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https://www.omnicalculator.com/chemistry/equilibrium-constant | [
"# Equilibrium Constant Calculator\n\nCreated by Dominika Śmiałek, MD, PhD candidate\nReviewed by Dominik Czernia, PhD and Jack Bowater\nLast updated: Jun 27, 2023\n\nThis equilibrium constant calculator will help you understand reversible chemical reactions, which are reactions in which both the forward and backward reactions occur simultaneously.\n\nAfter a certain amount of time, an equilibrium is formed, meaning that the rate of reactants being turned into products is the same as the rate of products being turned back into reactants. At this point, the reaction is considered stable. To determine the state of this equilibrium, the reaction quotient should remain constant. With this tool, you can calculate the value of an equilibrium constant for a reaction while learning how to calculate the equilibrium constant with ease!\n\nBelow you can find the reversible reaction and equilibrium constant equations:\n\na[A] + b[B] ⇌ c[C] + d[D]\n\nK = ([C]c × [D]d)/([B]b × [A]a),\n\nwhere [A] and [B] are the molar concentrations of the reactants, and [C] and [D] are the molar concentrations of the products. To understand those concepts better, take a look at the molarity calculator\n\nCalculating the value of the equilibrium constant for a reaction is helpful when determining the amount of each substance formed at equilibrium as a ratio of each other. The constant doesn't depend on the initial concentrations of the reactants and products, as the same ratio will always be reached after a certain period of time. However, the constant may be influenced by:\n\n• Temperature;\n• Solvent; and\n• Ionic strength.\n\nIf the reaction is still underway, with oscillations between reagents and products, you have to use the reaction quotient calculator instead. However, even if it applies in a different context, it is defined in the same way as the equilibrium constant!\n\n## When is equilibrium constant used?\n\nEquilibrium constants are useful if you want to understand biochemical processes such as oxygen transport by hemoglobin or acid-base homeostasis in humans. The changes in acid-base homeostasis are mainly reflected in changes in the arterial and venous blood pH. Doctors will also check the equilibrium constant of transferrin in the blood, as transferrin saturation is a symptom of iron deficiency anemia.\n\nThis equation helps explain what will be favored by the equilibrium – the reactants or the products. This can give important information about the nature of the reaction and its mechanism. You will find more on this topic below.\n\n## Equilibrium constant equation\n\nThe equilibrium constant of a reaction relates to all of the species present in the reaction. However, in this calculator, we assume that there is a maximum of two main reactants and two main products. For the hypothetical reaction:\n\na[A] + b[B] ⇌ c[C] + d[D]\n\nthe equilibrium constant equation has the following formula:\n\nK = ([C]c × [D]d)/([B]b × [A]a)\n\nThe constant K reflect two measurements of quantity:\n\n• Kc - represents concentration, molarity, expressed as moles per liter (M=mol/L)\n• Kp - a function of both reactants and products partial pressure, usually in atmospheres, useful for calculations in the gas phase\n\nIf K > 1 – equilibrium favors the products\n\nIf K < 1 – equilibrium favors the reactants\n\nIf K = 1 – the mixture contains similar amounts of both products and reactants at equilibrium\n\nIf you're not sure how to switch from moles to other units and the other way around, take a look at our mole calculator.\n\n## Let's calculate the value of the equilibrium constant for a reaction\n\nTo give you more insight into how this equation works in practice, we created this example.\n\nYou have a mixture of gaseous sulfur dioxide and oxygen, from which you can react to form sulfur trioxide. This is one of the steps in synthesizing sulphuric acid:\n\n2 SO₂ + O₂ ⇌ 2 SO₃\n\nTherefore the equilibrium constant equation for this reaction is:\n\nK = [SO₃]²/([SO₂]² × [O₂])\n\nThe reaction mixture is left for a while until an equilibrium is established. The reactants and products have the following concentrations:\n\n• SO₂: 0.03 mol/L\n• O₂: 0.035 mol/L\n• SO₃: 0.5 mol/L\n\nWhen you put these numbers into the equation, K is found to be:\n\nK = 0.05²/(0.03² × 0.035)\n\nK = 7.937 × 10³\n\nAs K > 1, the equilibrium favors the products.\n\nIn our example, the concentrations of reactants and products at equilibrium were provided. We then used this information to calculate the equilibrium constant. But what if you knew the equilibrium constant and the unknown was the initial concentration or coefficient of a component? Well, don't worry! Our calculator works in reverse – so it solves both kinds of problems. Just input all of the data you have, and the results will be computed for you in an instance.\n\n## How to calculate the equilibrium constant?\n\nThis paragraph mainly focuses on how the equilibrium constant is determined analytically. To calculate the value of the equilibrium constant for a reaction, you need to measure (maybe with our titration calculator) the concentrations of the reactants and/or products. There are both experimental and computational methods for constant evaluation. Among experimental methods, you can find:\n\n• Potentiometry\n• Spectrophotometry\n• NMR chemical shift\n• Calorimetry\n\nWithin computational methods, there are:\n\n• Chemical model\n• Speciation calculations\n• Refinement\n• Model selection.\n\nAlthough you already know how to calculate equilibrium constant, save yourself some time and make use of our calculator!\n\n## FAQ\n\n### What is an equilibrium constant?\n\nThe equilibrium constant, K, determines the ratio of products and reactants of a reaction at equilibrium.\n\nFor example, having a reaction a[A] + b[B] ⇌ c[C] + d[D], you should allow the reaction to reach equilibrium and then calculate the ratio of the concentrations of the products to the concentrations of the reactants:\n\nK = ([C]c × [D]d)/([B]b × [A]a).\n\n### How can I write an equilibrium constant expression?\n\nTo write out an equilibrium constant expression, follow these steps:\n\n1. Calculate the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation).\n\n2. Calculate the product of equilibrium concentrations of reactants (raised to their coefficients in the balanced chemical equation).\n\n3. Find the ratio of the two products.\n\nFor example, the equilibrium constant for the reaction Cl₂ + 2NO₂ ⇌ 2NO₂Cl will be [NO₂Cl]²/([Cl₂] × [NO₂]²).\n\n### How can I find the equilibrium constant?\n\nLet's say you have a reaction of synthesis of ammonia: N₂ + 3H₂ ⇌ 2NH₃, where the concentrations are: N₂ = 0.04 mol/L, H₂ = 0.125 mol/L, and NH₃ = 0.003 mol/L.\n\nTo find its equilibrium constant:\n\n1. Write the equilibrium constant equation:\n\nK = ([NH₃]2)/([H₂]3 × [N₂]1).\n\n2. Enter the concentration:\n\nK = ([0.003 mol/L]2)/([0.125 mol/L]3 × [0.04 mol/L]1) = 0.1152 = 1.152 × 10⁻¹.\n\n### What changes the equilibrium constant?\n\nThe equilibrium constant is changed by temperature, the direction of the writing equation, the stoichiometry of the chemical equation, and concentration units. In contrast, the equilibrium constant does not depend on changes in concentration or pressure or the presence of a catalyst.\n\n### What will be the concentration of reactant if the equilibrium constant is 0.03?\n\n1.33 mol/L, assuming that you have a reaction of [A] + [B] ⇌ [C] + [D], and concentration of A is 0.5 mol/L, C is 0.2 mol/L, and D is 0.1 mol/L. To find it:\n\n1. Write the equilibrium constant equation:\n\nK = ([C] × [D])/([B] × [A]).\n\n2. Solve for [B]:\n\n[B] = ([C] × [D])/(K × [A]).\n\n3. Enter data:\n\n[B] = (0.2 × 0.1)/(0.03 × 0.5) = 1.33 mol/L.\n\nDominika Śmiałek, MD, PhD candidate\na[A] + b[B] ⇌ c[C] + d[D]\nConcentration: [A]\nM\nCoefficient: a\nConcentration: [B]\nM\nCoefficient: b\nConcentration: [C]\nM\nCoefficient: c\nConcentration: [D]\nM\nCoefficient: d\nEquilibrium constant\nPeople also viewed…\n\n### Alien civilization\n\nThe alien civilization calculator explores the existence of extraterrestrial civilizations by comparing two models: the Drake equation and the Astrobiological Copernican Limits👽\n\n### Electronegativity\n\nThis electronegativity calculator is an efficient tool to calculate the type of bond formed between two atoms based on their electronegativities.\n\n### Molar mass\n\nCheck how many grams are contained in 1 mole of any element or chemical compound with this molar mass calculator.\n\n### Sunbathing\n\nDo you always remember to put on sunscreen before going outside? Are you sure that you use enough? The Sunbathing Calculator ☀ will tell you when's the time to go back under an umbrella not to suffer from a sunburn!"
] | [
null
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https://wiki.math.ucr.edu/index.php?title=009A_Sample_Final_A,_Problem_5&direction=next&oldid=280 | [
"# 009A Sample Final A, Problem 5\n\n5. Consider the function $h(x)={\\frac {x^{3}}{3}}-2x^{2}-5x+{\\frac {35}{3}}}.$",
null,
"(a) Find the intervals where the function is increasing and decreasing.\n(b) Find the local maxima and minima.\n(c) Find the intervals on which $f(x)$",
null,
"is concave upward and concave downward.\n(d) Find all inflection points.\n(e) Use the information in the above to sketch the graph of $f(x)$",
null,
".\n\nFoundations:\nWe learn a lot about the shape of a function's graph from its derivatives. When a first derivative is positive, the function is increasing (heading uphill). When the first derivative is negative, it is decreasing (heading downhill). Of particular interest is when the first derivative at a point is zero. If f '(z) = 0 at a point z, and the first derivative splits around it (either f '(x) < 0 for x < z and f '(x) > 0 for x > z or f '(x) > 0 for x < z and f '(x) < 0 for x > z), then the point (z,f(z)) is a local maximum or minimum, respectively.\nThe second derivative tells us how the first derivative is changing. If the second derivative is positive, the first derivative (the slope of the tangent line) is increasing. This is equivalent to the graph \"turning left\" if we consider moving from negative x-values to positive. We call this \"concave up\". The parabola y = x2 is an example of a purely concave up graph, and its second derivative is the constant function y \" = 2.\nIf the second derivative is negative, then the first derivative is decreasing. This means we are turning right as we move from negative x-values to positive. This is called \"concave down\". The inverted parabola y = -x2 is an example of a purely concave down graph.\nA point z where the second derivative is zero, and the sign of the second derivative splits around it (either f \"(x) < 0 for x < z and f \"(x) > 0 for x > z, or f \"(x) > 0 for x < z and f \"(x) < 0 for x > z), then the point (z,f(z) is an inflection point.\n\nOf course, there are tests we use to find local extrema (maxima and minima, which is the plural of maximum and minimum). We are assuming the function f is continuous and differentiable in an interval containing the point x0.\nFirst Derivative Test: If at a point x0, f '(x0) = 0, and f '(x) < 0 for x < x0 while f '(x) > 0 for x > x0, then f(x0) is a local minimum.\nOn the other hand, if f '(x0) = 0, and f '(x) > 0 for x < x0 while f '(x) < 0 for x > x0, then f(x0) is a local maximum.\nSecond Derivative Test: If at a point x0, f '(x0) = 0, and f \"(x0) > 0, then f(x0) is a local minimum. On the other hand, if f \"(x0) < 0, then f(x0) is a local maximum. If f \"(x0) = 0, the test is inconclusive.\n\nSolution:\n\nFind the Derivatives and Their Roots:\nNote that\n$f'(x)=x^{2}-4x-5=(x-5)(x+1),$",
null,
"while\n$f''(x)=2x-4.$",
null,
"The roots of f ' are -1 and 5, while the only root of f \" is 2.\nProduce Sign Charts and Evaluate:\nSince all of our tests rely on the signs of our derivatives, we need to produce sign charts. For the first derivative, we can test values below -1, between -1 and 5 and above 5. For example:\n$f'(-10)=(-)(-)=(+),\\quad f'(0)=(-)(+)=(-),\\quad f'(10)=(+)(+)=(+).$",
null,
"From this, we can build a sign chart:\n $x:$",
null,
"$x<-1$",
null,
"$x=-1$",
null,
"$-1",
null,
"$x=5$",
null,
"$x>5$",
null,
"$f'(x):$",
null,
"$(+)$",
null,
"$0$",
null,
"$(-)$",
null,
"$0$",
null,
"$(+)$",
null,
"This gives us the following answers:\n(a) The function is increasing on $(\\infty ,-1)$",
null,
"and $(5,\\infty )$",
null,
", and decreasing on $(-1,5)$",
null,
".\n(b) The first derivative test shows\n$f(-1)=-{\\frac {1}{3}}-2+5+{\\frac {35}{3}}=14\\,{\\frac {1}{3}}$",
null,
"is a local maximum, while\n$f(5)={\\frac {125}{3}}-50-25+{\\frac {35}{\\,3}}=-75+{\\frac {160}{3}}=-\\,{\\frac {65}{\\,3}}.$",
null,
"is a local minimum.\nThe second derivative has only the single root x = 2, so we need only look at values for x < 2 and x > 2. These values are clearly negative and positive, respectively, so we have a sign chart:\n $x:$",
null,
"$x<2$",
null,
"$x=2$",
null,
"$x>2$",
null,
"$f''(x):$",
null,
"$(-)$",
null,
"$0$",
null,
"$(+)$",
null,
"This gives us the following answers:\n(c) The function is concave downward on $(-\\infty ,2)$",
null,
"and concave upward on $(2,\\infty )$",
null,
".\n(d) The function has an inflection point at $(2,f(2))=\\left(2,-{\\frac {11}{3}}\\right).$",
null,
"Graph:\nThis is part (e) of the problem. We wish to use all our results. In the image, the dots represent the two local extrema at x = -1 and x = 5, as well as the inflection point at x = 2. The graph is drawn in blue where it is concave downward, and in red where it is concave upward."
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https://projecteuclid.org/Proceedings/berkeley-symposium-on-mathematical-statistics-and-probability/Proceedings-of-the-Fifth-Berkeley-Symposium-on-Mathematical-Statistics-and/toc/bsmsp/1200513615 | [
"Home > Proceedings > Berkeley Symp. on Math. Statist. and Prob. > Proceedings of the Fifth Berkeley Symposium on Mathematical Statistics and Probability, Volume 3: Physical Sciences\nFifth Berkeley Symposium on Mathematical Statistics and Probability\nJune 21-July 18, 1965 and December 27, 1965-January 7, 1966 | Statistical Laboratory of the University of California, Berkeley\nProceedings of the Fifth Berkeley Symposium on Mathematical Statistics and Probability, Volume 3: Physical Sciences\n\nEditor(s) Lucien M. Le Cam, Jerzy Neyman\n\nBerkeley Symp. on Math. Statist. and Prob., 5.3: 324pp. (1967).",
null,
"View All Abstracts +\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, - (1967).",
null,
"E. M. Burbidge , G. R. Burbidge\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 1-18 (1967).",
null,
"W. H. McCrea\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 19-29 (1967).",
null,
"Thornton Page\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 31-49 (1967).",
null,
"Beverly T. Lynds\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 51-60 (1967).",
null,
"W. C. Livingston\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 61-72 (1967).",
null,
"R. L. Dobrushin\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 73-87 (1967).",
null,
"J. M. Hammersley\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 89-117 (1967).",
null,
"KEYWORDS: 60.40\nHerbert Solomon\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 119-134 (1967).",
null,
"KEYWORDS: 65.10, 52.00\nM. S. Bartlett\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 135-153 (1967).",
null,
"KEYWORDS: 62.85\nBenoit Mandelbrot\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 155-179 (1967).",
null,
"John Bather , Herman Chernoff\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 181-207 (1967).",
null,
"KEYWORDS: 62.45, 93.00\nRichard Bellman\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 209-215 (1967).",
null,
"P. Whittle\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 217-227 (1967).",
null,
"KEYWORDS: 93.60\nRichard E. Barlow , A. W. Marshall\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 229-257 (1967).",
null,
"Yu. K. Belyaev , B. V. Gnedenko , A. D. Soloviev\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 259-270 (1967).",
null,
"Z. W. Birnbaum , J. D. Esary\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 271-283 (1967).",
null,
"KEYWORDS: 62.80\nB. V. Gnedenko\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 285-291 (1967).",
null,
"KEYWORDS: 62.80\nFrank Proschan , Ronald Pyke\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 293-312 (1967).",
null,
"KEYWORDS: 62.75\nA. D. Soloviev\nBerkeley Symposium on Mathematical Statistics and Probability Vol. 5.3, 313-324 (1967).",
null,
"KEYWORDS: 62.80"
] | [
null,
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null,
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https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-5-section-5-5-dividing-polynomials-exercise-set-page-386/60 | [
"Introductory Algebra for College Students (7th Edition)\n\n-2x$^2$+4x\nDivide the polynomial by the monomial by separating the polynomial into individual terms and dividing each one by the monomial. Then cancel out common factors of the coefficients and subtract the powers of the variables to simplify. $\\frac{10x^3-20x^2}{-5x}$=$\\frac{10x^3}{-5x}$+$\\frac{-20x^2}{-5x}$=-2x$^{(3-1)}$+4x$^{(2-1)}$=-2x$^2$+4x Check the answer. -5x(-2x$^2$+4x)=10x$^{(1+2)}$-20x$^{(1+1)}$=10x$^3$-20x$^2$"
] | [
null
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https://zh.m.wikipedia.org/wiki/%E6%9C%80%E5%B0%8F%E5%85%83 | [
"# 最小元\n\n$(A,\\leq )$",
null,
"偏序集$B\\subseteq A$",
null,
"$y\\in B$",
null,
",若对于所有的$x\\in B$",
null,
"都有$y\\leq x$",
null,
",则称$y$",
null,
"$B$",
null,
"最小元"
] | [
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/8dccbed0945476c5f14b21d76e3d936f4acf547b",
null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/fb8124cb68686ede7083aa2a5a821f262eb62954",
null,
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null,
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null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/7de6a6e4f44d9dfcbfaadbdcf388d4b8a6fed109",
null,
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null,
"https://wikimedia.org/api/rest_v1/media/math/render/svg/47136aad860d145f75f3eed3022df827cee94d7a",
null
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https://www.hunker.com/13416849/how-to-figure-out-a-45-degree-angle | [
"# How to Figure Out a 45-Degree Angle\n\nHunker may earn compensation through affiliate links in this story.\n\n#### Things You'll Need\n\n• Ruler\n\n• Right angle (such as a carpenter's square or the corner of factory-cut cardboard)\n\nIf you need to figure out a 45-degree angle and you don't have a protractor handy, you can create a workaround. A 45-degree angle is half the size of right angle, which is 90 degrees. Working with a ruler and a square or other right angle substitute, you can make the angle without any special tools. The 45-degree angle can be useful for projects like painting diagonals on walls, marking trim, or completing crafts and decoration projects. You may want to use the first 45-degree angle you make as a template so that you can use it to mark other angles easily.\n\n## Step 1\n\nMark a piece of paper (or whatever surface you are working on) with a right angle. Use the square or whatever substitute right angle you have to trace the angle. The back of a pad of paper or even a magazine will suffice.\n\n## Step 2\n\nMeasure off a distance (the length is unimportant) from the point of the angle, and mark it on one leg of the angle.\n\n## Step 3\n\nMeasure and mark the identical distance on the other leg.\n\n## Step 4\n\nDraw a diagonal line between the two points on the legs of the right angle.\n\n## Step 5\n\nMeasure the length of the diagonal line and make a mark at its center point. Divide the total distance by two to ascertain this measurement.\n\n## Step 6\n\nDraw a line from the corner of the original right angle to the center point on the diagonal line between the legs of the angle. This bisects the right angle, creating two 45-degree angles."
] | [
null
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https://forum.worldviz.com/showpost.php?s=bcb3782be764ac3b83b29ea0b223e669&p=19983&postcount=1 | [
"Thread: Mark & Print Oculus data View Single Post\n#1\n tianmoran",
null,
"Member Join Date: Nov 2016 Posts: 16\nMark & Print Oculus data\n\nHi there,\n\nI'm coding for an experiment but muddled with some trouble. My goal is to record the Oculus data(position, euler) every 50ms. At the same time, I need to mark some events, like below the events are 'cylinder is on', 'key pressed', 'cylinder is off'. So I put a variable called 'Event' and hope to change its value right after each event happens. Unfortunately, it didn't work.\n\nI kind of know why it's not working. But is there any solution on that?\n\nThank you so much if anyone would like to help me out!!!\n\nCode:\n```import viz\nimport vizact\nimport vizshape\nimport vizinput\nimport vizinfo\nimport time\nimport random\nimport oculus\nimport vizfx\nimport numpy\nfrom numpy import *\n\nviz.go()\n\n##### SET UP OCULUS\nhmd = oculus.Rift()\noculusView.setOffset([0,0.5,-1])\n\ndef Demo():\nglobal acc, RT, Keycode, trialNo, oculus, oculusView, Cond, FMM, Code_direct, Event, trial, block, unit, acc, RT, Keycode, trialNo, oculus, oculusView,oculus_timer, Group, Code_Type, Code_Letter, FirstTrial\nFirstTrial = 1\nInstr.setPosition(0,0,0)\nInstr.color(0.5,0,0)\nInstr.visible(viz.OFF)\n\nOculusdata = open('testtest'+ \"_Oculus.txt\",'a',1)\n\nfor trial in range(10):\n\n###### Calculating Reaction Time\ndef df():\nglobal startTime\nstartTime = viz.tick()\nprint str(trial), 'startTime', startTime\nInstr.visible(viz.ON)\nyield vizact.ontimer2(1,0,df)\nEvent = 0\n\ndef waitkeydown():\nglobal r, KeyPress, startTime\n\nif r.key == 'j':\nacc[trial] = 1\nKeycode[trial] = 2 \t### Key J is pressed\nelif r.key == 'f':\nacc[trial] = 0\nKeycode[trial]= 1 \t\t### Key F is pressed\n\nEvent = 1\nInstr.visible(viz.OFF)\nEvent = 2\n\ndef printOculusData():\nglobal oculus_timer\noculus_timer = viz.tick()\nOculusData = \"%s\\t%s\\t%s\\t%s\\t%s\\n\" %(trial, Event,oculus_timer,oculusView.getPosition(), oculusView.getEuler())\nOculusdata.write(OculusData)\nif FirstTrial == 1:\nvizact.ontimer(0.05,printOculusData)\nFirstTrial = 0"
] | [
null,
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https://lbartman.com/4th-grade-respect-worksheets/ | [
"## ↤ l\n\n👤 will chen 🗓 May 17, 2021, 11:49 pm ( Last Modified )\n\nWinter Word Puzzles. Now that I’ve gotten up our Christmas word puzzles, I wanted to add a ‘secular’ alternative with these winter word puzzles.These are great for 3rd, 4th and 5th grade students – maybe even 6th grade. Another set of worksheets that you can pair with these word puzzles are our winter mad libs found here..Need elementary math resources? Use Lesson Planet to find curriculum covering topics such as counting, shapes (like this shape-dice printable!), addition, subtraction, time (this time concept book is fantastic), measurement, and arithmetic.Try this interactive to help youngsters understand the relationship between addition and subtraction. For more advanced classes, find lessons on geometry ..We would like to show you a description here but the site won’t allow us..Worksheets. regents books. ai lesson plans. worksheet generators. extras. regents exam archives 1866-now. jmap resource archives ai/geo/aii (2015-now) ia/ge/a2 (2007-17) math a/b (1998-2010) regents resources. interdisciplinary exams. nyc teacher resources.\n\nTake A Sneak Peak At The Movies Coming Out This Week (8/12) #BanPaparazzi – Hollywood.com will not post paparazzi photos; New Movie Releases This Weekend: March 5th – March 7th..\n\nRelated to \"4th Grade Respect Worksheets\" ⤵\n\nName : __________________\n\nSeat Num. : __________________\n\nDate : __________________\n\n62 + 37 = ...\n\n49 + 68 = ...\n\n38 + 54 = ...\n\n21 + 48 = ...\n\n90 + 24 = ...\n\n71 + 61 = ...\n\n51 + 23 = ...\n\n47 + 22 = ...\n\n16 + 31 = ...\n\n32 + 41 = ...\n\n70 + 47 = ...\n\n60 + 38 = ...\n\n28 + 39 = ...\n\n96 + 61 = ...\n\n35 + 21 = ...\n\n96 + 70 = ...\n\n98 + 72 = ...\n\n88 + 51 = ...\n\n22 + 70 = ...\n\n75 + 69 = ...\n\n53 + 53 = ...\n\n66 + 97 = ...\n\n65 + 12 = ...\n\n92 + 49 = ...\n\n48 + 84 = ...\n\n53 + 92 = ...\n\n60 + 96 = ...\n\n53 + 79 = ...\n\n26 + 31 = ...\n\n74 + 85 = ...\n\n22 + 55 = ...\n\n48 + 33 = ...\n\n63 + 55 = ...\n\n78 + 41 = ...\n\n26 + 83 = ...\n\n47 + 93 = ...\n\n16 + 60 = ...\n\n83 + 44 = ...\n\n78 + 20 = ...\n\n64 + 60 = ...\n\n77 + 81 = ...\n\n38 + 93 = ...\n\n43 + 30 = ...\n\n97 + 66 = ...\n\n80 + 63 = ...\n\n36 + 32 = ...\n\n25 + 65 = ...\n\n68 + 92 = ...\n\n83 + 60 = ...\n\n64 + 99 = ...\n\n23 + 80 = ...\n\n11 + 99 = ...\n\n51 + 22 = ...\n\n29 + 86 = ...\n\n53 + 19 = ...\n\n22 + 21 = ...\n\n88 + 93 = ...\n\n97 + 88 = ...\n\n46 + 93 = ...\n\n34 + 47 = ...\n\n48 + 55 = ...\n\n46 + 66 = ...\n\n26 + 92 = ...\n\n95 + 11 = ...\n\n85 + 71 = ...\n\n51 + 65 = ...\n\n66 + 19 = ...\n\n25 + 90 = ...\n\n84 + 29 = ...\n\n19 + 23 = ...\n\n68 + 86 = ...\n\n30 + 55 = ...\n\n21 + 20 = ...\n\n38 + 34 = ...\n\n36 + 71 = ...\n\n93 + 55 = ...\n\n99 + 57 = ...\n\n20 + 62 = ...\n\n75 + 49 = ...\n\n79 + 66 = ...\n\n81 + 41 = ...\n\n47 + 20 = ...\n\n85 + 77 = ...\n\n63 + 63 = ...\n\n37 + 24 = ...\n\n93 + 79 = ...\n\n66 + 63 = ...\n\n80 + 46 = ...\n\n67 + 16 = ...\n\n93 + 48 = ...\n\n45 + 51 = ...\n\n35 + 64 = ...\n\n78 + 68 = ...\n\n43 + 82 = ...\n\n42 + 91 = ...\n\n83 + 23 = ...\n\n41 + 62 = ...\n\n15 + 16 = ...\n\n83 + 67 = ...\n\n31 + 82 = ...\n\n24 + 14 = ...\n\n78 + 34 = ...\n\n42 + 46 = ...\n\n74 + 12 = ...\n\n31 + 13 = ...\n\n46 + 90 = ...\n\n12 + 37 = ...\n\n76 + 71 = ...\n\n86 + 88 = ...\n\n47 + 76 = ...\n\n64 + 31 = ...\n\n59 + 97 = ...\n\n10 + 64 = ...\n\n26 + 76 = ...\n\n67 + 18 = ...\n\n28 + 41 = ...\n\n29 + 64 = ...\n\n92 + 83 = ...\n\n64 + 29 = ...\n\n56 + 63 = ...\n\n72 + 58 = ...\n\n22 + 25 = ...\n\n83 + 97 = ...\n\n59 + 22 = ...\n\n17 + 56 = ...\n\n78 + 37 = ...\n\n14 + 87 = ...\n\n47 + 67 = ...\n\n57 + 37 = ...\n\n98 + 49 = ...\n\n29 + 83 = ...\n\n91 + 17 = ...\n\n92 + 19 = ...\n\n87 + 24 = ...\n\n68 + 20 = ...\n\n87 + 42 = ...\n\n32 + 74 = ...\n\n82 + 50 = ...\n\n78 + 67 = ...\n\n64 + 87 = ...\n\n34 + 68 = ...\n\n86 + 28 = ...\n\n70 + 23 = ...\n\n10 + 55 = ...\n\n52 + 60 = ...\n\n11 + 28 = ...\n\n87 + 96 = ...\n\n78 + 76 = ...\n\n55 + 26 = ...\n\n96 + 71 = ...\n\n32 + 25 = ...\n\n89 + 50 = ...\n\n87 + 76 = ...\n\n45 + 89 = ...\n\n90 + 65 = ...\n\n79 + 64 = ...\n\n86 + 42 = ...\n\n22 + 49 = ...\n\n79 + 80 = ...\n\n65 + 28 = ...\n\n86 + 19 = ...\n\n29 + 83 = ...\n\n39 + 12 = ...\n\n48 + 89 = ...\n\n76 + 61 = ...\n\n87 + 34 = ...\n\n16 + 74 = ...\n\n85 + 63 = ...\n\n85 + 32 = ...\n\n95 + 25 = ...\n\n60 + 92 = ...\n\n86 + 68 = ...\n\n85 + 97 = ...\n\n91 + 89 = ...\n\n72 + 15 = ...\n\n13 + 20 = ...\n\n86 + 76 = ...\n\n44 + 15 = ...\n\n65 + 63 = ...\n\n73 + 13 = ...\n\n30 + 85 = ...\n\n33 + 86 = ...\n\n91 + 91 = ...\n\n24 + 79 = ...\n\n88 + 64 = ...\n\n82 + 14 = ...\n\n38 + 18 = ...\n\n27 + 83 = ...\n\n37 + 40 = ...\n\n53 + 81 = ...\n\n78 + 72 = ...\n\n81 + 29 = ...\n\n35 + 66 = ...\n\n35 + 94 = ...\n\n43 + 90 = ...\n\n38 + 46 = ...\n\n56 + 89 = ...\n\n73 + 57 = ...\n\n44 + 36 = ...\n\n78 + 60 = ...\n\nshow printable version !!!hide the show",
null,
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http://questions.instantgrades.com/in-the-figure-below-two-point-particles-are-fixed-on-an-x-axis-separated-by-a-distance-d-particle-a-has-mass-ma-and-particle-b-has-mass-2-00ma-a-third-particle-c-of-mass-55-0ma-is-to-be-placed-on/ | [
"Interested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality?\n\n# In the figure below, two point particles are fixed on an x axis separated by a distance d. Particle A has mass mA and particle B has mass 2.00mA. A third particle C, of mass 55.0mA, is to be placed on the x axis and near particles A and B. In terms of distance d, at what x coordinate should C be placed so that the net gravitational force on particle A from particles B and C is zero?\n\nIn the figure below, two point particles are fixed on an x axis separated by a distance d. Particle A has mass mA and particle B has mass 2.00mA. A third particle C, of mass 55.0mA, is to be placed on the x axis and near particles A and B. In terms of distance d, at what x coordinate should C be placed so that the net gravitational force on particle A from particles B and C is zero?\n\nInterested in a PLAGIARISM-FREE paper based on these particular instructions?...with 100% confidentiality?"
] | [
null
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https://openstax.org/books/introductory-business-statistics/pages/9-1-null-and-alternative-hypotheses | [
"# 9.1Null and Alternative Hypotheses\n\nIntroductory Business Statistics9.1 Null and Alternative Hypotheses\n\nThe actual test begins by considering two hypotheses. They are called the null hypothesis and the alternative hypothesis. These hypotheses contain opposing viewpoints.\n\nH0: The null hypothesis: It is a statement of no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.\n\nHa: The alternative hypothesis: It is a claim about the population that is contradictory to H0 and what we conclude when we cannot accept H0. The alternative hypothesis is the contender and must win with significant evidence to overthrow the status quo. This concept is sometimes referred to the tyranny of the status quo because as we will see later, to overthrow the null hypothesis takes usually 90 or greater confidence that this is the proper decision.\n\nSince the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.\n\nAfter you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are \"cannot accept H0\" if the sample information favors the alternative hypothesis or \"do not reject H0\" or \"decline to reject H0\" if the sample information is insufficient to reject the null hypothesis. These conclusions are all based upon a level of probability, a significance level, that is set my the analyst.\n\nTable 9.1 presents the various hypotheses in the relevant pairs. For example, if the null hypothesis is equal to some value, the alternative has to be not equal to that value.\n\nH0 Ha\nequal (=) not equal (≠)\ngreater than or equal to (≥) less than (<)\nless than or equal to (≤) more than (>)\nTable 9.1\n\n### Note\n\nAs a mathematical convention H0 always has a symbol with an equal in it. Ha never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test.\n\n### Example 9.1\n\nH0: No more than 30% of the registered voters in Santa Clara County voted in the primary election. p ≤ 30\nHa: More than 30% of the registered voters in Santa Clara County voted in the primary election. p > 30\n\n### Example 9.2\n\nWe want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are:\nH0: μ = 2.0\nHa: μ ≠ 2.0\n\n### Example 9.3\n\nWe want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:\nH0: μ ≥ 5\nHa: μ < 5"
] | [
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https://typeset.io/topics/multivariate-normal-distribution-3bbd5jb4 | [
"Topic\n\n# Multivariate normal distribution\n\nAbout: Multivariate normal distribution is a(n) research topic. Over the lifetime, 8304 publication(s) have been published within this topic receiving 324203 citation(s). The topic is also known as: multivariate Gaussian distribution & joint normal distribution.\n##### Papers\nMore filters\n\nBook\n01 Jan 1982-\nAbstract: (NOTE: Each chapter begins with an Introduction, and concludes with Exercises and References.) I. GETTING STARTED. 1. Aspects of Multivariate Analysis. Applications of Multivariate Techniques. The Organization of Data. Data Displays and Pictorial Representations. Distance. Final Comments. 2. Matrix Algebra and Random Vectors. Some Basics of Matrix and Vector Algebra. Positive Definite Matrices. A Square-Root Matrix. Random Vectors and Matrices. Mean Vectors and Covariance Matrices. Matrix Inequalities and Maximization. Supplement 2A Vectors and Matrices: Basic Concepts. 3. Sample Geometry and Random Sampling. The Geometry of the Sample. Random Samples and the Expected Values of the Sample Mean and Covariance Matrix. Generalized Variance. Sample Mean, Covariance, and Correlation as Matrix Operations. Sample Values of Linear Combinations of Variables. 4. The Multivariate Normal Distribution. The Multivariate Normal Density and Its Properties. Sampling from a Multivariate Normal Distribution and Maximum Likelihood Estimation. The Sampling Distribution of 'X and S. Large-Sample Behavior of 'X and S. Assessing the Assumption of Normality. Detecting Outliners and Data Cleaning. Transformations to Near Normality. II. INFERENCES ABOUT MULTIVARIATE MEANS AND LINEAR MODELS. 5. Inferences About a Mean Vector. The Plausibility of ...m0 as a Value for a Normal Population Mean. Hotelling's T 2 and Likelihood Ratio Tests. Confidence Regions and Simultaneous Comparisons of Component Means. Large Sample Inferences about a Population Mean Vector. Multivariate Quality Control Charts. Inferences about Mean Vectors When Some Observations Are Missing. Difficulties Due To Time Dependence in Multivariate Observations. Supplement 5A Simultaneous Confidence Intervals and Ellipses as Shadows of the p-Dimensional Ellipsoids. 6. Comparisons of Several Multivariate Means. Paired Comparisons and a Repeated Measures Design. Comparing Mean Vectors from Two Populations. Comparison of Several Multivariate Population Means (One-Way MANOVA). Simultaneous Confidence Intervals for Treatment Effects. Two-Way Multivariate Analysis of Variance. Profile Analysis. Repealed Measures, Designs, and Growth Curves. Perspectives and a Strategy for Analyzing Multivariate Models. 7. Multivariate Linear Regression Models. The Classical Linear Regression Model. Least Squares Estimation. Inferences About the Regression Model. Inferences from the Estimated Regression Function. Model Checking and Other Aspects of Regression. Multivariate Multiple Regression. The Concept of Linear Regression. Comparing the Two Formulations of the Regression Model. Multiple Regression Models with Time Dependant Errors. Supplement 7A The Distribution of the Likelihood Ratio for the Multivariate Regression Model. III. ANALYSIS OF A COVARIANCE STRUCTURE. 8. Principal Components. Population Principal Components. Summarizing Sample Variation by Principal Components. Graphing the Principal Components. Large-Sample Inferences. Monitoring Quality with Principal Components. Supplement 8A The Geometry of the Sample Principal Component Approximation. 9. Factor Analysis and Inference for Structured Covariance Matrices. The Orthogonal Factor Model. Methods of Estimation. Factor Rotation. Factor Scores. Perspectives and a Strategy for Factor Analysis. Structural Equation Models. Supplement 9A Some Computational Details for Maximum Likelihood Estimation. 10. Canonical Correlation Analysis Canonical Variates and Canonical Correlations. Interpreting the Population Canonical Variables. The Sample Canonical Variates and Sample Canonical Correlations. Additional Sample Descriptive Measures. Large Sample Inferences. IV. CLASSIFICATION AND GROUPING TECHNIQUES. 11. Discrimination and Classification. Separation and Classification for Two Populations. Classifications with Two Multivariate Normal Populations. Evaluating Classification Functions. Fisher's Discriminant Function...nSeparation of Populations. Classification with Several Populations. Fisher's Method for Discriminating among Several Populations. Final Comments. 12. Clustering, Distance Methods and Ordination. Similarity Measures. Hierarchical Clustering Methods. Nonhierarchical Clustering Methods. Multidimensional Scaling. Correspondence Analysis. Biplots for Viewing Sample Units and Variables. Procustes Analysis: A Method for Comparing Configurations. Appendix. Standard Normal Probabilities. Student's t-Distribution Percentage Points. ...c2 Distribution Percentage Points. F-Distribution Percentage Points. F-Distribution Percentage Points (...a = .10). F-Distribution Percentage Points (...a = .05). F-Distribution Percentage Points (...a = .01). Data Index. Subject Index.\n\n11,666 citations\n\nBook\n14 Sep 1984-\nAbstract: Preface to the Third Edition.Preface to the Second Edition.Preface to the First Edition.1. Introduction.2. The Multivariate Normal Distribution.3. Estimation of the Mean Vector and the Covariance Matrix.4. The Distributions and Uses of Sample Correlation Coefficients.5. The Generalized T2-Statistic.6. Classification of Observations.7. The Distribution of the Sample Covariance Matrix and the Sample Generalized Variance.8. Testing the General Linear Hypothesis: Multivariate Analysis of Variance9. Testing Independence of Sets of Variates.10. Testing Hypotheses of Equality of Covariance Matrices and Equality of Mean Vectors and Covariance Matrices.11. Principal Components.12. Cononical Correlations and Cononical Variables.13. The Distributions of Characteristic Roots and Vectors.14. Factor Analysis.15. Pattern of Dependence Graphical Models.Appendix A: Matrix Theory.Appendix B: Tables.References.Index.\n\n9,680 citations\n\nBook\n01 Jan 1976-\nAbstract: A text designed to make multivariate techniques available to behavioural, social, biological and medical students. Special features include an approach to multivariate inference based on the union-intersection and generalized likelihood ratio principles.\n\n5,797 citations\n\nJournal ArticleDOI\nAbstract: A common concern when faced with multivariate data with missing values is whether the missing data are missing completely at random (MCAR); that is, whether missingness depends on the variables in the data set. One way of assessing this is to compare the means of recorded values of each variable between groups defined by whether other variables in the data set are missing or not. Although informative, this procedure yields potentially many correlated statistics for testing MCAR, resulting in multiple-comparison problems. This article proposes a single global test statistic for MCAR that uses all of the available data. The asymptotic null distribution is given, and the small-sample null distribution is derived for multivariate normal data with a monotone pattern of missing data. The test reduces to a standard t test when the data are bivariate with missing data confined to a single variable. A limited simulation study of empirical sizes for the test applied to normal and nonnormal data suggests th...\n\n4,890 citations\n\nBook\n01 Jun 1970-\nAbstract: Foreword.Preface.PART ONE. SURVEY OF PROBABILITY THEORY.Chapter 1. Introduction.Chapter 2. Experiments, Sample Spaces, and Probability.2.1 Experiments and Sample Spaces.2.2 Set Theory.2.3 Events and Probability.2.4 Conditional Probability.2.5 Binomial Coefficients.Exercises.Chapter 3. Random Variables, Random Vectors, and Distributions Functions.3.1 Random Variables and Their Distributions.3.2 Multivariate Distributions.3.3 Sums and Integrals.3.4 Marginal Distributions and Independence.3.5 Vectors and Matrices.3.6 Expectations, Moments, and Characteristic Functions.3.7 Transformations of Random Variables.3.8 Conditional Distributions.Exercises.Chapter 4. Some Special Univariate Distributions.4.1 Introduction.4.2 The Bernoulli Distributions.4.3 The Binomial Distribution.4.4 The Poisson Distribution.4.5 The Negative Binomial Distribution.4.6 The Hypergeometric Distribution.4.7 The Normal Distribution.4.8 The Gamma Distribution.4.9 The Beta Distribution.4.10 The Uniform Distribution.4.11 The Pareto Distribution.4.12 The t Distribution.4.13 The F Distribution.Exercises.Chapter 5. Some Special Multivariate Distributions.5.1 Introduction.5.2 The Multinomial Distribution.5.3 The Dirichlet Distribution.5.4 The Multivariate Normal Distribution.5.5 The Wishart Distribution.5.6 The Multivariate t Distribution.5.7 The Bilateral Bivariate Pareto Distribution.Exercises.PART TWO. SUBJECTIVE PROBABILITY AND UTILITY.Chapter 6. Subjective Probability.6.1 Introduction.6.2 Relative Likelihood.6.3 The Auxiliary Experiment.6.4 Construction of the Probability Distribution.6.5 Verification of the Properties of a Probability Distribution.6.6 Conditional Likelihoods.Exercises.Chapter 7. Utility.7.1 Preferences Among Rewards.7.2 Preferences Among Probability Distributions.7.3 The Definitions of a Utility Function.7.4 Some Properties of Utility Functions.7.5 The Utility of Monetary Rewards.7.6 Convex and Concave Utility Functions.7.7 The Anxiomatic Development of Utility.7.8 Construction of the Utility Function.7.9 Verification of the Properties of a Utility Function.7.10 Extension of the Properties of a Utility Function to the Class ?E.Exercises.PART THREE. STATISTICAL DECISION PROBLEMS.Chapter 8. Decision Problems.8.1 Elements of a Decision Problem.8.2 Bayes Risk and Bayes Decisions.8.3 Nonnegative Loss Functions.8.4 Concavity of the Bayes Risk.8.5 Randomization and Mixed Decisions.8.6 Convex Sets.8.7 Decision Problems in Which ~2 and D Are Finite.8.8 Decision Problems with Observations.8.9 Construction of Bayes Decision Functions.8.10 The Cost of Observation.8.11 Statistical Decision Problems in Which Both ? and D contains Two Points.8.12 Computation of the Posterior Distribution When the Observations Are Made in More Than One Stage.Exercises.Chapter 9. Conjugate Prior Distributions.9.1 Sufficient Statistics.9.2 Conjugate Families of Distributions.9.3 Construction of the Conjugate Family.9.4 Conjugate Families for Samples from Various Standard Distributions.9.5 Conjugate Families for Samples from a Normal Distribution.9.6 Sampling from a Normal Distribution with Unknown Mean and Unknown Precision.9.7 Sampling from a Uniform Distribution.9.8 A Conjugate Family for Multinomial Observations.9.9 Conjugate Families for Samples from a Multivariate Normal Distribution.9.10 Multivariate Normal Distributions with Unknown Mean Vector and Unknown Precision matrix.9.11 The Marginal Distribution of the Mean Vector.9.12 The Distribution of a Correlation.9.13 Precision Matrices Having an Unknown Factor.Exercises.Chapter 10. Limiting Posterior Distributions.10.1 Improper Prior Distributions.10.2 Improper Prior Distributions for Samples from a Normal Distribution.10.3 Improper Prior Distributions for Samples from a Multivariate Normal Distribution.10.4 Precise Measurement.10.5 Convergence of Posterior Distributions.10.6 Supercontinuity.10.7 Solutions of the Likelihood Equation.10.8 Convergence of Supercontinuous Functions.10.9 Limiting Properties of the Likelihood Function.10.10 Normal Approximation to the Posterior Distribution.10.11 Approximation for Vector Parameters.10.12 Posterior Ratios.Exercises.Chapter 11. Estimation, Testing Hypotheses, and linear Statistical Models.11.1 Estimation.11.2 Quadratic Loss.11.3 Loss Proportional to the Absolute Value of the Error.11.4 Estimation of a Vector.11.5 Problems of Testing Hypotheses.11.6 Testing a Simple Hypothesis About the Mean of a Normal Distribution.11.7 Testing Hypotheses about the Mean of a Normal Distribution.11.8 Deciding Whether a Parameter Is Smaller or larger Than a Specific Value.11.9 Deciding Whether the Mean of a Normal Distribution Is Smaller or larger Than a Specific Value.11.10 Linear Models.11.11 Testing Hypotheses in Linear Models.11.12 Investigating the Hypothesis That Certain Regression Coefficients Vanish.11.13 One-Way Analysis of Variance.Exercises.PART FOUR. SEQUENTIAL DECISIONS.Chapter 12. Sequential Sampling.12.1 Gains from Sequential Sampling.12.2 Sequential Decision Procedures.12.3 The Risk of a Sequential Decision Procedure.12.4 Backward Induction.12.5 Optimal Bounded Sequential Decision procedures.12.6 Illustrative Examples.12.7 Unbounded Sequential Decision Procedures.12.8 Regular Sequential Decision Procedures.12.9 Existence of an Optimal Procedure.12.10 Approximating an Optimal Procedure by Bounded Procedures.12.11 Regions for Continuing or Terminating Sampling.12.12 The Functional Equation.12.13 Approximations and Bounds for the Bayes Risk.12.14 The Sequential Probability-ratio Test.12.15 Characteristics of Sequential Probability-ratio Tests.12.16 Approximating the Expected Number of Observations.Exercises.Chapter 13. Optimal Stopping.13.1 Introduction.13.2 The Statistician's Reward.13.3 Choice of the Utility Function.13.4 Sampling Without Recall.13.5 Further Problems of Sampling with Recall and Sampling without Recall.13.6 Sampling without Recall from a Normal Distribution with Unknown Mean.13.7 Sampling with Recall from a Normal Distribution with Unknown Mean.13.8 Existence of Optimal Stopping Rules.13.9 Existence of Optimal Stopping Rules for Problems of Sampling with Recall and Sampling without Recall.13.10 Martingales.13.11 Stopping Rules for Martingales.13.12 Uniformly Integrable Sequences of Random Variables.13.13 Martingales Formed from Sums and Products of Random Variables.13.14 Regular Supermartingales.13.15 Supermartingales and General Problems of Optimal Stopping.13.16 Markov Processes.13.17 Stationary Stopping Rules for Markov Processes.13.18 Entrance-fee Problems.13.19 The Functional Equation for a Markov Process.Exercises.Chapter 14. Sequential Choice of Experiments.14.1 Introduction.14.2 Markovian Decision Processes with a Finite Number of Stages.14.3 Markovian Decision Processes with an Infinite Number of Stages.14.4 Some Betting Problems.14.5 Two-armed-bandit Problems.14.6 Two-armed-bandit Problems When the Value of One Parameter Is Known.14.7 Two-armed-bandit Problems When the Parameters Are Dependent.14.8 Inventory Problems.14.9 Inventory Problems with an Infinite Number of Stages.14.10 Control Problems.14.11 Optimal Control When the Process Cannot Be Observed without Error.14.12 Multidimensional Control Problems.14.13 Control Problems with Actuation Errors.14.14 Search Problems.14.15 Search Problems with Equal Costs.14.16 Uncertainty Functions and Statistical Decision Problems.14.17 Sufficient Experiments.14.18 Examples of Sufficient Experiments.Exercises.References.Supplementary Bibliography.Name Index.Subject Index.\n\n4,255 citations\n\n##### Network Information\n###### Related Topics (5)\nTest statistic\n\n9K papers, 425.6K citations\n\n94% related\nNonparametric statistics\n\n19.9K papers, 844.1K citations\n\n93% related\nNonparametric regression\n\n7.6K papers, 354.4K citations\n\n93% related\nEfficient estimator\n\n6.5K papers, 228.9K citations\n\n93% related\nBias of an estimator\n\n6.1K papers, 171.9K citations\n\n93% related\n##### Performance\n###### Metrics\nNo. of papers in the topic in previous years\nYearPapers\n20227\n2021273\n2020326\n2019259\n2018241\n2017291\n\n###### Top Attributes\n\nShow by:\n\nTopic's top 5 most impactful authors\n\nNarayanaswamy Balakrishnan\n\n37 papers, 1.1K citations\n\nArjun K. Gupta\n\n37 papers, 989 citations\n\nMuni S. Srivastava\n\n35 papers, 1.3K citations"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.69898313,"math_prob":0.77100784,"size":15684,"snap":"2022-27-2022-33","text_gpt3_token_len":3599,"char_repetition_ratio":0.17321429,"word_repetition_ratio":0.029442692,"special_character_ratio":0.20798266,"punctuation_ratio":0.21218343,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9744244,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-18T20:39:05Z\",\"WARC-Record-ID\":\"<urn:uuid:67b59d1c-143b-48f3-925c-5157c0a39303>\",\"Content-Length\":\"246259\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:acd4e65d-fc9e-4c13-9016-2af22289f1a8>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd6a0b41-cd86-4cee-9c9b-43090e4c951e>\",\"WARC-IP-Address\":\"13.32.151.107\",\"WARC-Target-URI\":\"https://typeset.io/topics/multivariate-normal-distribution-3bbd5jb4\",\"WARC-Payload-Digest\":\"sha1:L2SOAV7DPWZFFKDMBSPOOOAGQJDQ326G\",\"WARC-Block-Digest\":\"sha1:QESPXCNPR37SVJEWDOVQ4256CUHGXN5X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573399.40_warc_CC-MAIN-20220818185216-20220818215216-00665.warc.gz\"}"} |
http://allenfleishmanbiostatistics.com/Articles/2012/04/5-a-accepting-the-null-hypothesis-a-bayesian-approach/ | [
"# 5.A Accepting the Null-Hypothesis a Bayesian approach\n\nThe following blog was written by Randy Gallistel, PhD of Rutgers. It presents a Bayesian approach to hypothesis testing. It was written on April 23, 2012, but will eventually appear to have an earlier date, to sort it immediately after my original (Frequentist) blog.\n\nThe Bayesian would say that the truism that one cannot prove the null is a consequence of a misformulation of the inference problem. If we agree that hypothesis-testing statistics is the mathematics of probabilistic inference and if we resort to probabilistic inference only when we are faced with some uncertainty as to which conclusion to draw, then the NHST formulation of the problem is ruled out because, given that formulation, we have no uncertainty: Only one possible conclusion is to be tested against the data, the null conclusion, and we are a priori certain that it cannot be true. Thus, there is no inference problem.\n\nOne might object that this is not so; the alternative is that there is “some” (positive!) effect. But until we specify what we understand by “some”, this is not a well-formulated alternative. For example, in a typical pharmacological clinical trial, “some” effect could mean that the drug had an effect anywhere between 0 and complete cure in every patient (maximum possible effect). If that is what we understand “some” effect to mean, then for most drugs, the null conclusion (no positive effect) has a greater likelihood than the “some” (positive) effect conclusion.\n\nThe Bayesian computation tells us how well each possible conclusion (aka hypothesis) predicts the data that we have gathered. The possible hypotheses are represented by prior distributions. These prior distributions may be thought of as bets made by each hypothesis before the data are examined. Each hypothesis has a unit mass of prior probability with which to bet. The null conclusion bets it all on 0. The unlimited “some” hypothesis spreads its unit mass of prior probability out over all possible effect sizes.The question then becomes which of these prior probability distributions does a better job of predicting the likelihood function.\n\nLikelihood is sometimes called the reverse probability. In forward probability, we assume that we know the distribution (that is, we know its form and the values of its parameters) and we use this knowledge to predict how probable different outcomes are. In reverse probability (likelihood), we assume we know the data and we use the data to compute how likely those data would be for various assumptions about the distribution from which they came (assumptions about the form and about the values of the parameters of the distribution from which the data may have come). The likelihood function tells us the likelihood for all different values of the parameters of an assumed distribution. The highly likely values are the ones that predict what we have observed; the highly unlikely ones are the ones that predict that we should not have observed what we have in fact observed\n\nThe possibilities for which probabilities are defined in a probability distribution are mutually exclusive and exhaustive, so their probabilities must sum (integrate) to one. Reverse probabilities (likelihoods), by contrast, are neither mutually exclusive nor exhaustive. It is possible to have two hypotheses that are distinct but overlapping and they may both either predict the data we have with absolute certainty (in which case, they both have a likelihood of 1) or not at all (in which case, they both have a likelihood of 0). Generally, however, one hypothesis does a better job of predicting our data than the other, in which case that hypothesis is more likely than the alternative. The Bayes Factor is the ratio of the likelihoods, in other words, the likelihood of the one hypothesis relative to the other.\n\nSuppose the data suggest only a weak positive effect. That means that we COULD have got those data with reasonable probability even if there is in fact no effect (the null hypothesis), whereas, we could not have got those data if the effect of the drug were so great as to completely cure every patient, which is one of the states of the world encompassed by the unbridled version of the “some” hypothesis. The marginal likelihood of an hypothesis is its average likelihood over each possible value of (say) its mean that is compassed by the associated prior probability function. Because weakly positive results are inconsistent with all the stronger forms of “some”, the marginal likelihood of the unbridled “some” hypothesis is low. The null places all its chips on a single value 0, so the “average” for this hypothesis is simply the likelihood at that value, and, as already noted, if the data are weak, then the likelihood that the true effect is 0 is substantial.\n\nThus, the Bayesian would argue that when we formulate the inference problem in such a way that there actually is some uncertainty–hence, something to be inferred–the data may very well favor the null hypothesis. The frequentist objects that when we frame the inference problem this way, our inference will depend on the upper limit that we put on what we understand by ‘some,’ and that is true. But there is no reason not to compute the Bayes Factor (the ratio of the marginal likelihoods) as a function of this upper limit. If the Bayes Factor in favor of the null approaches 1 from above as the upper limit goes to 0, that is, as “some effect” becomes indistinguishable for “no effect”, then we can conclude that the inference to the null is to be preferred over ANY (positive!) alternative to it.\n\nWhen the null is actually true, the data will yield such a function 50% of the time. And, when the pharmacological effect is actually slightly or strongly negative (deleterious), the data will yield such a function even more often. Moreover, this will be true no matter how small the N. Thus, we have a rational basis for favoring one conclusion over the other no matter how little data we have.\n\nWhen, by chance, the function relating the odds in favor of the null to the upper limit on “some positive effect” dips slightly below 1 for some non-zero assumption about the upper limit, it will not go very far below 1, that is, the “some effect” hypothesis cannot attain high relative likelihood when there is in fact no effect or when the effect is weak and we have little data. Therefore, if we insist that we want some reasonable odds (say 10:1) in favor of “some” (positive) effect before we put the drug on the market, we will more often than not conclude that there is no effect or none worth considering. And that is what we should conclude. Not because it is necessarily true–nothing is certain but death and taxes–but because that is what is consistent with the data we have and the principle that a drug should not be marketed unless the data we have make us reasonably confident that it will do good.\n\nThis entry was posted in Biostatistics, Uncategorized. Bookmark the permalink."
] | [
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https://formulasearchengine.com/wiki/Representation_theory | [
"# Representation theory\n\n{{#invoke:Hatnote|hatnote}}\n\nRepresentation theory is a branch of mathematics that studies abstract algebraic structures by representing their elements as linear transformations of vector spaces, and studies modules over these abstract algebraic structures. In essence, a representation makes an abstract algebraic object more concrete by describing its elements by matrices and the algebraic operations in terms of matrix addition and matrix multiplication. The algebraic objects amenable to such a description include groups, associative algebras and Lie algebras. The most prominent of these (and historically the first) is the representation theory of groups, in which elements of a group are represented by invertible matrices in such a way that the group operation is matrix multiplication.\n\nRepresentation theory is a powerful tool because it reduces problems in abstract algebra to problems in linear algebra, a subject that is well understood. Furthermore, the vector space on which a group (for example) is represented can be infinite-dimensional, and by allowing it to be, for instance, a Hilbert space, methods of analysis can be applied to the theory of groups. Representation theory is also important in physics because, for example, it describes how the symmetry group of a physical system affects the solutions of equations describing that system.\n\nA striking feature of representation theory is its pervasiveness in mathematics. There are two sides to this. First, the applications of representation theory are diverse: in addition to its impact on algebra, representation theory:\n\nThe second aspect is the diversity of approaches to representation theory. The same objects can be studied using methods from algebraic geometry, module theory, analytic number theory, differential geometry, operator theory, algebraic combinatorics and topology.\n\nThe success of representation theory has led to numerous generalizations. One of the most general is in category theory. The algebraic objects to which representation theory applies can be viewed as particular kinds of categories, and the representations as functors from the object category to the category of vector spaces. This description points to two obvious generalizations: first, the algebraic objects can be replaced by more general categories; second, the target category of vector spaces can be replaced by other well-understood categories.\n\nA representation should not be confused with a presentation.\n\n## Definitions and concepts\n\nLet V be a vector space over a field F. For instance, suppose V is Rn or Cn, the standard n-dimensional space of column vectors over the real or complex numbers respectively. In this case, the idea of representation theory is to do abstract algebra concretely by using n × n matrices of real or complex numbers.\n\nThere are three main sorts of algebraic objects for which this can be done: groups, associative algebras and Lie algebras.\n\nThis generalizes to any field F and any vector space V over F, with linear maps replacing matrices and composition replacing matrix multiplication: there is a group GL(V,F) of automorphisms of V, an associative algebra EndF(V) of all endomorphisms of V, and a corresponding Lie algebra gl(V,F).\n\n### Definition\n\nThere are two ways to say what a representation is. The first uses the idea of an action, generalizing the way that matrices act on column vectors by matrix multiplication. A representation of a group G or (associative or Lie) algebra A on a vector space V is a map\n\n$\\Phi \\colon G\\times V\\to V\\quad {\\text{or}}\\quad \\Phi \\colon A\\times V\\to V$",
null,
"with two properties. First, for any g in G (or a in A), the map\n\n{\\begin{aligned}\\varphi (g)\\colon V&\\to V\\\\v&\\mapsto \\Phi (g,v)\\end{aligned}}",
null,
"is linear (over F). Second, if we introduce the notation g · v for Φ (g, v), then for any g1, g2 in G and v in V:\n\n$(1)\\quad e\\cdot v=v$",
null,
"$(2)\\quad g_{1}\\cdot (g_{2}\\cdot v)=(g_{1}g_{2})\\cdot v$",
null,
"where e is the identity element of G and g1g2 is product in G. The requirement for associative algebras is analogous, except that associative algebras do not always have an identity element, in which case equation (1) is ignored. Equation (2) is an abstract expression of the associativity of matrix multiplication. This doesn't hold for the matrix commutator and also there is no identity element for the commutator. Hence for Lie algebras, the only requirement is that for any x1, x2 in A and v in V:\n\n$(2')\\quad x_{1}\\cdot (x_{2}\\cdot v)-x_{2}\\cdot (x_{1}\\cdot v)=[x_{1},x_{2}]\\cdot v$",
null,
"where [x1, x2] is the Lie bracket, which generalizes the matrix commutator MNNM.\n\nThe second way to define a representation focuses on the map φ sending g in G to a linear map φ(g): VV, which satisfies\n\n$\\varphi (g_{1}g_{2})=\\varphi (g_{1})\\circ \\varphi (g_{2})\\quad {\\text{for all }}g_{1},g_{2}\\in G\\,\\!$",
null,
"and similarly in the other cases. This approach is both more concise and more abstract. From this point of view:\n\n### Terminology\n\nThe vector space V is called the representation space of φ and its dimension (if finite) is called the dimension of the representation (sometimes degree, as in ). It is also common practice to refer to V itself as the representation when the homomorphism φ is clear from the context; otherwise the notation (V,φ) can be used to denote a representation.\n\nWhen V is of finite dimension n, one can choose a basis for V to identify V with Fn and hence recover a matrix representation with entries in the field F.\n\nAn effective or faithful representation is a representation (V,φ) for which the homomorphism φ is injective.\n\n### Equivariant maps and isomorphisms\n\nIf V and W are vector spaces over F, equipped with representations φ and ψ of a group G, then an equivariant map from V to W is a linear map α: VW such that\n\n$\\alpha (g\\cdot v)=g\\cdot \\alpha (v)$",
null,
"for all g in G and v in V. In terms of φ: G → GL(V) and ψ: G → GL(W), this means\n\n$\\alpha \\circ \\phi (g)=\\psi (g)\\circ \\alpha$",
null,
"for all g in G.\n\nEquivariant maps for representations of an associative or Lie algebra are defined similarly. If α is invertible, then it is said to be an isomorphism, in which case V and W (or, more precisely, φ and ψ) are isomorphic representations.\n\nIsomorphic representations are, for all practical purposes, \"the same\": they provide the same information about the group or algebra being represented. Representation theory therefore seeks to classify representations \"up to isomorphism\".\n\n### Subrepresentations, quotients, and irreducible representations\n\nIf (W,ψ) is a representation of (say) a group G, and V is a linear subspace of W that is preserved by the action of G in the sense that g · vV for all vV (Serre calls these V stable under G), then V is called a subrepresentation: by defining φ(g) to be the restriction of ψ(g) to V, (V, φ) is a representation of G and the inclusion of V into W is an equivariant map. The quotient space W/V can also be made into a representation of G.\n\nIf W has exactly two subrepresentations, namely the trivial subspace {0} and W itself, then the representation is said to be irreducible; if W has a proper nontrivial subrepresentation, the representation is said to be reducible.\n\nThe definition of an irreducible representation implies Schur's lemma: an equivariant map α: VW between irreducible representations is either the zero map or an isomorphism, since its kernel and image are subrepresentations. In particular, when V = W, this shows that the equivariant endomorphisms of V form an associative division algebra over the underlying field F. If F is algebraically closed, the only equivariant endomorphisms of an irreducible representation are the scalar multiples of the identity.\n\nIrreducible representations are the building blocks of representation theory: if a representation W is not irreducible then it is built from a subrepresentation and a quotient that are both \"simpler\" in some sense; for instance, if W is finite-dimensional, then both the subrepresentation and the quotient have smaller dimension.\n\n### Direct sums and indecomposable representations\n\nIf (V,φ) and (W,ψ) are representations of (say) a group G, then the direct sum of V and W is a representation, in a canonical way, via the equation\n\n$g\\cdot (v,w)=(g\\cdot v,g\\cdot w).$",
null,
"The direct sum of two representations carries no more information about the group G than the two representations do individually. If a representation is the direct sum of two proper nontrivial subrepresentations, it is said to be decomposable. Otherwise, it is said to be indecomposable.\n\nIn favourable circumstances, every representation is a direct sum of irreducible representations: such representations are said to be semisimple. In this case, it suffices to understand only the irreducible representations. In other cases, one must understand how indecomposable representations can be built from irreducible representations as extensions of a quotient by a subrepresentation.\n\n## Branches and topics\n\nRepresentation theory is notable for the number of branches it has, and the diversity of the approaches to studying representations of groups and algebras. Although, all the theories have in common the basic concepts discussed already, they differ considerably in detail. The differences are at least 3-fold:\n\n1. Representation theory depends upon the type of algebraic object being represented. There are several different classes of groups, associative algebras and Lie algebras, and their representation theories all have an individual flavour.\n2. Representation theory depends upon the nature of the vector space on which the algebraic object is represented. The most important distinction is between finite-dimensional representations and infinite-dimensional ones. In the infinite-dimensional case, additional structures are important (e.g. whether or not the space is a Hilbert space, Banach space, etc.). Additional algebraic structures can also be imposed in the finite-dimensional case.\n3. Representation theory depends upon the type of field over which the vector space is defined. The most important case is the field of complex numbers. The other important cases are the field of real numbers, finite fields, and fields of p-adic numbers. Additional difficulties arise for fields of positive characteristic and for fields that are not algebraically closed.\n\n### Finite groups\n\n{{#invoke:main|main}}\n\nGroup representations are a very important tool in the study of finite groups. They also arise in the applications of finite group theory to geometry and crystallography. Representations of finite groups exhibit many of the features of the general theory and point the way to other branches and topics in representation theory.\n\nOver a field of characteristic zero, the representation theory of a finite group G has a number of convenient properties. First, the representations of G are semisimple (completely reducible). This is a consequence of Maschke's theorem, which states that any subrepresentation V of a G-representation W has a G-invariant complement. One proof is to choose any projection π from W to V and replace it by its average πG defined by\n\n$\\pi _{G}(x)={\\frac {1}{|G|}}\\sum _{g\\in G}g\\cdot \\pi (g^{-1}\\cdot x).$",
null,
"πG is equivariant, and its kernel is the required complement.\n\nThe finite-dimensional G-representations can be understood using character theory: the character of a representation φ: G → GL(V) is the class function χφ: GF defined by\n\n$\\chi _{\\varphi }(g)=\\mathrm {Tr} (\\varphi (g))\\,$",
null,
"where $\\mathrm {Tr}$",
null,
"is the trace. An irreducible representation of G is completely determined by its character.\n\nMaschke's theorem holds more generally for fields of positive characteristic p, such as the finite fields, as long as the prime p is coprime to the order of G. When p and |G| have a common factor, there are G-representations that are not semisimple, which are studied in a subbranch called modular representation theory.\n\nAveraging techniques also show that if F is the real or complex numbers, then any G-representation preserves an inner product $\\langle \\cdot ,\\cdot \\rangle$",
null,
"on V in the sense that\n\n$\\langle g\\cdot v,g\\cdot w\\rangle =\\langle v,w\\rangle$",
null,
"for all g in G and v, w in W. Hence any G-representation is unitary.\n\nUnitary representations are automatically semisimple, since Maschke's result can be proven by taking the orthogonal complement of a subrepresentation. When studying representations of groups that are not finite, the unitary representations provide a good generalization of the real and complex representations of a finite group.\n\nResults such as Maschke's theorem and the unitary property that rely on averaging can be generalized to more general groups by replacing the average with an integral, provided that a suitable notion of integral can be defined. This can be done for compact groups or locally compact groups, using Haar measure, and the resulting theory is known as abstract harmonic analysis.\n\nOver arbitrary fields, another class of finite groups that have a good representation theory are the finite groups of Lie type. Important examples are linear algebraic groups over finite fields. The representation theory of linear algebraic groups and Lie groups extends these examples to infinite-dimensional groups, the latter being intimately related to Lie algebra representations. The importance of character theory for finite groups has an analogue in the theory of weights for representations of Lie groups and Lie algebras.\n\nRepresentations of a finite group G are also linked directly to algebra representations via the group algebra F[G], which is a vector space over F with the elements of G as a basis, equipped with the multiplication operation defined by the group operation, linearity, and the requirement that the group operation and scalar multiplication commute.\n\n### Modular representations\n\n{{#invoke:main|main}}\n\nModular representations of a finite group G are representations over a field whose characteristic is not coprime to |G|, so that Maschke's theorem no longer holds (because |G| is not invertible in F and so one cannot divide by it). Nevertheless, Richard Brauer extended much of character theory to modular representations, and this theory played an important role in early progress towards the classification of finite simple groups, especially for simple groups whose characterization was not amenable to purely group-theoretic methods because their Sylow 2-subgroups were \"too small\".\n\nAs well as having applications to group theory, modular representations arise naturally in other branches of mathematics, such as algebraic geometry, coding theory, combinatorics and number theory.\n\n### Unitary representations\n\n{{#invoke:main|main}}\n\nA unitary representation of a group G is a linear representation φ of G on a real or (usually) complex Hilbert space V such that φ(g) is a unitary operator for every gG. Such representations have been widely applied in quantum mechanics since the 1920s, thanks in particular to the influence of Hermann Weyl, and this has inspired the development of the theory, most notably through the analysis of representations of the Poincaré group by Eugene Wigner. One of the pioneers in constructing a general theory of unitary representations (for any group G rather than just for particular groups useful in applications) was George Mackey, and an extensive theory was developed by Harish-Chandra and others in the 1950s and 1960s.\n\nA major goal is to describe the \"unitary dual\", the space of irreducible unitary representations of G. The theory is most well-developed in the case that G is a locally compact (Hausdorff) topological group and the representations are strongly continuous. For G abelian, the unitary dual is just the space of characters, while for G compact, the Peter–Weyl theorem shows that the irreducible unitary representations are finite-dimensional and the unitary dual is discrete. For example, if G is the circle group S1, then the characters are given by integers, and the unitary dual is Z.\n\nFor non-compact G, the question of which representations are unitary is a subtle one. Although irreducible unitary representations must be \"admissible\" (as Harish-Chandra modules) and it is easy to detect which admissible representations have a nondegenerate invariant sesquilinear form, it is hard to determine when this form is positive definite. An effective description of the unitary dual, even for relatively well-behaved groups such as real reductive Lie groups (discussed below), remains an important open problem in representation theory. It has been solved for many particular groups, such as SL(2,R) and the Lorentz group.\n\n### Harmonic analysis\n\n{{#invoke:main|main}}\n\nThe duality between the circle group S1 and the integers Z, or more generally, between a torus Tn and Zn is well known in analysis as the theory of Fourier series, and the Fourier transform similarly expresses the fact that the space of characters on a real vector space is the dual vector space. Thus unitary representation theory and harmonic analysis are intimately related, and abstract harmonic analysis exploits this relationship, by developing the analysis of functions on locally compact topological groups and related spaces.\n\nA major goal is to provide a general form of the Fourier transform and the Plancherel theorem. This is done by constructing a measure on the unitary dual and an isomorphism between the regular representation of G on the space L2(G) of square integrable functions on G and its representation on the space of L2 functions on the unitary dual. Pontrjagin duality and the Peter–Weyl theorem achieve this for abelian and compact G respectively.\n\nAnother approach involves considering all unitary representations, not just the irreducible ones. These form a category, and Tannaka–Krein duality provides a way to recover a compact group from its category of unitary representations.\n\nIf the group is neither abelian nor compact, no general theory is known with an analogue of the Plancherel theorem or Fourier inversion, although Alexander Grothendieck extended Tannaka–Krein duality to a relationship between linear algebraic groups and tannakian categories.\n\nHarmonic analysis has also been extended from the analysis of functions on a group G to functions on homogeneous spaces for G. The theory is particularly well developed for symmetric spaces and provides a theory of automorphic forms (discussed below).\n\n### Lie groups\n\n{{#invoke:main|main}} Template:Lie groups\n\nA Lie group is a group that is also a smooth manifold. Many classical groups of matrices over the real or complex numbers are Lie groups. Many of the groups important in physics and chemistry are Lie groups, and their representation theory is crucial to the application of group theory in those fields.\n\nThe representation theory of Lie groups can be developed first by considering the compact groups, to which results of compact representation theory apply. This theory can be extended to finite-dimensional representations of semisimple Lie groups using Weyl's unitary trick: each semisimple real Lie group G has a complexification, which is a complex Lie group Gc, and this complex Lie group has a maximal compact subgroup K. The finite-dimensional representations of G closely correspond to those of K.\n\nA general Lie group is a semidirect product of a solvable Lie group and a semisimple Lie group (the Levi decomposition). The classification of representations of solvable Lie groups is intractable in general, but often easy in practical cases. Representations of semidirect products can then be analysed by means of general results called Mackey theory, which is a generalization of the methods used in Wigner's classification of representations of the Poincaré group.\n\n### Lie algebras\n\n{{#invoke:main|main}}\n\nA Lie algebra over a field F is a vector space over F equipped with a skew-symmetric bilinear operation called the Lie bracket, which satisfies the Jacobi identity. Lie algebras arise in particular as tangent spaces to Lie groups at the identity element, leading to their interpretation as \"infinitesimal symmetries\". An important approach to the representation theory of Lie groups is to study the corresponding representation theory of Lie algebras, but representations of Lie algebras also have an intrinsic interest.\n\nLie algebras, like Lie groups, have a Levi decomposition into semisimple and solvable parts, with the representation theory of solvable Lie algebras being intractable in general. In contrast, the finite-dimensional representations of semisimple Lie algebras are completely understood, after work of Élie Cartan. A representation of a semisimple Lie algebra g is analysed by choosing a Cartan subalgebra, which is essentially a generic maximal subalgebra h of g on which the Lie bracket is zero (\"abelian\"). The representation of g can be decomposed into weight spaces that are eigenspaces for the action of h and the infinitesimal analogue of characters. The structure of semisimple Lie algebras then reduces the analysis of representations to easily understood combinatorics of the possible weights that can occur.\n\n#### Infinite-dimensional Lie algebras\n\nThere are many classes of infinite-dimensional Lie algebras whose representations have been studied. Among these, an important class are the Kac–Moody algebras. They are named after Victor Kac and Robert Moody, who independently discovered them. These algebras form a generalization of finite-dimensional semisimple Lie algebras, and share many of their combinatorial properties. This means that they have a class of representations that can be understood in the same way as representations of semisimple Lie algebras.\n\nAffine Lie algebras are a special case of Kac–Moody algebras, which have particular importance in mathematics and theoretical physics, especially conformal field theory and the theory of exactly solvable models. Kac discovered an elegant proof of certain combinatorial identities, Macdonald identities, which is based on the representation theory of affine Kac–Moody algebras.\n\n#### Lie superalgebras\n\n{{#invoke:main|main}}\n\nLie superalgebras are generalizations of Lie algebras in which the underlying vector space has a Z2-grading, and skew-symmetry and Jacobi identity properties of the Lie bracket are modified by signs. Their representation theory is similar to the representation theory of Lie algebras.\n\n### Linear algebraic groups\n\nLinear algebraic groups (or more generally, affine group schemes) are analogues in algebraic geometry of Lie groups, but over more general fields than just R or C. In particular, over finite fields, they give rise to finite groups of Lie type. Although linear algebraic groups have a classification that is very similar to that of Lie groups, their representation theory is rather different (and much less well understood) and requires different techniques, since the Zariski topology is relatively weak, and techniques from analysis are no longer available.\n\n### Invariant theory\n\n{{#invoke:main|main}}\n\nInvariant theory studies actions on algebraic varieties from the point of view of their effect on functions, which form representations of the group. Classically, the theory dealt with the question of explicit description of polynomial functions that do not change, or are invariant, under the transformations from a given linear group. The modern approach analyses the decomposition of these representations into irreducibles.\n\nInvariant theory of infinite groups is inextricably linked with the development of linear algebra, especially, the theories of quadratic forms and determinants. Another subject with strong mutual influence is projective geometry, where invariant theory can be used to organize the subject, and during the 1960s, new life was breathed into the subject by David Mumford in the form of his geometric invariant theory.\n\nThe representation theory of semisimple Lie groups has its roots in invariant theory and the strong links between representation theory and algebraic geometry have many parallels in differential geometry, beginning with Felix Klein's Erlangen program and Élie Cartan's connections, which place groups and symmetry at the heart of geometry. Modern developments link representation theory and invariant theory to areas as diverse as holonomy, differential operators and the theory of several complex variables.\n\n### Automorphic forms and number theory\n\n{{#invoke:main|main}}\n\nAutomorphic forms are a generalization of modular forms to more general analytic functions, perhaps of several complex variables, with similar transformation properties. The generalization involves replacing the modular group PSL2 (R) and a chosen congruence subgroup by a semisimple Lie group G and a discrete subgroup Γ. Just as modular forms can be viewed as differential forms on a quotient of the upper half space H = PSL2 (R)/SO(2), automorphic forms can be viewed as differential forms (or similar objects) on Γ\\G/K, where K is (typically) a maximal compact subgroup of G. Some care is required, however, as the quotient typically has singularities. The quotient of a semisimple Lie group by a compact subgroup is a symmetric space and so the theory of automorphic forms is intimately related to harmonic analysis on symmetric spaces.\n\nBefore the development of the general theory, many important special cases were worked out in detail, including the Hilbert modular forms and Siegel modular forms. Important results in the theory include the Selberg trace formula and the realization by Robert Langlands that the Riemann-Roch theorem could be applied to calculate the dimension of the space of automorphic forms. The subsequent notion of \"automorphic representation\" has proved of great technical value for dealing with the case that G is an algebraic group, treated as an adelic algebraic group. As a result an entire philosophy, the Langlands program has developed around the relation between representation and number theoretic properties of automorphic forms.\n\n### Associative algebras\n\n{{#invoke:main|main}}\n\nIn one sense, associative algebra representations generalize both representations of groups and Lie algebras. A representation of a group induces a representation of a corresponding group ring or group algebra, while representations of a Lie algebra correspond bijectively to representations of its universal enveloping algebra. However, the representation theory of general associative algebras does not have all of the nice properties of the representation theory of groups and Lie algebras.\n\n#### Module theory\n\n{{#invoke:main|main}}\n\nWhen considering representations of an associative algebra, one can forget the underlying field, and simply regard the associative algebra as a ring, and its representations as modules. This approach is surprisingly fruitful: many results in representation theory can be interpreted as special cases of results about modules over a ring.\n\n#### Hopf algebras and quantum groups\n\n{{#invoke:main|main}}\n\nHopf algebras provide a way to improve the representation theory of associative algebras, while retaining the representation theory of groups and Lie algebras as special cases. In particular, the tensor product of two representations is a representation, as is the dual vector space.\n\nThe Hopf algebras associated to groups have a commutative algebra structure, and so general Hopf algebras are known as quantum groups, although this term is often restricted to certain Hopf algebras arising as deformations of groups or their universal enveloping algebras. The representation theory of quantum groups has added surprising insights to the representation theory of Lie groups and Lie algebras, for instance through the crystal basis of Kashiwara.\n\n## Generalizations\n\n### Set-theoretic representations\n\n{{#invoke:main|main}}\n\nA set-theoretic representation (also known as a group action or permutation representation) of a group G on a set X is given by a function ρ from G to XX, the set of functions from X to X, such that for all g1, g2 in G and all x in X:\n\n$\\rho (1)[x]=x$",
null,
"$\\rho (g_{1}g_{2})[x]=\\rho (g_{1})[\\rho (g_{2})[x]].$",
null,
"This condition and the axioms for a group imply that ρ(g) is a bijection (or permutation) for all g in G. Thus we may equivalently define a permutation representation to be a group homomorphism from G to the symmetric group SX of X.\n\n### Representations in other categories\n\nEvery group G can be viewed as a category with a single object; morphisms in this category are just the elements of G. Given an arbitrary category C, a representation of G in C is a functor from G to C. Such a functor selects an object X in C and a group homomorphism from G to Aut(X), the automorphism group of X.\n\nIn the case where C is VectF, the category of vector spaces over a field F, this definition is equivalent to a linear representation. Likewise, a set-theoretic representation is just a representation of G in the category of sets.\n\nFor another example consider the category of topological spaces, Top. Representations in Top are homomorphisms from G to the homeomorphism group of a topological space X.\n\nTwo types of representations closely related to linear representations are:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87891185,"math_prob":0.9468753,"size":36031,"snap":"2019-26-2019-30","text_gpt3_token_len":7982,"char_repetition_ratio":0.23218697,"word_repetition_ratio":0.06001864,"special_character_ratio":0.20132664,"punctuation_ratio":0.113803,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9902451,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,10,null,4,null,10,null,10,null,10,null,2,null,10,null,4,null,10,null,10,null,4,null,10,null,null,null,10,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T05:44:38Z\",\"WARC-Record-ID\":\"<urn:uuid:7ab2c373-9568-484e-9699-3e576d2bbfdd>\",\"Content-Length\":\"134122\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:de09d53e-b87b-42b8-b408-e8454fb2988d>\",\"WARC-Concurrent-To\":\"<urn:uuid:59aaa019-5f1f-4109-8465-e89f56b2e63c>\",\"WARC-IP-Address\":\"132.195.228.228\",\"WARC-Target-URI\":\"https://formulasearchengine.com/wiki/Representation_theory\",\"WARC-Payload-Digest\":\"sha1:KMRK5B6VV46NYGSFXXCBQ77CYZTLUTOJ\",\"WARC-Block-Digest\":\"sha1:4GD7EM5HOCLZUYFRFD7GIBUTRODRQUBX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527531.84_warc_CC-MAIN-20190722051628-20190722073628-00036.warc.gz\"}"} |
https://www.dataunitconverter.com/exabyte-to-petabit | [
"# EB to Pbit Calculator - Convert Exabytes to Petabits",
null,
"## Conversion History (Last 6)\n\nInput Exabyte - and press Enter\nEB\n\n## EB to Pbit - Conversion Formula and Steps\n\nExabyte and Petabit are units of digital information used to measure storage capacity and data transfer rate. Both are decimal units. One Exabyte is equal to 1000^6 bytes. One Petabit is equal to 1000^5 bits. There are 0.000125 Exabytes in one Petabit. - view the difference between both units",
null,
"Source Data UnitTarget Data Unit\nExabyte (EB)\nEqual to 1000^6 bytes\n(Decimal Unit)\nPetabit (Pbit)\nEqual to 1000^5 bits\n(Decimal Unit)\n\nThe formula of converting the Exabyte to Petabit is represented as follows :\n\nPbit = EB x (8x1000)\n\nBelow conversion diagram will help you to visualize the Exabyte to Petabit calculation steps in a simplified manner.\n\n÷ 8\n÷ 1000\nPetabit [Pbit]\nPetabyte [PB]\nExabyte [EB]\nx 8\nx 1000\n\nNow let us apply the above formula and, write down the steps to convert from Exabyte (EB) to Petabit (Pbit).\n\n1. STEP 1 → Petabit = Exabyte x (8x1000)\n2. STEP 2 → Petabit = Exabyte x 8000\n\nExample : If we apply the above steps, conversion from 10 EB to Pbit, will be processed as below.\n\n1. = 10 x (8x1000)\n2. = 10 x 8000\n3. = 80000\n4. i.e. 10 EB is equal to 80,000 Pbit.\n\n(Result rounded off to 40 decimal positions.)\n\nYou can use above formula and steps to convert Exabyte to Petabit using any of the programming language such as Java, Python or Powershell.\n\n#### Definition : Exabyte\n\nAn Exabyte (EB) is a unit of measurement for digital information storage. It is equal to 1,000,000,000,000,000,000 (one quintillion) bytes, It is commonly used to measure the storage capacity of large data centers, computer hard drives, flash drives, and other digital storage devices.\n\n#### Definition : Petabit\n\nA Petabit (Pb or Pbit) is a unit of measurement for digital information transfer rate. It is equal to 1,000,000,000,000,000 (one quadrillion) bits. It is commonly used to measure the speed of data transfer over computer networks, such as internet connection speeds.\n\n### Excel Formula to convert from EB to Pbit\n\nApply the formula as shown below to convert from Exabyte to Petabit.\n\nABC\n1Exabyte (EB)Petabit (Pbit)\n21=A2 * 8000\n3"
] | [
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"https://www.dataunitconverter.com/images/certificate_precise.png",
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"https://www.dataunitconverter.com/showimage.php/Exabyte_to_Petabit_Dataunitconverter.png",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.79457986,"math_prob":0.99199146,"size":2646,"snap":"2023-14-2023-23","text_gpt3_token_len":703,"char_repetition_ratio":0.16010597,"word_repetition_ratio":0.03524229,"special_character_ratio":0.27135298,"punctuation_ratio":0.11045365,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9975213,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-21T11:30:08Z\",\"WARC-Record-ID\":\"<urn:uuid:27b0fc25-6d4a-490d-85c7-d904efb937e1>\",\"Content-Length\":\"65824\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54efcf49-0784-4fec-b7ae-c85972c1ce88>\",\"WARC-Concurrent-To\":\"<urn:uuid:e34c5ee5-7653-40e5-8314-73d2b87141d1>\",\"WARC-IP-Address\":\"64.227.22.174\",\"WARC-Target-URI\":\"https://www.dataunitconverter.com/exabyte-to-petabit\",\"WARC-Payload-Digest\":\"sha1:U6ROQG3KVFNJN6NQTY7GHW25J2ICR7TT\",\"WARC-Block-Digest\":\"sha1:2PR3LR7TFCDKU2ENKEWCQGXDUBMWR5FF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943695.23_warc_CC-MAIN-20230321095704-20230321125704-00513.warc.gz\"}"} |
https://www.geeksforgeeks.org/atomiclong-doublevalue-method-in-java-with-examples/?ref=rp | [
"Related Articles\nAtomicLong doubleValue() method in Java with examples\n• Last Updated : 29 Jan, 2019\n\nThe Java.util.concurrent.atomic.AtomicLong.doubleValue() is an inbuilt method in java which returns the current value of the AtomicLong as a Double data-type after performing primitive conversion.\n\nSyntax:\n\n```public double doubleValue()\n```\n\nParameters: The function does not accepts any parameter.\n\nReturn value: The function returns the numeric value represented by this object after conversion to type double.\n\nBelow programs illustrate the above method:\n\nProgram 1:\n\n `// Java program that demonstrates``// the doubleValue() function`` ` `import` `java.util.concurrent.atomic.AtomicLong;`` ` `public` `class` `GFG {`` ``public` `static` `void` `main(String args[])`` ``{`` ` ` ``// Initially value as 0`` ``AtomicLong val = ``new` `AtomicLong(``0``);`` ` ` ``val.addAndGet(``7``);`` ` ` ``// Prints the updated value`` ``System.out.println(``\"Previous value: \"`` ``+ val);`` ` ` ``// Gets the double value`` ``double` `res = val.doubleValue();`` ` ` ``System.out.println(``\"Double value: \"`` ``+ res);`` ``}``}`\nOutput:\n```Previous value: 7\nDouble value: 7.0\n```\n\nProgram 2:\n\n `// Java program that demonstrates``// the doubleValue() function`` ` `import` `java.util.concurrent.atomic.AtomicLong;`` ` `public` `class` `GFG {`` ``public` `static` `void` `main(String args[])`` ``{`` ` ` ``// Initially value as 18`` ``AtomicLong val = ``new` `AtomicLong(``18``);`` ` ` ``val.addAndGet(``7``);`` ` ` ``// Gets the double value`` ``System.out.println(``\"Previous value: \"`` ``+ val);`` ` ` ``// Decreases the value by 1`` ``double` `res = val.doubleValue();`` ` ` ``System.out.println(``\"Double value: \"`` ``+ res);`` ``}``}`\nOutput:\n```Previous value: 25\nDouble value: 25.0\n```\n\nAttention reader! Don’t stop learning now. Get hold of all the important Java Foundation and Collections concepts with the Fundamentals of Java and Java Collections 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\nMy Personal Notes arrow_drop_up"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.65971446,"math_prob":0.6250629,"size":1881,"snap":"2021-04-2021-17","text_gpt3_token_len":430,"char_repetition_ratio":0.14864145,"word_repetition_ratio":0.296875,"special_character_ratio":0.23444976,"punctuation_ratio":0.19526628,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97416836,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T02:19:19Z\",\"WARC-Record-ID\":\"<urn:uuid:035dfb11-0add-4b35-ab91-44f1939fafdd>\",\"Content-Length\":\"90769\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e55a8c9-529e-48f5-b96f-a35fc1dc6c9d>\",\"WARC-Concurrent-To\":\"<urn:uuid:091e8b8a-d4ed-4cf9-9348-f0e72f09981a>\",\"WARC-IP-Address\":\"23.12.145.47\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/atomiclong-doublevalue-method-in-java-with-examples/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:LGTFLH63756CXKSBLCBBDAYLD2D4U625\",\"WARC-Block-Digest\":\"sha1:AKPGHJKCXOHXHLKA7ZFVTANVZFFIPYMC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038060603.10_warc_CC-MAIN-20210411000036-20210411030036-00001.warc.gz\"}"} |
https://codegolf.stackexchange.com/questions/6443/pythagorean-triplets | [
"# Pythagorean Triplets\n\nThe Pythagorean Theorem states that for a right triangle with sides a, b, and c, a2+b2=c2.\n\nA Pythagorean triplet is a set of three numbers, where a2+b2=c2, a To extend this even further, a primitive Pythagorean triplet is a Pythagorean triplet where gcd(a,b,c)=1.\n\nThe goal is to find 100 primitive Pythagorean triplets.\n\nInput: No input.\n\nOutput: 100 primitive Pythagorean triplets, in whatever order and shape, as long as the output contains only those triplets.\n\nShortest code wins. Good luck!\n\n• Is a < b or are (a,b,c) and (b,c,a) 2 solutions. – user unknown Jun 22 '12 at 23:44\n• @user unknown: a<b<c – beary605 Jun 23 '12 at 0:49\n• They're called \"triples\", not \"triplets\". – mbomb007 Oct 28 '16 at 15:04\n\n## Python, 42\n\ni=4\nwhile i<203:print 2*i,i*i-1,i*i+1;i+=2\n\n• i**2 -> i*i to save a couple of characters. – breadbox Jun 23 '12 at 3:17\n• I should mark this as cheating, but oh well. :) – beary605 Jun 23 '12 at 16:00\n• How is it cheating? – grc Jun 24 '12 at 1:09\n• \"hoping for brute force\" -- there's something you don't hear every day. – breadbox Jun 25 '12 at 16:14\n• @beary605 why have you accepted the answer that isn't the shortest? – Griffin Jun 27 '12 at 15:35\n\n# Javascript, 44 characters\n\nUpdate:\n\nfor(i=s=0;i^400;)s+=[i+=4,r=i*i/4-1,r+2]+' '\n\n\nI should have looked up an algorithm first. Well, can't beat python anyway.\n\nRun in a console to see the result.\n\nOld, slow, brute-force variant (140 characters)\n\nfor(q=a=[];563>++a;)for(b=a;--b;)(k=Math.sqrt(a*a+b*b))==~~k==function c(e,d,f){return f?c(c(f,e),d):d?c(d,e%d):e}(a,b,k)&&q.push([b,a,k]);q\n\n\n## J, 646038 49 characters\n\n100$}.~.(/:~*(0<*/*1=+./))\"1[4$.$.(=/+/~)2^~i.699 Should've realised that such a big gain was probably flawed. Still uses the brute-force method, but done a bit more efficiently. Explanation: x=.2^~i.699 generates a list of ints from 0 to 699 and squares 2^~ them (~ here reverses the order of the arguments). (=+/~) is a hook that generates an addition table and compares the result to the list. This gives me a three dimensional array with items which are either 1 or 0. A 1 means that a2+b2=c2. $. converts to a sparse array. For my smaller (9) example I get:\n\n $.(=/+/~)2^~i.9 0 0 0 | 1 1 0 1 | 1 1 1 0 | 1 2 0 2 | 1 2 2 0 | 1 3 0 3 | 1 3 3 0 | 1 4 0 4 | 1 4 4 0 | 1 5 0 5 | 1 5 3 4 | 1 5 4 3 | 1 5 5 0 | 1 6 0 6 | 1 6 6 0 | 1 7 0 7 | 1 7 7 0 | 1 8 0 8 | 1 8 8 0 | 1 4$. just gives me the left part of this list.\n\n(/:~*(0<*/*1=+./))\"1[ decides which rows meet the criteria. (verb)\"1 tells the verb to act on the individual rows rather than on the list. [ just separates the 1 and 4, otherwise J will think that 1 4 is a list of 2 numbers. '1=+./' gets the GCD of the three numbers and checks if it's 1. */ multiplies each triple together (getting 0 if it contains a 0) and these two are multiplied together. 0< turns the result into a boolean. * multiplies this result by each triple which eliminates all triples which contain a 0 or have a GCD which is not 1. /:~ sorts each triple.\n\n~. selects the unique items from the list.\n\n}. removes the first item (0 0 0) from the list.\n\n100\\$ takes the first 100 items from the list.\n\nThe above is my answer to this question, but as a matter of interest I implemented grc's method, (27 characters):\n\n2(*,.<:@^~,.>:@^~)2*2+i.100\n\n\n### Scala 180:\n\ndef g(a:Int,b:Int):Int=if(a==b)a else\nif(a>b)g(a-b,b)else\ng(b,a)\nfor(c<-(5 to 629);\na<-(1 to c-2);\nb<-(a to c-1);\nif((a*a+b*b==c*c)&&g(c,g(a,b))==1)&&a<400)println(a+\" \"+b+\" \"+c)\n\n\nUngolfed:\n\ndef gcd (a: Int, b: Int) : Int = if (a == b)\na else\nif (a > b) gcd (a-b, b) else\ngcd (b, a)\n\ndef gcd (a: Int, b: Int, c:Int): Int = gcd (c, gcd (a, b))\n\ndef pythagorean (a: Int, b: Int, c: Int) = (a * a + b * b == c * c)\n\nfor (c <- (5 to 629);\na <- (1 to c-2);\nb <- (a to c-1);\nif (pythagorean (a, b, c) && gcd (a, b, c) == 1)) yield {\nprintln (a + \"² + \" + b + \"² = \" + c + \"² => \" + (a*a) + \" + \" + (b*b) + \" = \" + (c*c))\nc*c\n}\n\n\n## APL (31)\n\n↑{(2×⍵),(A-1),1+A←⍵×⍵}¨2+2×⍳100\n\n\nIt uses basically the same approach as the Python program.\n\n# MATLAB 29\n\ni=2:2:200;[2*i;i.*i-1;i.*i+1]\n\n\nMathematica 33\n\n{2#,#^2-1,#^2+1}&/@Range[4,203,2]\n\n\n# APL(NARS), 26 chars, 52 bytes\n\n{k(4×⍵),2+k←¯1+×⍨2×⍵}¨⍳100\n\n\ntest:\n\n a←{k(4×⍵),2+k←¯1+×⍨2×⍵}¨⍳100\n≢a\n100\n+/{(x y z)←⍵⋄(z*2)=(y*2)+x*2}¨a\n100\n+/{∨/⍵}¨a\n100\n10↑a\n3 4 5 15 8 17 35 12 37 63 16 65 99 20 101 143 24 145 195 28 197 255 32 257 323 36 325 399 40 401\n\n\n## R, 49 characters\n\ntranslated from the Python solution:\n\ni=4;while(i<203){print(c(2*i,i*i-1,i*i+1));i=i+2}\n\n• You should take advantage of the functional nature of R i=seq(2,200,2);c(2*i,i*i-1,i*i+1) (though the output is a bit messy). – Griffin Jun 26 '12 at 13:55\n• Ops! I posted without taking a look at all the other answers! :-P – Paolo Jun 26 '12 at 14:01"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.80684614,"math_prob":0.9876371,"size":1695,"snap":"2020-45-2020-50","text_gpt3_token_len":641,"char_repetition_ratio":0.11709048,"word_repetition_ratio":0.0,"special_character_ratio":0.4141593,"punctuation_ratio":0.12926829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9952671,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T09:58:34Z\",\"WARC-Record-ID\":\"<urn:uuid:58bf43f4-c9c0-4b53-9e0a-11928acd9ef7>\",\"Content-Length\":\"231046\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:961a59b4-39af-4c7a-851a-c54d2d6f3cf3>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4d27e52-be9d-4105-b96b-284b4e3b3bf8>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://codegolf.stackexchange.com/questions/6443/pythagorean-triplets\",\"WARC-Payload-Digest\":\"sha1:CY7JNFL3GHJ7HC5VZS43YTU2GE5NUULK\",\"WARC-Block-Digest\":\"sha1:KNLLRREGP52OCHJKN4T5FZXGPDZVP3FQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141187753.32_warc_CC-MAIN-20201126084625-20201126114625-00622.warc.gz\"}"} |
https://scikit-image.org/docs/dev/auto_examples/transform/plot_register_rotation.html | [
"# Polar and Log-Polar Transformations¶\n\nRotation differences between two images can be converted to translation differences along the angular coordinate ($$\\theta$$) axis of the polar-transformed images. Scaling differences can be converted to translation differences along the radial coordinate ($$\\rho$$) axis if it is first log transformed (i.e., $$\\rho = \\ln\\sqrt{x^2 + y^2}$$). Thus, in this example, we use phase correlation (feature.register_translation) to recover rotation and scaling differences between two images that share a center point.\n\n## Recover rotation difference with a polar transform¶\n\nimport numpy as np\nimport matplotlib.pyplot as plt\n\nfrom skimage import data\nfrom skimage.feature import register_translation\nfrom skimage.transform import warp_polar, rotate\nfrom skimage.util import img_as_float\n\nangle = 35\nimage = data.retina()\nimage = img_as_float(image)\nrotated = rotate(image, angle)\n\nfig, axes = plt.subplots(2, 2, figsize=(8, 8))\nax = axes.ravel()\nax.set_title(\"Original\")\nax.imshow(image)\nax.set_title(\"Rotated\")\nax.imshow(rotated)\nax.set_title(\"Polar-Transformed Original\")\nax.imshow(image_polar)\nax.set_title(\"Polar-Transformed Rotated\")\nax.imshow(rotated_polar)\nplt.show()\n\nshifts, error, phasediff = register_translation(image_polar, rotated_polar)\nprint(\"Expected value for counterclockwise rotation in degrees: \"\nf\"{angle}\")\nprint(\"Recovered value for counterclockwise rotation: \"\nf\"{shifts}\")",
null,
"Out:\n\nExpected value for counterclockwise rotation in degrees: 35\nRecovered value for counterclockwise rotation: 35.0\n\n\n## Recover rotation and scaling differences with log-polar transform¶\n\nfrom skimage.transform import rescale\n\n# radius must be large enough to capture useful info in larger image\nangle = 53.7\nscale = 2.2\nimage = data.retina()\nimage = img_as_float(image)\nrotated = rotate(image, angle)\nrescaled = rescale(rotated, scale, multichannel=True)\nscaling='log', multichannel=True)\nscaling='log', multichannel=True)\n\nfig, axes = plt.subplots(2, 2, figsize=(8, 8))\nax = axes.ravel()\nax.set_title(\"Original\")\nax.imshow(image)\nax.set_title(\"Rotated and Rescaled\")\nax.imshow(rescaled)\nax.set_title(\"Log-Polar-Transformed Original\")\nax.imshow(image_polar)\nax.set_title(\"Log-Polar-Transformed Rotated and Rescaled\")\nax.imshow(rescaled_polar)\nplt.show()\n\n# setting upsample_factor can increase precision\ntparams = register_translation(image_polar, rescaled_polar, upsample_factor=20)\nshifts, error, phasediff = tparams\nshiftr, shiftc = shifts[:2]\n\n# Calculate scale factor from translation\nshift_scale = 1 / (np.exp(shiftc / klog))\n\nprint(f\"Expected value for cc rotation in degrees: {angle}\")\nprint(f\"Recovered value for cc rotation: {shiftr}\")\nprint()\nprint(f\"Expected value for scaling difference: {scale}\")\nprint(f\"Recovered value for scaling difference: {shift_scale}\")",
null,
"Out:\n\nExpected value for cc rotation in degrees: 53.7\nRecovered value for cc rotation: 53.75\n\nExpected value for scaling difference: 2.2\nRecovered value for scaling difference: 2.1981889915232165\n\n\nTotal running time of the script: ( 0 minutes 5.883 seconds)\n\nGallery generated by Sphinx-Gallery"
] | [
null,
"https://scikit-image.org/docs/dev/_images/sphx_glr_plot_register_rotation_001.png",
null,
"https://scikit-image.org/docs/dev/_images/sphx_glr_plot_register_rotation_002.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5656977,"math_prob":0.98788387,"size":3441,"snap":"2020-10-2020-16","text_gpt3_token_len":910,"char_repetition_ratio":0.14925808,"word_repetition_ratio":0.11948052,"special_character_ratio":0.2531241,"punctuation_ratio":0.16521738,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99592394,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-29T21:58:10Z\",\"WARC-Record-ID\":\"<urn:uuid:7fd71bac-2365-42fa-b91e-8c0f1d738eb5>\",\"Content-Length\":\"39659\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:699b60b7-3b79-4ce3-8c0e-1af743b44882>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4419a6b-1443-4d03-b224-1fdd7f25373d>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://scikit-image.org/docs/dev/auto_examples/transform/plot_register_rotation.html\",\"WARC-Payload-Digest\":\"sha1:DGVRXJIIBMXETJ62NAN5M6OJAI3WT3ZN\",\"WARC-Block-Digest\":\"sha1:BMZH5KQTWQU5GSOC6MPMSH3WL3LMGB6R\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370496227.25_warc_CC-MAIN-20200329201741-20200329231741-00500.warc.gz\"}"} |
https://www.esaral.com/q/mark-the-tick-against-the-correct-answer-in-the-following-76534 | [
"",
null,
"# Mark the tick against the correct answer in the following:\n\nQuestion:\n\nMark the tick against the correct answer in the following:\n\nRange of $\\operatorname{coses}^{-1} \\times$ is\n\nA. $\\left(\\frac{-\\pi}{2}, \\frac{\\pi}{2}\\right)$\n\nB. $\\left[\\frac{-\\pi}{2}, \\frac{\\pi}{2}\\right]$\n\nC. $\\left[\\frac{-\\pi}{2}, \\frac{\\pi}{2}\\right]-\\{0\\}$\n\nD. None of these\n\nSolution:\n\nTo Find: The range of $\\operatorname{cosec}^{-1}(x)$\n\nHere,the inverse function is given by $y=\\mathrm{f}^{-1}(x)$\n\nThe graph of the function $y=\\operatorname{cosec}^{-1}(x)$ can be obtained from the graph of\n\n$Y=\\operatorname{cosec} x$ by interchanging $x$ and $y$ axes.i.e, if $(a, b)$ is a point on $Y=\\operatorname{cosec} x$ then $(b, a)$ is the point on the function $y=\\operatorname{cosec}^{-1}(x)$\n\nBelow is the Graph of the range of $\\operatorname{cosec}^{-1}(x)$",
null,
"From the graph it is clear that the range of $\\operatorname{cosec}^{-1}(x)$ is restricted to interval\n\n$\\left[-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right]-\\{0\\}$"
] | [
null,
"https://www.facebook.com/tr",
null,
"https://www.esaral.com/media/uploads/2022/02/25/image68263.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.631105,"math_prob":0.99998677,"size":893,"snap":"2023-14-2023-23","text_gpt3_token_len":323,"char_repetition_ratio":0.1991001,"word_repetition_ratio":0.0,"special_character_ratio":0.36730123,"punctuation_ratio":0.08938547,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000075,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T19:51:05Z\",\"WARC-Record-ID\":\"<urn:uuid:1bd02ced-9b7b-4c7f-9f93-54f149171662>\",\"Content-Length\":\"25675\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76a06f34-de9d-4710-842c-4912905e9ecb>\",\"WARC-Concurrent-To\":\"<urn:uuid:62527b49-430c-4ef5-8950-3981947a765c>\",\"WARC-IP-Address\":\"172.67.213.11\",\"WARC-Target-URI\":\"https://www.esaral.com/q/mark-the-tick-against-the-correct-answer-in-the-following-76534\",\"WARC-Payload-Digest\":\"sha1:LLRKXUXW2S3J2QWYLEDX2UAFZSBCIIUH\",\"WARC-Block-Digest\":\"sha1:VNXHNUVAJU5UR4O6QH2C2SB2YXYJBHGO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653071.58_warc_CC-MAIN-20230606182640-20230606212640-00478.warc.gz\"}"} |
https://www.scec.org/publication/326 | [
"## Discrete Scale Invariance, Complex Fractal Dimensions and log-periodic fluctuations in Seismicity\n\nHubert Saleur, Charles G. Sammis, & Didier Sornette\n\nPublished August 10, 1996, SCEC Contribution #326\n\nWe discuss in detail the concept of discrete scale invariance and show how it leads to complex critical exponents and hence to the log-periodic corrections to scaling exhibited by various measures of seismic activity close to a large earthquake singularity. Discrete scale invariance is first illustrated on a geometrical fractal, the Sierpinsky gasket, which is shown to be fully described by a complex fractal dimension whose imaginary part is a simple function (inverse of the logarithm) of the discrete scaling factor. Then, a set of simple physical systems (spins and percolation) on hierarchical lattices is analyzed to exemplify the origin of the different terms in the discrete renormalization group formalism introduced to tackle this problem. As a more specific example of rupture relevant for earthquakes, we propose a solution of the hierarchical time-dependent fiber bundle of Newman et al. which exhibits explicitly a discrete renormalization group from which log-periodic corrections follow. We end by pointing out that discrete scale invariance does not necessarily require an underlying geometrical hierarchical structure. A hierarchy may appear “spontaneously” from the physics and/or the dynamics in a Euclidean (nonhierarchical) heterogeneous system. We briefly discuss a simple dynamical model of such mechanism, in terms of a random walk (or diffusion) of the seismic energy in a random heterogeneous system.\n\nCitation\nSaleur, H., Sammis, C. G., & Sornette, D. (1996). Discrete Scale Invariance, Complex Fractal Dimensions and log-periodic fluctuations in Seismicity. Journal of Geophysical Research, 101(B8), 17661-17677. doi: 10.1029/96JB00876."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82944095,"math_prob":0.8458644,"size":1922,"snap":"2020-45-2020-50","text_gpt3_token_len":428,"char_repetition_ratio":0.10583942,"word_repetition_ratio":0.043636363,"special_character_ratio":0.20343392,"punctuation_ratio":0.11598746,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96417236,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T22:15:41Z\",\"WARC-Record-ID\":\"<urn:uuid:70fbc9b1-1986-472e-9119-f95b8fcaf509>\",\"Content-Length\":\"19727\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:430a1ad1-1210-4927-b169-d5a7bc84d2a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:8e34f046-2f79-4ca9-94da-f03943d865e3>\",\"WARC-IP-Address\":\"52.36.134.75\",\"WARC-Target-URI\":\"https://www.scec.org/publication/326\",\"WARC-Payload-Digest\":\"sha1:KQM7LUYTXVNOSG7DV2I676GSHI3KXKTU\",\"WARC-Block-Digest\":\"sha1:E7TQ4MG3DU7RPV4PCKAFYVJ7SYUTJDOP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107878633.8_warc_CC-MAIN-20201021205955-20201021235955-00448.warc.gz\"}"} |
https://admin.clutchprep.com/organic-chemistry/practice-problems/51125/a-mixture-of-a-pair-of-enantiomers-has-30-ee-the-observed-rotation-of-this-mixtu | [
"# Problem: A mixture of a pair of enantiomers has 30% ee. The observed rotation of this mixture is +15º, and it is known by experiment that the (‐)‐enantiomer has the (R) configuration. (enantiomer excess (ee) = % one enantiomer – % the other enantiomer)(a) Calculate the percentage of (R) and (S) enantiomers of the natural product.(b) What is the optical rotation of a mixture of 20% (S) and 80% (R).\n\n###### Problem Details\n\nA mixture of a pair of enantiomers has 30% ee. The observed rotation of this mixture is +15º, and it is known by experiment that the (‐)‐enantiomer has the (R) configuration. (enantiomer excess (ee) = % one enantiomer – % the other enantiomer)\n\n(a) Calculate the percentage of (R) and (S) enantiomers of the natural product.\n\n(b) What is the optical rotation of a mixture of 20% (S) and 80% (R)."
] | [
null
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https://optovr.com/free-basic-pre-algebra-worksheets/ | [
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"https://optovr.com/images/free-printable-math-worksheets-kindergarten-grade-trigonometry-module-summer-multiplication-facts-book-algebra-fun-basic-pre.jpg",
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"https://optovr.com/images/helpful-tips-math-algebra-worksheets-untitled-design-games-level-2-grade-micro-lesson-free-basic-pre.jpg",
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"https://optovr.com/images/kindergarten-drawing-worksheets-free-math-tracing-decimal-games-play-easy-algebra-problems-answers-problem-solver-integrated-st-homework-school-readiness-assessment-printable-basic-pre.jpg",
null,
"https://optovr.com/images/pin-human-body-worksheets-senses-algebra-grade-math-standards-free-timetable-year-simple-logic-puzzles-easy-fraction-decimal-basic-pre.jpg",
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"https://optovr.com/images/free-time-worksheets-kindergarten-lemony-algebra-denominator-math-spring-multiple-choice-maker-array-fluency-games-basic-pre.jpg",
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https://www.clipsal.com/faq/fa174756 | [
"",
null,
"## Can a 60 Hz rated Transformer be energized with 50 Hz source?\n\nFA174756\n\n11 April 2022\n\nIssue:\nTransformers rated for 60 Hz ONLY are sometimes chosen to be installed in a location that has a 50 Hz Distribution system.\n\nProduct Line:\nLV Transformers\n\nEnvironment:\nApplies to MOST Transformers made by SquareD/Schneider Electric. Exceptions are 9070 series Control Transformers and Export Model Transformers which are dual rated for 50Hz and 60Hz.\n\nCause:\nDomestic (USA) power systems are rated 60 Hz. Many other power systems in the world are rated at 50 Hz.\n\nResolution:\nA 60 Hz rated transformer can only be used on a 60 Hz system (with certain exceptions -- read further.) Whereas a 50 or 50/60 Hz rated transformer can be used on either a 50 or a 60 Hz rated system.\nException: If the energizing voltage does not exceed 83% of the transformers nominal voltage, the transformer can be used. Remember that transformers are simple ratio devices and the output will also be lower. Another important factor to remember is that the transformer`s capacity will also have to be derated at the same ratio of the applied voltage divided by the nominal voltage.\n\nExample: An EXN30T3H, 30kVA 480 delta to 208Y/120 transformer 60 Hz, can be utilized on a 50 Hz system as long as the applied voltage does not exceed 50/60 or 5/6 of the nominal 480, or rather 400 volts. With 400 volts 50 Hz applied, the output voltage will be 400/480 * 208 = 173 volts line-to-line, and 100 volts line-to-neutral, or 173Y/100 50 Hz. As well, the capacity or kVA of the transformer will be 400/480 * 30 = 25kVA. Transformers cannot change frequency, thus frequency in = frequency out."
] | [
null,
"https://pixel.quantserve.com/pixel/p-yEy7R7_vSeq3Y.gif",
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https://numbermatics.com/n/5085/ | [
"# 5085\n\n## 5,085 is an odd composite number composed of three prime numbers multiplied together.\n\nWhat does the number 5085 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 12 divisors.\n\n5085 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twelve divisors.\n\n## Prime factorization of 5085:\n\n### 32 × 5 × 113\n\n(3 × 3 × 5 × 113)\n\nSee below for interesting mathematical facts about the number 5085 from the Numbermatics database.\n\n### Names of 5085\n\n• Cardinal: 5085 can be written as Five thousand and eighty-five.\n\n### Scientific notation\n\n• Scientific notation: 5.085 × 103\n\n### Factors of 5085\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 121\n\n### Divisors of 5085\n\n• Number of divisors d(n): 12\n• Complete list of divisors:\n• Sum of all divisors σ(n): 8892\n• Sum of proper divisors (its aliquot sum) s(n): 3807\n• 5085 is a deficient number, because the sum of its proper divisors (3807) is less than itself. Its deficiency is 1278\n\n### Bases of 5085\n\n• Binary: 10011110111012\n• Base-36: 3X9\n\n### Squares and roots of 5085\n\n• 5085 squared (50852) is 25857225\n• 5085 cubed (50853) is 131483989125\n• The square root of 5085 is 71.3091859441\n• The cube root of 5085 is 17.1961141393\n\n### Scales and comparisons\n\nHow big is 5085?\n• 5,085 seconds is equal to 1 hour, 24 minutes, 45 seconds.\n• To count from 1 to 5,085 would take you about twenty-four minutes.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 5085 cubic inches would be around 1.4 feet tall.\n\n### Recreational maths with 5085\n\n• 5085 backwards is 5805\n• The number of decimal digits it has is: 4\n• The sum of 5085's digits is 18\n• More coming soon!\n\nMLA style:\n\"Number 5085 - Facts about the integer\". Numbermatics.com. 2021. Web. 21 October 2021.\n\nAPA style:\nNumbermatics. (2021). Number 5085 - Facts about the integer. Retrieved 21 October 2021, from https://numbermatics.com/n/5085/\n\nChicago style:\nNumbermatics. 2021. \"Number 5085 - Facts about the integer\". https://numbermatics.com/n/5085/\n\nThe information we have on file for 5085 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 5085, math, Factors of 5085, curriculum, school, college, exams, university, Prime factorization of 5085, STEM, science, technology, engineering, physics, economics, calculator, five thousand and eighty-five.\n\nOh no. Javascript is switched off in your browser.\nSome bits of this website may not work unless you switch it on."
] | [
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-2-solving-equations-2-3-solving-multi-step-equations-lesson-check-page-97/8 | [
"## Algebra 1\n\n-19$\\frac{3}{4}$\n-$\\frac{2}{9}$$x - 4 = \\frac{7}{18} Multiply each side by 18 to get rid of the denominators. 18(-\\frac{2}{9}$$x$ - $4$) = 18($\\frac{7}{18}$) -4$x$-72 = 7 -4$x$ = 79 Divide each side by 4 to isolate the variable. x=-$\\frac{79}{4}$ = -19$\\frac{3}{4}$"
] | [
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https://nolver.net/home/pubtype/conference/ | [
"### Filter by type:\n\nSort by year:\n\n#### Majorizing measures for the optimizer\n\nConferencePreprintprobability\nSander Borst, Daniel Dadush, Neil Olver and Makrand Sinha,\nAccepted to ICTS 2021\n\nThe theory of majorizing measures, extensively developed by Fernique, Talagrand and many others, provides one of the most general frameworks for controlling the behavior of stochastic processes. In particular, it can be applied to derive quantitative bounds on the expected suprema and the degree of continuity of sample paths for many processes.\n\nOne of the crowning achievements of the theory is Talagrand’s tight alternative characterization of the suprema of Gaussian processes in terms of majorizing measures. The proof of this theorem was difficult, and thus considerable effort was put into the task of developing both shorter and easier to understand proofs. A major reason for this difficulty was considered to be theory of majorizing measures itself, which had the reputation of being opaque and\nmysterious. As a consequence, most recent treatments of the theory (including by Talagrand himself) have eschewed the use of majorizing measures in favor of a purely combinatorial approach (the generic chaining) where objects based on sequences of partitions provide roughly matching upper and lower bounds on the desired expected supremum.\n\nIn this paper, we return to majorizing measures as a primary object of study, and give a viewpoint that we think is natural and clarifying from an optimization perspective. As our main contribution, we give an algorithmic proof\nof the majorizing measures theorem based on two parts:\n\n• We make the simple (but apparently new) observation that finding the best majorizing measure can be cast as a convex program.\nThis also allows for efficiently computing the measure using off-the-shelf methods from convex optimization.\n• We obtain tree-based upper and lower bound certificates by rounding, in a series of steps, the primal and dual solutions to this convex program.\n\nWhile duality has conceptually been part of the theory since its beginnings, as far as we are aware no explicit link to convex optimization has been previously made.\n\n#### Long term behavior of dynamic equilibria in fluid queuing networks\n\nagtConferenceJournal\nRoberto Cominetti, Jose Correa and Neil Olver\nAccepted to Operations Research. Conference version: IPCO 2017\n\n#### Approximate multi-matroid intersection via iterative refinement\n\napproximationConferenceJournal\nAndré Linhares, Neil Olver, Chaitanya Swamy, Rico Zenklusen\nMathematical Programming. Conference version: IPCO 2019. arXiv:1811.09027\n\nWe introduce a new iterative rounding technique to round a point in a matroid polytope subject to further matroid constraints. This technique returns an independent set in one matroid with limited violations of the other ones. On top of the classical steps of iterative relaxation approaches, we iteratively refine/split involved matroid constraints to obtain a more restrictive constraint system, that is amenable to iterative relaxation techniques. Hence, throughout the iterations, we both tighten constraints and later relax them by dropping constrains under certain conditions. Due to the refinement step, we can deal with considerably more general constraint classes than existing iterative relaxation/rounding methods, which typically round on one matroid polytope with additional simple cardinality constraints that do not overlap too much.\n\nWe show how our rounding method, combined with an application of a matroid intersection algorithm, yields the first 2-approximation for finding a maximum-weight common independent set in 3 matroids. Moreover, our 2-approximation is LP-based, and settles the integrality gap for the natural relaxation of the problem. Prior to our work, no better upper bound than 3 was known for the integrality gap, which followed from the greedy algorithm. We also discuss various other applications of our techniques, including an extension that allows us to handle a mixture of matroid and knapsack constraints.\n\n#### Algorithms for Flows Over Time with Scheduling Costs\n\nagtConferencenetwork\nDario Frascaria and Neil Olver\nIPCO 2020\n\nFlows over time have received substantial attention from both an optimization and (more recently) a game-theoretic perspective. In this model, each arc has an associated delay for traversing the arc, and a bound on the rate of flow entering the arc; flows are time-varying. We consider a setting which is very standard within the transportation economic literature, but has received little attention from an algorithmic perspective. The flow consists of users who are able to choose their route but also their departure time, and who desire to arrive at their destination at a particular time, incurring a scheduling cost if they arrive earlier or later. The total cost of a user is then a combination of the time they spend commuting, and the scheduling cost they incur. We present a combinatorial algorithm for the natural optimization problem, that of minimizing the average total cost of all users (i.e., maximizing the social welfare). Based on this, we also show how to set tolls so that this optimal flow is induced as an equilibrium of the underlying game.\n\n#### Fixed-order scheduling on parallel machines\n\napproximationConference\nThomas Bosman, Dario Frascaria, Neil Olver, Rene Sitters and Leen Stougie\nIPCO 2019\n\nWe consider the following natural scheduling problem: Given a sequence of jobs with weights and processing times, one needs to assign each job to one of m identical machines in order to minimize the sum of weighted completion times. The twist is that for machine the jobs assigned to it must obey the order of the input sequence, as is the case in multi-server queuing systems. We establish a constant factor approximation algorithm for this (strongly NP-hard) problem. Our approach is necessarily very different from what has been used for similar scheduling problems without the fixed-order assumption. We also give a QPTAS for the special case of unit processing times\n\n#### A Simpler and Faster Strongly Polynomial Algorithm for Generalized Flow Maximization\n\nConferenceJournalnetwork\nNeil Olver and László Végh\nAccepted to J. ACM, subject to minor revisions. Conference version: STOC 2017.\n\nWe present a new strongly polynomial algorithm for generalized flow maximization. The first strongly polynomial algorithm for this problem was given in [Végh 2016]; our new algorithm is much simpler, and much faster. The complexity bound $O((m+n\\log n)mn\\log (n^2/m))$ improves on the previous estimate in [ Végh 2016] by almost a factor $O(n^2)$. Even for small numerical parameter values, our algorithm is essentially as fast as the best weakly polynomial algorithms. The key new technical idea is relaxing primal feasibility conditions. This allows us to work almost exclusively with integral flows, in contrast to all previous algorithms for the problem.\n\n#### The Itinerant List Update Problem\n\napproximationConference\nNeil Olver, Kirk Pruhs, Kevin Schewior, Rene Sitters and Leen Stougie\nProceedings of the 16th Workshop on Approximation and Online Algorithms (WAOA)\n\n#### Fast, Deterministic and Sparse Dimensionality Reduction\n\nConferenceprobability\nDaniel Dadush, Cristóbal Guzman, Neil Olver\nSODA 2018\n\n#### Chain-constrained spanning trees\n\nConferenceJournalnetwork\nNeil Olver and Rico Zenklusen\nMathematical Programming 167(2):293–314. Conference version in IPCO 2013.\n\nWe consider the problem of finding a spanning tree satisfying a family of additional constraints. Several settings have been considered previously, the most famous being the problem of finding a spanning tree with degree constraints. Since the problem is hard, the goal is typically to find a spanning tree that violates the constraints as little as possible.\n\nIterative rounding became the tool of choice for constrained spanning tree problems. However, iterative rounding approaches are very hard to adapt to settings where an edge can be part of a super-constant number of constraints. We consider a natural constrained spanning tree problem of this type, namely where upper bounds are imposed on a family of cuts forming a chain. Our approach reduces the problem to a family of independent matroid intersection problems, leading to a spanning tree that violates each constraint by a factor of at most 9.\n\nWe also present strong hardness results: among other implications, these are the first to show, in the setting of a basic constrained spanning tree problem, a qualitative difference between what can be achieved when allowing multiplicative as opposed to additive constraint violations.\n\n#### On the Integrality Gap of the Prize-Collecting Steiner Forest LP\n\napproximationConferencenetwork\nJochen Koenemann, Kanstantsin Pashkovich, Neil Olver, R. Ravi, Chaitanya Swamy, Jens Vygen\nAPPROX 2017\n\n#### Exploring the tractability of the capped hose model\n\nConferencenetwork\nThomas Bosman, Neil Olver\nProceedings of the 25th Annual European Symposium on Algorithms (ESA)\n\n#### On the Equivalence of the Bidirected and Hypergraphic Relaxations for Steiner Tree\n\nConferenceJournalnetwork\nAndreas E. Feldmann, Jochen Koenemann, Neil Olver and Laura Sanità\nMathematical Programming 160:379-406. Conference version: APPROX 2014\n\nThe bottleneck of the currently best (ln(4) + epsilon)-approximation algorithm for the NP-hard Steiner tree problem is the solution of its large, so called hypergraphic, linear programming relaxation (HYP). Hypergraphic LPs are NP-hard to solve exactly, and it is a formidable computational task to even approximate them sufficiently well. We focus on another well-studied but poorly understood LP relaxation of the problem: the bidirected cut relaxation (BCR). This LP is compact, and can therefore be solved efficiently. Its integrality gap is known to be greater than 1.16, and while this is widely conjectured to be close to the real answer, only a (trivial) upper bound of 2 is known. In this paper, we give an efficient constructive proof that BCR and HYP are polyhedrally equivalent in instances that do not have an (edge-induced) claw on Steiner vertices, i.e., they do not contain a Steiner vertex with 3 Steiner neighbors. This implies faster ln(4)-approximations for these graphs, and is a significant step forward from the previously known equivalence for (so called quasi-bipartite) instances in which Steiner vertices form an independent set. We complement our results by showing that even restricting to instances where Steiner vertices induce one single star, determining whether the two relaxations are equivalent is NP-hard.\n\nConferenceprobability\nJosé Correa, Marcos Kiwi , Neil Olver and Alberto Vera\nWINE 2015\n\nMotivated by the recent emergence of the so-called opportunistic communication networks, we consider the issue of adaptivity in the most basic continuous time (asynchronous) rumor spreading process. In our setting a rumor has to be spread to a population; the service provider can push it at any time to any node in the network and has unit cost for doing this. On the other hand, as usual in rumor spreading, nodes share the rumor upon meeting and this imposes no cost on the service provider. Rather than fixing a budget on the number of pushes, we consider the cost version of the problem with a fixed deadline and ask for a minimum cost strategy that spreads the rumor to every node. A non-adaptive strategy can only intervene at the beginning and at the end, while an adaptive strategy has full knowledge and intervention capabilities. Our main result is that in the homogeneous case (where every pair of nodes randomly meet at the same rate) the benefit of adaptivity is bounded by a constant. This requires a subtle analysis of the underlying random process that is of interest in its own right.\n\n#### Decentralized utilitarian mechanisms for scheduling games\n\nagtConferenceJournal\nR. Cole, J. Correa, V. Gkatzelis, V. Mirrokni and N. Olver\nGames and Economic Behavior, Volume 92, pp 306–326, 2015. Conference version: STOC 2011\n\nGame Theory and Mechanism Design are by now standard tools for studying and designing massive decentralized systems. Unfortunately, designing mechanisms that induce socially efficient outcomes often requires full information and prohibitively large computational resources. In this work we study simple mechanisms that require only local information. Specifically, in the setting of a classic scheduling problem, we demonstrate local mechanisms that induce outcomes with social cost close to that of the socially optimal solution. Somewhat counter-intuitively, we find that mechanisms yielding Pareto dominated outcomes may in fact enhance the overall performance of the system, and we provide a justification of these results by interpreting these inefficiencies as externalities being internalized. We also show how to employ randomization to obtain yet further improvements. Lastly, we use the game-theoretic insights gained to obtain a new combinatorial approximation algorithm for the underlying optimization problem.\n\n#### Pipage Rounding, Pessimistic Estimators and Matrix Concentration\n\nConferenceprobability\nN. Harvey, N. Olver\nSODA 2014\n\nPipage rounding is a dependent random sampling technique that has several interesting properties and diverse applications.\nOne property that has been useful in applications is negative correlation of the resulting vector. There are some further properties that would be interesting to derive, but do not seem to follow from negative correlation. In particular, recent concentration results for sums of independent random matrices are not known to extend to a negatively dependent setting.\n\nWe introduce a simple but useful technique called concavity of pessimistic estimators. This technique allows us to show concentration of submodular functions and concentration of matrix sums under pipage rounding. The former result answers a question of Chekuri et al. (2009). To prove the latter result, we derive a new variant of Lieb’s celebrated concavity theorem in matrix analysis.\n\nWe provide numerous applications of these results. One is to spectrally-thin trees, a spectral analog of the thin trees that played a crucial role in the recent breakthrough on the asymmetric traveling salesman problem. We show a polynomial time algorithm that, given a graph where every edge has effective conductance at least $\\kappa$, returns an $O(\\kappa^{-1} \\cdot \\log n / \\log \\log n)$-spectrally-thin tree. There are further applications to rounding of semidefinite programs and to a geometric question of extracting a nearly-orthonormal basis from an isotropic distribution.\n\n#### Approximability of robust network design\n\napproximationConferenceJournalnetwork\nNeil Olver and Bruce Shepherd\nMathematics of Operations Research 39(2):561–572, 2014. Conference version: SODA 2010\n\nWe consider robust network design problems where the set of feasible demands may be given by an arbitrary polytope or convex body more generally. This model, introduced by Ben-Ameur and Kerivin (2003), generalizes the well studied virtual private network (VPN) problem. Most research in this area has focused on finding constant factor approximations for specific polytope of demands, such as the class of hose matrices used in the definition of VPN. As pointed out in Chekuri (2007), however, the general problem was only known to be APX-hard (based on a reduction from the Steiner tree problem). We show that the general robust design is hard to approximate to within logarithmic factors. We establish this by showing a general reduction of buy-at-bulk network design to the robust network design problem. In the second part of the paper, we introduce a natural generalization of the VPN problem. In this model, the set of feasible demands is determined by a tree with edge capacities; a demand matrix is feasible if it can be routed on the tree. We give a constant factor approximation algorithm for this problem that achieves factor 8 in general, and 2 for the case where the tree has unit capacities.\n\n#### The VPN Conjecture is true\n\nConferenceJournalnetwork\nNavin Goyal, Neil Olver and Bruce Shepherd\nJournal of the ACM, 60(3), 2013. Conference version: STOC 2008\n\nWe consider the following network design problem. We are given an undirected graph $G=(V,E)$ with edges costs $c(e)$ and a set of terminal nodes $W$. A hose demand matrix for $W$ is any symmetric matrix $[D_{ij}]$ such that for each $i$, $\\sum_{j \\neq i} D_{ij} \\leq 1$. We must compute the minimum cost edge capacities that are able to support the oblivious routing of every hose matrix in the network. An oblivious routing template, in this context, is a simple path $P_{ij}$ for each pair $i,j \\in W$. Given such a template, if we are to route a demand matrix $D$, then for each $i,j$ we send $D_{ij}$ units of flow along each $P_{ij}$. Fingerhut et al. (1997) and Gupta et al. (2001) obtained a $2$-approximation for this problem, using a solution template in the form of a tree. It has been widely asked and subsequently conjectured that this solution actually results in the optimal capacity for the single path VPN design problem; this has become known as the VPN conjecture.\n\nThe conjecture has previously been proven for some restricted classes of graphs (Hurkens et al. 2005, Grandoni et al. 2007, Fiorini et al. 2007). Our main theorem establishes that this conjecture is true in general graphs. This proves that the single path VPN problem is solvable in polynomial time. We also show that the multipath version of the conjecture is false.\n\n#### Matroids and Integrality Gaps for Hypergraphic Steiner Tree Relaxations\n\nConferencenetwork\nMichel Goemans, Neil Olver, Thomas Rothvoss and Rico Zenklusen\nSTOC 2012\n\nUntil recently, LP relaxations have played a limited role in the design of approximation algorithms for the Steiner tree problem. In 2010, Byrka et al. presented a ln(4)+epsilon approximation based on a hypergraphic LP relaxation, but surprisingly, their analysis does not provide a matching bound on the integrality gap.\nWe take a fresh look at hypergraphic LP relaxations for the Steiner tree problem – one that heavily exploits methods and results from the theory of matroids and submodular functions – which leads to stronger integrality gaps, faster algorithms, and a variety of structural insights of independent interest. More precisely, we present a deterministic ln(4)+epsilon approximation that compares against the LP value and therefore proves a matching ln(4) upper bound on the integrality gap.\nSimilarly to Byrka et al., we iteratively fix one component and update the LP solution. However, whereas they solve an LP at every iteration after contracting a component, we show how feasibility can be maintained by a greedy procedure on a well-chosen matroid. Apart from avoiding the expensive step of solving a hypergraphic LP at each iteration, our algorithm can be analyzed using a simple potential function. This gives an easy means to determine stronger approximation guarantees and integrality gaps when considering restricted graph topologies. In particular, this readily leads to a 73/60 bound on the integrality gap for quasi-bipartite graphs.\nFor the case of quasi-bipartite graphs, we present a simple algorithm to transform an optimal solution to the bidirected cut relaxation to an optimal solution of the hypergraphic relaxation, leading to a fast 73/60 approximation for quasi-bipartite graphs. Furthermore, we show how the separation problem of the hypergraphic relaxation can be solved by computing maximum flows, providing a fast independence oracle for our matroids.\n\n#### Dynamic vs Oblivious Routing in Network Design\n\nConferenceJournalnetwork\nNavin Goyal, Neil Olver and Bruce Shepherd\nAlgorithmica 61(1): 161–173, 2011. Conference version: ESA 2009\n\nConsider the robust network design problem of finding a minimum cost network with enough capacity to route all traffic demand matrices in a given polytope. We investigate the impact of different routing models in this robust setting: in particular, we compare oblivious routing, where the routing between each terminal pair must be fixed in advance, to dynamic routing, where routings may depend arbitrarily on the current demand. Our main result is a construction that shows that the optimal cost of such a network based on oblivious routing (fractional or integral) may be a factor of $\\Omega(\\log{n})$ more than the cost required when using dynamic routing. This is true even in the important special case of the asymmetric hose model. This answers a question in Chekuri (2007), and is tight up to constant factors. Our proof technique builds on a connection between expander graphs and robust design for single-sink traffic patterns (Chekuri et al. 2007)."
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https://www.mail-archive.com/music-dsp@music.columbia.edu/msg02487.html | [
"# Re: [music-dsp] Time Varying BIBO Stability Analysis of Trapezoidal integrated optimised SVF v2\n\n```Hi Ross,\n\n```\nsince you opened this topic, I thought I'd try to share the intermediate results my findings, as much as I can remember them (that was a few years back). Most of them concern the continuous time case.\n```\n```\nFirst note regarding the continuous time case is that cutoff modulations do not affect the BIBO stability at all. More rigorously: - if the cutoff modulation is done by varying the gains *in front* (rather than behind) of *all* integrators in the system\n```- if the cutoff function w(t) is always positive\n- if the system is BIBO stable for some cutoff function w(t)\nthen the system is also BIBO stable for any other positive cutoff function\n\n```\nParticularly, if a linear system is BIBO-stable in time-invariant case (for the constant cutoff function), then it's also stable for varying cutoff.\n```\nThis is very easy to obtain from the state-space equation:\ndu/dt=w*F(u,x)\n```\nwhere u(t) is the state vector, x(t) is the input vector, w(t) is the cutoff scalar function and F(u,x,t) is the nonlinear time-varying version of A*u+B*x. Without reduction of generality we can assume w(t)=1 for the given stable case. Then, we simply rewrite the equation as\n```du/(w*dt)=F(u,x)\nand substitute the time parameter:\nd tau = w*dt\n```\nNow in \"tau\" time coordinates the modulated system is exactly the same as the unmodulated one in the \"t\" coordinates.\n```\nThe same doesn't seem to hold for the TPT discrete-time version, though.\n\n```\nIn a more general case for *linear* continuous time, IIRC, we have a sufficient (but it seems, not necessary) time-varying stability criterion: all eigenvalues of the matrix A+A^T must be \"uniformly negative\", that is they must be bounded by some negative number from above. It is essential to require this uniform negativity, otherwise the eigenvalues can get arbitrarily close to the self-oscillation case. This condition is simply obtained from the fact that in the absence of the input signal you want the absolute value of the state to decay with a relative speed, which is uniformly less that 1. This will make sure, that, whatever the bound of the input signal is, a large enough state will decay sufficiently fast, to win over the input vector B(t)*x(t). Indeed, ignoring the B*x term, we have\n```(d/dt) |u|^2=(d/dt)(u^T*u)=u'^T*u+u^T*u'=\n(A*u)^T*u+u^T*(A*u)=u^T*A^T*u+u^T*A*u=\nu^T*(A+A^T)*u<=|u|^2*max{lambda_i}\nwhere lambda_i are the eigenvalues of A+A^T.\nNow on the other hand\n(d/dt) |u|^2=2*|u|*(d/dt)|u|\nSo\n2*|u|*(d/dt)|u|<=|u|^2*max{lambda_i}\nand\n2*(d/dt)|u|<=|u|*max{lambda_i}\n\n```\nObviously, you don't have to satisfy the condition in the original state-space coordinates. Instead, you can satisfy it in any other coordinates, which corresponds to using P^T*A*P instead of A for some nonsingular matrix P.\n```\n```\nNow I didn't manage to get this condition satisfied for the continuous-time SVF. Reading your post, I admit, that I could have made a mistake there, but FWIW... First, I discarded the consideration of varying cutoff, as explained above and concentrated on the varying damping. Not managing to find a matrix P, I constructed an input signal, requiring the maximum possible growth of the state vector. The signal, IIRC was either sgn(s_1) or -sgn(s_1), where s_1 is the first of the state components (or it could have been s_2). Then I noticed that for low damping the state vector is moving in almost a circle, while for higher damping (but still with complex poles) is turns into an ellipse. This was exactly the problem: \"in principle\" the circle is having a bigger size, than the ellipse, but by switching the damping from low to high you could \"shoot\" the state point into a much \"higher orbit\". Much worse, in certain cases the system state can increase even in the full absence of the input signal!!! However, IIRC, I managed to show, that for a sufficiently large \"elliptic\" orbit (with high damping), (d/dt)|u|^2<=0 regardless of the current damping. Since we are already considering the \"worst possible\" input signal, the system state can't cross this boundary \"orbit\" to the outside.\n```\n```\nFor the discrete-time case the situation is more complicated, because we can't use the continuity of the state vector function. IIRC, I also didn't manage to build the \"worst-case\" signal, but there was the same problem of the state vector becoming larger in the absence of the input signal. That's why I was somewhat surprised that you simply managed to restrict the eigenvalues of the system matrix in some coordinates. Particularly suspicious is that your coordinate transformation matrix is \"built for the smallest damping\", while the more problematic case seems to occur \"at the larger damping\". But, as I said, I didn't finish that research and I could have been wrong. So just take my input FWIW.\n```\nRegards,\n\n--\nReaktor Application Architect\nNative Instruments GmbH\n+49-30-611035-0\n\nwww.native-instruments.com\n--\ndupswapdrop -- the music-dsp mailing list and website:\nsubscription info, FAQ, source code archive, list archive, book reviews, dsp"
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https://physics.stackexchange.com/questions/246454/are-gravitational-force-and-gravitational-time-dilation-proportional | [
"# Are gravitational force and gravitational time dilation proportional?\n\nParticles in gravitational fields are subject to gravitational time dilation. The closer a particle is near a gravitational source, the slower is running its clock. I would like to know more about the relation between gravity and gravitational time dilation.\n\nIn order to get a rough impression I used Newton's gravity equation (which may be used for weak fields, and I found that gravity and time dilation are (approximately) proportional: Can this result be confirmed on the base of Einstein's field equation (maybe even for stronger fields)?\n\ndτ = proper time of a particle in the gravitation field of Earth, dt = proper time of an observer in infinity, rs = Schwarzschild radius of Earth, r = distance particle - center of Earth\n\nGravitational time dilation:\n\n$\\frac{dτ}{dt}=\\sqrt{1- \\frac{r_s}{r}} ≈ 1- \\frac{r_s}{2r}$\n\nTime dilation (difference):\n\n$1-\\frac{dτ}{dt} ≈ \\frac{r_s}{2r}= \\frac{GM}{c^2 r}$\n\nGravitational force (Newton's equation):\n\n$F=G \\frac{mM}{r^2}$\n\n\\begin{equation} \\frac{Gravitational\\:force}{Time\\:dilation\\:(difference)} ≈ \\frac{G\\frac{mM}{r^2}}{\\frac{GM}{c^2 r}}=\\frac{mc^2}{r}=\\frac{rest\\:energy\\:(of\\:the\\:particle\\:subject\\:to\\:time\\:dilation)}{distance\\:(of\\:the\\:particle)} \\end{equation}\n\n(As a result, time dilation would be approximately gravitaty, divided by the rest energy of the particle, multiplied by its distance.)\n\n• Hi Moonraker. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. Mar 31, 2016 at 10:27\n• Hi @Qmechanic, thank you for your information, and I did read the page you indicated. - I don't know why you think that my question falls under this category. My question is clearly indicated in the title, and there is no other question: Is there some relation of proportionality between gravitational force and gravitational time dilation. I could even reformulate my question without the formulas I provided (if you wish so). I just need the information about Einstein field equations, nothing more. Mar 31, 2016 at 11:56\n• Also, for future reference, please don't delete and repost questions. Instead, if they're not well received, you should edit them to improve them. Mar 31, 2016 at 12:36\n• @David Z♦ : OK, I undeleted the former question. Thank you for this formal comment, I hope I fixed it. I would be keen on knowing your comment on the topic of my question? Sincerely Mar 31, 2016 at 13:23\n\nIf you have a look at my answer to Deriving a Schwarzschild radius using relativistic mass I discuss how the weak field approximation gives us an approximate metric for the Newtonian gravitational potential $\\phi$:\n\n$$ds^2 \\approx -\\left( 1 + \\frac{2\\phi}{c^2}\\right) c^2dt^2 + \\frac{1}{1 + 2\\phi/c^2}\\left(dx^2 + dy^2 + dz^2\\right)$$\n\nTo extract time dilation from this we take a stationary object, so $dx = dy = dz = 0$ and use the relationship between the line element and the proper time $ds^2 = -c^2d\\tau^2$ to get:\n\n$$\\frac{d\\tau}{dt} \\approx \\sqrt{ 1 + \\frac{2\\phi}{c^2}}$$\n\nTo clarify this, take two observers $A$ and $B$ with gravitational potential energies $\\phi_A$ and $\\phi_B$, then the equation tells us that the elapsed times recorded by $A$ and $B$ are related by:\n\n$$\\frac{dt_A}{dt_B} \\approx \\sqrt{ 1 + \\frac{2(\\phi_A - \\phi_B)}{c^2}}$$\n\nThis equation is only valid when $2\\Delta\\phi/c^2 \\ll 1$, in which case we can use the binomial expansion:\n\n$$\\frac{dt_A}{dt_B} \\approx 1 + \\frac{\\Delta\\phi}{c^2} + \\text{higher terms}$$\n\nand dropping the higher terms and rearranging:\n\n$$\\frac{dt_A - dt_B}{dt_B} \\approx \\frac{\\Delta\\phi}{c^2}$$\n\nAnd this is sort of what you describe. Remember that the potential energy $\\phi$ is the potential energy per unit mass, so if we multiply the top and bottom of the right side by the mass to get the total potential energy $\\Phi$ we get:\n\n$$\\frac{dt_A - dt_B}{dt_B} \\approx \\frac{\\Delta\\Phi}{mc^2}$$\n\nwhich is indeed the gravitational potential divided by the rest energy.\n\nBut this is an approximation that works (reliably) only in the weak field limit. As it happens the weak field expression works for any values of $r$ in the Schwarzschild metric, but as discussed in the linked question this is an accidental coincidence and can't be relied on.\n\n• Nice, interesting explanations, but you are far from answering my question! You confirmed that my rough calculation with Newton's equation yields approximately correct results for weak fields. I know this, even if I could not explain it with Schwarzschild metrics as you did. But my calculations (and yours also) induce the idea that gravitational force and gravitational time dilation are proportional, with a factor of proportionality equal to the rest energy of the particle divided by the distance. - Are they proportional or not? Mar 31, 2016 at 6:48\n• @Moonraker: I thought that was obvious from my final equation. In the weak field limit the relative time dilation is proportional to the gravitational potential. However this is only true in the weak field limit so it works for calculating the time dilation of geostationary satellites but not for calculating the time dilation near a black hole. Mar 31, 2016 at 7:08\n• My question is not for purposes of technological applications, but for exploring the nature of gravitational time dilation and gravity (even in stronger fields). Your final approximate equation does not include more insight than my final approximate equation, Newton's equation is appropriate for weak fields. Mar 31, 2016 at 8:09\n• I appreciate that you showed the Schwarzschild approximation, but there might be other insights deriving from Einstein's field equation which are putting some light on this supposed proportionality. Mass is producing gravity, and mass is producing gravitational time dilation - may be there is a direct relation between both of them?? Mar 31, 2016 at 8:09\n• @Moonraker: I have to confess that I don't understand what you are getting at. The linear proportionality is a weak field phenomenon and is an inevitable consequence of the fact that in the weak field limit GR must reproduce Newtonian gravity. Outside the weak field limit the time dilation is obviously related to the mass (more precisely the stress-energy tensor) but remember you can choose any coordinate system you want and the time dilation will depend on the coordinates you choose, so the relationship is a complicated one. Mar 31, 2016 at 8:22\n\nThey don't have to be related.\n\nFor instance if you have a hollow spherical shell of matter then the inside of the sphere is a flat spacetime region and it has the same time dilation as the shell.\n\nBut since the inside is flat, there is no gravitational force inside the shell. Yet there is time dilation."
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https://www.slideserve.com/yovela/sect-3-a-position-velocity-and-acceleration | [
"# Sect. 3-A Position , Velocity, and Acceleration - PowerPoint PPT Presentation",
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"Download Presentation",
null,
"Sect. 3-A Position , Velocity, and Acceleration\n\nSect. 3-A Position , Velocity, and Acceleration",
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"Download Presentation",
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"## Sect. 3-A Position , Velocity, and Acceleration\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Sect. 3-APosition, Velocity, and Acceleration\n\n2. Position function - gives the location of an object at time t, usually s(t), x(t) or y(t) Velocity - The rate of change( derivative) of position, usually v(t) Acceleration - The rate of change ( derivative) of velocity usually a(t) Average velocity -\n\n3. 1) If find v(t) and a(t).\n\n4. 2) The position of a particle moving along the x - axis at time t is given by . Find the particles velocity and acceleration at t = 5. Velocity Acceleration\n\n5. Speed – The absolute value of velocity otherwise known as the magnitude of velocity\n\n6. Velocity Positive - the particle is moving to the right Negative - the particle is moving to the left Zero - the particle has momentarily stopped or is changing direction ( must have a sign change)\n\n7. 3) Find where the object changes direction if Find where v(t) = 0\n\n8. Acceleration and Velocity • If the sign of acceleration is the same as velocity, the speed of the particle is ______________ (the two are working together) • If the sign of the acceleration is opposite that of velocity, the speed of the particle is _________________ (the two are working against each other)\n\n9. 4) Given the same position function as #3 find the interval during which the particle is slowing down.\n\n10. 5) Given find the interval during which the particle is speeding up.\n\n11. Distance vs. Displacement • Displacement- change in position ( final position minus original position) • Distance- the total distance travelled by an object in the time interval even if duplicated\n\n12. Consider two intervals 6) Find the DISTANCE traveled by the particle whose position is given by on the interval (0,4). Distance NOT displacement!!\n\n13. 7) If • find the DISTANCE traveled by the particle on the interval (2,4). • Find the DISPLACEMENT on the interval (1,5)\n\n14. The graph shows the position function of a radio controlled car a) Was the car going faster at B or at C? b) When was the car stopped? c) At which point was the car’s velocity the greatest? d) At which point was the car’s speed decreasing?\n\n15. Homework PVA Worksheet and worksheet 3-A"
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http://de.git.xonotic.org/?p=xonotic/darkplaces.git;a=blobdiff;f=draw.h;h=3857e98e303f53a4f188079d6be0474ff1b8a943;hp=bbd964ed3d9839518be6f8c67a5faaccae177a83;hb=9d6f77621de650ffaaecb65a3485d51378632925;hpb=c045878e6948f00e52f17c9df6beefa2dfdfd868 | [
"diff --git a/draw.h b/draw.h\nindex bbd964e..3857e98 100644 (file)\n--- a/draw.h\n+++ b/draw.h\n@@ -134,7 +134,15 @@ extern dp_fonts_t dp_fonts;\n#define STRING_COLOR_RGB_TAG_CHAR 'x'\n#define STRING_COLOR_RGB_TAG \"^x\"\n\n-// all of these functions will set r_defdef.draw2dstage if not in 2D rendering mode (and of course prepare for 2D rendering in that case)\n+// prepare for 2D rendering (sets r_refdef.draw2dstage = 1 and calls R_ResetViewRendering2D)\n+void DrawQ_Start(void);\n+// resets r_refdef.draw2dstage to 0\n+void DrawQ_Finish(void);\n+// batch draw the pending geometry in the CL_Mesh_UI() model and reset the model,\n+// to be called by things like DrawQ_SetClipArea which make disruptive state changes.\n+void DrawQ_FlushUI(void);\n+// use this when changing r_refdef.view.* from e.g. csqc\n+void DrawQ_RecalcView(void);\n\n// draw an image (or a filled rectangle if pic == NULL)\nvoid DrawQ_Pic(float x, float y, cachepic_t *pic, float width, float height, float red, float green, float blue, float alpha, int flags);\n@@ -163,12 +171,6 @@ void DrawQ_SetClipArea(float x, float y, float width, float height);\nvoid DrawQ_ResetClipArea(void);\n// draw a line\nvoid DrawQ_Line(float width, float x1, float y1, float x2, float y2, float r, float g, float b, float alpha, int flags);\n-// resets r_refdef.draw2dstage\n-void DrawQ_Finish(void);\n-void DrawQ_RecalcView(void); // use this when changing r_refdef.view.* from e.g. csqc\n-// batch draw the pending geometry in the CL_Mesh_UI() model and reset the model,\n-// to be called by things like DrawQ_SetClipArea which make disruptive state changes.\n-void DrawQ_FlushUI(void);\n\nconst char *Draw_GetPicName(cachepic_t *pic);\nint Draw_GetPicWidth(cachepic_t *pic);"
] | [
null
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https://sanaulla.info/2013/04/08/arrays-sort-versus-arrays-parallelsort/?amp=1 | [
"# Arrays.sort versus Arrays.parallelSort\n\nWe all have used Arrays.sort to sort objects and primitive arrays. This API used merge sort OR Tim Sort underneath to sort the contents as shown below:\n\n```public static void sort(Object[] a) {\nif (LegacyMergeSort.userRequested)\nlegacyMergeSort(a);\nelse\nComparableTimSort.sort(a);\n}\n```\n\nThis is all done sequentially, even though merge sort uses divide and conquer technique, its all done sequentially.\n\nCome Java 8, there is a new API introduced for sorting which is Arrays#parallelSort. This is does the sorting in parallel. Interesting right! Lets see how it does…\n\nArrays#parallelSort uses Fork/Join framework introduced in Java 7 to assign the sorting tasks to multiple threads available in the thread pool. This is called eating your own dog food. Fork/Join implements a work stealing algorithm where in a idle thread can steal tasks queued up in another thread.\n\n### An overview of Arrays#parallelSort:\n\nThe method uses a threshold value and any array of size lesser than the threshold value is sorted using the Arrays#sort() API (i.e sequential sorting). And the threshold is calculated considering the parallelism of the machine, size of the array and is calculated as:\n\n```private static final int getSplitThreshold(int n) {\nint p = ForkJoinPool.getCommonPoolParallelism();\nint t = (p > 1) ? (1 + n / (p << 3)) : n;\nreturn t < MIN_ARRAY_SORT_GRAN ? MIN_ARRAY_SORT_GRAN : t;\n}\n```\n\nOnce its decided whether to sort the array in parallel or in serial, its now to decide how to divide the array in to multiple parts and then assign each part to a Fork/Join task which will take care of sorting it and then another Fork/Join task which will take care of merging the sorted arrays. The implementation in JDK 8 uses this approach:\n– Divide the array into 4 parts.\n– Sort the first two parts and then merge them.\n– Sort the next two parts and then merge them.\nAnd the above steps are repeated recursively with each part until the size of the part to sort is not lesser than the threshold value calculated above.\n\n### Some interesting results:\n\nI tried to compare the time taken by the Arrays#sort and Arrays#parallelSort on a machine with 4 CPUs. The program which I used for this comparison is:\n\n```public class ArraysParallelDemo {\npublic static void main(String[] args) throws FileNotFoundException {\nList<Double> arraySource = new ArrayList<>();\n\ngetSystemResourceAsStream(\"java8demo/large_array_input\"));\nString[] strNums = line.split(\",\");\nfor ( String strN : strNums){\n}\n}\n\nSystem.out.println(arraySource.size());\n\nDouble [] myArray = new Double;\nmyArray = arraySource.toArray(myArray);\nlong startTime = System.currentTimeMillis();\nArrays.sort(myArray);\nlong endTime = System.currentTimeMillis();\nSystem.out.println(\"Time take in serial: \"+\n(endTime-startTime)/1000.0);\n\nDouble [] myArray2 = new Double;\nmyArray2 = arraySource.toArray(myArray);\nstartTime = System.currentTimeMillis();\nArrays.parallelSort(myArray2);\nendTime = System.currentTimeMillis();\nSystem.out.println(\"Time take in parallel: \"+\n(endTime-startTime)/1000.0);\n\n}\n}\n```\n\nThere is a similar implementation for Lists as well and lot of the operations on Lists have a parallel equivalent."
] | [
null
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http://maths-statistics-tutor.com/simple_linear_regression_pasw_spss.php | [
"# Performing simple linear regression in PASW (SPSS)\n\nWhen do we do simple linear regression?\nWe run simple linear regression when we want to access the relationship between two continuous variables.\n\nExample Scenario\nIn a statistics course, we want to see if there is any relationship between study time and scores in the mid-semester exam.\n\nIn this example, our null hypothesis is that there is no relationship between study time and exam scores. Our alternative hypothesis is that the more time students study, the higher the exam score.\n\nIn the data, the first column is exam scores and the second column is study time. The dataset can be obtained here.",
null,
"Before we perform the actual regression analysis, we can explore the relationship with a scatter plot.",
null,
"It appears that the more time students study, the higher the exam scores and the relationship looks linear. We now perform the regression analysis to see if there is an actual relationship between study time and exam scores. (We cannot make any definite conclusion until we do an appropriate statistical analysis.\n\nStep 1\nSelect \"Analyze -> Regression -> Linear\".",
null,
"A new window pops out.",
null,
"Step 2\nFrom the list on the left, select the variable \"Exam score\" as \"Dependent\" and the variable \"Hours\" as the \"Independent(s)\". Click \"OK\".",
null,
"Step 3\nThe results now pop out in the \"Output\" window.",
null,
"Step 4\nWe can now interpret the result.",
null,
"From B in the third table, since the p-value is 0, the relationship between study hours and exam scores is significant. From A in the second table, the correlation coefficient, R, is 0.827. Therefore, we can conclude that study hours is positively correlated with exam score and the relationship is very strong (R is positive and is very closed to 1). From C in the last table, we can conclude that on average, for every one hour a student study, he gets 2.391 more marks in the exam.",
null,
"© Maths-Statistics-Tutor.com 2010 Web Development Team."
] | [
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"http://maths-statistics-tutor.com/img/simple_linear_regression/1.jpg",
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"http://maths-statistics-tutor.com/img/simple_linear_regression/scatterplot.jpg",
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"http://maths-statistics-tutor.com/img/simple_linear_regression/2.jpg",
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"http://maths-statistics-tutor.com/img/simple_linear_regression/3.jpg",
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"http://maths-statistics-tutor.com/img/simple_linear_regression/4.jpg",
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"http://maths-statistics-tutor.com/img/simple_linear_regression/5.jpg",
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"http://maths-statistics-tutor.com/img/simple_linear_regression/6.jpg",
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"http://maths-statistics-tutor.com/img/contact_maths_statistics_button.jpg",
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http://beastie.cs.ua.edu/ACP-C/index_20.html | [
"",
null,
"",
null,
"",
null,
"Lists Contents\n\n# Lists\n\nYou can download the functions defined or used in this chapter with the following commands:\n\n``` wget troll.cs.ua.edu/ACP-C/ilist.c //integer-valued nodes\nwget troll.cs.ua.edu/ACP-C/ilist.h\nwget troll.cs.ua.edu/ACP-C/rlist.c //real-valued nodes\nwget troll.cs.ua.edu/ACP-C/rlist.h\nwget troll.cs.ua.edu/ACP-C/slist.c //string-valued nodes\nwget troll.cs.ua.edu/ACP-C/slist.h\nwget troll.cs.ua.edu/ACP-C/olist.c //'object'-valued nodes\nwget troll.cs.ua.edu/ACP-C/olist.h\nwget troll.cs.ua.edu/ACP-C/employeeList.c //employee linked list demo\n```\n\n## The List data structure\n\nWe could stop at this point and just use nodes and their operations to make true lists. But just as constructors, accessors, and mutators free us from being concerned how nodes and records are structured, we can design some list (sometimes called linked list) operators that free us from some of the details of the nodes themselves. If we do our design and implementation properly, a reader of our code will have few clues that lists are build from nodes.\n\nThe first task is to decide what will represent an empty list. The null pointer will work well for this role.\n\n``` Node *\nnewEmptyList(void)\n{\nreturn 0;\n}\n```\n\nNext, we define join, a function that will be used to add a value to a list, returning the new list. We will use our integer-valued nodes from the previous chapter:\n\n``` Node *\njoin(int value,Node *items)\n{\nreturn newNode(value,items);\n}\n```\n\nThe join function takes a value and a list as arguments and returns a new node that glues the two arguments together. The result is a list one value larger than the list that is passed in. Note that join is non-destructive with regards to the original list; if we want to modify the original list, we must reassign the original list with the return value of join:\n\n``` items = join(value,items);\n```\n\nWe can see from the definition of an empty list and from join that a list is either 0 (the null pointer) or a node. Two accessor functions are often defined for lists, head and tail. The head function returns the value stored in the first node in the list, while the tail returns the chain of nodes that follows the first node:\n\n``` int head(Node *items) { return getNodeValue(items); }\nNode *tail(Node *items) { return getNodeNext(items); }\n```\n\nTwo mutator functions are also commonly defined:\n\n``` void setHead(Node *items,int v) { setNodeValue(items,v); }\nvoid setTail(Node *items,Node *n) { setNodeNext(items,n); }\n```\n\nYou may be wondering at this point why we bother to distinguish lists and nodes. One reason is we can improve lists by keeping some more information around; we will save that improvement for later. Another reason is it is vitally important to practice the concept of abstraction. Modern software systems exhibit a large number of abstraction layers. With our lists, we can see for layers of abstractions: variables, which abstract locations in memory, structures, which abstract collections of variables, nodes, which abstract structures with two components, and lists, which abstracts a chain of `nodes`. Each level of abstraction frees us from thinking about the details of the underlying layers. We emphasize abstraction over and over, because it is so very important.\n\n### The join operator\n\nLet us now look at the join operation more closely. Here is a rewrite of join that uses a temporary variable, n, to hold the new node that will join the given value and list together:\n\n``` Node *\njoin(int v,Node *items)\n{\nNode *n = newEmptyNode(); //step 1\nsetNodeValue(n,v); //step 2\nsetNodeNext(n,items); //step 3\nreturn n;\n}\n\n```\n\nFirst, let us create an empty list to which we will join a value:\n\n``` a = newEmptyList();\n```\n\nGraphically, the situation looks like this:",
null,
"Now, we call join to join the value of 13 to the list a.\n\n``` a = join(13,a);\n```\n\nAt the very start of the join function, the formal parameters v and items have been set to the value 13 and a, respectively. The situation looks like this:",
null,
"The variables v and items, since they are local to the function join, are shown in blue35. After step 1, `n = newEmptyNode()`, we have:",
null,
"As with items and v, the variable n is shown in blue as it is a local variable. After step 2, `setNodeValue(v,None)`, the node n has its value set to v. the variable n. The situation changes to:",
null,
"After step 3, `setNodeNext(n,items)`, the next pointer of n, which is null, is changed to the value of items, which is also null. So the situation remains the same as before. Finally, n is returned and the variable a is reassigned to have the value of n. The variables v, items, and n go out of scope and disappear, leaving:",
null,
"If we add a second value to the list a, as in:\n\n``` a = join(7,a);\n```\n\nat the start of the join function, we have:",
null,
"After creating node n (step 1) and setting its value pointer (step 2), we have:",
null,
"Setting n's next pointer to items yields:",
null,
"At this point, we return the value of n and assign it to a:",
null,
"From this last drawing, we can see that list a now includes the value 7 at the front of the list, as intended. As before, at this point the variables local to join, are no longer in scope.\n\n### Chaining calls to join\n\nAs seen in the previous section, the join function is used to build a list:\n\n``` a = 0;\na = join(13,a);\na = join(7,a);\na = join(42,a);\n```\n\nThe above code builds the following list:",
null,
"An alternative way to build the exact same list is to chain together the series of calls to join:\n\n``` a = join(42,join(7,join(13,0)));\n```\n\nBoth methods are rather tedious for larger arrays, so we will make use of a list constructor that takes an array of integers and returns a list of those integers. This code:\n\n``` int numbers[] = {42,7,13};\nNode *a = arrayToList(numbers,sizeof(numbers)/sizeof(int));\n```\n\nproduces the same list. A first attempt at defining arrayToList might look like this:\n\n``` Node *\narrayToList(int *array,int size)\n{\nint i;\nNode *items = 0; //start items as the empty list\nfor (i = 0; i < size; ++i)\nitems = join(array[i],items);\nreturn items;\n}\n```\n\nUnfortunately, this produces a list whose values are found in the reverse order from that of the array. Why this is so is answered by the question, \"What is the last value joined to the growing list and where did it end up in the list?\". Obviously, the last value joined was the last value in the array and it ended up as the first value in the list, since join places its first argument at the head of the list. By a similar logic, the next-to-the-last value in the array ended up in the second position in the list, and so on.\n\nWe can fix this deficiency by defining a function that creates a new list in the reverse order of a given list. The code mimics our arrayToList function:\n\n``` Node *\nreverse(Node *a)\n{\nNode *items = 0;\nwhile (a != 0)\n{\na = tail(a);\n}\nreturn items;\n}\n```\n\nAgain, the last value in the incoming list a is placed in the first position of the new list items.\n\nThe arrayToList function becomes:\n\n``` Node *\narrayToList(int *array,int size)\n{\nint i;\nNode *items = 0; //start items as the empty list\nfor (i = 0; i < size; ++i)\n{\nitems = join(array[i],items);\n}\nNode *rev = reverse(items);\nfreeList(items);\nreturn rev;\n}\n```\n\nThis revised version of arrayToList is a bit inefficient in that the values to be placed in the list are traversed twice, once by accessing every value in the array (the for loop) and once by accessing every value in reverse's incoming list. The arrayToList functions found in the list modules mentioned in the start of this chapter cleverly make a single traversal. See if you can figure out how they work.\n\n### Reading a list from a file\n\nConsider a file of integers that we wish to read into a list. As always, we use the read pattern:\n\n``` do the initial read\n{\n}\n```\n\nConcretely, we have:\n\n``` Node *\n{\nint v;\nNode *items;\n\nitems = 0; //items points to an empty list\nfscanf(fp,\"%d\",&v);\nwhile (!feof(fp))\n{\nitems = join(v,items); //update items\nfscanf(fp,\"%d\",&v);\n}\nreturn items;\n}\n```\n\nLike the arrayToList function in the previous subsection, the readIntegers function also reverses the order of the integers as found in the file. If order must be maintained, one can reverse items before returning or use the same trick the version of arrayToList in the ilist module uses.\n\n### Displaying a list\n\nTo display a list, we need to walk the list. A walk of a list is comprised of visiting every node in the list from front to back. Typically, a loop is used to implement a walk. Here is a display function for a list of integer-valued nodes:\n\n``` void\ndisplayList(Node *items)\n{\nwhile (items != 0)\n{\nprintf(\"{%d}\",v);\nitems = tail(items); //take a step\n}\nprintf(\"\\n\");\n}\n```\n\nWith this function in place, we can see if join is really working. Here is some code that tests both join and display:\n\n``` //test (compile with ilist.c)\n#include \"ilist.h\"\nint numbers[] = {42,7,13};\nNode *a = arrayToList(numbers,sizeof(numbers)/sizeof(int));\ndisplayList(a);\n```\n\nRunning this code yields the following output, as expected:\n\n``` {42}{7}{13}\n```\n\n## Lists of other values\n\nWe can make lists of any kind of value. The modules rlist.c and slist.c contain analogs to the functions in ilist.c, except they work with real numbers (`double`) and strings (`char *`), respectively.\n\nAnother module, the olist.c module, allows for lists of pointers to arbitrary structures (or any pointers, other than function pointers). Consider the employee record structure from the chapter on structures. One might read a file of employee records, storing them into a linked list, this way:\n\n``` #include \"employee.h\"\n#include \"olist.h\"\n...\nNode *\n{\nNode *employees = 0; //zero is the empty list\nwhile (!feof(fp))\n{\nemployees = join(e,Employees);\n}\nreturn employees;\n}\n```\n\nA careful inspection of the olist module reveals that the value in a node has type `void *`. A variable of this type can store any pointer (except a function pointer) safely. However, once stored, the pointer loses any knowledge of the kind of thing to which it points (i.e. the kind of thing found at the address stored in the pointer). Another way to describe this loss of information is to say the pointer becomes generic. Once the pointer is genericized, it is up to the programmer to convert it back to the kind of pointer it once was. For example, this code fails:\n\n``` Node *emps = readEmployeeList(fp);\n```\n\nbecause the code writer has not converted the generic pointer returned by olist's head function into an Employee pointer. This code, on the other hand, succeeds:\n\n``` Node *emps = readEmployeeList(fp);\nEmployee *e = head(emps); //generic pointer converted to Employee pointer\nchar *name = e->name;\n```\n\nThe latter two lines can be combined into a single line with a cast:\n\n``` Node *emps = readEmployeeList(fp);\nchar *name = ((Employee *) head(emps))->name;\n```\n\nOne can also use the accessor function for the name field in the Employee structure:\n\n``` Node *emps = readEmployeeList(fp);\nThis gives the compiler the information that the generic pointer returned by `head(emps)` is really a pointer to an Employee structures, since getEmployeeName accepts Employee pointers.",
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https://abdulfatir.com/blog/2020/Langevin-Monte-Carlo/ | [
"Langevin Monte Carlo is a class of Markov Chain Monte Carlo (MCMC) algorithms that generate samples from a probability distribution of interest (denoted by $\\pi$) by simulating the Langevin Equation. The Langevin Equation is given by\n\n$\\lambda\\frac{dX_t}{dt} = -\\frac{\\partial V(x)}{\\partial x} + \\eta(t), \\tag{1}\\label{eq:langevin}$\n\nwhere $X_t$ is the position of a particle in a potential $V(x)$ and $\\eta(t)$ is a noise term. The dynamics in Eq. ($\\ref{eq:langevin}$) is also commonly written as the following Stochastic Differential Equation (SDE)\n\n$\\mathrm{d}X_t = \\underset{\\text{drift term}}{\\underbrace{-\\nabla V(x)\\mathrm{d}t}} + \\underset{\\text{diffusion term}}{\\underbrace{\\sqrt{2}\\mathrm{d}B_t}} \\tag{2}\\label{eq:itodiff}$\n\nwhich represents an Itô diffusion, where $\\mathrm{d}B_t$ denotes the time derivative of standard Brownian motion.\n\nIt can be shown that the SDE in Eq. ($\\ref{eq:itodiff}$) has a unique invariant measure (or simply, a steady-state distribution) that does not change along the trajectory ($X_t$) of the particle. This means that if $X_0$ is distributed according to some probability density function $p_\\infty$, then $X_t$ is also distributed according to $p_\\infty$ for all $t \\geq 0$. If we set the potential $V$ in Eq. ($\\ref{eq:itodiff}$) cleverly such that $p_\\infty = \\pi$, then we can simulate the SDE (Eq. $\\ref{eq:itodiff}$) to generate samples from $\\pi$.\n\n### The steady-state distribution: choosing the potential\n\nThe Fokker-Plank equation is a partial differential equation (PDE) that describes the evolution of a probability distribution over time under the effect of drift forces and random (or noise) forces. The equivalent Fokker-Plank equation for the SDE in Eq. ($\\ref{eq:itodiff}$) is given by\n\n$\\frac{\\partial p(x,t)}{\\partial t} = \\frac{\\partial}{\\partial x}\\left[\\frac{\\partial V(x)}{\\partial x}p(x,t)\\right] + \\frac{\\partial^2p(x,t)}{\\partial x^2}. \\tag{3}\\label{eq:fpe}$\n\nThe steady-state solution of the Fokker-Plank equation is given by $\\frac{\\partial p(x,t)}{\\partial t} = 0$. If $p_\\infty$ is the steady-state distribution, we have\n\n$\\frac{\\partial p(x,t)}{\\partial t} = \\frac{\\partial}{\\partial x}\\left[\\frac{\\partial V(x)}{\\partial x}p_\\infty(x) + \\frac{\\partial p_\\infty(x)}{\\partial x}\\right] = \\frac{\\partial}{\\partial x}J(x) = 0, \\tag{4}\\label{eq:steadystate}$\n\nwhere $J(x)$ denotes the probability “flux”. Eq. ($\\ref{eq:steadystate}$) implies that $J(x)$ must be a constant; however, $p_\\infty(x)$ and $\\frac{\\partial p_\\infty(x)}{\\partial x}$ must also satisfy certain boundary conditions. Specifically, the boundary condition that $J(x) = 0$ at infinity must be satisfied. Since $J(x) = 0$ at infinity and $J(x)$ is a constant, it must be equal to 0 everywhere. This leaves us with\n\n$J(x) = \\frac{\\partial V(x)}{\\partial x}p_\\infty(x) + \\frac{\\partial p_\\infty(x)}{\\partial x} = 0, \\tag{5}\\label{eq:zeroflux}$\n\nwhich has the solution\n\n$p_\\infty(x) \\propto \\exp(-V(x)). \\tag{6}\\label{eq:gibbs}$\n\nEq. ($\\ref{eq:gibbs}$) represents a Gibbs distribution. This means that we can sample from energy-based models (EBMs) of the form $\\pi(x) = \\frac{\\exp[-E(x)]}{Z}$, by setting $V(x) = E(x)$ in Eq. ($\\ref{eq:itodiff}$). We can also write the distribution $\\pi(x)$ as $\\exp[\\log\\pi(x)]$, which means that we can set $V(x) = -\\log\\pi(x)$. It must be noted that we do not really need to know the normalization constant $Z$ for this to work because Eq. ($\\ref{eq:itodiff}$) requires $\\nabla\\log\\pi(x)$ and $\\nabla Z = 0$ since $Z$ is a constant.\n\n### Simulating the SDE\n\nHaving derived the form of the potential $V(x)$, we are now interested in simulating the following SDE to sample from its steady state distribution, i.e., $\\pi(x)$,\n\n$\\mathrm{d}X_t = \\nabla \\log\\pi(x)\\mathrm{d}t + \\sqrt{2}\\mathrm{d}B_t. \\tag{7}\\label{eq:finalsde}$\n\nThe SDE can be discretized using a numerical method such as the Euler-Maruyama method. The Euler-Maruyama approximation of Eq. ($\\ref{eq:finalsde}$) can be written as\n\n$X_{t + \\tau} - X_{t} = \\tau\\nabla \\log\\pi(x) + \\sqrt{2}(B_{t+\\tau}-B_{t}), \\tag{8}\\label{eq:eulerapprox1}$\n\nwhere $\\tau$ is the step-size and $(B_{t+\\tau}-B_{t}) \\sim \\mathcal{N}(0,\\tau)$. This allows us to write Eq. ($\\ref{eq:eulerapprox1}$) as\n\n$X_{t + \\tau} = X_{t} + \\tau\\nabla \\log\\pi(x) + \\sqrt{2\\tau}\\xi, \\tag{9}\\label{eq:eulerapprox2}$\n\nwhere $\\xi \\sim \\mathcal{N}(0,1)$. The time-step $\\tau$ can also be changed over time.\n\nEq. ($\\ref{eq:eulerapprox2}$) gives us a method to sample from a probability distribution $\\pi(x)$ by setting an initial seed $X_0$ and simulating the dynamics which, after a burn-in phase, will generate samples from $\\pi(x)$. This algorithm is known as the Unadjusted Langevin Algorithm (ULA) which requires $\\nabla \\log\\pi(x)$ to be $L$-Lipschitz for stability.\n\nThe ULA always accepts the new sample proposed by Eq. ($\\ref{eq:eulerapprox2}$). Metropolis-adjusted Langevin Algorithm (MALA), on the other hand, uses the Metropolis-Hastings algorithm to accept or reject the proposed sample. Since $\\xi \\sim \\mathcal{N}(0,1)$, $X_{t + \\tau} \\sim \\mathcal{N}(X_{t} + \\tau\\nabla \\log\\pi(x),2\\tau)$ in Eq. ($\\ref{eq:eulerapprox2}$). This means that the proposal distribution is given by\n\n$q(x'|x) \\propto \\exp\\left(-\\frac{\\|x'-x-\\tau\\nabla \\log\\pi(x)\\|^2}{2\\cdot2\\tau}\\right).$\n\nThe sample ($\\tilde{X}_{k+1}$) proposed by Eq. ($\\ref{eq:eulerapprox2}$) is accepted with the following acceptance probability\n\n$\\alpha := \\min\\left(1, \\frac{\\pi(\\tilde{X}_{k+1})q(X_k|\\tilde{X}_{k+1})}{\\pi(X_{k})q(\\tilde{X}_{k+1}|X_k)}\\right).$\n\n### Visualizing Langevin Monte Carlo Sampling\n\nI set out to visualize these MCMC algorithms using matplotlib.animation to see how the distribution evolves over time. Unfortunately, writing to an MP4 file using matplotlib.animation is painfully slow and I could not find a simple way to speed it up. To solve this issue, I wrote a shell script to parallelize the generation of the chunks of the video and then combined them into one long video. The following video shows how samples are generated using MALA from a heart-shaped density given by\n\n$\\pi(\\mathbf{x}=\\begin{bmatrix}x_1 & x_2\\end{bmatrix}^\\top) \\propto \\exp\\left(-\\frac{0.8x_1^2 + \\left(x_2-\\sqrt{x_1^2}\\right)^2}{2^2}\\right).$\n\nThe code used to generate the video above can be found here.\n\n Working with the Langevin and Fokker-Planck equations (notes).\n Chapter 4, Stochastic Processes and Applications, Grigorios A. Pavliotis (book).\n Wikipedia articles (a, b, c, d)."
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https://whatisconvert.com/152-ounces-in-pounds | [
"# What is 152 Ounces in Pounds?\n\n## Convert 152 Ounces to Pounds\n\nTo calculate 152 Ounces to the corresponding value in Pounds, multiply the quantity in Ounces by 0.0625 (conversion factor). In this case we should multiply 152 Ounces by 0.0625 to get the equivalent result in Pounds:\n\n152 Ounces x 0.0625 = 9.5 Pounds\n\n152 Ounces is equivalent to 9.5 Pounds.\n\n## How to convert from Ounces to Pounds\n\nThe conversion factor from Ounces to Pounds is 0.0625. To find out how many Ounces in Pounds, multiply by the conversion factor or use the Mass converter above. One hundred fifty-two Ounces is equivalent to nine point five Pounds.",
null,
"## Definition of Ounce\n\nThe ounce (abbreviation: oz) is a unit of mass with several definitions, the most popularly used being equal to approximately 28 grams. The size of an ounce varies between systems. Today, the most commonly used ounces are the international avoirdupois ounce (equal to 28.3495231 grams) and the international troy ounce (equal to 31.1034768 grams).\n\n## Definition of Pound\n\nThe pound or pound-mass (abbreviations: lb, lbm, lbm, ℔) is a unit of mass with several definitions. Nowadays, the most common is the international avoirdupois pound which is legally defined as exactly 0.45359237 kilograms. A pound is equal to 16 ounces.\n\n### Using the Ounces to Pounds converter you can get answers to questions like the following:\n\n• How many Pounds are in 152 Ounces?\n• 152 Ounces is equal to how many Pounds?\n• How to convert 152 Ounces to Pounds?\n• How many is 152 Ounces in Pounds?\n• What is 152 Ounces in Pounds?\n• How much is 152 Ounces in Pounds?\n• How many lb are in 152 oz?\n• 152 oz is equal to how many lb?\n• How to convert 152 oz to lb?\n• How many is 152 oz in lb?\n• What is 152 oz in lb?\n• How much is 152 oz in lb?"
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https://discourse.mc-stan.org/t/possible-to-fit-this-time-series-model-in-brms/18168 | [
"# Possible to fit this time series model in brms?\n\nElection forecasting has, in recent years, come to rely on models like those described in Jackman (2005):\n\ny_{i} \\sim \\mathrm{Normal}(\\mu_{i}, \\sigma_{i}) \\\\ \\mu_{i} = \\alpha_{t[i]} +\\delta_{j[i]} \\\\ \\alpha_{t} \\sim \\mathrm{Normal}(\\alpha_{t-1}, \\omega), t = 2, ..., T\n\nHere, poll y is modelled as the function of some grand mean \\alpha that varies over each day in the data and a set of house effects, \\delta, due to persistent biases present in each polling company’s methods.\n\nImportantly, \\alpha_{t} is itself modelled as a function of \\alpha_{t-1}. I know that brms can handle autoregressive elements, but can it do so for anything other than the dependent variable?\n\nIt’s worth noting also that often there are missing days in the data too.\n\nHola,\n\nWell for those kind of models you can use atsar package. Still have some struggle with the documentation and performance but is a good start. If you look a bit in the repository you find the stan code , and tou can play a bit with it to get your model fit\n\nHope that helps you :)\n\n1 Like"
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https://codingtube.tech/2021/09/17/java/java-array/ | [
"# Array in java\n\nAn array in java is a data a structure used to process a collection of data that is all of the same type\n\nOther definition about Array in java\n\nAn array is a flexible structure for storing a sequence of values all of the same type.\n\nA structure that holds multiple values of the same type.\n\n• An array behaves behaves like a numbered numbered list of variables variables with a uniform naming mechanism\n• It has a part that does not change: the name of the array\n• It has a part that can change: an integer in square brackets\n\nExample\n\nscore , score , score , score , score , score, score, score, score, score\n\n## Creating and Accessing Arrays\n\n• An array that beheaves like this collection of variables, all of type double, can be created using one statement as follows\n\ndouble[] [ ]; score = new double;\n\n• Or using two statements:\n\ndouble[] score;\nscore = new double;\n\n• The individual variables that together make up the array are called indexed variables\n• They can also be called subscripted variables or elements of the array\n• The number in square brackets is called an index or subscript\n• In Java, indices must be numbered starting with 0, and nothing else\n\nscore , score, score , score , score , score, score, score, score, score\n\n• The number of indexed variables in an array is called the length or size of the array\n• When an array is created, the length of the array is given in square brackets after the array type\n• The indexed variables are then numbered starting with 0, and ending with the integer that is one less than the length of the array\n\nscore, score, score, score, score;\n\n• A variable may be used in place of the integer\n\nExample\n\ndouble[] score = new double;\n\n## Declaring and Creating an Array\n\n• An array is declared and created in almost the same way that objects are declared and created\n\nBaseType[] ArrayName = new BaseType[size];\n\n• The size may be given as an expression that evaluates to a nonnegative integer, for example, an int variable\n\nchar[] line = new char;\nPerson[] specimen = new Person;\n\n## Array Traversal in java\n\nProcessing each array element sequentially from the first to the last\n\n```for (int i = 0; i < <array>.length; i++) {\n<do something with array [i]>;\n}\n```\n\nA Complete Array Program in java\n\n``` import java.util.*;\npublic class Temperature1 {\npublic static void main(String[] args) {\nScanner console = new Scanner(System.in);\nSystem.out.print(\"How many days' temperatures? \");\nint numDays = console.nextInt();\nint sum = 0;\nfor (int i = 1; i <= numDays; i++) {\nSystem.out.print(\"Day \" + i + \"'s high temp: \");\nint next = console.nextInt();\nsum += next;\n}\ndouble average = (double) sum / numDays;\nSystem.out.println();\nSystem.out.println(\"Average = \" + average);\n}\n}\n\n```\n\n## Buffer Overruns\n\nOne of the earliest and still most common sources of computer security problems is a buffer overrun (also known as a buffer overflow). A buffer overrun is similar to an array index out of bounds exception. It occurs when a program writes data beyond the bounds of the buffer set aside for that data.\n\nBuffer overruns are often written as array code. You might wonder how such a malicious program could be written if the computer checks the bounds when you access an array. The answer is that older programming languages like C and C++ do not check bounds when you access an array. By the time Java was designed in the early 1990s, the danger of buffer overruns was clear and the designers of the language decided to include array-bounds checking so that Java would be more secure. Microsoft included similar bounds checking when it designed the language C# in the late 1990s\n\n## The For-Each Loop in java Array\n\nJava 5 introduced a new loop construct that simplifies certain array loops. It is known as the enhanced for loop, or the for-each loop. It can be used whenever you find yourself wanting to examine each value in an array.\n\nThe foreach loop, or enhanced for statement, as Java calls it, is used to enumerate the values in a collection of values\n\n```int sum= 0;\nfor (int v : b) {\nsum= sum + v;\n}\n```\n\nExample\n\n```public class ArrayExample {\npublic static void main(String[] args) {\nint[] numbers = {2, 4, 6, 8, 10};\nfor (int n: numbers) {\nSystem.out.println(n);\n}\n}\n}\n```\n\noutput\n\n2\n4\n6\n8\n10\n\n## Iterate over Java String Array using For-Each\n\n```public class ArrayExample {\npublic static void main(String[] args) {\nString names[] = {\"apple\", \"banana\", \"cherry\", \"mango\"};\nfor(String name: names) {\nSystem.out.println(\"Hello \" + name + \"!\");\n}\n}\n}\n\n```\n\nOutput\n\nHello apple!\nHello banana!\nHello cherry!\nHello mango!\n\n## Limitations of Arrays\n\nsome general limitations of arrays:\n\n• You can’t change the size of an array in the middle of program execution\n• You can’t compare arrays for equality using a simple == test. Remember that arrays are objects, so if you ask whether one array is == to another array, you are asking whether they are the same object, not whether they store the same values\n• You can’t print an array using a simple print or println statement. You will get odd output when you do so\n\nThe first limitation is more difficult to overcome. Because an array is allocated as a contiguous block of memory, it is not easy to make it larger. To make an array bigger, you’d have to construct a new array that is larger than the old one and copy values from the old to the new array. Java provides a class called ArrayList that does this growing operation automatically. It also provides methods for inserting values in and deleting values from the middle of a list\n\nMultidimensional Arrays\n\nArrays of more than one dimension are called multidimensional arrays\n\nRectangular Two-Dimensional Arrays\n\nThe most common use of a multidimensional array is a two-dimensional array of a certain width and height. For example, suppose that on three separate days you took a series of five temperature readings. You can define a two-dimensional array that has three rows and five columns as follows\n\ndouble[][] temps = new double;\n\nExample\n\n```public class Main {\npublic static void main(String[] args) {\nint[][] myNumbers = { {1, 2, 3, 4}, {5, 6, 7} };\nfor (int i = 0; i < myNumbers.length; ++i) {\nfor(int j = 0; j < myNumbers[i].length; ++j) {\nSystem.out.println(myNumbers[i][j]);\n}\n}\n}\n}\n\n```\n\nOutput\n\n1\n2\n3\n4\n5\n6\n7"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8188023,"math_prob":0.9036463,"size":6294,"snap":"2023-40-2023-50","text_gpt3_token_len":1513,"char_repetition_ratio":0.12384738,"word_repetition_ratio":0.047402006,"special_character_ratio":0.25961232,"punctuation_ratio":0.12123678,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9794123,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-05T21:52:30Z\",\"WARC-Record-ID\":\"<urn:uuid:e0e9709b-0f22-4d87-88cc-0e1d24aa57c8>\",\"Content-Length\":\"141248\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e952dcfe-a48b-4c85-a52f-a773a48226ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:a60ef9d2-77a2-4b16-844a-52c74481d923>\",\"WARC-IP-Address\":\"172.67.195.44\",\"WARC-Target-URI\":\"https://codingtube.tech/2021/09/17/java/java-array/\",\"WARC-Payload-Digest\":\"sha1:RSBI2U6TRAYFSIZ267U3EBMAHGSXC5W6\",\"WARC-Block-Digest\":\"sha1:PCVKXFOJAEFCRBATO7AZW6AJWSERN6IJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100568.68_warc_CC-MAIN-20231205204654-20231205234654-00104.warc.gz\"}"} |
http://serdis.dis.ulpgc.es/~lalvarez/research/demos/ColorStereoFlow/index.html | [
"3-D Geometry reconstruction using a color image\nstereo pair and partial differential equations\nby\nL.Alvarez and J.Sánchez\n\nIn this work, we present a new model for a dense disparity estimation and the 3-D geometry reconstruction using a\ncolor image stereo pair. First, we present a brief introduction to the 3-D Geometry of a camera system. Next, we\npropose a new model for the disparity estimation based on an energie functional. We look for the local minima of\nthe energy using the associated Euler-Lagrange partial differential equation. This model is a generalization to color\nimage of the model developed by L.Alvarez, R.Deriche, J.Sánchez and J.Weickert in the paper Dense Disparity Map\nEstimation Respecting Image Discontinuities with some changes in the strategy to avoid the irrelevant local minima.\nWe present some numerical experiences of \\$3-D\\$ reconstruction, using this method for some real stereo pairs.\n\nFor more details see the paper 3-D Geometry reconstruction using a color image stereo pair and partial\ndifferential equations\n\nNext we present several experiments to illustrate the method",
null,
"Figure 1. We present an stereo pair of the face of Javier Sánchez, one of author\nof this work, and 4 views of the 3-D reconstruction using the proposed method. If\nyou click on the image you will see a movie to ilustrate the 3-D reconstruction.",
null,
"Figure 2. We present an stereo pair of the face of Julio Esclarín, a member of the Comp.\nScience Depart. at the University of Las Palmas, and 4 views of the 3-D reconstruction\nusing the proposed method. If you click on the image you will see a movie to\nilustrate the 3-D reconstruction.",
null,
"Figure 3. We present an stereo pair of an outdoor scene of the laboratory Instituto Universitario de\nCiencias y Tecnologías Cibernéticas de la Universidad de Las Palmas de G.C., and 4 views of\nthe 3-D reconstruction using the proposed method. If you click on the image you will see a movie to\nilustrate the 3-D reconstruction.\n\nlalvarez@dis.ulpgc.es / updated March 17, 2000"
] | [
null,
"http://serdis.dis.ulpgc.es/~lalvarez/research/demos/ColorStereoFlow/JavierColor.JPG",
null,
"http://serdis.dis.ulpgc.es/~lalvarez/research/demos/ColorStereoFlow/JulioColor.JPG",
null,
"http://serdis.dis.ulpgc.es/~lalvarez/research/demos/ColorStereoFlow/InstitutoColor.JPG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73967755,"math_prob":0.74841255,"size":1595,"snap":"2019-13-2019-22","text_gpt3_token_len":382,"char_repetition_ratio":0.13073538,"word_repetition_ratio":0.20930232,"special_character_ratio":0.19623825,"punctuation_ratio":0.09508197,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9530805,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-21T20:19:14Z\",\"WARC-Record-ID\":\"<urn:uuid:3eeb6d60-7f92-4e45-8237-65304e03d2a4>\",\"Content-Length\":\"3962\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef92232e-aa30-4c17-bb18-7eb7649fa871>\",\"WARC-Concurrent-To\":\"<urn:uuid:b265eb5a-ebdd-42c6-b526-ba1d2853dcd3>\",\"WARC-IP-Address\":\"193.145.147.54\",\"WARC-Target-URI\":\"http://serdis.dis.ulpgc.es/~lalvarez/research/demos/ColorStereoFlow/index.html\",\"WARC-Payload-Digest\":\"sha1:TIIDNCKKA7Q4BSTXQ24AJ4KZ5LILTQKI\",\"WARC-Block-Digest\":\"sha1:GPRSNMIEI4SDA6VNT6ELJKKXDJ4BY7VI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912202572.29_warc_CC-MAIN-20190321193403-20190321215403-00475.warc.gz\"}"} |
https://tutorialspoint.dev/algorithm/mathematical-algorithms/modulo-2-binary-division | [
"# Cyclic Redundancy Check and Modulo-2 Division\n\nCRC or Cyclic Redundancy Check is a method of detecting accidental changes/errors in communication channel.\n\nCRC uses Generator Polynomial which is available on both sender and receiver side. An example generator polynomial is of the form like x3 + x + 1. This generator polynomial represents key 1011. Another example is x2 + 1 that represents key 101.\n\n```n : Number of bits in data to be sent\nfrom sender side.\nk : Number of bits in the key obtained\nfrom generator polynomial. ```\n\nSender Side (Generation of Encoded Data from Data and Generator Polynomial (or Key)):\n\n1. The binary data is first augmented by adding k-1 zeros in the end of the data\n2. Use modulo-2 binary division to divide binary data by the key and store remainder of division.\n3. Append the remainder at the end of the data to form the encoded data and send the same\n4. .\n\nReceiver Side (Check if there are errors introduced in transmission)\nPerform modulo-2 division again and if remainder is 0, then there are no errors.\n\nIn this article we will focus only on finding the remainder i.e. check word and the code word.\n\nModulo 2 Division:\nThe process of modulo-2 binary division is the same as the familiar division process we use for decimal numbers. Just that instead of subtraction, we use XOR here.\n\n• In each step, a copy of the divisor (or data) is XORed with the k bits of the dividend (or key).\n• The result of the XOR operation (remainder) is (n-1) bits, which is used for the next step after 1 extra bit is pulled down to make it n bits long.\n• When there are no bits left to pull down, we have a result. The (n-1)-bit remainder which is appended at the sender side.\n\nIllustration:\n\nExample 1 (No error in transmission):\n\n```Data word to be sent - 100100\nKey - 1101 [ Or generator polynomial x3 + x2 + 1]\n\nSender Side:",
null,
"Therefore, the remainder is 001 and hence the encoded\ndata sent is 100100001.",
null,
"Therefore, the remainder is all zeros. Hence, the\ndata received has no error. ```\n\nExample 2: (Error in transmission)\n\n```Data word to be sent - 100100\nKey - 1101\n\nSender Side:",
null,
"Therefore, the remainder is 001 and hence the\ncode word sent is 100100001.\n\nLet there be error in transmission media",
null,
"Since the remainder is not all zeroes, the error\nis detected at the receiver side.\n\nImplementation\nBelow is Python implementation for generating code word from given binary data and key.\n\n`# Returns XOR of 'a' and 'b' `\n`# (both of same length) `\n`def` `xor(a, b): `\n` `\n` ``# initialize result `\n` ``result ``=` `[] `\n` `\n` ``# Traverse all bits, if bits are `\n` ``# same, then XOR is 0, else 1 `\n` ``for` `i ``in` `range``(``1``, ``len``(b)): `\n` ``if` `a[i] ``=``=` `b[i]: `\n` ``result.append(``'0'``) `\n` ``else``: `\n` ``result.append(``'1'``) `\n` `\n` ``return` `''.join(result) `\n` `\n` `\n`# Performs Modulo-2 division `\n`def` `mod2div(divident, divisor): `\n` `\n` ``# Number of bits to be XORed at a time. `\n` ``pick ``=` `len``(divisor) `\n` `\n` ``# Slicing the divident to appropriate `\n` ``# length for particular step `\n` ``tmp ``=` `divident[``0` `: pick] `\n` `\n` ``while` `pick < ``len``(divident): `\n` `\n` ``if` `tmp[``0``] ``=``=` `'1'``: `\n` `\n` ``# replace the divident by the result `\n` ``# of XOR and pull 1 bit down `\n` ``tmp ``=` `xor(divisor, tmp) ``+` `divident[pick] `\n` `\n` ``else``: ``# If leftmost bit is '0' `\n` ``# If the leftmost bit of the dividend (or the `\n` ``# part used in each step) is 0, the step cannot `\n` ``# use the regular divisor; we need to use an `\n` ``# all-0s divisor. `\n` ``tmp ``=` `xor(``'0'``*``pick, tmp) ``+` `divident[pick] `\n` `\n` ``# increment pick to move further `\n` ``pick ``+``=` `1`\n` `\n` ``# For the last n bits, we have to carry it out `\n` ``# normally as increased value of pick will cause `\n` ``# Index Out of Bounds. `\n` ``if` `tmp[``0``] ``=``=` `'1'``: `\n` ``tmp ``=` `xor(divisor, tmp) `\n` ``else``: `\n` ``tmp ``=` `xor(``'0'``*``pick, tmp) `\n` `\n` ``checkword ``=` `tmp `\n` ``return` `checkword `\n` `\n`# Function used at the sender side to encode `\n`# data by appending remainder of modular divison `\n`# at the end of data. `\n`def` `encodeData(data, key): `\n` `\n` ``l_key ``=` `len``(key) `\n` `\n` ``# Appends n-1 zeroes at end of data `\n` ``appended_data ``=` `data ``+` `'0'``*``(l_key``-``1``) `\n` ``remainder ``=` `mod2div(appended_data, key) `\n` `\n` ``# Append remainder in the original data `\n` ``codeword ``=` `data ``+` `remainder `\n` ``print``(``\"Remainder : \"``, remainder) `\n` ``print``(``\"Encoded Data (Data + Remainder) : \"``, `\n` ``codeword) `\n` `\n`# Driver code `\n`data ``=` `\"100100\"`\n`key ``=` `\"1101\"`\n`encodeData(data, key) `\n\nOutput:\nRemainder : 001\nEncoded Data (Data + Remainder) : 100100001\n\nNote that CRC is mainly designed and used to protect against common of errors on communication channels and NOT suitable protection against intentional alteration of data (See reasons here)\nReferences:\nhttps://en.wikipedia.org/wiki/Cyclic_redundancy_check\n\n```"
] | [
null,
"https://tutorialspoint.dev/image/rational1.jpg",
null,
"https://tutorialspoint.dev/image/rational2.jpg",
null,
"https://tutorialspoint.dev/image/rational1.jpg",
null,
"https://tutorialspoint.dev/image/rational4.jpg",
null
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http://songcharts.tk/calculate-average-daily-range-forex-606653.html | [
"Calculate average daily range forex",
null,
"Average True Range Spreadsheet & Tutorial - Invest Excel\n\nAre you looking for an ADR indicator for MT4? Today, I'm going to show you what is, in my opinion, the best average daily range indicator for MetaTrader 4.",
null,
"Multi Range Calculator Metatrader 4 Indicator\n\nThe Average True Range The average range is also a good indicator to use for filtering which consists of news, opinions, daily and weekly forex analysis,",
null,
"5 Day Average Daily Range — Indicator by IamTheDudeMan\n\nHow to Calculate the Daily Range for Forex. we’re going to teach you all about the daily range, how to calculate the daily range, (Average Directional Movement)",
null,
"What Is Forex? - BabyPips.com\n\nDiscover how average true range is used as a stop-loss indicator in trading strategies, and learn how to calculate it in Excel. This volatility indicator is used to",
null,
"Average Daily Range Pro Calculator Metatrader 4 Indicator",
null,
"Average True Range (ATR) | Measure Market - FOREX.com\n\nHow to calculate ADR (Average Daily Range) This will give you the average daily range. Forex Average Pips: 250-300/month. Email: i_ipun[at]yahoo.com. Edward Revy,",
null,
"How to Use ATR in a Forex Strategy - Forex Trading News\n\nAverage Daily Range Pro Calculator. Scam or not - Click here to find out.",
null,
"Average True Range (ATR) - TradingView Wiki\n\nIf you are trading any kind of intraday forex system, then it's always a good idea to be fully aware of the average daily range of the pair(s) you are trading.",
null,
"Currencies Average daily range - Forex Market Hours\n\nAverage Daily Range Calculator . You can install and use the ADR Calculator with the MetaTrader 4.0 trading platform provided by any Forex broker.",
null,
"How to Trade Using the Average True Range Indicator\n\nHome Forex Blog Forex Tips – Be Wary of Average Daily Ranges. Forex Tips – Be Wary of Average Daily Ranges. where should you calculate the ADR range,",
null,
"How to Calculate the Daily Range for Forex | Investoo.com\n\n2019-03-12 · You can calculate the average high-low range on a piece of paper, The next day, the price opens gap up, but the daily range is the same \\$2.",
null,
"Forex Volatility - Mataf\n\nHow do we calculate the Average Daily Range? By taking the difference of the high and low of the trading day, you calculate the daily range. But you must be",
null,
"32# ADR (Average Day Range) Strategy Trading System. How to calculate ADR (Average Daily Range) Forex Indicator Average Range.",
null,
"CompassFX | Average Daily Range Calculator - ADR\n\nThe average true range indicator is an In order to calculate the average true range, you take the average of each true range value over a fixed period of",
null,
"Daily Range Calculator Metatrader 4 Indicator - fxtsp.com\n\nForex Volatility. Add our content on The volatility calculated on this page is called Average true range let's calculate the volatility of the Euro dollar",
null,
"The Average Daily Range Of The Major Forex Currency Pairs\n\n2018-07-04 · Forex, also known as Position Size Calculator; Check out the graph of the average daily trading volume for the forex market,",
null,
"How to Use the Average Daily Range When Trading Forex\n\nThe indicator I am referring to is called Average Daily Range How to Calculate ADR. The calculation of the daily range of a Average Daily Range of a Forex",
null,
"Foreign Exchange Volatility | Currency Movement | Forex\n\n0 Average Daily Range Pro Calculator Metatrader 4 Indicator. This indicator measures the average daily range (volatility) for the following time periods: 5 days",
null,
"I am working with daily data and 21 trading days corresponds to approximately one month. In column G we will calculate the AVERAGE of column F (true range).",
null,
"How to Analyze the Average Trading Range - dummies\n\n2008-09-23 · 100 pip movements and daily average movement. uses the average daily range of the past 10 days to calculate entry/exit and SL (look on the Forex Systems",
null,
"Calculating Average True Range (ATR) in Excel - Macroption\n\n2013-10-28 · How to Use ATR in a Forex Strategy. Average True Range Talking Points: Forex traders can use ATR to gauge market Get daily market analysis from our",
null,
"Measuring Volatility with Average True Range - DailyFX\n\nThis is an outside day that would use Method 1 to calculate and the first 14-day ATR is the average of the daily TR The Average True Range indicator can",
null,
"Average Daily Range 2010 - Currency Forecasts\n\nLearn about Average True Range and how to identify it on a forex chart.",
null,
"",
null,
""
] | [
null,
"https://c.mql5.com/31/36/average-daily-range-screen-9179.png",
null,
"http://www.billlions.com/wp-content/uploads/2016/03/ADR-bad-low-adr-200x200.png",
null,
"http://www.forex-tsd.com/attachments/metatrader-4/138268d1338419678-average-day-range-indicator-adr-daily-adr.png",
null,
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null,
"http://forexwinners.ru/wp-content/uploads/2014/01/Average-Daily-Range-Pro-Calculator-660x330.jpg",
null,
"https://media.giphy.com/media/l0HefOZVMXvIHC5ry/giphy.gif",
null,
"http://investexcel.net/wp-content/uploads/2012/04/Average-True-Range-Excel.png",
null,
"https://www.panduantrading.com/assets/uploads/2017/09/Daily-Average-Range-cpy-04.jpg",
null,
"http://www.billlions.com/wp-content/uploads/2016/03/ADR-bad-low-adr.png",
null,
"http://forex-indicators.net/files/indicators/truerange.png",
null,
"http://www.dolphintrader.com/wp-content/uploads/2014/01/average-daily-range-pro-indicator-mt4.png",
null,
"http://www.fxtsp.com/wp-content/uploads/2010/02/100-daily-range_calculator-metatrader-4-indicator.gif",
null,
"http://www.compassfx.com/images/adr_pro_2_lg.jpg",
null,
"http://forexwinners.net/wp-content/uploads/2013/05/Range.jpg",
null,
"http://mt4indicators.com/wp-content/uploads/2015/03/ADR_TheGreedyPig-640x439.png",
null,
"https://c.mql5.com/3/145/brentcrud-d1-fx-choice-limited.png",
null,
"https://i.ytimg.com/vi/15mvligjgoo/maxresdefault.jpg",
null,
"https://forex-station.com/download/file.php",
null,
"http://www.forex-indikatoren.com/wp-content/uploads/2017/04/average-daily-range.png",
null,
"http://www.billlions.com/wp-content/uploads/2016/03/ADR-good-high-adr-500x500.png",
null,
"https://c.mql5.com/forextsd/forum/63/hroutput01.jpg",
null,
"http://www.investoo.com/wp-content/uploads/2015/05/Screen-Shot-2015-05-11-at-14.37.23.png",
null,
"https://c.mql5.com/forextsd/forum/63/mp_mtf_pj9100_001com.jpg",
null,
"http://forexzz.com/wp-content/uploads/2015/11/adr-calc.png",
null,
"https://c.mql5.com/forextsd/forum/150/12-03-2015_6-34-59_am.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8285965,"math_prob":0.91635424,"size":4918,"snap":"2019-26-2019-30","text_gpt3_token_len":1086,"char_repetition_ratio":0.25600326,"word_repetition_ratio":0.057692308,"special_character_ratio":0.21207808,"punctuation_ratio":0.0706278,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97916216,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,null,null,3,null,3,null,null,null,3,null,3,null,3,null,3,null,7,null,4,null,3,null,null,null,3,null,3,null,null,null,null,null,null,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T20:00:24Z\",\"WARC-Record-ID\":\"<urn:uuid:d28bf4ea-56a1-409e-a6f0-0fb9bb7b21a1>\",\"Content-Length\":\"18618\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a5b576b8-edc7-4dca-8422-787f179e2a99>\",\"WARC-Concurrent-To\":\"<urn:uuid:008efc97-4790-4aa5-a659-0a153068b03d>\",\"WARC-IP-Address\":\"104.24.118.121\",\"WARC-Target-URI\":\"http://songcharts.tk/calculate-average-daily-range-forex-606653.html\",\"WARC-Payload-Digest\":\"sha1:3BDW5QHQA5GS66YEB2Q5KPPAFHYK3NX4\",\"WARC-Block-Digest\":\"sha1:LR3RG5OYKZDTORX6WFI2E5BNLXYIU7BT\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000545.97_warc_CC-MAIN-20190626194744-20190626220744-00368.warc.gz\"}"} |
https://huaibei.home.focus.cn/gonglue/5e84e6b4a91042e1.html | [
"|\n\n# 安装厨房推拉门 节省厨房空间\n\n安装厨房推拉门不但可以节省家居空间,还能起到一定的装饰效果,因此,厨房推拉门现在越来越受到人们的欢迎。那么,如何安装厨房推拉门呢?首先要根据房屋结构、空间大小和使用功能来设计。这可不是一件简单的事哦,下面就让我们来带您详细的了解一下。",
null,
"安装厨房推拉门\n\n1、上滑道安装\n\n上滑道安装时必须按门洞宽度和开启方向的要求将其同门梁或雨蓬梁进行固定一般以门洞宽度的中心为基准两分固定时要注意以下要点滑道同雨蓬梁或门梁连接处的左右标高一致滑道相对于门洞中心尺寸一致。滑道中心线与彩板外墙或砖墙外边的尺寸一致且其尺寸为门扇厚度的一半外加20mm的间隙如为彩板外墙尚需另加20mm的包边宽度。\n\n2、门扇安装\n\n门扇及滑轮系统由公司在车间内加工成型。上滑道安装结束后随即将滑轮成对放入滑道滑槽内然后用人工或其它吊装工具将门扇竖直运到其下方并将门扇上焊接的螺杆套入滑轮上的螺栓孔内用双螺帽临时紧固。同理将另一门扇先行挂起。\n\n3、门扇调整\n\n门扇均已挂好后将二个门扇均推到门洞中心位置并将其贴紧按室外地坪标高的要求将门扇下扁钢调到地坪上表面0~10mm同时调整两门扇连接处的缝隙到最小且侧边垂直于地面。标高调整用松紧上面的螺帽进行。门扇全部调整结束后将滑轮上的止推销用开口销固定在滑轮上。\n\n4、门上限位器\n\n安装门扇全部调整完后将门扇分别全部关闭和推开根据其位置将上滑道底部或内部用角钢上限位器焊接在距离滑轮边10mm的地方使门扇的开启区域限制在其有效范围内。角钢与滑轮接触处要求设置不小于20mm的硬质橡胶垫作为缓冲。",
null,
"5、导饼和门下限位器安装\n\n导饼和门下限位器安装要求同土建配合进行其要点是导饼中心线定位可按门扇下扁钢的实际位置进行定位导饼露出地面为10~15mm间距为500mm。下限位器定位时以将门扇推到距外边10~20mm位置外埋入砼中或砼施工完后用膨胀螺栓固定。\n\n6、门五金配件安装\n\n厨房推拉门的五金配件常用的有门拉手、锁扣可在门安装调整全部结束且无误后进行按各工程所先用的五金配件的安装要求分别进行应注意所的配件的平面位置和标高要求均在同一位置以影响美观。\n\n`声明:本文由入驻焦点开放平台的作者撰写,除焦点官方账号外,观点仅代表作者本人,不代表焦点立场错误信息举报电话: 400-099-0099,邮箱:jubao@vip.sohu.com,或点此进行意见反馈,或点此进行举报投诉。`",
null,
"A B C D E F G H J K L M N P Q R S T W X Y Z\nA - B - C - D - E\n• A\n• 鞍山\n• 安庆\n• 安阳\n• 安顺\n• 安康\n• 澳门\n• B\n• 北京\n• 保定\n• 包头\n• 巴彦淖尔\n• 本溪\n• 蚌埠\n• 亳州\n• 滨州\n• 北海\n• 百色\n• 巴中\n• 毕节\n• 保山\n• 宝鸡\n• 白银\n• 巴州\n• C\n• 承德\n• 沧州\n• 长治\n• 赤峰\n• 朝阳\n• 长春\n• 常州\n• 滁州\n• 池州\n• 长沙\n• 常德\n• 郴州\n• 潮州\n• 崇左\n• 重庆\n• 成都\n• 楚雄\n• 昌都\n• 慈溪\n• 常熟\n• D\n• 大同\n• 大连\n• 丹东\n• 大庆\n• 东营\n• 德州\n• 东莞\n• 德阳\n• 达州\n• 大理\n• 德宏\n• 定西\n• 儋州\n• 东平\n• E\n• 鄂尔多斯\n• 鄂州\n• 恩施\nF - G - H - I - J\n• F\n• 抚顺\n• 阜新\n• 阜阳\n• 福州\n• 抚州\n• 佛山\n• 防城港\n• G\n• 赣州\n• 广州\n• 桂林\n• 贵港\n• 广元\n• 广安\n• 贵阳\n• 固原\n• H\n• 邯郸\n• 衡水\n• 呼和浩特\n• 呼伦贝尔\n• 葫芦岛\n• 哈尔滨\n• 黑河\n• 淮安\n• 杭州\n• 湖州\n• 合肥\n• 淮南\n• 淮北\n• 黄山\n• 菏泽\n• 鹤壁\n• 黄石\n• 黄冈\n• 衡阳\n• 怀化\n• 惠州\n• 河源\n• 贺州\n• 河池\n• 海口\n• 红河\n• 汉中\n• 海东\n• I\n• J\n• 晋中\n• 锦州\n• 吉林\n• 鸡西\n• 佳木斯\n• 嘉兴\n• 金华\n• 景德镇\n• 九江\n• 吉安\n• 济南\n• 济宁\n• 焦作\n• 荆门\n• 荆州\n• 江门\n• 揭阳\n• 金昌\n• 酒泉\n• 嘉峪关\nK - L - M - N - P\n• K\n• 开封\n• 昆明\n• 昆山\n• L\n• 廊坊\n• 临汾\n• 辽阳\n• 连云港\n• 丽水\n• 六安\n• 龙岩\n• 莱芜\n• 临沂\n• 聊城\n• 洛阳\n• 漯河\n• 娄底\n• 柳州\n• 来宾\n• 泸州\n• 乐山\n• 六盘水\n• 丽江\n• 临沧\n• 拉萨\n• 林芝\n• 兰州\n• 陇南\n• M\n• 牡丹江\n• 马鞍山\n• 茂名\n• 梅州\n• 绵阳\n• 眉山\n• N\n• 南京\n• 南通\n• 宁波\n• 南平\n• 宁德\n• 南昌\n• 南阳\n• 南宁\n• 内江\n• 南充\n• P\n• 盘锦\n• 莆田\n• 平顶山\n• 濮阳\n• 攀枝花\n• 普洱\n• 平凉\nQ - R - S - T - W\n• Q\n• 秦皇岛\n• 齐齐哈尔\n• 衢州\n• 泉州\n• 青岛\n• 清远\n• 钦州\n• 黔南\n• 曲靖\n• 庆阳\n• R\n• 日照\n• 日喀则\n• S\n• 石家庄\n• 沈阳\n• 双鸭山\n• 绥化\n• 上海\n• 苏州\n• 宿迁\n• 绍兴\n• 宿州\n• 三明\n• 上饶\n• 三门峡\n• 商丘\n• 十堰\n• 随州\n• 邵阳\n• 韶关\n• 深圳\n• 汕头\n• 汕尾\n• 三亚\n• 三沙\n• 遂宁\n• 山南\n• 商洛\n• 石嘴山\n• T\n• 天津\n• 唐山\n• 太原\n• 通辽\n• 铁岭\n• 泰州\n• 台州\n• 铜陵\n• 泰安\n• 铜仁\n• 铜川\n• 天水\n• 天门\n• W\n• 乌海\n• 乌兰察布\n• 无锡\n• 温州\n• 芜湖\n• 潍坊\n• 威海\n• 武汉\n• 梧州\n• 渭南\n• 武威\n• 吴忠\n• 乌鲁木齐\nX - Y - Z\n• X\n• 邢台\n• 徐州\n• 宣城\n• 厦门\n• 新乡\n• 许昌\n• 信阳\n• 襄阳\n• 孝感\n• 咸宁\n• 湘潭\n• 湘西\n• 西双版纳\n• 西安\n• 咸阳\n• 西宁\n• 仙桃\n• 西昌\n• Y\n• 运城\n• 营口\n• 盐城\n• 扬州\n• 鹰潭\n• 宜春\n• 烟台\n• 宜昌\n• 岳阳\n• 益阳\n• 永州\n• 阳江\n• 云浮\n• 玉林\n• 宜宾\n• 雅安\n• 玉溪\n• 延安\n• 榆林\n• 银川\n• Z\n• 张家口\n• 镇江\n• 舟山\n• 漳州\n• 淄博\n• 枣庄\n• 郑州\n• 周口\n• 驻马店\n• 株洲\n• 张家界\n• 珠海\n• 湛江\n• 肇庆\n• 中山\n• 自贡\n• 资阳\n• 遵义\n• 昭通\n• 张掖\n• 中卫\n\n1室1厅1厨1卫1阳台\n\n1\n2\n3\n4\n5\n\n0\n1\n2\n\n1\n\n1\n\n0\n1\n2\n3",
null,
"",
null,
"",
null,
"报名成功,资料已提交审核",
null,
"A B C D E F G H J K L M N P Q R S T W X Y Z\nA - B - C - D - E\n• A\n• 鞍山\n• 安庆\n• 安阳\n• 安顺\n• 安康\n• 澳门\n• B\n• 北京\n• 保定\n• 包头\n• 巴彦淖尔\n• 本溪\n• 蚌埠\n• 亳州\n• 滨州\n• 北海\n• 百色\n• 巴中\n• 毕节\n• 保山\n• 宝鸡\n• 白银\n• 巴州\n• C\n• 承德\n• 沧州\n• 长治\n• 赤峰\n• 朝阳\n• 长春\n• 常州\n• 滁州\n• 池州\n• 长沙\n• 常德\n• 郴州\n• 潮州\n• 崇左\n• 重庆\n• 成都\n• 楚雄\n• 昌都\n• 慈溪\n• 常熟\n• D\n• 大同\n• 大连\n• 丹东\n• 大庆\n• 东营\n• 德州\n• 东莞\n• 德阳\n• 达州\n• 大理\n• 德宏\n• 定西\n• 儋州\n• 东平\n• E\n• 鄂尔多斯\n• 鄂州\n• 恩施\nF - G - H - I - J\n• F\n• 抚顺\n• 阜新\n• 阜阳\n• 福州\n• 抚州\n• 佛山\n• 防城港\n• G\n• 赣州\n• 广州\n• 桂林\n• 贵港\n• 广元\n• 广安\n• 贵阳\n• 固原\n• H\n• 邯郸\n• 衡水\n• 呼和浩特\n• 呼伦贝尔\n• 葫芦岛\n• 哈尔滨\n• 黑河\n• 淮安\n• 杭州\n• 湖州\n• 合肥\n• 淮南\n• 淮北\n• 黄山\n• 菏泽\n• 鹤壁\n• 黄石\n• 黄冈\n• 衡阳\n• 怀化\n• 惠州\n• 河源\n• 贺州\n• 河池\n• 海口\n• 红河\n• 汉中\n• 海东\n• I\n• J\n• 晋中\n• 锦州\n• 吉林\n• 鸡西\n• 佳木斯\n• 嘉兴\n• 金华\n• 景德镇\n• 九江\n• 吉安\n• 济南\n• 济宁\n• 焦作\n• 荆门\n• 荆州\n• 江门\n• 揭阳\n• 金昌\n• 酒泉\n• 嘉峪关\nK - L - M - N - P\n• K\n• 开封\n• 昆明\n• 昆山\n• L\n• 廊坊\n• 临汾\n• 辽阳\n• 连云港\n• 丽水\n• 六安\n• 龙岩\n• 莱芜\n• 临沂\n• 聊城\n• 洛阳\n• 漯河\n• 娄底\n• 柳州\n• 来宾\n• 泸州\n• 乐山\n• 六盘水\n• 丽江\n• 临沧\n• 拉萨\n• 林芝\n• 兰州\n• 陇南\n• M\n• 牡丹江\n• 马鞍山\n• 茂名\n• 梅州\n• 绵阳\n• 眉山\n• N\n• 南京\n• 南通\n• 宁波\n• 南平\n• 宁德\n• 南昌\n• 南阳\n• 南宁\n• 内江\n• 南充\n• P\n• 盘锦\n• 莆田\n• 平顶山\n• 濮阳\n• 攀枝花\n• 普洱\n• 平凉\nQ - R - S - T - W\n• Q\n• 秦皇岛\n• 齐齐哈尔\n• 衢州\n• 泉州\n• 青岛\n• 清远\n• 钦州\n• 黔南\n• 曲靖\n• 庆阳\n• R\n• 日照\n• 日喀则\n• S\n• 石家庄\n• 沈阳\n• 双鸭山\n• 绥化\n• 上海\n• 苏州\n• 宿迁\n• 绍兴\n• 宿州\n• 三明\n• 上饶\n• 三门峡\n• 商丘\n• 十堰\n• 随州\n• 邵阳\n• 韶关\n• 深圳\n• 汕头\n• 汕尾\n• 三亚\n• 三沙\n• 遂宁\n• 山南\n• 商洛\n• 石嘴山\n• T\n• 天津\n• 唐山\n• 太原\n• 通辽\n• 铁岭\n• 泰州\n• 台州\n• 铜陵\n• 泰安\n• 铜仁\n• 铜川\n• 天水\n• 天门\n• W\n• 乌海\n• 乌兰察布\n• 无锡\n• 温州\n• 芜湖\n• 潍坊\n• 威海\n• 武汉\n• 梧州\n• 渭南\n• 武威\n• 吴忠\n• 乌鲁木齐\nX - Y - Z\n• X\n• 邢台\n• 徐州\n• 宣城\n• 厦门\n• 新乡\n• 许昌\n• 信阳\n• 襄阳\n• 孝感\n• 咸宁\n• 湘潭\n• 湘西\n• 西双版纳\n• 西安\n• 咸阳\n• 西宁\n• 仙桃\n• 西昌\n• Y\n• 运城\n• 营口\n• 盐城\n• 扬州\n• 鹰潭\n• 宜春\n• 烟台\n• 宜昌\n• 岳阳\n• 益阳\n• 永州\n• 阳江\n• 云浮\n• 玉林\n• 宜宾\n• 雅安\n• 玉溪\n• 延安\n• 榆林\n• 银川\n• Z\n• 张家口\n• 镇江\n• 舟山\n• 漳州\n• 淄博\n• 枣庄\n• 郑州\n• 周口\n• 驻马店\n• 株洲\n• 张家界\n• 珠海\n• 湛江\n• 肇庆\n• 中山\n• 自贡\n• 资阳\n• 遵义\n• 昭通\n• 张掖\n• 中卫",
null,
"",
null,
"• 手机",
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"• 分享\n• 设计\n免费设计\n• 计算器\n装修计算器\n• 入驻\n合作入驻\n• 联系\n联系我们\n• 置顶\n返回顶部"
] | [
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"https://t1.focus-img.cn/sh740wsh/zx/duplication/5f12c893-6579-4423-ba0f-ea94e20e5d70.JPEG",
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"https://t1.focus-img.cn/sh740wsh/zx/duplication/648aa9ac-bfc2-4a4e-bec2-2a643c530a06.JPEG",
null,
"data:image/png;base64,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",
null,
"data:image/png;base64,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",
null,
"https://huaibei.home.focus.cn/gonglue/5e84e6b4a91042e1.html",
null,
"https://t1.focus-res.cn/front-pc/module/loupan-baoming/images/yes.png",
null,
"https://t1.focus-res.cn/front-pc/module/loupan-baoming/images/qrcode.png",
null,
"data:image/png;base64,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",
null,
"https://t.focus-res.cn/home-front/pc/img/qrcode.d7cfc15.png",
null,
"data:image/png;base64,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",
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.92394084,"math_prob":0.45310748,"size":870,"snap":"2020-24-2020-29","text_gpt3_token_len":1029,"char_repetition_ratio":0.045034643,"word_repetition_ratio":0.0,"special_character_ratio":0.11034483,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999534,"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,1,null,1,null,null,null,null,null,1,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-29T04:58:14Z\",\"WARC-Record-ID\":\"<urn:uuid:d6a112d3-9be7-4db2-9e42-5f4887521f9a>\",\"Content-Length\":\"144452\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f981bdff-d582-44ff-b2cb-c64c4702e42c>\",\"WARC-Concurrent-To\":\"<urn:uuid:3d13e39c-8e2d-4921-ab6c-7aa4e488aa11>\",\"WARC-IP-Address\":\"42.63.21.227\",\"WARC-Target-URI\":\"https://huaibei.home.focus.cn/gonglue/5e84e6b4a91042e1.html\",\"WARC-Payload-Digest\":\"sha1:EANU4SAAWLC4RELX6WQDBNN2EEG62LCU\",\"WARC-Block-Digest\":\"sha1:AK73IRAD77U4WCYQ3VPIAWRYE7LOOQKR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347401260.16_warc_CC-MAIN-20200529023731-20200529053731-00169.warc.gz\"}"} |
https://www.geeksforgeeks.org/spacing-between-boxplots-in-ggplot2-in-r/?ref=rp | [
"# Spacing between boxplots in ggplot2 in R\n\n• Last Updated : 18 Jul, 2021\n\nIn this article, we are going to see how to add space between the boxplots in ggplot2 using R programming language.\n\nDataset in use: Crop_recommendation\n\n### Method 1: Using width between boxplot\n\nHere we will use width attributes to define space between the boxplot. In this the value is passed to the attribute.\n\nSyntax: geom_boxplot(width)\n\nProgram:\n\n## R\n\n `library``(ggplot2)`` ` `# loading data set and storing it in ds variable``df <- ``read.csv``(``\"Crop_recommendation.csv\"``, header = ``TRUE``)`` ` `# create a boxplot by using geom_boxplot() function``# of ggplot2 package``plot = ``ggplot``(data=df,`` ``mapping=``aes``(`` ``x=label, y=temperature))+``geom_boxplot``(width = 0.5)``plot `\n\nOutput:",
null,
"### Method 2: Using position_dodge\n\nHere we will use position_dodge to define the vertical position of a geom while adjusting the horizontal position. position_dodge() requires the grouping variable to be specified position.\n\nSyntax:\n\ngeom_boxplot( position = position_dodge(width))\n\nProgram:\n\n## Python3\n\n `library(ggplot2)`` ` `# loading data set and storing it in ds variable``df <``-` `read.csv(``\"Crop_recommendation.csv\"``, header ``=` `TRUE)`` ` `# create a boxplot by using geom_boxplot() function``# of ggplot2 package``plot``=` `ggplot(data``=``df,`` ``mapping``=``aes(x``=``label, `` ``y``=``temperature))``+``geom_boxplot(width``=``0.1``, position ``=` `position_dodge(width``=``0.5``))``plot`\n\nOutput:",
null,
"My Personal Notes arrow_drop_up"
] | [
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210701115856/Capture-660x574.PNG",
null,
"https://media.geeksforgeeks.org/wp-content/uploads/20210701115856/Capture-660x574.PNG",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.55585873,"math_prob":0.9469237,"size":1367,"snap":"2022-40-2023-06","text_gpt3_token_len":359,"char_repetition_ratio":0.14966984,"word_repetition_ratio":0.16346154,"special_character_ratio":0.24725677,"punctuation_ratio":0.12719299,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98836446,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,8,null,8,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-04T01:38:33Z\",\"WARC-Record-ID\":\"<urn:uuid:6fcee1b3-8048-4ebb-9fdb-1d0d663001f2>\",\"Content-Length\":\"130244\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d646942c-b111-4d9c-9dcc-2c0c15485603>\",\"WARC-Concurrent-To\":\"<urn:uuid:faf74eae-b92b-4b96-a4e8-975cc5a2207d>\",\"WARC-IP-Address\":\"104.97.85.137\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/spacing-between-boxplots-in-ggplot2-in-r/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:75ZL7CIDIHGY4MS7CWQT2IIZRIHY4WQA\",\"WARC-Block-Digest\":\"sha1:RCRLCNYHHOXTRFS4OQBHQVMKRI7LZ3CV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337446.8_warc_CC-MAIN-20221003231906-20221004021906-00314.warc.gz\"}"} |
https://www.wilsonware.com/electronics/series_parallel_circuits.htm | [
"Series/Parallel Circuits Click one of the buttons above to move to that topic.\n Your browser does not support Java Applets Circuits consisting of both series and parallel elements are series-parallel circuits. RESISTANCE IN SERIES-PARALLEL CIRCUITS Equivalent circuits are used to find series-parallel resistance. REDRAWING CIRCUITS FOR CLARITY Circuits can be redrawn to make current flow understandable. Steps for Redrawing the Circuit for Clarity Trace the current paths in the circuit. Label the nodes (junctions) in the circuit. Recognize points that are the same voltage Visualize a rearrangement Redraw into a simpler form REDRAWING A SERIES-PARALLEL CIRCUIT Start with the negative terminal of the voltage source and follow the current flow through the components. SERIES-PARALLEL CIRCUIT ANALYSIS Reduce all parallel circuits to series-equivalent resistances. Combine all the branches containing more than one resistance in series into a single resistance. Redraw the resulting equivalent circuit and determine the equivalent resistance of the whole circuit. When the series-parallel circuit has been simplified, solve for the total current, voltage, or power, as required. To obtain a complete solution for the series-parallel circuit, find the individual component values by using the values obtained in the equivalent circuit and applying them to the original circuit. Solve for individual voltage, current, or power, as required. WHEATSTONE BRIDGE A series parallel circuit which is useful for rectifiers and measuring instruments is a bridge. A wheatstone bridge is a special measuring instrument. Balancing the bridge One of the resistances can be adjusted until there is zero current flow between the junction at the middle of each parallel branch. TROUBLESHOOTING SERIES-PARALLEL CIRCUITS Troubleshooting often involves isolating the circuit into smaller parallel or series elements."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81210154,"math_prob":0.9598569,"size":1043,"snap":"2019-51-2020-05","text_gpt3_token_len":233,"char_repetition_ratio":0.14051972,"word_repetition_ratio":0.0,"special_character_ratio":0.1581975,"punctuation_ratio":0.055555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9726689,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T21:36:15Z\",\"WARC-Record-ID\":\"<urn:uuid:6da7a5c8-8197-488b-a25c-9bf5b29afd1f>\",\"Content-Length\":\"8595\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a3d160e2-183d-4ec4-ba4a-0ef0e0d69a6d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9264f813-70e6-429a-9450-9bf33376a29d>\",\"WARC-IP-Address\":\"173.236.159.173\",\"WARC-Target-URI\":\"https://www.wilsonware.com/electronics/series_parallel_circuits.htm\",\"WARC-Payload-Digest\":\"sha1:J3CTCTVM6UR2A6VSZVFGPZ24OSVNYGGJ\",\"WARC-Block-Digest\":\"sha1:7HCIXJIJZ66PE3Q5IMG3PM47VBQ6UM3F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540514893.41_warc_CC-MAIN-20191208202454-20191208230454-00517.warc.gz\"}"} |
https://www.nationaltrustcollections.org.uk/results?ObjectType=saucer+dish | [
"## You searched , Object Type: “saucer dish”\n\nShow me:\nand\n\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 8 items Explore\n• Explore\n• Explore\n• Explore\n• 12 items Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 32 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• 4 items Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• 4 items\n• 8 items Explore\n• 3 items Explore\n• 5 items Explore\n• Explore\n• 8 items Explore\n• Explore\n• 5 items Explore\n• Explore\n• 8 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 6 items Explore\n• 2 items Explore\n• 3 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 7 items Explore\n• Explore\n• 1 items Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 9 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 11 items Explore\n• Explore\n• Explore\n• 3 items Explore\n• Explore\n• 1 items Explore\n• Explore\n• 2 items Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 3 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 1 items\n• Explore\n• 1 items Explore\n• 1 items Explore\n• Explore\n• Explore\n• 4 items\n• Explore\n• 4 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• 2 items Explore\n• Explore\n• Explore\n• Explore\n• 5 items Explore\n• 2 items Explore\n• Explore\n• 2 items Explore\n• Explore\n• 13 items Explore\n• Explore\n• Explore\n• Explore\n• 1 items Explore\n• Explore\n• Explore\n• Explore\n• Explore\n• Explore"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.59503293,"math_prob":0.980798,"size":598,"snap":"2022-27-2022-33","text_gpt3_token_len":181,"char_repetition_ratio":0.13636364,"word_repetition_ratio":0.0,"special_character_ratio":0.17892976,"punctuation_ratio":0.23333333,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98579216,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T16:56:22Z\",\"WARC-Record-ID\":\"<urn:uuid:7ead4c3e-9240-402f-8859-b3cb47f94748>\",\"Content-Length\":\"224610\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:caecfe72-4daa-49e3-b3b1-2c7710649941>\",\"WARC-Concurrent-To\":\"<urn:uuid:89d95be6-d01b-49f6-af79-0635459ee89a>\",\"WARC-IP-Address\":\"13.93.104.16\",\"WARC-Target-URI\":\"https://www.nationaltrustcollections.org.uk/results?ObjectType=saucer+dish\",\"WARC-Payload-Digest\":\"sha1:ECC5LUZKAUEAJJ65WNJBRGCYSWPY5A6G\",\"WARC-Block-Digest\":\"sha1:XW3GKGT64IXB43OI6E3AGEY23R4LAR6A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103850139.45_warc_CC-MAIN-20220630153307-20220630183307-00760.warc.gz\"}"} |
https://www.tutorialspoint.com/antenna_theory/antenna_theory_short_dipole.htm | [
"# Antenna Theory - Short Dipole\n\nA short dipole is a simple wire antenna. One end of it is open-circuited and the other end is fed with AC source. This dipole got its name because of its length.\n\n### Frequency range\n\nThe range of frequency in which short dipole operates is around 3KHz to 30MHz. This is mostly used in low frequency receivers.\n\n## Construction & Working of Short Dipole\n\nThe Short dipole is the dipole antenna having the length of its wire shorter than the wavelength. A voltage source is connected at one end while a dipole shape is made, i.e., the lines are terminated at the other end.",
null,
"The circuit diagram of a short dipole with length L is shown. The actual size of the antenna does not matter. The wire that leads to the antenna must be less than one-tenth of the wavelength. That is\n\n$$L < \\frac{\\lambda}{10}$$\n\nWhere\n\n• L is the length of the wire of the short dipole.\n\n• λ is the wavelength.\n\nAnother type of short dipole is infinitesimal dipole, whose length is far less than its wave length. Its constructiion is similar to it, but uses a capacitor plate.\n\n## Infinitesimal Dipole\n\nA dipole whose length is far less than wavelength is infitesimal dipole. This antenna is actually impractical. Here, the length of the dipole is less than even fiftith part of the wavelength.\n\nThe length of the dipole, Δl << λ. Where, λ is the wavelength.\n\n$$\\Delta l = \\frac{\\lambda}{50}$$\n\nHence, this is the infinitely small dipole, as the name implies.\n\nAs the length of these dipoles is very small, the current flow in the wire will be dI. These wires are generally used with capacitor plates on both sides, where low mutual coupling is needed. Because of the capacitor plates, we can say that uniform distribution of current is present. Hence the current is not zero here.\n\nThe capacitor plates can be simply conductors or the wire equivalents. The fields radiated by the radial currents tend to cancel each other in the far field so that the far fields of the capacitor plate antenna can be approximated by the infinitesimal dipole.\n\nThe radiation pattern of a short dipole and infinitesimal dipole is similar to a half wave dipole. If the dipole is vertical, the pattern will be circular. The radiation pattern is in the shape of “figure of eight” pattern, when viewed in two-dimensional pattern.\n\nThe following figure shows the radiation pattern of a short dipole antenna, which is in omni-directional pattern.",
null,
"The following are the advantages of short dipole antenna −\n\n• Ease of construction, due to small size\n\n• Power dissipation efficiency is higher\n\nThe following are the disadvantages of short dipole antenna −\n\n• High resistive losses\n• High power dissipation\n• Low Signal-to-noise ratio"
] | [
null,
"https://www.tutorialspoint.com/antenna_theory/images/short_dipole.jpg",
null,
"https://www.tutorialspoint.com/antenna_theory/images/omni_directional_pattern.jpg",
null
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https://panotbook.com/theory-of-consumer-behaviour/ | [
"# Theory Of Consumer Behaviour NCERT Textbook PDF\n\nTheory Of Consumer Behaviour NCERT Class 12 Economics Chapter 2 Solutions‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 12 Economics Chapter 2 Exercise Solution’ using the download button.\n\n## Theory Of Consumer Behaviour NCERT Textbook With Solutions Book PDF Free Download\n\n### Chapter 2: Theory Of Consumer Behaviour\n\nA consumer usually decides his demand for a commodity on the basis of utility (or satisfaction) that he derives from it.\n\nWhat is utility? The utility of a commodity is its want-satisfying capacity. The more the need of a commodity or the stronger the desire to have it, the greater is the utility derived from the commodity.\n\nThe utility is subjective. Different individuals can get different levels of utility from the same commodity.\n\nFor example, someone who likes chocolates will get much higher utility from chocolate than someone who is not so fond of chocolates, Also, a utility that one individual gets from the commodity can change with change in place and time.\n\nFor example, utility from the use of a room heater will depend upon whether the individual is in Ladakh or Chennai (place) or whether it is summer or winter (time).\n\nCardinal utility analysis assumes that level of utility can be expressed in numbers. For example, we can measure the utility derived from a shirt and say, this shirt gives me 50 units of utility.\n\nBefore discussing further, it will be useful to have a look at two important measures of utility.\nMeasures of Utility Total Utility: Total utility of a fixed quantity of a commodity (TU) is the total satisfaction derived from consuming the given amount of some commodity x.\n\nMore of commodity x provides more satisfaction to the consumer. TU depends on the quantity of the commodity consumed. Therefore, TUn refers to total utility derived from consuming n units of a commodity x.\n\nMarginal Utility: Marginal utility (MU) is the change in total utility due to consumption of one additional unit of a commodity.\n\nFor example, suppose 4 bananas give us 28 units of total utility and 5 bananas give us 30 units of total utility.\n\nClearly, consumption of the 5th banana has caused the total utility to increase by 2 units (30 units minus 28 units). Therefore, the marginal utility of the 5th banana is 2 units.\n\nCardinal utility analysis is simple to understand, but suffers from a major drawback in the form of quantification of utility in numbers. In real life, we never express utility in the form of numbers.\n\nAt the most, we can rank various alternative combinations in terms of having more or less utility. In other words, the consumer does not measure utility in numbers, though she often ranks\nvarious consumption bundles. This forms the starting point of this topic – Ordinal Utility Analysis.\n\nIt may be mentioned that the law of Diminishing Marginal Rate of Substitution causes an indifference curve to be convex to the origin.\n\nThis is the most common shape of an indifference curve. But in case of goods being perfect substitutes4, the marginal rate of substitution does not diminish. It remains the same.\n\n### NCERT Solutions Class 12 Economics Chapter 2 Theory Of Consumer Behaviour\n\n1. What do you mean by the budget set of a consumer?\n\nA budget set of a consumer is a bundle of two or more goods in certain quantities and combinations that is desirable and affordable for the consumer based on their price range. A budget set is also called as an opportunity set.\n\n2. What is budget line?\n\nA budget line is a graphical representation of a consumer’s constraints when buying a combination of two or more products with a given budget.\n\nA budget line will shift whenever there is a change in the prices, preferences or income. It is also called a consumption possibility line. Here, it is assumed that the customer spends the entire income on the bundle of products.\n\n3. Explain why the budget line is downward sloping.\n\nWith a limited income, the customer can increase the consumption of one good only by decreasing the consumption of the other good. This is why a budget line is downward sloping.\n\nNCERT Class 12 Economics Textbook Chapter 2 Theory Of Consumer Behaviour With Answer PDF Free Download"
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https://onlinecalculator.guru/calculus/derivative-calculator/ | [
"# Derivative Calculator\n\nFree Derivative Calculator helps you to find the differentiation of the given function, with steps shown. Make your calculations easier and faster by giving the input function and tap on the calculate button to get the output as derivative of the function in no time.\n\nDerivative Calculator: Struggling to calculate the derivative of a function? Then, here is the solution for your problem. Differentiation can be calculated easily by using the differentiation rules. You can understand and learn about the entire concept from here and also get the simple steps to solve the questions. Our handy calculator does all the required calculations and displays the exact output along with all the steps of calculations in fraction of seconds.\n\n## Find Derivative of Function\n\nHere are the easy steps that should be followed while finding the derivative. The steps are along the lines:\n\n• Take any function to calculate the derivative.\n• Have a look at the basic formulas or rules that are useful to solve the differentiation.\n• Apply those rules and solve the function easily.\n\n### Important Derivative Rules\n\nBelow mentioned are the some important derivate rules that are used while solving the derivative of any function. Have a look at them.\n\n1. d/dx (a) = 0 (where a is a constant)\n2. d/dx (x) = 1\n3. d/dx (xn) = nxn-1 [power rule]\n4. d/dx (ex) = ex [exponent rule]\n5. d/dx (log x) = 1/x\n6. d/dx (ax) = ax logx\n7. d/dx (f+g) = d/dx (f) + d/dx (g)\n8. d/dx (f-g) = d/dx (f) - d/dx (g)\n9. d/dx (ay) = a dy/dx\n10. (f.g)' = f'g + g'f [product rule]\n11. (f/g)' = (f'g - g'f) / (g2) [quotient rule]\n12. d/dx (f(g(x) = f'(g(x))g'(x) [chain rule]\n13. For Trigonometric Functions:\n• d/dx sin(x) = cos(x)\n• d/dx cos(x) = -sin(x)\n• d/dx tan(x) = sec2(x)\n• d/dx cot(x) = -cosec2(x)\n\nExample\n\nQuestion: Solve derivative of 6 / √z3 + 1 / (8z4) - 1 / (3z10)\n\nSolution:\n\nGiven function is 6 / √z3 + 1 / (8z4) - 1 / (3z10)\n\nd/ dz ( 6 / √z3 + 1 / (8z4) - 1 / (3z10)) = ?\n\n= d/ dz ( 6 z-3/2 + 1/8 (z-4) - 1/3 (z-10)\n\n= d/ dz (6 z-3/2) + d/ dz (1/8 (z-4)) - d/ dz (1/3 (z-10))\n\nApply the power rule i.e d/ dx xn = nxn-1\n\n= 6 (-3/2) z-3/2 - 1 + 1/8 (-4) z-4-1 -1/3(-10) z-10-1\n\n= -9z-5/2 - 1/2 z-5 + 10/3 z-11\n\nd/ dz ( 6 / √z3 + 1 / (8z4) - 1 / (3z10)) = -9z-5/2 - 1/2 z-5 + 10/3 z-11\n\nOnlinecalculator.guru has different algebra concepts calculators which are free to use & easy to understand and you can make any of your calculations easy & quick.",
null,
"### FAQs on Derivative Calculator\n\n1. What is derivative formula?\n\nDerivative is the fundamental tool of calculus. The derivative of a function of a real variable measures the sensitivity to change a quantity which is determined by another quantity. Derivative formula is\n\nf1(x) = lim△x→0 ([f(x) + △x) - f(x)] / △x\n\n2. What is the purpose of derivatives?\n\nDerivatives are the financial contracts whose value is linked to the value of an underlying asset.\n\n3. How do you find the derivative on a calculator?\n\nEnter the input function in the calculator and hit on the calculate button which is provided next to that input box to get the output instantly.\n\n4. Find the derivative of f(x). Where f(x) = 6x3 - 9x + 4?\n\nf'(x) = 6.3 x3-1 -9(1) + 0\n\n= 18 x2 -9"
] | [
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"https://onlinecalculator.guru/static/images/Calculus/Derivative-Calculator.jpeg",
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https://programmer.group/5e1f0182b5552.html | [
"# Common features of vue. Form, calculated properties, custom, listener, filter, lifecycle\n\nKeywords: Vue Attribute\n\n## Form operation\n\n### Form scope modifier\n\n1. Number: converts the form value to a number\n```data:{\ninputmsg:'123'\n}\n<input type=text v-model.number=\"inputmsg\">//The value of this form is output as data\n```\n1. trim: remove spaces at the beginning and end of the form\n```data:{\ninputmsg:' 123 '\n}\n<input type=text v-model.trim=\"inputmsg\">//The value of this form is' 123 ', without spaces\n```\n1. lazy: switch input event to change event (trigger only when cursor is lost)\n```data:{\ninputmsg:123\n}\n<input type=text v-model.lazy=\"inputmsg\">//When the form is input, the data in the corresponding data will not change immediately, but will change when the mouse leaves\n```\n\n### input single line text box\n\n```data:{\ninputmsg:'Single line text'\n}\n<input type=text v-model=\"inputmsg\">//This binds the input values in both directions\n```\n\n### textarea multiline text box\n\n```data:{\ntextareamsg:'Single line text'\n}\n<textarea type=text v-model=\"textarea\"></textarea>//In this way, the value of textarea is bound in both directions\n```\n\n### select drop-down list\n\n1. single election\n\n```data:{\nselect:1//Variable corresponds to the item selected by default, single variable definition\n}\n<select v-model=\"select\">//Bind data in both directions on select\n<option value=\"0\">0</option>//Each option is distinguished by value, which is also used to associate the corresponding binding variables.\n<option value=\"1\">1</option>\n<option value=\"2\">2</option>\n<option value=\"3\">3</option>\n</select>\n```\n\n1. more choices\n\n```data:{\nselect:['0']//If select is multiple, you need to define an array. The value in the array corresponds to the value item in the selected option\n}\n<select v-model=\"select\" multipel='true'>//Bind data in two directions on select. The multipel property sets select to multi select\n<option value=\"0\">0</option>//Each option is distinguished by value, which is also used to associate the corresponding binding variables.\n<option value=\"1\">1</option>\n<option value=\"2\">2</option>\n<option value=\"3\">3</option>\n</select>\n```\n\n```data:{\nmsg: 1 //The variable corresponds to the item selected by default,\n}\n//male<input type=\"radio\" name=\"sex\" value=\"0\" v-model=\"msg\">//The radio boxes of the same group are bound to the same variable, and value is used as the distinction\n```\n\n### checkbox check box\n\n```data:{\narr:[\"2\",'1'] //The data bound to the check box is an array. The values in the array correspond to the option to select the corresponding check box.\n}\n//eat<input type=\"checkbox\" name=\"checkbox\" value=\"1\" v-model=\"arr\">//The radio boxes of the same group are bound to the same variable, and value is used as the distinction\n//drink<input type=\"checkbox\" name=\"checkbox\" value=\"2\" v-model=\"arr\">\n//play<input type=\"checkbox\" name=\"checkbox\" value=\"3\" v-model=\"arr\">\n//Happy<input type=\"checkbox\" name=\"checkbox\" value=\"4\" v-model=\"arr\">\n```\n\n## directive custom instruction\n\n1. Global custom instruction\nDefine the custom instruction outside the vue instance, which can be used by all components.\n```Vue.directive('focus',{//Focus is the name of the user-defined instruction. You need to add V - to the reference. Such as: v-focus\ninserted:function(el,binding){//inserted is the hook function of the user-defined instruction, el is the element calling the instruction, and it is the native html element. binding parameter, which can be left blank, as follows\n//inserted:function(el){//inserted is the hook function of the custom instruction, el is the element of the calling instruction, and is the native html element\nel.focus();//Focus of attention\nconsole.log(binding)//Parameters can be output\n}\n})\n//Quote\ndata:{\nmsg:123\n}\n<input type='text' v-focus=\"msg\">\n<input type='text' v-focus >\n```\n1. Local custom instruction\n**The local custom instruction is defined in the directives parameter in the vue instance; it can only be used in its own components; * * as follows:\n```const vm = Vue({\ndata:{}\ndirectives:{//Custom instructions. Others are the same as global instructions.\nfocus:{\ninserted:function(el,binding){//inserted is the hook function of the user-defined instruction, el is the element calling the instruction, and it is the native html element. binding parameter, which can be left blank, as follows\n//inserted:function(el){//inserted is the hook function of the custom instruction, el is the element of the calling instruction, and is the native html element\nel.focus();//Focus of attention\nconsole.log(binding)//Parameters can be output\n}\n}\n}\n})\n//Same usage as global instruction\n```\n\n## Calculated calculation properties\n\nThe calculated attributes can depend on the data, and the data does not need to be re rendered. Based on data rendering in data\nAdd the computed parameter to the vue instance;\n\n```data:{msg:123}\ncomputed: {\ncomputed1: function(){//computed1 is the name of the calculation attribute, and the reference can directly refer to the name. Just like variables in data.\nreturn this.msg.split('').reverse().join('')//The calculation property needs to return the calculation result by return.\n}\n}\n//Quote\n<div>{{computed1}}</div>//The result is 321\n```\n```\tDifference between computed and methods methods methods.\ncomputed: there will be a cache. When the data in the data remains unchanged, it will not be re rendered.\nmethods: there is no cache, and the function will be called again every time.\n```\n\n## watch listener\n\n```\t**Handle some asynchronous and expensive events**\nAdd the watch parameter to the vue instance\n```\n```data:{\nfirstname:123,\nlastname:453,\nfullname:'123 453'\n},\nwatch:{\nfirstname: function(val){//Firstname is the data in the data to be monitored, val is the latest value of firstname\nthis.fullname = val+' '+this.lastname;//When the monitored variable changes, perform the corresponding operation.\n},\nlastname: function(val){//lastname is the data in the data to be monitored, val is the latest value of lastname\nthis.fullname = this.firstname+' '+val;//When the monitored variable changes, perform the corresponding operation.\n},\n}\n```\n\n## Filter filter\n\nData formatting, such as capitalization of the initial of a string.\n\n### Definition\n\nGlobal definition. It is defined outside the vue instance, similar to the custom instruction, and can be used by all components. as\n\n```//The filter can carry parameters, starting from the second bit, such as\n//**Vue.filter('filtername',function(value,arg1, arg2){//arg is the parameter passed by the filter**\nVue.filter('filtername',function(value){//filtername is the name of the filter, and value is the data that calls the filter.\nreturn value.charAt(0).toUpperCase()+value.slice(1)\n//The first letter in charAt(0) data, toUpperCase to uppercase, toLowerCase to lowercase, slice from the second bit\n})\n```\n\nLocal filter. Add the filters parameter in the Vue instance to define the filter in the parameter, which is only valid in this component. as\n\n```data:{msy:asd},\nfilters:{\n//The filter can carry parameters, starting from the second bit, such as\n//**filtername: function(value,arg1, arg2){//arg is the parameter passed by the filter**\nfiltername: function(value){\nreturn value.charAt(0).toUpperCase()+value.slice(1)\n}\n}\n```\n\n### usage\n\n```//It can be directly used in interpolation expressions, such as:\n//Usage of parameter passing, < div > {MSG | filtername (arg1,arg2)}} < / div >\n<div>{{msg | filtername}} </div>\n//The filter supports multiple functions, such as:\n<div>{{msg | filtername1 | filtername2}} </div>\n//It can also be used in property binding, such as:\n//The usage of parameter passing, < div: id = \"MSG | filtername (arg1, arg2)\" >\n<div :id=\"msg | filtername\"> </div>\n```\n\n## life cycle\n\nThese methods are the same as method and data.\n\n### Main stage\n\n```**Mount (initialize related properties)**\n```\n1. beforeCreate / / called after instance initialization, before data observation and event configuration\n2. Created / / called immediately after the instance is created.\n3. beforeMout / / called before the mount starts\n4. mouted //el is replaced by the newly created vm.\\$el. After mounting it to an instance, the hook is called. The function indicates that the initialization is complete. usage\n```mounted: function(){\n//operation\n}\n```\n\nUpdate (change operation of element or component)\n6. beforeUpdate / / called when updating data, which occurs before virtual DOM patching\n7. updated / / if the virtual DOM is re rendered and patched due to data update, the hook will be called after that\nDestroy (destroy related properties)\n\n1. Call before beforeDestroy / / instance destruction\n2. After destroyed / / instance destruction, call",
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"Published 8 original articles, praised 0, visited 83\n\nPosted by OuchMedia on Wed, 15 Jan 2020 04:11:00 -0800"
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https://ask.learncbse.in/t/write-an-equation-that-defines-the-exponential-function-with-base-b-0/63854 | [
"",
null,
"# Write an equation that defines the exponential function with base b > 0\n\na) Write an equation that defines the exponential function with base b > 0.\nb) What is the domain of this function?\nc) If b≠1, what is the range of this function?"
] | [
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http://forums.wolfram.com/mathgroup/archive/2004/Aug/msg00367.html | [
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"Hypergeometric function\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg50188] Hypergeometric function\n• From: jujio77 at yahoo.com (Scott)\n• Date: Thu, 19 Aug 2004 06:28:24 -0400 (EDT)\n• Sender: owner-wri-mathgroup at wolfram.com\n\n```I have a finite alternating series of hypergeometric (2F1) functions.\nThese functions have complex parameters. When I sum the series I get\nlarger and larger values. My question is this, does anyone know how\nprecise Mathematica is when calculating a hypergeometric fn\nnumerically ie, how many sig figs are correct?\n\nI have done various transformations on the hypergeometric fn and then\nsummed the series. Each time I arrive at the same result.\n\nThanks for any replies,\nScott\n\n```\n\n• Prev by Date: Label of Max[list]\n• Next by Date: GUIKit FileDialog Widget\n• Previous by thread: Re: Label of Max[list]\n• Next by thread: Re: Hypergeometric function"
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https://embed.tutorhub.com/question/algebra--34 | [
"Lesson hub\n\n## Can't find the answer? Try online tutoring",
null,
"We have the UK’s best selection of online tutors, when and for how long you need them.\n\nGetting 1-on-1 support is cheaper than you might think.\n\n### Participating users\n\nWelcome to our free-to-use Q&A hub, where students post questions and get help from other students and tutors.\n\nYou can ask your own question or look at similar Pure Mathematics questions.\n\nHi @luisxxi, thanks for taking your time to help me on this question. But as I've mentioned in my post, I need help on the 3rd part. I was able to manage the first two parts myself.\n\nAny help on this is highly appreciates.\n\nMany thanks!\n\nIn the first part you must transform the equation so that it fits you:\n\n((a+b+c)/a)+((a+b+c)/b)+((a+b+c)/c)>=9\n\n1+(b/a)+(c/a)+(a/b)+1+(c/b)+(a/c)+(b/c)+1>=9\n\nrearranging\n\n(a/b+b/a) + (a/c+c/a) + (b/c+c/b) + 3 >=9\n\nand you must develop the following remarkable product:\n\n(a-b)^2=a^2-2ab+b^2 if a=b then a^2-2ab+b^2=0\n\n=> a^2+b^2=2ab => (a^2+b^2)/ab=2 => a/b+b/a=2\n\nthe important thing here is that you make the assumption a = b and\n\na=c and b=c\n\nWith that assumption you have the problem done",
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https://codereview.stackexchange.com/questions/212825/discrete-event-simulation-for-production-defects/213209 | [
"# Discrete event simulation for production defects\n\nI have one machine, which produces parts. In machine_failure_rate% it produces faulty parts which need to be produced again. Thus, we end up with a simple queuing problem. Can the following code be futher functionalized? I have the feeling, I can get rid of time_parts, but all I have in mind deteriorates the code as I need further lookups in the production_df data frame to look for \"what was produced / what needs to be produced now?\". The following script is running:\n\ninput_rate <- 1/60 # input rate [1/min, 1/input_rate corresponds to interarrival time in min]\nn <- 1000 # number of parts\ndt <- 1 # timestep = time to transfer faulty parts back to production. [min]\n\nmachine_production_rate <- 1/40 # production rate [1/min]\nmachine_failure_rate <- 0.2 # machine failure rate\n\n# Sum all interarrival times\nset.seed(123456)\nt_event <- cumsum(rpois(n, 1/input_rate))\n\n# Create initial list of tasks. Produces parts will be cut off.\ntime_parts <- data.frame(id = c(1:n),\nt = t_event,\nstringsAsFactors = FALSE)\n\n# ========= Functions ==========================================================\ncreate_machine <- function(failure_rate, production_rate) {\nmachine <- list()\nmachine$failure_rate <- failure_rate machine$production_rate <- production_rate\nmachine$is_occupied <- FALSE return(machine); } update_machine <- function(ind_production_df, machine, production_df) { if (machine$is_occupied) {\nif (production_df$po_start[ind_production_df] + 1/machine$production_rate <= t) {\nmachine$is_occupied <- FALSE } } return(machine) } production_summary <- function(production_df, machine, input_rate) { no_of_failures <- sum(production_df$no_failures)\ntotal_production_time <- max(production_df$po_start) + 1/machine$production_rate\nuptime <- (no_of_failures + n)/machine$production_rate print(paste0(\"Estimated machine$failure_rate \",\nround(no_of_failures/(no_of_failures + n), 2),\n\" [theory \", round(machine$failure_rate, 2), \"]\")) print(paste0(\"Up-time \", uptime, \", of total time \", total_production_time, \". Auslastung \", round(uptime/total_production_time, 2), \" [theory \", round(input_rate/machine$production_rate*1/(1 - machine$failure_rate), 2), \"]\")) } # ========= DE simulation ====================================================== machine <- create_machine(machine_failure_rate, machine_production_rate) production_df <- data.frame(id = time_parts$id,\ntime = time_parts$t, production_start = rep(0, nrow(time_parts)), no_failures = rep(0, nrow(time_parts)), stringsAsFactors = FALSE) t <- 0 while (length(time_parts$t) > 0) {\nind_production_df <- which(production_df$id == time_parts$id)\n\nmachine <- update_machine(ind_production_df, machine, production_df)\n\nif (!machine$is_occupied & time_parts$t <= t) {\n# A machine is available and a part needs to be produced\nmachine$is_occupied <- TRUE production_df$po_start[ind_production_df] <- t\nif (runif(1) < machine$failure_rate) { # bad part time_parts$t <- time_parts$t + dt time_parts <- time_parts[sort(time_parts$t, index.return = TRUE)$ix, ] production_df$no_failures[ind_production_df] <-\nproduction_df$no_failures[ind_production_df] + 1 t <- t + min(time_parts$t, dt)\n} else {\n# good part\nif (production_df$po_start[ind_production_df] + 1/machine$production_rate >= t &&\nnrow(time_parts) >= 2) {\ntime_parts <- time_parts[2:(nrow(time_parts)), ]\n} else {\ntime_parts <- time_parts[FALSE, ]\n}\nt <- t + 1/machine$production_rate machine$is_occupied <- FALSE\n}\n} else {\n# machine is occupied or no part needs to be produced\nt <- t + min(time_parts$t, dt) } } # ========= Results ============================================================ production_summary(production_df, machine, input_rate) Backround: I think about a generalisation (more machines, more input-sources, more complex rules how/when/... parts a produces). I fear that I will end up with tons of unreadable and unmaintainable code-lines if I proceed like this. Edit: I think t <- t + min(time_parts$t, dt) is a bug and the correct version is t <- min(time_parts$t, t + dt). It only worked because the time difference dt was always the minimum. In the last case you could speed up using t <- max(time_parts$t, t + dt) as there is nothing to do in the time inbetween.\n\n• Could you explain what \"po_start\" represents - is this the time at which production starts for a given part? – Russ Hyde Feb 5 at 9:22\n• Yes, it means, when the production really starts - due to a queue, this can be delayed in contrast to the original t_event. I also made a tiny edit to the post... – Christoph Feb 5 at 9:48\n• I've been working on this, but can't post my code from work. I can restructure the code and get the same result but I was wondering whether there might be an error. Within your code t <- t + min(time_parts$t, dt). The contents of time_parts are the earliest-time a given part can be made and the current time must increase at each iteration, shouldn't you update current time to max(t + dt, time_parts$t)? – Russ Hyde Feb 6 at 10:02\n• @RussHyde No, I don't think so: if I increase t, I should take the earliest point in time where something can change. This is save. It might be, that sometimes a max would speed up... I am curious about your results :-) – Christoph Feb 6 at 15:39\n\nThis was a pretty difficult challenge - principally because R doesn't have a built-in priority queue data-structure, but also because the priority-queue-like data-frame (time_parts) was wrapped around the results-storing data-frame (production_df) and the main while loop contains code at a few different levels of abstraction.\n\n# Idiomatic R\n\nI did some simple stuff first: pulled all your functions to the start of the script, reformatted some code/comments.\n\nThere was a couple of things I changed for idiomatic reasons:\n\nwhich(production_df$id == time_parts$id)\n# -->\nmatch(time_parts$id, production_df$id)\n\n# time_parts[2:(nrow(time_parts)), ] # and\n# time_parts[FALSE, ] # when time_parts has only one row\n# can both be replaced with\ntime_parts[-1, ]\n# (which is the idiomatic way to drop the first row) so this allowed us to remove an if-else clause\n\n# You don't need to do rep(some_value, n) when you're adding a\n# constant column to a data-frame at construction:\nproduction_df <- data.frame(id = time_parts$id, time = time_parts$t,\nproduction_start = rep(0, nrow(time_parts)),\nno_failures = rep(0, nrow(time_parts)),\nstringsAsFactors = FALSE)\n# -->\nproduction_df <- data.frame(id = time_parts$id, time = time_parts$t,\nproduction_start = 0,\nno_failures = 0,\nstringsAsFactors = FALSE)\n\n# order(...) does the same thing as sort(..., index.return)$ix sort(time_parts$t, index.return = TRUE)$ix # --> order(time_parts$t)\n\n# nrow(x) is more idiomatic than length(x$some_column) while(length(time_parts$t) > 0){\n# -->\nwhile(nrow(time_parts) > 0) {\n# but I subsequently replaced this newer line as well\n\n\n# Explicit data-classes\n\nI converted your create_machine function so that it returns an object of class \"Machine\"; this wasn't really necessary.\n\ncreate_machine <- function(failure_rate, production_rate) {\nstructure(\nlist(\nfailure_rate = failure_rate,\nproduction_rate = production_rate,\nis_occupied = FALSE\n),\nclass = \"Machine\"\n)\n}\n\n\nI added a create_part function that similarly returns a Part object. There was a lot of repeats of 1 / machine$production_rate in your code; I replaced these with a call to part$production_duration. Also I thought your test to see whether a produced part was a failure should be associated with the produced part object (part$is_failure); with this, the while-loop logic becomes more explicit: create_part <- function(machine) { structure( list( is_failure = runif(1) < machine$failure_rate,\nproduction_duration = 1 / machine$production_rate ), class = \"Part\" ) } # then we can use this in the while-loop part <- create_part(machine) if (part$is_failure) {\n# bad part logic\n...\n} else {\n# good part logic\n...\n}\n\n\n# Restructuring the while loop\n\nI wanted to push that while-loop into a function - the less work you do in the global environment, the better.\n\nSince you want to extract data from production_df for your report, the function should return the production_df. During the while-loop, you access production_df, time_parts, t, dt (which I renamed dt_recovery based on your comments), n and machine. So we might want to pass all of those into that function. But we can compute some of those from the others:\n\n• n is the nrow of production_df,\n\n• t isn't needed outside of the while loop and\n\n• the data that initialises time_parts also initialises production_df.\n\nThe only thing we need to initialise both time_parts and production_df is the arrival-times or times at which the parts were ordered (which I renamed t_ordered).\n\nSo, we can put that while-loop into a function that takes arguments t_ordered, dt_recovery, machine.\n\nrun_event_simulation <- function(t_ordered, machine, dt_recovery) {\nn_parts <- length(t_ordered)\n\n# results data-frame\nproduction_df <- data.frame(\nid = seq(n_parts),\nt_ordered = t_ordered,\nt_started = 0,\nt_completed = 0,\nno_failures = 0,\nstringsAsFactors = FALSE\n)\n\ntime_parts <- ... # define in terms of production_df\n\n# while-loop logic\n\n# return the updated production_df\n\n\nI added the column t_completed into production_df so that you can more easily compute total_production_time from production_df in your report (this allows you to generalise the production rates)\n\n# in production_summary\n...\ntotal_production_time <- max(production_df$t_completed) ... # A functional priority queue The really big step: R doesn't have a native priority-queue, and it would be pretty hard to encode using the S3 or S4 classes since you can't update by reference in those classes. There is a priority-queue defined in the package liqueueR, but I've no experience of that. So I just wrote a simpler version of the priority queue (as an S3 class): this allows you to • peek: extract the element in the queue with the lowest priority value (without mutating the queue) • delete_min: remove that element with the lowest priority value from the queue and return the resulting queue • add: add a new element to the queue according to it's priority, returning the resulting queue • and provides a couple of helper methods (is_empty, nrow) However, this doesn't provide a pop_element(queue): typically, pop_element removes the leading element from the queue and returns that element. That is, it returns the leading element and updates the queue through a side-effect. This side-effect is problematic in R, so I didn't include a pop_element function. To achieve pop_element you have to peek and then delete_min. # Priority Queue class create_priority_queue <- function(x, priority_column) { structure( list( # note that we only order once - see add for how this is possible queue = x[order(x[[priority_column]]), ] ), class = \"PriorityQueue\", priority_column = priority_column ) } # generic methods for Priority Queue is_empty <- function(x, ...) UseMethod(\"is_empty\") peek <- function(x, ...) UseMethod(\"peek\") delete_min <- function(x, ...) UseMethod(\"delete_min\") add <- function(x, ...) UseMethod(\"add\") nrow <- function(x, ...) UseMethod(\"nrow\") nrow.default <- function(x, ...) { base::nrow(x) } # implemented methods for Priority Queue nrow.PriorityQueue <- function(x, ...) { nrow(x$queue)\n}\nis_empty.PriorityQueue <- function(x, ...) {\nnrow(x) == 0\n}\npeek.PriorityQueue <- function(x, ...) {\nx$queue[1, ] } delete_min.PriorityQueue <- function(x, ...) { x$queue <- x$queue[-1, ] x } add.PriorityQueue <- function(x, new_element, ...) { priority_column <- attr(x, \"priority_column\") # split the existing values by comparison of their priorities to # those of the new-element lhs <- which(x$queue[[priority_column]] <= new_element[[priority_column]])\nrhs <- setdiff(seq(nrow(x)), lhs)\nx$queue <- rbind(x$queue[lhs, ], new_element, x$queue[rhs, ]) x } Then I replaced your time_parts data-frame with a PriorityQueue: # inside run_event_simulation ... # Create initial list of tasks. Once produced, a part will be removed from the # queue. product_queue <- create_priority_queue( data.frame( id = production_df$id,\nt = production_df$t_ordered ), \"t\" ) ... I added a few other helpers. The final code looks like this: # ---- classes # Priority Queue class create_priority_queue <- function(x, priority_column) { structure( list( queue = x[order(x[[priority_column]]), ] ), class = \"PriorityQueue\", priority_column = priority_column ) } # A machine for producing Parts create_machine <- function(failure_rate, production_rate) { structure( list( failure_rate = failure_rate, production_rate = production_rate, is_occupied = FALSE ), class = \"Machine\" ) } # A manufactured part create_part <- function(machine) { structure( list( is_failure = runif(1) < machine$failure_rate,\nproduction_duration = 1 / machine$production_rate ), class = \"Part\" ) } # methods for Priority Queue is_empty <- function(x, ...) UseMethod(\"is_empty\") peek <- function(x, ...) UseMethod(\"peek\") delete_min <- function(x, ...) UseMethod(\"delete_min\") add <- function(x, ...) UseMethod(\"add\") nrow <- function(x, ...) UseMethod(\"nrow\") nrow.default <- function(x, ...) { base::nrow(x) } nrow.PriorityQueue <- function(x, ...) { nrow(x$queue)\n}\n\nis_empty.PriorityQueue <- function(x, ...) {\nnrow(x) == 0\n}\n\npeek.PriorityQueue <- function(x, ...) {\nx$queue[1, ] } delete_min.PriorityQueue <- function(x, ...) { x$queue <- x$queue[-1, ] x } add.PriorityQueue <- function(x, new_element, ...) { priority_column <- attr(x, \"priority_column\") lhs <- which(x$queue[[priority_column]] <= new_element[[priority_column]])\nrhs <- setdiff(seq(nrow(x)), lhs)\nx$queue <- rbind(x$queue[lhs, ], new_element, x$queue[rhs, ]) x } # ---- functions update_machine <- function(machine, ind_production_df, production_df, current_time) { if (machine$is_occupied) {\nif (\nproduction_df$t_started[ind_production_df] + 1 / machine$production_rate <= current_time\n) {\nmachine$is_occupied <- FALSE } } return(machine) } should_produce_part <- function(machine, earliest_production_time, current_time) { !machine$is_occupied &&\nearliest_production_time <= current_time\n}\n\nincrement_failures <- function(df, i) {\ndf[i, \"no_failures\"] <- 1 + df[i, \"no_failures\"]\ndf\n}\n\n# ---- format results\n\nproduction_summary <- function(production_df, machine, input_rate) {\nn_parts <- nrow(production_df)\nno_of_failures <- sum(production_df$no_failures) total_production_time <- max(production_df$t_completed)\nuptime <- (no_of_failures + n_parts) / machine$production_rate print(paste0( \"Estimated machine$failure_rate \",\nround(no_of_failures / (no_of_failures + n_parts), 2),\n\" [theory \", round(machine$failure_rate, 2), \"]\" )) print(paste0( \"Up-time \", uptime, \", of total time \", total_production_time, \". Auslastung \", round(uptime / total_production_time, 2), \" [theory \", round( input_rate / machine$production_rate * 1 / (1 - machine$failure_rate), 2 ), \"]\" )) } # ---- discrete-event simulation # run_event_simulation <- function(t_ordered, machine, dt_recovery) { n_parts <- length(t_ordered) # results data-frame production_df <- data.frame( id = seq(n_parts), t_ordered = t_ordered, t_started = 0, t_completed = 0, no_failures = 0, stringsAsFactors = FALSE ) # Create initial list of tasks. Once produced, a part will be removed from the # queue. product_queue <- create_priority_queue( data.frame( id = production_df$id,\nt = production_df$t_ordered ), \"t\" ) t <- 0 while (!is_empty(product_queue)) { queued_part <- peek(product_queue) ind_production_df <- match( queued_part$id, production_df$id ) machine <- update_machine(machine, ind_production_df, production_df, t) if ( should_produce_part(machine, earliest_production_time = queued_part$t,\ncurrent_time = t)\n) {\n# A machine is available and a part needs to be produced\n\n# - pop the scheduled part from the queue; add it back if it's production\n# fails\nproduct_queue <- delete_min(product_queue)\n\nmachine$is_occupied <- TRUE production_df$t_started[ind_production_df] <- t\npart <- create_part(machine)\n\nif (part$is_failure) { # bad part - add it back to the schedule queued_part$t <- queued_part$t + dt_recovery product_queue <- add(product_queue, queued_part) production_df <- increment_failures(production_df, ind_production_df) t <- t + min(peek(product_queue)$t, dt_recovery)\n} else {\n# good part\nt <- t + part$production_duration production_df$t_completed[ind_production_df] <- t\nmachine$is_occupied <- FALSE } } else { # machine is occupied or no part needs to be produced t <- t + min(peek(product_queue)$t, dt_recovery)\n}\n}\nproduction_df\n}\n\n# ---- script\nset.seed(123456)\n\n# Input rate [1/min, 1/input_rate corresponds to interarrival time in min]\ninput_rate <- 1 / 60\n\n# Number of parts\nn_parts <- 1000\n\n# timestep = time to transfer faulty parts back to production. [min]\ndt_recovery <- 1\n\n# Production rate [1/min]\nmachine_production_rate <- 1 / 40\n\n# Machine failure rate\nmachine_failure_rate <- 0.2\n\n# Sum all interarrival times\nt_ordered <- cumsum(rpois(n_parts, 1 / input_rate))\n\nmachine <- create_machine(machine_failure_rate, machine_production_rate)\n\n# ---- results\n\nproduction_df <- run_event_simulation(\nt_ordered, machine, dt_recovery\n)\n\nproduction_summary(production_df, machine, input_rate)\n\n\nWhy aren't S3 queues easy?\n\n(This is actually quite hard to explain). Well, the pop method on a priority-queue returns an element from the queue and moves the queue on by one step. (In R) Updating the queue might look like new_queue <- old_queue[-1] and obtaining the returned element might look like returned_element <- old_queue. So a pop function might look like\n\npop <- function(q) {\n# extract the head\nel <- q\n\n# In a reference-based language you could update the queue\n# using a side-effect like q.drop()\n# But in R, this creates a new queue: and if it isn't returned\n# explicitly, it is thrown away at the end of the pop function\nnew_q <- q[-1]\n\n# return the element that's at the head of the original queue\nel\n}\n\n# calling_env\nmy_q <- create_queue(...)\n\nBut the queue has not been altered by that pop. Now we could rewrite that function to do something dangerous like q <<- q[-1] and that would update the q in the calling environment. I consider this dangerous because q might not exist in the calling environment and that introduces side-effects, which are much harder to reason about.\n• @Christoph there was a couple of things I didn't feel comfortable restructuring. I couldn't work out why update_machine works the way it does: it seems to look into the future before it decides what to do now. It makes more sense to me for is_occupied to be set to FALSE at the end of each while-loop iteration. – Russ Hyde Feb 12 at 13:08\n• Updated. But it's pretty difficult to explain. I'm not particularly interested in stochastic simulation at present. I did have a look at your simmer documentation. Can you confirm that when you mclapply() over different simulation chains, different seeds are used for each chain? – Russ Hyde Feb 18 at 9:53\n• To your function pop <- function(q): why don't you just return(list(el=el, new_q=new_q)? Then you could work within one line r <- pop(q); el <- r$el; q <- r$new_q; rm(r);? But I still understand your comment that calling by reference would be smart... – Christoph Aug 7 at 14:35"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6955043,"math_prob":0.93471247,"size":13697,"snap":"2019-35-2019-39","text_gpt3_token_len":3417,"char_repetition_ratio":0.19696195,"word_repetition_ratio":0.22419028,"special_character_ratio":0.2833467,"punctuation_ratio":0.16697164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99092984,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-20T19:11:03Z\",\"WARC-Record-ID\":\"<urn:uuid:482fa957-edad-457a-a654-b22f759ae883>\",\"Content-Length\":\"169350\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:904caa97-99aa-40ce-ae61-3315aee0bc18>\",\"WARC-Concurrent-To\":\"<urn:uuid:e53548eb-a6a5-4123-9726-93e2ec672b5e>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/212825/discrete-event-simulation-for-production-defects/213209\",\"WARC-Payload-Digest\":\"sha1:KVHAEXAEJOIJ7GQNODSHCO3RWZDSN4JK\",\"WARC-Block-Digest\":\"sha1:2W5DPLNZJABJUYA6UTJVSJLFGBLJT46K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574058.75_warc_CC-MAIN-20190920175834-20190920201834-00170.warc.gz\"}"} |
https://www.colorhexa.com/2b28f8 | [
"# #2b28f8 Color Information\n\nIn a RGB color space, hex #2b28f8 is composed of 16.9% red, 15.7% green and 97.3% blue. Whereas in a CMYK color space, it is composed of 82.7% cyan, 83.9% magenta, 0% yellow and 2.7% black. It has a hue angle of 240.9 degrees, a saturation of 93.7% and a lightness of 56.5%. #2b28f8 color hex could be obtained by blending #5650ff with #0000f1. Closest websafe color is: #3333ff.\n\n• R 17\n• G 16\n• B 97\nRGB color chart\n• C 83\n• M 84\n• Y 0\n• K 3\nCMYK color chart\n\n#2b28f8 color description : Bright blue.\n\n# #2b28f8 Color Conversion\n\nThe hexadecimal color #2b28f8 has RGB values of R:43, G:40, B:248 and CMYK values of C:0.83, M:0.84, Y:0, K:0.03. Its decimal value is 2828536.\n\nHex triplet RGB Decimal 2b28f8 `#2b28f8` 43, 40, 248 `rgb(43,40,248)` 16.9, 15.7, 97.3 `rgb(16.9%,15.7%,97.3%)` 83, 84, 0, 3 240.9°, 93.7, 56.5 `hsl(240.9,93.7%,56.5%)` 240.9°, 83.9, 97.3 3333ff `#3333ff`\nCIE-LAB 35.61, 68.322, -98.377 18.695, 8.807, 89.516 0.16, 0.075, 8.807 35.61, 119.774, 304.78 35.61, -9.035, -129.307 29.677, 60.512, -158.067 00101011, 00101000, 11111000\n\n# Color Schemes with #2b28f8\n\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #f5f828\n``#f5f828` `rgb(245,248,40)``\nComplementary Color\n• #288df8\n``#288df8` `rgb(40,141,248)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #9328f8\n``#9328f8` `rgb(147,40,248)``\nAnalogous Color\n• #8df828\n``#8df828` `rgb(141,248,40)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #f89328\n``#f89328` `rgb(248,147,40)``\nSplit Complementary Color\n• #28f82b\n``#28f82b` `rgb(40,248,43)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #f82b28\n``#f82b28` `rgb(248,43,40)``\n• #28f5f8\n``#28f5f8` `rgb(40,245,248)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #f82b28\n``#f82b28` `rgb(248,43,40)``\n• #f5f828\n``#f5f828` `rgb(245,248,40)``\n• #0a07cd\n``#0a07cd` `rgb(10,7,205)``\n• #0b07e6\n``#0b07e6` `rgb(11,7,230)``\n• #130ff7\n``#130ff7` `rgb(19,15,247)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #4341f9\n``#4341f9` `rgb(67,65,249)``\n• #5c59fa\n``#5c59fa` `rgb(92,89,250)``\n• #7472fa\n``#7472fa` `rgb(116,114,250)``\nMonochromatic Color\n\n# Alternatives to #2b28f8\n\nBelow, you can see some colors close to #2b28f8. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #2859f8\n``#2859f8` `rgb(40,89,248)``\n• #2848f8\n``#2848f8` `rgb(40,72,248)``\n• #2836f8\n``#2836f8` `rgb(40,54,248)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #3c28f8\n``#3c28f8` `rgb(60,40,248)``\n• #4e28f8\n``#4e28f8` `rgb(78,40,248)``\n• #5f28f8\n``#5f28f8` `rgb(95,40,248)``\nSimilar Colors\n\n# #2b28f8 Preview\n\nText with hexadecimal color #2b28f8\n\nThis text has a font color of #2b28f8.\n\n``<span style=\"color:#2b28f8;\">Text here</span>``\n#2b28f8 background color\n\nThis paragraph has a background color of #2b28f8.\n\n``<p style=\"background-color:#2b28f8;\">Content here</p>``\n#2b28f8 border color\n\nThis element has a border color of #2b28f8.\n\n``<div style=\"border:1px solid #2b28f8;\">Content here</div>``\nCSS codes\n``.text {color:#2b28f8;}``\n``.background {background-color:#2b28f8;}``\n``.border {border:1px solid #2b28f8;}``\n\n# Shades and Tints of #2b28f8\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, #01000d is the darkest color, while #f9f9ff is the lightest one.\n\n• #01000d\n``#01000d` `rgb(1,0,13)``\n• #010120\n``#010120` `rgb(1,1,32)``\n• #020233\n``#020233` `rgb(2,2,51)``\n• #030246\n``#030246` `rgb(3,2,70)``\n• #040359\n``#040359` `rgb(4,3,89)``\n• #05046c\n``#05046c` `rgb(5,4,108)``\n• #06047f\n``#06047f` `rgb(6,4,127)``\n• #070592\n``#070592` `rgb(7,5,146)``\n• #0805a5\n``#0805a5` `rgb(8,5,165)``\n• #0906b8\n``#0906b8` `rgb(9,6,184)``\n• #0907cb\n``#0907cb` `rgb(9,7,203)``\n• #0a07de\n``#0a07de` `rgb(10,7,222)``\n• #0b08f1\n``#0b08f1` `rgb(11,8,241)``\n• #1815f7\n``#1815f7` `rgb(24,21,247)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\n• #3e3bf9\n``#3e3bf9` `rgb(62,59,249)``\n• #504ef9\n``#504ef9` `rgb(80,78,249)``\n• #6361fa\n``#6361fa` `rgb(99,97,250)``\n• #7674fa\n``#7674fa` `rgb(118,116,250)``\n• #8987fb\n``#8987fb` `rgb(137,135,251)``\n• #9b9afc\n``#9b9afc` `rgb(155,154,252)``\n``#aeadfc` `rgb(174,173,252)``\n• #c1c0fd\n``#c1c0fd` `rgb(193,192,253)``\n• #d4d3fe\n``#d4d3fe` `rgb(212,211,254)``\n• #e6e6fe\n``#e6e6fe` `rgb(230,230,254)``\n• #f9f9ff\n``#f9f9ff` `rgb(249,249,255)``\nTint Color Variation\n\n# Tones of #2b28f8\n\nA tone is produced by adding gray to any pure hue. In this case, #8f8e92 is the less saturated color, while #2b28f8 is the most saturated one.\n\n• #8f8e92\n``#8f8e92` `rgb(143,142,146)``\n• #86869a\n``#86869a` `rgb(134,134,154)``\n• #7e7da3\n``#7e7da3` `rgb(126,125,163)``\n• #7675ab\n``#7675ab` `rgb(118,117,171)``\n• #6d6cb4\n``#6d6cb4` `rgb(109,108,180)``\n• #6564bc\n``#6564bc` `rgb(101,100,188)``\n• #5d5bc5\n``#5d5bc5` `rgb(93,91,197)``\n• #5453cd\n``#5453cd` `rgb(84,83,205)``\n``#4c4ad6` `rgb(76,74,214)``\n• #4442de\n``#4442de` `rgb(68,66,222)``\n• #3c39e7\n``#3c39e7` `rgb(60,57,231)``\n• #3331ef\n``#3331ef` `rgb(51,49,239)``\n• #2b28f8\n``#2b28f8` `rgb(43,40,248)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #2b28f8 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":0.5199653,"math_prob":0.72056174,"size":3673,"snap":"2019-35-2019-39","text_gpt3_token_len":1703,"char_repetition_ratio":0.12237667,"word_repetition_ratio":0.011111111,"special_character_ratio":0.5567656,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9823716,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T09:02:23Z\",\"WARC-Record-ID\":\"<urn:uuid:626ecd54-f6c4-4d0d-822b-dd38728f40ff>\",\"Content-Length\":\"36256\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:30a7cb98-c643-4bab-af4c-85525f90d586>\",\"WARC-Concurrent-To\":\"<urn:uuid:03dc2f87-d258-472c-9602-25e6f2109723>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/2b28f8\",\"WARC-Payload-Digest\":\"sha1:PWDUW7WJ6BUQ2JTTHLTPILVJO3XZ7ZJL\",\"WARC-Block-Digest\":\"sha1:C7IPJBGYDVM4ER4FPQCM2VIAX7BGHK3J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573065.17_warc_CC-MAIN-20190917081137-20190917103137-00066.warc.gz\"}"} |
https://docs.w3cub.com/javascript/errors/invalid_array_length | [
"/JavaScript\n\n# RangeError: invalid array length\n\nThe JavaScript exception \"Invalid array length\" occurs when specifying an array length that is either negative, a floating number or exceeds the maximum supported by the platform (i.e. when creating an `Array` or `ArrayBuffer`, or when setting the `length` property).\n\nThe maximum allowed array length depends on the platform, browser and browser version. For `Array` the maximum length is 4GB-1 (2^32-1). For `ArrayBuffer` the maximum is 2GB-1 on 32-bit systems (2^31-1). From Firefox version 89 the maximum value of `ArrayBuffer` is 8GB on 64-bit systems (2^33).\n\nNote: `Array` and `ArrayBuffer` are independent data structures (the implementation of one does not affect the other).\n\n## Message\n\n```RangeError: invalid array length (V8-based & Firefox)\nRangeError: Array buffer allocation failed (V8-based)\nRangeError: Array size is not a small enough positive integer. (Safari)\n```\n\n## What went wrong?\n\nAn invalid array length might appear in these situations:\n\nIf you are creating an `Array`, using the constructor, you probably want to use the literal notation instead, as the first argument is interpreted as the length of the `Array`.\n\nOtherwise, you might want to clamp the length before setting the length property, or using it as argument of the constructor.\n\n## Examples\n\n### Invalid cases\n\n```new Array(Math.pow(2, 40))\nnew Array(-1)\nnew ArrayBuffer(Math.pow(2, 32)) // 32-bit system\nnew ArrayBuffer(-1)\n\nconst a = [];\na.length = a.length - 1; // set the length property to -1\n\nconst b = new Array(Math.pow(2, 32) - 1);\nb.length = b.length + 1; // set the length property to 2^32\nb.length = 2.5; // set the length property to a floating-point number\n\nconst c = new Array(2.5); // pass a floating-point number\n```\n\n### Valid cases\n\n```[ Math.pow(2, 40) ] // [ 1099511627776 ]\n[ -1 ] // [ -1 ]\nnew ArrayBuffer(Math.pow(2, 32) - 1)\nnew ArrayBuffer(Math.pow(2, 33)) // 64-bit systems after Firefox 89\nnew ArrayBuffer(0)\n\nconst a = [];\na.length = Math.max(0, a.length - 1);\n\nconst b = new Array(Math.pow(2, 32) - 1);\nb.length = Math.min(0xffffffff, b.length + 1);\n// 0xffffffff is the hexadecimal notation for 2^32 - 1\n// which can also be written as (-1 >>> 0)\n\nb.length = 3;\n\nconst c = new Array(3);\n```"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7573457,"math_prob":0.9747642,"size":666,"snap":"2023-14-2023-23","text_gpt3_token_len":158,"char_repetition_ratio":0.163142,"word_repetition_ratio":0.0,"special_character_ratio":0.23573573,"punctuation_ratio":0.096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9947877,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-29T20:22:34Z\",\"WARC-Record-ID\":\"<urn:uuid:38a2015b-69a7-4215-90ad-166ddd2e38b8>\",\"Content-Length\":\"19290\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dac5a3a3-bd9b-4e71-83c7-e71660a11296>\",\"WARC-Concurrent-To\":\"<urn:uuid:a51d188d-a877-4f80-8c24-ec3125626ae0>\",\"WARC-IP-Address\":\"104.21.8.155\",\"WARC-Target-URI\":\"https://docs.w3cub.com/javascript/errors/invalid_array_length\",\"WARC-Payload-Digest\":\"sha1:6VQI5DLKBQPYTWSDM6J3GKLFMAUOA6CC\",\"WARC-Block-Digest\":\"sha1:5PGSE4JMQLW2JSE2TL5EXQKNXALJLTUB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949025.18_warc_CC-MAIN-20230329182643-20230329212643-00353.warc.gz\"}"} |
https://www.ndt.net/article/v11n06/rosyidi/rosyidi.htm | [
"NDT.net • June 2006 • Vol. 11 No.6\n\n## Estimating G-Max & Field CBR of Soil Subgrade Using a Seismic Method\n\nS.A. Rosyidi1, K.A.M. Nayan2, M.R. Taha3 & A. ISMAIL4\n1Lecturer, Division of Transportation Eng., Department of Civil Engineering\nMuhammadiyah University of Yogyakarta, Lingkar Barat, 55183, Yogyakarta, Indonesia\nemail: atmaja_sri@umy.ac.id\n2Lecturer, 3Professor, 4Assoc.Prof., Department of Civil & Structural Engineering,\nUniversiti Kebangsaan Malaysia. 43600, Bangi, Malaysia\nemail: khairul@eng.ukm.my, drmrt@eng.ukm.my, abim@eng.ukm.my.\n\n### ABSTRACT\n\nThe knowledge of the in-situ strength and stiffness of the ground is normally required for the design monitoring and evaluation of highway pavement so as to ensure an adequate margin of safety. In order to estimate the stiffness of the pavement foundation a method called the spectral analysis of surface wave (SASW) has been developed. The method consists of generation, measurement and processing of the dispersive Rayleigh waves recorded from two vertical transducers. Subsequently, an inversion process is carried out to obtain the shear wave velocity versus depth profile of the site from the dispersive Rayleigh wave data. The results presented in this paper showed that the SASW method is able to determine reliably the shear modulus of the soil subgrade layer of the pavement profile. In situ subgrade bearing capacity test using the field CBR test was also carried out in the same location of SASW test. An empirical relationship between the CBR value and the dynamic soil stiffness from the SASW measurements was also established for practical applications for the purpose of pavement design and maintenance.\n\nKeywords: SASW, shear wave velocity, shear modulus, pavement subgrade layer\n\n### INTRODUCTION\n\nThe performance of pavement structures is strongly influenced by the stiffness of the soil subgrade layer. In order to establish the stiffness of subgrade structures of the existing roads, accurate information of the moduli of the various pavement layers is needed. The shear modulus parameter is used to calculate the bearing capacity and to characterize the mechanical behavior of the materials under different types of traffic loading in order to predict the performance, select and to design appropriate rehabilitation techniques. In order to effectively measure and evaluate the stiffness of soil subgrade layers, a non-destructive test (NDT) of the spectral analysis of surface wave (SASW) which is economic and fast is needed. The method is based on the dispersion of Rayleigh waves (R waves) to determine the shear wave velocity each layer of the pavement profile. The SASW method has been utilized in different applications over the past decade (Stokoe et al., 1994) after the advancement and improvement of the well-known steady-state (Jones, 1958) technique. These applications include detection of soil profile, evaluation of concrete structures, detection of anomalies, detection of the structural layer of cement mortar, assessing compaction of fills and the evaluation of railway ballast. The purpose of this paper is to estimate the dynamic shear modulus of the soil subgrade layer measured from the SASW test and to derive the empirical correlation between the dynamic shear modulus parameter and the CBR values from different sites.\n\n### LITERATURE REVIEW\n\nThe SASW method is based on the particles motion of R wave in heterogeneous media. The energy of R waves from the source propagates mechanically along the surface of media and their amplitude decrease rapidly with depth. Particle motions associated with R wave are composed of both vertical and horizontal components, which when combined, formed a retrogressive ellipse close to the surface. In homogenous, isotropic, elastic half-space, R wave velocity does not vary with frequency. However, R wave velocity varies with frequency in layered medium where there is a variation of stiffness with depth (Stokoe et al., 1994). This phenomenon is termed dispersion where the frequency is dependent on R wave velocity. The ability to detect and evaluate the depth of the medium is influenced by the wavelength and the frequency generated. The shorter wavelength of high frequency penetrates the shallower zone of the near surface and the longer wavelength of lower frequency penetrates deeper into the medium.\n\nThe range of wavelength to be used as a guide for the receiver spacing can be estimated from the shear wave velocities of the material anticipated at the site:",
null,
"where f is the frequency and VS is shear wave velocity. The higher and low frequency waves groups needed can be generated by various transient sources of different weights and shapes. Waves of low frequency for the base and subgrade layer could be generated from hammer weights of 3 to 5 kg (Rosyidi et al., 2005a, 2005b).\n\nThe experimental dispersion curve of phase velocity and wavelength may be developed from phase information of the transfer function at the frequency range satisfying the coherence criterion. In addition, most of researchers apply the filtering criteria (Heisey et al., 1982) with a wavelength greater than ½ and less than 3 receiver spacings. The time of travel between the receivers for each frequency can be calculated by:",
null,
"where f is the frequency, t(f) and ff are respectively the travel time and the phase difference in degrees at a given frequency. The distance of the receiver (d) is a known parameter. Therefore, R wave velocity, VR or the phase velocity at a given frequency is simply obtained by:",
null,
"and the corresponding wavelength of the R wave, LR may be written as:",
null,
"The actual shear wave velocity of the pavement profile is produced from the inversion of the composite experimental dispersion curve. In the inversion process, a profile of set of a homogeneous layer extending to infinity in the horizontal direction is assumed. The last layer is usually taken as a homogeneous half-space. Based on the initial profile, a theoretical dispersion curve is then calculated using an automated forward modeling analysis of the dynamic stiffness matrix method (Kausel & Röesset, 1981). The theoretical dispersion curve is ultimately matched to the experimental dispersion curve of the lowest RMS error with an optimization technique. Finally, the profile from the best-fitting (lowest RMS) of the theoretical dispersion curve to the experimental dispersion curve is used that represents the most likely pavement profile of the site. The dynamic shear moduli of the materials can be easily obtained using the shear wave velocity parameter of SASW from the following equation (Yoder & Witczak, 1975):",
null,
"where G is the dynamic shear modulus, VS the shear wave velocity, g the gravitational acceleration, ? the total unit weight of the material and µ the Poisson ratio. Nazarian & Stokoe (1986) explained that the shear modulus parameter of material obtained from the SASW test approaches to the maximum shear modulus at a strain below about 0.001 %. In this strain range, modulus of the subgrade materials is also taken as constant.\n\n### METHODOLOGY\n\nExperimental Set Up\n\nAn impact source on a pavement surface is used to generate R waves. These waves are detected using two accelerometers where the signals are recorded using an analog digital recorder and a notebook computer for post processing (Figure 1). Several configurations of the receiver and the source spacings are required in order to sample different depths. The best configuration in the SASW is the mid point receiver spacings (Heisey et al., 1982).\n\nIn this study, the short receiver spacings of 5 and 10 cm with a high frequency source (ball bearing) are used to sample the AC layers while the long receiver spacings of 20, 40 cm and 80, 160 cm with a set of low frequencies sources (a set of hammers) are used to sample the base and subgrade layers, respectively. The SASW tests were carried out at two sites which include 30 test locations on the main road in the campus of Universiti Kebangsaan Malaysia in Bangi, Selangor, Malaysia and 20 test locations on the State Road of Prambanan to Pakem and Piyungan to Gading, Yogyakarta Province, Indonesia. Data were collected together with the field CBR tests conducted on the same SASW measured centre points.",
null,
"Figure 1. (a) SASW equipments, (b) impact sources, (c) SASW experimental set up, and (d) SASW test conducted in field\n\nData Analysis\n\nAll the data collected from the recorder are transformed using the Fast Fourier Transform (FFT) to frequency domain by the dBFA32 software resident in the notebook computer. Two functions in the frequency domain between the two receivers are of great importance: (1) the coherence function and (2) the phase information of the transfer function. The coherence function is used to visually inspect the quality of signals being recorded in the field and have a real value between zero and one in the range of frequencies being measured. The value of one indicates a high signal-to-noise ratio (i.e., perfect correlation between the two signals) while values of zero represents no correlation between the two signals. The transfer function spectrum is used to obtain the relative phase shift between the two signals in the range of the frequencies being generated.\n\nFigure 2 shows a typical set of the coherence and the phase plot of the transfer function from the measurement of an 80 cm receiver spacing at the site of UKM's road. By unwrapping the data of the phase angle from the transfer function, a composite experimental dispersion curve of all the receiver spacings are generated. By repeating the procedure outlined above and using equation (2) through (4) for each frequency value, the R wave velocity corresponding to each wavelength is evaluated and the experimental dispersion curve is subsequently generated. Figure 3 shows the example of the composite experimental dispersion curve from measurements of all the receiver spacings.",
null,
"Figure 2. The coherence and the transfer function spectrum for an 80 cm receiver spacing on UKM's road.",
null,
"Figure 3. A typical dispersion curve from a set of SASW tests on the pavement showing the portion of the phase velocity for subgrade layer\n\n### RESULTS AND DISCUSSION\n\nBy unwrapping data of the phase angle from the transfer function (Fig.2 for site 1), the composite experimental dispersion curve obtained is as shown in Figure 3. The figure shows the horizontal dash lines for wavelengths ranging from 0.8 to 3 m, the subgrade layer with the minimum phase velocity of 180 m/s to the maximum phase velocity of 230 m/s are obtained.\n\nIn order to generate the actual shear wave velocity of the subgrade layer, an inversion process using 3 D forward modeling from the stiffness matrix method (Kausel & Röesset, 1981) and an optimization technique of the maximum likelihood method (Joh, 1996) were conducted. For developing the pavement profile from the inversion analysis, the starting model parameter is obtained from the cored road profile that was found to be consisting of an average asphalt concrete (AC) layer (70 mm thick), a base layer (400 mm thick) over a subgrade layer. Descriptions of model parameters and typical thickness of the pavement layers are shown in Figure 4.",
null,
"Figure 4. A typical pavement profile for starting model parameter\n\nThe pavement profile obtained from inversion process is shown in Figure 5. The profile is an example of the SASW result from the first location (UKM, Malaysia). Core drilling was also conducted in the same location after the SASW measurement. There is a reasonable agreement of the profile depth between the results from the SASW measurement and the core drilling. The average of inverted shear wave velocity for UKM's road measuring points is 178.419 m/s with a range of 116.44 to 263.226 m/s.\n\nThe dynamic shear modulus is then obtained from the shear wave velocity profile using the dynamic material equation (Equation 5). The average shear modulus of the subgrade material from the analysis is 69.779 MPa. Based on the shear modulus, the subgrade material maybe classified as a sandy soil material. The shear wave velocities and their corresponding shear modulus from this study were listed in Table 1 in comparison with the results of SASW testing obtained by other researchers such as Puri (1969) and Nazarian & Stokoe (1986). It is important to note that Puri (1969) obtained the dynamic properties of silty sand using the free vertical vibration stress waves measured at 1.0 x 10-4 % strain level.\n\nThe shear wave velocities from the SASW were then correlated to the CBR values for the evaluation of the bearing capacities of the subgrade materials. The relationship between the shear wave velocities and CBR values can also be derived as shown in Figure 6 for the subgrade layer.\n\n Table 1. Comparison of subgrade shear wave velocity and shear modulus. Compared parameter This study Puri (1969) Nazarian & Stokoe (1986) Shear wave velocity (m/s) 178.42 m/s --- 147.5 - 211.9 m/s Shear Modulus (MPa) 69.78 MPa Sandy soil with poorly graded 64.75 MPa Poorly graded fine silty sand 41.34 - 85.31 MPa Loose sand",
null,
"Figure 5. Comparison between actual pavement profile and the result from core drilling in the site\n\nFigure 6 also shows that the increased in the shear wave velocities correlates well with the increased in the CBR values. The coefficients of correlation obtained (Figure 6) indicate that the empirical equation derived between the shear wave velocities have significant correlations with the CBR value. The correlation coefficient, R2 of 0.938 was obtained for the subgrade layer. However, the empirical equations obtained only shows the best correlation of shear wave velocity to the CBR value that is not more than 400 m/s. Higher deviations obtained from Figure 6 for high values of CBR should be investigated. The derived equation of Vs and CBR can be written as:\n\nCBR = 0.0006 (VSS)1.99 (6)\n\nwhere, CBR is the field California Bearing Ratio in % and VS is the shear wave velocity in m/s. Figure 7 shows the empirical correlation between the CBR values to the dynamic shear modulus from SASW for the subgrade layer. The result shows a good agreement between the dynamic shear modulus from the SASW test and the CBR value with a deviation range of ± 20 %. The empirical equations obtained were summarized as below :\n\nCBR = 0.266 (G)1.0027, R2 = 0.947 (7)\n\nwhere G is an approach value of the maximum shear modulus in MPa obtained from the SASW analysis.",
null,
"Figure 6. Correlation between the shear wave velocity, DCP and CBR for the subgrade layer",
null,
"Figure 7. Empirical correlation between the CBR value and the dynamic elastic modulus from SASW for the subgrade layer\n\n### CONCLUSIONS\n\n1. Good agreements were obtained between the measured shear wave velocities and the corresponding dynamic shear modulus as compared to the work of Puri (1969) and Nazarian & Stokoe (1986). 2. This study has also managed to obtain good empirical correlations between the dynamic shear modulus and the field CBR values. 3. The SASW method is able to characterize the stiffness of the pavement subgrade layer in terms of shear wave velocity and its corresponding dynamic shear modulus satisfactorily for the propose of pavement design and evaluation.\n\n### ACKNOWLEDGEMENT\n\nThe authors would like to give our sincere appreciation to the Ministry of Science, Technology and Environmental of Malaysia for supporting this research through the IRPA Grant No.09-02-02-0055-EA151, and also to the Faculty of Engineering, Muhammadiyah University of Yogyakarta for supporting the study through Engineering Research Grant 2004. The authors also would like to thank Prof. Sung Ho Joh of the Chung Ang University, Korea for his assistance in using the WinSASW software in this study and Prof. Gucunski of Rutgers University, USA for his interested discussion.\n\n### REFERENCES\n\n1. Al-Hunaidi, M.O. 1998. Evaluation-Based Genetic Algorithms for Analysis of Non-Destructive Surface Waves Test on Pavements\". NDT&E international, Vol.31, No.4, 273-280.\n2. Heisey, J.S., Stokoe II, K.H. & Meyer, A.H. 1982. Moduli of Pavement Systems from Spectral Analysis of Surface Waves. Transportation Research Record (TRB) No.852, 22-31.\n3. Joh, S.H. 1996. Advance in Interpretation & Analysis Technique for Spectral Analysis of Surface Wave (SASW) Measurements. Ph.D Dissertation. The University of Texas at Austin.\n4. Jones, R.B. 1958. In-situ Measurement of the Dynamic Properties of Soil by Vibration Methods. Geotechnique, Vol.8, No.1, pp.1-21.\n5. Kausel, E. & Röesset, J.M. 1981. Stiffness Matrices for Layered Soils. Bulletin of the Seismological Society of America, Vol.71, No.6, pp.1743-1761.\n6. Nazarian, S. & Stokoe, K.H.II. 1986. In Situ Determination of Elastic Moduli of Pavement Systems by Spectral-Analysis-of-Surface-Wave Method (Theoretical Aspects). Research Report 437-2. Center of Transportation Research. Bureau of Engineering Research. The University of Texas at Austin.\n7. Stokoe, K.H.II, Wright, S.G., Bay, J.A. & Röesset, J.M. 1994. Characterization of Geotechnical Sites by SASW Method. GEOPHYSICAL CHARACTERIZATION OF SITES. XIII ICSMFE. 1994. Oxford & IBH Publishing Co.PVT.Ltd. New Delhi.\n8. Puri, V.K. 1969. Natural Frequency of Block Foundation under Free and Force Vibration. Master Thesis. University of Roorkee. India.\n9. Rosyidi, S.A., Taha, M.R., Nayan, K.A.M. & Ismail, A. 2005a. Predicting Soil Bearing Capacity of Pavement Subgrade System using SASW Method. Proceeding of the International Symposium of Geoline 2005, Lyon, France.\n10. Rosyidi, S.A., Taha, M.R. & Nayan, K.A.M. 2005b. Assessing In Situ Dynamic Stiffness of Pavement Layers with Simple Seismic Test. Proceeding of International Seminar and Exhibition on Road Constructions. Semarang, Indonesia. pp.15-24.\n11. Yoder, E.J. & Witczak, M.W. 1975. PRINCIPLES OF PAVEMENT DESIGN. New York: John Wiley & Son, Inc."
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https://physics.stackexchange.com/questions/263791/confusion-with-newtons-third-law | [
"# Confusion with Newton's third law\n\n“Every action has an equal and opposite reaction.”\n\nI have a query about the word every in that sentence.\n\nSuppose we have two objects A and B. A pushes B with a force of 5N and B will push A with a force of 5N. However, won’t the reaction that B has caused on A, serve as an action for A, causing A to again push B with its reaction and thus making a total of 10N? (And then, of course, B will also apply a force of 10N on A.)\n\n• You are counting the A on B force twice. There are not three (or more) forces, only two. – garyp Jun 21 '16 at 1:47\n• The reaction isn't really a separate thing from the action. They come as a pair. The 'action' is the reaction to the reaction (should that be rereaction?). – dmckee --- ex-moderator kitten Jun 21 '16 at 2:09\n• I think it means at every point of application, each force can be matched with a force of equal magnitude pointing in the opposite direction. – Emil Jun 21 '16 at 5:33\n• The law says every action HAS an equal and opposite reaction, not every action CREATES an equal and opposite reaction. If you start with A, then A's push on B will \"create\" the reaction, but B's push on A will already have its reaction, namely the original push. – João Mendes Jun 21 '16 at 10:52\n• Another analogy: \"every married person has a spouse\". I'm married, so I have a wife. But that means my wife is married, so she must have a husband, and that husband must have a wife, and ... – Harry Johnston Jun 21 '16 at 23:10\n\nThe way we are all taught Newton's Laws (by reciting them like mantras as children) is unfortunate because the traditional wording is misleading in many ways.\n\nA big problem (though not the only one) with the traditional wording of both Newton's second and third laws is that they incorrectly suggest cause and effect (and hence imply a chain of events, as you put it).\n\nNewton's second law, for example, suggests that a force 'causes' an acceleration, implying it happens first. It doesn't. The force and the acceleration occur jointly and concurrently, despite the persistent misconception and stubborn illusion of a temporal sequence.\n\nBut let's not get distracted with the second law right now, because you are understandably perplexed by the third ...\n\nAgain, the wording of the third law suggests that an 'action' happens first and then it 'causes' a 'reaction'. If this were literally true, you'd have every right to cry infinite regress!\n\nThe truth is, the forces occur jointly and simultaneously, and are not the causes of each other. If you want a better way to think about it, you can hardly do better than the way Newton himself came up with the third law. He argued for it as follows:\n\nSuppose you had a system of two objects interacting with each other, with no external forces acting on the system. Then you should be able to consider that system as a 'whole' if you want to, and from that perspective the system as a whole must not accelerate as it has no net force acting on it. But this can only be the case if the two objects making up the system have equal and opposite forces between them (i.e. all internal forces of the system must cancel out).\n\nDo you see how this argument does not involve any 'causal sequence' or 'chain' of forces? It is just an observation about what must be the case in order for Newton's force-based scheme to work consistently.\n\nNot convinced? Let me try an analogy. You and your friend each have a certain amount of money. You buy something from your friend. Your balance goes down and your friend's goes up. Was there a time-delayed causal sequence here? Nope. Your balance decreased concurrently (as you handed over the money) as your friend's balance increased. Looking at the system as a whole, we know that since no money flowed into or out of the system during the transaction, the net balance must be zero. Every payment entails a receipt and every receipt entails a payment, but, despite the illusion, there is no sequence (much less a perpetual one!).\n\nNote: You could also translate this argument into the language of momentum conservation, but I have tried to answer the question in the same language in which you phrased it.\n\nUPDATE: The 'infinite regress' problem highlighted here is not the only confusion that arises when we use the suggestive language of 'action' and 'reaction'. I've identified two other problems this language causes along with my proposed solution here.\n\n• Thanks for a great answer! But i did not get the crux! Whats the key point that you’re trying to highlight, that proves my statement wrong? – Aaryan Dewan Jun 21 '16 at 15:13\n• The gist is this: The popular idea of A pushing/pulling B and then B pushing/pulling A back in response is wrong. In the Newtonian model, at any given time each body in the system has a single force vector assigned to it (determined by whatever force-laws are in play). At any given time the sum of all those vectors is zero (assuming the system is isolated). If there are only two bodies in the system, this means their forces are equal and opposite. But this is not to be thought of as some sort of sequential tit-for-tat. There is just a background law assigning forces to bodies at each instant. – Physics Footnotes Jun 21 '16 at 15:25\n• Newton's language of Action/Reaction turns my clinical account into a compelling narrative that makes the story understandable. But if you take the narrative too seriously it achieves the exact opposite, leading as it has in your case, into the absurd idea that an object must remember that it has already pushed another object once so it had better not push again or else it will ruin all the equations. – Physics Footnotes Jun 21 '16 at 15:45\n• Please let me know if the above comments help, because if they do I'll transcribe them into my main answer for the benefit of others. – Physics Footnotes Jun 21 '16 at 16:10\n• Yeah I like the idea of \"the push\" being the causal agent of both the force and the acceleration. Both force and acceleration are technical parts of the abstraction, whereas \"the push\" is more colloquial, since we haven't decided what exactly this push is acting on. I think it's very dangerous to say both \"the force is not the cause of the acceleration\" and \"F=ma\" in the same sentence. The only correct form of N2 is $\\sum \\vec{F}=m\\vec{a}$, and then it's clear that \"THE force is the cause....\" is wrong. – levitopher Jun 24 '16 at 20:28\n\nIt does not matter which one you call action and which one reaction, what the law says is that two objects make the same force on each other. Thus if A pushed B with a force of 5 Newtons, B will also push A with the same force.\n\nYou need a better statement of Newton's Law. The one you are using is meaningless, because the word action is not defined. (In today's language of physics the word action is used in an entirely different context, sense, and meaning.) It's based on what Newton wrote, but is only half of what he wrote. Wikipedia gives us the whole thing\n\nTo every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.\n\nBut this statement still has the archaic language. We need to stop repeating Newton's use of the word action. It means nothing to us today.\n\nI intended to find a link to a good version of the law. I could not find one. I'm sure there are good ones out there, but a quick search came up with a bunch of bad ones, almost all of them using the words action and reaction. Here's a version based on the statement of the law in the book \"Integrated Science\" by Tillery.\n\nWhenever two objects interact, the force exerted on the first by the second is equal in strength and opposite in direction to the force exerted on the second by the first.\n\nBut no matter how it is phrased, it needs explanation and elaboration. The short statements of the law are not good ways to learn what the law is about. Hyperphysics (scroll to the bottom) does a pretty good job explaining it.\n\nYou're assuming \"action\" and \"reaction\" are the same things. Ok, there are lots of reasons why we might consider them to be either similar or exactly the same, but since the third law doesn't say \"for every action there is an equal and opposite action\", the third law is logically consistent.\n\nFor the purposes of the third law, reaction is NOT the same as action.\n\nThis might sound like a bit of a linguistic, rather then physical, argument, but the third law is motivated by the form of our analysis. We develop interactions as forces, and then we ask what happens under action of those forces. Well, for every force we 100% know exists (\"me pushing on the wall\"), the third law tells us that there is another, reactionary force (\"the wall pushing on me\"). \"reaction\" is a kind of force, not a name for all forces.\n\nHow about considering a specific force, such as the Newtonian gravitational force between two point masses $m_1$ and $m_2$. We could write the force that mass 1 exerts on mass 2 as follows:\n\n$$\\vec{F}_{12} =-G\\frac{m_1 m_2} {r_{12}^2} \\hat{r}_{12},$$\n\nwhere $r_{12}$ and $\\hat{r}_{12}$ are the distance 2 is from 1 and unit vector from 1 to 2, respectively. Does 2 exert a force on 1? According to Newton's Universal Law of Gravitation, it does (just as it's written above):\n\n$$\\vec{F}_{21} =-G\\frac{m_2 m_1} {r_{21}^2} \\hat{r}_{21}.$$\n\nHow are $\\vec{F}_{21}$ and $\\vec{F}_{21}$ similar? How are they different? Is there a difference in sign?\n\nIf you want to consider EVERY so-called action, then you need to consider different forces, keeping in mind the idea of \\it{force} being classical.\n\nRegarding (re)action, which force is exerted \"first\"? I don't think one can say. As mentioned in other answers, the wording is unfortunate and misleadingly.\n\nNote that Newton's Third Law is often (re)stated as follows:\n\nForces come in equal and opposite pairs.\n\nI might be a little late in answering this question, but I hope it'll help.\n\nPhysics Footnotes emphasised that action and reaction forces occur simultaneously at the same time. This is correct. The words action and reaction do create the misconception that one force is the reason for the other.\n\nLet me explain why these two forces happen simultaneously, with that knowledge you will understand why the reaction force(so called) does not act as yet another force. To understand this we must understand what causes a force.\n\nA force is always caused by the interaction of two bodies.(We are not considering fictitious forces here) These two objects can be touching each other or may be some distance apart(Eg - gravitational force)\n\nTaking your example, how would A have exerted the force of 5N, upon B? Possibly, A was in motion (relative to the earth) previously and would collide with B, thus exerting the force. (It is easy to visualise A as your hand and B as some object)\n\nBefore the collision, A would be in motion towards B,relative to the earth and thus, relative to B (B is in the same state of motion as the earth, which is the state of being still. So, as the earth and B share the same state of motion, a motion relative to the earth, is a motion relative to B as well) Also, relative to A, B would be in motion towards A.( If we regard A to be stationary, then B is the one that is moving)\n\n• So, relative to A, B is the moving object and A is the stationary object. So, when the two objects collide, relative to A, It is B that exerts the force. (Because relative to A, it is B that comes towards it and collides) Relative to A, A was just moving along happily until This B came along and bumped into him, exerting a force upon A\n• As in the traditional view, it is A that moves and collides with B. So, the collision would cause a force from A on B. Relative to B, B was sitting still, minding its own business until A came and bumped into him, thus exerting the force.\n\nIt is these two forces that are popularly regarded as action reaction force pairs.\n\nBoth forces exist simultaneously, relative to A, the acting force is a force of 5N, exerted upon it by B. Relative to B, the acting force is force of 5N exerted upon it by A.\n\nWith this understanding on how the forces concerned are created, it is easy understand that that the force exerted by B upon A (F1)does not lead to a whole new reaction force(so called) In fact, relative to A, the force exerted by it upon B can be regarded as the reaction force,to F1 (with the true sense of the third law, of course, which is to say that it is not a result of the first force)\n\nIf there is a certain force in some direction, the existence of another force is unavoidable( in the opposite direction, acting on the object which caused the force, to which we referred to first) Which is to say forces exist in pairs. Now, relative to A, it is B that exerts the force, So, relative to A, the action is the force exerted by B upon A.( Let's call it F1) Even as this action takes place relative to A, relative to B yet another action takes place. ( force exerted by A upon B, F2) For A, The action is F1. At the same time F2 is also exerted. For A, F2 is an indirect consequence of the collision. It can be considered to be the reaction force relative to A. ( The term \"reaction\" is not really suitable, but that is the commonly used term) Likewise For B, F1 is the indirect consequence; reaction. The creation of a whole new force as a reaction is not what Newton meant by the third law. So, every action has a reaction, as stated by the third law\n\n• Great answer! Thanks a lot. Can you please explain the last part, or the crux of your answer? – Aaryan Dewan Jul 5 '16 at 4:34\n• I edited my answer, see if it answers your problem – SNB Jul 5 '16 at 15:38",
null,
"It should be \"Every action has equal and opposite reaction of SAME TYPE\" , the best way to understand these contact force related questions is to draw a large clear free body diagram. Which will eventually lights up your problems.\n\nDraw all possible forces on both object(s) and or contact surface.\n\nPls see the image , the force you applied is not belong to the action, it is a separate force, action reaction forces are (Ra and Rb) their magnitude remains equal regardless of situation (acceleration, declaration, constant velocity etc) , and those are the action reaction pair, not the force you applied on the body.\n\nIf you are concerning about the reaction force due to 5N force applied on the object A , it will be acting on your hand which is still 5N acting opposite direction. Also to clarify your concern, newton pair of force always acts on different bodies not on the same body\n\n• please read my question again. – Aaryan Dewan Jun 21 '16 at 4:35\n• Pls see the image , the force you applied is not belong to the action, it is a separate force, action reaction forces are (Ra and Rb) their magnitude remains equal regardless of situation (acceleration, declaration, constant velocity etc) , and those are the action reaction pair, not the force you applied on the body. If you are concerning about the reaction force due to 5N force applied on the object A , it will be acting on your hand which is still 5N acting opposite direction. Also to clarify your concern, newton pair of force always acts on different bodies not on same body – Manoj Jun 21 '16 at 5:48"
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https://www.quantstart.com/articles/Cholesky-Decomposition-in-Python-and-NumPy/ | [
"# Cholesky Decomposition in Python and NumPy\n\nCholesky Decomposition in Python and NumPy\n\nFollowing on from the article on LU Decomposition in Python, we will look at a Python implementation for the Cholesky Decomposition method, which is used in certain quantitative finance algorithms.\n\nIn particular, it makes an appearance in Monte Carlo Methods where it is used to simulating systems with correlated variables. Cholesky decomposition is applied to the correlation matrix, providing a lower triangular matrix L, which when applied to a vector of uncorrelated samples, u, produces the covariance vector of the system. Thus it is highly relevant for quantitative trading.\n\nCholesky decomposition assumes that the matrix being decomposed is Hermitian and positive-definite. Since we are only interested in real-valued matrices, we can replace the property of Hermitian with that of symmetric (i.e. the matrix equals its own transpose). Cholesky decomposition is approximately 2x faster than LU Decomposition, where it applies.\n\nIn order to solve for the lower triangular matrix, we will make use of the Cholesky-Banachiewicz Algorithm. First, we calculate the values for L on the main diagonal. Subsequently, we calculate the off-diagonals for the elements below the diagonal:\n\n\\begin{eqnarray*} l_{kk} &=& \\sqrt{ a_{kk} - \\sum^{k-1}_{j=1} l^2_{kj}}\\\\ l_{ik} &=& \\frac{1}{l_{kk}} \\left( a_{ik} - \\sum^{k-1}_{j=1} l_{ij} l_{kj} \\right), i > k \\end{eqnarray*}\n\nAs with LU Decomposition, the most efficient method in both development and execution time is to make use of the NumPy/SciPy linear algebra (linalg) library, which has a built in method cholesky to decompose a matrix. The optional lower parameter allows us to determine whether a lower or upper triangular matrix is produced:\n\nimport pprint\nimport scipy\nimport scipy.linalg # SciPy Linear Algebra Library\n\nA = scipy.array([[6, 3, 4, 8], [3, 6, 5, 1], [4, 5, 10, 7], [8, 1, 7, 25]])\nL = scipy.linalg.cholesky(A, lower=True)\nU = scipy.linalg.cholesky(A, lower=False)\n\nprint \"A:\"\npprint.pprint(A)\n\nprint \"L:\"\npprint.pprint(L)\n\nprint \"U:\"\npprint.pprint(U)\n\n\nThe output from the code is given below:\n\nA:\narray([[ 6, 3, 4, 8],\n[ 3, 6, 5, 1],\n[ 4, 5, 10, 7],\n[ 8, 1, 7, 25]])\nL:\narray([[ 2.44948974, 0. , 0. , 0. ],\n[ 1.22474487, 2.12132034, 0. , 0. ],\n[ 1.63299316, 1.41421356, 2.30940108, 0. ],\n[ 3.26598632, -1.41421356, 1.58771324, 3.13249102]])\nU:\narray([[ 2.44948974, 1.22474487, 1.63299316, 3.26598632],\n[ 0. , 2.12132034, 1.41421356, -1.41421356],\n[ 0. , 0. , 2.30940108, 1.58771324],\n[ 0. , 0. , 0. , 3.13249102]])\n\n\nAs with LU Decomposition, it is unlikely that you will ever need to code up a Cholesky Decomposition in pure Python (i.e. without NumPy/SciPy), since you can just include the libraries and use the far more efficient implements found within. However, for completeness I have included the pure Python implementation of the Cholesky Decomposition so that you can understand how the algorithm works:\n\nfrom math import sqrt\nfrom pprint import pprint\n\ndef cholesky(A):\n\"\"\"Performs a Cholesky decomposition of A, which must\nbe a symmetric and positive definite matrix. The function\nreturns the lower variant triangular matrix, L.\"\"\"\nn = len(A)\n\n# Create zero matrix for L\nL = [[0.0] * n for i in xrange(n)]\n\n# Perform the Cholesky decomposition\nfor i in xrange(n):\nfor k in xrange(i+1):\ntmp_sum = sum(L[i][j] * L[k][j] for j in xrange(k))\n\nif (i == k): # Diagonal elements\n# LaTeX: l_{kk} = \\sqrt{ a_{kk} - \\sum^{k-1}_{j=1} l^2_{kj}}\nL[i][k] = sqrt(A[i][i] - tmp_sum)\nelse:\n# LaTeX: l_{ik} = \\frac{1}{l_{kk}} \\left( a_{ik} - \\sum^{k-1}_{j=1} l_{ij} l_{kj} \\right)\nL[i][k] = (1.0 / L[k][k] * (A[i][k] - tmp_sum))\nreturn L\n\nA = [[6, 3, 4, 8], [3, 6, 5, 1], [4, 5, 10, 7], [8, 1, 7, 25]]\nL = cholesky(A)\n\nprint \"A:\"\npprint(A)\n\nprint \"L:\"\npprint(L)\n\n\nThe output from the pure Python implementation is given below:\n\nA:\n[[6, 3, 4, 8], [3, 6, 5, 1], [4, 5, 10, 7], [8, 1, 7, 25]]\nL:\n[[2.449489742783178, 0.0, 0.0, 0.0],\n[1.2247448713915892, 2.1213203435596424, 0.0, 0.0],\n[1.6329931618554523, 1.414213562373095, 2.309401076758503, 0.0],\n[3.2659863237109046,\n-1.4142135623730956,\n1.5877132402714704,\n3.1324910215354165]]\n\n\nThe SciPy implementation and the pure Python implementation both agree, although we haven't calculated the upper version for the pure Python implementation. In production code you should use SciPy as it will be significantly faster at decomposing larger matrices."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.71294,"math_prob":0.99515694,"size":4332,"snap":"2019-51-2020-05","text_gpt3_token_len":1444,"char_repetition_ratio":0.115295745,"word_repetition_ratio":0.06865671,"special_character_ratio":0.39358264,"punctuation_ratio":0.23217922,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987934,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-20T15:17:54Z\",\"WARC-Record-ID\":\"<urn:uuid:913b25a1-fe43-4e09-bd51-e9b1fb149f7e>\",\"Content-Length\":\"17837\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3d210a4e-2da6-49e8-9b9f-168e4ee9a721>\",\"WARC-Concurrent-To\":\"<urn:uuid:3be7cdb2-f65b-4bc8-976c-4d221842d617>\",\"WARC-IP-Address\":\"23.23.50.20\",\"WARC-Target-URI\":\"https://www.quantstart.com/articles/Cholesky-Decomposition-in-Python-and-NumPy/\",\"WARC-Payload-Digest\":\"sha1:GZ44EOUEGL36CZXFXHVSBCADADFKKH7D\",\"WARC-Block-Digest\":\"sha1:FGROCOHZIDEXCTX24JJLLHDFLNBIUM5T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250598800.30_warc_CC-MAIN-20200120135447-20200120164447-00406.warc.gz\"}"} |
https://www.elsevier.com/books/feedback-systems-input-output-properties/desoer/978-0-12-212050-3 | [
"",
null,
"# Feedback Systems: Input-output Properties\n\n1st Edition - January 28, 1975\n\nWrite a review\n\n• Editor: C.A. Desoer\n• eBook ISBN: 9780323157797\n\n## Purchase options\n\nPurchase options\nDRM-free (PDF)\nSales tax will be calculated at check-out\n\n#### Institutional Subscription\n\nFree Global Shipping\nNo minimum order\n\n## Description\n\nFeedback Systems: Input-output Properties deals with the basic input-output properties of feedback systems. Emphasis is placed on multiinput-multioutput feedback systems made of distributed subsystems, particularly continuous-time systems. Topics range from memoryless nonlinearities to linear systems, the small gain theorem, and passivity. Norms and general theorems are also considered. This book is comprised of six chapters and begins with an overview of a few simple facts about feedback systems and simple examples of nonlinear systems that illustrate the important distinction between the questions of existence, uniqueness, continuous dependence, and boundedness with respect to bounded input and output. The next chapter describes a number of useful properties of norms and induced norms and of normed spaces. Several theorems are then presented, along with the main results concerning linear systems. These results are used to illustrate the applications of the small gain theorem to different classes of systems. The final chapter outlines the framework necessary to discuss passivity and demonstrate the applications of the passivity theorem. This monograph will be a useful resource for mathematically inclined engineers interested in feedback systems, as well as undergraduate engineering students.\n\n• Preface\n\nAcknowledgments\n\nList of Symbols\n\nI Memoryless Nonlinearities\n\n1 Sector Conditions\n\n2 Linear Feedback around a Nonlinearity (Memoryless Case)\n\n3 Multiple Nonlinearities\n\nNotes and References\n\nII Norms\n\n1 Norms: Definitions and Examples\n\n2 Equivalent Norms\n\n3 Relations between Normed Spaces\n\n4 Geometric Interpretation of Norms\n\n5 Induced Norms of Linear Maps\n\n6 Two Examples\n\n8 The Measure of a Matrix\n\nNotes and References\n\nIII General Theorems\n\n1 Setting of the Problem\n\n2 Small Gain Theorem\n\n3 Small Gain Theorem: Incremental Form\n\n4 A Boundedness Result\n\n5 An Existence and Uniqueness Theorem\n\n6 Loop Transformation Theorem\n\n7 L Stability\n\n8 General Feedback Formula\n\nNotes and References\n\nIV Linear Systems\n\n0 Introduction\n\n1 Linear Feedback Systems with Rational Transfer Functions\n\n2 Necessary and Sufficient Conditions: Factorization Method\n\n3 Linear Feedback Systems with Dynamics in the Feedback Path (Rational Transfer Functions Case)\n\n4 Convolution Feedback Systems\n\n5 Graphical Test\n\n6 Discrete-Time Systems\n\n7 Linear Time-Varying Systems\n\n8 Slowly Varying Systems\n\n9 Linearization\n\nNotes and References\n\nV Applications of the Small Gain Theorem\n\n1 Continuous-Time Systems—LP Stability\n\n2 L2 Stability—Circle Criterion\n\n3 Exponential Weighting—L∞ Stability\n\n4 Discrete-Time Systems—LP Stability\n\n5 Slowly-Varying Linear Systems\n\n6 Nonlinear Circuit Example\n\n7 Existence of Periodic Solutions\n\n8 Popov Criterion\n\n9 Instability\n\nNotes and References\n\nVI Passivity\n\n0 Introduction\n\n1 Motivation from Circuit Theory\n\n2 Scalar Products\n\n3 Formal Framework\n\n4 Passive Systems: Definition and Examples\n\n5 Passivity Theorem\n\n6 The Popov Criterion\n\n7 Discrete-Time Case\n\n8 Average Logarithmic Variation Criterion\n\n9 Multiplier Theory\n\n10 Relation between the Passivity Theorem and the Small Gain Theorem\n\n11 Invertibility of I + H\n\n12 Instability Theorems\n\nNotes and References\n\nAppendixes\n\nA Integrals and Series\n\nA.1 Regulated Functions\n\nA.2 Integrals\n\nA.3 Series\n\nB Fourier Transforms\n\nB.1 L1 Theory\n\nB.2 L2 Theory\n\nB.3 Laplace Transform\n\nC Convolution\n\nC.1 Introduction\n\nC.2 Convolution of Functions\n\nC.3 Convolution of a Measure and a Function\n\nC.4 Convolution of Sequences\n\nD Algebras\n\nD.1 Algebras\n\nD.2 Ideals\n\nD.3 Inverses in A\n\nE Bellman-Gronwall Lemma\n\nReferences\n\nIndex\n\n## Product details\n\n• No. of pages: 2\n• Language: English"
] | [
null,
"https://secure-ecsd.elsevier.com/covers/80/Tango2/large/9780122120503.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7829713,"math_prob":0.6830013,"size":4259,"snap":"2022-05-2022-21","text_gpt3_token_len":979,"char_repetition_ratio":0.12949471,"word_repetition_ratio":0.0,"special_character_ratio":0.19464663,"punctuation_ratio":0.07834758,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9741592,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-20T11:56:29Z\",\"WARC-Record-ID\":\"<urn:uuid:4f66e1f2-dcb1-42e9-ba6d-6abf63cad86f>\",\"Content-Length\":\"192782\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:133d73c6-5183-4350-aaa3-58fa0455b5d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:c72e9900-f9fd-4d15-b69d-34e2bb6d0455>\",\"WARC-IP-Address\":\"203.82.26.7\",\"WARC-Target-URI\":\"https://www.elsevier.com/books/feedback-systems-input-output-properties/desoer/978-0-12-212050-3\",\"WARC-Payload-Digest\":\"sha1:SSR4EVZOPTWQ5CLPCMQH4Y5KFOZATC4U\",\"WARC-Block-Digest\":\"sha1:LHGISTOMVJANAE3EQ7M4NCX6ZFMVITZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662531779.10_warc_CC-MAIN-20220520093441-20220520123441-00120.warc.gz\"}"} |
https://www.actucation.com/grade-6-maths/word-problems- | [
"Informative line\n\n# Word Problems on Fractions\n\n• To solve the word problems, we need to follow the given steps:\n\nStep 1: Read and understand the given situations.\n\n• In this step, we need to identify the keyword that will help us to determine which operation symbol to be used.\n\nStep 2: Choose the correct operation and solve the problem.\n\nTo understand the above steps in a better way, we should understand how problems involve the basic four operations:\n\n1. Addition: A problem in which two or more quantities are being combined.\n\nKeywords: Sum, in all, combined, altogether, added etc.\n\nFor example: Emma bought 2 cupcakes and Carl bought 5. How many cupcakes did they buy in all?\n\nKeyword = 'in all'\n\n2. Subtraction: A problem in which two quantities are given and one is being removed from another.\n\nKeywords: Left, left over, decreased by, take away, less, less than etc.\n\nFor example: Jose takes $$\\dfrac{1}{4}$$ minutes less than Marc to reach the staff-room from the corridor. If Marc takes $$\\dfrac{3}{4}$$ minutes to reach the staff-room, how many minutes does Jose take?\n\nIn this example, 'less than' is the keyword that helps to identify that subtraction is to be done.\n\n3. Multiplication: A problem in which repeated addition needs to be done.\n\nKeywords: Each, every, how much, of etc.\n\nFor example: What is $$\\dfrac{1}{2}$$ of $$4$$?\n\nIn this example, 'of' is the keyword that helps us to identify that we should use multiplication to solve this problem.\n\n4. Division: A problem in which a single quantity is divided into many equal parts.\n\nKeywords: Split, divide, shared, per etc.\n\nFor example: Kelvin teaches 8 students. He divides them equally into two groups. How many students are in each group?\n\nHere, the keyword is 'divided ', which means we need to divide.\n\n#### Lara runs a mile in an average of $$9 \\dfrac{3}{6}$$ minutes. She does a cool down jog in an average of $$3 \\dfrac{2}{3}$$ minutes. If her daily workout is running a mile and then doing the cool down jog, how long does it take her to do this workout?\n\nA $$\\dfrac{54}{5}$$ minutes\n\nB $$13\\dfrac{5}{6}$$ minutes\n\nC $$\\dfrac{58}{3}$$ minutes\n\nD $$13\\dfrac{1}{6}$$ minutes\n\n×\n\nAverage time taken by Lara to run a mile $$=9\\dfrac{3}{6}$$ minutes\n\n$$=\\dfrac{57}{6}$$ minutes\n\nAverage time taken by her for cool down jog $$=3\\dfrac{2}{3}$$ minutes\n\n$$=\\dfrac{11}{3}$$ minutes\n\nTotal time taken for workout\n\n$$=\\dfrac{57}{6}+\\dfrac{11}{3}$$\n\n$$=\\dfrac{57+22}{6} = \\dfrac{79}{6}$$\n\n$$=\\dfrac{79}{6}$$ minutes\n\n$$=13\\dfrac{1}{6}$$ minutes\n\nThus, the total time taken for workout is $$13\\dfrac{1}{6}$$ minutes.\n\nHence, option (D) is correct.\n\n### Lara runs a mile in an average of $$9 \\dfrac{3}{6}$$ minutes. She does a cool down jog in an average of $$3 \\dfrac{2}{3}$$ minutes. If her daily workout is running a mile and then doing the cool down jog, how long does it take her to do this workout?\n\nA\n\n$$\\dfrac{54}{5}$$ minutes\n\n.\n\nB\n\n$$13\\dfrac{5}{6}$$ minutes\n\nC\n\n$$\\dfrac{58}{3}$$ minutes\n\nD\n\n$$13\\dfrac{1}{6}$$ minutes\n\nOption D is Correct\n\n# Mathematical Modeling on Addition and Subtraction\n\nMany times in real life problems, we need to deal with more than one operations. Only addition or only subtraction would not help. We need to make use of the combination of operations in order to solve the problem.\n\nFor example:\n\nA town has a total population of $$20,000$$. There are $$2,000$$ children, $$1,250$$ youths, $$750$$ infants and rest are aged people. What fraction of the total population represents the aged people?\n\nTotal population of town $$=20,000$$\n\nNumber of children $$=2,000$$\n\nFraction of children to total population $$=\\dfrac {2,000}{20,000}$$\n\nNumber of youths $$=1,250$$\n\nFraction of youths to total population $$=\\dfrac {1,250}{20,000}$$\n\nNumber of infants $$=750$$\n\nFraction of infants to total population $$=\\dfrac {750}{20,000}$$\n\nTotal fraction of children, youths and infants\n\n$$=\\dfrac {2,000}{20,000}+\\dfrac {1,250}{20,000}+\\dfrac {750}{20,000}$$\n\n$$=\\dfrac {4,000}{20,000}=\\dfrac {1}{5}$$\n\n$$\\therefore\\;$$ Fraction of aged people $$=1-\\dfrac {1}{5}=\\dfrac {4}{5}$$\n\nThus, $$\\dfrac {4}{5}$$ of the total population represents the aged people.\n\n#### In a chutes and ladders game board, there are $$100$$ square blocks, $$9$$ chutes and $$10$$ ladders. What fraction of the total square blocks have the chutes and the ladders? Also find the fraction of the rest of the blocks.(chutes and ladders belong to the blocks where they begin)\n\nA $$\\dfrac {19}{100},\\dfrac {81}{100}$$\n\nB $$\\dfrac {18}{100},\\dfrac {82}{100}$$\n\nC $$\\dfrac {17}{100},\\dfrac {83}{100}$$\n\nD $$\\dfrac {16}{100},\\dfrac {84}{100}$$\n\n×\n\nTotal number of square blocks $$=100$$\n\nNumber of chutes $$=9$$\n\nNumber of ladders $$=10$$\n\nFraction of $$9$$ chutes to total square blocks $$=\\dfrac {9}{100}$$\n\nFraction of $$10$$ ladders to total square blocks $$=\\dfrac {10}{100}$$\n\nTotal fraction $$=\\dfrac {9}{100}+\\dfrac {10}{100}$$\n\n$$=\\dfrac {19}{100}$$\n\nFraction for remaining blocks $$=1-\\dfrac {19}{100}$$\n\n$$=\\dfrac {81}{100}$$\n\nThus, $$\\dfrac {19}{100}$$ part of the total square blocks have chutes and ladders while remaining blocks constitute $$\\dfrac {81}{100}$$ part of the total square blocks.\n\nHence, option (A) is correct.\n\n### In a chutes and ladders game board, there are $$100$$ square blocks, $$9$$ chutes and $$10$$ ladders. What fraction of the total square blocks have the chutes and the ladders? Also find the fraction of the rest of the blocks.(chutes and ladders belong to the blocks where they begin)\n\nA\n\n$$\\dfrac {19}{100},\\dfrac {81}{100}$$\n\n.\n\nB\n\n$$\\dfrac {18}{100},\\dfrac {82}{100}$$\n\nC\n\n$$\\dfrac {17}{100},\\dfrac {83}{100}$$\n\nD\n\n$$\\dfrac {16}{100},\\dfrac {84}{100}$$\n\nOption A is Correct\n\n# Mathematical Modeling on Addition and Division\n\nHere, we will use the combination of addition and division to solve the problem.\n\nFor example:\n\nCarl puts $$\\dfrac{4}{5}$$ kilogram of fruits in a basket. After a while he adds $$\\dfrac{1}{4}$$ kilogram of fruits more to it. He then distributes the total fruits equally among $$3$$ children. What quantity of fruits does each child get?\n\nQuantity of fruits added to the basket by Carl at first $$=\\dfrac{4}{5}$$ kg\n\nQuantity of fruits added later $$=\\dfrac{1}{4}$$ kg\n\nTotal quantity of fruits in the basket $$=\\dfrac{4}{5}+\\dfrac{1}{4}$$\n\n$$=\\dfrac{16+5}{20}$$\n\n$$=\\dfrac{21}{20}$$ kg\n\nCarl distributes the total fruits equally among $$3$$ children.\n\nSo, the quantity of fruits each child got $$=\\dfrac{21}{20}\\div3$$\n\n$$=\\dfrac{21}{20}\\times \\dfrac{1}{3}$$\n\n$$=\\dfrac{7}{20}$$ kg\n\nThus, each child got $$\\dfrac{7}{20}$$ kilogram of fruits.\n\n#### Sam has $$\\dfrac{5}{3}$$ pounds of tomatoes in a basket. He adds $$\\dfrac{3}{2}$$ pounds of tomatoes more to it and distributes them equally among $$4$$ kids. What quantity of tomatoes does each child get?\n\nA $$\\dfrac{19}{15}$$ pounds\n\nB $$\\dfrac{90}{19}$$ pounds\n\nC $$\\dfrac{20}{15}$$ pounds\n\nD $$\\dfrac{19}{24}$$ pounds\n\n×\n\nQuantity of tomatoes Sam already had in the basket $$=\\dfrac{5}{3}$$ pounds\n\nQuantity of tomatoes he added $$=\\dfrac{3}{2}$$ pounds\n\n$$\\therefore$$ Total quantity of tomatoes in the basket$$=\\dfrac{5}{3}+\\dfrac{3}{2}$$\n\n$$=\\dfrac{10+9}{6}$$\n\n$$=\\dfrac{19}{6}$$ pounds\n\nSam distributed the total quantity of tomatoes equally among $$4$$ kids.\n\nSo, the quantity of tomatoes that each kid got $$=\\dfrac{19}{6}\\div4$$\n\n$$=\\dfrac{19}{6}\\times\\dfrac{1}{4}$$\n\n$$=\\dfrac{19}{24}$$ pounds\n\nThus, each kid got $$\\dfrac{19}{24}$$ pounds of tomatoes.\n\nHence, option (D) is correct.\n\n### Sam has $$\\dfrac{5}{3}$$ pounds of tomatoes in a basket. He adds $$\\dfrac{3}{2}$$ pounds of tomatoes more to it and distributes them equally among $$4$$ kids. What quantity of tomatoes does each child get?\n\nA\n\n$$\\dfrac{19}{15}$$ pounds\n\n.\n\nB\n\n$$\\dfrac{90}{19}$$ pounds\n\nC\n\n$$\\dfrac{20}{15}$$ pounds\n\nD\n\n$$\\dfrac{19}{24}$$ pounds\n\nOption D is Correct\n\n# Mathematical Modeling on Division and Subtraction\n\nHere, we will use the combination of division and subtraction to solve the problem.\n\nFor example:\n\nChris and Sarah together had 4 apples. They divided all the apples equally among themselves. If Chris ate $$\\dfrac{1}{2}$$ of the apples he got, how many apples were left with him?\n\nTotal number of apples $$=4$$\n\nNumber of apples each got $$=4\\div 2$$\n\n$$=4\\times\\dfrac{1}{2}$$\n\n$$=2$$\n\nNumber of apples Chris ate $$=\\dfrac{1}{2}$$ of the apples he got $$=2\\times \\dfrac{1}{2}=1$$\n\nNumber of apples left with Chris $$=2-1$$\n\n$$=1$$\n\nThus, 1 apple was left with Chris.\n\n#### Keith and Kara have a $$5$$ liter juice can. They pour $$\\dfrac{2}{3}$$ fraction of juice in a bottle and divide it equally among themselves. If Keith drinks $$\\dfrac{1}{2}$$ of the juice he gets, how much quantity of juice is left with him?\n\nA $$\\dfrac{42}{5}$$ liters\n\nB $$\\dfrac{5}{6}$$ liters\n\nC $$\\dfrac{31}{2}$$ liters\n\nD $$\\dfrac{10}{3}$$ liters\n\n×\n\nTotal quantity of juice $$=5$$ liters\n\nQuantity of juice Kara and Keith poured $$=\\dfrac{2}{3}$$ of juice\n\n$$=5\\times \\dfrac{2}{3}$$\n\n$$=\\dfrac{10}{3}$$ liters\n\nQuantity of juice each got $$=\\dfrac{10}{3}\\div2$$\n\n$$=\\dfrac{10}{3}\\times\\dfrac{1}{2}$$\n\n$$=\\dfrac{5}{3}$$ liters\n\nQuantity of juice Keith drinks $$=\\dfrac{1}{2}$$ of juice he got\n\n$$=\\dfrac{1}{2}\\times\\dfrac{5}{3}=\\dfrac{5}{6}$$ liters\n\n$$\\therefore$$ Quantity of juice left with him\n\n$$=\\dfrac{5}{3}-\\dfrac{5}{6}$$\n\n$$=\\dfrac{30-15}{18}$$\n\n$$=\\dfrac{15}{18} =\\dfrac{5}{6}$$ liters\n\nThus, $$\\dfrac{5}{6}$$ liters of juice is left with Keith.\n\nHence, option (B) is correct.\n\n### Keith and Kara have a $$5$$ liter juice can. They pour $$\\dfrac{2}{3}$$ fraction of juice in a bottle and divide it equally among themselves. If Keith drinks $$\\dfrac{1}{2}$$ of the juice he gets, how much quantity of juice is left with him?\n\nA\n\n$$\\dfrac{42}{5}$$ liters\n\n.\n\nB\n\n$$\\dfrac{5}{6}$$ liters\n\nC\n\n$$\\dfrac{31}{2}$$ liters\n\nD\n\n$$\\dfrac{10}{3}$$ liters\n\nOption B is Correct\n\n#### Which one of the following options represents the mixed number which has the value between $$\\dfrac{7}{5} \\;and\\; \\dfrac{8}{5}$$?\n\nA $$2 \\dfrac{1}{5}$$\n\nB $$7 \\dfrac{1}{3}$$\n\nC $$1 \\dfrac{1}{2}$$\n\nD $$1 \\dfrac{1}{5}$$\n\n×\n\nGiven: $$\\dfrac{7}{5}$$ and $$\\dfrac{8}{5}$$\n\nFinding the equivalent fractions by multiplying numerator and denominator with the same non-zero number.\n\n$$\\dfrac{7}{5} = \\dfrac{7 \\times2}{5 \\times2} = \\dfrac{14}{10}$$\n\n$$\\dfrac{8}{5} = \\dfrac{8 \\times2}{5 \\times2} = \\dfrac{16}{10}$$\n\nThe two new fractions are:\n\n$$\\dfrac{14}{10}$$ and $$\\dfrac{16}{10}$$\n\nThus, we can say that the fraction between the above two fractions is:\n\n$$\\dfrac{15}{10}\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Bigg(\\because\\dfrac{14}{10},\\dfrac{15}{10},\\dfrac{16}{10}\\Bigg)$$\n\nThe greatest common factor of $$15$$ and $$10$$ is $$5$$. So, on simplifying the fraction $$\\dfrac{15}{10}$$, we get\n\n$$\\dfrac{15 \\div 5}{10 \\div 5}=\\dfrac{3}{2}$$\n\n$$3$$ and $$2$$ do not have any common factor other than 1. So, the fraction $$\\dfrac{3}{2}$$ is in its simplest form.\n\nOn dividing, we get the mixed fraction $$=1\\dfrac{1}{2}$$",
null,
"So, the mixed fraction between $$\\dfrac{7}{5}$$ and $$\\dfrac{8}{5}$$ is $$1\\dfrac{1}{2}$$.\n\nHence, option (C) is correct.\n\n### Which one of the following options represents the mixed number which has the value between $$\\dfrac{7}{5} \\;and\\; \\dfrac{8}{5}$$?\n\nA\n\n$$2 \\dfrac{1}{5}$$\n\n.\n\nB\n\n$$7 \\dfrac{1}{3}$$\n\nC\n\n$$1 \\dfrac{1}{2}$$\n\nD\n\n$$1 \\dfrac{1}{5}$$\n\nOption C is Correct\n\n# Mathematical Modeling on Addition and Multiplication\n\nMany times in real life problems, we need to deal with more than one operations. Only addition or only multiplication would not help. We need to make use of the combination of operations in order to solve the problem.\n\nFor example:\n\nCarl bought $$14$$ cookies at $$\\dfrac{2}{7}$$ each and $$20$$ candies at $$\\dfrac{4}{9}$$ each. What is the total amount he spent?\n\nNumber of cookies Carl bought $$=14$$\n\nPrice of $$1$$ cookie $$=\\dfrac{2}{7}$$\n\nPrice of $$14$$ cookies $$= 14 \\times \\dfrac{2}{7}$$ $$=4$$\n\nNumber of candies he bought $$=20$$\n\nPrice of $$1$$ candy $$=\\dfrac{4}{9}$$\n\nPrice of 20 candies $$=20 \\times \\dfrac{4}{9}$$ $$=\\dfrac{80}{9}$$\n\n$$\\therefore$$ Total money he spent $$=4+ \\dfrac{80}{9}$$\n\n$$=(\\dfrac{36+80}{9})$$\n\n$$=\\dfrac{116}{9}$$\n\n#### Sam bought $$13$$ chocolates and ate $$\\dfrac{1}{2}$$ of them. Casey bought $$10$$ chocolates and ate $$\\dfrac{3}{4}$$ of them. Find the total number of chocolates they both ate in all.\n\nA $$10$$\n\nB $$14$$\n\nC $$13$$\n\nD $$15$$\n\n×\n\nChocolates bought by Sam $$=13$$\n\nHe ate $$\\dfrac{1}{2}$$ of the chocolates $$=13\\times\\dfrac{1}{2}=\\dfrac{13}{2}$$\n\nChocolates bought by Casey $$=10$$\n\nShe ate $$\\dfrac{3}{4}$$ of the chocolates $$=\\dfrac{3}{4}\\times10 = \\dfrac{15}{2}$$\n\n$$\\therefore$$ Total number of chocolates they both ate $$=\\dfrac{13}{2}+\\dfrac{15}{2}$$\n\n$$=\\dfrac{13+15}{2}$$\n\n$$=\\dfrac{28}{2}$$\n\n$$=14$$\n\nThus , the total number of chocolates they both ate equals $$14$$.\n\nHence, option (B) is correct.\n\n### Sam bought $$13$$ chocolates and ate $$\\dfrac{1}{2}$$ of them. Casey bought $$10$$ chocolates and ate $$\\dfrac{3}{4}$$ of them. Find the total number of chocolates they both ate in all.\n\nA\n\n$$10$$\n\n.\n\nB\n\n$$14$$\n\nC\n\n$$13$$\n\nD\n\n$$15$$\n\nOption B is Correct\n\n# Mathematical Modeling on Subtraction and Multiplication\n\nHere, we will use the combination of subtraction and multiplication to solve the problem.\n\nFor example:\n\nMaria buys $$\\dfrac{1}{2}$$ kilogram(kg) of strawberries at the rate of $$10$$ per kg and Jacob buys the same amount of strawberries at the rate of $$12$$ per kg. Find who spends more and by how much.\n\nRate at which Maria buys strawberries $$=10$$ per kg\n\nQuantity of strawberries she buys $$=\\dfrac{1}{2}$$ kg\n\nTotal price of $$\\dfrac{1}{2}$$ kilogram of strawberries $$=\\dfrac{1}{2}\\times10$$ $$=5$$\n\nRate at which Jacob buys strawberries $$=12$$ per kg\n\nQuantity of strawberries he buys $$=\\dfrac{1}{2}$$ kg\n\nTotal price of $$\\dfrac{1}{2}$$ kilogram of strawberries $$=\\dfrac{1}{2}\\times12$$ $$=6$$\n\nThus, Jacob spends more.\n\nThe amount Jacob spends more $$= 6-5$$ $$=1$$\n\nHence, Jacob spends $$1$$ more than Maria.\n\n#### Kyle had $$3$$ bottles of milk. Each bottle contained $$\\dfrac{3}{5}$$ liters of milk. She emptied the milk of each bottle into a big jar, out of which she used $$\\dfrac{2}{5}$$ liters of milk for making coffee. How much quantity of milk was left in the jar?\n\nA $$\\dfrac{3}{4}\\;liters$$\n\nB $$\\dfrac{7}{5}\\;liters$$\n\nC $$\\dfrac{2}{3}\\;liters$$\n\nD $$\\dfrac{2}{5}\\;liters$$\n\n×\n\nTotal number of bottles of milk $$=3$$\n\nQuantity of milk in each bottle $$=\\dfrac{3}{5}$$ liters\n\nTotal quantity of milk in $$3$$ bottles $$=3\\times \\dfrac{3}{5}$$\n\n$$=\\dfrac{9}{5}$$ [Total quantity of milk in the jar before making coffee]\n\nQuantity of milk she used for coffee $$=\\dfrac{2}{5}$$ liters\n\n$$\\therefore$$ Quantity of milk left in the jar $$=\\dfrac{9}{5}-\\dfrac{2}{5}$$\n\n$$=\\dfrac{9-2}{5}$$\n\n$$=\\dfrac{7}{5}$$ liters\n\nThus, $$\\dfrac{7}{5}$$ liters of milk was left in the jar.\n\nHence, option (B) is correct.\n\n### Kyle had $$3$$ bottles of milk. Each bottle contained $$\\dfrac{3}{5}$$ liters of milk. She emptied the milk of each bottle into a big jar, out of which she used $$\\dfrac{2}{5}$$ liters of milk for making coffee. How much quantity of milk was left in the jar?\n\nA\n\n$$\\dfrac{3}{4}\\;liters$$\n\n.\n\nB\n\n$$\\dfrac{7}{5}\\;liters$$\n\nC\n\n$$\\dfrac{2}{3}\\;liters$$\n\nD\n\n$$\\dfrac{2}{5}\\;liters$$\n\nOption B is Correct\n\n# Mathematical Modeling on Multiplication and Division\n\nHere, we will use the combination of multiplication and division to solve the problem.\n\nFor example:\n\nFor a project, Isaac buys $$6$$ packets of pens for $$\\dfrac{20}{3}$$ each and decides to share the price among the $$5$$ project team members including him. How much does each member have to pay?\n\nPackets of pens Isaac buys $$=6$$\n\nPrice of $$1$$ packet $$=\\dfrac{20}{3}$$\n\nPrice of $$6$$ packets of pens $$=6\\times \\dfrac{20}{3}$$\n\n$$=40$$\n\nIsaac shares the price with his $$5$$ project team members $$=40\\div 5$$\n\n$$=40\\times \\dfrac{1}{5}$$\n\n$$=8$$\n\nThus, each member pays $$8$$.\n\n#### Julie has to clean her house, so she decides to do $$200\\,\\text{m}^2$$ area of her house today. If she cleans only $$\\dfrac{1}{5}$$ area of what she decided and asks her $$4$$ sisters to clean the rest of the area, find the area that each sister has to clean. Assume each sister has to clean equal amount of area.\n\nA $$25\\;\\text{m}^2$$\n\nB $$40\\;\\text{m}^2$$\n\nC $$30\\;\\text{m}^2$$\n\nD $$35\\;\\text{m}^2$$\n\n×\n\nTotal area to be cleaned $$=200\\;\\text{m}^2$$\n\nJulie cleans only $$\\dfrac{1}{5}$$ area of what she decided $$=200\\times\\dfrac{1}{5}$$\n\n$$=40\\;\\text{m}^2$$\n\nArea remained uncleaned by Julie $$=200\\;\\text{m}^2$$ $$-40\\;\\text{m}^2$$\n\n$$=160\\;\\text{m}^2$$\n\nShe divides the rest of her work equally among her $$4$$ sisters\n\n$$=160\\div 4=\\dfrac{160}{4}$$\n\n$$=40\\;\\text{m}^2$$\n\nThus, each sister has to clean $$40\\;\\text{m}^2$$ area.\n\nHence, option (B) is correct.\n\n### Julie has to clean her house, so she decides to do $$200\\,\\text{m}^2$$ area of her house today. If she cleans only $$\\dfrac{1}{5}$$ area of what she decided and asks her $$4$$ sisters to clean the rest of the area, find the area that each sister has to clean. Assume each sister has to clean equal amount of area.\n\nA\n\n$$25\\;\\text{m}^2$$\n\n.\n\nB\n\n$$40\\;\\text{m}^2$$\n\nC\n\n$$30\\;\\text{m}^2$$\n\nD\n\n$$35\\;\\text{m}^2$$\n\nOption B is Correct"
] | [
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"https://www.actucation.com/img/uploads/images/IllusSolG2L56M491.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.879529,"math_prob":0.9999914,"size":7012,"snap":"2021-31-2021-39","text_gpt3_token_len":2197,"char_repetition_ratio":0.14840183,"word_repetition_ratio":0.12268744,"special_character_ratio":0.37407303,"punctuation_ratio":0.109104045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999595,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T19:21:24Z\",\"WARC-Record-ID\":\"<urn:uuid:433b082f-5429-4625-9d97-74606cc8f48f>\",\"Content-Length\":\"290962\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8bd39ba9-ab33-427e-80e5-5125778adb8d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7a05cfb8-3bac-490a-af12-940be774e6e2>\",\"WARC-IP-Address\":\"166.62.28.118\",\"WARC-Target-URI\":\"https://www.actucation.com/grade-6-maths/word-problems-\",\"WARC-Payload-Digest\":\"sha1:YDWJMQAF5ZWT4UAF3AVI5SNQN5MUEWQM\",\"WARC-Block-Digest\":\"sha1:OZYQ5GA62CLAPO6NX7D6QM624QXGNV3D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153474.19_warc_CC-MAIN-20210727170836-20210727200836-00375.warc.gz\"}"} |
https://www.homeschoolingbooks.com/product-page/working-with-numbers-level-e-student-workbook | [
"`This arithmetic series is a survey program that offers solid practice in basic math skills at each grade level. The workbook is in two-color, with guided practice, regular exercises, and mixed review pages. Both computational and problem-solving skills are covered.The answers are in the corresponding teacher's guide, sold separately. Level E covers: place value through millions, review of arithmetic operations, multiplying and dividing multi-digit numbers, estimating, fraction, equivalent fractions, comparing and ordering fractions, higher terms and simplest terms, mixed numbers, arithmetic operations on fractions (like and unlike denominators), decimals, decimal place value, reading and writing decimals, comparing and ordering decimals, estimation and rounding, customary and metric measurement, comparing measurements, angles, perimeter and area, formulas, and problem-solving. 160 pp.`\n\n# Working with Numbers - Level E - Student Workbook\n\nSKU: 010489-1703E\n\\$24.13 Regular Price\n\\$11.49Sale Price\n• ISBN-10: 073989160X\nISBN-13: 9780739891605\nPublisher: Harcourt Achieve\nFormat: Paperback\nAge: 10, 11"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8427479,"math_prob":0.8977636,"size":1081,"snap":"2019-43-2019-47","text_gpt3_token_len":218,"char_repetition_ratio":0.11420613,"word_repetition_ratio":0.0,"special_character_ratio":0.21091582,"punctuation_ratio":0.21164021,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.981806,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-21T18:43:11Z\",\"WARC-Record-ID\":\"<urn:uuid:57cf57cd-3dbd-484b-ac5a-f6424a42ea98>\",\"Content-Length\":\"411405\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f07b8997-8b50-4684-ab3f-eb2580fe9969>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc6cf564-7025-4d13-abfe-b3bcdc182050>\",\"WARC-IP-Address\":\"185.230.60.211\",\"WARC-Target-URI\":\"https://www.homeschoolingbooks.com/product-page/working-with-numbers-level-e-student-workbook\",\"WARC-Payload-Digest\":\"sha1:C7INXXLLVCOFUXT73ZL2Z3HY5RZSP2XR\",\"WARC-Block-Digest\":\"sha1:E64MEWZHSHE2XNLVQRYTYZ2S3E5NJEQ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670948.64_warc_CC-MAIN-20191121180800-20191121204800-00435.warc.gz\"}"} |
https://www.vernier.com/experiment/chem-i-17_acid-base-titrations/ | [
"### Introduction\n\nA titration is a process used to determine the volume of a solution needed to react with a given amount of another substance. When titrating a solution of the strong acid hydrochloric acid, HCl, with a solution of the strong base sodium hydroxide, NaOH, the hydrogen ions from the HCl react with hydroxide ions from the NaOH in a one-to-one ratio to produce water in the overall reaction:",
null,
"${{\\text{H}}^{\\text{ + }}}{\\text{(aq) + C}}{{\\text{l}}^ - }{\\text{(aq) + N}}{{\\text{a}}^{\\text{ + }}}{\\text{(aq) + O}}{{\\text{H}}^ - }{\\text{(aq)}} \\to {{\\text{H}}_{\\text{2}}}{\\text{O(l) + N}}{{\\text{a}}^{\\text{ + }}}{\\text{(aq) + C}}{{\\text{l}}^ - }{\\text{(aq)}}$\n\nWhen an HCl solution is titrated with an NaOH solution, the pH of the acidic solution is initially low. As base is added, the change in pH is quite gradual until close to the equivalence point, when equimolar amounts of acid and base have been mixed. Near the equivalence point, the pH increases very rapidly. The change in pH then becomes more gradual again, before leveling off with the addition of excess base.\n\n### Objectives\n\nIn the Preliminary Activity, you will titrate a solution of the strong acid hydrochloric acid, HCl, with a solution of the strong base sodium hydroxide, NaOH. The concentration of the NaOH solution is given and you will determine the unknown concentration of the HCl.\n\nAfter completing the Preliminary Activity, you will first use reference sources to find out more about acids, bases, and acid-base titrations before you choose and investigate a researchable question utilizing acid-base titrations."
] | [
null,
"https://s0.wp.com/latex.php",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90631187,"math_prob":0.9837545,"size":1326,"snap":"2021-31-2021-39","text_gpt3_token_len":284,"char_repetition_ratio":0.1633888,"word_repetition_ratio":0.13953489,"special_character_ratio":0.18627451,"punctuation_ratio":0.10080645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9913221,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T17:48:08Z\",\"WARC-Record-ID\":\"<urn:uuid:a6dabf84-7678-45eb-8364-bd42ac583021>\",\"Content-Length\":\"313064\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:94f0264e-2c0c-4e9f-abbf-cbefd9d9e520>\",\"WARC-Concurrent-To\":\"<urn:uuid:4e9ab3b6-8269-4757-92a7-38a2564b18ad>\",\"WARC-IP-Address\":\"172.67.69.150\",\"WARC-Target-URI\":\"https://www.vernier.com/experiment/chem-i-17_acid-base-titrations/\",\"WARC-Payload-Digest\":\"sha1:KLEJYLLLYVUQO6O46EUXZGXQUTLUMJ2M\",\"WARC-Block-Digest\":\"sha1:LJBOQSXYCGHUKDIEYEMIWCLTVWTC6NK7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057733.53_warc_CC-MAIN-20210925172649-20210925202649-00595.warc.gz\"}"} |
https://emeritus.org/blog/data-analytics-what-is-logistic-regression/ | [
"# What is Logistic Regression? A Guide to Get Your Basics Right\n\nIf you’re trying to enter the field of data analytics, you must be looking for ways to grasp the various tools and techniques of the trade. One such particular type of analysis technique that data analysts commonly use is logistic regression. So, what is logistic regression? How does it work? What are its uses and advantages?\n\nThis guide will help you understand everything- from its importance to its uses and benefits. By the end of this post, you will know the difference between logistic and linear regression and become familiar with its different types.\n\n## What is Logistic Regression?\n\nAccording to Tech Target, it is a statistical analysis method to predict a binary outcome, such as yes or no, based on prior observations of a data set.\n\nIn other words, logistic regression models predict a dependent variable by analyzing the relationship between one or more independent variables. For instance, it predicts whether a student will get or not get admission into a particular university. These binary outcomes enable a straightforward decision between two alternative variables.\n\n## How Does Logistic Regression Work?\n\nThe analysis model considers multiple input criteria into the picture. In the above case of university acceptance, the logistic function can consider factors like grade point average, admission test score, and extracurricular activities. Based on the historical data about earlier outcomes with the same input criteria, the model scores the new cases on their chance of falling into one of the binary outcome categories.\n\n## Where is Logistic Regression Used?\n\nLet us now explore some real-world scenarios where this is relevant.\n\nCalculating the probability of a binary event occurring can help us classify whether an email is spam or not or whether a credit card transaction is a fraud or not. In the medical context, it can be used to predict a health outcome, such as whether a tumor is malignant or not.\n\nSo you can see logistic regression predicts the likelihood of all kinds of yes/no outcomes. By making predictions, the model helps data scientists make informed business decisions to minimize risks, optimize spending and maximize profits. And, isn’t that what all businesses are looking for?\n\nFor example, a credit card company cannot possibly issue a card to every person who applies. They need to analyze the two possible binary outcomes, the person “will default” or “not default” using a model to categorize customers who might be a good fit for the credit card.\n\n## What is Logistic Regression in Machine Learning?\n\nLogistic regression is a crucial technique in artificial intelligence and machine learning (AI/ML). Machine Learning (ML) models are software programs you can tune and train to perform complex data processing tasks without manual intervention. ML models built using logistic regression can provide actionable insights from business data and conduct predictive analysis to reduce operational expenses and increase productivity.\n\n## What are the Advantages of Logistic Regression?\n\nHere are some of the advantages of such analysis that bring value for data analysts:\n\n1. Simplicity: Models are mathematically less complex than other ML models and hence very useful in the machine learning context. Thus, you can build, train and deploy efficient ML models using logistic regression, even if you lack in-depth ML expertise.\n2. Speed: It processes large data sets at high speed because they require less computational capacities in terms of memory and processing power. This makes them ideal for programmers working on complex ML projects while aspiring for quick wins.\n3. Visibility: Analysis gives programmers greater visibility into the internal software processes than other data analysis techniques. Troubleshooting becomes easier as calculations using the equation are less complex.\n\n##",
null,
"What are the Different Types of Logistic Regression?\n\nIn this post, we’ve explained just one type so far- binary regression, where you have only two possible outcomes. However, there are three types:\n\n1. Binary: This statistical model predicts the relationship between the dependent variable (Y) and the independent variable X), where the dependent variable is binary. So the outcome is in terms of success/failure, 0/1, yes/no, or true/false.\n2. Multinomial: In this statistical analysis, you have one dependent variable with two or more discrete outcomes. For example, let’s say you want to predict the most popular transport type in 2040. Here the transport type will be a dependent variable with several possible outcomes like a train, bus, car, or bike.\n3. Ordinal: In this technique, the dependent variable (Y) has a meaningful order with two or more categorical variables. For example, let’s consider shirt size (S/M/L/XL) or an audience poll with (Agree/Disagree/Neutral) options.\n\n## Linear Regression vs Logistic Regression: What’s the Difference\n\nLinear regression equation uses continuous variables as targets, while in the logistic regression model, the target is a discrete variable or a binary value. The predicted value for linear regression is the mean of the target variables. Whereas, the predicted value is the probability of the target variables.\n\n## How to Learn About Logistic Regression\n\nEmeritus in association with the world’s best universities offers top data science and analytics courses. These completely online courses are designed to help you dive deep into the concepts and techniques used in data analysis, machine learning, and deep learning. You will also get to apply all the knowledge and skills to develop an ML model, build a job-ready portfolio, capture your prospective employer’s attention, and help land your dream job!\n\nBy Swet Kamal\n\nWrite to us content@emeritus.org",
null,
""
] | [
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"data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==",
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"data:image/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89294195,"math_prob":0.85513484,"size":5875,"snap":"2023-14-2023-23","text_gpt3_token_len":1111,"char_repetition_ratio":0.13251576,"word_repetition_ratio":0.0,"special_character_ratio":0.1853617,"punctuation_ratio":0.10400763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9870725,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-08T18:29:10Z\",\"WARC-Record-ID\":\"<urn:uuid:045098d9-ec37-4749-9e0e-81e5004f0429>\",\"Content-Length\":\"196193\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1efe3984-a838-4a74-9b39-6a2dcdae6797>\",\"WARC-Concurrent-To\":\"<urn:uuid:01a97724-f238-4401-a209-e053e6807695>\",\"WARC-IP-Address\":\"104.18.16.18\",\"WARC-Target-URI\":\"https://emeritus.org/blog/data-analytics-what-is-logistic-regression/\",\"WARC-Payload-Digest\":\"sha1:YRUW5ZAIN53DJBDTPJXJQW2QM3W5TM2N\",\"WARC-Block-Digest\":\"sha1:TB5K5J2XKCRVUGWMOE7PQJKJCTJ4AB26\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224655092.36_warc_CC-MAIN-20230608172023-20230608202023-00483.warc.gz\"}"} |
https://research.chalmers.se/publication/81492 | [
"Branching of some holomorphic representations of SO(2,n) Artikel i vetenskaplig tidskrift, 2007\n\nIn this paper we consider the analytic continuation of the weighted Bergman spaces on the Lie ball D = SO(2,n)/S(O(2) x O(n)) and the corresponding holomorphic unitary (projective) representations of SO(2, n) on these spaces. These representations are known to be irreducible. Our aim is to decompose them under the subgroup SO(1, n) which acts as the isometry group of a totally real submanifold H of D. We give a proof of a general decomposition theorem for certain unitary representations of semisimple Lie groups. In the particular case we are concerned with, we find an explicit formula for the Plancherel measure of the decomposition as the orthogonalising measure for certain hypergeometric polynomials. Moreover, we construct an explicit generalised Fourier transform that plays the role of the intertwining operator for the decomposition. We prove an inversion formula and a Plancherel formula for this transform. Finally we construct explicit realisations of the discrete part appearing in the decomposition and also for the minimal representation in this family.\n\nbounded symmetric domain\n\nintertwining operator\n\nspherical function\n\nKERNELS\n\nBEREZIN TRANSFORM\n\nLie group\n\nLie algebra\n\nrepresentation\n\nhypergeometric function\n\nBOUNDED SYMMETRIC DOMAINS\n\nunitary\n\nPOLYNOMIALS\n\nFörfattare\n\nHenrik Seppänen\n\nChalmers, Matematiska vetenskaper, Matematik\n\nGöteborgs universitet\n\nJournal of Lie Theory\n\n0949-5932 (ISSN)\n\nVol. 17 1 191-227\n\nMatematik\n\n2017-10-06"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84729975,"math_prob":0.9342882,"size":1368,"snap":"2019-43-2019-47","text_gpt3_token_len":304,"char_repetition_ratio":0.12536657,"word_repetition_ratio":0.0,"special_character_ratio":0.1754386,"punctuation_ratio":0.06726457,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97520256,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-18T07:12:31Z\",\"WARC-Record-ID\":\"<urn:uuid:14f260d1-72c8-4684-b0ae-5fc33d977a10>\",\"Content-Length\":\"30960\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:88647283-dbdf-4720-9a8b-d855d75a3a11>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd271d59-a8cc-45e1-9380-f3fdab3c67fc>\",\"WARC-IP-Address\":\"40.113.65.9\",\"WARC-Target-URI\":\"https://research.chalmers.se/publication/81492\",\"WARC-Payload-Digest\":\"sha1:UI2ILN7B23YAZQIC4ZAYTKGB4IPWJJ2H\",\"WARC-Block-Digest\":\"sha1:RHM4RMKXCV7VJ2X7XZICA63NYPQPXFD4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986677964.40_warc_CC-MAIN-20191018055014-20191018082514-00374.warc.gz\"}"} |
https://www.jiskha.com/questions/1424067/7-0-kilograms-of-ice-at-0-0c-and-are-melted-to-water-at-0-0c-find-the-increase-in | [
"# Physics\n\n7.0 kilograms of ice at 0 0C and are melted to water at 0 0C. Find the increase in internal energy of the ice as it is melted to water at 0 0C. Use 3.33x 105 J for Lf of ice. Express your answers in calories.\n\n1. 👍\n2. 👎\n3. 👁\n1. What is 7*Lf ?\n\nNow for calories, multiply Joules by .239\n\n1. 👍\n2. 👎\n👤\nbobpursley\n\n## Similar Questions\n\n1. ### Chemistry\n\ncalculate the final temperature (once the ice has melted) of a mixture made up of initially of 75.0mL liquid water at 29 degrees Celsius and 7.0 g of ice at 0.0 degrees Celsius\n\n2. ### physics\n\nFive kilograms of water at 40◦C is poured on a large block of ice at 0◦C. How much(in Kg) ice melts ? Hf = 335 kJ/kg cw = 4.184 kJ/(kg·◦C\n\n3. ### Chemistry\n\nAn ice cube at 0.00 degree celsius with a mass of 23.5 g is placed into 550.0 g of water, initially at 28.0 degree celsius, in an insulated container. Assuming that no heat is lost to the surroundings, what is the temperature of\n\n4. ### Chemistry - Heat of Fusion of Ice\n\nConclusion question(s) from a lab we did to find the heat of fusion of ice: Does the value obtained for the molar heat of fusion depend on the volume of water used? Does it depend on the mass of ice melted? Does it depend on the\n\n1. ### Chemistry - Energy: Phase Changes\n\nIf 13.4 kJ of energy are added to 1.00 kg of ice at 0 degrees Celsius, how much water at 0 degrees Celsius is produced? How much ice is left? The molar heat of melting is 6.01 kJ/mol. So... 6.01 kJ -> 1 mol (6.01 kJ of energy can\n\n2. ### chemistry\n\nDetermine the specific heat of a metal from the following data. A 75 g piece of the metal at 100.0 C was placed into an ice water bath. The heat loss from the metal melted 255g of ice. Some unmelted ice remained in the ice water\n\n3. ### Chemistry\n\nCalculate the final temperature (once the ice has melted) of a mixture made up initially of 75.0 mL liquid water at 29.0 Celsius and 7.0 g ice at 0.0 Celsius?\n\n4. ### physics\n\nYears ago, a block of ice with a mass of about 22kg was used daily in a home icebox. The temperature of the ice was 0.0 degrees Celsius when delivered. As it melted, how much heat did a block of ice that size absorb? The latent\n\n1. ### science\n\nan ice cube, a glass of cold water, and a swimming pool are each exposed to the same amount of sunlight on a hot day. after five minutes, the ice cube is melted and the cold drink is two degrees warmer, but the temperature of the\n\n2. ### physics\n\nAn insulated Thermos contains 140 cm3 of hot coffee at 89.0°C. You put in a 11.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted and equilibrium\n\n3. ### Chemistry\n\nHey, for chemistry class we are given days in the lab to conduct an experiment, then answer a number of questions that correlate with that experiment and how we can make use of it. This week we looked at the Molar Fusion of Ice\n\n4. ### Chemistry\n\nTo treat a burn on your hand, you decide to place an ice cube on the burned skin. The mass of the ice cube is 16.0 g, and its initial temperature is -11.4 °C. The water resulting from the melted ice reaches the temperature of"
] | [
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https://www.thefreelibrary.com/Non-Darcy+mixed+convective+heat+and+mass+transfer+flow+of+a+viscous+...-a0323408511 | [
"Non-Darcy mixed convective heat and mass transfer flow of a viscous electrically conducting fluid through a porous medium in a circular annulus in the presence of temperature gradient heat sources with Soret and Dufour effects--a finite element study.\n\nIntroduction\n\nConvection flows driven by temperature and concentration differences have been studied extensively in the past and various extensions of the problems have been reported in the literature with both temperature and concentration interacting simultaneously, the convection have become quite complex. Bejan and Khair have investigated the vertical free convection flow embedded in a porous medium resulting from the combined heat and mass transfer. Jang and Chang have used an implicit finite difference method to study the buoyancy induced inclined boundary layer in a porous medium resulting from the combined heat and mass buoyancy effects.\n\nHeat transfers in thermal insulation within vertical cylindrical annuli provide us insight into the mechanism of energy transport and enable engineers to use insulation more efficiently. In particular design engineers require relationship between heat transfer, geometry and boundary conditions which can be utilized cost-benefit analysis to determine the amount of insulation that will yield the maximum investment. An understanding of convective heat transfer in porous annuli is essential for its applications in packed-bed catalytic reactors, Geophysics, thermal insulation, design of regenerative heat exchangers, Geological disposal of high-level nuclear waste, petroleum resources and many other uses. Free convection in a vertical porous annulus has been extensively studied by Prasad , Prasad and Kulacki and Prasad et al., both theoretically and experimentally. Convection through annular regions under steady conditions has also been discussed with the two cylindrical surfaces kept at different temperatures . This work has been extended in temperature dependent convection flow as well as convection flow through horizontal porous channel whose inner surface is maintained at constant temperature, while the other surface is maintained at circumferentially varying sinusoidal temperature .\n\nThe applications of electromagnetic fields in controlling the heat transfer as in aerodynamic heating leads to the study of Magneto hydrodynamic heat transfer. This MHD heat transfer has gained significance owing to recent advancement of space technology. The MHD heat transfer can be divided into two sections. One contains problems in which the heating is an incidental by product of the electromagnetic fields as in MHD generators and pumps etc. and the second consist of problems in which the primary use of electromagnetic fields is to control the heat transfer(5). With fuel crisis deepening all over the world, there is a great concern to utilize the enormous power beneath the earth's crust in the geothermal region (10). Liquid in the geothermal region is an electrically conducting liquid because of high temperature and that they undergo the influence of magnetic field.\n\nIn many industrial applications of transient free convection flow problems, there occurs a heat source or a sink which is either a constant or temperature gradient or temperature dependent heat source. This heat source occurs in the form of a coil or a battery. Gokhale and Behnaz-Farman analyzed Transient free convection flow of a n incompressible fluid past an isothermal plate with temperature gradient dependent heat sources. Implicit finite difference scheme which is unconditionally stable has been used to solve the governing partial differential equations of the flow. Transient temperature and velocity profiles are plotted to show the effect of heat source. Muthukumara swamy et al has analyzed the radiation effect on moving vertical plate with variable temperature and mass diffusion. Sreevani has analyzed the Soret effect on convective heat and mass transfer flow of a viscous fluid in a cylindrical annulus with heat generating sources. Sivaiah has discussed the convective heat and mass transfer flow in a circular duct with Soret effect.\n\nLiterature suggests that the effect of viscous dissipation on heat transfer has been studied for different geometries. Brinkman have studied the viscous dissipation effect on natural convection in horizontal cylinder embedded in porous medium. Their study showed that the viscous dissipation effect might not be neglected. Saffman have studied the viscous dissipation effect on natural convection in a porous cavity and found that the heat transfer rate at hot surface decreases with increase of viscous dissipation parameter. Thermal radiation plays a significant role in the overall surface hear transfer where convective heat transfer is small. Verschoor et al have studied the effect of viscous dissipation and radiation on unsteady magneto hydro dynamic free convection flow fast vertical plate in porous medium. They found that the temperature profile increases when viscous dissipation increases. A good amount of work has been done to understand natural convection in porous cavity. In spite of endeavor efforts to study heat transfer in porous cavity, the combined effect of viscous dissipation and radiation on porous medium filled inside a square cavity has not received attention.\n\nThe Soret and Dufour effects have garnered considerable interest in both Newtonian and non-Newtonian convective heat and mass transfer. Such effects are significant when density differences exist in the flow regime. Soret and Dufour effects are important for intermediate molecular weight gases in coupled heat and mass transfer in binary systems, often encountered in chemical process engineering and also in high-speed aerodynamics. Soret and Dufour effects are also critical in various porous flow regimes occurring in chemical and geophysical systems. There are few studies about the Soret and Dufour effects in a Darcy or non-Darcy porous medium. Anghel et al has examined the composite Soret and Dufour effects on free convective heat and mass transfer in a Darcian porous medium with Soret and Dufour effects. Very recently, Barletta. A, Lazzari.S, and others have studied on Mixed convection with heating effects in a vertical porous annulus with a radially varying magnetic field. Emmunuel Osalusi, Jonathan Side, Robert Harris have discussed Thermal-diffusion and diffusion thermo effects on combined heat and mass transfer of a steady MHD convective and slip flow due to a rotating disk with viscous dissipation and ohmic heating.\n\nThe Weighted residual method is of the generalization of the Ritz-variational method where in we seek an approximate solution in the form of linear combination of suitable approximation functions. The parameters in the linear combination as determined by setting integral of a weighted residual of the approximation over the domain zero. A comprehensive description of a weighted residual method. In many situations the Galerkin method which is one of the important weighted residual methods is equivalent to the Ritz method for solving variational problems. The finite element method is piece--wise applications of weighted residual method in which the Ritz-Galerkin type methods are employed over each element of the domain. The finite element method was initially developed as an adhoc engineering procedure for constructing matrix solutions to stress and displacement calculations in structural analysis. Very few fluid dynamic problems can be expressed in a variational form. Consequently most of the finite element applications in fluid dynamics have used in Galerkin finite element formulation. The Galerkin finite element method has two important futures. Firstly the approximate solution is written directly as a linear combination of approximating functions in terms of the nodal unknowns. Secondly the approximating functions or the shape functions are chosen exclusively from low order piecewise polynomials restricted to contiguous elements.\n\nIn this paper we discuss the mixed convective viscous dissipative flow through a porous medium in a circular cylindrical annulus with Thermal-Diffusion and Diffusion-Thermo effects in the presence of temperature gradient heat source, where the inner wall is maintained constant temperature while the outer wall is maintained constant heat flux and the concentration is constant on the both walls. The coupled momentum, energy and diffusion equations are solved by using finite element analysis with quadratic interpolation polynomials. The effect of temperature gradient heat sources on the flow and heat transfer characteristics are analyzed. The stress, rate of heat transfer and the rate of mass transfer are discussed numerically for different variations of the governing parameters.\n\nFormulation of the Problem\n\nWe consider free and force convective flow of a viscous, electrically conducting fluid through a porous medium in a circular cylindrical annulus with Thermal-Diffusion and Diffusion-Thermo effects in the presence of temperature gradient heat source, whose inner wall is maintained at a constant temperature and the outer wall is maintained constant heat flux. Also the concentration is constant on the both walls. A uniform radial magnetic field is applied on the flow. The flow, temperature and concentration in the fluid are assumed to be fully developed. Both the fluid and porous region have constant physical properties and the flow is a mixed convection flow taking place under thermal and molecular buoyancies and uniform axial pressure gradient. The boussenissque approximation is invoked so that the density variation is confined to the thermal and molecular buoyancy forces. The Brinkman-Forchhimer-Extended Darcy model which accounts for the inertia and boundary effects has been used for the momentum equation in the porous region. In the momentum, energy and diffusion are coupled and non-linear. Also the flow in is unidirectional along the axial cylindrical annulus. Making use of the above assumptions the governing equations are Equation of linear momentum\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (1)\n\nEquation of Energy\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (2)\n\nEquation of diffusion\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (3)\n\nEquation of state\n\n[rho] - [[rho].sub.0] = -[beta][[rho].sub.0](T - [T.sub.0]) - [[beta].sup.*][[rho].sub.0](C - [C.sub.0]) (4)\n\nWhere u is the axial velocity in the porous region, T & C are the temperature and concentrations of the fluid, k is the permeability of porous medium, F is a function that depends on Reynolds number and the microstructure of the porous medium and [D.sub.1] is the Molecular diffusivity, [D.sub.m] is the coefficient of mass diffusitivity, [T.sub.m] is the mean fluid temperature, [K.sub.t] is the thermal diffusion, [C.sub.s] is the concentration susceptibility, [C.sub.p] is the specific heat, [rho] is density, g is gravity, [beta] is the coefficient of thermal expansion, [[beta].sup.*] is the coefficient of volume expansion. [sigma] is the electrical conductivity, [[mu].sub.e] is the magnetic permeability.\n\nu = 0, T = [T.sub.i], C = [C.sub.i] at r = a (6)\n\nu = 0, [partial derivative]T/[partial derivative]r = [Q.sub.1], C = Co at r = a + s (7)\n\nThe axial temperature gradient [partial derivative]T/[partial derivative]r and concentration gradient [partial derivative]C/[partial derivative]z are assumed to be constant say A and B respectively.\n\nUsing equations (2.5) and (2.6) equation (2.2) reduces to\n\n[[rho].sub.0][C.sub.p]uA = [lambda]([T.sub.rr] + [1/r] [T.sub.r]) + [[D.sub.m][K.sub.t]/[C.sub.s][C.sub.P]] ([C.sub.rr] + [1/r] [C.sub.r]) (8)\n\n[[rho].sub.0][C.sub.P]uB = [lambda]([C.sub.rr] + [1/r] [C.sub.r]) + [[D.sub.m][K.sub.t]/[C.sub.s][C.sub.P]] ([T.sub.rr] + [1/r] [T.sub.r]) (9)\n\nWe now define the following non-dimensional variables\n\n[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (10)\n\nIntroducing these non-dimensional variables, the governing equations in the non-dimensional form are (on removing the stars)\n\n[[d.sup.2]u/d[r.sup.2] + [1/r][du/dr]] = P + [delta]([D.sup.-1] + [[M.sup.2]/[r.sup.2]])u + [[delta].sup.2][LAMBDA] [u.sup.2] - [delta]G[theta] (11)\n\n[[d.sup.2][theta]/d[r.sup.2] + [1/r][d[theta]/dr] + [[alpha]/r] [d[theta]/dr]] = [P.sub.r][N.sub.T]u + Du[N.sub.t] ([d.sup.2]C/d[r.sup.2] + [1/r][d[theta]/dr]) + [P.sub.r][E.sub.c] [(du/dr).sup.2] (12)\n\n[d[r.sup.2]C/d[r.sup.2] + [1/r][dC/dr]] = Sc[N.sub.c]u + ScSr ([d.sup.2][theta]/d[r.sup.2] + [1/r][d[theta]/dr]) (13)\n\nwhere\n\n[LAMBDA] = F[D.sup.-1] (Forchheimer number)\n\n[P.sub.r] = [mu]/[C.sub.p]/[lambda] (Prandtl number)\n\nG = g[beta]([T.sub.1] - [T.sub.0])[a.sup.3]/[v.sup.2] (Grashof number)\n\n[D.sup.-1] = [a.sup.2]/k (Inverse Darcy parameter)\n\n[N.sub.t] = Aa/[T.sub.1] - [T.sub.0] (Temperature gradient)\n\n[N.sub.c] = Ba/[C.sub.1] - [C.sub.0] (Non-dimensional concentration gradient)\n\nDu = ([D.sub.m][K.sub.t][DELTA][ca.sup.2]/[C.sub.s][C.sub.p][DELTA]T][lambda]) (Dufour Number)\n\nSc = v/[D.sub.1] (Schmidt number)\n\nSr = ([D.sub.m][K.sub.t][DLETA]T/v[T.sub.m][DELTA]C) (Soret number)\n\n[alpha] = Q[L.sup.2]/[lambda][C.sub.p] (Heat source parameter)\n\n[M.sup.2] = [sigma][[mu].sup.2.sub.e][H.sup.2.sub.0][a.sup.2]/[[gamma].sup.2] (Hartman number)\n\n[E.sub.c] = [v.sup.2]/[a.sup.2](Aa)[c.sub.p] (Eckert number)\n\nWith the corresponding boundary conditions as:\n\nu = 0, [theta] = 0, C = 1 at r = 1 (14)\n\nu = 0, [partial derivative][theta]/[partial derivative]r = [Q.sub.1], C = 0 at r = 1 + s (15)\n\nNumerical Analysis\n\nThe finite element method has been implemented to obtain numerical solutions of equations (11) to (13) under boundary conditions (14) and (15). This technique is extremely efficient and allows robust solutions of complex coupled, nonlinear multiple degree differential equation systems. The fundamental steps comprising the method are now summarized:\n\nPhase 1] Discretization of the domain into elements\n\nPhase 2] Derivation of element equations\n\nPhase 3] Assembly of Element Equations\n\nPhase 4] Imposition of boundary conditions\n\nPhase 5] Solution of assembled equations\n\nThe shear stress are evaluated on the cylinder using the formula\n\n[tau] = [(du/dr).sub.r=1,1+s]\n\nThe rate of heat transfer (Nusselt number) are evaluated on the cylinder using the formula\n\nNu = [(d[theta]/dr).sub.r=1]\n\nThe rate of mass transfer (Sherwood Number) is evaluated using the formula\n\nSh = -[(dC/dr).sub.r=1,1+s]\n\nDiscussion of the Numerical Results\n\nIn this analysis we investigate thermo-Diffusion and Diffusion-Thermo effects on convective heat and mass transfer flow of a viscous conducting fluid through a porous medium in circular annulus in the presence of temperature gradient dependent heat source with viscous dissipation. The inner cylinder is maintained at constant temperature and the outer wall is maintained constant heat flux while the concentration is maintained constant on both the cylinders. The axial flow is in vertically downword direction, u > 0 indicates a reversal flow. The velocity, temperature and concentration distributions are shown in figures 1-27 for different values of the parameters G, [D.sup.-1], M, Sc, Sr, Du, N, [alpha] and Ec.\n\nThe variation of u with Grashof number G shows that the axial flow enhances with increase in G and the region of reversal flow enhances with increase in G (fig.1). With respect to the variation of u with [D.sup.-1] we find that lesser the permeability of porous medium smaller the magnitude of u and for further lowering of the permeability larger th-1e magnitude of u in the entire flow region, the region of reversal flow shrinks with [D.sup.-1] [less than or equal to] 2 X [10.sup.3] and enhances with higher [D.sup.-1] [greater than or equal to] 3 X [10.sup.3] (fig.2). From fig.3 we find that higher the Lorentz force larger the velocity in the flow region. Also the region of reversal flow enhances with increase in M. Fig.4 represents the variation of u with Sc. We notice that lesser the molecular diffusitivity smaller [absolute value of u] and for further lowering of molecular diffusitivity it experiences a depreciation in the entire flow region and it attains maximum at r = 1.5.\n\n[FIGURE 1 OMITTED]\n\n[FIGURE 2 OMITTED]\n\n[FIGURE 3 OMITTED]\n\n[FIGURE 4 OMITTED]\n\nThe variation of u with Soret parameter Sr shows that the velocity experiences an enhancement with increase in Sr [less than or equal to] 0.8 and for further increase in Sr [greater than or equal to] 1 it depreciates in its magnitude (fig.5). From fig.6 we observe that the region of reversal velocity enlarges with increase in Du and [absolute value of u] enhances with Du. The variation of u with N shows that when the molecular buoyancy force dominates over the thermal buoyancy force the actual axial velocity experiences a depreciation when the buoyancy forces act in the same direction while for the forces acting in the opposite directions it experiences an enhancement in the flow region (fig.7). The influence of heat source parameter [alpha] on u is shown in fig.8.\n\n[FIGURE 5 OMITTED]\n\n[FIGURE 6 OMITTED]\n\n[FIGURE 7 OMITTED]\n\n[FIGURE 8 OMITTED]\n\nAn increase in [alpha] < 0 enhances the actual axial velocity u while it depreciates with [alpha] > 0. This shows that in the presence of the temperature heat source depreciates the velocity in the flow region with maximum in the mid region. The influence of dissipative effect on u is shown in fig.9. We conclude that the axial velocity u experiences an enhancement with Ec.\n\n[FIGURE 9 OMITTED]\n\nThe non-dimensional temperature ([theta]) is shown in fig 10-18 for different values of the parameters. It is found that the non-dimensional temperature gradually increases from its prescribed value 0 on r = 1 to attain its prescribed value 1 at r = 2. An increase in G enhances the temperature (fig.10). The variation of [theta] with [D.sup.-1] shows that lesser the permeability of porous medium larger the actual temperature in the flow region (fig.11). From fig.12 we find that lesser the Lorentz force larger the temperature. With respect to Sc, we notice that lesser the molecular diffusitivity smaller the temperature in the flow region (fig.13).\n\n[FIGURE 10 OMITTED]\n\n[FIGURE 11 OMITTED]\n\n[FIGURE 12 OMITTED]\n\n[FIGURE 13 OMITTED]\n\nAn increase in Soret parameter Sr results in a depreciation in the actual temperature in the region (fig.14). The variation of [theta] with Dufour parameter Du shows that the actual temperature enhances gradually with increase in Du (fig.15). When the molecular buoyancy force dominates over the thermal buoyancy force the actual temperature decreases irrespective of the directions of the buoyancy forces (fig.16). The influence of temperature gradient heat source parameter [alpha] on [theta] is shown in fig.17. It found that the temperature is negative for [alpha] = 0, [alpha] < 0 and positive for [alpha] > 0. The actual temperature experiences a depreciation with increase in the strength of the heat sources. The variation of [theta] with Eckert number Ec is shown in fig.18. We find that the actual temperature enhances with increase in Ec.\n\n[FIGURE 14 OMITTED]\n\n[FIGURE 15 OMITTED]\n\n[FIGURE 16 OMITTED]\n\n[FIGURE 17 OMITTED]\n\n[FIGURE 18 OMITTED]\n\nThe non-dimensional concentration ([phi]) is shown in fig 19-27 for different values of the parameters G, [D.sup.-1], M, Sc, Sr, Du, N, [alpha] and Ec. It is found that the non-dimensional concentration gradually increases from its prescribed value 0 on r = 1 and attain its prescribed value 1 at r = 2. Fig.19. shows the variation of [phi] with G. It is noticed that the concentration depreciates with increase in the Grashof number G. The variation of [phi] with [D.sup.-1] shows that lesser the permeability of porous medium higher the actual concentration in the flow region and for further lowering of the permeability smaller the actual concentration in the flow region (fig.20). From fig.21 we find that higher the Lorentz force larger the concentration in the flow region. Lesser the molecular diffusitivity smaller the concentration in the flow field (fig.22). An increase in the Soret parameter Sr enhances the concentration everywhere in the flow region (fig.23). The variation of [phi] with Dufour parameter Du shows that the concentration experiences a marginal depreciation in the flow region (fig.24). The variation of [phi] with N shows that when the molecular buoyancy force dominates over the thermal buoyancy force the actual concentration experiences an enhancement when the buoyancy forces act in the same direction while for the forces acting in the opposite directions it experiences a depreciation in the flow region (fig.25). From fig.26, we observe that [phi] experiences a marginal enhancement with [alpha] > 0 and a depreciation with [alpha] < 0. The inclusion of the dissipation in the flow enhances the concentration in the flow region (fig.27).\n\n[FIGURE 19 OMITTED]\n\n[FIGURE 20 OMITTED]\n\n[FIGURE 21 OMITTED]\n\n[FIGURE 22 OMITTED]\n\n[FIGURE 23 OMITTED]\n\n[FIGURE 24 OMITTED]\n\n[FIGURE 25 OMITTED]\n\n[FIGURE 26 OMITTED]\n\n[FIGURE 27 OMITTED]\n\nReferences\n\n Angel M, Takhar HS, and Pop I (2000): Dofour and Soret effects on free convection boundary layer over a vertical surface embedded in a porous medium. Studia universities-Bolyai, Mathematica XLV, pp. 11-21.\n\n Barletta.A, Lazzari.S,(2008): Mixed convection with heating effects in a vertical porous annulus with a radially varying magnetic field.\n\n Bejan, A and Khair, K.R: Heat and mass transfer by natural convection in a pourous medium., Int.J. Heat and mass transfer, V.28.PP.908-818(1985).\n\n Brinkman H.C (1948): A Calculation of the viscous force eternal by a flowing fluid on a dense swarm of particles. Appl.Science Research, Ala, p81.\n\n Chandra sekhar. S: Hydrodynamic and Hydro magnetic stability, Clarandon press, oxford (1961).\n\n Emmunuel Osalusi, Jonathan Side, Robert Harris:\"Thermal-diffusion and diffusion thermo effects on combined heat and mass transfer of a steady MHD convective and slip flow due to a rotating disk with viscous dissipation and ohmic heating: Int.Communications in heat and mass transfer,Vol.35,PP.908-915(2008).\n\n Gokhale, M.Y, Behnaz-Farnam: \"Transient free convection flow on an isothermal plate with temperature dependent heat sources\", International review of Pure and Applied Mathematics, Vol.3, No.1, pp.129-136(2007).\n\n Jang J.Y and Chang W.J: The flow and vertex instability of horizontal natural convection in a porous medium resulting from combined heat and mass buoyancy effects. Int. Heat and Mass Transfer, V. 31, p.769-777 (1987).\n\n Muthukumara Swamy, Maheswari, J, Pandurangan, J: \"Study of MHD and Radiation effects on moving vertical plate with variable temperature and mass diffusion\", International Review of Pure and Applied Mathematics, Vol.3, No.1, pp.95-103(2007).\n\n Nanda, R.S and Mohan, M: Proc. Ind Acad.Sci., V.876 A,No.5,P.147(1978).\n\n Prasad, V: Natural convection in a vertical porous annulus, Int. J. Heat and Mass Transfer, V. 27, p 207-219 (1984).\n\n Prasad, V and Kulacki, F. A: Natural convection in porous media bounded by short concentric vertical cylinders, ASME J. Heat Transfer, V. 107, p 147-154 (1985).\n\n Prasad, V, Kulacki, F.A and Keyhani, M: Natural convection in porous media, J. fluid mech., V. 150, p 89-119 (1985).\n\n Saffman .P.G (1971): on the boundary conditions at the free surface of a porous medium. Stud.Appl.Maths, V.2, p 93.\n\n Sivaiah (2004): Thermo-diffusion effect on convective heat and mass transfer flow through a porous medium in ducts. Ph.D thesis, S.K. University. Anantapur.\n\n Sreevani, M: Mixed convection heat and mass transfer through a porous medium in channels with dissipative effects, Ph.D thesis, S, K.University, Anantpur, India(2003).\n\n Vasseur, P, Nguyen, T.H., Robillard and Thi. V.K.T: Int. Heat Mass Transfer, V. 27, p 337 (1984).\n\n Verschoor et al: Int. Heat and Mass Transfer, V. 31(1992).\n\nP. Sudarsan Reddy (1), G. Srinivas (2), P. Sreenivasa Rao (3), DRV. Prasada Rao (4)\n\n(1) Asst. Professor, Dept of Mathematics, RGM Engg. College, Nandyal, JNTU Anantapur, India Email: suda1983@gmail.com\n\n(2) Assoc. Professor, Dept of Mathematics, Vignan Engg. College, JNTU Hyderabad, India\n\n(3) Assoc. professor, Dept. of Physics, Jyothismathi Engg.College, Karimnagar, JNTU Kakinada, India.\n\n(4) Professor, Dept of Mathematics, SK University, Anantapur. India"
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https://www.sciences360.com/index.php/can-you-divide-by-0-yes-11886/ | [
"",
null,
"Mathematics\n\n# Can you Divide by 0 – Yes",
null,
"Hunter J.'s image for:\n\"Can you Divide by 0 - Yes\"\nCaption:\nLocation:\nImage by:",
null,
"Ah zero, such an interesting number. A number even today we do not know much about. One of the most popular questions asked today of course is: Is it really impossible for something to be divided by zero? My answer is, yes it is.\n\nTo understand zero we must understand the definition of a number. A number is simply something that can list a quantity of something and that we can do something to change its value. But then is zero even a number? After all, it doesn't list the quantity of anything because zero well, is simply nothing. However, we must note that nothing is still a quantity because it shows the value of something. Therefore it is first concluded that zero is infact a number.\n\nNext we must study zero in many ways. We know of course that any number multiplied by zero will equal zero. We also note that an number added or subtracted to zero will equal itself. We also know that anything to the zeroth power will always equal one. Now for each number property there is always a pattern. For example, we know that if we multiply a number by anything >1 then the number will go up. We also know that if we subtract a positive number from a number and add a negative number to a number the ending quantity will go down. Concurringly, if we add a positive number or subtract a negative number from any number, the number quantity will go up.\n\nNow let's look at the pattern of division. It seems that each time we divide a number closer to zero from another number, the larger the quantity becomes. For example 1/.25.1/.5. It seems however that no matter how close the number line is to 0, the matching result will always be another number. Because of this, it is safe to assume that any number divided by zero is infinity. This is because it seems that no matter how close you get to the number 0, there is never a matching result, so if you get a number infinitely close to zero that the number is zero, the matching result will of course have to be infinity.\n\nSo now we have figured that anything divided by 0 is infinity, we must understand what infinity is. Infinity, does not follow any mathematical pattern. If added, subtracted, multiplied it will equal itself. If divided it will equal a number infinitely close to 0. Therefore, it is safe to conclude that infinity is in fact not a number. Some argue that infinity does in fact represent a quantity, a quantity that is infinitely large. However, my argument is that although it is a quantity, we can do nothing to change its quantity. If we subtract, add, or multiply, the number with another number we get the same number. If you divide that number with something such as infinity/8, we reach inifinity still. If you divide infinity by a number, as previously stated, the number is still 1/infinity or 0. We can reach infinity, but we can never change infinity.\n\nTherefore, just like how infinity does not follow the rules of algebra, any number divided by 0 also does not follow the rules of algebra. In this way, we are unable to use anything divided by 0. This is exactly why we can conclude that anything divided by 0 is absolutely impossible.\n\nTweet"
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https://ixtrieve.fh-koeln.de/birds/litie/document/6257 | [
"# Document (#6257)\n\nAuthor\nTitle\n¬The digitisation of library material\nSource\nInformation management and technology. 26(1993) no.3, S.128-132\nYear\n1993\nAbstract\nConsiders conversion of materials currently held on paper or microfilm to electronic media using imaging technology. Briefly describes current research projects in the UK and USA. Explains the digitization process and describes document and microfilm scanners. Details scanning of print and microfilm, explaining the different problems and special needs each medium, outlines image enhancement and optical character recognition\n\n## Similar documents (content)\n\n1. 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null
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https://www.easycalculation.com/square-roots-325.html | [
"# What is Square Root of 325 ?\n\nThe square root is a number which results in a specific quantity when it is multiplied by itself. The square root of 325 is 18.02776\n\nSquare Root of 325\n √325 = √(18.028 x 18.028) 18.02776\n\nThe square root is a number which results in a specific quantity when it is multiplied by itself. The square root of 325 is 18.02776\n\nThe nearest previous perfect square is 324 and the nearest next perfect square is 361 ."
] | [
null
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https://www.colorhexa.com/1a7905 | [
"# #1a7905 Color Information\n\nIn a RGB color space, hex #1a7905 is composed of 10.2% red, 47.5% green and 2% blue. Whereas in a CMYK color space, it is composed of 78.5% cyan, 0% magenta, 95.9% yellow and 52.5% black. It has a hue angle of 109.1 degrees, a saturation of 92.1% and a lightness of 24.7%. #1a7905 color hex could be obtained by blending #34f20a with #000000. Closest websafe color is: #336600.\n\n• R 10\n• G 47\n• B 2\nRGB color chart\n• C 79\n• M 0\n• Y 96\n• K 53\nCMYK color chart\n\n#1a7905 color description : Dark lime green.\n\n# #1a7905 Color Conversion\n\nThe hexadecimal color #1a7905 has RGB values of R:26, G:121, B:5 and CMYK values of C:0.79, M:0, Y:0.96, K:0.53. Its decimal value is 1734917.\n\nHex triplet RGB Decimal 1a7905 `#1a7905` 26, 121, 5 `rgb(26,121,5)` 10.2, 47.5, 2 `rgb(10.2%,47.5%,2%)` 79, 0, 96, 53 109.1°, 92.1, 24.7 `hsl(109.1,92.1%,24.7%)` 109.1°, 95.9, 47.5 336600 `#336600`\nCIE-LAB 44.096, -46.591, 47.202 7.29, 13.905, 2.443 0.308, 0.588, 13.905 44.096, 66.324, 134.627 44.096, -38.511, 52.945 37.289, -30.357, 22.217 00011010, 01111001, 00000101\n\n# Color Schemes with #1a7905\n\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #640579\n``#640579` `rgb(100,5,121)``\nComplementary Color\n• #547905\n``#547905` `rgb(84,121,5)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #05792a\n``#05792a` `rgb(5,121,42)``\nAnalogous Color\n• #790554\n``#790554` `rgb(121,5,84)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #2a0579\n``#2a0579` `rgb(42,5,121)``\nSplit Complementary Color\n• #79051a\n``#79051a` `rgb(121,5,26)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #051a79\n``#051a79` `rgb(5,26,121)``\n• #796405\n``#796405` `rgb(121,100,5)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #051a79\n``#051a79` `rgb(5,26,121)``\n• #640579\n``#640579` `rgb(100,5,121)``\n• #0a3002\n``#0a3002` `rgb(10,48,2)``\n• #0f4803\n``#0f4803` `rgb(15,72,3)``\n• #156104\n``#156104` `rgb(21,97,4)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #1f9106\n``#1f9106` `rgb(31,145,6)``\n• #25aa07\n``#25aa07` `rgb(37,170,7)``\n• #2ac208\n``#2ac208` `rgb(42,194,8)``\nMonochromatic Color\n\n# Alternatives to #1a7905\n\nBelow, you can see some colors close to #1a7905. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #377905\n``#377905` `rgb(55,121,5)``\n• #2d7905\n``#2d7905` `rgb(45,121,5)``\n• #247905\n``#247905` `rgb(36,121,5)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #107905\n``#107905` `rgb(16,121,5)``\n• #077905\n``#077905` `rgb(7,121,5)``\n• #05790d\n``#05790d` `rgb(5,121,13)``\nSimilar Colors\n\n# #1a7905 Preview\n\nThis text has a font color of #1a7905.\n\n``<span style=\"color:#1a7905;\">Text here</span>``\n#1a7905 background color\n\nThis paragraph has a background color of #1a7905.\n\n``<p style=\"background-color:#1a7905;\">Content here</p>``\n#1a7905 border color\n\nThis element has a border color of #1a7905.\n\n``<div style=\"border:1px solid #1a7905;\">Content here</div>``\nCSS codes\n``.text {color:#1a7905;}``\n``.background {background-color:#1a7905;}``\n``.border {border:1px solid #1a7905;}``\n\n# Shades and Tints of #1a7905\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, #020800 is the darkest color, while #f6fff4 is the lightest one.\n\n• #020800\n``#020800` `rgb(2,8,0)``\n• #061b01\n``#061b01` `rgb(6,27,1)``\n• #0a2e02\n``#0a2e02` `rgb(10,46,2)``\n• #0e4003\n``#0e4003` `rgb(14,64,3)``\n• #125303\n``#125303` `rgb(18,83,3)``\n• #166604\n``#166604` `rgb(22,102,4)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #1e8c06\n``#1e8c06` `rgb(30,140,6)``\n• #229f07\n``#229f07` `rgb(34,159,7)``\n• #26b207\n``#26b207` `rgb(38,178,7)``\n• #2ac408\n``#2ac408` `rgb(42,196,8)``\n• #2ed709\n``#2ed709` `rgb(46,215,9)``\n• #32ea0a\n``#32ea0a` `rgb(50,234,10)``\n• #3bf512\n``#3bf512` `rgb(59,245,18)``\n• #4bf625\n``#4bf625` `rgb(75,246,37)``\n• #5af738\n``#5af738` `rgb(90,247,56)``\n• #6af84b\n``#6af84b` `rgb(106,248,75)``\n• #79f85d\n``#79f85d` `rgb(121,248,93)``\n• #89f970\n``#89f970` `rgb(137,249,112)``\n• #99fa83\n``#99fa83` `rgb(153,250,131)``\n• #a8fb96\n``#a8fb96` `rgb(168,251,150)``\n• #b8fba9\n``#b8fba9` `rgb(184,251,169)``\n• #c7fcbc\n``#c7fcbc` `rgb(199,252,188)``\n• #d7fdce\n``#d7fdce` `rgb(215,253,206)``\n• #e6fee1\n``#e6fee1` `rgb(230,254,225)``\n• #f6fff4\n``#f6fff4` `rgb(246,255,244)``\nTint Color Variation\n\n# Tones of #1a7905\n\nA tone is produced by adding gray to any pure hue. In this case, #3c443a is the less saturated color, while #177e00 is the most saturated one.\n\n• #3c443a\n``#3c443a` `rgb(60,68,58)``\n• #394935\n``#394935` `rgb(57,73,53)``\n• #364d31\n``#364d31` `rgb(54,77,49)``\n• #33522c\n``#33522c` `rgb(51,82,44)``\n• #305727\n``#305727` `rgb(48,87,39)``\n• #2d5c22\n``#2d5c22` `rgb(45,92,34)``\n• #29611d\n``#29611d` `rgb(41,97,29)``\n• #266618\n``#266618` `rgb(38,102,24)``\n• #236a14\n``#236a14` `rgb(35,106,20)``\n• #206f0f\n``#206f0f` `rgb(32,111,15)``\n• #1d740a\n``#1d740a` `rgb(29,116,10)``\n• #1a7905\n``#1a7905` `rgb(26,121,5)``\n• #177e00\n``#177e00` `rgb(23,126,0)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #1a7905 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://warrenadams.me/music-audio/how-to-make-a-double-box-plot.php | [
"Double box-and-whisker plots give you a quick visual comparison of 2 sets of data, as was also found with other double graph forms covered in. Read and learn for free about the following article: Box plot review. Let's make a box plot for the same dataset from above. Step 1: Scale and label an axis that. The box in the box-and-whisker plot contains, and thereby highlights, the middle portion of these data points. To create a box-and-whisker plot, we start by .\n\n## box and whisker plot problems\n\nQuickly make parallel box plots for your maths assignment. Then find out how to describe their spread and central tendency and compare them. A box and whisker plot is a diagram that shows the statistical distribution of a set of data. This makes it easy to see how data is distributed along. Note: After clicking Draw here, you can click the Copy to Clipboard button (in Internet Explorer), or right-click on the graph and choose Copy. In your Word.\n\nCreating box and whisker plots has never been so easy with Displayr's free online box and whisker plot maker. Generate professional area charts, customize . Box plots, or box-and-whisker plots, are fantastic little graphs that give you a lot of statistical information in a cute little square. Let's take a look. How to read a box plot/Introduction to box plots. Box plots are drawn for groups of [email protected] scale scores. They enable us to study the distributional characteristics of.\n\nMake box plots online with Excel, CSV, or SQL data. Make bar charts, histograms , box plots, scatter plots, line graphs, dot plots, and more. Free to get started!. Create a standard box plot to show the distribution of a set of data. Make a box-and-whisker plot for the ages of the members of the. U.S. Which statement is true about the double box-and-whisker plot?\n\n## box and whisker plot explained\n\nA box plot is a graphical rendition of statistical data based on the minimum, first quartile, median, third quartile, and maximum. The term box plot comes from the . In this article, you will learn to create whisker and box plot in R programming. You will also learn to draw multiple box plots in a single plot. However, there is no need to bother the double box plot for such as the following meteorological and astronomical phenomena is not easy. If you've never made one before, we'll show you how to create a box and whisker plot in Excel, then double-check the calculations, and. In this activity, you will create three parallel boxplots and compare them. Materials To change to the dot plot to a boxplot, select MENU > Plot Type >. Boxplot. An example of a formula is y~group where a separate boxplot for numeric variable y is generated for each value of group. Add varwidth=TRUE to make boxplot. This is actually more efficient because boxplot converts # a 2-D array into a list of the original sample, and a boxplot is one visual tool to make this assessment. A box and whisker plot (sometimes called a boxplot) is a graph that presents A breakdown of the various components that make up a box and whisker. The key information you want to get when reading box plots is: are these groups If two boxes do not overlap with one another, say, box A is. Box plot generator. Create AccountorSign In. enter your data into list A. enter your data into list A. 1. A = 5,27,7,2,8, \\$\\$ A. \\$\\$= 6 element list. 2. box plot."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8874269,"math_prob":0.93586016,"size":3692,"snap":"2020-10-2020-16","text_gpt3_token_len":846,"char_repetition_ratio":0.14560738,"word_repetition_ratio":0.015197569,"special_character_ratio":0.219935,"punctuation_ratio":0.12021136,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.987914,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-01T20:29:47Z\",\"WARC-Record-ID\":\"<urn:uuid:1062c267-829e-4e9e-9694-9482a6cd48d1>\",\"Content-Length\":\"13388\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98c4c273-3042-4b02-80d7-4063962c10a6>\",\"WARC-Concurrent-To\":\"<urn:uuid:bbc65faa-bb30-4233-ac58-19394eca6dde>\",\"WARC-IP-Address\":\"104.31.78.58\",\"WARC-Target-URI\":\"https://warrenadams.me/music-audio/how-to-make-a-double-box-plot.php\",\"WARC-Payload-Digest\":\"sha1:FKKPY32EKRWOOKCHGVPE5JPQWUTZECEU\",\"WARC-Block-Digest\":\"sha1:YIQUVJJ3LHIFEQKKS64ICPUETFQMDCUG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370506121.24_warc_CC-MAIN-20200401192839-20200401222839-00456.warc.gz\"}"} |
https://physics.stackexchange.com/questions/207115/harmonic-oscillator-relation-with-this-hamiltonian/207417 | [
"# harmonic oscillator relation with this hamiltonian\n\nI have studied the annihilation and creation operators and number operator $N$ in relation with the simple harmonic oscillator that is governed by: $\\ H = \\hbar\\omega(N+ \\frac{1}{2})$.\nI don't understand the relation between the harmonic oscillator and, for example, this Hamiltonian $\\ H = \\hbar\\omega_0a^{\\dagger}a+\\hbar\\omega_1a^{\\dagger}a^{\\dagger}aa$ that I have found in an example in a lecture notes. They calculate the energies of this system.\n\nThey use the annihilation operator that is defined from the simple harmonic oscillator to solve that system. what is physically this system? why I can use the SHO to calculate the energies? I feel that I am confused with the a operator. I thought that it was defined from the Hamiltonian of the simple harmonic oscillator ,,,, isn't it ?\n\n• First, do you understand the meaning of the Hamiltonian of a system? $H=\\hbar\\omega(N+\\frac{1}{2})$ is the reduced form of the Hamiltonian of a harmonic oscillator or electromagnetic field. – TBBT Sep 15 '15 at 6:07\n• Thanks for your response. Yes i understand , N here is the number operator equal to N=aa+ , but here they use the simple harmonic oscillator to solve this problem? why the a, a+ and N are the same ? i thought that those operators are ONLY for the simple harmonic oscillator. – Mati Sep 15 '15 at 6:13\n• That is not what I meant. – TBBT Sep 15 '15 at 6:14\n• TBBT please can you explain? thanks in advance – Mati Sep 15 '15 at 6:17\n• I am writing an answer as we speak. – TBBT Sep 15 '15 at 6:18\n\nLet's quickly review the quantum harmonic oscillator. We have a single particle moving in one dimension, so the Hilbert space is $L^2(\\mathbb{R})$: the set of square-integrable complex functions on $\\mathbb{R}$. The harmonic oscillator Hamiltonian is given by\n\n$$H= \\frac{P^2}{2m} + \\frac{m\\omega^2}{2}X^2$$\n\nwhere $X$ and $P$ are the usual position and momentum operators: acting on a wavefunction $\\psi(x)$ they are $X \\psi(x) = x\\psi(x)$ and $P \\psi(x) = -i\\hbar\\ \\partial \\psi / \\partial x$. Of course, we can also think of them as acting on an abstract vector $|\\psi\\rangle$.\n\nBy letting $P \\to -i\\hbar\\ \\partial/\\partial x$ we could solve the time independent Schrödinger equation $H \\psi = E \\psi$, but this is a bit of a drag. So instead we define operators $a$ and $a^\\dagger$ as in your post. Notice that the definition of $a$ and $a^\\dagger$ has nothing whatsoever to do with our Hamiltonian. It just so happen that these definitions are convenient because the Hamiltonian turns out to be $\\hbar \\omega (a^\\dagger a + 1/2)$.\n\nFor convenience we define the number operator $N = a^\\dagger a$; at this stage number is just a name with no physical interpretation. Using the commutation relation $[a,a^\\dagger] = 1$ and some algebra we notice that $N$ has a nondegenerate spectrum given by the natural numbers. In other words, the eigenvalues of $N$ are $\\{0,1,2,\\dots\\}$, and to each eigenvalue $n$ there corresponds a single state $|n\\rangle$ with $N|n\\rangle = n |n\\rangle$. Notice that, again, $N$ is independent of our Hamiltonian. However, because the Hamiltonian turns out to be $\\hbar \\omega (N+1/2)$ we immediately know that the states $|n\\rangle$ are its eigenvectors, with energies $\\hbar \\omega (n + 1/2)$.\n\nNow you are given a different Hamiltonian. The Hilbert space is still exactly the same, and so are $a$, $a^\\dagger$ and $N$, because their definition had nothing to do with the original Hamiltonian. You can still use their properties to find energies, eigenvectors, and so on. The states $|n\\rangle$ are still the eigenstates of $N$, though a priori they might not be eigenstates of the new $H$ (exercise 31 asks you to prove that they in fact are eigenstates of the new $H$). The important point here is that operators are (usually) defined independently of the Hamiltonian. They characterize the physical system. After all, you know that there are operators $X$ and $P$, and you have no qualms about using them with different Hamiltonians. The Hamiltonian gives the energy and the time evolution, but the observables and related operators are independent of your choice of Hamiltonian.\n\nAbout the physical interpretation... exercise 31 asks you to prove that $H=\\hbar\\omega_0 N + \\hbar \\omega_1 (N^2-N)$; notice that we have gotten rid of $\\hbar\\omega_0 /2$ since it is just a constant. I would usually expect $\\omega_1$ to be smaller than $\\omega_0$ so this is a small perturbation (for small $n$ at least), but we don't really care about that right now. You can see that $|n\\rangle$ are still the eigenstates of the Hamiltonian; all we did is shift the energies by an amount $\\hbar \\omega_1 (N^2-N)$.\n\n• Just a comment, i was born in BSAS :) – Mati Sep 19 '15 at 6:13\n\nThe Annihilation and Creation Operators are NOT specific to any particular Hamiltonian. They are defined through X and P which are position and momentum operators for ANY system that you are studying. It just so turned out that these operators facilitate the calculations in SHO specifically. (In fact, I believe these operators originated from Dirac's study of SHO)\n\nYour new Hamiltonian is different from SHO formally, but really the OPERATORs have NOT Changed! N is still the same N! Since arguments of H ONLY involve one operator N and arbitrary scalars, H and N commute and thus eigenstates of N are also eigenstates of H. So if H acts on an eigenstate of N, the output is just the energy. Here notice that although n no longer COUNTS the excitation level of a single oscillator, it is still a sufficiently good label for the energy levels of your new system.\n\nTo summarize, N is just a tool. You could have expressed your Hamiltonian in terms of other \"ingenious\" (or stupid) operators and solve the problem in terms of eigenstates of those operators! N is by NO MEANS unique. It can exist on its own without reference to SHO. I guess that is the biggest point to make.\n\n• Yes, this $a^\\dagger$ is related with the harmonic oscillator! The relation is very close as it creates eigenstates of this Hamiltonian. (Of course you can write other Hamiltonians in terms of these operators, but those Hamiltonians are in general not diagonalized by them). As you are generally interested in diagonalizing the Hamiltonian (as you can then read off the spectrum, ...) this counts for a lot. For another example, the operators creating Cooper pairs are related to the BCS Hamiltonian, they create the fundamental excitations of the system (as they diagonalize the Hamiltonian). – Sebastian Riese Sep 16 '15 at 18:36\n• Hi Sebastian Riese, I guess my answer was somewhat misleading. I am not saying that N is not associated with SHO. All I am saying is that it CAN exist on its own! And Mati's original confusion was \"why can we apply results of SHO to another Hamiltonian that seems unrelated.\" My answer is, we are not applying results of SHO, but rather just results derived from the operator N, which is universally applicable. – Zhengyan Shi Sep 17 '15 at 3:35\n• Your edit fixes the issues with the answer, so +1. – Sebastian Riese Sep 17 '15 at 10:58\n• @SebastianRiese , Zhengyan Shi thank you very much! – Mati Sep 19 '15 at 6:14\n\nHamiltonian of different systems is different. Hamiltonian of different systems do not have to be related. You are taught to study the dynamics of the harmonic oscillator in Quantum Mechanics is because it related to the electromagnetic field quantisation, which you will learn later on. Basically said, the Hamiltonian of the harmonic oscillator is the same with light field.\n\nAnyway, the exercise, which you are presented us, is asking you to study a quantum system with a total Hamiltonian of $H=\\hbar\\omega_{0}a^{\\dagger}a+\\hbar\\omega_{1}a^{\\dagger}a^{\\dagger}aa$. Therefore, you should be using this one to solve the exercise.\n\nThe annihilation operator is also given in the exercise. In case you wonder, that is the same as the relation you find in the harmonic oscillator system. I'll show you, $$a=\\sqrt{\\frac{m\\omega}{2\\hbar}}x+i\\sqrt{\\frac{1}{2m\\omega\\hbar}}p$$ and its Hermitian conjugate, $$a^{\\dagger}=\\sqrt{\\frac{m\\omega}{2\\hbar}}x-i\\sqrt{\\frac{1}{2m\\omega\\hbar}}p$$ From these to relations, you can be able to find $x$ and $p$.\n\nI can tell you are confusing with the fundamental concept of Quantum Mechanics. I suggest you to go back and read more materials. If you are not clear on the operators, as you said, you should learn more about Mathematics for Quantum Mechanics.\n\nI think this exercise is not very hard. Since you only asked us to clarify the confusion you have with the Hamiltonian, I won't go any further that this. If you find the exercise challenging, please ask us for specific assistant. Please note that this site is not for homework-solving but homework-helping. If you find my answer not helpful, please comment below.\n\n• Thanks. I understand that i need to use this Hamiltonian to solve the problem. I went back and read a lot before i wrote this question. Specifically i learned that a and a+ operators derive from the hamiltonian of the simple harmonic oscillator: H = p^2/2m + mw^2x^2/2 so why i can use them without adapting them to this problem ? – Mati Sep 15 '15 at 6:49"
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https://www.blackstonetutors.com/free-pre-registration-pharmacist-recruitment-numeracy-exam-practice-questions/ | [
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"# Free Pre-Registration Pharmacist Recruitment Numeracy Exam Practice Questions\n\nAdvice & Insight From Pre-Registration Recruitment Exam Specialists\n\n## Practice Question 1\n\nA 54 year old female patient has been admitted onto the ward after receiving a kidney transplant and the consultant would like to estimate creatinine clearance prior to administration of treatment. The patient’s weight is 64kg and he has a serum creatinine of 126 micromol/litre.\n\nYou may use the following equation to support you answering this question:\n\nEstimated CrCl (ml/min) = (140 – Age) x Weight x Constant\nSerum creatinine\n\nWhere:\n\nAge = years\nWeight = kg\nSerum creatinine = micromole/litre\nConstant = 1.23 for men or 1.04 for women\n\nEstimated CrCl (ml/min) = (140 – Age) x Weight x Constant\nSerum creatinine\n\n= (140 – 54) x 64 x 1.04\n126\n\n= 45.43ml/min\n\n## Practice Question 2\n\nA drug has an elimination half-life of 14 hours, assuming it is completely absorbed and 1200mg of the drug is administered as the dose, how many mg would you expect to remain in the body after 48 hours? Assume the drug in question has first order pharmacokinetics.\n\n1200mg starting off and after 14 hours it is halved to 600mg then after another 14 hours halved again to 300mg and after another 14 hours halved again to 150mg but this equals to 42 hours and we need to work out for 48 hours.\n\nTherefore, in the next halve of between 150mg to 75mg in 14 hours the drug is reduced by 75mg so in one hour the drug reduces by 5.36mg and you can then multiply by 6 to work out total drug depletion in 6 hours = 32.14mg.\n\n75mg – 32.14mg = 42.86mg is depleted in 8 hours so you would add this to 75mg to arrive at 117.86mg after 48 hours\n\n## Practice Question 3\n\nHow many litres of a 0.91% stock solution are required to make 2L of a 0.8% solution, using water as the solvent?\n\nNeed to apply formula: C1V1 = C2V2\n\nWe know C1 (0.91%) but not V1 and we know C2 (0.8%) and V2 (2L = 2000mls) so simply plug in the known values:\n0.91% x V1 = 0.8% x 2000mls\n\nV1 = 0.8% x 2000mls/0.91\nV1 = 1758mls/1000 = 1.76 Litres\n\n## Practice Question 4\n\nYou have been asked to calculate how many milligrams of sodium ions are found in a 200mg tablet of sodium chloride which is required to be used to treat a salt deficiency in a patient.\n\nYou would need to know the molecular weight of the two components before starting; Sodium (Na) = 23 and Chloride (Cl) = 35.5\nThese added together: 23 + 35.5 = 58.5\n\nAs you are required to calculate milligrams of sodium:\n23/58.5 x 200mg tablet = 78.63mg\n\n##### Optimise Your Pre-Reg Oriel Performance\n\nLearn the best Pre-Reg Oriel strategies and practice with reflective questions & worked solutions.\n\n## Practice Question 5\n\nYou have been asked to calculate the dose for an antibiotic for a 7 year old child of a drug that is being used unlicensed according to the adult dose. The adult dose is 650mg OD.\n\nYou may find the following equation useful:\n\nDose for child = age (years) x adult dose\nAge + 12\n\nDose for child = age (years) x adult dose\nAge + 12\n\nDose for child = 7 x 650mg\n7 + 12\n\n= 239.47 mg rounded to 240mg\n\n## Practice Question 6\n\nYou receive a prescription for Levetiracetam 100mg/1ml oral syrup for a 5 month old baby weighing 7.2kg. The dose is 7mg/kg once daily, then increased in steps of up to 7mg/kg twice daily (max. per dose 21mg/kg twice daily), dose to be increased every two weeks. Calculate the total quantity you would dispense in mls for a 4 week supply.\n\n7mg per 1kg\n50.4mg =x per 7.2kg\n\n100mg in 1ml\n50.4mg in x mls = 0.5mls per day\n\nFirst two weeks: 0.5mls x 14 days = 7mls\nLast two weeks: 0.5mls BD x 14 days = 14mls\n\n7ml + 14mls = 21mls\n\nShopping Cart\nScroll to Top"
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https://elo.mastermath.nl/course/info.php?id=171 | [
"### M1: Operator Algebras - 8EC\n\nPrerequisites\n\nStandard Bachelor courses on Analysis, plus some knowledge in Functional Analysis, as provided by the Mastermath course Functional Analysis (vGaans) or at least an introductory course like the \"Introduction to functional analysis\" as taught in Nijmegen. As an indication, you should be familiar with a reasonable part of the material in the first three chapters of Gert Pedersen's book \"Analysis NOW\", Springer-Verlag, 1989.\n\nAim of the course\n\nThe students are familiar with the basics of C*-algebras and von Neumann algebras, allowing them to specialize further or to apply operator algebras in the context of non-commutative geometry or the theory of infinite quantum systems. In 1929, J. von Neumann began studying what came to be called von Neumann algebras. C*-algebras were introduced by Gelfand and Nalmark in 1943. These two subjects together form the discipline of Operator Algebras, an important part of Functional Analysis with many applications in harmonic analysis and representation theory, quantum group theory, Connes' non-commutative geometry, and mathematical physics (quantum mechanics and field theory, statistical physics). The aim of this course is to lay the foundations for further studies of the subject and its applications.\n\nWe will cover at least the following subjects:\n\nBanach algebras, in particular spectral theory commutative C*-algebras ideals, quotients, homomorphisms states and representations weak topologies, density theorems von Neumann algebras Possible further subjects, time permitting:\n\ntensor products of C*-algebras some interesting examples of C*-algebras projections in von Neumann algebras and the type classification\n\nLecture Notes / Literature\n\nNecessary\n\nWe will mainly use the following book: Gerard J. Murphy: C*-algebras and operator theory, Academic Press, 1990. You need to find yourself a copy of this book to follow the course.\n\nOther literature\n\nFor some more advanced topics, we may also refer to the following book: Masamichi Takesaki: Theory of operator algebras I. Springer, 1979, 2001.\n\nA useful book giving many non-trivial examples of C*-algebras is: Kenneth R. Davidson: C*-Algebras by example. American Mathematical Society, 1996.\n\nLecturer\n\nM. Caspers (TUD)"
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https://kannte-koffer-glaube.com/pin/465489311459067364w4-63z250100jk | [
"Home\n\n# Graph function\n\n## Graph of a function - Wikipedi\n\nThere is a slider with \"a =\" on it. You can use \"a\" in your formula and then use the slider to change the value of \"a\" to see how it affects the graph. Practice: Graphs of logarithmic functions. Vertical asymptote of natural log. In exponential functions, what is the dependent variable and in logarithmic functions\n\nWe need to model the height at time t based on what we know about cones. We also need to assume several things. (We make life easy for ourselves as we go along. We are allowed to do this since we just need to come up with a basic graph for the height of the water at time t). In this HowTech written tutorial, we're going to show you how to graph functions in Excel 2016. Don't forget to check out our main channel.. When plotting inequalities, the \"monochrome shading\" checkbox can be used. If this is checked, the shaded areas for all three functions are all the same light gray. This allows you to more easily see where complex functions overlap, since the more overlap there is, the darker the shading. If left unchecked, each function is shaded in a different color.\n\n### GraphSketch Permanent link to this graph page\n\n• Function Grapher is a full featured Graphing Utility that supports graphing two functions together.\n• Example 2. Sketch graph of the function `f(x)=(x+2)^2(x-1)^3`. There is no simpler function that initial function is obtained from. Function is neither even nor odd and not periodic\n• Online Graph draw: plot function, plot parametric curves,plot polar curves. The online curve plotting software, also known as a graph plotter, is an online curve plotter that allows you to plot functions..\n• Since there is no limit to the possible number of points for the graph of the function, we will follow this procedure at first:\n• To graph a function, start by plugging in 0 for x and then solving the equation to find y. Then, mark that spot on the y-axis with a dot. Next, find the slope of the line, which is the number that's right before the variable. Once you know your slope, write it as a fraction over 1 and then use the rise over run to plot the rest of the points from the spot you marked on the y-axis. Finally, use a ruler to draw a line connecting all of the points on your graph. To learn how to graph complicated functions by hand, scroll down! Did this summary help you?YesNo\n• Teaching Resources & Lesson Plans | Teachers Pay Teachers. Quadratic Function Graph Transformations - Notes, Charts, and Quiz I have found that practice makes perfect when teaching..\n\n## Video: Draw Function Graphs - Plotte\n\nAdd edges to a graph. add.vertex.shape. Various vertex shapes when plotting igraph graphs. How igraph functions handle attributes when the graph changes We choose values `0.5` seconds apart (if we were to use `t = 1\\ \"s\"` intervals, we would not see enough detail on the graph). The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also Characteristics of Graphs. Consider the function f(x) = 2 x + 1. We recognize the equation y = 2 x + 1.. We can only take the square root of a positive number so `x ≥ 0`. The square root of a number can only be positive, so `y ≥ 0`. function User_name(user) { return user.getName(); } Once a GraphQL service is running (typically at a URL on a web service), it can receive GraphQL queries to validate and execute. A received query is..\n\n### Function Grapher and Calculato\n\n• imum over the same interval. In mathematics, the graph of a function f is, formally, the set..\n• The graph of a function is the set of all points whose co-ordinates (x, y) satisfy the function `y = f(x)`. This means that for each x-value there is a corresponding y-value which is obtained when we substitute into the expression for `f(x)`.\n• Note: Some programs cannot handle URLs beyond a certain length. (E.G. Microsoft Word has a 256 byte limit). Certain very complex charts may produce URLs that are longer than this.",
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"Note: This feature can mislead you. For example if you enter 1/2sin(x) GFE inserts a multiply between the 2 and the sin. Since there are no parentheses, it is executed from left to right so it sees it as one half of sin(x). You may have meant it as one over 2sin(x). In mathematics, the graph of a function f is the set of ordered pairs , where f = y. In the common case where x For faster navigation, this Iframe is preloading the Wikiwand page for Graph of a function Everything you're expected to graph on your own is based on a more basic graph (parent function) that you NEED to memorize. Look at the image above and check with your teacher to see which you..\n\nLearn about exponential functions graphing with free interactive flashcards. Choose from 500 different sets of flashcards about exponential functions graphing on Quizlet Graphical Function Explorer (GFE). Operating instructions. GFE is a free online function graphing tool that allows you to plot up to three functions on the same set of axes",
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"### Graphing Linear Functions\n\n1. Function graph drawing/rendering application for use with audio applications. Exports .wav files. Crispy Plotter is a function graph plotter for mathematical functions, featuring high drawing speed..\n2. (2) We could have written the function in this example with h(t) rather than just h. The following 2 equations mean the same thing.\n3. The tangent function has a parent graph just like any other function. Therefore, the graph of tangent has asymptotes, which is where the function is undefined, at each of these places\n\n### 3 Ways to Graph a Function - wikiHo\n\n1. Graph a linear function: a step by step tutorial with examples and detailed solutions. Free graph paper is available.\n2. The ball hits the ground at approx `t = 2.04\\ \"s\"` (we can see this from Example 1). The velocity when the ball hits the ground from the graph we just drew is about `-11\\ \"m/s\"`. The graph stops at this point.\n3. 12345-1-2-3-4-512345-1-2-3-4xyOpen image in a new pageGraph of `y=1+1/x`, a hyperbola. It's a discontinuous function.\n4. Function Grapher and Calculator. Description :: All Functions. Description. Function Grapher is a full featured Graphing Utility that supports graphing two functions together\n5. Since we've recognized it is a straight line, we only need to plot 2 points and join them. But we find 3 points, just to make sure we have the correct line.\n6. The features of a function graph can show us many aspects of the relationship represented by the function. Let's take a look at the more popular graphical features. Be sure to pay attention to the..\n7. Graph piecewise-defined functions. Sometimes, we come across a function that requires more Because this requires two different processes or pieces, the absolute value function is an example of..\n\nLearn more about frequency response, graphic System Identification Toolbox. I want to know if exist some MATLAB function that can give me a transfer function by analysis from points of a frequency.. (c) We know something strange will happen near `x = 0` (since the graph is not defined there). So we check what happens at some typical points between `x = -1` and `x = 1`: Safari users: Safari defaults to suppress popups. In theory, this should not prevent this feature working, but it does. If you do not see the dialog to get the link, adjust the browser preferences to allow popups.\n\nGraph can plot standard functions, parametric functions and polar functions. Given an x-coordinate Graph will calculate the function value and the first two derivatives for any given function require 'graph/function' Graph::Function.as_x11. If you don't want to output to x11, just set config.terminal to a different option. Two convenience methods exist for gif and canva (a) There are no restrictions on the values that x can take in this example, since it is a general question with no practical significance.",
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"### Graphical Function Explorer grapher (GFE) - Math Open Referenc\n\n• e if a graph shows a function, we see if there's any value of x for which a vertical line through that value will hit..\n• Different Functions and their graphs. Modulus Function. Last updated at July 12, 2018 by Teachoo\n• Here is a set of practice problems to accompany the Graphing Functions section of the Graphing and Functions chapter of the notes for Paul Dawkins Algebra course at Lamar University\n\n### Functions and Their Graphs\n\n1. Safari users: This will be ignored if the browser is set to block pop ups - the default in Safari. (This blocking should not strictly happen, since the pages come from the same domain).\n2. If a function (such as sin() ) is preceded by a number, GFE assumes you want to multiply them. For example 3cos(2.1) will be automatically treated as if you entered 3*cos(2.1): three times the cosine of 2.1. It will not work if the function is preceded by a variable name.\n3. For example, in the chart above, press 'reset'. Note that the first function is sin(a*x). This means that each time a point is plotted, it is the sine of the current value of x multiplied by the variable a. This variable is controlled by the a slider on the right, so as you move the slider you can see the effect of varying its value. *\n4. Electrical / Rotating Equipment & Mechanical Functions\n5. Plotting and graphing are methods of visualizing the behavior of mathematical functions. Use Wolfram|Alpha to generate plots of functions, equations and inequalities in one..\n6. Graphing Logarithmic Functions: Intro (page 1 of 3). By nature of the logarithm, most log graphs tend to have the same The graph of the square root starts at the point (0, 0) and then goes off to the right\n\n## An online tool to draw, display and investigate graphs of many different\n\nClick on \"full size\" under the chart window. A new window will open with a new instance of GFE in it that is as large as your monitor will allow. This can be useful in a classroom with a projector. Function Grapher is graph maker to create 2D, 2.5D, 3D and 4D function graphs, animations and table graphs. 2D functions can be in the form of explicit, parametric, piecewise, implicit and inequality (1) This graph is height against time. The ball went straight up, not forward. (Our graph may give the impression the ball moved in the x- direction as well as up, but this was not the case.) Online graph sketching app that can graph functions and numerically solve differential equations. Requires Javascript HTML 5 Rewrite the function as an equation. Graph the line using the slope and the y-intercept, or the points\n\nWhen you enter your equations, you can refer to up to four variables that are controlled by sliders. These are named a, b, c and d, and you can adjust the value of each variable by moving the slider up or down. You can also enter an exact value into the box at the top of the slider, followed by the GRAPH button or the Enter key. . The graphs of the original and inverse functions are symmetric about the line y=x. . Composite function Suppose that a function y=f(u). depends on an intermediate variable u Close Necessary Always Enabled Once you have the charts exactly as you want them, you can click on 'Make Link' below the applet. This will build a link to the chart that you can paste into a web page or Word document. When you later click on that link, the chart will come up exactly as you want it immediately. Also, by pasting the address back into the browser address bar and pressing Enter, you can then save the chart as a browser bookmark or favorite. Explore math with our beautiful, free online graphing calculator. Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more\n\n### 4. The Graph of a Function\n\n• However, you are encouraged to learn the general shapes of certain common curves (like straight line, parabola, trigonometric and exponential curves, which you'll come across in later chapters). It's much easier than plotting points and more useful for later!\n• The most sophisticated and comprehensive graphing calculator online. Includes all the functions and options you might need. Easy to use and 100% Free\n• GraphSketch.com. Click here to download this graph. Beyond simple math and grouping (like (x+2)(x-4)), there are some functions you can use as well\n• Graphing Exponential Functions. What is an Exponential Function? To begin graphing exponential functions we will start with two examples. We will graph the two exponential functions by making a..\n• function-graph. 0.1.2 • Public • Published 6 years ago. \\$ cd node_modules/function-graph. \\$ npm test\n\n### To plot a function just type it into the function box.\n\nGraphing Linear Functions. Graph a linear function: a step by step tutorial with examples and detailed solutions. Free graph paper is available Graphing Standard Function & Transformations. A few standard graphs. Graphing Standard Function & Transformations. The rules below take these standard plots and shift them horizontally.. Graph Individual (x,y) Points. The most basic plotting skill it to be able to plot x,y points. This page will help you to do that\n\n## Graph functions and relations (Algebra 2, How to graph functions\n\nEnter a formula into one of the three input boxes ( f(x), g(x), or h(x) ), then press GRAPH or the keyboard Enter key. For example: Press Clear, then in the top function box (f(x)) enter \"cos(x)\" then press GRAPH or the Enter key on the keyboard. The function will be plotted in the window above. The syntax rules are the same as for the typed-in expressions for the Math/Scientific Calculator. To zoom, use the zoom slider. To the left zooms in, to the right zooms out. When you let go of the slider it goes back to the middle so you can zoom more. (To convince yourself of this, plot points where `x = 0.4`, `x = 0.3`, `x = 0.2`, `x = 0.1` and even `x = 0.01`.) Graph functions and relations. In order to graph a linear equation we work in 3 steps: First we solve the equation for y. Second we make a table for our x- and y-values For the first `0.918\\ \"s\"`, the ball is going up (positive velocity - that is, the blue line is above the t-axis), but slowing down.\n\n### Graphs of Functions\n\n1. Function graph with dplyr. Ask Question. Asked 2 years ago. This is my function, it's walk when I try my code without function but it's not ok with function\n2. If the water drains out in `10` seconds, it means `100π/3\\ \"units\"^3` will drain out each second (This is just `1/10` of the volume). Thus the amount of water left after t seconds is given by\n3. Computations for graphing functions. Wanted: graphing procedure. Two young mathematicians discuss how to sketch the graphs of functions\n4. Syntax for Writing Functions in R. func_name <- function (argument) { statement }. Here, we can see that the reserved word function is used to declare a function in R. The statements within the curly..\n5. A graph of a function is a visual representation of a function's behavior on an x-y plane. Graphs help us understand different aspects of the function, which would be difficult to understand by just..\n\n### Zooming and Re-centering\n\nFunctions and Their Graphs. What you should learn. GOAL 2 Graph and evaluate linear functions, as applied in Exs. 55 and 56. Why you should learn it In these functions, a graph is represented either by a list of rules of the form {v_i _ 1->v_j _ 1 Graph Drawing Algorithms. The Wolfram Language provides functions for the aesthetic drawing of graphs © 2020 GeoGebra. Function Graph. Parent topic: Functions. Graphing The Derivative of a Function This lesson covers graphing functions by plotting points as well as finding the domain and range of a function after it has been graphed. Graphs are important in giving a visual representation of the.. Graphs of Functions. The coordinate plane can be used for graphing functions. To graph a function in the xy -plane, we represent each input x and its corresponding output f ( x ) as a point ( x..\n\nThe plot( ) function opens a graph window and plots weight vs. miles per gallon. click to view. Saving Graphs. You can save the graph in a variety of formats from the menu File -> Save As In mathematics, the graph of a function f is the set of ordered pairs (x, y), where f(x) = y. In the common case where x and f(x) are real numbers, these pairs are Cartesian coordinates of points in the Euclidean plane and thus form a subset of this plane A man who is `2\\ \"m\"` tall throws a ball straight up and its height at time t (in s) is given by h = 2 + 9t − 4.9t2 m.\n\n## Algebra - Graphing Functions (Practice Problems\n\nProvided is a method for creating a cardiac function graph of an atrial fibrillation or sinus arrhythmia patient and an analysis evaluation method for a pathophysiologic mechanism from thoracic.. Note the curve continues beyond what is shown in the graph. This is just a general question and there are no practical limits for either the x- or y-values. Mobile Version | Imprint & Privacy Instructions ← → | | | | | | | | | | | | | | | | | | | | | | | | |\n\n## Example 2: Graph the function using the given values of x\n\nTake your graph with you Share. Export as... Scalable Vector Graphics (.svg) Encapsulated PostScript (.eps) Portable Document Format (.pdf) Portable Function Polar Parametric Points. Add The graph of the function \\$f(x,y)=x^2+y^2\\$ is called an elliptic paraboloid. More information about applet. One can, of course, plot the graphs of all sorts of functions. Some of the most interesting..\n\n### Using \"a\" Values\n\nGraphs and coordinates. Functions and Limits. Coordinate Planes and Graphs. A rectangular coordinate system is a pair of perpendicular coordinate lines, called coordinate axes, which are placed.. The electric power P (in watts) delivered by a battery as a function of the resistance R (in ohms) is : A free graphing calculator - graph function, examine intersection points, find maximum and minimum and much more. Save Graph. Sorry, your browser does not support this application",
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"## Steps for Sketching the Graph of the Function on eMathHel\n\nIntuitively, we expect the water height to decrease slowly at first, then to drop more quickly near the end.(d) As the value of x gets closer to `0`, the points get closer to the y-axis, although they do not touch it. The y-axis is called an asymptote of the curve. GFE has the following built-in functions. The function names are not case sensitive. Example: sin(x) is the same as Sin(x). All trigonometric functions operate in radians. Plot the graphs of functions and their inverses by interchanging the roles of x and y. Find the relationship between the graph of a function and its inverse. Which functions inverses are also..\n\nGFE is a free online function graphing tool that allows you to plot up to three functions on the same set of axes. In the functions you can refer to up to four independent variables that are controlled by sliders. This allows you to easily see the effect of changes since the graphs change in real time as you drag the sliders. Is given the formula of function of two variables z = sin(x) + cos(y). Develop the application, which draws the graph of this function in a separate form. In additional, you need realize the rotation of.. A simple to use online function plotter with a lot of options for calculating and drawing graphs or charts of mathematical functions and their score tables is a quadratic function whose graph follows. and is shared by the graphs of all quadratic functions. Note that the graph is indeed a function as it passes the vertical line test",
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"## Graphs of Functions (examples, solutions, videos\n\nThe aspect ratio (ratio of width to height) of the graph window is exactly 4:3. The initial range of values on the x and y axes are in the same ratio, so a graph of y = x will be at 45°, and circles would be round, not squashed into ellipses. However, if you change the axis limits, this may no longer be true. Graphing Calculator which plots 2D graphs and 3D functions. Offers a handsome formula editor, exporting features and more. Multiple graphs per plot We take a cone with \"easy\" values, say `h = r = 10`. This has volume `1000π/3\\ \"units\"^3`.\n\nA graph of a function is a visual representation of a function's behavior on an x-y plane. Graphs help us understand different aspects of the function, which would be difficult to understand by just looking at the function itself. You can graph thousands of equations, and there are different formulas for each one. That said, there are always ways to graph a function if you forget the exact steps for the specific type of function. JSXGraph is a cross-browser JavaScript library for interactive geometry, function plotting, charting, and data visualization in the web browser GFE can be used to plot inequalities by changing the relational operator in the pull-down menu to the left of the function. There are five possible operators: .relops {margin-top:40px} .relops td {vertical-align:top} = Equals The default. The function will be plotted as a line as usual. Greater than As above, but the line is drawn dashed.\n\nHome | Sitemap | Author: Murray Bourne | About & Contact | Privacy & Cookies | IntMath feed | Page last modified: 18 March 2018 A graph is a pictorial representation of a set of objects where some pairs of objects are connected by Mathematical graphs can be represented in data structure. We can represent a graph using an.. * If you are curious: the sine curve shown is sometimes called a sine wave. The slider a is controlling the frequency of the wave. See Sine wave. Download Function graph stock vectors at the best vector graphic agency with millions of premium high quality, royalty-free stock vectors, illustrations and cliparts at reasonable prices There is another asymptote in this curve: `y = 1`, which is marked with a dashed line. Notice the curve does not pass through this value.\n\nImprove your math knowledge with free questions in Graph a quadratic function and thousands of other math skills At each end of the x and y axis is a box containing the end values. To change them, simply edit them in place and press GRAPH or the Enter key again. If you just click-and-release (without moving), then the spot you clicked on will be the new center\n\nfunctions derivatives polynomials graphing-functions intersection-theory. functions graphing-functions. asked Apr 30 at 11:58 Normally, when GFE starts up it displays a default chart. You can alter what is initially displayed by attaching parameters to the URL of the web page. You can override some or all the controls to display whatever initial chart you would like.\n\nThereafter, the ball is coming down towards the ground and getting faster (the portion where the blue line is below the t-axis). Note: The large version is a copy of the normal-size one. Any changes you make to the large one will not be copied back to the original when you close it.\n\nThe graph of a function f is the set of all ordered pairs ( x, f(x) ) where x is in the domain of f. A function is increasing on an open interval if the function rises (positive slope) on the interval as you.. GraphSketch.com. Click here to download this graph. Beyond simple math and grouping (like (x+2)(x-4)), there are some functions you can use as well\n\nCreates a kNN or saturated graph SpatialLinesDataFrame object. Distanced constrained spatial graph dist.graph <- knn.graph(ralu.site, row.names=ralu.site@data[,SiteName Q3. (Application) Water flows out of a tank in the shape of an inverted cone (i.e. the water flows through the pointy end of the cone and the widest part of the cone is at the top). The volume of the water is decreasing at a constant rate. Create graph online and use big amount of algorithms: find the shortest path, find adjacency matrix, find minimum spanning tree and others (a) Note: y is not defined for values of x less than `-1`. (Try some in your calculator, like `x = −4`.)\n\nHere graphs of numerous mathematical functions can be drawn, including their derivatives and integrals. Draw Function Graphs. Mathematics / Analysis - Plotter - Calculator 4.0 Then plot the graph: Applications of Quadratic Functions to Real-World Problems. Learning Objectives. Here you'll learn how to write and graph quadratic functions in intercept form In this mode, there is a gravitation pull that acts on the nodes and keeps them in the center of the drawing area. Also, the nodes exert a force on each other, making the whole graph look and act like.. Inverse Functions: Graphs. A feature of a pair of inverse function is that their ordered pairs are When graphed the functions will be a reflection of the other over the line y = x as shown below\n\nGraphing a recursive function (self.math). submitted 5 years ago by [deleted]. I have a recursive piecewise function that I'd like to graph, but I can't find any graphers that support recursion Algebra. When we look at a function such as. we call the variable that we are changing—in this case. --the independent variable. We assign the value of the function to a variable we call the dependent variable\n\n## Coordinate Planes and Graphs, Functions\n\nHow do you graph a function using the first and second derivatives to identify critical points and One more subtle note we can take from a function's graph is whether for increasing [math]x[/math]-values.. The graph of a function is the set of all points whose co-ordinates (x, y) satisfy the function `y = f Since there is no limit to the possible number of points for the graph of the function, we will follow.. To visualize the relationships within families of complex functions, parameterize them with the variables t, u, s, r, or n. The tool will render a range of complex functions for values of the parameter.. You can change the range of the slider by clicking on \"range\" below it. A dialog will appear that allows you to set the range of each slider separately. GFE will check to ensure that the lower value is at the bottom of the y axis or the left of the x-axis. Negative number are allowed.\n\n## Graphs of logarithmic functions (Algebra 2 level) Khan Academ\n\n(a) Negative values for R have no physical significance, hence P is not plotted for negative values of R.\n\n• Audi g tron.\n• Vapaaehtoistoiminta tutkimus.\n• Kuparisonni.\n• Jäsenmaksulaskuri talentia.\n• Elgiganten vitvaror.\n• Rtk palvelu oy.\n• Homeopaattinen apteekki tampere.\n• Spekti keikat tampere.\n• Mitä laittaa snäpissä.\n• Sound blaster drivers.\n• Yvonne catterfeld trennung.\n• Ligamentti selkäranka.\n• Kasvatuslaatikko.\n• Biltema träskruv utomhus.\n• Vr mobiililippu lähijuna.\n• Merkintänauhat.\n• Hallitilaa siilinjärvi.\n• 0 9 lautapeli hinta.\n• Astoria kuohuviini.\n• Konepellin kiveniskusuoja.\n• Edullisin et lehti.\n• Lobbaus esimerkki.\n• Kuinka usein vaippa vaihdetaan.\n• Contura i51 hinta.\n• Studio elite kuvaukset.\n• Koiran vienti ruotsiin.\n• Dr oberniedermayr.\n• Parts of manhattan.\n• Audi a5 sportback viat.\n• Helsinki melu valitus.\n• Liiga pistepörssi 2017 2018.\n• Toimiston nimikyltit.\n• 2 panzer division ss.\n• Red cups tokmanni.\n• Std error mean suomeksi.\n• Orvokki ruukussa.\n• Vauva kääntyy mahalleen nukkumaan.\n• Vantaan seurakuntayhtym.\n• Suomussalmen vuokratalot leena kela.\n• Suomen fanituote.\n• Myydään autotraileri."
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http://www.cut-the-knot.org/triangle/RelationsInTriangle.shtml | [
"# Relations between various elements of a triangle\n\n### 2S = ab sin(C)\n\nThis follows from 2S = aha because ha = b sin(C).",
null,
"### S = rp\n\nTriangle ABC is a union of three triangles ABI, BCI, CAI, with bases AB = c, BC = a, and AC = b, respectively. The altitudes to those bases all have the length of r.",
null,
"### r² = p-1(p - a)(p - b)(p - c)\n\nThis follows from S² = p(p - a)(p - b)(p - c) and S = rp.",
null,
"### 1/r = 1/ha + 1/hb + 1/hc\n\n2S = aha = bhb = chc. Therefore, a = 2S/ha, etc. On the other hand, S = rp, so that p = S/r, or (a + b + c) = 2S/r. From here, 2S/ha + 2S/hb + 2S/hc = 2S/r.",
null,
"### sin²(A/2) = (p - b)(p - c) / bc, etc.\n\nFirst of all, sin(A) = 2·sin(A/2)cos(A/2) = 2·sin²(A/2)/tan(A/2). Therefore,\n\n (1) sin²(A/2) = sin(A)·tan(A/2) /2.\n\nWe know that\n\n (2) sin(A) = 2S / bc\n\nand\n\n (3) tan(A/2) = r/(p - a).\n\nCombining (1)-(3) gives\n\nsin²(A/2) = 2S/bc · r/(p-a) · 1/2.\n\nTaking into account that S² = p(p - a)(p - b)(p - c) and r² = p-1(p - a)(p - b)(p - c), the latter leads to\n\nsin²(A/2) = (p - b)(p - c) / bc.",
null,
"### cos²(A/2) = p(p - a) / bc, etc.\n\nIndeed, from sin²(A/2) = (p - b)(p - c) / bc,\n\n cos²(A/2) = 1 - sin²(A/2) = 1 - (p - b)(p - c) / bc = (p(b + c) - p²) / bc = p((2p - a) - p) / bc = p(p - a) / bc.",
null,
"### cos²[(C-B)/2] = [(b+c)²(p-b)(p-c)] / [a²bc]\n\nThis is the consequence of the previous two. Indeed, cos²[(C+B)/2]=sin²(A/2).\n\ncos²[(C-B)/2]-cos²[(C+B)/2]=sin(C)sin(B)=4[ΔABC]²/(a²bc),\n\ni.e.,\n\ncos²[(C-B)/2]=sin²(A/2)+4p(p-a)(p-b)(p-c)/(a²bc).\n\nIn other words,\n\ncos²[(C-B)/2]=(p-b)(p-c)/bc + 4p(p-a)(p-b)(p-c)/(a²bc)=(b+c)²(p-b)(p-c)/(a²bc).",
null,
"### AI² = (p - a)bc/p\n\nSquare the obvious\n\nAI = r/sin(A/2).\n\nSubstitute there sin²(A/2) = (p - b)(p - c) / bc and r² = p-1(p - a)(p - b)(p - c):\n\n AI² = p-1(p - a)(p - b)(p - c)bc/(p - b)(p - c) = (p - a)bc/p.",
null,
"### bc·tan(B/2)·tan(C/2)\n\nSquaring AI = r/sin(A/2) and substituting sin²(A/2) = (p - b)(p - c) / bc, we obtain\n\nAI² = r²·bc/(p - b)(p - c).\n\nBy the incenter construction, tan(B/2) = r/(p - b) and also tan(C/2) = r/(p - c). Substituting these into the above gives the required\n\nAI² = bc·tan(B/2)·tan(C/2).",
null,
"### 1/r = 1/ra + 1/rb + 1/rc\n\nS = ra(p - a) = rb(p - b) = rc(p - c).\n\nTherefore\n\n 1/ra+ 1/rb + 1/rc = (p - a)/S + (p - b)/S + (p - c)/S = (3p - a - b - c)/S = (3p - 2p)/S = p/S = 1/r,\n\nsince S = rp.",
null,
"### ra + rb + rc = r + 4R\n\nS = rp,\n\nand also\n\nS = ra(p - a) = rb(p - b) = rc(p - c).\n\nFrom these we have\n\n (4) ra + rb + rc - r = S(1/(p - a) + 1/(p - b) + 1/(p - c) - 1/p).\n\nSimple algebra yields\n\n1/(p - a) + 1/(p - b) = c / (p - a)(p - b) and\n1/(p - c) - 1/p = c / p(p - c).\n\nAnd a little more effort makes a great payoff:\n\nc / (p - a)(p - b) + c / p(p - c) = abc / p(p - a)(p - b)(p - c) = abc / S²,\n\nby Heron's formula. To sum up, from (4)\n\n (5) ra + rb + rc - r = S·abc/S² = abc / S.\n\nHowever, abc = 4RS, so that (5) implies exactly what's needed:\n\nra + rb + rc - r = abc / S = 4R.",
null,
"### rarbrc = pS\n\nSince\n\nS = ra(p - a) = rb(p - b) = rc(p - c),\n\nwe immediatly obtain\n\n rarbrc = S3 / (p - a)(p - b)(p - c) = S3 / [S² / p],\n\nby Heron's formula. But\n\nS3 / [S² / p] = Sp.",
null,
"### r+R=R(cos(A)+cos(B)+cos(C))\n\nWe know that\n\nr + rc + rb - ra = 4Rcos(A), r + rb + ra - rc = 4Rcos(C), r + ra + rc - rb = 4Rcos(B)\n\nSo that 3r+(ra + rb rc=4R(cos(A)+cos(B)+cos(C)). But\n\nra + rb + rc = r+4R\n\nwhich combine into 4r+4R=4R(cos(A)+cos(B)+cos(C)), exactly as required.",
null,
"### r rarbrc = S²\n\nThis is an immediate consequence of rarbrc = pS and rp = S.",
null,
"### la = 4p(p-a)bc/(b+c)²\n\nFollows from la = 2bc cos(A/2)/2 and cos²(A/2) = p(p - a) / bc.",
null,
"### la = 2bc cos(A/2)/(b+c)\n\nApplying the sine area formula to triangles ABLa and ACLa and then to the entire ΔABC we see that\n\nblasin(A/2)/2 + clasin(A/2)/2 = bc sin(A)/2\n\nThis simplifies to\n\nla = bc sin(A)/ (b + c)sin(A/2) = 2bc cos(A/2) / (b + c).",
null,
"### ma² = (b² + c²)/2 - a²/4\n\nLet's use Stewart's theorem\n\nAB²·DC + AC²·BD - AD²·BC = BC·DC·BD\n\nwith D being the midpoint M of BC. Then AB = c, DC = a/2, AC = b, BD = a/2, AD = ma, BC = a. We have,\n\nc²·a/2 + b²·a/2 - ma²·a = a·a/2·a/2.\n\n(The above identity could be as easily obtained with the help of the Theorem of Cosines or the Parallelogram Law; here is an example.)",
null,
"### abc = 4RS",
null,
"Let AD be a diameter of the circumcircle of ΔABC and AH its altitude. Right triangles AHC and ABD are similar, for ∠ADB = ∠ACH. Therefore,\n\nIn other words,\n\n2R·AH = AB·AC = bc.\n\nAnd finally\n\nabc = 2R·AH·a = 4RS.",
null,
"### bc = 2Rha\n\nThis follows from the previous derivation or by substituting S = aha/2 into the final formula.",
null,
"### p = 4Rcos(A/2)·cos(B/2)·cos(C/2)\n\nBy the Law of Sines\n\na = 2R·sinA, b = 2R·sinB, c = 2R·sinC,\n\nso that\n\n p = R·(sinA + sinB + sinC) = R·(sinA + sinB + sin(180° - A - B) = R·(sinA + sinB + sin(A + B) = R·(sinA + sinB + sinA·cosB + cosA·sinB) = R·(sinA·(1 + cosB) + sinB·(1 + cosA)) = R·(2sin(A/2)cos(A/2)·2cos²(B/2) + 2sin(B/2)cos(B/2)·2cos²(A/2)) = 4R·cos(A/2)cos(B/2)(sin(A/2)cos(B/2) + sin(B/2)cos(A/2)) = 4R·cos(A/2)cos(B/2)sin((A + B)/2) = 4R·cos(A/2)cos(B/2)sin(90° - C/2) = 4R·cos(A/2)cos(B/2)cos(C/2).",
null,
"### S = 2R²sin(A)·sin(B)·sin(C)\n\nBy the Law of Sines\n\na = 2R·sinA, b = 2R·sinB,\n\nFor the area of the triangle we have\n\n 2S = ab·sinC = 2RsinA·2RsinB·sinC = 4R²·sinA·sinB·sinC.",
null,
"### r = 4Rsin(A/2)·sin(B/2)·sin(C/2)\n\nThis follows directly from",
null,
"### cot(A/2) + cot(B/2) + cot(C/2) = cot(A/2)·cot(B/2)·cot(C/2)\n\nThis is equivalent to showing that, for A + B + C = 180°,\n\ncos(A/2)sin(B/2)sin(C/2) + sin(A/2)cos(B/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2)\n= cos(A/2)cos(B/2)cos(C/2).\n\nLet's transform the left-hand side:\n\ncos(A/2)sin(B/2)sin(C/2) + sin(A/2)cos(B/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2)\n= sin((A+B)/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2).\n\nBut since (A + B)/2 C = 90° - C/2, this equals\n\ncos(C/2)sin(C/2) + sin(A/2)sin(B/2)cos(C/2) = cos(C/2)[sin(C/2) + sin(A/2)sin(B/2)].\n\nReversing the steps:\n\n sin(C/2) + sin(A/2)sin(B/2) = cos((A+B)/2) + sin(A/2)sin(B/2) = cos(A/2)cos(B/2) - sin(A/2)sin(B/2) + sin(A/2)sin(B/2) = cos(A/2)cos(B/2).\n\nCombining everything together we get the desired identity.",
null,
"### rR = abc / 4p\n\nr² = p-1(p - a)(p - b)(p - c) is equivalent to\n\nr = D / p,\n\nwhere D = p(p - a)(p - b)(p - c). Also,\n\nR = abc / 4D.\n\nMultiplying the two gives\n\nrR = abc / 4p.\n\nNote that the identity at hand also follows by combining S = rp with abc = 4RS.",
null,
"### AH = 2R·|cos(A)|\n\nIn ΔABH, if A < 90°, ∠ABH = 90° - ∠A. (This is because ΔABHb is right.) Applying the law of sines to ΔABH gives,\n\n AH / sin(∠ABH) = AB / sin(180° - ∠C) = AB / sin(∠C) = 2R\n\nfrom the lawa of sines applied in ΔABC. Thus\n\n 2R = AH / sin(∠ABH) = AH / sin(90° - ∠A) = AH / cos(∠A),\n\nwhich proves the assertion AH = 2R·|cos(A)| when A < 90°.\n\nFor the case where ∠A is obtuse, H falls outside ΔABC, ∠ABH = ∠A - 90° so at the end we'll get AH = -2R·cos(A), proving AH = 2R·|cos(A)| in this case also.",
null,
"### p² = rarb + rbrc + rcra\n\nAs we know, ra=S/(p-a). It follows that\n\nrarb + rbrc + rcra=S²(1/[(p-b)(p-c)]+1/[(p-c)(p-a)]+1/[(p-a)(p-b)]=S²p/[(p-a)(p-b)(p-c)]=p²,",
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"Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. 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Determine the number of full-time-equivalent students on a campus in which students registered for a total of 25,350 credits.\n• ti84maths program asolver\n• plotting points to make picture\n• high school math Simplifier that SHOWS WORK\n• \"arithmetic practice worksheets\"\n• math mcqs problem solving with solution\n• solving equations when multiplying by a negative number\n• simplyfiy (2w to the 2nd power) to the third power\n• problem involving system of linear equation using by the substitution with agons\n• lowell lynde intermediate algebra textbook 3rd edition\n• permutations and combinations elementarypowerpoint\n• 26.3% written as a decimal\n• college algebra special product formula\n• simplifying expressions calculator for square root\n• free boolean algebra simplifier\n• algebra worksheet packets\n• real life picture of a hyperbila\n• solve each equation or formula for the specified variable\n• ti-84 plus complex root solver\n• kumon solution book level I\n• free 8th grade math worksheets\n• mathematics investigatory project for grade 4\n• expanded notation with negitive\n• MULTIPLYCATION AND DIVISION OF FRACTION AND THEIR RULES\n• elementary algebra for dummies\n• website that solves math problems\n• combination and permutation for kids\n• sample lesson plan ratio intermediate algebra\n• concept of monimial,binomial and trinomial and adition and subtraction\n• singapore standard math for secondary worksheets\n• highest common factor calculator\n• factoring trinomials\n• how to do radicals in geometry\n• similar fraction worksheets\n• solving quadratic equation by extracting square root\n• mathematical trivias\n• logarithms for dummies\n• algebra worksheets for 7th graders\n• non linear differential equations\n• 11th grade printable math worksheets\n• flow square root formula\n• worksheets on square numbers\n• iowa algebra aptitude test\n• algebrator demos\n• simplifying complex fractions worksheet\n• Variable Expression Calculator\n• texas instruments 84 interpolation program\n• Solutions of Quadratic Equation by extracting the square roots\n• +The difference between the cubes of two consecutive numbers is 547. What are these numbers?\n• ged math worksheets\n• +evaluating expressions wiyth more than two operations\n• cubic equations in real life\n• Solve pair of equation by reducing them to a linear equation 1/2x +1/3y=2\n• trivias on numbers for grade 3\n• softmath.com]\n• hands-on trig workbook\n• Solve any system using elimination calculator\n• bar graph science worksheet\n• mathletics instant workbooks for 5th grades\n• derivative calculator roots\n• what is the difference beween ti83 plus and ti 83 plus silver\n• manual de algebrator\n• system of linear equation age problem worksheet\n• solution of a quadratic equation by extracting square roots\n• college algebra cheat sheet\n• list of math formulas algebra\n• y11 3u parabola parametric worksheets with solutions\n• Traverse this string just once and find if it is palindrome or not\n• Linear * problems for high school\n• extracting roots algebra\n• examples of solving problems using graphs and charts on the hanna orleans algebraic test\n• polynomial formula calculator hyperbola\n• multiplying decimals calculator\n• (8n+2)/7=6\n• www. discreatemathmatics codes and groups chap solution.ppt\n• www.uses of mathamatics.nic.in\n• az algebra pretest\n• clep test college algebra\n• fraction number line\n• orleans-hanna algebra prognosis test\n• problem solving of linear equations.FRACTIONS\n• questions & answers to reading compass practice test for college students\n• decimal worksheet greatest to lowest worksheet\n• online usable algaba I caculater\n• sample test papers in algebra answer sheet\n• simplifying rational expressions calculator and that shows the work\n• slope equations\n• multiplying and dividing radical expressions calculator\n• real life exampels of a hypebola\n• definition prealgebra\n• how to solve quotients of radicals\n• tagalog version mathematical poems\n• solution for an irrational number expression using newton's method\n• math trivia question and answer\n• simplifying variable expression worksheet\n• +14.7 convert to mix number\n• Rational Expression Games\n• rules of negatives and positives when simplifying equations\n• texas ged print out practice test\n• How is doing operations (adding, subtracting, multiplying, and dividing) with rational expressions similar to or different from doing operations with fractions?\n• Equation Writer from Creative Software Design for ti 89\n• algebra standardized tests\n• Free 8Th Grade Math Worksheets\n• multiplication property of equality\n• mathematical investigatory project\n• free college algebra tutorial software\n• exponential functions\n• Quantitative Ability: questions on fractions, percentages, decimals, log base 10, algebra, probability and statistics\n• binomial fraction inequalities\n• games to teach numbers in english to adolescents. 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] | [
null,
"https://www.softmath.com/images/video-pages/solver-top.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78439766,"math_prob":0.9889867,"size":64982,"snap":"2019-43-2019-47","text_gpt3_token_len":15121,"char_repetition_ratio":0.23374066,"word_repetition_ratio":0.0257121,"special_character_ratio":0.20639561,"punctuation_ratio":0.021374548,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999876,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T13:34:59Z\",\"WARC-Record-ID\":\"<urn:uuid:aeff26a1-740d-4d67-9be7-8f49c3bc58d1>\",\"Content-Length\":\"203895\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec820702-cccd-4c91-8f5b-cb9009a483e0>\",\"WARC-Concurrent-To\":\"<urn:uuid:b63cbb82-82d7-4572-8f23-1d9a07dbfe6a>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"https://www.softmath.com/math-com-calculator/function-range/simplify-expressions-and-use.html\",\"WARC-Payload-Digest\":\"sha1:JN2VKQYCXO64J7NJRGHMGWGKWX77GVQ2\",\"WARC-Block-Digest\":\"sha1:HVN5XH6G6QJU72CAERBLIUIYDH2UJXZC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671260.30_warc_CC-MAIN-20191122115908-20191122143908-00310.warc.gz\"}"} |
http://stp.clarku.edu/simulations/harddisks/metropolis/index.html | [
"### Monte Carlo simulation of hard disks\n\nIntroduction\n\nThe application of the Metropolis algorithm for a system of hard disks can be stated very simply:\n\n1. Choose a particle at random and generate trial a change in its x and y coordinates:\n\nx(i) = x(i) + (2r - 1)δ\n\ny(i) = y(i) + (2r - 1)δ,\n\nwhere r is a uniform random number in the unit interval and δ is the maximum displacement.\n\n2. Accept the trial move if the trial position of the disk does not overlap another disk. Otherwise, the move is rejected and the old configuration is retained. A reasonable, although not necessarily optimum choice for δ is to choose its value such that approximately 20% of the trial moves are accepted.\n\nThe program uses units such that the diameter σ = 1.\n\nProblems\n\n1. The main quantity of interest is the radial distribution function g(r). Describe its qualitative r-dependence. How does g(r) change with increasing density?\n\n2. How does the form of g(r) compare to that a system of particles interacting with the Lennard-Jones potential at the same density?\n\n3. The pressure P of a system of hard disks is related to the value of g(r) at contact by the expression",
null,
".\n\nBecause the hard disks rarely touch, it is difficult to obtain good statistics for g(r) at contact. Fit the values of g(r) close to r = σ to a second-order polynomial in r and extrapolate the values of g(r) for r greater than r = σ+.\n\n4. What does the phase diagram of a system of hard disks look like? Does the system become a solid at high densities? Start the system at a low density and slowly compress the system. We do so by first determining the minimum distance between the centers of any two disks. Because this distance cannot be less than σ, its value bounds the amount that we can compress the system. We multiply this distance by the parameter λ = scale lengths after every Monte Carlo step per particle. Start the system in a rectangular configuration and then compress the system by setting scale lengths = 0.95. Is the system a fluid or a solid at high densities? If its a solid, what is its symmetry?\n\nJava Classes\n\n• HD\n• HDMCApp\n\nUpdated 3 March 2009."
] | [
null,
"http://stp.clarku.edu/simulations/harddisks/metropolis/pressurehd.jpg",
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https://www.kylesconverter.com/time/weeks-to-sols | [
"# Convert Weeks to Sols\n\n### Kyle's Converter > Time > Weeks > Weeks to Sols\n\n Weeks (wk) Sols (sol) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18\nReverse conversion?\nSols to Weeks\n(or just enter a value in the \"to\" field)\n\n#### Please share if you found this tool useful:\n\nUnit Descriptions\n1 Week:\n1 Week is equal to 7 days. In SI units 1 week is 604,800 seconds.\n1 Sol:\n1 Sol is a martian solar day, one full rotation of Mars from the perspective of the Sun. Approximately 88775.24409 seconds or 24 hours, 39 minutes, 35.24409 seconds; slightly variable due to orbital eccentricity. 1 sol = 88775.24409 s.\n\nConversions Table\n1 Weeks to Sols = 6.812770 Weeks to Sols = 476.8897\n2 Weeks to Sols = 13.625480 Weeks to Sols = 545.0168\n3 Weeks to Sols = 20.438190 Weeks to Sols = 613.1439\n4 Weeks to Sols = 27.2508100 Weeks to Sols = 681.271\n5 Weeks to Sols = 34.0636200 Weeks to Sols = 1362.542\n6 Weeks to Sols = 40.8763300 Weeks to Sols = 2043.813\n7 Weeks to Sols = 47.689400 Weeks to Sols = 2725.084\n8 Weeks to Sols = 54.5017500 Weeks to Sols = 3406.355\n9 Weeks to Sols = 61.3144600 Weeks to Sols = 4087.626\n10 Weeks to Sols = 68.1271800 Weeks to Sols = 5450.1681\n20 Weeks to Sols = 136.2542900 Weeks to Sols = 6131.4391\n30 Weeks to Sols = 204.38131,000 Weeks to Sols = 6812.7101\n40 Weeks to Sols = 272.508410,000 Weeks to Sols = 68127.1008\n50 Weeks to Sols = 340.6355100,000 Weeks to Sols = 681271.0077\n60 Weeks to Sols = 408.76261,000,000 Weeks to Sols = 6812710.077"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.78743994,"math_prob":0.95890784,"size":1333,"snap":"2023-14-2023-23","text_gpt3_token_len":526,"char_repetition_ratio":0.37697518,"word_repetition_ratio":0.0,"special_character_ratio":0.49587396,"punctuation_ratio":0.1462585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9545448,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-25T20:36:04Z\",\"WARC-Record-ID\":\"<urn:uuid:43267336-8f04-4bae-a4eb-5791912cfd32>\",\"Content-Length\":\"18692\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02f5d92d-67f2-4464-9937-9a4953f1c049>\",\"WARC-Concurrent-To\":\"<urn:uuid:dc4bd1ea-6c72-44c7-a5a2-9be16fdcac6e>\",\"WARC-IP-Address\":\"18.160.10.41\",\"WARC-Target-URI\":\"https://www.kylesconverter.com/time/weeks-to-sols\",\"WARC-Payload-Digest\":\"sha1:Z2LMWCRBSGH5MIQHW3SUYXGTOSKXBEQB\",\"WARC-Block-Digest\":\"sha1:M5YDUF7636WQE6CX5Y3NOF2CSZJCMIZ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296945372.38_warc_CC-MAIN-20230325191930-20230325221930-00134.warc.gz\"}"} |
http://www.oalib.com/relative/3796171 | [
"Home OALib Journal OALib PrePrints Submit Ranking News My Lib FAQ About Us Follow Us+",
null,
"",
null,
"Title Keywords Abstract Author All\nSearch Results: 1 - 10 of 100 matches for \" \"\n Page 1 /100 Display every page 5 10 20 Item\n Viacheslav V. Nikulin Mathematics , 2007, Abstract: Using authors's methods of 1980, 1981, some explicit finite sets of number fields containing ground fields of arithmetic hyperbolic reflection groups are defined, and good bounds of their degrees (over Q) are obtained. For example, degree of the ground field of any arithmetic hyperbolic reflection group in dimension at least 6 is bounded by 56. These results could be important for further classification. We also formulate a mirror symmetric conjecture to finiteness of the number of arithmetic hyperbolic reflection groups which was established in full generality recently.\n Viacheslav V. Nikulin Mathematics , 2007, DOI: 10.1112/jlms/jdp003 Abstract: This paper continues arXiv.org:math.AG/0609256, arXiv:0708.3991 and arXiv:0710.0162 . Using authors's methods of 1980, 1981, some explicit finite sets of number fields containing all ground fields of arithmetic hyperbolic reflection groups in dimension at least 3 are defined, and explicit bounds of their degrees (over Q) are obtained. Thus, now, explicit bound of degree of ground fields of arithmetic hyperbolic reflection groups is known in all dimensions. Thus, now, we can, in principle, obtain effective finite classification of arithmetic hyperbolic reflection groups in all dimensions together.\n Viacheslav V. Nikulin Mathematics , 2009, Abstract: The transition constant was introduced in our 1981 paper and denoted as N(14). It is equal to the maximal degree of the ground fields of V-arithmetic connected edge graphs with 4 vertices and of the minimality 14. This constant is fundamental since if the degree of the ground field of an arithmetic hyperbolic reflection group is greater than N(14), then the field comes from very special plane reflection groups. In our recent paper (see also arXiv:0708.3991), we claimed its upper bound 56. Using similar but more difficult considerations, here we improve this bound. These results could be important for further classification.\n Mikhail Belolipetsky Mathematics , 2015, Abstract: This is a survey article about arithmetic hyperbolic reflection groups with an emphasis on the results that were obtained in the last ten years and on the open problems.\n Mathematics , 2006, Abstract: We prove that there are only finitely many conjugacy classes of arithmetic maximal hyperbolic reflection groups.\n Mathematics , 2012, Abstract: Following the previous work of Nikulin and Agol, Belolipetsky, Storm, and Whyte it is known that there exist only finitely many (totally real) number fields that can serve as fields of definition of arithmetic hyperbolic reflection groups. We prove a new bound on the degree \\$n_k\\$ of these fields in dimension 3: \\$n_k\\$ does not exceed 9. Combined with previous results of Maclachlan and Nikulin, this leads to a new bound \\$n_k \\le 25\\$ which is valid for all dimensions. We also obtain upper bounds for the discriminants of these fields and give some heuristic results which may be useful for the classification of arithmetic hyperbolic reflection groups.\n Mikhail Belolipetsky Mathematics , 2007, Abstract: We show that degrees of the real fields of definition of arithmetic Kleinian reflection groups are bounded by 35.\n Mathematics , 2007, Abstract: We describe a collection of computer scripts written in PARI/GP to compute, for reflection groups determined by finite-volume polyhedra in \\$\\mathbb{H}^3\\$, the commensurability invariants known as the invariant trace field and invariant quaternion algebra. Our scripts also allow one to determine arithmeticity of such groups and the isomorphism class of the invariant quaternion algebra by analyzing its ramification. We present many computed examples of these invariants. This is enough to show that most of the groups that we consider are pairwise incommensurable. For pairs of groups with identical invariants, not all is lost: when both groups are arithmetic, having identical invariants guarantees commensurability. We discover many ``unexpected'' commensurable pairs this way. We also present a non-arithmetic pair with identical invariants for which we cannot determine commensurability.\n Mahdi Izadi;Mohd Zainal Abidin Ab Kadir;Maryam Hajikhani PIER , 2013, DOI: 10.2528/PIER12112503 Abstract: In this paper, analytical field expressions are proposed to determine the electromagnetic fields due to an inclined lightning channel in the presence of a ground reflection at the striking point. The proposed method can support different current functions and models directly in the time domain without the need to apply any extra conversions. A set of measured electromagnetic fields associated with an inclined lightning channel from a triggered lightning experiment is used to evaluate the proposed field expressions. The results indicate that the peak of the electromagnetic fields is dependent on the channel angle, the observation point angle as well as the value of the ground reflection factor due to the difference between channel and ground impedances. Likewise, the effect of the channel parameters and the ground reflection on the values of the electromagnetic fields is considered and the results are discussed accordingly.\n Viacheslav V. Nikulin Mathematics , 2008, Abstract: An integral hyperbolic lattice is called reflective if its automorphism group is generated by reflections, up to finite index. Since 1981, it is known that their number is essentially finite. We show that K3 surfaces over C with reflective Picard lattices can be characterized in terms of compositions of their self-correspondences via moduli of sheaves with primitive isotropic Mukai vector: Their self-correspondences with integral action on the Picard lattice are numerically equivalent to compositions of a finite number of especially simple self-correspondences via moduli of sheaves. This relates two topics: Self-correspondences of K3 surfaces via moduli of sheaves and Arithmetic hyperbolic reflection groups. It also raises several natural unsolved related problems.\n Page 1 /100 Display every page 5 10 20 Item"
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https://dsp.stackexchange.com/questions/34424/question-about-ramp-filter-used-in-filtered-backprojection | [
"# Question about ramp filter used in filtered backprojection\n\nQuestion is this. First, a ramp filter (in frequency domain) is defined by $H(Q)=|Q|$. What are the responses of a ramp filter to (1) a constant function $f(r)=c$ and (2) a sinusoid function $f(r)=\\sin(wr)$? What does the response mean? Following is my work.\n\nMy work:\n\n1. First, take fourier transform of a function $f(r)=c$. It is $\\int_{-\\infty}^{\\infty}f(r)e^{-2i\\pi rQ}dr=c\\delta(Q)$. Then multiply ramp filter and take inverse fourier transform. It is $\\int_{-\\infty}^{\\infty}c\\delta(Q)|Q|e^{2i\\pi irQ}dQ=0$??\n\n2. Similarly, $\\int_{-\\infty}^{\\infty}\\sin(wr)e^{-2i\\pi rQ}dr=\\frac{\\delta(Q-w/2\\pi)-\\delta(Q+w/2\\pi)}{2i}$. So Applying the ramp filter and i.f.t gives $\\frac{w(e^{iwr}-e^{-iwr})}{4i\\pi}=\\frac{w\\sin(wr)}{2\\pi}$.\n\nIt this right?\n\n• Can you please clarify the math notation a little bit? Your first expression is like an integral equation not an evaluation of an integral. It is quite possible that your derivations are along the right track but with this notation it is unclear why. Also, what does your intuition say? What do you think might happen if you were to pass DC through a ramp filter? What is another name for the ramp filter? What does the ramp filter do at the end of the day? – A_A Sep 25 '16 at 17:01\n\nIf a filter has frequency response $H(Q)$, this means that its response to an input $e^{j2\\pi Q_0 r}$ is the signal $H(Q_0)e^{j2\\pi Q_0 r}$. In other words, a sinusoidal input of frequency $Q_0$ produces an output of the same frequency, but with amplitude $|H(Q_0)|$ and phase $\\angle H(Q_0)$.\nIn your first question, the input has frequency $Q_0=0$. The filter's response at that frequency is $|Q_0|=0$. Then, the filter's output will be 0: frequency $Q_0=0$ is completely absorbed by the filter and it does not appear at the output.\nIn your second question, the input has frequency $Q_0=w/2\\pi$. The filter's response at that frequency is $|Q_0|=w/2\\pi$. The output, then, should be $\\frac{w}{2\\pi}\\sin(wr)$."
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https://homes.cs.washington.edu/~jmschr/lectures/Parallel_Processing_in_Python.html | [
"# Parallel Processing in Python¶\n\nauthor: Jacob Schreiber\ncontact: jmschr@cs.washington.edu\n\nSimply put, parallel processing is splitting up a task between many CPUs to make it work faster or more efficiently. In the big data setting this often means speading up complex analyses by splitting up the task across the CPUs and getting a speedup bounded by the additional number of cores added. This is made easier because most tasks on big data are embarassingly parallel, which means having many independent tasks that can be distributed to many cores easily. For example, matrix-matrix multiplication can be time intensive for big matrices. However, each row-column dot product is independent from each other and so can be given to a core without the need to communicate between cores mid-task. This is great for parallel processing.\n\nWhile languages like C, Java, and R allow parallel processing fairly easily, life isn't easy being a Python programmer due to the Global Interpreter Lock (or GIL). This is a lock on a Python process which prevents it from executing multiple threads simultaneously.\n\nWhat does that even mean, though?\n\nA process in a computer can be depicted as the following:",
null,
"(ref: https://web.kudpc.kyoto-u.ac.jp/manual/en/parallel)\n\nYour computer will allocate some memory for each process, which can be thought of as basically a \"program.\" Every program is isolated from the others, and no process is allowed to infringe on the memory space of another process. If a process attempts to infringe on a neighbors space, this can lead to everyone's favorite error, the segfault.\n\nInside a process there can be multiple \"threads\" of execution. These threads share the underlying memory of the process, and can each be assigned to different cores for execution. This makes parallelization nice, because a process can load up some data and then process it using multiple cores much many people might eat a birthday cake.\n\nHowever, having shared memory can cause problems. For example, imagine trying to sum a trillion numbers and saving it to a local variable \"x\". At some point, one of your lines of code will look something like x = x + item. However, if multiple threads grab the old value of \"x\" at the same time, add some number to it, and try to update the variable, you will get mistakes called \"race conditions\" which will cause you to get an incorrect sum. What's even worse is that the simpler the task, the higher the probability of these happening due to the frequency that these variables are being called. Lets take a look at a simple code example (without focusing too much on what the code itself says.)\n\nIn :\n%load_ext Cython\n%pylab inline\nimport seaborn, time\nseaborn.set_style('whitegrid')\n\nPopulating the interactive namespace from numpy and matplotlib\n\nIn :\n%%cython --compile-args=-fopenmp --link-args=-fopenmp --force\nimport numpy\ncimport numpy\nfrom cython.parallel import prange\n\ncdef int x, i\n\nwith nogil:\nx = 0\nfor i in prange(10000000, schedule='guided', num_threads=4):\n(&x) = (&x) + 1\n\nprint x\n\n2505506\n\n\nLooks like we're getting a very wrong answer, and a different answer every time the code is run. This is where locks come in! Locks can be acquired by any thread and prevent other threads from executing. In this case, a thread could acquire a lock on x, update it, and then release the lock. This means only one thread is operating on x at a time. We can simulate that with the following code:\n\nIn :\n%%cython --compile-args=-fopenmp --link-args=-fopenmp --force\nimport numpy\ncimport numpy\nfrom cython.parallel import prange\n\ncdef int x, i\n\nwith nogil:\nx = 0\nfor i in prange(10000000, schedule='guided', num_threads=4):\nwith gil:\n(&x) = (&x) + 1\n\nprint x\n\n10000000\n\n\nHowever, there can still be race conditions if multiple threads can read a variable but then acquire the gil to update it sequentially. In fact, we have exactly the same problem as before, where multiple threads can read a variable and then overwrite each other's progress. We're just doing it in a more orderly fashion instead of a chaotic fashion. See the following code:\n\nIn :\n%%cython --compile-args=-fopenmp --link-args=-fopenmp --force\nimport numpy\ncimport numpy\nfrom cython.parallel import prange\n\ncdef int x, i, y\n\nwith nogil:\nx = 0\nfor i in prange(1000000, schedule='guided', num_threads=4):\ny = x + 1\nwith gil:\n(&x) = y\n\nprint x\n\n275824\n\n\nUsing the lock seemed to solve all of our issues, so what was the problem? Well, if we only have one thread running at a time, then we're not going to get any speed gain! Despite having distributed work to each of the seperate threads, only one of them is running at a time because we're constantly putting a lock on the work the threads are doing. We can see below roughly what was happening, where only one thread was allowed to do work at a time because it was acquiring and releasing a lock.",
null,
"(ref: www.tivix.com/blog/lets-go-python)\n\nNow, the GIL is like a lock on a variable, except it's a lock on the entire Python process. This allows only one thread to be running at a time, no matter how many threads you create during the process of your program. This is a hotly contended issue amongst Pythonistas, but the general argument is that the Python interpreter was built to not be thread-safe because several speed gains could be made on that assumption, and since Python is pretty slow natively these speed gains are important.\n\nIt is still possible to do parallel processing in Python. The most naive way is to manually partition your data into independent chunks, and then run your Python program on each chunk. A computer can run multiple python processes at a time, just in their own unqiue memory space and with only one thread per process. This will be tedious, and you'll have to use another script to aggregate your answers. Pretty much, this sounds awful and you shouldn't do it.\n\nThis process can be automated using the multiprocessing module.\n\n## Multiprocessing¶\n\nThe simplest way to include parallel processing in your code is through the multiprocessing module which is built into python. The way this works is through the built-in pickle module, which is a way of serializing data, functions, and objects. In essense, you start up a pool of processes which wait for instructions from the main process. The main process will then send serialized data, methods, and objects, to the new process, and the new process will perform the instructions and send the result back.\n\nThe module has a relatively simple interface, where you write a python method, create a pool of workers, and execute that function for different inputs. Lets see it in action for the simple task of taking in a series of numbers and returning the number of items and their sum.\n\nFirst, define the method:\n\nIn :\ndef summarize( X ):\n\"\"\"Summarize the data set by returning the length and the sum.\"\"\"\nreturn len(X), sum(X)\n\n\nNow lets run this on a large amount of data in a purely sequential way to see what the answer is.\n\nIn :\nX = numpy.random.randn(1e7) + 8.342\nx0, x1 = summarize(X)\nprint x1 / x0\n\n/home/jmschr/anaconda/lib/python2.7/site-packages/ipykernel/__main__.py:1: DeprecationWarning: using a non-integer number instead of an integer will result in an error in the future\nif __name__ == '__main__':\n\n8.34182145513\n\n\nLooks like we're recovering the mean fairly well, as would be expected given the large number of samples. Now lets time this:\n\nIn :\n%timeit summarize(X)\n\n100 loops, best of 3: 8.8 ms per loop\n\n\nSeems fairly fast given that we have 1e7 elements. But it is a simple operation!\n\nUsing multiprocessing in Python involves first setting up a pool of workers and then distributing the tasks to each worker involving a subsection of the dataset.\n\nIn :\nfrom multiprocessing import Pool\n\np = Pool(4)\nx = p.map( summarize, (X[i::4] for i in range(4)) )\nprint x\n\n[(2500000, 20853831.617965657), (2500000, 20855287.683435395), (2500000, 20856136.83326447), (2500000, 20852958.41664188)]\n\n\nThis returns a list with each element being the output of one of the jobs. We have to sum through each job now to get the total aggregate.\n\nIn :\nx0 = sum( y0 for y0, _ in x )\nx1 = sum( y1 for _, y1 in x )\nprint x1 / x0\n\n8.34182145513\n\n\nLooks like we're getting the same answer here! Now to time it:\n\nIn :\nn = len(X) / 4\n%timeit p.map( summarize, (X[i*n:(i+1)*n] for i in range(4)) )\n\nThe slowest run took 5.87 times longer than the fastest. This could mean that an intermediate result is being cached.\n1 loop, best of 3: 198 ms per loop\n\n\nThat looks like it's at least an order of magnitude slower! But we're using parallel processing, how could things ever be slower? The reason mostly revolves around the operation being too simple and the data being too small. Here are some of the more prominent reasons:\n\n(1) The cost of setting up the worker pool is very high given the cost of the original operation:\n\nIn :\n%timeit -n 3 Pool(4)\n\nThe slowest run took 14.03 times longer than the fastest. This could mean that an intermediate result is being cached.\n3 loops, best of 3: 23.7 ms per loop\n\n\n(2) The memory cost of using this is high, because you're copying your initial data at least twice. This is because one full copy lives on the original process, and you have to send a partition of the data to this new process, which all has to live in memory.\n\n(3) The cost of sending data across pipes can be higher than you want. Here is a speed comparison done by someone on StackOverflow:\n\nmpenning@mpenning-T61:~$python multi_pipe.py Sending 10000 numbers to Pipe() took 0.0369849205017 seconds Sending 100000 numbers to Pipe() took 0.328398942947 seconds Sending 1000000 numbers to Pipe() took 3.17266988754 seconds mpenning@mpenning-T61:~$ python multi_queue.py\nSending 10000 numbers to Queue() took 0.105256080627 seconds\nSending 100000 numbers to Queue() took 0.980564117432 seconds\nSending 1000000 numbers to Queue() took 10.1611330509 seconds\nmpnening@mpenning-T61:~$python multi_joinablequeue.py Sending 10000 numbers to JoinableQueue() took 0.172781944275 seconds Sending 100000 numbers to JoinableQueue() took 1.5714070797 seconds Sending 1000000 numbers to JoinableQueue() took 15.8527247906 seconds mpenning@mpenning-T61:~$\n\nThis can be made faster, but it shows that especially for large datasets the time spent just piping the data can be significant.\n\nSo, if a significant portion of time is being spent piping data, then we can get better improvement with more cores if we have a more complex task. Basically each core is speeding through the summation and gulping down data as fast as possible. If we had a more complex task then the core would be spending more time on it and not need to be requesting new data as often.\n\nLets look at the case of pairwise sums, where we take in a list of numbers $x_{1}, \\dots , x_{n}$ and return the sum of them all multiplied by each other, $\\sum\\limits_{i=1}^{n} \\sum\\limits_{j=1}^{n} x_{i}x_{j}$.\n\nIn :\ndef scalar_sum(X, y):\nreturn sum( x*y for x in X )\n\ndef pairwise_sum(X):\nreturn( sum( scalar_sum(X, y) for y in X ) )\n\n\nNow lets show the single process version time:\n\nIn :\nX = numpy.random.randn(1000)\n%timeit pairwise_sum(X)\n\n1 loop, best of 3: 221 ms per loop\n\n\nAnd a parallelized version. As a side note, we need a helper function to unwrap the arguments to the function since it has more than one function. It doesn't add a significant overhead to the program.\n\nIn :\ndef scalar_wrapper(args):\nreturn scalar_sum(*args)\n\np = Pool(4)\n%timeit p.map( scalar_wrapper, ((X, y) for y in X) )\n\n10 loops, best of 3: 84 ms per loop\n\n\nLooks like we're already getting a ~3x speed up when using 4 processes on this more complex operation!\n\nNow, lets go to a real world example which can be very sped up using parallelization: Doing Expectation-Maximization on a Gaussian Mixture Model.\n\nTo begin with, a multivariate gaussian $G$ of dimensionality $d$ is parameterized by $mu$, which is a vector of mean values for each dimension, and $\\Sigma$ which is the covariance matrix between all of these dimensions. The covariance matrix has a diagonal which is the variance of every dimension by itself, and each entry $\\Sigma_{i,j}$ represents the covariance between dimension $i$ and dimension $j$.\n\nA mixture model requires that its underlying components have two operations: (1) they can return probabilities of data given that component $P(D|M)$, and (2) they can be fit to data. The expectation-maximization procedure then iterates between these two operations, using Bayes rule to calculate the probability of each component producing each point, and then updating the components based on these weighted beliefs.\n\nMultivariate Gaussians have both of these operations. (1) The probability of a point under the multivariate Gaussian is as follows:\n\n\\begin{equation} P(x|\\mu, \\Sigma) = \\frac{1}{\\sqrt{(2\\pi)^{d}|\\Sigma|}}exp\\left(-\\frac{1}{2}(x-\\mu)^{T}\\Sigma^{-1}(x-\\mu)\\right) \\end{equation}\n\nFortunately it's also implemented in scipy as a simple function for us. We can use this to create the expectation part of EM, which is calculating $P(M|D)$ for every data point and every component. We then normalize this matrix so that each row sums to 1, which defines the probability that each component generated that sample. Lastly, we do weighted MLE estimates to update the distributions.\n\nWe can see an example of it at work here:",
null,
"(ref: http://twitwi.github.io/Presentation-2015-dirichlet-processes/gmm2d/difficult-soft.gif)\n\nThe colors added to the set are a bit confusing, but we can pretend that they don't matter. Roughly, we believe that the data was generated from three components, and want to be able to identify the underlying distributions which generated the data. We start off with rough, and very incorrect estimates, and then move towards better estimates iteratively until we converge at good solutions.\n\nIn :\nfrom scipy.stats import multivariate_normal\n\ndef expectation(X, mus, covs):\nr = numpy.hstack([multivariate_normal.pdf(X, mu, cov)[:, numpy.newaxis] for mu, cov in zip(mus, covs)])\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ndef covariance(X, weights):\nn, d = X.shape\nmu = numpy.average(X, axis=0, weights=weights)\ncov = numpy.zeros((d, d))\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\ndef maximization(X, r):\nmus = numpy.array([ numpy.average(X, axis=0, weights=r[:,i]) for i in range(r.shape) ])\ncovs = numpy.array([ covariance(X, r[:,i]) for i in range(r.shape) ])\nreturn mus, covs\n\n\nWe now have the two components which we need for the EM algorithm-- the expectation step and the maximization step. We just need to repeat these two steps until we achieve convergence. Instead of writing a convergence calculator we're just going to run the algorithm some fixed number of times until convergence,\n\nIn :\ndef EM(X, mu, cov):\nfor i in range(50):\nr = expectation(X, mu, cov)\nmu, cov = maximization(X, r)\n\nreturn mu, cov, r\n\n\nGreat. So we have a function now which will take in some initial values, and iterate until convergence, and return those values.\n\nLets generate some data with 4 underlying components\n\nIn :\nd, m = 2, 4\nX = numpy.concatenate([numpy.random.randn(100000, d)+i*4 for i in range(m)])\n\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c='c', linewidth=0 )\n\nOut:\n<matplotlib.collections.PathCollection at 0x7f1eb1a22050>",
null,
"Now lets run the sequential algorithm and plot the results.\n\nIn :\ninitial_mu = numpy.random.randn(m, d) + X.mean(axis=0)\ninitial_cov = numpy.array([numpy.eye(d) for i in range(m)])\n\nmu, cov = initial_mu.copy(), initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 26.227670908s\n\n\nHowever, the calculation of the responsibility matrix in the expectation step is embarassingly parallel. We can either use model parallelism, where we break the model up across processes and have it analyze the same data, or data parallelism, where we break up the data across processes and have the full model analyze the partition of the data.\n\nIf we try to do this using the multiprocessing module, we will usually run into a memory error. This is because we have to create a new pool of workers each process, and pipe the data to each worker. The module doesn't have great shutting down of idle processes, and so in an iterative algorithm like EM you will frequently get several hundreds of idle processes all with a huge slice of the data taking up memory. Since we are implementing model parallelism, this gets especially bad, because each process will have a full copy of the data stored on it.\n\nHow do we deal with this, then?\n\n## joblib¶\n\njoblib is a parallel processing library for python which was developed by many of the same people who work on scikit-learn, and is widely used inside scikit-learn to parallelize some of their algorithms. It is built on top of the multiprocessing and multithreading libraries in order to support both (multithreaded will be talked about later) but has a significant portion of additional features. One of the biggest ones is the ability to use a pool of workers like a context manager which can be reused across many tasks to be parallelized. This means each iteration of EM can use the same pool of workers. Newer versions will also actively time the duration of the tasks being dispatched to the processes to create an optimal schedule for splitting up data chunks. Lastly, if the number of jobs is set to 1, it will work in a purely sequential mode, with no overhead of setting up a pool or dispatching data.\n\nLets see how to use it below:\n\nIn :\nfrom joblib import Parallel, delayed\n\ndef _expectation(X, mu, cov):\nreturn multivariate_normal.pdf(X, mu, cov)[:, numpy.newaxis]\n\ndef _expectation_wrapper(args):\nreturn _expectation(*args)\n\ndef expectation(X, mus, covs, parallel):\ntasks = ((X, mu, cov) for mu, cov in zip(mus, covs))\nr = numpy.hstack( parallel( delayed(_expectation_wrapper)(t) for t in tasks ) )\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ndef EM(X, mu, cov):\nwith Parallel(n_jobs=4) as parallel:\nfor i in range(50):\nr = expectation(X, mu, cov, parallel)\nmu, cov = maximization(X, r)\n\nreturn mu, cov, r\n\nmu, cov = initial_mu.copy(), initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 39.5285749435s\n\n\nIt looks like we're not getting a speed increase despite using parallel processing. This is likely because of the overhead costs of setting up the pool and piping data between the processes can be quite high. It might be time to move back to multi-threading, since we're having limited success with multiprocessing.\n\nWe did not immediately use threading because the GIL prevented multiple threads from being executed at the same time. However, it is possible to release the GIL if we use cython, which is a compiler for both native Python code and its extension, also called Cython. Lets look into parallelizing the pairwise sum problem first in cython.\n\nIn :\nX = numpy.random.randn(5000)\n\n\nLets first take a look at the raw python code and see how long it takes.\n\nIn :\ndef scalar_sum(X, y):\nreturn sum( x*y for x in X )\n\ndef pairwise_sum(X):\nreturn( sum( scalar_sum(X, y) for y in X ) )\n\n%timeit pairwise_sum(X)\n\n1 loop, best of 3: 5.54 s per loop\n\n\nNow lets keep the code the same but turn cython on and see how long it takes.\n\nIn :\n%%cython\n\ndef scalar_sum(X, y):\nreturn sum( x*y for x in X )\n\ndef pairwise_sum(X):\nreturn( sum( scalar_sum(X, y) for y in X ) )\n\nIn :\n%timeit pairwise_sum(X)\n\n1 loop, best of 3: 4.74 s per loop\n\n\nLooks like we're getting a pretty good speed increase without even modifying the code!\n\nNow lets use some static typing of variables to utilize the Cython extension language. Since scalar_sum is the primary workhorse, lets focus on just improving that one for now.\n\nIn :\n%%cython\n\ncdef double scalar_sum(double[:] X, double y):\ncdef int i\ncdef double _sum = 0.0\n\nfor i in range(X.shape):\n_sum += X[i] * y\n\nreturn _sum\n\ndef pairwise_sum(X):\nreturn( sum( scalar_sum(X, y) for y in X ) )\n\nIn :\n%timeit pairwise_sum(X)\n\n10 loops, best of 3: 27.3 ms per loop\n\n\nLooks like a pretty good speed increase there! Static typing is one of the main boosts in speed you can get from Cython, because it can be very expensive to have to infer the type for numerics if you have to constantly repeat the checks.\n\nNow lets cythonize the pairwise_sum component as well, to make the entire thing written in the Cython extension language.\n\nIn :\n%%cython\n\ncdef double scalar_sum(double[:] X, double y):\ncdef int i\ncdef double _sum = 0.0\n\nfor i in range(X.shape):\n_sum += X[i] * y\n\nreturn _sum\n\ncpdef pairwise_sum(double[:] X):\ncdef int i\ncdef double _sum = 0.0\n\nfor i in range(X.shape):\n_sum += scalar_sum(X, X[i])\n\nreturn _sum\n\nIn :\n%timeit pairwise_sum(X)\n\n10 loops, best of 3: 23.7 ms per loop\n\n\nNow lets remove the GIL. In order to remove the GIL we need to make sure that the code with the GIL removed obeys several rules:\n\n(1) Only statically typed variasbles of C primitives (int, long, double...)\n\n(2) Arrays must be represented using pointers instead of numpy arrays or memoryviews\n\n(3) No python object or methods at all\n\n(4) All functions called must be tagged with nogil at the end\n\nWe then specify when we want to remove the GIL through the use of a context manager with gil and with nogil depending on what you want to do.\n\nPassing arrays of data to the C level form the Python level seems like it might be a daunting task at first. However, numpy has solved this problem relatively elegantly. You can simply use the following to extract a pointer from any numpy array:\n\ncdef dtype* X_ptr = <dtype*> X_ndarray.data\n\nThe .data attribute extracts the pointer to the underlying data, and dtype is whatever datatype is being stored. For example, most numerics are stored as a doubles, so we can do the following:\n\ncdef double* X_ptr = <double*> X_ndarray.data\n\nOne last consideration is that X_ndarray must be cast as a numpy array. If we haven't explicitly cast it we must explicitly cast it now:\n\ncdef double* X_ptr = <double*> (<numpy.ndarray> X_ndarray).data\nIn :\n%%cython\ncimport numpy\n\ncdef double scalar_sum(double* X, double y, int n) nogil:\ncdef int i\ncdef double _sum = 0.0\n\nfor i in range(n):\n_sum += X[i] * y\n\nreturn _sum\n\ncpdef pairwise_sum(numpy.ndarray X_ndarray):\ncdef int i, n = X_ndarray.shape\ncdef double* X = <double*> X_ndarray.data\ncdef double _sum = 0.0\n\nwith nogil:\nfor i in range(n):\n_sum += scalar_sum(X, X[i], n)\n\nreturn _sum\n\nIn :\n%timeit pairwise_sum(X)\n\n10 loops, best of 3: 21.2 ms per loop\n\n\nNow we can parallelize this function by changing the range to prange. There are a variety of great scheduling techniques to assign chunks of data to the various threads, but guided usually works the fastest in my experience.\n\nIn :\n%%cython --compile-args=-fopenmp --link-args=-fopenmp --force\ncimport numpy\nfrom cython.parallel cimport prange\n\ncdef double scalar_sum(double* X, double y, int n) nogil:\ncdef int i\ncdef double _sum = 0.0\n\nfor i in range(n):\n_sum += X[i] * y\n\nreturn _sum\n\ncpdef pairwise_sum(numpy.ndarray X_ndarray):\ncdef int i, n = X_ndarray.shape\ncdef double* X = <double*> X_ndarray.data\ncdef double _sum = 0.0\n\nwith nogil:\nfor i in prange(n, schedule='guided', num_threads=4):\n_sum += scalar_sum(X, X[i], n)\n\nreturn _sum\n\nIn :\n%timeit pairwise_sum(X)\n\n100 loops, best of 3: 6.49 ms per loop\n\n\nHowever, a major problem is that we can't use openmp on Windows machines, so if you want to write code which works cross-platform, or just on Windows, you need to use a different solution. This is where joblib comes in again, because it has both a multiprocessing and a multithreading backend.\n\nWe can use the same scalar_sum code from before exactly, but we need to make a few modifications since joblib works on the python level, but releasing the gil requires you to be on the C level.\n\nIn :\n%%cython\ncimport numpy\nfrom joblib import Parallel, delayed\n\ncdef double scalar_sum(double* X, double y, int n) nogil:\ncdef int i\ncdef double _sum = 0.0\n\nfor i in range(n):\n_sum += X[i] * y\n\nreturn _sum\n\ncpdef double scalar_wrapper(numpy.ndarray X_ndarray, double y):\ncdef double* X = <double*> X_ndarray.data\ncdef double _sum\n\nwith nogil:\n_sum = scalar_sum(X, y, X_ndarray.shape)\nreturn _sum\n\ncpdef pairwise_sum(numpy.ndarray X):\ncdef double _sum\n\n_sum = sum( parallel([delayed(scalar_wrapper, check_pickle=False)(X, y) for y in X ]) )\n\nreturn _sum\n\nIn :\n%timeit pairwise_sum(X)\n\n1 loop, best of 3: 2.12 s per loop\n\n\nThere is a lot of overhead here with going between python and cython as individual points are passed. So lets chunk the data and send all of the elements of $X$ which it is responsible for summing.\n\nIn :\n%%cython\ncimport numpy\nfrom joblib import Parallel, delayed\n\ncdef double scalar_sum(double* X, double* y, int n, int m) nogil:\ncdef int i, j\ncdef double _sum = 0.0\n\nfor i in range(n):\nfor j in range(m):\n_sum += X[i] * y[j]\n\nreturn _sum\n\ncpdef double scalar_wrapper(numpy.ndarray X_ndarray, numpy.ndarray y_ndarray):\ncdef double* X = <double*> X_ndarray.data\ncdef double* y = <double*> y_ndarray.data\ncdef int n = X_ndarray.shape, m = y_ndarray.shape\ncdef double _sum\n\nwith nogil:\n_sum = scalar_sum(X, y, n, m)\nreturn _sum\n\ncdef double _sum\n\n_sum = sum( parallel([ delayed(scalar_wrapper, check_pickle=False)(X, X[i::num_threads]) for i in range(num_threads) ]) )\n\nreturn _sum\n\nIn :\nX = numpy.random.randn(100000)\n%timeit pairwise_sum(X, 1)\n\n1 loop, best of 3: 8.55 s per loop\n\nIn :\n%timeit pairwise_sum(X, 4)\n\n1 loop, best of 3: 2.62 s per loop\n\n\nLooks like we're getting a near-linear speedup when we add more threads. Great!\n\n## Expectation-Maximization Revisited¶\n\nNow lets go back to the expectation-maximization algorithm on Gaussian Mixture models. Previously we found that using joblib and multiprocessing to parallelize jobs did not help significantly, likely due to the high overhead of setting up multiple processes and then pipe data between them. Lets see if we can lower this overhead using threads instead of processes.\n\nFirst, lets generate some data again, and create baseline parameter estimates to start at.\n\nIn :\nd, m = 2, 4\nX = numpy.concatenate([numpy.random.randn(1000000, d)+i*5 for i in range(m)]).astype('float64')\ninitial_mu = numpy.random.randn(m, d)*4 + X.mean(axis=0)\ninitial_cov = numpy.array([numpy.eye(d) for i in range(m)])\n\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c='c', linewidth=0 )\n\nOut:\n<matplotlib.collections.PathCollection at 0x7f1eb1d91850>",
null,
"Now, lets use the same code which we used before. We're going to need to get this code down to the cython level so that we can release the GIL and use multithreading. The most important function to ensure that the GIL is released for is the log probability function, because it takes up the most amount of time. This means that instead of using the convenient scipy function which we used before, we have to rewrite it ourselves. If the inner workings of log probability calculations don't interest you, ignore it! Otherwise, we basically need to pass in the inverse covariance matrix and the log of the determinant of the covariance matrix. Using these, we can easily calculate the log probability of each point under the given parameters.\n\nFirst lets define the functions we won't change this entire time, the maximization step and the EM iterator.\n\nIn :\ndef maximization(X, r):\nmus = numpy.array([ numpy.average(X, axis=0, weights=r[:,i]) for i in range(r.shape) ])\ncovs = numpy.array([ covariance(X, r[:,i], mus[i]) for i in range(r.shape) ])\nreturn mus, covs\n\ndef EM(X, mu, cov):\nfor i in range(25):\nr = expectation(X, mu, cov)\nmu, cov = maximization(X, r)\n\nreturn mu, cov, r\n\n\nNow lets define the expectation step functions which we will optimize, and the covariance calculator for the maximization step.\n\nIn :\nimport numpy\nLOG_2_PI = numpy.log(2*numpy.pi)\n\ndef log_probability(X, mu, inv_cov, log_det, d):\nlogp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i, j]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ndef _expectation(X, mu, cov ):\nn = X.shape\nd = X.shape\nlog_det = numpy.linalg.slogdet(cov)\ninv_cov_ndarray = numpy.linalg.inv(cov)\nr = numpy.zeros(n)\n\nfor i in range(n):\nr[i] = numpy.exp(log_probability(X[i], mu, inv_cov_ndarray, log_det, d ))\n\nreturn r\n\ndef expectation(X, mu, cov):\nr = numpy.hstack((_expectation(X, m, c)[:, numpy.newaxis] for m, c in zip(mu, cov)))\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ndef covariance(X, weights, mu):\nn, d = X.shape\ncov = numpy.zeros((d, d))\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 3808.46620393s\n\n\nAlright, looks like it's taking a fair amount of time to run this calculation on the data. Lets step through the steps we took to reduce this down to Cython that we did before. First, lets just turn on the Cython compiler and see what types of speed increases we can get from that.\n\nIn :\n%%cython\nimport numpy\nLOG_2_PI = numpy.log(2*numpy.pi)\n\ndef log_probability(X, mu, inv_cov, log_det, d):\nlogp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i, j]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ndef _expectation(X, mu, cov ):\nn = X.shape\nd = X.shape\nlog_det = numpy.linalg.slogdet(cov)\ninv_cov_ndarray = numpy.linalg.inv(cov)\nr = numpy.zeros(n)\n\nfor i in range(n):\nr[i] = numpy.exp(log_probability(X[i], mu, inv_cov_ndarray, log_det, d ))\n\nreturn r\n\ndef expectation(X, mu, cov):\nr = numpy.hstack((_expectation(X, m, c)[:, numpy.newaxis] for m, c in zip(mu, cov)))\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ndef covariance(X, weights, mu):\nn, d = X.shape\ncov = numpy.zeros((d, d))\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 2470.94735408s\n\n\nGreat! We didn't need to do anything differently but we still get a fairly nice speed gain. Almost all python scripts which have a lot of numeric calculations will experience a speed gain of, in my experience, around 30%.\n\nThe next step for cythonizing our code is to include static typing. This allows the cython compiler to turn for loops into C-for loops, which are significantly faster. As a note, despute xrange being faster and more memory efficient than range for large Python lists, use range when writing in cython. Lets also pull the pointer arrays out of the numpy arrays.\n\nIn :\n%%cython\nimport numpy\ncimport numpy\nLOG_2_PI = numpy.log(2*numpy.pi)\n\ncdef double log_probability(double* X, double* mu, double* inv_cov, double log_det, int d):\ncdef int i, j\ncdef double logp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i + j*d]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ncpdef numpy.ndarray _expectation( numpy.ndarray X_ndarray, numpy.ndarray mu_ndarray, numpy.ndarray cov_ndarray ):\ncdef int i, n = X_ndarray.shape, d = X_ndarray.shape\ncdef double log_det = numpy.linalg.slogdet(cov_ndarray)\ncdef numpy.ndarray inv_cov_ndarray = numpy.linalg.inv(cov_ndarray)\ncdef numpy.ndarray r = numpy.zeros(n)\n\ncdef double* inv_cov = <double*> inv_cov_ndarray.data\ncdef double* mu = <double*> mu_ndarray.data\ncdef double* X = <double*> X_ndarray.data\n\nfor i in range(n):\nr[i] = numpy.exp(log_probability(X + i*d, mu, inv_cov, log_det, d ))\n\nreturn r\n\ndef expectation(X, mu, cov):\nr = numpy.hstack((_expectation(X, m, c)[:, numpy.newaxis] for m, c in zip(mu, cov)))\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ncpdef numpy.ndarray covariance(numpy.ndarray X, numpy.ndarray weights, numpy.ndarray mu):\ncdef int i, j, n = X.shape, d = X.shape\ncdef numpy.ndarray cov = numpy.zeros((d, d))\ncdef double w_sum\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 718.266939878s\n\n\nThat's a pretty nice speed gain! Now lets release the GIL for the important functions to get it ready for multithreading.\n\nIn :\n%%cython\nimport numpy\ncimport numpy\nfrom libc.math cimport exp as cexp\n\nDEF LOG_2_PI = 0.79817986835\n\ncdef double log_probability(double* X, double* mu, double* inv_cov, double log_det, int d) nogil:\ncdef int i, j\ncdef double logp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i + j*d]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ncpdef numpy.ndarray _expectation( numpy.ndarray X_ndarray, numpy.ndarray mu_ndarray, numpy.ndarray cov_ndarray ):\ncdef int i, n = X_ndarray.shape, d = X_ndarray.shape\ncdef double log_det = numpy.linalg.slogdet(cov_ndarray)\ncdef numpy.ndarray inv_cov_ndarray = numpy.linalg.inv(cov_ndarray)\ncdef numpy.ndarray r = numpy.zeros(n)\n\ncdef double* inv_cov = <double*> inv_cov_ndarray.data\ncdef double* mu = <double*> mu_ndarray.data\ncdef double* X = <double*> X_ndarray.data\n\nfor i in range(n):\nr[i] = cexp(log_probability(X + i*d, mu, inv_cov, log_det, d ))\n\nreturn r\n\ndef expectation(X, mu, cov):\nr = numpy.hstack((_expectation(X, m, c)[:, numpy.newaxis] for m, c in zip(mu, cov)))\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ncpdef numpy.ndarray covariance(numpy.ndarray X, numpy.ndarray weights, numpy.ndarray mu):\ncdef int i, j, n = X.shape, d = X.shape\ncdef numpy.ndarray cov = numpy.zeros((d, d))\ncdef double w_sum\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 63.3489141464s\n\n\nThat is a pretty massive speed increase! Now lets try using multithreads with openmp and see what type of performance increase we can get.\n\nIn :\n%%cython --compile-args=-fopenmp --link-args=-fopenmp --force\nfrom cython.parallel cimport prange\nimport numpy\ncimport numpy\nfrom libc.math cimport exp as cexp\n\nDEF LOG_2_PI = 0.79817986835\n\ncdef double log_probability(double* X, double* mu, double* inv_cov, double log_det, int d) nogil:\ncdef int i, j\ncdef double logp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i + j*d]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ncpdef numpy.ndarray _expectation( numpy.ndarray X_ndarray, numpy.ndarray mu_ndarray, numpy.ndarray cov_ndarray ):\ncdef int i, n = X_ndarray.shape, d = X_ndarray.shape\ncdef double log_det = numpy.linalg.slogdet(cov_ndarray)\ncdef numpy.ndarray inv_cov_ndarray = numpy.linalg.inv(cov_ndarray)\ncdef numpy.ndarray r_ndarray = numpy.zeros(n)\n\ncdef double* inv_cov = <double*> inv_cov_ndarray.data\ncdef double* mu = <double*> mu_ndarray.data\ncdef double* X = <double*> X_ndarray.data\ncdef double* r = <double*> r_ndarray.data\n\nfor i in prange(n, nogil=True, num_threads=4, schedule='guided'):\nr[i] = cexp(log_probability(X + i*d, mu, inv_cov, log_det, d ))\n\nreturn r_ndarray\n\ndef expectation(X, mu, cov):\nr = numpy.hstack((_expectation(X, m, c)[:, numpy.newaxis] for m, c in zip(mu, cov)))\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ncpdef numpy.ndarray covariance(numpy.ndarray X, numpy.ndarray weights, numpy.ndarray mu):\ncdef int i, j, n = X.shape, d = X.shape\ncdef numpy.ndarray cov = numpy.zeros((d, d))\ncdef double w_sum\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 36.5230062008s\n\n\nIt's almost two times faster! That's better, but we aren't seeing nearly the linear gain we would expect to see. This is because we implemented a data parallel algorithm, where the only thing parallelized is calculating the log probabilities. This is mostly because it's easier to implement. What if, instead, we implemented a model parallel scheme where each component is parallelized instead of the data?\n\nIn :\n%%cython --compile-args=-fopenmp --link-args=-fopenmp --force\nfrom cython.parallel cimport prange\nimport numpy\ncimport numpy\nfrom libc.math cimport exp as cexp\n\nDEF LOG_2_PI = 0.79817986835\n\ncdef double log_probability(double* X, double* mu, double* inv_cov, double log_det, int d) nogil:\ncdef int i, j\ncdef double logp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i + j*d]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ncdef void _expectation( double* X, double* r, double* mu, double* inv_cov, double log_det, int n, int d ) nogil:\ncdef int i\nfor i in range(n):\nr[i] = cexp(log_probability(X + i*d, mu, inv_cov, log_det, d ))\n\ncpdef expectation(numpy.ndarray X_ndarray, numpy.ndarray mu_ndarray, numpy.ndarray cov_ndarray):\ncdef int i, n = X_ndarray.shape, d = X_ndarray.shape, m = mu_ndarray.shape\ncdef numpy.ndarray inv_cov_ndarrays = numpy.array([numpy.linalg.inv(c) for c in cov_ndarray])\ncdef numpy.ndarray log_dets_ndarray = numpy.array([numpy.linalg.slogdet(c) for c in cov_ndarray])\ncdef numpy.ndarray r_ndarray = numpy.zeros((4, n), dtype='float64')\n\ncdef double* inv_cov = <double*> inv_cov_ndarrays.data\ncdef double* log_dets = <double*> log_dets_ndarray.data\ncdef double* r = <double*> r_ndarray.data\n\ncdef double* X = <double*> X_ndarray.data\ncdef double* mu = <double*> mu_ndarray.data\n\nfor i in prange(4, num_threads=4, schedule='guided', nogil=True ):\n_expectation(X, r + i*n, mu + i*d, inv_cov + i*d*d, log_dets[i], n, d)\n\nr_ndarray = ( r_ndarray / r_ndarray.sum(axis=0) ).T\nreturn r_ndarray\n\ncpdef numpy.ndarray covariance(numpy.ndarray X, numpy.ndarray weights, numpy.ndarray mu):\ncdef int i, j, n = X.shape, d = X.shape\ncdef numpy.ndarray cov = numpy.zeros((d, d))\ncdef double w_sum\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)",
null,
"EM took 30.6642620564s\n\n\nLooks like we're doing a bit better with a model parallel scheme than a data parallel scheme. This makes sense, because we can easily have each thread scan the dataset rather than constantly chopping up the dataset for the threads to see. Oftentimes data parallel schemes are easier to implement, whereas model parallel schemes can be more efficient.\n\nLastly, lets implement this with joblib. We're going to implement the model parallel version.\n\nIn :\n%%cython\nimport numpy\ncimport numpy\nfrom libc.math cimport exp as cexp\nfrom joblib import Parallel, delayed\n\nDEF LOG_2_PI = 0.79817986835\n\ncdef double log_probability(double* X, double* mu, double* inv_cov, double log_det, int d) nogil:\ncdef int i, j\ncdef double logp = 0.0\n\nfor i in range(d):\nfor j in range(d):\nlogp += (X[i] - mu[i]) * (X[j] - mu[j]) * inv_cov[i + j*d]\n\nreturn -0.5 * (d * LOG_2_PI + log_det + logp)\n\ncpdef numpy.ndarray _expectation( numpy.ndarray X_ndarray, numpy.ndarray mu_ndarray, numpy.ndarray cov_ndarray ):\ncdef int i, n = X_ndarray.shape, d = X_ndarray.shape\ncdef double log_det = numpy.linalg.slogdet(cov_ndarray)\ncdef numpy.ndarray inv_cov_ndarray = numpy.linalg.inv(cov_ndarray)\ncdef numpy.ndarray r_ndarray = numpy.zeros(n)\n\ncdef double* inv_cov = <double*> inv_cov_ndarray.data\ncdef double* mu = <double*> mu_ndarray.data\ncdef double* X = <double*> X_ndarray.data\ncdef double* r = <double*> r_ndarray.data\n\nwith nogil:\nfor i in range(n):\nr[i] = cexp(log_probability(X + i*d, mu, inv_cov, log_det, d ))\n\nreturn r_ndarray\n\ndef expectation(X, mu, cov):\nr = parallel([ delayed(_expectation, check_pickle=False)(X, m, c) for m, c in zip(mu, cov) ])\nr = numpy.hstack([ a[:, numpy.newaxis] for a in r ])\nr = ( r.T / r.T.sum(axis=0) ).T\nreturn r\n\ncpdef numpy.ndarray covariance(numpy.ndarray X, numpy.ndarray weights, numpy.ndarray mu):\ncdef int i, j, n = X.shape, d = X.shape\ncdef numpy.ndarray cov = numpy.zeros((d, d))\ncdef double w_sum\n\nfor i in range(d):\nfor j in range(i+1):\ncov[i, j] = weights.dot( (X[:,i] - mu[i])*(X[:,j] - mu[j]) )\ncov[j, i] = cov[i, j]\n\nw_sum = weights.sum()\nreturn cov / w_sum\n\nIn :\nmu = initial_mu.copy()\ncov = initial_cov.copy()\n\ntic = time.time()\nmu, cov, r = EM(X, mu, cov)\ntoc = time.time() - tic\n\nr = r.argmax(axis=1)\nplt.figure( figsize=(14, 10) )\nplt.scatter( X[:,0], X[:,1], c=['cmgr'[i] for i in r], linewidth=0 )\nplt.show()\nprint \"EM took {}s\".format(toc)"
] | [
null,
"https://web.kudpc.kyoto-u.ac.jp/manual/sites/default/files/styles/large/public/thread_en.png",
null,
"https://lh4.googleusercontent.com/HXDr4afwx28XEZgogOWBMEcaU0updIy_BsRqOnq7kaGVq3kEyXMlwmrDTvi9ZlMRI7fdW4TT5sPO4z_9kSVxlhrUznOdvK_rHQtP6pfic8ABrVcm3lOWPEoMH8sDKK2fMhw1YLI",
null,
"http://twitwi.github.io/Presentation-2015-dirichlet-processes/gmm2d/difficult-soft.gif",
null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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null,
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https://sci2s.ugr.es/covariate-shift-cross-validation | [
"# A Study on the Impact of Partition-Induced Dataset Shift on k-fold Cross-Validation - Complementary Material\n\nThis Website contains additional material to the SCI2S research paper: A Study on the Impact of Partition-Induced Dataset Shift on k-fold Cross-Validation\n\nJ.G. Moreno-Torres, J.A. Sáez, and F. Herrera, A Study on the Impact of Partition-Induced Dataset Shift on k-fold Cross-Validation. IEEE Transactions on Neural Networks 23(8): 1304-1312 (2012)",
null,
"Summary:\n\n## Abstract\n\nJ.G. Moreno-Torres, J.A. Sáez, and F. Herrera, A Study on the Impact of Partition-Induced Dataset Shift on k-fold Cross-Validation.\n\nCross-validation is a very commonly employed technique to evaluate classifier performance. However, it can potentially introduce dataset shift, a harmful factor that is often not taken into account, and which can result in inaccurate performance estimation. This works analyzes both the prevalence and impact of partition-induced covariate shift on different k-fold cross-validation schemes.\n\nFrom the experimental results obtained we conclude that the degree of partition-induced covariate shift depends on the cross-validation scheme considered. In this way, worse schemes may harm the correctness of a single classifier performance estimation and also increase the needed number of repetitions of cross-validation to reach a stable performance estimation.\n\n## Single-experiment classifier performance analysis results\n\nThis section includes the results for the single-experiment classifier performance analysis experiment. There are 4 files, one for each type of partitioning studied (DOB-SCV, DB-SCV, SCV and MS-SCV). Inside each file, you can find the results divided in sheets. Each sheet corresponds to a different partition granularity: 10x1, 5x2 and 2x5. On each sheet, you can then find the test AUC obtained by each method on each dataset.\n\nWe also include here the results of the partitions created to obtain \"true\" classifier estimations. These are presented in a single file, since we use the same data as a reference when studiying all 4 methods. Remember the presented results are classifier performance measured as ROC AUC in the test set."
] | [
null,
"https://sci2s.ugr.es/sites/default/files/icons/pdf_download.png",
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http://www.weddslist.com/rmdb/pages/relations/cant.php | [
"# Cantellation\n\nCantellation is a non-symmetric relationship between some pairs of regular maps of the same genus. Any self-dual regular map can be cantellated.\n\nIf a self-dual regular map is described by\nM:{p,p} V V E\n(meaning, it is in manifold M, each face has p edges, each vertex has p edges, it has V vertices, V faces and E edges), then it can be cantellated. This yields a regular map described by\nM:{p,4} E 2V 2E.\n\nThis relationship is never symmetric: the cantellated regular map has twice as many edges as the original.\n\nFor example, if we cantellate the tetrahedron we get the octahedron.\n\nIf you have a regular map and want to cantellate it,\n\n• replace each edge by a vertex\n• retain each face as a face\n• replace each vertex by a face\n\nThe same procedure can be applied to a regular map which is not self-dual. However the result is not a regular map, it is semiregular. For example, if we cantellate the cube, we get the cuboctahedron.\n\nIf a regular map has Petrie polygons of size r, and we cantellate it, the result has holes of size r.\n\n### The name \"cantellation\"\n\nThe term \"cantellation\" was coined by Coxeter. It is further defined in the Wikipedia entry cantellation.\n\n### Halving\n\nThe book \"Abstract Regular Polytopes\"ARM, page 197, uses the term \"halving\" for an operation closely related to what we call \"cantellation\". Halving converts\nM:{4,p} 2V E 2E\nto\nM:{p,p} V V E\nand is the same as taking the dual and un-cantellating it.\n\nARM denotes halving by η."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.922447,"math_prob":0.9220573,"size":799,"snap":"2020-34-2020-40","text_gpt3_token_len":194,"char_repetition_ratio":0.19874214,"word_repetition_ratio":0.030075189,"special_character_ratio":0.19649562,"punctuation_ratio":0.10759494,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9556532,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-24T05:45:47Z\",\"WARC-Record-ID\":\"<urn:uuid:d0af5992-d517-40ca-a8b4-2ce95bf72659>\",\"Content-Length\":\"3296\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a5fa7534-9bd5-49f4-92c4-0ab067d5f1d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:1438372c-4437-4229-beeb-5bf59d5c375a>\",\"WARC-IP-Address\":\"209.68.18.39\",\"WARC-Target-URI\":\"http://www.weddslist.com/rmdb/pages/relations/cant.php\",\"WARC-Payload-Digest\":\"sha1:K7MDBX6W7YS2R55ZB6TVF2P6B3RCLTJN\",\"WARC-Block-Digest\":\"sha1:WRU3QZVFND26MR4YGF5PFGJXGXCMHAU6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400213454.52_warc_CC-MAIN-20200924034208-20200924064208-00614.warc.gz\"}"} |
https://calculator-converter.com/grams-to-grains.htm | [
"",
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"",
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"# Grams to Grains\n\nThis calculator provides conversion of grams to grains and backwards grains to grams (gr to g).\n\nEnter grams or grains for conversion:\n\nSelect conversion type:\n\nRounding options:\n\n## Conversion Table\n\n gram to grains Conversion Table: g to gr 1.0 = 15.43236 2.0 = 30.86472 3.0 = 46.29708 4.0 = 61.72943 5.0 = 77.16179 gram to grains 6.0 = 92.59415 7.0 = 108.02651 8.0 = 123.45887 9.0 = 138.89123 10 = 154.32358\n grains to grams Conversion Table: gr to g 1.0 = 0.06480 2.0 = 0.12960 3.0 = 0.19440 4.0 = 0.25920 5.0 = 0.32399 grains to grams 6.0 = 0.38879 7.0 = 0.45359 8.0 = 0.51839 9.0 = 0.58319 10 = 0.64799\n\nThe gram (British spelling: gramme, abbreviation: g) is a unit of mass in the SI system (metric system). One gram is equal to one thousandth of the kilogram (kg), which is the current SI (Metric system) base unit of mass. 1 gram (g) = 15.4323584 grains or \"troy grains\" (gr) = 1000000 microgram (mcg or µg) = 1000 milligram (mg) = 0.001 kilogram (kg) = 0.0352739619 ounces (oz) = 0.00220462262 pounds (lb). The grain or \"troy grain\" (gr) is the non-metric unit of mass used in the Imperial and U.S. customary units. Since 1958, 1 grain (gr) = 0.06479891 grams (g) = 64.79891 milligrams (mg) = 0.00006479891 kilogram (kg) = 0.00228571429 ounces (oz).\n\n## Weight Conversion\n\nOunces to grams (oz to gr)\nGrams to ounces (gr to oz)\nPounds to kilograms (lb to kg)\nKilograms to pounds (kg to lb)\nGrams to kilograms (g to kg)\nKilograms to grams (kg to g)\nKilograms to ounces (kg to oz)\nOunces to kilograms (oz to kg)\nKilograms to newtons (kg to N)\nNewtons to kilograms (N to kg)\nNewtons to pounds (N to lb)\nPounds to newtons (lb to N)\nMilligrams to kilograms (mg to kg)\nKilograms to milligrams (kg to mg)\nKilograms to tonnes (kg to t)\nTonnes to kilograms (t to kg)\nShort tons to kilograms (t to kg)\nKilograms to short tons (kg to t)\nMetric tons to short tons (t to ST)\nShort tons to tons [metric] (ST to t)\nKilograms to stones (kg to st)\nStones to kilograms (st to kg)\nMilligrams to grams (mg to g)\nGrams to milligrams (g to mg)\nMilligrams to micrograms (mg to mcg)\nMicrograms to milligrams (mcg to mg)\nMicrograms to grams (mcg to g)\nGrams to micrograms (g to mcg)\nGrains to grams (gr to g)\nGrams to grains (g to gr)\nOunces to grains (oz to gr)\nGrains to ounces (gr to oz)\nCarats to grams (ct to g)\nGrams to carats (g to ct)\nKarats to gold hallmarks (kt to hallmark)\nGold hallmarks to karats (hallmark to kt)\nCarats to ounces (ct to oz)\nOunces to carats (oz to ct)\nMilligrams to ounces (mg to oz)\nOunces to milligrams (oz to mg)\nMilligrams to pounds (mg to lb)\nPounds to milligrams (lb to mg)\nKilonewtons to tons (metric ton-force) (kN to t)\nTon-forces to kilonewtons (t to kN)\nNewtons to kilonewtons (N to kN)\nKilonewtons to newtons (kN to N)\nKilonewtons to pounds-force (kN to lb)\nPounds-force to kilonewtons (lbf to kN)\nTonnes to pounds (t to lb)\nPounds to tonnes (metric tons) (lb to t)\nShort tons to pounds (t to lb)\nPounds to short tons (US) (lb to t)\nPounds to grams (lb to g)\nGrams to pounds (g to lb)\nPounds to ounces (lb to oz)\nOunces to pounds (oz to lb)\nStones to pounds (st to lb)\nPounds to stones (lb to st)\nCups to fluid ounces (c to fl oz)\nFluid ounces to cups (fl oz to c)\nTroy ounces to ounces (ozt to oz)\nOunces to troy ounces (oz to ozt)\nTroy ounces to grams (ozt to g)\nGrams to troy ounces (g to ozt)"
] | [
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"https://calculator-converter.com/pics/calculator-converter.gif",
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https://stats.stackexchange.com/questions/127394/r-forecasting-flat-forecast | [
"# R forecasting, flat forecast\n\nI’m trying to produce a hourly, daily forecast for revenue in R. I set seasonal periods to 24, for 24 hours, and 365.25 for days in a year. I attached the fit vs actual plot and the forecast produced by R.\n\nI then fit the time series with the tbats model due to the high seasonal periods. I then try and forecast 8112 periods or just under 1 year.\n\nMy problem is that I keep getting a flat model\\$mean. However, the fitted vs actuals looks like its catching the seasonality.\n\nrev_ts <- msts(revenue_data, seasonal.periods=c(24,365.25))\nrev_fit <- tbats(rev_ts)\nrev_forecast <- forecast(rev_fit,h=8112)\nplot(rev_forecast )",
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"",
null,
"UPDATE:\n\nTrying to reference your write-up here Rob:\n\nhttp://robjhyndman.com/hyndsight/longseasonality/\n\nSo m=365.25? And n= # of observations? Sorry I'm a current student and new to R (and modeling for that matter). Where does this take into account hourly seasonality\n\nTrying to implement your code from post using these lines of code:\n\nm=365.25 (Where does this take into account hourly seasonality)\n\nn= 25656 (number of observations, historical data)\n\nrev_fit <- Arima(rev_ts, order=c(2,0,1), xreg=fourier(1:n,4,m))\n\nplot(forecast(rev_fit, h=2*m, xreg=fourier(n+1:(2*m),4,m)))\n\nAny explanation on the theory and what this is actually doing? Sorry for all the questions, but I'd love to understand this more fully.\n\nThanks!"
] | [
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https://qanda.ai/en/search/%5Cdfrac%7B%202%20%20%7D%7B%203%20%20%7D%20%20x%20%3D%20%20-2?search_mode=expression | [
"",
null,
"",
null,
"Calculator search results\n\nFormula\nSolve the equation",
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"",
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"",
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"",
null,
"Graph\n$y = \\dfrac { 2 } { 3 } x$\n$y = - 2$\n$x$Intercept\n$\\left ( 0 , 0 \\right )$\n$y$Intercept\n$\\left ( 0 , 0 \\right )$\n$\\dfrac{ 2 }{ 3 } x = -2$\n$x = - 3$\n Solve a solution to $x$\n$\\color{#FF6800}{ \\dfrac { 2 } { 3 } } \\color{#FF6800}{ x } = - 2$\n Calculate the multiplication expression \n$\\color{#FF6800}{ \\dfrac { 2 x } { 3 } } = - 2$\n$\\color{#FF6800}{ \\dfrac { 2 x } { 3 } } = \\color{#FF6800}{ - } \\color{#FF6800}{ 2 }$\n Multiply both sides by the least common multiple for the denominators to eliminate the fraction \n$\\color{#FF6800}{ 2 } \\color{#FF6800}{ x } = \\color{#FF6800}{ - } \\color{#FF6800}{ 6 }$\n$\\color{#FF6800}{ 2 } \\color{#FF6800}{ x } = \\color{#FF6800}{ - } \\color{#FF6800}{ 6 }$\n Divide both sides by the same number \n$\\color{#FF6800}{ x } = \\color{#FF6800}{ - } \\color{#FF6800}{ 3 }$\n 그래프 보기 \nGraph\nHave you found the solution you wanted?\nTry again"
] | [
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https://www.research.lancs.ac.uk/portal/en/publications/measurement-of-the-cjet-mistagging-efficiency-in-tbart-events-using-pp-collision-data-at-sqrts13-text-tev-collected-with-the-atlas-detector(ca07f6bc-80c1-478e-ad10-51d5fefea69d).html | [
"Home > Research > Publications & Outputs > Measurement of the c-jet mistagging efficiency ...\n\n## Measurement of the c-jet mistagging efficiency in $$t\\bar{t}$$ events using pp collision data at $$\\sqrt{s}=13$$ $$\\text {TeV}$$ collected with the ATLAS detector\n\nResearch output: Contribution to Journal/MagazineJournal articlepeer-review\n\nPublished\n• ATLAS Collaboration\nClose\nArticle number 95 31/01/2022 European Physical Journal C 1 82 27 Published English\n\n### Abstract\n\nA technique is presented to measure the efficiency with which c-jets are mistagged as b-jets (mistagging efficiency) using $$t\\bar{t}$$ t t ¯ events, where one of the W bosons decays into an electron or muon and a neutrino and the other decays into a quark–antiquark pair. The measurement utilises the relatively large and known $$W\\rightarrow cs$$ W → c s branching ratio, which allows a measurement to be made in an inclusive c-jet sample. The data sample used was collected by the ATLAS detector at $$\\sqrt{s} = 13$$ s = 13 $$\\text {TeV}$$ TeV and corresponds to an integrated luminosity of 139 fb$$^{-1}$$ - 1 . Events are reconstructed using a kinematic likelihood technique which selects the mapping between jets and $$t\\bar{t}$$ t t ¯ decay products that yields the highest likelihood value. The distribution of the b-tagging discriminant for jets from the hadronic W decays in data is compared with that in simulation to extract the mistagging efficiency as a function of jet transverse momentum. The total uncertainties are in the range 3–17%. The measurements generally agree with those in simulation but there are some differences in the region corresponding to the most stringent b-jet tagging requirement."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.90364873,"math_prob":0.94232273,"size":1218,"snap":"2023-14-2023-23","text_gpt3_token_len":284,"char_repetition_ratio":0.10214168,"word_repetition_ratio":0.0,"special_character_ratio":0.22577997,"punctuation_ratio":0.042056076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96763444,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T15:03:18Z\",\"WARC-Record-ID\":\"<urn:uuid:1ab863a5-5f4f-4f72-b247-e7d25e45a0c7>\",\"Content-Length\":\"32679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:931372a3-e8fd-419d-a66d-a382907cc93e>\",\"WARC-Concurrent-To\":\"<urn:uuid:316884a4-9089-4677-81fd-2960ccf9be46>\",\"WARC-IP-Address\":\"148.88.22.128\",\"WARC-Target-URI\":\"https://www.research.lancs.ac.uk/portal/en/publications/measurement-of-the-cjet-mistagging-efficiency-in-tbart-events-using-pp-collision-data-at-sqrts13-text-tev-collected-with-the-atlas-detector(ca07f6bc-80c1-478e-ad10-51d5fefea69d).html\",\"WARC-Payload-Digest\":\"sha1:TTWRM7CM5T3PPHOWCKDQ3SEFZXLQAKJ3\",\"WARC-Block-Digest\":\"sha1:HBFBXAWQPRVMBR3NXAW4T2BKKAKXYYTL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649986.95_warc_CC-MAIN-20230604125132-20230604155132-00282.warc.gz\"}"} |
https://www.vedantu.com/question-answer/the-radius-of-the-first-bohr-orbit-n1-of-class-11-chemistry-cbse-5fd6f1320cf84502ba7e82bc | [
"",
null,
"The radius of the first bohr orbit (n=1) of hydrogen atom is 53.4 pm. The radius of bohr orbit having n=3 in $L{i^{2 + }}$will be :A. 53.4 pmB. 106.8 pmC. 120.1 pmD. 160.2 pm",
null,
"Verified\n118.8k+ views\nHint: We know that neil bohr was the first to explain the general features of hydrogen atom structure and its spectrum. Bohr’s theory can be applied on the ions containing only one electron similar to that of hydrogen atom like $L{i^{2 + }}$, $B{e^{3 + }}$ and $H{e^ + }$, such species are also called hydrogen like species.\n\nFormula used: For hydrogen like species, the radii expression from bohr’s theory is given as:\n${r_n} = \\dfrac{{{a_ \\circ }({n^2})}}{Z}pm$\n\nComplete step-by-step solution:\nLet us understand the electron in the hydrogen atom can move around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus.\nThis energy and radius does not change with time but only as one moves from lower stationary state to higher stationary state when the electron absorbs the required amount of energy and emits the energy to move from higher state to lower state. These stationary states or orbits are denoted by an integral number known as principal quantum number which varies from n=1, 2, 3, 4 etc.\nThe radii of these stationary states can be expressed as:\n${r_n} = {n^2}{a_ \\circ }$\nwhere ${a_ \\circ }$ is the radius of the first stationary state and is called Bohr radius.\nIn the question above we have been given with first bohr orbit with n=1 therefore we can calculate the value of ${a_ \\circ }$by substituting the given values,\n${r_1} = {(1)^2}{a_ \\circ } \\\\ 53.4 = {a_ \\circ } \\\\$\nSimilarly for hydrogen like species such as lithium ion the radii expression from bohr’s theory is given as\n${r_n} = \\dfrac{{{a_ \\circ }({n^2})}}{Z}pm$\n$\\Rightarrow {r_n} = \\dfrac{{53.4({n^2})}}{Z}pm$\nNow we have been given n=3 and Z is the atomic number of lithium which is 3 therefore substituting we get,\n${r_3} = \\dfrac{{53.4({3^2})}}{3}pm = 160.2pm$\n\nHence, the correct option is D.\n\nNote: From the above we observe that the value of energy becomes more negative and that of radius becomes smaller with increase of atomic number Z, this means that the electron will be tightly bound to the nucleus."
] | [
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https://www.teachoo.com/3995/673/Example-41---If-R1--R2-are-equivalence-relations-in-set-A/category/Examples/ | [
"Examples\n\nChapter 1 Class 12 Relation and Functions\nSerial order wise",
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"Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class\n\n### Transcript\n\nExample 18 If R1 and R2 are equivalence relations in a set A, show that R1 ∩ R2 is also an equivalence relation. R1 is an equivalence relation 1. R1 is symmetric (a, a) ∈ R1, for all a ∈ A. 2. R1 is reflexive If (a, b) ∈ R1 , then (b, a) ∈ R1 3. R1 is transitive If (a, b) ∈ R1 & (b, c) ∈ R1 , then (a, c) ∈ R1 R2 is an equivalence relation 1. R2 is symmetric (a, a) ∈ R2, for all a ∈ A. 2. R2 is reflexive If (a, b) ∈ R2 , then (b, a) ∈ R2 3. R2 is transitive If (a, b) ∈ R2 & (b, c) ∈ R2 , then (a, c) ∈ R2 We have to prove R1 ∩ R2 is equivalence relation Check reflexive For all a ∈ A (a, a) ∈ R1, & (a, a) ∈ R2 Hence, (a, a) ∈ both R1 & R2 Hence, (a, a) ∈ R1 ∩ R2 ∴ R1 ∩ R2 is reflexive. Check symmetric R1 is symmetric ,hence If (a, b) ∈ R1 , then (b, a) ∈ R1 R2 is symmetric, hence If (a, b) ∈ R2 , then (b, a) ∈ R2 From (1) and (2) If (a, b) ∈ R1 ∩ R2, then (b, a) ∈ R1 ∩ R2 Hence , R1 ∩ R2 is symmetric. Checking transitive R1 is transitive, Hence, if (a, b) ∈ R1 & (b, c) ∈ R1 , then (a, c) ∈ R1 R2 is transitive, Hence, if (a, b) ∈ R2 & (b, c) ∈ R2 , then (a, c) ∈ R2 From (3) & (4) If (a, b) ∈ R1 ∩ R2 and (b, c) ∈ R1 ∩ R2 , then (a, c) ∈ R1 ∩ R2, ∴ R1∩ R2 is transitive. Thus, R1 ∩ R2 is an equivalence relation.",
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"https://www.teachoo.com/static/misc/Davneet_Singh.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8769599,"math_prob":0.9993765,"size":3147,"snap":"2023-40-2023-50","text_gpt3_token_len":1114,"char_repetition_ratio":0.25198856,"word_repetition_ratio":0.34115806,"special_character_ratio":0.35811883,"punctuation_ratio":0.10939907,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998621,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T21:13:24Z\",\"WARC-Record-ID\":\"<urn:uuid:a0e98783-9fff-4f41-bed7-ccda83a31bb3>\",\"Content-Length\":\"221554\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e5bcceb3-db72-4e2f-94cf-cbeca25eaad7>\",\"WARC-Concurrent-To\":\"<urn:uuid:f9a01a61-634d-47ea-9514-dc11a194566c>\",\"WARC-IP-Address\":\"52.21.227.162\",\"WARC-Target-URI\":\"https://www.teachoo.com/3995/673/Example-41---If-R1--R2-are-equivalence-relations-in-set-A/category/Examples/\",\"WARC-Payload-Digest\":\"sha1:V7UKCTSRMVTZ6QYIDXQC77OU5R5F6OA3\",\"WARC-Block-Digest\":\"sha1:X2H44JA7X5DTPBS2G74JHYH57TUFLOYB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102637.84_warc_CC-MAIN-20231210190744-20231210220744-00839.warc.gz\"}"} |
https://oeis.org/A343023 | [
"The OEIS is supported by the many generous donors to the OEIS Foundation.",
null,
"Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A343023 Number of cyclic cubic fields with discriminant n^2. 9\n 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,63 COMMENTS Equivalently, number of cubic fields with discriminant n^2. That is to say, it makes no difference if the word \"cyclic\" is omitted from the title. Let D be a discriminant of a cubic field F, then F is a cyclic cubic field if and only if D is a square. For D = k^2, k must be of the form (p_1)*(p_2)*...*(p_t) or 9*(p_1)*(p_2)*...*(p_{t-1}) with distinct primes p_i == 1 (mod 3), in which case there are exactly 2^(t-1) = 2^(omega(k)-1) (cyclic) cubic fields with discriminant D. See Page 17, Theorem 2.7 of the Ka Lun Wong link. Each term is 0 or a power of 2. The first occurrence of 2^t is 9*A121940(t) for t >= 1. LINKS Jianing Song, Table of n, a(n) for n = 1..16000 LMFDB, Cubic fields Wikipedia, Cubic field Ka Lun Wong, Maximal Unramified Extensions of Cyclic Cubic Fields, (2011), Theses and Dissertations, 2781. FORMULA a(n) = A160498(n)/2 for n > 1. EXAMPLE a(7) = 1 since there is only 1 (cyclic) cubic field with discriminant 7^2 = 49 is Q[x]/(x^3 - x^2 + x + 1). a(63) = 2 since there are 2 (cyclic) cubic fields with discriminant 63^2 = 3969: Q[x]/(x^3 - 21x - 28) and Q[x]/(x^3 - 21x - 35). a(819) = 4 since there are 4 (cyclic) cubic fields with discriminant 819^2 = 670761: Q[x]/(x^3 - 273x - 91), Q[x]/(x^3 - 273x - 728), Q[x]/(x^3 - 273x - 1547) and Q[x]/(x^3 - 273x - 1729). a(35) = 0 since it is not of form (p_1)*(p_2)*...*(p_t) or 9*(p_1)*(p_2)*...*(p_{t-1}) with distinct primes p_i == 1 (mod 3). Indeed, there are no (cyclic) cubic fields with discriminant 35^2 = 1225. PROG (PARI) a(n) = if(n<=1, 0, my(L=factor(n), w=omega(n)); for(i=1, w, if(!((L[i, 1]%3==1 && L[i, 2]==1) || L[i, 1]^L[i, 2] == 9), return(0))); 2^(w-1)) CROSSREFS Cf. A160498, A121940, A343000 (discriminants of cyclic cubic fields), A343001 (indices of positive terms). Sequence in context: A086071 A322212 A089813 * A337760 A037845 A037881 Adjacent sequences: A343020 A343021 A343022 * A343024 A343025 A343026 KEYWORD nonn,easy AUTHOR Jianing Song, Apr 02 2021 STATUS approved\n\nLookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam\nContribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent\nThe OEIS Community | Maintained by The OEIS Foundation Inc.\n\nLast modified January 18 07:55 EST 2022. Contains 350454 sequences. (Running on oeis4.)"
] | [
null,
"https://oeis.org/banner2021.jpg",
null
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https://www.convertunits.com/from/joule/metre/to/exanewton | [
"## ››Convert joule/metre to exanewton\n\n joule/metre exanewton\n\nHow many joule/metre in 1 exanewton? The answer is 1.0E+18.\nWe assume you are converting between joule/metre and exanewton.\nYou can view more details on each measurement unit:\njoule/metre or exanewton\nThe SI derived unit for force is the newton.\n1 newton is equal to 1 joule/metre, or 1.0E-18 exanewton.\nNote that rounding errors may occur, so always check the results.\nUse this page to learn how to convert between joules/meter and exanewtons.\nType in your own numbers in the form to convert the units!\n\n## ››Want other units?\n\nYou can do the reverse unit conversion from exanewton to joule/metre, or enter any two units below:\n\n## Enter two units to convert\n\n From: To:\n\n## ››Definition: Exanewton\n\nThe SI prefix \"exa\" represents a factor of 1018, or in exponential notation, 1E18.\n\nSo 1 exanewton = 1018 newtons.\n\nThe definition of a newton is as follows:\n\nIn physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.\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
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https://www.accountingformanagement.org/problem-4-pcs/ | [
"# Problem 4: Equivalent units for materials and conversion costs\n\n## Problem 4 (a) – weighted average vs FIFO method\n\nTK Company has two processing departments – Department X and Department Y. The quantity report of Department X reveals the following information:\n\nRequired: Using the quantity report of Department X given above, compute equivalent units for materials and conversion costs under:\n\n1. weighted average method\n2. FIFO method\n\n## Problem 4 (b) – weighted average method\n\nFor May 2020, the cost department of Abraham Company received the following data from one of its five producing departments:\n\n• Received from preceding department: 50,000 kilograms\n• Processed and transferred to next department: 38,000 kilograms\n• Work in process ending inventory: 12,000 kilograms\n\nIn this department, three different types of materials are added to the product received from previous department. The materials are added at three different stages of production as follows:\n\n• Material M-1 is added when production process starts\n• Material M-2 is added when the production process is 25% completed\n• Material M-3 is added when the production process is 75% completed\n\nThe conversion costs are incurred uniformly throughout the production process.\n\nAn examination of work in process ending inventory revealed that 3,000 kilograms were 85% processed; 6,000 kilograms were 50% processed; 3,000 kilograms were 15% processed. There was no unfinished work at the beginning of May.\n\nRequired:\n\n1. Compute equivalent units of production for each type of materials.\n2. Compute equivalent units of production for conversion costs.\n\n### Solution\n\n#### 1. Equivalent units for materials\n\nExplanation:\n\n• Material M-1 is included in all the units in ending inventory because it is added at the start of manufacturing process.\n• Material M-2 is included in both 85% and 50% processed units because it is added when the processing is 25% completed.\n• Material M-3 is included only in 85% processed units because it is added when the processing is 75% completed."
] | [
null
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http://mechatronics2018.dreslab.com/pages/190120 | [
"",
null,
"## Rui Shi & Ziyi Zhang\n\n### Description\n\nThe goal of the project is to make a robotic arm which can hold a pen or something else and is able to write in a sheet. To construct such a system, we used two large motors, two gyros, a medium motor with a gear mounted on it to adjust the height of the \"pen\". To ensure each rotate part has the largest degree of freedom so that it can cover as much part of the sheet as possible, we connected three motors in different horizontal level so they can rotate for almost 360°. Also, to avoid collision, one gyro was installed on the side of the motor rather than on top of it. Besides, we mounted a rod in the hole of the medium motor to make the arm capable to cover a larger area.",
null,
"",
null,
"",
null,
"",
null,
"### Challenge 1: Building the Robotic Arm\n\nAs shown in the pictures and video below, our motor is able to reach a much larger area than a 8.5*11 sheet.",
null,
"### Challenge 2: Calculate End Effector Positions\n\nFor challenge two, the code need to use the Gyro sensor to report the rotation angles of both parts of the arm and we need to make sure the report angles is the final angle of two parts. To achieve this, I wrote an if sentence in the while loop to achieve the function that the code can report values only when both parts of the arm already rotated and stopped rotating. When the rotation angle difference is smaller than 10, the code reports the value of two rotation angles for further calculation.\n\nIn the video, we used a green pad with grids as reference. The distance between two bold lines are one inch. The following picture shows the coordinate system we used to measure positions of the pen.",
null,
"### Challenge 3: Recording and Playing Back Positions\n\nFor challenge three, after getting the value of two rotation angles, we need to run the motor to return the arm to its origin shape. To achieve this function, first, we calculate the angle that two motors need to rotate. Then, we set two motors to rotate for that much in certain speed.\n\n```#!/usr/bin/env python3\nimport math\ndef main():\nev3 = Device('this')\n\n#Connect the #1 motor to port A\n#Connect 1the #2 motor to port B\nm1 = ev3.LargeMotor('outA')\nm2 = ev3.LargeMotor('outB')\n\n#Connect the #1 sensor to port 1\n#Connect 1the #2 motor to port 3\nsensor1 = ev3.GyroSensor('in1')\nsensor2 = ev3.GyroSensor('in3')\nsensor1.mode ='GYRO-ANG'\nsensor2.mode ='GYRO-ANG'\nangle1_origin = sensor1.value();\nangle2_origin = sensor2.value();\n\n#set the length of each part of the arm\nl1 = 8\nl2 = 9\nloop = 1\n\n#record the origin values of both sensors to prevent error\nprev1 = sensor1.value()\nprev2 = sensor2.value() - angle2_origin\n\n#this part use sensors to record the rotate angles of both parts of the arm\nwhile loop == 1:\nangle1 = sensor1.value()\nangle2 = sensor2.value() - angle2_origin\nprint(\"angle1 =%d, angle2 =%d\" % (angle1, angle2))\n\n#if both parts of the arm moved and already stoped moving, then break the loop and report the value of angle1 and angle2\nif abs(angle1 - angle1_origin) > 10 and abs(angle2 - angle2_origin) > 10 and abs(prev1 - angle1) < 10 and abs(prev2 - angle2) < 10:\nprev1 = angle1\nprev2 = angle2\nbreak\nsleep(1)\nprev1 = angle1\nprev2 = angle2\n\n#calculate angles that motor1 and motor2 should rotate, the angle motor2 should rotate equates to angle2 - angle1\ntheta1 = -(prev1)\ntheta2 = -(prev2 - prev1)\nprint(\"prev =%d, prev2 =%d\" % (theta1, theta2))\n\n#calculate x, y position\nx = l1 * math.cos(prev1*2*3.14/360) + l2 * math.cos(prev2*2*3.14/360)\ny = l1 * math.sin(prev1*2*3.14/360) + l2 * math.sin(prev2*2*3.14/360)\nprint(\"theta1 =%d, theta2 =%d\" % (theta1, theta2))\n\n#print x, y position\nprint(\"x =%d, y =%d\" % (x, y))]\n\n#run the motor to rotate back to its origin position\nm1.run_to_rel_pos(position_sp=theta1, speed_sp=100, stop_action=\"hold\")\nm2.run_to_rel_pos(position_sp=theta2, speed_sp=100, stop_action=\"hold\")\nsleep(2)\nm1.stop(stop_action=\"coast\")\nm2.stop(stop_action=\"coast\")\nif __name__ == '__main__':\nmain()```"
] | [
null,
"http://mechatronics2018.dreslab.com/thumbs/pages/190120/hero-5f544028f20bc41ed4475e15a8b7b689.jpg",
null,
"http://mechatronics2018.dreslab.com/pages/190120/22-5a84e7a5a2ca5.jpg",
null,
"http://mechatronics2018.dreslab.com/pages/190120/33-5a84e7c2450e4.jpg",
null,
"http://mechatronics2018.dreslab.com/pages/190120/44-5a84e7dc4c104.jpg",
null,
"http://mechatronics2018.dreslab.com/pages/190120/55-5a84e7e908098.jpg",
null,
"http://mechatronics2018.dreslab.com/pages/190120/11-5a84e67739498.jpg",
null,
"http://mechatronics2018.dreslab.com/pages/190120/wechat-image_20180216142423-5a8733286371c.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8060435,"math_prob":0.97847486,"size":3899,"snap":"2020-24-2020-29","text_gpt3_token_len":1061,"char_repetition_ratio":0.13838254,"word_repetition_ratio":0.02764977,"special_character_ratio":0.29263914,"punctuation_ratio":0.10966057,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9895163,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,1,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-03T10:22:54Z\",\"WARC-Record-ID\":\"<urn:uuid:5b289d3a-8203-400d-a506-cf18ad66f97f>\",\"Content-Length\":\"23473\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:191ddd8e-496d-4954-8c4b-dc7a4d73b971>\",\"WARC-Concurrent-To\":\"<urn:uuid:79a090a6-f298-4d84-a618-65ccb07a68d7>\",\"WARC-IP-Address\":\"130.64.112.173\",\"WARC-Target-URI\":\"http://mechatronics2018.dreslab.com/pages/190120\",\"WARC-Payload-Digest\":\"sha1:HL3T3LWU2BI6UUE3VUXZPFK2KENCLDML\",\"WARC-Block-Digest\":\"sha1:CCKFPJS6FIAQX7QIPBTI4755PMZP3JYD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655881984.34_warc_CC-MAIN-20200703091148-20200703121148-00181.warc.gz\"}"} |
http://grinebiter.com/Numbers/Cardinal/489-four-hundred-eighty-nine.html | [
"Four hundred eighty-nine\n\n Here is information about \"four hundred eighty-nine\" that you may find useful and interesting. Number Systems Four hundred eighty-nine is a decimal number and can be written with numbers: 489 Binary is a number system with only 0s and 1s. Four hundred eighty-nine in binary form is displayed below: 111101001 A Hexadecimal number has a base of 16 which means it includes the numbers 0 to 9 and A through F. Four hundred eighty-nine converted to hexadecimal is: 1E9 Roman Numerals is another number system. Below is four hundred eighty-nine in roman numerals: CDLXXXIX Scientific Notation Sometimes calculators and scientists shorten numbers using scientific notation. Here is four hundred eighty-nine as a scientific notation: 4.89E+02 Math Here are some math facts about Four hundred eighty-nine: Four hundred eighty-nine is a rational number and an integer. Four hundred eighty-nine is an odd number because it is not divisible by two. Four hundred eighty-nine is divisible by the following numbers: 1, 3, 163, 489 Four hundred eighty-nine is not a square number because no number multiplied by itself will equal four hundred eighty-nine. Number Lookup Four hundred eighty-nine is not the only number we have information about. Go here to look up other numbers.\n\n Translated Here we have translated four hundred eighty-nine into some of the most commonly used languages: Chinese: 四百八十九 French: quatre cent quatre-vingt-neuf German: vierhundert neunundachtzig Italian: quattrocento ottantanove Spanish: cuatrocientos ochenta y nueve\n\n Currency Here is four hundred eighty-nine written in different currencies: US Dollars: \\$489 Canadian Dollars: CA\\$489 Australian Dollars: A\\$489 British Pounds: £489 Indian Rupee: ₹489 Euros: €489\n\n Ordinal The cardinal number four hundred eighty-nine can also be written as an ordinal number: 489th Or if you want to write it with letters only: four hundred eighty-ninth.\n\n Four hundred ninety Go here for the next number on our list that we have information about."
] | [
null
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https://scfr.savingadvice.com/2008/06/09/retirement-gift-for-in-laws_39908/ | [
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119.155.20.164\n)\n\n => Array\n(\n => 103.121.43.68\n)\n\n => Array\n(\n => 5.62.58.6\n)\n\n => Array\n(\n => 203.122.40.154\n)\n\n => Array\n(\n => 222.254.96.203\n)\n\n => Array\n(\n => 103.83.148.167\n)\n\n => Array\n(\n => 103.87.251.226\n)\n\n => Array\n(\n => 123.24.129.24\n)\n\n => Array\n(\n => 137.97.83.8\n)\n\n => Array\n(\n => 223.225.33.132\n)\n\n => Array\n(\n => 128.76.175.190\n)\n\n => Array\n(\n => 195.85.219.32\n)\n\n => Array\n(\n => 139.167.102.93\n)\n\n => Array\n(\n => 49.15.198.253\n)\n\n => Array\n(\n => 45.152.183.172\n)\n\n => Array\n(\n => 42.106.180.136\n)\n\n => Array\n(\n => 95.142.120.9\n)\n\n => Array\n(\n => 139.167.236.4\n)\n\n => Array\n(\n => 159.65.72.167\n)\n\n => Array\n(\n => 49.15.89.2\n)\n\n => Array\n(\n => 42.201.161.195\n)\n\n => Array\n(\n => 27.97.210.38\n)\n\n => Array\n(\n => 171.241.45.19\n)\n\n => Array\n(\n => 42.108.2.18\n)\n\n => Array\n(\n => 171.236.40.68\n)\n\n => Array\n(\n => 110.93.82.102\n)\n\n => Array\n(\n => 43.225.24.186\n)\n\n => Array\n(\n => 117.230.189.119\n)\n\n => Array\n(\n => 124.123.147.187\n)\n\n => Array\n(\n => 216.151.184.250\n)\n\n => Array\n(\n => 49.15.133.16\n)\n\n => Array\n(\n => 49.15.220.74\n)\n\n => Array\n(\n => 157.37.221.246\n)\n\n => Array\n(\n => 176.124.233.112\n)\n\n => Array\n(\n => 118.71.167.40\n)\n\n => Array\n(\n => 182.185.213.161\n)\n\n => Array\n(\n => 47.31.79.248\n)\n\n => Array\n(\n => 223.179.238.192\n)\n\n => Array\n(\n => 79.110.128.219\n)\n\n => Array\n(\n => 106.210.42.111\n)\n\n => Array\n(\n => 47.247.214.229\n)\n\n => Array\n(\n => 193.0.220.108\n)\n\n => Array\n(\n => 1.39.206.254\n)\n\n => Array\n(\n => 123.201.77.38\n)\n\n => Array\n(\n => 115.178.207.21\n)\n\n => Array\n(\n => 37.111.202.92\n)\n\n => Array\n(\n => 49.14.179.243\n)\n\n => Array\n(\n => 117.230.145.171\n)\n\n => Array\n(\n => 171.229.242.96\n)\n\n => Array\n(\n => 27.59.174.209\n)\n\n => Array\n(\n => 1.38.202.211\n)\n\n => Array\n(\n => 157.37.128.46\n)\n\n => Array\n(\n => 49.15.94.80\n)\n\n => Array\n(\n => 123.25.46.147\n)\n\n => Array\n(\n => 117.230.170.185\n)\n\n => Array\n(\n => 5.62.16.19\n)\n\n => Array\n(\n => 103.18.22.25\n)\n\n => Array\n(\n => 103.46.200.132\n)\n\n => Array\n(\n => 27.97.165.126\n)\n\n => Array\n(\n => 117.230.54.241\n)\n\n => Array\n(\n => 27.97.209.76\n)\n\n => Array\n(\n => 47.31.182.109\n)\n\n => Array\n(\n => 47.30.223.221\n)\n\n => Array\n(\n => 103.31.94.82\n)\n\n => Array\n(\n => 103.211.14.45\n)\n\n => Array\n(\n => 171.49.233.58\n)\n\n => Array\n(\n => 65.49.126.95\n)\n\n => Array\n(\n => 69.255.101.170\n)\n\n => Array\n(\n => 27.56.224.67\n)\n\n => Array\n(\n => 117.230.146.86\n)\n\n => Array\n(\n => 27.59.154.52\n)\n\n => Array\n(\n => 132.154.114.10\n)\n\n => Array\n(\n => 182.186.77.60\n)\n\n => Array\n(\n => 117.230.136.74\n)\n\n => Array\n(\n => 43.251.94.253\n)\n\n => Array\n(\n => 103.79.168.225\n)\n\n => Array\n(\n => 117.230.56.51\n)\n\n => Array\n(\n => 27.97.187.45\n)\n\n => Array\n(\n => 137.97.190.61\n)\n\n => Array\n(\n => 193.0.220.26\n)\n\n => Array\n(\n => 49.36.137.62\n)\n\n => Array\n(\n => 47.30.189.248\n)\n\n => Array\n(\n => 109.169.23.84\n)\n\n => Array\n(\n => 111.119.185.46\n)\n\n => Array\n(\n => 103.83.148.246\n)\n\n => Array\n(\n => 157.32.119.138\n)\n\n => Array\n(\n => 5.62.41.53\n)\n\n => Array\n(\n => 47.8.243.236\n)\n\n => Array\n(\n => 112.79.158.69\n)\n\n => Array\n(\n => 180.92.148.218\n)\n\n => Array\n(\n => 157.36.162.154\n)\n\n => Array\n(\n => 39.46.114.47\n)\n\n => Array\n(\n => 117.230.173.250\n)\n\n => Array\n(\n => 117.230.155.188\n)\n\n => Array\n(\n => 193.0.220.17\n)\n\n => Array\n(\n => 117.230.171.166\n)\n\n => Array\n(\n => 49.34.59.228\n)\n\n => Array\n(\n => 111.88.197.247\n)\n\n => Array\n(\n => 47.31.156.112\n)\n\n => Array\n(\n => 137.97.64.180\n)\n\n => Array\n(\n => 14.244.227.18\n)\n\n => Array\n(\n => 113.167.158.8\n)\n\n => Array\n(\n => 39.37.175.189\n)\n\n => Array\n(\n => 139.167.211.8\n)\n\n => Array\n(\n => 73.120.85.235\n)\n\n => Array\n(\n => 104.236.195.72\n)\n\n => Array\n(\n => 27.97.190.71\n)\n\n => Array\n(\n => 79.46.170.222\n)\n\n => Array\n(\n => 102.185.244.207\n)\n\n => Array\n(\n => 37.111.136.30\n)\n\n => Array\n(\n => 50.7.93.28\n)\n\n => Array\n(\n => 110.54.251.43\n)\n\n => Array\n(\n => 49.36.143.40\n)\n\n => Array\n(\n => 103.130.112.185\n)\n\n => Array\n(\n => 37.111.139.202\n)\n\n => Array\n(\n => 49.36.139.108\n)\n\n => Array\n(\n => 37.111.136.179\n)\n\n => Array\n(\n => 123.17.165.77\n)\n\n => Array\n(\n => 49.207.143.206\n)\n\n => Array\n(\n => 39.53.80.149\n)\n\n => Array\n(\n => 223.188.71.214\n)\n\n => Array\n(\n => 1.39.222.233\n)\n\n => Array\n(\n => 117.230.9.85\n)\n\n => Array\n(\n => 103.251.245.216\n)\n\n => Array\n(\n => 122.169.133.145\n)\n\n => Array\n(\n => 43.250.165.57\n)\n\n => Array\n(\n => 39.44.13.235\n)\n\n => Array\n(\n => 157.47.181.2\n)\n\n => Array\n(\n => 27.56.203.50\n)\n\n => Array\n(\n => 191.96.97.58\n)\n\n => Array\n(\n => 111.88.107.172\n)\n\n => Array\n(\n => 113.193.198.136\n)\n\n => Array\n(\n => 117.230.172.175\n)\n\n => Array\n(\n => 191.96.182.239\n)\n\n => Array\n(\n => 2.58.46.28\n)\n\n)\n```\nRetirement Gift for In-Laws: scfr's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Retirement Gift for In-Laws",
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"June 9th, 2008 at 03:46 am\n\nMy in-laws are easing (or should I say lurching) in to retirement. FIL is 72 and MIL is 68. They both worked very hard all of their lives. FIL owned his own business (with several employees) and MIL worked with him part-time, running the office. FIL has had a series of health issues in the past few years, and after several years of talking about how he was going to retire soon, MIL decided to take charge and push him along. She talked him in to putting their house up for sale. It sold fairly quickly, and then they rented a place in a more rural area. MIL moved in to the rental in late-April, but FIL stayed behind (moving in to SIL's house) to wrap things up with the business. He goes and visits MIL on the weekends, and has promised he will have tied up all of the loose ends with the business on June 20th, when he will permanently moving to the country. The entire family is saying \"we'll believe it when we see it,\" as he really seems to be having a hard time actually taking the final steps.\n\nI'm so proud of my MIL. She has been really ready to retire for about 8 years now, and she has put up with many promises deferred. Because of her culture and her personality, she generally takes a fairly submissive role in the marriage, but I guess she decided she had had enough and she put her foot down! The rest of the family has really gotten a kick out of watching her put her assert herself and (hopefully) drag FIL kicking and screaming in to a well-deserved retirement.\n\nI think it's hard for FIL to deal with because at this stage in his life he really doesn't have interests outside of work (other than watching TV and reading the paper). Hopefully he won't turn in to a total couch potato. He is interested in gardening, and now that they have a place in the country hopefully he can get more involved in that. MIL on the other had is quite a social butterfly, especially with her religious group. She visits with her group members, makes friends quickly with the neighbors, keeps house, and likes to read. She will thrive in retirement, no doubt.\n\nAnyway ... DH & I want to give them a nice retirement gift. We have come up with 2 ideas, and would love to hear any other suggestions. If we ask them what they want, they will say they don't want anything. So we have to take the approach of offering them option A, B, or C and letting them choose.\n\nIf anyone who has already retired has suggestions of what they would have liked to have received when they retired, please share. If anyone has done something like this for their parents or have thought about doing something like this for their folks, or just dreams about being able to do something like this for their parents some day, please share suggestions.\n\nDon't be afraid to venture a bit into \"just dreaming\" sorts of ideas. FIL is not facing imminent death, but he is not going to be around forever, and this gift is our way of \"giving him roses while he is still alive.\" We are able to do something nice for them, and we very much want to.\n\nThe 2 ideas we have are:\n\n1. A 2-week cruise in their region (they live in another country) ... They have not cruised before, but they have enjoyed travelling in the past, and we thought they would feel more comfortable if they stuck close to home, especially given FIL's medical problems.\n\n2. Business Class tickets to visit us here in the USA, plus some domestic travel with us (perhaps via Amtrak in a sleeper car) ... DH & I always fly coach, as have my in-laws in the past, but we think at their age they deserve to travel in comfort.\n\nWe thought about replacing their old car for them, but MIL does not drive and we do not know how much longer FIL will be able to drive. DH thinks it would be best if they just drive the current car for as long as it lasts, and when it dies, to encourage them to just rely on public transportation (which is excellent where they live).\n\nI'd like to have at least another one or two options to toss out to them but am stumped, so any and all suggestions are welcome! Thank you!\n\n### 5 Responses to “Retirement Gift for In-Laws”\n\n1. Carolina Bound Says:\n\nI really appreciated getting a Barnes & Noble gift card when I retired.\n\nAnything related to travel is good, or something to help fund hobbies. I would have loved theater tickets, but of course, that depends on your in-laws.\n\n2. pretty cheap jewelry Says:\n\nI cruised once and it is not what all cracked up to be. (Of course, I am prone to sea sickness). It is but a temporary diversion and I am more in favor of lasting impacts.\n\nSort of the same opinion on train travel. There is a LOT of 'boring' time.\n\nOK now for the help!\nMy parents retired but only about a year ago, my mom so worried she wouldn't know what to do! Now? She has no free time! Lesson-it takes a little adjustment time, but they will find what they love and devote time toward.\n\nThey love to travel? Send a 'dream' basket of travel ideas with a coupon redeemable at your checking account for a portion of the trip. Include maps, food samples from various countries (ie Italian coffee, Hawaiian pineapple, etc), language translation booklets (small ones), a disposable camera, and more.\n\nNo grandkids? If so, there could be a 'playdate' themed gift with travel to the kids, and tickets to a zoo or somewhere.\n\nHow about throwing a surprise 'Welcome to Free Time' party (or some such fun title). The MIL sounds like she has plenty to invite. But if you are far away, it would be tough to arrange. Any best friend of hers that would work with you? Hold it at a community center, with photos and food. Let the guests write one line of a poem about the couple, or 'roast' them.\n\nhmmm, let me think some more and I'll come back if I can.\n\n3. pretty cheap jewelry Says:\n\nLast thoughts:\nSo they are going to be in the country, do they garden? do they need outdoor tools? Is there a nearby town with activities/groups of interest?\n\nI joined the local art association in our weekend property town. It is active and a big help in getting to know residents. Give them a membership or make a donation to such a group they will be near. Maybe there is a historic society, a Sierra Club chapter, or something you think will get them out of the house!\n\nGood luck!\n\n4. scfr Says:\n\nThanks for the great ideas!\n\nA party? Why didn't I think of that? I'll add that to the list. And we may add a gardening center certificate as well (we'll have to find out if their rental house allows them to make significant changes).\n\nPCJ - I hear you about the cruise. But the one I had picked out includes daily on-shore excursions (included in the price of the cruise) that sound really interesting, and there are daily after-dinner educational lectures. It's definitely aimed at the 55+ active learner crowd. Not much in the way of on-board activities. Sounds like the daily routine is\n\n- Breakfast on-board\n- Daily excursion in to town, including walking about (age-appropriate exercise)\n- Dinner on-board\n- After dinner lecture\n- Perhaps a bit of unwinding w/ TV (in each room) or reading\n- Sleep to get ready to do the same thing the next day\n\nOn this particular cruise, the boat seems to serve as a floating hotel & restaurant only, to keep you fed (without having to decide where to go) and get you from place to place in comfort (without having to pack & unpack every day).\n\nSince this particular cruise focuses on their home country, I thought it might introduce them to new places that they could go back and explore more in-depth on their own. Given their age & health, their long overseas travelling days are over. Just too grueling.\n\n5. pretty cheap jewelry Says:\n\noh yes yes, the educational cruises might be better\n\nFor a party wouldn't it be a hoot if each guest were to bring one 'Idea' for retirement activities? Some folks are not crafty/clever so you could even supply a 'form' for guests to fill out and make a binder with them.\n\nhave a good experience in whatever happens!\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4. [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]"
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"https://www.savingadvice.com/blogs/images/search/top_left.php",
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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-7-algebraic-fractions-7-4-addition-and-subtraction-of-algebraic-fractions-and-simplifying-complex-fractions-problem-set-7-4-page-300/71 | [
"## Elementary Algebra\n\n$96/b$\nWe use the equation for the area of a triangle: $$A=1/2 bh\\\\ 48 = 1/2 bh \\\\ 48 \\times 2 = bh \\\\ h = 96/b$$"
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https://encyclopediaofmath.org/wiki/Elliptic_genera | [
"# Elliptic genera\n\nThe name elliptic genus has been given to various multiplicative cobordism invariants taking values in a ring of modular forms. The following is an attempt to present the simplest case — level- $2$ genera in characteristic $\\neq 2$— in a unified way. It is convenient to use N. Katz's approach to modular forms (cf. [a7]) and view a modular form as a function of elliptic curves with a chosen invariant differential (cf. also Elliptic curve). A similar approach to elliptic genera was used by J. Franke [a3].\n\n## Jacobi functions.\n\nLet $K$ be any perfect field of characteristic $\\neq 2$ and fix an algebraic closure ${\\overline{K}\\; }$ of $K$( cf. Algebraically closed field). Consider a triple $( E, \\omega, \\alpha )$ consisting of:\n\ni) an elliptic curve $E$ over $K$, i.e. a smooth curve of genus $1$ with a specified $K$- rational base-point $O$;\n\nii) an invariant $K$- rational differential $\\omega$;\n\niii) a $K$- rational primitive $2$- division point $\\alpha$. Following J.I. Igusa [a6] (up to a point), one can associate to these data two functions, $x$ and $y$, as follows.\n\nThe set $E _ {4} \\subset E ( {\\overline{K}\\; } )$ of $4$- division points on $E$ can be described as follows. There are four $2$- division points $t$( $\\alpha$ is one of them), four primitive $4$- division points $r$ such that $2r = \\alpha$, and eight primitive $4$- division points $s$ such that $2s \\neq \\alpha$. Consider the degree- $0$ divisor $D = \\sum ( t ) - \\sum ( r )$. Since $\\sum t - \\sum r = 0$ in $E$ and since Galois symmetries transform $D$ into itself, Abel's theorem (cf., for example, [a11], III.3.5.1, or Abel theorem) implies that there is a function $x \\in K ( E ) ^ \\times$, uniquely defined up to a multiplicative constant, such that ${ \\mathop{\\rm div} } ( x ) = D$.\n\nThe function $x$ is odd, satisfies $x ( u + \\alpha ) \\equiv x ( u )$, and undergoes sign changes under the two other translations of exact order $2$. Moreover, if $r \\in E _ {4}$ satisfies $2r = \\alpha$, then translation by $r$ transforms $x$ into $Cx ^ {- 1 }$ for some non-zero constant $C$. This constant depends on the choice of $r$ but only up to sign. It follows that $x ^ {2} ( u + r ) x ^ {2} ( u )$ does not depend on the choice of $r$. This constant is written as $\\varepsilon ^ {- 1 }$, i.e.\n\n$$\\varepsilon \\equiv x ^ {- 2 } ( u + r ) x ^ {- 2 } ( u ) .$$\n\nOne also defines\n\n$$\\delta = { \\frac{1}{8} } \\sum x ^ {- 2 } ( s )$$\n\n(the summation is over the primitive $4$- division points $s$ such that $2s \\neq \\alpha$). If $a$ is one of the values of $x ( s )$, the other values are $\\pm a, \\pm \\varepsilon ^ {- {1 / 2 } } a ^ {- 1 }$, each taken twice. It follows that\n\n$$\\delta = { \\frac{1}{2} } ( a ^ {- 2 } + \\varepsilon a ^ {2} )$$\n\nand\n\n$$\\prod ( X - x ( s ) ) = \\varepsilon ^ {- 2 } ( 1 - 2 \\delta X ^ {2} + \\varepsilon X ^ {4} ) ^ {2} = \\varepsilon ^ {- 2 } R ( X ) ^ {2} .$$\n\nIt is now easy to see that\n\n$${ \\mathop{\\rm div} } ( R ( x ) ) = 2 \\left ( \\sum ( s ) - 2 \\sum ( r ) \\right ) .$$\n\nUsing once more Abel's theorem, one sees that there is a unique $y \\in K ( E ) ^ \\times$ such that ${ \\mathop{\\rm div} } ( y ) = \\sum ( s ) - 2 \\sum ( r )$, and $y ( O ) = 1$. Since $x ( O ) = 0$, one has $y ^ {2} = R ( x )$.\n\nThe differential $dx$ has four double poles $r$. Also, it is easy to see that $s$ is a double zero of $x - x ( s )$, hence a simple zero of $dx$. One concludes that\n\n$${ \\mathop{\\rm div} } ( dx ) = \\sum ( s ) - 2 \\sum ( r ) = { \\mathop{\\rm div} } ( y ) .$$\n\nand that ${ {dx } / y }$ is an invariant differential on $E$.\n\nA slight modification of the argument given in [a6] shows that the Jacobi elliptic functions satisfy the Euler addition formula\n\n$$x ( u + v ) ( 1 - \\varepsilon x ^ {2} ( u ) x ^ {2} ( v ) ) = x ( u ) y ( v ) + x ( v ) y ( u ) .$$\n\nAccordingly, one defines the Euler formal group law $F ( U,V ) \\in K [ [ U,V ] ]$ by\n\n$$F ( U,V ) = { \\frac{U \\sqrt {R ( V ) } + V \\sqrt {R ( U ) } }{1 - \\varepsilon U ^ {2} V ^ {2} } } .$$\n\nNotice that since ${ \\mathop{\\rm char} } K \\neq 2$, $F ( U,V )$ is defined over $K$.\n\n## The elliptic genus.\n\nAt this point, one normalizes $x$ over $K$ by requiring that ${ {dx } / y } = \\omega$( the given invariant differential). All the objects $x, y, \\delta, \\varepsilon$, and $F ( U,V )$ are now completely determined by the initial data. Replacing $\\omega$ by $\\lambda \\omega$( $\\lambda \\in K ^ \\times$) yields:\n\n$$\\tag{a1 } x \\asR \\lambda x, \\quad y \\asR y, \\quad \\delta \\asR \\lambda ^ {- 2 } \\delta,$$\n\n$$\\varepsilon \\asR \\lambda ^ {- 4 } \\varepsilon, \\quad F ( U,V ) \\asR \\lambda F ( \\lambda ^ {- 1 } U, \\lambda ^ {- 1 } V ) .$$\n\nAs any formal group law, $F ( U,V )$ is classified by a unique ring homomorphism\n\n$$\\psi : {\\Omega _ {*} ^ { { \\mathop{\\rm U} } } } \\rightarrow K$$\n\nfrom the complex cobordism ring. Since $F ( - U, - V ) = - F ( U,V )$, it is easy to see that $\\psi$ uniquely factors through a ring homomorphism\n\n$$\\varphi : {\\Omega _ {*} ^ { { \\mathop{\\rm SO} } } } \\rightarrow K$$\n\nfrom the oriented cobordism ring. By definition, $\\varphi$ is the level- $2$ elliptic genus. Suppose now that ${ \\mathop{\\rm char} } K = 0$. Define a local parameter $z$ near $O$ so that $z ( O ) = 0$ and $dz = \\omega$. Then $x$ can be expanded into a formal power series $x ( z ) \\in K [ [ z ] ]$ which clearly satisfies $x ( z ) = z + o ( z )$ and $x ( - z ) = - x ( z )$. In this case, the elliptic genus can be defined as the Hirzebruch genus (cf. [a4] or [a5]) corresponding to the series $P ( z ) = {z / {x ( z ) } }$. Since ${ {d x ( z ) } / {dz } } = y ( z )$, the logarithm $g ( z )$ of this elliptic genus is given by the elliptic integral\n\n$$\\tag{a2 } g ( z ) = \\int\\limits _ { 0 } ^ { z } { \\frac{dt }{\\sqrt {1 - 2 \\delta t ^ {2} + \\varepsilon t ^ {4} } } } ,$$\n\nwhich gives the original definition in [a9].\n\n## Modularity.\n\nFor any closed oriented manifold $M$ of dimension $4k$, $\\varphi ( M )$ is a function of the triple $( E, \\omega, \\alpha )$. As easily follows from (a1), multiplying $\\omega$ by $\\lambda$ results in multiplying $\\varphi ( M )$ by $\\lambda ^ {- 2k }$. Also, $\\varphi ( M )$ depends only on the isomorphism class of the triple $( E, \\omega, \\alpha )$ and commutes with arbitrary extensions of the scalar field $K$. In the terminology of Katz ([a7]; adapted here to modular forms over fields), $\\varphi ( M )$ is a modular form of level $2$ and weight $2k$. Let ${\\mathcal M} _ {*}$ be the graded ring of all such modular forms. Then $\\varphi ( M ) \\in {\\mathcal M} _ {2k }$, $\\delta \\in {\\mathcal M} _ {2}$, $\\varepsilon \\in {\\mathcal M} _ {4}$. Moreover, one can prove that ${\\mathcal M} _ {*} \\cong \\mathbf Z [ {1 / 2 } , \\delta, \\varepsilon ]$. If one identifies these two isomorphic rings, the elliptic genus becomes the Hirzebruch genus\n\n$$\\varphi : {\\Omega _ {*} ^ { { \\mathop{\\rm SO} } } } \\rightarrow {\\mathbf Z [ {1 / 2 } , \\delta, \\varepsilon ] }$$\n\nwith logarithm given by the formal integral (a2).\n\n## Integrality.\n\nConsider\n\n$${ {\\widetilde \\varphi } } : {\\Omega _ {*} ^ { { \\mathop{\\rm Spin} } } } \\rightarrow { {\\mathcal M} _ {*} } ,$$\n\ni.e., the composition of $\\varphi$ with the forgetful homomorphism $\\Omega _ {*} ^ { { \\mathop{\\rm Spin} } } \\rightarrow \\Omega _ {*} ^ { { \\mathop{\\rm SO} } }$. As is shown in [a2],\n\n$${\\widetilde \\varphi } ( \\Omega _ {*} ^ { { \\mathop{\\rm Spin} } } ) = \\mathbf Z [ 8 \\delta, \\varepsilon ] .$$\n\nThe ring $\\mathbf Z [ 8 \\delta, \\varepsilon ]$ agrees with the ring ${\\mathcal M} _ {*} ( \\mathbf Z )$ of modular forms over $\\mathbf Z$. Thus: If $M$ is a ${ \\mathop{\\rm Spin} }$- manifold of dimension $4k$, then $\\varphi ( M ) \\in {\\mathcal M} _ {2k } ( \\mathbf Z )$.\n\n## Example: the Tate curve.\n\nLet $K$ be a local field, complete with respect to a discrete valuation $v$, and let $q \\in K ^ \\times$ be any element satisfying $v ( q ) < 0$. Consider $E = K ^ \\times /q ^ {2 \\mathbf Z }$. It is well-known (cf. [a11], § C.14) that $E$ can be identified with the elliptic curve (known as the Tate curve)\n\n$$E _ {q ^ {2} } : Y ^ {2} + XY = X ^ {3} + a _ {4} X + a _ {6} ,$$\n\nwhere\n\n$$a _ {4} = \\sum _ {m \\geq 1 } ( - 5m ^ {3} ) { \\frac{q ^ {2m } }{1 - q ^ {2m } } } ,$$\n\n$$a _ {6} = \\sum _ {m \\geq 1 } \\left ( - { \\frac{5m ^ {3} + 7m ^ {5} }{12 } } \\right ) { \\frac{q ^ {2m } }{1 - q ^ {2m } } } .$$\n\n$E$ can be treated as an elliptic curve over $K$ with $O = 1$. Fix the invariant differential $\\omega = { {du } / u }$( $u \\in K ^ \\times$) on $E$( $\\omega$ corresponds to the differential $\\omega _ {\\textrm{ can } } = { {dX } / {( 2Y + X ) } }$ on the Tate curve). $E$ has three $K$- rational primitive $2$- division points: $- 1$, $q$ and $- q$. To describe the corresponding Jacobi function $x$, consider the theta-function\n\n$$\\Theta ( u ) = ( 1 - u ^ {- 2 } ) \\prod _ {n > 0 } ( 1 - q ^ {2n } u ^ {- 2 } ) ( 1 - q ^ {2n } u ^ {2} ) .$$\n\nThis is a \"holomorphic\" function on $K ^ \\times$ with simple zeros at points of $\\pm q ^ {\\mathbf Z}$( cf. [a10] for a justification of this terminology), satisfying\n\n$$\\Theta ( - u ) = \\Theta ( u ) , \\quad \\Theta ( q ^ {- 1 } u ) = - u ^ {2} \\Theta ( u ) .$$\n\nConsider the case where $\\alpha = - 1$. Let $i \\in {\\overline{K}\\; }$ be any square root of $- 1$, and let\n\n$$\\tag{a3 } f ( u ) = { \\frac{\\Theta ( u ) }{\\Theta ( iu ) } } =$$\n\n$$= { \\frac{u ^ {2} - 1 }{u ^ {2} + 1 } } \\prod _ {n > 0 } { \\frac{( 1 - q ^ {2n } u ^ {- 2 } ) ( 1 - q ^ {2n } u ^ {2} ) }{( 1 + q ^ {2n } u ^ {- 2 } ) ( 1 + q ^ {2n } u ^ {2} ) } } .$$\n\n$f$ is a meromorphic function on $E$ satisfying $f ( iu ) = {1 / {f ( u ) } }$ and\n\n$${ \\mathop{\\rm div} } ( f ) = ( 1 ) + ( - 1 ) + ( q ) + ( - q ) +$$\n\n$$- ( i ) - ( - i ) - ( iq ) - ( - iq ) ,$$\n\ni.e., $f$ is a multiple of the Jacobi function $x$ of $( E, \\omega, - 1 )$.\n\nNotice now that the normalization condition ${ {du } / u } = { {dx } / y }$ can be written as $y ( u ) = ux ^ \\prime ( u )$, where $x ^ \\prime ( u )$ is the derivative with respect to $u$. Since $y ( 1 ) = 0$, one has $x ^ \\prime ( 1 ) = 1$. Differentiating (a3), one obtains\n\n$$f ^ \\prime ( 1 ) = \\prod _ {n > 0 } \\left ( { \\frac{1 - q ^ {2n } }{1 + q ^ {2n } } } \\right ) ^ {2} ,$$\n\n$$x ( u ) = { \\frac{u ^ {2} - 1 }{u ^ {2} + 1 } } \\prod _ {n > 0 } { \\frac{( 1 - q ^ {2n } u ^ {- 2 } ) ( 1 - q ^ {2n } u ^ {2} ) ( 1 + q ^ {2n } ) ^ {2} }{( 1 + q ^ {2n } u ^ {- 2 } ) ( 1 + q ^ {2n } u ^ {2} ) ( 1 - q ^ {2n } ) ^ {2} } } ,$$\n\nand\n\n$$\\varepsilon = \\prod _ {n > 0 } \\left ( { \\frac{1 - q ^ {2n } }{1 + q ^ {2n } } } \\right ) ^ {8} .$$\n\nFinally, if ${ \\mathop{\\rm char} } K = 0$, the function $z = { \\mathop{\\rm log} } u$ satisfies $dz = { {du } / u }$. It follows that the generating series $P ( z ) = {z / {x ( z ) } }$ is given by\n\n$$P ( z ) =$$\n\n$$= { \\frac{z}{ { \\mathop{\\rm tanh} } z } } \\prod _ {n > 0 } { \\frac{( 1 + q ^ {2n } e ^ {- 2z } ) ( 1 + q ^ {2n } e ^ {2z } ) ( 1 - q ^ {2n } ) ^ {2} }{( 1 - q ^ {2n } e ^ {- 2z } ) ( 1 - q ^ {2n } e ^ {2z } ) ( 1 + q ^ {2n } ) ^ {2} } } .$$\n\nThe cases where $\\alpha = q$ or $\\alpha = - q$ are treated similarly, with\n\n$$f ( u ) = { \\frac{u \\Theta ( u ) }{\\Theta ( q ^ {- 1/2 } u ) } }$$\n\nand\n\n$$f ( u ) = { \\frac{u \\Theta ( u ) }{\\Theta ( iq ^ {- 1/2 } u ) } } ,$$\n\nrespectively.\n\n## Strict multiplicativity.\n\nThe following theorem, also known (in an equivalent form) as the Witten conjecture, was proven first by C. Taubes [a12], then by R. Bott and Taubes [a1]. Let $P$ be a principal $G$- bundle (cf. also Principal $G$- object) over an oriented manifold $B$, where $G$ is a compact connected Lie group, and suppose $G$ acts on a compact ${ \\mathop{\\rm Spin} }$- manifold $M$. Then\n\n$$\\varphi ( P \\times _ {G} M ) = \\varphi ( B ) \\varphi ( M ) .$$\n\nFor the history of this conjecture, cf. [a8].\n\nHow to Cite This Entry:\nElliptic genera. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Elliptic_genera&oldid=46811\nThis article was adapted from an original article by S. Ochanine (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article"
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https://chemistry.stackexchange.com/questions/73504/water-vapor-equilibrium-calculations | [
"Water vapor equilibrium calculations\n\nThe first part said:\n\nCalculate the amount of water required to saturate a vessel of volume 82 L at 300 K given that the vapor pressure of water is 38 torr at 300 K.\n\nThe solution went like this:\n\nSince we have to saturate the vessel, we have to find the minimum number of moles that will occupy the given volume (that is what I understood).\n\nNow, applying the gas equation\n\n$$PV=nRT$$\n\n$$\\frac{1}{20}*82 = n * 0.82 * 300$$\n\n$$n = \\frac{1}{6}; w=n * M = 3 gm$$\n\nThis part is fine.\n\nNow the second part:\n\nIf 5 gm of water is introduced in the vessel (initially evacuated), then, calculate the amount of water remaining.\n\nSo, my teacher said that, 3 gm will evaporate and 2 gm will remain. But, 3 gm will evaporate for 82 L. So, when I pour 5 gm of water, shouldn't the volume decrease?\n\n• You are correct that there will initially be a volume of 82L - 5cc. Then 3cc evaporates, so the volume of the container is 82 L - 0.002 L. Will this make a difference in your calculation that will show up in a 2 significant figure answer? Apr 28 '17 at 1:47\n• @airhuff then, in the first question also, the volume should be 82-0.003. right? Apr 28 '17 at 1:54\n• But the point for the second question, is would it matter if you just ignored the effect of the volume of the liquid water? In the first question there is no liquid water left because it all evaporated and you have nothing but water vapor. Back to the second question: 82L - 0.002L = 81.998L, which rounds to 82L (note that you are only given 2 significant figures in the problem), and for that matter it rounds to 82.00L! So, is the volume of the small amount of water insignificant compared to the size of the container in this question (hint, hint), such that you can just ignore it? Apr 28 '17 at 2:07\n• You are not supposed to write \"Thanks\" or \"Help is appreciated\" or any statements of weaknesses in your post. And fix that Vap0r in the title. [I edited his post, apparently he rejected it] Apr 28 '17 at 12:18"
] | [
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https://flylib.com/books/en/2.253.1/namespaces.html | [
"# namespaces\n\nA program includes many identifiers defined in different scopes. Sometimes a variable of one scope will \"overlap\" (i.e., collide) with a variable of the same name in a different scope, possibly creating a naming conflict. Such overlapping can occur at many levels. Identifier overlapping occurs frequently in third-party libraries that happen to use the same names for global identifiers (such as functions). This can cause compiler errors.\n\nGood Programming Practice 24.1",
null,
"Avoid identifiers that begin with the underscore character, which can lead to linker errors. Many code libraries use names that begin with underscores.\n\nThe C++ standard attempts to solve this problem with namespaces. Each namespace defines a scope in which identifiers and variables are placed. To use a namespace member, either the member's name must be qualified with the namespace name and the binary scope resolution operator (::), as in\n\nMyNameSpace::member\n\nor a using declaration or using directive must appear before the name is used in the program. Typically, such using statements are placed at the beginning of the file in which members of the namespace are used. For example, placing the following using directive at the beginning of a source-code file\n\n```using namespace MyNameSpace;\n```\n\nspecifies that members of namespace MyNameSpace can be used in the file without preceding each member with MyNameSpace and the scope resolution operator (::).\n\nA using declaration (e.g., using std::cout;) brings one name into the scope where the declaration appears. A using directive (e.g., using namespace std;) brings all the names from the specified namespace into the scope where the directive appears.\n\nSoftware Engineering Observation 24.1",
null,
"Ideally, in large programs, every entity should be declared in a class, function, block or namespace. This helps clarify every entity's role.\n\nError-Prevention Tip 24.2",
null,
"Precede a member with its namespace name and the scope resolution operator (::) if the possibility exists of a naming conflict.\n\nNot all namespaces are guaranteed to be unique. Two third-party vendors might inadvertently use the same identifiers for their namespace names. Figure 24.2 demonstrates the use of namespaces.\n\nFigure 24.2. Demonstrating the use of namespaces.\n\n(This item is displayed on pages 1204 - 1205 in the print version)\n\n``` 1 // Fig. 24.2: fig24_02.cpp\n2 // Demonstrating namespaces.\n3 #include\n4 using namespace std; // use std namespace\n5\n6 int integer1 = 98; // global variable\n7\n8 // create namespace Example\n9 namespace Example\n10 {\n11 // declare two constants and one variable\n12 const double PI = 3.14159;\n13 const double E = 2.71828;\n14 int integer1 = 8;\n15\n16 void printValues(); // prototype\n17\n18 // nested namespace\n19 namespace Inner\n20 {\n21 // define enumeration\n22 enum Years { FISCAL1 = 1990, FISCAL2, FISCAL3 };\n23 } // end Inner namespace\n24 } // end Example namespace\n25\n26 // create unnamed namespace\n27 namespace\n28 {\n29 double doubleInUnnamed = 88.22; // declare variable\n30 } // end unnamed namespace\n31\n32 int main()\n33 {\n34 // output value doubleInUnnamed of unnamed namespace\n35 cout << \"doubleInUnnamed = \" << doubleInUnnamed;\n36\n37 // output global variable\n38 cout << \"\n(global) integer1 = \" << integer1;\n39\n40 // output values of Example namespace\n41 cout << \"\nPI = \" << Example::PI << \"\nE = \" << Example::E\n42 << \"\ninteger1 = \" << Example::integer1 << \"\nFISCAL3 = \"\n43 << Example::Inner::FISCAL3 << endl;\n44\n45 Example::printValues(); // invoke printValues function\n46 return 0;\n47 } // end main\n48\n49 // display variable and constant values\n50 void Example::printValues()\n51 {\n52 cout << \"\nIn printValues:\ninteger1 = \" << integer1 << \"\nPI = \"\n53 << PI << \"\nE = \" << E << \"\ndoubleInUnnamed = \"\n54 << doubleInUnnamed << \"\n(global) integer1 = \" << ::integer1\n55 << \"\nFISCAL3 = \" << Inner::FISCAL3 << endl;\n56 } // end printValues\n```\n\n ``` doubleInUnnamed = 88.22 (global) integer1 = 98 PI = 3.14159 E = 2.71828 integer1 = 8 FISCAL3 = 1992 In printValues: integer1 = 8 PI = 3.14159 E = 2.71828 doubleInUnnamed = 88.22 (global) integer1 = 98 FISCAL3 = 1992 ```\n\nUsing the std Namespace\n\nLine 4 informs the compiler that namespace std is being used. The contents of header file are all defined as part of namespace std. [Note: Most C++ programmers consider it poor practice to write a using directive such as line 4 because the entire contents of the namespace are included, thus increasing the likelihood of a naming conflict.]\n\nThe using namespace directive specifies that the members of a namespace will be used frequently throughout a program. This allows the programmer to access all the members of the namespace and to write more concise statements such as\n\n```cout << \"double1 = \" << double1;\n```\n\nrather than\n\n```std::cout << \"double1 = \" << double1;\n```\n\nWithout line 4, either every cout and endl in Fig. 24.2 would have to be qualified with std::, or individual using declarations must be included for cout and endl as in:\n\n```using std::cout;\nusing std::endl;\n```\n\nThe using namespace directive can be used for predefined namespaces (e.g., std) or programmer-defined namespaces.\n\nDefining Namespaces\n\nLines 924 use the keyword namespace to define namespace Example. The body of a namespace is delimited by braces ({}). Namespace Example's members consist of two constants (PI and E at lines 1213), an int (integer1 at line 14), a function (printValues at line 16) and a nested namespace (Inner at lines 1923). Notice that member integer1 has the same name as global variable integer1 (line 6). Variables that have the same name must have different scopesotherwise compilation errors occur. A namespace can contain constants, data, classes, nested namespaces, functions, etc. Definitions of namespaces must occupy the global scope or be nested within other namespaces.\n\nLines 2730 create an unnamed namespace containing the member doubleInUnnamed. The unnamed namespace has an implicit using directive, so its members appear to occupy the global namespace, are accessible directly and do not have to be qualified with a namespace name. Global variables are also part of the global namespace and are accessible in all scopes following the declaration in the file.\n\nSoftware Engineering Observation 24.2",
null,
"Each separate compilation unit has its own unique unnamed namespace; i.e., the unnamed namespace replaces the static linkage specifier.\n\nAccessing Namespace Members with Qualified Names\n\nLine 35 outputs the value of variable doubleInUnnamed, which is directly accessible as part of the unnamed namespace. Line 38 outputs the value of global variable integer1. For both of these variables, the compiler first attempts to locate a local declaration of the variables in main. Since there are no local declarations, the compiler assumes those variables are in the global namespace.\n\nLines 4143 output the values of PI, E, integer1 and FISCAL3 from namespace Example. Notice that each must be qualified with Example:: because the program does not provide any using directive or declarations indicating that it will use members of namespace Example. In addition, member integer1 must be qualified, because a global variable has the same name. Otherwise, the global variable's value is output. Notice that FISCAL3 is a member of nested namespace Inner, so it must be qualified with Example::Inner::.\n\nFunction printValues (defined at lines 5056) is a member of Example, so it can access other members of the Example namespace directly without using a namespace qualifier. The output statement in lines 5255 outputs integer1, PI, E, doubleInUnnamed, global variable integer1 and FISCAL3. Notice that PI and E are not qualified with Example. Variable doubleInUnnamed is still accessible, because it is in the unnamed namespace and the variable name does not conflict with any other members of namespace Example. The global version of integer1 must be qualified with the unary scope resolution operator (::), because its name conflicts with a member of namespace Example. Also, FISCAL3 must be qualified with Inner::. When accessing members of a nested namespace, the members must be qualified with the namespace name (unless the member is being used inside the nested namespace).\n\nCommon Programming Error 24.1",
null,
"Placing main in a namespace is a compilation error.\n\nAliases for Namespace Names\n\nNamespaces can be aliased. For example the statement\n\n```namespace CPPHTP5E = CPlusPlusHowToProgram5E;\n```\n\ncreates the alias CPPHTP5E for CPlusPlusHowToProgram5E.",
null,
"C++ How to Program (5th Edition)\nISBN: 0131857576\nEAN: 2147483647\nYear: 2004\nPages: 627",
null,
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https://www.worldsrichpeople.com/what-number-is-12-of-150/ | [
"# What number is 12% of 150?\n\n## What number is 12% of 150?\n\n12 percent of 150 is 18.\n\n## How do you find 150 percent of a number?\n\n1. How to calculate percentage of a number. Use the percentage formula: P% * X = Y\n\n1. Convert the problem to an equation using the percentage formula: P% * X = Y.\n2. P is 10%, X is 150, so the equation is 10% * 150 = Y.\n3. Convert 10% to a decimal by removing the percent sign and dividing by 100: 10/100 = 0.10.\n\nHow do you calculate percentages in a worksheet?\n\nThe basic formula for calculating a percentage is =part/total. Say you want to reduce a particular amount by 25%, like when you’re trying to apply a discount. Here, the formula will be: =Price*1-Discount %. (Think of the “1” as a stand-in for 100%.)\n\n### How do you find the percentage of a number practice?\n\nWorksheet on Percentage of a Number\n\n1. We know, to find the percent of a number we obtain the given number and then multiply the number by the required percent i.e., x % of a = x100 × a.\n2. (x) 34 % of \\$ 780.\n3. (xiv) 35 % of 725 cm.\n\n### What number is 8% of 150?\n\n8 percent of 150 is 12.\n\nHow do we convert a percent to decimal decimal to percent?\n\nTo convert a percentage to a decimal, divide by 100. So 25% is 25/100, or 0.25. To convert a decimal to a percentage, multiply by 100 (just move the decimal point 2 places to the right). For example, 0.065 = 6.5% and 3.75 = 375%.\n\n## What is the percentage of 150?\n\nRelated Standard Percentage Calculations on 10 is what percent of 150\n\nX is Percentage(P) of Y\n97.5 65 150\n99 66 150\n100.5 67 150\n102 68 150\n\n## Can you have 150 percent?\n\nTechnically, “percent” should mean “for every hundred”. So, I would think that it’s perfectly fine to say “150%”. However, in common usage, people rarely say percentages greater than a hundred.\n\nHow do I convert to percentage in Excel?\n\nPercentages are calculated by using the equation amount / total = percentage. For example, if a cell contains the formula =10/100, the result of that calculation is 0.1. If you then format 0.1 as a percentage, the number will be correctly displayed as 10%.\n\n### How do I calculate 15% of a number in Excel?\n\nTo subtract 15%, use =1-15% as the formula….Here’s how to do it:\n\n1. Enter the numbers you want to multiply by 15% into a column.\n2. In an empty cell, enter the percentage of 15% (or 0.15), and then copy that number by pressing Ctrl-C.\n3. Select the range of cells A1:A5 (by dragging down the column).\n\n### How do you find the percentage of two numbers worksheet?\n\nAnswer: To find the percentage of a number between two numbers, divide one number with the other and then multiply the result by 100.\n\nHow do you work out 100 percent from a percentage?\n\nIn order to do this, we:\n\n1. Either add/subtract the percentage given in the problem from 100% to determine what percentage we have.\n2. Find 1% by dividing by percentage found in previous step.\n3. Find 100% (original amount) by multiplying your answer in step 2 by 100.\n\n## What percent is 12 of 150?\n\nSteps to solve “what percent is 12 of 150?” To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100 If you are using a calculator, simply enter 12÷150×100 which will give you 8 as the answer.\n\n## How do you write 12% as a percentage?\n\n“Percent” or “%” means “out of 100” or “per 100”, Therefore 12% can be written as 12 100. When dealing with percents the word “of” means “times” or “to multiply”.\n\nWhat is the value of 100% of 150?\n\n1. We assume, that the number 150 is 100% – because it’s the output value of the task. 2. We assume, that x is the value we are looking for. 3. If 100% equals 150, so we can write it down as 100%=150.\n\n### What is the value of X if 12 is 100%?\n\nWe assume, that the number 12 is 100% – because it’s the output value of the task. 2. We assume, that x is the value we are looking for. 3. If 12 is 100%, so we can write it down as 12=100%. 4. We know, that x is 150% of the output value, so we can write it down as x=150%. 5. Now we have two simple equations: 6.\n\nBegin typing your search term above and press enter to search. Press ESC to cancel."
] | [
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https://flylib.com/books/en/3.150.1.84/1/ | [
"# IIf",
null,
"We mentioned IIf earlier in the book (Chapter 4), and warned you of its dangers (all of the arguments are run), but there are some times when it can be really useful. For example, suppose we want to print our report out double sided, and would like the page numbers to appear at the outer edge on both odd and even numbered pages. This means that for some pages it must be on the left, and for the others it must be on the right. Note that as an alternative, we could write our own function to do this: we probably would if we wanted more flexibility than the IIf function offers.\n\n### Try It Out-The IIf Function\n\n1. Switch the form back into design mode.\n\n2. Move the page number field to the very left of the page, and set the text to be left aligned using the Align Left button on the toolbar.\n\n3. Modify the Control Source property of the page number field, so it now looks like this:\n\n` =IIf([Page] Mod 2 = 0, \"Page \" & [Page] & \" of \" & [Pages], \"\") `\n4. Add another textbox, this time at the right of the page and set the text to be right aligned using the Align Right button on the toolbar. You can delete the label again.\n\n5. Add the following code to this textbox's Control Source property:\n\n` =IIf([Page] Mod 2 = 1, \"Page \" & [Page] & \" of \" & [Pages], \"\") `\n6. Now switch to preview mode to see what effect the changes have had. Step forward a few pages to see what happens for odd and even pages.\n\nNotice that for the first page, and all odd numbered pages, the page numbers are on the right of the page. For all even numbers they are on the left.\n\nHow It Works\n\nLet's look again at the arguments for the IIf function:\n\n` IIf (Expression, TruePart, FalsePart) `\n\nThe arguments are:\n\n• Expression , which is the expression to test\n\n• TruePart , which is the value to return if Expression is True\n\n• FalsePart , which is the value to return if Expression is False\n\nSo, for the page numbers on the left we have this:\n\n` =IIf([Page] Mod 2 = 0, \"Page \" & [Page] & \" of \" & [Pages], \"\") `\n\nThat means the Expression we are testing is:\n\n` [Page] Mod 2 = 0 `\n\nThis uses Mod to return the integer remainder of dividing the page number by two. This will be if the page number is even, so the expression will only be True on even pages.\n\nIf the Expression is True , then the TruePart of the IIf function:\n\n` \"Page \" & [Page] & \" of \" & [Pages] `\n\nis returned.\n\nIf Expression is False , then the FalsePart of the IIf function is returned, which is empty.\n\nSo this whole field will only show up on even numbered pages, which produces the page count we are looking for.\n\nThe page number field for page numbers on the right is pretty similar. The only difference is in the expression to test:\n\n` [Page] Mod 2 = 1 `\n\nHere we check to see whether the page number is odd or not. If it is, then the same TruePart is returned.\n\nThis shows that with just one simple function you've made your report look much better than it did before.\n\nStill confused about Mod ? We need to use it because we don't have a programmatic concept of what is an odd or even page, and so we use the Mod operator to help us out. We know that if we set the Mod operator to 2 , that all page numbers (numerator) will be divided by 2 (denominator). It just so happens that whenever a page number is divided by two that if there is a remainder the page number is odd and whenever there is no remainder the page number is even.\n\nSo by using Mod 2 in this instance we can determine an odd from an even page number.\n\nYou can find more about the Mod operator by keying \"Mod\" into the help index when in the V-E (remember online help is context sensitive).",
null,
"",
null,
"Beginning Access 2002 VBA (Programmer to Programmer)\nISBN: 0764544020\nEAN: 2147483647\nYear: 2003\nPages: 256"
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