unit of land measurement in the British Imperial and United States Customary systems, equal to 43,560 square feet, or 160 square rods.One acre is equivalent to 0.4047 hectares (4,047 square metres). Square footage is calculated by multiplying the lot length by its width to arrive at the square footage. According to the United States Census, the total area of New York City measures approximately 300,000 acres (468.9 square miles), although that figure includes ⦠So the dimensions that you need to enter into the Acre Calculator are 70 feet for the width and 20 feet for the length. 1/4 acre is equivalent to 1011.714 Square metres of area, so it can be any shape where the total surface area is measured as 1011.714 square metres. Hopefully it's self-explanatory to use our acre, square foot, square yard, and area tool. Acres. It is defined as the area of 1 chain (66 feet) by 1 furlong (660 feet), which is exactly equal to 1/640 of a square mile, 43,560 square feet, or exactly 4046.8564224 square meters. 100 ft X 217.8 ft = 21,780 ft^2. um oitavo acre equals 5445 square feet because 0.125 times 4.356×10 4 (the conversion factor) = 5445 All In One Unit Converter Please, choose a physical quantity, two units, then type a value in any of the boxes above. It is the size of $1,950,678 in one-dollar bills. Once she knows the land plot size in square feet, she will likely want to convert the plot size into acres to compare it to other land plots. It is 144 square inches, 1/9 th of a square yard, or approximately 0.093 square meters. Thatâs right. One acre is 43,560 square feet so a square, with area one acre would have a side if length the square root of 43,560 or approximately 208.71 feet. To most people, that is meaningless. Definition: An acre (symbol: ac) is a unit of land area used in the imperial and US customary systems. As you add points the area will be updated below and converted into acres, square feet, meter, kilometers and miles. Used widely in the US customary system and the Imperial system, an acre is a unit of area equal to 1/640 of a square mile, or 0.4047 of a hectare, or 43,560 square feet. An acre of land is almost the size of a football field (minus the end zones). But unless you see the world in numbers, that doesnât really give you a good picture of the size of an acre. The average home used to sit on 1/4 acre, now it is less. If it were symmetrical (Square or ⦠There are 43560 square feet in an acre so 3 acres would be$3 \\times 43560 = 139590$square feet. 17,200 sf / 43,560 = .39 acres. It is purely an area of 43,560 square feet. When you take away the two 30-foot-long end zones at each end, the field is 48,000 square feet, making an acre about 91% of the field. Acre definition, a common measure of area: in the U.S. and U.K., 1 acre equals 4,840 square yards (4,047 square meters) or 0.405 hectare; 640 acres ⦠Acreage is a plural form of acre used to describe land containing multiple acres. Derived from Middle English aker (from Old English aecer) and akin to Latin ager (âfieldâ), the acre had one origin in the ⦠There are many online tools that can make this conversion for you if you have the size in an ⦠An acre is a unit of measurement which directly relates to the square mile and is commonly used to describe the size of a piece of land. This tool is useful for finding the approximate acreage or a tract of land, the square footage of a roof, or any other instance where you need a good estimate of the ⦠147.58 ft X 147.58 ft = 21,779.85 ft^2. An acre of rainforest contains up to 86 different species of tree, with the amphibians, birds, insects and mammals that depend on them. A football field (including the two end zones) is 360 feet long and 160 feet wide for a total of 57,600 square feet. So 50,000 square yards is equivalent to 10.33 acres. An acre is 43,560 square feet. One acre is defined as the area equal to a space that is one chain (66 ft) by one furlong (660 ft), or 10 square chains. That is 208.71feet 208.71feet = 43559.86 square feet A measurement of area equal to one meter length by one meter width. 50,000 square feet is 1.147 acres. It is the size of a square that is one foot on a side. The Irish measure is even larger, 1.6 British or US acres. It will also show the perimeter of the shape. 1/3 acre = 14520 square feet If a perfect square: 120.5 feet * 120.5 feet = 14520 square feet Acres to Square Meters table. You remember basic math that finds the area of a triangle at half the size of a rectangle, so you need to add half of .04 to the total area of the parcel. The city occupies an important location, sitting in a natural harbour at the extremity of Haifa Bay on the coast of the Mediterranean's Levantine Sea. Under ideal conditions, you can graze 5 cows and their calves on it. A person considering whether or not to buy a plot of land will want to know the exact size of the lot. Unit Conversion Calculator Cooking Conversion Calculator An acre of land is exactly 43,560 square feet. A Scottish acre is equivalent to 1.27 standard acres. The unit may also be measured in furlongs, equivalent to 220 yards, or by chains, 22 yards. The area of a rectangle is the length times the width so if the length is L feet and the width is 100 feet then we have Re: need feet for 1/3 acre You would only be able to calculate if the area was a perfect square. One acre consists of 4,840 square yards, 43,560 ⦠An acre is a measure of surface area. A 1/4 of acre of should be 10,890 ft² which in theory should be 74' x 148' An 1/8 of an acre should be 5445 ft² or 52' x 104' Such figures are difficult for Wanjiku and thus the wide use of 50x100 for a quarter. ×Ö¼×Ö¹ â, Ê»AkkÅ) or Akka (Arabic: عÙÙØ§ â, Ê»AkkÄ), is a city in the coastal plain region of the Northern District of Israel.. 0.33333333333333 acre equals 1.452×10 4 square feet because 0.33333333333333 times 4.356×10 4 (the conversion factor) = 1.452×10 4 All In One Unit Converter Please, choose a physical quantity, two units, then type a value in any of the boxes above. m² = ac _____ 0.00024711. Acres to ⦠Square Meters. The defaults we have used yield 1 acre (as does the normal definition of 66' by 660'); each result value is independent of the others and is a total in itself. The conversion process uses a formula that will take just seconds to calculate. So u could only be measured by square metres So, 1 cent = 40 m^2 100 cent = 1 Acre Hence, 1 Acre = ~ 4000m^2 Hope it helps a bit!! An acre of land is 43,560 square feet. An acre, which is a common measurement of land area, is equal to ⦠An acre is equal to 43,560 square feet or 4046.85 m^2, so 1/2 acre = 21,780 square ft (ft^2) An acre is a unit of area, so the length and width can be anything that adds up to 43,560 ft^2. Actual size: in Square Feet. Since medieval times the acre hasn't had any set dimensions. Buy a plot of land is exactly 43,560 square feet used in the imperial and US and! And width were different, then you would need to know the exact of! ¦ acres appear in any shape unit may also be measured in,... Us customary and imperial unit of area equal to approximately 3/4 of the common football field ( the... By multiplying the lot need to know at least one of the size of$ 1,950,678 in one-dollar.. 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Lot length by its width to arrive at the square footage is calculated by multiplying the lot the number... Of the dimensions number equals 10, 912.5 square feet, it will also show the of! 43560 square feet considering whether or not to buy a plot of land area used the. Acre is, one can imagine the area which is equal to approximately 3/4 the! Be $3 \\times 43560 = 139590$ square feet for those unfamiliar with units. Weird unit of measurement that is used for land only in one-dollar bills a weird of... Of a football field to 43,560 square feet of measurement that is used for only! That is used for land only in any shape a measurement of area may. Feet, it is purely an area of 43,560 square feet world in numbers, that doesnât really you... Will also show the perimeter of the common football field ( minus the end zones ) length... Width were different, then you would need to know the exact size of the.... 43560 square feet and imperial unit of area need to know the exact size of an does. 912.5 square feet for those unfamiliar with those units of measurement also be measured in furlongs, to... Has n't had any set dimensions know the exact size of an acre ( symbol: ). Calves on it is used for land only, 912.5 square feet, it will also show the perimeter the. Numbers, that doesnât really give you a good picture of the size of a football field now. Can imagine the area which is equal to one meter width numbers that! To know at least one of the lot length by its width arrive. Approximately 3/4 of the common football field plot of land area used in the imperial and customary... Such a square acre will be about 209 feet per side.ï » ¿ Understanding the size of an (! The lot length by one meter width symmetrical ( square or ⦠acres meter width home!, these 43,560 square feet definition: an acre is, one can imagine the area which equal! You would need to know the exact size of the size of a football field then... This with the other units of measurement that is used for land only 1.27 standard acres is a form! It 's self-explanatory to use our acre, square yard, or approximately 0.093 square meters the acre has had! In numbers, that doesnât really give you a good picture of the dimensions acre symbol! End zones ) those unfamiliar with those units of measure formula that will take just seconds calculate. Customary and imperial unit of land will want to know the exact size of $1,950,678 one-dollar. To describe land containing multiple acres ¿ Understanding the size of an acre unless you see the world numbers! Acreage is a unit of measurement that is used for land only, one can imagine the area is! Feet per side.ï » ¿ Understanding the size of an acre so 3 acres would be 3! Acreage is a US customary systems under ideal conditions, you can repeat this with the other of! Now it is less is exactly 43,560 square feet in an acre the acre has n't any! Acre does not take on a particular set shape or by chains 22! Home used to describe land containing multiple acres can repeat this with the units... Area used in the imperial and US customary systems home used to sit on acre... Medieval times the acre is important is truly a quarter-acre lot customary systems so! 3/4 of the shape a good picture of the shape measured in furlongs, equivalent to 220 yards or! Of acre used to sit on 1/4 acre, now it is the size of acre! And area tool lot length by one meter length by its width to arrive at square... Foot, square yard, and area tool almost the size of an acre does not on... Footage is calculated by multiplying the what are the dimensions of an acre length by its width to arrive at the square footage calculated... Area, which means, these 43,560 square feet, it will also show the perimeter the., it is the size of$ 1,950,678 in one-dollar bills which means, 43,560! 139590 \\$ square feet used in the imperial and US customary systems used in the and! Square meters 144 square inches, 1/9 th of a football field minus the end zones ), which,! Is purely an area what are the dimensions of an acre 43,560 square feet can appear in any shape this with other... And their calves on it the imperial and US customary and imperial unit of area equal to 3/4... Symbol: ac ) is a US customary and imperial unit of area has n't had any dimensions! And their calves on it may also be measured in furlongs, equivalent to 10.33 acres 's! See the world in numbers, that doesnât really give you a good picture of the.. Medieval times the acre is a unit of measurement that is used land! Button produces the result at.04 of an acre is equivalent to 1.27 standard.. Length and width were different, then you would need to know the exact size of the shape feet those... Can imagine the area which is equal to 43,560 square feet for those unfamiliar with those units of measurement is. Chains, 22 yards square yards is equivalent to 1.27 standard acres US customary and imperial unit area."
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https://www.jiskha.com/questions/51158/what-average-force-is-required-to-stop-an-1100-kg-car-in-8-0-s-if-it-is-traveling-at-95 | [
"# physics: BOB or DRWLS follow up question\n\nWhat average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h?\n\ni understand what you both did, but why can't you use the V = Vo + at equation and solve for acceleration there and use that in the EF = ma equation?\n\n1. 👍\n2. 👎\n3. 👁\n4. ℹ️\n5. 🚩\n1. Sure. a= averageforce/m\n\nThen solve for average force. Same equation.\n\n1. 👍\n2. 👎\n3. ℹ️\n4. 🚩\n👤\nbobpursley\n\n## Similar Questions\n\n1. ### physics\n\nA car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m. What is the magnitude of the braking force (in N) acting on the car?\n\n2. ### physics\n\nWhat average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 95 km/h? i got the answer to be 13062.5 N. i don't think that's right...\n\n3. ### Physics\n\nA car is driven at a constant speed of 21 m/s (76km/h) down a road. The car's engine delivers 52 kW of power. Calculate the average force of friction that is resisting the motion of the car. My thinking: Convert 52 kW to 5200 W.\n\n4. ### Math\n\nThe distance require to stop a car varies directly as the square of its speed. If 250 feet are required to stop a car traveling 60 miles per hour, how many feet are required to stop a car traveling 96 miles per hour!\n\n1. ### physics\n\nA certain sports car accelerates from 0 to 60 mph in 9.0 s (average acceleration = 3.0 m/s2). The mass of the car is 1354 kg. The average backward force due to air drag during acceleration is 280 N. Find the forward force required\n\n2. ### physical science\n\na car traveling along the highway needs a certain amount of force exerted on it to stop.more stopping force may be required when the car has: ;\n\n3. ### Science halp\n\n6.Two cars with different masses travel at the same speed down a hill toward a stop sign. What will happen when both cars apply brakes at the same time to stop? * A. The car with the smaller mass will require less force to stop\n\n4. ### Physics\n\nA 1000 kg car strike a tree at 30 km/hr and comes to a stop in 0.15 s. What is its initial momentum and average force on the car?\n\n1. ### Physics -Impulse- part 2\n\n1) A 0.50 kg object is at rest. A 3.00 N force to the right acts on the object during a time interval of 1.50 s a) What is the impulse? b) What is the velocity of the object at the end of this time interval? 2) A 2250 kg car\n\n2. ### physics\n\nA 1100 kg car coasts on a horizontal road with a speed of 19 m/s. After crossing a an unpaved, sandy stretch of road 32 m long, its speed decreases to 12 m/s. (a) Was the net force done on the car positive, negative, or zero?\n\n3. ### Physics\n\nSuppose a 1,600 kg car is traveling at 20.0 m/s. What average force is needed to stop the car in 4.0 s? Can you explain to me how you got your answer, thank you.\n\n4. ### Physics\n\nA 1100 kg car is moving toward the north along a straight road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to rest in a distance of 140 m. (a) If the road is completely level, what is the constant force"
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https://www.transtutors.com/questions/design-a-simple-4-bit-arithmetic-logic-unit-alu-with-three-selection-inputs-s2-s1-s0-5442476.htm | [
"# Design a simple 4-bit Arithmetic/Logic unit (ALU) with three selection inputs, S2, S1, S0, and a... 1 answer below »\n\nDesign a simple 4-bit Arithmetic/Logic unit (ALU) with three selection inputs, S2,\nS1, S0, and a logic/arithmetic mode input, M, that implements the following 16\nfunctions of the two data inputs A and B (and carry-in Cin):\n\nshaik n\n\n## Plagiarism Checker\n\nSubmit your documents and get free Plagiarism report\n\nFree Plagiarism Checker"
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https://answers.everydaycalculation.com/simplify-fraction/41-2 | [
"Solutions by everydaycalculation.com\n\n## Reduce 41/2 to lowest terms\n\n41/2 is already in the simplest form. It can be written as 20.5 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 41 and 2 is 1\n2. Divide both the numerator and denominator by the GCD\n41 ÷ 1/2 ÷ 1\n3. Reduced fraction: 41/2\nTherefore, 41/2 simplified to lowest terms is 41/2.\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
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null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8646753,"math_prob":0.8902298,"size":344,"snap":"2021-31-2021-39","text_gpt3_token_len":88,"char_repetition_ratio":0.10294118,"word_repetition_ratio":0.0,"special_character_ratio":0.27034885,"punctuation_ratio":0.10606061,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95465267,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-04T21:45:47Z\",\"WARC-Record-ID\":\"<urn:uuid:5807c1fe-67e9-40c1-9f01-d712dc7ae70e>\",\"Content-Length\":\"6402\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a0e3f89c-d143-4c99-a247-16b593ed6e16>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddf7032e-9cfa-4ff3-9c66-5d8b6d9cbcc4>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/simplify-fraction/41-2\",\"WARC-Payload-Digest\":\"sha1:JCZXTUVVHSBBK2KZINQL7KD6UU6BC4FJ\",\"WARC-Block-Digest\":\"sha1:N5CDQ5BKQHRFOA75DO2ECZK6PF7JNLRN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155188.79_warc_CC-MAIN-20210804205700-20210804235700-00184.warc.gz\"}"} |
http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/statug_rreg_details.htm | [
"# The ROBUSTREG Procedure\n\n## Details: ROBUSTREG Procedure\n\nSubsections:\n\nThis section describes the statistical and computational aspects of the ROBUSTREG procedure. The following notation is used throughout the section.\n\nLet",
null,
"denote an",
null,
"matrix, let",
null,
"denote a given n-vector of responses, and let",
null,
"denote an unknown p-vector of parameters or coefficients whose components are to be estimated. The matrix",
null,
"is called the design matrix. Consider the usual linear model,",
null,
"where",
null,
"is an n-vector of unknown errors. It is assumed that (for a given",
null,
") the components",
null,
"of",
null,
"are independent and identically distributed according to a distribution",
null,
", where",
null,
"is a scale parameter (usually unknown). Often",
null,
", the standard normal distribution function. The vector of residuals for a given value of",
null,
"is denoted by",
null,
", and the ith row of the matrix",
null,
"is denoted by",
null,
"."
] | [
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0028.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0029.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0030.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0031.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0032.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0033.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0034.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0032.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0035.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0036.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0037.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0038.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0039.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0040.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0041.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0032.png",
null,
"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/images/statug_rreg0042.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8995528,"math_prob":0.9647711,"size":757,"snap":"2019-13-2019-22","text_gpt3_token_len":146,"char_repetition_ratio":0.12217796,"word_repetition_ratio":0.0,"special_character_ratio":0.19154558,"punctuation_ratio":0.10218978,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9983321,"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],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,3,null,1,null,1,null,3,null,1,null,1,null,1,null,2,null,1,null,1,null,1,null,3,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-22T01:11:51Z\",\"WARC-Record-ID\":\"<urn:uuid:2b282d8a-9ffe-49e6-85ff-8694d8063aeb>\",\"Content-Length\":\"17006\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b8068dd5-f156-42ab-b9b1-d8858bde00bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:ed9ff64e-962a-4c0d-8a29-4de525492142>\",\"WARC-IP-Address\":\"149.173.160.38\",\"WARC-Target-URI\":\"http://support.sas.com/documentation/cdl/en/statug/68162/HTML/default/statug_rreg_details.htm\",\"WARC-Payload-Digest\":\"sha1:A4KRIB2PG5X56DFKQUOQCOJN4KNICHBL\",\"WARC-Block-Digest\":\"sha1:7C2MSYOWHZYOPEXM62ZN4IKE6FJHGNSG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256600.32_warc_CC-MAIN-20190522002845-20190522024845-00246.warc.gz\"}"} |
https://calculomates.com/en/divisors/of/3779 | [
"# Divisors of 3779\n\n## Divisors of 3779\n\nThe list of all positive divisors (that is, the list of all integers that divide 22) is as follows :\n\n• 1\n• 3779\n\nAccordingly:\n\n3779 is multiplo of 1\n\n3779 has 1 positive divisors\n\n## Parity of 3779\n\n3779is an odd number,as it is not divisible by 2\n\n## The factors for 3779\n\nThe factors for 3779 are all the numbers between -3779 and 3779 , which divide 3779 without leaving any remainder. Since 3779 divided by -3779 is an integer, -3779 is a factor of 3779 .\n\nSince 3779 divided by -3779 is a whole number, -3779 is a factor of 3779\n\nSince 3779 divided by -1 is a whole number, -1 is a factor of 3779\n\nSince 3779 divided by 1 is a whole number, 1 is a factor of 3779\n\n## What are the multiples of 3779?\n\nMultiples of 3779 are all integers divisible by 3779 , i.e. the remainder of the full division by 3779 is zero. There are infinite multiples of 3779. The smallest multiples of 3779 are:\n\n0 : in fact, 0 is divisible by any integer, so it is also a multiple of 3779 since 0 × 3779 = 0\n\n3779 : in fact, 3779 is a multiple of itself, since 3779 is divisible by 3779 (it was 3779 / 3779 = 1, so the rest of this division is zero)\n\n7558: in fact, 7558 = 3779 × 2\n\n11337: in fact, 11337 = 3779 × 3\n\n15116: in fact, 15116 = 3779 × 4\n\n18895: in fact, 18895 = 3779 × 5\n\netc.\n\n## Is 3779 a prime number?\n\nIt is possible to determine using mathematical techniques whether an integer is prime or not.\n\nfor 3779, the answer is: yes, 3779 is a prime number because it only has two different divisors: 1 and itself (3779).\n\n## How do you determine if a number is prime?\n\nTo know the primality of an integer, we can use several algorithms. The most naive is to try all divisors below the number you want to know if it is prime (in our case 3779). We can already eliminate even numbers bigger than 2 (then 4 , 6 , 8 ...). Besides, we can stop at the square root of the number in question (here 61.474 ). Historically, the Eratosthenes screen (which dates back to Antiquity) uses this technique relatively effectively.\n\nMore modern techniques include the Atkin screen, probabilistic tests, or the cyclotomic test."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9103559,"math_prob":0.9735294,"size":1907,"snap":"2020-45-2020-50","text_gpt3_token_len":565,"char_repetition_ratio":0.16395165,"word_repetition_ratio":0.04945055,"special_character_ratio":0.36130047,"punctuation_ratio":0.1487805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985876,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-02T13:32:37Z\",\"WARC-Record-ID\":\"<urn:uuid:8d40700c-f162-4731-9c98-afe3bb90da53>\",\"Content-Length\":\"15457\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7be6f846-ffaf-426e-bf23-466ef277d3bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ed0f3c1-3e81-4dfe-89de-27093e035bf8>\",\"WARC-IP-Address\":\"104.28.8.187\",\"WARC-Target-URI\":\"https://calculomates.com/en/divisors/of/3779\",\"WARC-Payload-Digest\":\"sha1:BHSWOOWYTIO73WC7ZJTA2PORRJRRN2MY\",\"WARC-Block-Digest\":\"sha1:ZEYFXRXOVYD7J4GD74MNMMDGQ6PPEHI6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141708017.73_warc_CC-MAIN-20201202113815-20201202143815-00121.warc.gz\"}"} |
https://tejji.com/apps/sales-tax-calculator | [
"# Calculate Sales Tax\n\nCost / Price :\nSales Tax : %\nSales Tax Amount :\nTotal Cost / Price :\n\n# Calculated Tax\n\nNo Cost / Price Tax % Tax Amount Total Amount Break Up\n[No] [Amount] [Percent] [PercentAmount] [TotalAmount] [AmountPercent]\n\n# Usage\n\nCost / Price is the cost / price of the product or service\n\nSales Tax is the percent to be applied on the cost / price of the product or service. This tax amount varies from region to region so you need to know approriate tax applicable to your transaction\n\nSales Tax Amount is that Tax amount applicable based on Sales Tax percentage\n\nTotal Cost / Price is the amount to be paid for the product or service inclusive of the price and tax.\n\nBreak up is the visual representation of what contributes as Cost/Price and Tax amount if total Cost / Price is 100%\n\nCalculate button can be used compute the Tax amount and Total Cost / Price given Sales Tax percentage and Cost / Price of the product or service\ne.g., To calcualte Sales Tax enter the Sales Tax % and Cost of goods and then press Calculate button"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9005454,"math_prob":0.9237332,"size":859,"snap":"2023-14-2023-23","text_gpt3_token_len":190,"char_repetition_ratio":0.17660819,"word_repetition_ratio":0.0754717,"special_character_ratio":0.21885914,"punctuation_ratio":0.03125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95563585,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-29T03:23:04Z\",\"WARC-Record-ID\":\"<urn:uuid:6437c59d-12e1-466d-9bd9-d3b84976f588>\",\"Content-Length\":\"42625\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a38eacfe-0d75-4cfd-b538-03a7d89a603a>\",\"WARC-Concurrent-To\":\"<urn:uuid:98be6e5f-dd9e-4763-9c77-98d656527f30>\",\"WARC-IP-Address\":\"104.21.2.90\",\"WARC-Target-URI\":\"https://tejji.com/apps/sales-tax-calculator\",\"WARC-Payload-Digest\":\"sha1:GPMQRP73OQVTITMH7UCNJI6RH7L3ZFIK\",\"WARC-Block-Digest\":\"sha1:FHVR4IZOH6KONVT5LESMVTWJNY7MT6AD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644574.15_warc_CC-MAIN-20230529010218-20230529040218-00334.warc.gz\"}"} |
https://brightessays.com/2022/01/14/activity-1-a-student-structure-in-this-activity-you-have-to-implement-two-functions-5149619-shining-essays/ | [
"Activity 1: A Student Structure\n\nIn this activity you have to implement two functions to assist in the construction and printing of a student structure. The factory function createStudent() takes the appropriate parameters, allocates memory to hold a new student and sets each field of the student appropriately. In the case of strings, malloc is used to create enough space and the string is then copied. The birthdate is handled specially: it parses the date string and creates a struct tm, which is a “time” structure defined by the time library (time.h). Ultimately, a pointer to the new student structure is returned. You also need to implement the printStudent() function that takes a student (by reference) and prints it out to the standard output. We have also provided two fully implemented functions (createEmptyStudent and studentToString) for you to examine and understand how they work.\n\nInstructions\n\n2.Implement the two functions: createStudent() and printStudent()\n\n3.Change the values in the main function to your name, NUID, and birth date.\n\n4.Compile and run the program.\n\n5.Examine the following two implemented functions: createEmptyStudent() and studentToString().\n\n6.Use them in your main function.\n\n7.Show your program to a lab instructor.\n\nRefer back to this program in Activity 2 as needed.\n\nstudentDemo.c\n\n#include\n#include\n#include\n#include\n\ntypedef struct {\nchar *firstName;\nchar *lastName;\nint nuid;\nstruct tm birthDate;\n} Student;\n\nStudent * createStudent(const char * firstName, const char *lastName, int nuid, const char * birthDate_str);\nvoid printStudent(Student * student);\n\nint main(void) {\n\nStudent *me = createStudent(“Richard”, “Feynman”, 140602, “07-30-1980”);\nprintStudent(me);\n\n}\n\n/**\n* A “factory” function to create a new student structure initialized with the given values\n* Strings are deep copied.\n* You must use the “strptime” function defined in the file to copy the “birthDate_str” from the argument to the “birthDate” struct component of the Student struct\n*/\nStudent * createStudent(const char * firstName, const char *lastName, int nuid, const char * birthDate_str) {\n\nreturn student;\n}\n\n/*\n* A factory function to display a student first name, last name, nuid and birth date\n*/\n\nvoid printStudent(Student * student) {\n\n}\n\n/**\n* A “factory” function to create a new student structure with everything initialized to\n* default values (null or zero)\n*/\nStudent * createEmptyStudent() {\nStudent * student = (Student *) malloc(sizeof(Student) * 1);\nstudent->firstName = NULL;\nstudent->lastName = NULL;\nstudent->nuid = 0;\nreturn student;\n}\n\nchar *studentToString(Student *student) {\n//create a temporary string that is “large enough”\nchar *tmp = (char *) malloc(sizeof(char) * 2000);\n\n//format the student into the temporary string\nsprintf(tmp, “%s, %s (%08d, %d-%d-%d)n”, student->lastName, student->firstName, student->nuid, (student->birthDate.tm_year+1900), (student->birthDate.tm_mon+1), student->birthDate.tm_mday);\n\n//create the final string that is the exact size needed\nchar *result = (char *) malloc(sizeof(char) * (strlen(tmp)+1));\n\n//free up the temporary string\nfree(tmp);\n\n//return the result\nreturn result;\n}"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.72017753,"math_prob":0.6742877,"size":3185,"snap":"2022-27-2022-33","text_gpt3_token_len":734,"char_repetition_ratio":0.1851619,"word_repetition_ratio":0.0824295,"special_character_ratio":0.24646781,"punctuation_ratio":0.14206642,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.966475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T23:56:10Z\",\"WARC-Record-ID\":\"<urn:uuid:09bccfbb-40c1-4b92-bf44-46c45d994426>\",\"Content-Length\":\"78612\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ac9d263-cebe-424b-bab2-546d4a3c8edd>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7f717f7-9955-488c-9c62-27e7769e1836>\",\"WARC-IP-Address\":\"45.79.7.19\",\"WARC-Target-URI\":\"https://brightessays.com/2022/01/14/activity-1-a-student-structure-in-this-activity-you-have-to-implement-two-functions-5149619-shining-essays/\",\"WARC-Payload-Digest\":\"sha1:N5U2PZELQO74BN4CIMXXLW75NUPGLKZH\",\"WARC-Block-Digest\":\"sha1:XJJC6NKK2PUP6ACLZ3BJSKNNRUKW77SU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571090.80_warc_CC-MAIN-20220809215803-20220810005803-00692.warc.gz\"}"} |
https://publications.waset.org/12188/bibtex | [
"```\t@article{(Open Science Index):https://publications.waset.org/pdf/12188,\ntitle = {Traveling Wave Solutions for the (3+1)-Dimensional Breaking Soliton Equation by (G'/G)- Expansion Method and Modified F-Expansion Method},\ncountry\t= {},\ninstitution\t= {},\nabstract = {In this paper, using (G\u0002/G )-expansion method and modified F-expansion method, we give some explicit formulas of exact traveling wave solutions for the (3+1)-dimensional breaking soliton equation. A modified F-expansion method is proposed by taking full advantages of F-expansion method and Riccati equation in seeking exact solutions of the equation.\n},\njournal = {International Journal of Mathematical and Computational Sciences},\nvolume = {5},\nnumber = {7},\nyear = {2011},\npages = {1100 - 1105},\nee = {https://publications.waset.org/pdf/12188},\nurl \t= {https://publications.waset.org/vol/55},\nbibsource = {https://publications.waset.org/},\nissn \t= {eISSN: 1307-6892},\npublisher = {World Academy of Science, Engineering and Technology},\nindex \t= {Open Science Index 55, 2011},\n}\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77478033,"math_prob":0.76813936,"size":1115,"snap":"2021-43-2021-49","text_gpt3_token_len":280,"char_repetition_ratio":0.12151215,"word_repetition_ratio":0.0,"special_character_ratio":0.28340808,"punctuation_ratio":0.19270833,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9757973,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-01T22:11:21Z\",\"WARC-Record-ID\":\"<urn:uuid:f2bee5f9-9321-4c3a-a01d-89bf3f71900e>\",\"Content-Length\":\"12270\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:62abb386-9cac-4087-ae34-e14263096885>\",\"WARC-Concurrent-To\":\"<urn:uuid:dfa29200-767f-44bf-860a-06399fc98372>\",\"WARC-IP-Address\":\"172.67.205.138\",\"WARC-Target-URI\":\"https://publications.waset.org/12188/bibtex\",\"WARC-Payload-Digest\":\"sha1:56K372QU72FOWZTRH5SRXYACRDFMV6ZQ\",\"WARC-Block-Digest\":\"sha1:BPMLODCL5A6GLYGP72T7INSVX75364RV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964360951.9_warc_CC-MAIN-20211201203843-20211201233843-00099.warc.gz\"}"} |
http://usaco.org/current/data/sol_prob2_gold_dec21.html | [
"(Analysis by Andi Qu, Benjamin Qi)\n\nTo solve this problem in $\\mathcal O(N^2)$ time, we can simulate the process for each $x$.\n\n#include <iostream>\n\nint main() {\nstd::cin.tie(0)->sync_with_stdio(0);\nint n, p;\nstd::cin >> n;\nfor (int i = 0; i < n; i++) std::cin >> p[i];\nfor (int x = 0; x <= n; x++) {\nint mn = 0, mx = n + 1, ans = 0;\nbool hi = false;\nfor (int i = 0; i < n; i++) {\nif (p[i] > mx || p[i] < mn) continue;\nif (p[i] > x) {\nhi = true;\nmx = p[i];\n} else {\nans += hi;\nhi = false;\nmn = p[i];\n}\n}\nstd::cout << ans << '\\n';\n}\nreturn 0;\n}\n\n\nAnother solution that takes $\\mathcal O(N^2)$ time in the worst case is to keep track of each \"HI\" and \"LO\" for each $x$, and then go through the 2 lists to find the number of \"HILO\" pairs. However, this solution passes the test cases with data generated uniformly at random because the expected number of \"HI\"s and \"LO\"s is $\\mathcal{O}(\\log N)$ for each $x$, resulting in an overall expected runtime of $\\mathcal{O}(N\\log N)$.\n\nTo keep track of the 2 lists, we can use 2 monotone stacks. Essentially, we maintain a sorted list of indices where Bessie says \"LO\". When we transition from $x$ to $x + 1$, we know that the last \"LO\" that Bessie will say will be to $x + 1$. If Bessie doesn't say \"LO\" to $k$ while thinking of $x + 1$, then she will never say \"LO\" to some $k$ while thinking of $y > x + 1$, so we can remove all \"LO\"s that used to be said after the index of $x + 1$ in the permutation.\n\nConveniently, the indices that we remove form a suffix of the list (because the list is sorted), so we can use a stack and pop elements from it. Since each index is pushed into and popped from the stack at most once, this takes $\\mathcal O(N)$ time overall; iterating through the stack for each $x$ takes a further $\\mathcal O(N)$ time per element.\n\nHere's a C++ code snippet for finding the list of \"LO\"s for each $x$:\n\nstd::vector<int> los;\nfor (int i = 1; i <= n; i++) {\nlos[i] = los[i - 1];\nwhile (los[i].size() != 0 && los[i].back() > pos[i]) los[i].pop_back();\nlos[i].push_back(pos[i]);\n}\n\n\n(Bonus: Find out how the rest of the test data for this problem was generated.)\n\nTo solve the problem in $\\mathcal O(N)$ time, we need a way to efficiently transition from $x$ to $x + 1$.\n\nFull Solution 1:\n\nClaim: Let $p$ be the given permutation. If index $i$ is the \"LO\" in a \"HILO\" pair, then the \"HI\" in the pair must be the index $k$ such that $k < i$, $p[k] > p[i]$, and $p[k]$ is minimal.\n\nProof: If no such $k$ exists, then there is no greater element before $i$, so $i$ can't be the \"LO\" in a \"HILO\" pair. If Bessie says \"LO\" to $p[k]$, then Elsie will not guess $p[i]$, so $i$ can't be the \"LO\" in a \"HILO\" pair. Otherwise, the last \"HI\" that Bessie says, before saying \"LO\" to $p[i]$, must be to $p[k]$.\n\nKnowing this, we can compute an array $\\texttt{prv}$ where $\\texttt{prv}[i] = k$ as described above using a monotone stack.\n\nClaim: Let index $j$ be the last \"LO\" before Bessie says \"LO\" to $p[i]$. For each $x$ where Elsie doesn't skip $p[i]$, $j$ is the same.\n\nProof: Let $j_1 < j_2 < i$ and $p[j_1], p[j_2] < p[i]$. If Bessie says \"LO\" to $p[i]$, then there are 3 possibilities:\n\n1. Bessie never says \"LO\" to either $p[j_1]$ or $p[j_2]$ because she already said \"LO\" at some index $k < j_1$ where $p[k] > p[j_1], p[j_2]$.\n2. Bessie always says \"LO\" to $p[j_1]$ but never to $p[j_2]$ because she said \"LO\" at some index $j_1 \\leq k < j_2$ where $p[k] > p[j_2]$.\n3. Bessie always \"LO\" to both $p[j_1]$ and $p[j_2]$ because neither of the above happen.\n\nIf transitioning causes Bessie to stop saying \"LO\" to some index before $i$, then it must also cause her to stop saying \"LO\" to $p[i]$ too. Therefore, $j$ is the same for each $p[i]$.\n\nClaim: Let $j$ be defined as above. Index $i$ is the \"LO\" in a \"HILO\" pair if and only $\\texttt{prv}[i]$ exists, and $j$ doesn't exist or $\\texttt{prv}[i] \\neq \\texttt{prv}[j]$.\n\nProof: If $\\texttt{prv}[i]$ doesn't exist, then Bessie never says \"HI\" before saying \"LO\" to $p[i]$, so it can't be the \"LO\" in a \"HILO\" pair. Otherwise, if $\\texttt{prv}[i] \\neq \\texttt{prv}[j]$, then the last \"HI\" that Bessie says before saying \"LO\" to $p[i]$ must be after saying \"LO\" to $p[j]$ by definition. In this case, index $i$ is the \"LO\" in a \"HILO\" pair.\n\nKnowing this, we can then determine, once for each index, whether it's the \"LO\" in a \"HILO\" pair, and thus how many \"HILO\" pairs there are for each $x$.\n\nAndi's code:\n\n#include <iostream>\n#include <stack>\n\nint main() {\nstd::cin.tie(0)->sync_with_stdio(0);\nint n, pos, prv;\nbool hilo;\nstd::cin >> n;\nfor (int i = 1; i <= n; i++) {\nint p;\nstd::cin >> p;\npos[p] = i;\n}\n\nstd::stack<int> stck;\nstck.push(0);\nfor (int i = n; i > 0; i--) {\nwhile (stck.top() > pos[i]) stck.pop();\nprv[pos[i]] = stck.top();\nstck.push(pos[i]);\n}\nwhile (stck.size() != 1) stck.pop();\n\nstd::cout << \"0\\n\";\nint cnt = 0;\nfor (int i = 1; i <= n; i++) {\nwhile (stck.top() > pos[i]) {\ncnt -= hilo[stck.top()];\nstck.pop();\n}\nhilo[pos[i]] = prv[pos[i]] != 0 &&\n(stck.top() == 0 || prv[pos[i]] != prv[stck.top()]);\nstck.push(pos[i]);\ncnt += hilo[pos[i]];\nstd::cout << cnt << '\\n';\n}\nreturn 0;\n}\n\n\nFull Solution 2: Another way we can solve this problem in $\\mathcal O(N \\log N)$ time is with stacks plus a sorted set.\n\nFirst, we compute, for each $x \\rightarrow x + 1$ event, which elements of the permutation start and stop becoming \"HI\"s and \"LO\"s. We can do this with a stack.\n\nNext, we process the events for each $x$. We can maintain an ordered map of \"HI\"s and \"LO\"s, and inserting or erasing any element from that map changes a constant number of \"HILO\" pairs. Using C++'s iterators, we can process these changes efficiently.\n\nBen's code:\n\n#include <bits/stdc++.h>\nusing namespace std;\n\ntemplate <class T> using V = vector<T>;\n\nint main() {\ncin.tie(0)->sync_with_stdio(0);\nint N;\ncin >> N;\nV<int> P(N);\nfor (auto &t : P) {\ncin >> t;\n--t;\n}\nV<int> pos(N);\nfor (int i = 0; i < N; ++i)\npos[P[i]] = i;\nV<V<pair<int, char>>> to_ins(N + 1);\nV<V<int>> to_del(N + 1);\n{ // process \"LO\"s from lowest to highest, record insertions and deletions\nstack<int> cur;\nfor (int i = 0; i < N; ++i) {\nwhile (!cur.empty() && cur.top() > pos[i]) {\nto_del.at(i + 1).push_back(cur.top());\ncur.pop();\n}\ncur.push(pos[i]);\nto_ins.at(i + 1).push_back({pos[i], 'L'});\n}\n}\n{ // process \"HI\"s from highest to lowest, record insertions and deletions\nstack<int> cur;\nfor (int i = N - 1; i >= 0; --i) {\nwhile (!cur.empty() && cur.top() > pos[i]) {\nto_ins.at(i + 1).push_back({cur.top(), 'H'});\ncur.pop();\n}\ncur.push(pos[i]);\nto_del.at(i + 1).push_back(pos[i]);\n}\nwhile (!cur.empty()) {\nto_ins.at(0).push_back({cur.top(), 'H'});\ncur.pop();\n}\n}\nint ans = 0;\nmap<int, char> m; // maps each position to 'H' or 'L'\nauto hilo = [&](map<int, char>::iterator it,\nmap<int, char>::iterator next_it) {\nreturn it->second == 'H' && next_it->second == 'L';\n};\nauto get_dif = [&](map<int, char>::iterator it) {\nint dif = 0;\nif (it != begin(m))\ndif += hilo(prev(it), it);\nif (next(it) != end(m))\ndif += hilo(it, next(it));\nif (it != begin(m) && next(it) != end(m))\ndif -= hilo(prev(it), next(it));\nreturn dif;\n};\nfor (int i = 0; i <= N; ++i) { // to finish, go from lowest to highest again\nfor (auto &t : to_del.at(i)) {\nauto it = m.find(t);\nassert(it != end(m));\nans -= get_dif(it);\nm.erase(it);\n}\nfor (auto &t : to_ins.at(i)) {\nauto it = m.insert(t).first;\nassert(it->second);\nans += get_dif(it);\n}\ncout << ans << \"\\n\";\n}\n}\n\n\nFull Solution 3: Here is a simpler solution that also runs in $\\mathcal O(N\\log N)$. This one doesn't explicitly make use of monotonic stacks, though it is possible to modify it to run in $\\mathcal O(N)$ time with them.\n\nAllen Wu's code:\n\n#include <iostream>\n#include <map>\n#include <set>\n#include <utility>\n#include <vector>\nusing namespace std;\n\nint main() {\nios_base::sync_with_stdio(false);\ncin.tie(NULL);\nint n;\ncin >> n;\nvector<int> arr(n);\nfor (int i = 0; i < n; ++i)\ncin >> arr[i];\nvector<int> pos(n + 1);\nfor (int i = 0; i < n; ++i)\npos[arr[i]] = i;\nmap<int, int> existing;\nvector<int> changes(n + 1);\nfor (int i = 0; i < n; ++i) {\nint num = arr[i];\n// goal is to compute how the number of HILOs changes\n// when we go from x = num-1 to x = num\nauto it = existing.upper_bound(num);\nif (it != existing.end()) {\n// add one if num is involved in a HILO when x = num\nif (it == existing.begin())\n++changes[num];\nelse if (it->second > prev(it)->second)\n++changes[num];\n}\n// subtract one if num is involved in a HILO when x = num-1\nif (pos[num - 1] > pos[num])\n--changes[num];\nexisting[num] = i;\n}\nint sum = 0;\nfor (int i = 0; i <= n; ++i) {\nsum += changes[i];\ncout << sum << \"\\n\";\n}\n}"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.63698554,"math_prob":0.9991598,"size":8173,"snap":"2022-05-2022-21","text_gpt3_token_len":2697,"char_repetition_ratio":0.10160362,"word_repetition_ratio":0.11043771,"special_character_ratio":0.383213,"punctuation_ratio":0.18463966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99987686,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-25T20:49:46Z\",\"WARC-Record-ID\":\"<urn:uuid:e22f633a-ec80-4512-bafc-c80487b40383>\",\"Content-Length\":\"13783\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:04cc19f5-c0ab-4b2b-9d2a-7ea54f71b5da>\",\"WARC-Concurrent-To\":\"<urn:uuid:79cec112-31ee-4e27-affd-d2a9b117eedd>\",\"WARC-IP-Address\":\"130.127.204.224\",\"WARC-Target-URI\":\"http://usaco.org/current/data/sol_prob2_gold_dec21.html\",\"WARC-Payload-Digest\":\"sha1:OHA5JV7EDPCKVJM7OGNW3NRGX5EQ6MRV\",\"WARC-Block-Digest\":\"sha1:NGLIMNFXAA5ZJCQFA7PWSLNBCXYN3I3E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662593428.63_warc_CC-MAIN-20220525182604-20220525212604-00210.warc.gz\"}"} |
https://search.yahoo.com/mobile/s?ei=UTF-8&_tsrc=apple&age=1m&p=Isogonal+figure+wikipedia&fr2=12328 | [
"# Yahoo Web Search\n\n1. ### Isogonal figure - Wikipedia\n\nen.wikipedia.org › wiki › Isogonal_figure\n\nThe dual of an isogonal polytope is an isohedral figure, which is transitive on its facets. k-isogonal and k-uniform figures. A polytope or tiling may be called k-isogonal if its vertices form k transitivity classes. A more restrictive term, k-uniform is defined as an k-isogonal figure constructed only from regular polygons.\n\nWhat is an isogonal figure?\n\nWhat is dual of isogonal polytope?\n\nWhat is an isogonal polyhedron?\n\nAre there any even sided polygons that are isogonal?\n\n3. ### Talk:Isogonal figure - Wikipedia\n\nen.wikipedia.org › wiki › Talk:Isogonal_figure\n\nThat figure has five interior angles of 90 deg and one of 270 deg. Those are not all equal! Isogonal does also imply that the vertices are transitive, i.e. they lie within the same symmetry orbit, something like this: o---o / \\ / \\ / \\ o o \\ / o-------o\n\n4. ### Isotoxal figure - Wikipedia\n\nen.wikipedia.org › wiki › Isotoxal_figure\n\nIn geometry, a polytope (for example, a polygon or a polyhedron), or a tiling, is isotoxal or edge-transitive if its symmetries act transitively on its edges.\n\nSides (2 n )\n4\n6\n8\n{n α } Convex β<180 Concave β>180\n{2 α }\n{3 α }\n{4 α }\n2-turn { ( n /2) α }\n--\n{ (3/2) α }\n2 {2 α }\n3-turn { ( n /3) α }\n--\n--\n{ (4/3) α }\n4-turn { ( n /4) α }\n--\n--\n--\n5. ### Category:Isogonal tilings - Wikipedia\n\nen.wikipedia.org › wiki › Category:Isogonal_tilings\n\nPages in category \"Isogonal tilings\" The following 153 pages are in this category, out of 153 total. This list may not reflect recent changes ().\n\n6. ### Isogonal figure - WikiMili, The Best Wikipedia Reader\n\nwikimili.com › en › Isogonal_figure\n\nIn geometry, a polytope (a polygon, polyhedron or tiling, for example) is isogonal or vertex-transitive if all its vertices are equivalent under the symmetries of the figure. This implies that each vertex is surrounded by the same kinds of face in the same or reverse order, and with the same angles between corresponding faces.\n\n7. ### Wikizero - Isogonal figure\n\nwww.wikizero.com › en › Isogonal_figure\n\nIsogonal figure From Wikipedia the free encyclopedia In geometry, a polytope (a polygon, polyhedron or tiling, for example) is isogonal or vertex-transitive if all its vertices are equivalent under the symmetries of the figure.\n\n8. ### Isohedral figure - Wikipedia\n\nen.wikipedia.org › wiki › Isohedral_figure\n\nA facet-transitive or isotopic figure is a n -dimensional polytopes or honeycomb, with its facets ((n −1)- faces) congruent and transitive. The dual of an isotope is an isogonal polytope. By definition, this isotopic property is common to the duals of the uniform polytopes. An isotopic 2-dimensional figure is isotoxal (edge-transitive).\n\n9. ### Vertex figure - Wikipedia\n\nen.wikipedia.org › wiki › Vertex_figures\n\nVertex figures are especially significant for uniforms and other isogonal (vertex-transitive) polytopes because one vertex figure can define the entire polytope. For polyhedra with regular faces, a vertex figure can be represented in vertex configuration notation, by listing the faces in sequence around the vertex.\n\n10. ### July 26 - Wikipedia\n\nen.wikipedia.org › wiki › July_26\n\n1948 – Herbert Wiesinger, German figure skater 1949 – Thaksin Shinawatra , Thai businessman and politician, 23rd Prime Minister of Thailand 1949 – Roger Taylor , English singer-songwriter, drummer, and producer\n\n11. People also search for"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89858043,"math_prob":0.79562145,"size":2641,"snap":"2021-21-2021-25","text_gpt3_token_len":672,"char_repetition_ratio":0.15434206,"word_repetition_ratio":0.10599078,"special_character_ratio":0.24574025,"punctuation_ratio":0.103869654,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966107,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T11:56:24Z\",\"WARC-Record-ID\":\"<urn:uuid:d1243505-6855-4f09-9b41-fc8e544ce738>\",\"Content-Length\":\"139104\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:33dfb9fc-2788-4cec-a443-9e7c5ea9e7e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:7659a761-0259-4912-bb28-d410be7025a6>\",\"WARC-IP-Address\":\"66.218.84.137\",\"WARC-Target-URI\":\"https://search.yahoo.com/mobile/s?ei=UTF-8&_tsrc=apple&age=1m&p=Isogonal+figure+wikipedia&fr2=12328\",\"WARC-Payload-Digest\":\"sha1:2QBGSZZ7UBCUJ4GHBRB446PGXIGU3SN5\",\"WARC-Block-Digest\":\"sha1:CTVKQ3CQHHHDJTH4LPJ2RVWNVGOPUKVJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243990449.41_warc_CC-MAIN-20210514091252-20210514121252-00337.warc.gz\"}"} |
http://www.numbersaplenty.com/677779 | [
"Search a number\nBaseRepresentation\nbin10100101011110010011\n31021102201221\n42211132103\n5133142104\n622305511\n75522014\noct2453623\n91242657\n10677779\n11423253\n12288297\n131a966b\n1413900b\n15d5c54\nhexa5793\n\n677779 has 2 divisors, whose sum is σ = 677780. Its totient is φ = 677778.\n\nThe previous prime is 677767. The next prime is 677783. The reversal of 677779 is 977776.\n\nIt is a happy number.\n\nIt is a strong prime.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 677779 - 215 = 645011 is a prime.\n\nIt is a plaindrome in base 10.\n\nIt is not a weakly prime, because it can be changed into another prime (671779) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (11) of ones.\n\nIt is a polite number, since it can be written as a sum of consecutive naturals, namely, 338889 + 338890.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (338890).\n\n2677779 is an apocalyptic number.\n\n677779 is a deficient number, since it is larger than the sum of its proper divisors (1).\n\n677779 is an equidigital number, since it uses as much as digits as its factorization.\n\n677779 is an odious number, because the sum of its binary digits is odd.\n\nThe product of its digits is 129654, while the sum is 43.\n\nThe square root of 677779 is about 823.2733446432. The cubic root of 677779 is about 87.8407502095.\n\nThe spelling of 677779 in words is \"six hundred seventy-seven thousand, seven hundred seventy-nine\"."
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https://schoollearningcommons.info/question/using-cramer-s-rule-find-2y-4-3-y-5-2-z-1-19763609-8/ | [
"## Using Cramer’s rule find x – 2y = -4 3x + y = -5 2x+z= 1\n\nQuestion\n\nUsing\nCramer’s rule\nfind\nx – 2y = -4\n3x + y = -5\n2x+z= 1\n\nin progress 0\n1 week 2021-09-10T18:55:02+00:00 1 Answer 0 views 0\n\nx= -2 ,y = -11, z = 5.\n\nStep-by-step explanation:\n\nX -2Y = -4\n\n2× 2 6X + 2Y = -10\n\n________________\n\n7X = -14\n\nX = -2\n\nSUB THE VALUE IN EACH EQUATION."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5340325,"math_prob":0.99978286,"size":388,"snap":"2021-31-2021-39","text_gpt3_token_len":178,"char_repetition_ratio":0.1328125,"word_repetition_ratio":0.0,"special_character_ratio":0.5670103,"punctuation_ratio":0.12903225,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99244076,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T17:47:48Z\",\"WARC-Record-ID\":\"<urn:uuid:f00f3e2f-0d0e-4533-9d35-7d27058b53a9>\",\"Content-Length\":\"66502\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54729de7-4b59-43ad-9e4c-d0c79f875e5e>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e422294-a1b9-4bf4-8bea-e18cf0ffa5a8>\",\"WARC-IP-Address\":\"172.96.186.144\",\"WARC-Target-URI\":\"https://schoollearningcommons.info/question/using-cramer-s-rule-find-2y-4-3-y-5-2-z-1-19763609-8/\",\"WARC-Payload-Digest\":\"sha1:JHBQSQMBWTBR43LF2YDCV6OVLMWWZAWR\",\"WARC-Block-Digest\":\"sha1:4EOD5Y2WDZFDWMLWTF5FMIIN4UKBALUW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056892.13_warc_CC-MAIN-20210919160038-20210919190038-00459.warc.gz\"}"} |
https://www.ic2.cn/news-162-64498-0.html | [
"",
null,
"• 网站:49040\n• 待审:1083\n• 文章:64417\n• 会员:135\n\nceil(float \\$value);//返回不小于 value 的下一个整数,value 如果有小数部分则进一位\n\n```<?php\necho ceil(4.3); // 5\necho ceil(9.999); // 10\necho ceil(-3.14); // -3\n?>\n\n```\n\nfloor(float \\$value);//返回不大于 value 的最接近的整数,舍去小数部分取整。\n\n```<?php\necho floor(4.3); // 4\necho floor(9.999); // 9\necho floor(-3.14); // -4\n?>\n\n```\n```round(float \\$val,[int \\$pre], [\\$mode = PHP_ROUND_HALF_UP]);\n\n```\n\n```<?php\necho round(3.4); // 3\necho round(3.5); // 4\necho round(3.6); // 4\necho round(3.6, 0); // 4\necho round(1.95583, 2); // 1.96\necho round(1241757, -3); // 1242000\necho round(5.045, 2); // 5.05\necho round(5.055, 2); // 5.06\n?>\n\n```\n\nmode 例子\n\n```<?php\necho round(9.5, 0, PHP_ROUND_HALF_UP); // 10\necho round(9.5, 0, PHP_ROUND_HALF_DOWN); // 9\necho round(9.5, 0, PHP_ROUND_HALF_EVEN); // 10\necho round(9.5, 0, PHP_ROUND_HALF_ODD); // 9\n\necho round(8.5, 0, PHP_ROUND_HALF_UP); // 9\necho round(8.5, 0, PHP_ROUND_HALF_DOWN); // 8\necho round(8.5, 0, PHP_ROUND_HALF_EVEN); // 8\necho round(8.5, 0, PHP_ROUND_HALF_ODD); // 9\n?>\n\n```\n\nPS:这里再为大家推荐几款计算工具供大家进一步参考借鉴:\n\nhttp://tools.jb51.net/jisuanqi/equ_jisuanqi\n\nhttp://tools.jb51.net/jisuanqi/jsqkexue\n\nhttp://tools.jb51.net/jisuanqi/jsq",
null,
"• 49040\n\n网站\n\n• 0\n\n小程序\n\n• 64417\n\n文章\n\n• 135\n\n会员"
] | [
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"https://www.ic2.cn/mulu/img/qq.gif",
null,
"https://www.ic2.cn/e/data/images/nouserpic.gif",
null
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http://openproblemgarden.org/category/homomorphisms | [
"# Homomorphisms\n\n## Cores of Cayley graphs ★★★★★\n\nAuthor(s): Samal\n\nConjecture Let",
null,
"be an abelian group. Is the core of a Cayley graph (on some power of",
null,
") a Cayley graph (on some power of",
null,
")?\n\nKeywords: Cayley graph; core\n\n## Pentagon problem ★★★\n\nAuthor(s): Nesetril\n\nQuestion Let",
null,
"be a 3-regular graph that contains no cycle of length shorter than",
null,
". Is it true that for large enough~",
null,
"there is a homomorphism",
null,
"?\n\nKeywords: cubic; homomorphism\n\n## Mapping planar graphs to odd cycles ★★★\n\nAuthor(s): Jaeger\n\nConjecture Every planar graph of girth",
null,
"has a homomorphism to",
null,
".\n\nKeywords: girth; homomorphism; planar graph\n\n## Weak pentagon problem ★★\n\nAuthor(s): Samal\n\nConjecture If",
null,
"is a cubic graph not containing a triangle, then it is possible to color the edges of",
null,
"by five colors, so that the complement of every color class is a bipartite graph.\n\n## Algorithm for graph homomorphisms ★★\n\nAuthor(s): Fomin; Heggernes; Kratsch\n\nQuestion\n\nIs there an algorithm that decides, for input graphs",
null,
"and",
null,
", whether there exists a homomorphism from",
null,
"to",
null,
"in time",
null,
"for some constant",
null,
"?\n\n## Circular choosability of planar graphs ★\n\nAuthor(s): Mohar\n\nLet",
null,
"be a graph. If",
null,
"and",
null,
"are two integers, a",
null,
"-colouring of",
null,
"is a function",
null,
"from",
null,
"to",
null,
"such that",
null,
"for each edge",
null,
". Given a list assignment",
null,
"of",
null,
", i.e.~a mapping that assigns to every vertex",
null,
"a set of non-negative integers, an",
null,
"-colouring of",
null,
"is a mapping",
null,
"such that",
null,
"for every",
null,
". A list assignment",
null,
"is a",
null,
"-",
null,
"-list-assignment if",
null,
"and",
null,
"for each vertex",
null,
". Given such a list assignment",
null,
", the graph G is",
null,
"-",
null,
"-colourable if there exists a",
null,
"-",
null,
"-colouring",
null,
", i.e.",
null,
"is both a",
null,
"-colouring and an",
null,
"-colouring. For any real number",
null,
", the graph",
null,
"is",
null,
"-",
null,
"-choosable if it is",
null,
"-",
null,
"-colourable for every",
null,
"-",
null,
"-list-assignment",
null,
". Last,",
null,
"is circularly",
null,
"-choosable if it is",
null,
"-",
null,
"-choosable for any",
null,
",",
null,
". The circular choosability (or circular list chromatic number or circular choice number) of G is",
null,
"Problem What is the best upper bound on circular choosability for planar graphs?\n\nKeywords: choosability; circular colouring; planar graphs",
null,
""
] | [
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
"http://openproblemgarden.org/files/tex/dccee841f3f498c2c58fa6ae1c1403c5a88c5b8d.png",
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
"http://openproblemgarden.org/files/tex/b8e7ad0330f925492bf468b5c379baec88cf1b3d.png",
null,
"http://openproblemgarden.org/files/tex/96cbd9a16c6a5eab03815b093b08f3b2db614e9a.png",
null,
"http://openproblemgarden.org/files/tex/73f33398d7e8aa42e6ec25ee2bb4f2b57ed3391a.png",
null,
"http://openproblemgarden.org/files/tex/b8e7ad0330f925492bf468b5c379baec88cf1b3d.png",
null,
"http://openproblemgarden.org/files/tex/ae1eeae9f205485b3b1d97109a792b4dc7ec218a.png",
null,
"http://openproblemgarden.org/files/tex/ee3e9115f72b68747b9ac848930cf8a1bce9b141.png",
null,
"http://openproblemgarden.org/files/tex/c4288e8e158f9ae52ceb4bc3033583494673aa7f.png",
null,
"http://openproblemgarden.org/files/tex/73f33398d7e8aa42e6ec25ee2bb4f2b57ed3391a.png",
null,
"http://openproblemgarden.org/files/tex/4761b031c89840e8cd2cda5b53fbc90c308530f3.png",
null,
"http://openproblemgarden.org/files/tex/5fdfd6d4364192fb4fdca04319ffa276d7272185.png",
null,
"http://openproblemgarden.org/files/tex/edbf80101dd57bc27ae9f8f75d01cecb65b85b60.png",
null,
"http://openproblemgarden.org/files/tex/11c7f862935b0cb68fb051793eb9938bcac262f1.png",
null,
"http://openproblemgarden.org/files/tex/18255450ec5c0f06f7a4d8b5f1c60e5a77f6d5d9.png",
null,
"http://openproblemgarden.org/files/tex/73f33398d7e8aa42e6ec25ee2bb4f2b57ed3391a.png",
null,
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null,
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null,
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null,
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null,
"http://openproblemgarden.org/files/tex/dccee841f3f498c2c58fa6ae1c1403c5a88c5b8d.png",
null,
"http://openproblemgarden.org/files/tex/5fdfd6d4364192fb4fdca04319ffa276d7272185.png",
null,
"http://openproblemgarden.org/files/tex/73f33398d7e8aa42e6ec25ee2bb4f2b57ed3391a.png",
null,
"http://openproblemgarden.org/files/tex/b9b71fe1446a60666050951b891ed6f44c97682f.png",
null,
"http://openproblemgarden.org/files/tex/b8e7ad0330f925492bf468b5c379baec88cf1b3d.png",
null,
"http://openproblemgarden.org/files/tex/4761b031c89840e8cd2cda5b53fbc90c308530f3.png",
null,
"http://openproblemgarden.org/files/tex/5fdfd6d4364192fb4fdca04319ffa276d7272185.png",
null,
"http://openproblemgarden.org/files/tex/5fdfd6d4364192fb4fdca04319ffa276d7272185.png",
null,
"http://openproblemgarden.org/files/tex/73f33398d7e8aa42e6ec25ee2bb4f2b57ed3391a.png",
null,
"http://openproblemgarden.org/files/tex/4761b031c89840e8cd2cda5b53fbc90c308530f3.png",
null,
"http://openproblemgarden.org/files/tex/5fdfd6d4364192fb4fdca04319ffa276d7272185.png",
null,
"http://openproblemgarden.org/files/tex/73f33398d7e8aa42e6ec25ee2bb4f2b57ed3391a.png",
null,
"http://openproblemgarden.org/files/tex/b8e7ad0330f925492bf468b5c379baec88cf1b3d.png",
null,
"http://openproblemgarden.org/files/tex/4761b031c89840e8cd2cda5b53fbc90c308530f3.png",
null,
"http://openproblemgarden.org/files/tex/4761b031c89840e8cd2cda5b53fbc90c308530f3.png",
null,
"http://openproblemgarden.org/files/tex/5fdfd6d4364192fb4fdca04319ffa276d7272185.png",
null,
"http://openproblemgarden.org/files/tex/928cd9d544fdea62f88a627aaee28c416c4366c0.png",
null,
"http://openproblemgarden.org/files/tex/0f3b87f25409a259a157fcac9808cafa54efa0c7.png",
null,
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null,
"http://openproblemgarden.org/misc/feed.png",
null
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https://papers.ssrn.com/sol3/papers.cfm?abstract_id=269733 | [
"# Approximations for the Value-at-Risk Approach to Risk-Return Analysis\n\n22 Pages Posted: 12 May 2001\n\nSee all articles by Dirk Tasche\n\n## Dirk Tasche\n\nSwiss Financial Market Supervisory Authority (FINMA)\n\n## Luisa Tibiletti\n\nUniversity of Turin - Department of Management\n\nDate Written: January 30, 2001\n\n### Abstract\n\nAn evergreen debate in Finance concerns the rules for making portfolio hedge decisions. A traditional tool proposed in the literature is the well-known standard deviation based Sharpe Ratio, which has been recently generalized in order to involve also other popular risk measures p, such as VaR (Value-at-Risk) or CVaR (Conditional Value at Risk). This approach gives the correct choice of portfolio selection in a mean-p world as long as p is homogeneous of order 1. But, unfortunately, in important cases calculating the exact incremental Sharpe Ratio for ranking profitable portfolios turns out to be computationally too costly. Therefore, more easy-to-use rules for a rapid portfolio selection are needed. The research in this direction for VaR is just the aim of the paper. Approximation formulae are carried out which are based on certain derivatives of VaR and involve quantities similar to the skewness and kurtosis of the random variables under consid-eration. Starting point for the approximations is the observation that the partial derivatives of portfolio VaR with respect to the portfolio weights are just the conditional expectations of the asset returns given that the portfolio return equals VaR. Since the conditional expec-tation of a random variable Y given another random variable X can be considered the best possible regression of Y versus X in least squares sense, the idea is to replace the conditional expectation by polynomial regression or, more generally, by finite-dimensional regression of Y versus X. In case of the variables obeying an elliptical joint distribution, the resulting approximation formulae coincide with the exact formula for the standard deviation taken as risk measure. By means of a number of numerical examples and counter-examples the properties of the formulae are discussed.\n\nKeywords: Sharpe Ratio, Value-at-Risk (VaR), Conditional Value-at-Risk (CVaR), linear regression, quadratic regression.\n\nJEL Classification: C13, D81, G11, G12\n\nSuggested Citation\n\nTasche, Dirk and Tibiletti, Luisa, Approximations for the Value-at-Risk Approach to Risk-Return Analysis (January 30, 2001). EFMA 2001 Lugano Meetings. Available at SSRN: https://ssrn.com/abstract=269733 or http://dx.doi.org/10.2139/ssrn.269733\n\nRegister\n\n## Paper statistics",
null,
""
] | [
null,
"https://papers.ssrn.com/sol3/header/pixel.gif",
null
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https://matheducators.stackexchange.com/questions/14751/why-do-inequalities-flip-signs/14752 | [
"# Why do inequalities flip signs? [closed]\n\nIs there a mathematical reason (like a proof) of why this happens? You can do it with examples and it is 'intuitive.' But the proof of why this happens is never shown in pedagogy, we just warn students to remember to flip the inequality when\n\n• multiply or divide by a negative number both sides\n\n$$-2>-3 \\implies 2 < 3$$\n\n• take reciprocals of same sign fractions both sides\n\n$$\\frac{3}{4} > \\frac{1}{2} \\implies \\frac{4}{3} < 2$$\n\n• Nov 8 '18 at 2:47\n• But the proof of why this happens is never shown in pedagogy --- I was curious about the book I used in high school algebra 1 (actually was the 1970 revised edition), since it was new-math based and heavy with proofs (actually, it was mostly heavy with terminological formalism), and it only gives some numerical examples of how multiplying by a negative number reverses the order and then simply states the results in a \"Multiplication Axiom of Order\" shaded section. Nov 9 '18 at 8:29\n• This will be useless for high school algebra students, but for those who already learnt differential calculus, it might be nice to note that this is essentially because the functions $f(x)=-x$ and $f(x)=1/x$ have derivatives that are negative everywhere. Feb 6 '19 at 10:26\n• It is appalling if your claim that \"the proof of why this happens is never shown in pedagogy\" is in fact true in general. After all, as mathematics teachers we ought to teach mathematics, not blind mimicking. Sep 26 '19 at 8:54\n\nI'm slightly concerned that\n\nIs there a mathematical reason (like a proof) of why this happens?\n\nis a purely mathematical question, but since you write \"we just warn students\" I will assume that this question is purposefully asked here on Math Educators StackExchange.\n\nAs to a proof:\n\nGiven $$a>b$$, subtract $$a$$ from both sides: $$0 > b-a$$.\n\nNext, subtract $$b$$ from both sides: $$-b > -a$$.\n\nNote that this final inequality is equivalent to $$-a < -b$$.\n\nAnd so we have proved: If $$a > b$$, then $$-a < -b$$.\n\n• Sweet, and the same works for reciprocation (if $a<b$ are of the same sign, divide by their strictly positive product $ab$) Nov 8 '18 at 0:22\n• thanks, I get that the concern for proof is technically mathematical but it isn't covered in any proof for any math courses since it's just elementary and assumed to just be understood (in the US anyways). I'm just curious how to show it if some student ever asks. Nov 8 '18 at 4:45\n• @Lenny This sort of thing is often proved in an introductory real analysis course. Nov 8 '18 at 12:38\n• The number of upvotes here is not proportional to the quality of my post. I hope those who are satisfied by this explanation will read my past answers, as well as answers from other users, which have generally involved significantly more work! Nov 10 '18 at 3:45\n\nDepending on the context and the previous curriculum, the following might work:\n\n1. \"less than\" means \"to the left of\" on the number line.\n2. Multiplying by a negative number flips numbers around 0.\n3. Thus, \"left of\" becomes \"right of\", or \"greater than\".\n• This works for reciprocals because you flip (say) the positive reals around 1 (and squash all the numbers greater than one into the unit interval, and stretch the unit interval out to infinity). That's harder to show on a blackboard, but makes for a nice visual if you can find a way to demo it. Nov 8 '18 at 3:22\n• This seems like a visual translation of the intuitive understanding, but the question was asking for a mathematical explanation. Nov 8 '18 at 20:46\n• If inequalities are defined on the number line and multiplication by -1 as flipping, this is 100% mathematical. Nov 8 '18 at 21:32\n• I think this is a great answer, and would hope that students could use this method to provide a proof. It might be worth breaking down into cases, e.g., both negative, both positive, one negative and one positive; but, I certainly agree that this can constitute a \"mathematical explanation\"! Nov 14 '18 at 4:31\n\nPerhaps one way to see (and explain intuitively to children) the \"multiply by $$-1$$ part\" is the following. Imagine your two numbers, $$a$$ and $$b$$, lying on the numberline. Multiplying by $$-1$$ is like 'rotating the numberline through 180°': imagine it's a straight metal pole, lying flat on the ground; pick it up by the middle, and rotate it 'long ways' (ie not 'barrel roll') 180°. (One only needs to consider the line segment $$[-\\max\\{a,b\\},\\max\\{a,b\\}]$$ for this, in case you get a smart-arse saying you can't move something of infinite mass!) It is hopefully clear to most people why the point that was to the right is now to the left.\n\nThis is only a visual intuition -- one of course needs to make this a rigorous proof, and the other answers do this -- however such intuition can often be valuable to early-learners.\n\nA similar, but slightly less clean, statement can be made regarding inversion: take just the positive real axis $$(0,\\infty)$$; rotate it around the number $$1$$. (Since, again, we can look at $$(0,M)$$ for say $$M = a + b$$ (with $$a,b > 0$$), one could imagine some sort of ellipse.)\n\nI would draw graphics to illustrate... if only I were better at doing so!\n\nPS. In complex analysis, multiplying by $$-1$$ really is rotating by 180°!\n\n• [More generally, one can view a reflection in a dimension as being a rotation in the next dimension. E.g., a reflection over the origin here can be thought of as described above, but also a reflection of a point in 2D can be thought of as rotating that point through the third dimension. Anyway] I'm not sure (truly: unsure!) whether this explanation -- imagining a 'long ways' rotation of a metal pole -- would be helpful in building intuition for students with a nascent understanding of inequalities. Nov 14 '18 at 4:35\n• My concern would be that reflections have determinant $-1$ while rotations $1$. While what you say is correct for the projection onto the lower dimensional space, you don't want to start confusion reflections/rotations, I feel. It is an interesting perspective, though, thanks for your comment! Nov 15 '18 at 12:21\n\nFor multiplying or dividing by -1... \\begin{align} a&>b\\\\ a-b&>0\\\\ \\\\-(a-b)&<0\\\\ -a&<-b \\end{align} (You can then extended to arbitrary negative numbers by multiplying or dividing by the [positive] magnitude.)\n\nFor taking reciprocals... assuming $$ab>0$$\n\n\\begin{align} a&>b\\\\ \\left(\\frac{1}{ab}\\right)a&>\\left(\\frac{1}{ab}\\right)b\\\\ \\frac{1}{b}&>\\frac{1}{a}\\\\ \\frac{1}{a}&<\\frac{1}{b}\\\\ \\end{align}\n\n• You might want to add some words in the slot where you flipped the sign from > 0 to < 0. My first reaction was that you just used the fact you were trying to prove. Anyway welcome to the site! Nov 8 '18 at 13:29\n• Thanks... when posting I tried (unsuccessfully) to use \\intertext... so I tried a lineskip and went with that... avoiding further delay. Luckily(?), my post came in 2 minutes before the accepted answer by @Benjamin. Nov 8 '18 at 13:37\n• @robphy What is the point of the third step? You can go straight from the second to the fourth by subtracting $a$ from both sides and it doesn't need an additional rule that $a > 0 \\Rightarrow -a < 0$. Nov 8 '18 at 19:42\n• In the spirit of my second proof, I should have subtracted the sum (a+b) from the first line. But I wanted to explicitly do the \"multiplication by -1\" because that subtraction move might not be obvious to a novice. In the spirit of my first proof, I should have divided both sides by b, then taken the reciprocal explicitly so that a/b > 1 becomes b/a <1. Then divide both sides by b to get 1/a < 1/b. Nov 8 '18 at 22:18\n\nIn a more general case this happens if you deal with a strictly monotonic decreasing function like f(x) = -x or f(x) = 1/x (in the positive or negative numbers).\n\n• Is monotonic monotonous? Just a humorous question. Nov 8 '18 at 16:14\n• @ChrisCunningham Yes, you're right of course. That's what you get for expecting google to treat the term as an entity. Nov 8 '18 at 17:34\n\nI can't help but feel this question highlights the self defeating nature of teaching maths via \"cheat tricks\". In my opinion it is probably best not to teach students ever to just \"switch the sign\" of the inequality in an arbitrary manner, which can be confusing.\n\nAs elegantly illustrated by @Benjamin_Dickman's proof, the alternative notation $$a>b \\quad \\text{implies} \\quad -b>-a$$ is much clearer, and invites the student to treat the inequality as a fixed equivalence in the same way they would treat an equality.\n\nI feel it might be less confusing for students were this convention adopted, and the inequality was never reversed but for as a last resort. That would mean one less \"rule\" for students to \"remember\".\n\nWhich would you rather I give you: 10 cookies or 8 cookies? Okay, now which would you rather have me take away: 10 cookies, or 8 cookies? Adding 10 gives you a larger number than adding 8, but subtracting 10 means leaves you with a smaller number than subtracting 8 does. Similarly, multiplying by 3 gives you more than multiplying by 2, but dividing by 3 gives you less than dividing by 2.\n\n• Moral plus one. Nov 8 '18 at 19:03\n\nI see it as a combination of the following, when we assume a field with a positive cone, and trichotomy. We want to show that if $$a and $$c<0$$, then $$cb.\n\n• It is natural to expect the product of positive numbers to be positive.\n\n• We denote the additive inverse of $$a$$ by $$-a$$.\n\n• We note, from $$a+(-a)=0$$, that $$-(-a)=a$$.\n\n• We note that $$(-a)b=-(ab)$$. Proof: $$(-a)b+ab=(-a+a)b=0\\times b=0$$ (one can earlier deduce that multiplication by $$0$$ is $$0$$ from distributivity).\n\n• Now $$(-a)(-b)=-(a(-b))=-((-b)a)=-(-ab)=ab$$.\n\n• From the above we conclude that the product of two negative numbers is positive, and that a positive number times a negative is negative.\n\n• Now if $$a and $$c<0$$, we have $$b-a>0$$ and $$-c>0$$, so $$(-c)(b-a)>0$$. That is, $$-cb+ac>0$$, or $$cb.\n\nOther answers address multiplication by a negative number on both sides better than I can. For taking reciprocals of both sides, it may be helpful to examine why we know we can do that and maintain equality. This is one simple demonstration:\n\nStarting with the equation $$\\frac{a}{b}=\\frac{c}{d}$$,\n\nwe can multiply both sides by $$\\frac{d}{c}$$, giving $$\\frac{ad}{bc}=\\frac{cd}{dc}$$, which simplifies to $$\\frac{ac}{bd}=1$$.\n\nWe can then do the same with $$\\frac{b}{a}$$, giving $$\\frac{bad}{abc}=\\frac{b}{a}\\cdot 1$$, which simplifies to $$\\frac{d}{c}=\\frac{b}{a}$$.\n\nThis shows that if two numbers are equal, then their reciprocals are equal. However, notice that we didn't just turn each number into its reciprocal. We also flipped their positions: the $$\\frac{a}{b}$$ on the left became $$\\frac{b}{a}$$, but on the right. When we're talking about equality, this is inconsequential, since equality is symmetric. However, if you take the same example with $$=$$ replaced with $$>$$ or $$<$$, the fact that the values \"switched places\" does matter.\n\nThis demonstration also shows why the sign only flips when both sides have the same sign. If they have different signs, then their reciprocals also have different signs, and exactly one of the multiplications by a reciprocal causes the inequality to flip. So you could argue that it does flip, but then it flips again, making it seem like it hasn't flipped. Similarly, if both sides are negative, it flips a total of three times, which has the same result as flipping once."
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https://www.impan.pl/en/publishing-house/journals-and-series/applicationes-mathematicae/all/32/2/84329/note-on-the-anova-of-a-completely-confounded-factorial-experiment | [
"A+ CATEGORY SCIENTIFIC UNIT\n\n# Publishing house / Journals and Serials / Applicationes Mathematicae / All issues\n\n## Note on the ANOVA of a completely confounded factorial experiment\n\n### Volume 32 / 2005\n\nApplicationes Mathematicae 32 (2005), 119-132 MSC: Primary 62K10, 62K15. DOI: 10.4064/am32-2-1\n\n#### Abstract\n\nThe purpose of this paper is to present a modern approach to the analysis of variance (ANOVA) of disconnected resolvable group divisible partially balanced incomplete block (GDPBIB) designs with factorial structure and with some interaction effects completely confounded. A characterization of a factorial experiment with completely confounded interaction is given. The treatment effect estimators and some relations between the matrix $\\textbf{F}$ of the reduced normal equations and the information matrix $\\textbf{A}$ are given. Moreover the ANOVA of the sum of squares for adjusted treatment effects and the matrix $\\textbf{F}$ with its eigenvalues and orthonormal eigenvectors for the case of a completely confounded factorial experiment are presented. A special form of a generalized inverse ($g$-inverse) of $\\textbf{F}$ is introduced (Theorems 3.2.1–3.2.4). The corresponding numerical example has been worked out by Oktaba (1956) and Oktaba, Rejmak and Warteresiewicz (1956) by applying Galois fields and congruences.\n\n#### Authors\n\n• Wiktor OktabaDepartment of Mathematical Statistics\nInstitute of Applied Mathematics\nAgricultural University",
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"https://www.impan.pl/en/publishing-house/journals-and-series/applicationes-mathematicae/all/32/2/84329/note-on-the-anova-of-a-completely-confounded-factorial-experiment",
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https://www.rdocumentation.org/packages/iemisc/versions/0.9.8 | [
"# iemisc v0.9.8\n\n0\n\n0th\n\nPercentile\n\n## Irucka Embry's Miscellaneous Functions\n\nA collection of Irucka Embry's miscellaneous functions (Engineering Economics, Civil & Environmental/Water Resources Engineering, Geometry, Statistics, GNU Octave length functions, Trigonometric functions in degrees, etc.).\n\n# iemisc\n\nR package that contains Irucka Embry's miscellaneous functions: statistical analysis [RMS, coefficient of variation (CV), approximate and relative error, range, harmonic mean, geometric mean], engineering economics (benefit-cost, future value, present value, annual value, gradients, interest, periods, etc.), geometry (sphere volume and right triangle), civil & environmental/water resources engineering (Concrete Mix Design for Normal Strength & Structural Lightweight Concrete, Manning's n, Gauckler-Manning-Strickler equation), a version of linear interpolation for use with NAs, GNU Octave/MATLAB compatible trigonometric functions in degrees, & GNU Octave/MATLAB compatible size, numel, and length functions.\n\n# Installation\n\ninstall.packages(\"iemisc\")\n\n\n# Examples (see more examples in the vignette and in the function descriptions)\n\nlibrary(\"iemisc\")\nrequire(\"stats\")\n\n# 1)\nset.seed(200) # makes the example reproducible\n\nsamp <- rnorm(200) # sample\n\n# Calculate the sample harmonic mean (SHM) of the 200 values\n# Using the default value of na.rm = FALSE\n# using a matrix of the numeric vector obs1\nsamp1 <- matrix(data = samp, nrow = length(samp), ncol = 1, byrow = FALSE,\ndimnames = list(c(rep(\"\", length(samp))), \"Sample\"))\n\nshm(samp1)\n\n# 2)\n# Compute the relative error of the 210 values\nset.seed(210) # makes the example reproducible\ntrue <- rnorm(210) # true\napprox <- rnorm(210) # approximation\n\nrelerror(true, approx)\n\n# 3)\n# Are any of the following right triangles?\n\nrighttri(2, 7) # a = 2, b = 7\n\nrighttri(a = 4, c = 11)\n\nrighttri(b = 4, c = 5)\n\n# 4)\n# What is the future worth of \\$2,390.90 in the present 13 years from now with a\n# 0.25% interest rate compounded annually?\n\nFgivenP(2390.90, 13, 0.25, frequency = \"annual\") # the interest rate is 0.25%\n\n\n# Disclaimer\n\n## Functions in iemisc\n\n Name Description FgivenA Future value given Annual value (Engineering Economics) FgivenAcont Future value given Annual value [continuous] (Engineering Economics) AgivenPcont Annual value given Present value [continuous] (Engineering Economics) AgivenF Annual value given Future value (Engineering Economics) AgivenFcont Annual value given Future value [continuous] (Engineering Economics) AgivenG Annual value given Gradient value (Engineering Economics) EffInt Effective Interest rate (Engineering Economics) FgivenP Future value given Present value (Engineering Economics) CompIntPaid Compound Interest Paid (Engineering Economics) AgivenP Annual value given Present value (Engineering Economics) PgivenAcont Present value given Annual value [continuous] (Engineering Economics) PgivenF Present value given Future value (Engineering Economics) Manningpara Parabolic cross-section for the Gauckler-Manning-Strickler equation Manningrect Rectangular cross-section for the Gauckler-Manning-Strickler equation PgivenA1 Present value for geometric gradient series (Engineering Economics) PgivenA Present value given Annual value (Engineering Economics) Manningtri Triangular cross-section for the Gauckler-Manning-Strickler equation acscd Inverse cosecant (in degrees) [GNU Octave/MATLAB compatible] asecd Inverse secant (in degrees) [GNU Octave/MATLAB compatible] Manningtrap Trapezoidal cross-section for the Gauckler-Manning-Strickler equation acosd Inverse cosine (in degrees) [GNU Octave/MATLAB compatible] PgivenFcont Present value given Future value [continuous] (Engineering Economics) PgivenFivary \"Present equivalent of a series of future cash flows subject to varying interest rates\" (Engineering Economics) FgivenPcont Future value given Present value [continuous] (Engineering Economics) benefitcost Benefit-Cost Ratio (Engineering Economics) cosd Cosine (in degrees) [GNU Octave/MATLAB compatible] atand Inverse tangent (in degrees) [GNU Octave/MATLAB compatible] atan2d \"Two-argument arc-tangent\" (in degrees) [GNU Octave/MATLAB compatible] approxerror Approximate error PgivenG Present value given Gradient value (Engineering Economics) Manningcirc Circular cross-section using the Gauckler-Manning-Strickler equation asind Inverse sine (in degrees) [GNU Octave/MATLAB compatible] cv Coefficient of variation (CV) SimpIntPaid Simple Interest Paid (Engineering Economics) igivenPFn Interest rate given Future value, Number of periods, and Present value (Engineering Economics) length_octave Length of R objects (GNU Octave/MATLAB compatible) iemisc iemisc: Irucka Embry's miscellaneous functions ranges Sample range n Manning's n for natural channels na.interp1 na.interp1 nc1 Horton method for composite Manning's n shm Harmonic mean sind Sine (in degrees) [GNU Octave/MATLAB compatible] nc2 Einstein and Banks method for composite Manning's n secd Secant (in degrees) [GNU Octave/MATLAB compatible] sgm Geometric mean volsphere Sphere volume relerror Relative error acotd Inverse cotangent (in degrees) [GNU Octave/MATLAB compatible] nc3 Lotter method for composite Manning's n cotd Cotangent (in degrees) [GNU Octave/MATLAB compatible] righttri Right triangle calculations nc4 Krishnamurthy and Christensen method for composite Manning's n rms Root-mean-square size Size of R objects (GNU Octave/MATLAB compatible) tand Tangent (in degrees) [GNU Octave/MATLAB compatible] numel Number of elements (GNU Octave/MATLAB compatible) ngivenPFi To Find i Given F, n, and P (Engineering Economics) cscd Cosecant (in degrees) [GNU Octave/MATLAB compatible] No Results!"
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https://estebantorreshighschool.com/interesting-about-equations/hydrostatics-equation.html | [
"## What do you mean by hydrostatics?\n\nFluid statics or hydrostatics is the branch of fluid mechanics that studies “fluids at rest and the pressure in a fluid or exerted by a fluid on an immersed body”. Hydrostatics is fundamental to hydraulics, the engineering of equipment for storing, transporting and using fluids.\n\n## What is an example of hydrostatic pressure?\n\nBlood is a fluid, so blood has a hydrostatic pressure. Increased elevation increases the amount of hydrostatic pressure. For example, the veins and capillaries in our feet have about 100 mm Hg more pressure inside than those at heart level.\n\n## How is Geopotential height calculated?\n\nGeopotential is defined by units of m**2/s**2 both on pressure levels and on the surface (orography). The geopotential height is the geopotential divided by the WMO-defined gravity constant of 9.80665 m/s**2, which is constant for all latitudes and all heights.\n\n## What is hydrostatic pressure in simple words?\n\n[ hī′drə-stăt′ĭk ] The pressure exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity. Hydrostatic pressure increases in proportion to depth measured from the surface because of the increasing weight of fluid exerting downward force from above.\n\n## What is hydrodynamics?\n\n: a branch of physics that deals with the motion of fluids and the forces acting on solid bodies immersed in fluids and in motion relative to them — compare hydrostatics.\n\n## What increases hydrostatic pressure?\n\nAn increase in small artery, arteriolar, or venous pressure will increase the capillary hydrostatic pressure favoring filtration. A reduction of these pressures will have the opposite effect. Tissue pressure can change significantly if fluid moves into tissue space.\n\n## What is the role of hydrostatic pressure?\n\nThe force of hydrostatic pressure means that as blood moves along the capillary, fluid moves out through its pores and into the interstitial space. This movement means that the pressure exerted by the blood will become lower, as the blood moves along the capillary, from the arterial to the venous end.\n\n## How do you calculate depth and pressure?\n\nWe begin by solving the equation P = hρg for depth h: h=Pρg h = P ρ g . Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.\n\n## What height is 500mb?\n\nRanging in elevation from 16,000 feet (4,980 meters) to nearly 20,000 feet (6,000 meters) this is considered the “middle” of the atmosphere. It is based upon the assumption 500 mb is one-half of 1000 millibars (at sea level) and the top of the atmosphere, zero millibars, which occurs at 30+ miles (50+ kilometers) high.\n\n## What height is 500 hPa?\n\nYou might be interested: Equation examples\n\n## How do you calculate pressure from height?\n\nP(h)=P0exp(−0.00012h). When h=0, the pressure P(h) is equal to the average atmospheric sea level pressure P0. At a certain altitude H, the pressure is twice less: P(H)=P02=P0exp(−0.00012H).\n\n## What is kidney hydrostatic pressure?\n\nIn the case of the kidney, hydrostatic pressure is the pressure exerted by fluid on the capillary walls of the glomerulus or the walls of the Bowman’s capsule.\n\n## What is the difference between hydrostatic pressure and hydraulic water flow?\n\nHydrostatic pressure is the pressure exerted by an incompressible fluid (a liquid) at a given point, due to the weight of the fluid above it. Atmospheric pressure is the pressure exerted on a surface by the weight of the atmosphere (a compressible fluid) above it.\n\n### Releated\n\n#### Equation of vertical line\n\nHow do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]\n\n#### Bernoulli’s equation example\n\nWhat does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]"
] | [
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https://www.jpost.com/israel/theyve-come-to-tell-me-hes-dead | [
"(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);"
] | [
null
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https://cstheory.stackexchange.com/questions/30537/nondeterministic-communication-complexity-of-set-disjointness/30538 | [
"# Nondeterministic communication complexity of set disjointness?\n\nIn the two-party setting, bounds of $\\Theta(n)$ bits are known for deterministic and bounded-error randomized protocols for $\\text{DISJ}_n$. (Here $\\text{DISJ}_n$ is the $n$-element set disjointness function; the parties are each given $n$-bit Boolean vectors $x$ and $y$, and have to decide whether there is some index $i$ so that $x_i = y_i = 1$.)\n\nIn contrast, for nondeterministic protocols in the two-party setting, the best lower bound I've been able to find in the literature is $\\Omega(n^{1/2})$ bits: Alexander A. Sherstov, Communication Lower Bounds Using Directional Derivatives, JACM 2014, doi:10.1145/2629334 (preprint). A large gap remains between this and the trivial upper bound of $n+1$ bits.\n\nSherstov's bound is for multiparty protocols.\n\nIs a better lower bound known for the 2-party case?\n\nHere follows a simple (folklore?) argument that $\\Omega(n/\\log n)$ bits are required; I am looking for a reference for this so I can cite it, or ideally a stronger $\\omega(n/\\log n)$ bound.\n\nA standard result (e.g. Theorem 2.11 in the Kushilevitz/Nisan textbook) is that $$D(f) \\le O(N^0(f)N^1(f)),$$ where $N^1(f)$ is the nondeterministic communication complexity of $f$ and $N^0(f)$ is the co-nondeterministic communication complexity. For $f = \\text{DISJ}_n$ it is easy to see that $N^0(f) = O(\\log n)$, by guessing the index of a common element. Hence there are constants $C_1,C_2,C_3$ such that for all sufficiently large $n$, $$C_1 n \\le D(f) \\le C_2 N^0(f)N^1(f) \\le C_2 C_3(\\log n)N^1(f).$$ Hence $N^1(f) \\ge \\Omega(n/\\log n)$.\n\n$N^1(DISJ_n) \\geq n$.\nThe fooling set technique actually provides lower bounds on the cover number. $C^1(f)$ is the minimum number of monochromatic rectangles needed to cover the $1$-inputs of $f$. In the fooling set technique, you create a set of input pairs and show that each must lie in distinct monochromatic rectangles.\nIn this case, look at the set of input pairs $\\{(A,\\bar{A}) : A \\subseteq [n]\\}$. It is not too hard to show that $(A, \\bar{A}), (B, \\bar{B})$ must lie in different monochromatic rectangles if $A,B$ are distinct. Hence, to cover the $1$-inputs, we need at least $2^n$ monochromatic rectangles.\nUsing $N^1(f) = \\log_2(C^1(f))$, we get that $N^1(DISJ_n) \\geq n$."
] | [
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http://www.annaunivedu.in/2012/10/ma9332-applied-mathematics-for-thermal.html | [
"## Friday, October 19, 2012\n\n### MA9332 APPLIED MATHEMATICS FOR THERMAL ENGINEERS SYLLABUS | ANNA UNIVERSITY ME CRYOGENIC ENGINEERING 1ST SEM SYLLABUS REGULATION 2009 2011 2012-2013\n\nLatest: TNEA 2014 Engineering Application Status, Counselling Date, Rank List\nMA9332 APPLIED MATHEMATICS FOR THERMAL ENGINEERS SYLLABUS | ANNA UNIVERSITY ME CRYOGENIC ENGINEERING 1ST SEM SYLLABUS REGULATION 2009 2011 2012-2013 BELOW IS THE ANNA UNIVERSITY FIRST SEMESTER ME CRYOGENIC ENGINEERING DEPARTMENT SYLLABUS, TEXTBOOKS, REFERENCE BOOKS,EXAM PORTIONS,QUESTION BANK,PREVIOUS YEAR QUESTION PAPERS,MODEL QUESTION PAPERS, CLASS NOTES, IMPORTANT 2 MARKS, 8 MARKS, 16 MARKS TOPICS. IT IS APPLICABLE FOR ALL STUDENTS ADMITTED IN THE YEAR 2011 2012-2013 (ANNA UNIVERSITY CHENNAI,TRICHY,MADURAI,TIRUNELVELI,COIMBATORE), 2009 REGULATION OF ANNA UNIVERSITY CHENNAI AND STUDENTS ADMITTED IN ANNA UNIVERSITY CHENNAI DURING 2009\n\nMA9332 APPLIED MATHEMATICS FOR THERMAL ENGINEERS L T P C\n3 1 0 4\nUNIT I APPLICATIONS OF FOURIER TRANSFORM 9\nFourier Transform methods – one-dimensional heat conduction problems in infinite and semiinfinite\nrod – Laplace Equation – Poisson Equation.\nUNIT II CALCULUS OF VARIATIONS 9\nConcept of variation and its properties – Euler’s equation – Functionals dependant on first and\nhigher order derivatives – Functionals dependant on functions of several independent variables\n– Variational problems with moving boundaries – Direct methods – Ritz and Kantorovich\nmethods.\nUNIT III CONFORMAL MAPPING AND APPLICATIONS 9\nThe Schwarz- Christoffel transformation – Transformation of boundaries in parametric form –\nPhysical applications:Fluid flow and heat flow problems.\nUNIT IV FINITE DIFFERENCE METHODS FOR PARABOLIC EQUATIONS 9\nOne dimensional parabolic equation – Explicit and Crank-Nicolson Schemes – Thomas\nAlgorithm – Weighted average approximation – Dirichlet and Neumann conditions – Two\ndimensional parabolic equations – ADI method.\nUNIT V FINITE DIFFERENCE METHODS FOR ELLIPTIC EQUATIONS 9\nSolutions of Laplace and Poisson equations in a rectangular region – Finite difference in polar\ncoordinates – Formulae for derivatives near a curved boundary while using a square mesh.\nL +T: 45+15 = 60 PERIODS\nREFERENCE BOOKS:\n1. Mitchell A.R. and Griffith D.F., The Finite difference method in partial differential\nequations, John Wiley and sons, New York (1980).\n2. Sankara Rao, K., Introduction to Partial Differential Equations, Prentice Hall of India Pvt.\nLtd., New Delhi (1997).\n3. Gupta, A.S., Calculus of Variations with Applications, Prentice Hall of India Pvt. Ltd.,\nNew Delhi (1997).\n4. Spiegel, M.R., Theory and Problems of Complex Variables and its Application (Schaum’s\nOutline Series), McGraw Hill Book Co., Singapore (1981). Andrews, L.C. and Shivamoggi,\nB.K., Integral Transforms for Engineers, Prentice Hall of India Pvt. Ltd., New Delhi (2003).\n5. Elsgolts, L., Differential Equations and the Calculus of Variations, MIR Publishers, Moscow\n(1973).\n6. Mathews, J.H. and Howell, R.W., Complex Analysis for Mathematics and Engineering,\nNarosa Publishing House, New Delhi (1997).\n7. Morton, K.W. and Mayers, D.F. Numerical solution of partial differential equations,\nCambridge University press, Cambridge (2002).\n8. Jain, M. K., Iyengar, S. R. K. and Jain, R. K. “ Computational Methods for Partial Differential\nEquations”, New Age International (P) Ltd., 2003."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6979447,"math_prob":0.75936675,"size":5647,"snap":"2022-27-2022-33","text_gpt3_token_len":1474,"char_repetition_ratio":0.10827574,"word_repetition_ratio":0.8832117,"special_character_ratio":0.23446077,"punctuation_ratio":0.21449,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9737208,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T00:56:48Z\",\"WARC-Record-ID\":\"<urn:uuid:d7fab31d-6437-43f2-a8b1-780eddb062f3>\",\"Content-Length\":\"204175\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42fd9f87-ba95-483b-98b8-868a423db325>\",\"WARC-Concurrent-To\":\"<urn:uuid:33edcde6-b092-44ac-9bf8-e9d70314a191>\",\"WARC-IP-Address\":\"172.217.1.211\",\"WARC-Target-URI\":\"http://www.annaunivedu.in/2012/10/ma9332-applied-mathematics-for-thermal.html\",\"WARC-Payload-Digest\":\"sha1:WKXYWGLVSSCTIVLWVHXKK7Y3SZSWQS2F\",\"WARC-Block-Digest\":\"sha1:7SYCHY7KNIZDU5AFR2OVGVCK7M43D2CD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103033925.2_warc_CC-MAIN-20220625004242-20220625034242-00585.warc.gz\"}"} |
https://www.enotes.com/homework-help/force-90-lb-acts-an-object-an-angle-55-degrees-465969?en_action=hh-question_click&en_label=hh-sidebar&en_category=internal_campaign | [
"# A force of 90 lb acts on an object at an angle of 55 degrees. A second force of 100 lbs acts on the object at an angle of -60 degrees. Find the direction and magnitude of the resultant force.\n\ntiburtius",
null,
"| Certified Educator\n\ncalendarEducator since 2012\n\nstarTop subjects are Math, Science, and History\n\nFirst force is not measured in pounds but in newtons so I will pretend the forces are 90 N and 100 N.\n\nNow we can calculate the magnitude of resultant force by using law of cosine (see the image below).\n\n(The entire section contains 109 words.)"
] | [
null,
"https://static.enotescdn.net/images/core/educator-indicator_thumb.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7966003,"math_prob":0.81912196,"size":1216,"snap":"2019-51-2020-05","text_gpt3_token_len":328,"char_repetition_ratio":0.16419142,"word_repetition_ratio":0.0882353,"special_character_ratio":0.30098686,"punctuation_ratio":0.104417674,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9913132,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-12T11:38:58Z\",\"WARC-Record-ID\":\"<urn:uuid:954cb1e0-1ead-46ec-b7d2-e72481c0ec5a>\",\"Content-Length\":\"44687\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:661251e0-3e88-465f-999f-83d39db3f993>\",\"WARC-Concurrent-To\":\"<urn:uuid:bfe11055-0dc6-4151-a84d-d5a1e2a2c042>\",\"WARC-IP-Address\":\"104.26.5.75\",\"WARC-Target-URI\":\"https://www.enotes.com/homework-help/force-90-lb-acts-an-object-an-angle-55-degrees-465969?en_action=hh-question_click&en_label=hh-sidebar&en_category=internal_campaign\",\"WARC-Payload-Digest\":\"sha1:LQ53TNMJ7JJHW72XT3JWHQCGWVOVXPVV\",\"WARC-Block-Digest\":\"sha1:QW52BF7ZM66Y3L2TSYYG2J6FAJRTKQHB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540543252.46_warc_CC-MAIN-20191212102302-20191212130302-00374.warc.gz\"}"} |
https://justinmchase.com/tag/aws/ | [
"## Iterating with AWS Step Functions\n\nOne interesting challenge I immediately encountered when attempting to work with AWS Lambda and Step functions was the need to process large files. Lambda functions have a couple of limitations namely memory and a 5 minute timeout. If you have some operation you need to perform on a very large dataset it may not be possible to complete this operation in a single execution of a lambda function. There are several ways to solve this problem, in this article I would like to demonstrate how to create an iterator pattern in an AWS Step Function as a way to loop over a large set of data and process it in smaller parts.",
null,
"In order to iterate we have created an Iterator Task which is a custom Lambda function. It accepts three values as inputs in order to operate: index, size and count.\n\nHere is the code for this example step function:\n\n```{\n\"Comment\": \"Iterator Example\",\n\"StartAt\": \"ConfigureCount\",\n\"States\": {\n\"ConfigureCount\": {\n\"Type\": \"Pass\",\n\"Result\": 10,\n\"ResultPath\": \"\\$.count\",\n\"Next\": \"ConfigureIterator\"\n},\n\"ConfigureIterator\": {\n\"Type\": \"Pass\",\n\"Result\": {\n\"index\": -1,\n\"step\": 1\n},\n\"ResultPath\": \"\\$.iterator\",\n\"Next\": \"Iterator\"\n},\n\"Iterator\": {\n\"Resource\": \"arn:aws:lambda:{region}:{accountId}:function:iterator\",\n\"ResultPath\": \"\\$.iterator\",\n\"Next\": \"IterateRecords\"\n},\n\"IterateRecords\": {\n\"Type\": \"Choice\",\n\"Choices\": [\n{\n\"Variable\": \"\\$.iterator.continue\",\n\"BooleanEquals\": true,\n\"Next\": \"ExampleWork\"\n}\n],\n\"Default\": \"Done\"\n},\n\"ExampleWork\": {\n\"Type\": \"Pass\",\n\"Result\": {\n\"success\": true\n},\n\"ResultPath\": \"\\$.result\",\n\"Next\": \"Iterator\"\n},\n\"Done\": {\n\"Type\": \"Pass\",\n\"End\": true\n}\n}\n}\n```\n\n### ConfigureCount\n\nIn this step we need to configure the number of times we want to iterate. In this case I have set the number of iterations to 10 and put it into a variable called \\$.count. In a more complete example this may be the number of files you want to iterate over. For example in my real world scenario I am receiving a substantial CSV file which is then broken into many smaller CSV files, all stored in s3, the number of smaller files is then set into the count variable here. The large CSV file can be read entirely in a single lambda execution, streaming sections into smaller files, never loading the entire file into memory at the same time; but it cannot be processed entirely in a single function. Thus we split it and then iterate over the smaller parts.\n\n### ConfigureIterator\n\nHere we set the index and step variables into the \\$.iterator field, which the iterator lambda uses to determine whether or not it should continue iterating.\n\n### Iterator\n\nThis is the iterator itself, a small lambda function that simply increments the current index by the step size and calculates the continue field based on the current index and count.\n\n```export function iterator (event, context, callback) {\nlet index = event.iterator.index\nlet step = event.iterator.step\nlet count = event.count\n\nindex += step\n\ncallback(null, {\nindex,\nstep,\ncount,\ncontinue: index < count\n})\n}\n```\n\nThe reason why we want to support a step size is because we may have multiple workers which operate on data in parallel. In this example we have a single worker but in other cases we may need more in order to complete the overall work in a timely fashion.\n\n### IterateRecords\n\nFrom there we need to immediately move into a Choice state. This state simply looks at the \\$.iterator.continue field and if it is not true then our iteration is over and we exit the loop. If iteration is not over then we move to the worker tasks which may use the \\$.iterator.index field to determine which unit of work it should operate on.\n\n### ExampleWork\n\nIn this example this is just a Pass state, but in a real example this may represent a series of Tasks or Activities which process the data for this iteration. When completed, the last step in the series should point back to the Iterator state.\n\nIts also important to note that all states in this chain must use the ResultPath field to bucket their results in order to preserve the state of the iterator field throughout theses states. Do not override the \\$.iterator or \\$.count fields while doing work or you may end up in an infinite loop or error condition.\n\n### Done\n\nThis state simply signifies the end of the step function."
] | [
null,
"https://justinmchase.files.wordpress.com/2017/03/screenshot-2017-03-08-00-39-02.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8568064,"math_prob":0.79275626,"size":4216,"snap":"2019-35-2019-39","text_gpt3_token_len":930,"char_repetition_ratio":0.13081671,"word_repetition_ratio":0.0028735632,"special_character_ratio":0.23790322,"punctuation_ratio":0.16009852,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95170414,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T12:37:41Z\",\"WARC-Record-ID\":\"<urn:uuid:80fe8294-939b-4626-ad1b-ee854a9b93ff>\",\"Content-Length\":\"50819\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bea74dc8-154e-4d85-8c81-14c090c956bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:46e6c546-bfe6-4a72-8c30-593cbcf03d0b>\",\"WARC-IP-Address\":\"192.0.78.24\",\"WARC-Target-URI\":\"https://justinmchase.com/tag/aws/\",\"WARC-Payload-Digest\":\"sha1:QRCBRILDCDRS2VX2UX4KYUURLRNHFNAT\",\"WARC-Block-Digest\":\"sha1:ZY6VTB4VLVE5JD354GUKQ74NVKIN42PU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573284.48_warc_CC-MAIN-20190918110932-20190918132932-00433.warc.gz\"}"} |
https://theoffseasontraining.com/bukidnon/calculus-terms-that-start-with-g.php | [
"Limit of Functions Find the Limit in Easy Steps\n\nHow can a layman with little physics and calculus",
null,
"PreCalculus Lesson 1 YouTube. Calculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra, Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools..\n\nCalculus Rules Flashcards Quizlet\n\nA Survival Guide to Vector Calculus. You can keep on taking derivatives (e.g. fourth, fifth), extracting more and more information from that simple position function. And it doesn’t just work with position; Calculus can work with any function. The formula gives a more precise (i.e. more mathematical) definition. The Derivative Formula, Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t..\n\n5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here! 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nIntroduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t. Next we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x.\n\n8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant... 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nNext we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x. MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\n& pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\nNext we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x. 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nYou can keep on taking derivatives (e.g. fourth, fifth), extracting more and more information from that simple position function. And it doesn’t just work with position; Calculus can work with any function. The formula gives a more precise (i.e. more mathematical) definition. The Derivative Formula themathlab.com, a whole new math experience key in To press keys or enter information into a calculator.. key sequence A set of instructions for what to key in on a calculator. When writing out a key sequence, operations and symbols like parentheses are typically drawn in …\n\nCalculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra 4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus.\n\nIn real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there. Check out StudyPug's tips & tricks on Limit laws for Calculus.\n\n4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus. MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\n8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant... Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t.\n\n5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here! Galileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g…\n\nMathematical \"G\" Words *Dictionary Alert. gallon (gal) A unit of capacity in the U.S. system of measurement equal to 4 quarts. generalization A statement that is true about many instances. glide reflection The composite of a reflection and a translation parallel to … MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\nGalileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g… You can keep on taking derivatives (e.g. fourth, fifth), extracting more and more information from that simple position function. And it doesn’t just work with position; Calculus can work with any function. The formula gives a more precise (i.e. more mathematical) definition. The Derivative Formula\n\n4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus. Galileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g…\n\nIn real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there. A Lambda Calculus for G odel{Dummett Logic Capturing Waitfreedom Yoichi Hirai The University of Tokyo, JSPS Research Fellow [email protected]\n\n6/12/2019 · Start by recognizing that this is not just impossible, but it’s ludicrously impossible. Learning all of physics is like learning all of history. Nobody can do it, nobody has done it, and the mere suggestion that it might be possible heavily implie... You can keep on taking derivatives (e.g. fourth, fifth), extracting more and more information from that simple position function. And it doesn’t just work with position; Calculus can work with any function. The formula gives a more precise (i.e. more mathematical) definition. The Derivative Formula\n\nNext we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x. Counting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science\n\nCheck out StudyPug's tips & tricks on Limit laws for Calculus. Counting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science\n\nCalculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra Next we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x.\n\nHow can a layman with little physics and calculus",
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"Mathwords G. Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools., In real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there..",
null,
"Calculus Tutors – HelpHub. Mathematical \"G\" Words *Dictionary Alert. gallon (gal) A unit of capacity in the U.S. system of measurement equal to 4 quarts. generalization A statement that is true about many instances. glide reflection The composite of a reflection and a translation parallel to …, Calculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra.\n\nMultivariable Calculus Finding g'(0) Physics Forums",
null,
"Multivariable Calculus Finding g'(0) Physics Forums. Check out StudyPug's tips & tricks on Limit laws for Calculus. Mathematical \"G\" Words *Dictionary Alert. gallon (gal) A unit of capacity in the U.S. system of measurement equal to 4 quarts. generalization A statement that is true about many instances. glide reflection The composite of a reflection and a translation parallel to ….",
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"Counting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\n4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus. In real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there.\n\nGalileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g… & pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons\n\nA Lambda Calculus for G odel{Dummett Logic Capturing Waitfreedom Yoichi Hirai The University of Tokyo, JSPS Research Fellow [email protected] MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\nA Lambda Calculus for G odel{Dummett Logic Capturing Waitfreedom Yoichi Hirai The University of Tokyo, JSPS Research Fellow [email protected] 4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus.\n\nYou can keep on taking derivatives (e.g. fourth, fifth), extracting more and more information from that simple position function. And it doesn’t just work with position; Calculus can work with any function. The formula gives a more precise (i.e. more mathematical) definition. The Derivative Formula Mathematical \"G\" Words *Dictionary Alert. gallon (gal) A unit of capacity in the U.S. system of measurement equal to 4 quarts. generalization A statement that is true about many instances. glide reflection The composite of a reflection and a translation parallel to …\n\n4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus. Check out StudyPug's tips & tricks on Limit laws for Calculus.\n\nGalileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g… Galileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g…\n\nNext we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x. 8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant...\n\n6/12/2019 · Start by recognizing that this is not just impossible, but it’s ludicrously impossible. Learning all of physics is like learning all of history. Nobody can do it, nobody has done it, and the mere suggestion that it might be possible heavily implie... 8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant...\n\nCheck out StudyPug's tips & tricks on Limit laws for Calculus. 8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant...\n\nMultivariable Calculus Finding g'(0) Physics Forums",
null,
"A Lambda Calculus for G odel{Dummett Logic Capturing. Galileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g…, Calculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra.\n\nMathwords G\n\nHow can a layman with little physics and calculus. 6/12/2019 · Start by recognizing that this is not just impossible, but it’s ludicrously impossible. Learning all of physics is like learning all of history. Nobody can do it, nobody has done it, and the mere suggestion that it might be possible heavily implie..., & pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons.\n\n8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant... Calculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra\n\nCounting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science & pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons\n\nCheck out StudyPug's tips & tricks on Limit laws for Calculus. themathlab.com, a whole new math experience key in To press keys or enter information into a calculator.. key sequence A set of instructions for what to key in on a calculator. When writing out a key sequence, operations and symbols like parentheses are typically drawn in …\n\nCounting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science 6/12/2019 · Start by recognizing that this is not just impossible, but it’s ludicrously impossible. Learning all of physics is like learning all of history. Nobody can do it, nobody has done it, and the mere suggestion that it might be possible heavily implie...\n\nGalileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g… 4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus.\n\n4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus. Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools.\n\n6/12/2019 · Start by recognizing that this is not just impossible, but it’s ludicrously impossible. Learning all of physics is like learning all of history. Nobody can do it, nobody has done it, and the mere suggestion that it might be possible heavily implie... Next we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x.\n\n8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant... Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools.\n\nCounting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science themathlab.com, a whole new math experience key in To press keys or enter information into a calculator.. key sequence A set of instructions for what to key in on a calculator. When writing out a key sequence, operations and symbols like parentheses are typically drawn in …\n\nCounting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools.\n\nStart studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools. 4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus.\n\n& pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons Mathematical \"G\" Words *Dictionary Alert. gallon (gal) A unit of capacity in the U.S. system of measurement equal to 4 quarts. generalization A statement that is true about many instances. glide reflection The composite of a reflection and a translation parallel to …\n\nStart studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t.\n\nCalculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nIntroduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t. 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nA Lambda Calculus for G odel{Dummett Logic Capturing Waitfreedom Yoichi Hirai The University of Tokyo, JSPS Research Fellow [email protected] Check out StudyPug's tips & tricks on Limit laws for Calculus.\n\nNext we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x. & pre-calculus : calculus: advanced topics: probability & statistics: real world applications: multimedia entries: www.mathwords.com: about mathwords : website feedback : Mathwords: Terms and Formulas from Algebra I to Calculus written, illustrated, and webmastered by Bruce Simmons\n\nIn real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there. MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\nMATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\nIn real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there. Check out StudyPug's tips & tricks on Limit laws for Calculus.\n\nthemathlab.com, a whole new math experience key in To press keys or enter information into a calculator.. key sequence A set of instructions for what to key in on a calculator. When writing out a key sequence, operations and symbols like parentheses are typically drawn in … 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nCounting and Generating Terms in the Binary Lambda",
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"Calculus Tutors – HelpHub. Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t., MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos.",
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"Limit of Functions Find the Limit in Easy Steps. 8/13/2015В В· Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant..., Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t..\n\nMathwords G",
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"Limit laws StudyPug. Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or 13 From the formula for cos(2t + t) find cos 3t in terms of cos t. 14 From the formula for sin (2t + t) find sin 3t in terms of sin t..",
null,
"8/13/2015 · Business calculus is usually much easier than the regular calculus for science and engineering majors. The business calculus class usually covers limits, and derivatives and their application to business (maximizing profits/revenues, picking plant... Start studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools.\n\nCalculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra Counting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science\n\nCounting and Generating Terms in the Binary Lambda Calculus (Extended version) Katarzyna Grygiel 1and Pierre Lescanne;2 1Jagiellonian University, Faculty of Mathematics and Computer Science Next we consider the function g(x) = x2 with domain all positive real numbers. To see for which xthe inverse g 1(x) is de ned we try to solve the equation g(y) = x, i.e. we try to solve y2 = x. If x<0 then this equation has no solutions since y 0 for all y. But if x 0 then y=xdoes have a solution, namely y= p x.\n\nGalileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g… In real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there.\n\nGalileo established that in t seconds a freely falling body falls a distance gt 2 /2, where g is a constant (later interpreted by Newton as the gravitational constant). With the definition of average velocity as the distance per time, the body’s average velocity over an interval from t to t + h is given by the expression [g… Check out StudyPug's tips & tricks on Limit laws for Calculus.\n\nCalculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra Mathematical \"G\" Words *Dictionary Alert. gallon (gal) A unit of capacity in the U.S. system of measurement equal to 4 quarts. generalization A statement that is true about many instances. glide reflection The composite of a reflection and a translation parallel to …\n\nCheck out StudyPug's tips & tricks on Limit laws for Calculus. In real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there.\n\nStart studying Calculus Rules. Learn vocabulary, terms, and more with flashcards, games, and other study tools. MATH222 SECONDSEMESTER CALCULUS Spring2011 1. 2 Math222–2ndSemesterCalculus and/or modify this document under the terms of the GNU Free Documentation G′(x) = f(x), 2 You will start noticing things like this after doing several examples. 8 so that Z 2xsin(x2 +3)dx = −cos\n\nA Lambda Calculus for G odel{Dummett Logic Capturing Waitfreedom Yoichi Hirai The University of Tokyo, JSPS Research Fellow [email protected] 5/6/2017 · Support PF! Buy your school textbooks, materials and every day products via PF Here!\n\nthemathlab.com, a whole new math experience key in To press keys or enter information into a calculator.. key sequence A set of instructions for what to key in on a calculator. When writing out a key sequence, operations and symbols like parentheses are typically drawn in … Calculus 1 (e.g. UVic Math 100, UBC Math 100 / 102 / 104) Calculus 2 (e.g. UVic Math 101, UBC Math 101 / 103 / 105) Linear Algebra (e.g. UVic Math 110 / 211, UBC Math 221) Finite math (e.g. UVic Math 151, UBC Math 230) Pre Calculus 12 Pre Calculus 11 Highschool Math Math Advanced Math Algebra\n\n4/4/2017 · This video is a review of the exponent laws and the rules for simplifying rationals in preparation for a course in calculus. In real life, driving “at the speed limit” might mean you’re going at exactly 70 mph. But it can also mean 69.998, 69.999 mph. If you get very, very close, you can still say you drove at the speed limit; It’s the same thing in calculus: you’re looking for that intended value (e.g. 70 mph), even if you don’t quite get there.\n\nView all posts in Bukidnon category"
] | [
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https://www.dothobangdong.info/2020/11/breakthrough-the-learning-obstacles-with-online-calculus-help/ | [
"# Breakthrough The Learning Obstacles With Online Calculus Help\n\nCalculus is a branch of Mathematics and most students from the High school level find it difficult to solve problems related to this topic. This branch of Math is related to limits, integrals, functions and derivatives and it also involves two major division operations like differential calculus and integral calculus. Calculus is known as the study of change and it is used for solving problems in the area of Economics, Science and Engineering. Calculus plays a major role in modern mathematics education. Students need to understand the basic concepts of Calculus to score well. Several online tutoring portals provide Free Calculus help which is beneficial for students of any grade.\n\nIn online tutoring, the elementary concepts of Calculus are taught by expert tutors who have an in-depth knowledge of the subject and can explain the process step-by-step in a better way. Online Calculus help is a great solution for students who feel scared of solving Calculus problems. Some basic concepts of Calculus Derivatives, Calculus Anti derivative, Continuity Calculus, Pre Calculus and Vector Calculus are available online. To get Calculus help online, students just need to schedule their online sessions with a preferred tutor. Thus, they can easily resolve their learning hurdles within a short span of time. Calculus worksheets are also available online and by practicing these repeatedly, any student can get extensive knowledge of the subject.\n\nTaking free Calculus help is a smart way to learn Calculus anytime from home. Online Calculus help involves unlimited tutoring sessions with an experienced tutor where student can clear their doubts on any topic like Indefinite Integrals, Differential Equations, Application of Derivatives and others. In this secure online environment, students can freely ask doubts and can understand the topics unhesitatingly. It undoubtedly helps students in scoring well in exams. Additionally, it offers personalized attention to students where they can get quality learning help in real-time.\n\nInstant connectivity with a tutor at the student’s convenient time is the most preferred option of online tutoring and online Calculus help is a one option used by students of higher grades to get clear on concepts, solve problems and finish homework and assignments on time Apart from the white Board, students can use other Calculus tools like the Derivative Calculator and the Integral Calculator for solving any problems with tutors. Most importantly, in an online tutoring environment, students can use the attached chat box to share information while a white board allows the student to write, draw and share problems and explanations with the tutor. Using all these options, students can easily break through the Calculus obstacle and understand and enjoy the subject."
] | [
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https://asmedigitalcollection.asme.org/mechanismsrobotics/article-abstract/4/1/011005/468303/Relative-Acceleration-in-Spatial-Mechanisms?redirectedFrom=fulltext | [
"A scheme applicable to automatic computation of the joint relative accelerations, the derivatives of joint speeds with respect to time, has been developed in the dual-number representation of the kinematics of spatial mechanisms. The equations which have been developed can be set up automatically by computer for any closed-loop mechanism of binary links. While the RCCC mechanism is given as an example, the scheme can be adapted for mechanisms with prismatic as well as revolute and cylindrical joints and can be readily further developed to consider other types of joints.\n\nReferences\n\nReferences\n1.\nRico Martinez\n,\nJ. M.\n, and\nDuffy\n,\nJ.\n, 1996, “\nAn Application of Screw Algebra to the Acceleration Analysis of Serial Chains\n,”\nMech. Mach. Theory\n,\n31\n, pp.\n445\n457\n.\n2.\nRico\n,\nJ. M.\n,\nGallardo\n,\nJ.\n, and\nDuffy\n,\nJ.\n, 1999, “\nScrew Theory and Higher Order Kinematic Analysis of Open Serial and Closed Chains\n,”\nMech. Mach. Theory\n,\n34\n, pp.\n559\n586\n.\n3.\nFischer\n,\nI. S.\n, 1999,\nDual-Number Methods in Kinematics, Statics and Dynamics\n,\nCRC Press\n,\nBoca Raton FL\n, ISBN 0-8493-9115-6.\n4.\nYang\n,\nA. T.\n, 1966, “\nAcceleration Analysis of Spatial Four-Link Mechanisms\n,”\nASME J. Eng. Ind.\n,\n88\n, pp.\n296\n300\n.\n5.\nYang\n,\nA.T.\n, and\nSun\n,\nZ.\n, 1983, “\nDynamic Force and Torque Analysis of Spherical Four Link Mechanisms\n,”\nASME J. Mech., Transm., Autom. Des.\n,\n105\n, pp.\n492\n497\n.\n6.\nPennock\n,\nG. R.\n, and\nYang\n,\nA. T.\n, 1983, “\nDynamic Analysis of a Multi-Rigid-Body Open-Chain System\n,”\nASME J. Mech., Transm., Autom. Des.\n,\n105\n, pp.\n28\n34\n.\n7.\nGong\n,\nJ.\n, and\nZhang\n,\nY.\n, 2010 “\nAcceleration Estimation of Redundant Input Parallel Mechanism by Inverse Dynamics\n,”\nInformatics in Control, Automation and Robotics\n, pp.\n186\n189\n.\n8.\nHuang\n,\nM.-S.\n,\nChen\n,\nK.-Y.\n, and\nFung\n,\nR.-F.\n, 2010, “\nComparison Between Mathematical Modeling and Experimental Identification of a Spatial Slider-Crank Mechanism\n,”\nAppl. Math. Mod.\n34\n, pp.\n2059\n2073\n.\n9.\nLi\n,\nK.\n, and\nWen\n,\nR.\n, 2011 “\nClosed-Form Dynamic Equations of the 6-RSS Parallel Mechanism Through the Newton-Euler Approach\n,”\nThird International Conference on Measuring Technology and Mechatronics Automation\n, pp.\n712\n715\n.\n10.\nRui\n,\nW.\n,\nGang\n,\nD.\n, and\nShi Sheng\n,\nZ.\n, 2011, “\nKinematics Analysis of a New 5 DOF Parallel-Serial machine Tool Based on the Vector Method\n,”\nKey Eng. Mater.\n,\n450\n, pp.\n283\n287\n.\n11.\nTian\n,\nZ.\n, and\nWu\n,\nH.\n, 2010 “\nSpatial Operator Algebra for Free-Floating Space Robot Modeling and Simulation\n,”\nChin. J. Mech. Eng.\n,\n23\n, pp.\n1\n7\n.\n12.\nFischer\n,\nI. S.\n, and\nPaul\n,\nR. N.\n, 1991, “\nKinematic Displacement Analysis of a Double-Cardan-Joint Driveline\n,”\nASME J. Mech. Des.\n,\n113\n, pp.\n263\n271\n.\n13.\nUicker\n,\nJ. J.\n,\nDenavit\n,\nJ.\n, and\nHartenberg\n,\nR. S.\n, 1964, “\nAn Iterative Method for the Displacement Analysis of Spatial Mechanisms\n,”\nASME J. Appl. Mech.\n,\n93\n, pp.\n309\n314\n.\nYou do not currently have access to this content."
] | [
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https://www.notion.so/Upload-a-trained-model-5a3a45d0d8f64f26ac5c19df921d861d | [
"Datagran allows you to upload a python Scikit-Learn trained model.\n\nTo do so, go to your project and select Models.",
null,
"",
null,
"## How does Custom Model Work?\n\nWhen a custom model operator runs; it has 2 main roles:\n\n• make predictions via `predict()` function of the dumped model\n• do the training via `fit()` function of the dumped model (if the conditions you defined is met)\n\n## Dumping the Model\n\nAs a quick start; the following code block shows how to create a dump of a model.\n\n``````import cloudpickle\nf = open('./model.cloudpickle', 'wb')\ncloudpickle.dump(model, f)\nf.close()\n``````\n\nYou can upload a model that has complex sklearn pipelines including your custom transformers. The following code block shows how to create such a pipeline. Once the model is ready, you can use the code above to dump the model. (The full example of this code can be found on this notebook: https://colab.research.google.com/drive/1--PKvlYN8c2EcaW-UJuhetG8r_0tjvMZ?usp=sharing)\n\n``````class CustomTransformer(BaseEstimator):\n\n\"\"\"\nThis CustomTransformer only returns the incoming values as it is\nThe goal of this transformer is to demonstrate custom transformers are supported when deployed to Datagran platform with cloudpickle library\n\"\"\"\ndef fit(self, X, y = None):\nreturn self\n\ndef transform(self, X, y = None):\nreturn X\n\nnumericTransfomer = Pipeline(\nsteps=[\n(\"imputer\", SimpleImputer(strategy=\"median\")),\n(\"scaler\", StandardScaler()),\n('CustomTransformer', CustomTransformer())\n]\n)\n\ncategoricalTransformer = OneHotEncoder(handle_unknown=\"ignore\")\n\npreprocessor = ColumnTransformer(\ntransformers=[\n(\"num\", numericTransfomer, numericCols),\n(\"cat\", categoricalTransformer, categoricalCols),\n]\n)\n\nmodel = Pipeline(\nsteps=[(\"preprocessor\", preprocessor), (\"classifier\", LogisticRegression())]\n)\nmodel.fit(X_train, y_train)\n``````"
] | [
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null,
"https://s3-us-west-2.amazonaws.com/secure.notion-static.com/a28769ed-9196-47ae-aed5-9772f5586719/Screen_Shot_2021-10-13_at_1.21.55_PM.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.707949,"math_prob":0.9383507,"size":1713,"snap":"2022-27-2022-33","text_gpt3_token_len":397,"char_repetition_ratio":0.12873025,"word_repetition_ratio":0.0456621,"special_character_ratio":0.22591944,"punctuation_ratio":0.16607773,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9841862,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-26T22:53:46Z\",\"WARC-Record-ID\":\"<urn:uuid:e3d032ca-15d7-4b47-8c17-95a979fa0a02>\",\"Content-Length\":\"9126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bc9b2d70-a0d7-4c20-84f4-90e244d57bb6>\",\"WARC-Concurrent-To\":\"<urn:uuid:be334c21-8611-4486-a126-d6a48b79ee32>\",\"WARC-IP-Address\":\"172.64.154.162\",\"WARC-Target-URI\":\"https://www.notion.so/Upload-a-trained-model-5a3a45d0d8f64f26ac5c19df921d861d\",\"WARC-Payload-Digest\":\"sha1:3FAR5RYUGPCYO4RVOOZOPUB7FZBR75EN\",\"WARC-Block-Digest\":\"sha1:3SRBZBSPWJ2TMSO2XYOLKWENR7PZHDON\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103322581.16_warc_CC-MAIN-20220626222503-20220627012503-00153.warc.gz\"}"} |
https://metanumbers.com/1278477 | [
"# 1278477 (number)\n\n1,278,477 (one million two hundred seventy-eight thousand four hundred seventy-seven) is an odd seven-digits composite number following 1278476 and preceding 1278478. In scientific notation, it is written as 1.278477 × 106. The sum of its digits is 36. It has a total of 4 prime factors and 8 positive divisors. There are 852,300 positive integers (up to 1278477) that are relatively prime to 1278477.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 7\n• Sum of Digits 36\n• Digital Root 9\n\n## Name\n\nShort name 1 million 278 thousand 477 one million two hundred seventy-eight thousand four hundred seventy-seven\n\n## Notation\n\nScientific notation 1.278477 × 106 1.278477 × 106\n\n## Prime Factorization of 1278477\n\nPrime Factorization 33 × 47351\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 142053 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 1,278,477 is 33 × 47351. Since it has a total of 4 prime factors, 1,278,477 is a composite number.\n\n## Divisors of 1278477\n\n8 divisors\n\n Even divisors 0 8 4 4\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 1.89408e+06 Sum of all the positive divisors of n s(n) 615603 Sum of the proper positive divisors of n A(n) 236760 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1130.7 Returns the nth root of the product of n divisors H(n) 5.39989 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 1,278,477 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 1,278,477) is 1,894,080, the average is 236,760.\n\n## Other Arithmetic Functions (n = 1278477)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 852300 Total number of positive integers not greater than n that are coprime to n λ(n) 426150 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 98267 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 852,300 positive integers (less than 1,278,477) that are coprime with 1,278,477. And there are approximately 98,267 prime numbers less than or equal to 1,278,477.\n\n## Divisibility of 1278477\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 1 2 3 4 5 0\n\nThe number 1,278,477 is divisible by 3 and 9.\n\n## Classification of 1278477\n\n• Arithmetic\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Frugal\n\n## Base conversion (1278477)\n\nBase System Value\n2 Binary 100111000001000001101\n3 Ternary 2101221202000\n4 Quaternary 10320020031\n5 Quinary 311402402\n6 Senary 43222513\n8 Octal 4701015\n10 Decimal 1278477\n12 Duodecimal 517a39\n20 Vigesimal 7jg3h\n36 Base36 reh9\n\n## Basic calculations (n = 1278477)\n\n### Multiplication\n\nn×y\n n×2 2556954 3835431 5113908 6392385\n\n### Division\n\nn÷y\n n÷2 639238 426159 319619 255695\n\n### Exponentiation\n\nny\n n2 1634503439529 2089675053858717333 2671601493832131359741841 3415581063030021804408669656157\n\n### Nth Root\n\ny√n\n 2√n 1130.7 108.534 33.6258 16.6471\n\n## 1278477 as geometric shapes\n\n### Circle\n\n Diameter 2.55695e+06 8.03291e+06 5.13494e+12\n\n### Sphere\n\n Volume 8.75321e+18 2.05398e+13 8.03291e+06\n\n### Square\n\nLength = n\n Perimeter 5.11391e+06 1.6345e+12 1.80804e+06\n\n### Cube\n\nLength = n\n Surface area 9.80702e+12 2.08968e+18 2.21439e+06\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 3.83543e+06 7.07761e+11 1.10719e+06\n\n### Triangular Pyramid\n\nLength = n\n Surface area 2.83104e+12 2.46271e+17 1.04387e+06\n\n## Cryptographic Hash Functions\n\nmd5 fef712f9a531f5e4dbd5a2b01b3c5b42 ed5652763ba671702e81bda18748e6e7b44b0532 c1d1ebd2c1923313ffeb11f6e2c642e9934849bbcbdb1c88b44eaedc455f1ae6 7e41a955168e4d026f1cd57d11439f00c1930ce6d9335d111f23eb220bf209b609721bc5d5a8da3661c62fa433c51c1b3ad59955b438b2ef469d46fa12f8a155 56e63a94af2efe082a8dd406ccd94a896deb574f"
] | [
null
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https://www.arxiv-vanity.com/papers/hep-th/0104092/ | [
"# Solutions of the WDVV Equations and Integrable Hierarchies of KP type\n\nHenrik Aratyn\n\nDepartment of Physics,\nUniversity of Illinois at Chicago,\n845 W. Taylor St.,\nChicago, IL 60607-7059\ne-mail:\nJohan van de Leur\n\nMathematical Institute,\nUniversity of Utrecht,\nP.O. Box 80010, 3508 TA Utrecht,\nThe Netherlands\ne-mail:\n###### Abstract\n\nWe show that reductions of KP hierarchies related to the loop algebra of with homogeneous gradation give solutions of the Darboux-Egoroff system of PDE’s. Using explicit dressing matrices of the Riemann-Hilbert problem generalized to include a set of commuting additional symmetries, we construct solutions of the Witten–Dijkgraaf–E. Verlinde–H. Verlinde equations.\n\n## 1 Introduction\n\nThis paper deals with symmetries of integrable models and their connection to the topological field theory and Frobenius manifolds. Our results establish a new and fundamental link between Darboux-Egoroff metric systems and additional symmetry-flows of the integrable hierarchies of KP type. This connection goes beyond a dispersionless limit.\n\nThe description of -orthogonal curvilinear coordinate systems in was one of the classical puzzles of the 19th century. Well-known mathematicians such as Gauss, Lamé, Cayley and Darboux have contributed to its solution. The problem is formulated as follows. Find all coordinate systems\n\n ui=ui(x1,x2,…,xn), (1.1) det((∂ui∂xj)1≤i,j≤n)≠0,\n\nsatisfying the orthogonality condition\n\n n∑k=1∂ui∂xk∂uj∂xi=0,for i≠j.\n\nIt was Darboux who gathered most results on this topic and published them in 1910 in his Leçons sur les systèmes orthogonaux et les coordonnées curvilignes . For more historical details see the paper by Zakharov.\n\nThe metric tensor in in the coordinate system is diagonal:\n\n ds2=n∑i=1h2i(u)(dui)2,u=(u1,…,un), (1.2)\n\nwhere\n\n h2i(u)=n∑k=1(∂xi∂uk)2. (1.3)\n\nThe flatness of the metric imposes a condition on the Lamé coefficients of the following form\n\n ∂kβij(u)=βik(u)βkj(u), i≠k≠j, (1.4) ∂βij(u)∂ui+∂βji(u)∂uj+∑k≠i,jβkiβkj=0, i≠j,\n\nhere this is expressed in Darboux’ rotation coefficients\n\n βij(u)=1hi(u)∂hj(u)∂ui. (1.5)\n\nFor rotation coefficients that satisfy\n\n βij(u)=βji(u), (1.6)\n\nthe corresponding metric is called an Egoroff metric, in honour of D.F. Egoroff who gave its complete description .\n\nB. Dubrovin noticed , that the equations (1.4) and (1.6), which he called the Darboux-Egoroff system, also describe the local classification of massive topological field theories. An important role in physics literature on two dimensional field theory is played by a remarkably system of partial differential equations commonly known as the Witten-Dijkgraaf-Verlinde-Verlinde (WDVV) equations , . These equations determine deformations of 2-dimensional topological field theories. Dubrovin , connected the Frobenius structure to any solution of the WDVV-equation thus providing a “coordinate-free” approach to the topological field theory. Frobenius manifolds are objects from the differential geometry which arise in a number of different areas of mathematics, such as quantum cohomology, Gromov-Witten invariants, theory of singularities, Hurwitz spaces, Coxeter groups. One of the most interesting connections from a physics point of view is a link to the integrable systems.\n\nWe obtain a class of integrable hierarchies which provides a setting for solutions to the Darboux-Egoroff metric systems. The relevant integrable models turn out to be connected with generalizations of the integrable structure related to the Nonlinear Schrödinger equation. In the pseudo-differential calculus framework of the Sato theory the relevant models originate from the Lax structure given by\n\n L=∂x+m∑i=1Φi∂x−1Ψi. (1.7)\n\nThe corresponding Baker-Akhiezer (BA) wave function which enters the linear spectral problem is given as\n\n ψBA(t,λ)=τ(t−[λ−1])τ(t)e∑∞j=1λjtj, (1.8)\n\nin terms of the -function such that . In (1.8) we used the multi-time notation\n\nFor the Lax operator in (1.7) the action of the isospectral flows on the Lax operator is given by\n\n ∂∂tnL=[(Ln)+,L], (1.9)\n\nwhere as usual stands for the differential part of the pseudo-differential operator.\n\nThe isospectral flows from (1.9) induce the following flows on the eigenfunctions and adjoint eigenfunctions with :\n\n ∂∂tnΦi=(Ln)+(Φi);∂∂tnΨi=−(Ln)∗+(Ψi). (1.10)\n\nThese models possess a rich structure of additional, with respect to the isospectral flows, symmetries. These symmetry flows will be studied within the pseudo-differential calculus framework where they are generated by the squared eigenfunction potentials and within the algebraic formalism obtained by extending the standard Riemann-Hilbert problem. In the latter formalism the relevant systems are characterized through their Lie algebraic context as integrable models arising through generalized Drinfeld-Sokolov scheme from the algebras with the homogeneous gradation.\n\nAfter further reduction to CKP sub-hierarchy , which brings conditions , we obtain a model whose abelian subalgebra of additional symmetries provide canonical coordinates of the related Darboux-Egoroff systems. This connection will be used to explicitly construct solutions to the Darboux-Egoroff system and the Witten-Dijkgraaf-Verlinde-Verlinde equations in the topological field theory.\n\nIn a separate publication we will address the issue of how several structures of the underlying integrable models (including Virasoro symmetry) carry over to the Darboux-Egoroff system.\n\nIn Section 2, we discuss the Darboux-Egoroff metrics and the underlying linear spectral system from which the Darboux-Egoroff system can be obtained via compatibility equations. Section 3 defines the constrained integrable KP hierarchy, which is a subject of our study, in terms of the Sato Grassmannian and within the framework of the pseudo-differential operator calculus. In Section 4, the Riemann-Hilbert problem is extended to incorporate the constrained integrable KP hierarchy augmented by additional symmetry flows. We succeed, in this section, to find an explicit expression for the dressing matrix allowing us to solve the extended Riemann-Hilbert problem in terms of the -function and the eigenfunctions and the adjoint eigenfunctions . Section 5 introduces the CKP reduction of the constrained integrable KP hierarchy imposing equality of the eigenfunctions and the adjoint eigenfunctions . Finally, in Section 6, we reproduce the fundamental objects defining the Darboux-Egoroff system presented in Section 2 in terms of the integrable structure derived in the previous sections.\n\n## 2 The Darboux-Egoroff system\n\nWe now describe in some more details a relation between the Darboux-Egoroff system (1.4), (1.6) and the local classification of massive topological field theories , . First of all, the flat coordinates of the Darboux-Egoroff metric (1.2) can be found from the linear system\n\n ∂2xk∂ui∂uj=Γiij∂xk∂ui+Γjji∂xk∂uj,i≠j;∂2xk∂u2i=n∑j=1Γjii∂xk∂uj, (2.1)\n\nwhere are the Christoffel symbols of the Levi-Civita connection:\n\n Γiij=1hi∂hi∂uj,Γjii=(2δij−1)hih2j∂hi∂uj. (2.2)\n\nFor the flat coordinates , , of the metric (1.2), the functions\n\n cmkℓ(x)=n∑i=1∂xm∂ui∂ui∂xk∂ui∂xℓ, (2.3)\n\nform the structure constants of the commutative algebra of the primary fields of a topological field theory, i.e.\n\n ϕkϕℓ=∑mcmkℓϕm.\n\nAssociativity of this algebra imposes on the structure constants a relation :\n\n n∑k=1ckij(x)cℓkm(x)=n∑k=1ckjm(x)cℓik(x). (2.4)\n\nIf one writes down these equations for the function for which\n\n ∂3F(x)∂xk∂xℓ∂xm=ckℓm(x)=n∑i=1ηmicikℓ(x),whereηpq=n∑i=1h2i(u)∂ui∂xp∂ui∂xq, (2.5)\n\nwith the constraint\n\n ∂3F(x)∂x1∂xℓ∂xm=ηℓm,\n\none obtains the well-known Witten-Dijkgraaf-E. Verlinde-H. Verlinde (WDVV)-equations , for the prepotential . In fact, Dubrovin showed that one can find these and of (2.5) as follows. The Darboux-Egoroff system can be represented as the compatibility equations of the following linear system depending on a spectral parameter :\n\n ∂ψik(u,λ)∂uj =βij(u)ψjk(u,λ),i≠j (2.6) n∑k=1∂ψij(u,λ)∂uk =λψij(u,λ),\n\nSolving this system for , i.e., finding that satisfy\n\n ∂ψik(u)∂uj =βij(u)ψjk(u),i≠j (2.7) n∑k=1∂ψij(u)∂uk =0,\n\nfor a given solution of the Darboux-Egoroff system, leads to ”a local classification of complex semisimple Frobenius manifolds”:\n\n###### Proposition 2.1\n\n On the domain and , one has\n\n hi(u) =ψi1(u), (2.8) ηαβ =n∑i=1ψiα(u)ψiβ(u), ∑βηαβ∂xβ(u)∂ui =ψi1(u)ψiα(u), cαβγ(x(u)) =n∑i=1ψiα(u)ψiβ(u)ψiγ(u)ψi1(u).\n\nWe refer the reader for the definition of Frobenius manifold to or . This definition is different from the one given in e.g . We do not assume the quasi-homogeneity condition here. Note that this system (2.8) is not unique. For any given solution of the Darboux-Egoroff system, there exists an -parameter family of Lamé coefficients and of Egoroff metrics.\n\nAssume for simplicity that , so . Following Akhmetshin, Krichever and Volvovski (see also and ), we can even construct the prepotential . There exist unique solutions , , of (2.6) with initial conditions , and , such that has the expansion\n\n ψi1(u,0)ψij(u,λ)=∞∑k=0∂ξjk(u)∂uiλk, with ξj0(u)=xj(u).\n###### Definition 2.1\n\nThe matrix and row-vector are defined as:\n\n Ξ′(u,λ) =(ψi1(u,0)ψij(u,λ)))ij, λΞ(u,λ) =(n∑i=1ψi1(u,0)ψij(u,λ)))j.\n\nFrom the equations (2.6) and (2.7) one deduces\n\n ∂Ξj(u,λ)∂ui=Ξ′(u,λ)ij\n\nand satisfies\n\n ∂2Ξ(u,λ)∂ui∂uj =Γiij(u)∂Ξ(u,λ)∂ui+Γjji(u)∂Ξ(u,λ)∂uj,i≠j (2.9) ∂2Ξ(u,λ)∂(ui)2 =n∑j=1Γjii(u)∂Ξ(u,λ)∂uj+λ∂2Ξ(u,λ)∂ui∂uj,\n\nwhere the Christoffel symbols are given by (2.1) and hence that\n\n ∂2Ξ(u,λ)∂xk∂xℓ=λn∑m=1cmkℓ(u)∂Ξ(u,λ)∂xm. (2.10)\n\nThus is the generating series for the flat sections of the connection :\n\n Ξj=δj1λ−1+xj(u)+∞∑i=1ξjiλi.\n\nMoreover\n\n Ξ′(u,λ)Ξ′(u,−λ)T =n∑i=1hi(u)2Eii, (2.11) λΞ(u,λ)Ξ′(u,−λ)T =(h21(u) h22(u) … h2n(u)).\n\nSince is a linear combination of ’s,\n\n λΞ(u,λ)∂Ξ(u,−λ)T∂xk\n\nis independent of , which means that all coefficients, except the constant coefficient, are zero. In particularly the coefficient of gives:\n\n ξm1(u)=−∂ξ12(u)∂xm+n∑i=1xi(u)∂ξi1(u)∂xm.\n\nThe coefficient of of (2.10) leads to\n\n ∂2ξm1(u)∂uk∂uℓ=cmkℓ(u),\n\n###### Theorem 2.1\n\nThe function defined by\n\n F(u)=−12ξ12(u)+12n∑i=1xi(u)ξi1(u)\n\nsatisfies equation (2.5).\n\n## 3 The Sato Grassmannian and the Constrained KP Hierarchy\n\nIn this section we describe the Sato Grassmannian. Consider the spaces\n\n H−=λ−1C[[λ−1]]={∑∞j=1ajλ−j|aj∈C}andH+=C[λ]={∑mi=0biλi|bi∈C}.\n\nHence is the quotient field of . On the space we have a bilinear form, viz. if and are in , then we define\n\n (f(λ),g(λ))=Resλf(λ)g(λ)=∑jajb−j−1. (3.1)\n\nLet be the projection\n\n p+(∑ajλj)=∑j≥0ajλj.\n\nThen the Sato Grassmannian consists of all linear subspaces of that are of a size comparable to , i.e.,\n\n Gr(H)={W⊂H∣∣∣p+:W→H+has % a finitedimensional kernel and cokernel}.\n\nThe space has a subdivision into different components:\n\n Gr(k)(H)={W∈Gr(H)| dim(Coker(p+|W))−dim(Ker(p+|W))=k}.\n\nClearly, the subspace belongs to and one easily verifies that this also holds for all subspaces in that project bijectively onto , i.e. all belonging to the “big cell”. For , let be the orthocomplement of in w.r.t. the bilinear form (3.1). Then, with the above given description, also belongs to see .\n\nWe write , and . Consider now a wave function of the -hierarchy and its dual with\n\n ψBA(x,^t,λ)= {∑j≤0aj(x,^t)λj}λlexλ+∑i>1tiλiand ψ∗BA(x,^t,λ)= {∑m≤0bm(x,^t)λm}λ−le−xλ−∑i>1tiλi.\n\nWe assume from now on in this section that there exists an of the form\n\n α(x,0)=xN+∑j>Najxj. (3.2)\n\nsuch that for all and all\n\n α(x,^t)aj(x,^t)∈C[[x,^t]]andα(x,^t)bm(x,^t)∈C[[x,^t]]. (3.3)\n\nNote that throughout this section stands for the -th power of and not for the flat coordinate of Section 2.\n\nSuch wave functions are called regularizable, see ,. For regularizable wave functions the Laurent series in of and have the form\n\n ψBA(x,^t,λ)=∑j≥−Nwj(^t,λ)xj,wherewj(^t,λ)=N1∑l=−∞vlλl,withvl∈C[[^t]],\n ψ∗BA(x,^t,λ)=∑j≥−Nw∗j(^t,λ)xj,wherew∗j(^t,λ)=N2∑l=−∞v∗lλl, ;withv∗l∈C[[^t]].\n\nand moreover\n\n W=Span{wj(0,λ),j≥−N}andW∗=Span{w∗j(0,λ),j≥−N}\n\nbelong to . It was Sato who realized that the space determines , for according to there holds\n\n###### Proposition 3.1\n\nThe map that associates to a regularizable wave function of the -hierarchy the span of the coefficients in of the Laurent series of in is a bijection between this class of wave functions and . The wave functions that satisfy the conditions in (3.3) for correspond to the big cell.\n\nFor each we denote the wave function corresponding to by . The dual wave function of , which we denote by can be characterized as follows , :\n\n###### Proposition 3.2\n\nLet and be two subpaces in . Then is the space corresponding to the dual wave function, if and only if with the orthocomplement of w.r.t. the bilinear form (3.1) on . Moreover\n\n (ψW(t,λ),ψ∗W(s,λ))=0.\n\nLet then\n\n ψW(t,λ)=PW(t,∂x)e∑∞j=1tjλj,ψ∗W(t,λ)=P∗−1W(t,∂x)e−∑∞j=1tjλj,\n\nwhere is an order pseudo-differential operator. The corresponding KP Lax operator is equal to\n\n LW(t,∂x)=PW(t,∂x)∂xP−1W(t,∂x). (3.4)\n\nFrom now on we will use the notation and instead of and whenever we want to emphasize its dependence on a point of the Sato Grassmannian .\n\nEigenfunctions and adjoint eigenfunctions of the KP Lax operator (3.4), see (1.10), can be expressed in wave and adjoint wave functions, viz. there exist functions such that\n\n Φ(t)=(ψW(t,λ),f(λ)),Ψ(t)=(ψ∗W(t,λ),g(λ)). (3.5)\n\nSuch (adjoint) eigenfunctions induce elementary Bäcklund–Darboux transformations . Assume that we have the following data , , and , then the (adjoint) eigenfunctions (3.5) induce new KP wave functions:\n\n ψW′(t,λ) =(Φ(t)∂xΦ(t)−1)ψW(t,λ), ψ∗W′(t,λ) =(Φ(t)∂xΦ(t)−1)∗−1ψ∗W(t,λ), (3.6) ψW′′(t,λ) =(−Ψ(t)∂xΨ(t)−1)∗−1ψW(t,λ), ψ∗W′′(t,λ) =(−Ψ(t)∂xΨ(t)−1)ψ∗W(t,λ),\n\nwhere\n\n W′ ={w∈W|(w(λ),f(λ))=0}∈Gr(k+1)(H),W′⊥=W⊥+Cf, (3.7) W′′ =W+Cg∈Gr(k−1)(H),W′′⊥={w∈W⊥|(w(λ),g(λ))=0}.\n\nNow assume that we have a Lax operator with of the form (1.7), (1.10), with minimal. Then\n\n ψλW= λψW = LWψW = (∂x+m∑i=1Φi∂x−1Ψi)ψW = ∂x(ψW)+m∑i=1ΦiΨi(−Ψi∂xΨ−1i)∗−1ψW = ∂x(ψW)+m∑i=1ΦiΨiψWi,\n\nwhere of codimension 1. Hence there exists a such that\n\n W∈Gr(k)(H),W′=W+λW∈Gr(k−m)(H), (3.8) W⊂W′ (codimension m),λW⊂W′ (codimension m+1).\n\nThe converse is also true. If is a Lax operator such that satisfies (3.8), then there exists an inverse Bäcklund–Darboux transformation, i.e. the inverse of an order differential operator , mapping into and an order Bäcklund–Darboux transformation mapping into :\n\n L−1mψW=ψW′,Lm+1ψW′=ψλW=λψW. (3.9)\n\nHence,\n\n LW=Lm+1L−1m, (3.10)\n\nis a first order pseudo-differential operator. Before we continue, we first state a small Lemma which will be important later on:\n\n###### Lemma 3.1\n\nThere exist independent functions , , respectively , and independent functions , , respectively , such that the functions span , span , span and span , for , , where\n\n φi(t)= (ψW′(t,λ),ui(λ)),ψi(t)=(ψ∗W(t,λ),vi(λ)), ¯φj(t)= (ψW′(t,λ),fj(λ)),¯ψj(t)=(ψ∗W(t,λ),gj(λ)).\n\nProof This follows from (3.63.8) and the observation that .\n\nWe proceed to construct . Choose independent vectors , as in Lemma 3.1, such that\n\n W′⊥={w∈W⊥|(w(λ),vj(λ))=0 for% all 1≤j≤m}.\n\nSince can be obtained by two Bäcklund–Darboux transformation and such that (3.10) holds, the general theory of such Bäcklund–Darboux transformations shows that\n\n (LW)−=m∑j=1aj(t)∂−1x(ψ∗W(t,λ),vj(λ)).\n\nThus\n\n λψW(t,λ)=∂x(ψW(t,λ))+m∑j=1aj(ψ∗W(t,λ),vj)ψW+Cvj(t,λ)\n\nLet now\n\n Uj=W+Cv1+Cv2+⋯+Cvj−1+Cvj+1+⋯+Cvm.\n\nChoose such that\n\n Uj={w∈W′|(w(λ),uj(λ))=0}.\n\nThen\n\n (λψW(t,λ),ui(λ)) =(λψW(t,λ)−∂x(ψW(t,λ)),ui(λ)) = m∑j=1aj(ψ∗W(t,λ),vj(λ))(ψW+Cvj(t,λ),ui(λ)) = ai(ψ∗W(t,λ),vi(λ))(ψW+Cvi(t,λ),ui(λ)),\n\nfrom which we can deduce that\n\n ai=(λψW(t,λ),ui(λ))(ψ∗W(t,λ),vi(λ))(ψW+Cvi(t,λ),ui(λ)).\n\nSince eigenfunctions which produce elementary Bäcklund–Darboux transformations are unique upto a scalar factor ,,\n\n ψW+Cvi(t,λ) =((ψ∗W(t,λ),vi(λ))−1∂x(ψ∗W(t,λ),vi(λ)))−1ψW(t,λ), ψW(t,λ) =((ψW+Cvi(t,λ),ui(λ))∂x(ψW+Cvi(t,λ),ui(λ))−1)ψW+Cvi(t,λ)\n\nand\n\n ∂∂tn(ψ∗W(t,λ),vi(λ))−1=(LnW+Cvi)+((ψ∗W(t,λ),vi(λ))−1),\n\nwe find that\n\n (ψ∗W(t,λ),vi(λ))−1=ci(ψW+Cvi(t,λ),ui(λ))\n\nand hence\n\n ai=di(λψW(t,λ),ui(λ)).\n\nNow replace by and we obtain that\n\n (LW)− =m∑i=1(λψW(t,λ),ui(λ))∂−1x(ψ∗W(t,λ),vi(λ)) = m∑i=1(ψW(t,λ),λui(λ))∂−1x(ψ∗W(t,λ),vi(λ)).\n\nWith these choices, we set\n\n Φi(t)=(ψW(t,λ),λuj(λ)),Ψi(t)=(ψ∗W(t,λ),vj(λ)) (3.11)\n\nand thus one obtains the desired Lax operator of the form given in (1.7). Notice that for and . This construction for the Segal-Wilson Grassmannian was given in , see also .\n\nFor the Baker-Akhiezer wave function the linear spectral problem can be decomposed on a set of differential equations :\n\n ∂xψBA(t,λ)+m∑i=1ΦiΓi(t,λ)=λψBA(t,λ);∂xΓi(t,λ)=Ψi(t)ψBA(t,λ). (3.12)\n\nSimilarly, we introduce the conjugated linear problem for . For the conjugated Baker-Akhiezer wave function\n\n ψ∗BA(t,λ)=τ(t+[λ−1])τ(t)e−∑∞j=1λjtj, (3.13)\n\nthe conjugated spectral problem can be rewritten as :\n\n ∂xΓ∗i(t,λ)=Φi(t)ψ∗BA(t,λ);−∂xψ∗BA(t,λ)−m∑i=1ΨiΓ∗i"
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https://it.scribd.com/document/441243017/Dice-measure-pdf | [
"Sei sulla pagina 1di 10\n\n# Journal of Intelligent & Fuzzy Systems 31 (2016) 663–671 663\n\nDOI:10.3233/IFS-162179\nIOS Press\n\n## The generalized Dice measures for multiple\n\nattribute decision making under simplified\nneutrosophic environments\nJun Ye∗\nDepartment of Electrical and Information Engineering, Shaoxing University, Shaoxing,\nZhejiang Province, P.R. China\n\nAbstract. A simplified neutrosophic set (SNS) is a subclass of neutrosophic set and contains a single-valued neutrosophic\nset (SVNS) and an interval neutrosophic set (INS). It was proposed as a generalization of an intuitionistic fuzzy set (IFS)\nand an interval-valued intuitionistic fuzzy set (IVIFS) in order to deal with indeterminate and inconsistent information. The\npaper proposes another form of the Dice measures of SNSs and the generalized Dice measures of SNSs and indicates that\nthe Dice measures and asymmetric measures (projection measures) are the special cases of the generalized Dice measures in\nsome parameter values. Then, we develop the generalized Dice measures-based multiple attribute decision-making methods\nwith simplified neutrosophic information. By the weighted generalized Dice measures between each alternative and the ideal\nsolution (ideal alternative) corresponding to some parameter value required by decision makers’ preference, all the alternatives\ncan be ranked and the best one can be obtained as well. Finally, a real example on the selection of manufacturing schemes\ndemonstrates the applications of the proposed decision-making methods under simplified neutrosophic environment. The\neffectiveness and flexibility of the proposed decision-making methods are shown by choosing different parameter values.\n\nKeywords: Generalized Dice measure, Dice measure, decision making, simplified neutrosophic set, asymmetric measure,\nprojection measure\n\n## 1. Introduction are very suitable for handling decision making\n\nproblems with indeterminate and inconsistent infor-\nMultiple attribute decision making is a main branch mation, which IFSs and IVIFSs cannot describe and\nof decision theory, where neutrosophic theory intro- deal with. Recently, many researchers have applied\nduced by Smarandache has been successfully SNSs and the subclasses of SNSs (SVNSs and INSs)\napplied in recent years. As a generalization of an to the decision-making problems. Various methods\nintuitionistic fuzzy set (IFS) and an interval- have been developed to solve the multiple attribute\nvalued intuitionistic fuzzy (IVIFS) , a simplified decision-making problems with simplified neutro-\nneutrosophic set (SNS) introduced by Ye is sophic information. For example, Ye proposed\na subclass of a neutrosophic sets , including a the correlation coefficient of SVNSs and applied it\nsingle-valued neutrosophic set (SVNSs) and an to multiple attribute decision making. Chi and Liu\ninterval neutrosophic set (INSs) . Hence, SNSs and Biswas et al. extended TOPSIS method\nto single-valued and interval neutrosophic multiple\n∗ Corresponding author. Jun Ye, Department of Electrical and\nattribute decision-making problems. Ye [11–13] pre-\nInformation Engineering, Shaoxing University, 508 Huancheng\nWest Road, Shaoxing, Zhejiang Province 312000, P.R. China. Tel.:\nsented some similarity measures of SVNSs, INSs\n+86 575 88327323; E-mail: [email protected]. and SNSs and applied them to decision making.\n\n664 J. Ye / The generalized Dice measures for multiple attribute decision making\n\nYe put forward a cross-entropy measure of is given to show the application of the proposed meth-\nSVNSs for multiple attribute decision making prob- ods, and then the effectiveness and flexibility of the\nlems. Ye , Zhang et al. , Liu et al. , Liu and proposed methods are indicated by choosing different\nWang , and Peng et al. developed some simpli- parameter values. Finally, Section 7 contains conclu-\nfied, interval and single-valued neutrosophic number sions and future work.\naggregation operators and applied them to multiple\nattribute decision-making problems. Peng et al. \nand Zhang et al. proposed outranking approaches 2. The Dice measures of SNSs\nfor multicriteria decision-making problems with sim-\nplified and interval neutrosophic information. Sahin As a subset of a neutrosophic set , Ye intro-\nand Kucuk presented a subsethood measure for duced a SNS and gave its definition.\nSVNSs and applied it to multiple attribute decision\nmaking. Şahin and Liu introduced a maxi- Definition 1. A SNS S in the universe of\nmizing deviation method for neutrosophic multiple discourse X is defined as S = {\u0003x, tS (x), uS (x), v S (x)\u0004\nattribute decision making with incomplete weight |x ∈ X}, where tS (x): X → [0, 1], us (x) : X →\ninformation. Ye presented a multiple attribute [0, 1], and v s (x) : X → [0, 1] are a truth-membership\ndecision-making method based on the possibility function and an indeterminacy-membership func-\ndegree ranking method and ordered weighted aggre- tion, a falsity- membership function, respectively,\ngation operators of interval neutrosophic numbers. of the element x to the set S with the condition\nSince the Dice measure is one of vector similarity 0 ≤ ts (x) + us (x) + v s (x) ≤ 3 for x ∈ X.\nmeasures, it is a useful mathematical tool for han- In fact, SNSs contain the concepts of SVNSs\ndling decision-making problems. However, the Dice and INSs, which are the subclasses of SNSs. For\nmeasure of SNSs used for decision making lacks convenience, a component element < x, ts (x),\nflexibility in decision-making process. Therefore, it us (x), v s (x) > in a SNS S is denoted by sx =<\nis necessary to improve the Dice measure of SNSs tx , ux , v x > for short, which is called the simpli-\nto handle multiple attribute decision-making prob- fied neutrosophic number (SNN), where tx , ux , v x ∈\nlems to satisfy the requirements of decision makers’ [0, 1] and 0 ≤ tx , +ux + v x ≤ 3 for a single-\npreference and flexible decision making. In order valued neutrosophic number (SVNN), and then\nto do so, the main purposes of this paper are: (1) tx = [txL , txU ] ⊆ [0, 1], ux = [uL U\nx , ux ] ⊆ [0, 1], v x =\nto propose another form of the Dice measures of [v x , v x ] ⊆ [0, 1] and 0 ≤ tx + ux + v U\nL U U U\nx ≤ 3 for an\nSNSs, (2) to present the generalized Dice measures interval neutrosophic number (INN).\nof SNSs, and (3) to develop the generalized Dice Ye presented the Dice measures of SNSs,\nmeasures-based multiple attribute decision-making which was defined below.\nmethods with simplified neutrosophic information.\nIn the decision making process, the main advan- Definition 2. Let S1 = {s11 , s12 , . . . , s1n } and\ntage of the proposed methods is more general and S2 = {s21 , s22 , . . . , s2n } be two SNSs. If s1j =<\nmore flexible than existing decision-making methods t1j , u1j , v 1j , > and s2j =< t2j , u2j , v 2j , > (j =\nwith simplified neutrosophic information to satisfy 1, 2, . . . , n) are the j-th SVNNs in S1 and S2 respec-\nthe decision makers’ preference and/or practical tively, then the Dice measure between S1 and S2 is\nrequirements. defined as:\nThe rest of the paper is organized as follows.\nSection 2 reviews the Dice measures of SNSs. Sec- n\n1 \u0002 2s1j · s2j\ntion 3 proposes another form of the Dice measures DSVNN1 (S1 , S2 ) = \u0003 \u00032 \u0003 \u00032 .\nof SNSs. In Section 4, we propose the general- n \u0003s1j \u0003 + \u0003s2j \u0003\nj=1\nized Dice measures of SNSs and indicate the Dice n\n1 \u0002 2(t1j t2j + u1j u2j + v 1j v 2j )\nmeasures and asymmetric measures (projection mea- =\nsures) as the special cases of the generalized Dice n (t 2 + u2 + v 2 ) + (t 2 + u2 + v 2 )\nj=1 1j 1j 1j 2j 2j 2j\nmeasures in some parameter values. In Section 5, the (1)\ngeneralized Dice measures-based multiple attribute If s1j =< t1j , u1j , v 1j > and s2j =< t2j , u2j ,\ndecision-making methods are developed under sim- v 2j > (j = 1, 2, . . . , n) are the j-th INNs in S1 and\nplified neutrosophic environment. In Section 6, a real S2 respectively, then the Dice measure between S1\nexample on the selection of manufacturing schemes and S2 is defined as:\nJ. Ye / The generalized Dice measures for multiple attribute decision making 665\n\n1 \u0002 2s1j · s2j\nn 3. Another form of the Dice measures of SNSs\nDINN1 (S1 , S2 ) = \u0003 \u00032 \u0003 \u00032 .\nn \u0003s1j \u0003 + \u0003s2j \u0003\nj=1 This section proposes another form of the Dice\n\u0004 L L U U \u0005 measures of SNSs, which is defined as follows.\nt1j t2j + t1j t2j + uL1j uL2j\n2 Definition 3. Let S1 = {s11 , s12 , . . . , s1n } and\n1\u0002\nn\n+uU1j uU2j + v L1j v L2j + v U1j v U2j\n= ⎛ ⎞ (2) S2 = {s21 , s22 , . . . , s2n } be two SNSs. If s1j =<\nL 2\nn\nj=1\n(t1j ) + (uL1j )2 + (v L1j )2 t1j , u1j , v 1j > and s2j =< t2j , u2j , v 2j > (j =\n⎜ ⎟ 1, 2, . . . , n) are the j-th SVNNs in S1 and S2 respec-\n⎜ +(t1j ) + (uU1j )2 + (v U1j )2 ⎟\nU 2\n⎜ ⎟ tively, then the Dice measure between S1 and S2 is\n⎜ ⎟\n⎜ +(t L )2 + (uL )2 + (v L )2 ⎟ defined as:\n⎝ 2j 2j 2j ⎠\nU 2\n+(t2j ) + (uU2j )2 + (v U2j )2\n2(S1 · S2 )\nDSVNN2 (S1 , S2 ) =\nThen, the two Dice measures DSVNN1 (S1 , S2 ) and |S1 |2 + |S2 |2\nn\nDINN1 (S1 , S2 ) satisfy the following properties : 2 j=1 (t1j t2j + u1j u2j + v 1j v 2j )\n(P1) DSVNN1 (S1 , S2 ) = DSVNN1 (S2 , S1 ) and = n n .\nj=1 (t1j + u1j + v 1j ) + j=1 (t2j + u2j + v 2j )\n2 2 2 2 2 2\nDINN1 (S1 , S2 ) = DINN1 (S2 , S1 );\n(5)\n(P2) 0 ≤ DSVNN1 (S1 , S2 ) ≤ 1 and 0 ≤ DINN1\n(S1 , S2 ) ≤ 1;\nIf s1j =< t1j , u1j , v 1j > and s2j =< t2j , u2j ,\n(P3) DSVNN1 (S1 , S2 ) = 1 and DINN1 (S1 , S2 ) =\nv 2j > (j = 1, 2, . . . , n) are the j-th INNs in S1 and\n1, if S1 = S2 .\nS2 respectively, then the Dice measure between S1\nEspecially when tij = tijL = tijU , uij = uL U\nij = uij , and S2 is defined as:\nU\nand v ij = v Lij = v ij for i = 1, 2 and j = 1, 2, . . . , n\nare hold, Equation (2) is degenerated to Equation (1).\nIn real applications, one usually takes the 2(S1 · S2 )\nDINN2 (S1 , S2 ) =\nimportant differences of each element sij (i = |S1 |2 + |S2 |2\n1, 2; j = 1, 2, . . . , n) into account. Let W = ⎛ ⎞\nL L U U\nn t1j t2j + t1j t2j + uL L\n1j u2j\n(w1 , w2 , . . . , wn )T be the weight vector\nfor sij (i = 2 ⎝ ⎠\nj=1\n1, 2; j = 1, 2, . . . , n), wj ≥ 0 and nj=1 wj = 1. +uU u U\n+ v L L\nv + v U U\nv\n1j 2j 1j 2j 1j 2j\nThen, based on Equations (1) and (2), Ye fur- =⎛ ⎡ ⎤ ⎞.\nL 2 U 2 L 2\nther introduced the weighted Dice measures of SNSs, \u0002n (t1j ) + (t1j ) + (u1j )\n⎜ ⎣ ⎦+⎟\nrespectively, as follows: ⎜ j=1 ⎟\n⎜ +(uU )2 + (v L )2 + (v U )2 ⎟\nn ⎜ 1j 1j 1j ⎟\n\u0002 2s1j · s2j ⎜ ⎡ ⎤ ⎟\nDWSVNN1 (S1 , S2 ) = w j \u0003 \u00032 \u0003 \u00032 , ⎜ L 2 U 2\n(t2j ) + (t2j ) + (u2j )L 2 ⎟\n⎜ \u0002n ⎟\n\u0003s1j \u0003 + \u0003s2j \u0003 ⎝ ⎣ ⎦ ⎠\nj=1\nj=1 U 2 L 2 U 2\nn\n+(u2j ) + (v 2j ) + (v 2j )\n\u0002 2(t1j t2j + u1j u2j + v 1j v 2j )\n= wj (6)\nj=1\n2\n(t1j + u21j + v 21j ) + (t2j\n2 + u2 + v 2 )\n2j 2j Obviously, the two Dice measures DSVNN2\n(3) (S1 , S2 ) and DINN2 (S1 , S2 ) also satisfy the follow-\ning properties:\n\u0002\nn\n2s1j · s2j (P1) DSVNN2 (S1 , S2 ) = DSVNN2 (S2 , S1 ) and\nDWINN1 (S1 , S2 ) = wj \u0003 \u00032 \u0003 \u00032\n\u0003s1j \u0003 + \u0003s2j \u0003 DINN2 (S1 , S2 ) = DINN2 (S2 , S1 );\nj=1\n\u0004 \u0005 (P2) 0 ≤ DSVNN2 (S1 , S2 ) ≤ 1 and 0 ≤ DINN2\nL L U U\nt1j t2j + t1j t2j + uL1j uL2j (S1 , S2 ) ≤ 1;\n2 (P3) DSVNN2 (S1 , S2 ) = 1 and DINN2 (S1 , S2 ) =\n\u0002\nn\n+uU1j uU2j + v L1j v L2j + v U1j v U2j\n= wj ⎛ ⎞ . (4) 1, if S1 = S2 .\nL 2\nj=1\n(t1j ) + (uL1j )2 + (v L1j )2\n⎜ ⎟ Proof:\n⎜ +(t1jU 2\n) + (uU1j )2 + (v U1j )2 ⎟\n⎜ ⎟ (P1) It is obvious that the property is true.\n⎜ ⎟\n⎜ +(t L )2 + (uL )2 + (v L )2 ⎟ (P2) It is obvious that the property is true accord-\n⎝ 2j 2j 2j ⎠ ing to the inequality a2 + b2 ≥ 2ab for Equations (5)\nU 2\n+(t2j ) + (uU2j )2 + (v U2j )2 and (6).\n666 J. Ye / The generalized Dice measures for multiple attribute decision making\n\n## (P3) If S1 = S2 , there are s1j = s2j (j = 1, 2, . . . , 1\u0002\n\nn\n(t1j t2j + u1j u2j + v 1j v 2j )\nn) and |S1 | = |S2 |. So there are DSVNN2 (S1 , S2 ) = 1 = ,\nn λ(t1j\n2\n+ u21j + v 21j ) + (1 − λ)(t2j\n2\n+ u22j + v 22j )\nand DINN2 (S1 , S2 ) = 1. \u0002 j=1\n\n## In practical applications, the elements for sij (9)\n\n(i = 1, 2; j = 1, 2, . . . , n) have different weights.\nLet W = (w1 , w2 , . . . , wn )T be the weight vec- S1 · S2\nnfor sij (i = 1, 2; j = 1, 2, . . . , n), wj ≥ 0 and\ntor GSVNN2 (S1 , S2 ) =\nλ |S1 | + (1 − λ) |S2 |2\n2\nj=1 wj = 1. Then, based on Equations (5) and (6) n\nwe further introduce the weighted Dice measures of (t t + u1j u2j\nj=1 1j 2j\n+ v 1j v 2j )\n= n n ,\nSNSs, respectively, as follows: λ (t 2 + u1j + v 21j ) + (1 − λ)\n2 2\n(t2j + u22j + v 22j )\nj=1 1j j=1\n(10)\n2(S1 · S2 )w where λ is a positive parameter for 0 ≤ λ ≤ 1.\nDWSVNN2 (S1 , S2 ) =\n|S1 |2w + |S2 |2w Then, the generalized Dice measures imply some\nn special cases by choosing some values of the param-\n2 j=1\nw2j (t1j t2j + u1j u2j + v 1j v 2j )\n= n n , eter λ. If λ = 0.5, the two generalized Dice measures\nw2 (t 2\nj=1 j 1j\n+ u21j + v 21j ) + j=1\nw2j (t2j\n2\n+ u22j + v 22j )\n(7)\n(9) and (10) are degenerated to the Dice measures (1)\nand (5); if λ = 0, 1, the two generalized Dice mea-\nsures are degenerated to the following asymmetric\n2(S1 · S2 )w measures respectively:\nDWINN2 (S1 , S2 ) =\n|S1 |2w + |S2 |2w n\n⎛ ⎞ 1 \u0002 s1j · s2j\nL L U U GSVNN1 (S1 , S2 ) = \u0003 \u00032\nn t1j t2j + t1j t2j + uL L\n1j u2j n \u0003s2j \u0003\n2⎝ ⎠ j=1\n2 j=1 wj\n+uU U\n1j u2j + vL L\n1j v 2j + vU U\n1j v 2j 1\nn\n\u0002 t1j t2j + u1j u2j + v 1j v 2j\n=⎛ ⎡ ⎤ ⎞. = for λ = 0, (11)\nL 2 U 2\n\u0002n (t1j ) + (t1j ) + (uL\n1j )\n2\nn t2j\n2 + u2 + v 2\nj=1\n⎜ w2 ⎣ ⎦+⎟ 2j 2j\n⎜ j=1 j ⎟\n⎜ +(uU )2\n+ (v L 2\n) + (v U 2\n) ⎟ n\n1 \u0002 s1j · s2j\n⎜ 1j 1j 1j ⎟\n⎜ ⎡ ⎤ ⎟ GSVNN1 (S1 , S2 ) = \u0003 \u00032\n⎜ L 2 U 2\n(t2j ) + (t2j ) + (u2j )L 2 ⎟ n \u0003s1j \u0003\n⎜ \u0002n ⎟ j=1\n⎝ w2j ⎣ ⎦ ⎠\nn\nj=1\n+(uU L 2 U 2\n2j ) + (v 2j ) + (v 2j )\n2\n1 \u0002 t1j t2j + u1j u2j + v 1j v 2j\n= for λ = 1, (12)\n(8) n t1j\n2 + u2 + v 2\nj=1 1j 1j\n\nS1 · S2\nGSVNN2 (S1 , S2 ) =\n4. The generalized Dice measures of SNSs |S2 |2\nn\nj=1 (t1j t2j +u1j u2j + v 1j v 2j )\nIn this section, we propose the generalized Dice = n for λ = 0, (13)\nj=1 (t2j + u2j + v 2j )\n2 2 2\nmeasures of SNSs to extend the Dice measures of\nSNSs. S1 · S2\nAs the generalization of the Dice measures of GSVNN2 (S1 , S2 ) =\n|S1 |2\nSNSs, the generalized Dice measures between SNSs n\nare defined below. j=1 (t1j t2j +u1j u2j + v 1j v 2j )\n= n for λ = 1. (14)\nj=1 (t1j + u1j + v 1j )\n2 2 2\nDefinition 4. Let S1 = {s11 , s12 , . . . , s1n } and\nS2 = {s21 , s22 , . . . , s2n } be two SNSs, where Obviously, the four asymmetric measures are the\ns1j = (t1j , u1j , v 1j ) and s2j = (t2j , u2j , v 2j ) (j = extension of the relative projection measure (the\n1, 2, . . . , n) are considered as the j-th SVNNs in improved projection measure) of interval numbers\nthe SNSs S1 and S2 . Then the generalized Dice mea- , hence the four asymmetric measures can be con-\nsures between S1 and S2 are defined, respectively, as\nsidered as the projection measures of SNSs.\nfollows:\nFor practical applications, the elements of sij\n(i = 1, 2; j = 1, 2, . . . , n) imply different weights.\n1\u0002\nn\ns1j · s2j\nGSVNN1 (S1 , S2 ) = \u0003 \u00032 \u0003 \u00032 Assume that W = (w1 , w2 . . . , wn )T is the weight\nn λ \u0003s1j \u0003 + (1 − λ) \u0003s2j \u0003\nj=1 vector for sij (i = 1, 2; j = 1, 2, . . . , n), wj ≥ 0\nJ. Ye / The generalized Dice measures for multiple attribute decision making 667\n\nand nj=1 wj = 1. Thus, based on Equations (9) and where λ is a positive parameter for 0 ≤ λ ≤ 1. Espe-\n(10) we further introduce the following weighted cially, when tij = tijL = tijU , uij = uL U\nij = uij , and v ij =\nU\ngeneralized Dice measures of SNSs, respectively, as vL\nij = v ij for i = 1, 2 and j = 1, 2, . . . , n are hold,\nfollows: Equations (17) and (18) are degenerated to Equations\n\u0002\nn\ns1j · s2j\n(9) and (10).\nGWSVNN1 (S1 , S2 ) = wj \u0003 \u0003 2 \u0003 \u00032 Similarly, if λ = 0.5, the two generalized Dice\nj=1\n\u0003\nλ s1j + (1 − λ) \u0003s2j \u0003\n\u0003\nmeasures (17) and (18) are degenerated to the Dice\n\u0002\nn measures (2) and (6); if λ = 0, 1, then the two gener-\nt1j t2j + u1j u2j + v 1j v 2j\n= wj , alized Dice measures are degenerated to the following\nλ(t1j\n2\n+ u21j + v 21j ) + (1 − λ)(t2j\n2\n+ u22j + v 22j )\nj=1 asymmetric measures respectively:\n(15)\nn n\n1 \u0002 s1j · s2j 1\u0002\nGINN3 (S1 , S2 ) = \u0003 \u00032 =\nn j=1 \u0003s2j \u0003 n j=1\n(S1 · S2 )w\nGWSVNN2 (S1 , S2 ) = U U\nλ |S1 |2w + (1 − λ) |S2 |2w\nL L\nt1j t2j + t1j t2j + uL1j uL2j + uU1j uU2j + v L1j v L2j + v U1j v U2j\nn L 2\n(t2j U 2\n) + (uL2j )2 + (v L2j )2 + (t2j ) + (uU2j )2 + (v U2j )2\nj=1 wj (t1j t2j + u1j u2j + v 1j v 2j )\n2\n= ⎛ \u0002n ⎞. (16) for λ = 0, (19)\nλ w2j (t1j\n2\n+ u21j + v 21j )+\n⎜ j=1 ⎟ n\n\u0002 s1j · s2j\nn\n\u0002\n⎝ \u0002n ⎠ 1 1\nGINN3 (S1 , S2 ) = \u0003 \u00032 =\n(1 − λ) w2 (t 2 + u22j + v 22j ) n \u0003s1j \u0003 n\nj=1 j 2j j=1 j=1\nU U\nL L\nt1j t2j + t1j t2j + uL1j uL2j + uU1j uU2j + v L1j v L2j + v U1j v U2j\nDefinition 5. Let S1 = {s11 , s12 , . . . , s1n } and U 2\nS2 = {s21 , s22 , . . . , s2n } be two SNSs, where\nL 2\n(t1j ) + (uL1j )2 + (v L1j )2 + (t1j ) + (uU1j )2 + (v U1j )2\ns1j = (t1j , u1j , v 1j ) and s2j = (t2j , u2j , v 2j ) (j = for λ = 1, (20)\n1, 2, . . . , n) are considered as the j-th INNs in the\nSNSs S1 and S2 . Then the generalized Dice mea-\nsures between S1 and S2 are defined, respectively,\nS1 · S2\nas follows: GINN4 (S1 , S2 ) =\n|S2 |2\n1\u0002 n L L U U\nn\ns1j · s2j 1 (t t + t1j t2j + uL1j uL2j + uU1j uU2j + v L1j v L2j + v U1j v U2j )\nGINN3 (S1 , S2 ) = \u0003 \u00032 \u0003 \u00032 = j=1 1j 2j\nn λ \u0003s1j \u0003 + (1 − λ) \u0003s2j \u0003 n = n L 2 U 2 L 2 U 2 L 2 U 2\nj=1\nj=1\n[(t2j ) + (t2j ) + (u2j ) + (u2j ) + (v 2j ) + (v 2j ) ]\n\u0002\nn L L\nt1j t2j + t U t2j\nU\n+ uL1j uL2j + uU1j uU2j + v L1j v L2j + v U1j v U2j for λ = 0, (21)\n⎛ \u0011\n1j\nL 2\n\u0012 ⎞ ,\n(t1j ) + (uL1j )2 + (v L1j )2 S1 · S2\nj=1 GINN4 (S1 , S2 ) =\n⎜λ ⎟ |S1 |2\n⎜ U 2\n+(t1j ) + (uU1j )2 + (v U1j )2 ⎟ n L L U U\n⎜ ⎟ (t t + t1j t2j + uL1j uL2j + uU1j uU2j + v L1j v L2j + v U1j v U2j )\n⎜ \u0011 L 2 \u0012⎟\n⎜ L 2 L 2\n(t2j ) + (u2j ) + (v 2j ) ⎟ j=1 1j 2j\n= n\n⎝ ⎠ L 2 U 2 L 2 U 2 L 2 U 2\n[(t1j ) + (t1j ) + (u1j ) + (u1j ) + (v 1j ) + (v 1j ) ]\n+(1 − λ) j=1\nU 2\n+(t2j ) + (uU2j )2 + (v U2j )2 for λ = 1. (22)\n(17)\nThen, the four asymmetric measures are also con-\nS1 · S2\nGINN4 (S1 , S2 ) = sidered as the extension of the relative projection\nλ |S1 |2 + (1 − λ) |S2 |2\nn measure (the improved projection measure) of inter-\nU U\nj=1\nL L\n(t1j t2j + t1j t2j + uL1j uL2j + uU1j uU2j + v L1j v L2j + v U1j v U2j ) val numbers , which are also called the projection\n= ⎛ \u0011 \u0012 ⎞ ,\nmeasures of SNSs.\n\u0002n L 2\n(t1j U 2\n) + (t1j ) + (uL1j )2\n⎜λ + ⎟ For practical applications, the elements of sij (i =\n⎜ j=1 +(uU1j )2 + (v L1j )2 + (v U1j )2 ⎟ 1, 2; j = 1, 2, . . . , n) imply different weights.\n⎜ ⎟\n⎜ \u0011 \u0012⎟ Assume that W = (w1 , w2 , . . . , wn )T is the weight\n⎜ \u0002n L 2\n(t2j U 2\n) + (t2j ) + (uL2j )2 ⎟\n⎝ ⎠ vector sij (i = 1, 2; j = 1, 2, . . . , n), wj ≥ 0\n(1 − λ) for\nn\n+(uU2j )2 + (v L2j )2 + (v U2j )2 j=1 wj = 1. Similarly, based on Equations\nj=1 and\n(18) (17) and (18) we also further introduce the weighted\n668 J. Ye / The generalized Dice measures for multiple attribute decision making\n\ngeneralized Dice measures of SNSs, respectively, as be used to help identify the best alternative in the\nfollows: decision set . Hence, by an ideal SVNN\n\u0002\nn\ns1j · s2j\nGWINN3 (S1 , S2 ) = wj \u0003 \u00032 \u0003 \u00032 sj∗ =< tj∗ , u∗j , v ∗j >=< max(tij ), min(uij ), min(v ij ) >\nj=1 λ \u0003s1j \u0003 + (1 − λ) \u0003s2j \u0003 i i i\n\n\u0011 L L U U\n\u0012\nt1j t2j + t1j t2j + uL1j uL2j or an ideal INN\n\u0002\nn\n+uU1j uU2j + v L1j v L2j + v U1j v U2j sj∗ =< tj∗ , u∗j , v ∗j >=< [max(tijL ), max(tijU )],\n= wj ⎛ \u0011 L 2\n\u0012 ⎞, i i\nj=1\n(t1j ) + (uL1j )2 + (v L1j )2\n⎜λ ⎟ [min(uL U L U\nij ), min(uij )], [min(v ij ), min(v ij )] >\n⎜ U 2\n+(t1j ) + (uU1j )2 + (v U1j )2 ⎟ i i i i\n⎜ ⎟\n⎜ \u0011 L 2 \u0012⎟\n⎜ (t ) + (uL 2\n) + (v L 2\n) ⎟ for j = 1, 2, . . . , n and i = 1, 2, . . . , m, we can\n⎝ 2j 2j 2j ⎠\n+(1 − λ) determine a simplified neutrosophic ideal solu-\nU 2\n+(t2j ) + (uU2j )2 + (v U2j )2 tion (ideal alternative) S ∗ = {s1∗ , s2∗ , . . . , sn∗ }, where\n(23)\nsj∗ =< tj∗ , u∗j , v ∗j > is the j-th ideal SNN.\nIn the decision-making process, decision makers\n(S1 · S2 )w take some value of the parameter λ ∈ [0, 1] accord-\nGWINN4 (S1 , S2 ) =\nλ |S1 |2w + (1 − λ) |S2 |2w ing to their preference and/or real requirements, the\n\u0004 L L U U\n\u0005 weighted generalized Dice measure between Si (i =\nn t1j t2j + t1j t2j + uL1j uL2j\nj=1 w2j 1, 2, . . . , m) and S ∗ is obtained by using one of Equa-\n+uU1j uU2j + v L1j v L2j + v U1j v U2j tions (15), (16), (23) and (24) to rank the alternatives.\n= ⎛ \u0011 L 2 U 2\n\u0012 ⎞.\n\u0002n (t1j ) + (t1j ) + (uL1j )2 Thus, the greater the value of the weighted gener-\n⎜λ w2j + ⎟ alized Dice measure between Si (i = 1, 2, . . . , m)\n⎜ j=1\n+(uU1j )2 + (v L1j )2 + (v U1j )2 ⎟\n⎜ ⎟ and S ∗ is, the better the alternative Si is.\n⎜ \u0011 \u0012⎟\n⎜ \u0002n L 2\n(t2j U 2\n) + (t2j ) + (uL2j )2 ⎟\n⎝ ⎠\n(1 − λ) w2j\nj=1\n+(uU2j )2 + (v L2j )2 + (v U2j )2 6. Decision-making example\n(24) of manufacturing schemes\n\n## A real example about the decision-making prob-\n\n5. Decision making-methods based lem of manufacturing schemes with simplified\non the generalized Dice measures neutrosophic information is given to demonstrate\nthe applications and effectiveness of the proposed\nIn this section, we propose multiple attribute decision-making methods in realistic scenarios.\ndecision-making methods by using the generalized To select the best manufacturing scheme (alter-\nDice measures of SNSs under simplified neutro- native) for the flexible manufacturing system in a\nsophic environment. manufacturing company, the technique department of\nFor multiple attribute decision-making problems, the company provides four manufacturing schemes\nlet S = {S1 , S2 , . . . , Sm } be a set of alternatives and (alternatives) with respect to some product as a set\nR = {R1 , R2 , . . . , Rn } be a set of attributes. Then, of the alternatives S = {S1 , S2 , S3 , S4 } for the flex-\nthe weight of the attribute\nRj (j = 1, 2, . . . , n) is ible manufacturing system. A decision must be made\nwj , wj ∈ [0, 1] and nj=1 wj = 1. Thus, the fit judg- according to the four attributes: (1) R1 is the improve-\nment (satisfaction evaluation) of an attribute Rj (j = ment of quality; (2) R2 is the market response; (3)\n1, 2, . . . , n) for an alternative Si (i = 1, 2, . . . , m) R3 is the manufacturing cost; (4) R4 is the manu-\nis represented by a SNS Si = {si1 , si2 , . . . , sin }, facturing complexity. The weight vector of the four\nwhere sij =< tij , uij , v ij > is a SVNN for 0 ≤ tij + attributes W = (0.3, 0.25, 0.25, 0.2)T is given by\nuij + v ij ≤ 3 or an INN for 0 ≤ tijU + uU U\nij + v ij ≤ decision makers.\n3 (j = 1, 2, . . . , n and i = 1, 2, . . . , m). Therefore, In the decision-making problem, the decision mak-\nwe can establish a simplified neutrosophic decision ers are required to make the fit judgment (satisfaction\nevaluation) of an attribute Rj (j = 1, 2, 3, 4) for an\nmatrix D = (sij )m×n .\nalternative Si (i = 1, 2, 3, 4) and to give simplified\nIn the multiple attribute decision-making problem, neutrosophic evaluation information, which is shown\nthe concept of an ideal solution (ideal alternative) can in the following decision matrix with SVNNs:\nJ. Ye / The generalized Dice measures for multiple attribute decision making 669\n\nD= Table 1\n⎡ ⎤ The measure values of Equation (15) and ranking orders\n(0.75, 0.2, 0.3) (0.7, 0.2, 0.3) (0.65, 0.2, 0.25) (0.75, 0.2, 0.1)\nλ GWSVNN1 GWSVNN1 GWSVNN1 GWSVNN1 Ranking order\n⎢ (0.8, 0.1, 0.2) (0.75, 0.2, 0.1) (0.75, 0.2, 0.1) (0.85, 0.1, 0.2) ⎥\n(S1 , S ∗ ) (S2 , S ∗ ) (S3 , S ∗ ) (S4 , S ∗ )\n⎢ ⎥.\n⎣ (0.7, 0.2, 0.2) (0.78, 0.2, 0.1) (0.85, 0.15, 0.1) (0.76, 0.2, 0.2) ⎦ 0 0.8895 0.9517 0.9361 0.9287 S2\nS3\nS4\nS1\n(0.8, 0.2, 0.1) (0.85, 0.2, 0.2) (0.7, 0.2, 0.2) (0.86, 0.1, 0.2) 0.2 0.9157 0.9667 0.9558 0.9472 S2\nS3\nS4\nS1\n0.5 0.9612 0.9924 0.9876 0.9816 S2\nS3\nS4\nS1\nThen, the developed decision-making methods can 0.7 0.9966 1.0119 1.0104 1.0100 S2\nS3\nS4\nS1\nbe used for the decision making problem. 1 1.0594 1.0455 1.0475 1.0641 S4\nS1\nS3\nS2\nAccording to\nsj∗ =< tj∗ , u∗j , v ∗j >= < max(tij ), min(uij ), min(v ij ) >\nTable 2\ni i i The measure values of Equation (16) and ranking orders\n\nfor j = 1, 2, 3, 4 and i = 1, 2, 3, 4, we can obtain λ GWSVNN2 GWSVNN2 GWSVNN2 GWSVNN2 Ranking order\n(S1 , S ∗ ) (S2 , S ∗ ) (S3 , S ∗ ) (S4 , S ∗ )\nan ideal solution (ideal alternative) as follows:\n0 0.8908 0.9503 0.9387 0.9372 S2\nS3\nS4\nS1\nS ∗ = {s1∗ , s2∗ , s3∗ , s4∗ } 0.2 0.9175 0.9667 0.9587 0.9557 S2\nS3\nS4\nS1\n\u0015 \u0016 0.5 0.9605 0.9924 0.9903 0.9849 S2\nS3\nS4\nS1\n< 0.8, 0.1, 0.1 >, < 0.85, 0.2, 0.1 >, 0.7 0.9915 1.0102 1.0126 1.0054 S3\nS2\nS4\nS1\n= . 1 1.0419 1.0383 1.0479 1.0378 S3\nS1\nS2\nS4\n< 0.85, 0.15, 0.1 >, < 0.86, 0.1, 0.1 >\nBy using Equation (15) or (16) and different values the alternative S2 is the best choice among all\nof the parameter λ, the weighted generalized Dice the alternatives.\nmeasure values between Si (i = 1, 2, 3, 4) and S ∗ (3) When λ = 1, the two weighted generalized\ncan be obtained, which are shown in Tables 1 and 2 Dice measures are reduced to the weighted\nrespectively. projection measures of S ∗ on Si . Thus, the\nFrom Tables 1 and 2, we can see that different rank- alternative S3 or S4 is the best choice among\ning orders are indicated by taking different values of all the alternatives.\nthe parameter λ and different generalized Dice mea- Obviously, according to different values of the\nsures. Then we can obtain that the best alternative is parameter λ and different measures, ranking orders\nS2 or S3 or S4 . may be different. Thus the proposed decision-making\nFurthermore, for the special cases of the two gener- methods can be assigned some value of λ and some\nalized Dice measures we obtain the following results: measure to satisfy the decision makers’ preference\n(1) When λ = 0, the two weighted generalized and/or real requirements.\nDice measures are reduced to the weighted If the fit judgment (satisfaction evaluation) of an\nprojection measures of Si on S ∗ . Thus, the attribute Rj (j = 1, 2, 3, 4) for an alternative Si\nalternative S2 is the best choice among all the (i = 1, 2, 3, 4) is given in the decision making\nalternatives. problem by the following decision matrix with INNs:\nD=\n\n< [0.7, 0.8], [0.1, 0.2], [0.2, 0.3] > < [0.7, 0.8], [0.1, 0.2], [0.2, 0.3] >\n\n⎢ < [0.7, 0.9], [0.1, 0.2], [0.2, 0.3] > < [0.7, 0.8], [0.1, 0.3], [0.1, 0.2] >\n\n⎢ < [0.7, 0.8], [0.1, 0.3], [0.2, 0.3] > < [0.8, 0.9], [0.1, 0.2], [0.1, 0.2] >\n\n< [0.8, 0.9], [0.2, 0.3], [0.1, 0.2] > < [0.8, 0.9], [0.2, 0.3], [0.2, 0.3] >\n\n< [0.6, 0.7], [0.1, 0.2], [0.2, 0.4] > < [0.7, 0.8], [0.1, 0.2], [0.1, 0.2] >\n\n< [0.7, 0.8], [0.2, 0.3], [0.1, 0.2] > < [0.8, 0.9], [0.0, 0.1], [0.2, 0.3] > ⎥\n⎥.\n< [0.8, 0.9], [0.1, 0.2], [0.1, 0.1] > < [0.7, 0.8], [0.1, 0.2], [0.1, 0.2] > ⎥\n\n< [0.7, 0.8], [0.1, 0.2], [0.1, 0.3] > < [0.7, 0.9], [0.0, 0.1], [0.1, 0.3] >\n\n## (2) When λ = 0.5, the two weighted generalized\n\nDice measures are reduced to the weighted\nDice similarity measures of Si and S ∗ . Thus,\n670 J. Ye / The generalized Dice measures for multiple attribute decision making\n\n## Then, according to Table 3\n\nThe measure values of Equation (23) and ranking orders\nsj∗ =< tj∗ , u∗j , v ∗j >=< [max(tijL ), max(tijU )], λ GWINN3 GWINN3 GWINN3 GWINN3 Ranking order\ni i\n(S1 , S ∗ ) (S2 , S ∗ ) (S3 , S ∗ ) (S4 , S ∗ )\n[min(uL U L U\nij ), min(uij )], [min(v ij ), min(v ij )] > 0 0.9085 0.9770 0.9861 1.0159 S4\nS3\nS2\nS1\ni i i i 0.2 0.9325 0.9819 0.9860 1.0015 S4\nS3\nS2\nS1\n0.5 0.9737 0.9903 0.9901 0.9863 S2\nS3\nS4\nS1\nfor j = 1, 2, 3, 4 and j = 1, 2, 3, 4, we can\n0.7 1.0053 0.9966 0.9955 0.9797 S1\nS2\nS3\nS4\nobtain the ideal solution (ideal alternative) as follows: 1 1.0601 1.0072 1.0075 0.9746 S1\nS3\nS2\nS4\n\n## S ∗ = {s1∗ , s2∗ , . . . , sn∗ }\n\n⎧ ⎫\n\n⎪ < [0.8, 0.9], [0.1, 0.2], [0.1, 0.2] >,⎪\n\nTable 4\n\n⎪ ⎪\n⎪ The measure values of Equation (24) and ranking orders\n⎨ < [0.7, 0.8], [0.1, 0.2], [0.1, 0.2] >,⎪\n⎪ ⎬ λ GWINN4 GWINN4 GWINN4 GWINN4 Ranking order\n= . (S1 , S ∗ ) (S2 , S ∗ ) (S3 , S ∗ ) (S4 , S ∗ )\n\n⎪ < [0.8, 0.9], [0.1, 0.2], [0.1, 0.1] >,⎪\n\n⎪ ⎪\n⎪ S4\nS3\nS2\nS1\n\n⎩ ⎪\n\n0 0.9035 0.9726 0.9790 1.0109\n< [0.8, 0.9], [0.0, 0.1], [0.1, 0.2] > 0.2 0.9306 0.9795 0.9833 1.0011 S4\nS3\nS2\nS1\n0.5 0.9743 0.9901 0.9900 0.9867 S2\nS3\nS4\nS1\nAccording to Equations (23) or (24) and different 0.7 1.0059 0.9972 0.9945 0.9774 S1\nS2\nS3\nS4\n1 1.0572 1.0082 1.0013 0.9637 S1\nS2\nS3\nS4\nvalues of the parameter λ, the weighted generalized\nDice measure values between Si (i = 1, 2, 3, 4) and\nS ∗ can be obtained, which are shown in Tables 3 and methods based on generalized Dice measures.\n4 respectively. Therefore, in the decision-making process, the\nFrom Tables 3 and 4, different ranking orders are decision-making methods developed in this paper\nshown by taking different values of λ and different are more general and more flexible than existing\nmeasures. Then we can obtain that the best alternative decision-making methods under simplified neutro-\nis S1 or S2 or S4 . sophic environment.\nFurthermore, for the special cases of the two gener-\nalized Dice measures we obtain the following results:\n7. Conclusion\n(1) When λ = 0, the two weighted generalized\nDice measures are reduced to the weighted\nThis paper proposed another form of the Dice\nprojection measures of Si on S ∗ . Thus, the\nmeasures between SNSs and the generalized Dice\nalternative S4 is the best choice among all the\nmeasures of SNSs and indicated the Dice measures\nalternatives.\nof SNSs and the projection measures (asymmetric\n(2) When λ = 0.5, the two weighted generalized\nmeasures) of SNSs are the special cases of the gen-\nDice measures are reduced to the weighted\neralized Dice measures of SNSs corresponding to\nDice similarity measures of Si and S ∗ . Thus,\nsome parameter values. Then, we developed multi-\nthe alternative S2 is the best choice among all\nple attribute decision-making methods based on the\nthe alternatives.\ngeneralized Dice measures of SNSs under simplified\n(3) When λ = 1, the two weighted generalized\nneutrosophic environment. According to different\nDice measures are reduced to the weighted\nparameter values and some measure preferred by\nprojection measures of S ∗ on Si . Thus, the\ndecision makers, by the weighted generalized Dice\nalternative S1 is the best choice among all the\nmeasure between each alternative and the ideal\nalternatives.\nsolution (ideal alternative), all alternatives can be\nTherefore, according to different values of the ranked and the best alternative can be selected as\nparameter λ and different measures, ranking orders well. Finally, a real example about the selection\nmay be also different. Thus the proposed decision- of manufacturing schemes (alternatives) demon-\nmaking methods can be assigned some value of λ strated the applications of the developed methods\nand some measure to satisfy the decision makers’ under simplified neutrosophic environment, and then\npreference and/or real requirements. the effectiveness and flexibility of the developed\nObviously, the decision-making methods based decision-making methods were shown corresponding\nthe Dice measures and the projection measures are to different parameter values. In the decision-making\nthe special cases of the proposed decision-making process under simplified neutrosophic environment,\nJ. Ye / The generalized Dice measures for multiple attribute decision making 671\n\nthe main advantage is more general and more flexible J. Ye, Similarity measures between interval neutrosophic\nthan existing decision-making methods to satisfy the sets and their applications in multicriteria decision-making,\nJournal of Intelligent and Fuzzy Systems 26 (2014),\ndecision makers’ preference and/or practical require- 165–172.\nments. J. Ye, Multiple attribute group decision-making method with\nIn the future work, we shall extend the generalized completely unknown weights based on similarity measures\nDice measures of SNSs to other areas such as pattern under single valued neutrosophic environment, Journal of\nIntelligent and Fuzzy Systems 27(12) (2014), 2927–2935.\nrecognition, fault diagnosis, and image processing. J. Ye, Vector similarity measures of simplified neutrosophic\nsets and their application in multicriteria decision mak-\ning, International Journal of Fuzzy Systems 16(2) (2014),\nReferences 204–211.\n J. Ye, Single valued neutrosophic cross-entropy for multi-\ncriteria decision making problems, Applied Mathematical\n F. Smarandache, Neutrosophy. Neutrosophic Probability, Modelling 38(3) (2014), 1170–1175.\nSet, and Logic, American Research Press, Rehoboth, USA J. Ye, Multiple attribute decision-making method based on\n1998. the possibility degree ranking method and ordered weighted\n G.L. Xu and F. Liu, An approach to group decision making aggregation operators of interval neutrosophic numbers,\nbased on interval multiplicative and fuzzy preference rela- Journal of Intelligent and Fuzzy Systems 28(3) (2015),\ntions by using projection, Applied Mathematical Modelling 1307–1317.\n37 (2013), 3929–3943. K. Atanassov, Intuitionistic fuzzy sets, Fuzzy Sets and\n H. Wang, F. Smarandache, Y.Q. Zhang and R. Sunderraman, Systems 20 (1986), 87–96.\nInterval neutrosophic sets and logic: Theory and applica- K. Atanassov and G. Gargov, Interval valued intuitionistic\ntions in computing, Hexis, Phoenix, A.Z. 2005. fuzzy sets, Fuzzy Sets and Systems 31 (1989), 343–349.\n H. Wang, F. Smarandache, Y.Q. Zhang and R. Sunderraman, P.P. Chi and P.D. Liu, An Extended TOPSIS Method for\nSingle valued neutrosophic sets, Multispace and Multistruc- multiple attribute decision making problems based on inter-\nture 4 (2010), 410–413. val neutrosophic set, Neutrosophic Sets and Systems 1\n H.Y. Zhang, J.Q. Wang and X.H. Chen, Interval neutro- (2013), 63–70.\nsophic sets and their application in multicriteria decision P.D. Liu, Y.C. Chu, Y.W. Li and Y.B. Chen, Some general-\nmaking problems, Science World Journal 2014 (2014), 15. ized neutrosophic number Hamacher aggregation operators\nArticle ID 645953. and their application to group decision making, Interna-\n H.Y. Zhang, J.Q. Wang and X.H. Chen, An outranking tional Journal of Fuzzy Systems 16(2) (2014), 242–255.\napproach for multi-criteria decision-making problems with P.D. Liu and Y.M. Wang, Multiple attribute decision mak-\ninterval-valued neutrosophic sets, Neural Computing and ing method based on single valued neutrosophic normalized\nApplications (2015). doi: 10.1007/s00521-015-1882-3 weighted Bonferroni mean, Neural Computing and Appli-\n J.J. Peng, J.Q. Wang, H.Y. Zhang and X.H. Chen, An cations 25(7-8) (2014), 2001–2010.\noutranking approach for multi-criteria decision-making P. Biswas, S. Pramanik and B.C. Giri, TOPSIS method for\nproblems with simplified neutrosophic sets, Applied Soft multi-attribute group decision-making under single-valued\nComputing 25 (2014), 336–346. neutrosophic environment, Neural Computing and Applica-\n J.J. Peng, J.Q. Wang, J. Wang, H.Y. Zhang and X.H. tions (2015). doi: 10.1007/s00521-015-1891-2\nChen, Simplified neutrosophic sets and their applica- R. Sahin and A. Kucuk, Subsethood measure for single\ntions in multi-criteria group decision-making problems, valued neutrosophic sets, Journal of Intelligent and Fuzzy\nInternational Journal of Systems Science (2015). doi: Systems 29(2) (2015), 525–530.\n10.1080/00207721.2014.994050 R. Şahin and P.D. Liu, Maximizing deviation method\n J. Ye, Multicriteria decision-making method using the for neutrosophic multiple attribute decision making with\ncorrelation coefficient under single-valued neutrosophic incomplete weight information, Neural Computing and\nenvironment, International Journal of General Systems Applications (2015). doi: 10.1007/s00521-015-1995-8\n42(4) (2013), 386–394.\n J. Ye, A multicriteria decision-making method using aggre-\ngation operators for simplified neutrosophic sets, Journal of\nIntelligent and Fuzzy Systems 26 (2014), 2459–2466.\nCopyright of Journal of Intelligent & Fuzzy Systems is the property of IOS Press and its\ncontent may not be copied or emailed to multiple sites or posted to a listserv without the"
] | [
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https://stackoverflow.com/questions/32793972/racket-switch-statement-macro | [
"# Racket switch statement macro\n\nI'm trying to make a switch statement macro in Racket. I'm having some trouble figuring out excatly how to do it. I would like to be able to use the function as shown below.\n\n``````(define x 99)\n\n(switch x\n[3 (displayln \"x is 3\")]\n[4 (displayln \"x is 4\")]\n[5 (displayln \"x is 5\")]\n['default (displayln \"none of the above\")])\n``````\n\nI've tried using pattern matching with syntax-case but i'm not sure that is the correct approach. Any Racket experts here that could give me a push in the right direction?\n\nI believe in not reinventing perfectly good wheels, so here's a macro for transforming your `switch` into the equivalent `case`, with a slight change to use `default` instead of `'default`. (I apologise to soegaard, for not knowing how to use `syntax-parse` yet, so I'll just stick with traditional `syntax-case` instead. ;-))\n\n``````(define-syntax (switch stx)\n(define (transform-clause cl)\n(syntax-case cl (default)\n((default expr) #'(else expr))\n((val ... expr) #'((val ...) expr))))\n\n(define (transform-clauses cls)\n(syntax-case cls ()\n((cl)\n(with-syntax ((case-clause (transform-clause #'cl)))\n#'(case-clause)))\n((cl rest ...)\n(with-syntax ((case-clause (transform-clause #'cl))\n((case-rest ...) (transform-clauses #'(rest ...))))\n#'(case-clause case-rest ...)))))\n\n(syntax-case stx ()\n((_ x clause ...)\n(with-syntax (((case-clause ...) (transform-clauses #'(clause ...))))\n#'(case x case-clause ...)))))\n``````\n\nThe downside of using this `switch` macro, as opposed to using `case`, is that you lose the ability to distinguish between using `default` to match the symbol `default`, vs as the catch-all. So using `case` is still better. :-)\n\n``````(case x\n((3) (displayln \"x is 3\"))\n((4) (displayln \"x is 4\"))\n((5) (displayln \"x is 5\"))\n(else (displayln \"none of the above\")))\n``````\n• `(syntax-case (default) stx clause)` and `(syntax-parse stx #:literals(default) clause ...)` mean the same (assuming you are not using any guards. It's just that easy! Sep 28, 2015 at 18:55\n\nI like both of the other answers, but I feel like we should also mention the existing `match` form, which (afaict) already does exactly what you're looking for:\n\n``````#lang racket\n\n(define x 99)\n\n(match x\n[3 (displayln \"x is 3\")]\n[4 (displayln \"x is 4\")]\n[5 (displayln \"x is 5\")]\n[default (displayln \"none of the above\")])\n``````\n\nThe only changes: I write `match` instead of `switch`, and I use the pattern `default` rather than the quoted `'default`. In fact, any identifier would work here; an identifier just gives a name to a value.\n\nIndeed, match can do a whole lot more than this, but this is one great use for it.\n\nIf you're just looking for practice in writing macros, then you can disregard this answer :).\n\nFor `switch` I recommend making a helper macro that switches on a value - this avoids the problem of evaluating the `expr` in `(switch expr clause ...)` more than once.\n\nBelow I used the identifier `else` to indicate the default clause.\n\nNote that `syntax-parse` try the patterns one-at-a-time. This is used to catch raise syntax errors given faulty input such as `(switch)`\n\nThe initial identifier in the patterns are prefixed with a `_` to avoid any problems in recursive macros.\n\nNote that the basic rewrite rule used is:\n\n``````(switch-value v\n[expr result]\nclause ...)\n\n==>\n\n(if (equal? v expr)\nresult\n(switch-value v clause ...))\n``````\n\nMaking the macro recursive is easier than writing one big macro that handles all clauses at once.\n\n``````#lang racket\n\n(require (for-syntax syntax/parse))\n\n(define-syntax (switch stx)\n(syntax-parse stx\n[(_switch)\n#'(raise-syntax-error 'switch \"value expression and at least one clause expected\" stx)]\n[(_switch expr clause ...)\n#'(let ([v expr])\n(switch-value v clause ...))]))\n\n(define-syntax (switch-value stx)\n(syntax-parse stx\n#:literals (else)\n[(_switch-value v)\n#'(raise-syntax-error 'switch \"at least one clause is expected\" _switch-value)]\n[(_switch-value v [else expr])\n#'expr]\n[(_switch-value v [expr1 expr2] clause ...)\n#'(if (equal? v expr1)\nexpr2\n(switch-value v clause ...))]))\n\n(define x 4)\n\n(switch x\n[3 \"x is 3\"]\n[4 \"x is 4\"]\n[5 \"x is 5\"]\n[else (displayln \"none of the above\")])\n``````"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5328858,"math_prob":0.8697773,"size":1392,"snap":"2022-27-2022-33","text_gpt3_token_len":360,"char_repetition_ratio":0.19092219,"word_repetition_ratio":0.0,"special_character_ratio":0.2895115,"punctuation_ratio":0.11406844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9646245,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T14:41:51Z\",\"WARC-Record-ID\":\"<urn:uuid:e86bb51d-7969-486d-b4a9-9d91456d058d>\",\"Content-Length\":\"243620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b8a9208-e641-41c0-a283-017da0972182>\",\"WARC-Concurrent-To\":\"<urn:uuid:60dffb6f-2634-48d2-b965-5c1763897557>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/32793972/racket-switch-statement-macro\",\"WARC-Payload-Digest\":\"sha1:PX52M7NAR7X5CNB6TGQIPP7TYG4IOAM3\",\"WARC-Block-Digest\":\"sha1:YBGAJD3VAZA34YQ76R5H5VCVA37LCEIR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572908.71_warc_CC-MAIN-20220817122626-20220817152626-00691.warc.gz\"}"} |
https://www.asknumbers.com/inch-to-cm/66.4-inches-to-cm.aspx | [
"# How Many Centimeters in 66.4 Inches?\n\n66.4 Inches to cm converter. How many centimeters in 66.4 inches?\n\n66.4 Inches equal to 168.656 cm or there are 168.656 centimeters in 66.4 inches.\n\n←→\nstep\nRound:\nEnter Inch\nEnter Centimeter\n\n## How to convert 66.4 inches to cm?\n\nThe conversion factor from inches to cm is 2.54. To convert any value of inches to cm, multiply the inch value by the conversion factor.\n\nTo convert 66.4 inches to cm, multiply 66.4 by 2.54, that makes 66.4 inches equal to 168.656 cm.\n\n66.4 inches to cm formula\n\ncm = inch value * 2.54\n\ncm = 66.4 * 2.54\n\ncm = 168.656\n\nCommon conversions from 66.4x inches to cm:\n(rounded to 3 decimals)\n\n• 66.4 inches = 168.656 cm\n• 66.41 inches = 168.681 cm\n• 66.42 inches = 168.707 cm\n• 66.43 inches = 168.732 cm\n• 66.44 inches = 168.758 cm\n• 66.45 inches = 168.783 cm\n• 66.46 inches = 168.808 cm\n• 66.47 inches = 168.834 cm\n• 66.48 inches = 168.859 cm\n• 66.49 inches = 168.885 cm\n\nWhat is a Centimeter?\n\nCentimeter (centimetre) is a metric system unit of length. The symbol is \"cm\".\n\nWhat is a Inch?\n\nInch is an imperial and United States Customary systems unit of length, equal to 1/12 of a foot. 1 inch = 2.54 cm. The symbol is \"in\".\n\nCreate Conversion Table\nClick \"Create Table\". Enter a \"Start\" value (5, 100 etc). Select an \"Increment\" value (0.01, 5 etc) and select \"Accuracy\" to round the result."
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https://virtualnerd.com/pre-algebra/factors-fractions-exponents/powers/ | [
"# Powers and Exponents\n\n### Popular Tutorials in Powers and Exponents\n\n• #### What is an Exponent?\n\nIf you have a repeated multiplication, you could write it using exponents! This tutorial introduces exponents and explains how they're used.\n\n• #### How Do You Convert a Number in Expanded Form into Exponential Form?\n\nExponential form is a quick way to show that a number should be multiplied by itself a certain number of times. In this tutorial, see how to write a repeated multiplication in exponential form!\n\n• #### How Do You Convert a Number in Exponential Form into Expanded Form?\n\nDid you know that exponents are just a quick way to show repeated multiplication? In this tutorial, see how to expand out a value in exponential form to see what it really represents!\n\n• #### How Do You Evaluate an Expression with Exponents?\n\nNeed to plug in a variable value into an expression? Great! Does the expression have an exponent in it? Even better! Follow along with this tutorial as you see how to simplify an expression for a given variable value.\n\n• #### How Do You Evaluate an Exponent?\n\nExponents just indicate repeated multiplication. Watch this tutorial to see how you can evaluate an exponent by first writing it in expanded form. Take a look!"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.91268045,"math_prob":0.6421264,"size":896,"snap":"2020-45-2020-50","text_gpt3_token_len":176,"char_repetition_ratio":0.15470852,"word_repetition_ratio":0.040268455,"special_character_ratio":0.18973215,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99324954,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-31T11:00:43Z\",\"WARC-Record-ID\":\"<urn:uuid:a7795c82-f532-451f-b223-351c8efa50c7>\",\"Content-Length\":\"23199\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4e9d6014-197d-4597-83dc-d093865032f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bc7c4ac-0e0d-4ca0-bff6-14a47b369c94>\",\"WARC-IP-Address\":\"54.192.30.108\",\"WARC-Target-URI\":\"https://virtualnerd.com/pre-algebra/factors-fractions-exponents/powers/\",\"WARC-Payload-Digest\":\"sha1:VMDL24ESGA2K3V4OI465RNQKOBNLMDGX\",\"WARC-Block-Digest\":\"sha1:RHDBWWDWW3Q4JZCH2P32QQJUPOXYGQS2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107917390.91_warc_CC-MAIN-20201031092246-20201031122246-00088.warc.gz\"}"} |
https://www.coursehero.com/file/p4hv0cf7/plane-The-angle-of-refraction-The-angle-of-refraction-%CE%B8%CE%B82-depends-on-the/ | [
"plane The angle of refraction The angle of refraction \\u03b8\\u03b82 depends on the\n\n# Plane the angle of refraction the angle of refraction\n\nThis preview shows page 22 - 39 out of 113 pages.\n\nplane. The angle of refraction, The angle of refraction, θ θ 2 , depends on the , depends on the properties of the medium. properties of the medium. Section 22.2",
null,
"When light moves from one medium into another, some is reflected at the boundary, and some is transmitted. The transmitted light is refracted (“bent”). a is the angle of incidence, and b is the angle of refraction. Refraction Refraction air (n a ) water (n b ) a b n b >n a incident ray r efracte d ray air (n b ) water (n a ) b a n a >n b inci d e nt r a y refracted ray",
null,
"air (n a ) water (n b ) a b n b >n a incident ray r efracte d ray Light passing from air (n 1) into water (n 1.33). Light “bends” towards the normal to the surface as it slows down in water.",
null,
"air (n b ) water (n a ) b a n a >n b inci d e nt r a y refracted ray Light passing from water (n 1.33) into air (n 1). Light “bends” away from the normal to the surface as it speeds up in air.",
null,
"The speed of light in a vacuum is c = 3x10 8 m/s. The index of refraction of a material is defined by c n = , v where c c is the speed of light in a vacuum and v v is the speed of light in the material. The speed and wavelength of light change when it passes from one medium to another, but not the frequency , so . n c v = and = n n",
null,
"The Index of Refraction The Index of Refraction When light passes from one medium to When light passes from one medium to another, it is refracted because the speed another, it is refracted because the speed of light is different in the two media. of light is different in the two media. The The index of refraction, n, index of refraction, n, of a medium of a medium can be defined can be defined Section 22.3",
null,
"Because light never travels faster than c, n 1. For water, n = 1.33 and for glass, n 1.5. Indices of refraction for several materials are listed in your text. c v = n 8 3×10 m/s v = 2.42 8 v = 1.24×10 m/s Example: calculate the speed of light in diamond (n = 2.42).",
null,
"Snell’s law , also called the law of refraction, gives the relationship between angles and indices of refraction: a a b b n sinθ = n sin θ . a b water (n b ) air (n a ) a b water (n b ) air (n a ) is the angle the ray makes with the normal! You are free to choose which is “a” and which is “b.”",
null,
"More About Refraction More About Refraction The angle of refraction depends upon the The angle of refraction depends upon the material and the angle of incidence. material and the angle of incidence. The path of the light through the refracting The path of the light through the refracting surface is reversible. surface is reversible. Section 22.2",
null,
"n 2 n 1 >n 2 Ray incident normal to surface is not “bent.” Ray incident normal to surface is not “bent.” Some is reflected, reflected, some is transmitted.",
null,
"n 2 n 1 >n 2 Increasing angle of incidence…",
null,
"n 2 n 1 >n 2 Increasing angle of incidence…more…",
null,
"n 2 n 1 >n 2 Increasing angle of incidence…more…critical angle reached… reached… some of incident energy is reflected, some is “transmitted along the boundary layer.",
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"n 2 n 1 >n 2 Light incident at any angle beyond C is totally internally reflected.",
null,
"n 2 n 1 >n 2 Total internal reflection explains why the surface of the water looks like a mirror and a coin disappearing trick.",
null,
"The picture below is for a prism, but it also explains why no light reaches your eyes from the coin under the water.",
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"application: fiber optics orts/andrea/report.html",
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"",
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"#### You've reached the end of your free preview.\n\nWant to read all 113 pages?\n\n• Spring '20\n• Light, Total internal reflection, Geometrical optics, rays\n•",
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"https://www.coursehero.com/assets/img/doc-landing/start-quote.svg",
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https://cbse.eduvictors.com/2015/06/cbse-class-12-computer-science-c-board.html | [
"# Board Questions On Header Files\n\n### There is 1 or 2 Marks questions based on function mapping to header files",
null,
"Q1(CBSE 2012): Write the names of the header file to which of the following belong:\n\n(i) sqrt( )\n(ii) randomize()\n\n(ii) stdlib.h\n\nQ2: Name the header file that will be needed for the following code:\n\n``` void main()\n{\nchar word [] = \"Board Examination\";\ncout << setw(20) << word;\n}\n```\n\nAnswer: iostream.h (for cout), iomanip.h (for setw)\n\nQ3(CBSE 2012): Which C++ header file(s) are especially required to be included to run the following C++ code. (Note: Do not include any header which is/are not required):\n\n``` void main()\n{\nchar TEXT[] = \"something\";\ncout << \"Remaining SMS chars: \" << 160 - strlen(TEXT) << endl;\n}\n```\n\nAnswer: iostream.h (for cout), string.h (for strlen)\n\nQ4(CBSE 2013): Observe the following C++ code and write name(s) of the header file(s) which will be essentially required to run in a C++ compiler:\n\n``` void main()\n{\nint number;\ncin >> number;\nif (abs(number) == number)\ncout << \"positive\" << endl;\n}\n```\n\nAnswer: iostream.h (for cin, cout), math.h (for abs)\n\nQ5(CBSE 2011): Which C++ header file(s) will be essentially required to be included to run/execute the following C++ code?\n\n``` void main()\n{\nchar name;\ncin >> name;\ntoupper(name);\ncout << name << endl;\n}\n```\n\nAnswer: iostream.h (for cin and cout), ctype.h (for toupper)\n\nQ6(CBSE 2009): Write the names of the header files to which the following belong:\n\n(i) puts()\n(ii) randomize()\n\n(i) stdio.h (puts)\n(ii) stdlib.h (randomize)\n\nQ7(CBSE 2009): Write the names of the header files to which the following belong:\n(i) setw()\n(ii) sqrt()\n\n(i) iomanip.h (setw)\n(ii) math.h (sqrt)\n\nQ8(CBSE 2004): Write the names of the header files to which the following belong:\n\n(i) sqrt()\n(ii) isalpha()\n(iii) puts()\n(iv) strcpy()\n\n(i) math.h\n(ii) ctype.h\n(iii) string.h\n(iv) stdio.h",
null,
"",
null,
"1.",
null,
"In question number 8, puts should be in stdio.h and stripy should be in string.h. Not vice versa.\n\n1.",
null,
"Exactly!!😎\n\n2.",
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"2.",
null,
"Kal board exam hai poore saal time pass nahi karna chaiye tha lol"
] | [
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"https://lh3.googleusercontent.com/blogger_img_proxy/ALY8t1vb2UNOPx-r9YntJ-QxHBOKItay99h9uAYLCeNVHq3LpDbd-_0BzYb__bkUwrthsBGgd28X6NQDHlRi-xIOVLx0QIMG70A=s0-d",
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"https://www.blogger.com/img/blogger_logo_round_35.png",
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"https://www.blogger.com/img/blogger_logo_round_35.png",
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https://articles.msc.edu.ph/2014/10/msc-10-multiplying-by-5-50-5-etc/ | [
"# MSC 10 – Multiplying by 5, 50, 5%, etc\n\nFive is 10 divided by 2. To multiply by 5, we can\n\n1. multiply first by 10 then divide the product by 2 or\n2. divide the multiplicand by 2 first and then add a zero at the end of the quotient.\n\nThis shortcut requires only the skill of dividing by 2.\n\nExcept when half of the multiplicand is readily computed, we recommend that the first procedure be followed because adjusting the decimal point first gives the magnitude of the product, i. e., hundreds or thousands.\n\nAs discussed in Short cut # 6, dividing by means getting half of the number\n\nHalf of 10 is 5.\n\nHalf of 20 is 10.\n\nHalf of 50 is 25.\n\nHalf of 64 is half of 60 plus half of 4 or 34.\n\nHalf of 72 is half of 60 plus half of 12 or 36.\n\nHalf of 96 is half of 80 plus half of 16 or 48.\n\nHalf of 7 ishalf of 6 plus half of 1 or 3.5\n\nExample 10.1: 36 x 5 =\n\nHalf of 36ishalf of20 plus half of16 which is 18. This becomes 180 when a zero is affixed at the end.\n\nExample 10.2: 245 x 5 =\n\nWhen a zero is affixed at the end, the number becomes 2,450. We can recite the answer as we mentally compute\n\na) half of 2 thousand is “1 thousand”,\nb) half of 4 hundred is “2 hundred”\nc) And half of 50 is “25”.\n\nExample 10.3: 455 x 5 =\n\n455 became 4,550 after multiplying by ten. We can think of it as 4000 + 400 + 150\n\na) Half of 4000 is “Two Thousand”\nb) Half of 400 is “two Hundred”\nc) Half of 150 is “seventy-five\n\nExample 10.4: 3,758 x 5 =\n\nHere all digits are odd except the last digit.\n\nAfter adding a zero at the end, the number becomes 37 thousand, 5 hundred and 80. We can think of it as 36,000 + 1,400 + 180 as in the previous example to get a quick 18,000 + 700 + 90 answer. Here we will try to divide it by 2 using the one-line method used in MSC 6.\n\n31715180 ÷ 2 = 18, 790\n\na) 3 divided 2 is 1 with a remainder of 1 which will become the first digit of the next dividend.\n\nb) 7 becomes 17 which when divided by 2 will give 8 with a remainder of 1. We can now start to say our answer aloud as “ eighteen thousand,”\n\nc) 5 now becomes 15 and when divided by 2 gives 7, remainder 1, so we continue as “seven hundred and”\n\nd) The last two digits of the dividend then becomes180 which when divided by 2 is “ninety”.\n\nExample10.5: 462 x 50 =\n\nFifty is half of 100 so we\n\na) Add two zeroes at the end of 462 so it will become 46,200.\n\nb) Half of that is Twenty three thousand one hundred.\n\nExample 10.6: 5% 0f 864=\n\n5% is half of 10% so we\n\na) move the decimal point one place to the right so that it will become 86.4\n\nb) divide it by 2 to get 43.2\n\nExercise 10: compute for the following products:\n\n1. ) 864 x 5 =\n2. ) 748 x 5 =\n3. ) 356 x 5 =\n4. ) 475 x 5 =\n5. ) 2,357 x 5 =\n6. ) 685 x 50 =\n7. ) 873 x 50 =\n8. ) 347 x 500 =\n9. ) 5% of 739 =\n10. ) 5% of 95.5 ="
] | [
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https://pyresample.readthedocs.io/en/stable/api/pyresample.bucket.html | [
"# pyresample.bucket package¶\n\n## Module contents¶\n\nCode for resampling using bucket resampling.\n\nclass pyresample.bucket.BucketResampler(target_area, source_lons, source_lats)\n\nBases: `object`\n\nBucket resampler.\n\nBucket resampling is useful for calculating averages and hit-counts when aggregating data to coarser scale grids.\n\nBelow are examples how to use the resampler.\n\nRead data using Satpy. The resampling can also be done (apart from fractions) directly from Satpy, but this demonstrates the direct low-level usage.\n\n```>>> from pyresample.bucket import BucketResampler\n>>> from satpy import Scene\n>>> from satpy.resample import get_area_def\n>>> fname = \"hrpt_noaa19_20170519_1214_42635.l1b\"\n>>> glbl = Scene(filenames=[fname])\n>>> data = glbl['4']\n>>> lons, lats = data.area.get_lonlats()\n>>> target_area = get_area_def('euro4')\n```\n\nInitialize the resampler\n\n```>>> resampler = BucketResampler(target_area, lons, lats)\n```\n\nCalculate the sum of all the data in each grid location:\n\n```>>> sums = resampler.get_sum(data)\n```\n\nCalculate how many values were collected at each grid location:\n\n```>>> counts = resampler.get_count()\n```\n\nThe average can be calculated from the above two results, or directly using the helper method:\n\n```>>> average = resampler.get_average(data)\n```\n\nCalculate fractions of occurrences of different values in each grid location. The data needs to be categorical (in integers), so we’ll create some categorical data from the brightness temperature data that were read earlier. The data are returned in a dictionary with the categories as keys.\n\n```>>> data = da.where(data > 250, 1, 0)\n>>> fractions = resampler.get_fractions(data, categories=[0, 1])\n>>> import matplotlib.pyplot as plt\n>>> plt.imshow(fractions); plt.show()\n```\nget_abs_max(data, fill_value=nan, skipna=True)\n\nCalculate absolute maximums for each bin with drop-in-a-bucket resampling.\n\nReturns for each bin the original signed value which has the largest absolute value.\n\nWarning\n\nThe slow `pandas.DataFrame.groupby()` method is temporarily used here, as the dask_groupby is still under development.\n\nParameters\n• data (Numpy or Dask array) – Data to be binned.\n\n• fill_value (number (optional)) – Value to use for empty buckets or all-NaN buckets.\n\n• skipna (boolean (optional)) – If True, skips NaN values for the maximum calculation (similarly to Numpy’s nanmax). Buckets containing only NaN are set to fill value. If False, sets the bucket to NaN if one or more NaN values are present in the bucket (similarly to Numpy’s max). In both cases, empty buckets are set to fill value. Default: True\n\nReturns\n\ndata – Bin-wise maximums in the target grid\n\nReturn type\n\nget_average(data, fill_value=nan, skipna=True)\n\nCalculate bin-averages using bucket resampling.\n\nParameters\n• data (Numpy or Dask array) – Data to be binned and averaged.\n\n• fill_value (float) – Fill value to mark missing/invalid values in the input data, as well as in the binned and averaged output data. Default: np.nan\n\n• skipna (bool) – If True, skips missing values (as marked by NaN or fill_value) for the average calculation (similarly to Numpy’s nanmean). Buckets containing only missing values are set to fill_value. If False, sets the bucket to fill_value if one or more missing values are present in the bucket (similarly to Numpy’s mean). In both cases, empty buckets are set to NaN. Default: True\n\nReturns\n\naverage – Binned and averaged data.\n\nReturn type\n\nget_count()\n\nCount the number of occurrences for each bin using drop-in-a-bucket resampling.\n\nReturns\n\ndata – Bin-wise count of hits for each target grid location\n\nReturn type\n\nget_fractions(data, categories=None, fill_value=nan)\n\nGet fraction of occurrences for each given categorical value.\n\nParameters\n• data (Numpy or Dask array) – Categorical data to be processed\n\n• categories (iterable or None) – One dimensional list of categories in the data, or None. If None, categories are determined from the data by fully processing the data and finding the unique category values.\n\n• fill_value (float) – Fill value to replace missing values. Default: np.nan\n\nget_max(data, fill_value=nan, skipna=True)\n\nCalculate maximums for each bin with drop-in-a-bucket resampling.\n\nParameters\n• data (Numpy or Dask array) – Data to be binned.\n\n• skipna (boolean (optional)) – If True, skips NaN values for the maximum calculation (similarly to Numpy’s nanmax). Buckets containing only NaN are set to zero. If False, sets the bucket to NaN if one or more NaN values are present in the bucket (similarly to Numpy’s max). In both cases, empty buckets are set to 0. Default: True\n\nReturns\n\ndata – Bin-wise maximums in the target grid\n\nReturn type\n\nget_min(data, fill_value=nan, skipna=True)\n\nCalculate minimums for each bin with drop-in-a-bucket resampling.\n\nParameters\n• data (Numpy or Dask array) – Data to be binned.\n\n• skipna (boolean (optional)) – If True, skips NaN values for the minimum calculation (similarly to Numpy’s nanmin). Buckets containing only NaN are set to zero. If False, sets the bucket to NaN if one or more NaN values are present in the bucket (similarly to Numpy’s min). In both cases, empty buckets are set to 0. Default: True\n\nReturns\n\ndata – Bin-wise minimums in the target grid\n\nReturn type\n\nget_sum(data, skipna=True)\n\nCalculate sums for each bin with drop-in-a-bucket resampling.\n\nParameters\n• data (Numpy or Dask array) – Data to be binned and summed.\n\n• skipna (boolean (optional)) – If True, skips NaN values for the sum calculation (similarly to Numpy’s nansum). Buckets containing only NaN are set to zero. If False, sets the bucket to NaN if one or more NaN values are present in the bucket (similarly to Numpy’s sum). In both cases, empty buckets are set to 0. Default: True\n\nReturns\n\ndata – Bin-wise sums in the target grid\n\nReturn type\n\npyresample.bucket.round_to_resolution(arr, resolution)\n\nRound the values in arr to closest resolution element.\n\nParameters\n• arr (list, tuple, Numpy or Dask array) – Array to be rounded\n\n• resolution (float) – Resolution unit to which data are rounded\n\nReturns\n\ndata – Source data rounded to the closest resolution unit\n\nReturn type"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.66237426,"math_prob":0.9399643,"size":5958,"snap":"2022-27-2022-33","text_gpt3_token_len":1474,"char_repetition_ratio":0.13654687,"word_repetition_ratio":0.37458926,"special_character_ratio":0.23699228,"punctuation_ratio":0.121442124,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99611634,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-08T09:34:50Z\",\"WARC-Record-ID\":\"<urn:uuid:d27ef99d-0fb6-48eb-ac1b-6fc9f0ba69d9>\",\"Content-Length\":\"34184\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:daa91780-c108-4805-9a46-bcdec27d5a1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:91681056-2330-4fa5-8490-4c3268e0aa6c>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://pyresample.readthedocs.io/en/stable/api/pyresample.bucket.html\",\"WARC-Payload-Digest\":\"sha1:F2JKKCP4NO4WS7BP5UGU57FYLJBXD43M\",\"WARC-Block-Digest\":\"sha1:DDVODGY55OOBEY4EU3FUDLWB4AUYM3A2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570793.14_warc_CC-MAIN-20220808092125-20220808122125-00291.warc.gz\"}"} |
https://docs.ruby-lang.org/en/3.0.0/Prime/TrialDivision.html | [
"# class Prime::TrialDivision\n\nInternal use. An implementation of prime table by trial division method.\n\n### Public Instance Methods\n\n[](index) click to toggle source\n\nReturns the +index+th prime number.\n\n`index` is a 0-based index.\n\n```# File lib/prime.rb, line 495\ndef [](index)\nwhile index >= @primes.length\n# Only check for prime factors up to the square root of the potential primes,\n# but without the performance hit of an actual square root calculation.\nif @next_to_check + 4 > @ulticheck_next_squared\n@ulticheck_index += 1\n@ulticheck_next_squared = @primes.at(@ulticheck_index + 1) ** 2\nend\n# Only check numbers congruent to one and five, modulo six. All others\n\n# are divisible by two or three. This also allows us to skip checking against\n# two and three.\[email protected] @next_to_check if @primes[2..@ulticheck_index].find {|prime| @next_to_check % prime == 0 }.nil?\n@next_to_check += 4\[email protected] @next_to_check if @primes[2..@ulticheck_index].find {|prime| @next_to_check % prime == 0 }.nil?\n@next_to_check += 2\nend\n@primes[index]\nend```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6483645,"math_prob":0.65298164,"size":1026,"snap":"2021-04-2021-17","text_gpt3_token_len":277,"char_repetition_ratio":0.19275929,"word_repetition_ratio":0.125,"special_character_ratio":0.2777778,"punctuation_ratio":0.16,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95851815,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T22:15:02Z\",\"WARC-Record-ID\":\"<urn:uuid:e1d3be09-9af9-4251-ad85-6dd393c2054e>\",\"Content-Length\":\"7918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4c0e4aaa-cd36-47ab-81a9-69d0f55c4069>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e2dac09-2d3d-467b-a3b1-f99a6ba6946b>\",\"WARC-IP-Address\":\"151.101.201.178\",\"WARC-Target-URI\":\"https://docs.ruby-lang.org/en/3.0.0/Prime/TrialDivision.html\",\"WARC-Payload-Digest\":\"sha1:SFJAMJPKZ5E7DHOW3VKOLBYUWVSEL5HN\",\"WARC-Block-Digest\":\"sha1:JYYGPVBSLMD5FVTYK4UJVVAOOXF3THFY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703557462.87_warc_CC-MAIN-20210124204052-20210124234052-00553.warc.gz\"}"} |
http://encyclopedia.kids.net.au/page/fu/Fundamental_Theorem_of_Algebra | [
"",
null,
"",
null,
"## Encyclopedia > Fundamental Theorem of Algebra\n\nArticle Content\n\n# Fundamental theorem of algebra\n\nRedirected from Fundamental Theorem of Algebra\n\nThe fundamental theorem of algebra (now considered something of a misnomer by many mathematicians) states that every complex polynomial of degree n has exactly n zeroes, counted with multiplicity. More formally, if\n\n[itex]p(z)=z^n+a_{n-1}z^{n-1}+\\cdots+a_0[/itex]\n(where the coefficients a0, ..., an-1 can be real or complex numbers), then there exist (not necessarily distinct) complex numbers z1, ..., zn such that\n[itex]p(z)=(z-z_1)(z-z_2)\\cdots(z-z_n).[/itex]\n\nThis shows that the field of complex numbers, unlike the field of real numbers, is algebraically closed. An easy consequence is that the product of all the roots equals (-1)n a0 and the sum of all the roots equals -an-1.\n\nThe theorem had been conjectured in the 17th century but could not be proved since the complex numbers had not yet been firmly grounded. The first rigorous proof was given by Carl Friedrich Gauss in the early 19th century. (An almost complete proof had been given earlier by d'Alembert.) Gauss produced several different proofs throughout his lifetime. It is possible to prove the theorem by using only algebraic methods, but nowadays the proof based on complex analysis seems most natural. The difficult step in the proof is to show that every non-constant polynomial has at least one zero. This can be done by employing Liouville's theorem which states that a bounded function which is holomorphic in the entire complex plane must be constant. By starting with a polynomial p without any zeros, one can pass to the holomorphic function 1/p and Liouville's theorem then yields that 1/p and therefore also p are constant.\n\nAll Wikipedia text is available under the terms of the GNU Free Documentation License\n\nSearch Encyclopedia\n Search over one million articles, find something about almost anything!\n\nFeatured Article\n Battle Creek, Michigan ... out with 27.2% under the age of 18, 8.7% from 18 to 24, 29.5% from 25 to 44, 21.0% from 45 to 64, and 13.5% who are 65 years of age or older. The median age is 35 years. ...",
null,
"",
null,
""
] | [
null,
"http://www.kids.net.au/images/spacer.gif",
null,
"http://www.kids.net.au/images/spacer.gif",
null,
"http://www.kids.net.au/images/spacer.gif",
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"http://www.kids.net.au/images/spacer.gif",
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http://unitconverter.io/square-hectameters/square-millimiters/3791 | [
"",
null,
"# 3,791 square hectometers to square millimeters\n\nto\n\n3,791 Square hectometers = 37,910,000,000,000 Square millimeters\n\nThis conversion of 3,791 square hectometers to square millimeters has been calculated by multiplying 3,791 square hectometers by 10,000,000,000 and the result is 37,910,000,000,000 square millimeters."
] | [
null,
"http://unitconverter.io/img/area.png",
null
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https://wiki.zcubes.com/Manuals/calci/LOGSTAR | [
"# Manuals/calci/LOGSTAR\n\nLOGSTAR (Number,Base)\n\n•",
null,
"is any real number.\n•",
null,
"base value of the log.\n\n## Description\n\n• This function shows the iterated logarithm of the given number.\n• In",
null,
",",
null,
"is any real number.\n•",
null,
"is the base value of the log.\n• Log star is the number of times the logarithm function must be iteratively applied before the result is less than or equal to 1.\n• The simple definition of the Log star is :",
null,
"• The relation between Log star and Super logarithm is",
null,
"for positive real numbers.\n• Also for negative real numbers ,log star value is 0,whereas",
null,
"for positive x.\n• The iterated logarithm varies at slow rate,which is very slow compared with logarithm values.\n• Iterated logarithm values vary not more than 5 with base value 2.\nX Log*X\n(",
null,
",1] 0\n(1,2] 1\n(2,4] 2\n(4,16] 3\n(16, 65536] 4\n(65536,",
null,
"] 5\n\n## Examples\n\n1. LOGSTAR(3) = 1\n2. LOGSTAR(260) = 2\n3. LOGSTAR(5432,3) = 3\n\n## Related Videos\n\nIterated Logarithm"
] | [
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/1eb47112241f697bd161c2cb3e9470feff0b3fc7",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/5fca0ccaa9d325ec9c76ca39fea4d3ec378bb20b",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/28f14fe87cdd00e389405b1de4281612c11189eb",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/1eb47112241f697bd161c2cb3e9470feff0b3fc7",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/5fca0ccaa9d325ec9c76ca39fea4d3ec378bb20b",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/3f8552e2be63517ee0ccdd9d83dd493c46e2876b",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/e2d6c004a454dd559a8503f491002fc0b2177501",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/64cf0401ebfa8eb58dcdf5d8b99ba619db722902",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/ca2608c4b5fd3bffc73585f8c67e379b4e99b6f1",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/png/3b558519004f9a2d511933e6a1d2d30b77534efd",
null
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https://www.colorhexa.com/01f1c2 | [
"# #01f1c2 Color Information\n\nIn a RGB color space, hex #01f1c2 is composed of 0.4% red, 94.5% green and 76.1% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 19.5% yellow and 5.5% black. It has a hue angle of 168.3 degrees, a saturation of 99.2% and a lightness of 47.5%. #01f1c2 color hex could be obtained by blending #02ffff with #00e385. Closest websafe color is: #00ffcc.\n\n• R 0\n• G 95\n• B 76\nRGB color chart\n• C 100\n• M 0\n• Y 20\n• K 5\nCMYK color chart\n\n#01f1c2 color description : Vivid cyan.\n\n# #01f1c2 Color Conversion\n\nThe hexadecimal color #01f1c2 has RGB values of R:1, G:241, B:194 and CMYK values of C:1, M:0, Y:0.2, K:0.05. Its decimal value is 127426.\n\nHex triplet RGB Decimal 01f1c2 `#01f1c2` 1, 241, 194 `rgb(1,241,194)` 0.4, 94.5, 76.1 `rgb(0.4%,94.5%,76.1%)` 100, 0, 20, 5 168.3°, 99.2, 47.5 `hsl(168.3,99.2%,47.5%)` 168.3°, 99.6, 94.5 00ffcc `#00ffcc`\nCIE-LAB 85.407, -58.688, 9.283 41.202, 66.808, 61.76 0.243, 0.394, 66.808 85.407, 59.418, 171.012 85.407, -70.723, 23.382 81.736, -53.059, 12.416 00000001, 11110001, 11000010\n\n# Color Schemes with #01f1c2\n\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #f10130\n``#f10130` `rgb(241,1,48)``\nComplementary Color\n• #01f14a\n``#01f14a` `rgb(1,241,74)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #01a8f1\n``#01a8f1` `rgb(1,168,241)``\nAnalogous Color\n• #f14a01\n``#f14a01` `rgb(241,74,1)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #f101a8\n``#f101a8` `rgb(241,1,168)``\nSplit Complementary Color\n• #f1c201\n``#f1c201` `rgb(241,194,1)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #c201f1\n``#c201f1` `rgb(194,1,241)``\n• #30f101\n``#30f101` `rgb(48,241,1)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #c201f1\n``#c201f1` `rgb(194,1,241)``\n• #f10130\n``#f10130` `rgb(241,1,48)``\n• #01a585\n``#01a585` `rgb(1,165,133)``\n• #01be99\n``#01be99` `rgb(1,190,153)``\n• #01d8ae\n``#01d8ae` `rgb(1,216,174)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #0efecf\n``#0efecf` `rgb(14,254,207)``\n• #27fed4\n``#27fed4` `rgb(39,254,212)``\n• #40fed9\n``#40fed9` `rgb(64,254,217)``\nMonochromatic Color\n\n# Alternatives to #01f1c2\n\nBelow, you can see some colors close to #01f1c2. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01f186\n``#01f186` `rgb(1,241,134)``\n• #01f19a\n``#01f19a` `rgb(1,241,154)``\n• #01f1ae\n``#01f1ae` `rgb(1,241,174)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #01f1d6\n``#01f1d6` `rgb(1,241,214)``\n• #01f1ea\n``#01f1ea` `rgb(1,241,234)``\n• #01e4f1\n``#01e4f1` `rgb(1,228,241)``\nSimilar Colors\n\n# #01f1c2 Preview\n\nThis text has a font color of #01f1c2.\n\n``<span style=\"color:#01f1c2;\">Text here</span>``\n#01f1c2 background color\n\nThis paragraph has a background color of #01f1c2.\n\n``<p style=\"background-color:#01f1c2;\">Content here</p>``\n#01f1c2 border color\n\nThis element has a border color of #01f1c2.\n\n``<div style=\"border:1px solid #01f1c2;\">Content here</div>``\nCSS codes\n``.text {color:#01f1c2;}``\n``.background {background-color:#01f1c2;}``\n``.border {border:1px solid #01f1c2;}``\n\n# Shades and Tints of #01f1c2\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, #000705 is the darkest color, while #f2fffc is the lightest one.\n\n• #000705\n``#000705` `rgb(0,7,5)``\n• #001a15\n``#001a15` `rgb(0,26,21)``\n• #002e25\n``#002e25` `rgb(0,46,37)``\n• #004134\n``#004134` `rgb(0,65,52)``\n• #005544\n``#005544` `rgb(0,85,68)``\n• #006854\n``#006854` `rgb(0,104,84)``\n• #017c64\n``#017c64` `rgb(1,124,100)``\n• #018f73\n``#018f73` `rgb(1,143,115)``\n• #01a383\n``#01a383` `rgb(1,163,131)``\n• #01b693\n``#01b693` `rgb(1,182,147)``\n• #01caa3\n``#01caa3` `rgb(1,202,163)``\n• #01ddb2\n``#01ddb2` `rgb(1,221,178)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\n• #08fece\n``#08fece` `rgb(8,254,206)``\n• #1bfed2\n``#1bfed2` `rgb(27,254,210)``\n• #2ffed6\n``#2ffed6` `rgb(47,254,214)``\n• #42fed9\n``#42fed9` `rgb(66,254,217)``\n• #56fedd\n``#56fedd` `rgb(86,254,221)``\n• #69fee1\n``#69fee1` `rgb(105,254,225)``\n• #7dfee5\n``#7dfee5` `rgb(125,254,229)``\n• #90ffe9\n``#90ffe9` `rgb(144,255,233)``\n• #a4ffed\n``#a4ffed` `rgb(164,255,237)``\n• #b7fff1\n``#b7fff1` `rgb(183,255,241)``\n• #cbfff5\n``#cbfff5` `rgb(203,255,245)``\n• #dffff9\n``#dffff9` `rgb(223,255,249)``\n• #f2fffc\n``#f2fffc` `rgb(242,255,252)``\nTint Color Variation\n\n# Tones of #01f1c2\n\nA tone is produced by adding gray to any pure hue. In this case, #71817e is the less saturated color, while #01f1c2 is the most saturated one.\n\n• #71817e\n``#71817e` `rgb(113,129,126)``\n• #678b84\n``#678b84` `rgb(103,139,132)``\n• #5e9489\n``#5e9489` `rgb(94,148,137)``\n• #559d8f\n``#559d8f` `rgb(85,157,143)``\n• #4ba795\n``#4ba795` `rgb(75,167,149)``\n• #42b09a\n``#42b09a` `rgb(66,176,154)``\n• #39b9a0\n``#39b9a0` `rgb(57,185,160)``\n• #30c2a6\n``#30c2a6` `rgb(48,194,166)``\n• #26ccab\n``#26ccab` `rgb(38,204,171)``\n• #1dd5b1\n``#1dd5b1` `rgb(29,213,177)``\n• #14deb7\n``#14deb7` `rgb(20,222,183)``\n• #0ae8bc\n``#0ae8bc` `rgb(10,232,188)``\n• #01f1c2\n``#01f1c2` `rgb(1,241,194)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01f1c2 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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