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https://virtualnerd.com/algebra-1/linear-inequalities/solve-by-multiplication-division/solve-by-multiplication/word-problem-multiply-negative-example | [
"# How Do You Use Multiplication with Negative Numbers to Solve an Inequality Word Problem?\n\n### Note:\n\nThis tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality! That's one of the big differences between solving equalities and solving inequalities.\n\n### Keywords:\n\n• problem\n• word problem\n• word\n• translate words to algebra\n• inequality\n• solve\n• solve by multiplication\n• multiplication\n• multiplication property of inequality\n• single variable\n• 1 variable\n• 1 step\n• single step\n• negative numbers\n• negative\n• flip inequality"
] | [
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https://samuelstevens.me/writing/python-multiprocessing | [
"# Behold, My Stuff\n\n[Home] [Writing] [CV] [GitHub] [Email]\n\n# `multiprocessing.Queue` in Python\n\nI tried to use the `multiprocessing` version of `Queue` rather than the threaded version (`queue.Queue`) and found that without `task_done()` and `Queue.join()` I didn’t understand how to actually end a queue. This is my approach, followed by how the docs do it.\n\nThe key understanding that I missed: processees exit once their target finishes. If you know that, this probably won’t be useful.\n\nI didn’t see how to exit a queue without `task_done` and `join`. However, if you know that a process exits once it finishes it’s target, it becomes clearer. For example, assume we have 2 producers and 2 consumers:\n\n``````import time\nfrom multiprocessing import Queue, Process\n\ndef produce(q: \"Queue[int]\", length: int) -> None:\nfor _ in range(length):\nq.put(3)\n\ndef consume(q: \"Queue[int]\") -> None:\nwhile True:\nnum = q.get()\nprint(f\"Sleeping for {num} seconds.\")\ntime.sleep(num) # expensive work\n\n# q.task_done() would go here! How do we know to exit?\n\ndef main() -> None:\nq: \"Queue[int]\" = Queue()\n\nfor _ in range(2):\nc = Process(target=consume, args=(q,))\nc.start()\n\nfor _ in range(2):\np = Process(target=produce, args=(q, 5))\np.start()\n\nfor _ in range(2):\np.join()\n\nfor _ in range(2):\nc.join()\n\nif __name__ == \"__main__\":\nmain()``````\n\nRunning this program will never exit because of `while True:` in `consume`.\n\nTo satisfy `mypy`, `Queue` needs a type parameter. However, `Queue` cannot be indexed via `[]`, so you can put the whole type in quotes so that it’s not evaluated at runtime.\n\nHow does `consume` know to exit? It can’t check if `q.empty` because according to the docs it’s not reliable. We could just use `get_nowait()` and exit if it raises an exception. But can the queue ever be empty without it being finished? Yes, so just checking if it’s empty isn’t a reliable way to end the program either.\n\nThe solution is to send some sort of “stop”-value that tells the consumer that it’s done working. For example, if we were to use negative values as a stop value:\n\n``````def consume(q: \"Queue[int]\") -> None:\nwhile True:\nnum = q.get()\nif num < 0: # sentinel value\nbreak\n\nprint(f\"Sleeping for {num} seconds.\")\ntime.sleep(num) # expensive work``````\n\nNow if we modify `produce`:\n\n``````def produce(q: \"Queue[int]\", length: int) -> None:\nfor _ in range(length):\nq.put(3)\n\nq.put(-1) # stop-value``````\n\nNow we can run `time python queue_demo.py` and see that it takes less than 30 seconds (3 seconds * 5 elements produced * 2 producers). It’s not perfect (should be exactly 15 seconds), but it’s definitely faster than in a single process.\n\nHere’s the final program, licensed under GNU AGPLv3. If you have any improvements/suggestions, I can be reached at samuel.robert.stevens@gmail.com\n\n``````import time\nfrom multiprocessing import Process, Queue\n\ndef produce(q: \"Queue[int]\", length: int) -> None:\nfor _ in range(length):\nq.put(3)\n\nq.put(-1) # stop-value\n\ndef consume(q: \"Queue[int]\") -> None:\nwhile True:\nnum = q.get()\nif num < 0: # sentinel value\nbreak\n\nprint(f\"Sleeping for {num} seconds.\")\ntime.sleep(num) # expensive work\n\ndef main() -> None:\nq: \"Queue[int]\" = Queue()\n\nfor _ in range(2):\nc = Process(target=consume, args=(q,))\nc.start()\n\nfor _ in range(2):\np = Process(target=produce, args=(q, 5))\np.start()\n\nfor _ in range(2):\np.join()\n\nfor _ in range(2):\nc.join()\n\nif __name__ == \"__main__\":\nmain()``````"
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https://www.geeksforgeeks.org/program-to-check-if-n-is-a-centered-hexagonal-number/ | [
"Related Articles\n\n# Program to check if N is a Centered Hexagonal Number\n\n• Last Updated : 16 Jul, 2021\n\nGiven an integer N, the task is to check if N is a Centered Hexagonal Number or not. If the number N is a Centered Hexagonal Number then print “Yes” else print “No”.\n\nCentered hexagonal number are figurate numbers and are in the form of the Hexagon. The Centered Hexagonal number is different from Hexagonal Number because it contains one element at the center..The first few Centered hexagonal numbers are 1, 7, 19, 37, 61, 91, 127 …\n\nExamples:\n\nInput: N = 7\nOutput: Yes\nExplanation:\nSecond Centered hexagonal number is 7.\n\nInput: N = 20\nOutput: No\n\nApproach:\n\n• The Kth term of the Centered hexagonal number is given as",
null,
"• As we have to check that the given number can be expressed as a Centered hexagonal number or not. This can be checked as:\n\n=>",
null,
"=>",
null,
"• If the value of K calculated using the above formula is an integer, then N is a Centered Hexagonal Number.\n• Else the number N is not a Centered Hexagonal Number.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n // C++ program for the above approach#include using namespace std; // Function to check that the// number is a Centered hexagonal numberbool isCenteredhexagonal(int N){ float n = (3 + sqrt(12 * N - 3)) / 6; // Condition to check if the // number is a Centered hexagonal number return (n - (int)n) == 0;} // Driver Codeint main(){ int N = 7; // Function call if (isCenteredhexagonal(N)) { cout << \"Yes\"; } else { cout << \"No\"; } return 0;}\n\n## Java\n\n // Java program for the above approachclass GFG{ // Function to check that the// number is a Centered hexagonal numberstatic boolean isCenteredhexagonal(int N){ float n = (float)((3 + Math.sqrt(12 * N - 3)) / 6); // Condition to check if the // number is a Centered hexagonal number return (n - (int)n) == 0;} // Driver Codepublic static void main(String[] args){ int N = 7; // Function call if (isCenteredhexagonal(N)) { System.out.print(\"Yes\"); } else { System.out.print(\"No\"); }}} // This code is contributed by sapnasingh4991\n\n## Python3\n\n # Python3 program for the above approachimport math # Function to check that the number# is a centered hexagonal numberdef isCenteredhexagonal(N): n = (3 + math.sqrt(12 * N - 3)) / 6 # Condition to check if the number # is a centered hexagonal number return (n - int(n)) == 0 # Driver CodeN = 7 if isCenteredhexagonal(N): print(\"Yes\")else : print(\"No\") # This code is contributed by ishayadav181\n\n## C#\n\n // C# program for the above approachusing System; class GFG{ // Function to check that the number// is a centered hexagonal numberstatic bool isCenteredhexagonal(int N){ float n = (float)((3 + Math.Sqrt(12 * N - 3)) / 6); // Condition to check if the number // is a centered hexagonal number return (n - (int)n) == 0;} // Driver Codepublic static void Main(String[] args){ int N = 7; // Function call if (isCenteredhexagonal(N)) { Console.Write(\"Yes\"); } else { Console.Write(\"No\"); }}} // This code is contributed by amal kumar choubey\n\n## Javascript\n\n \nOutput:\nYes\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"
] | [
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"https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-c78758d64ecdc5cca0b05023c410d2c3_l3.png",
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"https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-7accc3126af69a9a257fa49e3b6844d9_l3.png",
null,
"https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-9105632ea520d0c9aa5d088062adc313_l3.png",
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https://www.quantum-espresso.stage.24h.it/Doc/INPUT_BANDS.html | [
"# Input File Description\n\n## Program: bands.x / PWscf / Quantum Espresso (version: 7.0)\n\nINTRODUCTION\n\n&BANDS\n\nprefix | outdir | filband | spin_component | lsigma | lp | filp | lsym | no_overlap | plot_2d | firstk | lastk\n\n### INTRODUCTION\n\nPurpose of bands.x:\nRe-order bands, computes band-related properties. Currently,\nre-ordering can be done with two different algorithms:\n(a) by maximising the overlap with bands at previous k-point\n(b) by computing symmetry properties of each wavefunction\nBands-related properties that can be computed are currently\n(a) The expectation value of the spin operator on each spinor\nwave-function (noncolinear case only)\n(b) The expectation value of p\n\nThe input data can be read from standard input or from file using\ncommand-line options \"bands.x -i file-name\" (same syntax as for pw.x)\n\nOutput files:\n- file filband containing the band structure, in a format\nsuitable for plotting code \"plotband.x\"\n- file \"filband\".rap (if lsym is .t.) with symmetry information,\nto be read by plotting code \"plotband.x\"\n- if (lsigma(i)): file \"filband\".i, i=1,2,3, with expectation values\nof the spin operator in the noncolinear case\n- file \"filband\".gnu with bands in eV, directly plottable using gnuplot\n- file filp with matrix elements of p (including the nonlocal potential\ncontribution i*m*[V_nl,x])\n\nStructure of the input data:\n============================\n\n&BANDS\n...\n/\n\n\n## Namelist: &BANDS\n\n prefix CHARACTER Default: 'pwscf' prefix of files saved by program pw.x \n outdir CHARACTER Default: value of the ESPRESSO_TMPDIR environment variable if set; current directory ('./') otherwise directory containing the input data, i.e. the same as in pw.x \n filband CHARACTER Default: 'bands.out' file name for band output (to be read by \"plotband.x\") \nspin_component INTEGER In the lsda case select: 1 = spin-up 2 = spin-down \nlsigma(i), i=1,3 LOGICAL If true computes expectation values of the spin operator on the spinor wave-functions (only in the noncollinear case), writes them to a file \"filband\".i, i=1,2,3 \n lp LOGICAL Default: .false. If .true. matrix elements of the momentum operator p between conduction and valence bands are computed and written to file specified in filp. The matrix elements include the contribution from the nonlocal potential, i*m*[V_nl, x]. In other words, the calculated matrix elements are those of the velocity operator i*m*[H, x] times mass, not those of the true momentum operator. \n filp CHARACTER Default: 'p_avg.dat' If lp is set to .true., file name for matrix elements of p \n lsym LOGICAL Default: .true. If .true. the bands are classified according to the irreducible representations of the small group of k. A file \"filband\".rap with the same format of \"filband\" is written, for usage by \"plotband.x\" \n no_overlap LOGICAL Default: .true. If .false., and if lsym is .false., writes the eigenvalues in the order that maximises overlap with the neighbor k-points \n plot_2d LOGICAL Default: .false. If .true. writes the eigenvalues in the output file in a 2D format readable by gnuplot. Band ordering is not changed. Each band is written in a different file called filband.# with the format: xk, yk, energy xk, yk, energy .. .. .. energies are written in eV and xk in units 2\\pi/a. \n firstk, lastk INTEGER if lsym=.true. makes the symmetry analysis only for k points between firstk to lastk \nThis file has been created by helpdoc utility on Sat Dec 18 20:08:46 CET 2021."
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https://www.vexforum.com/t/how-do-i-make-tank-control-with-fixed-speed-percentage/77508 | [
"# How do I make tank control with fixed speed percentage?\n\nint main() {\n// Initializing Robot Configuration. DO NOT REMOVE!\nvexcodeInit();\n\n// Deadband stops the motors when Axis values are close to zero.\n\nwhile (true) {\n// Get the velocity percentage of the left motor. (Axis3)\nint leftMotorSpeed = Controller1.Axis3.position();\n// Get the velocity percentage of the right motor. (Axis2)\nint rightMotorSpeed = Controller1.Axis1.position();\n\n``````// Set the speed of the left motor. If the value is less than the deadband,\n// set it to zero.\n// Set the speed to zero.\nLeftMotor.setVelocity(0, percent);\n} else {\n// Set the speed to leftMotorSpeed\nLeftMotor.setVelocity(leftMotorSpeed, percent);\n}\n\n// Set the speed of the right motor. If the value is less than the deadband,\n// set it to zero.\n// Set the speed to zero\nRightMotor.setVelocity(0, percent);\n} else {\n// Set the speed to rightMotorSpeed\nRightMotor.setVelocity(rightMotorSpeed, percent);\n}\n\nelse if(x != 1){\n\n}\n\n// Spin both motors in the forward direction.\nLeftMotor.spin(forward);\nRightMotor.spin(forward);\n\nwait(25, msec);\n``````\n\n}\n}\n\nSo you see, I want to be able to move the robot at a fixed speed, instead of it being proportional to the joystick. For example, if you see whats inside the first else statement for leftMotorSpeed, it averages the speed of the robot depending on how far the joystick is. But I dont want that to happen, no matter how far the joystick is, I want the robot to go at a fixed speed of, lets say, 50 percent. How do I do that?\n\nYou can simply do this…\n\nEdit: this is wrong check below\n\n``````if (abs(leftMotorSpeed) < deadband) {\n// Set the speed to zero.\nLeftMotor.setVelocity(0, percent);\n} else {\n// Set the speed to 50\nLeftMotor.setVelocity(50, percent);\n}\n``````\n\nNo matter what the joystick value is if it is greater than 5 the velocity will be set to 50.\n\n1 Like\n\nbut you see I tried that, but for some reason it wont go backwards\n\nOhh yeah sorry, I forget that part too…\n\n``````if (abs(leftMotorSpeed) < deadband) {\n// Set the speed to zero.\nLeftMotor.setVelocity(0, percent);\n} else if(leftMotorSpeed > 0) {\n// Set the speed to 50\nLeftMotor.setVelocity(50, percent);\n} else {\n// Set the speed to -50\nLeftMotor.setVelocity(-50, percent);\n}\n``````\n\nIf the joystick value is negative the percentage assigned to the motor should be negative too.\n\n5 Likes\n\nwait wait wait, im confused, why didnt you add “abs” behind leftMotorSpeed in the else if statement?\n\nIt is to check weather the value is positive or negative. You want to be able to go forwards and backwards.\n\n2 Likes\n\nthank you so much!!!"
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https://brilliant.org/problems/a-truefalse-question/ | [
"# A true/false question\n\nAlgebra Level 1\n\nIf x and y are real numbers, $(x+y)^{2}$ is always equal to $(x^{2} + y^{2})$\n\n×"
] | [
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https://www.geeksforgeeks.org/class-9-rd-sharma-solutions-chapter-8-introduction-to-lines-and-angles-exercise-8-2-set-2/?ref=leftbar-rightbar | [
"# Class 9 RD Sharma Solutions – Chapter 8 Introduction to Lines and Angles- Exercise 8.2 | Set 2\n\n• Last Updated : 04 May, 2021\n\n### Question 11. In Figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x.",
null,
"Solution:\n\nAttention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.\n\nThe students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.\n\nIt is given that ACB is a line in the figure given below.\n\nThus, ∠ACD and ∠BCD form a linear pair.\n\nTherefore, their sum must be equal toh 180°.\n\nOr we can say that\n\n∠ACD + ∠BCD = 180°\n\nAlso, ∠ACD = 4x and ∠BCD = 5x.\n\nThis further simplifies to:\n\n4x + 5x = 180\n\n9x = 180\n\nx =",
null,
"x = 20°\n\n### Question 12. In the given figure, ∠POR = 3x and ∠QOR = 2x + 10, find the value of x for which POQ will be a line.",
null,
"Solution:\n\nHere we have POQ as a line\n\nSo, ∠POR and ∠QOR form a linear pair.\n\nTherefore, their sum must be equal to 180°\n\nOr\n\n∠POR + ∠QOR = 180°\n\nIt is given that ∠POR = (3x)° and ∠QOR = (2x + 10)°. On substituting these values above we get,\n\n3x + (2x + 10) = 180°\n\n3x + 2x + 10 = 180°\n\n5x + 10 = 180°\n\n5x = 180 – 10\n\n5x = 170\n\nx =",
null,
"x = 34°\n\n### Question 13. In the given figure, a is greater than b by one-third of a right-angle. Find the values of a and b.",
null,
"Solution:\n\nIt is given that in the figure given below; a is greater than b by one-third of a right angle.\n\nOr we can say that, the difference between a and b is",
null,
"i.e.\n\na – b =",
null,
"a – b = 30° ……..(i)\n\nAlso a and b forn a linear pair. Therefore, their sum must be equal to 180°.\n\nWe can say that:\n\na + b = 180° ……….(ii)\n\nOn adding (i) and (ii), we get:\n\n2a = 180 + 30\n\n2a = 210\n\na =",
null,
"a = 105°\n\nOn putting a = 105 in (i)\n\n105 – b = 30\n\n-b = 30 – 105\n\n-b = -75\n\nb = 75°\n\nHence, a = 105° and b = 75°\n\n### Question 14. What value of y would make AOB a line in the given figure, if ∠AOC = 4y and ∠BOC = (6y + 30)",
null,
"Solution:\n\nLet us assume, AOB as a straight line.\n\nThis makes ∠AOC and ∠BOC to form a linear pair. Therefore, their sum must be equal to 180°.\n\nWe can say that:\n\n∠AOC + ∠BOC = 180°\n\nAlso, ∠AOC = 4y and ∠BOC = 6y + 30. This further simplifies to:\n\n4y + (6y + 30) = 180\n\n10y + 30 = 180\n\n10y = 180 – 30\n\n10y = 150\n\ny =",
null,
"y = 15°\n\nThus, the value of y = 15° make AOB as a line.\n\n### Question 15. If the given figure, ∠AOF and ∠FOG form a linear pair.\n\n∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°",
null,
"(i) Find the measure of ∠FOE, ∠COB and ∠DOE.\n\n(ii) Name all the right angles.\n\n(iii) Name three pairs of adjacent complementary angles.\n\n(iv) Name three pairs of adjacent supplementary angles.\n\n(v) Name three pairs of adjacent angles.\n\nSolution:\n\nThe given figure is as follows:\n\n(i) It is given that ∠AOB, ∠FOE, ∠EOB and ∠FOG form a linear pair.\n\nTherefore, their sum must be equal to 180°\n\ni.e.\n\n∠AOB + ∠FOE + ∠EOB + ∠FOG = 180°\n\nIt is given that:\n\n∠FOG = 30°\n\n∠AOB = 30°\n\n∠EOB = 90° in equation above, we get:\n\n∠AOB + ∠FOE + ∠EOB + ∠FOG = 180°\n\n30° + ∠FOE + 90° + 30° = 180°\n\n∠FOE + 150° = 180°\n\n∠FOE = 180° – 150°\n\n∠FOE = 30°\n\nIt is given that\n\n∠FOC = 90°\n\nFrom the above figure:\n\n∠FOE + ∠DOE + ∠COD = 90°\n\n30° + ∠DOE + 30° = 90°\n\n∠DOE + 60° = 90°\n\n∠DOE = 90° – 60°\n\n∠DOE = 30°\n\nSimilarly, we have:\n\n∠EOB = 90°\n\nFrom the above figure:\n\n∠DOE + ∠DOC + ∠COB = 90°\n\n30° + 30° + ∠COB = 90°\n\n∠COB + 60° = 90°\n\n∠COB = 90° – 60°\n\n∠COB = 30°\n\n(ii) We have:\n\n∠FOG = 30°\n\n∠FOE = 30°\n\n∠EOD = 30°\n\n∠COD = 30°\n\n∠COB = 30°\n\n∠AOB = 30°\n\nFrom the figure above and the measurements of the calculated angles we get two right angles as ∠DOG and ∠AOD.\n\nTwo right angles are already given as ∠FOC and ∠EOB\n\n(iii) We have to find the three pair of adjacent complementary angles.\n\nWe know that ∠EOB is a right angle.\n\nTherefore,\n\n∠EOC and ∠COB are complementary angles.\n\nSimilarly, ∠AOD is a right angle.\n\nTherefore,\n\n∠AOC and ∠COD are complementary angles.\n\n(iv) We have to find the three pair of adjacent supplementary angles.\n\nSince, ∠AOG is a straight line.\n\nTherefore, following are the three linear pair, which are supplementary:\n\n∠AOB and ∠BOG\n\n∠AOC and ∠COG\n\n∠AOD and ∠DOG\n\n(v) We have to find pair of adjacent angles, which are as follows:\n\n∠AOB and ∠BOC\n\n∠COD and ∠DOE\n\n∠EOF and ∠FOG\n\n### Question 16. In the given figure, OP, OQ , OR, and OS are four rays. Prove that:\n\n∠POQ + ∠QOR + ∠SOR + ∠POS = 360°.",
null,
"Solution:\n\nLet us draw TOP as a straight line.",
null,
"Since, TOP is a line, therefore, ∠POQ, ∠QOR and ∠ROT form a linear pair.\n\nAlso, ∠POS and ∠SOT form a linear pair.\n\nThus, we have:\n\n∠POQ + ∠QOR + ∠ROT = 180° ……(i)\n\nand\n\n∠POS + ∠SOT = 180° …….(ii)\n\nOn adding (i) and (ii), we get;\n\n(∠POQ + ∠QOR + ∠ROT) + (∠POS + ∠SOT) = 180° + 180°\n\n∠POQ + ∠QOR + (∠ROT + ∠SOT) + ∠POS = 360°\n\n∠POQ + ∠QOR + ∠SOR + ∠POS = 360°\n\nHence proved.\n\n### Question 17. In the given figure, ray OS stand on a line POQ, Ray OR and ray OT are angle bisectors of ∠POS and∠SOQ respectively. If ∠POS = x, find ∠ROT.",
null,
"Solution:\n\nIn the figure given below, we have\n\nRay OR as the bisector of ∠POS\n\nTherefore,\n\n∠POR = ∠ROS\n\nor,\n\n∠POS = 2∠ROS ………..(i)\n\nSimilarly, ray OT as the bisector of ∠SOQ\n\nTherefore,\n\n∠TOQ = ∠TOS\n\nor,\n\n∠QOS = 2∠TOS ……….(ii)\n\nAlso, Ray OS stand on a line POQ. Therefore, ∠POS and ∠QOS form a linear pair.\n\nThus,\n\n∠POS + ∠QOS = 180°\n\nFrom (i) and (ii)\n\n2∠ROS + 2∠TOS = 180°\n\n2(∠ROS + ∠TOS) = 180°\n\n∠ROS + ∠TOS =",
null,
"∠ROT = 90°\n\n### Question 18. In the given figure, lines PQ and RS intersect each other at point O. If ∠POR: ∠ROQ = 5:7, find all the angles.",
null,
"Solution:\n\nLet ∠POR and ∠ROQ be 5x and 7x respectively.\n\nSince, Ray OR stand on line POQ. Thus, ∠POR and ∠ROQ form a linear pair.\n\nTherefore, their sum must be equal to 180°.\n\nOr,\n\n∠POR + ∠ROQ = 180°\n\n5x + 7x = 180°\n\n12x = 180°\n\nx =",
null,
"x = 15° …….(i)\n\nThus,\n\n∠POR = 5x\n\n= 5(15)\n\n= 75\n\n∠POR = 75°\n\nThus,\n\n∠ROQ = 7x\n\n= 7(15)\n\n=105\n\n∠ROQ = 105°\n\nIt is evident from the figure, that ∠QOS and ∠POR are vertically opposite angles.\n\nAnd we know that vertically opposite angles are equal.\n\nTherefore,\n\n∠QOS = ∠POR\n\n∠QOS = 75°\n\nSimilarlly, ∠POS and ∠ROQ are vertically opposite angles.\n\nAnd we know that vertically opposite angles are equal.\n\nTherefore,\n\n∠POS = ∠ROQ\n\n∠POS = 105°\n\n### Question 19. In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that\n\n∠ROS = 1212 (∠QOS − POS).",
null,
"Solution:\n\nThe given figure shows:\n\nWe have POQ as a line. Ray OR is perpendicular to line PQ. Therefore,\n\n∠ROQ = 90°\n\n∠POR = 90°\n\nFrom the figure above, we get\n\n∠ROS + ∠POS = 90° ………(i)\n\n∠POS and ∠QOS form a linear pair.\n\nTherefore,\n\n∠QOS + ∠POS = 180° ……(ii)\n\nFrom (i) and (ii) equation we get:\n\n∠QOS + ∠POS = 2 × 90 ∠QOS + ∠POS = 2 × 90\n\n∠QOS + ∠POS = 2(∠ROS + ∠POS)\n\n2∠ROS = ∠QOS – ∠POS\n\n∠ROS =",
null,
"(∠QOS – ∠POS)\n\nHence, proved.\n\nMy Personal Notes arrow_drop_up"
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https://docs.opencv.org/4.5.2/d4/d93/group__img__hash.html | [
"",
null,
"OpenCV 4.5.2 Open Source Computer Vision\nThe module brings implementations of different image hashing algorithms.\n\n## Classes\n\nclass cv::img_hash::AverageHash\nComputes average hash value of the input image. More...\n\nclass cv::img_hash::BlockMeanHash\nImage hash based on block mean. More...\n\nclass cv::img_hash::ColorMomentHash\nImage hash based on color moments. More...\n\nclass cv::img_hash::ImgHashBase\nThe base class for image hash algorithms. More...\n\nclass cv::img_hash::MarrHildrethHash\nMarr-Hildreth Operator Based Hash, slowest but more discriminative. More...\n\nclass cv::img_hash::PHash\npHash More...\n\nImage hash based on Radon transform. More...\n\n## Enumerations\n\nenum cv::img_hash::BlockMeanHashMode {\ncv::img_hash::BLOCK_MEAN_HASH_MODE_0 = 0,\ncv::img_hash::BLOCK_MEAN_HASH_MODE_1 = 1\n}\n\n## Functions\n\nvoid cv::img_hash::averageHash (cv::InputArray inputArr, cv::OutputArray outputArr)\nCalculates img_hash::AverageHash in one call. More...\n\nvoid cv::img_hash::blockMeanHash (cv::InputArray inputArr, cv::OutputArray outputArr, int mode=BLOCK_MEAN_HASH_MODE_0)\nComputes block mean hash of the input image. More...\n\nvoid cv::img_hash::colorMomentHash (cv::InputArray inputArr, cv::OutputArray outputArr)\nComputes color moment hash of the input, the algorithm is come from the paper \"Perceptual Hashing for Color Images Using Invariant Moments\". More...\n\nvoid cv::img_hash::marrHildrethHash (cv::InputArray inputArr, cv::OutputArray outputArr, float alpha=2.0f, float scale=1.0f)\nComputes average hash value of the input image. More...\n\nvoid cv::img_hash::pHash (cv::InputArray inputArr, cv::OutputArray outputArr)\nComputes pHash value of the input image. More...\n\nvoid cv::img_hash::radialVarianceHash (cv::InputArray inputArr, cv::OutputArray outputArr, double sigma=1, int numOfAngleLine=180)\nComputes radial variance hash of the input image. More...\n\n## Detailed Description\n\nProvide algorithms to extract the hash of images and fast way to figure out most similar images in huge data set.\n\nNamespace for all functions is cv::img_hash.\n\n### Supported Algorithms\n\n• Average hash (also called Different hash)\n• PHash (also called Perceptual hash)\n• Marr Hildreth Hash\n• Block Mean Hash (modes 0 and 1)\n• Color Moment Hash (this is the one and only hash algorithm resist to rotation attack(-90~90 degree))\n\nYou can study more about image hashing from following paper and websites:\n\n• \"Implementation and benchmarking of perceptual image hash functions\" \n• \"Looks Like It\" \n\n### Code Example\n\n#include \"opencv2/core.hpp\"\n#include <iostream>\nusing namespace cv;\nusing namespace cv::img_hash;\nusing namespace std;\ntemplate <typename T>\ninline void test_one(const std::string &title, const Mat &a, const Mat &b)\n{\ncout << \"=== \" << title << \" ===\" << endl;\nTickMeter tick;\nMat hashA, hashB;\nfunc = T::create();\ntick.reset(); tick.start();\nfunc->compute(a, hashA);\ntick.stop();\ncout << \"compute1: \" << tick.getTimeMilli() << \" ms\" << endl;\ntick.reset(); tick.start();\nfunc->compute(b, hashB);\ntick.stop();\ncout << \"compute2: \" << tick.getTimeMilli() << \" ms\" << endl;\ncout << \"compare: \" << func->compare(hashA, hashB) << endl << endl;;\n}\nint main(int argc, char **argv)\n{\nif (argc != 3)\n{\ncerr << \"must input the path of input image and target image. ex : hash_samples lena.jpg lena2.jpg\" << endl;\nreturn -1;\n}\ntest_one<AverageHash>(\"AverageHash\", input, target);\ntest_one<PHash>(\"PHash\", input, target);\ntest_one<MarrHildrethHash>(\"MarrHildrethHash\", input, target);\ntest_one<BlockMeanHash>(\"BlockMeanHash\", input, target);\nreturn 0;\n}\n\n### Performance under different attacks",
null,
"Performance chart\n\n### Speed comparison with PHash library (100 images from ukbench)",
null,
"Hash Computation chart",
null,
"Hash comparison chart\n\nAs you can see, hash computation speed of img_hash module outperform PHash library a lot.\n\nPS : I do not list out the comparison of Average hash, PHash and Color Moment hash, because I cannot find them in PHash.\n\n### Motivation\n\nCollects useful image hash algorithms into opencv, so we do not need to rewrite them by ourselves again and again or rely on another 3rd party library(ex : PHash library). BOVW or correlation matching are good and robust, but they are very slow compare with image hash, if you need to deal with large scale CBIR(content based image retrieval) problem, image hash is a more reasonable solution.\n\nYou can learn more about img_hash modules from following links, these links show you how to find similar image from ukbench dataset, provide thorough benchmark of different attacks(contrast, blur, noise(gaussion,pepper and salt), jpeg compression, watermark, resize).\n\n### Contributors\n\nTham Ngap Wei, thamn.nosp@m.gapw.nosp@m.ei@gm.nosp@m.ail..nosp@m.com\n\n## ◆ BlockMeanHashMode\n\n#include <opencv2/img_hash/block_mean_hash.hpp>\n\nEnumerator\nBLOCK_MEAN_HASH_MODE_0\nPython: cv.img_hash.BLOCK_MEAN_HASH_MODE_0\n\nuse fewer block and generate 16*16/8 uchar hash value\n\nBLOCK_MEAN_HASH_MODE_1\nPython: cv.img_hash.BLOCK_MEAN_HASH_MODE_1\n\nuse block blocks(step sizes/2), generate 31*31/8 + 1 uchar hash value\n\n## ◆ averageHash()\n\n void cv::img_hash::averageHash ( cv::InputArray inputArr, cv::OutputArray outputArr )\nPython:\noutputArr=cv.img_hash.averageHash(inputArr[, outputArr])\n\n#include <opencv2/img_hash/average_hash.hpp>\n\nCalculates img_hash::AverageHash in one call.\n\nParameters\n inputArr input image want to compute hash value, type should be CV_8UC4, CV_8UC3 or CV_8UC1. outputArr Hash value of input, it will contain 16 hex decimal number, return type is CV_8U\n\n## ◆ blockMeanHash()\n\n void cv::img_hash::blockMeanHash ( cv::InputArray inputArr, cv::OutputArray outputArr, int mode = BLOCK_MEAN_HASH_MODE_0 )\nPython:\noutputArr=cv.img_hash.blockMeanHash(inputArr[, outputArr[, mode]])\n\n#include <opencv2/img_hash/block_mean_hash.hpp>\n\nComputes block mean hash of the input image.\n\nParameters\n inputArr input image want to compute hash value, type should be CV_8UC4, CV_8UC3 or CV_8UC1. outputArr Hash value of input, it will contain 16 hex decimal number, return type is CV_8U mode the mode\n\n## ◆ colorMomentHash()\n\n void cv::img_hash::colorMomentHash ( cv::InputArray inputArr, cv::OutputArray outputArr )\nPython:\noutputArr=cv.img_hash.colorMomentHash(inputArr[, outputArr])\n\n#include <opencv2/img_hash/color_moment_hash.hpp>\n\nComputes color moment hash of the input, the algorithm is come from the paper \"Perceptual Hashing for Color Images Using Invariant Moments\".\n\nParameters\n inputArr input image want to compute hash value, type should be CV_8UC4, CV_8UC3 or CV_8UC1. outputArr 42 hash values with type CV_64F(double)\n\n## ◆ marrHildrethHash()\n\n void cv::img_hash::marrHildrethHash ( cv::InputArray inputArr, cv::OutputArray outputArr, float alpha = 2.0f, float scale = 1.0f )\nPython:\noutputArr=cv.img_hash.marrHildrethHash(inputArr[, outputArr[, alpha[, scale]]])\n\n#include <opencv2/img_hash/marr_hildreth_hash.hpp>\n\nComputes average hash value of the input image.\n\nParameters\n inputArr input image want to compute hash value, type should be CV_8UC4, CV_8UC3, CV_8UC1. outputArr Hash value of input, it will contain 16 hex decimal number, return type is CV_8U alpha int scale factor for marr wavelet (default=2). scale int level of scale factor (default = 1)\n\n## ◆ pHash()\n\n void cv::img_hash::pHash ( cv::InputArray inputArr, cv::OutputArray outputArr )\nPython:\noutputArr=cv.img_hash.pHash(inputArr[, outputArr])\n\n#include <opencv2/img_hash/phash.hpp>\n\nComputes pHash value of the input image.\n\nParameters\n inputArr input image want to compute hash value, type should be CV_8UC4, CV_8UC3, CV_8UC1. outputArr Hash value of input, it will contain 8 uchar value\n\n void cv::img_hash::radialVarianceHash ( cv::InputArray inputArr, cv::OutputArray outputArr, double sigma = 1, int numOfAngleLine = 180 )\n#include <opencv2/img_hash/radial_variance_hash.hpp>"
] | [
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"https://docs.opencv.org/4.5.2/attack_performance.JPG",
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"https://docs.opencv.org/4.5.2/hash_computation_chart.JPG",
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"https://docs.opencv.org/4.5.2/hash_comparison_chart.JPG",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5747758,"math_prob":0.7533462,"size":6508,"snap":"2021-31-2021-39","text_gpt3_token_len":1833,"char_repetition_ratio":0.20741083,"word_repetition_ratio":0.3,"special_character_ratio":0.2556853,"punctuation_ratio":0.26807472,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9763436,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-02T00:27:01Z\",\"WARC-Record-ID\":\"<urn:uuid:f5203209-4774-4681-8301-bef5b61f74ac>\",\"Content-Length\":\"40717\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b414393-5fd5-411c-913c-030e70d10669>\",\"WARC-Concurrent-To\":\"<urn:uuid:d7208651-bfc6-40e6-9ebf-6a28b837f1ea>\",\"WARC-IP-Address\":\"207.38.86.214\",\"WARC-Target-URI\":\"https://docs.opencv.org/4.5.2/d4/d93/group__img__hash.html\",\"WARC-Payload-Digest\":\"sha1:ID76VIEHAOFADMV5MT6SAAL7THEUJBQQ\",\"WARC-Block-Digest\":\"sha1:67XIRGVQD26QUS5XLLKN7XZABKHSETJG\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154277.15_warc_CC-MAIN-20210801221329-20210802011329-00512.warc.gz\"}"} |
https://metanumbers.com/255085500000001 | [
"## 255085500000001\n\n255,085,500,000,001 (two hundred fifty-five trillion eighty-five billion five hundred million one) is an odd fifteen-digits composite number following 255085500000000 and preceding 255085500000002. In scientific notation, it is written as 2.55085500000001 × 1014. The sum of its digits is 31. It has a total of 5 prime factors and 24 positive divisors. There are 217,515,703,860,000 positive integers (up to 255085500000001) that are relatively prime to 255085500000001.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 15\n• Sum of Digits 31\n• Digital Root 4\n\n## Name\n\nShort name 255 trillion 85 billion 500 million 1 two hundred fifty-five trillion eighty-five billion five hundred million one\n\n## Notation\n\nScientific notation 2.55085500000001 × 1014 255.085500000001 × 1012\n\n## Prime Factorization of 255085500000001\n\nPrime Factorization 11 × 192 × 101 × 636010831\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 13425552631579 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 255,085,500,000,001 is 11 × 192 × 101 × 636010831. Since it has a total of 5 prime factors, 255,085,500,000,001 is a composite number.\n\n## Divisors of 255085500000001\n\n24 divisors\n\n Even divisors 0 24 12 12\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 2.966e+14 Sum of all the positive divisors of n s(n) 4.15143e+13 Sum of the proper positive divisors of n A(n) 1.23583e+13 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59714e+07 Returns the nth root of the product of n divisors H(n) 20.6408 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 255,085,500,000,001 can be divided by 24 positive divisors (out of which 0 are even, and 24 are odd). The sum of these divisors (counting 255,085,500,000,001) is 296,599,835,438,208, the average is 12,358,326,476,592.\n\n## Other Arithmetic Functions (n = 255085500000001)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 217515703860000 Total number of positive integers not greater than n that are coprime to n λ(n) 120842057700 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7935040922259 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 217,515,703,860,000 positive integers (less than 255,085,500,000,001) that are coprime with 255,085,500,000,001. And there are approximately 7,935,040,922,259 prime numbers less than or equal to 255,085,500,000,001.\n\n## Divisibility of 255085500000001\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 1 1 1 6 1 4\n\n255,085,500,000,001 is not divisible by any number less than or equal to 9.\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n## Base conversion (255085500000001)\n\nBase System Value\n2 Binary 111001111111111110111000100111010110011100000001\n3 Ternary 1020110011221021002020110202011\n4 Quaternary 321333332320213112130001\n5 Quinary 231413310101000000001\n6 Senary 2302304354433052521\n8 Octal 7177767047263401\n10 Decimal 255085500000001\n12 Duodecimal 247393419b7141\n20 Vigesimal 14i45aif0001\n36 Base36 2if4nsl5hd\n\n## Basic calculations (n = 255085500000001)\n\n### Multiplication\n\nn×i\n n×2 510171000000002 765256500000003 1020342000000004 1275427500000005\n\n### Division\n\nni\n n⁄2 1.27543e+14 8.50285e+13 6.37714e+13 5.10171e+13\n\n### Exponentiation\n\nni\n n2 65068612310250510171000000001 16598059505466471580836930750765256500000001 4233924307981684234493084365495911673861501020342000000001 1080012699063666147722093653598941870367163414436123102501275427500000001\n\n### Nth Root\n\ni√n\n 2√n 1.59714e+07 63420.3 3996.42 760.917\n\n## 255085500000001 as geometric shapes\n\n### Circle\n\n Diameter 5.10171e+14 1.60275e+15 2.04419e+29\n\n### Sphere\n\n Volume 6.95258e+43 8.17676e+29 1.60275e+15\n\n### Square\n\nLength = n\n Perimeter 1.02034e+15 6.50686e+28 3.60745e+14\n\n### Cube\n\nLength = n\n Surface area 3.90412e+29 1.65981e+43 4.41821e+14\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 7.65257e+14 2.81755e+28 2.20911e+14\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.12702e+29 1.9561e+42 2.08276e+14\n\n## Cryptographic Hash Functions\n\nmd5 9da0bad7a1f7314ee9bbc63cc3e3c1ef c56c93a4b221b6bdcdd5f50b84eac55b80d4a8e5 8eeddce064a91410b65d8a4903df4ee50bbb35dc04d7a90f4a211cb8e2af96f8 d0f0dae71a95baa212479fcfa115cefa5e62038261f5fb7ac69d6d7338c60733ecf03fc80c12f4d5562665c838f35c0e64352529f1c361f315b00e2ecb891b8c 794e68e2fe74907e5482745e316dda24f077b04b"
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https://flipcode.net/archives/String_Matching_In_Linear-Time.shtml | [
"",
null,
"This section of the archives stores flipcode's complete Developer Toolbox collection, featuring a variety of mini-articles and source code contributions from our readers.",
null,
"String Matching In Linear-Time Submitted by",
null,
"Hi, I noticed the queue seemed to be empty so here's a string matching implementation that runs in linear time, O(n+m) to be exact, where n is the length of the text to be searched and m is the length of the text to search for. It's based on the Knuth-Morris-Pratt algorithm (or my understanding of it at least :)). Although not an explicit game-programming issue, I hope this can be helpful for some people.\n\nThe single file actually compiles into a program that searches a file for a string. However here's an example:\n ```// ex1 string_search my_search; int pos = my_search.find_first(\"badfbhfbsdcbaca\", \"ba\"); // ex2 std::vector\n\nThe first example will return 0, and the other 0 and ... eh, something else. You may use it as you like, but if you're going to give credits, give them where they are due.\n\n/Andreas Magnusson\nhttp://www.mdstud.chalmers.se/~md7amag/\n\n ```/* String search based on the Knuth-Morris-Pratt algorithm for linear running-time, O(m+n) where m=length of pattern and n=length of text. Written by Andreas Magnusson in November 2001 You may do as you please with this code, but don't blame me if anything goes wrong... */ #include #include #include #include typedef std::vector int_vec;class string_search { int_vec shifts; void compute_shifts(const std::string &pattern); public: int find_first(const std::string &text, const std::string &pattern); int_vec find_all(const std::string &text, const std::string &pattern);};// create the shift-lookup-table void string_search::compute_shifts(const std::string &pattern) { int next_shift = 0; shifts.clear(); shifts.push_back(0); // shift to the first character // start with the second character, since the shift to the first is always 0 for(int i = 1; i < pattern.length(); i++) { while(next_shift > 0 && pattern[next_shift] != pattern[i]) next_shift = shifts[next_shift]; if(pattern[next_shift] == pattern[i]) next_shift++; shifts.push_back(next_shift); } }// search the string and return when the first occurrence is found int string_search::find_first(const std::string &text, const std::string &pattern) { int next_shift = 0; compute_shifts(pattern); for(int i = 0; i < text.length(); i++) { while(next_shift > 0 && pattern[next_shift] != text[i]) next_shift = shifts[next_shift - 1]; if(pattern[next_shift] == text[i]) next_shift++; if(next_shift == pattern.length()) return i - (pattern.length() - 1); // found the first so return } return -1; }// search the string and put every occurence in a vector int_vec string_search::find_all(const std::string &text, const std::string &pattern) { int next_shift = 0; int_vec positions; compute_shifts(pattern); for(int i = 0; i < text.length(); i++) { while(next_shift > 0 && pattern[next_shift] != text[i]) next_shift = shifts[next_shift - 1]; if(pattern[next_shift] == text[i]) next_shift++; if(next_shift == pattern.length()) { positions.push_back(i - (pattern.length() - 1)); // found one, put in list next_shift = shifts[next_shift - 1]; // restart pattern with last shift } } return positions; }int main(int argc, char **argv) { if(argc <= 2){ cout << \"Usage: \" << argv << \" filename searchpattern\" << endl; return 0; } std::string pattern = argv; // read the file. Since the file is read like this all white-characters // are eaten, so a search including white-characters will fail... fstream fs; std::string text, temp; fs.open(argv, ios::in); while(!fs.eof()){ fs >> temp; text += temp; } fs.close(); // search the file string_search search; int_vec pos_list = search.find_all(text, pattern); // print out result std::vector::iterator it; cout << \"Found \" << pos_list.size() << \" occurrences\" << endl; for(it = pos_list.begin(); it != pos_list.end(); it++){ temp = text.substr(*it, pattern.length()); cout << \"Pos=\" << *it << \" == \" << temp << endl; } } ```"
] | [
null,
"https://flipcode.net/archives/logo_toolbox.png",
null,
"https://flipcode.net/archives/icon_articles.png",
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"https://flipcode.net/archives/line_grey.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6455306,"math_prob":0.77184564,"size":4124,"snap":"2023-14-2023-23","text_gpt3_token_len":1076,"char_repetition_ratio":0.175,"word_repetition_ratio":0.14046822,"special_character_ratio":0.30237633,"punctuation_ratio":0.22010177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98428386,"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\":\"2023-03-23T04:10:44Z\",\"WARC-Record-ID\":\"<urn:uuid:ae3b6531-ddb3-4c2b-8864-1e66e6916db6>\",\"Content-Length\":\"12429\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ad800807-930c-483a-8033-9f41007d7dcf>\",\"WARC-Concurrent-To\":\"<urn:uuid:9661a94a-fa95-42b3-b8f1-c4abf3921493>\",\"WARC-IP-Address\":\"52.4.205.129\",\"WARC-Target-URI\":\"https://flipcode.net/archives/String_Matching_In_Linear-Time.shtml\",\"WARC-Payload-Digest\":\"sha1:KIDK4PRZ63P4XATTKKYJ3RDP7CZCT63Q\",\"WARC-Block-Digest\":\"sha1:TB2BSQYY54ENKCSWW5BMWRMKF2YUC5TT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296944996.49_warc_CC-MAIN-20230323034459-20230323064459-00039.warc.gz\"}"} |
https://programmingpraxis.com/2016/04/22/gcd-sum/ | [
"## GCD Sum\n\n### April 22, 2016\n\nToday’s exercise is inspired by A018804: Find the sum of the greatest common divisors gcd(k, n) for 1 ≤ kn.\n\nYour task is to write a function that finds the sum of the greatest common divisors. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.\n\nPages: 1 2 3\n\n### 9 Responses to “GCD Sum”\n\n1. Rutger said\n\nIn Python, bruteforce solution:\n\n```from fractions import gcd\n\nfor n in range(20):\nprint n, \":\", sum([gcd(n,k) for k in range(1,n+1)])\n\n# Result\n# 0 : 0\n# 1 : 1\n# 2 : 3\n# 3 : 5\n# 4 : 8\n# 5 : 9\n# 6 : 15\n# 7 : 13\n# 8 : 20\n# 9 : 21\n# 10 : 27\n# 11 : 21\n# 12 : 40\n# 13 : 25\n# 14 : 39\n# 15 : 45\n# 16 : 48\n# 17 : 33\n# 18 : 63\n# 19 : 37\n# [Finished in 0.2s]\n```\n2. Paul said\n\nIn Python, td_factors is from Programming Praxis primes essay and divisor_gen generates all divisors.\n\n```def totient(n):\nproduct = n\nfor prime in set(td_factors(n)):\nproduct -= product // prime\nreturn product\n\ndef sumgcd(n):\nreturn sum(d * totient(n // d) for d in divisor_gen(n))\n```\n3. matthew said\n\nhttp://oeis.org/A018804 also tells us that sumgcd (or Pillai’s arithmetic function) is multiplicative (f is multiplicative if f(nm) = f(n)f(m) for coprime n and m) and that its value for p^e is p^(e-1)*((p-1)e+p), for prime p – this suffices to calculate the function for any n, given a prime factorization. Here we use the Python primefac library, converting the output to prime-exponent pairs, and use that to define a generic multiplicative function. Fairly snappy, even for “colossally abundant” numbers with lots of divisors (has anyone seen the Ramanujan film yet?):\n\n```from primefac import primefac\nfrom itertools import groupby\nfrom operator import mul\n\ndef factors(n): return ((s, len(list(s))) for s in groupby(primefac(n)))\ndef multiplicative(f): return lambda n: reduce(mul, (f(p,e) for (p,e) in factors(n)))\ngcdsum = multiplicative(lambda p,e: p**(e-1)*((p-1)*e+p))\n\nprint gcdsum(991*997)\nprint gcdsum(224403121196654400) # 22nd colossally abundant number\n```\n```\\$ ./gcdsum.py\n3948133\n4812563181512005920000\n```\n4. Zack said\n\nHere is an alternative in Julia, without invoking a bunch of packages to do all the hard work:\n\nfunction gcd_(x::Int64, y::Int64)\nm = min(x,y)\n\nfor d in m:-1:1\nif (x%d == 0) && (y%d == 0)\nreturn d\nend\nend\nend\n\nfunction gcd_sum(n::Int64)\ns = n + 1\n\nfor k = 2:(n-1)\ns += gcd(k,n)\nend\n\nreturn s\nend\n\n5. Zack said\n\n@matthew: This CSS add-on doesn’t support formatting for Julia…\n\n6. Daniel said\n\nHere’s a Java solution that is similar in spirit to Sieve of Eratosthenes.\n\n```static long gcdSum(int n) {\nint[] array = new int[n+1]; // array ignored\n\nfor (int i = 1; i <= n/2; i++) {\nif (n % i == 0) {\nfor (int j = i; j <= n; j += i) {\narray[j] = i;\n}\n}\n}\n\nlong sum = n;\nfor (int i = 1; i < n; i++) {\nsum += array[i];\n}\n\nreturn sum;\n}\n```\n7. matthew said\n\n@Zack: you can just leave out the language specification, which will preserve indentation at least, eg “\n\n`\" at the start, \"`\n\n” at the end:\n\n```function gcd_(x::Int64, y::Int64)\nm = min(x,y)\n\nfor d in m:-1:1\nif (x%d == 0) && (y%d == 0)\nreturn d\nend\nend\nend\n\nfunction gcd_sum(n::Int64)\ns = n + 1\n\nfor k = 2:(n-1)\ns += gcd(k,n)\nend\n\nreturn s\nend\n```\n8. matthew said\n\nGah, I was attempting to show the formatting commands `[code]` and `[/code]` – I had a notion they only applied when on a new line, but evidently not. Using HTML entities for the square brackets here."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.85801154,"math_prob":0.9664932,"size":1888,"snap":"2021-43-2021-49","text_gpt3_token_len":509,"char_repetition_ratio":0.097133756,"word_repetition_ratio":0.0127795525,"special_character_ratio":0.26271185,"punctuation_ratio":0.13032581,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.997599,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T03:37:09Z\",\"WARC-Record-ID\":\"<urn:uuid:bb453665-8402-4982-8a48-af131f3df1d7>\",\"Content-Length\":\"122095\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:952c8b21-8ed7-42f2-b71f-95563fe87c7e>\",\"WARC-Concurrent-To\":\"<urn:uuid:af3be3b4-616e-4d33-b9f8-8d8c6a1fc9cb>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://programmingpraxis.com/2016/04/22/gcd-sum/\",\"WARC-Payload-Digest\":\"sha1:5TL75BTEOOXNSKFUMK66XNDJHJMQVRSM\",\"WARC-Block-Digest\":\"sha1:BB42ISOWCL6FIXNIKK6F5YTGKO2OP3YQ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362589.37_warc_CC-MAIN-20211203030522-20211203060522-00361.warc.gz\"}"} |
https://mathindemand.com/products/math-coloring-worksheets-bundle | [
"# Math Coloring Worksheets Bundle\n\nThis is a growing bundle! Currently there are 11 coloring worksheets. This includes:\n\n(1) Adding and Subtracting Integers Halloween Coloring Worksheet\n(2) Multiplying and Dividing Integers Halloween Coloring Worksheet\n(3) Adding and Subtracting Fractions Coloring Worksheet\n(4) One-Step Equations Halloween Coloring Worksheet\n(5) Two-Step Equations Halloween Coloring Worksheet\n(6) Slope Between Two Points Coloring Worksheet\n(7) Slope of a Linear Equation Coloring Worksheet\n(8) Comparing and Ordering Rationals & Irrationals Coloring Worksheet\n(9) Number Sense Vocabulary Coloring Worksheet for 7th Grade\n(10) Ratios & Proportional Relationships Vocabulary Coloring Worksheet for 7th Grade\n(11) Multi-Step Equations Coloring Worksheet\n\nThis growing bundle will have at least 20 coloring worksheets. You get all future coloring worksheets for no additional cost."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7280517,"math_prob":0.61679304,"size":901,"snap":"2023-40-2023-50","text_gpt3_token_len":198,"char_repetition_ratio":0.27759197,"word_repetition_ratio":0.033613447,"special_character_ratio":0.19866815,"punctuation_ratio":0.046153847,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95362425,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T21:51:37Z\",\"WARC-Record-ID\":\"<urn:uuid:dd70c730-2ad2-4eda-bf45-e6b7a65de87e>\",\"Content-Length\":\"109642\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66f61322-3021-4f32-bb37-df986db60a25>\",\"WARC-Concurrent-To\":\"<urn:uuid:c3962bde-7782-441d-82c1-da1297859d5e>\",\"WARC-IP-Address\":\"23.227.38.65\",\"WARC-Target-URI\":\"https://mathindemand.com/products/math-coloring-worksheets-bundle\",\"WARC-Payload-Digest\":\"sha1:DOL2MADKETJSRSXTM7ICNLTWZGL2J6PS\",\"WARC-Block-Digest\":\"sha1:TVELI7QBEW3D5EKOG2DPS3MI2P2IFLAC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510225.44_warc_CC-MAIN-20230926211344-20230927001344-00462.warc.gz\"}"} |
https://www.asvabtestbank.com/arithmetic-reasoning/t/76/q/practice-test/25134/5 | [
"## ASVAB Arithmetic Reasoning Operations on Exponents Practice Test 25134\n\n Questions 5 Focus Operations on Exponents Topics Defining Exponents Question Type Questions\n\n#### Study Guide\n\n###### Defining Exponents\n\nAn exponent (cbe) consists of coefficient (c) and a base (b) raised to a power (e). The exponent indicates the number of times that the base is multiplied by itself. A base with an exponent of 1 equals the base (b1 = b) and a base with an exponent of 0 equals 1 ( (b0 = 1)."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8387243,"math_prob":0.96616346,"size":373,"snap":"2021-21-2021-25","text_gpt3_token_len":100,"char_repetition_ratio":0.17073171,"word_repetition_ratio":0.03076923,"special_character_ratio":0.2707775,"punctuation_ratio":0.04347826,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9831276,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T12:17:56Z\",\"WARC-Record-ID\":\"<urn:uuid:9b169adb-9a99-42da-9aab-49b042fe85d8>\",\"Content-Length\":\"9660\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2fb6daff-b16d-42ee-a51f-34e4b7af6b02>\",\"WARC-Concurrent-To\":\"<urn:uuid:676b40fd-6c68-4e60-8142-c500538ae981>\",\"WARC-IP-Address\":\"68.183.116.70\",\"WARC-Target-URI\":\"https://www.asvabtestbank.com/arithmetic-reasoning/t/76/q/practice-test/25134/5\",\"WARC-Payload-Digest\":\"sha1:K6VXPWOFEYQXXYMPWEODWJCZM54SG25D\",\"WARC-Block-Digest\":\"sha1:MJGVI7KDS4CSWHV52JP5XYJZAXLS5OZ6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991269.57_warc_CC-MAIN-20210516105746-20210516135746-00571.warc.gz\"}"} |
http://www.monkeybreadsoftware.net/example-chartdirector-surfaceperspective.shtml | [
"Platforms to show: All Mac Windows Linux Cross-Platform\n\n/ChartDirector/surfaceperspective\nFunction:\nRequired plugins for this example: MBS ChartDirector Plugin\nYou find this example project in your Plugins Download as a Xojo project file within the examples folder: /ChartDirector/surfaceperspective\nThis example is the version from Sun, 17th Mar 2012.\nProject \"surfaceperspective.rbp\"\nClass App Inherits Application\nConst kEditClear = \"&Löschen\"\nConst kFileQuit = \"Beenden\"\nConst kFileQuitShortcut = \"\"\nEventHandler Sub Open() createChart(0, \"surfaceperspective0\") createChart(1, \"surfaceperspective1\") createChart(2, \"surfaceperspective2\") createChart(3, \"surfaceperspective3\") createChart(4, \"surfaceperspective4\") createChart(5, \"surfaceperspective5\") End EventHandler\nProtected Sub CreateChart(img as integer, filename as string) dim w as new PicWindow w.Title=filename // The x and y coordinates of the grid dim dataX(-1) as double = array(0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0) dim dataY(-1) as double = array(0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0) // The values at the grid points. In this example, we will compute the values // using the formula z = sin((x - 0.5) * 2 * pi) * sin((y - 0.5) * 2 * pi) dim dataZ(120) as double for yIndex as integer = 0 to UBound(datay) dim y as double = (dataY(yIndex) - 0.5) * 2 * 3.1416 for xIndex as integer = 0 to UBound(Datax) dim x as double = (dataX(xIndex) - 0.5) * 2 * 3.1416 dataZ(yIndex * 11 + xIndex) = sin(x) * sin(y) next next // the perspective level dim perspective as integer = img * 12 // Create a SurfaceChart object of size 360 x 360 pixels, with white (ffffff) // background and grey (888888) border. dim c as new CDSurfaceChartMBS(360, 360, &hffffff, &h888888) // Set the perspective level c.setPerspective(perspective) call c.addTitle(\"Perspective = \"+str(perspective)) // Set the center of the plot region at (195, 165), and set width x depth x // height to 200 x 200 x 150 pixels c.setPlotRegion(195, 165, 200, 200, 150) // Set the plot region wall thichness to 5 pixels c.setWallThickness(5) // Set the elevation and rotation angles to 30 and 30 degrees c.setViewAngle(30, 30) // Set the data to use to plot the chart c.setData(dataX, dataY, dataZ) // Spline interpolate data to a 40 x 40 grid for a smooth surface c.setInterpolation(40, 40) // Use smooth gradient coloring. c.colorAxis.setColorGradient // Output the chart w.Backdrop=c.makeChartPicture w.top=50+380*(img\\2) w.left=400*(img mod 2) w.show End Sub\nEnd Class\nClass PicWindow Inherits Window\nEnd Class\nEnd Project",
null,
"",
null,
""
] | [
null,
"http://www.monkeybreadsoftware.net/images/xojoplugin.png",
null,
"http://www.monkeybreadsoftware.net/images/olark.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5188909,"math_prob":0.97567266,"size":3353,"snap":"2019-51-2020-05","text_gpt3_token_len":1033,"char_repetition_ratio":0.1403404,"word_repetition_ratio":0.199187,"special_character_ratio":0.30181926,"punctuation_ratio":0.16534181,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9784374,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T04:37:03Z\",\"WARC-Record-ID\":\"<urn:uuid:c152b7fe-e18b-45e9-a302-8ff74db2a77b>\",\"Content-Length\":\"30013\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb89e06c-a6c7-4e29-824e-fc77df698b23>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1f56e6a-96e3-4de3-a36b-d54bde20ccde>\",\"WARC-IP-Address\":\"109.239.53.23\",\"WARC-Target-URI\":\"http://www.monkeybreadsoftware.net/example-chartdirector-surfaceperspective.shtml\",\"WARC-Payload-Digest\":\"sha1:NE773YS3TQ62UXNAQEKKLP3VPZTVEO44\",\"WARC-Block-Digest\":\"sha1:7ZFOO6B2WMOICWO3ASVQRCM3EKLOKELX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251788528.85_warc_CC-MAIN-20200129041149-20200129071149-00001.warc.gz\"}"} |
http://dondun.com/information/mathematics/index.html | [
"# Useful Calculations\n\nIs Primen:r:\nNext Prime\nPrime Factors\nHCF/LCM\nFactorial\nnCr\nnPr\nPoisson Mean: Number:\nBinomial Number:Selection: Prob:\nNormal From x = to x = Mean:St Devn:\nPolynomial Coeffs * Value of x:\nPolygons\nDtaw of coloured ballsNumber and colour of balls to be drawn Red:Green: Blue:\nSubmit\n\n*Polynomial coefficients should be entered in descending order and separated by spaces\nExample: 4x4+2x3-x2-5x+2 would be entered as 4 2 -1 -5 2\n\n†Enter the number required in the space for n on the right\n‡Enter the two numbers in the spaces for n and r on the right"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8732719,"math_prob":0.9696079,"size":264,"snap":"2021-31-2021-39","text_gpt3_token_len":77,"char_repetition_ratio":0.13846155,"word_repetition_ratio":0.0,"special_character_ratio":0.26136363,"punctuation_ratio":0.018518519,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9776413,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T08:12:39Z\",\"WARC-Record-ID\":\"<urn:uuid:04ac79ea-60a7-40be-8c1c-98afcab16ee4>\",\"Content-Length\":\"4273\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:be5e09d4-8f33-428c-a01b-2bf72e08e05c>\",\"WARC-Concurrent-To\":\"<urn:uuid:d4baab64-20da-4613-a20a-ea420ca6b2ed>\",\"WARC-IP-Address\":\"217.160.0.241\",\"WARC-Target-URI\":\"http://dondun.com/information/mathematics/index.html\",\"WARC-Payload-Digest\":\"sha1:VZMHYJN6VXAQFAOAO7NUSXFOKREQZPKN\",\"WARC-Block-Digest\":\"sha1:QRKJBYJZ53LZIQQW5YCUR4IRYBWW7KGU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153934.85_warc_CC-MAIN-20210730060435-20210730090435-00212.warc.gz\"}"} |
https://zbmath.org/?q=an:1038.47501 | [
"# zbMATH — the first resource for mathematics\n\nCovering dimension and differential inclusions. (English) Zbl 1038.47501\nLower semicontinuous multivalued mappings $$\\Phi$$, $$\\Psi : X \\to 2^Y$$ are considered, where $$X$$, $$Y$$ are Banach spaces. Under some assumptions (closedness and convexity of the values of $$\\Phi$$ and $$\\Psi$$, surjectivity of $$\\Phi$$, a certain compactness of $$\\Psi$$, etc.), it is shown that $\\dim (\\{x \\in X;\\;\\Phi (x) \\cap \\Psi (x) \\neq \\emptyset \\}) \\geq \\dim (\\Phi ^{-1}(0)),$ where $$\\dim$$ denotes the covering dimension. Two applications to differential inclusions are given.\n##### MSC:\n 47H04 Set-valued operators 26E25 Set-valued functions\nFull Text:"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.755253,"math_prob":0.99985766,"size":973,"snap":"2021-43-2021-49","text_gpt3_token_len":313,"char_repetition_ratio":0.090815276,"word_repetition_ratio":0.04477612,"special_character_ratio":0.34326825,"punctuation_ratio":0.24598931,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999753,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T04:15:56Z\",\"WARC-Record-ID\":\"<urn:uuid:83452d23-02d3-4e3b-8fa7-7663585f5d7a>\",\"Content-Length\":\"46657\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ed2e96e-c5de-4db2-ab4f-858249ec94e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd80ef4b-87cf-4fe2-a8b6-89f8f6b3ac4c>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an:1038.47501\",\"WARC-Payload-Digest\":\"sha1:Z6JR3SSJQ4EPNUFLRGDYOMQ7BX66DOXM\",\"WARC-Block-Digest\":\"sha1:GL3IKZ5DBMKEVQISRMZE4UJO4EHRA7TT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363659.21_warc_CC-MAIN-20211209030858-20211209060858-00068.warc.gz\"}"} |
https://braingenie.ck12.org/skills/103391/learn | [
"### Sample Problem\n\nThe perimeter of the un-shaded part of the figure below is cm.",
null,
"#### Solution\n\nThe perimeter is the sum of AB, BC, CD, DE, EF, FG, GH and HA. So it is (3 + 2) + (2 + 2 + 4) + 3 + 4 + (4 - 2) + 2 + 4 + 2 = 30 cm."
] | [
null,
"https://s3.amazonaws.com/ck12bg.ck12.org/curriculum/103391/1.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8607245,"math_prob":0.99984646,"size":195,"snap":"2020-45-2020-50","text_gpt3_token_len":85,"char_repetition_ratio":0.13612565,"word_repetition_ratio":0.0,"special_character_ratio":0.4974359,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99579334,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T04:18:57Z\",\"WARC-Record-ID\":\"<urn:uuid:fae5788b-64d2-4f79-a032-c3d72709d86f>\",\"Content-Length\":\"2893\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2bc100b3-34e6-4bae-8752-ab481b698a3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:293c9831-f2dc-4c63-ae34-72af153b93ac>\",\"WARC-IP-Address\":\"54.236.247.140\",\"WARC-Target-URI\":\"https://braingenie.ck12.org/skills/103391/learn\",\"WARC-Payload-Digest\":\"sha1:LHJPPAGHAQ2OQD3WXNBESUANH6DGT5UK\",\"WARC-Block-Digest\":\"sha1:SVMZI24ETH25NRFWBNY2IHA6NMR6JAAD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878879.33_warc_CC-MAIN-20201022024236-20201022054236-00625.warc.gz\"}"} |
http://share.web-tinker.com/SimpleWebGL.Matrix.1.0.html | [
"• Matrix\n• 版本：1.0\n• 程序链接：http://www.web-tinker.com/files/SimpleWebGL.Matrix.1.0.js\n• 描述：一个SimpleWebGL的矩阵插件，用于处理一些简单的矩阵计算。\n• 继承于：Array\n• 构造器\n• new Matrix(Number dimension)\n• 描述: 生成一个指定矩阵阶数的单位矩阵。\n• 参数说明:\n• dimension: 一个大于1的整数。\n• new Matrix(Array data)\n• 描述: 使用现有数据生成一个矩阵。\n• 参数说明:\n• data: 存放数据的数组，该数组长度必须是n²，数据类型为数字。\n• 静态成员\n• Matrix projection(Number angle,Number proportion,Number near,Number far)\n• 描述: 生成一个投射矩阵。\n• 返回: Matrix对象（4阶）。\n• 参数说明:\n• angle: 视角大小。\n• proportion: 显示区域的长宽比（width/height）。\n• near: 近平面与原点的距离。\n• far: 远平面与原点的距离。\n• Matrix view(Array position[,Number yaw[,Number pitch[,Number roll]]])\n• 描述: 生成一个视图矩阵。\n• 返回: Matrix对象（4阶）。\n• 参数说明:\n• position: 一个数组[x,y,z]，描述摄像机位置。\n• yaw: 摄像机绕y轴旋转的度数，可选，不使用可为null。\n• pitch: 摄像机绕x轴旋转的度数，可选，不使用可为null。\n• roll: 摄像机绕z轴旋转的度数，可选，不使用可为null。\n• Matrix model(Array position[,Number yaw[,Number pitch[,Number roll]]])\n• 描述: 生成一个模型矩阵。\n• 返回: Matrix对象（4阶）。\n• 参数说明:\n• position: 一个数组[x,y,z]，描述模型位置。\n• yaw: 模型绕y轴旋转的度数，可选，不使用可为null。\n• pitch: 模型绕x轴旋转的度数，可选，不使用可为null。\n• roll: 模型绕z轴旋转的度数，可选，不使用可为null。\n• 实例成员\n• Number dimension\n• 描述: 标识当前Matrix实例的阶数。\n• Number data(Number row,Number column)\n• 描述: 取出矩阵中指定行列的数据，注意下标从0开始。\n• 返回: 一个实数。\n• 参数说明:\n• row: 一个整数，如果越界将取余数计算。\n• column: 一个整数，如果越界将取余数计算。\n• Self data(Number row,Number column,Number value)\n• 描述: 设置矩阵中指定行列的数据，注意下标从0开始。\n• 返回: 调用对象。\n• 参数说明:\n• row: 一个整数，如果越界将取余数计算。\n• column: 一个整数，如果越界将取余数计算。\n• value: 将要写入指定位置的数据。\n• Self equate(Array data);\n• 描述: 传入一个数组来设置矩阵中的每一项。\n• 返回: 调用对象。\n• 参数说明:\n• data: 一个数组，长度应为矩阵阶数的平方。\n• Self plus(Number value);\n• 描述: 为矩阵中的每一项加上一个实数。\n• 返回: 调用对象。\n• 参数说明:\n• value: 需要加到矩阵每一项上的数值。\n• Self plus(Matrix matrix);\n• 描述: 把传入的矩阵的每一项加到当前矩阵的相应位置。\n• 返回: 调用对象。\n• 参数说明:\n• matrix: 传入一个和当前矩阵阶数相同的矩阵。\n• Self multiply(Number value);\n• 描述: 为矩阵中的每一项乘上一个实数。\n• 返回: 调用对象。\n• 参数说明:\n• value: 需要乘到矩阵每一项上的数值。\n• Self multiply(Matrix matrix);\n• 描述: 把当前矩阵乘上一个矩阵，或者说对当前矩阵做指定的矩阵变换（当前×传入）。\n• 返回: 调用对象。\n• 参数说明:\n• matrix: 传入一个和当前矩阵阶数相同的矩阵。\n• Number determinant();\n• 描述: 计算当前矩阵的行列式值。\n• 返回: 一个实数。\n• Self inverse();\n• 描述: 对当前矩阵求逆。\n• 返回: 调用对象。\n• Self yaw(Number angle);\n• 描述: 把一个绕y轴旋转的旋转矩阵乘上当前矩阵，需要当前矩阵的阶数大于或等于3。\n• 返回: 调用对象。\n• 参数说明:\n• angle: 旋转的角度。\n• Self pitch(Number angle);\n• 描述: 把一个绕x轴旋转的旋转矩阵乘上当前矩阵，需要当前矩阵的阶数大于或等于3。\n• 返回: 调用对象。\n• 参数说明:\n• angle: 旋转的角度。\n• Self roll(Number angle);\n• 描述: 把一个绕z轴旋转的旋转矩阵乘上当前矩阵，需要当前矩阵的阶数大于或等于3。\n• 返回: 调用对象。\n• 参数说明:\n• angle: 旋转的角度。\n• Self move(Number x,Number y,Number z);\n• 描述: 让当前矩阵移动一段距离(无视旋转)，需要当前矩阵的阶数大于或等于3。\n• 返回: 调用对象。\n• 参数说明:\n• x: x方向的移动距离。\n• y: y方向的移动距离。\n• z: z方向的移动距离。\n• Self moveTo(Number x,Number y,Number z);\n• 描述: 让当前矩阵移动到一个绝对位置(无视旋转)，需要当前矩阵的阶数大于或等于3。\n• 返回: 调用对象。\n• 参数说明:\n• x: 目标位置的x坐标。\n• y: 目标位置的y坐标。\n• z: 目标位置的z坐标。"
] | [
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.9291811,"math_prob":0.9954531,"size":2427,"snap":"2019-26-2019-30","text_gpt3_token_len":1519,"char_repetition_ratio":0.2150227,"word_repetition_ratio":0.17554858,"special_character_ratio":0.31479192,"punctuation_ratio":0.2740047,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996694,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T04:35:46Z\",\"WARC-Record-ID\":\"<urn:uuid:061a4c62-4192-4673-b278-5ab4b7680469>\",\"Content-Length\":\"10448\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54b50524-c443-43b4-8479-293c1687c785>\",\"WARC-Concurrent-To\":\"<urn:uuid:21c2c4de-938f-441a-b483-a62f63872b46>\",\"WARC-IP-Address\":\"47.90.22.219\",\"WARC-Target-URI\":\"http://share.web-tinker.com/SimpleWebGL.Matrix.1.0.html\",\"WARC-Payload-Digest\":\"sha1:JAJ7DWKZNYTC2WCCV5HUTI2ULVSYYY2C\",\"WARC-Block-Digest\":\"sha1:TEZ6ZRRCP6YMYSUM75COSMJJOQFYAAGT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525500.21_warc_CC-MAIN-20190718042531-20190718064531-00216.warc.gz\"}"} |
https://au.mathworks.com/help/wavelet/ug/continuous-wavelet-analysis-of-modulated-signals.html | [
"# Continuous Wavelet Analysis of Modulated Signals\n\nThis example shows how to use the continuous wavelet transform (CWT) to analyze modulated signals.\n\nLoad a quadratic chirp signal. The signal's frequency begins at approximately 500 Hz at t = 0, decreases to 100 Hz at t=2, and increases back to 500 Hz at t=4. The sampling frequency is 1 kHz.\n\n```load quadchirp; fs = 1000;```\n\nObtain a time-frequency plot of this signal using the CWT with a bump wavelet. The bump wavelet is a good choice for the CWT when your signals are oscillatory and you are more interested in time-frequency analysis than localization of transients.\n\n```[cfs,f] = cwt(quadchirp,'bump',fs); helperCWTTimeFreqPlot(cfs,tquad,f,'surf','CWT of Quadratic Chirp','Seconds','Hz')```",
null,
"The CWT clearly shows the time evolution of the quadratic chirp's frequency. The quadratic chirp is a frequency-modulated signal. While that signal is synthetic, frequency and amplitude modulation occur frequently in natural signals as well. Use the CWT to obtain a time-frequency analysis of an echolocation pulse emitted by a big brown bat (Eptesicus Fuscus). The sampling interval is 7 microseconds. Use the bump wavelet with 32 voices per octave. Thanks to Curtis Condon, Ken White, and Al Feng of the Beckman Center at the University of Illinois for the bat data and permission to use it in this example.\n\n```load batsignal t = 0:DT:(numel(batsignal)*DT)-DT; [cfs,f] = cwt(batsignal,'bump',1/DT,'VoicesPerOctave',32); helperCWTTimeFreqPlot(cfs,t.*1e6,f./1e3,'surf','Bat Echolocation (CWT)',... 'Microseconds','kHz')```",
null,
"For the final example, obtain a time-frequency analysis of some seismograph data recorded during the 1995 Kobe earthquake. The data are seismograph (vertical acceleration, nm/sq.sec) measurements recorded at Tasmania University, Hobart, Australia on 16 January 1995 beginning at 20:56:51 (GMT) and continuing for 51 minutes at 1 second intervals. Use the default analytic Morse wavelet.\n\n```load kobe; dt = 1; cwt(kobe,1); title('CWT of 1995 Kobe Earthquake Seismograph Data');```",
null,
"##### Support",
null,
"Get trial now"
] | [
null,
"https://au.mathworks.com/help/examples/wavelet/win64/TimeFrequencyAnalysisWithTheCWTExample_01.png",
null,
"https://au.mathworks.com/help/examples/wavelet/win64/TimeFrequencyAnalysisWithTheCWTExample_02.png",
null,
"https://au.mathworks.com/help/examples/wavelet/win64/TimeFrequencyAnalysisWithTheCWTExample_03.png",
null,
"https://au.mathworks.com/images/responsive/supporting/apps/doc_center/bg-trial-arrow.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84958655,"math_prob":0.96766603,"size":1974,"snap":"2020-24-2020-29","text_gpt3_token_len":516,"char_repetition_ratio":0.11370558,"word_repetition_ratio":0.0071174377,"special_character_ratio":0.24164134,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97128415,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-08T05:03:47Z\",\"WARC-Record-ID\":\"<urn:uuid:11c02f9a-19c9-456d-ae58-0e361f568ed4>\",\"Content-Length\":\"68662\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa9c64c1-0253-429e-b9f0-a00f8c3126d8>\",\"WARC-Concurrent-To\":\"<urn:uuid:5666c613-790d-46b8-95aa-e7a77792d634>\",\"WARC-IP-Address\":\"23.66.56.59\",\"WARC-Target-URI\":\"https://au.mathworks.com/help/wavelet/ug/continuous-wavelet-analysis-of-modulated-signals.html\",\"WARC-Payload-Digest\":\"sha1:ONM4NZF4RXLHG6OHTCBBD6MFGVJPOCT2\",\"WARC-Block-Digest\":\"sha1:EFBDW7AEISFT62ZE6ZOORPLCK57W6O5A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655896374.33_warc_CC-MAIN-20200708031342-20200708061342-00000.warc.gz\"}"} |
https://docs.lucee.org/reference/functions/log10.html | [
"# Log10\n\nCalculates the logarithm of number, to base 10.\n\n```Log10( number )\n```\n\nReturns: Number\n\nArgument Description\nnumber\nnumber, required\n\nPositive real number for which to calculate the logarithm\n\n#### Examples\n\nThe following code example will output an example number returned from the log10() function.\n\n```<cfoutput>#log10(10)#</cfoutput>\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59403753,"math_prob":0.9732813,"size":329,"snap":"2019-13-2019-22","text_gpt3_token_len":80,"char_repetition_ratio":0.16307692,"word_repetition_ratio":0.0,"special_character_ratio":0.24012157,"punctuation_ratio":0.09803922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976475,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-24T10:51:26Z\",\"WARC-Record-ID\":\"<urn:uuid:ce2cfce9-44a4-4e6c-b4bd-b979c1132e23>\",\"Content-Length\":\"45765\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43ef9545-4b2d-45b3-ac74-19b5b139b089>\",\"WARC-Concurrent-To\":\"<urn:uuid:b99ea62a-2f4b-4094-ab98-28c7f3288450>\",\"WARC-IP-Address\":\"99.84.104.119\",\"WARC-Target-URI\":\"https://docs.lucee.org/reference/functions/log10.html\",\"WARC-Payload-Digest\":\"sha1:R7WIWFTIUYIDOBT4JTBQKQ5MBQQ4U6YL\",\"WARC-Block-Digest\":\"sha1:UDDABSEL4VGD3J56H4BHJO67VVSPJM4Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257605.76_warc_CC-MAIN-20190524104501-20190524130501-00544.warc.gz\"}"} |
https://cheatography.com/fanthom1234/cheat-sheets/python-note-sheet/ | [
"Cheatography\n\n# Python Note Sheet by fanthom1234\n\n### Function\n\n input() Ask for information from user. print() Show information that we want on the scree. int() Change number to be number integer. float() Change number to be decimal. str() String. A list of number, letter, and symbol.\n\n### Math\n\n == equal to != not equal to < less than > more than <= less than or equal to >= more than or equal to\n\n### IF..ELSE..ELIF\n\n IF ELSE ELIF\n\n### mylist\n\nmylist = range(5)\nmylist = [0, 1, 2, 3, 4]\n\n Str + Str Combine together. Like \"1\" + \"2\" = 12 Str + number Crash Number + Number Sum, addition\n\n### Subtraction\n\n str - number crash str - str crash number - number Subtract\n\n### Division\n\n str / str crash str / number crash number / number division\n\n### Multiplication\n\n str * number combine that string. \"6\" * 2 = '66' str * str crash number * number multiply str ** str crash number ** number exponent, power str ** number crash\n\n### Vocab\n\n Variable Something that can change String a list of character like number, letter and symbols. Integer Number Whole number or Counting number Float Number The number into decimal number Syntax Structure of language Loop while Input Gain information from user Print Show information that we want on the screen.\n\n### Random\n\n random.choice() random word in the list. need to put 'import random' at the top first.\n\n### True and False\n\ny = true\nprint (not y or 2<3)\nnot true or 2<3\nFalse or True\nTrue\n\nTrue or anything == True\nFalse or anything == False\n\n### Help Us Go Positive!\n\nWe offset our carbon usage with Ecologi. Click the link below to help us!",
null,
""
] | [
null,
"https://api.ecologi.com/badges/cpw/5f87619570394d001d816113",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59816736,"math_prob":0.9880576,"size":1579,"snap":"2020-45-2020-50","text_gpt3_token_len":452,"char_repetition_ratio":0.14222223,"word_repetition_ratio":0.021201413,"special_character_ratio":0.25712475,"punctuation_ratio":0.10309278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97440803,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-22T09:07:15Z\",\"WARC-Record-ID\":\"<urn:uuid:576bab53-1297-4bd4-a458-4ab611724618>\",\"Content-Length\":\"65190\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c8a837f3-0e44-43c5-ad49-255760cc031d>\",\"WARC-Concurrent-To\":\"<urn:uuid:e84c0c48-6c61-4158-9805-4a6f1ba91f88>\",\"WARC-IP-Address\":\"178.79.154.177\",\"WARC-Target-URI\":\"https://cheatography.com/fanthom1234/cheat-sheets/python-note-sheet/\",\"WARC-Payload-Digest\":\"sha1:PK52L4HMEHDKIKHD5GPXV7NVEIES24OT\",\"WARC-Block-Digest\":\"sha1:A56K2CL7RSVZYDMBHQOB43L2JE6ATIZE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107879362.3_warc_CC-MAIN-20201022082653-20201022112653-00638.warc.gz\"}"} |
https://matheducators.stackexchange.com/questions/414/where-can-i-find-example-of-lagrange-multipliers-and-constrained-optimization | [
"# Where can I find example of Lagrange multipliers and constrained optimization?\n\nI want give some extra homework about Lagrange multipliers and constrained optimization. I am interested in examples that are complex (for example, those having more than two variables and more than one restriction). Where I can find sufficient examples for students' practice?\n\n• I need any webpage address that contains these examples. Mar 18 '14 at 7:00\n• Just curious, is this for a calculus class? I'm tutoring multivariable calculus and this could be of help. Mar 18 '14 at 10:54\n• Yes @Fantini . I need more examples for Economics students in Calculus class (Math II). Mar 18 '14 at 15:16\n• Ask the Economics faculty for relevant, simple to explain, examples? You might substitute \"realistic\" functions by simpler ones with roughly the right behaviour. Mar 19 '14 at 1:56\n• @vonbrand I need exapmles. Mar 19 '14 at 6:45\n\nSome simple examples you can try:\n\n1) Maximize the product of $n$ positive numbers, given their sum is one.\n\n2)Maximize entropy of $p_1,\\dots, p_n$ (all $p_i>0$, summing to one) that is maximizing $-\\sum p_i \\log p_i$.\n\n3) More conditions: Maximizing entropy, but with the extra condition that $\\sum a_i p_i = \\mu$. You can also try introducing more conditions (variance?). The $a_i$s are known constants.\n\n4) You can find a few more good problems here: https://www.math.ucdavis.edu/~thomases/W11_16C1_lec_2_4_11.pdf\n\n5) A few more good problems here: http://homepages.math.uic.edu/~dcabrera/practice_exams/topics/lagrangemultipliers.html\n\n• Second link was exactly which I had needed. Mar 19 '14 at 19:57"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8246835,"math_prob":0.86955,"size":1636,"snap":"2021-43-2021-49","text_gpt3_token_len":454,"char_repetition_ratio":0.11580882,"word_repetition_ratio":0.017167382,"special_character_ratio":0.27139366,"punctuation_ratio":0.15050167,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987833,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-17T09:46:27Z\",\"WARC-Record-ID\":\"<urn:uuid:e3c5f633-289a-4040-94f0-005024e2caa0>\",\"Content-Length\":\"171205\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f78866f7-83b8-4247-a418-344de64d0d5e>\",\"WARC-Concurrent-To\":\"<urn:uuid:197807df-dffd-41b0-a346-0521445d6ae7>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://matheducators.stackexchange.com/questions/414/where-can-i-find-example-of-lagrange-multipliers-and-constrained-optimization\",\"WARC-Payload-Digest\":\"sha1:YLG5P3YBCZ4KUGTU6FFIHOBICFPRV62T\",\"WARC-Block-Digest\":\"sha1:BC3JFUY7YOYZWBDM3TJP3EGDK4GYPL63\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585171.16_warc_CC-MAIN-20211017082600-20211017112600-00432.warc.gz\"}"} |
https://www.global-sci.org/intro/article_detail/jcm/9366.html | [
"Volume 10, Issue 4\nOptimal Interpolation of Scattered Data on a Circular Domain with Boundary Conditions\n\nJ. Comp. Math., 10 (1992), pp. 339-347\n\nPublished online: 1992-10\n\nPreview Full PDF 189 1870\nExport citation\n\nCited by\n\n• Abstract\n\nOptimal interpolation problems of scattered data on a circular domain with two different types of boundary value conditions are studied in this paper. Closed-form optimal solutions, a new type of spline functions defined by partial differential operators, are obtained. This type of new splines is a generalization of the well-known $L_g$-splines and thin-plate splines. The standard reproducing kernel structure of the optimal solutions is demonstrated. The new idea and technique developed in this paper are finally generalized to solve the same interpolation problems involving a more general class of partial differential operators on a general region.\n\n• Keywords\n\n@Article{JCM-10-339, author = {}, title = {Optimal Interpolation of Scattered Data on a Circular Domain with Boundary Conditions}, journal = {Journal of Computational Mathematics}, year = {1992}, volume = {10}, number = {4}, pages = {339--347}, abstract = { Optimal interpolation problems of scattered data on a circular domain with two different types of boundary value conditions are studied in this paper. Closed-form optimal solutions, a new type of spline functions defined by partial differential operators, are obtained. This type of new splines is a generalization of the well-known $L_g$-splines and thin-plate splines. The standard reproducing kernel structure of the optimal solutions is demonstrated. The new idea and technique developed in this paper are finally generalized to solve the same interpolation problems involving a more general class of partial differential operators on a general region. }, issn = {1991-7139}, doi = {https://doi.org/}, url = {http://global-sci.org/intro/article_detail/jcm/9366.html} }\nTY - JOUR T1 - Optimal Interpolation of Scattered Data on a Circular Domain with Boundary Conditions JO - Journal of Computational Mathematics VL - 4 SP - 339 EP - 347 PY - 1992 DA - 1992/10 SN - 10 DO - http://doi.org/ UR - https://global-sci.org/intro/article_detail/jcm/9366.html KW - AB - Optimal interpolation problems of scattered data on a circular domain with two different types of boundary value conditions are studied in this paper. Closed-form optimal solutions, a new type of spline functions defined by partial differential operators, are obtained. This type of new splines is a generalization of the well-known $L_g$-splines and thin-plate splines. The standard reproducing kernel structure of the optimal solutions is demonstrated. The new idea and technique developed in this paper are finally generalized to solve the same interpolation problems involving a more general class of partial differential operators on a general region."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7119485,"math_prob":0.918088,"size":3074,"snap":"2021-21-2021-25","text_gpt3_token_len":883,"char_repetition_ratio":0.17687297,"word_repetition_ratio":0.39814815,"special_character_ratio":0.3458035,"punctuation_ratio":0.08203125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95410365,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-05T22:27:09Z\",\"WARC-Record-ID\":\"<urn:uuid:1ff7751e-ae4e-42e7-b06e-4ebcac92b461>\",\"Content-Length\":\"75863\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d58f7fb6-4abd-498a-bae9-011c8432f81e>\",\"WARC-Concurrent-To\":\"<urn:uuid:60c0c5c5-3b18-40bf-8e53-8168d9b6ffe9>\",\"WARC-IP-Address\":\"47.52.67.10\",\"WARC-Target-URI\":\"https://www.global-sci.org/intro/article_detail/jcm/9366.html\",\"WARC-Payload-Digest\":\"sha1:ETIZ7BT5THEYO3BVPQUG2754WC24RDP4\",\"WARC-Block-Digest\":\"sha1:GUMH72LHYP5NNKEU4EARUDADFMU35XS3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988696.23_warc_CC-MAIN-20210505203909-20210505233909-00373.warc.gz\"}"} |
https://documen.tv/use-the-number-line-below-where-rs-5y-5-st-2y-5-and-rt-66-a-what-is-the-value-of-y-b-find-rs-and-28292952-91/ | [
"Question\n\nUse the number line below, where RS=5y +5, ST = 2y + 5, and RT= 66.\na. What is the value of y?\nb. Find RS and ST.\nR\na. What is the value of y?\n(Type an integer or a decimal.)\n\n1.",
null,
"Jezebel\nUse the number line below, where RS=5y +5, ST = 2y + 5, and RT= 66.The value of y=8,the value of RS=45, and ST=21.\nWhat is Number Line?\n• In math, a number line can be defined as a straight line with numbers placed at equal intervals or segments along its length.\n• A number line can be extended infinitely in any direction and is usually represented horizontally.\n• The numbers on the number line increase as one moves from left to right and decrease on moving from right to left.\n• A number line can also be used to represent both positive and negative integers as well.\n• Writing numbers on a number line make comparing numbers easier. The numbers on the left are smaller than the numbers on the right of the number line.\n• A number line can also be used to carry out addition, subtraction, and multiplication. We always move right to add, move left to subtract, and skip count to multiply.\nRS=5y +5\nST = 2y + 5\nRT= 66\nRS+ST=RT\n(5y +5)+(2y + 5)=66\n7y+10=66\n7y=66-10\n7y=56\ny=8\nRS=5y +5,\nput value of y in these question.\nRS=5(8)+5\nRS=45\nST = 2y + 5,\nput value of y in these question.\nST = 2(8)+ 5\nST=21"
] | [
null,
"https://documen.tv/wp-content/litespeed/avatar/02ae93fd6ec71ade8be79bb94062ad66.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.69203234,"math_prob":0.9995646,"size":242,"snap":"2023-14-2023-23","text_gpt3_token_len":92,"char_repetition_ratio":0.11344538,"word_repetition_ratio":0.5555556,"special_character_ratio":0.3966942,"punctuation_ratio":0.2112676,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998387,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-21T23:19:54Z\",\"WARC-Record-ID\":\"<urn:uuid:c1dd0c60-0b3a-4098-bea3-810bd6f03f91>\",\"Content-Length\":\"65647\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7b6d9059-7792-46c8-906c-66b47a116a8e>\",\"WARC-Concurrent-To\":\"<urn:uuid:560ac428-fb9f-4148-acb7-6e50bc0f996e>\",\"WARC-IP-Address\":\"103.56.160.25\",\"WARC-Target-URI\":\"https://documen.tv/use-the-number-line-below-where-rs-5y-5-st-2y-5-and-rt-66-a-what-is-the-value-of-y-b-find-rs-and-28292952-91/\",\"WARC-Payload-Digest\":\"sha1:SZAWWMPNYRXCOPGVN3CKSLBEZFJVXBU6\",\"WARC-Block-Digest\":\"sha1:BJSF6SBMSKNTDK4MFCUNZSBGUV2VZZHW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943747.51_warc_CC-MAIN-20230321225117-20230322015117-00160.warc.gz\"}"} |
https://optovr.com/number-patterns-grade-math-worksheets-combination-set-large-quadrilateral-3-test-ordering-decimals-converting-fractions-worksheet-square-root-4-patterning/ | [
"# Number Patterns Grade Math Worksheets Combination Set Large Quadrilateral 3 Test Ordering Decimals Converting Fractions Worksheet Square Root 4 Patterning",
null,
"Number Patterns Grade Math Worksheets Combination Set Large Quadrilateral 3 Test Ordering Decimals Converting Fractions Worksheet Square Root 4 Patterning.\n\nPattern recognition and prediction skills lay an important foundation for subjects like math, poetry, and more. with our patterns worksheets and, students of all ages and levels can explore patterns, use their reasoning skills to complete them, and even create their own.\n\nGrade 4 number patterns worksheets worksheet answer math problem free exam templates algebra questions answers learn match collection patterning. Worksheet grade number patterns specialists image ideas 4 worksheets jolly phonics essential math 3 digit 1 division patterning. Number patterns worksheets teachers pay grade 4 patterning. Grade 3 maths worksheets geometry geometric patterns shapes numbers lets share knowledge 4 patterning. Grade patterns worksheets fun teaching 4 patterning. Free repeating patterns worksheet foundation stage grade 4 patterning worksheets. Pattern worksheets printable activities grade plural nouns build number worksheet math practice book 4 patterning."
] | [
null,
"https://optovr.com/images/number-patterns-grade-math-worksheets-combination-set-large-quadrilateral-3-test-ordering-decimals-converting-fractions-worksheet-square-root-4-patterning.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77522737,"math_prob":0.8134035,"size":1270,"snap":"2021-04-2021-17","text_gpt3_token_len":224,"char_repetition_ratio":0.22748815,"word_repetition_ratio":0.17751479,"special_character_ratio":0.15826772,"punctuation_ratio":0.07978723,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9568775,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-16T07:58:24Z\",\"WARC-Record-ID\":\"<urn:uuid:cb9f0018-e174-47f9-bf3f-25458ac69fb9>\",\"Content-Length\":\"22073\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c2ce4ff2-7b65-449d-9b2d-d758bca468b4>\",\"WARC-Concurrent-To\":\"<urn:uuid:bb08d069-8c6d-4605-929b-8e094fb908cf>\",\"WARC-IP-Address\":\"172.67.133.61\",\"WARC-Target-URI\":\"https://optovr.com/number-patterns-grade-math-worksheets-combination-set-large-quadrilateral-3-test-ordering-decimals-converting-fractions-worksheet-square-root-4-patterning/\",\"WARC-Payload-Digest\":\"sha1:2HNBRPASGKQX4DQ6KL6LBDVMC6GWD7LE\",\"WARC-Block-Digest\":\"sha1:CB66R7ELG5HDZVOZP7ZXXGTE5UMHC3PA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038088731.42_warc_CC-MAIN-20210416065116-20210416095116-00609.warc.gz\"}"} |
https://asastandards.org/terms/interaural-cross-correlation/ | [
"Select Page\n\n7.29 cross correlation; interaural cross correlation. Covariance of signals vs. time shift; i.e., the degree of similarity of a reference signal and a time-shifted signal, as a function of time shift or delay.\n\nAnnotation 1 The cross correlation is calculated as",
null,
"where T is the observation period, and a(t) and b(t) are the respective signals.\n\nAnnotation 2 The coefficient of correlation varies from +1 (perfectly correlated; i.e., the two signals are identical), through zero (completely uncorrelated; i.e., the two signals are completely independent or orthogonal), to –1 (perfectly correlated, but with a complete 180° phase inversion between the signals).\n\nAnnotation 3 Auto correlation is the degree of similarity of a signal and a time-shifted version of itself.\n\nAnnotation 4 Interaural cross correlation describes the correlation between signals received at the two ears of a listener.\n\n« Back to Standards Terminolgy Index"
] | [
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"https://p2e2d2i6.rocketcdn.me/wp-content/uploads/2016/09/7.29-300x76.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8992806,"math_prob":0.9815331,"size":942,"snap":"2023-40-2023-50","text_gpt3_token_len":200,"char_repetition_ratio":0.17910448,"word_repetition_ratio":0.04347826,"special_character_ratio":0.20700637,"punctuation_ratio":0.14772727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9797691,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T06:47:48Z\",\"WARC-Record-ID\":\"<urn:uuid:8bf9dc3c-bad8-4947-bb4c-e1d83f715574>\",\"Content-Length\":\"201139\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1312eec5-f23b-48fd-bb73-ea42ce0ecff3>\",\"WARC-Concurrent-To\":\"<urn:uuid:939ad085-eddc-442d-b5e4-155acf732e5f>\",\"WARC-IP-Address\":\"72.167.34.78\",\"WARC-Target-URI\":\"https://asastandards.org/terms/interaural-cross-correlation/\",\"WARC-Payload-Digest\":\"sha1:6P6JUU6VZNJMJEBZ7QEQIRXZ72T3U7BO\",\"WARC-Block-Digest\":\"sha1:6TQN5HFRZM4XER4VLEB45E4Z3H7WSTJ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511055.59_warc_CC-MAIN-20231003060619-20231003090619-00596.warc.gz\"}"} |
https://discuss.codechef.com/t/digsmpar-editorial/100348 | [
"",
null,
"# DIGSMPAR Editorial\n\nSetter: Srikkanth R\nTester: Harris Leung\nEditorialist: Pratiyush Mishra, Jakub Safin\n\nCakewalk\n\nBasic Arithmetic\n\n# PROBLEM:\n\nFor a positive integer M, let 𝚍𝚒𝚐𝚒𝚝𝚂𝚞𝚖(𝙼) be the sum of digits of the number M (when written in decimal).\n\nFor example, 𝚍𝚒𝚐𝚒𝚝𝚂𝚞𝚖(𝟷𝟶𝟸𝟹)=1+0+2+3=6.\n\nGiven a positive integer N, find the smallest integer X greater than N such that\n\n𝚍𝚒𝚐𝚒𝚝𝚂𝚞𝚖(𝙽) and 𝚍𝚒𝚐𝚒𝚝𝚂𝚞𝚖(𝚇) have different parity, i.e. one of them is odd and the other is even\n\n# EXPLANATION:\n\nLet I be the given integer, S be the digit sum, x be the last digit.\n\nCase 1: S+1 has different parity as S\nAns: (I+1)\n\nCase 2: S+1 has same parity as S\n\\implies x = 9\n\\implies x+1 = 10\n\\therefore x for S+1 will be 0 and previous digits would be affected\n\\implies x+1 = 1\n\\therefore x for S+2 will be 1 and previous digits won’t be affected\n\\implies S+2 will have different parity than S+1 and hence, S\nAns: (I+2)\n\n# TIME COMPLEXITY:\n\nThe above computation can be done in constant time. Hence, the solution has a time complexity of O(1).\n\n# SOLUTION:\n\nfor calculating the sum of digits of n,there would be time complexity of o(log n) , but u said O\n(1) , how ??\n\n1 Like\n\nIt is O(length of n) that is const approx\n\n1 Like\n\nokkk ,thnxx i got your point\n\n``````#include <iostream>\nusing namespace std;\n\nlong int sum(long int n){\nlong int temp = n;\nlong int sum1 = 0;\nwhile(temp>0){\nsum1 = sum1 + temp%10;\ntemp = temp/10;\n}\nreturn sum1;\n}\n\nint main() {\nint t;\ncin>>t;\nwhile(t--){\nlong int n;\ncin>>n;\nif(n%10 == 9){\nif(sum(n)%2 == 0){\nif(sum(n+1)%2==0){\ncout<<n+2<<endl;\n}\nelse{\ncout<<n+1<<endl;\n}\n}\nelse{\ncout<<n+1<<endl;\n}\n}\nelse{\ncout<<n+1<<endl;\n}\n}\n\nreturn 0;\n}\n``````\n\nCan someone please tell me , what’s wrong with this solution I am getting correct o/p\n\nHey, there is a problem in the logic of your code.\nHere is a test cases on which I found your code’s output wrong wrong\n\n3\n9\n10\n99\n\nHere I have made few changes in your code and it passed, please go through it to understand:\n\n``````#include <iostream>\nusing namespace std;\n\nlong int sum(long int n){\nlong int temp = n;\nlong int sum1 = 0;\nwhile(temp>0){\nsum1 = sum1 + temp%10;\ntemp = temp/10;\n}\nreturn sum1;\n}\n\nint main() {\nint t;\ncin>>t;\nwhile(t--){\nlong int n;\ncin>>n;\nif(n%10 == 9){\nif(sum(n)%2 == 0){\nif(sum(n+1)%2==0){\ncout<<n+2<<endl;\n}\nelse{\ncout<<n+1<<endl;\n}\n}\nelse{\nif(sum(n+1)%2==1){\ncout<<n+2<<endl;\n}\nelse{\ncout<<n+1<<endl;\n}\n}\n}\nelse{\ncout<<n+1<<endl;\n}\n}\n\nreturn 0;\n}\n``````\n\n``````using namespace std;\n\nint main() {\nint t;\ncin >> t;\n\nwhile(t--){\nlong long n;\ncin >> n;\n\nif(n % 10 == 9){\ncout << n + 2 << \"\\n\";\n} else{\ncout << n + 1 << \"\\n\";\n}\n\n}\nreturn 0;\n}\n``````\n\nIt works on test cases. But shows wrong on submit. Can anyone help me find the error?\n\n``````#include<iostream>\nusing namespace std;\n\nint main() {\nint l;\ncin>>l;\nint n, s;\nwhile(l--){\ns=0;\ncin>>n;\nint x=n;\nwhile(s%2==0){\nx=n;\nwhile(x>0){\ns+=x%10;\nx/=10;\n}\nn++;\n}\ncout<<--n<<endl;\n}\nreturn 0;\n}\n``````\n\nCongrats on posting your first query!\n\n``````#include<iostream>\nusing namespace std;\nint digitSum(int n){\nint sum=0;\nwhile(n>0){\nsum+=n%10;\nn=n/10;\n}\nreturn sum;\n}\nint main() {\nint l;\ncin>>l;\nint n, s;\nwhile(l--){\ncin>>n;\nint x=n;\nwhile(digitSum(n)%2==digitSum(x)%2){// increase the number x until the parity of digitSum of n is not same as parity of digitSum of x\nx++;\n\n}\ncout<<x<<endl;\n}\nreturn 0;\n}\n``````\n\nHey @mannshah1211 , @kuldeep_singh",
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","
] | [
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"https://s3.amazonaws.com/discourseproduction/original/3X/7/f/7ffd6e5e45912aba9f6a1a33447d6baae049de81.svg",
null,
"https://discuss.codechef.com/images/emoji/apple/wave.png",
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https://www.afbcash.live/live-casino/winning-for-roulette/ | [
"# Winning for Roulette | AFBCash.live",
null,
"## Roulette winning formula The best analytical techniques to do is get rich 2019\n\nThe success formula winning for roulette can be enriched every day. There are many different technologies for uploading to social media for us to choose to learn and apply. Betting formulas in roulette can be used and earn real money. In this article will introduce a lot of formulas. Let’s see what should not be done in step betting.",
null,
"## Winning for roulette\n\n### Formula 1: The best roulette technique to play\n\nThis formula will focus on the numbers bet. Which must bet on numbers in a distributed form Which must be distributed as much as possible Dooya must select a total of 25 numbers from 37 numbers and bet RM 50 each. The initial cost will be RM 1250. If you correctly guess 1 number, you will get RM 1800 money minus the cost. There will be RM 550 profit and then take The remaining capital will be further expanded.\n\n### Formula 2: Roulette formula for beginners\n\nThis formula is known as the James Bond formula. It will cost 2 thousand ringgit. The bet divided as Place a high bet of RM 1400 (numbers 19 – 36), staggering 6 numbers, RM 500 (numbers 13-18) and finally, stabbing. Zero number RM 100. If correctly guessed, if the iron ball falls at the numbers 19 – 36, we will profit RM 800. If falling 13 – 18 will get a RM 100 profit. Which is a simple profit of RM 1,600. If the iron ball falls to 1 – 12, our capital will be lost.\n\n### Formula 3: How to play roulette to make money\n\nFocus on the edge of the table Because edge numbers will come out very often, but just getting the bet amount is not equal to the middlebox\n\n### Formula 4: Board for placing bets on roulette games\n\nRoulette winning formula This formula is the final formula Which will be suitable only for heavy bag people Which if any gambler applies this formula In playing baccarat I want to say that this formula is the fastest because the simplest is that this formula will be Make more investment plans by using this formula Will be compounded from the lost capital\n\nWhich this formula will be used only when the bettor Would like to get the capital back From the previous loss In a very easy way Just the bettor will have to increase the funds to bet. When losing the first eye Will increase in the next eye to double. How to apply AFBCASH Malaysia\n\nA minimum bet of RM 10 with AFBCASH Malaysia returns a commission of 0.5% for every bet placed, withdrawal and withdrawal services. Without limit And there is no minimum 2-hour limit."
] | [
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"https://www.afbcash.live/wp-content/uploads/2019/09/Afbcash-10-Bonus-01_Small-Banner_040719-300x216.jpg",
null,
"https://www.afbcash.live/wp-content/uploads/2019/09/Afbcash-Online-Casino-02_Web-Banner-280619-1024x417.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8908238,"math_prob":0.89388984,"size":2471,"snap":"2021-31-2021-39","text_gpt3_token_len":566,"char_repetition_ratio":0.15403324,"word_repetition_ratio":0.0,"special_character_ratio":0.2335087,"punctuation_ratio":0.07474747,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.955767,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T18:57:03Z\",\"WARC-Record-ID\":\"<urn:uuid:a5362c9b-6169-44d8-94d9-1376513a8ca5>\",\"Content-Length\":\"33766\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2c8f242-e303-4eb1-b378-f24e713c4d55>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab2d33ba-1990-4199-8536-544f1415e89a>\",\"WARC-IP-Address\":\"172.67.223.26\",\"WARC-Target-URI\":\"https://www.afbcash.live/live-casino/winning-for-roulette/\",\"WARC-Payload-Digest\":\"sha1:C53TRY6EU2OF6WESO7OOPB36NYMJZW7Z\",\"WARC-Block-Digest\":\"sha1:3CMZNVMBG7VRVMJBFKCVE6MMRKAYA6W4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780055775.1_warc_CC-MAIN-20210917181500-20210917211500-00451.warc.gz\"}"} |
http://blog-studia-historiae-scientiarum.pau.krakow.pl/foundations-of-mathematics-and-mathematical-practice-the-case-of-polish-mathematical-school/ | [
"",
null,
"",
null,
"# Foundations of Mathematics and Mathematical Practice. The Case of Polish Mathematical School\n\nThe foundations of mathematics cover mathematical as well as philosophical problems. At the turn of the 20th century logicism, formalism and intuitionism, main foundational schools were developed. A natural problem arose, namely of how much the foundations of mathematics influence the real practice of mathematicians. Although mathematics was and still is declared to be independent of philosophy, various foundational controversies concerned some mathematical axioms, e.g. the axiom of choice, or methods of proof (particularly, non-constructive ones) and sometimes qualified them as admissible (or not) in mathematical practice, relatively to their philosophical (and foundational) content. Polish Mathematical School was established in the years 1915–1920. Its research program was outlined by Zygmunt Janiszewski (the Janiszewski program) and suggested that Polish mathematicians should concentrate on special branches of studies, including set theory, topology and mathematical logic. In this way, the foundations of mathematics became a legitimate part of mathematics. In particular, the foundational investigations should be conducted independently of philosophical assumptions and apply all mathematically accepted methods, finitary or not, and the same concerns other branches of mathematics. This scientific ideology contributed essentially to the remarkable development of logic, set theory and topology in Poland.\n\nsee the article on the SHS website"
] | [
null,
"http://blog-studia-historiae-scientiarum.pau.krakow.pl/wp-content/uploads/2022/12/michael-dziedzic-nbW-kaz2BlE-unsplash-scaled.jpg",
null,
"http://blog-studia-historiae-scientiarum.pau.krakow.pl/wp-content/uploads/2022/12/michael-dziedzic-nbW-kaz2BlE-unsplash-scaled.jpg",
null
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https://nl.mathworks.com/matlabcentral/cody/problems/44950-calculate-inner-product/solutions/2237485 | [
"Cody\n\n# Problem 44950. Calculate Inner Product\n\nSolution 2237485\n\nSubmitted on 26 Apr 2020 by Andrzej Mamczura\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = [1 2]; y = [2;3]; z_correct = 8; assert(isequal(in_prod(x,y),z_correct))\n\nz = 8\n\n2 Pass\nx = -5; y = 100; z_correct = -500; assert(isequal(in_prod(x,y),z_correct))\n\nz = -500\n\n3 Pass\nx = [1 2 3]; y = [2;3]; assert(isStringScalar(in_prod(x,y)))\n\nz = \"The inner dimensions are 1 and 1Matrix multiplication not possibke\"\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.63782394,"math_prob":0.99381787,"size":708,"snap":"2020-45-2020-50","text_gpt3_token_len":212,"char_repetition_ratio":0.12215909,"word_repetition_ratio":0.0,"special_character_ratio":0.3375706,"punctuation_ratio":0.125,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99613124,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-25T14:39:46Z\",\"WARC-Record-ID\":\"<urn:uuid:03008de6-3ab1-41d8-a33b-926ddb0e7429>\",\"Content-Length\":\"80076\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8b96cd8-a27f-4a39-9370-1c367b5bb55e>\",\"WARC-Concurrent-To\":\"<urn:uuid:d6ff5978-de8c-4052-bd2e-a5c62c9f9d3f>\",\"WARC-IP-Address\":\"23.194.181.89\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/44950-calculate-inner-product/solutions/2237485\",\"WARC-Payload-Digest\":\"sha1:DA3MNT4ORMCAQOJ3VM7ICJG5HG6JMJZY\",\"WARC-Block-Digest\":\"sha1:Q2MO6HJEN5BN4J64CUZXH3PONJY5SZRM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107889173.38_warc_CC-MAIN-20201025125131-20201025155131-00267.warc.gz\"}"} |
https://math.stackexchange.com/questions/615937/converting-from-spherical-coordinates-to-cartesian-around-arbitrary-vector-n | [
"# Converting from spherical coordinates to cartesian around arbitrary vector $N$\n\nSo if I'm given an arbitrary unit vector $N$ and another vector $V$ defined in spherical coordinates $\\theta$ (polar angle between $N$ and $V$) and $\\phi$ (azimuthal angle) and $r = 1$. How do I convert vector $V$ into cartesian coordinates?\n\nNow, I know that in general the conversion from spherical to cartesian is as follows:\n\n$$x = r \\sin \\theta \\cos \\phi$$ $$y = r \\sin \\theta \\sin \\phi$$ $$z = r \\cos \\theta$$\n\nHowever, since the angles $\\theta$ and $\\phi$ are defined respective to the vector $N$ and not the axes, the above conversion wouldn't work, yes? So how would I go about modifying the conversion?\n\nThis is not well defined. You really need three vectors to define the spherical coordinate system. With only $n$, how do you decide where $\\phi = 0$ is?\nAnyway, suppose that you do have an orthonormal basis $e_1, e_2, e_3 = n$. Then you can write V as $x e_1 + y e_2 + z e_3$, where $x,y,z$ are given by the formula you gave. If you have the Cartesian coordinates of $e_1, e_2, e_3$, then you are done."
] | [
null
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http://ixtrieve.fh-koeln.de/birds/litie/document/34780 | [
"# Document (#34780)\n\nAuthor\nBehrens-Neumann, R.\nTitle\nAus der 55. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 5. und 6. November 2008 in Frankfurt am Main\nSource\nBibliotheksdienst. 43(2009) H.2, S.139-181\nYear\n2009\nSeries\nThemen: Gremien\nAbstract\nAm 5. und 6. November 2008 fand die 55. Sitzung der Arbeitsgemeinschaft der Verbundsysteme auf Einladung der Deutschen Nationalbibliothek in Frankfurt am Main statt. Zusammenarbeit der Verbundsysteme In der diesjährigen Herbstsitzung der Arbeitsgemeinschaft der Verbundsysteme wurde einmal mehr die intensive Zusammenarbeit der Verbünde in ihrer Rolle als Dienstleister für die Bibliotheken deutlich. Die vielfältigen Aktivitäten der regionalen Verbundzentralen sichern die Funktionalität der einzelnen Bibliotheken vor Ort und gleichen strukturelle Unterschiede aus, sei es zwischen öffentlichen und wissenschaftlichen Bibliotheken, sei es zwischen Bibliotheken, Museen und Archiven. Solche Dienstleistungen zu bündeln und Kooperationen einzugehen ist unter den Mitgliedern der Arbeitsgemeinschaft der Verbundsysteme längst schon Alltag. Hierzu ein kurzer Überblick über den aktuellen Stand.\nContent\nDarin auch: \"Crisscross Ziel des von der DFG geförderten Projekts ist die Schaffung eines multilingualen, thesaurusbasierten und benutzerorientierten Recherchevokabulars zu heterogen erschlossenen Dokumenten für die Nutzer des deutschen wissenschaftlichen Bibliothekswesens. Dazu werden die Sachschlagwörter der SWD mit den Notationen der Dewey Dezimalklassifikation verbunden sowie mit ihren Äquivalenten in LCSH und Rameau verknüpft. Im Berichtszeitraum wurde der Verlängerungsantrag bewilligt, der eine Laufzeit von 24 Monaten umfasst. Projektende ist jetzt ca. Januar 2010. In der Vergangenheit wurde der Schwerpunkt auf die Vergabe der DDC-Notationen für die Sachschlagwörter der SWD gelegt. Bisher nicht berücksichtigt wurden die Verknüpfungen der Sachschlagwörter mit den äquivalenten LCSH- und Rameau-Termen. Arbeitstechnische Mängel in der MACS-Datenbank (u.a. nicht aktuelle Datenbasis, Dubletten, fehlende Identifier) haben die Verlinkungsarbeit behindert und konnten im Berichtszeitraum in Zusammenarbeit mit den MACS-Partnern, insbesondere der Schweizerischen Nationalbibliothek, gelöst werden. Initiert wurde u.a. ein Update-Verfahren, das nun regelmäßig die Original-Thesauri und die vorhandenen Pärchen aktualisiert, so dass die konkrete Verlinkungsarbeit - unmittelbar in der LMI-Umgebung - demnächst beginnen kann.\" (S.161)\nTheme\nKatalogfragen allgemein\nObject\nCrissCross\nLocation\nD\n\n## Similar documents (author)\n\n1. Behrens-Neumann, R.: Aus der 53. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 6. und 7. November 2007 in München (2008) 5.83\n```5.830306 = sum of:\n5.830306 = sum of:\n2.8977642 = weight(author_txt:neumann in 4218) [ClassicSimilarity], result of:\n2.8977642 = score(doc=4218,freq=1.0), product of:\n0.70428926 = queryWeight, product of:\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.08558725 = queryNorm\n4.114452 = fieldWeight in 4218, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.5 = fieldNorm(doc=4218)\n2.9325418 = weight(author_txt:behrens in 4218) [ClassicSimilarity], result of:\n2.9325418 = score(doc=4218,freq=1.0), product of:\n0.7099131 = queryWeight, product of:\n1.0039847 = boost\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.08558725 = queryNorm\n4.1308465 = fieldWeight in 4218, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.5 = fieldNorm(doc=4218)\n```\n2. Behrens-Neumann, R.: Aus der 54. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 22. und 23. April 2008 in Konstanz (2008) 5.83\n```5.830306 = sum of:\n5.830306 = sum of:\n2.8977642 = weight(author_txt:neumann in 4220) [ClassicSimilarity], result of:\n2.8977642 = score(doc=4220,freq=1.0), product of:\n0.70428926 = queryWeight, product of:\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.08558725 = queryNorm\n4.114452 = fieldWeight in 4220, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.5 = fieldNorm(doc=4220)\n2.9325418 = weight(author_txt:behrens in 4220) [ClassicSimilarity], result of:\n2.9325418 = score(doc=4220,freq=1.0), product of:\n0.7099131 = queryWeight, product of:\n1.0039847 = boost\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.08558725 = queryNorm\n4.1308465 = fieldWeight in 4220, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.5 = fieldNorm(doc=4220)\n```\n3. Behrens-Neumann, R.: Aus der 56. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 23. April 2009 in Wien : ein Bericht (2009) 5.83\n```5.830306 = sum of:\n5.830306 = sum of:\n2.8977642 = weight(author_txt:neumann in 42) [ClassicSimilarity], result of:\n2.8977642 = score(doc=42,freq=1.0), product of:\n0.70428926 = queryWeight, product of:\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.08558725 = queryNorm\n4.114452 = fieldWeight in 42, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.5 = fieldNorm(doc=42)\n2.9325418 = weight(author_txt:behrens in 42) [ClassicSimilarity], result of:\n2.9325418 = score(doc=42,freq=1.0), product of:\n0.7099131 = queryWeight, product of:\n1.0039847 = boost\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.08558725 = queryNorm\n4.1308465 = fieldWeight in 42, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.5 = fieldNorm(doc=42)\n```\n4. Behrens-Neumann, R.: Aus der 62. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 17. und 18. April 2012 in München (2012) 5.83\n```5.830306 = sum of:\n5.830306 = sum of:\n2.8977642 = weight(author_txt:neumann in 344) [ClassicSimilarity], result of:\n2.8977642 = score(doc=344,freq=1.0), product of:\n0.70428926 = queryWeight, product of:\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.08558725 = queryNorm\n4.114452 = fieldWeight in 344, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.5 = fieldNorm(doc=344)\n2.9325418 = weight(author_txt:behrens in 344) [ClassicSimilarity], result of:\n2.9325418 = score(doc=344,freq=1.0), product of:\n0.7099131 = queryWeight, product of:\n1.0039847 = boost\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.08558725 = queryNorm\n4.1308465 = fieldWeight in 344, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.5 = fieldNorm(doc=344)\n```\n5. Behrens-Neumann, R.: Aus der 57. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 4. und 5. November 2009 in Köln (2010) 5.83\n```5.830306 = sum of:\n5.830306 = sum of:\n2.8977642 = weight(author_txt:neumann in 497) [ClassicSimilarity], result of:\n2.8977642 = score(doc=497,freq=1.0), product of:\n0.70428926 = queryWeight, product of:\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.08558725 = queryNorm\n4.114452 = fieldWeight in 497, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.5 = fieldNorm(doc=497)\n2.9325418 = weight(author_txt:behrens in 497) [ClassicSimilarity], result of:\n2.9325418 = score(doc=497,freq=1.0), product of:\n0.7099131 = queryWeight, product of:\n1.0039847 = boost\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.08558725 = queryNorm\n4.1308465 = fieldWeight in 497, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.261693 = idf(docFreq=29, maxDocs=42740)\n0.5 = fieldNorm(doc=497)\n```\n\n## Similar documents (content)\n\n1. Oehlschläger, S.: Aus der 51. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 29. und 30. November 2006 in Frankfurt am Main (2007) 0.77\n```0.76613474 = sum of:\n0.76613474 = product of:\n2.7361956 = sum of:\n0.15850887 = weight(abstract_txt:einladung in 2040) [ClassicSimilarity], result of:\n0.15850887 = score(doc=2040,freq=1.0), product of:\n0.12327969 = queryWeight, product of:\n1.0966012 = boost\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.0136615755 = queryNorm\n1.2857662 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.228904 = idf(docFreq=30, maxDocs=42740)\n0.15625 = fieldNorm(doc=2040)\n0.056284457 = weight(abstract_txt:main in 2040) [ClassicSimilarity], result of:\n0.056284457 = score(doc=2040,freq=1.0), product of:\n0.07788578 = queryWeight, product of:\n1.2326707 = boost\n4.6249847 = idf(docFreq=1138, maxDocs=42740)\n0.0136615755 = queryNorm\n0.72265387 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.6249847 = idf(docFreq=1138, maxDocs=42740)\n0.15625 = fieldNorm(doc=2040)\n0.1905286 = weight(abstract_txt:november in 2040) [ClassicSimilarity], result of:\n0.1905286 = score(doc=2040,freq=1.0), product of:\n0.1755925 = queryWeight, product of:\n1.8508489 = boost\n6.9443917 = idf(docFreq=111, maxDocs=42740)\n0.0136615755 = queryNorm\n1.0850612 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.9443917 = idf(docFreq=111, maxDocs=42740)\n0.15625 = fieldNorm(doc=2040)\n0.19277214 = weight(abstract_txt:frankfurt in 2040) [ClassicSimilarity], result of:\n0.19277214 = score(doc=2040,freq=1.0), product of:\n0.17696823 = queryWeight, product of:\n1.8580853 = boost\n6.971543 = idf(docFreq=108, maxDocs=42740)\n0.0136615755 = queryNorm\n1.0893036 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n6.971543 = idf(docFreq=108, maxDocs=42740)\n0.15625 = fieldNorm(doc=2040)\n0.26091495 = weight(abstract_txt:sitzung in 2040) [ClassicSimilarity], result of:\n0.26091495 = score(doc=2040,freq=1.0), product of:\n0.21653685 = queryWeight, product of:\n2.055341 = boost\n7.711647 = idf(docFreq=51, maxDocs=42740)\n0.0136615755 = queryNorm\n1.2049448 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.711647 = idf(docFreq=51, maxDocs=42740)\n0.15625 = fieldNorm(doc=2040)\n0.8675706 = weight(title_txt:arbeitsgemeinschaft in 2040) [ClassicSimilarity], result of:\n0.8675706 = score(doc=2040,freq=1.0), product of:\n0.48565406 = queryWeight, product of:\n4.3530784 = boost\n8.166383 = idf(docFreq=32, maxDocs=42740)\n0.0136615755 = queryNorm\n1.7863963 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.166383 = idf(docFreq=32, maxDocs=42740)\n0.21875 = fieldNorm(doc=2040)\n1.0096158 = weight(title_txt:verbundsysteme in 2040) [ClassicSimilarity], result of:\n1.0096158 = score(doc=2040,freq=1.0), product of:\n0.5788036 = queryWeight, product of:\n5.3131685 = boost\n7.974011 = idf(docFreq=39, maxDocs=42740)\n0.0136615755 = queryNorm\n1.7443149 = fieldWeight in 2040, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.974011 = idf(docFreq=39, maxDocs=42740)\n0.21875 = fieldNorm(doc=2040)\n0.28 = coord(7/25)\n```\n2. Behrens-Neumann, R.: Aus der 59. 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Behrens-Neumann, R.: Aus der 61. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 23. und 24. 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Oehlschläger, S.: Aus der 49. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 23. und 24. 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Behrens-Neumann, R.: Aus der 53. Sitzung der Arbeitsgemeinschaft der Verbundsysteme am 6. und 7. 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https://www.nuclear-power.net/nuclear-power/reactor-physics/atomic-nuclear-physics/fundamental-interactions-fundamental-forces/electroweak-interaction/ | [
"# Electroweak Interaction – Electroweak Force\n\n## Electroweak Interaction\n\nAlthough weak force is considered one of the four fundamental forces, at high energy the weak force and electromagnetic force are unified as a single electroweak force. The unification energy is on the order of 100 GeV. At low energies electromagnetic and weak interaction appear to be very different. Electroweak theory is very important for modern cosmology. In the history of the universe, during the quark epoch (shortly after the Big Bang) the unified force broke into the two separate forces as the universe cooled.\n\nFor contributions to the unification of the weak and electromagnetic interaction between elementary particles, Abdus Salam, Sheldon Glashow and Steven Weinberg were awarded the 1979 Nobel Prize in Physics.\n\n## Weak Interaction",
null,
"The weak interaction or weak force is one of the four fundamental forces and involves the exchange of the intermediate vector bosons, the W and the Z. Since these bosons are very massive (on the order of 80 GeV, the uncertainty principle dictates a range of about 10-18 meters which is less than the diameter of a proton. As a result, the weak interaction takes place only at very small, sub-atomic distances.\n\nThe weak interaction responsible for some nuclear phenomena such as beta decay, which can be understood in terms of the weak force operating on the quarks within the neutron. One of two down quarks changes into an up quark by emitting a W boson (carries away a negative charge). The W boson then decays into a beta particle and an antineutrino. This process is equivalent to the process, in which a neutrino interacts with a neutron.",
null,
"## Electromagnetic Interaction\n\nThe electromagnetic force is the force responsible for all electromagnetic processes. It acts between electrically charged particles. It is infinite-ranged force, much stronger than gravitational force, obeys the inverse square law, but neither electricity nor magnetism adds up in the way that gravitational force does. Since there are positive and negative charges (poles), these charges tend to cancel each other out. Electromagnetism includes the electrostatic force acting between charged particles at rest, and the combined effect of electric and magnetic forces acting between charged particles moving relative to each other.\n\nThe photon, the quantum of electromagnetic radiation, is an elementary particle, which is the force carrier of the electromagnetic force. Photons are gauge bosons having no electric charge or rest mass and one unit of spin. Common to all photons is the speed of light, the universal constant of physics. In empty space, the photon moves at c (the speed of light – 299 792 458 metres per second).\n\nReferences:\nNuclear and Reactor Physics:\n1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).\n2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.\n3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.\n4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317\n5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467\n6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965\n7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.\n8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.\n9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414."
] | [
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"https://www.nuclear-power.net/wp-content/uploads/weak-and-electromagnetic-interaction-253x300.png",
null,
"https://www.nuclear-power.net/wp-content/uploads/weak-interaction-weak-force-example-1024x262.png",
null
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http://ewyz.spectehnika-rb02.ru/calculate-bmi-centile-children.asp | [
"# Calculate bmi centile children\n\n## A simple guide to classifying body mass index in children https://khub.net › documents › A+simple+guide+to+classifying+body+mas...\n\nGirls with both parents overweight had the most rapid increases in bmi from 5. Bmi percentiles were used to determine the prevalence of overweight in girls in.There is also a bmi centile lookup on the standard 218 chart for less bmi is calculated by dividing weight (in kg) by the square of height (in metres e.Nov 28, 2017 misclassification was calculated comparing the frequency distributions for bmi categories defined by rounded percentiles and zscore.Apr 16, 2019 use our bmi calculator to find out if you or your child are a healthy weight.May 15, 2017 using bmi percentiles, researchers noted that adolescents are status and obesity based on bmi levels specific to a childs age and gender.The calculation is the same as for the bmi for adults, but the results are interpreted differently. The younger the child, the lower are the scores, those for girls are different to those for boys.Sep 14, 2015 you can use the bmi calculator below to find your childs bmi, but its bmi percentiles show how kids measurements compare with others the.This bmi calculator provides initial information about the weight status of the child. In any case, it is not intended as a substitute for professional bmi is a parameter calculated from height and we.\n\ngeneric viagra direct\n\n## Consideration of issues around the use of BMI centile ... https://assets.publishing.service.gov.uk › uploads › attachment_data › file\n\nThe way that bmi is calculated in children is different from the way it is calculated in adults. when a child’s body mass index is being configured, their weight and height are compared to a national.Related: causes of child obesity. Once the bmi value has been calculated, it is plotted on a bmiforage percentile chart. Separate charts are used for boys and.Bmi decreases during the preschool years, then increases into adulthood. The percentile curves show this pattern of growth. If your child has a bodymass index.2, introductionthe bmi centile calculator uses the who growth standard and the who growth reference to calculate the exact bmi centile for boys and girls.Dec 31, 2015 the bmi was calculated from measured height and weight as. Age and percentile in a sample of 1167 norwegian children between.Trust docs id: 9239 author: paediatric dietitians date calculation: bmi = (weight in kg) (height in meters)². In adults : o overweight (91st 98th bmi centile).Bmi stands for body mass index. The normal range is considered to be between 18. the same formula is used to calculate bmi in children and adults. However, the definitions of normal weight.You cannot measure childrens weight using solely bmi. You must calculate bmi and then plot it on the bmiforage percentile graph to find the childs weight.Concluded that bmi is a reasonable measure of body adiposity in children. 38 as in previous 95th (obese) bmi centiles of the 1990 reference population.\n\n## Calculator: Body mass index (BMI) for boys (Patient education ... https://www.uptodate.com › contents › calculator-body-mass-index-bmi-fo...\n\nCalculate and compare your bmi quickly and free of charge online. Whether adult or child. the calculator for children is designed for boys and girls from 5 to 18 years. We do not recommend its use for.Calculate your bmi. Questionnaire. This calculator computes the body mass index and rates it appropriately for men, women, children, juveniles and seniors. The sbmi – an index that has been developed f.Because children are actively growing, the same bmi from different ages can indicate a different severity of being overweight or you first calculate the bmi per individual and compare it to the percen.Bmi is calculated for children and adults in the same way; however the results should be the age and gender specific growth centiles (zones) are based on 17.Jul 27, 2017 a child who has a large upward change in bmi percentile, even if not considered overweight, should be evaluated to determine the cause.Calculating children’s bmi is much simpler than interpreting the answer because unlike adults, children are not done growing. to help interpret children’s bmi, you can graph the information on a gende.Sep 14, 2015 you can use the bmi calculator below to find your childs bmi, but its bmi percentiles show how kids measurements compare with others the.This calculator provides bmi and the corresponding bmiforage percentile based on the cdc growth charts for children and teens (ages 2 through 19 if the child is under 2 years old, bmi cannot be calcul.Bmi for adults. Calculate bmi ( body mass index ). in children and teens, bmi is based on age and sex, thus referred to as bmi for age. A high composition of body fat in children can lead to weightrel.Feb 14, 2018 bmi is a number calculated from your childs height and weight, bmiforagepercentile charts provide measures for boys and girls ages 2 to.\n\n## BMI, BMI %, BMI z-score or BMI centile? - CiteSeerX citeseerx.ist.psu.edu › viewdoc › download\n\nCalculating body mass index (bmi) is the most widely accepted method of bmi cutoffs in children are determined by age and sexspecific percentiles based.Body mass index (bmi) is an important measurement used to determine whether your child is overweight, underweight, or at an ideal weight for his bmi doesn't measure body fat but rather the volume.Childrens body mass index percentile calculator childs body mass index (bmi) calculator which calculates not only your childs body mass index, but also.How to calculate body mass index (bmi) in microsoft excel itfriend exceltricks продолжительность: 3:37 itfriend 37 029 просмотров. bmi & children продолжительность: 2:04 drmdk 5 805 просмотров.The percentiles generated for bmiforage were: 5th, 10th, 25th, 50th, 75th, figure 1 body mass index (bmi) curves for male children and adolescents with.Use this calculator to check your body mass index (bmi) and find out if youre a a childs bmi is expressed as a centile to show how their bmi compares with.Body mass index (bmi) is used to assess whether or not children are considered obese, overweight,underweight. for children, bmi is plotted on a growth chart that uses percentile lines to tell whether.Percentiles for children and adolescents. Methodology. B) to develop weight, height and bmi percentiles the bmi was calculated by applying the formula.Jun 29, 2018 child & teen bmi calculator · bmi percentile calculator for child and teen: results on a growth chart · childrens bmi tool for schools.Nov 12, 2018 once the bmi is calculated, the next step is to plot it onto the gender specific body mass indexforage percentiles: girls, 2 to 20 years (pdf.Paediatric bmi centile calculator (220 years), new zealand. This is the healthy weight for the childs age and height using a bmi centile range of 5 to 85.\n\n## Body Mass Index (BMI) - Kids Health https://kidshealth.org › parents › bmi-charts\n\nFeb 1, 2009 and when you calculate their childs bmi and compare it for age, you create whats called a bmi or body mass index percentile.A ruby gem for calculating important medical child health parameters such as growth charts and bmicentile. Important: this library does not yet use interpolation to enable it to return accurate centile.Use this body mass index (bmi) calculator to estimate your body fat and to see if you’re considered underweight, normal, overweight or obese. the body mass index, or bmi, is a number calculated from a.This web page allows you to calculate the body mass index (bmi) of your patients between the ages of 2 and 20 years, as well as the exact bmi percentile and.Bmi is calculated the same way for children as for adults, but the criteria for determining a healthy weight is different. the current bmi charts for children should be used as a guide to indicate whe..Oct 5, 2018 and older plotted on bmiforage percentile charts (for boys or girls) as an o bmiforage calculation using height and weight measurements.For use with children 2 years of age up to those 20 years of age to determine whether a child is at a healthy weight for hisher height, age and gender.Body mass index is calculated using weight and height measurements and is an the group is divided into percentiles that reflect whether a child is at a.You can use the bmi calculator below to find your childs bmi, but its also bmi percentiles show how kids measurements compare with others the same.Heres how to calculate bmi and understand what the numbers mean. Bmi percentiles show how kids measurements compare with others the same gender.Percentile calculator for ages 2 to 20 yrs. The bmi percentile chart works for all ages of children and teens and avoids having to have a separate bmi chart for.Jump to bmi percentiles for children the bmi percentile is a tool to categorize a childs bmi determine the correct bmi percentile by doing the.The body mass index percentiles for boys calculation allows you to evaluate if a patient is at an appropriate weight based on their age, height and weight.What should be my bmi? try this bmi calculator for kids and adults to determine ideal body weight for your height, age and gender.The paediatric weight management service is a new lothian wide service the healthy weight calculator to calculate bmi centile; child and parentcarer must.\n\nprice fixing levitra\n\n## BMI Calculator India - Calculate Your Body Mass Index\n\nherbal cure impotent\n\n## Child BMI: Body Mass Index Calculator for Children - Disabled ... https://www.disabled-world.com › calculators-charts › child-bmi\n\nJump to bmi for children bmi centile is a good way of telling whether a child is a once your childs bmi centile has been calculated, they will be.This bmi calculator calculates your bmi (body mass index) by entering your height and weight. note that this index should only be used to give you a general idea of where body weight should be, so mak.How to determine body mass index (bmi). Percentiles for children and teens. Body mass index: body mass index (bmi) is a value that is calculated by dividing.Height and body mass indexforage : methods and development. Coordinating. Who lengthforage percentiles for boys from birth to 24 months.Also calculate the bmi if the weight and height centiles appear very different. in a child over 2 years of age, the bmi centile is a better indicator of overweight or underweight than the weight centi.Cdc: “about bmi for children and teens,” “overweight and obesity: health consequences,” “tips for parents – ideas to help management of eating disorders in children and adolescents,” “associations bet.Apr 30, 2013 centile. Healthy weight. Bmi greater than 2 nd centile and less than 85 th centile. Each childs bmi is calculated then converted into.Body mass index (bmi) is a physical measurement used to assess an individual’s total amount of body fat. bmi is a simple, inexpensive screening tool used to identify possible weight problems for both."
] | [
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https://geotec.tongji.edu.cn/geotec_en/6a/e6/c25371a223974/page.htm | [
"Research",
null,
"Large In Small\n Analytical Solving Method to Steady-State Temperature Field of Artificial Ground Freezing Published：2015-01-02 Hits：1614 At present, AGF (artificial ground freezing) technique is a mature construction method, which has been widely used in such fields as mine engineering, tunneling engineering, temporary foundation, reinforcement, ground water pollution control, waste landfills. During construction, significant engineering parameters such as the width and mechanical properties of frozen soil wall, are determined by the temperature field of frozen soil wall, thus making the theory of temperature field the basis for AGF theory. Among the main calculating methods of temperature field of ground freezing (analytical method, simulation and numerical analysis), analytical method is always a key part in the study of temperature field because of its theoretical advantages. So far, a series of classical analytical solutions of steady-state temperature field have been made. But the solving method to steady-state temperature field of artificial ground freezing has not been summarized systematically and there is no explicit solving method to direct the subsequent freezing problem. The Analytical Solving Method to Steady-State Temperature Field of Artificial Ground Freezing should be studied and obtained. Objective: Obtains the Solving Method to Steady-State Temperature Field of Artificial Ground Freezing and gives solutions to some freezing problems that the arrangement of freezing pipe is in common such as multi-pipe freezing, row-pipe freezing and circle-pipe freezing. Approach: Based on the theory of analogy between thermal and hydraulic problems, the means of superposition of potential is introduced in artificial ground freezing and it is the based approach to solve freezing problems. Then conformal mapping and mirror reflection method is introduced to direct complex freezing problems. Based on the separability of boundary conditions for Laplace equations, the means of decomposition of function as an expended approach is introduced in artificial ground freezing Significant Results and Potential Impact: Means of superposition of potential: many classic solutions are derived again and many spilt-new solutions are obtained such as few-pipe freezing solution, three-row-pipe freezing solution. This means offers the possibility to solve the pending problem. Means of decomposition of function: the circle-pipe freezing solution is given perfectly. As an expended approach, this means makes the more freezing problems solved. Principal Investigator: Xiangdong HU Multi-pipe freezing:",
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"",
null,
"Three freezing pipes arranged at random in the infinite region",
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"",
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"Three freezing pipes arranged near a right angle adiabatic boundary Multi-row-pipe freezing:",
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"",
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"The aligned three-row freezing pipes and their constant potential boundaries Multi-circle-pipe freezing:",
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"",
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"Model of single- and double-circle-pipe freezing",
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"",
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null,
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null,
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https://mgtoml.com/mg-to-ml-conversion/ | [
"# MG to ML Conversion\n\nBy |\n\n## mL to mg conversion – ml to mg calculator\n\nTo perform the reverse conversion, i.e., how many mg in ml, we rearrange the above equation in terms of mg, which gives:\n\n• `mg = 1000 * mL`\n\n## Example of mg to mL conversion\n\nLet’s say you have 5 grams of water, which is 5,000 mg. Dividing by 1,000, as shown in the mg to mL formula, gives an answer of 5 mL.",
null,
"How many mg in an mL?\nmL to mg conversion – ml to mg calculator\nHow to use this mg to mL conversion calculator?\nExample of mg to mL conversion\nLiquid concentrations – how many milligrams in a milliliter"
] | [
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http://c.biancheng.net/view/6116.html | [
"# Java递归算法\n\n```public class Factorial {\nint fact(int n) {\nint result;\nif (n == 1) {\nreturn 1;\n}\nresult = fact(n - 1) * n;\nreturn result;\n}\n}\n\nclass Recursion {\npublic static void main(String args[]) {\nFactorial f = new Factorial();\nSystem.out.println(\"3的阶乘是 \" + f.fact(3));\nSystem.out.println(\"4的阶乘是 \" + f.fact(4));\nSystem.out.println(\"5的阶乘是 \" + f.fact(5));\n}\n}```\n\n3的阶乘是 6\n4的阶乘是 24\n5的阶乘是 120\n\n```class RecTest {\nint values[];\n\nRecTest(int i) {\nvalues = new int[i];\n}\n\nvoid printArray(int i) {\nif (i == 0){\nreturn;\n} else {\nprintArray(i - 1);\n}\nSystem.out.println(\"[\" + (i - 1) + \"] \" + values[i - 1]);\n}\n}\n\nclass Recursion2 {\npublic static void main(String args[]) {\nRecTest ob = new RecTest(10);\nint i;\nfor (i = 0; i < 10; i++) {\nob.values[i] = i;\n}\nob.printArray(10);\n}\n}```\n\n 0\n 1\n 2\n 3\n 4\n 5\n 6\n 7\n 8\n 9",
null,
""
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null,
"http://c.biancheng.net/templets/new/images/material/qrcode_original.png",
null
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https://xmonkeys360.com/2020/02/23/how-to-use-oxyplot-chart-in-xamarin-ios/ | [
"# How To Use Oxyplot Chart in Xamarin iOS\n\n### Introduction\n\nShowing data in a chart/graph is much easier to understand. In Xamarin iOS no default option to show the chart. We must use a third-party plugin to show charts. A lot of third-party plugins are paid like Syncfusion, Telerik. I recommend using Oxyplot charts because it’s full of open-source and easy to use this in your project. In this article, we are going to learn how to use and understand the Oxyplot chart in Xamarin.iOS Project.\n\n### Let’s start,\n\nCreate Xamarin iOS Single View Application in Visual Studio for Mac\n\nBefore start designing, we must install an Oxyplot chart plugin from NuGet Manager in your project.\n\nNow, open Main.Storyboard in Visual Studio designer and drag the View from Toolbox and place as per your requirement and apply class as PlotView in the properties window and set identity Name [Eg – plotview]\n\nNext, write your code in ViewController.cs file. Here the many different types of chart API will be available for use like Bar, Stack, Function, Pie Chart and etc,.\n\nHere, Create PlotModel and set required properties as per your requirement and create the X, Y-axis, position, labels, zoom, angle, Max, Min, and other properties.\n\n• The Zoom property for the zoom level of the chart.\n• The Max, Min properties for avoiding more scrolling in the non-data representation in the chart, I mean you set min level is a zero, it will avoid minus(-) side-scrolling, because, we don’t have data representation on minus side.\n• The Position property used to set the position of chart in the top, bottom, left or right side\n\nAfterward, create Axis and Data Column Series. In the Column series also have different properties like color, width and etc,. Apply this X, Y-Axis and Column series to this PlotModel as we created first. Finally set this plotModel to PlotView as we designed in the Storyboard scene. Here is the full source code.\n\n``````using System;\nusing System.Collections.ObjectModel;\nusing CoreGraphics;\nusing OxyPlot;\nusing OxyPlot.Axes;\nusing OxyPlot.Series;\nusing UIKit;\n\nnamespace chartsDemo\n{\npublic partial class ViewController : UIViewController\n{\npublic ViewController(IntPtr handle) : base(handle)\n{\n}\n\npublic override void ViewDidLoad()\n{\n// Perform any additional setup after loading the view, typically from a nib.\n\nvar model = new PlotModel()\n{\nTitle = \"Column\",\nPlotType = PlotType.XY,\nLegendSymbolLength = 5,\nLegendPlacement = LegendPlacement.Outside,\nLegendOrientation = LegendOrientation.Vertical,\n};\n\nCategoryAxis xaxis = new CategoryAxis();\nxaxis.Position = AxisPosition.Bottom;\nxaxis.AbsoluteMinimum = -.5;\nxaxis.AbsoluteMaximum = 6;\nxaxis.Zoom(0, 4.5);\nxaxis.Angle = 45;\n\nLinearAxis yaxis = new LinearAxis();\nyaxis.Position = AxisPosition.Left;\nyaxis.AbsoluteMinimum = 0;\nyaxis.AbsoluteMaximum = 100;\n\nColumnSeries s1 = new ColumnSeries();\ns1.IsStacked = true;\ns1.ColumnWidth = 20;\ns1.FillColor = OxyColor.FromRgb(255, 0, 0);\n\nColumnSeries s2 = new ColumnSeries();\ns2.IsStacked = true;\ns2.ColumnWidth = 20;\n\nthis.plotview.Model = model;\nplotview.Frame = new CGRect(0, 0, this.View.Frame.Width + 20, this.View.Frame.Height);\n\n}\n\npublic override void DidReceiveMemoryWarning()\n{\n// Release any cached data, images, etc that aren't in use.\n}\n}\n}``````\n\nNow, you can execute the project and the output like below.\n\nNext, create another type of chart. Create a new Model named item.cs and write the below-given code.\n\n``````namespace chartsDemo\n{\ninternal class Item\n{\npublic string Label { get; set; }\npublic int Value1 { get; set; }\npublic int Value2 { get; set; }\npublic int Value3 { get; set; }\n}\n}``````\n\nAs usual use bar Charts and apply item model values to this plot chart. The code is given below.\n\n``````using System;\nusing System.Collections.ObjectModel;\nusing CoreGraphics;\nusing OxyPlot;\nusing OxyPlot.Axes;\nusing OxyPlot.Series;\nusing UIKit;\n\nnamespace chartsDemo\n{\npublic partial class ViewController : UIViewController\n{\npublic ViewController(IntPtr handle) : base(handle)\n{\n}\n\npublic override void ViewDidLoad()\n{\n// Perform any additional setup after loading the view, typically from a nib.\n\nvar model = new PlotModel()\n{\nTitle = \"Column\",\nPlotType = PlotType.XY,\nLegendSymbolLength = 5,\nLegendPlacement = LegendPlacement.Outside,\nLegendOrientation = LegendOrientation.Vertical,\n};\n\nCategoryAxis xaxis = new CategoryAxis();\nxaxis.Position = AxisPosition.Bottom;\nxaxis.AbsoluteMinimum = -.5;\nxaxis.AbsoluteMaximum = 6;\nxaxis.Zoom(0, 4.5);\nxaxis.Angle = 45;\n\nLinearAxis yaxis = new LinearAxis();\nyaxis.Position = AxisPosition.Left;\nyaxis.AbsoluteMinimum = 0;\nyaxis.AbsoluteMaximum = 100;\n\nColumnSeries s1 = new ColumnSeries();\ns1.IsStacked = true;\n\ns1.ColumnWidth = 20;\ns1.FillColor = OxyColor.FromRgb(255, 0, 0);\n\nColumnSeries s2 = new ColumnSeries();\ns2.IsStacked = true;\ns2.ColumnWidth = 20;\n\nvar Items = new Collection<Item>\n{\nnew Item {Label = \"Apples\", Value1 = 37, Value2 = 12},\nnew Item {Label = \"Pears\", Value1 = 7, Value2 = 21},\nnew Item {Label = \"Bananas\", Value1 = 23, Value2 = 2}\n};\n\nmodel.Axes.Add(new CategoryAxis { ItemsSource = Items, LabelField = \"Label\", AbsoluteMinimum = -0.5 });\nmodel.Axes.Add(new LinearAxis { Position = AxisPosition.Left, MinimumPadding = 0, AbsoluteMinimum = 0 });\nmodel.Series.Add(new ColumnSeries { Title = \"2009\", ItemsSource = Items, ValueField = \"Value1\", ColumnWidth = 20 });\nmodel.Series.Add(new ColumnSeries { Title = \"2010\", ItemsSource = Items, ValueField = \"Value2\", ColumnWidth = 20 });\nmodel.Series.Add(new ColumnSeries { Title = \"2011\", ItemsSource = Items, ValueField = \"Value3\" });\n\nthis.plotview.Model = model;\nplotview.Frame = new CGRect(0, 0, this.View.Frame.Width + 20, this.View.Frame.Height);\n\n}\n\npublic override void DidReceiveMemoryWarning()\n{\n// Release any cached data, images, etc that aren't in use.\n}\n}\n}``````\n\nRun your project by pressing F5, you will get output like below.\n\nThe Full source code in Github\n\nSummary\n\n1. Install Xamarin.iOS.Oxtplot Plugin\n2. Place View and apply PlotView class\n3. Create PlotModel, Column Series, X and Y-Axis\n4. Apply to PlotView"
] | [
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https://tex.stackexchange.com/questions/439244/using-answers-with-tasks | [
"I'm writing a book of maths problems. I need a way to present both the problems and their solutions at the end of the book.\n\nI worked out a way to get the problems and answers to look just as I would like them to, except for one thing.\n\nI like to use tasks for some problems that have multiple questions. Tasks gives a nice way to work with horizontal lists.\n\nBut using them with the answers package gives strange results.",
null,
"The problems look just the way I want them.",
null,
"But the numbering and spacing goes wrong in the solutions when I use tasks, in problem 1.2.2.\n\nMy question is simply if anyone knows why this happens, knows way to fix the numbering or has a good substitute for tasks.\n\n\\documentclass[a4paper,12pt,twoside,openright]{book}\n\n% -- For nice horizontal lists -- %\n\n% -- For solutions -- %\n\\Newassociation{solution}{Sol}{solutions}\n\\renewenvironment{Sol}{\\begin{enumerate}\\item[\\bfseries #1.]}{\\end{enumerate}}\n\\Newassociation{secnr}{SectionNumbering}{solutions}\n\\renewenvironment{SectionNumbering}{\\begin{trivlist}\\item \\bfseries #1}{\\end{trivlist}}\n\n\\begin{document}\n\n\\Opensolutionfile{solutions}\n\\chapter{Mathematics}\n\\begin{secnr}Bls. \\pageref{section_1A}\\end{secnr}\n\n\\begin{enumerate}\n\\item What is $2+3$?\n\\begin{solution}\n5\n\\end{solution}\n\\item Calculate the following:\n\\begin{enumerate}\n\\item $8+2$\n\\item $9+3$\n\\item $10+4$\n\\end{enumerate}\n\\begin{solution}\n\\begin{enumerate}\n\\item $10$\n\\item $12$\n\\item $14$\n\\end{enumerate}\n\\end{solution}\n\\end{enumerate}\n\n\\section{Subtraction}\\label{section_1B}\n\\begin{secnr}Bls. \\pageref{section_1B}\\end{secnr}\n\n\\begin{enumerate}\n\\item What is $100 - 3$\n\\begin{solution}\n97\n\\end{solution}\n\\item Calculate the following:\n\\task $10 - 3$\n\\task $10 - 4$\n\\task $20 - 4$\n\\task $20 - 5$\n\\begin{solution}\n\\task $7$\n\\task $6$\n\\task $16$\n\\task $15$\n\\end{solution}\n\\item How are you doing?\n\\begin{solution}\nFine.\n\\end{solution}\n\\end{enumerate}\n\n\\Closesolutionfile{solutions}\n\n\\chapter*{Solutions}\n\\input{solutions}\n\n\\end{document}\n\n• I think both of these packages use \\@currentlabel. So the last \\task (in question not answer) has label d) which is then written to the answer. I don't know how to make them compatible but I assume that's not too easy. You probably need to create a new custom label to be used by one of those packages. Sorry I cannot help. – nox Jul 4 '18 at 17:42\n\nA crude way of approaching the matter could be copying those macros that hold info that usually is actualized by \\refstepcounter and that gets written to .aux-file by the \\label-command, i.e., the macros \\@currentlabel, \\@currentHref (in case the hyperref package is in use) and \\@currentlabelname (in case the hyperref package is in use), right before starting the tasks-environment, and having them restored after ending the tasks-environment, right before starting the solution-environment.\n\nAs the user nox pointed out, this can be automatized by using the commands \\BeforeBeginEnvironment and \\AfterEndEnvironment from the etoolbox package.\n\nBesides this I recommend using the command \\Readsolutionfile from the answers package for inputting the file with the solutions rather than using \\input.\n\nThe former command also takes into account that one might wish to load the package answers with its option nosolutionfiles which means that no external solutions-file will be generated.\n\nI also recommend placing the sectioning-command for the section-header of the solutions-section into the solution-file itself. This can be done using the environment Filesave from the answers package. This takes into account that such a section-header is not needed in case the package answers is loaded with the option nosolutionfiles which means that solutions are not written into an extra file and laterwards inserted as an extra section but each solution is placed right by the question.\n\nWith the following example you can choose whether either to generate an extra section for solutions or to have each solution printed near the question by turning into a comment/not turning into a comment the line \\PassOptionsToPackage{nosolutionfiles}{answers}:\n\n\\documentclass[a4paper,12pt,twoside,openright]{book}\n\n% -- For patching environments --\n\n\\usepackage{etoolbox}\n\n% -- For nice horizontal lists --\n\n% Unlike enumerate environment the tasks environment does not\n% change info needed for placing labels etc locally but globally.\n% Thus patch the tasks environment to save these things at the\n% start of the environment and to reset these things at\n% the end of the environment:\n\\makeatletter\n\\newcommand\\savelabelinfo{%\n\\let\\MyNicecurrentlabel=\\@currentlabel\n\\let\\MyNicecurrenthref=\\@currentHref\n\\let\\MyNicecurrentlabelname=\\@currentlabelname\n}%\n\\newcommand\\restorelabelinfo{%\n\\global\\let\\@currentlabel=\\MyNicecurrentlabel\n\\global\\let\\@currentHref=\\MyNicecurrenthref\n\\global\\let\\@currentlabelname=\\MyNicecurrentlabelname\n}%\n\\makeatother\n\n% -- For solutions --\n\n% In case you wish solutions writtem directly after questions,\n% enable the following line:\n\n\\Newassociation{solution}{Sol}{solutions}\n\\Newassociation{secnr}{SectionNumbering}{solutions}\n\n\\makeatletter\n% \\InCaseSolutionsViaFile{<Tokens to deliver in case option\n% \"nosolutionfiles\" is not provided>}%\n% {<Tokens to deliver in case option\n% \"nosolutionfiles\" is provided>}%\n\\newcommand\\InCaseSolutionsViaFile{}%\n\\let\\InCaseSolutionsViaFile=\\@secondoftwo\n}{%\n\\let\\InCaseSolutionsViaFile=\\@firstoftwo\n}%\n\\InCaseSolutionsViaFile{%\n\\renewenvironment{Sol}{\\begin{enumerate}\\item[\\bfseries #1.]}{\\end{enumerate}}%\n\\renewenvironment{SectionNumbering}{\\begin{trivlist}\\item \\bfseries #1}{\\end{trivlist}}%\n}{%\n% Turn sol-environment into something that prints the body\n% within something like a deflist environment with label \"Solution(s):\"\n\\renewenvironment{Sol}{%\n\\begin{list}{}{%\n\\renewcommand\\makelabel{\\textbf{##1}\\hfil}%\n\\settowidth\\labelwidth{\\textbf{Solution(s):}}%\n\\setlength\\leftmargin{\\labelwidth}%\n}%\n\\item[Solution(s):]%\n}{\\end{list}}%\n% Turn secnr-environment into something like the verbatim-environment\n% but without printing the body, thus gobbling its own content:\n\\renewenvironment{secnr}{%\n\\@bsphack\n\\let\\do=\\@makeother\n\\dospecials\n\\obeylines\n\\secnrgobble\n}{\\@Esphack}%\n}%\n\\newcommand\\secnrgobble{}%\n\\begingroup\n\\catcode|=0 \\catcode[= 1 \\catcode]=2 %\n\\catcode\\{=12 \\catcode\\}=12 \\catcode\\\\=12 %\n|@firstofone[%\n|endgroup\n|def|secnrgobble#1\\end{secnr}[|end[secnr]]%\n]%\n\\makeatother\n\n% \\usepackage{hyperref}\n\n\\begin{document}\n\n\\Opensolutionfile{solutions}\n\\begin{Filesave}{solutions}\n\\chapter*{Solutions}\n\\end{Filesave}\n%\n\\chapter{Mathematics}\n\\begin{secnr}Bls. \\pageref{section_1A}\\end{secnr}\n\n\\begin{enumerate}\n\\item What is $2+3$?\n\\begin{solution}\n5\n\\end{solution}\n\\item Calculate the following:\n\\begin{enumerate}\n\\item $8+2$\n\\item $9+3$\n\\item $10+4$\n\\end{enumerate}\n\\begin{solution}\n\\begin{enumerate}\n\\item $10$\n\\item $12$\n\\item $14$\n\\end{enumerate}\n\\end{solution}\n\\end{enumerate}\n\n\\section{Subtraction}\\label{section_1B}\n\\begin{secnr}Bls. \\pageref{section_1B}\\end{secnr}\n\n\\begin{enumerate}\n\\item What is $100 - 3$\n\\begin{solution}\n97\n\\end{solution}\n\\begin{samepage}\n\\item Calculate the following:\n\\task $10 - 3$\n\\task $10 - 4$\n\\task $20 - 4$\n\\task $20 - 5$\n\\end{samepage}\n\\begin{solution}\n\\task $7$\n\\task $6$\n\\task $16$\n\\task $15$\n\\end{solution}\n\\item How are you doing?\n\\begin{solution}\nFine.\n\\end{solution}\n\\end{enumerate}\n\n\\Closesolutionfile{solutions}\n\n\nIn case the line \\PassOptionsToPackage{nosolutionfiles}{answers} is turned into a comment, you get:\nIn case the line \\PassOptionsToPackage{nosolutionfiles}{answers} is not turned into a comment, you get:\n• Nice. By the way you can patch the tasks environment in order to not have to type \\savelabelinfo and \\restorelabelinfo around it by hand: \\usepackage{etoolbox} \\BeforeBeginEnvironment{tasks}{\\savelabelinfo} \\AfterEndEnvironment{tasks}{\\restorelabelinfo} – nox Jul 4 '18 at 21:22"
] | [
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"https://i.stack.imgur.com/C8x5V.png",
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"https://i.stack.imgur.com/BpqYk.png",
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https://arduino.stackexchange.com/questions/4451/counting-pulses-with-interrupt | [
"# Counting pulses with interrupt\n\nI have been trying to count pulses from a 12,500 Hz square wave to trigger an output. Here's the code I have so far. When the Arduino is reset it prints 315 to the serial over a 25 ms sample. 315 x 40 = 12600. Which seems to me it's working perfectly.\n\nMy only problem is it only returns this number once upon reset of the board. Now if I move that same code down into `void loop`, it counts consecutively giving me inconstant returns.\n\nI am not understanding what I need to put in the loop section so I can repeatedly and accurately count how many toggles of the input pin I am getting over a period of time so I can do something to the output based off the presence of the 12,500 Hz signal or not.\n\n``````volatile int IRQcount;\nint pin = 2;\nint pin_irq = 0; //IRQ that matches to pin 2\n\nvoid setup() {\n// Put your setup code here, to run once:\nSerial.begin (9600);\nattachInterrupt(pin_irq, IRQcounter, RISING);\ndelay(25);\ndetachInterrupt(pin);\nSerial.print(F(\"Counted = \"));\nSerial.println(IRQcount);\n}\n\nvoid IRQcounter() {\nIRQcount++;\n}\n\nvoid loop() {\n// Put your main code here, to run repeatedly:\n}\n``````\n\nUsing the above code, every time I press the reset button I get one line in the serial window.\n\n``````Counted = 441\nCounted = 442\nCounted = 441\nCounted = 441\nCounted = 441\n``````\n\nNow I want to get the same result, but repeating over and over. That way if the signal drops out I can trigger an output to turn off (LOW). When the signal is present the output will go high.\n\nMy attempt was to move the attach interrupt down into `void loop`, so it would repeat. Here's what it looks like.\n\n``````volatile int IRQcount;\nint pin = 2;\nint pin_irq = 0; //IRQ that matches to pin 2\n\nvoid setup() {\n// Put your setup code here, to run once:\nSerial.begin (9600);\n}\n\nvoid IRQcounter() {\nIRQcount++;\n}\n\nvoid loop() {\n// Put your main code here, to run repeatedly:\n\nattachInterrupt(pin_irq, IRQcounter, RISING);\ndelay(25);\ndetachInterrupt(pin);\nSerial.print(F(\"Counted = \"));\nSerial.println(IRQcount);\n}\n``````\n\nThe return I get is self-updating, but the \"count\" instead of starting from 0 each time starts from the previous count. So it gets larger and larger. I am looking to return a constant value that represents my 12500 Hz signal so that, and only that, will trigger my output.\n\n``````Counted = 442\nCounted = 886\nCounted = 1330\nCounted = 177\nCounted = 2221\nCounted = 2667\nCounted = 3112\nCounted = 3557\nCounted = 4002\nCounted = 4448\nCounted = 4893\nCounted = 5338\nCounted = 5784\nCounted = 6229\nCounted = 6674\nCounted = 7120\nCounted = 7565\nCounted = 8010\nCounted = 8456\nCounted = 8901\nCounted = 9347\nCounted = 9792\nCounted = 10237\nCounted = 10683\nCounted = 11130\nCounted = 11576\nCounted = 12022\nCounted = 12469\nCounted = 12915\nCounted = 13361\nCounted = 13808\nCounted = 14254\nCounted = 14700\nCounted = 15147\nCounted = 15593\nCounted = 16040\nCounted = 16486\nCounted = 16932\nCounted = 17378\nCounted = 17825\nCounted = 18271\nCounted = 18717\nCounted = 19164\nCounted = 19610\nCounted = 20056\nCounted = 20503\nCounted = 20949\nCounted = 21395\nCounted = 21842\nCounted = 22288\nCounted = 22735\nCounted = 23169\nCounted = 23616\nCounted = 24062\nCounted = 24508\nCounted = 24955\nCounted = 25401\nCounted = 25730\nCounted = 25756\nCounted = 26200\nCounted = 26646\nCounted = 27093\nCounted = 27539\nCounted = 27985\nCounted = 28432\nCounted = 28878\nCounted = 29324\nCounted = 29770\nCounted = 30217\nCounted = 30663\nCounted = 31110\nCounted = 31556\nCounted = 32002\nCounted = 32449\nCounted = -32641\nCounted = -32195\nCounted = -31748\nCounted = -31302\nCounted = -30855\nCounted = -30408\nCounted = -29962\nCounted = -29515\nCounted = -29069\nCounted = -28622\n``````\n• \"Now if i move that same code down into void loop it counts consecutively giving me inconstant returns.\" means what exactly? – Ignacio Vazquez-Abrams Oct 4 '14 at 15:39\n• I edited my question to try and better explain myself – Brandon Whosville Oct 4 '14 at 16:11\n\nYou need to reset IRQCount back to `0` before attaching the interrupt again. Otherwise it will just continue counting from where it stopped last time.\n\nI would actually keep the interrupt attached and just reset the variable just before the delay. That way the overhead of attach/detachinterrupt doesn't get added to the 25ms delay.\n\n``````volatile int IRQcount;\nint pin = 2;\nint pin_irq = 0; //IRQ that matches to pin 2\n\nvoid setup() {\n// put your setup code here, to run once:\nSerial.begin (9600);\nattachInterrupt(pin_irq, IRQcounter, RISING);\n}\n\nvoid IRQcounter() {\nIRQcount++;\n}\n\nvoid loop() {\n// put your main code here, to run repeatedly:\nIRQcount = 0;\ndelay(25);\nint result = IRQcount;\nSerial.print(F(\"Counted = \"));\nSerial.println(result);\n}\n``````\n\nSince an int is 2 bytes an interrupt might occur in the middle of setting/reading those two bytes. This might result in an occasional wrong value. To prevent that you should disable interrupt while setting/reading the value\n\n``````volatile int IRQcount;\nint pin = 2;\nint pin_irq = 0; //IRQ that matches to pin 2\n\nvoid setup() {\n// put your setup code here, to run once:\nSerial.begin (9600);\nattachInterrupt(pin_irq, IRQcounter, RISING);\n}\n\nvoid IRQcounter() {\nIRQcount++;\n}\n\nvoid loop() {\n// put your main code here, to run repeatedly:\n\ncli();//disable interrupts\nIRQcount = 0;\nsei();//enable interrupts\n\ndelay(25);\n\ncli();//disable interrupts\nint result = IRQcount;\nsei();//enable interrupts\n\nSerial.print(F(\"Counted = \"));\nSerial.println(result);\n}\n``````\n• Thanks Gerben! Using that code i get the occasional junk return. What would be the easiest way to denounce this? Make an out take lets say 3 readings before making a decision on what to do. Or averaging pulses over a span instead of a raw count? Heres an example of the return, i get the anomaly every few seconds. Counted = 439, Counted = 438, Counted = 430, Counted = 48, Counted = 318, Counted = 438, – Brandon Whosville Oct 4 '14 at 16:16\n• I lengthened the delay to give a larger sample size. This gives me a workable return from my noisy input source. This seems to work perfect! Thank you! – Brandon Whosville Oct 4 '14 at 16:54\n• I assume you used the code with the `cli`s and `sei`s in it. Rather strange having two consecutive wrong values. – Gerben Oct 5 '14 at 12:32\n• Gerben, Yes i used the code with the cli and sei in it. Do you mean its strange getting the two wrong returns in a row? It seems the returns are correct, it is the incoming signal that is not stable giving junk returns. If i sample over lets say 100ms it gives plenty of pulses per sample to safely activate or kill the circuit. I really appreciate your help! – Brandon Whosville Oct 6 '14 at 15:13\n• But the 48 and 318 you got are both lower that the average of about 435. I don't know the connected circuit, so it could just as well be the circuit, instead of the arduino code. Anyways, as long as you are happy with the end result... Glad to have helped. – Gerben Oct 6 '14 at 17:41"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9143863,"math_prob":0.8580088,"size":3556,"snap":"2020-45-2020-50","text_gpt3_token_len":1149,"char_repetition_ratio":0.27195945,"word_repetition_ratio":0.15821813,"special_character_ratio":0.3844207,"punctuation_ratio":0.102564104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99182755,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T02:12:31Z\",\"WARC-Record-ID\":\"<urn:uuid:684c4b7a-870c-4389-83b7-1c9f466e7a02>\",\"Content-Length\":\"143006\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd523828-9ebb-4847-95f0-71fa04d76c10>\",\"WARC-Concurrent-To\":\"<urn:uuid:fae56ca8-7ac6-4aff-94a2-fd6641510d2e>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://arduino.stackexchange.com/questions/4451/counting-pulses-with-interrupt\",\"WARC-Payload-Digest\":\"sha1:3T5TBRKRAX63E3QFCGMTEPBZXVRQU2S5\",\"WARC-Block-Digest\":\"sha1:VIA67H4WINFAHDI6PB2NTE77LKC7YR7Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141204453.65_warc_CC-MAIN-20201130004748-20201130034748-00111.warc.gz\"}"} |
http://herrstrathmann.de/learning-deep-kernels-for-exponential-family-densities/ | [
"# Learning deep kernels for exponential family densities\n\nAn ICML 2019 paper, together with Kevin Li, Dougal Sutherland, and Arthur Gretton. Code by Kevin.\n\nMore score matching for estimating gradients using the infinite dimensional kernel exponential family (e.g. for gradient-free HMC)! This paper tackles one of the most limiting practical characteristics of using the kernel infinite dimensional exponential family model in practice: the smoothness assumptions that come with the standard “swiss-army-knife” Gaussian kernel. These smoothness assumptions are blessing and curse at the same time: they allow for strong statistical guarantees, yet they can quite drastically restrict the expressiveness of the model.\n\nTo see this, consider a log-density model of the form (the “lite” estimator from our HMC paper) $$\\log p(x) = \\sum_{i=1}^n\\alpha_ik(x,z_i)$$for “inducing points” $z_i$ (the points that “span” the model, could be e.g. the input data) and the Gaussian kernel $$k(x,y)=\\exp\\left(-\\Vert x-y\\Vert^2/\\sigma\\right)$$Intuitively, this means that the log-density is simply a set of layered Gaussian bumps — (infinitely) smooth, with equal degrees of variation everywhere. As the paper puts it\n\nThese kernels are typically spatially invariant, corresponding to a uniform smoothness assumption across the domain. Although such kernels are sufficient for consistency in the infinite-sample limit, the\ninduced models can fail in practice on finite datasets, especially if the data takes differently-scaled shapes in different parts of the space. Figure 1 (left) illustrates this problem when fitting a simple mixture of Gaussians. Here there are two “correct” bandwidths, one for the broad mode and one for the narrow mode. A translation-invariant kernel must pick a single one, e.g. an average between the two, and any choice will yield a poor fit on at least part of the density\n\nHow can we learn a kernel that locally adapts to the density? Deep neural networks! We construct a non-stationary (i.e. location dependent) kernel using a deep network $\\phi(x)$ on top a Gaussian kernel, i.e. $$k(x,y) = \\exp\\left(-\\Vert \\phi(x)-\\phi(y)\\Vert^2 / \\sigma\\right)$$ The network $\\phi(x)$ is fully connected with softplus nonlinearity, i.e. $$\\log(1+e^x)$$Softplus gives us some nice properties such as well-defined loss functions and a normalizable density model (See Proposition 2 in the paper for details).\n\nHowever, we need to learn the parameters of the network. While kernel methods typically have nice closed-form solution with guarantees (and so does the original kernel exponential family model, see my post). Optimizing the parameters of ϕ(x)$\\varphi \\left(x\\right)$$\\phi(x)$ obviously makes things way more complicated: Whereas we could use a simple grid-search or black-box optimizer for a single kernel parameter, this approach here fails due to the number of parameters in 12p0(x)xlogp(x)xlogp0(x)2dx$\\varphi \\left(x\\right)$$\\phi(x)$.\n\nCould we use naive gradient descent? Doing so on our score-matching objectivewith $\\log p(x)=\\sum_{i=1}^n \\alpha_i k(z_i, x)$ and $k(x,y)=\\exp\\left(-\\Vert x-y\\Vert ^2 / \\sigma \\right)$ will always overfit to the training data as the score (gradient error loss) can be made arbitrarily good by moving ${\\mathrm{zi}}_{}$$z_i$ towards a data point and making $\\sigma$ go to zero. Stochastic gradient descent (the swiss-army-knife of deep learning) on the score matching objective might help, but would indeed produce very unstable updates.\n\nInstead, we employed a two-stage training procedure that is conceptually motivated by cross-validation: we first do a closed-form update for the kernel model coefficients αi${\\mathrm{\\alpha i}}^{}$$\\alpha_i$ on one half of the dataset, then we perform a gradient step on the parameters of the deep kernel on the other half. We make use of auto-diff — extremely helpful here as we need to propagate gradients through a quadratic-form-style score matching loss, the closed-form kernel solution, and the network. This seems to work quite well in practice (the usual deep trickery to make it work applies). Take away: By using a two-stage procedure, where each gradient step involves a closed form (linear solve) solutions for the kernel coefficients ${\\mathrm{\\alpha i}}^{}$$\\alpha_i$ we can fit this model reliably. See Algorithm 1 in the paper for more nitty-gritty details.\n\nA cool extension of the paper would be to get rid of the sub-sampling/partitioning of the data and instead auto-diff through the leave-one-out error, which is closed form for these type of kernel models, see e.g. Wittawat’s blog post.\n\nWe experimentally compared the deep kernel exponential family to a number of other approaches based on likelihoods, normalizing flows, etc, and the results are quite positive, see the paper!\n\nNaturally, as I have worked on using these gradients in HMC where the exact gradients are not available, I am very interested to see whether and how useful such a more expressive density model is. The case that comes to mind (and in fact motivated one of the experiments in the this paper) is the Funnel distribution (e.g. Neal 2003, picture by Stan), e.g.$$p(y,x) = \\mathsf{normal}(y|0,3) * \\prod_{n=1}^9 \\mathsf{normal}(x_n|0,\\exp(y/2)).$$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89069825,"math_prob":0.98801374,"size":5452,"snap":"2021-43-2021-49","text_gpt3_token_len":1240,"char_repetition_ratio":0.11049926,"word_repetition_ratio":0.0,"special_character_ratio":0.21386647,"punctuation_ratio":0.09620991,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988183,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-30T20:20:50Z\",\"WARC-Record-ID\":\"<urn:uuid:e8c80195-5ffd-478d-b45a-b26036ac44e1>\",\"Content-Length\":\"68160\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0382c3d2-3c0a-4284-bba5-35769b2739b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:14cfbfb4-2c1b-4c27-9e95-d621dd55401a>\",\"WARC-IP-Address\":\"85.13.139.98\",\"WARC-Target-URI\":\"http://herrstrathmann.de/learning-deep-kernels-for-exponential-family-densities/\",\"WARC-Payload-Digest\":\"sha1:LRZJS62VKZ4AIHGAEK6A4L4D3JLVXEIF\",\"WARC-Block-Digest\":\"sha1:K6QFVJVP72VRPQ6RNIKWHDRW7G3NUVP2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964359073.63_warc_CC-MAIN-20211130201935-20211130231935-00503.warc.gz\"}"} |
https://homeguides.sfgate.com/order-mulch-cubic-yard-52848.html | [
"# How to Order Mulch by the Cubic Yard\n\nMost gardeners understand the value of mulch in the landscape. Many gardeners' mulch needs can be satisfied by taking home a few bags of the product from the garden center. If you have a large yard and require mulch by the cubic yard, however, you'll have to order it from a local organic products supplier.\n\n#### 1\n\nDetermine how deep you intend to apply the mulch. Factor in about 1 to 2 inches for flower beds and 2 to 3 inches for trees and shrubs. If you have a mix of trees, shrubs and flowers, estimate and average of 2 to 2 1/2 inches.\n\n#### 2\n\nMultiply the length and width of the mulched areas of your yard, to determine the square footage. For example, 25-by-50-foot border measures 1250 square feet.\n\n#### 3\n\nDivide the square footage by 9, to determine the square yardage. For example, 1250 square feet converts to 138.88 square yards.\n\n#### 4\n\nDivide the square yardage of your garden by 18 for each 2 inches of mulch depth. If you only need 1 inch of mulch, divide by 36. If you need 3 inches, divide by 12. For example, 138.88 square yards divided by 18, for 2-inch coverage, equals 7.72 cubic yards of mulch.\n\n#### 5\n\nVisit your local organic products supplier to investigate their products first hand before ordering. For a 1250-square-foot garden in the example, order 8 cubic yards of mulch, rounding up from 7.72."
] | [
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https://www.enjoytutorials.com/c-programming/c-functions-complete-tutorial/ | [
"# C Functions Complete Tutorial\n\n• Post author:\n• Post category:C\n\nIn this section, we will learn what the Functions are and how to use them in C.\n\n## C Function Analogy:\n\nThe definition of the word `Function` According to the Longman dictionary is: “the purpose that something has, or the job that someone or something does.”\n\nWhen you go to a restaurant, probably the first person you’ll see is the Host that he/she will greet you and show you a table to seat. This is the function of a Host to `greet and show a place for customers to seat`.\n\nThen you have restaurant’s server (waiter and waitress) that he/ she will take your orders and probably give it to the chief in order to make them ready. So the function of waiter/ waitress is to `take orders and bring food to customers`.\n\nOf course there are other people working in a restaurant like a dishwasher that he/ she should wash the dishes and sometimes even the floor. So we can say the functions of a dishwasher are `washing the dishes and the floors`.\n\nEach person who works in this restaurant has a name. For example, Ellen is the Host, Jack is the Restaurant’s server and Bob is the dishwasher.\n\nLet’s go a little deeper:\n\nThe waiter/ waitress, in order to function correctly first and foremost, should be called by the customer (she can’t just go on their table over and over to see if they need anything). After the waitress is called, she needs input (orders) from the customer to proceed to its function! But she can’t just receive any type of input! For example, the customer of a restaurant can’t order a Car!!! So there are some limits on the types of orders that the customer can give. These limits become clear when the waitress puts a menu in front of the customer!\n\nAfter the waitress took the correct inputs, she’ll call another person (let’s say the chief) and give her order to that chief. The waitress should be careful as well because the chief also can’t take any types of order!\n\nDuring this time, the customer can’t proceed to eat because she is waiting for her order to be received.\n\nAlso, the waitress is waiting for the food so we can say that her function is not completed yet.\n\nNow the ordered food is ready and the chief `returns` the actual food to the waitress and waitress passes this food to the customer.\n\nThe customer can now proceed to eat and the function of the waitress and the chief are completed at this time.\n\nNote: the customer had one function here and that was eating. She started by calling a waitress which is equal to start someone else’s function, and then waited for that function to be finished. The function of the waitress finished when she returned the food to the customer. At this point, the customer can proceed and run the rest of her function (eating).\n\n## What is Function in C?\n\nIn the world of C programming, we have the concept of functions as well!\n\nOne function can call another function (like the way customer called the waitress), one function can pass inputs to another function (just like the way the customer passed order to the waitress), the caller function can wait until the called function finished its work and the called function can return a value to the caller function (just like the way the waitress passed the actual food to the customer).\n\nNote: In C programming, we name the function that starts a communication the `caller` function and the target function as the `called` or `callee` function. In our example, the caller function is the customer and the called or callee function is the waitress.\n\nAlso, a function can be `called/ callee ` and `caller ` at the same time! In our example, the waitress has these adjectives! The customer `called` the waitress and so now she becomes the `called/ callee` function. Then she calls the chief and passes the order and so now she is in the position of the `caller` function.\n\n## Function Creation in C:\n\nThere are two steps in order to create a function in C:\n\n1. Function Declaration\n2. Function Definition\n\n## 1- Declaration of a Function in C:\n\n1. Let’s back to the restaurant; the declared function of the waitress is to take an order from any customer in the restaurant and return food relative to the order she took.\n\nDeclaring is about what the inputs of a function are and what the output is. At this stage, we don’t care about the logics that should be defined within the body of that function. All we care is the `name of the function` and `the type` of `inputs` and `output` of the function.\n\nIn C programming, we call function declaration as `prototype` of that function.\n\nNotes:\n\n• The usual place of declaring a function is above the `main` function. But that doesn’t mean we can’t put the prototype inside a function. The point is that, before a function is called, there should be a prototype already declared somewhere that the compiler knows about.\n• We can skip the declaration of a function. In this case, the function should be defined above the `main` function.\n\nThere are two structures by which we can declare the prototype of a function in C:\n\n```Data_type function_name();\n\nData_type function_name(data_type parameter_1, double parameter_2, double parameter_N );```\n\nNote: If the function that we are going to build does not need inputs in order to function correctly, then we can use the first structure, otherwise, if inputs are necessary, the second structure should be used instead.\n\nData_type: Here, we set the type of output that is expected from the function.\n\nThe types can be those that are covered in the data-type section as well as enum, union and structure types.\n\nNote: if the function does not return any output, we replace the `data_type` with the keyword `void` which means empty or basically a signal to the compiler that after running the function, there won’t be any returning value.\n\nExample:\n\n`void sum();`\n\n## Function Identifier in C(function_name):\n\nAny function should have a unique name in order to differentiate functions from each other and be able to call them. This name is known as the identifier of the function.\n\nRules for creating a name for a function are the same as those rules that apply to the variable names.\n\n## C Functions Open and close parentheses ()\n\nThere are two reasons why we use parentheses here:\n\nFirst= using parentheses helps us and the compilers to differentiate a function from a variable! Think about it, how could we figure out which name is for a function and which name is for a variable if there weren’t any parentheses?\n\nSecond= if a function also takes inputs, we declare the type of each input within the parentheses when we declare a function. This way, we know what type of arguments (inputs) we should pass to the target function when that function is getting called.\n\nNote: we also use parentheses in order to call a function which is covered later in this section.\n\n## Function Parameters in C:\n\nAs mentioned before, a function can take inputs from its caller. These input values should be stored in variables in order to be able to use those values in a function at later times. We store these values in a variable that is known as function-parameter.\n\nThese parameters are declared within the parentheses that come after the name of a function.\n\nNote: multiple parameter should be separated from each other using comma `,`.\n\n## C Function Prototype Semicolon `;`:\n\nWhen a function is declared (prototyped), we end that function declaration with a semicolon `; `.\n\n## Example: function declaration in C\n\n`int sum(int valueOne, int valueTwo );`\n\nIn the example above, we’ve `declared` or `prototyped` a function that outputs a value of type `int`, the name of the function is `sum` and it takes `two inputs of type int`.\n\nEach input of the function is called the `parameter` of that function, we separate them via comma `,` and each parameter is considered the local variable of that function.\n\nSo in the example above, we’ve declared two parameters for the `sum` function with the names `valueOne` and `valueTwo`. These two parameters are also the local variables of the `sum` function.\n\nNote: if a function has inputs, the type of those inputs should be declared in the function prototype, but the names of variables are optional. In this case, when we defined the body of that function, we will set the name of those parameters as well.\n\n## Example: declaring a function in C\n\n`double maxNumber(int , double, float, long);`\n\nIn the example above, the function’s name is `maxNumber`, the output’s type of this function is `double` and it has 4 parameters with the types of `int`, `double`, `float`, and `long`, respectively.\n\nNote that these parameters don’t have a name for now, but we will set their name when we defined the function.\n\nWhen defining a function, setting the names for parameters is necessary because these parameters are actually the local variables of the function and any variable should have a name!\n\nAs the last note: the order of parameters is important. When calling a function, we can’t just put a value randomly in a function or put less or more inputs than what actually is needed for that function! These actions will cause the compiler to return an error instead of compiling our source code.\n\nFor example, the prototype mentioned above needs 4 inputs (also called `arguments`) and the order is `int`, `double`, `float`, `long` respectively. So the correct way of calling this function would be:\n\n`maxNumber(1 , 3.2, 2.4f, 2235);`\n\nThese are some examples of the wrong way of calling this function:\n\n```maxNumber(1 , 3.2);// less arguments than is expected.\nmaxNumber(1); // less arguments than is expected.\nmaxNumber(1 , 3.2, 2.4f, 2235,23,233); // inserting more arguments than what it can take.```\n\nAll of these function calls will cause the compiler to return an error.\n\nAgain, don’t worry about how to call a function at this stage, as you’ll learn about how to call them later in this section.\n\n## 2- How to define a Function in C:\n\nNow that the function is declared, we need to define the logic within the body of that function.\n\nThese are the structures we can use to define a function:\n\n```Data_type function_name(){\n// logic of the function goes here between open and close brackets.\nreturn value; // the return keyword is only used when the function expected output.\n\n}\n\nData_type function_name(data_type parameter_1, double parameter_2, double parameter_N ){\n// logic of the function goes here between open and close brackets.\nreturn value; // the return keyword is only used when the function expected output.\n\n}\n```\n\nData_type: the data-type of a defined function should be the same as its prototype. For example, if the prototype is `double` then the defined function should also set the return data type to `double`.\n\nFunction_name: the name of a defined function should be the same as its prototype. For example, if the prototype’s name is set to `sum` then the defined function should also be named as `sum`.\n\nParameters: the `type`, `number` and `order` of the parameters of a defined function should be the same as its prototype. But the `variable-name` for each parameter in the defined function can be different from what is set in the prototype of that function.\n\nOpen and close brackets {}: The logic of a function is defined within its body and the body of a function starts from the open brace `{` and ends right at the close brace `}`.\n\nNote: there’s no need to put `;` after the close bracket.\n\nreturn: if the defined function returns an output, we use the keyword `return` in order to set the value that the defined function should return. (The `return` keyword creates a statement and this statement must end with the semicolon `;`). (In the C Function return statement section, we’ve explained this statement in greater details.)\n\nNote: if the defined function does not output any value (the output type is set to `void`) then there’s no need to use the `return` keyword.\n\n## Example: Defining a function in C\n\n```#include <stdio.h>\n\nint sum(int v1 , int v2);\n\nint main() {\n\nint result= sum(1,2);\n\nprintf(\"result is: %d\\n\", result);\nreturn 0;\n}\n\nint sum(int valueOne , int valueTwo ){\n\nreturn valueOne + valueTwo;\n}\n```\n\nOutput:\n\n`result is: 3`\n\nIn the example above, we first declared a function named `sum` right above the `main` function. The return type of this function is `int`, the name as mentioned is `sum` and it takes two parameters both of type `int`.\n\nAfter that, right below the body of the `main` function, the `sum` function is defined. As you can see, the return type + the name and the number of parameters of this function is exactly the same as its prototype. But parameters’ names are intentionally different, which is OK.\n\nAlso, because this function outputs a value of type `int`, within the body of the function, the keyword `return` is used, which in this case it returns the sum of two values (valueOne and valueTwo).\n\nNote: it’s a compiler error if we try to define a function within the body of another function.\n\n## Example: C Void Function/ void keyword\n\nLet’s run another example and this time the function won’t return any output, so the type is `void`:\n\n```#include <stdio.h>\n\nvoid sum(int v1 , int v2);\n\nint main() {\n\nsum(1,2);\n\nreturn 0;\n}\n\nvoid sum(int valueOne , int valueTwo ){\nint result = valueOne + valueTwo;\nprintf(\"Result is: %d\\n\", (result));\n}\n```\n\nOutput:\n\n`Result is: 3`\n\nIn this example because the `sum` function doesn’t return any output, we used `void` as the return type of this function which means `empty` and of course there’s no `return` keyword in the body of this function then.\n\n## Calling a Function in C: How to Call a Function in C?\n\nIn order to call a function, write the name of that function followed by a pair of parentheses () and end it with a semicolon `; `.\n\nStructure:\n\n`Function_name();`\n\nIf the called function takes arguments (inputs) then we need to put the inputs within the parentheses as well. This means we’re assigning values to the parameters of the function. The parameters, as mentioned before, are also the local variables of that function.\n\nStructure:\n\n`Function_name( value1, value2, valueN);`\n\nNote: each input is separated via comma `, `.\n\nYou should know that calling a function can only happen within the body of another function or the body of itself, which in this case is called `recursion` and you can learn about it in the C recursion section.\n\nNote: calling a function outside any function will cause a compile time error.\n\n## Example: calling a function in C\n\n```#include <stdio.h>\n\nint multiply(int valueOne , int valueTwo);\n\nint main() {\n\nint result = multiply(300,21);\nprintf(\"Result is: %d\\n\", result);\n\nreturn 0;\n}\n\nint multiply(int valueOne , int valueTwo ){\n\nint result = valueOne * valueTwo;\nreturn result;\n}\n```\n\nIn the example above, we’ve called the `multiply` function within the body of the `main` function and because this function returns an `int` value, we’ve created a variable of this type in the body of `main` function and assigned the returned value of the `multiply` function to this variable.\n\nThen in the next line we simply passed the value of the variable to another function called `printf` and so if you run the example, the result will be printed on the screen because of calling this `printf` function.\n\nThis last example might raise the question about how the instructions in a program run and where the entry point in a program is! You’ll get the answer to these questions and some others in later sections where we dive deep into the concept of `stack`.\n\n## What is argument in C?\n\nThe value we put as the input for a function when it’s being called is known as the arguments of that function.\n\n## Example: arguments in C\n\n```#include <stdio.h>\n\nvoid printAge(int);\nint main() {\n\nprintAge(1000);\nreturn 0;\n}\nvoid printAge(int age){\nprintf(\"%s%d\\n\", \"My name is John Doe and my age is: \",age);\n}\n```\n\nOutput:\n\n`My name is John Doe and my age is: 1000`\n\nHere the `printAge()` function takes one argument and that should be of type int. So the value we’ve passed to this function in the `main` function is 1000 and this value is considered as the argument of the `printAge()` function. Note that this argument will be assigned to the `age` parameter and we can use this parameter in the body of the printAge() function to access the value then.\n\n## Return Type of a Function in C\n\nIf the data-type of a function is not `void` then that function returns a value of a specific data-type.\n\nFor example, the data type of a function could be `int`, `double`, `float`, `short`, a pointer, etc.!\n\nThe point is, the moment a function designed to return a value, in that function we should use the `return` statement in order to return a value.\n\nA return statement is a combination of the `return` keyword and a value that we put on the right side of this operator.\n\nFor example:\n\n`return 10;`\n\nThis statement sits inside the body of a function and it means when the execution engine reached to this statement in that function, it should terminate the work of the function and return the value 10 as a result.\n\nNote: the data type of the value we put for the `return` statement should match the return data type of the enclosing function. For example, if the return data type of a function is set to double, then we need to return a double value!\n\n## Example: Functions with different return types in C\n\n```#include <stdio.h>\n\ndouble multiply(double, double);\nint sum(int, int);\nint main() {\n\ndouble res = multiply(2.36, 56.4);\nint s = sum(14,15);\nprintf(\"The value of the res variable is: %f\\n\",res);\nprintf(\"The value of the sum variable is: %d\\n\",s);\nreturn 0;\n}\ndouble multiply(double val1, double val2){\nreturn val1 * val2;\n}\nint sum(int val1, int val2){\nreturn val1+ val2;\n}\n```\n\nOutput:\n\n```The value of the res variable is: 133.104000\n\nThe value of the sum variable is: 29```\n\n## Multiple return statements in a Function\n\nIn a function, there can be multiple return statements as well.\n\nThese statements mostly used when there are conditional statements like `if-else` or loops like `while` and `for` in a function and we want to terminate the work of a function in those statements if needed.\n\n## Example: using multiple return statements in a function\n\n```#include <stdio.h>\n\nint sum(int, int);\n\nint main() {\n\nint res = sum(100, 200);\nprintf(\"The result is: %d\\n\",res);\nreturn 0;\n}\n\nint sum(int val1, int val2){\nint res = val2 + val1;\nif (res>100){\nreturn 150;\n}else{\nreturn 23;\n}\n}\n```\n\nOutput:\n\n`The result is: 150`\n\n## C Local variables\n\nA variable that is created within the body of a function is known as the local variable of that function.\n\nA local variable, as the name suggest is local to its enclosing function! That means we can’t use such variable in other functions!\n\nNote: the body of a function starts from the opening brace `{` and ends right at its closing brace `}`.\n\nFor example, let’s say a function named A and it has a local variable named `age`. Now if there’s another function (Let’s say the name of this second function is B), this function can’t access the `age` variable that belongs to the function A! Basically, to the eyes of the B function, there’s no such variable as `age` at all. So calling a variable named `age` in the body of the `B` function would result a compile time error.\n\n## Example: creating local variables in C\n\n```#include <stdio.h>\n\nvoid printName(void);\nint main() {\n\nprintName();\nreturn 0;\n}\nvoid printName(void){\nint age =2000;\nprintf(\"%s%d\\n\", \"My name is John Doe and my age is: \",age);\n}\n```\n\nOutput:\n\n`My name is John Doe and my age is: 2000`\n\nHere, in the body of the `printName()` function, there’s a variable named `age`. This is a local variable! That means we can’t access the age variable in the body of other functions like the `main`. If we do so, that would be a compile time error.\n\n## Types of Function in C\n\nIn short, there are two types of functions in C.\n\n## – standard library functions in C:\n\nThese are the built-in functions that creators of C added to this language and we can use them whenever needed. For example, printf() function is a built in function in C where we can use to send data to the output.\n\n## – User defined functions in C:\n\nThese are the functions that we as developers build in order to cover the tasks we want to be done in our programs.\n\n## Difference between Parameter and Argument\n\nA parameter is the variable that we declare for a function and they are the local variables of the target function. Also, when we call a function, the input values we put for that function will be stored in these parameters.\n\nOn the other hand, arguments are those input values we put for a function when the function is being called."
] | [
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https://practiceadvices.com/data/advice/read/1030-are-all-rotation-matrices-orthogonal | [
"# Are all rotation matrices orthogonal?",
null,
"### Are all rotation matrices orthogonal?\n\nThus rotation matrices are always orthogonal. Now that was not your question. You asked why orthogonal matrices represent rotations. The columns (and rows) of an orthogonal matrix are unit vectors that are orthogonal to each other.\n\n### How do you check if a matrix is a rotation matrix?\n\nRotation matrices are square matrices, with real entries. More specifically, they can be characterized as orthogonal matrices with determinant 1; that is, a square matrix R is a rotation matrix if and only if RT = R−1 and det R = 1.\n\n### Do rotation matrices in 2d always commute?\n\nAs leftaroundabout said in a comment to the question itself, rotations not commuting is not really anything noteworthy. The fact that they do commute in two dimensions is notable, but asking why they do not commute in general is not very fruitful apart from a concrete demonstration.\n\n### What is an inverse rotation?\n\nThe inverse of a rotation matrix is the rotation matrix's transpose. The inverse of a matrix product is the product of the inverse matrices ordered in reverse.\n\n### What is rotation in simple words?\n\n1a(1) : the action or process of rotating on or as if on an axis or center. (2) : the act or an instance of rotating something. b : one complete turn : the angular displacement required to return a rotating body or figure to its original orientation.\n\n### Are rotation matrices unique?\n\nAre rotation matrices unique? Yes they are, as this answer that Francesco quoted explains well. If they were not unique, then Qv = Rv and thus (Q-R)*v = 0 would be true for any vector. The latter is only true for the null matrix, however.\n\n### How do I turn a picture 90 degrees?\n\nRotate 90 degrees\n\n1. Click the object that you want to rotate.\n2. Under Drawing Tools (or Picture Tools if you're rotating a picture), on the Format tab, in the Arrange group, click Rotate, and then: To rotate the object 90 degrees to the right, click Rotate Right 90°.\n\n### Are 3D rotation matrices commutative?\n\nNow we're going to talk about one of the more perplexing elements of rotations in 3D. I've got three coordinate frames here and they're all initially parallel to each other. The x-axes, the y-axes and the z-axes are all aligned.\n\n### Why are 3D rotations not commutative?\n\nRotations in three dimensions are generally not commutative, so the order in which rotations are applied is important even about the same point. ... Rotations about the origin have three degrees of freedom (see rotation formalisms in three dimensions for details), the same as the number of dimensions.\n\n### When is a matrix called an invertible matrix?\n\nA matrix A of dimension n x n is called invertible if and only if there exists another matrix B of the same dimension, such that AB = BA = I, where I is the identity matrix of the same order. Matrix B is known as the inverse of matrix A. Inverse of matrix A is symbolically represented by A -1.\n\n### Is there an inverse of the rotation matrix?\n\nMy problem is to find an inverse of the rotation matrix so that I can later “undo” the rotation performed on the vector so that I get back the original vector. The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix.\n\n### Which is the correct equation for matrix inversion?\n\nIf A and B are matrices of the same order and are invertible, then (AB) -1 = B -1 A -1. Matrix inversion is the method of finding the other matrix, say B that satisfies the previous equation for the given invertible matrix, say A. Matrix inversion can be found using the following methods:\n\n### Can a rotation matrix be a parametric matrix?\n\nThe rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Provide details and share your research!"
] | [
null,
"https://i.ytimg.com/vi/kR9rO-6Y2Zk/hq720.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8895561,"math_prob":0.9673268,"size":4611,"snap":"2021-43-2021-49","text_gpt3_token_len":1003,"char_repetition_ratio":0.20338616,"word_repetition_ratio":0.14214464,"special_character_ratio":0.21665582,"punctuation_ratio":0.11386697,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99640256,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T07:08:50Z\",\"WARC-Record-ID\":\"<urn:uuid:d4c26e70-f0c1-48e6-8f06-aa5f71dae7d8>\",\"Content-Length\":\"47173\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6a4d9bda-6742-4ec1-9efd-33b91070edb5>\",\"WARC-Concurrent-To\":\"<urn:uuid:52d97a62-579a-4336-b4d1-28756f2379d9>\",\"WARC-IP-Address\":\"95.216.182.30\",\"WARC-Target-URI\":\"https://practiceadvices.com/data/advice/read/1030-are-all-rotation-matrices-orthogonal\",\"WARC-Payload-Digest\":\"sha1:WTP6MGU43DFSZDYSPQCMJOIRBFDXW3TU\",\"WARC-Block-Digest\":\"sha1:X6SDOCLTLXESKRBNLVO2KCWBONQ5JJMM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363149.85_warc_CC-MAIN-20211205065810-20211205095810-00248.warc.gz\"}"} |
https://www.tutorialgateway.org/c-sqrt-function/ | [
"# C sqrt Function\n\nThe C sqrt function is one of the Math Functions which is used to find the square root of a number or a specified expression. The syntax of the sqrt in this language is\n\n`double sqrt(double number);`\n\n## C sqrt Function Example\n\nThe math square root Function allows you to find the square root of a given number. In this program, We are going to find the square root and display the output.\n\n```#include <stdio.h>\n#include <math.h>\n\nint main()\n{\nprintf(\"\\n The Square Root Value of 0 = %.4f \", sqrt(0));\nprintf(\"\\n The Square Root Value of 1 = %.4f \", sqrt(1));\n\nprintf(\"\\n The Square Root Value of 125 = %.4f \", sqrt(125));\nprintf(\"\\n The Square Root Value of 500.36 = %.4f \", sqrt(500.36));\n\nprintf(\"\\n The Square Root Value of -27.32 = %.4f \", sqrt(-27.32));\nprintf(\"\\n The Square Root Value of -90 = %.4f \", sqrt(-90));\n\nreturn 0;\n}```\n\n## C Square root Example 2\n\nIn this C Programming example, we are allowing the user to enter their own value. Next, this program uses the sqrt function to find the square root of the user given number.\n\n```#include <stdio.h>\n#include <math.h>\n\nint main()\n{\nfloat number, sqrtValue;\n\nprintf(\" Please Enter any Numeric : \");\nscanf(\"%f\", &number);\n\nsqrtValue = sqrt(number);\n\nprintf(\"\\n The Square Root of %.2f = %.4f \", number, sqrtValue);\n\nreturn 0;\n}```\n`````` Please Enter any Numeric : 64\n\nThe Square Root of 64.00 = 8.0000``````"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.64299893,"math_prob":0.999343,"size":1322,"snap":"2023-40-2023-50","text_gpt3_token_len":354,"char_repetition_ratio":0.17526555,"word_repetition_ratio":0.12608695,"special_character_ratio":0.33358547,"punctuation_ratio":0.17993079,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998116,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T01:41:49Z\",\"WARC-Record-ID\":\"<urn:uuid:54fb34aa-ba76-4cda-9648-a8b9b27aadaa>\",\"Content-Length\":\"61634\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12bebdf8-5b21-4346-99db-d15882aaf4b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:ec38afb1-567d-4680-909b-d6ad3349924a>\",\"WARC-IP-Address\":\"172.67.73.221\",\"WARC-Target-URI\":\"https://www.tutorialgateway.org/c-sqrt-function/\",\"WARC-Payload-Digest\":\"sha1:R2FDODLC3IMPUU33BAT2V2TS5WGKBNPV\",\"WARC-Block-Digest\":\"sha1:YM3XXDOASSD36J6XKLKDOUXU7FZ3FR2O\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100540.62_warc_CC-MAIN-20231205010358-20231205040358-00062.warc.gz\"}"} |
https://uniteasy.com/solver/cubicequation/3x%5E3-9x%5E2-8x%2B2%3D0/ | [
" Solve 3x^3-9x^2-8x+2=0 | Uniteasy.com\n\n# Solve the cubic equation:\n\n## $$3x^3-9x^2-8x+2=0$$\n\nSince the discriminant $$\\Delta <0$$, the cubic equation has three distinct real roots.\n\n$$\\Delta=-2.7393689986283$$\n\n$$\\begin{cases} x_1=\\dfrac{2}{3}\\sqrt{17}\\cos \\bigg[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\bigg]+1 \\\\ x_2=\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2\\pi}{3}\\bigg]+1 \\\\ x_3=\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{4\\pi}{3} \\bigg]+1 \\end{cases}$$\n\nIn decimals,\n\n$$\\begin{cases} x_1=3.6760776527123 \\\\ x_2=-0.881751033463 \\\\ x_3=0.20567338075069 \\end{cases}$$\n\nDetailed Steps on Solution\n\n## 1. Convert to depressed cubic equation\n\nThe idea is to convert general form of cubic equation\n\n$$ax^3+bx^2+cx+d = 0$$\n\nto the form without quadratic term.\n\n$$t^3+pt+q = 0$$\n\nBy substituting $$x$$ with $$t - \\dfrac{b}{3a}$$, the general cubic equation could be transformed to\n\n$$t^3+\\dfrac{3ac-b^2}{3a^2}t+\\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0$$\n\nCompare with the depressed cubic equation. Then,\n\n$$p = \\dfrac{3ac-b^2}{3a^2}$$\n\n$$q = \\dfrac{2b^3-9abc+27a^2d}{27a^3}$$\n\nSubstitute the values of coefficients, $$p, q$$ is obtained as\n\n$$p = \\dfrac{3\\cdot 3\\cdot (-8)-(-9)^2}{3\\cdot 3^2}=-\\dfrac{17}{3}$$\n\n$$q = \\dfrac{2\\cdot (-9)^3-9\\cdot3\\cdot (-9)\\cdot (-8)+27\\cdot 3^2\\cdot2}{27\\cdot 3^3}=-4$$\n\n### Use the substitution to transform\n\nLet $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.\n\n$$t^3 +pt+q=0$$\n\nLet $$x=t+1$$\n\nThe cubic equation $$3x³ - 9x² - 8x + 2=0$$ is transformed to\n\n$$t^3 -\\dfrac{17}{3}t-4=0$$\n\n## 2. Cardano's solution\n\nLet $$t=u-v$$\n\nCube both sides and extract common factor from two middle terms after expanding the bracket.\n\n\\begin{aligned} \\\\t^3&=(u-v)^3\\\\ & =u^3-3u^2v+3uv^2-v^3\\\\ & =-3uv(u-v)+u^3-v^3\\\\ \\end{aligned}\n\nSince $$u-v=t$$, substitution gives a linear term for the equation. Rearrange terms.\n\n$$x^3+3uvx-u^3+v^3=0$$\n\nCompare the cubic equation with the original one (1)\n\n$$\\begin{cases} 3uv=-\\dfrac{17}{3}\\quad\\text{or}\\quad v=-\\dfrac{17}{9u}\\\\ v^3-u^3=-4\\\\ \\end{cases}$$\n\n$$v=-\\dfrac{17}{9u}$$ gives relationship between the two variables. Substitute the value of $$v$$ to the second equation\n\n$$\\Big(-\\dfrac{17}{9u}\\Big)^3-u^3=-4$$\n\nSimplifying gives,\n\n$$u^3+\\dfrac{4913}{729}\\dfrac{1}{u^3}-4=0$$2\n\nLet $$m=u^3$$, then the equation is transformed to a quadratic equation in terms of $$m$$. Once the value of $$m$$ is determined, $$v^3$$ could be determined by $$v^3=-4+u^3$$.\n\n$$m^2-4m+\\dfrac{4913}{729}=0$$\n\nSovling the quadratic euqation will give two roots (some may be equal). Here we only cosider one case with positive sign before the square root radical since the negative case will produce the same result.\n\n\\begin{aligned} \\\\u^3=m&=2+\\dfrac{1}{2}\\sqrt{\\Big(-4^2\\Big)-4\\cdot \\dfrac{4913}{729}}\\\\ & =2+\\dfrac{1}{2}\\sqrt{16-\\dfrac{19652}{729}}\\\\ & =2+\\dfrac{1}{2}\\sqrt{\\dfrac{7988}{729}}i\\\\ & =2+\\dfrac{\\sqrt{1997}}{27}i\\\\ \\end{aligned}\n\n$$v^3$$ can be determined by the equation we deduced $$v^3-u^3=-4$$. Then,\n\n\\begin{aligned} \\\\v^3&=-4+u^3\\\\ & =-4+2+\\dfrac{\\sqrt{1997}}{27}i\\\\ & =-2+\\dfrac{\\sqrt{1997}}{27}i\\\\ \\end{aligned}\n\nNow we have,\n\n$$u^3=2+\\dfrac{\\sqrt{1997}}{27}i$$ and $$v^3=-2+\\dfrac{\\sqrt{1997}}{27}i$$\n\nEvaluating the simplest cubic equation $$x^3-A=0$$, it has 3 roots, in which the first root is a real number or a complex number. The second and third are expressed in the product of cubic root of unity and the first one.\n\nIf $$ω = \\dfrac{-1+i\\sqrt{3}}{2}$$, then its reciprocal is equal to its conjugate, $$\\dfrac{1}{ω}=\\overline{ω}$$.\n\n$$\\begin{cases} r_1=\\sqrt{A}\\\\ r_2=\\dfrac{-1+i\\sqrt{3}}{2}\\cdot \\sqrt{A}\\\\ r_3=\\dfrac{-1-i\\sqrt{3}}{2}\\cdot \\sqrt{A}\\\\ \\end{cases}$$\n\nSimilary, taking cubic root for $$u^3$$ and $$v^3$$ also gives 3 roots.\n\n$$\\begin{cases} u_1=\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}\\\\ u_2=\\dfrac{-1+i\\sqrt{3}}{2}\\cdot \\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}\\\\ u_3=\\dfrac{-1-i\\sqrt{3}}{2}\\cdot \\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}\\\\ \\end{cases}$$\n\nFor $$v_2$$ and $$v_3$$, the complex numbers before radicals are the conjugates of those for $$u_2$$ and $$u_3$$, which can be verified by the reciprocal property of the cubic root of unity from the equation $$v=-\\dfrac{17}{9u}$$. The radicand can be taken as the negative conjugate of that in $$u_1$$, $$u_2$$ and $$u_3$$, which is the same in value.\n\n$$\\begin{cases} v_1=\\sqrt{-2+\\dfrac{\\sqrt{1997}}{27}i}\\\\ v_2=\\dfrac{-1-i\\sqrt{3}}{2}\\cdot \\sqrt{-2+\\dfrac{\\sqrt{1997}}{27}i}\\\\ v_3=\\dfrac{-1+i\\sqrt{3}}{2}\\cdot \\sqrt{-2+\\dfrac{\\sqrt{1997}}{27}i}\\\\ \\end{cases}$$\n\nSince $$t=u-v$$, the firt root $$t_1$$ can be expressed as the sum of cubic root of two conjugate complex numbers\n\n$$t_1=\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}+\\sqrt{2-\\dfrac{\\sqrt{1997}}{27}i}$$\n\nLet $$u^3_1=z=2+\\dfrac{\\sqrt{1997}}{27}i$$, and $$z$$ can be expressed in trigonomic form $$r(\\cos θ + i \\sin θ)$$, where $$r$$ and $$θ$$ are the modulus and principle argument of the complex number.\n\nThen $$-v^3_1$$ is the conjugate $$\\overline{z}=2-\\dfrac{\\sqrt{1997}}{27}i$$, or $$\\overline{z} = r(\\cos θ - i \\sin θ)$$\n\nNow let's calculate the value of $$r$$ and $$θ$$.\n\n\\begin{aligned} \\\\r&=\\sqrt{\\Big(2\\Big)^2+\\Big(\\dfrac{\\sqrt{1997}}{27}\\Big)^2}\\\\ & =\\dfrac{17}{27}\\sqrt{17}\\\\ \\end{aligned}\n\n$$\\cosθ=\\dfrac{2}{\\dfrac{17}{27}\\sqrt{17}}=\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}$$\n\nThe argument is the inverse function of the cosθ\n\n$$θ=\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)$$\n\nUsing de Moivre’s formula, the cubic root of $$z$$ could be determined.\n\n\\begin{aligned} \\\\u_1=\\sqrt{z}&=\\sqrt{r}(\\cos\\dfrac{θ}{3}+i\\sin\\dfrac{θ}{3})\\\\ & =\\sqrt{\\dfrac{17}{27}\\sqrt{17}}\\Big[\\cos\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\sin\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\Big]\\\\ & =\\dfrac{\\sqrt{17}}{3}\\Big[\\cos\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\sin\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\Big]\\\\ \\end{aligned}\n\nSince $$-v^3$$ is the conjugate of $$z$$ as we mentioned above,\n\n\\begin{aligned} \\\\v_1=-\\sqrt{\\overline{z}}&=-\\sqrt{r}(\\cos\\dfrac{θ}{3}-i\\sin\\dfrac{θ}{3})\\\\ & =\\dfrac{\\sqrt{17}}{3}\\Big[-\\cos\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\sin\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\Big]\\\\ \\end{aligned}\n\nThe first root is the difference of $$u_1$$ and $$v_1$$.\n\n\\begin{aligned} \\\\t_1&=u_1-v_1\\\\ & =2\\cdot \\dfrac{\\sqrt{17}}{3}\\cos\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\\\ & =\\dfrac{2}{3}\\sqrt{17}\\cos\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\\\ \\end{aligned}\n\nThe second root is the difference of $$u_2$$ and $$v_2$$, in which $$u_2$$ is the product of the cubic root of $$z$$ and the cubic root of unity, $$v_2$$ is the product of the negative of the conjugate of $$z$$ and the conjugate of the cubic root of unity.\n\n\\begin{aligned} \\\\t_2&=u_2-v_2\\\\ & =\\dfrac{-1+i\\sqrt{3}}{2}\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}+\\dfrac{-1-i\\sqrt{3}}{2}\\sqrt{2-\\dfrac{\\sqrt{1997}}{27}i}\\\\ & =\\dfrac{-1+i\\sqrt{3}}{2}\\sqrt{r}(\\cos\\dfrac{θ}{3}+i\\sin\\dfrac{θ}{3})+\\dfrac{-1-i\\sqrt{3}}{2}\\sqrt{r}(\\cos\\dfrac{θ}{3}-i\\sin\\dfrac{θ}{3})\\\\ & =-\\sqrt{r}(\\cos\\dfrac{ θ}{3} + \\sqrt{3} \\sin\\dfrac{ θ}{3} ) \\\\ & =2\\sqrt{r}\\cos\\Big(\\dfrac{θ}{3}+ \\dfrac{4\\pi}{3}\\Big)\\\\ & =\\dfrac{2}{3}\\sqrt{17}\\cos\\Big(\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+ \\dfrac{4\\pi}{3}\\Big)\\\\ \\end{aligned}\n\nThe third root is the difference of $$u_3$$ and $$v_3$$, in which $$u_3$$ is the product of the cubic root of $$z$$ and the conjugate of the cubic root of unity, $$v_3$$ is the product of the negative of the conjugate of $$z$$ and the cubic root of unity.\n\n\\begin{aligned} \\\\t_3&=u_3-v_3\\\\ & =\\dfrac{-1-i\\sqrt{3}}{2}\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}+\\dfrac{-1+i\\sqrt{3}}{2}\\sqrt{2-\\dfrac{\\sqrt{1997}}{27}i}\\\\ & =\\dfrac{-1-i\\sqrt{3}}{2}\\sqrt{r}(\\cos\\dfrac{θ}{3}+i\\sin\\dfrac{θ}{3})+\\dfrac{-1+i\\sqrt{3}}{2}\\sqrt{r}(\\cos\\dfrac{θ}{3}-i\\sin\\dfrac{θ}{3})\\\\ & =\\sqrt{r}(-\\cos\\dfrac{θ}{3}+\\sqrt{3} \\sin \\dfrac{ θ}{3})\\\\ & =2\\sqrt{r}\\cos\\Big(\\dfrac{θ}{3}+ \\dfrac{2\\pi}{3}\\Big)\\\\ & =\\dfrac{2}{3}\\sqrt{17}\\cos\\Big(\\dfrac{1}{3}\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+ \\dfrac{2\\pi}{3}\\Big)\\\\ \\end{aligned}\n\n## 3. Vieta's Substitution\n\nIn Cardano' solution, $$t$$ is defined as the difference of $$u$$ and $$v$$. If we substitute the value of $$v$$ (4) into (2), we get the equation. $$t=u+\\dfrac{17}{9u}$$. And then substitute the equation to the cubic equation $$t^3-\\dfrac{17}{3}t-4=0$$. This method is called Vieta's Substitution for solving a cubic equation, which simplied the Cardano' solution. The substitution expression can be obtained by the following formula directly.\n\n$$t=u-\\dfrac{p}{3u}$$\n\nSubstitute the expression $$t=u+\\dfrac{17}{9u}$$ to the cubic equation\n\n$$\\Big(u+\\dfrac{17}{9u}\\Big)^3-\\dfrac{17}{3}\\Big(u+\\dfrac{17}{9u}\\Big)-4=0$$\n\nExpand brackets and cancel the like terms\n\n$$u^3+\\cancel{\\dfrac{17}{3}u^2\\dfrac{1}{u}}+\\cancel{\\dfrac{289}{27}u\\dfrac{1}{u^2}}+\\dfrac{4913}{729}\\dfrac{1}{u^3}-\\cancel{\\dfrac{17}{3}u}-\\cancel{\\dfrac{289}{27}\\dfrac{1}{u}}-4=0$$\n\nThen we get the same equation as (2)\n\n$$u^3+\\dfrac{4913}{729}\\dfrac{1}{u^3}-4=0$$\n\nThe rest of the steps will be the same as those of Cardano's solution\n\n## $$t^3-\\dfrac{17}{3}t-4=0$$\n\nMove the linear term and constant of (1) to its right hand side. We get the following form of the equation.\n\n$$t^3=\\dfrac{17}{3}t+4$$3\n\nLet the root of the cubic equation be the sum of two cubic roots\n\n$$t=\\sqrt{r_1}+\\sqrt{r_2}$$4\n\nin which $$r_1$$ and $$r_2$$ are two roots of a quadratic equation\n\n$$z^2-\\alpha z+ β=0$$5\n\nUsing Vieta's Formula, the following equations are established.\n\n$$r_1+r_2 = \\alpha \\quad \\text{and} \\quad r_1r_2 = β$$\n\nTo determine $$\\alpha$$, $$β$$, cube both sides of the equation (4)\n\n$$t^3=3\\sqrt{r_1r_2}(\\sqrt{r_1}+\\sqrt{r_2})+r_1+r_2$$\n\nSubstituting, the equation is simplified to\n\n$$t^3=3\\sqrt{β}t+\\alpha$$\n\nCompare the cubic equation with (3), the following equations are established\n\n$$\\begin{cases} 3\\sqrt{β}=\\dfrac{17}{3}\\\\ \\alpha=4\\\\ \\end{cases}$$\n\nSolving for $$β$$ gives\n\n$$β=\\dfrac{4913}{729}$$\n\nSo the quadratic equation (5) is determined as\n\n$$z^2-4z+\\dfrac{4913}{729}=0$$6\n\n$$\\begin{cases} r_1=2+\\dfrac{\\sqrt{1997}}{27}i\\approx2+1.6551039238151i\\\\ r_2=2-\\dfrac{\\sqrt{1997}}{27}i\\approx2-1.6551039238151i\\\\ \\end{cases}$$\n\nTherefore, one of the roots of the cubic equation could be obtained from (4).\n\n$$t_1=\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}+\\sqrt{2-\\dfrac{\\sqrt{1997}}{27}i}$$\n\nin decimals,\n\n$$t_1=2.6760776527123$$\n\nHowever, since the cube root of a quantity has triple values,\n\nThe other two roots could be determined as,\n\n$$t_2=\\dfrac{-1+i\\sqrt{3}}{2}\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}+\\dfrac{-1-i\\sqrt{3}}{2}\\sqrt{2-\\dfrac{\\sqrt{1997}}{27}i}$$\n\n$$t_3=\\dfrac{-1-i\\sqrt{3}}{2}\\sqrt{2+\\dfrac{\\sqrt{1997}}{27}i}+\\dfrac{-1+i\\sqrt{3}}{2}\\sqrt{2-\\dfrac{\\sqrt{1997}}{27}i}$$\n\nSince the expression involes cubic root of complex number, the final result can be deduced by using trigonometric method as shown in Cardano's solution.\n\nFor the equation $$t^3 -\\dfrac{17}{3}t-4$$, we have $$p=-\\dfrac{17}{3}$$ and $$q = -4$$\n\n### Calculate the discriminant\n\nThe nature of the roots are determined by the sign of the discriminant.\n\nSince $$p$$ is negative, the discriminant will be less than zero if the absolute value of $$p$$ is large enough.\n\n\\begin{aligned} \\\\\\Delta&=\\dfrac{q^2}{4}+\\dfrac{p^3}{27}\\\\ & =\\dfrac{(-4)^2}{4}+\\dfrac{\\Big(-\\dfrac{17}{3}\\Big)^3}{27}\\\\ & =4-\\dfrac{4913}{729}\\\\ & =\\dfrac{4\\cdot 729-4913\\cdot 1}{729}\\\\ & =-2.7393689986283\\\\ \\end{aligned}\n\n### 4.1 Use the root formula directly\n\nIf $$\\Delta < 0$$, then there are 3 distinct real roots for the cubic equation\n\n$$\\begin{cases} t_1 &= 2\\sqrt{r} \\cos\\dfrac{ θ}{3} \\\\ t_2 & = 2\\sqrt{r}\\cos\\Big( \\dfrac{ θ}{3}+\\dfrac{2\\pi}{3} \\Big) \\\\ t_3&= 2\\sqrt{r}\\cos\\Big( \\dfrac{ θ}{3}+\\dfrac{4\\pi}{3} \\Big) \\end{cases}$$\n\nwhere\n\n$$\\theta = \\arccos(\\dfrac{3q}{2p}\\sqrt{-\\dfrac{3}{p} } )$$ and $$\\sqrt{r} = \\sqrt{\\dfrac{-p}{3} }$$\n\nSubstitute the values of $$p$$ and $$q$$ to determine the value of $$\\theta$$\n\n\\begin{aligned} \\\\\\theta&= \\arccos(\\dfrac{3q}{2p}\\sqrt{-\\dfrac{3}{p} } )\\\\ & =\\arccos\\Big(\\dfrac{3\\cdot -4}{2 \\cdot -\\dfrac{17}{3}}\\sqrt{-\\dfrac{3}{-\\dfrac{17}{3}} }\\Big)\\\\ & = \\arccos\\Big(\\dfrac{1}{2}\\cdot 3\\cdot 4\\cdot \\dfrac{3}{17}\\sqrt{3\\cdot \\dfrac{3}{17}}\\Big)\\\\ & =\\arccos\\Big(\\dfrac{1}{2}\\cdot 3\\cdot 4\\cdot \\dfrac{3}{17}\\cdot 3\\sqrt{\\dfrac{1}{17}}\\Big)\\\\ & = \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\\\ \\end{aligned}\n\nThen we can determine one third of $$\\theta$$\n\n$$\\dfrac{\\theta}{3} = \\dfrac{1}{3} \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)$$\n\nSubstitute the value of $$p$$ to determine the cubic root of $$r$$\n\n\\begin{aligned} \\\\\\sqrt{r}& =\\sqrt{\\dfrac{-p}{3}}\\\\ & =\\sqrt{\\frac{-(-17)}{3 \\cdot 3}} \\\\ & =\\sqrt{\\dfrac{17}{9}}\\\\ & =\\dfrac{\\sqrt{17}}{3}\\\\ \\end{aligned}\n\nSubstitute the value of $$\\dfrac{\\theta}{3}$$ and $$\\sqrt{r}$$ to the root formulas. Then we get\n\n\\begin{aligned} \\\\t_1&= 2\\sqrt{r} \\cos\\dfrac{ θ}{3}\\\\ & =2\\cdot \\dfrac{\\sqrt{17}}{3}\\cdot \\cos \\bigg[ \\dfrac{1}{3}\\cdot\\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\bigg]\\\\ & =\\dfrac{2}{3}\\sqrt{17}\\cos \\bigg[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\bigg]\\\\ & \\approx \\dfrac{2}{3}\\sqrt{17} \\cos 0.23043886792429\\\\ & \\approx 2.6760776527123\\\\ \\end{aligned}\n\n\\begin{aligned} \\\\t_2&=\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2\\pi}{3}\\bigg]\\\\ & \\approx -1.881751033463\\\\ \\end{aligned}\n\n\\begin{aligned} \\\\t_3&=\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{4\\pi}{3} \\bigg]\\\\ & \\approx -0.79432661924931\\\\ \\end{aligned}\n\n### 4.2 Trigonometric Method\n\nFor a depressed cubic equation, another method is to convert the equation to the form of triple angle identity. By comparing the trigonometric value of the triple angle, the value of the single angle is determined. And then the roots of the cubic equation can be found.\n\nCompare the given equation with the standard depressed cubic equation $$t^3 +pt+q=0$$, we get $$p=-\\dfrac{17}{3}$$ and $$q = -4$$\n\nUsing the formula\n\n$$t= 2\\sqrt{\\dfrac{-p}{3}}u$$\n\nto introduce an intermediate variable $$u$$ so that the given equation is transformed to a new equation in terms of $$u$$, which is analogous to a trignometric triple angle identity.\n\nThen,\n\n$$t= \\dfrac{2}{3}\\sqrt{17}u$$\n\nSubstitute to the cubic equation and simplify.\n\n$$3\\Big(\\dfrac{2}{3}\\sqrt{17}u\\Big)^3 -17\\Big(\\dfrac{2}{3}\\sqrt{17}u\\Big)-12=0$$\n\nExpand and simplify coefficients\n\n$$3\\cdot \\dfrac{8}{27}\\cdot17\\sqrt{17}u^3 -\\dfrac{34}{3}\\sqrt{17}u-12=0$$\n\nContinute simplifying\n\n$$136\\sqrt{17}u^3-102\\sqrt{17}u-108=0$$\n\nCancel the common factors of coefficients of each term\n\n$$68\\sqrt{17}u^3-51\\sqrt{17}u-54=0$$\n\nDividing the equation by $$17\\sqrt{17}$$ gives\n\n$$4u^3-3u-\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}=0$$\n\nThe equation becomes the form of a triple angle identity for cosine function.\n\n$$4\\cos^3θ-3\\cos θ-\\cos3θ =0$$\n\nComparing the equation with triple angle identity gives\n\n$$u =\\cos θ$$\n\nand\n\n$$\\cos3θ =\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}$$\n\nThe fact that the cosine function is periodic with period 2π implies the following equation.\n\n$$\\cos(3θ-2πk) =\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}$$\n\nSolving for θ yields\n\n$$θ =\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2πk}{3} ,$$\n\nin witch $$k=0,1,2$$.\n\nThen we can determine the value of $$u$$, that is $$\\cos θ$$\n\n\\begin{aligned} \\\\u&= \\cos θ\\\\ & = \\cos\\Big[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2πk}{3}\\Big]\\\\ \\end{aligned}\n\nand subsiquently,\n\n\\begin{aligned} \\\\t&= \\dfrac{2}{3}\\sqrt{17}u\\\\ & = \\dfrac{2}{3}\\sqrt{17} \\cos\\Big[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2πk}{3}\\Big]\\\\ \\end{aligned}\n\nSubstituting $$k=0,1,2$$ gives the roots of solution set.\n\n\\begin{aligned} \\\\t_1&= \\dfrac{2}{3}\\sqrt{17}\\cos\\Big[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\Big]\\\\ & \\approx2.6760776527123\\\\ \\end{aligned}\n\n\\begin{aligned} \\\\t_2&= \\dfrac{2}{3}\\sqrt{17}\\cos\\Big[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2π}{3}\\Big]\\\\ & \\approx-1.881751033463\\\\ \\end{aligned}\n\n\\begin{aligned} \\\\t_3&= \\dfrac{2}{3}\\sqrt{17}\\cos\\Big[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{4π}{3}\\Big]\\\\ & \\approx-0.79432661924931\\\\ \\end{aligned}\n\nwhich shows the trigonometric method gets the same solution set as that by using roots formula.\n\n## Roots of the general cubic equation\n\nSince $$x = t - \\dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives\n\n\\begin{aligned} \\\\x_1&=t_1-\\dfrac{b}{3a}\\\\ & =\\dfrac{2}{3}\\sqrt{17}\\cos \\bigg[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\bigg]+1\\\\ \\end{aligned}\n\n\\begin{aligned} \\\\x_2&=t_2-\\dfrac{b}{3a}\\\\ & =\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2\\pi}{3}\\bigg]+1\\\\ \\end{aligned}\n\n\\begin{aligned} \\\\x_3&=t_3-\\dfrac{b}{3a}\\\\ & =\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{4\\pi}{3} \\bigg]+1\\\\ \\end{aligned}\n\n## 5. Summary\n\nIn summary, we have tried the method of cubic root formula, trigonometric to explore the solutions of the equation. The cubic equation $$3x³ - 9x² - 8x + 2=0$$ is found to have three real roots . Exact values and approximations are given below.\n\n$$\\begin{cases} x_1=\\dfrac{2}{3}\\sqrt{17}\\cos \\bigg[\\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)\\bigg]+1 \\\\ x_2=\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{2\\pi}{3}\\bigg]+1 \\\\ x_3=\\dfrac{2}{3}\\sqrt{17} \\cos \\bigg[ \\dfrac{1}{3}\\cdot \\arccos\\Big(\\dfrac{54}{17}\\sqrt{\\dfrac{1}{17}}\\Big)+\\dfrac{4\\pi}{3} \\bigg]+1 \\end{cases}$$\n\nConvert to decimals,\n\n$$\\begin{cases} x_1=3.6760776527123 \\\\ x_2=-0.881751033463 \\\\ x_3=0.20567338075069 \\end{cases}$$\n\n## 6. Graph for the function $$f(x) = 3x³ - 9x² - 8x + 2$$\n\nSince the discriminat is less than zero, the curve of the cubic function $$f(x) = 3x³ - 9x² - 8x + 2$$ has 3 intersection points with horizontal axis.\n\nScroll to Top"
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"SGPLOT - REG Graph\n\nQ\n\nHow to generate REG graphs with SAS SGPLOT procedure?\n\n✍: FYIcenter.com\n\nA",
null,
"If you have a set of X-Y data pairs, you can present the data with a REG graph, which displays locations of data pairs, and its straight regression line.\n\nThe following SAS program shows a REG graph on the example data set, sashelp.class, provided in the SAS University Edition:\n\n```proc sgplot data=sashelp.class;\nreg x=height y=weight;\n```\n\nThe result looks like the following:"
] | [
null,
"http://faq.fyicenter.com/SAS/_icon_SAS.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81136113,"math_prob":0.7098253,"size":1264,"snap":"2023-40-2023-50","text_gpt3_token_len":322,"char_repetition_ratio":0.08888889,"word_repetition_ratio":0.0,"special_character_ratio":0.25949368,"punctuation_ratio":0.17120622,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9768057,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T14:52:23Z\",\"WARC-Record-ID\":\"<urn:uuid:3572554d-9168-4a46-ae61-feecd8ef68af>\",\"Content-Length\":\"24758\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:12454956-d9a6-4a10-be27-ed39339c632e>\",\"WARC-Concurrent-To\":\"<urn:uuid:a7b62d9c-7d07-43c5-b3c6-f744f62d4744>\",\"WARC-IP-Address\":\"74.208.236.35\",\"WARC-Target-URI\":\"http://faq.fyicenter.com/1062_SGPLOT-REG_Graph.html\",\"WARC-Payload-Digest\":\"sha1:G6PRYBJIQMYCDTUG5GOJF5M4X6JDZQAT\",\"WARC-Block-Digest\":\"sha1:DA2ZUT5PS7XLWENW7GRTNOGDPEYO6OSM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510903.85_warc_CC-MAIN-20231001141548-20231001171548-00240.warc.gz\"}"} |
https://pureportal-staging.strath.ac.uk/en/studentTheses/stabilization-of-hybrid-systems-by-discrete-time-feedback-control-2 | [
"# Stabilization of hybrid systems by discrete-time feedback controls\n\n• Jianqiu Lu\n\nStudent thesis: Doctoral Thesis\n\n### Abstract\n\nHybrid stochastic differential equations (SDEs) (also known as SDEs with Markovian switching) have been used to model many practical systems where they may experience abrupt changes in their structure and parameters. One of the important issues in the study of hybrid systems is the automatic control, with consequent emphasis being placed on the asymptotic analysis of stability. One classical topic in this field is the problem of stabilization. The stability of hybrid systems by feedback control based on continuous-time observations has been studied extensively in the past decades. Recently, Mao initiates the study on the mean-square exponential stabilization of continuous-time hybrid stochastic differential equations by feedback controls based on discrete-time state observations. Mao also obtains an upperbound on the duration between two consecutive state observations. However, it is due to the general technique used there that the bound on is not very sharp. In this thesis, we will consider a couple of important classes of hybrid SDEs. Making full use of their special features, we will be able to establish a better bound on. Our new theory enables us to observe the system state less frequently (so costless) but still to be able to design the feedback control based on the discrete-timestate observations to stabilize the given hybrid SDEs in the sense of mean-square exponential stability. Moreover, we will be able to establish a better bound on making use of Lyapunov functionals. By this method, we will not only discuss the stabilization in the sense of exponential stability but also in other sense of H1 stability or asymptotic stability as well. We will not only consider the mean square stability but also the almost sure stability. It is easy to observe that the feedback control there still depends on the continuous-time observations of the mode. However, it usually costs to identify the current mode of the system in practice. So we can further improve the control to reduce the control cost by identifying the mode at discrete times when we make observations for the state. Therefore, we will also design such a type of feedback control based on the discrete-time observations of both state and mode to stabilize the given hybrid stochastic differential equations (SDEs) in the sense of mean-square exponential stability in this thesis. Similarly, we can extend our discussion to the stabilization of continuous-time hybrid stochastic differential equations by feedback controls based on not only discrete-time state observations but also discrete-time mode observations by Lyapunov method. At last, we will investigate stability of Stochastic differential delay equations with Markovian switching by feedback control based on discrete-time State and mode observations by using Lyapunov functional. Hence, we will get the upperbound on the duration between two consecutive state and mode observations.\nDate of Award 26 Sep 2018 English University Of Strathclyde Xuerong Mao (Supervisor) & Wei Yang (Supervisor)\n\n'"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.93962926,"math_prob":0.8812639,"size":3114,"snap":"2022-05-2022-21","text_gpt3_token_len":599,"char_repetition_ratio":0.15305467,"word_repetition_ratio":0.093023255,"special_character_ratio":0.17886962,"punctuation_ratio":0.06153846,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95242083,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T12:24:10Z\",\"WARC-Record-ID\":\"<urn:uuid:8e6b6749-aca8-412c-85f3-be8e9ce87c1f>\",\"Content-Length\":\"25570\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:930e07a7-223b-4e31-8996-2a7dd0caf0b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:31ea5547-f58e-440f-a4c8-3b0002ea552e>\",\"WARC-IP-Address\":\"34.248.98.230\",\"WARC-Target-URI\":\"https://pureportal-staging.strath.ac.uk/en/studentTheses/stabilization-of-hybrid-systems-by-discrete-time-feedback-control-2\",\"WARC-Payload-Digest\":\"sha1:KLPFTECZK2WMGFEZE7JT65J7NCDYX2GL\",\"WARC-Block-Digest\":\"sha1:UDZ3CA5YN4ELFCG3AUC3Y4ACFJEPFS5C\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662522270.37_warc_CC-MAIN-20220518115411-20220518145411-00203.warc.gz\"}"} |
https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0200209 | [
"Browse Subject Areas\n?\n\nClick through the PLOS taxonomy to find articles in your field.\n\n# Network-based risk measurements for interbank systems\n\n• Yongli Li,\n\nRoles Formal analysis, Writing – original draft\n\nAffiliation School of Business Administration, Northeastern University, Shenyang, P.R.China\n\n• Guanghe Liu,\n\nRoles Formal analysis, Writing – original draft\n\nAffiliations School of Business Administration, Northeastern University, Shenyang, P.R.China, Business School, Sun Yat-Sen University, Guangzhou, P.R.China\n\n• Paolo Pin\n\nRoles Formal analysis, Writing – original draft\n\npaolo.pin@unibocconi.it\n\nAffiliation Department of Decision Sciences, Innocenzo Gasparini Institute for Economic Research, and Bocconi Institute for Data Science and Analytics, University of Bocconi, Milano, Italy\n\n## Abstract\n\nThis paper focuses on evaluating the systemic risk in interbank networks, proposing a series of measurements: risk distance, risk degree and m-order risk degree. The proposed measurements are formally proven to have good basic and extended properties that are able to reflect the effect of bank size, liability size, liability distribution, and the discount factor on the default risk, not only of a single bank, but also of the entire system. Additionally, the abovementioned properties and the relationship between risk distance and financial contagion indicate the rationality embodied in the proposed measurements. This paper also provides some implications on how to decrease or prevent the systemic risk in an interbank system.\n\n## 1 Introduction\n\nSince the Asian financial crisis of 1997, special attention has been paid to the role of the growing interconnectedness between financial institutions among the many factors that affect financial contagion [1,2]. Particularly after the global crisis of 2007–08, the architecture of financial system building on the abovementioned interconnectedness was viewed as being crucial for its central role in the financial contagion . In fact, the abovementioned interconnectedness between financial institutes constitutes the edge of a financial network and the corresponding financial institutes are regarded as the nodes. Particularly, following numerous studies in this field such as and , we also focus on the interbank system that can be considered as a fundamental structure for complex financial systems. Note that interbank borrowing and loans, if any, form the abovementioned interconnectedness that link the corresponding banks in the interbank network . The network representation allows to study propagation of failures: recalling the two mentioned financial crisis, for example, one bank’s insolvency may lead to the default cascades in the interbank network. Here, two periods are considered: several banks are assumed to default in the first period and the set of these banks is named initial default set, and then in the second period, some of the remaining banks may be induced to default because of the existing borrowing and loan links. Facing this phenomenon, we want to explore two problems. The first one is when one bank’s insolvency occur, which bank will be the next victim? The second one is which initial default set will cause the largest amount of banks to default in the second period? This paper will inherit the idea of risk propagation in networks to cope with the two problems by way of providing a series of convenient measurements.\n\nMore concretely, if the answer of the first question is known in advance of financial contagion, we can inject liquidity into the more susceptive banks to avoid the spread of the crisis. As the first contribution, we provide a new measurement named risk distance, with the property that a shorter risk distance with the given initial default set means a higher likelihood of default. Among a growing literature on risk analysis in financial networks, the harmonic distance presented by is noteworthy because it captures the susceptibility of each bank to the distress of any other, so that it functions similarly to our proposed risk distance. However, the risk distance that we define is different from the harmonic distance in two aspects: one is that our risk distance considers the cash and marketable assets carried by the banks so that the analysed banks can be heterogeneous, the other is that the risk distance defined here does not assume that the initial default set only contains a single bank. As a result, we prove that our newly proposed risk distance has several different properties with the famous harmonic distance in the following parts of this paper. Overall, our risk distance is a node-level (or say microscopic) indicator that reflects the default likelihood of the remaining banks given an initial default set.\n\nBesides, the second question aims to find the “important” banks from the perspective of financial system risk. To that end, this paper further provides a second new measurement that can reflect the amount of “damage” caused by the initial default set, which is the second contribution. Then, the “damages” caused by different initial default sets of the same size can be compared to find which initial default set is most harmful. In particular, when the initial default set contains only one bank, the above problem can be simplified into finding the critical node and measuring its influence on causing financial contagion. Intuitively, the famous Katz-Bonacich centrality can provide the basic idea concerning how to address such a problem [11,12]. With Katz-Bonacich centrality’s becoming conventional wisdom [13,14], we attempt to inherit the basic framework of this centrality and to further develop it to solve the new problem, now that the classical Katz-Bonacich centrality cannot be directly adopted here . The key challenge of solving this problem is how to establish a new matrix that fully reflects the financial system in order to replace the classical adjacent matrix used in Katz-Bonacich centrality when an initial default set is given. Fortunately, has established a Markov transfer matrix with absorbing states, which inspires us on how to obtain a system matrix with the information of the initial default set. As a result, this paper successfully proposes the risk degree of a given initial default set and the risk degrees of an m-order initial default set, which reflects the default risk caused by a given initial default set and the maximal system risk caused by all possible default sets with any m banks, respectively. Overall, the newly provided measurements, called risk degree and m-order risk degree, belong to the type of system-level (i.e., macroscopic) indicators that reflect the collapsing force of the default sets.\n\nFurthermore, apart from the three newly established measurements, this paper also focuses on uncovering these measurements’ properties and demonstrating their rationale, which constitutes the third contribution. Specifically, our provided measurements are considered as a function of liability size, liability distribution, bank size and the discount factor, and we further find a much deeper relationship between the provided measurements and the number of failed banks caused by financial contagion. By extending the stylized setup used in papers such as those by , we also represent interbank systems that consists of n banks that are linked via unsecured debt contracts. Finally, abstracting from the background of finance, the nodes of our model have analogies with nodes of information in processes where knowledge is shared instead of risk . So, in the future, managers or government officers monitoring the financial system may be inspired by more general results that fit this analogy.\n\nThe rest of this paper is organized as follows, to present the abovementioned ideas and contributions. Section 2 starts from the balance sheet of inter-banks and then defines the three new measurements based on the payment balance of these banks. Section 3 provides the basic properties of the proposed measurements, and Section 4 further proves the extended properties, which are useful to validate the rationality of the proposed measurements. Section 5 concludes and discusses the managerial implications.\n\n## 2 Network-based risk measurements\n\n### 2.1 Balance sheet and payment equilibrium\n\nAn interbank system is considered here, which consists of several banks that are linked by interbank lending via unsecured debt contracts signed at the initial period. To better clarify the risk contagion process, we start with the balance sheet of a stylized commercial bank within the interbank system. As illustrated in Fig 1, bank i’s total assets Ai are composed of liquidities ci, securities si, interbank loans li and other assets pi; thus, the following equation holds:",
null,
"(1)\n\nFig 1. Balance sheet of a stylized commercial bank.\n\nhttps://doi.org/10.1371/journal.pone.0200209.g001\n\nMeanwhile, bank i’s total liabilities Hi consist of shareholder equity ei, deposits di, interbank borrowing bi and other liabilities qi:",
null,
"(2)\n\nIn fact, the risk in interbank payment systems can originate from different factors; for example, discussed how a shock to deposits could lead to the bank’s default on interbank borrowing or even part of its retail deposits, and considered that the uncertain returns on banks’ securities and other assets were likely to cause a system risk. Unlike the abovementioned studies, this paper does not distinguish the reasons that cause the risk, but rather focuses on the bank’s default induced by any possible initial shocks. Moreover, we further assume that each bank embedded in the interbank system have borrowing or loans at least with the other one bank, namely bi > 0 and li > 0, because the banks without any interbank lending and borrowing are isolated nodes and should not be considered into the interbank system.\n\nNote that the liabilities displayed in the balance sheet can be divided into the senior type and the junior type, and therefore they should be repaid in a different order when shocks occur. Here, the deposits di have seniority relative to the bank’s other liabilities; in other words, the available liquidities of bank i should repay di first and then bi as well as the other liabilities. Furthermore, the other liabilities generally contain long-term borrowing, bonds and sub debts that really exist as a commercial bank’s liabilities as displayed in Fig 1. Although interbank borrowings and other liabilities mentioned above both belong to the junior liabilities, interbank borrowings are always short-term and other liabilities are often long-term. Thus, we consider that interbank borrowings should be repaid firstly and immediately compared to other liabilities.\n\nLet ali denote the total available liquidities of bank i, and then regardless of the reasons that cause the shocks, two cases may exist: (1) if ali < di, bank i is called complete default. In this case, the total available liquidities of bank i are not enough to pay its senior liabilities; (2) if diali < di + bi, bank i is called part default, which means that, in this case, senior liabilities can be paid in full and the junior creditors are repaid in part. Furthermore, let xij (ji) denote the amount of money borrowed by bank j from bank i. When bank j is in part default, the amount of money repaid by bank j to bank i (denoted as yij) is in proportion to their contract xij. In mathematics, we have",
null,
"(3)\n\nSummarizing the two cases above, yij can be further expressed as",
null,
"(4) where xij implies the network structure imbedded in the interbank payments, and therefore their interdependence may induce a cascade of defaults when one or more banks default. Then, Eq (4) will be useful in determining the payment equilibrium because part default actually exists when financial contagion occurs in interbank payments.\n\nFurthermore, because rapid liquation is costly in most cases, bank i can only recover a fraction ηi < 1 of the securities si and other assets pi, where the defined fraction ηi is defined as the discount factor of bank i. According to the expression of repaid money shown in Eq (4), the total available liquidities of bank i can be expressed as",
null,
"(5)\n\nHere, ∑jyij denotes the realised payments made by all the other banks.\n\nTo sum up, Eq (4) is a rule that a bank returns the interbank borrowings when facing default, and Eq (5) measures the total available liquidity of a bank. By considering Eqs (4) and (5) together, it seems that the fix-point method can directly be adopted by using the framework of , but we do not adopt it in our model setting because the fix-point method ignores the time process of risk contagion to some extent and potentially assumes that all the banks can make the optimal decisions at the same time when the default risk appears in the system. Parallel to these studies related to DebtRank [23,24], this paper also pays attention to the dynamic process of risk contagion and provides the dynamic mechanism as displayed in the next subsection. Besides, this paper defines and validates several risk distances without needing all the banks make the optimal decisions at the same time, and therefore the method of this paper is quite different with fix-point method.\n\n### 2.2 Mechanism of risk contagion\n\nIn order to make clear the mechanism of risk contagion captured by this paper, Fig 2 provides a visual and simple representation, where the interbank system consists of three banks with different bank sizes and these banks are linked by their lending-borrowing relationship. Here, the process of risk contagion can be roughly divided into three successive phases.\n\nFig 2. Sketch map of risk contagion.\n\nhttps://doi.org/10.1371/journal.pone.0200209.g002\n\nAs Phase 1 displays, one of the banks suffers a sufficiently large negative shock so as to completely default, and thus the other banks’ loans to this bank become unrecoverable. As a result, the unrecoverable loads cause the bank with bigger size distressed and the one with the smaller size completely default in Phase 2. Subsequently, the complete default of the smaller bank causes the remaining bank’s loan to it unrecoverable so that the remaining bank can also default at this time as shown in Phase 3. Although the real interbank system always contain more banks than our simple example, the mechanism of risk contagion is similar. Note that when part default appears, Eqs (4) and (5) are useful to determine how much repaid money can be received within the mechanism of risk contagion displayed in Fig 2.\n\n### 2.3 Risk distance, risk degree and m-order risk degree\n\nThe considered interbank system consists of n banks indexed by N = {1, 2, ⋯, n}, and the lending- borrowing matrix, denoted as X, is expressed as X = [xij]i,jN, where xij (ji) denote the amount of money borrowed by bank j from bank i (we pose xii = 0). Based on the defined matrix X, the deduced system matrix G = [gij]i,jN is defined as below.",
null,
"(6)\n\nHere, gij (ij) represents the ratio of the amount of money lent from bank i to bank j to bank i’s total the total available liquidities. Then, it is not difficult to find that gij ≥ 0 (jN) and ∑jNgij = 1 so that the deduced matrix G can be understood as a Markov transfer matrix. Note that the above designed matrix does not contain yij expressed in Eq (4), but is dependent on xij, which implies that the designed matrix G does not change once the balance sheet is given. In fact, the unchanged G facilitates the calculations because we do not need to change G frequently with the change of yij (yij is endogenous variable of our model). Besides, in order to check the designed G is good or not, we test the designed G whether to satisfy much more desirable properties. In other words, if the designed G can meet much more good properties in measuring the system risk of risk contagion, it is a good design; otherwise, we should try the other forms of G. To be honest, the reported G in Eq (6) is selected from numerous possible forms and is found to meet much more desirable properties compared to the other tried forms, which will be presented in details in the latter part of this paper.\n\nLet S denote the set of initial default banks; for example, if the initial default set only contains bank j, then S = {j}, and if the initial default set contains banks i and j, then S = {i, j}. The risk distance from a non-default bank to the initial default set S is given in Definition 1.\n\n[Definition 1] (risk distance). The risk distance from each non-default bank to the initial default set S is expressed in the vector R−s:",
null,
"(7) Where 0 < β < 1 guaranteeing that the inverse matrix exists, I is an #(NS) × #(NS) identity matrix, 1 is an #(NS) vector consisting of 1, and G−s is G deleting the rows and the columns of the banks contained in the set S. In addition, the ith element of R−s is denoted as ri,S that represents the risk distance from ith bank to the default bank set S.\n\nBy recalling the defined Eq (6), the normalization is a useful procedure, whose theoretical basis is Markov chain, by noting that the normalized matrix can be understood or regarded as a Markov transfer matrix. If the status of default is regarded as the absorbing state, the defined risk distance in Eq (7) can be similarly understood as a likelihood that each remaining bank does not reach the absorbing state, according to the principle of finite-state Markov chain. Thus, a longer risk distance means a lower likelihood of default, which accords with our common sense. Interestingly, this form is quite similar to the famous Katz-Bonacich centrality in the field of network science .\n\nFurther, taking gij (ij) as an example without loss of generality, the mechanism of risk contagion explained in Section 2.2 means the following statements: when bank i face the risk of complete default, bank j should first returned the money borrowed from bank i, so gij = 0 at this time; and then even if bank i has received all the money that was lent out to the other banks, bank i can also completely default so that it appears in the initial default set S, and at this time, gij = 0 because complete default means that bank i cannot return the money borrowed from bank j (∀ji). By considering the above mechanism, we should delete the rows and the columns of the banks in the initial default set S, which interprets why we adopt the G−s to calculate the risk distance in Eq (7).\n\nAs a result, the normalization is important here for three reasons. First, the normalization guarantees that the sum of elements in each row equals 1, which accords with the definition of Markov transfer matrix, and therefore, the principle of Markov transfer matrix can be adopted directly as we have explained. Second, the normalization makes each element in the normalized matrix reflect the ratio of the corresponding liabilities, so that the bank size is potentially considered. In fact, the bank size, defined as the total amount of assets (or liabilities) of one bank, is an important factor influencing the risk contagion. To make it clear, if two banks borrow the same amount of money from another bank but the two banks have different sizes, the normalization will shows different ratios of the borrowed money in the two banks, but without the normalization, the effect of bank size cannot be reflected. Moreover, the normalization can also reflect the effect from each bank’s discount factor ηi (∀iN), which is also important in the process of risk contagion.\n\nSubsequently, let rv(S) denote the risk degree of the whole interbank system given the initial default set S. Based on the defined risk distance (see Definition 1), rv(S) is defined as follows.\n\n[Definition 2] (risk degree). Given the initial default set S, the risk degree rvs is defined as",
null,
"(8) where 1 is an #(NS) vector whose elements are all 1. In other words,1R−s equals the sum of all of the elements in R−s\n\nRecalling the explanations of Definition 1, the longer risk distance means a lower likelihood of default. Here, we first sum all the distances from the remaining banks to the given initial default set, and accordingly the mathematical expression is 1R−s. Note that a larger value of 1R−s means a lower system risk degree, which does not accord with our common sense. Thus, we use (1R−s)−1 to measure the system risk degree to avoid this problem.\n\nFurthermore, from the perspective of the system, we can compare the risk degrees relative to all possible initial default sets and determine the maximal risk degree and the corresponding default set. Considering the amount of default banks that will affect the risk degree, we keep this amount identical for fair comparison. As a result, let rvm(G) denote the m-order risk degree of the given system G, where m is the number of banks contained in the given initial default set S. For example, when m = 2, rv2(G) means the maximum sum of risk distance from each non-default bank to any possible default bank sets which contain two banks. Then, the rvm(G) is defined as below, where m = 1, 2, ⋯, n.\n\n[Definition 3] (m-order risk degree). For all initial default sets containing m banks, the maximal risk degree among all initial default sets is defined as the m-order risk degree, whose mathematical expression is",
null,
"(9) Where i1, i1, ⋯, im are the different banks contained in the initial default set and, therefore, the total number of their combination is n!/(m!(nm)!).\n\nKeeping the number of banks in initial default set identical, the defined m-order risk degree identifies not only the maximal risk degree for all possible initial default sets containing m banks but also the corresponding default set that leads to the highest system risk.\n\n## 3 Basic properties\n\nThis section aims to provide the basic properties of the proposed three measurements. In detail, these basic properties include the non-negativity, the asymmetry and the monotonicity of the defined risk distance, the monotonicity of the risk degree, and the heterogeneity of different-order risk degrees. Although these basic properties seem natural at first glance, they can not only deepen our understanding of the proposed measurements but also illustrate the rationality of these measurements.\n\n[Property B1] (Non-negativity of risk distance). For any given system G and the initial default set S, the risk distance defined in Eq (7) and the corresponding risk degree defined in Eq (8) are both non-negative.\n\nProof. Because each element of GS1T is no more than 1 and 0<β<1, it holds that",
null,
"(10)\n\nEq (10) clearly shows that all of the elements in [IβGS]−1 are non-negative since all of the elements in G−s are non-negative. Thus, Property B1 is immediately obtained.\n\n[Property B2] (Asymmetry of risk distance). The risk distance from the bank i to the initial default set {j} must not be equal to that from bank j to the initial default set {i}.\n\nProof. Here, we provide an example to show that ri,{j}rj,{i}. Given β = 0.9, the corresponding system matrix of the three banks is set as below:",
null,
"As a result, r2,{1} = 3.36 and r1,{2} = 2.59, which immediately obtains the result.\n\n[Property B3] (Monotonicity of risk distance). For any given system G, it holds that ri,{s}ri,{s,t}, where is and ti, s.\n\nProof. According to Eqs (7) and (10), we have",
null,
"(11)\n\nIf only considering the path starting from t and ending at i, we can further have",
null,
"(12) where",
null,
"means the product of all of the elements in the path starting from i and ending at t with the length of l in G−s. Then, because gij (i, jN) are all non-negative, it is not difficult to determine that ri,{s}ri,{s,t}.\n\nThe above three basic properties of the defined risk distance show that (1) the defined risk distance is non-negative, which accords with our common sense; (2) banks takes different effect on the system risk or financial contagion according to the different positions in the networks and thus have an asymmetrical risk distance; and (3) adding one more bank to the initial default set will not increase the risk distance, which is true because much greater bank defaulting in the first period often implies much greater fragility for the entire system.\n\n[Property B4] (Monotonicity of risk degree). For any given financial system G, it holds that rvm+1(G) ≥ rvm(G), where m = 1, 2, ⋯, n.\n\nProof. For any {i1, i2, ⋯, im} and {i1, i2, ⋯, im}\\{ik}(k = 1, 2, ⋯, m), it holds that",
null,
"(13a) because of the proven monotonicity of risk distance in Property B3. Then, we have",
null,
"(13b) because their maximum values inherit the inequality relation. Then, Inequality (13b) guarantees that rvm+1(G) ≥ rvm(G).\n\n[Property B5] (Heterogeneity of different-order risk degrees). For two given financial systems G1 and G2 whose node number is n1 and n2, respectively, if rv1(G1) ≤ rv1(G2), then it must not hold that rvm(G1) ≤ rvm(G2), where 2 ≤ m ≤ min{n1, n2}.\n\nProof. Similar to the proof of Property B2, we here provide an example to illustrate that rv1(G1) < rv1(G2) and rv2(G1) > rv2(G2) can exist simultaneously. Additionally, given β = 0.9, the two system matrices are shown below. According to Definition 3, we have rv1(G1) = 0.0286, rv1(G2) = 0.0342, rv2(G1) = 0.10, and rv2(G2) = 0.0950, meaning that rv1(G1) < rv1(G2) and rv2(G1) > rv2(G2) can exist simultaneously. Thus, Property B5 holds.",
null,
"Properties B4 and B5 focus on the defined risk degree and m-order risk degree, respectively. Here, Property B4 implies that adding one more bank to the initial default set S will not decrease the risk degree of the financial system, and Property B5 indicates that if one system’s 1-order risk degree is larger than another system’s 1-order risk degree, we can’t take for that the system’s higher order risk degree is still larger. In fact, two conflicting perspectives exist in the existing literature: one perspective supports that a more connected network structure enhances the system’s resilience , whereas the other suggests that dense interconnections will cause a destabilizing force to decrease the system’s risk . The illustrated Property B5 can explain the conflicting findings from the defined different order risk degree.\n\n## 4 Extended properties\n\nAlmost every central bank needs deposit reserves and limits the reserve requirement, i.e. ci / di, according to the specific law of its nation. Although the deposit reserve ratio of each bank is slightly different, we can assume a constant ratio of di to Hi since empirical analysis finds that such a ratio is almost the same for different banks within one nation . Thus, we make the following assumption.\n\n[Assumption 1] For any given system G, we assume that di = γHi for all iN, where 0 < γ < 1.\n\nBased on Assumption 1, for any given initial default set S, if the default set S leads to the complete default of bank i (iS) with the risk contagion, then the following inequality will hold:",
null,
"(14) Where ali has been defined in Eq (5), meaning that the numerator is the available asset after the risk contagion caused by the initial default set S and the denominator is the initial available asset. Furthermore, we next provide several extended properties to demonstrate the rationality of the proposed measurements; in other words, to check whether the proposed measurements can correctly reflect the system risk as the function of liability size, liability distribution, bank size and the discount factor. Additionally, the relationship between financial contagion and risk distance is also explored, which is significant and meaningful for validating the rationality of the proposed measurements.\n\n### 4.1 Relationship between financial contagion and risk distance\n\nFirst, we explore whether the defined risk distance can reflect the risk level of banks by controlling the effect of the discount factor. To this end, this subsection sets ηi = 1 (iN) to focus on the relationship between financial contagion and risk distance. Specifically, given an initial default set S, if bank i defaults with the financial contagion, then do all the banks that are shorter to the set S than bank i? The following Property E1 gives the answer.\n\n[Property E1] For any given initial default set S and two banks i and j (i, jS) in a given system G with ηi = 1 for any iN, if bank j completely defaults with the financial contagion and ri,S < rj,S for any β ∈ (0, 1), then bank i must completely default. In addition, there exists Δr, such that bank w defaults if rw,S < Δr.\n\nProof. For any given initial default set S and ηi = 1 for any iN, Eq (5) and Assumption 1 guarantee that the necessary and sufficient condition of bank j completely defaulting is",
null,
"(15) Recalling Eqs (3) and (5), Inequality (15) equals",
null,
"(16)\n\nOn the other hand, ri,S < rj,S can be further expressed as",
null,
"(17) and the Taylor expansion guarantees that",
null,
"(18) which holds for any β ∈ (0, 1). Then, let β → 0, and we immediately obtain that",
null,
"(19) which equals 1i(IGS)1T > 1j(IGS)1T > 1 − γ. Thus, the first part of Property E2 holds.\n\nAs a deduction, let Δr = max(rk,S | bank k defaults), and then the second part of Property E2 is obtained based on the result of the first part.\n\nProperty E1 provides the answer of the question is YES. Again, given an initial default set S, if bank i defaults with the financial contagion, all the banks shorter to the set S than bank i will default. Thus, the defined risk distance is demonstrated to be reasonable because it can fully reflect the insolvency contagion of banks when banks can liquate their assets freely. Note that the effect of the discount factor will be discussed in Subsection 4.5 to enrich our illustrations.\n\n### 4.2 Risk measurements as a function of liability size\n\nIn this subsection, we first check whether the increase in all pairwise liabilities raises the system risk, and the following Lemma 1 gives the answer.\n\n[Lemma 1] For any given financial system G and an initial default set S, let",
null,
"for all ij and α > 1 in the new financial system denoted as",
null,
". If bank i (iS) completely defaults with the risk contagion in G, then it must completely default in",
null,
".\n\nProof. Here, we first investigate the relationship of the available assets of bank i (iS) between G and",
null,
", and its results are as follows:",
null,
"(20) because of Eq (5) and the fact that",
null,
". Then, because α > 1 and ∑jxij > ∑jyij, it is not difficult to obtain that",
null,
"(21)\n\nThen, according to Inequality (14), for the given default set S and bank i (iS), it holds that",
null,
"(22) and Inequalities (20) and (21) guarantee that",
null,
"(23)\n\nOverall, Lemma 1 holds.\n\nThe proven Lemma 1 gives us a hint that increasing all pairwise liabilities in the network will raise the systemic risk, because the banks that completely default in G also completely default in",
null,
". Note that, here, we use the number of banks that completely default to depict the level of system risk. Then, we further check whether the defined risk measurements can correctly reflect the change in the system risk when all pairwise liabilities rise, as discussed above. As a result, by keeping all liquidities ci, securities si and other assets pi unchanged, the following Property E2 gives the answer.\n\n[Property E2] For any given financial system G, if",
null,
"for all ij and α > 1 in the new financial system denoted as",
null,
", for any given initial default set S, the defined risk distance is shorter, and the defined risk degree and the m-order risk degree are larger in the new financial system.\n\nProof. Based on the definition of G = [gij]i,jN in Eq (5), the relationship between",
null,
"and G is",
null,
"(24) and",
null,
"(25) whose elements are all more than 1. Then,",
null,
"(26)\n\nNow,",
null,
"is considered the function of α, and we further have",
null,
"(27)\n\nOn one hand, all of the elements in",
null,
"are non-negative, which has been proven in Property B1. On the other hand,",
null,
"(28) and therefore, Eq (27) is equivalent to",
null,
"(29)\n\nNote that",
null,
"and because all of the elements in",
null,
"are less than 0. Thus, all of the elements in",
null,
"are decreasing functions of α, meaning that the defined risk distance obtains its maximal value when α → 1 and obtains its minimum value when α → +∞. Furthermore, because the inequality of risk degrees holds for any initial default set S, the inequality is naturally inherited by the defined m-order risk degree. Thus, Property E2 holds, and we also obtain the boundary of the defined measurements when the liability size changes.\n\nOverall, Property E2 reflects that the rise in pairwise liabilities will increase the systemic risk, regardless of the type of network structure. Hence, controlling the level of total credits in the system is a method of managing systemic risk. From another perspective, the proposed measurements can properly reflect the increase in systemic risk, which indicates that the definition of the proposed measurement is rational.\n\n### 4.3 Risk measurements as a function of liability distribution\n\nIntuitively, a bank with positive net interbank lending is more likely to be harmed by the risk contagion because it will get loss from its counterparties’ default. Accordingly, this subsection focuses on the effect of liability distributions and explores how to arrange the liability distribution among banks to obtain a smaller system risk degree. To this end, consider any given system G, and suppose that each bank’s liability, on one hand, is kept unchanged, (that is, keeping ∑ijxij unchanged for any iN to avoid the effect from the amount of liability that has been proven influential), and, on the other hand, each bank’s total lending amount is kept equal to its borrowing amount (that is, ∑ixij = ∑jxij to ignore the effect of the difference between the loan and borrowing amounts). This subsection focuses on the effect of liability distributions and explores how to arrange the liability distribution among banks to obtain a smaller system risk degree. The following Property E3 gives the answer.\n\n[Property E3] Given any system G with the precondition that ∑ixij = ∑jxij for any i, jN, let",
null,
"for any i, jN in the newly generated system",
null,
"so the operation does not change each bank’s liability. Then, if the same default set S is given for the two systems, we have",
null,
"and the equation holds if and only if xij = xji holds in the given system G.\n\nProof. Here, system",
null,
"changes the liability distribution of the given system G and does not change each bank’s liability, noting that the precondition guarantees the following equation:",
null,
"(30)\n\nThus, the above operation, which changes the liability distribution, does not change each bank’s liability. In addition, the above operation does not change ci + ηi ⋅ (pi + si) for any iN, meaning that",
null,
". Moreover, we have",
null,
"(31)\n\nAccordingly, it holds that",
null,
".\n\nKnowing rvS(G) = (1RS(G))−1, we next focus on the item 1RS(G) whose detailed form is 1 ⋅ (IβGS)−11T / n. Accordingly, the problem of proving",
null,
"is changed into the problem of proving",
null,
", where",
null,
". The following part is to prove the above inequality. First, we have",
null,
"(32)\n\nThen, because 0.5([IβGS]−1 + [IβGS]−T) and",
null,
"are both symmetric matrices, the theorem of Rayleigh-Ritz guarantees that",
null,
"(33)\n\nSince |λ ([IβGS]−1[IβGS]T)| = |λ([IβGS]−T[IβGS])| = 1 and ([IβGS]−1[IβGS]T)−1 = ([IβGS]−T[IβGS], let exp(iw) and exp(−iw) be the eigenvalue of [IβGS]−1[IβGS]T and [IβGS]−T[IβGS], respectively, where w ∈ [0, 2π). As a result, it holds that",
null,
"(34) and if and only if w = 0, the equality holds.\n\nTo summarize, if",
null,
"(or",
null,
"),",
null,
", and if and only if",
null,
"(or",
null,
"),",
null,
". Property E3 holds.\n\nAs Property E2 has proven that liability size is an influencing factor of systemic risk, we here fix the liability size and keep each bank’s total loan and borrowing amounts unchanged in order to precisely study the influence of liability distribution on system risk. Although various liability distributions could exist in real interbank system, Property E3 uncovers that pairwise liabilities symmetrically distributed among the banks, without changing the liability size, can lead to a smaller system risk degree irrespective of the variety of liability distributions.\n\n### 4.4 Risk measurements as a function of bank size\n\nHere we define the bank size as the amount of total liquidity a bank holds. Considering the bank size as the main factor and keeping the lending-borrowing matrix X unchanged, when bank i does not belong to the initial default set S and its size becomes larger, this subsection further checks whether the defined risk degree can reflect the change of bank size. The following Property E4 provides the result.\n\n[Property E4] For any given system G, a new system",
null,
"is generated by keeping the lending-borrowing matrix X unchanged and the size of bank i becomes larger, that is,",
null,
"and",
null,
". Then, for any given default set S that does not contain bank i, it holds that",
null,
"and",
null,
".\n\nProof. Based on the operation described here, the relationship between",
null,
"and GS is",
null,
"(35)\n\nThen, we have",
null,
"and, further,",
null,
", where Ti is a #(NS) × #(NS) diagonal matrix whose ith element is",
null,
"and the remaining elements are all 1. Then,",
null,
"can be expressed as a function of Ti as follows, where 1i is an #(NS) vector whose ith element is 1 and the remaining elements are 0. Then, we further have",
null,
"(36)\n\nHere, (IβGS) + β(TiI) ⋅ (IGS) = (IβGS)[I + β(IβGS)−1(TiI) ⋅ (IGS)], and note that",
null,
"(37)\n\nThus, 1i ⋅ [(IβGS) + β(TiI) ⋅ (IGS)]−11T > 1i ⋅ (IβGS)−11T. That is,",
null,
". In addition, by replacing",
null,
"with 1T in Eqs (36) and (37), we can immediately obtain that",
null,
"(38) which equals",
null,
".\n\nProperty E4 demonstrates the intuition that a larger bank size decreases the systemic risk, which means larger size is a good buffer for insolvency contagion, and indicates that the proposed measurements can well reflect the effect of bank size on system risk.\n\n### 4.5 Risk measurements as a function of the discount factor\n\nThis subsection focuses on how each bank’s discount factor ηi affects the defined risk distance and risk degree. Intuitively, an increase in ηi means a stronger ability to liquidate assets; thus, systemic risk should decrease. Here, for any given default set S, we check whether the risk distance from bank i to S increases and the corresponding risk degree rvS decreases. The following Property E5 provides the answer.\n\n[Property E5] For any given default set S, if there exists a bank set Ξ such that Ξ ∩ S = Φ and",
null,
"for any i ∈ Ξ in the new system",
null,
", then the risk distance ri,S is longer in",
null,
"and the risk degree",
null,
"is smaller compared to the original G, where i ∈ Ξ.\n\nProof. The relationship between",
null,
"and G is",
null,
", and further,",
null,
", where Qi is a #(NS) × #(NS)diagonal matrix whose ith element is",
null,
"for i ∈ Ξ and the remaining elements are all 1. Then, this problem has the same structure with Property E4, and therefore, according to the conclusion provided in Property E4, Property E5 holds.\n\nProperty E5 validates the intuition that a rise in the discount factor of some banks will increase the risk distance and decrease the level of system risk. More importantly, the proposed measurements make it possible to reflect the effect of the discount factor, which implies that the effective liquidity management of every bank will benefit the safety of a financial system.\n\n## 5 Conclusions, discussions and future work\n\nThis paper belongs to the growing literature that focuses on designing appropriate measurements to evaluate systemic risk in financial networks . Specifically, we aim to measure two kinds of risks: one is the susceptibility of each bank to the distress of an initial default set from the microscopic angle, and the other is the the degree of the systemic failures due to contagion of counterparty risk from the macroscopic angle. To this end, this paper proposed a series of computationally tractable measurements that are risk distance, risk degree and m-order risk degree, among which the first one captures the first kind of risk and the latter two capture the second kind. Furthermore, this paper also uncovers their basic and extended properties. Regarding the basic properties, we show that the risk distance satisfies non-negativity, asymmetry and monotonicity, and that the risk degree and the m-order risk degree follows monotonicity and heterogeneity. In particular, heterogeneity implies that our measurements support neither “too central to fail” nor “too diverse to fail” , owing that different orders reflect different dimensions. On the other hand, the proven extended properties show that our measurements are able to reflect the effect of bank size, liability size, liability distribution and the discount factor on the default risk of one bank and also on the failures of the entire system. Moreover, the rationality of our measurements is embodied not only in the proven basic and extended properties but also in the relationship between the risk distance and financial contagion. From the perspective of methodology, the proposed risk distance is a node-level microscopic indicator that reflects the default likelihood of each remaining bank when an initial default set is given., while the proposed risk degree and m-order risk degree is a system-level macroscopic indicator that reflects the collapsing force of the given default sets. Both of them inherit the basic framework of the classical Katz-Bonacich centrality and establish a Markov transfer matrix with absorbing states.\n\nBased on the proposed measurements, this paper also provides some implications for guiding how to decrease or prevent the systemic risk of interbank systems: (1) since liability size influences systemic risk, it is an effective method for controlling the level of total credits within normal levels, such as deleveraging; (2) a symmetric liability distribution between pairwise banks will create a safer system under the precondition that the borrowing amount of each bank is equal to its loan amount; (3) the targeted liquidity injection is useful because banks with large bank size or liquidities are good buffers in the way of insolvency contagion; (4) enhancing the ability of banks to liquidate their assets or raise their capital adequacy ratios will decrease systemic risk.\n\nTwo issues are further discussed here: one is related to the problem of missing information and the other is related to varieties of measures of controlling the system risk. With regards to the first issue, one precondition of this paper is to know all the information about each bank’s balance sheet. However, granular data on financial networks is often lacking and the limitedness of the information available always exists in real practice. Thus, how to achieve the missing information and how to make decisions based on partial information also become two potential problem, although they are not deeply discussed in our work. Fortunately, and provided some feasible approaches to cope with the problem. Based on their work, once the missing information is estimated, our approach and the main results can also be adopted to analyze the risk contagion of interbank system. With regards to the second issue, apart from the above suggested measures of controlling the system risk, many others are also potentially useful in the context of network-based interbank system. For example, uncovered how the topological features of network structures influence the risk contagion and suggested avoiding the measures such as market integration and diversification to decrease system stability, and studied how types of debt contracts affected the system risk and suggested a more suitable type of contract that guaranteed for a unique Pareto efficient clearing payment vector. Accordingly, the mentioned two papers enrich the measures to reduce the system risk, which guides our future work to extend our proven properties.\n\nIn our opinion, it is meaningful to explore the relationship between the network structure and the proposed m-order system value because doing so may provide specific guidance on how to design a robust network structure under certain conditions. To this end, we suggest plotting all the different-order risk degrees on one line graph for finding out a sudden change on risk degrees in the process of adding banks into the given initial default set, which is a visual way to identify the influence of financial contagion. Moreover, note that the fix-point method captures the system risk under another mechanism of risk contagion different with ours , and therefore it will be interesting to check whether the fix-point method also shares the similar properties with ours. In addition, since this paper focuses on only interbank interactions with a balance-sheet mechanism, it would be interesting to study a more general framework that extends the interbank system to a generalized financial system that includes more types of financial institutions and interaction.\n\n## Acknowledgments\n\nThis work was supported by the National Natural Science Foundation of China under Grant 71501034 and 71771041, by China Postdoctoral Science Foundation under Grant 2016M590230 and 2017T100183, and by Italian Ministry of Education Progetti di Rilevante Interesse Nazionale (PRIN) Grant 2015592CTH.\n\n## References\n\n1. 1. Eisenberg L, Noe TH. Systemic risk in financial systems. Management Science. 2001; 47(2): 236–249.\n2. 2. Elliott M, Golub B, Jackson MO. Financial networks and contagion. American Economic Review. 2014; 104(10): 3115–3153.\n3. 3. Haldane AG, May RM. Systemic risk in banking ecosystems. Nature. 2011; 469(7330): 351–355. pmid:21248842\n4. 4. Chen N, Liu X, Yao DD. An optimization view of financial systemic risk modeling: Network effect and market liquidity effect. Operations Research. 2016; 64(5): 1089–1108.\n5. 5. Bekiros S, Nguyen DK, Junior LS, Uddin GS. Information diffusion, cluster formation and entropy-based network dynamics in equity and commodity markets. European Journal of Operational Research. 2017; 256(3): 945–961.\n6. 6. Summer M. Financial Contagion and Network Analysis. Annual Review of Financial Economics. 2013; 5: 277–97.\n7. 7. Sachs A. Completeness, Interconnectedness and Distribution of Interbank Exposures—A Parameterized Analysis of the Stability of Financial Networks. Quantitative Finance. 2014; 14(9): 1677–92.\n8. 8. Boss M, Helmut E, Martin S, Stefan T. Network Topology of the Interbank Market. Quantitative Finance. 2004; 4(6): 677–84.\n9. 9. Glasserman P, Young HP. Contagion in financial networks. Journal of Economic Literature. 2016; 54(3): 779–831.\n10. 10. Acemoglu D, Ozdaglar A, Tahbaz-Salehi A. Systemic risk and stability in financial networks. American Economic Review. 2015; 105(2): 564–608.\n11. 11. Bonacich P. Power and centrality: A family of measures. American Journal of Sociology. 1987; 92(5): 1170–1182.\n12. 12. Siebenbrunner C. Clearing algorithms and network centrality. SSRN: 2959680. 2017.\n13. 13. Ballester C, Calvó-Armengol A, Zenou Y. Who’s who in networks. Wanted: The key player. Econometrica. 2006; 74(5): 1403–1417.\n14. 14. Li Y, Wu C, Wang X, Luo P. A network-based and multi-parameter model for finding influential authors. Journal of Informetrics. 2014; 8(3): 791–799.\n15. 15. Cohen-Cole E, Patacchini E, Zenou Y. Static and dynamic networks in interbank markets. Network Science. 2015; 3(01): 98–123.\n16. 16. Soramäki K, Cook S. SinkRank: An algorithm for identifying systemically important banks in payment systems. Economics: The Open-Access, Open-Assessment E-Journal. 2013; 7(2013–28): 1–27.\n17. 17. Georg CP. The effect of the interbank network structure on contagion and common shocks. Journal of Banking & Finance. 2013; 37(7): 2216–2228.\n18. 18. Van Lelyveld I. Finding the core: Network structure in interbank markets. Journal of Banking & Finance. 2014; 49: 27–40.\n19. 19. Glasserman P, Young HP. How likely is contagion in financial networks?. Journal of Banking & Finance. 2015; 50: 383–399.\n20. 20. Basole RC, Bellamy MA. Supply network structure, visibility, and risk diffusion: a computational approach. Decision Sciences. 2014; 45(4): 753–789.\n21. 21. Lux T. Emergence of a core-periphery structure in a simple dynamic model of the interbank market. Journal of Economic Dynamics and Control. 2015; 52: A11–A23.\n22. 22. Chen C, Iyengar G, Moallemi CC. An axiomatic approach to systemic risk. Management Science. 2013; 59(6): 1373–1388.\n23. 23. Bardoscia M, Battiston S, Caccioli F, Caldarelli G. DebtRank: A microscopic foundation for shock propagation. PloS one. 2015; 10(6): e0130406. pmid:26091013\n24. 24. Bardoscia M, Battiston S, Caccioli F, Caldarelli G. Pathways towards instability in financial networks. Nature Communications. 2016; 8: 14416.\n25. 25. Allen F, Gale D. Financial contagion. Journal of Political Economy. 2000; 108(1): 1–33.\n26. 26. Blume L, Easley D, Kleinberg J, Kleinberg R, Tardos É. Network formation in the presence of contagious risk. ACM Transactions on Economics and Computation. 2013; 1(2): 6.\n27. 27. Allen F, Carletti E, Marquez R. Deposits and bank capital structure. Journal of Financial Economics. 2015; 118(3): 601–619.\n28. 28. Bae KH, Karolyi GA, Stulz RM. A new approach to measuring financial contagion. Review of Financial studies. 2003; 16(3): 717–763.\n29. 29. Elsinger H, Lehar A, Summer M. Risk assessment for banking systems. Management science. 2006; 52(9): 1301–1314.\n30. 30. Billio M, Getmansky M, Lo AW, Pelizzon L. Econometric measures of connectedness and systemic risk in the finance and insurance sectors. Journal of Financial Economics. 2012; 104(3): 535–559.\n31. 31. Battiston S, Puliga M, Kaushik R, Tasca P, Caldarelli G. Debtrank: too central to fail? financial networks, the fed and systemic risk. Scientific Reports. 2012; 2(8): 541.\n32. 32. Caccioli F, Shrestha M, Moore C, Farmer JD. Stability analysis of financial contagion due to overlapping portfolios. Journal of Banking & Finance. 2014; 46(3): 233–245.\n33. 33. Anand K, Lelyveld IV, Banai Ádám, Friedrich S, Garratt R, Hałaj G, et al. The missing links: a global study on uncovering financial network structures from partial data. Journal of Financial Stability. 2018; 35: 107–119.\n34. 34. Cimini G, Squartini T, Garlaschelli D, Gabrielli A. Systemic risk analysis on reconstructed economic and financial networks. Scientific Reports. 2015; 5: 15758. pmid:26507849\n35. 35. Bardoscia M, Caccioli F, Perotti JI, Vivaldo G, Caldarelli G. Distress propagation in complex networks: the case of non-linear DebtRank. PloS one. 2016; 11(10): e0163825. pmid:27701457\n36. 36. Schuldenzucker S, Seuken S, Battiston S. Clearing Payments in Financial Networks with Credit Default Swaps. ACM Conference on Economics and Computation. 2016: 759–759."
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https://old.liu.se/mai/foutb/kurser/mai0137-applied-statistical-methods-part-1?l=en | [
"# MAI0137 Applied Statistical Methods - Part I (Tillämpade statistiska metoder – del I)\n\nNumber of credits: 5 hp\n\nExaminer: Martin Singull\n\nCourse literature: Two books: 1) “Introduction to the Theory of Statistical Inference” by Hannelore Liero and Silvelyn Zwanzig, and 2) “An Introduction to Generalized Linear Models” by Annette J. Dobson.\n\nAim: The course is intended to give basic knowledge of the theory and methods of statistical inference, i.e. how to use observed data to draw conclusions about phenomena influenced by random factors. By the end of the course, the student should be able to:\n\n• use an appropriate probability model to describe and analyse observed data and draw conclusions concerning interesting parameters;\n• understand the principles of statistical inference;\n• explore the nature of the relationships between two or several variables by using different kinds of linear models and discuss the adequacy of the models;\n• use nonparametric methods to analyse data of different types and discuss the applicability of the methods;\n• find probability models and statistical methods in applications from engineering, economy and science and evaluate the results;\n• use suitable software (e.g., R, Matlab) for certain types of statistical analyses.\n\nCourse contents:\n\n• Point estimation, properties of estimators, different methods, e.g., maximum likelihood, method of moments and least squares\n• Confidence intervals and tests of hypotheses for one or several samples\n• The multivariate normal distribution\n• General linear models\n• Multiple linear regression\n• Single-factor (including random effects model), two-factor and multifactor experiments in theory and practice\n• Analysis of variance table (ANOVA)\n• Generalized linear models; Logistic regression, Poisson regression\n• Variable selection and model building\n• Nonparametric methods; sign test, Wilcoxon's tests, Kruskal-Wallis test, Friedmann's test\n• Introduction to Bayesian statistical inference\n• Analysis of data by using appropriate statistical software, e.g., R, Matlab.\n\nSome of these topics may be omitted, and/or other topics included, depending on the students need and other circumstances.\n\nOrganisation: Lectures and Computer example classes.\n\nExamination: Hand-in assignments\n\nPrerequisites: Knowledge of probability, calculus and algebra is assumed and some familiarity with matrix algebra.\n\nCourse web page\n\nPage manager: karin.johansson@liu.se\nLast updated: 2017-04-03"
] | [
null
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https://www.colorhexa.com/0149fd | [
"# #0149fd Color Information\n\nIn a RGB color space, hex #0149fd is composed of 0.4% red, 28.6% green and 99.2% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 71.1% magenta, 0% yellow and 0.8% black. It has a hue angle of 222.9 degrees, a saturation of 99.2% and a lightness of 49.8%. #0149fd color hex could be obtained by blending #0292ff with #0000fb. Closest websafe color is: #0033ff.\n\n• R 0\n• G 29\n• B 99\nRGB color chart\n• C 100\n• M 71\n• Y 0\n• K 1\nCMYK color chart\n\n#0149fd color description : Vivid blue.\n\n# #0149fd Color Conversion\n\nThe hexadecimal color #0149fd has RGB values of R:1, G:73, B:253 and CMYK values of C:1, M:0.71, Y:0, K:0.01. Its decimal value is 84477.\n\nHex triplet RGB Decimal 0149fd `#0149fd` 1, 73, 253 `rgb(1,73,253)` 0.4, 28.6, 99.2 `rgb(0.4%,28.6%,99.2%)` 100, 71, 0, 1 222.9°, 99.2, 49.8 `hsl(222.9,99.2%,49.8%)` 222.9°, 99.6, 99.2 0033ff `#0033ff`\nCIE-LAB 40.995, 52.324, -92.272 20.121, 11.862, 94.152 0.16, 0.094, 11.862 40.995, 106.075, 299.556 40.995, -16.17, -131.19 34.441, 44.012, -137.975 00000001, 01001001, 11111101\n\n# Color Schemes with #0149fd\n\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #fdb501\n``#fdb501` `rgb(253,181,1)``\nComplementary Color\n• #01c7fd\n``#01c7fd` `rgb(1,199,253)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #3701fd\n``#3701fd` `rgb(55,1,253)``\nAnalogous Color\n• #c7fd01\n``#c7fd01` `rgb(199,253,1)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #fd3701\n``#fd3701` `rgb(253,55,1)``\nSplit Complementary Color\n• #49fd01\n``#49fd01` `rgb(73,253,1)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #fd0149\n``#fd0149` `rgb(253,1,73)``\n• #01fdb5\n``#01fdb5` `rgb(1,253,181)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #fd0149\n``#fd0149` `rgb(253,1,73)``\n• #fdb501\n``#fdb501` `rgb(253,181,1)``\n• #0133b1\n``#0133b1` `rgb(1,51,177)``\n• #013aca\n``#013aca` `rgb(1,58,202)``\n• #0142e4\n``#0142e4` `rgb(1,66,228)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #195bfe\n``#195bfe` `rgb(25,91,254)``\n• #336dfe\n``#336dfe` `rgb(51,109,254)``\n• #4c7ffe\n``#4c7ffe` `rgb(76,127,254)``\nMonochromatic Color\n\n# Alternatives to #0149fd\n\nBelow, you can see some colors close to #0149fd. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0188fd\n``#0188fd` `rgb(1,136,253)``\n• #0173fd\n``#0173fd` `rgb(1,115,253)``\n• #015efd\n``#015efd` `rgb(1,94,253)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #0134fd\n``#0134fd` `rgb(1,52,253)``\n• #011ffd\n``#011ffd` `rgb(1,31,253)``\n• #010afd\n``#010afd` `rgb(1,10,253)``\nSimilar Colors\n\n# #0149fd Preview\n\nThis text has a font color of #0149fd.\n\n``<span style=\"color:#0149fd;\">Text here</span>``\n#0149fd background color\n\nThis paragraph has a background color of #0149fd.\n\n``<p style=\"background-color:#0149fd;\">Content here</p>``\n#0149fd border color\n\nThis element has a border color of #0149fd.\n\n``<div style=\"border:1px solid #0149fd;\">Content here</div>``\nCSS codes\n``.text {color:#0149fd;}``\n``.background {background-color:#0149fd;}``\n``.border {border:1px solid #0149fd;}``\n\n# Shades and Tints of #0149fd\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, #000513 is the darkest color, while #fefeff is the lightest one.\n\n• #000513\n``#000513` `rgb(0,5,19)``\n• #000b26\n``#000b26` `rgb(0,11,38)``\n• #00113a\n``#00113a` `rgb(0,17,58)``\n• #00164d\n``#00164d` `rgb(0,22,77)``\n• #001c61\n``#001c61` `rgb(0,28,97)``\n• #002274\n``#002274` `rgb(0,34,116)``\n• #012788\n``#012788` `rgb(1,39,136)``\n• #012d9b\n``#012d9b` `rgb(1,45,155)``\n• #0132af\n``#0132af` `rgb(1,50,175)``\n• #0138c2\n``#0138c2` `rgb(1,56,194)``\n• #013ed6\n``#013ed6` `rgb(1,62,214)``\n• #0143e9\n``#0143e9` `rgb(1,67,233)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\n• #1457fe\n``#1457fe` `rgb(20,87,254)``\n• #2765fe\n``#2765fe` `rgb(39,101,254)``\n• #3b73fe\n``#3b73fe` `rgb(59,115,254)``\n• #4e80fe\n``#4e80fe` `rgb(78,128,254)``\n• #628efe\n``#628efe` `rgb(98,142,254)``\n• #759cfe\n``#759cfe` `rgb(117,156,254)``\n• #89aaff\n``#89aaff` `rgb(137,170,255)``\n• #9cb8ff\n``#9cb8ff` `rgb(156,184,255)``\n• #b0c6ff\n``#b0c6ff` `rgb(176,198,255)``\n• #c3d4ff\n``#c3d4ff` `rgb(195,212,255)``\n• #d7e2ff\n``#d7e2ff` `rgb(215,226,255)``\n• #eaf0ff\n``#eaf0ff` `rgb(234,240,255)``\n• #fefeff\n``#fefeff` `rgb(254,254,255)``\nTint Color Variation\n\n# Tones of #0149fd\n\nA tone is produced by adding gray to any pure hue. In this case, #767b88 is the less saturated color, while #0149fd is the most saturated one.\n\n• #767b88\n``#767b88` `rgb(118,123,136)``\n• #6c7792\n``#6c7792` `rgb(108,119,146)``\n• #63739b\n``#63739b` `rgb(99,115,155)``\n• #596fa5\n``#596fa5` `rgb(89,111,165)``\n• #4f6aaf\n``#4f6aaf` `rgb(79,106,175)``\n• #4566b9\n``#4566b9` `rgb(69,102,185)``\n• #3c62c2\n``#3c62c2` `rgb(60,98,194)``\n• #325ecc\n``#325ecc` `rgb(50,94,204)``\n``#285ad6` `rgb(40,90,214)``\n• #1e56e0\n``#1e56e0` `rgb(30,86,224)``\n• #1551e9\n``#1551e9` `rgb(21,81,233)``\n• #0b4df3\n``#0b4df3` `rgb(11,77,243)``\n• #0149fd\n``#0149fd` `rgb(1,73,253)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0149fd 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"
] | [
null
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https://www.mlpack.org/doc/mlpack-3.3.2/doxygen/anntutorial.html | [
"Neural Network tutorial\n\n# Introduction\n\nThere is vast literature on neural networks and their uses, as well as strategies for choosing initial points effectively, keeping the algorithm from converging in local minima, choosing the best model structure, choosing the best optimizers, and so forth. mlpack implements many of these building blocks, making it very easy to create different neural networks in a modular way.\n\nmlpack currently implements two easy-to-use forms of neural networks: Feed-Forward Networks (this includes convolutional neural networks) and Recurrent Neural Networks.\n\nThis tutorial is split into the following sections:\n\n# Model API\n\nThere are two main neural network classes that are meant to be used as container for neural network layers that mlpack implements; each class is suited to a different setting:\n\n• `FFN:` the Feed Forward Network model provides a means to plug layers together in a feed-forward fully connected manner. This is the 'standard' type of deep learning model, and includes convolutional neural networks (CNNs).\n• `RNN:` the Recurrent Neural Network model provides a means to consider successive calls to forward as different time-steps in a sequence. This is often used for time sequence modeling tasks, such as predicting the next character in a sequence.\n\nBelow is some basic guidance on what should be used. Note that the question of \"which algorithm should be used\" is a very difficult question to answer, so the guidance below is just that—guidance—and may not be right for a particular problem.\n\n• Feed-forward Networks allow signals or inputs to travel one way only. There is no feedback within the network; for instance, the output of any layer does only affect the upcoming layer. That makes Feed-Forward Networks straightforward and very effective. They are extensively used in pattern recognition and are ideally suitable for modeling relationships between a set of input and one or more output variables.\n• Recurrent Networks allow signals or inputs to travel in both directions by introducing loops in the network. Computations derived from earlier inputs are fed back into the network, which gives the recurrent network some kind of memory. RNNs are currently being used for all kinds of sequential tasks; for instance, time series prediction, sequence labeling, and sequence classification.\n\nIn order to facilitate consistent implementations, the `FFN` and `RNN` classes have a number of methods in common:\n\n• `Train()`: trains the initialized model on the given input data. Optionally an optimizer object can be passed to control the optimization process.\n• `Predict()`: predicts the responses to a given set of predictors. Note the responses will reflect the output of the specified output layer.\n• `Add()`: this method can be used to add a layer to the model.\nNote\nTo be able to optimize the network, both classes implement the OptimizerFunction API. In short, the `FNN` and `RNN` class implement two methods: `Evaluate()` and `Gradient()`. This enables the optimization given some learner and some performance measure.\n\nSimilar to the existing layer infrastructure, the `FFN` and `RNN` classes are very extensible, having the following template arguments; which can be modified to change the behavior of the network:\n\n• `OutputLayerType:` this type defines the output layer used to evaluate the network; by default, `NegativeLogLikelihood` is used.\n• `InitializationRuleType:` this type defines the method by which initial parameters are set; by default, `RandomInitialization` is used.\ntemplate<\ntypename OutputLayerType = NegativeLogLikelihood<>,\ntypename InitializationRuleType = RandomInitialization\n>\nclass FNN;\n\nInternally, the `FFN` and `RNN` class keeps an instantiated `OutputLayerType` class (which can be given in the constructor). This is useful for using different loss functions like the Negative-Log-Likelihood function or the `VRClassReward` function, which takes an optional score parameter. Therefore, you can write a non-static OutputLayerType class and use it seamlessly in combination with the `FNN` and `RNN` class. The same applies to the `InitializationRuleType` template parameter.\n\nBy choosing different components for each of these template classes in conjunction with the `Add()` method, a very arbitrary network object can be constructed.\n\nBelow are several examples of how the `FNN` and `RNN` classes might be used. The first examples focus on the `FNN` class, and the last shows how the `RNN` class can be used.\n\nThe simplest way to use the FNN<> class is to pass in a dataset with the corresponding labels, and receive the classification in return. Note that the dataset must be column-major – that is, one column corresponds to one point. See the matrices guide for more information.\n\nThe code below builds a simple feed-forward network with the default options, then queries for the assignments for every point in the `queries` matrix.",
null,
"Note\nThe number of inputs in the above graph doesn't match with the real number of features in the thyroid dataset and are just used as an abstract representation.\n#include <mlpack/core.hpp>\nusing namespace mlpack;\nusing namespace mlpack::ann;\nint main()\n{\n// Load the training set and testing set.\narma::mat trainData;\narma::mat testData;\n// Split the labels from the training set and testing set respectively.\narma::mat trainLabels = trainData.row(trainData.n_rows - 1);\narma::mat testLabels = testData.row(testData.n_rows - 1);\ntrainData.shed_row(trainData.n_rows - 1);\ntestData.shed_row(testData.n_rows - 1);\n// Initialize the network.\nFFN<> model;\n// Train the model.\nmodel.Train(trainData, trainLabels);\n// Use the Predict method to get the predictions.\narma::mat predictionTemp;\nmodel.Predict(testData, predictionTemp);\n/*\nSince the predictionsTemp is of dimensions (3 x number_of_data_points)\nwith continuous values, we first need to reduce it to a dimension of\n(1 x number_of_data_points) with scalar values, to be able to compare with\ntestLabels.\nThe first step towards doing this is to create a matrix of zeros with the\ndesired dimensions (1 x number_of_data_points).\nIn predictionsTemp, the 3 dimensions for each data point correspond to the\nprobabilities of belonging to the three possible classes.\n*/\narma::mat prediction = arma::zeros<arma::mat>(1, predictionTemp.n_cols);\n// Find index of max prediction for each data point and store in \"prediction\"\nfor (size_t i = 0; i < predictionTemp.n_cols; ++i)\n{\n// we add 1 to the max index, so that it matches the actual test labels.\nprediction(i) = arma::as_scalar(arma::find(\narma::max(predictionTemp.col(i)) == predictionTemp.col(i), 1)) + 1;\n}\n/*\nCompute the error between predictions and testLabels,\nnow that we have the desired predictions.\n*/\nsize_t correct = arma::accu(prediction == testLabels);\ndouble classificationError = 1 - double(correct) / testData.n_cols;\n// Print out the classification error for the testing dataset.\nstd::cout << \"Classification Error for the Test set: \" << classificationError << std::endl;\nreturn 0;\n}\n\nNow, the matrix prediction holds the classification of each point in the dataset. Subsequently, we find the classification error by comparing it with testLabels.\n\nIn the next example, we create simple noisy sine sequences, which are trained later on, using the RNN class in the `RNNModel()` method.\n\nvoid GenerateNoisySines(arma::mat& data,\narma::mat& labels,\nconst size_t points,\nconst size_t sequences,\nconst double noise = 0.3)\n{\narma::colvec x = arma::linspace<arma::Col<double>>(0,\npoints - 1, points) / points * 20.0;\narma::colvec y1 = arma::sin(x + arma::as_scalar(arma::randu(1)) * 3.0);\narma::colvec y2 = arma::sin(x / 2.0 + arma::as_scalar(arma::randu(1)) * 3.0);\ndata = arma::zeros(points, sequences * 2);\nlabels = arma::zeros(2, sequences * 2);\nfor (size_t seq = 0; seq < sequences; seq++)\n{\ndata.col(seq) = arma::randu(points) * noise + y1 +\narma::as_scalar(arma::randu(1) - 0.5) * noise;\nlabels(0, seq) = 1;\ndata.col(sequences + seq) = arma::randu(points) * noise + y2 +\narma::as_scalar(arma::randu(1) - 0.5) * noise;\nlabels(1, sequences + seq) = 1;\n}\n}\nvoid RNNModel()\n{\nconst size_t rho = 10;\n// Generate 12 (2 * 6) noisy sines. A single sine contains rho\n// points/features.\narma::mat input, labelsTemp;\nGenerateNoisySines(input, labelsTemp, rho, 6);\narma::mat labels = arma::zeros<arma::mat>(rho, labelsTemp.n_cols);\nfor (size_t i = 0; i < labelsTemp.n_cols; ++i)\n{\nconst int value = arma::as_scalar(arma::find(\narma::max(labelsTemp.col(i)) == labelsTemp.col(i), 1)) + 1;\nlabels.col(i).fill(value);\n}\nLinear<> lookup(1, 4);\nSigmoidLayer<> sigmoidLayer;\nLinear<> linear(4, 4);\nRecurrent<> recurrent(add, lookup, linear, sigmoidLayer, rho);\nRNN<> model(rho);\nStandardSGD opt(0.1, 1, input.n_cols /* 1 epoch */, -100);\nmodel.Train(input, labels, opt);\n}\n\nFor further examples on the usage of the ann classes, see mlpack models.\n\n# Layer API\n\nIn order to facilitate consistent implementations, we have defined a LayerType API that describes all the methods that a `layer` may implement. mlpack offers a few variations of this API, each designed to cover some of the model characteristics mentioned in the previous section. Any `layer` requires the implementation of a `Forward()` method. The interface looks like:\n\ntemplate<typename eT>\nvoid Forward(const arma::Mat<eT>& input, arma::Mat<eT>& output);\n\nThe method should calculate the output of the layer given the input matrix and store the result in the given output matrix. Next, any `layer` must implement the Backward() method, which uses certain computations obtained during the forward pass and should calculate the function f(x) by propagating x backward through f:\n\ntemplate<typename eT>\nvoid Backward(const arma::Mat<eT>& input,\nconst arma::Mat<eT>& gy,\narma::Mat<eT>& g);\n\nFinally, if the layer is differentiable, the layer must also implement a Gradient() method:\n\ntemplate<typename eT>\nconst arma::Mat<eT>& error,\n\nThe Gradient function should calculate the gradient with respect to the input activations `input` and calculated errors `error` and place the results into the gradient matrix object `gradient` that is passed as an argument.\n\nNote\nNote that each method accepts a template parameter InputType, OutputType or GradientType, which may be arma::mat (dense Armadillo matrix) or arma::sp_mat (sparse Armadillo matrix). This allows support for both sparse-supporting and non-sparse-supporting `layer` without explicitly passing the type.\n\nIn addition, each layer must implement the Parameters(), InputParameter(), OutputParameter(), Delta() methods, differentiable layer should also provide access to the gradient by implementing the Gradient(), Parameters() member function. Note each function is a single line that looks like:\n\nOutputDataType const& Parameters() const { return weights; }\n\nBelow is an example that shows each function with some additional boilerplate code.\n\nNote\nNote this is not an actual layer but instead an example that exists to show and document all the functions that mlpack layer must implement. For a better overview of the various layers, see mlpack::ann. Also be aware that the implementations of each of the methods in this example are entirely fake and do not work; this example exists for its API, not its implementation.\n\nNote that layer sometimes have different properties. These properties are known at compile-time through the mlpack::ann::LayerTraits class, and some properties may imply the existence (or non-existence) of certain functions. Refer to the LayerTraits layer_traits.hpp for more documentation on that.\n\nThe two template parameters below must be template parameters to the layer, in the order given below. More template parameters are fine, but they must come after the first two.\n\n• `InputDataType:` this defines the internally used input type for example to store the parameter matrix. Note, a layer could be built on a dense matrix or a sparse matrix. All mlpack trees should be able to support any Armadillo- compatible matrix type. When the layer is written it should be assumed that MatType has the same functionality as arma::mat. Note that\n• `OutputDataType:` this defines the internally used input type for example to store the parameter matrix. Note, a layer could be built on a dense matrix or a sparse matrix. All mlpack trees should be able to support any Armadillo- compatible matrix type. When the layer is written it should be assumed that MatType has the same functionality as arma::mat.\ntemplate<typename InputDataType = arma::mat,\ntypename OutputDataType = arma::mat>\nclass ExampleLayer\n{\npublic:\nExampleLayer(const size_t inSize, const size_t outSize) :\ninputSize(inSize), outputSize(outSize)\n{\n/* Nothing to do here */\n}\n}\n\nThe constructor for `ExampleLayer` will build the layer given the input and output size. Note that, if the input or output size information isn't used internally it's not necessary to provide a specific constructor. Also, one could add additional or other information that are necessary for the layer construction. One example could be:\n\nExampleLayer(const double ratio = 0.5) : ratio(ratio) {/* Nothing to do here*/}\n\nWhen this constructor is finished, the entire layer will be built and is ready to be used. Next, as pointed out above, each layer has to follow the LayerType API, so we must implement some additional functions.\n\ntemplate<typename InputType, typename OutputType>\nvoid Forward(const InputType& input, OutputType& output)\n{\noutput = arma::ones(input.n_rows, input.n_cols);\n}\ntemplate<typename InputType, typename ErrorType, typename GradientType>\nvoid Backward(const InputType& input, const ErrorType& gy, GradientType& g)\n{\ng = arma::zeros(gy.n_rows, gy.n_cols) + gy;\n}\ntemplate<typename InputType, typename ErrorType, typename GradientType>\nErrorType& error,\n{\ngradient = arma::zeros(input.n_rows, input.n_cols) * error;\n}\n\nThe three functions `Forward()`, `Backward()` and `Gradient()` (which is needed for a differentiable layer) contain the main logic of the layer. The following functions are just to access and manipulate the different layer parameters.\n\nOutputDataType& Parameters() { return weights; }\nInputDataType& InputParameter() { return inputParameter; }\nOutputDataType& OutputParameter() { return outputParameter; }\nOutputDataType& Delta() { return delta; }\n\nSince some of this methods return internal class members we have to define them.\n\nprivate:\nsize_t inSize, outSize;\nInputDataType inputParameter;\n\nNote some members are just here so `ExampleLayer` compiles without warning. For instance, `inputSize` is not required to be a member of every type of layer.\n\nThere is one last method that is especially interesting for a layer that shares parameter. Since the layer weights are set once the complete model is defined, it's not possible to split the weights during the construction time. To solve this issue, a layer can implement the `Reset()` method which is called once the layer parameter is set.\n\n# Model Setup & Training\n\nOnce the base container is selected (`FNN` or `RNN`), the `Add` method can be used to add layers to the model. The code below adds two linear layers to the model—the first takes 512 units as input and gives 256 output units, and the second takes 256 units as input and gives 128 output units.\n\nFFN<> model;\n\nThe model is trained on Armadillo matrices. For training a model, you will typically use the `Train()` function:\n\narma::mat trainingSet, trainingLabels;\nmodel.Train(trainingSet, trainingLabels);\n\nYou can use mlpack's `Load()` function to load a dataset like this:\n\narma::mat trainingSet;\n\\$ cat dataset.csv\n0, 1, 4\n1, 0, 5\n1, 1, 1\n2, 0, 2\n\nThe type does not necessarily need to be a CSV; it can be any supported storage format, assuming that it is a coordinate-format file in the format specified above. For more information on mlpack file formats, see the documentation for mlpack::data::Load().\n\nNote\nIt’s often a good idea to normalize or standardize your data, for example using:\nfor (size_t i = 0; i < dataset.n_cols; ++i)\ndataset.col(i) /= norm(dataset.col(i), 2);\n\nAlso, it is possible to retrain a model with new parameters or with a new reference set. This is functionally equivalent to creating a new model.\n\nUsing `boost::serialization` (for more information about the internals see Serialization - Boost C++ Libraries), mlpack is able to load and save machine learning models with ease. To save a trained neural network to disk. The example below builds a model on the `thyroid` dataset and then saves the model to the file `model.xml` for later use.\n\narma::mat dataset;\n// Split the labels from the training set.\narma::mat trainData = dataset.submat(0, 0, dataset.n_rows - 4,\ndataset.n_cols - 1);\n// Split the data from the training set.\narma::mat trainLabelsTemp = dataset.submat(dataset.n_rows - 3, 0,\ndataset.n_rows - 1, dataset.n_cols - 1);\n// Initialize the network.\nFFN<> model;\n// Train the model.\nmodel.Train(trainData, trainLabels);\n// Use the Predict method to get the assignments.\narma::mat assignments;\nmodel.Predict(trainData, assignments);\ndata::Save(\"model.xml\", \"model\", model, false);\n\nAfter this, the file model.xml will be available in the current working directory.\n\nNow, we can look at the output model file, `model.xml:`\n\n\\$ cat model.xml\n<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\" ?>\n<!DOCTYPE boost_serialization>\n<boost_serialization signature=\"serialization::archive\" version=\"15\">\n<model class_id=\"0\" tracking_level=\"0\" version=\"0\">\n<parameter class_id=\"1\" tracking_level=\"1\" version=\"0\" object_id=\"_0\">\n<n_rows>66</n_rows>\n<n_cols>1</n_cols>\n<n_elem>66</n_elem>\n<vec_state>0</vec_state>\n<item>-7.55971528334903642e+00</item>\n<item>-9.95435955058058930e+00</item>\n<item>9.31133928948225353e+00</item>\n<item>-5.36784434861701953e+00</item>\n...\n</parameter>\n<width>0</width>\n<height>0</height>\n<currentInput object_id=\"_1\">\n<n_rows>0</n_rows>\n<n_cols>0</n_cols>\n<n_elem>0</n_elem>\n<vec_state>0</vec_state>\n</currentInput>\n<network class_id=\"2\" tracking_level=\"0\" version=\"0\">\n<count>3</count>\n<item_version>0</item_version>\n<item class_id=\"3\" tracking_level=\"0\" version=\"0\">\n<which>18</which>\n<value class_id=\"4\" tracking_level=\"1\" version=\"0\" object_id=\"_2\">\n<inSize>21</inSize>\n<outSize>3</outSize>\n</value>\n</item>\n<item>\n<which>2</which>\n<value class_id=\"5\" tracking_level=\"1\" version=\"0\" object_id=\"_3\"></value>\n</item>\n<item>\n<which>20</which>\n<value class_id=\"6\" tracking_level=\"1\" version=\"0\" object_id=\"_4\"></value>\n</item>\n</network>\n</model>\n</boost_serialization>\n\nAs you can see, the `<parameter>` section of `model.xml` contains the trained network weights. We can see that this section also contains the network input size, which is 66 rows and 1 column. Note that in this example, we used three different layers, as can be seen by looking at the `<network>` section. Each node has a unique id that is used to reconstruct the model when loading.\n\nThe models can also be saved as .bin or .txt; the .xml format provides a human-inspectable format (though the models tend to be quite complex and may be difficult to read). These models can then be re-used to be used for classification or other tasks.\n\nSo, instead of saving or training a network, mlpack can also load a pre-trained model. For instance, the example below will load the model from `model.xml` and then generate the class predictions for the `thyroid` test dataset.\n\narma::mat testData = dataset.submat(0, 0, dataset.n_rows - 4,\ndataset.n_cols - 1);\narma::mat predictions;\nmodel.Predict(testData, predictions);\n\nThis enables the possibility to distribute a model without having to train it first or simply to save a model for later use. Note that loading will also work on different machines.\n\n# Extracting Parameters\n\nTo access the weights from the neural network layers, you can call the following function on any initialized network:\n\nmodel.Parameters();\n\nwhich will return the complete model parameters as an armadillo matrix object; however often it is useful to not only have the parameters for the complete network, but the parameters of a specific layer. Another method, `Model()`, makes this easily possible:\n\nmodel.Model().Parameters();\n\nIn the example above, we get the weights of the second layer.\n\n# Further documentation\n\nFor further documentation on the ann classes, consult the complete API documentation."
] | [
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"https://www.mlpack.org/doc/mlpack-3.3.2/doxygen/dot_inline_dotgraph_1.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8543428,"math_prob":0.9307551,"size":10971,"snap":"2020-34-2020-40","text_gpt3_token_len":2204,"char_repetition_ratio":0.13029999,"word_repetition_ratio":0.070414536,"special_character_ratio":0.19223407,"punctuation_ratio":0.11072665,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98821145,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T04:42:20Z\",\"WARC-Record-ID\":\"<urn:uuid:96721070-86ff-4353-9181-dea0245da5f3>\",\"Content-Length\":\"48310\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d6375998-7568-49f6-a3ab-459af5fa2d74>\",\"WARC-Concurrent-To\":\"<urn:uuid:1ee56566-0503-4d14-bec5-0021e65f6f84>\",\"WARC-IP-Address\":\"216.24.161.178\",\"WARC-Target-URI\":\"https://www.mlpack.org/doc/mlpack-3.3.2/doxygen/anntutorial.html\",\"WARC-Payload-Digest\":\"sha1:J4JHCKVLANN5CG5QO2VUQMLLAPQLTH3D\",\"WARC-Block-Digest\":\"sha1:C2XWX7QVERIOVSLJQ5NDQ7DDKOUL5XBA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735909.19_warc_CC-MAIN-20200805035535-20200805065535-00028.warc.gz\"}"} |
https://answers.everydaycalculation.com/multiply-fractions/2-4-times-63-36 | [
"Solutions by everydaycalculation.com\n\n## Multiply 2/4 with 63/36\n\n1st number: 2/4, 2nd number: 1 27/36\n\nThis multiplication involving fractions can also be rephrased as \"What is 2/4 of 1 27/36?\"\n\n2/4 × 63/36 is 7/8.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 2/4 × 63/36 = 2 × 63/4 × 36 = 126/144\n3. After reducing the fraction, the answer is 7/8\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8823212,"math_prob":0.9838518,"size":426,"snap":"2021-31-2021-39","text_gpt3_token_len":164,"char_repetition_ratio":0.14454976,"word_repetition_ratio":0.0,"special_character_ratio":0.42723006,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9781682,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-25T07:17:56Z\",\"WARC-Record-ID\":\"<urn:uuid:f1402bb6-c110-4bb2-abbc-209486cfbf57>\",\"Content-Length\":\"7080\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c268d0a8-7860-4070-af2c-ef0291f092f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:2d0a3352-b8cb-4483-a83e-622204efd097>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/2-4-times-63-36\",\"WARC-Payload-Digest\":\"sha1:PUOTCUL6ARLZHZMB552LQDVMJSZSZUR3\",\"WARC-Block-Digest\":\"sha1:677GSNCPD7M7LXY7GSSK2YRRXDTQTTWY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046151638.93_warc_CC-MAIN-20210725045638-20210725075638-00158.warc.gz\"}"} |
https://docs.microsoft.com/en-us/dotnet/api/system.windows.forms.numericupdown.accelerations?view=netframework-3.5 | [
"# NumericUpDown.Accelerations Property\n\n## Definition\n\nGets a collection of sorted acceleration objects for the NumericUpDown control.\n\n``````public:\nproperty System::Windows::Forms::NumericUpDownAccelerationCollection ^ Accelerations { System::Windows::Forms::NumericUpDownAccelerationCollection ^ get(); };``````\n``````[System.ComponentModel.Browsable(false)]\npublic System.Windows.Forms.NumericUpDownAccelerationCollection Accelerations { get; }``````\n``member this.Accelerations : System.Windows.Forms.NumericUpDownAccelerationCollection``\n``Public ReadOnly Property Accelerations As NumericUpDownAccelerationCollection``\n\n#### Property Value\n\nA NumericUpDownAccelerationCollection containing the sorted acceleration objects for the NumericUpDown control\n\nAttributes\n\n## Examples\n\nThe following code example demonstrates how to use the Accelerations property. To run this example, paste the following code into a form and call the `InitializeAcceleratedUpDown` method from the form's constructor or Load event-handling method. While the code is running, press and hold the up or down arrow to see the acceleration occur.\n\n`````` private NumericUpDown numericUpDown1;\nprivate void InitializeAcceleratedUpDown()\n{\nnumericUpDown1 = new NumericUpDown();\nnumericUpDown1.Location = new Point(40, 40);\nnumericUpDown1.Maximum = 40000;\nnumericUpDown1.Minimum = -40000;\n\n// Add some accelerations to the control.\n\n}\n``````\n``````Private numericUpDown1 As NumericUpDown\n\nPrivate Sub InitializeAcceleratedUpDown()\nnumericUpDown1 = New NumericUpDown()\nnumericUpDown1.Location = New Point(40, 40)\nnumericUpDown1.Maximum = 40000\nnumericUpDown1.Minimum = - 40000\n\n' Add some accelerations to the control."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5715366,"math_prob":0.5774887,"size":2679,"snap":"2019-43-2019-47","text_gpt3_token_len":531,"char_repetition_ratio":0.324486,"word_repetition_ratio":0.029850746,"special_character_ratio":0.18402389,"punctuation_ratio":0.20052083,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97692466,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T09:58:37Z\",\"WARC-Record-ID\":\"<urn:uuid:48c8a713-28a9-4382-9e1c-f246508cd8af>\",\"Content-Length\":\"37287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eca92aa7-fc3a-42f0-ad3f-8232a9fa8bd5>\",\"WARC-Concurrent-To\":\"<urn:uuid:4cb2835e-39ca-4192-ab01-69e28106d686>\",\"WARC-IP-Address\":\"104.108.100.37\",\"WARC-Target-URI\":\"https://docs.microsoft.com/en-us/dotnet/api/system.windows.forms.numericupdown.accelerations?view=netframework-3.5\",\"WARC-Payload-Digest\":\"sha1:2ESIB7BAWXFQONX2W4T2YYWBBVQNN5PP\",\"WARC-Block-Digest\":\"sha1:3PKJZFGXXZJH2ADD6K4T3XEXVWJPJS3E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667177.24_warc_CC-MAIN-20191113090217-20191113114217-00255.warc.gz\"}"} |
https://www.convertunits.com/from/grav/to/hectometer/square+second | [
"## ››Convert grav to hectometre/square second\n\n grav hectometer/square second\n\nHow many grav in 1 hectometer/square second? The answer is 10.197162129779.\nWe assume you are converting between grav and hectometre/square second.\nYou can view more details on each measurement unit:\ngrav or hectometer/square second\nThe SI derived unit for acceleration is the meter/square second.\n1 meter/square second is equal to 0.10197162129779 grav, or 0.01 hectometer/square second.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between grav and hectometers/square second.\nType in your own numbers in the form to convert the units!\n\n## ››Quick conversion chart of grav to hectometer/square second\n\n1 grav to hectometer/square second = 0.09807 hectometer/square second\n\n10 grav to hectometer/square second = 0.98067 hectometer/square second\n\n20 grav to hectometer/square second = 1.96133 hectometer/square second\n\n30 grav to hectometer/square second = 2.942 hectometer/square second\n\n40 grav to hectometer/square second = 3.92266 hectometer/square second\n\n50 grav to hectometer/square second = 4.90333 hectometer/square second\n\n100 grav to hectometer/square second = 9.80665 hectometer/square second\n\n200 grav to hectometer/square second = 19.6133 hectometer/square second\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from hectometer/square second to grav, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Metric conversions and more\n\nConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3\", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7700268,"math_prob":0.9919216,"size":1989,"snap":"2022-27-2022-33","text_gpt3_token_len":503,"char_repetition_ratio":0.37027708,"word_repetition_ratio":0.027681662,"special_character_ratio":0.24585219,"punctuation_ratio":0.1292876,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9846552,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-27T05:08:21Z\",\"WARC-Record-ID\":\"<urn:uuid:f5eec01b-2877-46e4-ad1d-0f03e8abe6ee>\",\"Content-Length\":\"40777\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:418b5972-e934-485e-82aa-32bb6f1bb955>\",\"WARC-Concurrent-To\":\"<urn:uuid:c29cbe83-c6a7-4258-9a97-5e6386f84f97>\",\"WARC-IP-Address\":\"34.236.206.159\",\"WARC-Target-URI\":\"https://www.convertunits.com/from/grav/to/hectometer/square+second\",\"WARC-Payload-Digest\":\"sha1:J7QAMJT4N6RYZUFWKGUHHOOIOIRHM7AE\",\"WARC-Block-Digest\":\"sha1:U62JYUMZ4ZXBYXP4Q6H23RHMQDZU2VJH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103328647.18_warc_CC-MAIN-20220627043200-20220627073200-00332.warc.gz\"}"} |
https://dodgerocksgasmonkey.com/college-186 | [
"# Distance time acceleration calculator\n\nHere, we will be discussing about Distance time acceleration calculator.",
null,
"Do mathematic problems",
null,
"",
null,
"## Acceleration Calculator\n\nOur acceleration calculator is a tool that helps you to find out how fast the speed of an object is changing. It works in three different ways, based on: Difference between velocities at two distinct points in time. Distance traveled\n\n## Acceleration to Distance Calculator\n\nCalculator Use. This Displacement Calculator finds the distance traveled or displacement (s) of an object using its initial velocity (u), acceleration (a), and time (t) traveled. The equation used is s = ut + ½at 2; it is manipulated below to",
null,
"You have questions and we have answers!",
null,
"Get calculation assistance online\n\nNo need to be a math genius, our online calculator can do the work for you.",
null,
"Top Specialists\n\nWe have the best specialists in the business.\n\n## Calculate Distance at Acceleration\n\nWhat Is Acceleration? “The rate of changing velocity with respect to time is called acceleration” How to Calculate Acceleration? The following example will help you in calculating\n\nTimely Delivery\n\nTimely delivery is important to us.\n\nYou can improve your scholarly performance by following some simple tips.\n\nGet help from expert tutors\n\nIf you need help with your homework, our expert writers are here to assist you.\n\nDo mathematic problems\n\nI can't do math equations.\n\nmath is the study of numbers, shapes, and patterns. It is used in everyday life, from counting to measuring to more complex calculations.\n\nLearning math can be tough, but there are ways to make it easier.\n\n## Acceleration Calculator, Time, Speed, Velocity\n\nAcceleration calculator. This function calculates acceleration as a function of distance. The acceleration or change in speed over a certain distance is calculated. To perform the\n\nFigure out math\n\nMath is a subject that can be difficult for some students to grasp. However, with a little practice and perseverance, anyone can learn to love math!\n\nExplain mathematic problem\n\nMath is the study of numbers, shapes, and patterns.\n\nSolve word questions too\n\nIn addition to solving math problems, students should also be able to answer word questions.\n\nClear up mathematic\n\nAlthough math may seem daunting at first, with a little practice it can be easy to clear up any confusion and get better at solving problems."
] | [
null,
"https://dodgerocksgasmonkey.com/images/b1b8ab1e83584cf3/pmcodhbeqangfijkl-pic-boy.webp",
null,
"data:image/svg+xml,%3csvg%20xmlns=%27http://www.w3.org/2000/svg%27%20version=%271.1%27%20width=%27500%27%20height=%27400%27/%3e",
null,
"https://dodgerocksgasmonkey.com/images/b1b8ab1e83584cf3/maklnqoebfhidcjgp-snap.png",
null,
"https://dodgerocksgasmonkey.com/images/b1b8ab1e83584cf3/hpbfjdnqkleogmaic-best-writers.svg",
null,
"https://dodgerocksgasmonkey.com/images/b1b8ab1e83584cf3/hpbfjdnqkleogmaic-top-quality.svg",
null,
"https://dodgerocksgasmonkey.com/images/b1b8ab1e83584cf3/hpbfjdnqkleogmaic-always-on-time.svg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94216603,"math_prob":0.96820354,"size":2744,"snap":"2022-40-2023-06","text_gpt3_token_len":568,"char_repetition_ratio":0.10875913,"word_repetition_ratio":0.11304348,"special_character_ratio":0.19752187,"punctuation_ratio":0.11494253,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99716604,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,null,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T13:22:08Z\",\"WARC-Record-ID\":\"<urn:uuid:dc2d8004-0d2b-442b-bd2f-b73ba37b03c0>\",\"Content-Length\":\"22009\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:139f10de-534a-4171-974a-8f9bd95035a2>\",\"WARC-Concurrent-To\":\"<urn:uuid:00b9fef8-38cf-4949-b398-70e8f03be1d1>\",\"WARC-IP-Address\":\"170.178.164.167\",\"WARC-Target-URI\":\"https://dodgerocksgasmonkey.com/college-186\",\"WARC-Payload-Digest\":\"sha1:QGP7WMPYU6DJXSEUO6DH3PCMIMCUMNT7\",\"WARC-Block-Digest\":\"sha1:WE73MFWWV6UDYKQ7MJRQAT3XCKTNFZKJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499713.50_warc_CC-MAIN-20230129112153-20230129142153-00146.warc.gz\"}"} |
https://feet-to-meters.appspot.com/723-feet-to-meters.html | [
"Feet To Meters\n\n# 723 ft to m723 Foot to Meters\n\nft\n=\nm\n\n## How to convert 723 foot to meters?\n\n 723 ft * 0.3048 m = 220.3704 m 1 ft\nA common question is How many foot in 723 meter? And the answer is 2372.04724409 ft in 723 m. Likewise the question how many meter in 723 foot has the answer of 220.3704 m in 723 ft.\n\n## How much are 723 feet in meters?\n\n723 feet equal 220.3704 meters (723ft = 220.3704m). Converting 723 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 723 ft to m.\n\n## Convert 723 ft to common lengths\n\nUnitUnit of length\nNanometer2.203704e+11 nm\nMicrometer220370400.0 µm\nMillimeter220370.4 mm\nCentimeter22037.04 cm\nInch8676.0 in\nFoot723.0 ft\nYard241.0 yd\nMeter220.3704 m\nKilometer0.2203704 km\nMile0.1369318182 mi\nNautical mile0.1189904968 nmi\n\n## What is 723 feet in m?\n\nTo convert 723 ft to m multiply the length in feet by 0.3048. The 723 ft in m formula is [m] = 723 * 0.3048. Thus, for 723 feet in meter we get 220.3704 m.\n\n## 723 Foot Conversion Table",
null,
"## Alternative spelling\n\n723 ft in Meters, 723 ft to Meter, 723 Feet in Meter, 723 ft to m, 723 Feet in m, 723 Foot to Meters, 723 Foot in Meters, 723 Foot to Meter, 723 Foot in Meter,"
] | [
null,
"https://feet-to-meters.appspot.com/image/723.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8163735,"math_prob":0.8696158,"size":643,"snap":"2023-14-2023-23","text_gpt3_token_len":204,"char_repetition_ratio":0.20031299,"word_repetition_ratio":0.0,"special_character_ratio":0.41835147,"punctuation_ratio":0.15189873,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98306113,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-09T08:39:35Z\",\"WARC-Record-ID\":\"<urn:uuid:082e2fe7-605e-4d0d-be1f-47b16e2c1628>\",\"Content-Length\":\"27787\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0ff59d8-a9a8-44e6-8e1a-9afede853d3e>\",\"WARC-Concurrent-To\":\"<urn:uuid:9bef2f93-4286-45a4-8992-2d0a59a03a74>\",\"WARC-IP-Address\":\"142.251.167.153\",\"WARC-Target-URI\":\"https://feet-to-meters.appspot.com/723-feet-to-meters.html\",\"WARC-Payload-Digest\":\"sha1:U4B3JDDVNVAVR6MJL5WBRFEQOEG6U7FN\",\"WARC-Block-Digest\":\"sha1:DREVR6QVGXWR7FKZCDE4VY2RAZV2D74L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655446.86_warc_CC-MAIN-20230609064417-20230609094417-00781.warc.gz\"}"} |
https://www.colorhexa.com/0085f5 | [
"# #0085f5 Color Information\n\nIn a RGB color space, hex #0085f5 is composed of 0% red, 52.2% green and 96.1% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 45.7% magenta, 0% yellow and 3.9% black. It has a hue angle of 207.4 degrees, a saturation of 100% and a lightness of 48%. #0085f5 color hex could be obtained by blending #00ffff with #000beb. Closest websafe color is: #0099ff.\n\n• R 0\n• G 52\n• B 96\nRGB color chart\n• C 100\n• M 46\n• Y 0\n• K 4\nCMYK color chart\n\n#0085f5 color description : Pure (or mostly pure) blue.\n\n# #0085f5 Color Conversion\n\nThe hexadecimal color #0085f5 has RGB values of R:0, G:133, B:245 and CMYK values of C:1, M:0.46, Y:0, K:0.04. Its decimal value is 34293.\n\nHex triplet RGB Decimal 0085f5 `#0085f5` 0, 133, 245 `rgb(0,133,245)` 0, 52.2, 96.1 `rgb(0%,52.2%,96.1%)` 100, 46, 0, 4 207.4°, 100, 48 `hsl(207.4,100%,48%)` 207.4°, 100, 96.1 0099ff `#0099ff`\nCIE-LAB 55.447, 11.823, -64.221 24.865, 23.365, 89.581 0.18, 0.17, 23.365 55.447, 65.3, 280.431 55.447, -31.296, -102.244 48.338, 7.23, -76.041 00000000, 10000101, 11110101\n\n# Color Schemes with #0085f5\n\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #f57000\n``#f57000` `rgb(245,112,0)``\nComplementary Color\n• #00f5eb\n``#00f5eb` `rgb(0,245,235)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #000bf5\n``#000bf5` `rgb(0,11,245)``\nAnalogous Color\n• #f5eb00\n``#f5eb00` `rgb(245,235,0)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #f5000b\n``#f5000b` `rgb(245,0,11)``\nSplit Complementary Color\n• #85f500\n``#85f500` `rgb(133,245,0)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #f50085\n``#f50085` `rgb(245,0,133)``\n• #00f570\n``#00f570` `rgb(0,245,112)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #f50085\n``#f50085` `rgb(245,0,133)``\n• #f57000\n``#f57000` `rgb(245,112,0)``\n• #005ba9\n``#005ba9` `rgb(0,91,169)``\n• #0069c2\n``#0069c2` `rgb(0,105,194)``\n• #0077dc\n``#0077dc` `rgb(0,119,220)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #1092ff\n``#1092ff` `rgb(16,146,255)``\n• #299dff\n``#299dff` `rgb(41,157,255)``\n• #43a9ff\n``#43a9ff` `rgb(67,169,255)``\nMonochromatic Color\n\n# Alternatives to #0085f5\n\nBelow, you can see some colors close to #0085f5. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00c2f5\n``#00c2f5` `rgb(0,194,245)``\n• #00aef5\n``#00aef5` `rgb(0,174,245)``\n• #0099f5\n``#0099f5` `rgb(0,153,245)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #0071f5\n``#0071f5` `rgb(0,113,245)``\n• #005cf5\n``#005cf5` `rgb(0,92,245)``\n• #0048f5\n``#0048f5` `rgb(0,72,245)``\nSimilar Colors\n\n# #0085f5 Preview\n\nThis text has a font color of #0085f5.\n\n``<span style=\"color:#0085f5;\">Text here</span>``\n#0085f5 background color\n\nThis paragraph has a background color of #0085f5.\n\n``<p style=\"background-color:#0085f5;\">Content here</p>``\n#0085f5 border color\n\nThis element has a border color of #0085f5.\n\n``<div style=\"border:1px solid #0085f5;\">Content here</div>``\nCSS codes\n``.text {color:#0085f5;}``\n``.background {background-color:#0085f5;}``\n``.border {border:1px solid #0085f5;}``\n\n# Shades and Tints of #0085f5\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, #00050a is the darkest color, while #f5faff is the lightest one.\n\n• #00050a\n``#00050a` `rgb(0,5,10)``\n• #00101d\n``#00101d` `rgb(0,16,29)``\n• #001b31\n``#001b31` `rgb(0,27,49)``\n• #002544\n``#002544` `rgb(0,37,68)``\n• #003058\n``#003058` `rgb(0,48,88)``\n• #003a6c\n``#003a6c` `rgb(0,58,108)``\n• #00457f\n``#00457f` `rgb(0,69,127)``\n• #005093\n``#005093` `rgb(0,80,147)``\n• #005aa7\n``#005aa7` `rgb(0,90,167)``\n• #0065ba\n``#0065ba` `rgb(0,101,186)``\n• #0070ce\n``#0070ce` `rgb(0,112,206)``\n• #007ae1\n``#007ae1` `rgb(0,122,225)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\n• #0a8fff\n``#0a8fff` `rgb(10,143,255)``\n• #1d98ff\n``#1d98ff` `rgb(29,152,255)``\n• #31a1ff\n``#31a1ff` `rgb(49,161,255)``\n• #44aaff\n``#44aaff` `rgb(68,170,255)``\n• #58b3ff\n``#58b3ff` `rgb(88,179,255)``\n• #6cbcff\n``#6cbcff` `rgb(108,188,255)``\n• #7fc5ff\n``#7fc5ff` `rgb(127,197,255)``\n• #93ceff\n``#93ceff` `rgb(147,206,255)``\n• #a7d7ff\n``#a7d7ff` `rgb(167,215,255)``\n• #bae0ff\n``#bae0ff` `rgb(186,224,255)``\n• #cee8ff\n``#cee8ff` `rgb(206,232,255)``\n• #e1f1ff\n``#e1f1ff` `rgb(225,241,255)``\n• #f5faff\n``#f5faff` `rgb(245,250,255)``\nTint Color Variation\n\n# Tones of #0085f5\n\nA tone is produced by adding gray to any pure hue. In this case, #717b84 is the less saturated color, while #0085f5 is the most saturated one.\n\n• #717b84\n``#717b84` `rgb(113,123,132)``\n• #687c8d\n``#687c8d` `rgb(104,124,141)``\n• #5e7d97\n``#5e7d97` `rgb(94,125,151)``\n• #557ea0\n``#557ea0` `rgb(85,126,160)``\n• #4b7faa\n``#4b7faa` `rgb(75,127,170)``\n• #427fb3\n``#427fb3` `rgb(66,127,179)``\n• #3980bc\n``#3980bc` `rgb(57,128,188)``\n• #2f81c6\n``#2f81c6` `rgb(47,129,198)``\n• #2682cf\n``#2682cf` `rgb(38,130,207)``\n• #1c83d9\n``#1c83d9` `rgb(28,131,217)``\n• #1383e2\n``#1383e2` `rgb(19,131,226)``\n• #0984ec\n``#0984ec` `rgb(9,132,236)``\n• #0085f5\n``#0085f5` `rgb(0,133,245)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0085f5 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"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5192081,"math_prob":0.8644438,"size":3684,"snap":"2022-27-2022-33","text_gpt3_token_len":1667,"char_repetition_ratio":0.14619565,"word_repetition_ratio":0.0073664826,"special_character_ratio":0.55456024,"punctuation_ratio":0.23015873,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9858842,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-13T12:58:33Z\",\"WARC-Record-ID\":\"<urn:uuid:e38ed173-71ec-45cb-bf2c-a09d96c3fe7d>\",\"Content-Length\":\"36143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:373b7f3b-6a44-4e7e-b4c9-76e7a04679e4>\",\"WARC-Concurrent-To\":\"<urn:uuid:d44a78ad-d22c-40ee-ab71-e50a793aae9d>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0085f5\",\"WARC-Payload-Digest\":\"sha1:QKYRVSSLPCTUQ4HWZ2JJXDLCI2OKHEQ5\",\"WARC-Block-Digest\":\"sha1:NPN7RXRRXJVMGNH6YDMDPZ4ZOD46RV76\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571950.76_warc_CC-MAIN-20220813111851-20220813141851-00134.warc.gz\"}"} |
https://discuss.codechef.com/t/queens-necklace-editorial-rdc1/87302 | [
"# Queen's Necklace Editorial RDC1\n\nAuthor - Yash Dwivedi\nTester - Shubham Kushwaha\nEditorialist - Yash Dwivedi\n\nEasy\n\nC++ Stl\n\n# Problem :\n\nGiven an array of different types of pearls you need to print Yes if all the types of pearls are present in a unique quantity else print No.\n\n# Quick Explanation\n\nCount the frequency of each a[i] in a map and then insert all those count in a map then if size of map and set are same print Yes else print No\n\n# Explanation\n\nThe logic for this Question was simple: you needed to count the number of occurrences for all types of pearls and then put them in a map as well after that check if both have the same size then it would mean that all the pearls are present in a unique quantity.\nCount the frequency of each a[i] in a map and then insert all those count in a map then if size of map and set are same print Yes else print No.\n\n# Solution\n\nSetter's Solution\n\n#include<bits/stdc++.h>\nusing namespace std;\nint main()\n{\nios::sync_with_stdio(NULL);\ncin.tie(NULL);\ncout.tie(NULL)\nint t, n;\ncin >> t;\nwhile (t–)\n{\ncin >> n;\nint a[n];\nmap<int, int> mp;\nset < int> s;\nfor(int i=0; i<n; i++)\n{\ncin >> a[i];\nmp[a[i]]++;\n}\nfor (auto i : mp)\n{\ns.insert(i.second);\n}\nif (s.size() == m.size())\ncout<<“Yes”;\nelse\ncout<<“No”;\ncout<<endl;\n}\nreturn 0;\n}\n\n2 Likes"
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https://www.wiki.eigenvector.com/index.php?title=Faq_where_to_find_regression_vector_and_regression_intercept_for_a_PLS_or_PCR_model | [
"# Faq where to find regression vector and regression intercept for a PLS or PCR model\n\n### Issue:\n\nWhere do I find the regression vector (B) and regression intercept (B0) for a PLS or PCR model?\n\n### Possible Solutions:\n\nNOTE: If you are trying to implement a model in an on-line prediction environment, please see our Model_Exporter and Solo_Predictor products or contact us regarding use of on-line prediction engines.\n\nRegression models store the regression vector information in units of the preprocessed data. In order to use this, you will either need to use additional information stored in the model to preprocess new data first (before using the regression vector) or use one of a couple of functions to convert the regression vector into one that can be applied to raw data.\n\nFor some simple preprocessing situations, you may be able to convert a regression model over to a simple slope (a) and intercept (b) form:\n\n```y = ax+b\n```\n\nusing the Analysis menu item: \"File > Export Model > To Regression Vector\" which uses the PLS_Toolbox regcon function:\n\n``` >> regcon\nREGCON Converts regression model to y = ax + b form.\n'regcon help' gives extended help\nI/O: [a,b] = regcon(mod);\n```\n\nHowever, this function will only operate on models with simple autoscaling or mean centering.\n\nMore advanced preprocessing methods require either the Model_Exporter add-on product, or custom programming to convert into a simple equation format.\n\nManually Extracting and Applying\n\nTo manually extract the regression vector (B) and the intercept (B0), these items would need to be extracted from the model object containing the regression model. The following describes these fields and the preprocessing steps needed to be performed.\n\nThe regression vector is stored in the `model.reg` field of the model:\n\n```>> b = model.reg\n```\n\nB0 is more subtle. In PLS_Toolbox, the regression vector and intercept are stored in terms of preprocessed data and the intercept is effectively zero.\n\nPreprocessing information in general can be found in:\n\n```>> model.detail.preprocessing{1}\n```\n\nfor the x-block and\n\n```>> model.detail.preprocessing{2}\n```\n\nfor the y-block. If you are using ONLY mean centering or autoscaling (i.e. no other preprocessing steps), the y-block mean is stored in:\n\n```>> model.detail.preprocessing{2}(1).out{1}\n```\n\nThis may look like a terribly complicated command, but basically, it is saying that from preprocessing for the y-block \"{2}\" pull the first step \"(1)\" and return the first stored value \".out{1}\". This will be the mean from your y-block (or means for multi-column y). If you are using autoscaling, there will also be a vector of standard deviations:\n\n```>> model.detail.preprocessing{2}(1).out{2}\n```\n\nHere are several simple situations and the equations you can use this information in:\n\n• If you have mean-centered both X and Y prior to building the model (or applying it to new data), you can calculate a y prediction (`y_hat`) using:\n```>> b = model.reg;\n>> y_hat = x*b\n```\n\nThis returns `y_hat` in preprocessed y-block units.\n\n• If your model used ONLY mean-centering on ONLY the y-block (the model contained NO preprocessing on the X-block), the following will make a prediction:\n```>> b = model.reg;\n>> y_mn = model.detail.preprocessing{2}.out{1};\n>> y_hat = x*b + y_mn;\n```\n\nThis returns `y_hat` in original y-block units.\n\n• If your model used ONLY autoscaling on ONLY the y-block (the model contained NO preprocessing of the X-block), this will make a prediction:\n```>> b = model.reg;\n>> y_mn = model.detail.preprocessing{2}.out{1};\n>> y_sc = model.detail.preprocessing{2}.out{2};\n>> y_hat = (x*b)*diag(y_sc) + y_mn;\n```\n\nThis also returns `y_hat` in original y-block units.\n\nFor other preprocessing methods, you will need to use the `preprocess` function to apply the preprocessing."
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https://forum.allaboutcircuits.com/threads/computing-the-performance-improvement-with-parallelization.120174/ | [
"# Computing the performance improvement with parallelization\n\n#### HunterDX77M\n\nJoined Sep 28, 2011\n104\nSo I was doing a textbook problem for homework, and after a few parts, I got an answer that didn't really make practical sense.\n\nProblem:\nYour company's internal studies show that a single core system is sufficient for the demand on your processing power; however, you are exploring whether you could save power by using two cores.\n(a) Assume your application is 80% parallelizable. By how much could you decrease the frequency and get the same performance?\n\nApplying the following formula:\n$$Speedup_{overall} = (1 - Fraction_{enhanced} + \\frac{Fraction_{enhanced}}{Speedup_{enhanced}})^{-1}$$\nThe fraction enhanced would be 0.8 since 80% can be parallelized.\nThe Speedup enhanced would be 2 since parallelizing on two cores would double the speed of that portion.\n$$Speedup_{overall} = (1 - 0.8+ \\frac{0.8}{2})^{-1}$$\n\nSo the overall speed up would be 5/3. Therefore the frequency can be multiplied by the reciprocal, 3/5 (or 0.6) to have the same performance as the single core processor. Does that make sense?\n\n(b) Assume that voltage may be decreased linearly with the frequency. How much dynamic power would the dual core system require compared to the single core system?\n\nPower is proportional to\n$$C \\times V^2 \\times f$$\n\nC is capacitive load, V is the operating voltage and f is the frequency. Since frequency is multiplied by 0.6, so is voltage, since it tracks linearly, giving a cubic reduction in power:\n$$2 \\times C \\times (0.6 \\times V)2 \\times (0.6f) = 0.432 \\times C \\times V^2 \\times f$$\n\nI am multiplying by 2 at the beginning there since I figure there would be two cores using power. So this comes out to using less than half the power of the original single core system. Now I am suspicious something is wrong.\n\n(c) Now assume the voltage may not decrease below 25% of the original voltage. This is referred to as the voltage floor and any voltage lower than that will lose the state. What percentage of parallelization gives you a voltage at the voltage floor.\n\n1/4 of the voltage would mean 1/4 of the frequency. If the frequency can by cut down to just 1/4 of the original frequency, then the speed up must have been by 4 times.\n\n$$Speedup_{overall} = (1 - Fraction_{enhanced} + \\frac{Fraction_{enhanced}}{Speedup_{enhanced}})^{-1}$$\n$$Speedup_{overall} = 4 = (1 - x+ \\frac{x}{2})^{-1}$$\n$$0.25 = 1 - x+ \\frac{x}{2} = 1 - \\frac{x}{2}$$\n\nThis comes out to x = 1.5, where x is the fraction enhanced. That would mean 150% of the application would have to be parallelizable! That just doesn't add up (pun intended, sorry)! Is there a mistake in my logic here? Was it just an arithmetic error?\n\n#### sailorjoe\n\nJoined Jun 4, 2013\n363\nI think your answer to part a, Amdahl's Law, is good, and so is part b. It looks bad, but remember that the voltage is squared, so the power goes down exponentially.\nIn part c, I think your interpretation of the question may be off. If you reduced the voltage to .25, did you get a 4x speedup? Did you get that without adding cores? Explore that further.\n\nLast edited:"
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https://englishlangkan.com/question/two-automobiles-left-simultaneously-from-cities-a-and-b-heading-towards-each-other-and-met-in-5-12789623-52/ | [
"## Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobile that left\n\nQuestion\n\nTwo automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobile that left city A was 10 km/hour less than the speed of the other automobile. If the first automobile had left city A 4 1/2 hours earlier than the other automobile left city B, then the two would have met 150 km away from B. Find the distance between A and B.\n\nin progress 0\n2 weeks 2021-10-13T04:06:27+00:00 2 Answers 0\n\n450km\n\nStep-by-step explanation:\n\nTake it that each automobile travels at 30 km an hour, for 150 km, meaning it will be 450 km apart.\n\n450 km\n\nStep-by-step explanation:\n\nEquations\n\nWe can define 3 variables: a, b, d. Let “a” and “b” represent the speeds of the cars leaving cities A and B, respectively. Let “d” represent the distance between the two cities. We can write three equations in these three variables:\n\n1. The relation between “a” and “b”:\n\na = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B\n\n2. The relation between speed and distance when the cars leave at the same time:\n\nd = (a +b)·5 . . . . . . distance = speed × time\n\n3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is …\n\ndistance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance\n\n4.5a + (150/b)(a +b) = d\n\n__\n\nSolution\n\nSubstituting for d, we have …\n\n4.5a + 150/b(a +b) = 5(a +b)\n\n4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses\n\n5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side\n\nNow, we can substitute for “a” and solve for b.\n\n5b^2 + 0.5b(b-10) -150(b -10 +b) = 0\n\n5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses\n\n11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2\n\n(11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor\n\nThe solutions to this equation are …\n\nb = 60/11 = 5 5/11 . . . and . . . b = 50\n\nSince b must be greater than 10, the first solution is extraneous, and the values of the variables are …\n\n• b = 50\n• a = b-10 = 40\n• d = 5(a+b) = 5(90) = 450\n\nThe distance between A and B is 450 km.\n\n_____\n\nCheck\n\nWhen the cars leave at the same time, their speed of closure is the sum of their speeds. They will cover 450 km in …\n\n(450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h\n\n__\n\nWhen car A leaves 4 1/2 hours early, it covers a distance of …\n\n(4.5 h)(40 km/h) = 180 km\n\nbefore car B leaves. The distance remaining to be covered is …\n\n450 km – 180 km = 270 km\n\nWhen car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in …\n\n(270 km)/(90 km/h) = 3 h\n\nIn that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required."
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https://gcc.gnu.org/ml/gcc-patches/2013-09/msg01026.html | [
"This is the mail archive of the gcc-patches@gcc.gnu.org mailing list for the GCC project.\n\nIndex Nav: Message Nav: [Date Index] [Subject Index] [Author Index] [Thread Index] [Date Prev] [Date Next] [Thread Prev] [Thread Next] [Raw text]\n\n# [PATCH] Handle loops with control flow in loop-distribution\n\n• From: Richard Biener <rguenther at suse dot de>\n• To: gcc-patches at gcc dot gnu dot org\n• Date: Fri, 13 Sep 2013 14:23:55 +0200 (CEST)\n• Subject: [PATCH] Handle loops with control flow in loop-distribution\n• Authentication-results: sourceware.org; auth=none\n\n```The following patch makes loop-distribution able to distribute loops\nthat have control flow. It does so by adding control dependence\nedges to the RDG (removing the need for special-casing the loop\nexit condition and its dependencies). With this we can now\ndistribute the loop of the testcase\n\nfor (i = 0; i < 1024; ++i)\n{\na[i] = 0;\nif (i > 100)\nb[i] = i;\n}\n\nand generate a memset for zeroing a. The patch does not in any\nway add or adjust a cost model, so it remains necessary that\nthere is a write in each partition (a cost model could also be\nthat if we were able to factor out a control dependence from\nat least one partition then that is worthwhile on its own).\n\nIn theory this should expose a greater number of loops to\nvectorization (if you'd enable -ftree-loop-distribution, that is).\n\nBootstrapped (with -O2 -g -ftree-loop-distribution) on\nx86_64-unknown-linux-gnu, testing in progress.\n\nI'll leave it for comments until Monday (if anyone cares).\n\nThe next restriction to go is that of us only distributing innermost\nloops.\n\nThanks,\nRichard.\n\n2013-09-13 Richard Biener <rguenther@suse.de>\n\n* tree-loop-distribution.c (enum rdg_dep_type): Add control_dd.\n(dot_rdg_1): Handle control_dd.\n(create_edge_for_control_dependence): New function.\n(build_rdg): Likewise.\n(generate_loops_for_partition): If there are not necessary\ncontrol stmts remove all their dependencies.\n(collect_condition_stmts, rdg_flag_loop_exits): Remove.\n(distribute_loop): Pass on control dependences.\n(tree_loop_distribution): Compute control dependences and remove\nrestriction on number of loop nodes.\n\n* gcc.dg/tree-ssa/ldist-22.c: New testcase.\n\nIndex: gcc/tree-loop-distribution.c\n===================================================================\n*** gcc/tree-loop-distribution.c.orig\t2013-09-13 13:34:49.000000000 +0200\n--- gcc/tree-loop-distribution.c\t2013-09-13 13:44:09.771379182 +0200\n*************** enum rdg_dep_type\n*** 92,98 ****\noutput_dd = 'o',\n\n! input_dd = 'i'\n};\n\n/* Dependence information attached to an edge of the RDG. */\n--- 92,101 ----\noutput_dd = 'o',\n\n! input_dd = 'i',\n!\n! /* Control dependence (execute conditional on). */\n! control_dd = 'c'\n};\n\n/* Dependence information attached to an edge of the RDG. */\n*************** dot_rdg_1 (FILE *file, struct graph *rdg\n*** 218,223 ****\n--- 221,230 ----\nfprintf (file, \"%d -> %d [label=anti] \\n\", i, e->dest);\nbreak;\n\n+ \t case control_dd:\n+ fprintf (file, \"%d -> %d [label=control] \\n\", i, e->dest);\n+ break;\n+\ndefault:\ngcc_unreachable ();\n}\n*************** create_rdg_edges_for_scalar (struct grap\n*** 325,334 ****\n}\n}\n\n/* Creates the edges of the reduced dependence graph RDG. */\n\nstatic void\n! create_rdg_edges (struct graph *rdg, vec<ddr_p> ddrs)\n{\nint i;\nstruct data_dependence_relation *ddr;\n--- 332,369 ----\n}\n}\n\n+ /* Creates an edge for the control dependences of BB to the vertex V. */\n+\n+ static void\n+ create_edge_for_control_dependence (struct graph *rdg, basic_block bb,\n+ \t\t\t\t int v, control_dependences *cd)\n+ {\n+ bitmap_iterator bi;\n+ unsigned edge_n;\n+ EXECUTE_IF_SET_IN_BITMAP (cd->get_edges_dependent_on (bb->index),\n+ \t\t\t 0, edge_n, bi)\n+ {\n+ basic_block cond_bb = cd->get_edge (edge_n)->src;\n+ gimple stmt = last_stmt (cond_bb);\n+ if (stmt && is_ctrl_stmt (stmt))\n+ \t{\n+ \t struct graph_edge *e;\n+ \t int c = rdg_vertex_for_stmt (rdg, stmt);\n+ \t if (c < 0)\n+ \t continue;\n+\n+ \t e = add_edge (rdg, c, v);\n+ \t e->data = XNEW (struct rdg_edge);\n+ \t RDGE_TYPE (e) = control_dd;\n+ \t RDGE_RELATION (e) = NULL;\n+ \t}\n+ }\n+ }\n+\n/* Creates the edges of the reduced dependence graph RDG. */\n\nstatic void\n! create_rdg_edges (struct graph *rdg, vec<ddr_p> ddrs, control_dependences *cd)\n{\nint i;\nstruct data_dependence_relation *ddr;\n*************** create_rdg_edges (struct graph *rdg, vec\n*** 345,350 ****\n--- 380,400 ----\nFOR_EACH_PHI_OR_STMT_DEF (def_p, RDG_STMT (rdg, i),\niter, SSA_OP_DEF)\ncreate_rdg_edges_for_scalar (rdg, DEF_FROM_PTR (def_p), i);\n+\n+ if (cd)\n+ for (i = 0; i < rdg->n_vertices; i++)\n+ {\n+ \tgimple stmt = RDG_STMT (rdg, i);\n+ \tif (gimple_code (stmt) == GIMPLE_PHI)\n+ \t {\n+ \t edge_iterator ei;\n+ \t edge e;\n+ \t FOR_EACH_EDGE (e, ei, gimple_bb (stmt)->preds)\n+ \t create_edge_for_control_dependence (rdg, e->src, i, cd);\n+ \t }\n+ \telse\n+ \t create_edge_for_control_dependence (rdg, gimple_bb (stmt), i, cd);\n+ }\n}\n\n/* Build the vertices of the reduced dependence graph RDG. Return false\n*************** free_rdg (struct graph *rdg)\n*** 465,471 ****\nscalar dependence. */\n\nstatic struct graph *\n! build_rdg (vec<loop_p> loop_nest)\n{\nstruct graph *rdg;\nvec<gimple> stmts;\n--- 515,521 ----\nscalar dependence. */\n\nstatic struct graph *\n! build_rdg (vec<loop_p> loop_nest, control_dependences *cd)\n{\nstruct graph *rdg;\nvec<gimple> stmts;\n*************** build_rdg (vec<loop_p> loop_nest)\n*** 497,503 ****\nfree_rdg (rdg);\nreturn NULL;\n}\n! create_rdg_edges (rdg, dependence_relations);\ndependence_relations.release ();\ndatarefs.release ();\n\n--- 547,553 ----\nfree_rdg (rdg);\nreturn NULL;\n}\n! create_rdg_edges (rdg, dependence_relations, cd);\ndependence_relations.release ();\ndatarefs.release ();\n\n*************** generate_loops_for_partition (struct loo\n*** 699,710 ****\n&& !is_gimple_debug (stmt)\n&& !bitmap_bit_p (partition->stmts, gimple_uid (stmt)))\n{\n! \t gsi_remove (&bsi, true);\n! \t release_defs (stmt);\n}\n! \t else\n! \t gsi_next (&bsi);\n}\n}\n\n--- 749,776 ----\n&& !is_gimple_debug (stmt)\n&& !bitmap_bit_p (partition->stmts, gimple_uid (stmt)))\n{\n! \t /* Choose an arbitrary path through the empty CFG part\n! \t\t that this unnecessary control stmt controls. */\n! \t if (gimple_code (stmt) == GIMPLE_COND)\n! \t\t{\n! \t\t gimple_cond_make_false (stmt);\n! \t\t update_stmt (stmt);\n! \t\t}\n! \t else if (gimple_code (stmt) == GIMPLE_SWITCH)\n! \t\t{\n! \t\t gimple_switch_set_index\n! \t\t (stmt, CASE_LOW (gimple_switch_label (stmt, 1)));\n! \t\t update_stmt (stmt);\n! \t\t}\n! \t else\n! \t\t{\n! \t\t gsi_remove (&bsi, true);\n! \t\t release_defs (stmt);\n! \t\t continue;\n! \t\t}\n}\n! \t gsi_next (&bsi);\n}\n}\n\n*************** rdg_flag_vertex_and_dependent (struct gr\n*** 1031,1092 ****\nnodes.release ();\n}\n\n- /* Initialize CONDS with all the condition statements from the basic\n- blocks of LOOP. */\n-\n- static void\n- collect_condition_stmts (struct loop *loop, vec<gimple> *conds)\n- {\n- unsigned i;\n- edge e;\n- vec<edge> exits = get_loop_exit_edges (loop);\n-\n- FOR_EACH_VEC_ELT (exits, i, e)\n- {\n- gimple cond = last_stmt (e->src);\n-\n- if (cond)\n- \tconds->safe_push (cond);\n- }\n-\n- exits.release ();\n- }\n-\n- /* Add to PARTITION all the exit condition statements for LOOPS\n- together with all their dependent statements determined from\n- RDG. */\n-\n- static void\n- rdg_flag_loop_exits (struct graph *rdg, partition_t partition,\n- \t\t bitmap processed)\n- {\n- unsigned i;\n- bitmap_iterator bi;\n- vec<gimple> conds;\n- conds.create (3);\n-\n- EXECUTE_IF_SET_IN_BITMAP (partition->loops, 0, i, bi)\n- collect_condition_stmts (get_loop (cfun, i), &conds);\n-\n- while (!conds.is_empty ())\n- {\n- gimple cond = conds.pop ();\n- int v = rdg_vertex_for_stmt (rdg, cond);\n- \t{\n- \t bitmap saved_loops = BITMAP_ALLOC (NULL);\n- \t bitmap_copy (saved_loops, partition->loops);\n- \t rdg_flag_vertex_and_dependent (rdg, v, partition, processed);\n- \t EXECUTE_IF_AND_COMPL_IN_BITMAP (partition->loops, saved_loops,\n- \t\t\t\t\t 0, i, bi)\n- \t collect_condition_stmts (get_loop (cfun, i), &conds);\n- \t BITMAP_FREE (saved_loops);\n- \t}\n- }\n-\n- conds.release ();\n- }\n-\n/* Returns a partition with all the statements needed for computing\nthe vertex V of the RDG, also including the loop exit conditions. */\n\n--- 1097,1102 ----\n*************** build_rdg_partition_for_vertex (struct g\n*** 1096,1102 ****\npartition_t partition = partition_alloc (NULL, NULL);\nbitmap processed = BITMAP_ALLOC (NULL);\nrdg_flag_vertex_and_dependent (rdg, v, partition, processed);\n- rdg_flag_loop_exits (rdg, partition, processed);\nBITMAP_FREE (processed);\nreturn partition;\n}\n--- 1106,1111 ----\n*************** partition_contains_all_rw (struct graph\n*** 1463,1469 ****\nReturns the number of distributed loops. */\n\nstatic int\n! distribute_loop (struct loop *loop, vec<gimple> stmts)\n{\nstruct graph *rdg;\nvec<loop_p> loop_nest;\n--- 1472,1479 ----\nReturns the number of distributed loops. */\n\nstatic int\n! distribute_loop (struct loop *loop, vec<gimple> stmts,\n! \t\t control_dependences *cd)\n{\nstruct graph *rdg;\nvec<loop_p> loop_nest;\n*************** distribute_loop (struct loop *loop, vec<\n*** 1479,1485 ****\nreturn 0;\n}\n\n! rdg = build_rdg (loop_nest);\nif (!rdg)\n{\nif (dump_file && (dump_flags & TDF_DETAILS))\n--- 1489,1495 ----\nreturn 0;\n}\n\n! rdg = build_rdg (loop_nest, cd);\nif (!rdg)\n{\nif (dump_file && (dump_flags & TDF_DETAILS))\n*************** tree_loop_distribution (void)\n*** 1631,1636 ****\n--- 1641,1647 ----\nloop_iterator li;\nbool changed = false;\nbasic_block bb;\n+ control_dependences *cd = NULL;\n\nFOR_ALL_BB (bb)\n{\n*************** tree_loop_distribution (void)\n*** 1660,1669 ****\nif (!optimize_loop_for_speed_p (loop))\ncontinue;\n\n- /* Only distribute loops with a header and latch for now. */\n- if (loop->num_nodes > 2)\n- \tcontinue;\n-\n/* Initialize the worklist with stmts we seed the partitions with. */\nbbs = get_loop_body_in_dom_order (loop);\nfor (i = 0; i < loop->num_nodes; ++i)\n--- 1671,1676 ----\n*************** out:\n*** 1697,1703 ****\nfree (bbs);\n\nif (work_list.length () > 0)\n! \tnb_generated_loops = distribute_loop (loop, work_list);\n\nif (nb_generated_loops > 0)\nchanged = true;\n--- 1704,1718 ----\nfree (bbs);\n\nif (work_list.length () > 0)\n! \t{\n! \t if (!cd)\n! \t {\n! \t calculate_dominance_info (CDI_POST_DOMINATORS);\n! \t cd = new control_dependences (create_edge_list ());\n! \t free_dominance_info (CDI_POST_DOMINATORS);\n! \t }\n! \t nb_generated_loops = distribute_loop (loop, work_list, cd);\n! \t}\n\nif (nb_generated_loops > 0)\nchanged = true;\n*************** out:\n*** 1714,1719 ****\n--- 1729,1737 ----\nwork_list.release ();\n}\n\n+ if (cd)\n+ delete cd;\n+\nif (changed)\n{\nmark_virtual_operands_for_renaming (cfun);\nIndex: gcc/testsuite/gcc.dg/tree-ssa/ldist-22.c\n===================================================================\n*** /dev/null\t1970-01-01 00:00:00.000000000 +0000\n--- gcc/testsuite/gcc.dg/tree-ssa/ldist-22.c\t2013-09-13 14:06:48.135500474 +0200\n***************\n*** 0 ****\n--- 1,32 ----\n+ /* { dg-do run } */\n+ /* { dg-options \"-O3 -fdump-tree-ldist-details\" } */\n+\n+ extern void abort (void);\n+\n+ int a, b;\n+\n+ void __attribute__((noinline,noclone))\n+ foo (void)\n+ {\n+ int i;\n+ for (i = 0; i < 1024; ++i)\n+ {\n+ a[i] = 0;\n+ if (i > 100)\n+ \tb[i] = i;\n+ }\n+ }\n+\n+ int main()\n+ {\n+ b = 1;\n+ foo ();\n+ if (b != 1 || b != 101)\n+ abort ();\n+ if (a != 0 || a != 0)\n+ abort ();\n+ return;\n+ }\n+\n+ /* { dg-final { scan-tree-dump \"generated memset zero\" \"ldist\" } } */\n+ /* { dg-final { cleanup-tree-dump \"ldist\" } } */\n\n```\n\nIndex Nav: Message Nav: [Date Index] [Subject Index] [Author Index] [Thread Index] [Date Prev] [Date Next] [Thread Prev] [Thread Next]"
] | [
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https://omics.sbs/blog/prs/prs_calculation.html | [
"# Polygenic Risk Score Calculation\n\nYou can find the interactive app here.\n\n## 1. Intro",
null,
"",
null,
"These 4 nucleotides in sequence make up our DNA.\n\nOur observable traits like height, hair, and eye color as well as diseases manifest according to our biological heritage. For all humans, this heritage comes in the form of ~3.25 billion bases long DNA sequence from both parents chunked and packaged as chromosomes. What makes this heritage unique is the collection of differences in these bases. Every person carries a different set of these variations made unique by the shuffling of DNA segments in germ cells. These variations can be large or small; it could be the addition or deletion of bases or the change of bases from one to another. For some locations in the genome, multiple versions of these changes will be present in the population.\n\nSome of these variations have a high enough impact to project their effect on phenotype by themself. Some of the variations affect the phenotype in concert with each other showing their effect in an additive way. We model this additive effect of multiple variations on a given phenotype with a Polygenic Risk Score (PRS). This writing aims to give a basic understanding of PRS and some concepts related to it in literature.\n\n### 1.1. Polygenic\n\nPolygenic means multiple genes are involved in the emergence of the given phenotype. Gene is a fuzzy concept with evolving definitions and boundaries that are used as a catch-all phrase. (Cipriano and Ballarino 2018) In our case what we are interested in are loci in the genome that carries the variation that effect the same phenotype. This variation can be in a gene that shows its effect through its impact on the protein like a change from one amino acid to another. However, regions outside of the genes also play a role in biological processes and the emergence of phenome which we will see in the association studies. Since association studies show correlations and not causation a variant like rs1421085 found on FTO affects ARID5B millions of bases away. (Claussnitzer et al. 2015)\n\nYou can create this ideogram at: https://www.ncbi.nlm.nih.gov/genome/tools/gdp\n\nWith the following intervals:\n\n 98956422 98956423 chr5 21538259 21538260 chr6 50805979 50805980 chr7 99345973 99345974 chr8 95138636 95138637 chr9 68394717 68394718 chr16 30338345 30338346 chr17 33257441 33257442 chr19 38617616 38617617 chr20 34444167 34444168 chr22 30818468 30818469",
null,
"Figure 1: Variations are marked with pink circles.\n\n### 1.2. Risk\n\nOxford English Dictionary defines risk as:\n\nthe possibility of something bad happening at some time in the future.\n\nIn statistics, there exist two types of risk. (“Understanding Statistics: Risk | BMJ Best Practice” n.d.)\n\n• Absolute Risk (AR) ratio of the events to the total number in a given group\n• Relative Risk (RR) ratio of the event happening in the case group to an event happening in the control group.\n\nHere the term risk refers to a relative risk since we are comparing the case group with the control group. (“Polygenic Risk Scores” n.d.) RR should be evaluated with absolute risk on an individual basis. (Andrade 2015)\n\n### 1.3. Score\n\nWe can look at two different equations that calculate PRS. They both calculate",
null,
"by summing the effect size (denoted as",
null,
"in Collister et al. and",
null,
"in PLINK formula) and dosage (denoted as",
null,
"in Collister et al. and",
null,
"in PLINK formula) of the alleles. Differently, the PLINK formula has",
null,
"in the denominator. Where",
null,
"is the ploidy and",
null,
"is the number of non-missing Single Nucleotide Polymorphisms (SNPs). If a sample missing SNPs their score would be lower than samples with full genotypes. Dividing by non-missing number of SNPs and ploidy makes the score comparable between individuals with missing genotypes. (Collister, Liu, and Clifton 2022) With enough samples, these scores would approach a normal distribution.",
null,
"1",
null,
"2\n\n## 2. Calculating Polygenic Risk Scores\n\nWe are going to mock a Genome-Wide Association (GWAS) results and create a mock study of our own. We are going to use the summary statistics from the GWAS data. The idea here is that most of the time we don’t have access to every genotype from a GWAS. Instead, we use the most significant variation and investigate their effect in our cohort.\n\n### 2.1. Genome-wide Association Studies\n\nA great resource for viewing studies and their results is the GWAS catalog. A more hands-on example can be found at Hail’s site.\n\nGWAS performs statistical tests on thousands of genomic loci in thousands of genomes to see if any of those regions are associated with the given trait. (Uffelmann et al. 2021) As a result, we get p-values for how significant that association is and the effect size of the variation.\n\nLet’s say we have found 10 SNPs significant in a study with 50000 cases and 50000 controls. First, let’s go over the SNPs.\n\nsnp_number <- 10\n\ngwas <- data.frame(\nSNP_id = paste(\"SNP\", seq(snp_number), sep=\"\")\n)\ngwas[c(\n\"non_effect_allele\",\n\"effect_allele\"\n)] <- t(replicate(snp_number,\nsample(c(\"A\", \"T\", \"C\", \"G\"), 2)\n))\n\n\nSNP_id non_effect_allele effect_allele\nSNP1 C A\nSNP2 T C\nSNP3 G C\nSNP8 G C\nSNP9 A T\nSNP10 T G\n\nI selected some genotypes randomly. However, we don’t need to know the non-effect allele and effect allele nucleotides, that’s why they will be simply named A or 0 for the non-effect allele and B or 1 for the effect allele.\n\ngwas$non_effect_allele = \"A\" gwas$effect_allele = \"B\"\n\nSNP_id non_effect_allele effect_allele\nSNP1 A B\nSNP2 A B\nSNP3 A B\nSNP8 A B\nSNP9 A B\nSNP10 A B\n\nAnother thing we need from GWAS summary statistics is the effect sizes of these SNPs. We are going to use the log of OR as the effect size.\n\nI want to set it up so every SNP is observed 2 % more incrementally. The lines with step_case and obs_case just generalize this instead of hard-coding so it applies to any number of SNPs and sample size.\n\ngwas_control_number = 50000\ngwas_case_number = 50000\nincrement_percent = 0.02\ngwas_sample_number = gwas_control_number + gwas_case_number\n\nstep_case = round(gwas_case_number * increment_percent)\ngwas$obs_case = seq(gwas_case_number / 2 + step_case, gwas_case_number / 2 + step_case * nrow(gwas), step_case) step_control = round(gwas_control_number * increment_percent) gwas$obs_control = seq(gwas_control_number / 2 - step_control,\ngwas_control_number / 2 - step_control * nrow(gwas),\n-step_control)\n\ngwas$log_OR <- signif(log( (gwas$obs_case / (gwas_case_number - gwas$obs_case)) / (gwas$obs_control / (gwas_control_number - gwas$obs_control)) ), 2) print(rbind(head(gwas, 3), tail(gwas, 3))) SNP_id non_effect_allele effect_allele obs_case obs_control log_OR SNP1 A B 26000 24000 0.16 SNP2 A B 27000 23000 0.32 SNP3 A B 28000 22000 0.48 SNP8 A B 33000 17000 1.3 SNP9 A B 34000 16000 1.5 SNP10 A B 35000 15000 1.7 We get effect sizes for the first genotype and for the last genotype. A BB genotype would have score of",
null,
"",
null,
"for the first genotype and",
null,
"",
null,
". ### 2.2. Toy dataset We are screening our population concerning a phenotype. We aim to determine risk scores using the GWAS summary statistics and see if we can make a correlation between PRS and phenotype. We have genotyped the same 10 SNPs in 2000 people for our study. The set.seed() in the first line ensure get_genotype function randoms the same values every time. set.seed(42) sample_number <- 2000 snp_number <- 10 samples <- data.frame( sample_id = paste(\"Sample\", seq(sample_number), sep=\"\"), status = c( rep(\"Control\", sample_number / 2), rep(\"Case\", sample_number / 2)) ) get_genotype <- function(prob=c(.5, .5)) { paste(sample(c(\"A\", \"B\"), 2, replace=T, prob = prob), collapse=\"\") } control_prob <- c(0.55, 0.45) case_prob <- c(0.45, 0.55) samples[gwas$SNP_id] <- rbind(\nmatrix(\nreplicate(\nsnp_number * sample_number / 2,\nget_genotype(control_prob)),\nnrow=sample_number / 2),\nmatrix(\nreplicate(\nsnp_number * sample_number / 2,\nget_genotype(case_prob)),\nnrow=sample_number / 2)\n)\n\nsample_id status SNP1 SNP2 SNP3 SNP4 SNP5 SNP6 SNP7 SNP8 SNP9 SNP10\nSample1 Control BB BA BB AA AA AB BA BA AA AB\nSample2 Control AB BB AB AA AA BA AB BA AB AB\nSample3 Control BA BB AA AA AB BB AB BA AA AB\nSample1998 Case BB AB BA AA BB BB BA BB BA AA\nSample1999 Case BB BB AB AB AA AA BB BA BA AB\nSample2000 Case BA BB AB BB AB AB AA AA AA AB\n\nWe get random genotypes for controls for alleles A, and B with probabilities 0.55, 0.45 ; cases for alleles A, and B with probabilities 0.45, 0.55 This way cases are more prone to having the effect allele.\n\n### 2.3. Calculate the PRS\n\nWe will be using the simpler version of the equations. Mainly because we don’t have any missing genotypes and secondly because it’s simpler.\n\nWe implement the formula by using an apply function to multiply dosage and the effect size then doing a row sum would give us the score.\n\nWe have both AB and BA genotypes in the dosage list because our get_genotype function returns both of these. Since there are no haplotypes or variant phasing both AB and BA are the same.\n\ndosage <- list(\"AA\" = 0, \"AB\" = 1, \"BA\" = 1, \"BB\" = 2)\n\nsamples$PRS <- colSums(as.data.frame( apply( samples[gwas$SNP_id],\n1,\nfunction(x) as.numeric(dosage[x]) * gwas$log_OR))) print(rbind(head(samples, 3), tail(samples, 3))) sample_id status SNP1 SNP2 SNP3 SNP4 SNP5 SNP6 SNP7 SNP8 SNP9 SNP10 PRS Sample1 Control BB BA BB AA AA AB BA BA AA AB 6.78 Sample2 Control AB BB AB AA AA BA AB BA AB AB 7.96 Sample3 Control BA BB AA AA AB BB AB BA AA AB 7.77 Sample1998 Case BB AB BA AA BB BB BA BB BA AA 10 Sample1999 Case BB BB AB AB AA AA BB BA BA AB 8.99 Sample2000 Case BA BB AB BB AB AB AA AA AA AB 6.07 ### 2.4. Ploting the normal curves What we will be seeing around in papers and around the web is a plot of normal distributions comparing cases to controls. To create the plot we will need ggplot2. library(ggplot2) We shift the scores by subtracting the mean of the controls to make the controls center around 0. We get the mean and standard deviations to plot the curves with the stat_function. samples$status <- as.factor(with(samples, reorder(status, PRS)))\nsamples$PRS <- samples$PRS - mean(samples[which(samples$status==\"Control\"), \"PRS\"]) mean_control <- mean(samples[which(samples$status==\"Control\"), \"PRS\"])\nsd_control <- sd(samples[which(samples$status==\"Control\"), \"PRS\"]) mean_case <- mean(samples[which(samples$status==\"Case\"), \"PRS\"])\nsd_case <- sd(samples[which(samples$status==\"Case\"), \"PRS\"]) ggplot(samples, aes(x=PRS, color=status, fill=status)) + geom_histogram(aes(y=..density..), position=\"dodge\", alpha=.5, bins=30) + stat_function(fun = dnorm, color=\"darkred\", args = list(mean = mean_control, sd = sd_control)) + stat_function(fun = dnorm, color=\"darkblue\", args = list(mean = mean_case, sd = sd_case )) + ggtitle(sprintf( \"PRS distributions Controls μ %.2f σ %.2f Cases μ %.2f σ %.2f\", mean_control, sd_control, mean_case, sd_case))",
null,
"When you visit www.pgscatalog.org or image search “polygenic risk score” you would be seeing the plot of these two peaks. Cases have more effect alleles with higher totals on average this is why their curve is slightly shifted to the right. The case curve also has a larger standard deviation because effect alleles have increasing effect sizes. ### 2.5. Logistic regression Logistic regression is a model used in statistics that fit an S curve to data based on probabilities. Details of the subject are out of the scope of this post. I would suggest StatQuest’s logistic regression playlist to anyone interested. Logistic regressions are used for prediction. However, one thing I would note in context to PRS calculations is that, unlike this common use, it’s not used for prediction in genetic studies. Because the scores and the risk calculated here are relative. The effect becomes tangible only in the tails of the curves but as the number of cases and controls drops the significance of the effect also drops. • Because most of the totals are the same they overlap and we plot only a few points as a result. position_jitter with w=0.5 spreads the points around a bit allowing us the plot all of the points. ggplot(samples, aes(x=PRS, y=as.numeric(samples$status) -1, color=status)) +\ngeom_point(shape = \"|\", position = position_jitter(w = 0.5, h = 0)) +\ngeom_hline(yintercept = c(0,1), linetype = \"dashed\", color = \"grey\") +\nscale_y_discrete(name =\"Status\", labels=c(\"Control\",\"Case\"), limits=c(0,1))",
null,
"model <- glm(status~PRS, family=\"binomial\", data=samples)\nsummary(model)\n\n\nCall:\nglm(formula = status ~ PRS, family = \"binomial\", data = samples)\n\nCoefficients:\nEstimate Std. Error z value Pr(>|z|)\n(Intercept) -0.30654 0.07353 -4.169 3.06e-05 ***\nPRS 0.34170 0.03149 10.852 < 2e-16 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\nNull deviance: 1386.3 on 999 degrees of freedom\nResidual deviance: 1242.5 on 998 degrees of freedom\nAIC: 1246.5\n\nNumber of Fisher Scoring iterations: 4\n\n\nIn the model summary, we can find coefficients\n\nfor the control group and\n\nfor the case group. We can write our equation as",
null,
"•",
null,
".\n\nInserting the sample’s score into this equation would give us the log odds of having the phenotype. One unit increase in the",
null,
"would increase the log odds by units.\n\nWe can translate the log of odds to probabilities. Because it’s log we can turn them back to odds with exponential function. Odds are the event happening (let’s say",
null,
") ratio to the event not happening (let’s say",
null,
") which would be",
null,
".",
null,
"gives us the log of the odds;",
null,
"gives us the odds. We can write the odds as",
null,
"; this would be",
null,
"which would simplify to",
null,
". It would give us the probability of the event happening over the total events.",
null,
"3\n\n### 2.6. Plotting the regression curve\n\nWe can plot our model’s S-shaped curve that shows the probabilities.\n\nggplot(samples, aes(x=PRS, y=as.numeric(samples$status) -1)) + geom_point(shape = \"|\", position = position_jitter(w = 0.5, h = 0), aes(color=status)) + geom_smooth(method = \"glm\", method.args = list(family = \"binomial\"), se = FALSE) + geom_hline(yintercept = c(0,1), linetype = \"dashed\", color = \"grey\") + scale_y_discrete(name =\"Status\", labels=c(\"Control\",\"Case\"), limits=c(0,1))",
null,
"### 2.7. Measuring goodness of fit One of the measures of how good or model is",
null,
".",
null,
"in linear regression is calculated using the sum of squares of residuals. This calculation does not apply to the logistic regression because we have a probability curve. There are several approaches to calculating",
null,
"and the most used one I have seen is the Nagelkerke method (also the most optimistic). The closer it gets to 1 better the fit. rcompanion has the",
null,
"calculations modules when the Nagelkerke function is called other methods results are also returned. library(rcompanion) nagelkerke(model) $Models\n\nModel: \"glm, status ~ PRS, binomial, samples\"\nNull: \"glm, status ~ 1, binomial, samples\"\n\n$Pseudo.R.squared.for.model.vs.null Pseudo.R.squared McFadden 0.103711 Cox and Snell (ML) 0.133917 Nagelkerke (Cragg and Uhler) 0.178556$Likelihood.ratio.test\nDf.diff LogLik.diff Chisq p.value\n-1 -71.887 143.77 3.9798e-33\n\n$Number.of.observations Model: 1000 Null: 1000$Messages\n \"Note: For models fit with REML, these statistics are based on refitting with ML\"\n\n$Warnings \"None\" ### 2.8. Enrichment in the highest PRS burden percentiles This is the table-2 from the paper (Leu et al. 2019). Here it looks at the most extreme tails of the curve to see if the effect is larger. We get the samples bigger and smaller than our percent thresholds with the quantile function. We then pull the values used in the table from the summary function. Differently, we use the confusionMatrix from the caret library to calculate the sensitivity and specificity. What we are looking at in this table is the increase in the odds in the more extreme percentages. One odd thing I notice is that in my simulations sensitivity goes up and specificity goes down as we get to the edge of the tails where the opposite is seen in the paper. library(caret) my_sum_rows = list() percents = c(20, 10, 5, 1, 0.5) for (percent in percents) { samples$percentile <- ifelse(\nsamples$PRS >= quantile(samples$PRS,prob=1-percent/100),\n\"Above\", \"Below\")\nsamples$percentile <- as.factor(with(samples, reorder(percentile, PRS))) model <- glm(status~percentile, family=\"binomial\", data=samples) threshold <- 0.5 cm <- confusionMatrix( as.factor(ifelse(predict(model,type=\"response\")> threshold, 1, 0)), as.factor(model$y)\n)\n\nabove_control <- nrow(\nsamples[which(samples$status==\"Control\" & samples$percentile==\"Above\"),]\n)\nbelow_control <- nrow(\nsamples[which(samples$status==\"Control\" & samples$percentile==\"Below\"),]\n)\nabove_case <- nrow(\nsamples[which(samples$status==\"Case\" & samples$percentile==\"Above\"),]\n)\nbelow_case <- nrow(\nsamples[which(samples$status==\"Case\" & samples$percentile==\"Below\"),]\n)\n\nmy_sum_row <- data.frame(\n\"Reference Group\" = sprintf(\"Remaining %s%% \", 100-percent),\n\"Cases/Controls Above PRS%\" = sprintf(\"%s/%s\", above_case, above_control),\n\"Cases/Controls Below PRS%\" = sprintf(\"%s/%s\", below_case, below_control),\n\"Odds Ratio\" = round(exp(coef(model)), 2),\n\"95% CI\" = do.call(sprintf, c(\n\"%.2f - %.2f\",\nas.list(exp(confint(model, level=.90)[2,])))),\n\"P value\" = sprintf(\"%.2e\",\nsummary(model)$coefficients['percentileAbove', 'Pr(>|z|)'] ), \"Sensitivity / Specificity\" = sprintf( \"%.3f / %.3f\", as.list(cm$byClass)$Sensitivity, as.list(cm$byClass)$Specificity ) ) my_sum_rows[[sprintf(\"Top %s%% of distribution\", percent)]] <- my_sum_row } my_sum_table <- do.call(rbind, my_sum_rows) colnames(my_sum_table) <- c( \"Reference group\", \"Cases/Controls above PRS%\", \"Cases/Controls below PRS%\", \"Odds ratio\", \"95% CI\", \"P value\", \"Sensitivity / Specificity\" ) my_sum_table Reference group Cases/Controls above PRS% Cases/Controls below PRS% Odds ratio 95% CI P value Sensitivity / Specificity Top 20% of distribution Remaining 80% 290/110 710/890 3.3 2.71 - 4.05 2.1e-22 0.890 / 0.290 Top 10% of distribution Remaining 90% 149/52 851/948 3.19 2.43 - 4.23 4.68e-12 0.948 / 0.149 Top 5% of distribution Remaining 95% 78/23 922/977 3.59 2.44 - 5.43 1.2e-07 0.977 / 0.078 Top 1% of distribution Remaining 99% 15/6 985/994 2.52 1.18 - 5.93 0.0565 0.994 / 0.015 Top 0.5% of distribution Remaining 99.5% 9/1 991/999 9.07 2.14 - 90.13 0.0366 0.999 / 0.009 ### 2.9. PRS deciles This graph is from the paper (Kloeve-Mogensen et al. 2021). Here they compare scores from each decile. The fifth decile is used as the reference that’s why we are seeing a stair pattern where the decile before the fifth gets closer and the deciles after go up incrementally. I have found this answer which sets the fifth decile as a reference differently. (socialscientist 2022) I just set the levels so the fifth is the first one which the generalized linear models (glm) uses as the reference. Resulting coefficients are considered as odds ratio.(Noah 2018) samples$deciles <- factor(.bincode(samples$PRS, breaks = quantile(samples$PRS, seq(0, 1, 0.1)),\ninclude.lowest = TRUE\n), levels = c(5, 1, 2, 3, 4, 6, 7, 8, 9, 10))\n\nmodel <- glm(status~deciles, family=\"binomial\", data=samples)\nsummary(model)\n\n\nCall:\nglm(formula = status ~ deciles, family = \"binomial\", data = samples)\n\nCoefficients:\nEstimate Std. Error z value Pr(>|z|)\n(Intercept) -0.19867 0.14141 -1.405 0.16006\ndeciles1 -1.32376 0.23203 -5.705 1.16e-08 ***\ndeciles2 -0.88014 0.21511 -4.092 4.28e-05 ***\ndeciles3 -0.36885 0.20435 -1.805 0.07108 .\ndeciles4 -0.03352 0.20092 -0.167 0.86749\ndeciles6 0.50095 0.20114 2.491 0.01276 *\ndeciles7 0.58733 0.20242 2.902 0.00371 **\ndeciles8 1.02227 0.20877 4.897 9.75e-07 ***\ndeciles9 1.07696 0.20974 5.135 2.82e-07 ***\ndeciles10 1.26406 0.21532 5.871 4.34e-09 ***\n---\nSignif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1\n\n(Dispersion parameter for binomial family taken to be 1)\n\nNull deviance: 2772.6 on 1999 degrees of freedom\nResidual deviance: 2484.1 on 1990 degrees of freedom\nAIC: 2504.1\n\nNumber of Fisher Scoring iterations: 4\n\nmy_sum_data <- as.data.frame(cbind(\nexp(confint(model, level=.90)[2:10,]),\n\"Odds ratio\"=round(exp(coef(model)[2:10]), 2),\n\"P value\"=summary(model)$coefficients[2:10, 'Pr(>|z|)'] )) colnames(my_sum_data) <- c( \"lower\", \"upper\", \"odds_ratio\", \"p_val\" ) my_sum_data <- rbind(my_sum_data, c(1,1,1,1)) my_sum_data$decile <- factor(c(1,2,3,4,6,7,8,9,10,5))\n\nggplot(my_sum_data, aes(decile, odds_ratio)) +\ngeom_errorbar(aes(ymin = lower, ymax = upper), width=.2) +\ngeom_hline(yintercept = 1, linetype = \"dashed\", color = \"grey\") +\ngeom_point(shape=22, size=6, color=\"black\", fill=\"orange\", stroke=.8)",
null,
"## 3. Conclusion\n\nPRS is a method for estimating the underlying genetic risk that common variation create for complex phenotypes. As of pgscatalog.org hosts 3,688 calculations for 619 traits from 480 publications. Their interpretation in the diseases risk and their role in understanding disease etiology still needs further work but there are also some proposals for other uses. Lu et al. concludes that it can be useful for prioritizing patients with low scores for monogenic testing. (Lu et al. 2022) Campbell et al. suggests polygenic burden plays role in the severity and penetrance of the developmental epileptic encephalopathies. (Campbell et al. 2022) Another approach taken by Darst et al. and Hassanin et. al is to calculate cancer risk with PRS combined with rare variants. (Hassanin et al. 2023; Darst et al. 2021) Considering papers above, I think investigating PRS with variant of unknown significance with low allele frequency might give an insight to some of the complex diseases.\n\nThis has been my introduction to polygenic risk scores and genetic burden. This is not a tutorial on how to perform the analyses but just to understand basic underlying concept. This post also never goes into more finickier subjects like population stratification, better effect size estimation, controlling linkage disequilibrium.\n\n## 4. References\n\nAndrade, Chittaranjan. 2015. “Understanding Relative Risk, Odds Ratio, and Related Terms: As Simple as It Can Get.” The Journal of Clinical Psychiatry 76 (7): 21865. https://doi.org/10.4088/JCP.15f10150.\n“An Introduction to Statistical Learning.” n.d. An Introduction to Statistical Learning. https://www.statlearning.com. Accessed August 7, 2023.\nCampbell, Ciarán, Costin Leu, Yen-Chen Anne Feng, Stefan Wolking, Claudia Moreau, Colin Ellis, Shiva Ganesan, et al. 2022. “The Role of Common Genetic Variation in Presumed Monogenic Epilepsies.” Ebiomedicine 81 (July): 104098. https://doi.org/10.1016/j.ebiom.2022.104098.\nChoi, Shing Wan, Timothy Shin-Heng Mak, and Paul F. O’Reilly. 2020. “Tutorial: A Guide to Performing Polygenic Risk Score Analyses.” Nature Protocols 15 (9): 2759–72. https://doi.org/10.1038/s41596-020-0353-1.\nCipriano, Andrea, and Monica Ballarino. 2018. “The Ever-Evolving Concept of the Gene: The Use of RNA/Protein Experimental Techniques to Understand Genome Functions.” Frontiers in Molecular Biosciences 5.\nClaussnitzer, Melina, Simon N. Dankel, Kyoung-Han Kim, Gerald Quon, Wouter Meuleman, Christine Haugen, Viktoria Glunk, et al. 2015. “FTO Obesity Variant Circuitry and Adipocyte Browning in Humans.” New England Journal of Medicine 373 (10): 895–907. https://doi.org/10.1056/NEJMoa1502214.\nCollister, Jennifer A., Xiaonan Liu, and Lei Clifton. 2022. “Calculating Polygenic Risk Scores (PRS) in UK Biobank: A Practical Guide for Epidemiologists.” Frontiers in Genetics 13.\nDarst, Burcu F., Xin Sheng, Rosalind A. Eeles, Zsofia Kote-Jarai, David V. Conti, and Christopher A. Haiman. 2021. “Combined Effect of a Polygenic Risk Score and Rare Genetic Variants on Prostate Cancer Risk.” European Urology 80 (2): 134–38. https://doi.org/10.1016/j.eururo.2021.04.013.\nHassanin, Emadeldin, Isabel Spier, Dheeraj R. Bobbili, Rana Aldisi, Hannah Klinkhammer, Friederike David, Nuria Dueñas, et al. 2023. “Clinically Relevant Combined Effect of Polygenic Background, Rare Pathogenic Germline Variants, and Family History on Colorectal Cancer Incidence.” Bmc Medical Genomics 16 (1): 42. https://doi.org/10.1186/s12920-023-01469-z.\nKloeve-Mogensen, Kirstine, Palle Duun Rohde, Simone Twisttmann, Marianne Nygaard, Kristina Magaard Koldby, Rudi Steffensen, Christian Møller Dahl, et al. 2021. “Polygenic Risk Score Prediction for Endometriosis.” Frontiers in Reproductive Health 3: 793226. https://doi.org/10.3389/frph.2021.793226.\nLeu, Costin, Remi Stevelink, Alexander W Smith, Slavina B Goleva, Masahiro Kanai, Lisa Ferguson, Ciaran Campbell, et al. 2019. “Polygenic Burden in Focal and Generalized Epilepsies.” Brain 142 (11): 3473–81. https://doi.org/10.1093/brain/awz292.\nLu, Tianyuan, Vincenzo Forgetta, John Brent Richards, and Celia M. T. Greenwood. 2022. “Polygenic Risk Score as a Possible Tool for Identifying Familial Monogenic Causes of Complex Diseases.” Genetics in Medicine: Official Journal of the American College of Medical Genetics 24 (7): 1545–55. https://doi.org/10.1016/j.gim.2022.03.022.\nNoah. 2018. “Answer to ‘Why Are Exponentiated Logistic Regression Coefficients Considered ‘Odds Ratios’?’.” Cross Validated.\n“Polygenic Risk Scores.” n.d. Genome.Gov. https://www.genome.gov/Health/Genomics-and-Medicine/Polygenic-risk-scores. Accessed July 29, 2023.\nsocialscientist. 2022. “Answer to ‘How to Calculate and Plot Odds-Ratios and Their Standard Errors from a Logistic Regression in R?’.” Stack Overflow.\nUffelmann, Emil, Qin Qin Huang, Nchangwi Syntia Munung, Jantina de Vries, Yukinori Okada, Alicia R. Martin, Hilary C. Martin, Tuuli Lappalainen, and Danielle Posthuma. 2021. “Genome-Wide Association Studies.” Nature Reviews Methods Primers 1 (1): 1–21. https://doi.org/10.1038/s43586-021-00056-9.\n“Understanding Statistics: Risk | BMJ Best Practice.” n.d. Accessed July 29, 2023.\n\n## 5. Acronyms\n\nAR Absolute Risk 1\n\nGWAS Genome-Wide Association 1, 2, 3, 4\n\nGlm generalized linear models 1\n\nPRS Polygenic Risk Score 1, 2, 3, 4, 5, 6, 7, 8\n\nRR Relative Risk 1, 2\n\nSNP Single Nucleotide Polymorphism 1, 2, 3, 4, 5\n\n## 6. Glossary\n\nAllele frequency The ratio of given allele to total number of alleles in a given population 1\n\nAllele One of the alternatives of a genomic locus. Human autosomal chromosomes have two alleles for every locus each inherited from a parent 1, 2, 3\n\nDosage Number of effect allele. It would be 0 if both alleles are non-effect; 1 if one allele is an effect allele; and 2 if both alleles are the effect allele 1, 2\n\nEffect size Weight of the effect allele. It can be odds ratio, hazard ratio, allele frequency 1, 2, 3, 4\n\nEffect allele The alternative allele associated with the phenotype 1, 2, 3\n\nOdds ratio ratio of the odds of the two events 1\n\nOdds ratio of number of times event happening to event not happening 1, 2, 3, 4, 5, 6, 7, 8, 9\n\nPenetrance Probability of an variants effect is seen in the phenotype 1\n\nVariant of unknown significance The variants we can’t decide whether its pathogenic or benign 1"
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null,
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null,
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https://www.aqua-calc.com/one-to-all/dynamic-viscosity/preset/pound-per-foot-hour/1 | [
"# Convert pounds per foot-hour to other units of dynamic viscosity\n\n## pounds/foot-hour [lb/ft/h] dynamic viscosity conversions\n\n 1 lb/ft/h = 0.004 gram per centimeter-second lb/ft/h to g/cm/s 1 lb/ft/h = 1.49 kilograms per meter-hour lb/ft/h to kg/m/h 1 lb/ft/h = 0.0004 kilogram per meter-second lb/ft/h to kg/m/s 1 lb/ft/h = 0.41 millipascal-second lb/ft/h to mPa·s 1 lb/ft/h = 0.0004 newton-second per square meter lb/ft/h to N·s/m² 1 lb/ft/h = 0.0004 pascal-second lb/ft/h to Pa·s 1 lb/ft/h = 0.0003 pound per foot-second lb/ft/h to lb/ft/s 1 lb/ft/h = 0.08 pound per inch-hour lb/ft/h to lb/in/h 1 lb/ft/h = 2.31 × 10-5 pound per inch-second lb/ft/h to lb/in/s 1 lb/ft/h = 6 × 10-8 reyn lb/ft/h to reyn 1 lb/ft/h = 0.41 centipoise lb/ft/h to cP 1 lb/ft/h = 0.04 decipoise lb/ft/h to dP 1 lb/ft/h = 0.004 dyne-second per square centimeter lb/ft/h to dyn·s/cm² 1 lb/ft/h = 4.22 × 10-6 gram-force second per square centimeter lb/ft/h to gf·s/cm² 1 lb/ft/h = 4.22 × 10-5 kilogram-force second per square meter lb/ft/h to kgf·s/m² 1 lb/ft/h = 4 133.79 micropoises lb/ft/h to µP 1 lb/ft/h = 0.004 poise lb/ft/h to P 1 lb/ft/h = 0.0004 Poiseuille lb/ft/h to Pl 1 lb/ft/h = 8.63 × 10-6 pound-force second per square foot lb/ft/h to lbf·s/ft² 1 lb/ft/h = 6 × 10-8 pound-force second per square inch lb/ft/h to lbf·s/in² 1 lb/ft/h = 0.0003 poundal second per square foot lb/ft/h to pdl·s/ft² 1 lb/ft/h = 8.63 × 10-6 slug per foot-second lb/ft/h to sl/ft/s\n\n#### Foods, Nutrients and Calories\n\nSAVOY, COCONUT CREAM, UPC: 016229004019 weigh(s) 190 grams per metric cup or 6.3 ounces per US cup, and contain(s) 222 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]\n\n5029 foods that contain Histidine. List of these foods starting with the highest contents of Histidine and the lowest contents of Histidine\n\n#### Gravels, Substances and Oils\n\nCaribSea, Freshwater, Super Naturals, Snowy River weighs 1 457.68 kg/m³ (90.99999 lb/ft³) with specific gravity of 1.45768 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]\n\nHydrochloric acid, solid [HCl] weighs 1 507 kg/m³ (94.07894 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]\n\nVolume to weightweight to volume and cost conversions for Refrigerant R-32, liquid (R32) with temperature in the range of -40°C (-40°F) to 65.56°C (150.008°F)\n\n#### Weights and Measurements\n\nThe long ton per metric cup density measurement unit is used to measure volume in metric cups in order to estimate weight or mass in long tons\n\nThe length measurement was introduced to measure distance between any two objects.\n\ngr/mm² to t/ft² conversion table, gr/mm² to t/ft² unit converter or convert between all units of surface density measurement.\n\n#### Calculators\n\nCalculate Maximum Heart Rate (MHR) based on age and gender"
] | [
null
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https://www.kaysonseducation.co.in/questions/p-span-sty_3700 | [
" A small drop of water of surface tension is squeezed between two clean glass plates so that a thin layer of thickness d and area A is formed between them. If the angle of contact is zero, the force required to pull the plates apart is : Kaysons Education\n\n# A Small Drop Of Water Of Surface Tension is Squeezed Between Two Clean Glass Plates So That A Thin Layer Of Thickness d and Area A is Formed Between Them. If The Angle Of Contact Is Zero, The Force Required To Pull The Plates Apart Is\n\n#### Video lectures\n\nAccess over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.\n\n#### Online Support\n\nPractice over 30000+ questions starting from basic level to JEE advance level.\n\n#### Live Doubt Clearing Session\n\nAsk your doubts live everyday Join our live doubt clearing session conducted by our experts.\n\n#### National Mock Tests\n\nGive tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.\n\n#### Organized Learning\n\nProper planning to complete syllabus is the key to get a decent rank in JEE.\n\n#### Test Series/Daily assignments\n\nGive tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.\n\n## Question\n\n### Solution\n\nCorrect option is",
null,
"An extremely thin layer of a liquid can be regarded as a collection of a large number of semi-spherical drops. Hence the excess pressure across a thin layer of a liquid is",
null,
"as in the case of a spherical drop, where r = d/2. Therefore, excess pressure is",
null,
"",
null,
"Force due to surface tension pushing the two plates together is\n\nF = excess pressure",
null,
"area of layer",
null,
".\n\n#### SIMILAR QUESTIONS\n\nQ1\n\nTwo identical cylindrical vessels, each of base area A, have their bases at the same horizontal level. They contain a liquid of density",
null,
". In one vessel the height of the liquid is h1 and in the other h2 > h1. When the two vessels are connected, the work done by gravity in equalizing the levels is\n\nQ2\n\nA cylindrical jar has radius r. To what height h should it be filled with a liquid so that the force exerted by the liquid on the sides of the jar equals the force exerted on the bottom?\n\nQ3\n\nA rectangular tank is filled to the brim with water. When a hole at its bottom is unplugged, the tank is emptied in time T. If the tank is half-filled with water, it will be emptied in time\n\nQ4\n\nTwo capillary tubes A and B of radii ra and rb and lengths la and lbrespectively are held horizontally. The volume of water flowing per second through tube A is Qa when the pressure difference across its ends is maintained at P. When the same pressure difference is maintained across tube B, the volume of water flowing per second through it is Qb. The ratio",
null,
"is\n\nQ5",
null,
". If a pressure difference P is maintained across tube A, the rate of flow of water in it is Q. If the tubes are connected in series and the same pressure difference P is maintained across the combination, the rate of flow through the combination will be\n\nQ6\n\nIf the surface tension of soap solution is",
null,
", What is the work done in blowing soap bubble of radius r?\n\nQ7\n\nThe work done to break up a drop of a liquid of radius R and surface tension",
null,
"into eight drops, all of the same size, is\n\nQ8\n\nA soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If",
null,
"is the surface tension of soap solution, the energy spent in doing so is\n\nQ9\n\nIf the air bubble is formed at a depth h inside the container of soap solution of density",
null,
", the total pressure inside the bubble is (here P0denotes the atmospheric pressure)\n\nQ10\n\nEight drop of water, each of radius r, coalesce to form a single big drop. The energy released is used up in raising the temperature of the big drop. If",
null,
"is the surface tension of water and",
null,
"its density and J the mechanical equivalent of heat (all expressed in SI units), the rise in the temperature of the drop is"
] | [
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image2183.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image2175.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image2177.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image1184.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image1186.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image2179.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image024.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image2073.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image2085.png",
null,
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null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image026.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image026.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image024.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image026.png",
null,
"http://kaysonseducation.co.in/Question/IIT/physics/Waves%20&%20Fluid%20Mechanics/Fluid/Fluid_files/image024.png",
null
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https://cffmv.blogspot.com/2010/04/multiplying-on-grid.html | [
"## Sunday, April 25, 2010\n\n### Multiplying on the Grid\n\nLearning to multiply whole numbers using a grid makes far more sense than the traditional method which has been utilized since well before any currently living person was born!\n\nFurthermore, multiplying on a grid can help students with mental math, but more importantly, make it easier to understand multiplying and factoring polynomials later in algebra.\n\nAs seen in the first two images below, the method is remarkably simple. First sketch a grid to accommodate the number of digits in each factor. For instance, if multiplying a one digit factor by a two digit factor, you'll need a 1 by 2 grid. Use a 2 by 2 grid if the factors are each two digit numbers, and so forth.\n\nPlace the digits of the factors on the outside of the grid, each digit having been multiplied by its place value. For example, if one of the factors is 64, put 60 (6x10) and 4 outside the grid.\n\nNext, just as in a common \"times table\", fill the grid spaces with the products of the numbers on the outside. Finally, add the numbers in the grid to get the product.\n\nImage 1",
null,
"Image 2",
null,
"The following images illustrate how grid multiplication can facilitate polynomial multiplication and factoring as students progress through mathematics.\n\nThe size of the grid is determined by the number of terms in each polynomial factor. Multiplying a binomial by a trinomial would require a grid 2 by 3. Then each term of the factors is placed outside the grid just as the digits of the whole number factors were placed. The spaces of the grid are filled with the products of the monomial terms outside the grid. Finally, again just as with the whole numbers, the product of the two polynomials is the sum of the terms inside the grid.\n\nImage 3",
null,
"Image 4",
null,
"Image 5",
null,
"Image 6 shows how the grid concept can be used to help factor a binomial. Place the first and last terms of the trinomial in the first and last grid spaces. Then find numbers for the sun ☀ and the cloud ☁ whose product is the constant term (20) and whose sum is the coefficient of the linear term (9), 4 and 5. Therefore the factors are (x+4) and (x+5).\n\nImage 6",
null,
"I have long thought that just making this simple change in the way we teach a basic skill might have a subtle but significant impact on how students learn and understand higher level skills later. Why do we continue to teach the traditional method of whole number multiplication? Because that's the way we've always done it. That's what we know. That doesn't mean it's the best way.\n\nAdd-ons:. Try applying the grid method to mutiplying mixed numbers.\nYou can put a tech spin on this method by having students implement it in a spreadsheet.\n\nChristiansen said...\n\nThis makes even more sense from an application standpoint if you consider multiplication as finding the area of a rectangle.\n\nThomas Boito said...\n\nYou make an excellent point\n\nMrs. Tenkely said...\n\nWhy, why wasn't I taught this way? Mental math is infinitely easier with this model! Now to teach students with it. Thanks!\n\nEmily Starr, CEO StarrMatica.com said...\n\nThis is a great method for teaching students to multiply that I think many teachers aren't even aware of! (In the US at least) On StarrMatica, we share 3 different ways of approaching multi-digit mutliplication problems--one of the ways is the grid method.\n\nThe BBC has a nice online activity that practices solving problems with this strategy: http://www.bbc.co.uk/skillswise/numbers/wholenumbers/multiplication/written/flash2.shtml\n\nNOTICE TO SPAM COMMENTERS\nAll comments to this blog are reviewed before being published. The chances of you getting a comment including ridiculously obvious \"hidden\" hyperlinks to porn sites or other spam published is virtually zero. So, save your time as well as mine, and take your tawdry business elsewhere."
] | [
null,
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null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9394603,"math_prob":0.9188134,"size":3346,"snap":"2021-31-2021-39","text_gpt3_token_len":737,"char_repetition_ratio":0.14841412,"word_repetition_ratio":0.0034722222,"special_character_ratio":0.21129707,"punctuation_ratio":0.092987806,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97865486,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T13:34:20Z\",\"WARC-Record-ID\":\"<urn:uuid:c61b0ae9-9192-478c-be21-463e86374e71>\",\"Content-Length\":\"90029\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0209582f-7ed3-4a40-9f80-5d58e3201fba>\",\"WARC-Concurrent-To\":\"<urn:uuid:101040c5-ba8e-4833-8624-5666f32f2c43>\",\"WARC-IP-Address\":\"172.217.15.65\",\"WARC-Target-URI\":\"https://cffmv.blogspot.com/2010/04/multiplying-on-grid.html\",\"WARC-Payload-Digest\":\"sha1:2RF2NMF3LYIIEXPV5IEG3T5QXOAVMJX5\",\"WARC-Block-Digest\":\"sha1:7UIW7USZMZPJDAG4HCV3GL6L2YMFLR5I\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060803.2_warc_CC-MAIN-20210928122846-20210928152846-00595.warc.gz\"}"} |
https://mathoverflow.net/questions/359616/continuity-lipschitz-regularity-of-exponential-map-from-c-c-to-operatorname | [
"# Continuity/Lipschitz regularity of exponential map from $C_c$ to $\\operatorname{Diff}_c$?\n\nFor finite-dimensional Lie algebras, see this for a nice example, the exponential map is smooth and in particular, it is locally-Lipschitz onto its image. However, things are different when moving to the infinite-dimensional setting, as discussed for example in the answer to this post.\n\nLet $$G=\\operatorname{Diff}_c(M)$$ the space of compactly supported diffeomorphisms on $$M$$, say $$M$$ is Riemannian manifold diffeomorphic to $$\\mathbb{R}^k$$, and let $$\\mathfrak{g}=C_c^{\\infty}(M,M)$$ be the $$C^{\\infty}$$-vector fields on $$M$$. Again making use of comments in the answer to this post the exponential map taking a vector field $$V$$ in $$\\mathfrak{g}$$ to its flow $$\\Phi^V:x\\to x_1^x$$ where $$t\\to x_t$$ is well-defined by $$\\partial x_t^x = V(x_t^x), \\, x_0^x=x .$$\n\n• Is this map continuous?\n• Moreover, when is it locally-Lipschitz, in the sense that, for every $$\\emptyset \\subset K \\subset M$$ compact there exists some $$L_K>0$$ such that for every $$U,V \\in C^{\\infty}_c(M,M)$$ $$\\sup_{x \\in K} d_M\\left( \\Phi^V(x),\\Phi^U(x) \\right)\\leq L_K \\sup_{x \\in X} d_M\\left( V(x),U(x) \\right)$$ where $$d_M$$ is the metric induced by the Riemannian metric on $$M$$?\n\nThere is no need to be so specific in the question, you can indeed answer this in general for $$M$$ a (paracompact, finite-dimensional) manifold. It is well known, that in this setup $$\\mathrm{Diff}_c(M)$$ is an infinite-dimensional Lie group with Lie algebra $$\\mathfrak{X}_c(M)$$ (this was your space of $$C^\\infty_c(M,M)$$. Before we continue, it needs to be mentioned that there are at least two settings in which the Lie group statement makes sense and is true:\nHow does this relate to your question? Well $$\\exp$$ is convenient smooth according to 1, but we can not use that! Convenient smooth mappings are in general NOT continuous with respect to the original topology. Note that this phenomenon happens only outside of Frechet spaces and some other nice classes of spaces. However, the model space of compactly supported vector fields is not one of the nice spaces, so we can not deduce it from this result. In the Bastiani setting I know several sources wich mention Bastiani smoothness of the exponential map (this is a consequence of regularity of the Lie group, here regularity means that the flow map for curves in the Lie algebra exists and is smooth, in your picture this means that you take the flow map of time dependent vector fields) but only feature proofs in more restricted settings (i.e. $$M$$ compact). To my knowledge a proof of the regularity of $$\\mathrm{Diff}(M)$$ in the Bastiani setting for paracompact $$M$$ appeared for the first time in print in my thesis 3 (as a special case of the orbifold diffeomorphism group discussed there). Regularity of a Lie group (in the Bastiani setting) guarantees smoothness (in Bastiani sense) of the exponential map. Now as Bastiani smooth implies continuity, this answers your first question affirmatively.\nFirst of all the second question seems to be off, or at least contain some non-trivial identification since the vector fields map into the tangent bundle and are still comared using the metric induced by the Riemannian metric on $$M$$. Anyway, a closer analysis of the problem should show that whatever you want is probably a finite-dimensional problem anyway. Therefore it might be misleading to think about your second property as Lipschitzness of the exponential.\nTo see this, note that the source of the differentiability of the exponential map is the so called exponential law (for function spaces, has nothing to do with Lie group exponentials). Here this translates roughly to the insight that you do not need to solve the defining equation on an infinite-dimensional manifold, but can place yourself in a situation where you are solving a differential equation on the underlying manifold $$M$$ (see e.g. 4 for more information on this topic). This is usually highly technical but since your question fixes the compact set, it should be doable (I would have to think more closely about how to do this though)."
] | [
null
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https://earth-planets-space.springeropen.com/articles/10.1186/s40623-021-01421-4 | [
"# The dynamics of dual-magma-chamber system during volcanic eruptions inferred from physical modeling\n\n## Abstract\n\nA magma plumbing system with dual magma chambers beneath active volcanoes is commonly observed through petrological and geophysical measurements. This paper developed a physical model for the dynamics of a dual-magma-chamber system during volcanic eruptions. The model consists of the plumbing system where two elastically deformable magma chambers are connected in series with non-deformable conduits. Based on this model, we obtained an analytical solution that describes temporal changes in pressures at the two chambers accompanied by the eruption. The analytical solution showed that the feature of the chamber pressure changes is mainly controlled by two non-dimensional numbers $$C'$$ and $$\\Omega '$$. Here, $$C'$$ is the ratio of the parameter controlling the magnitude of pressure change in the shallower chamber to that in the deeper chamber, and $$\\Omega '$$ is the ratio of conduit’s conductivity (inverse of resistivity to magma flow) between the shallower chamber and the surface to that between the chambers. For smaller $$C'$$ and $$\\Omega '$$, the shallower chamber’s pressure is kept constant during the decrease in the pressure at the deeper chamber in the initial phase of the eruption. This corresponds to a deformation pattern commonly observed in some eruptions, in which the deflation of the deeper chamber was predominant. The estimation of $$C'$$ and $$\\Omega '$$ based on the parameters related to magma properties and geometries of the chambers and the conduits revealed that the smaller $$C'$$ and $$\\Omega '$$ conditions are satisfied under realistic magmatic and geological parameters. This indicates that the magma dynamics in the dual-chamber system generally cause the dominance of the deeper chamber’s deflation.\n\n## Introduction\n\nA magma plumbing system beneath an active volcano shows a wide variation in size, geometry, and magma properties inside it. This variation causes complex features of geodetic and petrological observations during volcanic eruptions (e.g., Dvorak and Dzurisin 1997; Cashman et al. 2017; Edmonds et al. 2019). To retrieve information about the eruption dynamics from the observations, it is necessary to understand the relationship between the magma plumbing system and observed signals. The simplest magma plumbing system is composed of a single and elastically deformable magma chamber connected to the surface by a volcanic conduit, in which inflation or deflation of the magma chamber is caused by a balance between a magma influx from depth to the chamber and a magma outflux from the chamber to the conduit. In this case, the inflation/deflation of the magma chamber, which can be detected by the geodetic observations of the surface deformation, is straightforwardly linked with the magma activity in the plumbing system (e.g., Mogi 1958; Dzurisin 2003; Segall 2013). When additional effects such as magma lateral migration from the magma chamber and viscoelastic deformation of the chamber are considered, the relationship between the chamber deformation and the magma activity becomes more complex (e.g., Ueda et al. 2005; Segall 2016; Anderson et al. 2019).\n\nOne of the complex magma plumbing systems observed in the volcanoes with silicic eruptions is the system composed of dual magma chambers. In some petrological studies, the dual-chamber system has been proposed to explain the magma mixing process: two types of magma with lower and higher degrees of fractionation (i.e., lower and higher $$\\hbox {SiO}_{2}$$) are stored in the deeper and shallower chambers, respectively, and the magma injection from the deeper to shallower chambers leads to the magma mixing (Sparks et al. 1977). The magma chamber’s depth is deduced from the magma storage pressure that is estimated from thermodynamic constraints for petrological data. In geodetic analyses for volcanic activities such as based on GPS, tiltmeter, strainmeter, and InSAR, multiple deformation sources are sometimes detected to explain the surface deformation patterns (e.g., Kohno et al. 2008; Foroozan et al. 2011; Reverso et al. 2014). In this case, the chamber depth is directly estimated from the depth of the deformation source.\n\nIn the dual-magma-chamber system’s observations, there is a concern that the petrological imaging of the chamber system does not always agree with the geodetic imaging. In the observations during some well-monitored eruptions, there is a common feature that the dual chambers at the shallower and the deeper depths were inferred from the petrological observations, whereas the geodetic observations mainly detected the deformation of the deeper chamber. At Mount Usu, Hokkaido, Japan, two magma chambers with storage pressures (depths) of 200–250 MPa (8–10 km) and 100–150 MPa (4–6 km) were inferred from petrological experiments (Tomiya et al. 2010). On the other hand, geodetic observations during the first few days of the 2000 eruption report a deflation of a deeper magma chamber at a depth of 10 km (Murakami et al. 2001). In the 2011 eruption at Kirishima Shinmoe-dake volcano, Kyushu, Japan, petrological observations revealed that basaltic andesite magma and silicic andesite magma were stored in deeper and shallower magma chambers with pressures (depths) of 250 MPa (10 km) and 125 MPa (5 km), respectively (Suzuki et al. 2013). The erupted magma originates from the mixing of these magmas. Precise geodetic observations based on borehole-type tiltmeter and GPS measurements detected a deflation of a spherical source at a depth of 9.8 km that was accompanied with three sub-Plinian eruptions with the duration of 2–3 h in each event and lava extrusion with two days duration, whereas a shallower source was not detected (Ueda et al. 2013). At Sakurajima volcano, Kyushu, Japan, petrological observations based on melt inclusion data revealed that magma storage depth immediately before the 1914 Plinian eruption is 0.9–3.2 km, and a few data show a storage depth of 4.1 km (Araya et al. 2019). A geodetic deformation source accompanied by recent Vulcanian explosions is estimated to be located at a depth of 4 km beneath the crater (Iguchi et al. 2013), implying a shallower magma chamber. In contrast, a precise leveling survey revealed that a sudden deflation source during the 1914 eruption was deeper, located at a depth of 10 km beneath Aira caldera (Mogi 1958). These observations at three volcanoes show a common feature: a deeper chamber deflation was predominant during the eruption in the dual-magma-chamber system.\n\nThe dual-chamber system dynamics have been investigated in some previous studies, mainly focusing on the magma plumbing system’s long-term evolution. At Soufrière Hills volcano on Montserrat, West Indies, a magma transfer between the dual chambers connected by a vertical conduit has been imaged by inverting magma efflux data and surface deformation data by GPS during eruptive and inter-eruptive periods for about 12 years (1995–2007) (Elsworth et al. 2008; Foroozan et al. 2011). These results also showed that the eruptive episodes deplete the deeper chamber only (Elsworth et al. 2008), whereas whether the shallower chamber’s deflation is apparent or not depends on model assumptions of the plumbing system and the GPS data quality (Foroozan et al. 2011). Melnik and Costa (2014) developed a numerical model for the plumbing system composed of two magma chambers connected by a dyke and a shallower conduit with dyke-to-cylindrical shape by referring to Soufrière Hills volcano. The model showed that connectivity between the chambers significantly influences the dynamics, such as the timing of chamber pressure change and the period of cyclic behavior of discharge rate corresponding to 2–3 years eruption cyclicity. At Grímsvötn Volcano, Iceland, posteruptive displacements measured by dense GPS network after the 1998, 2004, 2011 eruptions were explained by an analytical model that has two connected magma chambers (Reverso et al. 2014). More recently, Bato et al. (2018) applied a sequential data assimilation technique (Ensemble Kalman Filter) to the GPS data after the 2011 eruption based on the two-chamber model and revealed the temporal evolution of the basal magma inflow and a possible interplay between the activities of Grímsvötn volcano and its adjacent volcano (Bárðarbunga).\n\nRecently, the short-term geodetic deformations accompanied with the eruptions were successfully measured by high-resolution observations such as based on tiltmeter and strainmeter (e.g., Bonaccorso et al. 2013; Iguchi et al. 2013; Ueda et al. 2013). As shown in the previous studies for the long-term geodetic data, an appropriate physical model is also necessary to observe the short-term phenomena. In this study, we try to develop a physical model for the dual-magma-chamber system dynamics during the eruption. Using this model, we obtain an analytical solution for temporal changes in pressures at the two chambers accompanied by the eruption, which describes the relationship between a pattern of the chamber pressure changes and magmatic and geological conditions. We discuss whether a deformation pattern in which the pressure decrease in a deeper magma chamber is predominant, commonly observed in some eruptions, can occur under realistic conditions.\n\n## A model for dual-magma-chamber system\n\nWe consider a magma plumbing system in which two elastically deformable magma chambers are connected in series with non-deformable conduits (Fig. 1). A pressure change in each magma chamber is controlled by the balance between magma influx from the deeper conduit and outflux to the shallower conduit. In this case, the temporal changes in the chamber pressures are formulated as\n\n\\begin{aligned} \\dfrac{dp_\\text {1}}{dt} = C_\\text {1} (Q_{\\text {in1}}-Q_{\\text {out1}}), \\end{aligned}\n(1)\n\nand\n\n\\begin{aligned} \\dfrac{dp_\\text {2}}{dt} = C_\\text {2} (Q_\\text {in2}-Q_{\\text {out2}}), \\end{aligned}\n(2)\n\nwhere\n\n\\begin{aligned} C_ {1} = \\dfrac{1}{\\rho _{1} V_{1} (\\kappa _\\text {m1}+\\kappa _{\\text {ch1}})} \\quad {\\text {and}} \\quad C_{2} = \\dfrac{1}{\\rho _{2} V_ {2} (\\kappa _\\text {m2}+\\kappa _{\\text {ch2}})}, \\end{aligned}\n(3)\n\np is the chamber pressure, t is time, $$Q_{\\text {in}}$$ and $$Q_{\\text {out}}$$ are the magma influx and outflux, respectively (in kg s$$^{-1}$$), $$\\rho$$ is the magma density, V is the chamber volume, $$\\kappa _\\text {m}$$, is the magma compressibility in the chamber, and $$\\kappa _{\\text {ch}}$$ is the compressibility of the magma chamber. The number subscripts “1” and “2” correspond to the deeper and shallower magma chambers, respectively (Fig. 1). Here, $$C_\\text {1}$$ and $$C_\\text {2}$$ are the parameters controlling a variability of the chamber pressure: a larger $$C_\\text {1}$$ or $$C_\\text {2}$$ leads to a larger magnitude of the chamber pressure change for given magma influx and outflux.\n\nThe magma flow in the conduit is modeled as a Poiseuille flow. To set the conduit geometry as general as possible, we consider a dyke-like conduit with an elliptical cross section (Costa et al. 2007, Fig. 1). Under the assumption that a temporal change in the magma density in the conduit is negligible, the mass conservation in the conduit between the two chambers is expressed as\n\n\\begin{aligned} \\rho _\\text {12} w_\\text {12} \\pi a_\\text {12} b_\\text {12} = Q_{\\text {out1}} = Q_\\text {in2}, \\end{aligned}\n(4)\n\nwhere w is the ascent velocity, a and b are the semi-major and semi-minor axes of the ellipse, respectively, and the subscript “12” represents the conduit between the two chambers (Fig. 1). In the momentum conservation, we assume a linear pressure gradient and that the inertia term is negligible. In this case, we obtain:\n\n\\begin{aligned} \\dfrac{p_\\text {2}-p_\\text {1}}{L_\\text {12}} = - \\rho _\\text {12} g - \\frac{4 \\eta _\\text {12} w_\\text {12} (a_\\text {12}^2 + b_\\text {12}^2)}{a_\\text {12}^2 b_\\text {12}^2}, \\end{aligned}\n(5)\n\nwhere L is the conduit length, g is the gravity acceleration, and $$\\eta$$ is the magma viscosity. Similarly, the mass and momentum conservations in the conduit between the shallower chamber and the surface are approximated as\n\n\\begin{aligned} \\rho _\\text {2s} w_\\text {2s} \\pi a_\\text {2s} b_\\text {2s} = Q_{\\text {out2}}, \\end{aligned}\n(6)\n\nand\n\n\\begin{aligned} \\dfrac{p_\\text {s}-p_\\text {2}}{L_\\text {2s}} = - \\rho _\\text {2s} g - \\frac{4 \\eta _\\text {2s} w_\\text {2s}(a_\\text {2s}^2 + b_\\text {2s}^2)}{a_\\text {2s}^2 b_\\text {2s}^2}, \\end{aligned}\n(7)\n\nwhere $$p_\\text {s}$$ is the pressure at the surface, and “$$\\text {2s}$$” represents the conduit between the shallower chamber and the surface (Fig. 1). From Eqs. (4)–(7), we obtain\n\n\\begin{aligned} Q_{\\text {out1}} = \\Omega _\\text {12} (p_\\text {1}-p_\\text {2} -\\rho _\\text {12} g L_\\text {12}), \\end{aligned}\n(8)\n\nand\n\n\\begin{aligned} Q_{\\text {out2}} = \\Omega _\\text {2s} (p_\\text {2}-p_\\text {s} -\\rho _\\text {2s} g L_\\text {2s}), \\end{aligned}\n(9)\n\nwhere\n\n\\begin{aligned} \\Omega _\\text {12} = \\dfrac{\\rho _\\text {12} \\pi a_\\text {12}^3 b_\\text {12}^3}{4 \\eta _\\text {12} L_\\text {12}(a_\\text {12}^2 + b_\\text {12}^2)} \\; {\\text {and}} \\; \\Omega _\\text {2s} = \\dfrac{\\rho _\\text {2s} \\pi a_\\text {2s}^3 b_\\text {2s}^3}{4 \\eta _\\text {2s} L_\\text {2s}(a_\\text {2s}^2 + b_\\text {2s}^2)}. \\end{aligned}\n(10)\n\nThe parameters $$\\Omega _\\text {12}$$ and $$\\Omega _\\text {2s}$$ correspond to “conduit conductivity” (Slezin 2003): a larger $$\\Omega _\\text {12}$$ or $$\\Omega _\\text {2s}$$ leads to a higher magma flux for a given pressure gradient in the conduit due to lower viscous resistance. Here, the chamber pressure is expressed using the overpressure ($$\\Delta p$$) with respect to lithostatic pressure as\n\n\\begin{aligned} p_\\text {1} = \\rho _\\text {c} g (L_\\text {12}+L_\\text {2s}) + \\Delta p_\\text {1} \\quad {\\text {and}} \\quad p_\\text {2} = \\rho _\\text {c} g L_\\text {2s} + \\Delta p_\\text {2}, \\end{aligned}\n(11)\n\nwhere $$\\rho _\\text {c}$$ is the crust density ($$\\sim 2500$$ kg m$$^{-3}$$).\n\nFrom Eqs. (1), (2), (8), (9), and (11) and under the assumption that $$p_\\text {s} \\ll p_\\text {2}$$, we obtain\n\n\\begin{aligned} \\dfrac{d \\Delta p_\\text {1}}{dt} = C_\\text {1} \\left[ Q_{\\text {in1}}-\\Omega _\\text {12} \\left\\{ \\Delta p_\\text {1}-\\Delta p_\\text {2} +(\\rho _\\text {c} -\\rho _\\text {12})g L_\\text {12} \\right\\} \\right] , \\end{aligned}\n(12)\n\nand\n\n\\begin{aligned} \\dfrac{d \\Delta p_\\text {2}}{dt} = C_\\text {2} \\left[ \\Omega _\\text {12} \\left\\{ \\Delta p_\\text {1}-\\Delta p_\\text {2} +(\\rho _\\text {c} -\\rho _\\text {12})g L_\\text {12} \\right\\} -\\Omega _\\text {2s} \\left\\{ \\Delta p_\\text {2} +(\\rho _\\text {c} -\\rho _\\text {2s})g L_\\text {2s} \\right\\} \\right] . \\end{aligned}\n(13)\n\nEqs. (12) and (13) are normalized by substituting $$t = \\bar{t} t'$$, $$\\Delta p = \\bar{p} \\Delta p'$$, and $$Q = \\bar{Q} Q'$$, where $$\\bar{t}$$, $$\\bar{p}$$, and $$\\bar{Q}$$ are\n\n\\begin{aligned} \\bar{t} = \\frac{1}{C_\\text {1} \\Omega _\\text {12}}, \\quad \\bar{p} = (\\rho _\\text {c}-\\rho _\\text {12}) g L_\\text {12}, \\quad {\\text {and}} \\quad \\bar{Q} = \\Omega _\\text {12} (\\rho _\\text {c}-\\rho _\\text {12}) g L_\\text {12}. \\end{aligned}\n(14)\n\nAs a result, we get\n\n\\begin{aligned} \\dfrac{d \\Delta p'_\\text {1}}{dt'} = - \\Delta p'_\\text {1} + \\Delta p'_\\text {2} - 1 + Q'_{\\text {in1}} \\end{aligned}\n(15)\n\nand\n\n\\begin{aligned} \\dfrac{d \\Delta p'_\\text {2}}{dt'} = C' \\left[ \\Delta p'_\\text {1}-(\\Omega '+1) \\Delta p'_\\text {2} + 1 - \\Omega ' \\rho ' L' \\right] , \\end{aligned}\n(16)\n\nwhere\n\n\\begin{aligned} \\rho ' = \\dfrac{1-\\dfrac{\\rho _\\text {2s}}{\\rho _\\text {c}}}{1-\\dfrac{\\rho _\\text {12}}{\\rho _\\text {c}}}, \\end{aligned}\n(17)\n\nand\n\n\\begin{aligned} L' = \\frac{L_\\text {2s}}{L_\\text {12}}. \\end{aligned}\n(18)\n\nIt is noted that when $$\\rho _\\text {12}$$ and $$\\rho _\\text {2s}$$ in Eq. (17) are approximated as $$(1-\\phi _\\text {12}) \\rho _\\text {lc}$$ and $$(1-\\phi _\\text {2s}) \\rho _\\text {lc}$$, respectively, where $$\\phi$$ is the magma porosity in the conduit and $$\\rho _\\text {lc}$$ is the liquid-crystals density which is close to $$\\rho _\\text {c}$$, $$\\rho '$$ corresponds to the ratio of the porosity in the shallower conduit to that in the deeper conduit, $$\\phi _\\text {2s}/\\phi _\\text {12}$$. Here, the non-dimensional numbers $$C'$$ and $$\\Omega '$$ are the ratio of $$C_{2}$$ to $$C_{1}$$ and that of $$\\Omega _{2s}$$ to $$\\Omega _{12}$$, respectively, expressed as\n\n\\begin{aligned} C' \\equiv \\frac{C_\\text {2}}{C_\\text {1}} = \\dfrac{\\rho _\\text {1} V_\\text {1}}{\\rho _\\text {2} V_\\text {2}} \\kappa '^{-1}, \\end{aligned}\n(19)\n\nand\n\n\\begin{aligned} \\Omega ' \\equiv \\frac{\\Omega _\\text {2s}}{\\Omega _\\text {12}} = \\dfrac{\\rho _\\text {2s} \\eta _\\text {12}}{\\rho _\\text {12} \\eta _\\text {2s}} \\left( \\dfrac{b_\\text {2s}}{b_\\text {12}} \\right) ^{4} \\dfrac{ r_\\text {2s} (r_\\text {12}^{-2} + 1 ) }{ r_\\text {12} (r_\\text {2s}^{-2} + 1 ) } L'^{-1}, \\end{aligned}\n(20)\n\nwhere $$\\kappa '$$ is the ratio of the sum of the magma and the chamber compressibilities in the shallower chamber to that in the deeper chamber:\n\n\\begin{aligned} \\kappa ' = \\dfrac{\\kappa _\\text {m2}+\\kappa _{\\text {ch2}}}{\\kappa _\\text {m1}+\\kappa _{\\text {ch1}}}, \\end{aligned}\n(21)\n\nand r is the ratio of the semi-major to semi-minor axes of the ellipse in the conduit cross section:\n\n\\begin{aligned} r_\\text {12} = \\dfrac{a_\\text {12}}{b_\\text {12}} \\quad {\\text {and}} \\quad r_\\text {2s} = \\dfrac{a_\\text {2s}}{b_\\text {2s}}. \\end{aligned}\n(22)\n\nFrom the simultaneous differential equations (15) and (16) and under the assumption that $$\\rho '$$, $$L'$$, $$C'$$, $$\\Omega '$$, and $$Q'_{\\text {in1}}$$ are constant during the pressure changes, we obtain an analytical solution for $$\\Delta p'_\\text {1}$$ and $$\\Delta p'_\\text {2}$$ as\n\n\\begin{aligned} \\Delta p'_\\text {1} = A_\\text {+} \\text {e}^{\\lambda _\\text {+}t'} + A_\\text {-} \\text {e}^{\\lambda _\\text {-}t'} -1 -\\rho ' L' +\\dfrac{\\Omega ' + 1}{\\Omega '} Q'_{\\text {in1}}, \\end{aligned}\n(23)\n\nand\n\n\\begin{aligned} \\Delta p'_\\text {2} = (\\lambda _\\text {+}+1) A_\\text {+} \\text {e}^{\\lambda _\\text {+}t'} +(\\lambda _\\text {-}+1) A_\\text {-} \\text {e}^{\\lambda _\\text {-}t'} - \\rho ' L' +\\dfrac{Q'_{\\text {in1}}}{\\Omega '}, \\end{aligned}\n(24)\n\nwhere\n\n\\begin{aligned} \\lambda _{\\pm } = \\frac{1}{2} \\left[ - \\left\\{ C'(\\Omega '+1)+1 \\right\\} \\pm \\sqrt{ \\left\\{ C'(\\Omega '+1)+1 \\right\\} ^2 - 4 C' \\Omega ' } \\right] . \\end{aligned}\n(25)\n\nWhen we use the boundary conditions that $$\\Delta p'_\\text {1}=\\Delta p'_\\text {1,0}$$ and $$\\Delta p'_\\text {2}=\\Delta p'_\\text {2,0}$$ for $$t'=0$$, $$A_\\text {+}$$ and $$A_\\text {-}$$ are expressed as\n\n\\begin{aligned} A_{\\pm } = \\mp \\frac{\\lambda _{\\mp } \\left[ 1 +\\rho ' L' -\\dfrac{\\Omega '+1}{\\Omega '} Q'_{\\text {in1}} + \\Delta p'_\\text {1,0} \\right] + 1 - Q'_{\\text {in1}} + \\Delta p'_{1,0} - \\Delta p'_\\text {2,0}}{\\sqrt{ \\left\\{ C'(\\Omega '+1)+1 \\right\\} ^2 - 4 C' \\Omega ' }}. \\end{aligned}\n(26)\n\n## Results\n\nWe investigate temporal changes in the chamber pressures during an eruption using the analytical solution (Eqs. (23) and (24)). For simplicity, the magma influx to the deeper chamber $$Q_{\\text {in1}}$$ is set to be 0 under the assumption that it is substantially smaller than the magma flux in the conduits during the eruption, and $$\\Delta p_\\text {1,0}'$$ and $$\\Delta p_\\text {2,0}'$$ are set to be 0. We first show the typical results for temporal changes in $$\\Delta p_\\text {1}'$$ and $$\\Delta p_\\text {2}'$$ for $$\\rho ' L' = 3$$ (Fig. 2). This value of $$\\rho ' L'$$ is obtained under realistic conditions in which, for example, $$\\rho _\\text {12}/\\rho _\\text {c}=0.9$$, $$\\rho _\\text {2s}/\\rho _\\text {c}=0.7$$, and $$L_\\text {2s}/L_\\text {12}=1$$. These conditions correspond to the case that the average magma porosities in the deeper and the shallower conduits ($$\\phi _\\text {12}$$ and $$\\phi _\\text {2s}$$) are 0.1 and 0.3, respectively, and the deeper magma chamber is two times deeper than the shallower chamber. The temporal changes in $$\\Delta p_\\text {1}'$$ and $$\\Delta p_\\text {2}'$$ strongly depend on the non-dimensional parameters $$C'$$ and $$\\Omega '$$. When $$C'=\\Omega '=1$$, both $$\\Delta p_\\text {1}'$$ and $$\\Delta p_\\text {2}'$$ decrease simultaneously (Fig. 2e). For larger $$C'$$ and $$\\Omega '$$, the decrease in $$\\Delta p_\\text {2}'$$ is followed by the decrease in $$\\Delta p_\\text {1}'$$ (Fig. 2c). On the other hand, for smaller $$C'$$ and $$\\Omega '$$, the decrease in $$\\Delta p_\\text {1}'$$ is followed by the decrease in $$\\Delta p_\\text {2}'$$ (Fig. 2g). In this case, $$\\Delta p_\\text {2}'$$ is kept constant during the decrease in $$\\Delta p_\\text {1}'$$ in the initial phase. This feature reflects the fact that a larger $$C_\\text {1}$$ (i.e., higher variability of the pressure in the deeper chamber) or a larger $$\\Omega _\\text {12}$$ (i.e., higher conductivity of the deeper conduit) leads to an efficient pressure decrease in the deeper chamber. For larger $$C'$$ and smaller $$\\Omega '$$, $$\\Delta p_\\text {2}'$$ increases during the decrease in $$\\Delta p_\\text {1}'$$ (Fig. 2i). The deformation pattern observed during some eruptions, in which the deeper magma chamber’s deflation was predominant, corresponds to the case of smaller $$C'$$ and $$\\Omega '$$ (Fig. 2g).\n\nThe dependence of the features of the temporal changes in the chamber pressures on the parameters can be evaluated by checking the magnitude of $$\\Delta p_\\text {2}'$$ when $$\\Delta p_\\text {1}'$$ reaches a critical value (Fig. 3). To investigate the features of the pressure changes during the initial phase of the eruption, we set the critical value of $$\\Delta p_\\text {1}'$$ as $$-0.5$$ or $$-1$$. Figure 3 shows the contour map for $$\\Delta p_\\text {2}'$$ at the critical value of $$\\Delta p_\\text {1}'$$ in the parameter space of $$C'$$ and $$\\Omega '$$ for $$\\rho ' L'$$ ranging from 1 to 10. The range of $$\\rho ' L'$$ was set under the assumptions that $$\\phi _\\text {12}=0.1$$, $$\\phi _\\text {2s}=0.2$$–0.5, and $$L'=0.5$$–2. The region of negative $$\\Delta p_\\text {2}'$$ smaller than the critical value of $$\\Delta p_\\text {1}'$$ corresponds to the case in which the decrease in $$\\Delta p_\\text {2}'$$ is followed by the decrease in $$\\Delta p_\\text {1}'$$ (i.e., Fig. 2c). On the other hand, the region of $$\\Delta p_\\text {2}' \\sim 0$$ corresponds to the case in which the decrease in $$\\Delta p_\\text {1}'$$ is followed by the decrease in $$\\Delta p_\\text {2}'$$ (i.e., Fig. 2g). The region of positive $$\\Delta p_\\text {2}'$$ at $$\\Delta p_\\text {1}'=-0.5$$ (Fig. 3a, c, and e) reflects the case in which $$\\Delta p_\\text {2}'$$ increases during the decrease in $$\\Delta p_\\text {1}'$$ (i.e., Fig. 2i). Figure 3 indicates that the typical patterns of the temporal changes in the overpressures shown in Fig. 2 exist for a wide range of $$\\rho ' L'$$. From the quantitative point of view, the boundary of the region depends on $$\\rho ' L'$$: As $$\\rho ' L'$$ increases, the region of negative $$\\Delta p_\\text {2}'$$ expands to the smaller $$\\Omega '$$ range, whereas the region of positive $$\\Delta p_\\text {2}'$$ contracts.\n\nIn the region of smaller $$C'$$ and $$\\Omega '$$, we can specify a condition for $$\\Delta p_\\text {2}'$$ to be close to 0 during the decrease in $$\\Delta p_\\text {1}'$$. As an example, we now specify the condition of the ranges of $$C'$$ and $$\\Omega '$$ for $$\\Delta p_\\text {2}'$$ to be in the range of $$-0.1$$ to 0.1 both at $$\\Delta p_\\text {1}' = -0.5$$ and $$-1$$ (referred to as “small-$$\\Delta p_\\text {2}'$$ condition”). When $$\\rho ' L' =1$$, the small-$$\\Delta p_\\text {2}'$$ condition is satisfied for $$C' < 0.200$$ and $$\\Omega '<0.493$$ (Fig. 3a, b). As $$\\rho ' L'$$ increases, the range of $$C'$$ is unchanged, whereas that of $$\\Omega '$$ decreases: $$\\Omega '<0.048$$ at $$\\rho ' L' =10$$ (Fig. 3e, f). Therefore, under a wide range of $$\\rho ' L'$$, the small-$$\\Delta p_\\text {2}'$$ condition is satisfied when $$C' < 0.200$$ and $$\\Omega '<0.048$$.\n\n## Discussion\n\nWe have specified the ranges of $$C'$$ and $$\\Omega '$$ for the small-$$\\Delta p_\\text {2}'$$ condition in the previous section. This condition is interpreted as the case in which the deflation of the deeper magma chamber is predominant. Here, we discuss whether the small-$$\\Delta p_\\text {2}'$$ condition is satisfied under realistic magma properties and geological conditions. Among the parameters included in $$C'$$ and $$\\Omega '$$ (Eqs. (19) and (20)), $$V_\\text {2}/V_\\text {1}$$, $$L'$$, $$b_\\text {2s}/b_\\text {12}$$, $$r_\\text {2s}$$, and $$r_\\text {12}$$ are related with the geometry of the magma plumbing system such as the chamber volume and the conduit length and depth, and $$\\rho _\\text {2}/\\rho _\\text {1}$$, $$\\rho _\\text {2s}/\\rho _\\text {12}$$, and $$\\eta _\\text {2s}/\\eta _\\text {12}$$ are determined by the magma properties such as the density and the viscosity. The parameter $$\\kappa '$$, which is defined as the ratio of the sum of the magma compressibility $$\\kappa _\\text {m}$$ and the chamber compressibility $$\\kappa _{\\text {ch}}$$ (Eq. (21)), is a complex function of the magmatic and geological parameters: $$\\kappa _\\text {m}$$ is expressed by a function of the magma density $$\\rho$$ as $$(\\partial \\rho /\\partial p)/\\rho$$, and $$\\kappa _{\\text {ch}}$$ is controlled by the chamber shape and the rigidity of the surrounding rocks.\n\nTo obtain the relationship between $$\\kappa '$$ and the magmatic and geological parameters, we first calculate the magma compressibility by formulating the magma density as the density of the gas-liquid-crystals mixture, in which the gas density follows the equation of state for ideal gas and the liquid-crystals density is constant ($$=2500$$ kg $$\\hbox {m}^{-3}$$). The equilibrium volatile exsolution is assumed to follow the $$\\hbox {H}_2\\hbox {O}$$-$$\\hbox {CO}_2$$ solubility model by Newman and Lowenstern (2002). We also assume that the magma temperature is 1000 $$^\\text {o}\\hbox {C}$$, the melt-crystals compressibility is $$10^{-10}$$ $$\\hbox {Pa}^{-1}$$, and the pressure and the crystal content in the deeper and the shallower chambers are 250 MPa and 0 vol.%, and 125 MPa and 30 vol.%, respectively. Under these formulations and assumptions, we can calculate the magma compressibilities in the deeper and the shallower chambers ($$\\kappa _\\text {m1}$$ and $$\\kappa _\\text {m2}$$) as a function of initial $$\\hbox {H}_2\\hbox {O}$$ and $$\\hbox {CO}_2$$ contents (Fig. 4a). Furthermore, the compressibility of the magma chamber is expressed by $$(3 S_\\text {c})/(4\\mu )$$ where $$\\mu$$ is the rigidity of the surrounding rocks set to be 10 GPa, and $$S_\\text {c}$$ is a non-dimensional parameter used for evaluating the effect of the shape change in the magma chamber on the compressibility. The value of $$S_\\text {c}$$ is equal to 1 when the magma chamber is purely spherical, and it increases as the chamber changes to a penny-shaped sill. When the ratio of the longer to shorter axes of the sill is 10, $$S_\\text {c}$$ reaches up to about 10 (Anderson and Segall 2011). We defined $$S_\\text {c}$$ for the deeper and the shallower chambers as $$S_\\text {c1}$$ and $$S_\\text {c2}$$, respectively.\n\nIn the case that both the deeper and the shallower chambers are spherical (i.e., $$S_\\text {c1}=S_\\text {c2}=1$$), the variation of $$\\kappa '$$ ranges from 1 to about 20 (Fig. 4b), which originates from the effect of volatile exsolution on the magma compressibility (Fig. 4a). For higher H$$_2$$O, the volatiles are exsolved and the gas phase is generated, leading a drastic increase in the magma compressibility. On the other hand, for lower H$$_2$$O, the volatiles are not exsolved and the magma compressibility is equal to the melt-crystals compressibility. For given volatile contents, $$\\kappa _\\text {m2}$$ is larger than $$\\kappa _\\text {m1}$$ because lower pressure and higher crystallinity lead to more efficient volatile exsolution in the shallower chamber. Especially, $$\\kappa '$$ drastically increases under the conditions of larger $$\\kappa _\\text {m2}$$ due to volatile exsolution and constant $$\\kappa _\\text {m1}$$ without volatile exsolution (i.e., $$\\kappa _\\text {m1}$$ is equal to the melt-crystals compressibility), as shown in the case of $$\\hbox {H}_2\\hbox {O} = 4$$ wt % and $$\\hbox {CO}_2 = 500$$ ppm (Fig. 4b). Figure 4c shows the effects of the change in the combination of the chamber shape on $$\\kappa '$$, in which $$S_\\text {c1}$$ and/or $$S_\\text {c2}$$ are set as 10 to take into account the penny-shaped chamber. Compared to the case of $$S_\\text {c1}=S_\\text {c2}=1$$, $$\\kappa '$$ is larger for $$S_\\text {c1}=1$$ and $$S_\\text {c2}=10$$, whereas it is smaller for $$S_\\text {c1}=10$$ and $$S_\\text {c2}=1$$. When $$S_\\text {c1}=S_\\text {c2}=10$$, the increase in the magnitude of $$\\kappa _{\\text {ch}}$$ in $$\\kappa '$$ weakens the effect of the contrast between $$\\kappa _\\text {m1}$$ and $$\\kappa _\\text {m2}$$, leading to a smaller $$\\kappa '$$ than the case of $$S_\\text {c1}=S_\\text {c2}=1$$. Nevertheless, $$\\kappa '$$ is larger than 1 except for the case of extremely small H$$_2$$O. When $$S_\\text {c1}=1$$ and $$S_\\text {c2}=10$$, which corresponds to the combination of the deeper spherical chamber and the shallower penny-shaped chamber, $$\\kappa '$$ reaches up to 20.\n\nNow we investigate the assemblage of the magmatic and geological parameters which satisfies the small-$$\\Delta p_\\text {2}'$$ condition (i.e., $$C'<0.2$$ and $$\\Omega ' < 0.048$$). Figure 5a shows the dependence of $$C'$$ on the ratio of the chamber volume ($$V_\\text {2}/V_\\text {1}$$) and $$\\kappa '$$. If the volumes of the two chambers are comparable (i.e., $$V_\\text {2}/V_\\text {1} \\sim 1$$), a larger $$\\kappa '$$ due to the effect of the compressibility contrast between the chambers plays a key role in satisfying the condition of $$C'<0.2$$. Under an extreme condition that $$\\kappa ' = 1$$, in which the two chambers with the same shape are filled with incompressible magma, a larger $$V_\\text {2}/V_\\text {1}$$ up to about 10 is needed to satisfy the condition of $$C'<0.2$$. On the other hand, when $$\\kappa '$$ reaches up to 20 because of larger magma compressibility due to efficient volatile exsolution and larger chamber compressibility due to penny shape in the shallower chamber (Fig. 3b, c), the condition of $$C'<0.2$$ is satisfied even in the case of $$V_\\text {2}/V_\\text {1} < 1$$. Figure 5b shows the dependence of $$\\Omega '$$ on the ratio of the magma viscosity in the conduit ($$\\eta _\\text {2s}/\\eta _\\text {12}$$) and conduit geometry. In the case that the conduit is entirely cylindrical and keeping the same size ($$r_\\text {12} = r_\\text {2s} = 1$$ and $$b_\\text {2s}/b_\\text {12} = 1$$), the condition of $$\\Omega ' < 0.048$$ is satisfied when $$\\eta _\\text {2s}/\\eta _\\text {12}$$ is larger than about 20. Here, we consider a combination of the deeper dyke-like conduit and the shallower cylindrical conduit ($$r_\\text {12} > 1$$ and $$r_\\text {2s} = 1$$) under the constraint that the cross-sectional area of the conduit is constant (i.e., $$a_\\text {12} b_\\text {12} = a_\\text {2s} b_\\text {2s}$$). In this case, a larger $$b_\\text {2s}/b_\\text {12}$$ is set for a larger $$r_\\text {12}$$, which corresponds to a thinner dyke with a larger aspect ratio. Because the thinning of the dyke leads to the decrease in the conductivity of the deeper conduit and the increase in $$\\Omega '$$, the minimum value of $$\\eta _\\text {2s}/\\eta _\\text {12}$$ satisfying the condition of $$\\Omega ' < 0.048$$ is larger than the case of the entirely cylindrical conduit, and it reaches $$10^3$$ for $$r_\\text {12} = 100$$ and $$b_\\text {2s}/b_\\text {12} = 10$$. Generally, the magma viscosity increases during magma ascent because of crystallization and the evolution of melt composition such as the decrease in H$$_2$$O dissolved in the melt. Furthermore, the basaltic magma injection from the deeper to the shallower chambers has been inferred from many petrological studies, implying a lower magma viscosity in the deeper conduit. Therefore, $$\\eta _\\text {2s}/\\eta _\\text {12}$$ is estimated to be much greater than 1, and the condition of $$\\Omega ' < 0.048$$ is easily satisfied. These results indicate that the deformation pattern characterized by the dominance of the deeper chamber’s deflation can occur under realistic magmatic and geological conditions.\n\nFinally, we evaluate the timescale and the magnitude of the volume change of the magma chamber analyzed in this study. The non-dimensional timescale for $$\\Delta p_\\text {1}'$$ to reach $$-1$$ is of the order of 1 (i.e., $$t' \\sim 1$$; Fig. 2). Therefore, the dimensional timescale is approximated as $$\\bar{t} = 1/(C_\\text {1} \\Omega _\\text {12})$$ (Eq. (14)), which is a function of magmatic and geological parameters included in $$C_\\text {1}$$ and $$\\Omega _\\text {12}$$ ($$a_\\text {12}$$, $$b_\\text {12}$$, $$L_\\text {12}$$, $$V_\\text {1}$$, $$\\kappa _\\text {m1}$$, $$\\kappa _{\\text {ch1}}$$, $$\\eta _\\text {12}$$, $$\\rho _\\text {1}$$, and $$\\rho _\\text {12}$$; see Eqs. (3) and (10)). Under the assumptions that $$a_\\text {12} \\sim b_\\text {12} \\sim 10$$ m, $$L_\\text {12} \\sim 5000$$ m, $$V_\\text {1} \\sim 10^8$$$$10^{10}$$ m$$^3$$, $$\\kappa _\\text {m1} \\sim \\kappa _{\\text {ch1}} \\sim 10^{-10}$$ Pa$$^{-1}$$, $$\\eta _\\text {12} \\sim 10^{3}$$$$10^{5}$$ Pa s$$^{-1}$$, and $$\\rho _\\text {1} \\sim \\rho _\\text {12}$$, $$\\bar{t}$$ is estimated as about 10–$$10^5$$ s. From the definition of the chamber compressibility, the volume change of the deeper magma chamber ($$\\delta V_{1}$$) is related with the pressure change ($$\\delta p_\\text {1}$$) as $$\\delta V_{1} = \\kappa _{\\text {ch1}} V_{1} \\delta p_\\text {1}$$. Because the dimensional magnitude of the pressure change in the deeper chamber ($$|\\delta p_\\text {1}|$$) at $$\\Delta p_\\text {1}' = -1$$ is equal to $$|\\bar{p}| = (\\rho _\\text {c}-\\rho _\\text {12}) g L_\\text {12}$$ (Eq. (14)), the magnitude of the volume change ($$|\\delta V_{1}|$$) is expressed as $$\\kappa _{\\text {ch1}} V_{1} (\\rho _\\text {c}-\\rho _\\text {12}) g L_\\text {12}$$. By setting the values of $$\\kappa _{\\text {ch1}}$$, $$V_{1}$$, and $$L_\\text {12}$$ same as those for the estimation of $$\\bar{t}$$ and assuming that $$\\rho _\\text {12} \\sim 2000$$–2250 kg m$$^{-3}$$ (i.e., the porosity is 0.1–0.2), $$|\\delta V_{1}|$$ is estimated as about $$10^5$$$$10^7$$ m$$^3$$. The estimated values of $$\\bar{t}$$ and $$|\\delta V_{1}|$$ cover the duration of the crustal deformation and the magnitude of the change in the chamber volume observed during the 2000 Usu and the 2011 Shinmoe-dake eruptions (Murakami et al. 2001; Ueda et al. 2013). Although we recognize that both $$\\bar{t}$$ and $$|\\delta V_{1}|$$ strongly depend on unconstrained parameters such as the chamber volume, a rough agreement between the analytical results and the observations implies that the analytical solution has the potential to be combined with the observation data during the eruptions.\n\n## Conclusions\n\nWe developed the physical model of the co-eruptive magma dynamics in the plumbing system, where the two elastically deformable magma chambers are connected in series with the conduits. Based on the physical model, we obtained the analytical solution for the temporal changes in the two chambers’ pressures accompanied by the eruption. The solution showed that the feature of the pressure changes is mainly controlled by the non-dimensional numbers $$C'$$ and $$\\Omega '$$: $$C'$$ is the ratio of the parameter controlling the magnitude of the pressure change in the shallower chamber to that in the deeper chamber, and $$\\Omega '$$ is the ratio of the conduit’s conductivity between the shallower chamber and the surface to that between the chambers. We found that for smaller $$C'$$ and $$\\Omega '$$, the shallower chamber’s pressure is kept constant during the decrease in the pressure at the deeper chamber in the initial phase of the eruption. This corresponds to the deformation pattern commonly observed in some eruptions, in which the deflation of the deeper magma chamber was predominant. The estimation of $$C'$$ and $$\\Omega '$$ based on the parameters related to the magma properties and the geometries of the chambers and the conduits revealed that the smaller $$C'$$ and $$\\Omega '$$ conditions are satisfied under realistic magmatic and geological parameters. This indicates that the magma dynamics in the dual-chamber system generally cause the dominance of the deeper chamber’s deflation. In the dual-chamber system modeling, additional effects should be considered in future studies, such as elastic deformation of conduits, inclined conduits, magma temperature change, and viscoelastic deformation of magma chambers. Nevertheless, because the analytical solution obtained in this study can reproduce various chamber pressure changes for wide ranges of the parameters, it can be used for sophisticated geodetic data analyses such as based on data assimilation techniques and Bayesian inversions.\n\nNot applicable.\n\n## Abbreviations\n\nGPS:\n\nGlobal positioning system\n\nInSAR:\n\n## References\n\n1. Anderson K, Segall P (2011) Physics-based models of ground deformation and extrusion rate at effusively erupting volcanoes. J Geophys Res 116:B07204. https://doi.org/10.1029/2010JB007939\n\n2. Anderson KR, Johanson IA, Patrick MR, Gu M, Segall P, Poland MP, Montgomery-Brown EK, Miklius A (2019) Magma reservoir failure and the onset of caldera collapse at Kīlauea Volcano in 2018. Science. https://doi.org/10.1126/science.aaz1822\n\n3. Araya N, Nakamura M, Yasuda A, Okumura S, Sato T, Iguchi M, Miki D, Geshi N (2019) Shallow magma pre-charge during repeated Plinian eruptions at Sakurajima volcano. Sci Rep 9(1):1–10. https://doi.org/10.1038/s41598-019-38494-x\n\n4. Bato MG, Pinel V, Yan Y, Jouanne F, Vandemeulebrouck J (2018) Possible deep connection between volcanic systems evidenced by sequential assimilation of geodetic data. Sci Rep 8(1):1–13. https://doi.org/10.1038/s41598-018-29811-x\n\n5. Bonaccorso A, Currenti G, Linde A, Sacks S (2013) New data from borehole strainmeters to infer lava fountain sources (Etna 2011–2012). Geophys Res Lett 40(14):3579–3584. https://doi.org/10.1002/grl.50692\n\n6. Cashman KV, Sparks RSJ, Blundy JD (2017) Vertically extensive and unstable magmatic systems: a unified view of igneous processes. Science. https://doi.org/10.1126/science.aag3055\n\n7. Costa A, Melnik O, Sparks RSJ (2007) Controls of conduit geometry and wallrock elasticity on lava dome eruptions. Earth Planet Sci Lett 260(1–2):137–151. https://doi.org/10.1016/j.epsl.2007.05.024\n\n8. Dvorak JJ, Dzurisin D (1997) Volcano geodesy: The search for magma reservoirs and the formation of eruptive vents. Rev Geophys 35(3):343–384. https://doi.org/10.1029/97RG00070\n\n9. Dzurisin D (2003) A comprehensive approach to monitoring volcano deformation as a window on the eruption cycle. Rev Geophys. https://doi.org/10.1029/2001RG000107\n\n10. Edmonds M, Cashman KV, Holness M, Jackson M (2019) Architecture and dynamics of magma reservoirs. Philos Trans R Soc A 377:20180298. https://doi.org/10.1098/rsta.2018.0298\n\n11. Elsworth D, Mattioli G, Taron J, Voight B, Herd R (2008) Implications of magma transfer between multiple reservoirs on eruption cycling. Science 322(5899):246–248. https://doi.org/10.1126/science.1161297\n\n12. Foroozan R, Elsworth D, Voight B, Mattioli GS (2011) Magmatic-metering controls the stopping and restarting of eruptions. Geophys Res Lett. https://doi.org/10.1029/2010GL046591\n\n13. Iguchi M, Tameguri T, Ohta Y, Ueki S, Nakao S (2013) Characteristics of volcanic activity at Sakurajima volcano’s Showa crater during the period 2006 to 2011. Bull Volcanol Soc Japan 58:115–135\n\n14. Kohno Y, Matsushima T, Shimizu H (2008) Pressure sources beneath unzen volcano inferred from leveling and GPS data. J Volcanol Geotherm Res 175(1–2):100–109. https://doi.org/10.1016/j.jvolgeores.2008.03.022\n\n15. Melnik O, Costa A (2014) Dual-chamber-conduit models of non-linear dynamics behaviour at Soufrière Hills Volcano, Montserrat. In: Wadge G, Robertson R, Voight B (eds) The Eruption of Soufrière Hills Volcano, Montserrat, from 2000 to 2010, Geological Society, London, Memoirs 39, pp 61–69. https://doi.org/10.1144/M39.3\n\n16. Mogi K (1958) Relations between the eruptions of various volcanoes and the deformations of the ground surfaces around them. Bull Earthq Res Inst 36:99–134\n\n17. Murakami M, Ozawa S, Nishimura T, Tada T (2001) A model of magma movements associated with the 2000 eruption of Usu volcano inferred by crustal deformation detected by continuous GPS and other geodetic measurements. Bull GSI 95:99–105 (in Japanese)\n\n18. Newman S, Lowenstern JB (2002) Volatilecalc: a silicate melt-H$$_{2}$$O-CO$$_{2}$$ solution model written in visual basic for excel. Comput Geosci 28(5):597–604. https://doi.org/10.1016/S0098-3004(01)00081-4\n\n19. Reverso T, Vandemeulebrouck J, Jouanne F, Pinel V, Villemin T, Sturkell E, Bascou P (2014) A two-magma chamber model as a source of deformation at Grímsvötn Volcano. J Geophys Res 119(6):4666–4683. https://doi.org/10.1002/2013JB010569\n\n20. Segall P (2013) Volcano deformation and eruption forecasting. In: Pyle DM, Mather TA, Biggs J (eds) Remote sensing of volcanoes and volcanic processes: integrating observation and modelling, Geological Society, London, Memoirs 380, pp 85–106. https://doi.org/10.1144/SP380.4\n\n21. Segall P (2016) Repressurization following eruption from a magma chamber with a viscoelastic aureole. J Geophys Res 121(12):8501–8522. https://doi.org/10.1002/2016JB013597\n\n22. Slezin YB (2003) The mechanism of volcanic eruptions (a steady state approach). J Volcanol Geotherm Res 122(1–2):7–50. https://doi.org/10.1016/S0377-0273(02)00464-X\n\n23. Sparks SRJ, Sigurdsson H, Wilson L (1977) Magma mixing: a mechanism for triggering acid explosive eruptions. Nature 267(5609):315–318\n\n24. Suzuki Y, Yasuda A, Hokanishi N, Kaneko T, Nakada S, Fujii T (2013) Syneruptive deep magma transfer and shallow magma remobilization during the 2011 eruption of Shinmoe-dake, Japan-Constraints from melt inclusions and phase equilibria experiments. J Volcanol Geotherm Res 257:184–204. https://doi.org/10.1016/j.jvolgeores.2013.03.017\n\n25. Tomiya A, Takahashi E, Furukawa N, Suzuki T (2010) Depth and evolution of a silicic magma chamber: melting experiments on a low-k rhyolite from usu volcano, japan. J Petrol 51(6):1333–1354. https://doi.org/10.1093/petrology/egq021\n\n26. Ueda H, Fujita E, Ukawa M, Yamamoto E, Irwan M, Kimata F (2005) Magma intrusion and discharge process at the initial stage of the 2000 activity of Miyakejima, Central Japan, inferred from tilt and GPS data. Geophys J Int 161(3):891–906. https://doi.org/10.1111/j.1365-246X.2005.02602.x\n\n27. Ueda H, Kozono T, Fujita E, Kohno Y, Nagai M, Miyagi Y, Tanada T (2013) Crustal deformation associated with the 2011 Shinmoe-dake eruption as observed by tiltmeters and GPS. Earth Planets Space 65(6):517–525. https://doi.org/10.5047/eps.2013.05.001\n\n## Acknowledgements\n\nThe author thanks two anonymous reviewers for insightful comments and suggestions that significantly improved the manuscript.\n\n## Funding\n\nThis work was supported by JSPS KAKENHI Grant Numbers 19K04029, 18H01296, and 16H06348, the Ministry of Education, Culture, Sports, Science and Technology (MEXT) of Japan through its “Integrated Program for Next Generation Volcano Research and Human Resource Development” (Grant Number JPJ005391) and “Earthquake and Volcano Hazards Observation and Research Program”, and ERI JURP 2020-B-07.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nTK developed the physical model, conducted the simulations, and wrote the manuscript. The author read and approved the manuscript.\n\n### Corresponding author\n\nCorrespondence to Tomofumi Kozono.\n\n## Ethics declarations\n\n### Competing interests\n\nThe author declares that there are no competing interests.",
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"https://earth-planets-space.springeropen.com/track/article/10.1186/s40623-021-01421-4",
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https://encyclopediaofmath.org/wiki/Nash_theorems_(in_differential_geometry) | [
"# Nash theorems (in differential geometry)\n\n1. Nash’s theorem on $C^{1}$-imbeddings and $C^{1}$-immersions. A $C^{1}$-immersion (-imbedding) $f: V^{n} \\to E^{m}$ of an $n$-dimensional Riemannian space $V^{n}$ with a $C^{0}$-metric $g$ into an $m$-dimensional Euclidean space $E^{m}$ is called short if and only if the metric $g_{f}$ induced by it on $V^{n}$ is such that the quadratic form $g - g_{f}$ is positive definite. If $V^{n}$ has a short immersion (imbedding) into $E^{m}$, where $m \\geq n + 1$, then $V^{n}$ also has an isometric $C^{1}$-immersion (-imbedding) into $E^{m}$. Under the restriction $m \\geq n + 2$, this theorem was proved in , and in the form stated above in . This theorem implies, in particular, that if a compact Riemannian manifold $V^{n}$ has a $C^{1}$-imbedding (-immersion) into $E^{m}$, where $m \\geq n + 1$, then $V^{n}$ also has an isometric $C^{1}$-imbedding (-immersion) into $E^{m}$. Another consequence of Nash’s theorem is that every point of $V^{n}$ has a sufficiently small neighborhood that admits an isometric $C^{1}$-imbedding into $E^{n + 1}$.\n2. Nash’s theorem on regular imbeddings. Every compact Riemannian manifold $V^{n}$ of class $C^{r}$, where $3 \\leq r \\leq \\infty$, has an isometric $C^{r}$-imbedding into $E^{m}$, where $m = \\dfrac{3 n^{2} + 11 n}{2}$. If $V^{n}$ is not compact, then it has an isometric $C^{r}$-imbedding into $E^{m_{1}}$, where $m_{1} = \\dfrac{(3 n^{2} + 11) (n + 1)}{2}$.\nNash’s theorem on regular imbeddings results from an application of Nash’s implicit-function theorem on the inversion of a broad class of differential operators. The meaning of this theorem is that when a certain linear algebraic system of equations connected naturally with a differential operator $L$ is solvable, and when a reasonable topology is introduced in the image and inverse image, then the operator in question is an open mapping, i.e., $L$ is locally invertible near any point of its range. For the equations of an imbedding of a Riemannian manifold into a Euclidean space, this reduces to the fact that the first and second derivatives of the mapping $f: V^{n} \\to E^{m}$ with respect to the intrinsic coordinates of $V^{n}$ must be linearly independent. Such imbeddings were first considered in ; they are called free. Nash’s implicit-function theorem implies that a compact Riemannian manifold $V^{n}$ that is sufficiently close to another one $W^{n}$ having a free imbedding into $E^{m}$ also has a free imbedding into $E^{m}$. This fact, and the original method of extension with respect to a parameter, lead to Nash’s theorem on regular imbeddings (see ). By extending Nash’s method to non-compact manifolds and analytic imbeddings, and also by a principal refinement of the process of extension with respect to a parameter, it has been proved that every infinitely differentiable (analytic) Riemannian manifold $V^{n}$ has an isometric differentiable (analytic) imbedding into $E^{m}$, where $m = \\dfrac{n (n + 1)}{2} + 3 n + 5$ (see )."
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http://scholarsmine.mst.edu/math_stat_facwork/316/ | [
"",
null,
"## Mathematics and Statistics Faculty Research & Creative Works\n\n#### Title\n\nThe Beverton-Holt Dynamic Equation\n\n#### Abstract\n\nThe Cushing-Henson conjectures on time scales are presented and verified. The central part of these conjectures asserts that based on a model using the dynamic Beverton-Holt equation, a periodic environment is deleterious for the population. The proof technique is as follows. First, the Beverton-Holt equation is identified as a logistic dynamic equation. The usual substitution transforms this equation into a linear equation. Then the proof is completed using a recently established dynamic version of the generalized Jensen inequality.\n\n#### Department(s)\n\nMathematics and Statistics\n\n#### Keywords and Phrases\n\nComplex Variables; Computational Numerical Analysis; Differential Equations; Fourier Analysis; Functional Analysis; Integral Transforms & Equations; Mathematical Analysis; Mathematical Numerical Analysis; Operator Theory; Real Functions; Sequences & Series; Special Functions\n\n0003-6811\n\n#### Document Type\n\nArticle - Journal\n\nCitation\n\ntext\n\nEnglish"
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http://forums.wolfram.com/mathgroup/archive/2008/Apr/msg00416.html | [
"",
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"Re: Just primitive ColorFunction\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg87521] Re: Just primitive ColorFunction\n• From: ucervan at gmail.com\n• Date: Sat, 12 Apr 2008 06:57:58 -0400 (EDT)\n• References: <ftfej7\\$bu7\\$1@smc.vnet.net> <ftfk8g\\$fab\\$1@smc.vnet.net>\n\n```> 1. Is there a more appropriate way to force Plot to calculate the\n> function value at certain points?\n\nHow about (not very clean, but does the work if you really need to use\nColorFunction for this to make it work with Filling):\n\nep = 0.0001;\nPlot[Sin[x], {x, 0, 4 Pi}, PlotStyle -> Thick,\nColorFunction -> (If[Sin[#] >= 0, RGBColor[1, 0, 0],\nRGBColor[0, 0, 1]] &), ColorFunctionScaling -> False,\nMesh -> {{Pi, Pi + ep, 2 Pi, 2 Pi + ep, 3 Pi, 3 Pi + ep}},\nMeshStyle -> None]\n\nand if you still need to add mesh points in the x direction, use:\n\nep = 0.0001;\nPlot[Sin[x], {x, 0, 4 Pi}, PlotStyle -> Thick,\nColorFunction -> (If[Sin[#] >= 0, RGBColor[1, 0, 0],\nRGBColor[0, 0, 1]] &), ColorFunctionScaling -> False,\nMesh -> {{Pi, Pi + ep, 2 Pi, 2 Pi + ep, 3 Pi, 3 Pi + ep}, 10},\nMeshStyle -> {None, Black}, MeshFunctions -> {#1 &, #1 &}]\n\n>\n> 2. Is there a way to avoid having to find the zeros of the function\n> manually? (More generally: avoid having to calculate the points where\n> the colouring changes abruptly.)\n\nLook at these two cases:\n\n1)\n\nPlot[x, {x, 0, 4 Pi}, PlotStyle -> Thick,\nColorFunction -> (If[Mod[IntegerPart[30 #], 2] == 0,\nRGBColor[1, 0, 0], RGBColor[0, 0, 1]] &)]\n\n2)\n\nPlot[x, {x, 0, 4 Pi}, PlotStyle -> Thick, Mesh -> 20,\n\nSubdividing 1) by color will succeed only if the initial sampling gets\nthe colored regions right. Then a color based \"find root\" will need to\nbe computed for every single \"color jump\". We will also need to define\nwhat a \"jump\" or color based Exclusion is.\n\n2) is the way of dealing with subdividing curves and surfaces via the\n\nAn Automatic intersection method can be attained in many cases (only\nif the mesh functions evaluate to +/- values, so no min/max tangential\nintersections) with something like:\n\nPlot[Sin[x], {x, 0, 4 Pi}, PlotStyle -> Thick, Mesh -> {{0.}},\nMeshShading -> {Red, Blue}, MeshFunctions -> {(Sin[#]) &}]\n\nPlot[{x, Sin[x] + x}, {x, 0, 4 Pi}, PlotStyle -> Thick,\nMesh -> {{0.}}, MeshShading -> {Red, Blue},\nMeshFunctions -> {(Sin[#]) &}]\n\nPlot[{Cos[x] + x, Sin[x] + x}, {x, 0, 4 Pi}, PlotStyle -> Thick,\nMesh -> {{0.}}, MeshShading -> {Red, Blue},\nMeshFunctions -> {(Sin[#] - Cos[#]) &}]\n\nFinally, using all ColorFunction/Mesh/MeshFunctions/Filling options\ntogether:\n\nPlot[{x, Sin[x] + x}, {x, 0, 4 Pi}, PlotStyle -> Thick,\nMesh -> {{0.}}, MeshFunctions -> {(Sin[#]) &},\nColorFunction -> (If[Sin[#] >= 0, RGBColor[1, 0, 0],\nRGBColor[0, 0, 1]] &), ColorFunctionScaling -> False,\nFilling -> {1 -> {2}}]\n\n-Ulises Cervantes\nWRI\n\n```\n\n• Prev by Date: Re: Problem for using \"Epilog\" to plot legend\n• Next by Date: Re: Documentation - what is the big secret?\n• Previous by thread: Re: Just primitive ColorFunction\n• Next by thread: basic questions"
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https://www.brilliantmaths.com/lessons/lesson-1-identifying-odd-and-even-numbers/ | [
"# Lesson 1 – Identifying Odd and Even Numbers\n\nOBJECTIVE:\n\nAt the end of this lesson, students should be able to identify odd and even as well as work with them comfortably.\n\nODD NUMBERS\n\nWhen a number is divided by 2 and it gives a remainder of 1, the number is said to be an odd number.\n\nOdd numbers are not seen in the two times table. They end in 1, 3, 5, 7 or 9.\n\nExamples are 1, 3, 5, 7, 9, 11, 13, 15, 45, 127, 2369, 33033, 101 . . .\n\nEVEN NUMBERS\n\nWhen an even number is divided by 2, there is no remainder. Even numbers are numbers in the two times table. They end in 2, 4, 6, 8 or 0.\n\nExamples are 2, 4, 6, 8, 10, 12, 14, 16, 46, 932, 10258, 5620 . . .\n\nLesson Content\nerror: Content is protected !!"
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https://nrich.maths.org/2095/solution | [
"#### You may also like",
null,
"### Giant Holly Leaf\n\nFind the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr).",
null,
"Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the area enclosed by PQRS.",
null,
"### Get Cross\n\nA white cross is placed symmetrically in a red disc with the central square of side length sqrt 2 and the arms of the cross of length 1 unit. What is the area of the disc still showing?\n\n# Triangles and Petals\n\n##### Age 14 to 16Challenge Level\n\nHerschel from the European School of Varese sent us this solution:\n\nThe first flower has 3 petals corresponding to the 3 corners of the triangle. The completed animation shows that each petal is a semicircle, so the perimeter of the flower is $3\\times \\pi \\times \\text{ radius } = 3 \\times \\pi \\times \\text{ (side of triangle) }= 3\\pi r$.\n\nThe second flower has 4 petals. This time, each petal is a sector of a circle rather than a simple semicircle. The angle of this sector is 360 - (2 triangle corners) - (1 square corner) = $360 - 2 \\times 60 - 90 = 150^\\circ$.\nTherefore, the total perimeter of this petal is $4 \\times \\frac{150}{360}\\times (2\\times \\pi \\times \\text{ radius }) =\\frac{10}{3} \\times \\pi \\times \\text{ (side of the square) }= \\frac{10}{3} \\pi r$.\n\nIn general, we need to know 3 key bits of data to work out the perimeter of the flower.\nThey are:\n• The number of sides of the central shape; we'll call this $n$.\n• The length of each side in the central shape; we'll call this $r$. (Note that this is equal to the radius of the petals).\n• The angle at the centre of each petal. This can be derived from $n$:\n$\\text{Angle } = 360 - 2 \\times 60 - \\text{( Corner of shape)}$\n$\\text{Angle } = 360 - 120 - \\frac{180(n-2)}{n}$\n$\\text{Angle } = 240 - 180 - \\frac{360}{n}$\n$\\text{Angle } = 60 + \\frac{360}{n}$\nGiven these data, we can proceed to work out a general formula:\n\nPerimeter= (number of petals) $\\times$ (perimeter of a full circle) $\\times \\frac{\\text{angle at centre of petal}}{360}$\n$\\text{Perimeter }= n \\times 2 \\times \\pi \\times r \\times \\frac{(60+ \\frac{360}{n})}{360}$\n$\\text{Perimeter }= 2 \\times \\pi \\times n \\times r \\times (\\frac{1}{6}+\\frac{1}{n})$\n$\\text{Perimeter }= 2 \\times \\pi \\times n \\times r \\times \\frac{6+n}{6n}$\n$\\text{Perimeter }= \\pi \\times r \\times \\frac{6+n}{3}$\n\nUsing this formula, we find the following results:\n$n=3$ (Triangle): Perimeter = $\\pi \\times r \\times \\frac{9}{3} = 3 \\pi r$\n$n=4$ (Square): Perimeter = $\\pi \\times r \\times \\frac{10}{3}$\n$n=5$ (Pentagon): Perimeter = $\\pi \\times r \\times \\frac{11}{3}$\n$n=6$ (Hexagon): Perimeter = $\\pi \\times r \\times \\frac{12}{3} = 4\\pi r$\n$n=7$ (Heptagon): Perimeter = $\\pi \\times r \\times \\frac{13}{3}$\n$n=8$ (Octagon): Perimeter = $\\pi \\times r \\times \\frac{14}{3}$\n...\n$n=100$: Perimeter = $\\pi \\times r \\times \\frac{106}{3}$\n\nSo a shape with 100 sides will produce a flower with a perimeter of $\\pi \\times r \\times \\frac{106}{3}$.\nIf each edge of the central shape has a length of 1, the perimeter of the flower will be $35.333 \\times \\pi$, which is 111.00 to two decimal places.\n\n• Well done to Saif from Havering Sixth Form College, Nina, Jure and Kristjan from Elementary school Loka Crnomelj, Slovenia, Chi from Raynes Park, Rajeev from Haberdashers' Aske's Boys' School, Yun Seok Kang, and Cameron, who also sent in correct solutions. Click here to read Nina, Jure and Kristjan's thoughts.\n\n•\n•\n•"
] | [
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"https://nrich.maths.org/content/99/12/15plus1/icon.png",
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"https://nrich.maths.org/content/01/07/15plus4/icon.gif",
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"https://nrich.maths.org/content/01/12/15plus4/icon.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6990214,"math_prob":0.99988985,"size":2896,"snap":"2022-05-2022-21","text_gpt3_token_len":950,"char_repetition_ratio":0.2257953,"word_repetition_ratio":0.08179959,"special_character_ratio":0.3646409,"punctuation_ratio":0.089788735,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999542,"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\":\"2022-05-25T15:41:23Z\",\"WARC-Record-ID\":\"<urn:uuid:1f4003cb-2b7d-4436-9aa1-e663532333a0>\",\"Content-Length\":\"16243\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:309d9cd9-0cc2-4cfe-829f-6ffe8f1a4375>\",\"WARC-Concurrent-To\":\"<urn:uuid:cd687fcb-fdd0-4746-8eb9-f5b9313acf53>\",\"WARC-IP-Address\":\"131.111.18.195\",\"WARC-Target-URI\":\"https://nrich.maths.org/2095/solution\",\"WARC-Payload-Digest\":\"sha1:7ZUGJMSJPMUUOIFGPFZUGIXOF7FEXCAL\",\"WARC-Block-Digest\":\"sha1:WVYSG5D5XJIDKSB63WQRBSKSL4IROBCC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662588661.65_warc_CC-MAIN-20220525151311-20220525181311-00140.warc.gz\"}"} |
https://www.otomas.co.id/chili-oil-tevm/page.php?page=f5ddee-latex-multiple-equations | [
"Align In Latex Equation Editor Doesn T Work Properly. In LaTeX, amsmath package facilitates many useful features for displaying and representing equations. LaTeX is obviously pretty good at typesetting mathsâit was one of the chief aims of the core TeX system that LaTeX extends. Latex allows two writing modes for mathematical expressions. To create a 3x3 matrix equation in the LaTeX format, type the following into a math zone: A=\\{\\matrix{a&b&c\\\\d&e&f\\\\g&h&j}\\} This will build into the following professional equation: Here are some other examples of LaTeX expressions that can be built-up into a professional format. Multiline Formulae In Latex . IEEEtrantools.sty [2015/08/26 V1.5 by Michael Shell]: package needed for the That can be achieve in plain LaTeX without any specific package. LaTeX.org. Using the multiline, aligned packages. The % sign tells $\\mathrm{\\LaTeX}$ to ignore the rest of the current line. LaTeX equation editing supports most of the common LaTeX mathematical keywords. Thanks for contributing an answer to TeX - LaTeX Stack Exchange! Logout. Most LaTeX ⦠Can I write a LaTeX equation over multiple lines? Explaining Equation Steps Multiple Line Text Tex Latex. Systems of linear equations involving more than two variables work similarly, having either one solution, no solutions or infinite solutions (the latter in the case that all component equations are equivalent). Latex Basic Code. LaTeX Features . LaTeX forum â Math & Science â Multi-line equation alignment. Related articles. Alirezakn Posts: 3 Joined: Sat Nov 17, 2018 6:23 am. The align and align* environments, available through the amsmath package, are used for arranging equations of multiple lines. I used Libreoffice 4.4.3.2 in Linux Mint 16.2 and the TexMaths ( ⦠Grouping and centering equations. Post by Alirezakn » Thu Nov 22, 2018 7:02 am . share | improve this question | follow | asked Apr 8 '10 at 12:36. Also that each equation is separated from the one before by an. This tutorial talks about the usage of multiple columns in LaTeX. This text should show what a printed text will look like at this place. Putting multiple \\label s inside it will result in errors â Martijn Feb 6 '16 at 10:50 Latex reset equation numbering. 3 Special Characterulti Line Equations. Add Mathematical Equations In Keynote On Mac Apple Support Again, use * to toggle the equation numbering. How to Typeset Equations in LATEX 3 typeset_equations.tex: LATEX source le of this manual. â
Convert multiple equations at the same time, using any function supported by LaTeX! Open an example of the amsmath package in Overleaf. Information and discussion about LaTeX's math and science related features (e.g. In large equations or derivations which span multiple lines, we can use the \\begin{align} and \\end{align} commands to correctly display the aligned mathematics. Equations In Multiple Lines Tex Latex Stack Exchange. This is the 17th video in a series of 21 by Dr Vincent Knight of Cardiff University. L a T e X allows two writing modes for mathematical expressions: the inline mode and the display mode. How to align the beginning and = sign of equations of different length. Fractions in Equations . so it didn't seem to make much sense to them when it was suggested that they might like to use O365 tools with no equation support. The Moodle XML file can then be used to bulk import multiple questions into a question bank in cuLearn âFind out more about Moodle XML file formats Indexes and bibliographies are generated automatically. Sometimes a long equation needs to be broken over multiple lines, especially if using a double column export style. 4 posts ⢠Page 1 of 1. Overview of basic math features, with live-rendering and sandbox in your browser. for example: \\begin{align} y_1(x)=&x^2\\\\ y_2(x)=&2x+1 \\end{align} Thanks in advance . What is LaTeX . The {and } characters are used to surround multiple characters that $\\mathrm{\\LaTeX}$ should treat as a single character. Hello, here is some text without a meaning. Art Of Problem Solving. Work well with technical reports, journal articles, slide presentations, and books. This is best used at the start of a line to mark that line as a âcommentâ. When numbering is allowed, you can label each row individually. \\documentclass{article} \\begin{document} This is your only binary choices \\begin{math} \\left\\{ \\begin{array}{l} 0\\\\ 1 \\end{array} \\right. However, it can't always be relied upon to accurately interpret formulas in the way you did. \\end{math} \\end{document} This code produces something which looks what you seems to need. Have Metafont or PostScript fonts. First of all, you probably don't want the align environment if you have only one column of equations. 73. dot_emacs: commands to be included in the preference le of Emacs (.emacs) (see Section7). Logout Latex Support In Pages Macs Chemistry. \\begin{equation*} \\systeme{ x+y+z = 1, x+y+z = \\frac{5}{2}, x+y+z = 5 } \\end{equation*} or \\begin{equation*} \\systeme{ 3x +7z = 20, y - 17z = -3, 24x + 15y = 7 } \\end{equation*} which may or may not suit your taste. LaTeX is good for complex documents. Recently I plan to transplant one of my old post from other site to this new site (you can find the new post here). The first ⦠The result tends to be slightly incorrect horizontal spacing. Open an example in Overleaf. latex equation alignment. Information and discussion about LaTeX's math and science related features (e.g. As with matrices and tables, \\\\ specifies a line break, and & is used to indicate the point at which the lines should be aligned. When numbering is allowed, you can label each row individually. The old post contains quite a lot of equations in LaTeX format. This command forces LaTeX to give an equation the full height it needs to display as if it were on its own line. The right-hand side of the equation is an array with an opening brace sized to fit on the left. This environment takes one argument, the number of âequation columnsâ: count the maximum number of &s in any row, add 1 and divide by 2. The bracket can removed by specifying empty delimiters by preceding the \\systeme command with \\sysdelim.. (. Mine ponds amplify mercury risks in Peru's Amazon; Melting ice patch in Norway reveals large collection of ancient arrows ; Comet 2019 ⦠With so much equations in one post, I find it really hard to refer to some equations without using proper numbering. â
Always render in the highest quality! The first ⦠says to make the matching container on the right side empty. With the Moodle LaTex package, one can generate a Moodle XML file containing quiz questions from a document authored using LaTex. 1. Cham ... You can choose the layout that better suits your document, even if the equations are really long, or if you have to include several equations in the same line. If the paper is very short, or if there are only a few numbered equations, this is fine, but once the numbers get into the twenties and higher, a scheme that numbers equations by section, as in (1.1), (1.2), ..., (2.1), etc., is preferable. multiple alignment in an equation. The align environment is used for two or more equations when vertical alignment is desired; usually binary relations such as equal signs are aligned. Go to website. Recently I had to do a homework assignment using linear regression in OLS equations and LaTex. How Can I Split An Equation Over Two Or More Lines Tex Latex. Skip to content. Braces are special characters and so the opening brace needs to be escaped with a backslash. Multiple lines of equations on either side of left brace with alignment. If you want to create a document with more than two columns, use the package multicol, which has a set of commands for the same. LaTeX assumes that each equation consists of two parts separated by a &; also that each equation is separated from the one before by an &. Multiple Equations. LaTeX requires a right for every left but the dot in right. Be careful in using it as it can make a document due to variable line height. Multi-line equation alignment. Adding right brace and equation number. \\documentclass {article} \\usepackage [utf8] {inputenc} \\usepackage [english] {babel} \\usepackage {multicol} \\begin {document} \\begin {multicols}{3} [ \\section {First Section} All human things are subject to decay. Mathematical modes. Grouping and centering equations. Answers and Replies Related MATLAB, Maple, Mathematica, LaTeX News on Phys.org. How can I write two differents equations at the same line using Latex? Open an example in Overleaf. LaTeX forum â Document Classes â multiple alignment in an equation. By default, LaTeX will number equations consecutively, as (1), (2), etc., assuming you use the automatic equation numbering mechanism. ... Aligning an equation at multiple points, with both left and right alignment, as well as equals sign alignment. It has to make certain assumptions when there are ambiguous expressions. The scientists are all used to equations being accessible in web pages, LaTeX, jupyter notebooks, mediawiki, PowerPoint, Word etc. Inline math; Equations; Fractions; Matrices; Scaling of Parentheses, Brackets etc. Tim Tim. L a T e X allows two writing modes for mathematical expressions: the inline mode and the display mode. Open an example of the amsmath package in Overleaf. â
By using automatic sizing, the rendered image will exactly match the font size of the equation! formulas, graphs). Again, use * to toggle the equation numbering. LaTex is a popular document typesetting system for typesetting math. How To Linebreak A Equation In Latex Dealing Long You. As you see, the way the equations are displayed depends on the delimiter, in this case and . A variant environment alignat allows the horizontal space between equations to be explicitly specified. Aligning Multiline Equation To The Left With Only One. For e.g., you can include multiple equations within the same line and select the layout that best suits your document. Auto Latex Equations Google Workspace Marketplace. So I decide to figure out how to use LaTeX equation numbering properly in my post. LaTeX assumes that each equation consists of two parts separated by a &; also that each equation is separated from the one before by an &. This package allows you to choose the layout for your document that best suits your requirements. In fact, your example is probably best with the cases environment. Post by daviddoria » Sun Oct 05, 2008 7:38 pm . 9 3 Multiple Lines Of Displayed Maths. Latex equation in multiple lines. And when fate summons, Monarchs must obey. ] How do I reference my LaTeX tables or equations? Equations. Text with two or double columns can be created by passing the parameter \\twocolumn to the document class statement. LaTeX assumes that each equation consists of two parts separated by a &; also that each equation is separated from the one before by an &. There are several ways to format multiple equations and the amsmath package adds several more. There are two major modes of typesetting math in LaTeX one is embedding the math directly into your text by encapsulating your ⦠To change size, simply drag the box or change the original font size. LaTeX math and equations Learn to typeset and align equations, matrices and fractions in LaTeX. Writing. formulas, graphs). LaTeX Multiple Columns. More general systems involving nonlinear functions are possible as well. LaTeX is used to prepare a document with high-quality typesetting. When numbering is allowed you can label each row individually. Mathematical modes. Jupyter Notebook LaTeX equation example. 6 posts ⢠Page 1 of 1. daviddoria Posts: 60 Joined: Tue Sep 30, 2008 7:24 pm. 1. add a comment | 2 Answers Active Oldest Votes. Description below mirrored brace. 7 posts ⢠Page 1 of 1. Information and discussion about specific document classes and how to create your own document classes. How To Have A Single Vertically Centred Equation Number For. We love good questions. As you see, the way the equations are displayed depends on the delimiter, in this case and .\n\n## latex multiple equations\n\nDynamic Web Page Tutorial, Rosa Glauca Invasive, Poinsettia Plant Images, Application Architecture Examples, Silica Gel Vs Dehumidifier, Egg Pie Quiche, Just Enough Planning, How To Do Strawberry Tiktok Challenge, City Park New Orleans Jobs, Texas Tech Emergency Medicine Residency, Upper Hutt City Council Xplorer,"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.54382336,"math_prob":0.97561663,"size":370,"snap":"2021-04-2021-17","text_gpt3_token_len":82,"char_repetition_ratio":0.1010929,"word_repetition_ratio":0.0,"special_character_ratio":0.17567568,"punctuation_ratio":0.17741935,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9943831,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-13T00:46:02Z\",\"WARC-Record-ID\":\"<urn:uuid:40aff5dd-01b3-4c57-b3d2-69399771b7b0>\",\"Content-Length\":\"17267\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bd82c86e-7371-443b-824c-d02223a7e1e3>\",\"WARC-Concurrent-To\":\"<urn:uuid:148fd57a-2a5f-4343-94a3-8e6c19bea94e>\",\"WARC-IP-Address\":\"103.253.107.97\",\"WARC-Target-URI\":\"https://www.otomas.co.id/chili-oil-tevm/page.php?page=f5ddee-latex-multiple-equations\",\"WARC-Payload-Digest\":\"sha1:BD4QGOWYYBKTBDQTGTHZ72AZ24OGJHVF\",\"WARC-Block-Digest\":\"sha1:S7RZKSQ4PSVQXSGHSXOK67T346H6TEFS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038071212.27_warc_CC-MAIN-20210413000853-20210413030853-00095.warc.gz\"}"} |
http://www.arduinoos.com/2010/10/fast-fourier-transform-fft-cont-3/ | [
"## Fast Fourier Transform (FFT) (Part 4)\n\nPart 1234567891011\n\nOnce the FFT algorithm is executed, we get complex values, made of imaginary values and real values. We need to apply some maths in order to convert them in magnitude values. This is quite simply done thanks to the following routine which converts the complex (so as to say rectangular) values into a polar value:\n\n```void PlainFFT::complexToMagnitude(double *vR, double *vI, uint8_t samples) {\n// vM is half the size of vR and vI\nfor (uint8_t i = 0; i < samples; i++) {\nvR[i] = sqrt(sq(vR[i]) + sq(vI[i]));\n}\n}```"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8368344,"math_prob":0.9946601,"size":984,"snap":"2022-05-2022-21","text_gpt3_token_len":279,"char_repetition_ratio":0.096938774,"word_repetition_ratio":0.0,"special_character_ratio":0.2804878,"punctuation_ratio":0.13366337,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9948835,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T10:06:48Z\",\"WARC-Record-ID\":\"<urn:uuid:fcb786c8-3e5d-441b-953c-03fdc111bdd5>\",\"Content-Length\":\"34116\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:712f0121-862e-41b1-943c-f2b127890af2>\",\"WARC-Concurrent-To\":\"<urn:uuid:6bbb39c2-c663-4e23-a9cd-551a86fb52cb>\",\"WARC-IP-Address\":\"213.186.33.3\",\"WARC-Target-URI\":\"http://www.arduinoos.com/2010/10/fast-fourier-transform-fft-cont-3/\",\"WARC-Payload-Digest\":\"sha1:7EVMM6XRTTFEEVDXXRBDFUV5YJVKS2VL\",\"WARC-Block-Digest\":\"sha1:KWKU5Q6SIOWVUEZ6QKHG6GJJUCXLEZUS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531779.10_warc_CC-MAIN-20220520093441-20220520123441-00325.warc.gz\"}"} |
https://mclee.foolme.net/category/work | [
"# DEMO China 在北京",
null,
"",
null,
"",
null,
"",
null,
"# 一塊破紀錄，一塊送愛心~",
null,
"# 基本功果然很重要…\n\n#!/bin/sh\nmon=`date -d “-1 day” “+%m”`\n\nif [ \\$mon = 01 ]; then\nmon=’Jan’;\nelif [ \\$mon = 02 ]; then\nmon=’Feb’;\nelif [ \\$mon = 03 ]; then\nmon=’Mar’;\nelif [ \\$mon = 04 ]; then\nmon=’Apr’;\nelif [ \\$mon = 05 ]; then\nmon=’Mar‘;\nelif [ \\$mon = 06 ]; then\nmon=’Jun’;\nelif [ \\$mon = 07 ]; then\nmon=’Jul’;\nelif [ \\$mon = 08 ]; then\nmon=’Aug’;\nelif [ \\$mon = 09 ]; then\nmon=’Sep’;\nelif [ \\$mon = 10 ]; then\nmon=’Oct’;\nelif [ \\$mon = 11 ]; then\nmon=’Nov’;\nelif [ \\$mon = 12 ]; then\nmon=’Dec’;\nelse exit;\nfi\n\nyesdate=`date -d “-1 day” “+%d/\\$mon/%Y”`\n\nyesdate=`date -d yesterday “+%d/%b/%Y”`\n\n[Note] 這邊的 date 參數是給 GNU 版本的 date 吃，和 BSD family 的 date 不一樣。"
] | [
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"http://farm4.static.flickr.com/3257/2901276054_c2311885c9.jpg",
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"http://farm4.static.flickr.com/3165/2901522964_901aa05478.jpg",
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"http://farm4.static.flickr.com/3099/2900605809_2eee7f487c.jpg",
null,
"http://farm4.static.flickr.com/3245/2901454254_a7e35b55a8.jpg",
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"http://static.pe.recordcup.com/event/anti-seismic/sticker.jpg",
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.91273123,"math_prob":0.98429775,"size":1547,"snap":"2019-43-2019-47","text_gpt3_token_len":1318,"char_repetition_ratio":0.21062864,"word_repetition_ratio":0.0,"special_character_ratio":0.30575308,"punctuation_ratio":0.101960786,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9758401,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T11:30:33Z\",\"WARC-Record-ID\":\"<urn:uuid:c04be075-30d7-4e8f-a00b-f3408aceb7e4>\",\"Content-Length\":\"44350\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3c518dfd-1486-44d1-83fe-d86647970221>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcafbbc6-64ea-48dc-a858-0e9e04331ac1>\",\"WARC-IP-Address\":\"172.104.84.76\",\"WARC-Target-URI\":\"https://mclee.foolme.net/category/work\",\"WARC-Payload-Digest\":\"sha1:RZYCGGTHKI5AO3I6XFE5FBYQMQIZZDCY\",\"WARC-Block-Digest\":\"sha1:VRBOXTAPMKN7HDJDFJI3QTPGDRK25YJA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670135.29_warc_CC-MAIN-20191119093744-20191119121744-00456.warc.gz\"}"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-p-section-p-1-algebraic-expressions-mathematical-models-and-real-numbers-exercise-set-page-18/121 | [
"## Precalculus (6th Edition) Blitzer\n\n$x-(x+4)$ $-4$\nSince the number is being decreased by the sum of it and four, it would be $x-(x+4)$. The sum is the answer you get when you add numbers, and decreased implies subtraction. This then simplifies into $x-x-4$, which is just $-4$."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9386352,"math_prob":0.999739,"size":267,"snap":"2020-24-2020-29","text_gpt3_token_len":78,"char_repetition_ratio":0.117870726,"word_repetition_ratio":0.0,"special_character_ratio":0.29962546,"punctuation_ratio":0.1,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997768,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-04T11:31:55Z\",\"WARC-Record-ID\":\"<urn:uuid:259754a3-4990-4762-ac17-1be462264fcb>\",\"Content-Length\":\"90304\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ba534127-175e-4812-a576-c2704309fa55>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f1d4d9e-2064-4a86-b258-cb6329582e2f>\",\"WARC-IP-Address\":\"52.87.139.33\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-p-section-p-1-algebraic-expressions-mathematical-models-and-real-numbers-exercise-set-page-18/121\",\"WARC-Payload-Digest\":\"sha1:SV2GICG3PBTA46SHGYPYGQKCM7BU5SDF\",\"WARC-Block-Digest\":\"sha1:OBGN7YPOTCWRA4BTGGT5BX7KTUZCA4EX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655886121.45_warc_CC-MAIN-20200704104352-20200704134352-00274.warc.gz\"}"} |
https://timmousk.com/blog/typescript-reduce/ | [
"# How Does The Reduce Function Work In TypeScript?",
null,
"Tim Mouskhelichvili\n\nThe reduce function lets the developer iterate over an array and execute a reducer on each array element. This function returns a single result. Here is a simple example of the reduce function in TypeScript.\n\n```typescript```const arr = [ 1, 5, 7, 3 ];\n\nconst sum = arr.reduce((a: number, b: number) => {\nreturn a + b;\n}, 0);\n\n// Outputs: 16\nconsole.log(sum);``````\n\nIn this article, I will explain the reduce function in detail, how to use it in TypeScript, and give different examples of typical use cases.\n\nLet's get to it 😎.\n\nPage content\n\n## The definition\n\nThe reduce function is a built-in array function added in ES5.\n\nThe reduce function executes a reducer on each array's element and returns the accumulated result in a single result.\n\nHere is the reduce function's syntax.\n\n`typescript`array.reduce(function(total, value, index, array), initialValue)``\n\n### Return value\n\nThe reduce function returns the accumulated result from the last call of the function.\n\nNote: In TypeScript, you can specify the return type of this function by using a generic argument.\n\n### Browser support\n\nThe reduce function works on all major browsers, including IE9, IE10, and IE11 🥳.\n\n## How to type the reduce function in TypeScript?\n\nSometimes, the TypeScript compiler cannot infer the reduce function return type.\n\nThis is when you need to specify a return type.\n\nThe reduce function accepts an optional generic argument to specify the return type.\n\nHere is an example of this:\n\n```typescript```const arr = [\n{ firstName: 'Tim' },\n{ lastName: 'Mousk' }\n];\n\ntype Person = {\nfirstName: string;\nlastName: string;\n}\n\nconst person = arr.reduce<Person>((acc, current) => {\nreturn { ...acc, ...current };\n}, {} as Person);\n\n// Outputs: { firstName: \"Tim\", lastName: \"Mousk\" }\nconsole.log(person);``````\n\nIn this case, we create a person object by reducing each array item and specifying a return type.\n\nIn this example, we are using the spread operator to merge objects.\n\nThe best Web Development course in 2023! 👉 Learn Web Development\n\n## How to find the sum with reduce?\n\nTo find the sum of array elements, use the reduce function like so:\n\n```typescript```const arr = [ 5, 2, 5, 12 ];\n\nconst sum = arr.reduce((a: number, b: number) => {\nreturn a + b;\n}, 0);\n\n// Outputs: 24\nconsole.log(sum);``````\n\n## How to calculate the average with reduce?\n\nTo calculate the average with the reduce function, you need to get the accumulated result sum and divide it by the array's length.\n\nHere is an example of how to do it:\n\n```typescript```const arr = [ 10, 4, 7, 3 ];\n\nconst sum = arr.reduce((a: number, b: number) => {\nreturn a + b;\n}, 0);\n\nconst average = sum / arr.length;\n\n// Outputs: 6\nconsole.log(average);``````\n\n## How to reduce an array of objects?\n\nThe reduce function works very well on an array of objects.\n\nHere is an example of a reducer on an array of objects:\n\n```typescript```const arr = [\n{ value: 32 },\n{ value: 12 },\n{ value: 4 }\n];\n\nconst sum = arr.reduce((a, b) => {\nreturn a + b.value;\n}, 0);\n\n// Outputs: 48\nconsole.log(sum);``````\n\n## How to use reduce with a condition?\n\nTo execute a reducer conditionally, you need to use the filter function with the reduce function.\n\nHere is an example of a reduce with a condition:\n\n```typescript```const persons = [\n{\nvalue: 22\n},\n{\nrole: 'manager',\nvalue: 30\n},\n{\nvalue: 31\n}\n];\n\nconst valueSum = persons\n.filter(({ role }) => role === 'admin')\n.reduce<number>((sum: number, person) => {\nreturn sum + person.value\n}, 0);\n\n// Outputs: 53\nconsole.log(valueSum);``````\n\nIn this example, we get the sum of all the admin's values in the array.\n\n## Final thoughts\n\nIn conclusion, the reduce function is a great way to simplify your TypeScript code and eliminate unnecessary calls to the forEach function."
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"https://timmousk.com/wp-content/uploads/2022/03/author_tim.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6278947,"math_prob":0.9498602,"size":4167,"snap":"2023-40-2023-50","text_gpt3_token_len":1022,"char_repetition_ratio":0.16478501,"word_repetition_ratio":0.10407876,"special_character_ratio":0.2711783,"punctuation_ratio":0.2045728,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9901212,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T20:40:18Z\",\"WARC-Record-ID\":\"<urn:uuid:0cab271b-790a-4c9c-9286-0beb258c6a80>\",\"Content-Length\":\"146014\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:536a74c3-eb3c-4292-a18d-7212c063ce1d>\",\"WARC-Concurrent-To\":\"<urn:uuid:60e5b5e3-699d-41fc-bd6e-5682c01642b8>\",\"WARC-IP-Address\":\"199.241.136.156\",\"WARC-Target-URI\":\"https://timmousk.com/blog/typescript-reduce/\",\"WARC-Payload-Digest\":\"sha1:FHN6QDXUNC5ZKW2BOLE76HMICDCMSYOL\",\"WARC-Block-Digest\":\"sha1:GZOCRYDFUI6EHX4VD52PGCVIA3VYO2KD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506528.3_warc_CC-MAIN-20230923194908-20230923224908-00021.warc.gz\"}"} |
https://www.perlmonks.org/?node_id=1215373 | [
"",
null,
"Clear questions and runnable codeget the best and fastest answer PerlMonks\n\n### Re: difference in regex\n\nby haukex (Bishop)\n on May 29, 2018 at 14:03 UTC ( #1215373=note: print w/replies, xml ) Need Help??\n\nin reply to difference in regex\n\nYou will find the answer to your question in \"Regexp Quote-Like Operators\" in perlop - basically, different regex operations have different return values in different contexts. See also perlretut for a tutorial.\n\nOperation Context () Capturing\nGroups\nReturn Value on Match\n(and notes on behavior)\nReturn Value on Failure Example\nm// scalar - true false\n```my \\$x = \"foobar\"=~/[aeiou]/; # => \\$x is true\nmy \\$y = \"foobar\"=~/[xyz]/; # => \\$y is false\nm//g scalar - true\n(each execution of m//g finds the next match,\nsee \"Global matching\" in perlretut)\nfalse if there is no further match\n```my \\$str = \"foobar\";\nmy \\$x = \\$str=~/[aeiou]/g;\n# matches first \"o\" => \\$x is true, pos(\\$str) is 2\n\\$x = \\$str=~/[aeiou]/g;\n# matches second \"o\" => \\$x is true, pos(\\$str) is 3\n\\$x = \\$str=~/[aeiou]/g;\n# matches \"a\" => \\$x is true, pos(\\$str) is 5\n\\$x = \\$str=~/[aeiou]/g;\n# no more matches => \\$x is false, pos(\\$str) is undef\nm// list no the list (1) the empty list ()\n```my (\\$x) = \"foobar\"=~/[aeiou]/; # => \\$x is 1\nm//g list no a list of all the matched strings, as if there were parentheses around the whole pattern the empty list ()\n```my (\\$x,\\$y,\\$z) = \"foobar\"=~/[aeiou]/g;\n# => \\$x is \"o\", \\$y is \"o\", \\$z is \"a\"\nm// list yes a list consisting of the subexpressions matched by the parentheses in the pattern, that is, (\\$1, \\$2, \\$3...) the empty list ()\n```my (\\$x,\\$y) = \"foobar\"=~/([aeiou])(.)/;\n# => \\$x is \"o\", \\$y is \"o\"\nm//g list yes a list of the substrings matched by any capturing parentheses in the regular expression, that is, (\\$1, \\$2...) repeated for each match the empty list ()\n```my (\\$w,\\$x,\\$y,\\$z) = \"foobar\"=~/([aeiou])(.)/g;\n# => \\$w is \"o\", \\$x is \"o\", \\$y is \"a\", \\$z is \"r\"\ns/// - - the number of substitutions made false\n```my \\$x = \"foobar\";\nmy \\$y = \\$x=~s/[aeiou]/x/g; # => \\$y is 3\ns///r - - a copy of the original string with substitution(s) applied\n(available since Perl 5.14)\nthe original string\n```my \\$x = \"foobar\"=~s/[aeiou]/x/gr;\n# => \\$x is \"fxxbxr\"\n\nIn this table, \"true\" and \"false\" refer to Perl's notion of Truth and Falsehood. Remember not to rely on any of the capture variables like \\$1, \\$2, etc. unless the match succeeds!\n\nIn my \\$foo = \"bar\"=~/a/;, the right-hand side of the assignment (\"bar\"=~/a/) is in scalar context. In my (\\$foo) = \"bar\"=~/a/; or my @foo = \"bar\"=~/a/;, the right-hand side is in list context. That's why, in your example, you need those parens in (\\$value): because you want the matching operation to return the contents of the capture group.\n\nNote that your expressions can be slightly simplified, not all the parens you showed are needed:\n\n```my (\\$value) = \\$row =~ /.*,(.*)/;\n# and\n\\$row =~ s/,[^,]*\\$//;\n\n• (\\$row =~ s/,[^,]*\\$//); # gets substring before the last comma - this comment isn't quite right or at least potentially misleading, since it deletes the string before after and including the last comma.\n• /.*,(.*)/ matches any comma anywhere in the string, for simple input strings it may behave correctly, but I'd strongly recommend coding more defensively and writing it like your second expression: my (\\$value) = \\$row=~/,([^,]*)\\$/; - the \\$ anchor makes sure that the regex only matches the last comma and what follows it (unless you use the /m modifier, since it changes the meaning of \\$).\n• While the use of Scalar::Util's looks_like_number is often a good idea, note that if you don't mind being a little more restrictive, Regexp::Common (or a hand-written regex) would allow you to combine the two regular expressions:\n```use Regexp::Common qw/number/;\n\nmy \\$row = \"a,b,c,d,15\";\n\nif ( \\$row=~s/,(\\$RE{num}{real})\\$// ) {\nprint \"matched <\\$1>\\n\";\n}\nprint \"row is now <\\$row>\\n\";\n\n__END__\n\nmatched <15>\nrow is now <a,b,c,d>\n• If this is a CSV file, consider using Text::CSV (also install Text::CSV_XS for speed)\n\nUpdate: Added s///r to the table and added a few more doc links. A few other edits and updates. 2019-02-16: Added \"Return Value on Failure\" column to table, and a few other small updates. 2019-08-17: Updated the link to \"Truth and Falsehood\".\n\nReplies are listed 'Best First'.\nRe^2: difference in regex\nby ovedpo15 (Pilgrim) on May 29, 2018 at 14:11 UTC\nThank you for the replay!\nAs I mentioned on one of the posts on this thread - I would like to split it somehow into two scalars. I can use my (\\$a,\\$b) = (\\$row=~ /(.*),(.*)/); But if \\$row doesn't have commas it won't work. how do I make always put a string into \\$path\nfor example:\nif \"abc\" it will be \\$path = \"abc\" and \\$value is undefined.\nif \"abc,5\" it will be \\$path = \"abc\" and \\$value = 5\nif \"a,b,c,5\" it will be \\$path = \"a,b,c\" and \\$value = 5\n\nAlthough personally I'd still use a conditional, of course it's possible to do it all in one regex. One way is by making the comma optional by putting a ? on a group, in this case I'm using a non-capturing (?:...) group, and I had to make the first part of the regex non-greedy so that it doesn't swallow an existing comma:\n\n```use warnings;\nuse strict;\nuse Test::More;\n\nmy \\$regex = qr/ ^ (.*?) (?: , ([^,]*) )? \\$ /x;\n\nok \"abc\"=~\\$regex;\nis \\$1, \"abc\";\nis \\$2, undef;\n\nok \"abc,5\"=~\\$regex;\nis \\$1, \"abc\";\nis \\$2, 5;\n\nok \"a,b,c,5\"=~\\$regex;\nis \\$1, \"a,b,c\";\nis \\$2, 5;\n\ndone_testing;\n\nUpdate: An alternative that says a little more explicitly: either match a string with no commas in it, or, if there are commas, I want to match the thing after the last one: /^ (?| ([^,]*) | (.*) , ([^,]*) ) \\$/x Update 2: And it turns out this regex is much faster than the above! (try using it in this benchmark)\n\nCreate A New User\nDomain Nodelet?\nNode Status?\nnode history\nNode Type: note [id://1215373]\nhelp\nChatterbox?\nand the web crawler heard nothing...\n\nHow do I use this? | Other CB clients\nOther Users?\nOthers taking refuge in the Monastery: (2)\nAs of 2021-07-26 17:18 GMT\nSections?\nInformation?\nFind Nodes?\nLeftovers?\nVoting Booth?\n\nNo recent polls found\n\nNotices?"
] | [
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"https://promote.pair.com/i/pair-banner-current.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8366055,"math_prob":0.8523191,"size":5517,"snap":"2021-31-2021-39","text_gpt3_token_len":1719,"char_repetition_ratio":0.10429893,"word_repetition_ratio":0.039553754,"special_character_ratio":0.35254666,"punctuation_ratio":0.17472434,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9715523,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-26T17:19:31Z\",\"WARC-Record-ID\":\"<urn:uuid:49c013c3-ed48-4935-84ef-154e7496c94b>\",\"Content-Length\":\"29815\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d627e4ef-7b9e-4680-acff-64282aeb97d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:129900fd-5dc6-46d2-b51b-99c7d802e2ed>\",\"WARC-IP-Address\":\"66.39.54.27\",\"WARC-Target-URI\":\"https://www.perlmonks.org/?node_id=1215373\",\"WARC-Payload-Digest\":\"sha1:OEP42CQITGL6PW4U43XPL2PXVLE5ZI2X\",\"WARC-Block-Digest\":\"sha1:7RWNBWBH3WID6UNIYGX7XSKE6OZH3L47\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046152144.81_warc_CC-MAIN-20210726152107-20210726182107-00673.warc.gz\"}"} |
https://unix.stackexchange.com/questions/519129/repairing-the-headers-of-phylip-bioinformatics-files-to-accurately-reflect-the-u | [
"# Repairing the headers of phylip bioinformatics files to accurately reflect the updated number of samples in the file(s)\n\nI have a dataset that I am working with made up of phylip files that I have been editing. Phylip format is a bioinformatics format that contains as a header the number of samples and the sequence length, followed by each sample and its sequence. for example:\n\n5 10\nsample_1 gaatatccga\nsample_2 gaatatccga\nsample_3 gaatatcgca\nsample_4 caatatccga\nsample_5 gaataagcga\n\n\nMy issue is that in trimming these datasets, the sample number in the header no longer is accurate (e.g. in above example might say five, but I've since trimmed to have only three samples). What I need to do is to replace that sample count with the new, accurate sample count but I cannot figure out how to do so without losing the sequence length number (e.g. the 10).\n\nI have 550 files so simply doing this by hand is not an option. I can for-loop the wc but again I need to retain that sequence length information and somehow combine it with a new, accurate wc.\n\n• Please don't post pictures of text but post the actual text in a code block. May 15 '19 at 17:49\n• Also, it's unclear how the number should be changed. Always to 3? May 15 '19 at 17:55\n• ok thank you, will do in the future. Not always to three as the sample number is different across files but most of the files have been edited and so the new number of samples is often fewer than what is currently stated in the header May 15 '19 at 18:18\n\nIf I understand your requirement correctly you can use the following awk command:\n\nawk -v samples=\"$(($(grep -c . input)-1))\" 'NR == 1 { $1=samples }1' input samples will be set to the number of lines in the input file minus one (since you aren't counting the header line). awk will then change the first column of the first line to the new sample number and print everything. $ cat input\n5 10\nsample_1 gaatatccga\nsample_2 gaatatccga\nsample_3 gaatatccga\n$awk -v samples=\"$(($(grep -c . input)-1))\" 'NR == 1 {$1=samples }1' input\n3 10\nsample_1 gaatatccga\nsample_2 gaatatccga\nsample_3 gaatatccga\n\n\nWith GNU awk you can use the -i flag to modify the files in place but I would prefer to make a second set of modified files to ensure the correct changes have been made.\n\nSomething like:\n\nfor file in *.phy; do\nawk -v samples=\"$(($(grep -c . \"$file\")-1))\" 'NR == 1 {$1=samples }1' \"$file\" > \"${file}.new\"\ndone\n\n\nAnother option would be to use ed (of course!):\n\nfor f in input*\ndo\nprintf '1s/[[:digit:]][[:digit:]]*/%d\\nw\\nq' $(($(wc -l < \"$f\") - 1 )) | ed -s \"$f\"\ndone\n\n\nThis loops over the files (named, for example input-something) and sends a simple ed-script to ed:\n\n• on line 1, search and replace (s//) one or more digits at the beginning of the line with another number -- that replacement number being the result of computing the line length of the input minus one\n• after that, w write the file out and\n• then q quit ed\n\nIn Vim, run:\n\n:execute '1s/^[0-9]\\+/' . (line('\\$')-1) . '/'\n\n\n(Thanks also to this answer for pointing me in the right direction.)\n\nYou can also do this in a loop, e.g. using :bufdo or just a shell for loop."
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https://wiki.met.no/aerocom/indirect?rev=1380754185 | [
"# Aerocom wiki\n\n## AerChemMIP\n\naerocom:indirect\n\nThis is an old revision of the document!\n\n## Indirect Effect Experiment Remarks\n\n### Indirect forcing experiment\n\n* Submission of results by 1 December 2013\n\n##### Simulation setup\n\n* Simulation start 1 October 2005\n* Forcing by AMIP2 sea surface temperature and sea-ice extent\n* Preferred: Nudge toward ECMWF reanalysis winds (not temperature) through 2010\n* Acceptable: Nudge toward winds from one baseline simulation by your model\n* Less acceptable: No nudging\n* Greenhouse gas concentrations for year 2000\n* Aerosol direct, semi-direct, and indirect effects taken into account.\n\n* all_2000: simulation PD (present-day): year 2000 IPCC aerosol emissions\n* all_1850: simulation PI (pre-industrial): year 1850 IPCC aerosol emissions (year 2000 GHG concentration)\n* hom_2000: present day emissions no heterogeneous nucleation of ice in cirrus clouds with T<-37 C\n* hom_1850: as in hom_2000, but for pre-industrial emissions\n* fix_2000: present day emissions fixed ice nucleation for T<-37 C using a constant ice number of 383.6 /L, which is from Cooper (1986) at T=-37C\n* fix_1850: as in fix_2000, but for pre-industrial emissions\n\n#### Diagnostics\n\n* All data except COSP diagnostics is to be collected at the AEROCOM server.\n* Groups hold COSP diagnostics and contact Kenta Suzuki (Kentaro.Suzuki@jpl.nasa.gov) for analysis\n* follow the aerocom data protocol (http://aerocom.met.no/protocol.html)\n* Data in NetCDF format, one variable and year per file with CMOR variable names\n* All data are 3-dimensional ( lon x lat x time )\n* filenames aerocom_<ModelName>_<ExperimentName>_<VariableName>_<VerticalCoordinateType>_<Period>_<Frequency>.nc\n\n``` where <ModelName> can be chosen such that Model Name, Model version and possibly the institution can be identified. No underscores (_) are allowed in <ModelName>. Use (-) instead. Max 20 characters.\n<ExperimentName> = all_2000, all_1850, hom_2000, hom_1850, fix_2000, or fix_1850\n<VariableName> see list below\n<VerticalCoordinateType> => \"Surface\", \"Column\", \"ModelLevel\"\n<Period> => \"2008\", \"2010\", ...\n<Frequency> => \"timeinvariant\",\"hourly\", ,\"3hourly\", \"daily\", \"monthly\" ```\n\n* CFMIP COSP diagnostics provided by COSP do not need to be run through cmor because the names are the same,\n* but please separate files for each variable\n\n``` In addition to the diagnostics below, it is highly recommended to store the AEROCOM standard and forcing diagnostics,\nso that the simulations can be analysed for the direct forcing as well, and future more in-depth analyses are possible. ```\n\n(1) 2D diagnostics for evaluation with satellite data\n\n5 years (years 2006-2010) of 3-hourly data from the PD run\n\nname long_name (CF if possible) units description\nod550aer atmosphere_optical_thickness_due_to_aerosol 1 Aerosol optical depth (@ 550 nm)\nangstrm AOD_Angstrom_exponent 1\naerindex aerosol_index 1 od550aer*angstrm\ncdr liquid_cloud-top_droplet_effective_radius m Grid cell mean droplet effective radius at top of liquid water clouds\ncdnc liquid_cloud_droplet_number_concentration m-3 Grid cell mean droplet number concentration in top layer of liquid water clouds\ncdnum column_cloud_droplet_number_concentration m-2 grid cell mean column total\nicnum column_ice_crystal_number_concentration m-2 grid cell mean column total\nclt cloud_area_fraction 1 Fractional cover by all clouds\nlcc liquid_cloud_area_fraction 1 Fractional cover by liquid water clouds\nlwp atmosphere_cloud_ice_content kg m-2 grid cell mean liquid water path for liquid water clouds\niwp atmosphere_cloud_ice_content kg m-2 grid cell mean ice water path for ice clouds\nicr cloud-top_ice_crystal_effective_radius m grid cell mean effective radius of crystals at top of ice clouds\nicc ice_cloud_area_fraction 1 Fractional cover by ice clouds\ncod cloud_optical_depth 1 Grid cell mean cloud optical depth\ncodliq cloud_optical_depth_due_to_liquid 1 Grid cell mean cloud optical depth\ncodice cloud_optical_depth_due_to_ice 1 Grid cell mean cloud optical depth\nccn0.1bl cloud_condensation_nuclei_0.1_pbl m-3 CCN number concentration at S=0.1% at 1 km above the surface\nccn0.3bl cloud_condensation_nuclei_0.3_pbl m-3 CCN number concentration at S=0.3% at 1 km above the surface\ncolccn.1 column_cloud_condensation_nuclei_0.1 m-2 column-integrated CCN number concentration at S=0.1%\ncolccn.3 column_cloud_condensation_nuclei_0.3 m-2 column-integrated CCN number concentration at S=0.3%\nrsut toa_upward_shortwave_flux W m-2 TOA upward SW flux, all-sky\nrsutcs toa_upward_shortwave_flux_assuming_clear_sky W m-2 TOA upward SW flux, clear-sky\nrsutnoa toa_upward_shortwave_flux_no_aerosol W m-2 TOA upward SW flux, all-sky, aerosol removed from calculation\nrsutcsnoa toa_upward_shortwave_flux_clear_sky_no_aerosol W m-2 TOA upward SW flux, clear-sky, aerosol removed from calculation\nrlut toa_upward_longwave_flux W m-2 TOA upward LW flux, all-sky\nrlutcs toa_upward_longwave_flux_assuming_clear_sky W m-2 TOA upward LW flux, clear-sky\nhfls surface_upward_latent_heat_flux W m-2 Surface latent heat flux\nhfss surface_upward_sensible_heat_flux W m-2 Surface sensible heat flux\nrls surface_net_downward_longwave_flux_in_air W m-2 Net surface LW downward flux\nrss surface_net_downward_shortwave_flux W m-2 Net surface SW downward flux\nrsds surface_downwelling_shortwave_flux_in_air W m-2 Surface SW downward flux (to estimate the model's 'true' surface albedo)\nttop air_temperature_at_cloud_top K Temperature at top of clouds, weighted by cloud cover\nlts lower_tropospheric_stability K Difference in potential temperature between 700 hPa and 1000 hPa\nw500 vertical_velocity_dpdt_at_500_hPa hPa s-1\nsprecip stratiform_precipitation_rate kg m-2 s-1 grid cell mean at surface\nautoconv column_autoconversion_rate kg m-2 s-1 grid cell mean column total\naccretn column_accretion_rate kg m-2 s-1 grid cell mean column total\n\n(2) For forcing estimates: as in (1), but monthly-mean fields for both PD and PI simulations, plus a land-ocean mask (0 land, 1 ocean).\n\n(3) 3D monthly mean diagnostics\n\nname long_name (CF if possible) units description\nt temperature K each layer\nhus specific_humidity kg/kg each layer\nz altitude m each layer\nairmass atmosphere_mass_content_of_air kg m-2 each layer\nccn0.1 cloud_condensation_nuclei_0.1 m-3 each layer (S=0.1%)\nccn0.3 cloud_condensation_nuclei_0.3 m-3 each layer (S=0.3%)\nnc liquid_cloud_droplet_number_concentration m-3 grid cell mean each layer\nlwc cloud_liquid_water_content kg m-3 grid cell mean each layer\nrel droplet_effective_radius m grid cell mean each layer\nlccl liquid_cloud_fraction 1 Fractional cover by liquid water clouds each layer\nwsubc subgrid_vertical_velocity_for_stratiform m s-1\nautocl autoconversion_rate kg m-2 s-1 layer total in grid cell\naccretl accretion_rate kg m-2 s-1 layer total in grid cell\nni ice_cloud_crystal_number_concentration m-3 grid cell mean each layer\niwc cloud_ice_water_content kg m-3 grid cell mean each layer\nrei Ice_effective_radius m grid cell mean each layer\niccl ice_cloud_fraction 1 Fractional cover by ice water clouds each layer\nsati ice_supersaturation 1 Supersaturation with respect to ice\nwsubi subgrid_vertical_velocity_for_cirrus m s-1\ncirrus_nso4 sulfate_aerosol_number_for_homogeneous m-3 grid cell mean sulfate aerosol number used for homogeneous aerosol freezing for T←37C\ncirrus_ndust dust_aerosol_number_for_heterogeneous m-3 grid cell mean dust aerosol number used for heterogeneous aerosol freezing for T←37C\ncirrus_nbc BC_aerosol_number_for_heterogeneous m-3 grid cell mean BC aerosol number used for heterogeneous aerosol freezing for T←37C\ncirrus_nihom homogeneous_nucleation_number m-3 grid cell mean ice crystal number production from homogeneous aerosol freezing for T←37C during one model time step\ncirrus_nihet heterogeneous_nucleation_number m-3 grid cell mean ice crystal number production from heterogeneous aerosol freezing for T←37C during one model time step\ncirrus_freqhom homogeneous_nucleation_frequency 1 frequency counter of homogeneous aerosol freezing for T←37C. For each time step, freqhom = 1 if homogeneous ice nucleation happens; otherwise freqhom = 0. Monthly average of this value indicates the homogeneous nucleation frequency.\ncirrus_freqhet heterogeneous_nucleation_frequency 1 frequency counter of heterogeneous aerosol freezing for T←37C. At each model time step, set freqhom = 1 if heterogeneous ice nucleation happens; otherwise freqhom = 0. Monthly average of this value indicates the heterogeneous nucleation frequency.\nmp_hetnuc droplet_freezing_rate_by_heterogeneous m-3 s-1 grid cell mean freezing rate of cloud droplets in mixed-phase clouds for T>-37C\nmp_homnuc droplet_freezing_rate_by_homogeneous m-3 s-1 grid cell mean instantaneous freezing rate of cloud droplets for T⇐-37C\n\n(4) Optional CFMIP COSP diagnostics. Highly desirable for models with COSP\n3-hr snapshots and daily means for January-March 2008 PD simulation only.\n(a) 2D\n\nname long_name (CF if possible) units description comment notes\nclwmodis modis_liquid_cloud_fraction 1 Column fractional cover by liquid water clouds from modis simulator\nreffclwmodis modis_droplet_effective_radius*clwmodis m grid cell mean from modis simulator\nclimodis modis_ice_cloud_fraction 1 Column fractional cover by ice water clouds from modis simulator\nreffclimodis modis_ice_effective_radius*climodis m grid cell mean from modis simulator\ntauwmodis modis_liquid_cloud_optical_thickness*clwmodis 1 grid cell mean from modis simulator\ntauimodis modis_ice_cloud_optical_thickness*climodis 1 grid cell mean from modis simulator\nparasolRefl toa_bidirectional_reflectance 1 PARASOL Reflectance Simulated reflectance from PARASOL as seen at the top of the atmosphere for 5 solar zenith angles. Valid only over ocean and for one viewing direction (viewing zenith angle of 30 degrees and relative azimuth angle 320 degrees).\ncltcalipso cloud_area_fraction % CALIPSO Total Cloud Fraction\ncllcalipso cloud_area_fraction_in_atmosphere_layer % CALIPSO Low Level Cloud Fraction\nclmcalipso cloud_area_fraction_in_atmosphere_layer % CALIPSO Middle Level Cloud Fraction\nclhcalipso cloud_area_fraction_in_atmosphere_layer % CALIPSO High Level Cloud Fraction\n\n(b) 3D\n\nname long_name (CF if possible) units description comment notes\nt temperature K each layer\nz altitude m each layer\npressure atmospheric_pressure Pa each layer\nairmass atmosphere_mass_content_of_air kg m-2 each layer\nccn0.1 cloud_condensation_nuclei_0.1 m-3 each layer (S=0.1%)\nccn0.3 cloud_condensation_nuclei_0.3 m-3 each layer (S=0.3%)\nnc liquid_cloud_droplet_number_concentration m-3 grid cell mean each layer\nlwc cloud_liquid_water_content kg m-3 grid cell mean each layer stratiform cld only\nrel droplet_effective_radius m grid cell mean each layer stratiform cld only\nlccl layer_liquid_cloud_fraction 1 Fractional cover by liquid water stratiform clouds each layer\nni ice_cloud_crystal_number_concentration m-3 grid cell mean each layer\niwc cloud_ice_water_content kg m-3 grid cell mean each layer stratiform cld only\nrei ice_effective_radius m grid cell mean each layer stratiform cld only\niccl layer_ice_cloud_fraction 1 Fractional cover by ice water stratiform clouds each layer\nfracout fracout_cloud_flag_subcolumn 1 subcolumn cloud flag each model layer in 100 subcolumns 0 clear, 1 strat 2 conv\nclcalipso cloud_area_fraction_in_atmosphere_layer % CALIPSO Cloud Area Fraction at 40 height levels\nclcalipso2 cloud_area_fraction_in_atmosphere_layer % CALIPSO Cloud Fraction Undetected by CloudSat Clouds detected by CALIPSO but below the detectability threshold of CloudSat at 40 height levels\ncfadLidarsr532 histogram_of_backscattering_ratio_over_height_above_reference_ellipsoid 1 CALIPSO Scattering Ratio CFAD CFADs (Cloud Frequency Altitude Diagrams) are joint height - lidar scattering ratio distributions. 40 levels x 15 bins\n##### Sampling of cloud-top quantities\n\nThe idea is to use the cloud overlap assumption (maximum, random, or maximum-random) to estimate which part of the cloud in a\nlayer can be seen from above.\n\nNote: For the CCN, whether to sample it in the same way as CDNC, or use a similar approach (going from bottom up)\nto sample it at cloud base depends on your parameterization of the activation.\n\n``` let i=1,2,...,nx be the index for the horizontal grid-points\nlet k=1,2,...,nz be the index for the vertial levels, with 1 being the uppermost level, and nz the surface level ```\n\nnaming convention for the 3D input fields:\n\n``` iovl is the flag to select the overlap hypothesis\ncod3d(nx,nz) cloud optical thickness\nf3d(nx,nz) cloud fraction\nt3d(nx,nz) temperature\nphase3d(nx,nz) cloud thermodynamic phase (0: entire cloud consists of ice, 1: entire cloud consists of liquid water, between 0 and 1: mixed-phase)\ncdnc3d(nx,nz) cloud droplet number concentration ```\n\nthres_cld = 0.001\nthres_cod = 0.3\nIF ( iovl = random OR iovl = maximum-random ) THEN\n\n`clt(i) = 1.`\n\nELSE\n\n`clt(:) = 0`\n\nENDIF\nicc(:) = 0\nlcc(:) = 0\nttop(:) = 0\ncdr(:) = 0\nicr(:) = 0\ncdnc(:) = 0\n\nDO i=1,nx\n\n```DO k=2,nz ! assumption: uppermost layer is cloud-free (k=1)\nIF ( cod3d(i,k) > thres_cod and f3d(i,k) > thres_cld ) THEN ! visible, not-too-small cloud\n! flag_max is needed since the vertical integration for maximum overlap is different from the two others: for maximum, clt is the actual cloud cover in the level, for the two others, the actual cloud cover is 1 - clt\n! ftmp is total cloud cover seen from above down to the current level\n! clt is ftmp from the level just above\n! ftmp - clt is thus the additional cloud fraction seen from above in this level```\n```\t\tIF ( iovl = maximum ) THEN\nflag_max = -1.\nftmp(i) = MAX( clt(i), f3d(i,k)) ! maximum overlap\nELSEIF ( iovl = random ) THEN\nflag_max = 1.\nftmp(i) = clt(i) * ( 1 - f3d(i,k) ) ! random overlap\nELSEIF ( iovl = maximum-random ) THEN\nflag_max = 1.\nftmp(i) = clt(i) * ( 1 - MAX( f3d(i,k), f3d(i,k-1) ) ) / &\n( 1 - MIN( f3d(i,k-1), 1 - thres_cld ) ) ! maximum-random overlap\nENDIF\nttop(i) = ttop(i) + t3d(i,k) * ( clt(i) - ftmp(i) )*flag_max ```\n```\t\t! ice clouds\nicr(i) = icr(i) + icr3d(i,k) * ( 1 - phase3d(i,k) ) * ( clt(i) - ftmp(i) )*flag_max\nicc(i) = icc(i) + ( 1 - phase3d(i,k) ) * ( clt(i) - ftmp(i) )*flag_max\n\n! liquid water clouds\ncdr(i) = cdr(i) + cdr3d(i,j) * phase3d(i,k) * ( clt(i) - ftmp(i) )*flag_max\ncdnc(i) = cdnc(i) + cdnc3d(i,j) * phase3d(i,k) * ( clt(i) - ftmp(i) )*flag_max\nlcc(i) = lcc(i) + phase3d(i,k) * ( clt(i) - ftmp(i) )*flag_max\n\nclt(i) = ftmp(i)\nENDIF ! is there a visible, not-too-small cloud?\nENDDO ! loop over k```\n```IF ( iovl = random OR iovl = maximum-random ) THEN\nclt(i) = 1. - clt(i)\nENDIF```\n\nENDDO ! loop over I\n\nnaming convention for the input variables:\n\n``` utctime current time of the day in UTC in seconds\ntime_step_len length of model time-step\nlon(nx) longitude in degrees from 0 to 360 ```\n\n### Q/A\n\n* 2D cloud fields (lwp, iwp, cdr, cdnc, ttop, cod): Please compute them from grid-box mean values at each level but DO NOT divide by the total (2D) cloud cover, which will be done in analysis after averaging in time and space.\n\n* The three months 1 October - 31 December 2005 are thought as spin-up, which can of course be longer. Please choose as overlap assumption the one you use in the radiation scheme.\n\n* ATTENTION: clt(i) has to be initialized to 1 for random or maximum-random overlap assumptions in the “satellite simulator”\n\n* CCN definition: Compute CCN using Kohler theory at 0.1 and 0.3 % supersaturation.",
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"https://wiki.met.no/lib/exe/indexer.php",
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https://number.academy/38206 | [
"# Number 38206\n\nNumber 38,206 spell 🔊, write in words: thirty-eight thousand, two hundred and six . Ordinal number 38206th is said 🔊 and write: thirty-eight thousand, two hundred and sixth. The meaning of number 38206 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 38206. What is 38206 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 38206.\n\n## What is 38,206 in other units\n\nThe decimal (Arabic) number 38206 converted to a Roman number is (X)(X)(X)(V)MMMCCVI. Roman and decimal number conversions.\n\n#### Weight conversion\n\n38206 kilograms (kg) = 84228.9 pounds (lbs)\n38206 pounds (lbs) = 17330.1 kilograms (kg)\n\n#### Length conversion\n\n38206 kilometers (km) equals to 23741 miles (mi).\n38206 miles (mi) equals to 61487 kilometers (km).\n38206 meters (m) equals to 125347 feet (ft).\n38206 feet (ft) equals 11646 meters (m).\n38206 centimeters (cm) equals to 15041.7 inches (in).\n38206 inches (in) equals to 97043.2 centimeters (cm).\n\n#### Temperature conversion\n\n38206° Fahrenheit (°F) equals to 21207.8° Celsius (°C)\n38206° Celsius (°C) equals to 68802.8° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n38206 seconds equals to 10 hours, 36 minutes, 46 seconds\n38206 minutes equals to 3 weeks, 5 days, 12 hours, 46 minutes\n\n### Zip codes 38206\n\n• Zip code 38206 San Cristobal De La Laguna, Canarias, Santa Cruz de Tenerife, Spain a map\n• Zip code 38206 Cerca Larga, Guanajuato, Comonfort, Mexico a map\n• Zip code 38206 Morales, Guanajuato, Comonfort, Mexico a map\nZip code areas 38206\n\n### Codes and images of the number 38206\n\nNumber 38206 morse code: ...-- ---.. ..--- ----- -....\nSign language for number 38206:",
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"More images, other sizes, codes and colors ...\n\n#### Number 38206 infographic",
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"## Share in social networks",
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"## Mathematics of no. 38206\n\n### Multiplications\n\n#### Multiplication table of 38206\n\n38206 multiplied by two equals 76412 (38206 x 2 = 76412).\n38206 multiplied by three equals 114618 (38206 x 3 = 114618).\n38206 multiplied by four equals 152824 (38206 x 4 = 152824).\n38206 multiplied by five equals 191030 (38206 x 5 = 191030).\n38206 multiplied by six equals 229236 (38206 x 6 = 229236).\n38206 multiplied by seven equals 267442 (38206 x 7 = 267442).\n38206 multiplied by eight equals 305648 (38206 x 8 = 305648).\n38206 multiplied by nine equals 343854 (38206 x 9 = 343854).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 38206\n\nHalf of 38206 is 19103 (38206 / 2 = 19103).\nOne third of 38206 is 12735,3333 (38206 / 3 = 12735,3333 = 12735 1/3).\nOne quarter of 38206 is 9551,5 (38206 / 4 = 9551,5 = 9551 1/2).\nOne fifth of 38206 is 7641,2 (38206 / 5 = 7641,2 = 7641 1/5).\nOne sixth of 38206 is 6367,6667 (38206 / 6 = 6367,6667 = 6367 2/3).\nOne seventh of 38206 is 5458 (38206 / 7 = 5458).\nOne eighth of 38206 is 4775,75 (38206 / 8 = 4775,75 = 4775 3/4).\nOne ninth of 38206 is 4245,1111 (38206 / 9 = 4245,1111 = 4245 1/9).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 38206\n\n#### Is Prime?\n\nThe number 38206 is not a prime number. The closest prime numbers are 38201, 38219.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 38206 are 2 * 7 * 2729\nThe factors of 38206 are 1 , 2 , 7 , 14 , 2729 , 5458 , 19103 , 38206\nTotal factors 8.\nSum of factors 65520 (27314).\n\n#### Powers\n\nThe second power of 382062 is 1.459.698.436.\nThe third power of 382063 is 55.769.238.445.816.\n\n#### Roots\n\nThe square root √38206 is 195,463552.\nThe cube root of 338206 is 33,680396.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 38206 = loge 38206 = 10,550748.\nThe logarithm to base 10 of No. log10 38206 = 4,582132.\nThe Napierian logarithm of No. log1/e 38206 = -10,550748.\n\n### Trigonometric functions\n\nThe cosine of 38206 is -0,460942.\nThe sine of 38206 is -0,88743.\nThe tangent of 38206 is 1,925253.\n\n### Properties of the number 38206\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 38206 in Computer Science\n\nCode typeCode value\n38206 Number of bytes37.3KB\nUnix timeUnix time 38206 is equal to Thursday Jan. 1, 1970, 10:36:46 a.m. GMT\nIPv4, IPv6Number 38206 internet address in dotted format v4 0.0.149.62, v6 ::953e\n38206 Decimal = 1001010100111110 Binary\n38206 Decimal = 1221102001 Ternary\n38206 Decimal = 112476 Octal\n38206 Decimal = 953E Hexadecimal (0x953e hex)\n38206 BASE64MzgyMDY=\n38206 MD511f58f0c21cb2dae75e392564078002b\n38206 SHA1d997058b5da49ecaaaf7deb6301da9f064f95e40\n38206 SHA22487f12f86b6a45eb8e04df271a5393af8b2415a8e83a6f918276ae8d6\n38206 SHA384f2e2893b75b608a516c1bfd4306989b5fb15a4382c4fe235eaac5b944325d65dd14fd5fe312aba7d5c4bd713e29995e4\nMore SHA codes related to the number 38206 ...\n\nIf you know something interesting about the 38206 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 38206\n\n### Character frequency in number 38206\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 3 1 8 1 2 1 0 1 6 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 38206, the numbers 3+8+2+0+6 = 1+9 = 1+0 = 1 are added and the meaning of the number 1 is sought.\n\n## Interesting facts about the number 38206\n\n### Asteroids\n\n• (38206) 1999 ML1 is asteroid number 38206. It was discovered by LONEOS from Anderson Mesa on 6/20/1999.\n\n## № 38,206 in other languages\n\nHow to say or write the number thirty-eight thousand, two hundred and six in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 38.206) treinta y ocho mil doscientos seis German: 🔊 (Anzahl 38.206) achtunddreißigtausendzweihundertsechs French: 🔊 (nombre 38 206) trente-huit mille deux cent six Portuguese: 🔊 (número 38 206) trinta e oito mil, duzentos e seis Chinese: 🔊 (数 38 206) 三万八千二百零六 Arabian: 🔊 (عدد 38,206) ثمانية و ثلاثون ألفاً و مئتان و ستة Czech: 🔊 (číslo 38 206) třicet osm tisíc dvěstě šest Korean: 🔊 (번호 38,206) 삼만 팔천이백육 Danish: 🔊 (nummer 38 206) otteogtredivetusinde og tohundrede og seks Dutch: 🔊 (nummer 38 206) achtendertigduizendtweehonderdzes Japanese: 🔊 (数 38,206) 三万八千二百六 Indonesian: 🔊 (jumlah 38.206) tiga puluh delapan ribu dua ratus enam Italian: 🔊 (numero 38 206) trentottomiladuecentosei Norwegian: 🔊 (nummer 38 206) tretti-åtte tusen, to hundre og seks Polish: 🔊 (liczba 38 206) trzydzieści osiem tysięcy dwieście sześć Russian: 🔊 (номер 38 206) тридцать восемь тысяч двести шесть Turkish: 🔊 (numara 38,206) otuzsekizbinikiyüzaltı Thai: 🔊 (จำนวน 38 206) สามหมื่นแปดพันสองร้อยหก Ukrainian: 🔊 (номер 38 206) тридцять вiсiм тисяч двiстi шiсть Vietnamese: 🔊 (con số 38.206) ba mươi tám nghìn hai trăm lẻ sáu Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 38206 or any natural number (positive integer) please write us here or on facebook."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.539677,"math_prob":0.960619,"size":6941,"snap":"2022-27-2022-33","text_gpt3_token_len":2559,"char_repetition_ratio":0.15165056,"word_repetition_ratio":0.010810811,"special_character_ratio":0.41939202,"punctuation_ratio":0.1624521,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9925053,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T17:45:55Z\",\"WARC-Record-ID\":\"<urn:uuid:8e384dea-3cc4-4840-85b2-0513dec75447>\",\"Content-Length\":\"41163\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c2fcec2-4cfa-47a2-a031-a97cf65b76f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:70f58c1b-2b05-4ce6-86f0-77a5bc58956b>\",\"WARC-IP-Address\":\"162.0.227.212\",\"WARC-Target-URI\":\"https://number.academy/38206\",\"WARC-Payload-Digest\":\"sha1:DFEYVU6AO5JDURGQWZY6TCR6YP747PUO\",\"WARC-Block-Digest\":\"sha1:ISQWP5AJKVKZZNHMP3O2FAUUGQJFL37Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571056.58_warc_CC-MAIN-20220809155137-20220809185137-00763.warc.gz\"}"} |
https://swift-online.de/en/vac-virtual-analog-channel/ | [
"Search\n• Measuring, Analysing, Storing\n\n# VAC – Virtual Analog Channel",
null,
"Virtual channels are measurement channels that are not directly connected to a physical sensor but determine the actual measurement value through mathematical calculations from analogue and digital channels. The source channels (analogue and digital) are selectable. Additional coefficients (constants) can be defined for some of the formulas. The mathematical operations are performed based on the physical scaling defined for the channels to be calculated. The user can adjust the actually used values margin within certain limits. Like any real channel, the result of a virtual channel can be transformed into another physical unit by specifying the sensitivity and offset. To operate a virtual channel you should have this software. The price for the respective formula must be added on.\n\n## Formel „Power Calculation“: A1*A2\n\nThe formula “Power Calculation” calculates the value of a virtual channel as product of two analogue input channels (A1, A2). For example, using analogue inputs for the engine speed and torque, this formula calculates the engine power output.\n\n## Formel „Ratio“: A1/A2\n\nThe formula “Ratio” divides two analogue input channels (A1, A2). For example, the ratio of vehicle speed to engine speed gives the gear transmission ratio and hence suggests the actually selected gear.\n\n## Formel „weighted sums“:k1*A1 + k2*A2 + k3*A3 + k4 * A4\n\nThe formula “Weighted Sums” calculates the sum of up to four analogue input channels (A1 to A4) with consideration of their real quality rating. In addition, a weighting factor (k1 to k4) can be associated to each input channel. For example, this formula can be used to determine the total force resulting from up to four independent force inputs.\n\n## Formel „Digital Gate“: A1*D1( 0 := k1; 1:= k2) + A2*D2( 0 := k3; 1:= k4)\n\nThe formula “Digital Gate” establishes multiple connections between two analogue signals (A1, A2) and two digital channels (D1, D2). For each of the two states (0 or 1) of the digital channels, the user can assign a given value (k1 and k2 for D1 and k3 and k4 for D2). Depending on the state of the digital input the corresponding analogue channel is multiplied with the associated value. This can be done for two selectable analogue channels. The result is obtained by summation of the two products calculated before. Thus, this formula can select one of two analogue channels depending on one digital channel (D1 = D2). For example, given the vehicle speed signal (A1) and the directional information (D1), this formula enables the change of the algebraic sign (+ or -) for the vehicle speed depending on the driving direction.\n\n## Formel „Derive“: d(A1)/dt\n\nThe formula “Derive” calculates the derivative of an analogue input signal (A1). The calculation is based on the difference between the actual value of the input signal and this value delayed by a constant time, i.e. in mathematical terms corresponding to the differential quotient. The user can select the delay of the signal up to 1 second. The output derivative signal can be transformed into another user-defined unit and re-scaled in meaningful values. The effect of the selected time difference is automatically considered. A typical application of this formula is deriving the power of a shock absorber from its covered distance per time unit. The deflection of the shock absorber can be easily recorded by a displacement sensor. Thus, the formula “Derive” can online determine the derivative of the distance signal over time which corresponds to a value proportional to the shock absorber power.\n\n## Formel “Low Pass”\n\nThis formula reproduces a first order digital low pass filter. The cut-off frequency of the filter and the source channel can be selected by the user. The cut-off frequency may range from 1/40 to 1/400 000 of the system sampling rate. Based on the standard sampling rate of 2kHz the corresponding cut-off frequency ranges between 0.005Hz to 50Hz. Since low and very low cut-off frequencies can be set, this formula is best suited to measure the quasi static part of any dynamic value."
] | [
null,
"data:image/svg+xml,%3Csvg%20xmlns=%22http://www.w3.org/2000/svg%22%20viewBox=%220%200%20630%20454%22%3E%3C/svg%3E",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8579113,"math_prob":0.97936916,"size":4032,"snap":"2019-51-2020-05","text_gpt3_token_len":875,"char_repetition_ratio":0.14721946,"word_repetition_ratio":0.015337423,"special_character_ratio":0.21750993,"punctuation_ratio":0.09016393,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99755865,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T18:07:48Z\",\"WARC-Record-ID\":\"<urn:uuid:dff36e24-a3c2-40f7-bf9b-8ff26c9810f2>\",\"Content-Length\":\"51984\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cd31f641-49eb-498f-a1de-55c73ced2c1f>\",\"WARC-Concurrent-To\":\"<urn:uuid:57c2323b-bcd2-432a-8a05-87e85dcf9d0b>\",\"WARC-IP-Address\":\"85.199.141.61\",\"WARC-Target-URI\":\"https://swift-online.de/en/vac-virtual-analog-channel/\",\"WARC-Payload-Digest\":\"sha1:OAWARAZHCDDRDOQY36EHFUIMKVS4ZKFZ\",\"WARC-Block-Digest\":\"sha1:MLQUVUNAOMECZY244WYZBYKS7NF4EDPR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540531974.7_warc_CC-MAIN-20191211160056-20191211184056-00393.warc.gz\"}"} |
https://getremedyonline.com/31-excel-vba-chart-vertical-line/ | [
"# Unique 31 Sample Excel Vba Chart Vertical Line\n\nBy | April 13, 2019",
null,
"Unique 31 Sample Excel Vba Chart Vertical Line. Good day dear readers. In the current modern era, all information about the expansion of technologies is very simple to get. One can find a number of reports, tips, content, from any location in only a few moments. And also specifics about your ideal house can be accessed from lots of free places online.\nJust like now, you are researching for more knowledge about Unique 31 Sample Excel Vba Chart Vertical Line, arent you? Just sit in front of your beloved laptop or computer that is definitely linked to the Internet, you may get a variety of interesting new tips and you will apply it for your needs.\nDo you know The concept of Unique 31 Sample Excel Vba Chart Vertical Line that we give you here relates to the desire report about Unique 31 Sample Excel Vba Chart Vertical Line. We found that most people look for Unique 31 Sample Excel Vba Chart Vertical Line on search engines like bing. We tend to present a most recent picture to suit your needs.\nAlthough within our viewpoint, which we have provided the best Unique 31 Sample Excel Vba Chart Vertical Line picture, but your thought may be little bit different with us. Okay, You can use it as your research content only. And Unique 31 Sample Excel Vba Chart Vertical Line has been uploaded at 22 by [author] in excel field.\n\nUnique 31 Sample Excel Vba Chart Vertical Line\nquick tip vertical line chart in excel goodly the horizontal line chart is pretty standard stuff why dont we try making a vertical line chart well there isnt a standard vertical line chart in excel so well tweak some how to add a vertical line to a horizontal bar chart excels built in chart types are great for quickly visualizing your data the horizontal bar chart is a great example of an easy to use graph type how to add a vertical line to a horizontal bar chart learn how to add a vertical line to a horizontal bar chart in excel the tutorial walks through adding an average value line to a new series on the graph todays date line to gantt chart free excelvba help forum test xlsxi have a gantt chart that i would like to add a vertical line to that would indicate todays date to show progress standings of the project draw a connecting line between specified data points on chart ive tried using addline to place a straight line between to points on a chart the problem with addline is that it appears to reference the top left corner of the chart for its coordinates adding an excel chart target line detailed instructions a target line on excel charts can be used to visually compare a budget or target to actual expenses see different ways to create this kind of chart how to create waterfall chart in excel in 2 minutes a waterfall chart is used to represent a set of figures when they all impact the same derived number a waterfall chart helps to link the individual values to a whole excel waterfall chart 8 reasons why it still sucks in july 2015 microsoft announced that the then upcoming office 2016 would introduce 6 new charts to their line of charts the one that was most highly anticipated in the financial community was defini scatter chart in excel easy excel tutorial use a scatter chart xy chart in excel to show scientific xy data scatter charts are often used to find out if theres a relationship between variable x and y to create a scatter chart in excel ex how to create a timeline milestone chart in excel steps to create milestone chart in excel get the data in place to create this i have two columns of data date in b3b10 and activity in c3c10 and three helper columns Unique 31 Sample Excel Vba Chart Vertical Line\n\nExcel Vba Chart Vertical Line Excel Vba Set X Axis Maximum Excel Vba Chart X Axis Max\n\nExcel Vba Chart Vertical Line Adding Colored Vertical Band to Excel Chart\n\nExcel Vba Chart Vertical Line Excel Vba Chart Y Axis Range Excel Vba Set Chart Axis\n\nExcel Vba Chart Vertical Line Excel Vba Horizontal Axis Labels Excel Vba Chart X Axis\n\nExcel Vba Chart Vertical Line Excel Vba Set X Axis Maximum Excel Vba Change X Axis\n\nexcel vba zeilen ausblenden, excel vba vlookup, ex, excel vba bild ndern, excel vba lernen, excel vba msgbox zeilenumbruch, excel vba insert row, excel vba cells, excel vba value of cell, excel vba querytable refresh, excel vba json gemb, excel vba youtube, excel vba get cell value from another sheet,"
] | [
null,
"https://getremedyonline.com/wp-content/uploads/2019/02/excel-vba-chart-vertical-line-excel-vba-chart-vertical-line-line-and-bar-graph-excel-how-to-make-a-bar-graph-in-excel.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8706825,"math_prob":0.7005396,"size":4826,"snap":"2019-26-2019-30","text_gpt3_token_len":1023,"char_repetition_ratio":0.19701369,"word_repetition_ratio":0.1146789,"special_character_ratio":0.19995856,"punctuation_ratio":0.051948052,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.95103866,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T09:43:48Z\",\"WARC-Record-ID\":\"<urn:uuid:fe19d105-9a7a-49c7-9ad9-8a49242231fb>\",\"Content-Length\":\"162940\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67b4f50c-92c2-4108-bc2f-a22939c3923f>\",\"WARC-Concurrent-To\":\"<urn:uuid:b477bd40-ac8c-4cfd-8552-03813fc4478a>\",\"WARC-IP-Address\":\"104.18.35.10\",\"WARC-Target-URI\":\"https://getremedyonline.com/31-excel-vba-chart-vertical-line/\",\"WARC-Payload-Digest\":\"sha1:X2N2VQYBLIBRVKJMOXOHK53PVSCVI5LN\",\"WARC-Block-Digest\":\"sha1:JLGK3W5EDBAZJCQ7Y7ZT5GP7C4UCY7QO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527907.70_warc_CC-MAIN-20190722092824-20190722114824-00045.warc.gz\"}"} |
https://www.biaodianfu.com/detect-trend-increasing-or-decreasing-in-time-series.html | [
"# 时间序列分析之趋势判断\n\n## 方案二：斜率法\n\n• k > 0.1763 表示上升\n• k < -0.1763 表示下降\n• 其他，则表示平稳或震荡\n\nimport numpy as np\ndef trendline(index,data, order=1):\ncoeffs = np.polyfit(index, list(data), order)\nslope = coeffs[-2]\nreturn float(slope)\n\nindex=[1,2,3,4]\nList=[1043,6582,5452,7571]\nresultent=trendline(index,List)\nprint(resultent)\n\n\n## 方案三：Cox-Stuart趋势检验\n\nCox-Stuart是一种不依赖趋势结构的快速判断趋势是否存在的方法。Cox-Stuart趋势存在性检验的理论基础是符号检验。它的检验思想是：直接考虑数据的变化趋势，若数据有上升趋势，那么排在后面的数据的值要比排在前面的数据的值显著的大，反之，若数据有下降趋势，那么排在后面的数据的值要比排在前面的数据的值显著的小，利用前后两个时期不同数据的差值正负来判断数据总的变化趋势。为保证数对同分布，前后两个数的间隔应固定。这就意味着将数据一分为二，自然形成前后数对。Cox-Staut提出最优的拆分点是数列中位于中间位置的数。\n\n• 取$x_i$和$x_{i+c}$组成一对$(x_i, x_{i+c})$。这里如果n为偶数，则$c=n/2$，如果n是奇数，则c=(n+1)/2。当n为偶数时，共有n’=c对，而n是奇数时，共有 n’=c-1对。\n• 用每一对的两元素差$D_i= x_i-x_{i+c}$的符号来衡量增减。令$S^+$为正的$D_i$的数目，$S^-$为负的$D_i$的数目。显然当正号太对时有下降趋势，反之有增长趋势。在没有趋势的零假设下他们因服从二项分布b(n’,0.5)。\n• 用p(+)表示取到正数的概率，用p(-)表示取到负数的概率，这样我们就得到符号检验方法来检验序列是否存在趋势性。\n 双侧检验 H0：无趋势 H1：有增长或减少趋势 左侧检验 H0：无减少趋势 H1：有减少趋势 右侧检验 H0：无增加趋势 H1：有增加趋势",
null,
"import scipy.stats as stats\ndef cos_staut(list_c,debug=False):\nlst=list_c.copy()\nraw_len=len(lst)\nif raw_len%2==1:\ndel lst[int((raw_len-1)/2)]\nc=int(len(lst)/2)\nn_pos=n_neg=0\nfor i in range(c):\ndiff=lst[i+c]-lst[i]\nif diff>0:\nn_pos+=1\nelif diff<0:\nn_neg+=1\nelse:\ncontinue\nnum=n_pos+n_neg\nk=min(n_pos,n_neg)\np_value=2*stats.binom.cdf(k,num,0.5)\nif debug:\nprint('fall:%i, rise:%i, p-value:%f'%(n_neg, n_pos, p_value))\nif n_pos>n_neg and p_value<0.05:\nreturn 'increasing'\nelif n_neg>n_pos and p_value<0.05:\nreturn 'decreasing'\nelse:\nreturn 'no trend'\n\n## 方案四：Mann-Kendall趋势检验法\n\nMann-Kendall(MK)检验(test)(Mann 1945, Kendall 1975, Gilbert 1987) 的目的是统计评估我们所感兴趣的变量，随着时间变化，是否有单调上升或下降的趋势。单调上升（下降）的趋势意味着该变量随时间增加（减少），但此趋势可能是、也可能不是线性的。MK test可替代参数线性回归分析——线性回归可检验线性拟合直线的斜率是否不为零。回归分析要求拟合回归线的残差是正态分布的，MK检验不需要这种假设，MK检验是非参数检验（不要求服从任何分布-distribution free）\n\n• 当没有趋势时，随时间获得的数据是独立同分布的。独立的假设是说数据随着时间不是连续相关的。\n• 所获得的时间序列上的数据代表了采样时的真是条件。（样本具有代表性）\n• 样本的采集、处理和测量方法提供了总体样本中的无偏且具有代表性的观测值。\n\nMK检验不要求数据是正态分布，也不要求变化趋势——如果存在的话——是线性的。如果有缺失值或者值低于一个或多个检测限制，是可以计算MK检测的，但检测性能会受到不利影响。独立性假设要求样本之间的时间足够大，这样在不同时间收集的测量值之间不存在相关性。\n\nMK检验是检验是否拒绝零假设(null hypothesis: $H_0$)，并接受替代假设（alternative hypothesis: $H_a$）：\n\n• $H_0$：没有单调趋势\n• $H_a$：存在单调趋势\n\n• 将数据按采集时间列出：$x_1,x_2,…,x_n$，即分别在时间1,2,…,n得到的数据。\n• 确定所有n(n-1)/2个$x_j-x_k$差值的符号，其中j > k\n• 令$sgn(x_j-x_k)$作为指示函数，依据$x_j-x_k$的正负号取值为1,0或-1\n• 计算$S = \\sum_{k-1}^{n-1}\\sum_{j-k+1}^{n}sgn(x_j-x_k)$。即差值为正的数量减去差值为负的数量。如果S是一个正数，那么后一部分的观测值相比之前的观测值会趋向于变大；如果S是一个负数，那么后一部分的观测值相比之前的观测值会趋向于变小。\n• 如果$n\\leq 10$，依据Gilbert (1987, page 209, Section 16.4.1)中所描述，要在概率表 (Gilbert 1987, Table A18, page 272) 中查找S。如果此概率小于$\\alpha$(认为没有趋势时的截止概率)，那就拒绝零假设，认为趋势存在。如果在概率表中找不到n(存在结数据——tied data values——会发生此情况)，就用表中远离0的下一个值。比如S=12，如果概率表中没有S=12，那么就用S=13来处理也是一样的。如果n > 10，则依以下步骤6-10来判断有无趋势。这里遵循的是Gilbert (1987, page 211, Section 16.4.2)中的程序。\n• 计算S的方差如下：$\\text{VAR}(S)=\\frac{1}{18}[n(n-1)(2n+5)-\\sum_{p-1}^{g}t_p(t_p-1)(2t_p+5)]$。其中g是结组（tied groups）的数量，$t_p$是第p组的观测值的数量。例如：在观测值的时间序列{23, 24, 29, 6, 29, 24, 24, 29, 23}中有g = 3个结组，相应地，对于结值(tiied value)23有$t_1= 2$、结值24有$t_2=3$、结值29有$t_3=S3$。当因为有相等值或未检测到而出现结时，VAR(S)可以通过Helsel (2005, p. 191)中的结修正方法来调整。\n• 计算MK检验统计量Z_{MK}:\n\n$$Z_{M K}=\\left\\{\\begin{array}{cl}{\\frac{S-1}{\\sqrt{V A R(S)}},} & {S>0} \\\\{0} & {,\\quad S=0} \\\\{\\frac{S+1}{\\sqrt{V A R(S)}},} & {S<0}\\end{array}\\right.$$\n\n• 设想我们要测试零假设。$H_0$(没有单调趋势)对比替代假设$H_a$(有单调增趋势)，其1型错误率为$\\alpha$，$0<\\alpha<0.50$（注意$\\alpha$是MK检验错误地拒绝了零假设时可容忍的概率——即MK检验拒绝了零假设是错误地，但这个事情发生概率是$\\alpha$，我们可以容忍这个错误）。如果$Z_{MK}\\geq Z_{1-\\alpha}$，就拒绝零假设$H_0$，接受替代假设$H_a$，其中$Z_{1-\\alpha}$是标准正态分布的$100(1-\\alpha)^{th}$百分位。\n• 测试上面的$H_0$与$H_a$（有单调递减趋势），其1型错误率为$alpha$，$0<\\alpha<0.5$，如果$Z_{MK}\\leq – Z_{1-\\alpha}$，就拒绝零假设$H_0$，接受替代假设$H_a$\n• 测试上面的$H_0$与$H_a$（有单调递增或递减趋势），其1型错误率为$alpha$，$0<\\alpha<0.5$，如果$|Z_{MK}|\\geq Z_{1-\\frac{\\alpha}{2}}$，就拒绝零假设$H_0$，接受替代假设$H_a$，其中竖线代表绝对值。\nimport math\nfrom scipy.stats import norm, mstats\ndef mk_test(x, alpha=0.05):\n\"\"\"\nThis function is derived from code originally posted by Sat Kumar Tomer\n(satkumartomer@gmail.com)\nThe purpose of the Mann-Kendall (MK) test (Mann 1945, Kendall 1975, Gilbert\n1987) is to statistically assess if there is a monotonic upward or downward\ntrend of the variable of interest over time. A monotonic upward (downward)\ntrend means that the variable consistently increases (decreases) through\ntime, but the trend may or may not be linear. The MK test can be used in\nplace of a parametric linear regression analysis, which can be used to test\nif the slope of the estimated linear regression line is different from\nzero. The regression analysis requires that the residuals from the fitted\nregression line be normally distributed; an assumption not required by the\nMK test, that is, the MK test is a non-parametric (distribution-free) test.\nHirsch, Slack and Smith (1982, page 107) indicate that the MK test is best\nviewed as an exploratory analysis and is most appropriately used to\nidentify stations where changes are significant or of large magnitude and\nto quantify these findings.\nInput:\nx: a vector of data\nalpha: significance level (0.05 default)\nOutput:\ntrend: tells the trend (increasing, decreasing or no trend)\nh: True (if trend is present) or False (if trend is absence)\np: p value of the significance test\nz: normalized test statistics\nExamples\n--------\n>>> x = np.random.rand(100)\n>>> trend,h,p,z = mk_test(x,0.05)\n\"\"\"\nn = len(x)\n\n# calculate S\ns = 0\nfor k in range(n-1):\nfor j in range(k+1, n):\ns += np.sign(x[j] - x[k])\n\n# calculate the unique data\nunique_x, tp = np.unique(x, return_counts=True)\ng = len(unique_x)\n\n# calculate the var(s)\nif n == g: # there is no tie\nvar_s = (n*(n-1)*(2*n+5))/18\nelse: # there are some ties in data\nvar_s = (n*(n-1)*(2*n+5) - np.sum(tp*(tp-1)*(2*tp+5)))/18\n\nif s > 0:\nz = (s - 1)/np.sqrt(var_s)\nelif s < 0:\nz = (s + 1)/np.sqrt(var_s)\nelse: # s == 0:\nz = 0\n\n# calculate the p_value\np = 2*(1-norm.cdf(abs(z))) # two tail test\nh = abs(z) > norm.ppf(1-alpha/2)\n\nif (z < 0) and h:\ntrend = 'decreasing'\nelif (z > 0) and h:\ntrend = 'increasing'\nelse:\ntrend = 'no trend'\n\nreturn trend",
null,
"##### 网站与APP开发中的字体设置",
null,
"##### 用户体系搭建之ID-Mapping",
null,
"## 4 Replies to “时间序列分析之趋势判断”\n\n1. Ding Zhaohai说道：\n\n序列的趋势判断，看上去是个简单地不能再简单地需求，不过像通过稳健的方法对陌生的序列进行判断，需要有强大的理论支撑。\n\n2. Ding Zhaohai说道：\n\n最后一段代码：\n“h = abs(z) > norm.ppf(1-alpha/2)”，是否出错了。\n修改意见：“h = abs(z) > norm.ppf((1-alpha)/2)”\n\n3. Ding Zhaohai说道：\n\n是公式错误，不是代码错误。\n\n4. Xuance说道：\n\nCox-Stuart趋势检验中，单边检验p_value是否不用*2\np_value=2*stats.binom.cdf(k,num,0.5)"
] | [
null,
"https://www.biaodianfu.com/wp-content/uploads/2020/09/Cox-Stuart.png",
null,
"https://www.biaodianfu.com/wp-content/uploads/2021/09/shape.png",
null,
"https://www.biaodianfu.com/wp-content/uploads/2021/09/id-mapping.png",
null,
"https://www.biaodianfu.com/wp-content/uploads/2021/09/lvs.png",
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.55818516,"math_prob":0.99816364,"size":7026,"snap":"2021-43-2021-49","text_gpt3_token_len":4317,"char_repetition_ratio":0.0669325,"word_repetition_ratio":0.0063492064,"special_character_ratio":0.3182465,"punctuation_ratio":0.12132921,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99931705,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T00:24:27Z\",\"WARC-Record-ID\":\"<urn:uuid:0d960056-b45c-4cb3-8786-cc8fda8f9331>\",\"Content-Length\":\"45636\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:eb0aa67a-5ef2-4ff9-9dab-e51634134339>\",\"WARC-Concurrent-To\":\"<urn:uuid:046e508c-1ee9-4c38-b5c8-8824771cca6f>\",\"WARC-IP-Address\":\"42.192.37.225\",\"WARC-Target-URI\":\"https://www.biaodianfu.com/detect-trend-increasing-or-decreasing-in-time-series.html\",\"WARC-Payload-Digest\":\"sha1:S5FG3PPPLDOAYNDLNFDBDUE3XS5KWJDX\",\"WARC-Block-Digest\":\"sha1:WIZGLJEYFDXFCDQBW63UTISR67XAEGR5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585186.33_warc_CC-MAIN-20211018000838-20211018030838-00399.warc.gz\"}"} |
https://math.stackexchange.com/questions/912480/definition-of-multiplication-in-grothendieck-ring | [
"# Definition of multiplication in Grothendieck ring\n\nLet $X$ be a smooth variety over an algebraically closed field $k$ of dimension $n$. Consider the Grothendieck Group $K(X)$ of coherent sheaves on $X$, i.e. the free abelian group generated by expressions $[\\mathscr F]$ with $\\mathscr F$ a coherent sheaf on $X$, modulo the relation $[\\mathscr F]=[\\mathscr K]+[\\mathscr Q]$ whenever there is an exact sequence $$0\\to\\mathscr K\\to\\mathscr F\\to\\mathscr Q\\to 0.$$\n\nYou can turn $K(X)$ into a ring with multiplication induced by the tensor product, i.e. $[\\mathscr F]\\cdot[\\mathscr F]:=[\\mathscr F\\otimes \\mathscr G]$. However, I saw the product defined as $$[\\mathscr F]\\cdot[\\mathscr G] := \\sum_{i=0}^n (-1)^i [\\operatorname{Tor}_i(\\mathscr F,\\mathscr G)]$$ in Positivity in the Grothendieck group of complex flag varieties by Brion, it's on page 5 in the arXiv version.\n\nNow this should come out of the very definition of $\\operatorname{Tor}_i$ as the left derived of the tensor product. However, it does not work out that way for me, so I am looking for the mistake in my calculation.\n\nChoose a projective resolution of $\\mathscr G$, say $0\\to\\mathscr R_n\\to\\cdots\\to\\mathscr R_1\\to \\mathscr R_0 := \\mathscr G\\to 0$.\n\nApply $\\mathscr F\\otimes (-)$ to this sequence and with $\\mathscr P_i := \\mathscr F\\otimes \\mathscr R_i$, we get the exact sequence $$0\\xrightarrow{d_{n+1}=0} \\mathscr P_n \\xrightarrow{d_n} \\cdots \\xrightarrow{d_2} \\mathscr P_1 \\xrightarrow{d_1} \\mathscr P_0 = \\mathscr F\\otimes \\mathscr G\\xrightarrow{d_0=0} 0$$ Now set $\\mathscr K_i := \\ker(d_i)$, $\\mathscr I_i := \\operatorname{im}(d_i)$ and $\\mathscr T_i := \\operatorname{Tor}_i(\\mathscr F,\\mathscr G)= \\mathscr K_i / \\mathscr I_{i+1}$. We then have two exact sequences, \\begin{align*} 0&\\to \\mathscr I_{i+1} \\to \\mathscr K_i \\to \\mathscr T_i \\to 0 &&\\text{and}& 0&\\to \\mathscr K_{i} \\to \\mathscr P_i \\to \\mathscr I_i \\to 0 \\end{align*} Note that $\\mathscr I_{n+1}=0$ and $\\mathscr K_0=\\mathscr P_0 = \\mathscr F\\otimes\\mathscr G$, so we have \\begin{align*} \\sum_{i=0}^n (-1)^i [\\mathscr T_i] &= \\sum_{i=0}^n (-1)^i ([\\mathscr K_i]-[\\mathscr I_{i+1}]) = [\\mathscr K_0] + \\sum_{i=1}^{n} (-1)^i ([\\mathscr K_i]+[\\mathscr I_i]) \\\\ &= [\\mathscr P_0] + \\sum_{i=1}^n (-1)^i [\\mathscr P_i] = \\sum_{i=0}^n (-1)^i [\\mathscr P_i] = \\sum_{i=0}^n (-1)^i [\\mathscr F\\otimes \\mathscr R_i] \\\\ &= \\sum_{i=0}^n (-1)^i [\\mathscr F]\\cdot[\\mathscr R_i] = [\\mathscr F]\\cdot \\sum_{i=0}^n (-1)^i [\\mathscr R_i] \\end{align*} But by basically the same calculation, by exactness of the complex $\\mathscr R_\\bullet$, we have $\\sum_{i=0}^n (-1)^i [\\mathscr R_i] = 0$.\n\nWhat am I doing wrong?\n\n• A resolution of a module doesn't include the module itself. Put it another way, a resolution is an exact sequence if and only if it is a resolution of $0$. – Zhen Lin Aug 28 '14 at 23:50\n• @ZhenLin: Yea, I think I see now. I should replace $\\mathscr P_0$ and $d_1$ by $0$ in my notation. I will get roughly the same, but will end up with $[\\mathscr F]\\cdot\\sum_{i=1}^n (-1)^{i+1} [\\mathscr R_i]$ which is precisely $[\\mathscr F]\\cdot[\\mathscr G]$. – Jesko Hüttenhain Aug 29 '14 at 0:11\n\n## 1 Answer\n\nI think it's only a matter of indexing. For example, according to the notation in Weibel's book we would have $\\cdots\\to \\mathscr R_1\\to \\mathscr R_0\\to \\mathscr G\\to 0$ in the projective resolution, rather than defining $\\mathscr R_0 = \\mathscr G$. That is, in a projective (or free) resolution, the final map $\\mathscr R_1\\to\\mathscr R_0$ is typically defined so that it has cokernel isomorphic to the resolved module/sheaf, rather than incorporating the module/sheaf directly into the complex as a term.\n\nAlternatively, you can just start your sum at $i=1$. (This statement is a little misleading, since we would also need to re-index; see comments below.)\n\n• Hm. I actually thought about this? But starting the sum at $i=1$ doesn't seem to change anything, since $[\\mathscr T_0]=[\\mathscr K_0]-[\\mathscr I_1]=[\\mathscr P_0]-[\\mathscr P_0]=0$ in my notation. – Jesko Hüttenhain Aug 28 '14 at 23:57\n• You should add Zhen Lin's comment to your answer: The problem is really that I treated $\\mathscr G$ as part of the resolution. It is not enough to just start the sum at $1$, one needs to look at a different complext $\\mathscr P_\\bullet$ than I did. – Jesko Hüttenhain Aug 29 '14 at 0:13\n• Dear @JeskoHüttenhain, it seems to me that starting at $i=1$ doesn't change the Tor-zero term, but it changes the final product so you get $[\\mathscr F]\\cdot[\\mathscr G]$. (We also might want to change $(-1)^i$ to $(-1)^{i-1}$....) – Andrew Aug 29 '14 at 0:30\n• Well, changing that $(-1)^i$ to $(-1)^{i-1}$ is pretty much the same as losing the Tor-zero term. – Jesko Hüttenhain Aug 29 '14 at 0:34\n• Ah, I see, perhaps saying we can \"start at $i=1$\" is a little sloppy. Sorry about that. – Andrew Aug 29 '14 at 0:41"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71675223,"math_prob":0.99905115,"size":2579,"snap":"2019-35-2019-39","text_gpt3_token_len":952,"char_repetition_ratio":0.25165048,"word_repetition_ratio":0.016759777,"special_character_ratio":0.35866615,"punctuation_ratio":0.07429719,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999988,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T20:45:54Z\",\"WARC-Record-ID\":\"<urn:uuid:66782c58-4066-46b3-8a08-55d575d5292c>\",\"Content-Length\":\"144923\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68d8e3c1-1649-4fe4-b37d-fe69df689440>\",\"WARC-Concurrent-To\":\"<urn:uuid:f8c12ef4-9e81-48e4-a144-a026dfe05a40>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/912480/definition-of-multiplication-in-grothendieck-ring\",\"WARC-Payload-Digest\":\"sha1:STBJTVU6OPJ47CKXAOXSKA4YLWOV5V5B\",\"WARC-Block-Digest\":\"sha1:IYQ4MPO2XH7C4ELO33I2M6KWPCS2UXUP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315618.73_warc_CC-MAIN-20190820200701-20190820222701-00517.warc.gz\"}"} |
https://testmaxprep.com/lsat/community/100007048-help | [
"# Which one of the following could be a pair of people whose arrangements are identical?\n\nErikia on April 26, 2020\n\nHelp\n\nCan someone help me figure out how to answer this question?\n\nSkylar on April 26, 2020\n\n@Edunn, happy to help!\n\nWe can set up the game as follows:\nGHLR\n\nS: R R __ __\nT: H G __ __\nU: R H H __\nW: G G __ __\nZ: H R __ __\n\nWe get this setup from transcribing the rules. We also assigned a second H to U because of the 5th rule which states only Z has exactly one H and exactly one R. Since U originally had exactly one R and at least one H, we added another H so that it would not break Rule 5 by copying Z.\n\nTo build on this setup, we also know that each person must have exactly one pair of the same type of flower (because each has four spots and three types of flowers). T and Z are the only two arrangements missing their pair, so the double lily discussed in Rule 6 will be assigned to one of them.\n\nIn this specific question, we are tasked with identifying a pair of people who could have matching arrangements.\n- (A) is incorrect because Rule 1 tells us that S is the only person with two Rs.\n- (B) is incorrect for the same reason that (A) is.\n- (C) is correct. T and W could both have an arrangement of: H G G L without violating any rules.\n- (D) is incorrect because we have already established that U = R R H, so it only has one open spot left. Rule 4 tells us that W already has G G in two of its four spots, so for these two people to match, their arrangements would need to be R R H G G, which requires more than four spots.\n- (E) is incorrect because in order for W and Z to match, their arrangements would both have to be G G H R. This would violate Rule 5, which states that Z is the only person with exactly one H and exactly one R.\n\nIf you are having trouble with any of the other questions from this game, a walkthrough is posted on the message board of the last question.\n\nDoes that make sense? Please let us know if you have any additional questions!"
] | [
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https://studysoup.com/tsg/239736/calculus-early-transcendentals-2-edition-chapter-8-5-problem-10 | [
"×\nGet Full Access to Calculus: Early Transcendentals - 2 Edition - Chapter 8.5 - Problem 10\nGet Full Access to Calculus: Early Transcendentals - 2 Edition - Chapter 8.5 - Problem 10\n\n×\n\n# Solution: 918. The Ratio Test Use the Ratio Test to",
null,
"ISBN: 9780321947345 167\n\n## Solution for problem 10 Chapter 8.5\n\nCalculus: Early Transcendentals | 2nd Edition\n\n• Textbook Solutions\n• 2901 Step-by-step solutions solved by professors and subject experts\n• Get 24/7 help from StudySoup virtual teaching assistants",
null,
"Calculus: Early Transcendentals | 2nd Edition\n\n4 5 1 247 Reviews\n10\n1\nProblem 10\n\n918. The Ratio Test Use the Ratio Test to determine whether the following series converge. a _ k = 1 2k k!\n\nStep-by-Step Solution:\nStep 1 of 3\n\nStep 2 of 3\n\nStep 3 of 3\n\n##### ISBN: 9780321947345\n\nThis full solution covers the following key subjects: . This expansive textbook survival guide covers 128 chapters, and 9720 solutions. The full step-by-step solution to problem: 10 from chapter: 8.5 was answered by , our top Calculus solution expert on 12/23/17, 04:24PM. The answer to “918. The Ratio Test Use the Ratio Test to determine whether the following series converge. a _ k = 1 2k k!” is broken down into a number of easy to follow steps, and 22 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 2. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321947345. Since the solution to 10 from 8.5 chapter was answered, more than 241 students have viewed the full step-by-step answer.\n\nUnlock Textbook Solution"
] | [
null,
"https://studysoup.com/cdn/77cover_2630001",
null,
"https://studysoup.com/cdn/77cover_2630001",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.94386095,"math_prob":0.60955364,"size":11859,"snap":"2021-04-2021-17","text_gpt3_token_len":2326,"char_repetition_ratio":0.13606073,"word_repetition_ratio":0.20908593,"special_character_ratio":0.19360822,"punctuation_ratio":0.11800948,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98715615,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T00:23:41Z\",\"WARC-Record-ID\":\"<urn:uuid:00e20ca9-5c63-47db-8edb-bbda643b5ff0>\",\"Content-Length\":\"113235\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67faaaa8-1f9c-415c-bb85-73e53800e62a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6762ba7b-a81b-49d4-ac00-61406d873746>\",\"WARC-IP-Address\":\"54.189.254.180\",\"WARC-Target-URI\":\"https://studysoup.com/tsg/239736/calculus-early-transcendentals-2-edition-chapter-8-5-problem-10\",\"WARC-Payload-Digest\":\"sha1:KGR63EETS4QVDEXRK3WCX6RH26LV2LLC\",\"WARC-Block-Digest\":\"sha1:UMEK6ZKDS6SQ2TJA6QUZULCSPHGMRXSI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703538741.56_warc_CC-MAIN-20210123222657-20210124012657-00519.warc.gz\"}"} |
https://techcommunity.microsoft.com/t5/sql-server-blog/maintaining-unique-indexes/ba-p/383318 | [
"Maintaining Unique Indexes\nBy\nPublished Mar 23 2019 05:07 AM 469 Views\n\n# Maintaining Unique Indexes\n\nFirst published on MSDN on Sep 06, 2007\n\nConsider the following schema:\n\nCREATE TABLE T (PK INT PRIMARY KEY, A INT, B INT)\nCREATE INDEX TA ON T(A)\nCREATE UNIQUE INDEX TB ON T(B)\n\nINSERT T VALUES (0, 0, 0)\nINSERT T VALUES (1, 1, 1)\n\nNow suppose we run the following update statement:\n\nUPDATE T SET A = 1 - A\n\nThis update statement affects the clustered index and the non-clustered index TA. The plan is pretty much what you might expect:\n\n|--Clustered Index Update(OBJECT:([T].[PK__T__15502E78]), OBJECT:([T].[TA]), SET:([T].[A] = [Expr1003]))\n|--Compute Scalar(DEFINE:([Expr1016]=[Expr1016]))\n|--Compute Scalar(DEFINE:([Expr1016]=CASE WHEN [Expr1004] THEN (1) ELSE (0) END))\n|--Compute Scalar(DEFINE:([Expr1003]=(1)-[T].[A], [Expr1004]=CASE WHEN [T].[A] = ((1)-[T].[A]) THEN (1) ELSE (0) END))\n|--Top(ROWCOUNT est 0)\n|--Clustered Index Scan(OBJECT:([T].[PK__T__15502E78]))\n\nThis is a typical narrow update plan. The single clustered index update operator maintains both the clustered index and the non-clustered index TA. The plan includes compute scalars that detect whether each row of the non-clustered index really needs to be updated. I wrote about plans just like this one in this post .\n\nNow suppose we run the same update statement but this time we modify column B:\n\nUPDATE T SET B = 1 - B\n\nSuddenly the plan is much more complex:\n\nRows Executes\n2 1 |--Index Update(OBJECT:([T].[TB]), SET:([PK1022] = [T].[PK],[B1023] = [T].[B]))\n2 1 |--Collapse(GROUP BY:([T].[B]))\n4 1 |--Sort(ORDER BY:([T].[B] ASC, [Act1021] ASC))\n4 1 |--Filter(WHERE:(NOT [Expr1019]))\n4 1 |--Split\n2 1 |--Clustered Index Update(OBJECT:([T].[PK__T__15502E78]), SET:([T].[B] = [Expr1003]))\n2 1 |--Compute Scalar(DEFINE:([Expr1019]=[Expr1019]))\n0 0 |--Compute Scalar(DEFINE:([Expr1019]=CASE WHEN [Expr1004] THEN (1) ELSE (0) END))\n0 0 |--Compute Scalar(DEFINE:([Expr1003]=(1)-[T].[B], [Expr1004]=CASE WHEN [T].[B] = ((1)-[T].[B]) THEN (1) ELSE (0) END))\n2 1 |--Top(ROWCOUNT est 0)\n2 1 |--Clustered Index Scan(OBJECT:([T].[PK__T__15502E78]))\n\nWhat's going on? This time we updated a unique index. The SQL Server storage engine enforces the uniqueness of the index at all times. It does not permit the query processor to insert any duplicate rows into the index at any time. The query processor, on the other hand, must ensure that an update statement succeeds so long as it does not leave behind any uniqueness violations upon its completion. Based on this rule, the above update statement must succeed.\n\nLet's consider what would happen if SQL Server simply tried to update the non-clustered index TB using the same simple plan that it used when we updated column A. To execute this plan the server must scan the rows in the table and update them one at a time. Suppose the server chose to update the row with PK=0 first. In that case, it would attempt to change the value of B from 0 to 1. Unfortunately, there is already a row in the index with B=1. The storage engine would enforce the uniqueness of the index and the update would fail. The update would similarly fail if the server chose to update the row with PK=1 first. It would seem that there is no way to execute this update!\n\nFortunately, SQL Server has a solution. The basic idea is simple. Instead of updating the key columns of a unique index which can cause \"false\" uniqueness violations, the query processor reorganizes the update such that the non-key columns are updated instead. This reorganization is implemented by the split, sort, and collapse operators. Let's take a closer look at how this works.\n\nIn our example, we begin with two rows to update:\n\n PK B_old B_new 0 0 1 1 1 0\n\nThe split operator transforms the updates into deletes followed by inserts:\n\n Action PK B Delete 0 0 Insert 0 1 Delete 1 1 Insert 1 0\n\nNotice that as indicated by the above STATISTICS PROFILE output, we now have 4 rows.\n\nThe sort operator reorders the inserts and deletes by the key columns of the non-clustered index (in this case by column B). If there is a delete and an insert that share the same key value, the delete sorts before the insert. The results of the sort are:\n\n Action PK B Delete 0 0 Insert 1 0 Delete 1 1 Insert 0 1\n\nThe collapse operator combines adjacent delete and insert pairs that share the same key value into a single update:\n\n Action PK_old PK_new B Update 0 1 0 Update 1 0 1\n\nIn this example, the 4 rows collapse back into 2 rows leaving only updates. Notice that the updates no longer change column B (which could lead to a false uniqueness violation) but rather change column PK which is not a unique key of index TB and which cannot fail due to a uniqueness violation. Note also that in general it is not necessarily the case that all deletes and inserts collapse into updates. The results of the collapse operator may include any combination of inserts, updates, and deletes.\n\nFinally, the index update operator executes the update operations output by the collapse operator.\n\nKeep in mind that just because a query plan includes split, sort, and collapse operators does not mean that it may not result in an actual uniqueness violation. It simply ensures that there will be no false uniqueness violations. Moreover, SQL Server generates plans with split, sort, and collapse operators anywhere that there is a risk of a false uniqueness violation. There may or may not be an actual uniqueness violation. For example, the following update statement which (with the current data set) would not result in a false uniqueness violation generates a nearly identical plan:\n\nUPDATE T SET B = B + 10\n\nOn the other hand, the following update statements can only update a single row. SQL Server is clever enough to recognize that these statements cannot generate false uniqueness violations and generates simpler plans without the split, sort, and collapse operators:\n\nUPDATE T SET B = B + 10 WHERE PK = 0\nUPDATE TOP (1) T SET B = B + 10\n\nVersion history\nLast update:\nMar 23 2019 05:07 AM\nUpdated by:"
] | [
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https://eng.ichacha.net/m/arithmetic.html | [
"×\n\n# arithmetic meaning in Chinese\n\n[ ə'riθmətik ] Pronunciation: \"arithmetic\" in a sentence \"arithmetic\" meaning\n• n.\n1.算术，算法；计算。\n2.算术书。\nPhrases\n+More...\n\n### Examples\n\nMore: Next\n1. She grounded her pupils well in arithmetic .\n她给自己的学生打下良好的算术基础。\n2. An arithmetic or an algebraic identity is an equation .\n算数或代数恒等式是方程。\n3. Subsistence only increases in an arithmetic ratio .\n生活资料只能按算术级数增长。\n4. He has a knack of teaching arithmetic .\n他教算术有诀窍。\n5. One should therefore take the arithmetic mean as the sum .\n必须取算术平均作为和。\n\n### Related Words\n\nPC Version한국어简体繁體日本語DefinitionHindi"
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https://www.jiskha.com/search?query=I+solved+%282x-8%29%2Fz%3Dy%2C+and+got+2yz-8%3Dx+as+an+answer%2C+which+my+teacher+said+was+wrong.+What+would+the+right+answer+be+then%3F+What+did+I+do+wrong%3F+Thanks%21 | [
"# I solved (2x-8)/z=y, and got 2yz-8=x as an answer, which my teacher said was wrong. What would the right answer be then? What did I do wrong? Thanks!\n\n83,117 results\n1. ## Math - AP Calculus\n\nThe rate at which water flows into a tank, in gallons per hour, is given by a differentiable, increasing function R of time t. The table below gives the rate as measured at various times in an 8-hour time period. t (hours) 0 2 3 7 8 R(t) (gallons per hour)\n\n2. ## Calculus\n\nFind the rate of change of f(x,y,z) = xyz in the direction normal to the surface yx^2 + xy^2 + yz^2 = 75 at (1,5,3). I keep getting this wrong and I don't know what I did right. The answer's 147/sqrt(101) but I got 357/sqrt(1361). Is my method right? Or\n\n3. ## Math - Ms. Sue\n\n1) Which is M = 9pn solved for p? a. p = M/n b. p = M - 9n c. p = 9n/M d. p = M/9n 2) Which is M = 300J/P solved for J? a. J = 300PM b. J = 300M/P c. J = 3OO/PM d. J = PM/300 I don't really understand this but my answer are 1. d 2. b\n\n4. ## calc-kinematics\n\nif a particle moves along a line so that at time t where t [0,pi] its position is given by s(t)=-4cost-t^2/2+10 what is the velocity of the particle when its acceleration is zero? so i found v(t) = 4sint-t a(t)=4cost-1 = 0 solved for t plugged it into v(t)\n\n5. ## Physics MCAT, Help please\n\nA box of mass m = 6kg slides with speed v = 4 m/s across a frictionless floor. It suddenly explodes into two pieces. One piece, with mass m1 = 2kg moves in the same direction with speed v1 = 8m/s. Find the velocity of the second piece. Its answer is 2 m/s\n\n6. ## calc 1\n\nIf h(x) = the square root of 4 + 3f(x) where f(2) = 4 and f '(2) = 2, find h'(2). i got the derivative as 1/2(4+3f(x))^-1/2 then i solved for the inside which is (0+3(8)) then multiplied the the two eating and i got 3.... but answer is wrong =(\n\n7. ## math\n\nWhat two fractions, add up to 1/3? i know its a very very stupid question, but my teacher said adding 5/30 + 5/30 is wrong (Which i did), but maybe he read the question wrong. Answer, Maybe?\n\n8. ## Health of young children\n\nA teacher is in a prime position to observe children because A.the teacher sees children in relation to other children of the same age. B. the teacher sees each child for fewer hours than the parent and can, therefore be objective. C.he or she is often\n\n9. ## Science/Math/HElP\n\nScience- A 25N object is 3 meters up. How much potential energy does it have? our teacher said that the formula for finding out potential energy is PE = Mass * Gravity * Height. The gravity on Earth is always 9.8, but my problem is that instead of mass the\n\n10. ## Math\n\nEvaluate the following definite Integral ∫x * ln(x) dx from 1 to e^(2) I used the DI method which gave me D I + ln(x) x - 1/x (x^2)/2 = (ln(x)*x^(2))/2 - ∫ 1/x * x^(2)/2 = (ln(x)*x^(2))/2 - ∫ x/2 = (ln(x)*x^(2))/2 - x^(2)/4 + C (ln(x)*x^(2))/2 -\n\n11. ## Algebra\n\nsimplify (6x^-2)^2(0.5x)^4 I think I know the answer is 2.25 but my work is wrong my teacher said and I need to get this right so plz any help is fine\n\n12. ## Math\n\nEvaluate the integral of (e^2x)*sin^3 x dx I let u = e^2x, du = (1/2)e^2x dx v= (-1/3)cos^3 x , dv =sin^3 x dx When I used integration by parts and solved it all out I got: (37/36)intgral of (e^2x)*sin^3 x dx = (-1/3)(e^2x)*cos^3 x + (1/18)(e^2x)*sin^3 x\n\n13. ## math\n\nSolve the quadratic equation by completing the square. Verify your answer graphically. (Enter your answers as a comma-separated list.) 16x2 − 24x − 23 = 0 my answer was: x =16(x−3/4)^2−32 but online website (where i do my assignments that is posted\n\n14. ## calculus; implicit differentiation\n\nUse implicit differentiation to find the slope of the tangent line to the curve at the point (4,1) -1x^2 - 4xy + 3y^3 = -29 I differentiated both sides and solved for dy/dx and got 2x-4y/9y^2-4x. Then I plugged in X and Y and got -4/7 but when i enter my\n\n15. ## math\n\nI'm having a little trouble solving inverse variation. I have four problems - 2 are solved using proportions, 2 are not --- why can they not all be solved using proportions? 1. y varies inversely with x. If y = 40 when x = 16, find x when y = -5 If I use a\n\n16. ## math\n\nReduce the fraction to lowest terms. Do not use spaces in answer. 38x^2yz^2 __________= -19xy^2z^3\n\n17. ## chemistry\n\n6.2(+_ 0.2) - 4.1 (+_ 0.1 ) + ? [6.2( +_ 0.2) - 4.1(+_ 0.1)] : 9.43(+_0.05) find absolut uncertainty and relative uncertainty. PLEASE HELP:I solved them and got the exact absolute uncertainty like the answer key, but I didn't get right answer of relative\n\n18. ## algebra\n\nSolve for s in this equation. Depreciation. D=C-s/n This is my answer D-C=C-s/n-c D-C=-s/n -n(D-C)=-s/n*(-n) -nD+nC=s s=n(C-D) Please chack to see if this is correct Looks fine to me. A slightly shorter (by 2 lines) rearragement is: D=C-s/n D+s/n=C s/n=C-D\n\n19. ## Supportive learning environment\n\nIt was the wrong question...Sorry Scott is just beginning to learn the names for the letters of the alphabet. Which of the following would be an example of the teacher scaffolding his learning? A. The teacher tells Scott, \"You need to learn all your\n\n20. ## math\n\nFor the pair of numbers (-1/5),(-1/2) My teacher wants us to find the quadratic equation with integral coefficients that has these numbers as solutions. I am not sure if I have the right equation for this. Could someone let me know if I am right or wrong\n\n21. ## Math integrals\n\nWhat is the area of the region bounded by y=x^2, the tangent to this parabola at (1, 1) and the x-axis? Since it says that the parabola passes through 1,1 can I assume that the line is y = x or is that completely wrong? Doing that I got ∫ x - x^2 dx\n\n22. ## Physics\n\nA capacitor is made from two 1.2 cm diameter coins separated by a 0.18 mm thick piece of paper (K = 3.7). A 6 V battery is connected to the capacitor. How much charge is on each coin? My solution: C= K * Permitivity * A/d = q/v I solved for q and got 1.2 X\n\n23. ## math:solving for an exponent.\n\nI can't get the right answer, where am I going wrong? 7.0x10^9= (6.7x10^9)(2.72^r*3) 1.04 = 2.72^r*3 When I solve for this, I get ln (1.40) = 0.03922 then 0.03922 / 3 = 0.0130 So my r equals 0.0130 BUT the teacher says the answer is 0.05 = r *3 0.02 = r\n\n24. ## Math\n\nlim as x approaches negative infinity (3x^5 + 6x + - 3) / (8x^4 + 10) a. Does Not Exist b. 0 c. -3/8 d. 3/8 I solved this and got negative infinity as the answer, but none of the answers are listed above. I finally chose Choice C, but my teacher said the\n\n25. ## chemistry/math\n\nThe density of air at ordinary atmospheric pressure and 25°C is 1.19 g/L. What is the mass, in kilograms, of the air in a room that measures 12.0 ft 12.5 ft 7.3 ft? i made ft^3 into Liters and got 8,346,211.799 then I made the grams into kilograms and set\n\n26. ## Social Studies 7R (answer quick)\n\nI got my Native Americans test back today during ss and I got a 77 (4 qs. wrong there was 10 mulitple and 13 reponse but my teacher think it will take for ever for us to do the test so we had to do all the multiple qs. and for the reponse qs. do 9 qs. and\n\n27. ## math,help\n\non one of last week homeworks i got this one wrong can someone explain to me what is the correct format. The demand supply equations for a certain item are given by: D= -5p+40 S=-p^2+30p-8 The correct answer was: \\$1.43 how do you get to this answer because\n\n28. ## math\n\nCould someone please show me what I did worng in my math questions? any help will be greatly appreciated! I put my teachers response under the questions 4. Solve: 5|2x + 1 | = 55 Divide both side by 5: |2x+1|=11 2x+1=11 and 2x+1= -1 +1 +1 +1 +1 2x = 12 2x\n\n29. ## math\n\nThe graph of the equation 8x–5y=14 passes through a point with an abscissa of 1.2. Find the ordinate of the point. I got .8 repeating but my teacher said that it was wrong. I then got 9.6 and it was also wrong. what am I doing wrong and can you put the\n\n30. ## Chemistry\n\nso i got a test and like usual my teacher didn't teach me half of the stuff that's going to be on it so he gave us a pretest of what is going to be on it and an answer key and i got several questions 13.) what is the pH of a formic acid if the\n\n31. ## MATH-Stuck :-(\n\nI really need help factoring these special Trinomials. This question really puzzled me. I thought I had it correct, but when I checked the answer at the back of the book, I was wrong, and so I would like to know what was it that I had done wrong. This was\n\n32. ## English\n\nI wrote, \"I like to swim in summer.\" My teacher marked it wrong and said it should read \"I like to swim in THE summer.\" Can anyone offer an explanation, please, why my way is wrong? I don't know why it was marked wrong. Both with and without \"the\" are\n\n33. ## Please Check My Math Work\n\nHi. - I solved this problem,but I am not sure if it is correct.If you can please tell me if it is right or if it is wrong what is wrong with it.: 6y^2-54= (3y-6)(2y+9) *Many of you may be are wondering why my title is,\"[C]razy Baby. - [Iloveyou]\"it is\n\n34. ## Chemistry\n\nAn aqueous solution of ammonium nitrite decomposes when heated to give off nitrogen gas and water. This reaction can be used to prepare pure nitrogen gas. How many grams of ammonium nitrite must have reacted if 3.75 liters of nitrogen gas was collected\n\n35. ## English\n\n1. He is a social science teacher. 2. He is teacher. 3. She is a home economics teacher. 4. She is a HE teacher. 5. She is a technical arts and homemaking teacher. 6. She is a TAH teacher. 7. She is a healthe education teacher. 8. She is a HE teacher. 9.\n\n36. ## General Chemistry\n\nSo I've been working on this online homework problem and got it wrong 8 times. I need help. Here's the question: 1,2-diaminoethane, H2N-CH2-CH2-NH2, is used extensively in the synthesis of compounds containing transition-metal complexes in water. If pK_b1\n\n37. ## math\n\nkelly solved the following problem on her recent math test 121/11+4*8=120 the answer was marked wrong by your teacher and kelly is asking you to explain why she missed the promblem.whicthsolutions would you tell her A.divide 121 by 11,multipy 4 by 8 then\n\n38. ## Calculus\n\nWhich of the following is a point at which the tangent to y=(x-2)^2(x+3) is horizontal? a) (-2,16) b) (-1, 18) c) (1,4) d) (2,0) I found the derivative of y=(x-2)^2(x+3) which is 3x^2-2x-8 if I solved for 0, it wouldn't give me the answer that are listed\n\n39. ## Math\n\nlim as x approaches negative infinity (3x^5 + 6x + - 3) / (8x^4 + 10) a. Does Not Exist b. 0 c. -3/8 d. 3/8 I solved this and got negative infinity as the answer, but none of the answers are listed above. I finally chose Choice C, but my teacher said the\n\n40. ## Introduction to Educational Theory\n\nWe have to analyze this situation: A teacher demeans her students. The teacher calls her students lazy and says they will end up in remedial classes. She states that anyone with less than a B \"obviously doesn't care.\" The teacher sometimes refuses to\n\n41. ## algebra\n\nThere was a question on my quiz asking 5/6x=25 what is x. I got the answer 30.12 but my teacher says it is wrong because the answer is 30. I don't understand how to get just 30\n\n42. ## physics\n\nA 0.40kg piece of putty is dropped from a height of 2.9m above a flat surface. When it hits the surface, the putty comes to rest in 0.32s . What is the magnitude of the average force exerted on the putty by the surface? I keep getting 9.4 as the answer,\n\n43. ## physics\n\na 1500 kg car travelling at 16.7 m/s to the south collides with a 6300 kg truck that is initially at rest at a stop light. the car and the truck sticks togetherand moves together after the collision. what is the final velocity of the two-vehicle mass?\n\n44. ## math\n\nCan someone help me answer this question? It's combining like terms, my teacher said that the first term is right but the rest is wrong. Question~~ 2x^2y–5xy–7xy–8x^2y+9xy^2 –xy^2 My answer ~~ 2x^2+-8x^2 (2-8)x^2 -6x^2 y+y+y+y (1+1+1+1)y 4y\n\n45. ## Math\n\nHi, I am getting very frustrated. I am doing an online homework for chemistry and it's telling me to round to the nearest WHOLE number. So the answer I got was -1435.650.... And I typed in -1436. and said I was wrong and so I tried it again and typed 1.4 x\n\n46. ## Algebra 1\n\nRandy, Sam, Tim and Will counted the number of puzzles they had solved in two months. Sam has solved twelve more than Tim. Randy had solved seventeen less than Sam.Will had solved twice as many as Randy. They have solved a total of 267 problems. How many\n\n47. ## Chemistry\n\nIGNORE MY FIRST POST. I CORRECTED THIS ONE Write IUPAC name of the following alkanes: 1. (CH3)3 -C-CH2-CH2-CH3 all the numbers are subscripts. So there are 3 CH3's, C,CH2,CH2, and CH3 This is what my teacher has in my homework. The parenthesis really throw\n\n48. ## Algebra\n\n(−2x^2yz^2+3x^2y)·(−5y^2z^3−4z)·(xy^2+z)\n\n49. ## Writing:Quotes, Journal\n\nPlease may you check my answers for a quote that says---> Knowledge speaks, Wisdom listens. This quote indicates that people with knowledge will give aedvices to people, and people that listens to the individuals are wise to do so. For an example, let's\n\n50. ## pre-algebra\n\nit says use equivalent ratio's to solve the proportion!!! ok is there a different way to do them kind of proportions because i did it like that yesterday and my teacher said they were all wrong Give an example of what you did and how you came up with your\n\n51. ## Science math\n\nCan no one help me? I posted a question earlier I don't need answer I need to understand what I did wrong my thinking is that if I have a 2:1 ratio And my total is 99 that if I divide by 3 then I can get the amounts 33x2 is 66, then 33x1 is 33 so it would\n\n52. ## math\n\nCould someone please check my math questions? any help will be greatly appreciated! 4. Solve: 5|2x + 1 | = 55 Divide both side by 5: |2x+1|=11 2x+1=11 and 2x+1= -1 +1 +1 +1 +1 2x = 12 2x = -10 X=5 x=-6 What my teacher said was wrong~~ 2x+1=11 and 2x+1= -11\n\n53. ## math\n\nCan you tell me how you would answer this question? My teacher says that one answer is correct but I think that my answer is correct. I want a second opinion. A soccer coach purchases 12 uniforms for 17.50 each. He buys 6 soccer balls for 8.58. How much\n\n54. ## Math\n\nI Don't Know What Is This Subject Called In English But It's About Complex Numbers it says 25 = 16 - 9 i^2 = (4 ± 3i) But My Teacher Says 25 = 25 - 0 i^2 = (5 ± 0i) I Think My Teacher Is Wrong , The Result Is 25 But There Will Not Be Complex Number\n\n55. ## Algebra\n\nOn my quiz for (3x+2)^2, I wrote the answer as 21x=4. My teacher gave me a minus one, which meant it was half wrong. What is wrong about my answer? Thanks\n\n56. ## Physics\n\nAn 85kg person bounces on a bathroom scale having a force constant of 1.50*10^6 N/m. The amplitude of the bounce is 0.200 cm. What is the maximum energy stored in the spring? I use the formula mg = kx, solved for x, and got 5.55333 * 10^-4. Then I added\n\n57. ## Physics 11\n\nThe brakes on a car permit it to decelerate at the rate of -8.0m/s^2. How much distance is required to stop this car when it is travelling 60.0km/hr? The answer given is 173.6m I solved it this way 60km/h = 16.667m/s A=V/T 16.667m/s / -.80m/s^2 = 20.834\n\n58. ## MATH-STUCK :-(\n\nI really need help factoring these special Trinomials. This question really puzzled me. I thought I had it correct, but when I checked the answer at the back of the book, I was wrong, and so I would like to know what was it that I had done wrong. This was\n\n59. ## Geometry\n\nI need help solving for the height of a trapezoid and finding the correct final answer for a semi circle (a half circle).In area form Here's what I have so far: Trapezoid Formula: 1/2(b1+b2)h My Answer:1/2(5+15)=10 (My bases are 5 and 15) I need help\n\nThe problem is: (8.86 + 1.0 X 10^-3) / 3.610 X 10^-3 Well I got 2.73 X 10^-3 as an answer, but I know for a fact that this is wrong because my teacher said it was. Your teacher is right. What do you get when you add 8.86 + 1.0 X 10^-3 = 8.86 + .001 ? What\n\n61. ## chemistry\n\nSF4 ---> SF2 + F2 rate constant (k) = .011 l/molxS [SF4]2 How many minutes will it take for the concentration of SF4 to be reduced from 2.5 M to .25 M? I said this was a 2nd order reaction and used : (1/final)= kt + (1/initial) and solved for t. the answer\n\n62. ## Physics HELP! QUICK!\n\nA banked circular highway curve is designed for traffic moving at 55 km/h. The radius of the curve is 220 m. Traffic is moving along the highway at 38 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow\n\n63. ## Geometery\n\nA water storage tank is shaped like a circular cylinder with half of a sphere at each end. Find the volume of the tank if the cylindrical section is 100 cm long with a 40cm diameter. Use 3.14 for ~~ Be sure to round the final answer to two decimal places.\n\n64. ## English\n\n1. I stayed there for two weeks. 2. I stayed there two weeks. Are both grammatical? -------------------------- 3. This is the way how he solved the problem. 4. This is the way that he solved the problem. 5. This is the way he solved the problem. 6. This is\n\n65. ## Calculus\n\nFind the area under the curve of y=-2x+6 from x=0 to x=3 using 3 strips of equal length using right endpoints. I tried this equation before and my answer was 9 but my teacher said it was wrong? and I also posted this before and Someone who helped me got\n\n66. ## Chemistry\n\nCan somebody explain how to do this? I am stuck. I've tried solving for H+ concentration. I got .00024M and subtracted it from [OH-]. Then I solved for pOH and used that to get pH but still got the wrong answer. Somone plz explain this to me.\n\n67. ## Physics!\n\nTwo identical isolated particles, each of mass 1.55 kg, are separated by a distance of 30.6 cm. What is the magnitude of the gravitational force exerted by one particle on the other? --------------- I solved it as: Fg = G(m^2/r2) = G(m/r)2 = 6.673x10^−11\n\n68. ## Due in an hour help please\n\nSimplify: x^3+x^3 My answer was : 2x^6 is this incorrect because my test the teacher had marked it wrong as if he was saying that it. Was going to be. 2x^3 instead\n\n69. ## plz help english\n\nmy question is:- Complete the sentence below by interrupting it with two parallel if clauses: The problem of race relations, if………., and if………. must be solved. question is i wrote:- the problem of race relation, if it goes on increasing, and if\n\n70. ## Chemistry\n\n@DrBob222 I got the last question that I posted wrong and I solved it the same way. Could you show me the answer and how you got it? My test is Friday and this is a practice problem. Suppose that 30 mL of 0.5 M CsOH and 55 mL of 0.1 M NaHS are mixed. What\n\n71. ## Early Child Ed.\n\nWould you mind please, checking these questions and my answers? 1. Which of the following is the best example of a teachable moment? A. A teacher telling a young child it is time to use the potty. B. A child struggling to put on his own shoes. C. A child\n\n72. ## College Mathematics. 3\n\n(3t+1)/5=(t+7)/10+(t-5)/10 I solved this question through the linear equation simplification timing both sides by the least common denominator. 5 Once I did the the result was > 3t+1= 2t+14+2t-10 T= 3 However the real answer is t=0 Where am I going wrong?\n\n73. ## Algebra\n\nRandy, Sam, Tim and Will counted the number of puzzles they had solved in two months. Sam has solved twelve more than Tim. Randy had solved seventeen less than Sam.Will had solved twice as many as Randy. They have solved a total of 267 problems. How many\n\n74. ## math\n\nI'm trying to multiply this and keep getting the wrong answer: (x - 5)(-0.025x + 220) I have used the FOIL method and I have multiplied (x - 5) -0.025x and then multiplied (x - 5) 220. Both methods I get the same answer. But... the answer book and the\n\n75. ## Chemistry\n\nWhat is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HC2H3O2(aq)? Can somebody explain how to do this? I am stuck. I've tried solving for H+ concentration. I got .00024M and subtracted it from [OH-].\n\n76. ## algebra- integer problems\n\nHi! i have this problem, I'm stuck on. it says find 2 consecutive multiples of 9 whose sum is -153. I started of my saying the 1st number is 9x and the 2nd is 9x+9. I continued on to get, 9x+(9x+9)=-153 and solved it. I got an irrelevant answer: -9. I was\n\n77. ## Calculator Help\n\nIs there a difference between TI-89 Titanium calculator and TI-83 calculator? Because I'm taking a final exam and my teacher will only allow us to use a TI-83 calculator and I've never used it... like for putting in the parentheses with the log when\n\n78. ## English Grammar\n\nAs many of you know, I'm an ESL Teacher in Taiwan. I came across a sentence that does not sound right, but I am wondering if it's grammatically correct. \"I like to have cats.\" I've NEVER heard it said like that before, but I cannot think of any reason why\n\n79. ## algebra hsa\n\n5x-3=2(x-6) i got -3 when i solved it but when i check my work its not correct what am i doing wrong?\n\n80. ## Calculus\n\nsif xyz=0 (-3x^2yz)^3 p=27x^7y^4z^6 then p=?\n\n81. ## French\n\nmy teacher marked these phrases as wrong but i don't know what about them is wrong: 1. je revelerais sans prevenir 2. j'ai vendu mes robes plus jolies both of them need \"un mot a ajouter\"\n\n82. ## Algebra\n\nI solved (2x-8)/z=y, and got 2yz-8=x as an answer, which my teacher said was wrong. What would the right answer be then? What did I do wrong? Thanks!\n\n83. ## tere\n\nwhat did i do wrong? i need to find lowest vapor pressure. a 0.13 m FeCl3 b 0.50 m sucrose c 0.24 m NaNO3 d 0.15 m Na2PO4 so i solved. a 4 x 0.13 = 0.52 b 1 x 0.50 = 0.50 c 5 x 0.24 = 1.2 d 7 x 0.15 = 1.05 i chose C but the answer is A. did i do something\n\n84. ## Math\n\nCan you tell me how you would answer this question? My teacher says that one answer is correct but I think that my answer is correct. I want a second opinion. A soccer coach purchases 12 uniforms for 17.50 each. He buys 6 soccer balls for 8.58. How much\n\n85. ## math\n\n(4 − 2x)−2 x=-2 y=4 z=-3 how do you solve this? i have solved it and gotten 6 for my answer many times and it says it's wrong. other equations are (y − z)−3 and (x − 5y)0. i cant get them right\n\n86. ## cacl\n\nintegral of x^3+65/X^+7x+12 dx when i solved i got x^2/2-7x+64ln(x+4)-111ln(x+3)+c however this isn't the right answer....what have i done wrong??\n\n87. ## Physics\n\nA rectangular piece of gold is 3.1 cm long and 3 cm wide. Resistivity of gold is 2.4*10^-8. How deep is the piece when it displays a resistance of R = 1.65 μω for a current that flows along the depth-direction? I used R=(p*l)/A and solved for l. I got\n\n88. ## grammar\n\nMs Sue your a good pesron and a good teacher to admitt that you was wrong and I admire you for that. I am still confused. I don't know what is right or wrong on this and I am getting freaked out on this. So please send me some explanation on the answer\n\n89. ## math\n\nhow can I solve: 6³ + 6 ÷ 2¹ - 4² I got the answer wrong on the test and the teacher told me the answer was 203 but I want to know how to do it. I was to embarrassed to ask because all the other students knew how to do it\n\n90. ## Ap algebra\n\n2+3X^2-X^3=0 by use quadratic formula. The answer of textbook is 3.19 So I solved and came out a wrong answer What I did factor out (-X+3)X^2+2=0 Then By quadratic formula, a=1,b=0,c=2. Thus I got sqrt(8)/2\n\n91. ## Calculus\n\n1.) Consider the graphs x+5y=17 and x+7=y^2 where a=7 b=2 c=57 f(x) = ? g(x) = ? I have solved for a,b, and c but I can't figure out f(x) and g(x) For f(x) I thought it was (17-x)/5 but it's not the correct answer. For g(x) I thought it was the sqrt(x+7)\n\n92. ## math\n\nthe tripled product of 6a and b^2. I got 216a^3b^6, but my teacher says that this is wrong. my equation is \\(6ab^2)^3 is this wrong?\n\n93. ## math\n\nrounding to the nearest ten 2,123 my child put 2,120 and the teacher said it was wrong. what is the correct answer?\n\n94. ## Algebra\n\nI need to rewrite the following as an equivalent logarithmic yt = x I wrote logy=logt=logx. My teacher said that was wrong. Can someone guide me on the right answer?\n\n95. ## Math\n\nSolve the indefinite integral of 1/sqrt(x^2+2x+5). I solved it all out by completing the square and then trig sub and then drawing a triangle to go back to x variable and got: ln |(sqrt(x^2+2x+5)/2 + (X+1)/2 ) | but the book answer is: ln |(sqrt(x^2+2x+5)\n\n96. ## Math\n\nSolve the indefinite integral of 1/sqrt(x^2+2x+5). I solved it all out by completing the square and then trig sub and then drawing a triangle to go back to x variable and got: ln |(sqrt(x^2+2x+5)/2 + (X+1)/2 ) | but the book answer is: ln |(sqrt(x^2+2x+5)\n\n97. ## Math\n\nThe problem is ; 10x+8-y X= 2 Y=5 Every time I do it I get 23 but my teacher says I'm wrong . where am I wrong ? someone help me please .\n\n98. ## Algebra\n\nWhat is the best estimate for 3 + (-14.9)(-2.3)? a 36 b 31 c 48 d 33 I came up with 36 but my teacher counted it wrong. I reworked it but still come up with A-36. What am I doing wrong?\n\n99. ## Math\n\nFind the point (x,y) on the unit cirlce that corresponds to the real number t. t= 11 pie / 6 My answer was (1/2 , -sqrt3/2) but my teacher said it is wrong!\n\n100. ## Algebra\n\nRemove the parentheses from the following expression: (2y + x) ÷ z. A. 2yz + xz B. 2y/z + x/z C. Z/2y + x D. 2y + x + z"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95398897,"math_prob":0.8944268,"size":24592,"snap":"2022-05-2022-21","text_gpt3_token_len":7571,"char_repetition_ratio":0.13636734,"word_repetition_ratio":0.13137174,"special_character_ratio":0.31929082,"punctuation_ratio":0.10361613,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982001,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-23T01:26:11Z\",\"WARC-Record-ID\":\"<urn:uuid:5a1e82ee-7182-4698-84ee-2fd6fa2d59c1>\",\"Content-Length\":\"68790\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8ea86204-199b-44c1-a908-81fc70e361a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b773102-50a6-4d61-848f-637124de8ec0>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/search?query=I+solved+%282x-8%29%2Fz%3Dy%2C+and+got+2yz-8%3Dx+as+an+answer%2C+which+my+teacher+said+was+wrong.+What+would+the+right+answer+be+then%3F+What+did+I+do+wrong%3F+Thanks%21\",\"WARC-Payload-Digest\":\"sha1:HC4L4DJ7Q2BLBQ2X4B5OK2HGVYQAQR5C\",\"WARC-Block-Digest\":\"sha1:6GH6JO6T7IVZCLNDQLKNEMBB6XMDPAHB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662552994.41_warc_CC-MAIN-20220523011006-20220523041006-00211.warc.gz\"}"} |
https://stat.ethz.ch/pipermail/r-devel/2005-January/031803.html | [
"# [Rd] How to connect R with a C code which includes Atlas\n\nProf Brian Ripley ripley at stats.ox.ac.uk\nWed Jan 5 07:37:42 CET 2005\n\n```On Wed, 5 Jan 2005 xt_wang at cse.concordia.ca wrote:\n\n> Hello, everybody,\n>\n> Happy New Year!\n>\n> I am a graduate student from Concordia University. I meet a problem when\n> I am working on my thesis. I hope I will get help from you.\n>\n> I have a mathematical model which was already implemented by using R\n> language. However, part of this model includes fixed point iteration\n> algorithm and calculation of large linear equations which n will get to\n> 5000. Because of limitation of memory, R is not enough to support this\n> kind of calculation. Now, I use C language with Matlab C library to\n> implement this part of model and solve this problem. But the problem is\n> that I can not connect my C program with R code since C program include\n> Matlab.h.\n>\n> The function I used from matlab is Cholesky algorithm which is applied\n> to calculate inv(C)*b where C is a symmetric positive definite matrix.\n> Since R is not easy to connect with C which include Maltab C library,\n> and I find Atlas is optimized linear algebra library and possible to be\n> used by R, do you think it is possible for me to replace matlab by using\n> Atlas function in my C code, and finally the C code can be connected\n> with R dynamically? That means I can run a R program -> which include C\n> code-> which include Atlas automatically. If it is possible, could you\n\nCertainly. Just put your C code into a package and include \\$(BLAS_LIBS)\nwhen you build the package: see `Writing R Extensions'.\n\nNote that ATLAS is an optimized BLAS, the B standing for Basic. For a\nCholeski decomposition you could use LAPACK as the R sources do and that\nwill in turn use ATLAS if you built R against ATLAS.\n\n--\nBrian D. Ripley, ripley at stats.ox.ac.uk\nProfessor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/\nUniversity of Oxford, Tel: +44 1865 272861 (self)\n1 South Parks Road, +44 1865 272866 (PA)\nOxford OX1 3TG, UK Fax: +44 1865 272595\n\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92597157,"math_prob":0.49022502,"size":2137,"snap":"2019-51-2020-05","text_gpt3_token_len":554,"char_repetition_ratio":0.09048289,"word_repetition_ratio":0.0,"special_character_ratio":0.26438934,"punctuation_ratio":0.12162162,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9591911,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T20:11:03Z\",\"WARC-Record-ID\":\"<urn:uuid:3e28633d-155e-4e86-a699-46cc46d4149c>\",\"Content-Length\":\"5169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a4fb54f-4a15-4415-add4-0a09f637a3bb>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ddddae3-824e-463e-9801-4232865384e6>\",\"WARC-IP-Address\":\"129.132.119.195\",\"WARC-Target-URI\":\"https://stat.ethz.ch/pipermail/r-devel/2005-January/031803.html\",\"WARC-Payload-Digest\":\"sha1:7GZZEXDSBESYXOBFTISDJNXG3QB6KUT6\",\"WARC-Block-Digest\":\"sha1:3NWWUXGEENWE7XREGJ7FAUORJUZZWHW4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250599789.45_warc_CC-MAIN-20200120195035-20200120224035-00054.warc.gz\"}"} |
https://republicofsouthossetia.org/question/calculate-the-instantaneous-rate-of-change-of-3-2-2-at-1-16346766-20/ | [
"## Calculate the instantaneous rate of change of () = 3x^2 − 2 at x = 1.\n\nQuestion\n\nCalculate the instantaneous rate of change of () = 3x^2 − 2 at x = 1.\n\nin progress 0\n4 days 2021-09-14T20:29:50+00:00 1 Answer 0"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5886294,"math_prob":0.99925184,"size":345,"snap":"2021-31-2021-39","text_gpt3_token_len":121,"char_repetition_ratio":0.09970675,"word_repetition_ratio":0.42105263,"special_character_ratio":0.4173913,"punctuation_ratio":0.15189873,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99355596,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T05:57:01Z\",\"WARC-Record-ID\":\"<urn:uuid:815dc247-fd4b-413d-bb39-097756098e9d>\",\"Content-Length\":\"67626\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7c94b77f-0307-45f8-8370-296edeb4ae8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:5f67bc7f-c684-4987-baa0-731788894b3a>\",\"WARC-IP-Address\":\"198.252.99.154\",\"WARC-Target-URI\":\"https://republicofsouthossetia.org/question/calculate-the-instantaneous-rate-of-change-of-3-2-2-at-1-16346766-20/\",\"WARC-Payload-Digest\":\"sha1:4UD55AJPTTUT335HYOU4WM3HYOKY3FSP\",\"WARC-Block-Digest\":\"sha1:LSSQSPORCLRP7HC7DBJA553FQMHJ2K6B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056711.62_warc_CC-MAIN-20210919035453-20210919065453-00665.warc.gz\"}"} |
http://uoj.ac/problem/316 | [
"# #316. 【NOI2017】泳池\n\n• 必须保证安全性。即游泳场中的每一个单位海域都是安全的。\n• 必须是矩形。即游泳场必须是整个网格中的一个 $a \\times b$ 的子网格。\n• 必须和海滩相邻。即游泳场的下边界必须紧贴网格的下边界。",
null,
"### 样例一\n\n#### input\n\n10 5 1 2\n\n\n\n#### output\n\n342025319\n\n\n\n### 提示\n\n$x^{p-1} \\equiv 1 \\mod p$，其中 $p$ 为质数，$x \\in [1,p)$。\n\n### 限制与约定\n\n3$\\leq 10$$\\leq 8 4\\leq 10$$\\leq 9$\n5$\\leq 10$$\\leq 10 6\\leq 1000$$\\leq 7$\n7$\\leq 1000$$\\leq 8 8\\leq 1000$$\\leq 9$\n9,10,11$\\leq 1000$$\\leq 100 12,13,14\\leq 1000$$\\leq 1000$\n15,16$\\leq 10^9$$\\leq 10 17,18\\leq 10^9$$\\leq 100$\n19,20$\\leq 10^9$$\\leq 1000$"
] | [
null,
"http://img.uoj.ac/problem/316/1.png",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.94792575,"math_prob":1.0000097,"size":1587,"snap":"2019-26-2019-30","text_gpt3_token_len":1369,"char_repetition_ratio":0.18066962,"word_repetition_ratio":0.0,"special_character_ratio":0.4221802,"punctuation_ratio":0.046875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998012,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-17T17:23:18Z\",\"WARC-Record-ID\":\"<urn:uuid:012f2e32-1163-4bae-8172-2eac84d5e89a>\",\"Content-Length\":\"16162\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a882940b-612c-4052-bc05-c1090ecda863>\",\"WARC-Concurrent-To\":\"<urn:uuid:b8be9666-8c1f-44f8-b1fa-424b293f1f03>\",\"WARC-IP-Address\":\"114.215.147.61\",\"WARC-Target-URI\":\"http://uoj.ac/problem/316\",\"WARC-Payload-Digest\":\"sha1:OJ4D27BVL6M77KWVM3H5AE3O2GKL43V7\",\"WARC-Block-Digest\":\"sha1:DPARRM7QQJHFVXPUBGNHIQRH37Y7DSBE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525355.54_warc_CC-MAIN-20190717161703-20190717183703-00522.warc.gz\"}"} |
https://k12.libretexts.org/Bookshelves/Mathematics/Geometry/09%3A_Solid_Figures/9.19%3A_Surface_Area_of_Cylinders | [
"# 9.19: Surface Area of Cylinders\n\n•",
null,
"• Contributed by CK12\n• CK12\n\n$$SA=2 \\pi r(h+r)$$",
null,
"Figure $$\\PageIndex{1}$$\n\nMrs. Johnson is wrapping a cylindrical package in brown paper so that she can mail it to her son. Based on the dimensions shown, how much paper will she need to cover the package?\n\nIn this concept, you will learn to find the surface area of cylinders.\n\n## Surface Area\n\nA cylinder has two parallel congruent circular bases with a curved rectangle as its side. One way is to use a net.\n\nA net is a two-dimensional diagram of a three-dimensional figure. If you could unfold a cylinder (like a can) so that it is completely flat, you would have something that looks like this.",
null,
"Figure $$\\PageIndex{2}$$\n\nWith the net of a cylinder, you would need to calculate the area of each circle and the area of the curved side of the cylinder. Then you could add these values together to find the k.\n\nThe formula $$A&= \\pi r^{2}$$ can be used to find the area of a circle bases. Look closely again at the cylinder above. The two circular faces are congruent, so they must have the same radius and diameter.\n\nLet’s look at an example.",
null,
"Figure $$\\PageIndex{3}$$\n\nFirst, calculate the area of the circular bases.\n\n\\begin{aligned} A&= \\pi r^{2}\\\\ A&= \\pi (4)^{2} \\\\ A&= \\pi (16) \\\\ A&=50.3\\end{aligned}\n\nNext, find the length of the cylinder’s side. You know the width is 8 cm but the length is not shown. The cylinder was unwrapped to form the net. Therefore, the circumference of the circle would be the length of the side.\n\n\\begin{aligned}C&=2 \\pi r \\\\ C&=2 \\pi \\times 4 \\\\ C&=25.1\\end{aligned}\n\nThen, calculate the area of the side.\n\n\\begin{aligned}A&=l\\times w \\\\ A&=25.1\\times 8 \\\\ A&=200.8\\end{aligned}\n\nThen, find the surface area of the cylinder by adding the side area to the top and bottom area.\n\n\\begin{aligned}SA&=bottom+top+side \\\\ SA&=50.3+50.3+200.8 \\\\ SA&=301.4\\end{aligned}\n\nThe surface area of the cylinder is $$301.4 \\text{ cm}^{2}$$.\n\nPutting this all together, you can use the following formula to find the surface area of the cylinder:\n\n$$SA=2 \\pi r^{2}+2 \\pi rh$$\n\nThe formula $$2 \\pi r^{2}$$ represents the area of the top and bottom circles of the cylinder. The $$2 \\pi r h$$ represents the perimeter ($$2 \\pi r$$) multiplied by the height, $$h$$.\n\nLet’s look at an example.\n\nWhat is the surface area of the figure below?",
null,
"Figure $$\\PageIndex{4}$$\n\nFirst, substitute what you know into the surface area formula.\n\n\\begin{aligned}SA&=2 \\pi r^{2}+2 \\pi r h \\\\ SA&=2 \\pi (3.5)^{2}+2 \\pi (3.5)(28)\\end{aligned}\n\nNext, calculate the surface area.\n\n\\begin{aligned}SA&=2 \\pi (3.5)^{2}+2 \\pi (3.5)(28) \\\\ SA&=2 \\pi (12.25)+2 \\pi (98) \\\\ SA&=76.97+615.75 \\\\ SA&=692.72\\end{aligned}\n\nThe surface area of the cylinder is $$692.7 \\text{ cm}^{2}$$.\n\nSometimes, you can have a cylinder that has been cut. This is called a truncated cylinder. This is where you only see a section of the cylinder and will need to figure out the surface area of what you see.",
null,
"Figure $$\\PageIndex{5}$$\n\nExample $$\\PageIndex{1}$$\n\nEarlier, you were given a problem about Mrs. Johnson’s cylindrical wrapping.",
null,
"Figure $$\\PageIndex{6}$$\n\nSolution\n\nFirst, you need to find the radius. Remember, the radius is half the measure of the diameter.\n\n\\begin{aligned}r&=\\dfrac{d}{2} \\\\ r&=\\dfrac{11}{2} \\\\ r&=5.5v\\end{aligned}\n\nNext, substitute what you know into the surface area formula.\n\n\\begin{aligned}SA&=2 \\pi r^{2}+2 \\pi rh \\\\ SA&=2 \\pi (5.5)^{2}+2 \\pi (5.5)(22)\\end{aligned}\n\nThen, calculate the surface area.\n\n\\begin{aligned}SA&=2 \\pi (5.5)^{2}+2 \\pi (5.5)(22) \\\\ SA&=2 \\pi (30.25)+2 \\pi (121) \\\\ SA&=190.1+760.3 \\\\ SA&=950.4\\end{aligned}\n\nMrs. Johnson needs $$950.4 \\: cm^{2}$$ of brown paper to wrap her parcel.\n\nExample $$\\PageIndex{2}$$\n\nWhat is the surface area of the figure below?",
null,
"Figure $$\\PageIndex{7}$$\n\nSolution\n\nFirst, you need to find the radius. Remember that the radius is half the measure of the diameter.\n\n\\begin{aligned}r&=\\dfrac{d}{2} \\\\ r&=\\dfrac{13}{2} \\\\ r&=6.5\\end{aligned}\n\nNext, substitute what you know into the surface area formula.\n\n\\begin{aligned}SA&=2 \\pi r^{2}+2 \\pi rh \\\\ SA&=2 \\pi (6.5)^{2}+2 \\pi (6.5)(11)\\end{aligned}\n\nThen, calculate the surface area.\n\n\\begin{aligned}SA&=2 \\pi (6.5)^{2}+2 \\pi (6.5)(11) \\\\ SA&=2 \\pi (42.25)+2 \\pi (71.5) \\\\ SA&=265.5+449.2 \\\\ SA&=714.7\\end{aligned}\n\nThe surface area of the cylinder is $$714.7 \\: ft^{2}. Example \\(\\PageIndex{3}$$\n\nFind the surface area of a cylinder with a radius of 6 inches and a height of 5 inches.\n\nSolution\n\nFirst, substitute what you know into the surface area formula.\n\n\\begin{aligned}SA&=2 \\pi r^{2}+2 \\pi rh \\\\ SA&=2 \\pi (6)^{2}+2 \\pi (6)(5)\\end{aligned}\n\nNext, calculate the surface area.\n\n\\begin{aligned} SA&=2 \\pi (6)^{2}+2 \\pi (6)(5) \\ SA&=2 \\pi (36)+2 \\pi (30) \\\\ SA&=226.2+188.5 \\\\ SA&=414.7\\end{aligned}\n\nThe surface area of the cylinder is $$414.7 \\: in^{2}$$.\n\nExample $$\\PageIndex{4}$$\n\nFind the surface area of a cylinder with a radius of 4 cm and a height of 12 cm.\n\nSolution\n\nFirst, substitute what you know into the surface area formula.\n\nSA&=2 \\pi r^{2}+2 \\pi rh \\\\ SA&=2 \\pi (4)^{2}+2 \\pi (4)(12)\\end{aligned}\n\nNext, calculate the surface area.\n\n\\begin{aligned} SA&=2 \\pi (4)^{2}+2 \\pi (4)(12) \\\\ SA&=2 \\pi (16)+2 \\pi (48) \\\\ SA&=100.5+301.6 \\\\ SA&=402.1\\end{aligned}\n\nThe surface area of the cylinder is $$402.1 \\: cm^{2}$$.\n\nExample $$\\PageIndex{5}$$\n\nFind the surface area of a cylinder with a diameter of 10 meters and a height of 15 meters.\n\nSolution\n\nFirst, you need to find the radius. Remember that the radius is half the measure of the diameter.\n\n\\begin{aligned} r&=\\dfrac{d}{2} \\\\ r&=\\dfrac{10}{2} \\\\ r&=5\\end{aligned}\n\nNext, substitute what you know into the surface area formula.\n\n\\begin{aligned} SA&=2 \\pi r^{2}+2 \\pi r h \\\\ SA&=2 \\pi (5)^{2}+2 \\pi (5)(15)\\end{aligned}\n\nThen, calculate the surface area.\n\n\\begin{aligned}SA&=2 \\pi (5)^{2}+2 \\pi (5)(15) \\\\ SA&=2 \\pi (25)+2 \\pi (75) \\\\ SA&=157.1+471.2 \\\\ SA&=628.3\\end{aligned}\n\nThe surface area of the cylinder is $$628.3 \\: m^{2}$$.\n\n## Review\n\nUse the diagrams to answer the questions below each one.",
null,
"Figure $$\\PageIndex{8}$$\n1. What is the name of this figure?\n2. What is the shape of the base of this figure?\n3. How many bases are there?\n4. What is the surface area of this figure?\n5. Which measurement is needed the radius or the diameter?",
null,
"Figure $$\\PageIndex{9}$$\n1. What is the name of this figure?\n2. Which measurement is given the radius or the diameter?\n3. What is the surface area of the figure?\n\nUse what you have learned to answer each question.\n\n1. A cylindrical water tank is 35 long and 10 feet across. How much sheet metal is the tank made of?\n2. Did you use area or surface area to solve this problem?\n3. True or false. You can only find the surface area if you know the volume.\n4. True or false. Surface area and volume measure the same thing.\n5. True or false. Surface area measures the outside of a cylinder.\n6. True or false. You need the radius to find the surface area of a cylinder.\n7. True or false. The radius is one-half of the diameter.\n\n## Vocabulary\n\nTerm Definition\nCylinder A cylinder is a solid figure with two parallel congruent circular bases.\nNet A net is a diagram that shows a “flattened” view of a solid. In a net, each face and base is shown with all of its dimensions. A net can also serve as a pattern to build a three-dimensional solid.\nPrism A prism is a three-dimensional object with two congruent parallel bases that are polygons.\nSurface Area Surface area is the total area of all of the surfaces of a three-dimensional object.\nThree – Dimensional A figure drawn in three dimensions is drawn using length, width and height or depth.\nTruncated Cylinder A cylinder that is cut in part from a complete cylinder."
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https://edurev.in/studytube/Previous-Year-Questions--2014-19--The-p-Block-Elem/7931dc18-6dd3-4f3d-b19f-53f71359c6d9_t | [
"Courses\n\n# Previous Year Questions (2014-19) - The p-Block Elements (Group 15, 16, 17 and 18) Notes | EduRev\n\n## NEET : Previous Year Questions (2014-19) - The p-Block Elements (Group 15, 16, 17 and 18) Notes | EduRev\n\nThe document Previous Year Questions (2014-19) - The p-Block Elements (Group 15, 16, 17 and 18) Notes | EduRev is a part of the NEET Course Chemistry 28 Years Past year papers for NEET/AIPMT Class 12.\nAll you need of NEET at this link: NEET\n\nQ.1. Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code: (2019)",
null,
"",
null,
"A: (1)\nB: (2)\nC: (3)\nD: (4)\nAns:\nB",
null,
"",
null,
"",
null,
"",
null,
"Q.2. Which is the correct thermal stability order for H2E (E = O, S, Se, Te and Po)? (2019)\nA: H2S < H2O < H2Se < H2Te < H2Po\nB: H2O < H2S < H2Se < H2Te < H2Po\nC: H2Po < H2Te < H2Se < H2S < H2O\nD: H2Se < H2Te < H2Po < H2O < H2S\nAns:\nC\nOn going down the group thermal stability order for H2E decreases because H–E bond energy decreases\n∴ Order of stability would be:-\nH2Po < H2Te < H2Se < H2S < H2O\n\nQ.3. Which of the following statements is not true for halogens ? (2018)\nA: All form monobasic oxyacids.\nB: All are oxidizing agents.\n\nC: All but fluorine show positive oxidation states.\nD: Chlorine has the highest electron-gain enthalpy.\nAns:\nC\nDue to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid. Oxidation number of F is +1 in HOF.\n\nQ.4. In the structure of ClF3, the number of lone pairs of electrons on central atom 'Cl' is (2018)\nA: one\nB: two\nC: three\nD: four\n\nAns: B",
null,
"2 lone pairs at equitorial position.\n\nQ.5. Match the interhalogen compounds of column-I with the geometry in column II and assign the correct code. (2017)",
null,
"",
null,
"A: (1)\nB: (2)\nC: (3)\nD: (4\nAns:\nA",
null,
"Q.6. In which pair of ions both the species contain S–S bond? (2017)\nA:",
null,
"B:",
null,
"C:",
null,
"D:",
null,
"Ans:\nA",
null,
"Q.7. When copper is heated with conc. HNO3 it produces : (2016)\nA: Cu(NO3)2 and N2O\nB: Cu(NO3)2 and NO2\nC: Cu(NO3)2 and NO\nD: Cu(NO3)2 , NO and NO2\nAns:\nB\nNitric acid acts as an oxidising agent while reacting with copper.\ni) when copper reacts with dilute nitric acid it forms,\n3Cu + 4HNO3(dilute) → 3Cu(NO3)2 +2NO +4H2O\nii) When copper reacts with concentrated nitric acid it forms,\nCu +4HNO3(conc.) → Cu(NO3)2 +NO2 +2H2O\n\nQ.8. Which is the correct statement for the given acids? (2016)\nA: Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid.\nB: Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.\nC: Both are diprotic acids.\nD: Both are triprotic acids.\nAns:\nB\nPhosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.\nPhosphinic acid-",
null,
"Phosphonic acid-",
null,
"Due to the presence of one replaceable proton in phosphinic acid, it is monoprotic acid and due to the presence of two replaceable proton in phosphinic acid, it is diprotic acid.\n\nQ.9. Among the following, the correct order of acidity is : (2016)\nA: HCIO4 < HCIO2 < HCIO < HCIO3\nB: HCIO3 < HCIO4 < HCIO2 < HCIO\nC: HCIO < HCIO2 < HCIO3 < HCIO4\nD: HCIO2 < HCIO < HCIO3 < HCIO4\nAns:\nC\nAs oxidation number of central atom increases, acidic nature increases.\nHClO < HClO2 < HClO3 < HClO4\n\nQ.10. The product obtained as a result of a reaction of nitrogen with CaCis : (2016)\nA: Ca2CN\nB: Ca(CN)2\nC: CaCN\nD: CaCN3\nAns:\nB\nWhen calcium carbide reacts with nitrogen under high temperature, it forms calcium cyanamide which is also called nitrolim.",
null,
"Q.11. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (2016)\nA: F2 > Cl2 > Br2 > I2\nB: I2 > Br2 > CI2 > F2\nC: CI2 > Br2 > F2 > I2\nD: Br2 > I2 > F2 > CI2\nAns:\nC\nBond dissociation energies of halogen family decrease down the group as the size of the atom increases. The bond dissociation energy of fluorine, is, however, lower than those of chlorine and bromine because of interelectronic repulsions present in the small atom of fluorine.\nHence bond energy decreases in the order Cl2 > Br2 > F2 > I2\n\nQ.12. Match the compound given in column I with the hybridization and shape given in column II and mark the correct option. (2016)\nColumn-I Column-II\n(a) XeF6 (i) distorted octahedral\n(b) XeO3 (ii) square planar\n(c) XeOF4 (iii) pyramidal\n(d) XeF4 (iv) square pyramidal\nCode :\n(a) (b) (c) (d)\nA: (iv) (i) (ii) (iii)\nB: (i) (iii) (iv) (ii)\nC: (i) (ii) (iv) (iii)\nD: (iv) (iii) (i) (ii)\nAns:\nB\n(i) (iii) (iv) (ii)",
null,
"Q.13. Which of the following species contain equal number of σ - and π - bonds ? (2015)\nA: CH2(CN)2\nB:",
null,
"C: XeO4\nD:(CN)2\nAns:\nC",
null,
"Q.14. Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other ? (2015)\nA: is used as a food-preservative\nB: forms 'acid-rain'\nC: is a reducing agent\nD: is soluble in water\nAns:\nA\nSO2 is used in the manufacture of sodium bisulphate (NaHSO3) which is used as preservatives for jams, jellies and squashes.\n\nQ.15. Which of the following pairs of ions are isoelectronic and isostructural ? (2015)\nA:",
null,
"B:",
null,
"C:",
null,
"D:",
null,
"Ans:\nA",
null,
"Q.16. Maximum bond angle at nitrogen is present in which of the following ? (2015)\nA:",
null,
"B: NO2\nC:",
null,
"D:",
null,
"Ans: D\nIn all of the four molecules NO2- and NO2 have one lone pair thus bond angle is less than 1200 in the case of NO2- and more than 120in the case of NO2 but in the NO3- there is no lone pair hence all are bonding pairs leading to an ideal bond angle of 1200.\nIn NO2+, the is no lone pair only bond pair exist thus leading a bond angle is 1800.\n\nQ.17. Acidity of diprotic acids in aqueous solutions increases in the order : (2014)\nA: H2Te < H2S < H2Se\nB: H2Se < H2Te < H2S\nC: H2S < H2Se < H2Te\nD: H2Se < H2S < H2Te\nAns:\nC\nAcidic strength of hydrides increases as the size\nof central atom increases which weakens the\nM- H bond. Since, the size increases from S to Te\nthus acidic strength follows the order.\nH2S<H2Se<H2Te\nAcidic nature ∝",
null,
"S to Te size increases, bond dissociation enthalpy\ndecreases and acidic nature increases.\n\nOffer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n,\n\n;"
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https://encyclopediaofmath.org/wiki/Klein%E2%80%93Gordon_equation | [
"# Klein-Gordon equation\n\n(Redirected from Klein–Gordon equation)\n\nThe relativistically-invariant quantum equation describing spinless scalar or pseudo-scalar particles, for example, $\\pi$-, and $K$- mesons. The equation was established by O. Klein and somewhat later by V.A. Fock [V.A. Fok] as a wave equation under the conditions of cyclicity in the fifth coordinate and was shortly afterwards deduced by several authors (for example, W. Gordon ) without this requirement on the fifth coordinate.\n\nThe subsequent application of the Klein–Gordon equation as a relativistic quantum equation proved possible in quantum field theory but not in quantum mechanics. In an interpretation of the Klein–Gordon equation was given as an equation for fields of particles of zero spin. The Klein–Gordon equation is applied in the description of $\\pi$- mesons and corresponding fields; it plays the role of one of the fundamental equations of quantum field theory.\n\nThe Klein–Gordon equation is a linear homogeneous second-order partial differential equation with constant coefficients:\n\n$$\\tag{1 } \\left ( \\frac{\\partial ^ {2} }{\\partial x ^ {2} } + \\frac{\\partial ^ {2} }{\\partial y ^ {2} } + \\frac{\\partial ^ {2} }{\\partial z ^ {2} } - \\frac{\\partial ^ {2} }{c ^ {2} \\partial t ^ {2} } - \\mu ^ {2} \\right ) \\phi = 0 ,$$\n\nwhere $\\phi ( \\mathbf x , t )$ is a (pseudo-) scalar function, in the general case — complex, $\\mu = m c / \\hbar$ and $m$ is the rest mass of the particle. If $\\phi$ is a real function, then the Klein–Gordon equation describes neutral (pseudo-) scalar particles, while when $\\phi$ is complex it describes charged particles.\n\nIn the latter case equation (1) is supplemented by the equation for the complex-conjugate scalar function $\\phi ^ {*}$:\n\n$$\\tag{2 } \\left ( \\frac{\\partial ^ {2} }{\\partial x ^ {2} } + \\frac{\\partial ^ {2} }{\\partial y ^ {2} } + \\frac{\\partial ^ {2} }{\\partial z ^ {2} } - \\frac{\\partial ^ {2} }{c ^ {2} \\partial t ^ {2} } - \\mu ^ {2} \\right ) \\phi ^ {*} = 0 .$$\n\nThe interaction of (pseudo-) scalar particles with the electromagnetic field is described by the minimal substitution $\\partial / {\\partial x ^ \\alpha } \\rightarrow ( \\partial / {\\partial x ^ \\alpha } ) - i e A _ \\alpha / \\hbar$. Each component of the wave function of particles of any spin also satisfies the Klein–Gordon equation, but only for the case where the spin is 0 is the function invariant with respect to the Lorentz–Poincaré group.\n\nThe Klein–Gordon equation can be obtained by means of the relationship between the energy $E$ and the momentum $\\mathbf p$ of the particle in special relativity theory,\n\n$$\\frac{1}{c ^ {2} } E ^ {2} - p _ {x} ^ {2} - p _ {y} ^ {2} - p _ {z} ^ {2} = \\ m ^ {2} c ^ {2} ,$$\n\nby replacing quantities by operators (see , ):\n\n$$E \\rightarrow - \\frac \\hbar {i} \\frac \\partial {\\partial t } ,\\ \\ \\mathbf p \\rightarrow \\frac \\hbar {i} \\frac \\partial {\\partial \\mathbf x } .$$\n\nAs for all relativistic equations, the Klein–Gordon equation can be expressed in the form of the Dirac equation, that is, it can be reduced to a first-order linear equation:\n\n$$\\tag{3 } \\left ( \\Gamma ^ \\alpha \\frac \\partial {\\partial x ^ \\alpha } - \\mu \\right ) \\psi = 0 ,$$\n\nwhere the coefficients $\\Gamma ^ \\alpha$ are matrices similar to the Dirac matrices $\\gamma ^ \\alpha$. In the case of the Klein–Gordon equation the matrices $\\Gamma ^ \\alpha$ satisfy the commutation relations:\n\n$$\\tag{4 } \\Gamma _ \\mu \\Gamma _ \\nu \\Gamma _ \\rho + \\Gamma _ \\rho \\Gamma _ \\nu \\Gamma _ \\mu = \\ \\eta _ {\\mu \\nu } \\Gamma _ \\rho + \\eta _ {\\rho \\nu } \\Gamma _ \\mu .$$\n\nFor example, $( \\Gamma _ \\alpha ) ^ {3} = \\eta _ {\\alpha \\alpha } \\Gamma _ \\alpha$( Kemmer–Duffin matrices). Here $\\eta _ {\\mu \\nu }$ is the metric tensor of Minkowski space. All the $\\Gamma ^ \\alpha$ are singular matrices $( \\mathop{\\rm det} \\Gamma _ \\alpha = 0 )$. Hence they do not have inverses.\n\nApart from the trivial solution $\\Gamma _ \\alpha = 0$, $\\psi = 0$ to (4) and a solution in the form of five-row matrices, describing the scalar field $\\phi$ itself and the four components of its gradient, equation (4) has a further solution in the form of ten-row matrices. The corresponding ten-component function contains the four components of the potential $A _ \\alpha$ and the six components of the stress $F _ {\\alpha \\beta } = 2 \\partial _ {[ \\alpha{} } A _ { {}\\beta ] }$, that is, equations (3) and (4) can simultaneously give a representation for the Proca equation describing vector particles with spin 1; for $\\mu = 0$ and real $\\phi$ they give a representation of the Maxwell equations.\n\nWhen taking into account the interaction of the (pseudo-) scalar particles with a gravity field in accordance with the general theory of relativity, the Klein–Gordon equation is generalized onto an arbitrary Riemannian space as:\n\n$$\\tag{5 } \\frac{1}{\\sqrt - g } \\frac \\partial {\\partial x ^ \\alpha } \\left ( \\sqrt - g g ^ {\\alpha \\beta } \\frac{\\partial \\phi }{\\partial x ^ \\beta } \\right ) - \\mu ^ {2} \\phi = 0 ,$$\n\nwhere $g _ {\\alpha \\beta }$ is the metric tensor and $g$ is the determinant of the matrix $\\| g _ {\\alpha \\beta } \\|$. In equation (5) the term $R \\phi / 6$ is frequently added, where $R$ is the scalar curvature, as a result of which, when $\\mu = 0$, the general relativistic Klein–Gordon equation\n\n$$\\frac{1}{\\sqrt - g } \\frac \\partial {\\partial x ^ \\alpha } \\left ( \\sqrt - g g ^ {\\alpha \\beta } \\frac{\\partial \\phi }{\\partial x ^ \\beta } \\right ) - \\frac{R \\phi }{6} = 0$$\n\nbecomes conformally invariant.\n\nHow to Cite This Entry:\nKlein–Gordon equation. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Klein%E2%80%93Gordon_equation&oldid=22652"
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https://www.ginac.de/reference/classGiNaC_1_1matrix.html | [
"GiNaC 1.8.6\nGiNaC::matrix Class Reference\n\nSymbolic matrices. More...\n\n`#include <matrix.h>`\n\nInheritance diagram for GiNaC::matrix:",
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"## Public Member Functions\n\nmatrix (unsigned r, unsigned c)\nVery common ctor. More...\n\nmatrix (unsigned r, unsigned c, const lst &l)\nConstruct matrix from (flat) list of elements. More...\n\nmatrix (std::initializer_list< std::initializer_list< ex > > l)\nConstruct a matrix from an 2 dimensional initializer list. More...\n\nsize_t nops () const override\nnops is defined to be rows x columns. More...\n\nex op (size_t i) const override\nreturns matrix entry at position (i/col, icol). More...\n\nexlet_op (size_t i) override\nreturns writable matrix entry at position (i/col, icol). More...\n\nex evalm () const override\nEvaluate sums, products and integer powers of matrices. More...\n\nex subs (const exmap &m, unsigned options=0) const override\nSubstitute a set of objects by arbitrary expressions. More...\n\nex eval_indexed (const basic &i) const override\nAutomatic symbolic evaluation of an indexed matrix. More...\n\nex add_indexed (const ex &self, const ex &other) const override\nSum of two indexed matrices. More...\n\nex scalar_mul_indexed (const ex &self, const numeric &other) const override\nProduct of an indexed matrix with a number. More...\n\nbool contract_with (exvector::iterator self, exvector::iterator other, exvector &v) const override\nContraction of an indexed matrix with something else. More...\n\nex conjugate () const override\nComplex conjugate every matrix entry. More...\n\nex real_part () const override\n\nex imag_part () const override\n\nvoid archive (archive_node &n) const override\nSave (a.k.a. More...\n\nvoid read_archive (const archive_node &n, lst &syms) override\n\nunsigned rows () const\nGet number of rows. More...\n\nunsigned cols () const\nGet number of columns. More...\n\nmatrix add (const matrix &other) const\nSum of matrices. More...\n\nmatrix sub (const matrix &other) const\nDifference of matrices. More...\n\nmatrix mul (const matrix &other) const\nProduct of matrices. More...\n\nmatrix mul (const numeric &other) const\nProduct of matrix and scalar. More...\n\nmatrix mul_scalar (const ex &other) const\nProduct of matrix and scalar expression. More...\n\nmatrix pow (const ex &expn) const\nPower of a matrix. More...\n\nconst exoperator() (unsigned ro, unsigned co) const\noperator() to access elements for reading. More...\n\nexoperator() (unsigned ro, unsigned co)\noperator() to access elements for writing. More...\n\nmatrixset (unsigned ro, unsigned co, const ex &value)\n\nmatrix transpose () const\nTransposed of an m x n matrix, producing a new n x m matrix object that represents the transposed. More...\n\nex determinant (unsigned algo=determinant_algo::automatic) const\nDeterminant of square matrix. More...\n\nex trace () const\nTrace of a matrix. More...\n\nex charpoly (const ex &lambda) const\nCharacteristic Polynomial. More...\n\nmatrix inverse () const\nInverse of this matrix, with automatic algorithm selection. More...\n\nmatrix inverse (unsigned algo) const\nInverse of this matrix. More...\n\nmatrix solve (const matrix &vars, const matrix &rhs, unsigned algo=solve_algo::automatic) const\nSolve a linear system consisting of a m x n matrix and a m x p right hand side by applying an elimination scheme to the augmented matrix. More...\n\nunsigned rank () const\nCompute the rank of this matrix. More...\n\nunsigned rank (unsigned solve_algo) const\nCompute the rank of this matrix using the given algorithm, which should be a member of enum solve_algo. More...\n\nbool is_zero_matrix () const\nFunction to check that all elements of the matrix are zero. More...",
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"Public Member Functions inherited from GiNaC::basic\nvirtual ~basic ()\nbasic destructor, virtual because class ex will delete objects of derived classes via a basic*. More...\n\nbasic (const basic &other)\n\nconst basicoperator= (const basic &other)\nbasic assignment operator: the other object might be of a derived class. More...\n\nvirtual basicduplicate () const\nCreate a clone of this object on the heap. More...\n\nvirtual ex eval () const\nPerform automatic non-interruptive term rewriting rules. More...\n\nvirtual ex evalf () const\nEvaluate object numerically. More...\n\nvirtual ex evalm () const\nEvaluate sums, products and integer powers of matrices. More...\n\nvirtual ex eval_integ () const\nEvaluate integrals, if result is known. More...\n\nvirtual ex eval_indexed (const basic &i) const\nPerform automatic symbolic evaluations on indexed expression that contains this object as the base expression. More...\n\nvirtual void print (const print_context &c, unsigned level=0) const\nOutput to stream. More...\n\nvirtual void dbgprint () const\nLittle wrapper around print to be called within a debugger. More...\n\nvirtual void dbgprinttree () const\nLittle wrapper around printtree to be called within a debugger. More...\n\nvirtual unsigned precedence () const\nReturn relative operator precedence (for parenthezing output). More...\n\nvirtual bool info (unsigned inf) const\n\nvirtual size_t nops () const\nNumber of operands/members. More...\n\nvirtual ex op (size_t i) const\nReturn operand/member at position i. More...\n\nvirtual ex operator[] (const ex &index) const\n\nvirtual ex operator[] (size_t i) const\n\nvirtual exlet_op (size_t i)\nReturn modifiable operand/member at position i. More...\n\nvirtual exoperator[] (const ex &index)\n\nvirtual exoperator[] (size_t i)\n\nvirtual bool has (const ex &other, unsigned options=0) const\nTest for occurrence of a pattern. More...\n\nvirtual bool match (const ex &pattern, exmap &repls) const\nCheck whether the expression matches a given pattern. More...\n\nvirtual ex subs (const exmap &m, unsigned options=0) const\nSubstitute a set of objects by arbitrary expressions. More...\n\nvirtual ex map (map_function &f) const\nConstruct new expression by applying the specified function to all sub-expressions (one level only, not recursively). More...\n\nvirtual void accept (GiNaC::visitor &v) const\n\nvirtual bool is_polynomial (const ex &var) const\nCheck whether this is a polynomial in the given variables. More...\n\nvirtual int degree (const ex &s) const\nReturn degree of highest power in object s. More...\n\nvirtual int ldegree (const ex &s) const\nReturn degree of lowest power in object s. More...\n\nvirtual ex coeff (const ex &s, int n=1) const\nReturn coefficient of degree n in object s. More...\n\nvirtual ex expand (unsigned options=0) const\nExpand expression, i.e. More...\n\nvirtual ex collect (const ex &s, bool distributed=false) const\nSort expanded expression in terms of powers of some object(s). More...\n\nvirtual ex series (const relational &r, int order, unsigned options=0) const\nDefault implementation of ex::series(). More...\n\nvirtual ex normal (exmap &repl, exmap &rev_lookup, lst &modifier) const\nDefault implementation of ex::normal(). More...\n\nvirtual ex to_rational (exmap &repl) const\nDefault implementation of ex::to_rational(). More...\n\nvirtual ex to_polynomial (exmap &repl) const\n\nvirtual numeric integer_content () const\n\nvirtual ex smod (const numeric &xi) const\nApply symmetric modular homomorphism to an expanded multivariate polynomial. More...\n\nvirtual numeric max_coefficient () const\nImplementation ex::max_coefficient(). More...\n\nvirtual exvector get_free_indices () const\nReturn a vector containing the free indices of an expression. More...\n\nvirtual ex add_indexed (const ex &self, const ex &other) const\n\nvirtual ex scalar_mul_indexed (const ex &self, const numeric &other) const\nMultiply an indexed expression with a scalar. More...\n\nvirtual bool contract_with (exvector::iterator self, exvector::iterator other, exvector &v) const\nTry to contract two indexed expressions that appear in the same product. More...\n\nvirtual unsigned return_type () const\n\nvirtual return_type_t return_type_tinfo () const\n\nvirtual ex conjugate () const\n\nvirtual ex real_part () const\n\nvirtual ex imag_part () const\n\ntemplate<class T >\nvoid print_dispatch (const print_context &c, unsigned level) const\nLike print(), but dispatch to the specified class. More...\n\nvoid print_dispatch (const registered_class_info &ri, const print_context &c, unsigned level) const\nLike print(), but dispatch to the specified class. More...\n\nvirtual void archive (archive_node &n) const\nSave (serialize) the object into archive node. More...\n\nvirtual void read_archive (const archive_node &n, lst &syms)\nLoad (deserialize) the object from an archive node. More...\n\nex subs_one_level (const exmap &m, unsigned options) const\nHelper function for subs(). More...\n\nex diff (const symbol &s, unsigned nth=1) const\nDefault interface of nth derivative ex::diff(s, n). More...\n\nint compare (const basic &other) const\nCompare objects syntactically to establish canonical ordering. More...\n\nbool is_equal (const basic &other) const\nTest for syntactic equality. More...\n\nconst basichold () const\nStop further evaluation. More...\n\nunsigned gethash () const\n\nconst basicsetflag (unsigned f) const\nSet some status_flags. More...\n\nconst basicclearflag (unsigned f) const\nClear some status_flags. More...",
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"Public Member Functions inherited from GiNaC::refcounted\nrefcounted () noexcept\n\nunsigned int remove_reference () noexcept\n\nunsigned int get_refcount () const noexcept\n\nvoid set_refcount (unsigned int r) noexcept\n\n## Protected Member Functions\n\nmatrix (unsigned r, unsigned c, const exvector &m2)\nCtor from representation, for internal use only. More...\n\nmatrix (unsigned r, unsigned c, exvector &&m2)\n\nbool match_same_type (const basic &other) const override\nReturns true if the attributes of two objects are similar enough for a match. More...\n\nunsigned return_type () const override\n\nex determinant_minor () const\nRecursive determinant for small matrices having at least one symbolic entry. More...\n\nstd::vector< unsigned > echelon_form (unsigned algo, int n)\n\nint gauss_elimination (const bool det=false)\nPerform the steps of an ordinary Gaussian elimination to bring the m x n matrix into an upper echelon form. More...\n\nint division_free_elimination (const bool det=false)\nPerform the steps of division free elimination to bring the m x n matrix into an upper echelon form. More...\n\nint fraction_free_elimination (const bool det=false)\nPerform the steps of Bareiss' one-step fraction free elimination to bring the matrix into an upper echelon form. More...\n\nstd::vector< unsigned > markowitz_elimination (unsigned n)\n\nint pivot (unsigned ro, unsigned co, bool symbolic=true)\nPartial pivoting method for matrix elimination schemes. More...\n\nvoid print_elements (const print_context &c, const char *row_start, const char *row_end, const char *row_sep, const char *col_sep) const\n\nvoid do_print (const print_context &c, unsigned level) const\n\nvoid do_print_latex (const print_latex &c, unsigned level) const\n\nvoid do_print_python_repr (const print_python_repr &c, unsigned level) const",
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"Protected Member Functions inherited from GiNaC::basic\nbasic ()\n\nvirtual ex eval_ncmul (const exvector &v) const\n\nvirtual bool match_same_type (const basic &other) const\nReturns true if the attributes of two objects are similar enough for a match. More...\n\nvirtual ex derivative (const symbol &s) const\nDefault implementation of ex::diff(). More...\n\nvirtual int compare_same_type (const basic &other) const\nReturns order relation between two objects of same type. More...\n\nvirtual bool is_equal_same_type (const basic &other) const\nReturns true if two objects of same type are equal. More...\n\nvirtual unsigned calchash () const\nCompute the hash value of an object and if it makes sense to store it in the objects status_flags, do so. More...\n\nvoid ensure_if_modifiable () const\nEnsure the object may be modified without hurting others, throws if this is not the case. More...\n\nvoid do_print (const print_context &c, unsigned level) const\nDefault output to stream. More...\n\nvoid do_print_tree (const print_tree &c, unsigned level) const\nTree output to stream. More...\n\nvoid do_print_python_repr (const print_python_repr &c, unsigned level) const\nPython parsable output to stream. More...\n\n## Protected Attributes\n\nunsigned row\nnumber of rows More...\n\nunsigned col\nnumber of columns More...\n\nexvector m\nrepresentation (cols indexed first) More...",
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"Protected Attributes inherited from GiNaC::basic\nunsigned flags\nof type status_flags More...\n\nunsigned hashvalue\nhash value More...\n\n## Detailed Description\n\nSymbolic matrices.\n\nDefinition at line 37 of file matrix.h.\n\n## ◆ matrix() [1/5]\n\n GiNaC::matrix::matrix ( unsigned r, unsigned c )\n\nVery common ctor.\n\nInitializes to r x c-dimensional zero-matrix.\n\nParameters\n r number of rows c number of cols\n\nDefinition at line 71 of file matrix.cpp.\n\nReferences GiNaC::status_flags::not_shareable, and GiNaC::basic::setflag().\n\nReferenced by add(), conjugate(), determinant(), imag_part(), mul(), mul_scalar(), real_part(), sub(), subs(), and transpose().\n\n## ◆ matrix() [2/5]\n\n GiNaC::matrix::matrix ( unsigned r, unsigned c, const lst & l )\n\nConstruct matrix from (flat) list of elements.\n\nIf the list has fewer elements than the matrix, the remaining matrix elements are set to zero. If the list has more elements than the matrix, the excessive elements are thrown away.\n\nDefinition at line 80 of file matrix.cpp.\n\nReferences c, m, GiNaC::status_flags::not_shareable, r, GiNaC::basic::setflag(), and x.\n\n## ◆ matrix() [3/5]\n\n GiNaC::matrix::matrix ( std::initializer_list< std::initializer_list< ex > > l )\n\nConstruct a matrix from an 2 dimensional initializer list.\n\nThrows an exception if some row has a different length than all the others.\n\nDefinition at line 99 of file matrix.cpp.\n\nReferences c, col, m, GiNaC::status_flags::not_shareable, r, row, and GiNaC::basic::setflag().\n\n## ◆ matrix() [4/5]\n\n GiNaC::matrix::matrix ( unsigned r, unsigned c, const exvector & m2 )\nprotected\n\nCtor from representation, for internal use only.\n\nDefinition at line 119 of file matrix.cpp.\n\nReferences GiNaC::status_flags::not_shareable, and GiNaC::basic::setflag().\n\n## ◆ matrix() [5/5]\n\n GiNaC::matrix::matrix ( unsigned r, unsigned c, exvector && m2 )\nprotected\n\nDefinition at line 124 of file matrix.cpp.\n\nReferences GiNaC::status_flags::not_shareable, and GiNaC::basic::setflag().\n\n## ◆ nops()\n\n size_t GiNaC::matrix::nops ( ) const\noverridevirtual\n\nnops is defined to be rows x columns.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 206 of file matrix.cpp.\n\nReferences col, and row.\n\nReferenced by let_op(), and op().\n\n## ◆ op()\n\n ex GiNaC::matrix::op ( size_t i ) const\noverridevirtual\n\nreturns matrix entry at position (i/col, icol).\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 212 of file matrix.cpp.\n\nReferences GINAC_ASSERT, m, and nops().\n\nReferenced by contract_with().\n\n## ◆ let_op()\n\n ex & GiNaC::matrix::let_op ( size_t i )\noverridevirtual\n\nreturns writable matrix entry at position (i/col, icol).\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 220 of file matrix.cpp.\n\nReferences GiNaC::basic::ensure_if_modifiable(), GINAC_ASSERT, m, and nops().\n\n## ◆ evalm()\n\n ex GiNaC::matrix::evalm ( ) const\ninlineoverridevirtual\n\nEvaluate sums, products and integer powers of matrices.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 55 of file matrix.h.\n\n## ◆ subs()\n\n ex GiNaC::matrix::subs ( const exmap & m, unsigned options = `0` ) const\noverridevirtual\n\nSubstitute a set of objects by arbitrary expressions.\n\nThe ex returned will already be evaluated.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 228 of file matrix.cpp.\n\nReferences c, col, m, matrix(), options, r, row, and subs().\n\nReferenced by fraction_free_elimination(), and subs().\n\n## ◆ eval_indexed()\n\n ex GiNaC::matrix::eval_indexed ( const basic & i ) const\noverridevirtual\n\nAutomatic symbolic evaluation of an indexed matrix.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 320 of file matrix.cpp.\n\n ex GiNaC::matrix::add_indexed ( const ex & self, const ex & other ) const\noverridevirtual\n\nSum of two indexed matrices.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 397 of file matrix.cpp.\n\nReferences add(), col, GINAC_ASSERT, GiNaC::ex::is_equal(), GiNaC::ex::nops(), GiNaC::ex::op(), row, and transpose().\n\n## ◆ scalar_mul_indexed()\n\n ex GiNaC::matrix::scalar_mul_indexed ( const ex & self, const numeric & other ) const\noverridevirtual\n\nProduct of an indexed matrix with a number.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 433 of file matrix.cpp.\n\nReferences GINAC_ASSERT, mul(), GiNaC::ex::nops(), and GiNaC::ex::op().\n\n## ◆ contract_with()\n\n bool GiNaC::matrix::contract_with ( exvector::iterator self, exvector::iterator other, exvector & v ) const\noverridevirtual\n\nContraction of an indexed matrix with something else.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 448 of file matrix.cpp.\n\nReferences GiNaC::_ex1, col, GINAC_ASSERT, GiNaC::is_dummy_pair(), mul(), GiNaC::ex::op(), op(), row, and transpose().\n\n## ◆ conjugate()\n\n ex GiNaC::matrix::conjugate ( ) const\noverridevirtual\n\nComplex conjugate every matrix entry.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 239 of file matrix.cpp.\n\nReferences GiNaC::are_ex_trivially_equal(), col, m, matrix(), row, and x.\n\n## ◆ real_part()\n\n ex GiNaC::matrix::real_part ( ) const\noverridevirtual\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 264 of file matrix.cpp.\n\nReferences col, m, matrix(), and row.\n\n## ◆ imag_part()\n\n ex GiNaC::matrix::imag_part ( ) const\noverridevirtual\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 273 of file matrix.cpp.\n\nReferences col, m, matrix(), and row.\n\n## ◆ archive()\n\n void GiNaC::matrix::archive ( archive_node & n ) const\noverridevirtual\n\nSave (a.k.a.\n\nserialize) object into archive.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 152 of file matrix.cpp.\n\nReferences col, m, n, and row.\n\n void GiNaC::matrix::read_archive ( const archive_node & n, lst & syms )\noverridevirtual\n\ndeserialize) object from archive.\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 134 of file matrix.cpp.\n\nReferences col, m, n, and row.\n\n## ◆ match_same_type()\n\n bool GiNaC::matrix::match_same_type ( const basic & other ) const\noverrideprotectedvirtual\n\nReturns true if the attributes of two objects are similar enough for a match.\n\nThis function must not match subexpressions (this is already done by basic::match()). Only attributes not accessible by op() should be compared. This is also the reason why this function doesn't take the wildcard replacement list from match() as an argument: only subexpressions are subject to wildcard matches. Also, this function only needs to be implemented for container classes because is_equal_same_type() is automatically used instead of match_same_type() if nops() == 0.\n\nbasic::match\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 309 of file matrix.cpp.\n\nReferences col, cols(), GINAC_ASSERT, row, and rows().\n\n## ◆ return_type()\n\n unsigned GiNaC::matrix::return_type ( ) const\ninlineoverrideprotectedvirtual\n\nReimplemented from GiNaC::basic.\n\nDefinition at line 71 of file matrix.h.\n\nReferences GiNaC::return_types::noncommutative.\n\n## ◆ rows()\n\n unsigned GiNaC::matrix::rows ( ) const\ninline\n\nGet number of rows.\n\nDefinition at line 75 of file matrix.h.\n\nReferences row.\n\n## ◆ cols()\n\n unsigned GiNaC::matrix::cols ( ) const\ninline\n\nGet number of columns.\n\nDefinition at line 77 of file matrix.h.\n\nReferences col.\n\n matrix GiNaC::matrix::add ( const matrix & other ) const\n\nSum of matrices.\n\nExceptions\n logic_error (incompatible matrices)\n\nDefinition at line 555 of file matrix.cpp.\n\nReferences col, m, matrix(), and row.\n\n## ◆ sub()\n\n matrix GiNaC::matrix::sub ( const matrix & other ) const\n\nDifference of matrices.\n\nExceptions\n logic_error (incompatible matrices)\n\nDefinition at line 572 of file matrix.cpp.\n\nReferences col, m, matrix(), and row.\n\n## ◆ mul() [1/2]\n\n matrix GiNaC::matrix::mul ( const matrix & other ) const\n\nProduct of matrices.\n\nExceptions\n logic_error (incompatible matrices)\n\nDefinition at line 589 of file matrix.cpp.\n\nReferences c, col, cols(), GiNaC::is_zero(), m, matrix(), row, and rows().\n\nReferenced by charpoly(), contract_with(), GiNaC::ncmul::evalm(), pow(), and scalar_mul_indexed().\n\n## ◆ mul() [2/2]\n\n matrix GiNaC::matrix::mul ( const numeric & other ) const\n\nProduct of matrix and scalar.\n\nDefinition at line 610 of file matrix.cpp.\n\nReferences c, col, m, matrix(), r, and row.\n\n## ◆ mul_scalar()\n\n matrix GiNaC::matrix::mul_scalar ( const ex & other ) const\n\nProduct of matrix and scalar expression.\n\nDefinition at line 623 of file matrix.cpp.\n\nReferences c, col, GiNaC::return_types::commutative, m, matrix(), r, GiNaC::ex::return_type(), and row.\n\n## ◆ pow()\n\n matrix GiNaC::matrix::pow ( const ex & expn ) const\n\nPower of a matrix.\n\nCurrently handles integer exponents only.\n\nDefinition at line 639 of file matrix.cpp.\n\n## ◆ operator()() [1/2]\n\n const ex & GiNaC::matrix::operator() ( unsigned ro, unsigned co ) const\n\noperator() to access elements for reading.\n\nParameters\n ro row of element co column of element\nExceptions\n range_error (index out of range)\n\nDefinition at line 686 of file matrix.cpp.\n\nReferences col, m, and row.\n\n## ◆ operator()() [2/2]\n\n ex & GiNaC::matrix::operator() ( unsigned ro, unsigned co )\n\noperator() to access elements for writing.\n\nParameters\n ro row of element co column of element\nExceptions\n range_error (index out of range)\n\nDefinition at line 700 of file matrix.cpp.\n\nReferences col, GiNaC::basic::ensure_if_modifiable(), m, and row.\n\n## ◆ set()\n\n matrix & GiNaC::matrix::set ( unsigned ro, unsigned co, const ex & value )\ninline\n\nDefinition at line 87 of file matrix.h.\n\nReferences value.\n\n## ◆ transpose()\n\n matrix GiNaC::matrix::transpose ( ) const\n\nTransposed of an m x n matrix, producing a new n x m matrix object that represents the transposed.\n\nDefinition at line 712 of file matrix.cpp.\n\nReferences c, cols(), m, matrix(), r, and rows().\n\n## ◆ determinant()\n\n ex GiNaC::matrix::determinant ( unsigned algo = `determinant_algo::automatic` ) const\n\nDeterminant of square matrix.\n\nThis routine doesn't actually calculate the determinant, it only implements some heuristics about which algorithm to run. If all the elements of the matrix are elements of an integral domain the determinant is also in that integral domain and the result is expanded only. If one or more elements are from a quotient field the determinant is usually also in that quotient field and the result is normalized before it is returned. This implies that the determinant of the symbolic 2x2 matrix [[a/(a-b),1],[b/(a-b),1]] is returned as unity. (In this respect, it behaves like MapleV and unlike Mathematica.)\n\nParameters\n algo allows to chose an algorithm\nReturns\nthe determinant as a new expression\nExceptions\n logic_error (matrix not square)\ndeterminant_algo\n\nDefinition at line 737 of file matrix.cpp.\n\nReferenced by charpoly(), and GiNaC::tensepsilon::contract_with().\n\n## ◆ trace()\n\n ex GiNaC::matrix::trace ( ) const\n\nTrace of a matrix.\n\nThe result is normalized if it is in some quotient field and expanded only otherwise. This implies that the trace of the symbolic 2x2 matrix [[a/(a-b),x],[y,b/(b-a)]] is recognized to be unity.\n\nReturns\nthe sum of diagonal elements\nExceptions\n logic_error (matrix not square)\n\nDefinition at line 866 of file matrix.cpp.\n\nReferenced by charpoly(), and eval_indexed().\n\n## ◆ charpoly()\n\n ex GiNaC::matrix::charpoly ( const ex & lambda ) const\n\nCharacteristic Polynomial.\n\nFollowing mathematica notation the characteristic polynomial of a matrix M is defined as the determinant of (M - lambda * 1) where 1 stands for the unit matrix of the same dimension as M. Note that some CASs define it with a sign inside the determinant which gives rise to an overall sign if the dimension is odd. This method returns the characteristic polynomial collected in powers of lambda as a new expression.\n\nReturns\ncharacteristic polynomial as new expression\nExceptions\n logic_error (matrix not square)\nmatrix::determinant()\n\nDefinition at line 894 of file matrix.cpp.\n\nReferences c, col, GiNaC::ex::collect(), determinant(), GiNaC::basic::ex, m, mul(), GiNaC::info_flags::numeric, poly, r, row, and trace().\n\n## ◆ inverse() [1/2]\n\n matrix GiNaC::matrix::inverse ( ) const\n\nInverse of this matrix, with automatic algorithm selection.\n\nDefinition at line 939 of file matrix.cpp.\n\nReferences GiNaC::solve_algo::automatic, and inverse().\n\nReferenced by inverse(), and pow().\n\n## ◆ inverse() [2/2]\n\n matrix GiNaC::matrix::inverse ( unsigned algo ) const\n\nInverse of this matrix.\n\nParameters\n algo selects the algorithm (one of solve_algo)\nReturns\nthe inverted matrix\nExceptions\n logic_error (matrix not square) runtime_error (singular matrix)\n\nDefinition at line 950 of file matrix.cpp.\n\nReferences GiNaC::_ex1, c, col, r, row, and solve().\n\n## ◆ solve()\n\n matrix GiNaC::matrix::solve ( const matrix & vars, const matrix & rhs, unsigned algo = `solve_algo::automatic` ) const\n\nSolve a linear system consisting of a m x n matrix and a m x p right hand side by applying an elimination scheme to the augmented matrix.\n\nParameters\n vars n x p matrix, all elements must be symbols rhs m x p matrix algo selects the solving algorithm\nReturns\nn x p solution matrix\nExceptions\n logic_error (incompatible matrices) invalid_argument (1st argument must be matrix of symbols) runtime_error (inconsistent linear system)\nsolve_algo\n\nDefinition at line 995 of file matrix.cpp.\n\nReferenced by inverse(), GiNaC::lsolve(), and GiNaC::sqrfree_parfrac().\n\n## ◆ rank() [1/2]\n\n unsigned GiNaC::matrix::rank ( ) const\n\nCompute the rank of this matrix.\n\nDefinition at line 1058 of file matrix.cpp.\n\nReferences GiNaC::solve_algo::automatic, and rank().\n\nReferenced by rank().\n\n## ◆ rank() [2/2]\n\n unsigned GiNaC::matrix::rank ( unsigned solve_algo ) const\n\nCompute the rank of this matrix using the given algorithm, which should be a member of enum solve_algo.\n\nDefinition at line 1065 of file matrix.cpp.\n\nReferences col, echelon_form(), GINAC_ASSERT, m, r, and row.\n\n## ◆ is_zero_matrix()\n\n bool GiNaC::matrix::is_zero_matrix ( ) const\n\nFunction to check that all elements of the matrix are zero.\n\nDefinition at line 1676 of file matrix.cpp.\n\nReferences m.\n\n## ◆ determinant_minor()\n\n ex GiNaC::matrix::determinant_minor ( ) const\nprotected\n\nRecursive determinant for small matrices having at least one symbolic entry.\n\nThe basic algorithm, known as Laplace-expansion, is enhanced by some bookkeeping to avoid calculation of the same submatrices (\"minors\") more than once. According to W.M.Gentleman and S.C.Johnson this algorithm is better than elimination schemes for matrices of sparse multivariate polynomials and also for matrices of dense univariate polynomials if the matrix' dimension is larger than 7.\n\nReturns\nthe determinant as a new expression (in expanded form)\nmatrix::determinant()\n\nDefinition at line 1096 of file matrix.cpp.\n\nReferences GiNaC::_ex0, GiNaC::_ex1, c, cols(), GiNaC::ex::expand(), GiNaC::ex::is_zero(), GiNaC::is_zero(), m, n, and r.\n\nReferenced by determinant().\n\n## ◆ echelon_form()\n\n std::vector< unsigned > GiNaC::matrix::echelon_form ( unsigned algo, int n )\nprotected\n\nDefinition at line 1197 of file matrix.cpp.\n\nReferenced by rank(), and solve().\n\n## ◆ gauss_elimination()\n\n int GiNaC::matrix::gauss_elimination ( const bool det = `false` )\nprotected\n\nPerform the steps of an ordinary Gaussian elimination to bring the m x n matrix into an upper echelon form.\n\nThe algorithm is ok for matrices with numeric coefficients but quite unsuited for symbolic matrices.\n\nParameters\n det may be set to true to save a lot of space if one is only interested in the diagonal elements (i.e. for calculating determinants). The others are set to zero in this case.\nReturns\nsign is 1 if an even number of rows was swapped, -1 if an odd number of rows was swapped and 0 if the matrix is singular.\n\nDefinition at line 1269 of file matrix.cpp.\n\nReferenced by determinant(), and echelon_form().\n\n## ◆ division_free_elimination()\n\n int GiNaC::matrix::division_free_elimination ( const bool det = `false` )\nprotected\n\nPerform the steps of division free elimination to bring the m x n matrix into an upper echelon form.\n\nParameters\n det may be set to true to save a lot of space if one is only interested in the diagonal elements (i.e. for calculating determinants). The others are set to zero in this case.\nReturns\nsign is 1 if an even number of rows was swapped, -1 if an odd number of rows was swapped and 0 if the matrix is singular.\n\nDefinition at line 1441 of file matrix.cpp.\n\nReferences GiNaC::_ex0, c, cols(), GiNaC::basic::ensure_if_modifiable(), GINAC_ASSERT, m, n, pivot(), r, and rows().\n\nReferenced by determinant(), and echelon_form().\n\n## ◆ fraction_free_elimination()\n\n int GiNaC::matrix::fraction_free_elimination ( const bool det = `false` )\nprotected\n\nPerform the steps of Bareiss' one-step fraction free elimination to bring the matrix into an upper echelon form.\n\nFraction free elimination means that divide is used straightforwardly, without computing GCDs first. This is possible, since we know the divisor at each step.\n\nParameters\n det may be set to true to save a lot of space if one is only interested in the last element (i.e. for calculating determinants). The others are set to zero in this case.\nReturns\nsign is 1 if an even number of rows was swapped, -1 if an odd number of rows was swapped and 0 if the matrix is singular.\n\nDefinition at line 1495 of file matrix.cpp.\n\nReferenced by determinant(), and echelon_form().\n\n## ◆ markowitz_elimination()\n\n std::vector< unsigned > GiNaC::matrix::markowitz_elimination ( unsigned n )\nprotected\n\nDefinition at line 1326 of file matrix.cpp.\n\nReferences GiNaC::_ex0, c, col, GINAC_ASSERT, GiNaC::ex::is_zero(), GiNaC::is_zero(), k, m, n, GiNaC::basic::normal(), r, row, and std::swap().\n\nReferenced by echelon_form().\n\n## ◆ pivot()\n\n int GiNaC::matrix::pivot ( unsigned ro, unsigned co, bool symbolic = `true` )\nprotected\n\nPartial pivoting method for matrix elimination schemes.\n\nUsual pivoting (symbolic==false) returns the index to the element with the largest absolute value in column ro and swaps the current row with the one where the element was found. With (symbolic==true) it does the same thing with the first non-zero element.\n\nParameters\n ro is the row from where to begin co is the column to be inspected symbolic signal if we want the first non-zero element to be pivoted (true) or the one with the largest absolute value (false).\nReturns\n0 if no interchange occurred, -1 if all are zero (usually signaling a degeneracy) and positive integer k means that rows ro and k were swapped.\n\nDefinition at line 1636 of file matrix.cpp.\n\nReferenced by division_free_elimination(), and gauss_elimination().\n\n## ◆ print_elements()\n\n void GiNaC::matrix::print_elements ( const print_context & c, const char * row_start, const char * row_end, const char * row_sep, const char * col_sep ) const\nprotected\n\nDefinition at line 168 of file matrix.cpp.\n\nReferences c, col, m, and row.\n\nReferenced by do_print(), do_print_latex(), and do_print_python_repr().\n\n## ◆ do_print()\n\n void GiNaC::matrix::do_print ( const print_context & c, unsigned level ) const\nprotected\n\nDefinition at line 184 of file matrix.cpp.\n\nReferences c, and print_elements().\n\n## ◆ do_print_latex()\n\n void GiNaC::matrix::do_print_latex ( const print_latex & c, unsigned level ) const\nprotected\n\nDefinition at line 191 of file matrix.cpp.\n\nReferences c, col, and print_elements().\n\n## ◆ do_print_python_repr()\n\n void GiNaC::matrix::do_print_python_repr ( const print_python_repr & c, unsigned level ) const\nprotected\n\nDefinition at line 198 of file matrix.cpp.\n\nReferences c, and print_elements().\n\n## ◆ row\n\n unsigned GiNaC::matrix::row\nprotected\n\nnumber of rows\n\nDefinition at line 115 of file matrix.h.\n\n## ◆ col\n\n unsigned GiNaC::matrix::col\nprotected\n\nnumber of columns\n\nDefinition at line 116 of file matrix.h.\n\n## ◆ m\n\n exvector GiNaC::matrix::m\nprotected\n\nrepresentation (cols indexed first)\n\nDefinition at line 117 of file matrix.h.\n\nThe documentation for this class was generated from the following files:\n\nThis page is part of the GiNaC developer's reference. It was generated automatically by doxygen. For an introduction, see the tutorial."
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