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[ "GCSE Maths Algebra\n\nTypes of Graphs\n\n# Types Of Graphs\n\nHere we will learn about types of graphs, including straight line graphs, quadratic graphs, cubic graphs, reciprocal graphs, exponential graphs and circle graphs.\n\nThere are also types of graphs worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.\n\n## What are types of graphs?\n\nTypes of graphs include different types of straight and curved graphs.\n\nWe need to be able to recognise and distinguish between the main types of graphs.\n\n### What are types of graphs?", null, "### Straight line graphs\n\nStraight line graphs are graphs of linear functions and are of the form:\n\ny=mx+c\n\nWhere m is the gradient and c is the y -intercept (where the line crosses the y -axis).\n\nThe graphs look like this:\n\nStep-by-step guide: Straight line graphs\n\nQuadratic graphs are graphs of a quadratic function and can be recognised as they include an squared term.\nE.g.\nAn x^2 term.\n\nThe shape of a quadratic graph is a parabola.\nThe graphs have one tuning point – a minimum point or a maximum point.\n\n### Cubic graphs\n\nCubic graphs are graphs of a cubic function and can be recognised as they include a cubed term.\nE.g.\nAn x^3 term.\n\nThe graphs often have two turning points – a minimum point and a maximum point.\n\nStep-by-step guide: Cubic graphs\n\n### Exponential graphs\n\nExponential graphs are graphs of an exponential function and can be recognised as they include a k^x term where k is the base and x is the exponent (power).\nThe graphs can be a growth curve when k is greater than 1 or a decay curve when k is less than 1 .\n\nMore complex exponential curves are of the form:\n\ny=ab^x\n\nStep-by-step guide: Exponential graphs\n\n### Reciprocal graphs\n\nReciproccal graphs are graphs of a reciprocal function and can be recognised as they include a \\frac{1}{x} term.\nThe graphs can be recognised as they have two separate parts, often in different quadrants of the coordinate grid.\n\nA simple reciprocal graph\n\ny=\\frac{1}{x}\n\nA more complex reciprocal graph\n\ny=\\frac{1}{x} +x\n\nStep-by-step guide: Reciprocal graphs\n\n### Circle graphs\n\nCircle graphs at GCSE are graphs of a circle with centre (0,0)\nThey are of the form:\n\nx^2+y^2=r^2\nWhere r is the radius of the circle.\n\nE.g.\nThis circle graph has:\n\ncentre (0,0)\n\nIts equation is:\n\nx^2+y^2=3^2\n\nWhich can be simplified to:\n\nx^2+y^2=9", null, "Step-by-step guide: Circle graphs\n\n### Plotting graphs\n\nAt GCSE you may be asked to plot graphs. You may be asked to plot a straight line graph, a quadratic graph or a cubic graph. This involves finding coordinates and plotting these on a x-y axes and joining them.\n\nYou can find out more about plotting graphs here:\n\nStep-by-step guide: Plotting graphs\n\n### Sketching graphs\n\nWe need to be able to sketch the different types of graphs and provide enough information on the coordinate grid to be able to distinguish the graph from others. This includes providing a rough drawing of the type of graph (a straight line, a parabola, a cubic etc), labelling key coordinates / axis values, labelling the line with an equation, and drawing any asymptotes.\n\nStep-by-step guide: Sketching graphs (coming soon)\n\n## Graphs of real-world contexts\n\nThe graphs of algebraic equations can also be used to model real-world situations. For example, part of an exponential curve may be used to model how a population of rabbits in a field is growing over time.\n\nExponential curves may also be used to model how a hot drink cools over time.\n\n## Interpret graph\n\nA question may ask you to interpret a graph. Make sure you look at the key information on the graph. You may be asked to read a value from the graph or find the gradient at a point.\n\nIf the graph is a real-life graph, the interpretation needs to include the situation.\n\nFor example,\n\nThis graph represents the calculation of the cost of a taxi ride.\n\nThe 4 on the vertical axis can be interpreted as the fixed charge when hiring a taxi. We have to pay £4 to hire the taxi, no matter what distance we travel.\n\n## Graphical misrepresentation\n\nGraphs can be misleading depending on subtle changes in the axes.\n\nThe standard set of axes for most graphical representations of a function are plus-shaped with a horizontal x-axis, and a vertical y-axis, both containing the same sized scale and intersecting at the origin (0,0).\n\nE.g.\n\nThis is the graph of y=2x-3 plotted onto a standard set of axes.\n\nIf the scale of the y-axis is changed so that each square represents 2, the gradient of the line stays the same. However visually the graph looks shallower than on the set of axes above.\n\nThis graph could therefore be considered to be misleading as the scale on the x-axis is different to the scale on the y-axis.\n\nStep-by-step guide: Representing data\n\n## How to recognise types of graphs\n\nIn order to recognise and discriminate the types of graphs:\n\n2. Identify if there is a circle graph.\n3. Identify other curves by looking at the features such as growth, or vertices.\n4. Identify all the graphs clearly.\n\n## Recognising types of graphs examples\n\n### Example 1: recognise the types of graphs\n\nMatch the graph with its equation", null, "", null, "", null, "", null, "Equation 1\n\ny=2^x\n\nEquation 3\n\nx^2+y^2=4\n\nEquation 2\n\ny=x^2+x-2\n\nEquation 4\n\ny=x^3-x-2\n\nThere is no straight line graph, so there is no linear function.\n\nThere is a parabola graph, so this is the graph of a quadratic function which has an x^2 term.\n\nGraph C is Equation 2\n\n2Identify if there is circle graph.\n\nThere is a circle graph.  It has centre (0,0) and radius 2 ; so its equation would be:\n\nx^2+y^2=2^2\n\nGraph D is Equation 3\n\n3Identify other curves by looking at the features such as growth, or vertices.\n\nGraph A has 2 vertices; it is very likely to be a cubic function.\nEquation 4 is a cubic function with a y -intercept at −2 which is at the end of the equation for when x = 0 .\n\nGraph B is a growth curve so its equation will have a term with x as an exponent (power); equation 1 has a term with x as an exponent.\n\n4Identify all the graphs clearly.", null, "", null, "", null, "", null, "### Example 2: recognise the types of graphs\n\nMatch the graph with its equation", null, "", null, "", null, "", null, "Equation 1\n\nx+y=5\n\nEquation 3\n\nx^2+y^2=25\n\nEquation 2\n\ny=\\frac{5}{x}\n\nEquation 4\n\ny=0.5^x\n\nThere is a straight line graph so this is the graph of a linear function which has no visible powers.\n\nThere is no parabola graph so there is no quadratic function.\n\nGraph B is Equation 1\n\nThere is a circle graph.  It has centre (0,0) so its equation would be:\n\nx^2+y^2=r^2\n\nGraph A is Equation 3\n\nGraph C is not a growth curve but it is a decay curve so its equation will have a term with x as an exponent (power).  Equation 4 has a term with x as an exponent.\n\nGraph D is a curve which has 2 sections in different quadrants of the coordinate grid.  It is likely to be a reciprocal graph with a \\frac{1}{x} term.  Equation 2 has a term with with x as a denominator of a fraction.", null, "", null, "", null, "", null, "## How to use types of graphs\n\nIn order to use different types of graph to solve an equation:\n\n1. Add a line to the coordinate grid.\n2. See where the line crosses the curve.\n3. Draw a straight vertical line from the curve to the x -axis.\n4. Read off the value on the x -axis.\n\n## Using types of graphs examples\n\n### Example 3: using types of graphs\n\nUse the graph of y=1.5^x to find an approximate solution to the equation 1.5^x=7\n\nSince the y in the equation of the curve has been replaced by the 7 , add the horizontal line y=7 to the coordinate grid.\n\nMake sure you use a ruler and be as accurate as you can.\n\nThe line y=7 crosses the curve at one point.\n\nContinue to use a ruler and try to be as accurate as you can.\n\nBe careful with the scale on x -axis.  The solution to the equation is only approximate, but try to be as accurate as you can.\n\nx=4.8\n\n### Example 4: using types of graphs\n\nUse the graph of y=\\frac{1}{x} to find an approximate solution to the equation \\frac{1}{x}=x+2\n\nSince the y in the equation of the curve has been replaced by the x + 2 , add the horizontal line y = x + 2 to the coordinate grid.\n\nMake sure you use a ruler and be as accurate as you can.\n\nThe line y = x + 2 crosses the curve at two points.\n\nContinue to use a ruler and try to be as accurate as you can.\n\nBe careful with the scale on x -axis.  The solution to the equation is only approximate, but try to be as accurate as you can.\n\n## How to plot types of graphs\n\nIn order to plot different types of graphs:\n\n1. Complete the table of values.\n2. Plot the coordinates.\n3. Draw a smooth curve through the points.\n\n## Plotting types of graphs examples\n\n### Example 5: Plot types of graphs\n\nDraw the curve for -2\\leq{x}\\leq2 :\n\ny=x^3-2x+3\n\nSubstitute the values of x into the equation.  Write the values of y in the table.\n\nIt is best practice to use a pencil and plot the coordinates using small crosses.\n\nThe coordinates would be (-2,-1), (-1,4) and so on.  Note that the y -interceptis +3 which is also at the end of the equation for when x = 0.\n\nUse a pencil and turn the paper around if it makes it easier for you.\n\n### Example 6: Plot types of graphs\n\nDraw the curve for 1\\leq{x}\\leq18 :\n\ny=\\frac{18}{x}\n\nSubstitute the values of x into the equation.  Write the values of y in the table.\n\nIt is best practice to use a pencil and plot the coordinates using small crosses.\n\nThe coordinates would be (1,18), (2,9) and so on.\n\nUse a pencil and turn the paper around if it makes it easier for you.\n\n### Common misconceptions\n\n• Smooth curve\n\nPoints should be joined with a smooth curve – NOT line segments like the example below.\n\n• Points on the curve\n\nAll points should be on the curve. If there is a point that does not join up with the other points when you draw a smooth curve – go back and check the coordinate.  In the example below there is a point that does not lie on the smooth curve so needs checking.\n\n### Practice types of graphs questions\n\n1. Identify the correct equation for this graph:", null, "y=5^x", null, "y=x^3+5x", null, "y=\\frac{5}{x}", null, "x^2+y^2=25", null, "The graph shows a growth curve. We need to look for an equation which has a term like k^x where x is the exponent (power).\n\n2. Identify the correct equation for this graph:", null, "y=\\frac{1}{x+1}", null, "y=x^3-5x+1", null, "x^2+y^2=10", null, "y=(\\frac{1}{5})^x", null, "The graph shows a curve with 2 vertices – a minimum point and a maximum point.\n\nWe need to look for an equation which has a x^3 term.\n\nAlso the y -intercept is positive, and the equation ends +1 for when x=0.\n\n3. Identify the correct graph for this equation:\n\nx^2+y^2=9", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "The equation is of the form:\n\nx^2+y^2=r^2\n\nSo we are looking for a circle with centre (0,0).\n\n\\begin{aligned} r^2 &= 9 \\\\\\\\ r &= \\sqrt{9} = 3 \\end{aligned}\n\nThe radius of the circle is 3 .\n\nSo the circle will pass through 3 on the x-axis and the y-axis.\n\n4. Identify the correct graph for this equation:\n\ny=-x^3+2x+3", null, "", null, "", null, "", null, "", null, "", null, "The equation is a cubic with a negative x^3 term.\n\nSo we look for a cubic graph which shows as x increases so does y but negatively.\n\nSo this is the approximate shape:", null, "The equation also ends with +3 for when x=0.\n\nSo this tells us the y -intercept is also +3\n\n5. Use the graph of y=4^x to find an approximate solution to:\n\n4^x=3", null, "1.8", null, "0.8", null, "0.4", null, "1.4", null, "Draw a horizontal line at y=3 and see where it crosses the line.\n\nThen draw a vertical line down and read the value of the x-axis.", null, "6. Use the graph of y=\\frac{1}{x-2} to find an approximate solution to:\n\n\\frac{1}{x-2}=1.4", null, "2.7", null, "3.7", null, "3.2", null, "2.2", null, "Draw a horizontal line at y=1.4 and see where it crosses the line.\n\nThen draw a vertical line down and read the value of the x-axis.", null, "### Types of graphs GCSE questions\n\n1.  On the grid, sketch the curve with the equation\n\ny=3^x\n\nGive the coordinates of any points of intersection with the axes.", null, "(2 marks)", null, "for drawing a growth curve\n\n(1)\n\nfor the y -intercept (0,1)\n\n(1)\n\n2. Here are three graphs:", null, "Here are three equations:\n\nMatch each graph to the correct equation\n\nGraph A and y = ………………\n\nGraph B and y = ………………\n\nGraph C and y = ……………….\n\n(3 marks)\n\nGraph A and y=x^3\n\n(1)\n\nGraph B and y=x^{2}+2\n\n(1)\n\nGraph C and y=\\frac{1}{x}\n\n(1)\n\n3. (a)  Complete the table of values for y=x^{3}-6x+1\n\n(b)      On the grid, draw the graph of y=x^{3}-6x+1 for -3\\leq{x}\\leq3", null, "(c) Use the graph to find an approximate solution for\n\nx^{3}-6x+1=7\n\n(5 marks)\n\n(a)\n\nCorrect Values are: 5, 1 and -4\n\nFor 1 correct y -values\n\n(1)\n\nFor all correct y -values\n\n(1)\n\n(b)", null, "for plotting the points correctly\n\n(1)\n\nfor the smooth curve\n\n(1)\n\n(c)\n\nx=2.8\n\napproximate solution\n\n(1)\n\n## Learning checklist\n\nYou have now learned how to:\n\n• Recognise and sketch different types of curved graphs\n• Plot different types of curved graphs\n• Use and interpret different types of curved graphs\n\n## Still stuck?\n\nPrepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.\n\nFind out more about our GCSE maths revision programme.\n\nx\n\n#### FREE GCSE maths practice papers (Edexcel, AQA & OCR)\n\n8 sets of free exam practice papers written by maths teachers and examiners for Edexcel, AQA and OCR.\n\nEach set of exam papers contains the three papers that your students will expect to find in their GCSE mathematics exam." ]

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[ "Sciences‎ > ‎\n\n### Introduction to A. I.\n\nIntelligent agent\n\nTerminology\n• Fully vs. Partially Observable visibility (chess vs. poker)\n• Deterministic (outcome consistent with action, e.g. chess) vs. Stochastic(random factor, e.g. dice)\n• Discrete vs. Continuous (whether finite/infinite possibilities spacewise, e.g. chess vs. dart)\n• Benign vs. Adversarial motivation to bother or counteractiveness\nSources of uncertainty: Stochastic environments, sensor limits, adversaries, laziness, ignorance\n\n### Problem Solving\n\nComparison of frontier and explored set\nBreadth First Search: scans all possible paths at the same step\nDepth First Search: scans one particular full path at a time\nUniform Cost Search: propagates with the lowest distance first\nA* Search: proceeds with lowest (distance to destination(h) + distance travelled) on condition that h < true cost (admissible)\n\nState Spaces: product of multi-dimensional conditions\n• admissible: describing a heuristic that never overestimates the cost of reaching a goal\n• guaranteed to work when fully observable, known, deterministic, discrete and static\n\n### Statistics Uncertainty, and Bayes networks\n\nBayes Rule: P(A|B) = P(B|A) * P(A)/P(B)\nComplex Bayes network: multiply all conditional probabilities, and analyze the provided distribution\n\nD-Separation/Reachability\nConditional independence: disjoint relationship, linked by known cause, linked by unknown effect\nConditional dependence: direct causal relationship, linked by unknown cause, linked by known effect, or its successor\n\nMinimum number of parameters necessary to specify joint probability  = ∑ 2^(number of causes for each nodes)\n\n### Machine Learning\n\nMaximum likelihood: proportionate ratio\nLaplace smoothing: P = (occurrence + k) / (data + #variables)\nLinear Regression: minimize the sum of errors between real value-calculated value\nPerceptron Algorithm: linear separation by taking the majority class label of k(recolarizer) nearest neighbors\nUnsupervised Learning\n\n### Hidden Markov models and Bayes filters\n\nPropositional Logic\n\n### Adversarial planning (games) and belief space planning (POMDPs)\n\nLogic and Logical Problem Solving\n\nImage Processing and Computer Vision\n\nRobotics and robot motion planning\n\nNatural Language Processing and Information Retrieval" ]

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[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Astron. Astrophys. 343, 325-332 (1999)\n\n## 2. Stationary and axisymmetric field\n\nIn Euclidean space and in ellipsoidal coordinates", null, ", related to the Cartesian coordinates", null, "by", null, "the kinetic energy of a particle of mass m is", null, "where a is a non-negative constant of the dimension of length and the dot denotes differentiation with respect to the time parameter", null, "(normalized so that the momentum is", null, "). The motion of the particle in a stationary axisymmetric potential", null, "is described by equations", null, "where we denote, for example,", null, "", null, "", null, "", null, "", null, "© European Southern Observatory (ESO) 1999\n\nOnline publication: March 1, 1999", null, "" ]

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[ "# I'm trying to solve Bernoulli equation in differential form.\n\n1 view (last 30 days)\nsuresh berike on 3 Aug 2021\nCommented: suresh berike on 3 Aug 2021\nI'm trying to solve this bernoulli equation.\n(∂u/∂t)+ρ(u∂u/∂x)=-∂p/∂x+(μ∂^2 u)/(∂x^2 )\nafter assuming the variation of u with time is zero and ignoring viscous effects, I got to ρ(u∂u/∂x)=-∂p/∂x\nby solving this, i wish to get value of u for different values of dx( variation of u along a straight line ) for some constant value of p. But to do this i need to convert this equation to the form required by matlab, which im not able to do.At the end I need to plot the variation of u against dx. How can i achieve this?\n\nAlan Stevens on 3 Aug 2021\nIf p is constant then dp/dx = 0, hence du/dx = 0, which means u is constant.\nIf you meant the pressure gradient is constant then u*du/dx = -(p/dx)/rho = constant, or du^2/dx = -2(dp/dx)/rho, so u^2 = (-2(dp/dx)/rho)*x + uo^2, where uo is the value of u when x = 0.\nsuresh berike on 3 Aug 2021\nOk thanks a lot for the help." ]

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[ "", null, "Level 1\n\n0% to", null, "Level 2\n\nTip /\n\nto gain points, level up, and earn exciting badges like the new\nMission!\n\nView all\n\n# Adobe LiveCycle (Archived)\n\nSOLVED\n\n## Auto-Fill in of minimum value", null, "", null, "Level 3\n\nI have two cells being calculated. I want the the third cell to automatically fill-in based upon the lesser of those two values.\n\nExample: If cell1 = 5 and cell2 = 7, cell3 should autofill with 5, the lessor of the two\n\nIf cell1 = 5 and cell2 =3, cell3 sould fill with 3.\n\nFormula so far in layout:\n\nif (MaxLoan > MaxLoanLTV) then // NewLoan\n\nMaxLoan.presence = \"visible\"\n\nMaxLoanLTV.presence = \"hidden\"\n\nelseif (MaxLoanLTV > MaxLoan) then\n\nMaxLoan.presence = \"hidden\"\n\nMaxLoanLTV.presence = \"visible\"\n\nendif\n\nFormula in calculate:\n\n\\$ = sum(MaxLoan) // NewLoan\n\n\\$ = sum(MaxLoanLTV) // NewLoan\n\nIt does reflect a number but not the minimum value.\n\nThank you.\n\n1 Accepted Solution", null, "", null, "Level 5\n\nHi,\n\nyou have t copy this script in the calculate event of your target field, use Javascript.\n\n`this.rawValue =  Math.min(num_1.rawValue, num_2.rawValue);`", null, "Hope it will helps you,\n\nMandy\n\n0 Replies", null, "", null, "Level 5\n\nHi,\n\nyou have t copy this script in the calculate event of your target field, use Javascript.\n\n`this.rawValue =  Math.min(num_1.rawValue, num_2.rawValue);`", null, "Hope it will helps you,\n\nMandy", null, "", null, "Level 3\n\nThank you, worked great!", null, "" ]

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[ "# Protocol math (Ethereum)\n\nInterest rate models in use for EVM protocol\n\n## Utilisation\n\nThis current state of supply and demand for liquidity is measured with the utilisation rate, in other words, how much of the supplied liquidity is being borrowed (utilised) by borrowers. The higher the utilisation rate, the higher the interest rate in a lending market.\nUt = utilisation rate at time t Uoptimal = optimal utilisation rate Rv = variable borrow rate Rv0 = base variable borrow rate (interest when utilisation = 0%) Rslope1 = constant which determines the progression of the interest rate until Uoptimal Rslope2 = constant which determines the progression of the interest rate after Uoptimal\nThe protocol has built in incentives in the interest rate model. To be capitally efficient, it sets an optimal utilisation rate. Below this rate, the protocol will incentivise utilisation, above this rate and it will disincentivise utilisation.\nTwo different slopes are used to measure interest rates, one for when utilisation is below optimal, and one for when it is above the optimal rate.\nWhen not enough borrowers are borrowing available liquidity, the interest rate will be calculated as such:\n$R_v =R_{v0} + (U_t \\div U_{optimal}) \\times R_{slope1}$\nWhen too many borrowers are borrowing available liquidity, the interest rate will be calculated as such:\n$R_v = R_{v0} + R_{slope1}+(U_t - U_{optimal})\\div(1-U_{optimal})\\times R_{slope2}$\n\n## Protocol parameters\n\nHoney Finance lending markets use the following parameters as default:\nOptimal utilisation: 80% Borrow APR at Uoptimal: 40% Base borrow APR: 10% Rslope1 constant: 0.3 Rslope2 constant: 1\nYou can try these parameters out for yourself and test models here." ]

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[ "# Can a nonabelian group $G$ have a normal abelian subgroup $H$ with $[G:H]=3$?\n\nIf $$H$$ is a cyclic group, for instance, and $$[G:H］=2$$, then $$G$$ is a dihedral group with elements equal to the possible states of a regular polygon upon which rotations and reflections are applied. In this case, the quotient group $$G/H$$ of order $$2$$ can be interpreted as the group whose elements are the states \"not reflected\" and \"reflected.\"\n\nI'm having lots of trouble imagining what it would mean if $$H$$ were a cyclic normal subgroup of $$G$$ and $$[G:H]=3$$, or any odd number, for that matter. For even numbers, however, I can imagine the group whose elements are states of a regular polygon upon which rotations and reflections are applied, and whose vertices change color whenever two reflections are applied. In this case, if $$H$$ is the subgroup of rotations, then $$[G:H］$$ could be made to be any even numbers. However, I can't imagine any such thing for odd numbers.\n\nAny hints or advice would be greatly appreciated. For context, I'm doing research about Cayley graphs, and my group theory knowledge isn't as good as it should be. I tried looking through textbooks to find the answer to this question, but I haven't been successful.\n\n• Here's a simple infinite example: take $\\mathbb Z^3 \\rtimes (\\mathbb Z/3\\mathbb Z)$, where $t$ is a generator of the $\\mathbb Z/3\\mathbb Z$ factor and conjugation by $t$ cyclically permutes the factors of $\\mathbb Z^3$, i.e. $t^{-1}(a,b,c)t = (c,a,b)$ for $(a,b,c) \\in \\mathbb Z^3$. $\\mathbb Z^3$ is normal and has index $3$. – Rylee Lyman Apr 26 '19 at 1:35\n\n## 2 Answers\n\nYes. The Klein subgroup $$\\{1, (1~2)(3~4), (1~3)(2~4), (1~4)(2~3)\\} \\triangleleft A_4$$. This subgroup is abelian, as the title of the question requests, but it's not cyclic.\n\nMore generally, let $$p$$ be any prime. If one has a finite group $$G$$ of order $$p^n$$, then it has a composition series with quotients of order $$p$$, that is there is a sequence $${e}=G_0\\triangleleft G_1\\triangleleft G_2\\triangleleft \\cdots \\triangleleft G_{n-1} \\triangleleft G_n=G$$ with each $$|G_j|=p^j$$. Each $$|G_j:G_{j-1}|=p$$. If $$G$$ is non-Abelian, then somewhere the sequence flips from Abelian to non-Abelian, so there is some $$j$$ with $$G_{j-1}$$ Abelian, and normal of index $$p$$ in the non-Abelian group $$G_j$$.\n\nADDED IN EDIT\n\nHere's a construction for the original problem with $$H$$ cyclic. Let $$H$$ be a cyclic group of prime order $$p$$ with $$p\\equiv 1\\pmod3$$ (for instance $$p=7$$). Then $$H$$ has an automorphism of order $$3$$. Now let $$G$$ be the semidirect product of $$H$$ by a cyclic group of order $$3$$ acting on $$H$$ via this automorphism.\n\nThis construction also works for other values of $$3$$....." ]

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[ "• API\n• FAQ\n• Tools\n• Archive\nSHARE\nTWEET", null, "# Untitled", null, "a guest", null, "Sep 18th, 2012", null, "46", null, "Never\nNot a member of Pastebin yet? Sign Up, it unlocks many cool features!\n1.\n2.  And player getting 2 first, looses\n3.\n4. User turn --- Press 2 to roll\n5. 2\n6.\n7. The user rolled        Dice 1 = 5 and Dice 2 = 3\n8. Total = 8\n9.\n10. User turn --- Press 2 to roll\n11. 2\n12.\n13. The user rolled        Dice 1 = 1 and Dice 2 = 2\n14. Total = 3\n15.\n16. User turn --- Press 2 to roll\n17. 2\n18.\n19. The user rolled        Dice 1 = 2 and Dice 2 = 2\n20. Total = 4\n21.\n22. User turn --- Press 2 to roll\n23. 2\n24.\n25. The user rolled        Dice 1 = 4 and Dice 2 = 6\n26. Total = 10\n27.\n28. User turn --- Press 2 to roll\n29. 2\n30.\n31. The user rolled        Dice 1 = 2 and Dice 2 = 2\n32. Total = 4\n33.\n34. User turn --- Press 2 to roll\n35. 2\n36.\n37. The user rolled        Dice 1 = 6 and Dice 2 = 2\n38. Total = 8\n39.\n40. User turn --- Press 2 to roll\n41. 2\n42.\n43. The user rolled        Dice 1 = 1 and Dice 2 = 5\n44. Total = 6\n45.\n46. User turn --- Press 2 to roll\nRAW Paste Data\nWe use cookies for various purposes including analytics. By continuing to use Pastebin, you agree to our use of cookies as described in the Cookies Policy.\nTop" ]

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[ "You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.\n\n#### 93 lines 1.9 KiB Raw Normal View History Unescape Escape", null, "20 years ago `.TH array 3` `.SH NAME` `array \\- The array library interface` `.SH SYNTAX` `.B #include ` ``` ``` `.SH DESCRIPTION` `An \\fBallocated\\fR array variable keeps track of` ``` ``` `.sp 1` `.IP \\(bu` `a (nonzero) pointer to a dynamically allocated region of memory;` `.IP \\(bu` `the number of bytes allocated (always positive); and` `.IP \\(bu` `the number of bytes initialized (between 0 and the number of bytes` `allocated).` `.PP` ``` ``` `There are two other possibilities for the state of an array variable:` `\\fBunallocated\\fR and \\fIfailed\\fR. In both cases, there is no` `dynamically allocated region of memory.` ``` ``` `A new array variable is normally created as a static variable:` ``` ``` ` #include \"array.h\"` ``` ``` ` static array x;` ``` ``` `At this point it is unallocated. The array library provides various` `allocation and inspection functions.` ``` ``` `A new array variable can also be created dynamically. It must be` `initialized to all-0, meaning unallocated, before it is given to any of` `the array functions. It must be returned to the unallocated (or failed)` `state, for example with array_reset, before it is destroyed. These rules` `prevent all memory leaks.` `.SH \"Expansion and inspection\"` ``` ``` ` array x;` ``` ``` ` t* p1 = array_allocate(&x,sizeof(t),pos);` ``` ``` ` t* p2 = array_get(&x,sizeof(t),pos);` ``` ``` ` t* p3 = array_start(&x);` ``` ``` ` int64 len = array_length(&x,sizeof(t));` ``` ``` ` int64 bytes = array_bytes(&x);` ``` ``` `.SH \"Truncation and deallocation\"` ``` ``` ` array x;` ``` ``` ` array_truncate(&x,sizeof(t),len);` ``` ``` ` array_trunc(&x);` ``` ``` ` array_reset(&x);` ``` ``` ` array_fail(&x);` ``` ``` `.SH \"Comparison\"` ``` ``` ` array x;` ` array y;` ``` ``` ` if (array_equal(&x,&y))` ` /* arrays are equal... */` ``` ``` `.SH \"Concatenation\"` ``` ``` ` array x;` ` array y;` ``` ``` ` array_cat(&x,&y);` ``` ``` ` array_catb(&x,\"fnord\",5);` ``` ``` ` array_cats(&x,\"fnord\");` ``` ``` ` array_cats0(&x,\"fnord\"); /* also append the \\\\0 */` ``` ``` ` array_cat0(&x); /* append \\\\0 */` ``` ``` ` array_cate(&x,\"fnord\",1,4); /* append \"nor\" */` ``` ``` `.SH \"ORIGINAL API DEFINITION\"` `http://cr.yp.to/lib/array.html` `.SH \"SEE ALSO\"` `array_get(3), array_start(3), array_fail(3)`" ]

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[ "# 487 is 74 Percent of what?\n\n## 487 is 74 Percent of 658.11\n\n%\n\n487 is 74% of 658.11\n\nCalculation steps:\n\n487 ÷ ( 74 ÷ 100 ) = 658.11\n\n### Calculate 487 is 74 Percent of what?\n\n• F\n\nFormula\n\n487 ÷ ( 74 ÷ 100 )\n\n• 1\n\nPercent to decimal\n\n74 ÷ 100 = 0.74\n\n• 2\n\n487 ÷ 0.74 = 658.11 So 487 is 74% of 658.11\n\nExample" ]

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[ "At the end of each semester, a scholarship index grade point average (GPA) is computed for each student.\n\n A = 4.00 points for each credit B+ = 3.33 points for each credit C = 2.00 points for each credit D+ = 1.33 points for each credit F = 0.00 points C+ = 2.33 points for each credit A- = 3.67 points for each credit B = 3.00 points for each credit C- = 1.67 points for each credit D = 1.00 point for each credit B- = 2.67 points for each credit\n\nHF (HONOR CODE VIOLATION) = 0.00 points. Thus, a grade of A in a three-credit course would be assigned 12 points.\n\nThe GPA is determined by dividing the total number of earned quality points by the number of hours towards the GPA. Grades of AU, I and W are excluded from the calculations. Other symbols in use (not included in computation of average):\n\n• W Officially Withdrawn from the Course\n• AU Audit\n• I Incomplete\n• P Pass\n• NP Not Passed\n• T Transfer credit from another institution\n• PC Credit for Prior Learning and Experience\n• CC Credit for externally recognized credentials" ]

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[ "# K3 surfaces with a non-symplectic automorphism and product-quotient surfaces with cyclic groups\n\n• ### Alice Garbagnati\n\nUniversità degli Studi di Milano, Italy\n• ### Matteo Penegini\n\nUniversità degli Studi di Milano, Italy", null, "## Abstract\n\nWe classify all the K3 surfaces which are minimal models of the quotient of the product of two curves $C_1 \\times C_2$ by the diagonal action of either the group $\\mathbb Z/p\\mathbb Z$ or the group $\\mathbb Z/2p\\mathbb Z$ where $p$ is an odd prime. These K3 surfaces admit a non-symplectic automorphism of order $p$ induced by an automorphism of one of the curves $C_1$ or $C_2$. We prove that most of the K3 surfaces admitting a non-symplectic automorphism of order $p$ (and in fact a maximal irreducible component of the moduli space of K3 surfaces with a non-symplectic automorphism of order $p$) are obtained in this way.\n\nIn addition, we show that one can obtain the same set of K3 surfaces under more restrictive assumptions namely one of the two curves, say $C_2$, is isomorphic to a rigid hyperelliptic curve with an automorphism $\\delta_p$ of order $p$ and the automorphism of the K3 surface is induced by $\\delta_p$.\n\nFinally, we describe the variation of the Hodge structures of the surfaces constructed and we give an equation for some of them." ]

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[ "# Number Systems\n\nNumbers are everywhere and used in many places – financial institution, statistics, engineering and so on. Without number system our daily life would be difficult.\n\nHumans use a number system called the decimal number system and everyone can easily understand this system. You go to a store to buy something, you need a number system to count your money. Even a small thing such as making a list of its need numbers.\n\n### What is a number system?\n\nA number belonging to a number system with:\n\n1. A base ‘r’\n2. Coefficient ‘a’ that is between 0 to r-1.\n\nExamples\n\n``````\nDecimal with r=10 and coefficient a = 0 to 9\nBinary with r=2 and coefficient a = 0 and 1\nOctal with r= 8 and coefficient a = 0 to 7\nHexadecimal with r=16 and coefficient a = 0 to 9, A, B, C, D, E, F``````\n\n### Converting from any number system to decimal system\n\nIf the base ‘r’ and coefficients are given then any number can be converted to decimal system using following\n\nan * rn + an-1 * rn-1 + … + a1 * r + a0 + a-1 * r-1 + a-2 * r-2 + a-n-1 * r-n-1 + rn\n\nExample\nTo convert binary number 1101 into decimal number, do following\n\n``````\n1011 = 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 2^0\n\n= 8 + 0 + 2 + 1\n\n= 11``````\n\n11 is decimal equivalent of 1101.\n\nExample\nConvert 1101.0112 into decimal equivalent.\nSolution:\n\n``````Integer part -> 1101\n\n1 * 2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0\n\n8 + 4 + 0 + 1\n\n13\n\nDecimal part -> .011\n\n0 * 2^-1 + 1 * 2^-2 + 1 * 2^-3\n\n0 + 1/4 + 1/8\n\n3/8\n\nAnswer = 13 + 3/8 = (13 * 8 + 3)/8\n\n= 107/8\n\n= 13.37``````\n\n### Conversions from Decimal to Other Number Systems\n\nThe decimal number system have a base r = 10 and coefficients ranging from 0 –to- 9.\n\nDecimal to Binary\n\nTo convert from decimal from binary, divide the decimal number repeatedly until no longer possible to divide. Each time you get a quotient and remainder, divide the quotient and note the remainder. The remainders of this division put together in reverse order is the answer. See the example below.\n\nFor the fractional part, you need to multiply the decimal fraction with base r repeatedly until you reach 0 or close to zero for the fractional part. When you multiply a fraction with base r, you get two thing – an integer and a fraction.\n\nIf the fraction has not reached zero or real close to zero, multiply it with base in next iteration and check the integer and fraction part. The integer part put together is the binary equivalent for the decimal fraction. See the example below\n\nProblem:\nConvert 25.13 to binary equivalent number.\nSolution:\n\n``````25/2 quotient = 12, remainder = 1\n2/2 quotient = 6, remainder = 0\n6/2 quotient = 3, remainder = 0\n3/2 quotient = 1, remainder = 1\n1/2 quotient = 0, remainder = 1\n\nNow we find the fractional part binary equivalent.\n\n0.13 * 2 = 0.26 => 0 + .26\n0.26 * 2 = 0.52 => 0 + .52\n0.52 * 2 = 1.04 => 1 + .04\n\nAnswer: - 11001 + .001 = 11001.001``````\n\nDecimal to Octal\n\nThe decimal to octal conversion is done in the same way as decimal to binary except the base r = 8. For example\n\nProblem:\nConvert 235 to its octal equivalent.\nSolution:\n\n``````235/8 quotient = 29, remainder = 3\n29/8 quotient = 3, remainder = 5\n3/8 quotient = 0, remainder = 3\n\nThe hexadecimal number has a base r = 16 and we need to divide decimal number with base r to get the hexadecimal equivalent, but hexadecimal digits 10 – 15 are A – F.\n\n``````2344/16 quotient = 146, remainder = 8\n146/16 quotient = 9, remainder = 2\n2/16 quotient = 0, remainder = 2\n\n### Conversions to Binary\n\nOctal to Binary\n\nTo convert octal to binary, convert each digit of the octal number to its binary equivalent, but you are only allowed to use 3-bit positions because of 111 = 7 which is the maximum range of octal coefficient. For example\n\nProblem:\nConvert 256 into its binary equivalent.\n\nSolution:\n\n``````2 -> 0 1 0\n5 -> 1 0 1\n6 -> 1 1 0\n\nAnswer: The binary equivalent is 010 101 110.``````\n\nWe use 4 bits to convert each digit in a hexadecimal number to get the binary equivalent. This is because 1111 = 15 = F in decimal, the maximal value allowed in hexadecimal number. For example\n\nProblem:\nConvert 2CD3 into binary equivalent.\n\nSolution:\n\n``````2 C D 3\n0010 1100 1101 0011``````\n\nThe binary equivalent of 2CD3 is 0010 1100 1101 0011.\n\n### Binary to Other Number Systems\n\nBinary to Octal\n\nTo convert from binary to octal use 3-bit position and convert it into its decimal equivalent.\n\n1. For integer part 3 bit from left to right\n2. For fractional part 3 bit from right to left\n\nThis will give you the octal equivalent of binary number. For example,\n\n``````1001.0101 => 001 110 . 010 100\n1 6 . 2 1\n\nAnswer: - The octal equivalent is 16.21.``````\n\nGroup the given binary number into a group of four and then convert the each of the group into the decimal equivalent to get the hexadecimal digit and group all the digits found to get the hexadecimal number. For example\n\nProblem:\nConvert the given binary number 110101011100 into hexadecimal equivalent.\n\nSolution:\n\n``````Given binary number can be separated into group of four.\n\n110101011100 => 1101 0101 1100\n13 5 12\nD 5 C\n\nTherefore, the hexadecimal equivalent number is D5C.``````\n\n### Other Number Systems\n\nWe use a very simple procedure to convert octal to hexadecimal number. You can do this two easy steps.\n\n1. Convert Octal to Binary number.\n2. Convert the Binary number obtained into hexadecimal.\n\nIn the previous section, we learned about converting octal to binary and binary into hexadecimal. In this section, you have to use that knowledge in converting octal to hexadecimal. For example,\n\nProblem:\nConvert the given octal number 234 into Hexadecimal equivalent.\n\nSolution:\n\n``````Step1: Convert the octal into binary number.\n\n2 3 4\n010 011 100\n\nStep2: Convert the binary number into a hexadecimal number.\n\n010011100\n\nGroup the binary number into groups of four bits from left to right.\n\n0000 1001 1100\n0 9 12\n\nConvert the result into a hexadecimal number.\n\nTherefore, the hexadecimal equivalent is 09C.``````" ]

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[ "+0\n\n# Find angle P in the Parallelogram LMNP\n\n0\n174\n1\n\nFind angle P in the Parallelogram LMNP", null, "Mar 19, 2021\n\n#1\n0\n\nm∠L  +  m∠P   =   180°\n\n(14x - 22)°  +  (4x + 4)°   =   180°\n\n14x - 22  +  4x + 4   =   180\n\n18x + 18   =   180\n\n18x   =   180 - 18\n\n18x   =   162\n\nx   =   162/18\n\nx   =   9\n\nNow that we know  x  we can find the measure of angle  P\n\nm∠P   =   (4x + 4)°   =   (4(9) + 4)°   =   (36 + 4)°   =   40°\n\nMar 19, 2021" ]

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[ "", null, "## Lesson Tutor : Lesson Plans : Mathematics : Algebra : Grade 9\n\n/  Lesson Tutor : Lesson Plans : Mathematics : Algebra : Grade 9\n\nAlgebra Lesson 8 – The Equation Behind an Old Trick\nby Elaine Ernst Schneider\n\nObjective(s): By the end of this lesson the student will be able to:\n\nPre-Class Assignment: Review/ completion of Algebra Lesson 7\n\nResources/Equipment/Time Required:\n\nOutline:\n\nIn the last lesson, you learned that equations can be used to solve puzzles. For years, mathematicians have been solving the following puzzle to the amazement of their students. Not only is it a fun “trick,” but it is based on a sound mathematical principle that makes it work:\n\nHere’s how it works:\n\n3.Multiply by 50.\n5.Add the number of days in a year (365).\n6.Take that number and subtract 615.\n7.Pretend it is an amount of money and position the decimal point for dollars and cents." ]

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[ "Page images PDF EPUB\n .flow { margin: 0; font-size: 1em; } .flow .pagebreak { page-break-before: always; } .flow p { text-align: left; text-indent: 0; margin-top: 0; margin-bottom: 0.5em; } .flow .gstxt_sup { font-size: 75%; position: relative; bottom: 0.5em; } .flow .gstxt_sub { font-size: 75%; position: relative; top: 0.3em; } .flow .gstxt_hlt { background-color: yellow; } .flow div.gtxt_inset_box { padding: 0.5em 0.5em 0.5em 0.5em; margin: 1em 1em 1em 1em; border: 1px black solid; } .flow div.gtxt_footnote { padding: 0 0.5em 0 0.5em; border: 1px black dotted; } .flow .gstxt_underline { text-decoration: underline; } .flow .gtxt_heading { text-align: center; margin-bottom: 1em; font-size: 150%; font-weight: bold; font-variant: small-caps; } .flow .gtxt_h1_heading { text-align: center; font-size: 120%; font-weight: bold; } .flow .gtxt_h2_heading { font-size: 110%; font-weight: bold; } .flow .gtxt_h3_heading { font-weight: bold; } .flow .gtxt_lineated { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; } .flow .gtxt_lineated_code { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; font-family: monospace; } .flow .gtxt_quote { margin-left: 2em; margin-right: 2em; margin-top: 1em; margin-bottom: 1em; } .flow .gtxt_list_entry { margin-left: 2ex; text-indent: -2ex; } .flow .gimg_graphic { margin-top: 1em; margin-bottom: 1em; } .flow .gimg_table { margin-top: 1em; margin-bottom: 1em; } .flow { font-family: serif; } .flow span,p { font-family: inherit; } .flow-top-div {font-size:83%;}", null, "By def. 27. sect. 4. The sine of 96°=the sine of 84°, which is the supplement thereof ; therrtore instead of the sine of 96°, look in the tables for the sine of 84.", null, "Extend from 84° (which is the supplement of 969) to 46° 30' on the sines; that distance will reach from 230 to 168, on the line of numbers, for BC. Extend from 840 to 370.30', on the sines ; that extent will reach from 230 to 141, on the line of numbers, for AC. CASE III. Two sides and a contained angle given ; to find the other ana gles and side. PL. 5. fig. 16. In the triangle ABC, there is AR 240, the angle A 36° 404 and AC 180, given ; to find the angles C and B, and the side BC. Ist. B, Construction. Draw a blank line, on which from a scale of equal parts, lay AB 240; at the point A of the line AB, make an angle of 36° 40', by a blank line ; on which from A, lay AC 180, from the same scale of equal parts ; measure the angles C and B, and the the side BC, as before ; and you have the answers required. 2d. By Calculation. By cor 1. theo. 5. sect. 4. 180°—the angle A 36° 40' =143o. 20/ the sum of the angles C and B: therefore half of 143°. 20', will be half the sum of the two required angles, C and B. By theo. 2. of this sect. As the sum of the two sides AB and AC = 420 is to their difference, = 60 So is the tangent of half the sum of = 71° 404 the two unknown angles C and B to the tangent of half their difference - 23o 20% By theo. 4. To half the sum of the angles Cand B=71° 40 Add half their difference as now found = 23 20. The sum is the greatest angle, or ang. C=95 00 Subtract, and you have the least angle, or B=48 20 The angle C and B being found ; BC is had, as before, by theo. l. of this sect. Thus, S. B: AC::S: A: BC. 48° 206 : 180: : 36° 40: 143. 9. 3d. By Gunter's Scale. Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers ; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 71o. 40' ; that distance laid from 45° on the same line will reach to 23o. 30', the half difference of the required angles. Whence the angles are obtained, as before. The second proportion may be easily extended, from what has been already said. CASE IV. Pl. 5. fig. 17. The three sides given, to find the angles. In the triangle ABC, there is given, AB 64, AC 47, BC 34: the angles A, B, C, are required. Ist. By Construction. The construction of this triangle must be manifest, from prob. I. sect. 4. 2d. By Calculation. From the point C, let fall the perpendicular CD on the base AB; and it will divide the triangle into two right angled oness ADC and CBD ; as well as the base AB, into the two segments, AD and DB.", null, "As the base or the longest side, AB 64 is to the sum of the other sides, AC and BC, 81 So is the difference of those sides 13 to the difference of the segments of 16.46 the base AD, DB. By theo. 4. of this sect. To half the base, or to half the sum of the segments AD and DB. Add half their difference, now found, }", null, "", null, "Subtract, and their difference will be the least segment DB, be} 23.77 In the right angled triangle ADC, there is AC 47, and AD 40. 28, given, to find the angle A. This is resolved by case 4. of right angled plane trigonometry, thus, AD: R::AC: Sec. A. 40. 23: 90° : : 47: 31° 08 Or it may be bad by finding the angle ACD, the complement of the angle A ; without a secant, thus, AC: R::AD: S. ACD. 44 : 900 : : 40.23: 580 527 90—58° 52' =319.08, the angle A. Then by theo. 1 of this sect. BC: S. A:: AC: S. B. 34: 21° 08'::47: 45° 37. By cor. 1. theo. 5. sect. 4. 180o--the sum of A and Bec. A 314.08' B 45. 37 180°-76. 45=1039.15', the angle C. « PreviousContinue »" ]

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[ "# Euler's theorem: $1808^{80085^{2014}} \\mod 17$\n\nThere are many examples on math SE on how to use Euler's theorem to compute modulos of large powers. Yet, I can't seem to find a solution to the following problem:\n\nCompute $1808^{80085^{2014}} \\mod 17$\n\nEuler's theorem is $a^{\\phi(n)} \\equiv 1 \\mod n \\iff gcd(a,n) = 1$\n\nThe solution is as follows:\n\n\\begin{align} &1808^{80085^{2014}} & \\mod 17 \\\\ =& 6^{80085^{2014} mod \\phi(17)} & \\mod 17 \\\\ =&6^{5^{2014 \\mod \\phi(16)} \\mod 16} & \\mod 17 \\\\ =&6^{5^{2014 \\mod 8} \\mod 16} & \\mod 17 \\\\ =&6^{5^{6} \\mod 16} & \\mod 17 \\\\ =&6^{9} & \\mod 17 \\\\ =&11 \\end{align}\n\nI know how to compute $\\phi(n)$, I know that $1808 \\mod 17 = 6$, but how does $6^{80085^{2014} mod \\phi(17)} \\mod 17$ follow from $1808^{80085^{2014}} \\mod 17$?\n\n• Do you mean 80085 or 8085? (although both ≡ 5 (mod 16)) – kennytm Aug 23 '16 at 8:01\n• @kennytm 80085. Fixed the question. Thank you! – Auberon Aug 23 '16 at 8:03\n\nThat is because $n^m \\equiv (n-17)^m \\mod 17$ thus\n\n$$n^m \\equiv (n \\mod 17)^m \\mod 17$$\n\n$$1808^m \\equiv(1808 \\mod 17)^m \\equiv 6^m \\mod 17$$\n\nthen you have\n\n$$6^m \\equiv 6^{m \\mod \\phi(17)} \\mod 17$$ due to Euler's theorem.\n\nDetail:\n\n$$6^{\\phi(17)} \\equiv 1\\mod 17$$ thus $$6^m = 6^{m-\\phi(17)}\\cdot 6^{\\phi(17)} \\equiv 6^{m-\\phi(17)}\\cdot1 \\equiv 6^{m-\\phi(17)}\\mod 17$$ so $$6^m \\equiv 6^{m \\mod \\phi(17)} \\mod 17$$\n\n• Excellent! Blew my mind a little. – Auberon Aug 23 '16 at 8:20\n\nWe want to calculate $1808^{80085^{2014}}\\bmod{17}$.\n\nLet's observe $1808^{k}\\pmod{17}$:\n\n• By Euler's theorem: $\\gcd(1808,17)=1\\implies1808^{\\phi(17)}\\equiv1\\pmod{17}$\n• Since $17$ is prime, $\\phi(17)=17-1=16$, hence $1808^{16}\\equiv1\\pmod{17}$\n\nLet's observe $80085^{k}\\pmod{16}$:\n\n• By Euler's theorem: $\\gcd(80085,16)=1\\implies80085^{\\phi(16)}\\equiv1\\pmod{16}$\n• Since $16$ is a power of two, $\\phi(16)=16/2=8$, hence $80085^{8}\\equiv1\\pmod{16}$\n\nLet's compute $80085^{2014}\\bmod{16}$:\n\n• $80085^{2014}\\equiv80085^{8\\cdot251+6}\\equiv(80085^{8})^{251}\\cdot80085^{6}\\equiv1^{251}\\cdot80085^{6}\\equiv80085^{6}\\pmod{16}$\n• $80085^{6}\\bmod{16}=(80085\\bmod{16})^{6}\\bmod{16}=5^{6}\\bmod{16}=15625\\bmod{16}=9$\n\nLet's compute $1808^{80085^{2014}}\\bmod{17}$:\n\n• $1808^{80085^{2014}}\\equiv1808^{16n+9}\\equiv(1808^{16})^{n}\\cdot1808^{9}\\equiv1^{n}\\cdot1808^{9}\\equiv1808^{9}\\pmod{17}$\n• $1808^{9}\\bmod{17}=(1808\\bmod{17})^{9}\\bmod{17}=6^{9}\\bmod{17}=10077696\\bmod{17}=11$" ]

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[ "You could take a comparison with the Memory Manager embedded with the FreePascalCompiler.\nIt has also a per-thread heap, with another implementation design. And it is now pretty stable and cross-platform!\nIt uses some nice FPC compiler tricks, like a `prefetch()` function which is quite unique and powerful when dealing with such low-level stuff like a memory manager.\nThe FPC guys did great job at the compiler level, and they do not forget to optimize the RTL in their work, which is pretty reassuring for the future - do you follow my mind?", null, "First of all, we run all our mORMot regression tests, which consists on a whole range of low-level and high-level tests, including stress tests and concurrent tests.\n\n```1. SapMMTime elapsed for all tests: 27.98s\n```\n```Peak Private Bytes = 233 408 KBVirtual Size = 286 112 KBPeak Working Set = 102 788 KB2. FastMM4Time elapsed for all tests: 27.45s\n```\n```Peak Private Bytes = 39 668 KBVirtual Size = 92 968 KBPeak Working Set = 35 236 KB\n```\n\nSo, a small speed penalty, but almost all those tests are mono-threaded, therefore it does make sense that FastMM4 is still the winner here.\nMemory consumption of SapMM is much bigger, since during those tests, several thread pools were allocated and released, so SapMM did have to manage several heaps.\n\nBut SapMM is very close - congrats!\n\nNow, some data from our multi-thread tests.\nWe took the tests included in the above tests, but made it 10 times longer (2000 insertions+reads in the DB instead of 200).\n\nIt was run on a 4 cores i7 CPU, and compiled with Delphi XE4 targeting Win32 platform (the only one supported by SapMM by now):\n\n```1. SapMM - Create thread pool: 1 assertion passed 2.77ms - TSQLRestServerDB: 120,022 assertions passed 1.80s 1=38460/s 2=37022/s 5=35766/s 10=37032/s 30=36353/s 50=35737/s - TSQLRestClientDB: 120,022 assertions passed 1.89s 1=35718/s 2=35437/s 5=33373/s 10=35740/s 30=33346/s 50=34342/s - TSQLRestClientURINamedPipe: 60,011 assertions passed 3.04s 1=9621/s 2=10212/s 5=12503/s - TSQLRestClientURIMessage: 80,002 assertions passed 1.91s 1=19494/s 2=22776/s 5=23490/s 10=20981/s - TSQLHttpClientWinHTTP_HTTPAPI: 119,861 assertions passed 7.63s 1=5156/s 2=7578/s 5=8701/s 10=9314/s 30=9507/s 50=10730/s - TSQLHttpClientWinSock_WinSock: 119,971 assertions passed 4.74s 1=10990/s 2=12606/s 5=13702/s 10=13840/s 30=14196/s 50=14210/s Total failed: 0 / 619,890 - Multi thread process PASSED 21.05s\nPeak Private Bytes = 198 628 KB\nVirtual Size = 248 160 KB\nPeak Working Set = 85 328 KB\n```\n```2. FastMM4 - Create thread pool: 1 assertion passed 4.44ms - TSQLRestServerDB: 120,022 assertions passed 2.21s 1=40017/s 2=25030/s 5=31335/s 10=24284/s 30=31308/s 50=28787/s - TSQLRestClientDB: 120,022 assertions passed 2.23s 1=35795/s 2=34538/s 5=31646/s 10=25018/s 30=27057/s 50=24016/s - TSQLRestClientURINamedPipe: 60,012 assertions passed 3.20s 1=9607/s 2=10242/s 5=10418/s - TSQLRestClientURIMessage: 80,004 assertions passed 2.98s 1=18035/s 2=19944/s 5=11601/s 10=9950/s - TSQLHttpClientWinHTTP_HTTPAPI: 119,970 assertions passed 8.02s 1=5156/s 2=7249/s 5=8265/s 10=8135/s 30=8983/s 50=10190/s - TSQLHttpClientWinSock_WinSock: 119,995 assertions passed 5.76s 1=10814/s 2=10875/s 5=12501/s 10=11365/s 30=12040/s 50=8079/s Total failed: 0 / 620,026 - Multi thread process PASSED 24.42s\nPeak Private Bytes = 23 956 KB\nVirtual Size = 79 388 KB\nPeak Working Set = 20 356 KB\n```\n\nOf course, due to its per-thread heap, SapMM eats much more memory than FastMM4, even with a thread pool like in those tests.\nIn the above tests, each protocol create up to 50 threads for its concurrent client access - and both HTTP servers (http.sys or WinSock) do also use a thread pool to handle its clients.\nSo memory grows a lot. SapMM consumes more than 4 times the same amount of memory than FastMM4. This is worth considering!\nIt may be even worse if not thread pool were used, but a new thread created by client request (e.g. with a DataSnap/Indy server).\n\n### Conclusion: it depends on what you need!\n\nFor the main mORMot use case (which is to serve HTTP content, preferred using http.sys), it is only 7.63 seconds for SapMM, and 8.02 seconds for FastMM4.\nSo we would perhaps stick to FastMM4 here, since memory allocation is not the bottleneck.\nFor direct server-side process (i.e. TSQLRestServerDB and TSQLRestClientDB), SapMM does an amazing job, and let concurrent access be linear, whereas FastMM4 performance was almost cut in half with 50 client threads.\n\n### Apologizes for false positive alert\n\nIn fact, those tests helped to (re) identify an issue in our own test code of weak references.\nWe first thought it was SapMM's fault, but it was not.\nIn fact, when you use manual weak references - `pointer(@fMyInterfaceField) := pointer(aInterface)` - you must set it to zero at destroying - `pointer(@fMyInterfaceField) := nil` - otherwise you may get an access violation, which was the case with SapMM and with FastMM4 in fulldebug mode (whereas default mode was tolerant enough to continue the tests).\n\nSome months ago (probably after a long night of work), we already identified the error, but, by laziness, we just disabled the test when fulldebugg mode was on...", null, "Shame on us!\nNever disable a test which does not pass.\nNever!\n\nOur test case is fixed now.\nThere was no problem on our implementation of zeroing weak references - just an incorrect test case implementation.\nSorry Anton for the misunderstanding on our side!" ]

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[ "# The factors of the expression ( x + 3 ) 2 − 4 .", null, "### Precalculus: Mathematics for Calcu...\n\n6th Edition\nStewart + 5 others\nPublisher: Cengage Learning\nISBN: 9780840068071", null, "### Precalculus: Mathematics for Calcu...\n\n6th Edition\nStewart + 5 others\nPublisher: Cengage Learning\nISBN: 9780840068071\n\n#### Solutions\n\nChapter 1.3, Problem 76E\nTo determine\n\n## To calculate: The factors of the expression (x+3)2−4 .\n\nExpert Solution\n\nThe factors of the expression (x+3)24 are x+5 and x+1 .\n\n### Explanation of Solution\n\nGiven information:\n\nThe expression (x+3)24 .\n\nFormula used:\n\nThe special factoring formula of difference of squares which is expressed as,\n\nA2B2=(A+B)(AB)\n\nCalculation:\n\nConsider the given expression (x+3)24 .\n\nRewrite it as,\n\n(x+3)24=(x+3)2(2)2\n\nRecall the special factoring formula of difference of squares which is expressed as,\n\nA2B2=(A+B)(AB)\n\nHere, the values of A and B are A=(x+3) and B=2 .\n\n(x+3)24=(x+3+2)(x+32)=(x+5)(x+1)\n\nThus, the factors of expression (x+3)24 are x+5 and x+1 .\n\n### Have a homework question?\n\nSubscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!", null, "" ]

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[ "", null, "Physics\n\n# 10.2Consequences of Special Relativity\n\nPhysics10.2 Consequences of Special Relativity\n\n### Section Learning Objectives\n\nBy the end of this section, you will be able to do the following:\n\n• Describe the relativistic effects seen in time dilation, length contraction, and conservation of relativistic momentum\n• Explain and perform calculations involving mass-energy equivalence\n\n### Teacher Support\n\n#### Teacher Support\n\n• (4) Science concepts. The student knows and applies the laws governing motion in a variety of situations. The student is expected to:\n• (F) Identify and describe motion relative to different frames of reference.\n• (8) Science concepts. The student knows simple examples of atomic, nuclear, and quantum phenomena. The student is expected to:\n• (C) Describe the significance of mass-energy equivalence and apply it in explanations of phenomena such as nuclear stability, fission, and fusion.\n\n### Section Key Terms\n\n binding energy length contraction mass defect time dilation proper length relativistic relativistic momentum relativistic energy relativistic factor rest mass\n\n### Teacher Support\n\n#### Teacher Support\n\nIn this section, you will see how the postulates lead to the theory of special relativity and see how that theory predicts effects on time, distance, momentum, and energy at velocities approaching the speed of light.\n\n[BL] Begin a discussion by asking if students have ever seen a science fiction movie where space travelers age more slowly than the people left behind on Earth. Tell them there is some basis in fact to these stories. Discuss nuclear power. Ask if they know the basic difference between the nuclear power and combustion power.\n\n[OL] Explain that Newton’s laws are valid for everyday mechanics but break down at speeds approaching the speed of light. Discuss the relationship between relativity theory and Newton’s laws. Briefly describe the changes predicted for measurements of time, length, momentum, and energy. See how much they know about energy derived from nuclear reactions.\n\n[AL] Ask what the students already know about relativity theory. See if they know that relative motion is an old idea and ask for examples of relative motion in everyday situations. Explain that special relativity is similar but describes unexpected results at speeds approaching the speed of light. Ask if anyone can explain why this statement is true: “The original source of all the energy we use is the conversion of matter into energy.”\n\n### Relativistic Effects on Time, Distance, and Momentum\n\nConsideration of the measurement of elapsed time and simultaneity leads to an important relativistic effect. Time dilation is the phenomenon of time passing more slowly for an observer who is moving relative to another observer.\n\nFor example, suppose an astronaut measures the time it takes for light to travel from the light source, cross her ship, bounce off a mirror, and return. (See Figure 10.5.) How does the elapsed time the astronaut measures compare with the elapsed time measured for the same event by a person on the earth? Asking this question (another thought experiment) produces a profound result. We find that the elapsed time for a process depends on who is measuring it. In this case, the time measured by the astronaut is smaller than the time measured by the earth bound observer. The passage of time is different for the two observers because the distance the light travels in the astronaut’s frame is smaller than in the earth bound frame. Light travels at the same speed in each frame, and so it will take longer to travel the greater distance in the earth bound frame.\n\n### Teacher Support\n\n#### Teacher Support\n\n[OL] Discuss the expression for the relativistic factor. Explain that this is involved in all relativistic effects. Show how to tell when relativistic effects are significant and when they are negligible by plugging in values of v and c.\n\nFigure 10.5 (a) An astronaut measures the time $Δ t 0 Δ t 0$ for light to cross her ship using an electronic timer. Light travels a distance $2D 2D$ in the astronaut’s frame. (b) A person on the earth sees the light follow the longer path $2s 2s$ and take a longer time $Δt . Δt .$\n\n### Teacher Support\n\n#### Teacher Support\n\n[AL]Figure 10.5, like Figure 10.4, may be hard for some students to grasp. Refer back to the previous figure. The animation in the discussion of length contraction further on should also be some help.\n\nThe relationship between Δt and Δto is given by\n\n$Δt=γΔ t 0 , Δt=γΔ t 0 ,$\n\nwhere $γ γ$ is the relativistic factor given by\n\n$γ= 1 1− v 2 c 2 , γ= 1 1− v 2 c 2 ,$\n\nand v and c are the speeds of the moving observer and light, respectively.\n\n### Tips For Success\n\nTry putting some values for v into the expression for the relativistic factor ( $γ γ$ ). Observe at which speeds this factor will make a difference and when $γ γ$ is so close to 1 that it can be ignored. Try 225 m/s, the speed of an airliner; 2.98 × 104 m/s, the speed of Earth in its orbit; and 2.990 × 108 m/s, the speed of a particle in an accelerator.\n\n### Teacher Support\n\n#### Teacher Support\n\nTry putting some values for v into the expression for the relativistic factor. Observe at which speeds this factor will make a difference and when it is so close to 1 that it can be ignored.\n\nNotice that when the velocity v is small compared to the speed of light c, then v/c becomes small, and $γ γ$ becomes close to 1. When this happens, time measurements are the same in both frames of reference. Relativistic effects, meaning those that have to do with special relativity, usually become significant when speeds become comparable to the speed of light. This is seen to be the case for time dilation.\n\nYou may have seen science fiction movies in which space travelers return to Earth after a long trip to find that the planet and everyone on it has aged much more than they have. This type of scenario is a based on a thought experiment, known as the twin paradox, which imagines a pair of twins, one of whom goes on a trip into space while the other stays home. When the space traveler returns, she finds her twin has aged much more than she. This happens because the traveling twin has been in two frames of reference, one leaving Earth and one returning.\n\nTime dilation has been confirmed by comparing the time recorded by an atomic clock sent into orbit to the time recorded by a clock that remained on Earth. GPS satellites must also be adjusted to compensate for time dilation in order to give accurate positioning.\n\nHave you ever driven on a road, like that shown in Figure 10.6, that seems like it goes on forever? If you look ahead, you might say you have about 10 km left to go. Another traveler might say the road ahead looks like it is about 15 km long. If you both measured the road, however, you would agree. Traveling at everyday speeds, the distance you both measure would be the same. You will read in this section, however, that this is not true at relativistic speeds. Close to the speed of light, distances measured are not the same when measured by different observers moving with respect to one other.\n\nFigure 10.6 People might describe distances differently, but at relativistic speeds, the distances really are different. (Corey Leopold, Flickr)\n\n### Teacher Support\n\n#### Teacher Support\n\n[OL] Discuss the relationship between time dilation and length contraction. If observers agree on speed, but not on time, they must also disagree on length because v = d/t.\n\nOne thing all observers agree upon is their relative speed. When one observer is traveling away from another, they both see the other receding at the same speed, regardless of whose frame of reference is chosen. Remember that speed equals distance divided by time: v = d/t. If the observers experience a difference in elapsed time, they must also observe a difference in distance traversed. This is because the ratio d/t must be the same for both observers.\n\nThe shortening of distance experienced by an observer moving with respect to the points whose distance apart is measured is called length contraction. Proper length, L0, is the distance between two points measured in the reference frame where the observer and the points are at rest. The observer in motion with respect to the points measures L. These two lengths are related by the equation\n\n$L= L 0 γ . L= L 0 γ .$\n\nBecause $γ γ$ is the same expression used in the time dilation equation above, the equation becomes\n\n$L= L 0 1− v 2 c 2 . L= L 0 1− v 2 c 2 .$\n\nTo see how length contraction is seen by a moving observer, go to this simulation. Here you can also see that simultaneity, time dilation, and length contraction are interrelated phenomena.\n\nThis link is to a simulation that illustrates the relativity of simultaneous events.\n\nIn classical physics, momentum is a simple product of mass and velocity. When special relativity is taken into account, objects that have mass have a speed limit. What effect do you think mass and velocity have on the momentum of objects moving at relativistic speeds; i.e., speeds close to the speed of light?\n\nMomentum is one of the most important concepts in physics. The broadest form of Newton’s second law is stated in terms of momentum. Momentum is conserved in classical mechanics whenever the net external force on a system is zero. This makes momentum conservation a fundamental tool for analyzing collisions. We will see that momentum has the same importance in modern physics. Relativistic momentum is conserved, and much of what we know about subatomic structure comes from the analysis of collisions of accelerator-produced relativistic particles.\n\nOne of the postulates of special relativity states that the laws of physics are the same in all inertial frames. Does the law of conservation of momentum survive this requirement at high velocities? The answer is yes, provided that the momentum is defined as follows.\n\nRelativistic momentum, p, is classical momentum multiplied by the relativistic factor $γ . γ .$\n\n$p=γmu, p=γmu,$\n10.3\n\nwhere $m m$ is the rest mass of the object (that is, the mass measured at rest, without any $γ γ$ factor involved), $u u$ is its velocity relative to an observer, and $γ, γ,$ as before, is the relativistic factor. We use the mass of the object as measured at rest because we cannot determine its mass while it is moving.\n\nNote that we use $u u$ for velocity here to distinguish it from relative velocity $v v$ between observers. Only one observer is being considered here. With $p p$ defined in this way, $p tot p tot$ is conserved whenever the net external force is zero, just as in classical physics. Again we see that the relativistic quantity becomes virtually the same as the classical at low velocities. That is, relativistic momentum $γmu γmu$ becomes the classical $mu mu$ at low velocities, because $γ γ$ is very nearly equal to 1 at low velocities.\n\nRelativistic momentum has the same intuitive feel as classical momentum. It is greatest for large masses moving at high velocities. Because of the factor $γ, γ,$ however, relativistic momentum behaves differently from classical momentum by approaching infinity as $u u$ approaches $c . c .$ (See Figure 10.7.) This is another indication that an object with mass cannot reach the speed of light. If it did, its momentum would become infinite, which is an unreasonable value.\n\nFigure 10.7 Relativistic momentum approaches infinity as the velocity of an object approaches the speed of light.\n\n### Teacher Support\n\n#### Teacher Support\n\n[OL] Discuss the graph. Explain how it shows that objects that have mass cannot reach the speed of light. Have the students analyze the equation for relativistic momentum and see how this supports this conclusion. Explain that light can travel at the speed of light because it has no rest mass.\n\nRelativistic momentum is defined in such a way that the conservation of momentum will hold in all inertial frames. Whenever the net external force on a system is zero, relativistic momentum is conserved, just as is the case for classical momentum. This has been verified in numerous experiments.\n\n### Mass-Energy Equivalence\n\nLet us summarize the calculation of relativistic effects on objects moving at speeds near the speed of light. In each case we will need to calculate the relativistic factor, given by\n\n$γ= 1 1− v 2 c 2 , γ= 1 1− v 2 c 2 ,$\n\nwhere v and c are as defined earlier. We use u as the velocity of a particle or an object in one frame of reference, and v for the velocity of one frame of reference with respect to another.\n\n#### Time Dilation\n\nElapsed time on a moving object, $Δ t 0 , Δ t 0 ,$ as seen by a stationary observer is given by $Δt=γΔ t 0 , Δt=γΔ t 0 ,$ where $Δ t 0 Δ t 0$ is the time observed on the moving object when it is taken to be the frame or reference.\n\n#### Length Contraction\n\nLength measured by a person at rest with respect to a moving object, L, is given by\n\n$L= L 0 γ , L= L 0 γ ,$\n\nwhere L0 is the length measured on the moving object.\n\n#### Relativistic Momentum\n\nMomentum, p, of an object of mass, m, traveling at relativistic speeds is given by $p=γmu , p=γmu ,$ where u is velocity of a moving object as seen by a stationary observer.\n\n#### Relativistic Energy\n\nThe original source of all the energy we use is the conversion of mass into energy. Most of this energy is generated by nuclear reactions in the sun and radiated to Earth in the form of electromagnetic radiation, where it is then transformed into all the forms with which we are familiar. The remaining energy from nuclear reactions is produced in nuclear power plants and in Earth’s interior. In each of these cases, the source of the energy is the conversion of a small amount of mass into a large amount of energy. These sources are shown in Figure 10.8.\n\nFigure 10.8 The sun (a) and the Susquehanna Steam Electric Station (b) both convert mass into energy. ((a) NASA/Goddard Space Flight Center, Scientific Visualization Studio; (b) U.S. government)\n\nThe first postulate of relativity states that the laws of physics are the same in all inertial frames. Einstein showed that the law of conservation of energy is valid relativistically, if we define energy to include a relativistic factor. The result of his analysis is that a particle or object of mass m moving at velocity u has relativistic energy given by\n\n$E=γm c 2 . E=γm c 2 .$\n\nThis is the expression for the total energy of an object of mass m at any speed u and includes both kinetic and potential energy. Look back at the equation for $γ γ$ and you will see that it is equal to 1 when u is 0; that is, when an object is at rest. Then the rest energy, E0, is simply\n\n$E 0 =m c 2 . E 0 =m c 2 .$\n\nThis is the correct form of Einstein’s famous equation.\n\nThis equation is very useful to nuclear physicists because it can be used to calculate the energy released by a nuclear reaction. This is done simply by subtracting the mass of the products of such a reaction from the mass of the reactants. The difference is the m in $E 0 =m c 2 . E 0 =m c 2 .$ Here is a simple example:\n\nA positron is a type of antimatter that is just like an electron, except that it has a positive charge. When a positron and an electron collide, their masses are completely annihilated and converted to energy in the form of gamma rays. Because both particles have a rest mass of 9.11 × 10–31 kg, we multiply the mc2 term by 2. So the energy of the gamma rays is\n\n10.4\n\nwhere we have the expression for the joule (J) in terms of its SI base units of kg, m, and s. In general, the nuclei of stable isotopes have less mass then their constituent subatomic particles. The energy equivalent of this difference is called the binding energy of the nucleus. This energy is released during the formation of the isotope from its constituent particles because the product is more stable than the reactants. Expressed as mass, it is called the mass defect. For example, a helium nucleus is made of two neutrons and two protons and has a mass of 4.0003 atomic mass units (u). The sum of the masses of two protons and two neutrons is 4.0330 u. The mass defect then is 0.0327 u. Converted to kg, the mass defect is 5.0442 × 10–30 kg. Multiplying this mass times c2 gives a binding energy of 4.540 × 10–12 J. This does not sound like much because it is only one atom. If you were to make one gram of helium out of neutrons and protons, it would release 683,000,000,000 J. By comparison, burning one gram of coal releases about 24 J.\n\n### Teacher Support\n\n#### Teacher Support\n\n[BL] In regards to the change in the law of conservation of energy to the law of conservation of mass-energy, it may help to think of mass as simply a very concentrated form of energy.\n\n[OL] Impress upon the students the enormous amount of energy derived from the conversion of a small amount of mass. Have them note that c2 is a very large number. Students try to understand new concepts by using previous knowledge, and that may result in a misconception here. They are comfortable with chemical reactions and may try to relate this to the burning of a piece of wood. Tell them that burning the wood chemically might provide energy for a single room in a house, but converting the mass of the wood completely to energy according to E = mc2 would provide power for thousands of houses.\n\n[AL] Ask students if they know the difference between fission and fusion and where examples of each of these occur.\n\n### Boundless Physics\n\n#### The RHIC Collider\n\nFigure 10.9 shows the Brookhaven National Laboratory in Upton, NY. The circular structure houses a particle accelerator called the RHIC, which stands for Relativistic Heavy Ion Collider. The heavy ions in the name are gold nuclei that have been stripped of their electrons. Streams of ions are accelerated in several stages before entering the big ring seen in the figure. Here, they are accelerated to their final speed, which is about 99.7 percent the speed of light. Such high speeds are called relativistic. All the relativistic phenomena we have been discussing in this chapter are very pronounced in this case. At this speed $γ γ$ = 12.9, so that relativistic time dilates by a factor of about 13, and relativistic length contracts by the same factor.\n\nFigure 10.9 Brookhaven National Laboratory. The circular structure houses the RHIC. (energy.gov, Wikimedia Commons)\n\nTwo ion beams circle the 2.4-mile long track around the big ring in opposite directions. The paths can then be made to cross, thereby causing ions to collide. The collision event is very short-lived but amazingly intense. The temperatures and pressures produced are greater than those in the hottest suns. At 4 trillion degrees Celsius, this is the hottest material ever created in a laboratory\n\nBut what is the point of creating such an extreme event? Under these conditions, the neutrons and protons that make up the gold nuclei are smashed apart into their components, which are called quarks and gluons. The goal is to recreate the conditions that theorists believe existed at the very beginning of the universe. It is thought that, at that time, matter was a sort of soup of quarks and gluons. When things cooled down after the initial bang, these particles condensed to form protons and neutrons.\n\nSome of the results have been surprising and unexpected. It was thought the quark-gluon soup would resemble a gas or plasma. Instead, it behaves more like a liquid. It has been called a perfect liquid because it has virtually no viscosity, meaning that it has no resistance to flow.\n\n#### Teacher Support\n\n##### Teacher Support\n\nDiscuss particle colliders such as the relatively new Large Hadron Collider built by CERN. Students may want to know more about this project and the God particle. Explain why there are so many applications of special relativity theory in the field of particle physics.\n\n### Grasp Check\n\nCalculate the relativistic factor γ, for a particle traveling at 99.7 percent of the speed of light.\n\n1. 0.08\n2. 0.71\n3. 1.41\n4. 12.9\n\n### Worked Example\n\n#### The Speed of Light\n\nOne night you are out looking up at the stars and an extraterrestrial spaceship flashes across the sky. The ship is 50 meters long and is travelling at 95 percent of the speed of light. What would the ship’s length be when measured from your earthbound frame of reference?\n\n### Strategy\n\nList the knowns and unknowns.\n\nKnowns: proper length of the ship, L0 = 50 m; velocity, v, = 0.95c\n\nUnknowns: observed length of the ship accounting for relativistic length contraction, L.\n\nChoose the relevant equation.\n\nDiscussion\n\nCalculations of $γ γ$ can usually be simplified in this way when v is expressed as a percentage of c because the c2 terms cancel. Be sure to also square the decimal representing the percentage before subtracting from 1. Note that the aliens will still see the length as L0 because they are moving with the frame of reference that is the ship.\n\n### Teacher Support\n\n##### Teacher Support\n\nIdentify the variables, the knowns and unknowns, and the relevant equation. Understand clearly which length applies to your frame of reference and which applies to the ship’s frame of reference; that is, which is proper length.\n\n### Practice Problems\n\n7.\n\nCalculate the relativistic factor, γ, for an object traveling at 2.00×108 m/s.\n\n1. 0.74\n2. 0.83\n3. 1.2\n4. 1.34\n8.\n\nThe distance between two points, called the proper length, L0, is 1.00 km. An observer in motion with respect to the frame of reference of the two points measures 0.800 km, which is L. What is the relative speed of the frame of reference with respect to the observer?\n\n1. 1.80×108 m/s\n2. 2.34×108 m/s\n3. 3.84×108 m/s\n4. 5.00×108 m/s\n9.\n\nConsider the nuclear fission reaction $n+ U 92 235 → C 55 137 s+ R 37 97 b+2n+E n+ U 92 235 → C 55 137 s+ R 37 97 b+2n+E$ . If a neutron has a rest mass of 1.009u, $U 92 235 U 92 235$ has a rest mass of 235.044u, $C 55 137 s C 55 137 s$ has rest mass of 136.907u, and $R 37 97 b R 37 97 b$ has a rest mass of 96.937u, what is the value of E in joules?\n\n1. $1.8× 10 −11 1.8× 10 −11$ J\n2. $2.9× 10 −11 2.9× 10 −11$ J\n3. $1.8× 10 −10 1.8× 10 −10$ J\n4. $2.9× 10 −10 2.9× 10 −10$ J\n10.\n\nConsider the nuclear fusion reaction $H 1 2 + H 1 2 → H 1 3 + H 1 1 +E H 1 2 + H 1 2 → H 1 3 + H 1 1 +E$ . If $H 1 2 H 1 2$ has a rest mass of 2.014u, $H 1 3 H 1 3$ has a rest mass of 3.016u, and $H 1 1 H 1 1$ has a rest mass of 1.008u, what is the value of E in joules?\n\n1. $6× 10 −13 6× 10 −13$ J\n2. $6× 10 −12 6× 10 −12$ J\n3. $6× 10 −11 6× 10 −11$ J\n4. $6× 10 −10 6× 10 −10$ J\n\n11.\nDescribe time dilation and state under what conditions it becomes significant.\n1. When the speed of one frame of reference past another reaches the speed of light, a time interval between two events at the same location in one frame appears longer when measured from the second frame.\n2. When the speed of one frame of reference past another becomes comparable to the speed of light, a time interval between two events at the same location in one frame appears longer when measured from the second frame.\n3. When the speed of one frame of reference past another reaches the speed of light, a time interval between two events at the same location in one frame appears shorter when measured from the second frame.\n4. When the speed of one frame of reference past another becomes comparable to the speed of light, a time interval between two events at the same location in one frame appears shorter when measured from the second frame.\n12.\n\nThe equation used to calculate relativistic momentum is p = γ · m · u. Define the terms to the right of the equal sign and state how m and u are measured.\n\n1. γ is the relativistic factor, m is the rest mass measured when the object is at rest in the frame of reference, and u is the velocity of the frame.\n2. γ is the relativistic factor, m is the rest mass measured when the object is at rest in the frame of reference, and u is the velocity relative to an observer.\n3. γ is the relativistic factor, m is the relativistic mass $( i.e., m 1− u 2 c 2 ) ( i.e., m 1− u 2 c 2 )$ measured when the object is moving in the frame of reference, and u is the velocity of the frame.\n4. γ is the relativistic factor, m is the relativistic mass $( i.e., m 1− u 2 c 2 ) ( i.e., m 1− u 2 c 2 )$ measured when the object is moving in the frame of reference, and u is the velocity relative to an observer.\n13.\nDescribe length contraction and state when it occurs.\n1. When the speed of an object becomes the speed of light, its length appears to shorten when viewed by a stationary observer.\n2. When the speed of an object approaches the speed of light, its length appears to shorten when viewed by a stationary observer.\n3. When the speed of an object becomes the speed of light, its length appears to increase when viewed by a stationary observer.\n4. When the speed of an object approaches the speed of light, its length appears to increase when viewed by a stationary observer.\n\n### Teacher Support\n\n#### Teacher Support\n\nUse the Check Your Understanding questions to assess students’ achievement of the section’s learning objectives. If students are struggling with a specific objective, the Check Your Understanding will help identify which and direct students to the relevant content.\n\nOrder a print copy\n\nAs an Amazon Associate we earn from qualifying purchases." ]

[ null, "data:image/svg+xml;base64,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", null ]

[ "# how to right a program that find kth number in two sorted array?\n\nPrintable View\n\n• 04-15-2008, 09:24 PM\nfireball2008\nhow to right a program that find kth number in two sorted array?\nThere is two sorted array A, and B, how to write a program that runs in log(n)^2 that compute the kth value in the union of this two array.\nAny idea is welcome.\nI know it has something to do with using binary search.\n• 04-16-2008, 11:06 AM\nsukatoa\nWhat have you done so far!!!!\n\nkind regards,\nsukatoa\n• 04-16-2008, 11:24 AM\nEranga\nYes, using binary search you can find the union of two arrays. So, as sukatoa says, what you have done up to now?\n• 04-16-2008, 09:04 PM\nfireball2008\nI only know how to do it in linear time, I have no idea how to do it in log(n)^2\nfor linear time:\npublic int find(int k, int j, int i){\nint temp = A[i]\nint temp2 = B[j]\nif(A[i]>B[j])\nj++;\nelse\ni++;\nif((i+j)==k)\n{if(A[i]>B[j])\nreturn A[i];\nelse\nreturn B[j];}\nelse\nreturn find(k, i, j);}}\n\ncall find(k, 0, 0);\n\nbut that is linear time I need log(n)^2 time.\n• 04-17-2008, 03:41 PM\nsukatoa\nTake a look at this....\n\nHave some experiments on it,\n\nupdate us,\nsukatoa\n• 04-18-2008, 04:48 AM\nEranga\nFor sorted array you can make a simple binary search as follows. I've try it and seems working fine. But you have to use a sorted array ;)\n\nCode:\n\n```public class BunarySearch {     /**     * @param args the command line arguments     */     public static void main(String[] args) {         // TODO code application logic here         int[] myArray = {1, 3, 5, 6, 8};         System.out.println(\"5 is found at \" + binarySearch(myArray, 8));     }     private static int binarySearch(int[] array, int lookFor) {                 int high = array.length;         int low = -1;         int temp;                 while((high - low) > 1) {             temp = (high + low) >>> 1;             if(array[temp] > lookFor) {                 high = temp;             }             else {                 low = temp;             }         }         if(low == -1 || array[low] != lookFor)             return -1;         else             return low;     } }```\n• 04-18-2008, 05:09 PM\nfireball2008\nI know the binary search, but how to search two arrays at same time without combine the elements together.\n\nI think I got some ideas\n\nfirst pick a value in array A and note down it's index, and binary search in array B.\nand some how the returning index of B + index of A = K, if this is true, then return the larger element is that right?\n• 04-21-2008, 03:44 AM\nEranga\nQuote:\n\nI know the binary search, but how to search two arrays at same time without combine the elements together.\n\nwhy don't you impossible. I mean If you know the binary search, as I do select the comparing value form one array and do the binary search with other array. Do it for the length of the first array.\n\nQuote:\n\nfirst pick a value in array A and note down it's index, and binary search in array B.\nand some how the returning index of B + index of A = K, if this is true, then return the larger element is that right?\n\nYou are adding index of two array elements, then what is variable K?\n• 04-22-2008, 03:21 AM\nfireball2008\nQuote:\n\nwhy don't you impossible. I mean If you know the binary search, as I do select the comparing value form one array and do the binary search with other array. Do it for the length of the first array.\n\nbut would that run in n log n runtime. each array has n elements. I need (log n)^2 runtime." ]

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[ "", null, "Mechanical Energy Printer Friendly Version\nEnergy is defined as the ability to do work and is a scalar quantity. Energy has no direction, only magnitude. Mechanical energy comes in two varieties: kinetic energy (KE) and potential energy (PE).\n\nKinetic Energy\n\nKinetic energy represents the energy caused by an object's motion\n\nKE = ½mv2\n\nwhere v is the object's actual speed, that is, the magnitude of the object's instantaneous resultant velocity. In this formula, m must be measured in kg and v must be measured in m/sec. Note that this collection of units\n\nkg (m/sec)2 = kg m2/sec2 is called a joule (J).\n\nRefer to the following information for the next three questions.\n\nSuppose the skater shown below has a mass of 25 kg and is moving at a speed of 8 m/sec along a level surface.", null, "1. How much KE would he possess?\n\n 2. How would his kinetic energy change if his speed doubled?\n\n 3. How would the kinetic energy of a second skater having twice as much mass but still moving at 8 m/sec compare with the kinetic energy of our original skater?\n\nNotice that the kinetic energy of an object is a scalar quantity whose value is directly proportional to the object's mass and is also directly proportional to the square of the object's velocity.\n\nRefer to the following information for the next question.", null, "4. How do the kinetic energies of these two carts compare?\n\nRefer to the following information for the next question.\n\nThe following three 1-kg projectiles are all released at the same speed, 20 m/sec. However, the first one is released vertically, the second one is released at 37º, and the third one is released horizontally.", null, "5. How do the initial kinetic energies of these three projectiles compare?\n\nGravitational Potential Energy\n\nThis type of potential energy represents the energy an object possesses by virtue of its location in a gravitational field.\n\nPE = mgh\n\nwhere h is the height above an arbitrary zero level that is convenient for the solution of the problem. For instance, the zero level might be assigned as the base of a cliff for a projectile being thrown from the top of the cliff or the zero level might be the floor for a marble rolling off a table onto the ground.\n\nIn this formula, m must be measured in kg, g equals 9.8 m/sec2, and h is in meters. Note that this collection of units\n\nkg (m/sec2)(m) = kg (m2/sec2) is also a joule (J).\n\nThe expression mg represents the objects weight or the force of gravitational attraction between the earth and the object. Forces are measured in newtons. Thus the expression\n\nNm also equals a joule (J).\n\nRefer to the following information for the next two questions.\n\nThe table is 1-meter tall and the apple has a mass of 100 grams.", null, "What is the potential energy of the apple with respect to the top of the table?\n\n What is the potential energy of the apple with respect to the floor at the base of the table?\n\nNotice that the potential energy of an object is a scalar quantity whose value is directly proportional to the object's mass and is also directly proportional to the object's height above an arbitrarily chosen zero level.", null, "Related Documents" ]

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[ "", null, "# RD Sharma Solutions Class 12 Chapter 29 The Plane\n\n## RD Sharma Solutions for Class 12 Maths Chapter 29 – Free PDF Download\n\nThe RD Sharma Solutions for Class 12 Maths Chapter 29 is given here. To achieve an excellent academic score in Class 12 Maths, students should solve the questions of each and every exercise in RD Sharma Class 12 Maths Textbook. RD Sharma Solutions for Class 12 is fabricated by a team of Maths experts to help the students excel in their exams.\n\nRD Sharma Solutions for Class 12 are curated in a stepwise manner in simple language for the ease of understanding by students. Referring to these solutions while solving can strengthen the conceptual knowledge of students. Students can score high marks in the exams by solving all the questions and cross-checking the answers with the RD Sharma Solutions of BYJU’S.\n\n### Highlights of RD Sharma Class 12 Solutions Chapter 29 The Plane\n\nA plane is a flat, 2-dimensional surface having infinite dimensions, but having zero thickness. Such that if any two points are taken on it, the line segment joining them lies completely on the surface. A plane in 3-dimensional space has the following equation\n\nax + by + cz + d = 0,\n\nWhere at least one of the values of a,b,c, must be a non-zero.\n\nThe three possible planes in the 3-D coordinate system are\n\n1. XY plane, where the value of z coordinates is zero.\n2. Y-Z plane, where the value of x coordinate is zero.\n3. X-Z plane, where the value of y coordinate is zero." ]

[ null, "https://www.facebook.com/tr", null ]

[ "# CONFLUENTES MATHEMATICI\n\nA posteriori error estimates for a fully discrete approximation of Sobolev equations\nConfluentes Mathematici, Tome 11 (2019) no. 1, pp. 3-28.\n\nThe paper presents an a posteriori error estimator for a (piecewise linear) conforming finite element approximation of some (linear) Sobolev equations in ${ℝ}^{d}$, $d=2$ or 3, using implicit Euler’s scheme. For this discretization, we derive a residual indicator, which uses a spatial residual indicator based on the jumps of conormal derivatives of the approximations and a time residual indicator based on the jump (in an appropriated norm) of the successive solutions at each time step. Lower and upper bounds are obtained with minimal assumptions on the meshes. Numerical experiments that confirm and illustrate the theoretical results are given.\n\nReçu le :\nRévisé le :\nAccepté le :\nPublié le :\nDOI : https://doi.org/10.5802/cml.53\nClassification : 65M15,  65M50,  65M60\nMots clés : Sobolev equations, a posteriori error analysis\n@article{CML_2019__11_1_3_0,\nauthor = {Serge Nicaise and Fatiha Bekkouche},\ntitle = {A posteriori error estimates for a fully discrete approximation of {Sobolev} equations},\njournal = {Confluentes Mathematici},\npages = {3--28},\npublisher = {Institut Camille Jordan},\nvolume = {11},\nnumber = {1},\nyear = {2019},\ndoi = {10.5802/cml.53},\nlanguage = {en},\nurl = {https://cml.centre-mersenne.org/articles/10.5802/cml.53/}\n}\nNicaise, Serge; Bekkouche, Fatiha. A posteriori error estimates for a fully discrete approximation of Sobolev equations. Confluentes Mathematici, Tome 11 (2019) no. 1, pp. 3-28. doi : 10.5802/cml.53. https://cml.centre-mersenne.org/articles/10.5802/cml.53/\n\n D. N. Arnold, J. Douglas, Jr., and V. Thomée. Superconvergence of a finite element approximation to the solution of a Sobolev equation in a single space variable. Math. Comp., 36(153):53–63, 1981.\n\n G. I. Barenblatt, V. M. Entov, and V. M. Ryzhik. Theory of fluid flow through natural rocks. Dordrecht: Kluwer, 1990.\n\n F. Bekkouche, W. Chikouche, and S. Nicaise. Fully discrete approximation of general nonlinear Sobolev equations. Afr. Mat., Sep 2018.\n\n A. Bergam, C. Bernardi, and Z. Mghazli. A posteriori analysis of the finite element discretization of some parabolic equations. Math. Comp., 74(251):1117–1138, 2005.\n\n C. Bernardi and B. Métivet. Indicateurs d’erreur pour l’équation de la chaleur. Revue Européenne des éléments finis, 9(4):425–438, 2000.\n\n C. Bernardi and R. Verfürth. A posteriori error analysis of the fully discretized time-dependent Stokes equations. M2AN Math. Model. Numer. Anal., 38(3):437–455, 2004.\n\n J. M. Cascón, L. Ferragut, and M. I. Asensio. Space-time adaptive algorithm for the mixed parabolic problem. Numer. Math., 103(3):367–392, 2006.\n\n P. J. Chen and M. E. Gurtin. On a theory of heat conduction involving two temperatures. Z. Angew. Math. Phys., 19:614–627, 1968.\n\n P. Ciarlet. The finite element method for elliptic problems. North Holland, 1996.\n\n P. Clément. Approximation by finite element functions using local regularization. Rev. Française Automat. Informat. Recherche Opérationnelle Sér., 9(R-2):77–84, 1975.\n\n B. D. Coleman and W. Noll. An approximation theorem for functionals, with applications in continuum mechanics. Arch. Rational Mech. Anal., 6:355–370 (1960), 1960.\n\n C. Cuesta, C. J. van Duijn, and J. Hulshof. Infiltration in porous media with dynamic capillary pressure: travelling waves. European J. Appl. Math., 11(4):381–397, 2000.\n\n R. E. Ewing. Time-stepping Galerkin methods for nonlinear Sobolev partial differential equations. SIAM J. Numer. Anal., 15(6):1125–1150, 1978.\n\n S. M. Hassanizadeh and W. G. Gray. Thermodynamic basis of capillary pressure in porous media. Water Resour. Res., 29:858–879, 1993.\n\n P. Houston and E. Süli. Adaptive Lagrange-Galerkin methods for unsteady convection-diffusion problems. Math. Comp., 70(233):77–106, 2001.\n\n C. Johnson, Y. Y. Nie, and V. Thomée. An a posteriori error estimate and adaptive timestep control for a backward Euler discretization of a parabolic problem. SIAM J. Numer. Anal., 27(2):277–291, 1990.\n\n T. Liu, Y.-p. Lin, M. Rao, and J. R. Cannon. Finite element methods for Sobolev equations. J. Comput. Math., 20(6):627–642, 2002.\n\n S. Nicaise and N. Soualem. A posteriori error estimates for a nonconforming finite element discretization of the heat equation. M2AN Math. Model. Numer. Anal., 39(2):319–348, 2005.\n\n M. R. Ohm and H. Y. Lee. ${L}^{2}$-error analysis of fully discrete discontinuous Galerkin approximations for nonlinear Sobolev equations. Bull. Korean Math. Soc., 48(5):897–915, 2011.\n\n M. R. Ohm, H. Y. Lee, and J. Y. Shin. ${L}^{2}$-error estimates of the extrapolated Crank-Nicolson discontinuous Galerkin approximations for nonlinear Sobolev equations. J. Inequal. Appl., pages Art. ID 895187, 17, 2010.\n\n M. Picasso. Adaptive finite elements for a linear parabolic problem. Comput. Methods Appl. Mech. Engrg., 167(3-4):223–237, 1998.\n\n M. Picasso. An anisotropic error indicator based on Zienkiewicz-Zhu error estimator: Application to elliptic and parabolic problems. SIAM J. Sci. Comput., 24(4):1328–1355, 2003.\n\n R. E. Showalter. The Sobolev equation. II. Applicable Anal., 5(2):81–99, 1975.\n\n T. W. Ting. Certain non-steady flows of second-order fluids. Arch. Rational Mech. Anal., 14:1–26, 1963.\n\n T. W. Ting. A cooling process according to two-temperature theory of heat conduction. J. Math. Anal. Appl., 45:23–31, 1974.\n\n T. Tran and T.-B. Duong. A posteriori error estimates with the finite element method of lines for a Sobolev equation. Numer. Methods Partial Differential Equations, 21(3):521–535, 2005.\n\n R. Verfürth. A review of a posteriori error estimation and adaptive mesh-refinement techniques. Wiley-Teubner, Chichester; Stuttgart, 1996.\n\n R. Verfürth. Error estimates for some quasi-interpolation operators. M2AN Math. Model. Numer. Anal., 33(4):695–713, 1999.\n\n R. Verfürth. A posteriori error estimates for finite element discretization of the heat equation. Calcolo, 40:195–212, 2003." ]

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[ "# pcaTransform¶\n\n## Purpose¶\n\nReduces the dimension of a matrix using principal component vectors previously returned by pcaFit().\n\n## Format¶\n\nX_transform = pcaTransform(X, mdl)\nParameters: X (NxP matrix) – The matrix to transform. This matrix must have the same number of columns as the matrix passed to pcaFit(). mdl (pcaModel structure) – This must have been previously been filled in by a call to pcaFit(). X_transform (Nxn_components matrix) – The input matrix projected on the component vectors returned by a previous call to pcaFit().\n\n## Examples¶\n\nnew;\nlibrary gml;\n\n// Get file name with full path\nfname = getGAUSSHome() \\$+ \"pkgs/gml/examples/winequality.csv\";\n\nX_train = X[1:1000,.];\nX_test = X[1001:rows(X),.];\n\nn_components = 3;\n\nstruct pcaModel mdl;\nmdl = pcaFit(X_train, n_components);\n\nX_transform = pcaTransform(X_test, mdl);\n\n\nAfter the above code, the first 5 rows of X_transform will be:\n\n 37.441282 1.2145282 -1.5416867\n-2.0454164 -15.738950 1.0084994\n21.315231 -2.4328631 0.15655108\n41.776957 2.2901582 -2.2804431\n0.73984770 -12.260074 -0.68265628" ]

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[ "Cody\n\n# Problem 43298. Calculate area of sector\n\nSolution 1809237\n\nSubmitted on 7 May 2019 by Gareth Davies\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nr=1 seta=pi/2 y_correct = 0.7854; assert(abs(sectorarea(r,seta)-y_correct)<0.001)\n\nr = 1 seta = 1.5708\n\n2   Pass\nr=2 seta=pi/2 y_correct = pi; assert(abs(sectorarea(r,seta)-y_correct)<0.001)\n\nr = 2 seta = 1.5708" ]

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[ "# 10.1 Wave Basics\n\nWaves generally begin as a disturbance of some kind, and the energy of that disturbance gets propagated in the form of waves. We are most familiar with the kind of waves that break on shore, or rock a boat at sea, but there are many other types of waves that are important to oceanography:\n\n• Internal waves form at the boundaries of of different densities (i.e. at a ), and propagate at depth. These generally move more slowly than surface waves, and can be much larger, with heights exceeding 100 m. However, the height of the deep wave would be unnoticeable at the surface.\n• Tidal waves are due to the movement of the tides. What we think of as tides are basically enormously long waves with a that may span half the globe (see section 11.1). Tidal waves are not related to , so don’t confuse the two.\n• Tsunamis are large waves created as a result of earthquakes or other seismic disturbances. They are also called seismic sea waves (section 10.4).\n• Splash waves are formed when something falls into the ocean and creates a splash. The giant wave in Lituya Bay that was described in the introduction to this chapter was a splash wave.\n• Atmospheric waves form in the sky at the boundary between air masses of different densities. These often create ripple effects in the clouds (Figure 10.1.1).", null, "Figure 10.1.1 Wake patterns in cloud cover over Possession Island, East Island, Ile aux Cochons, Ile de Pingouins. The ripple pattern is a result of internal waves in the atmosphere (NASA [Public domain], via Wikimedia Commons).\n\nThere are several components to a basic wave (Figure 10.1.2):\n\n• Still water level: where the water surface would be if there were no waves present and the sea was completely calm.\n• Crest: the highest point of the wave.\n• Trough: the lowest point of the wave.\n• Wave height: the distance between the crest and the trough.\n• Wavelength: the distance between two identical points on successive waves, for example crest to crest, or trough to trough.\n• Wave steepness: the ratio of wave height to length (H/L). If this ratio exceeds 1/7 (i.e. height exceeds 1/7 of the wavelength) the wave gets too steep, and will break.", null, "Figure 10.1.2 Components of a basic wave (Modified by PW from Steven Earle “Physical Geology”).\n\nThere are also a number of terms used to describe wave motion:\n\n• Period: the time it takes for two successive crests to pass a given point.\n• Frequency: the number of waves passing a point in a given amount of time, usually expressed as waves per second. This is the inverse of the period.\n• Speed: how fast the wave travels, or the distance traveled per unit of time. This is also called (c), where\n\nc = wavelength  x  frequency\n\nTherefore, the longer the , the faster the wave.\n\nAlthough waves can travel over great distances, the water itself shows little horizontal movement; it is the energy of the wave that is being transmitted, not the water. Instead, the water particles move in circular orbits, with the size of the orbit equal to the (Figure 10.1.3). This orbital motion occurs because water waves contain components of both longitudinal (side to side) and transverse (up and down) waves, leading to circular motion. As a wave passes, water moves forwards and up over the wave , then down and backwards into the , so there is little horizontal movement. This is evident if you have ever watched an object such as a seabird floating at the surface. The bird bobs up and down as the wave pass underneath it; it does not get carried horizontally by a single wave crest.", null, "Figure 10.1.3 Animation showing the orbital motion of particles in a surface wave (By Kraaiennest (Own work) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC BY-SA 4.0], via Wikimedia Commons).\n\nThe circular orbital motion declines with depth as the wave has less impact on deeper water and the diameter of the circles is reduced. Eventually at some depth there is no more circular movement and the water is unaffected by surface wave action. This depth is the and is equivalent to half of the (Figure 10.1.4). Since most ocean waves have wavelengths of less than a few hundred meters, most of the deeper ocean is unaffected by surface waves, so even in the strongest storms marine life or submarines can avoid heavy waves by submerging below the wave base.", null, "Figure 10.1.4 Orbital motion of water within a wave, extending down to the wave base at a depth of half of the wavelength (Modified by PW from Steven Earle, “Physical Geology”).\n\nWhen the water below a wave is deeper than the wave base (deeper than half of the wavelength), those waves are called . Most open ocean waves are deep water waves. Since the water is deeper than the wave base, deep water waves experience no interference from the bottom, so their speed only depends on the wavelength:", null, "where g is gravity and L is wavelength in meters. Since g and π are constants, this can be simplified to:", null, "occur when the depth is less than 1/20 of the wavelength. In these cases, the wave is said to “touch bottom” because the depth is shallower than the so the orbital motion is affected by the seafloor. Due to the shallow depth, the orbits are flattened, and eventually the water movement becomes horizontal rather than circular just above the bottom. The speed of shallow water waves depends only on the depth:", null, "where g is gravity and d is depth in meters. This can be simplified to:", null, "or transitional waves are found in depths between ½ and 1/20 of the wavelength. Their behavior is a bit more complex, as their speed is influenced by both wavelength and depth. The speed of an intermediate wave is calculated as:", null, "which contains both depth and wavelength variables.", null, "" ]

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[ "# Simulation time\n\nHi folks -\n\nI was wondering what we can do to increase simulation speed. Its taking ridecolus amount of time to simulate 1000 subjects.\n\n``````Sub = map(i-> Subject(;id=i, events=event, covariates=(dose=dose, wt=70, sex=1,)), 1:1000)\n__sim = simobs(model, Sub, final_estimates,\nsimulate_error=false, obstimes=0:0.01:288)\n``````\n\nThis is the code.\n\n``````\nmodel = @model begin\n@param begin\nθ ∈ VectorDomain(7, lower=zeros(7))\nΩ ∈ PDiagDomain(5)\nσ ∈ RealDomain(;lower=0.001)\nend\n\n@random begin\nη ~ MvNormal(Ω)\nend\n@covariates dose wt\n\n@pre begin\nCLi = exp(log(θ) + 0.75 * log(wt/70) + η)\nVi = exp(log(θ) + log(wt/70) + η)\nkai = exp(log(θ) + η)\nQi = exp(log(θ)+ 0.75 * log(wt/70))\nVpi = exp(log(θ) + log(wt/70) + η)\nend\n\n@dosecontrol begin\nf50i = exp(log(θ) + η)\nfmax = exp(log(θ))\nbio = dose < 50 ? 1 : 1 - fmax * (dose - 50)/ (f50i + (dose - 50) )\n\nbioav = (y1=bio, y2=1, y3 = 1,)\nend\n\n@dynamics begin\ny1' = -kai * y1\ny2' = kai * y1 - (CLi/Vi) * y2 + (Qi/Vpi) * y3 - (Qi/Vi) * y2\ny3' = - (Qi/Vpi) * y3 + (Qi/Vi) * y2\ny4' = y2/Vi\nend\n\n@derived begin\nauc = @. 1_000 * y4\ncp := @. 1_000 * y2 / Vi\ndv ~ @. Normal(cp, cp * σ)\nend\nend\n``````\n\nThe final estimates:\n\n``````(θ = [0.04389707139276508, 7.642248641569024, 0.5063665507090237, 55.583019021299606, 0.7764954718388033, 0.7811025178500393, 6.995697749715079],\nΩ = Diagonal(ones(5) * 0.15),\nσ = 0.01)\n``````\n\nIf I did few subjects, It’s fast, the estimation, is fast, but when simulating this much, it takes more than 30 minutes. Any ideas? I think the issue is coming from `DataFrame`\n\n@mjaber by default the `simobs` is not parallelized by default. Can you just add\n\n``````simobs(...., ensemblealg = EnsembleThreads())\n``````\n\nif you want this to be reproducible, you can also add a random seed within the function. Please check `? simobs` for more information. Let us know if that helps. In the meanwhile, we will test it at our end.\n\nThanks Vijay for your suggestion. I already tried that before and nothing much gained. After I did some testing. It looks like `DataFrame` function when converting simobs output is taking that much time. I think thats where attention should be.\n\nNot sure which `DataFrame` call are you referring to? There is nothing in your code. Also, can you please share your `events` set up for the `Sub`. I just assumed a 100 mg dose using `events = DosageRegimen(100)` and covariate value for `dose = 100` and I get the following results\n\n``````julia> @time sims = simobs(model, Sub, final_estimates,\nsimulate_error=false, obstimes=0:0.01:288)\n135.055649 seconds (965.08 M allocations: 77.680 GiB, 7.06% gc time, 5.82% compilation time)\nSimulated population (Vector{<:Subject})\nSimulated subjects: 1000\nSimulated variables: auc, dv\n\n``````\n\nOpps. my bad. Here you go:\n\n``````dose = 450\nSub = map(i-> Subject(;id=i, events=event, covariates=(dose=dose, wt=70,)), 1:1000)\n@time __sim = simobs(model, Sub, final_estimates,\n@time SIM_450 = DataFrame(__sim)\n``````\n1. The slowdown is not in the simulation as that takes less than 3 minutes (single threaded)\n2. Your simulation is generating 28000 datapoints per subject and you have 1000 subjects resulting in 28 million data points that has to be converted to a dataframe.\n3. 28 million dataframe will obviously take time, but my question to you is if you really need that fine a time grid. Also, what is the purpose of the simulation and what do you want to do as post-processing.\n\nThanks Vijay; The aim is to simulate and get some endpoints including parameters such as cmax, tmax, AUC with different time points (partials) and assess concentrations at specific points. The workflow, is to simulate with this much followed by calculating some statistics. Does that make sense?\n\nYou can use `NCA` within a model to extract them easily. Since you set `simulate_error` to `false`, you can replace the `derived` block with `observed` block and then do the simulation. You can also directly convert a simulation to a `NCAPopulation` and perform NCA. I have a fully reproducible example on just 10 subjects. Go through it and let me know if you have any questions.\n\n``````using Pumas\nusing PumasUtilities\nusing Random\ndose = 450\nevent = DosageRegimen(dose, time=0, ii=24, addl=1, route=NCA.EV)\nSub = map(i -> Subject(; id=i, events=event, covariates=(dose=dose, wt=70,)), 1:10)\n# Sub = map(i -> Subject(; id=i, events=DosageRegimen(100), covariates=(dose=100, wt=70, sex=1,)), 1:1000)\n\n# integrated NCA within the model\nmodel = @model begin\n@param begin\nθ ∈ VectorDomain(7, lower=zeros(7))\nΩ ∈ PDiagDomain(5)\nσ ∈ RealDomain(; lower=0.001)\nend\n\n@random begin\nη ~ MvNormal(Ω)\nend\n@covariates dose wt\n\n@pre begin\nCLi = exp(log(θ) + 0.75 * log(wt / 70) + η)\nVi = exp(log(θ) + log(wt / 70) + η)\nkai = exp(log(θ) + η)\nQi = exp(log(θ) + 0.75 * log(wt / 70))\nVpi = exp(log(θ) + log(wt / 70) + η)\nend\n\n@dosecontrol begin\nf50i = exp(log(θ) + η)\nfmax = exp(log(θ))\nbio = dose < 50 ? 1 : 1 - fmax * (dose - 50) / (f50i + (dose - 50))\n\nbioav = (y1=bio, y2=1, y3=1,)\nend\n\n@dynamics begin\ny1' = -kai * y1\ny2' = kai * y1 - (CLi / Vi) * y2 + (Qi / Vpi) * y3 - (Qi / Vi) * y2\ny3' = -(Qi / Vpi) * y3 + (Qi / Vi) * y2\n#y4' = y2 / Vi\nend\n\n@observed begin\n# auc = @. 1_000 * y4\ncp := @. 1_000 * y2 / Vi\nsimnca := @nca cp\ncmax = NCA.cmax(simnca)\nauc = NCA.auc(simnca)\nauc04 = NCA.auc(simnca, interval=(0, 4))\n# dv ~ @. Normal(cp, cp * σ)\nend\nend\n\nfinal_estimates = (θ=[0.04389707139276508, 7.642248641569024, 0.5063665507090237, 55.583019021299606, 0.7764954718388033, 0.7811025178500393, 6.995697749715079],\nΩ=Diagonal(ones(5) * 0.15),\nσ=0.01)\n#\n@time sims = simobs(model, Sub, final_estimates,\nsimulate_error=false, obstimes=0:0.5:288, rng = Random.seed!(1234))\nsimdf = DataFrame(sims, include_events=false, include_covariates=false)\n#\n\n## converting simulations directly to NCAPopulation\nmodel = @model begin\n@param begin\nθ ∈ VectorDomain(7, lower=zeros(7))\nΩ ∈ PDiagDomain(5)\nσ ∈ RealDomain(; lower=0.001)\nend\n\n@random begin\nη ~ MvNormal(Ω)\nend\n@covariates dose wt\n\n@pre begin\nCLi = exp(log(θ) + 0.75 * log(wt / 70) + η)\nVi = exp(log(θ) + log(wt / 70) + η)\nkai = exp(log(θ) + η)\nQi = exp(log(θ) + 0.75 * log(wt / 70))\nVpi = exp(log(θ) + log(wt / 70) + η)\nend\n\n@dosecontrol begin\nf50i = exp(log(θ) + η)\nfmax = exp(log(θ))\nbio = dose < 50 ? 1 : 1 - fmax * (dose - 50) / (f50i + (dose - 50))\n\nbioav = (y1=bio, y2=1, y3=1,)\nend\n\n@dynamics begin\ny1' = -kai * y1\ny2' = kai * y1 - (CLi / Vi) * y2 + (Qi / Vpi) * y3 - (Qi / Vi) * y2\ny3' = -(Qi / Vpi) * y3 + (Qi / Vi) * y2\nend\n\n@observed begin\ndv = @. 1_000 * y2 / Vi\nend\nend\n\nfinal_estimates = (θ=[0.04389707139276508, 7.642248641569024, 0.5063665507090237, 55.583019021299606, 0.7764954718388033, 0.7811025178500393, 6.995697749715079],\nΩ=Diagonal(ones(5) * 0.15),\nσ=0.01)\n#\n@time sims = simobs(model, Sub, final_estimates,\nsimulate_error=false, obstimes=0:0.5:288, rng = Random.seed!(1234))\nncapop = NCAPopulation(Subject.(sims))\nNCA.cmax(ncapop)\nNCA.auc(ncapop)\nNCA.auc(ncapop, interval=(0,4))\n#\n``````\n\nNote that the results between the two methods are identical and generate results by each dose per subject. So you don’t really have to take these into `DataFrame`s which are always expensive\n\n1 Like\n\nThis is amazing. I will go through your code and let you know if I have any questions/comments. Appreciate the response, Vijay!\n\n@vijay I have had a similar experience in the past. Converting simulations to DataFrame is very slow. In some cases, it was required for me to convert a simulation to a DataFrame for the sake of post processing. My dataset had approximately 6000 subjects and took approximately 10 minutes for this conversion.\n\nFor the case study you shared here, I used your code to simulate 5000 subjects instead of 10. Converting by doing `DataFrame(sims)` took > 1 minute than converting the required columns to a DataFrame on my own that took ~1 second.\n\n``````julia> Sub = map(i -> Subject(; id=i, events=event, covariates=(dose=dose, wt=70,)), 1:5000)\nPopulation\nSubjects: 5000\nCovariates: dose, wt\nObservations:\n\njulia> @time sims = simobs(model, Sub, final_estimates,\nsimulate_error=false, obstimes=0:0.5:288, rng = Random.seed!(1234))\n12.803235 seconds (87.68 M allocations: 7.893 GiB, 13.31% gc time)\nSimulated population (Vector{<:Subject})\nSimulated subjects: 5000\nSimulated variables: cmax, auc, auc04\n\njulia> @time simdf = DataFrame(sims, include_events=false, include_covariates=false)\n75.476109 seconds (173.89 M allocations: 221.364 GiB, 6.89% gc time)\n2885000×19 DataFrame\nRow │ id time cmax auc auc04 evid η_1 η_2 η_3 η_4 η_5 y1 y2 y3 CLi Vi ⋯\n│ String? Float64 Float64? Float64? Float64? Int64? Float64? Float64? Float64? Float64? Float64? Float64? Float64? Float64? Float64? Float ⋯\n─────────┼─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────\n1 │ 1 0.0 9743.63 1.37404e6 25207.5 0 -0.139322 0.421074 -0.162506 0.278433 0.162761 154.772 0.0 0.0 0.0381881 11.64 ⋯\n2 │ 1 0.5 9743.63 1.37404e6\n\njulia> @time sims_df = vcat(map(sims) do z\nobs = DataFrame(id = z.subject.id, cmax = z.observations.cmax, auc = z.observations.auc, auc04 = z.observations.auc04)\ncovs = DataFrame([(z.subject.covariates.u..., id = z.subject.id)])\neves = vcat(DataFrame.(z.subject.events)...)\nobs_eves = hcat(obs, eves)\nsubj_sims = outerjoin(covs, obs_eves, on=[:id])\nend...)\n0.966107 seconds (5.02 M allocations: 327.896 MiB, 30.04% compilation time)\n``````\n1 Like\n\nOne additional tip that is not the bottleneck here but might be a good thing to know in the future, especially if you are doing heavy `map` stuff (such as in tons of simulations).\n\nYou can use the `ThreadsX.map` instead of `map`. It will run a multithreaded parallel `map`.\n\n2 Likes\n\nThis is really helpful folks! Appreciate that.\n\nSorry for being late to the party, but if any of you could try to replicate this on v2.4 (the latest version of Pumas) I would appreciate it. On an internal project we found a similar bottleneck, and realized that we could improve the DataFrame constructor significantly, so now it should run much faster.\n\n1 Like" ]

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[ "# Number 13209\n\nNumber 13,209 spell 🔊, write in words: thirteen thousand, two hundred and nine . Ordinal number 13209th is said 🔊 and write: thirteen thousand, two hundred and ninth. The meaning of number 13209 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 13209. What is 13209 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 13209.\n\n## What is 13,209 in other units\n\nThe decimal (Arabic) number 13209 converted to a Roman number is (X)MMMCCIX. Roman and decimal number conversions.\n\n#### Weight conversion\n\n13209 kilograms (kg) = 29120.6 pounds (lbs)\n13209 pounds (lbs) = 5991.6 kilograms (kg)\n\n#### Length conversion\n\n13209 kilometers (km) equals to 8208 miles (mi).\n13209 miles (mi) equals to 21258 kilometers (km).\n13209 meters (m) equals to 43337 feet (ft).\n13209 feet (ft) equals 4027 meters (m).\n13209 centimeters (cm) equals to 5200.4 inches (in).\n13209 inches (in) equals to 33550.9 centimeters (cm).\n\n#### Temperature conversion\n\n13209° Fahrenheit (°F) equals to 7320.6° Celsius (°C)\n13209° Celsius (°C) equals to 23808.2° Fahrenheit (°F)\n\n#### Time conversion\n\n(hours, minutes, seconds, days, weeks)\n13209 seconds equals to 3 hours, 40 minutes, 9 seconds\n13209 minutes equals to 1 week, 2 days, 4 hours, 9 minutes\n\n### Zip codes 13209\n\n• Zip code 13209 Zapotitlán, Ciudad de México, Tláhuac, Mexico a map\n\n### Codes and images of the number 13209\n\nNumber 13209 morse code: .---- ...-- ..--- ----- ----.\nSign language for number 13209:", null, "", null, "", null, "", null, "", null, "Number 13209 in braille:", null, "Images of the number\nImage (1) of the numberImage (2) of the number", null, "", null, "More images, other sizes, codes and colors ...\n\n#### Number 13209 infographic", null, "## Share in social networks", null, "## Mathematics of no. 13209\n\n### Multiplications\n\n#### Multiplication table of 13209\n\n13209 multiplied by two equals 26418 (13209 x 2 = 26418).\n13209 multiplied by three equals 39627 (13209 x 3 = 39627).\n13209 multiplied by four equals 52836 (13209 x 4 = 52836).\n13209 multiplied by five equals 66045 (13209 x 5 = 66045).\n13209 multiplied by six equals 79254 (13209 x 6 = 79254).\n13209 multiplied by seven equals 92463 (13209 x 7 = 92463).\n13209 multiplied by eight equals 105672 (13209 x 8 = 105672).\n13209 multiplied by nine equals 118881 (13209 x 9 = 118881).\nshow multiplications by 6, 7, 8, 9 ...\n\n### Fractions: decimal fraction and common fraction\n\n#### Fraction table of 13209\n\nHalf of 13209 is 6604,5 (13209 / 2 = 6604,5 = 6604 1/2).\nOne third of 13209 is 4403 (13209 / 3 = 4403).\nOne quarter of 13209 is 3302,25 (13209 / 4 = 3302,25 = 3302 1/4).\nOne fifth of 13209 is 2641,8 (13209 / 5 = 2641,8 = 2641 4/5).\nOne sixth of 13209 is 2201,5 (13209 / 6 = 2201,5 = 2201 1/2).\nOne seventh of 13209 is 1887 (13209 / 7 = 1887).\nOne eighth of 13209 is 1651,125 (13209 / 8 = 1651,125 = 1651 1/8).\nOne ninth of 13209 is 1467,6667 (13209 / 9 = 1467,6667 = 1467 2/3).\nshow fractions by 6, 7, 8, 9 ...\n\n### Calculator\n\n 13209\n\n#### Is Prime?\n\nThe number 13209 is not a prime number. The closest prime numbers are 13187, 13217.\n\n#### Factorization and factors (dividers)\n\nThe prime factors of 13209 are 3 * 7 * 17 * 37\nThe factors of 13209 are\n1 , 3 , 7 , 17 , 21 , 37 , 51 , 111 , 119 , 259 , 357 , 629 , 777 , 1887 , 4403 , 13209\nTotal factors 16.\nSum of factors 21888 (8679).\n\n#### Powers\n\nThe second power of 132092 is 174.477.681.\nThe third power of 132093 is 2.304.675.688.329.\n\n#### Roots\n\nThe square root √13209 is 114,930414.\nThe cube root of 313209 is 23,638685.\n\n#### Logarithms\n\nThe natural logarithm of No. ln 13209 = loge 13209 = 9,488654.\nThe logarithm to base 10 of No. log10 13209 = 4,12087.\nThe Napierian logarithm of No. log1/e 13209 = -9,488654.\n\n### Trigonometric functions\n\nThe cosine of 13209 is -0,172816.\nThe sine of 13209 is 0,984954.\nThe tangent of 13209 is -5,699438.\n\n### Properties of the number 13209\n\nIs a Friedman number: No\nIs a Fibonacci number: No\nIs a Bell number: No\nIs a palindromic number: No\nIs a pentagonal number: No\nIs a perfect number: No\n\n## Number 13209 in Computer Science\n\nCode typeCode value\n13209 Number of bytes12.9KB\nUnix timeUnix time 13209 is equal to Thursday Jan. 1, 1970, 3:40:09 a.m. GMT\nIPv4, IPv6Number 13209 internet address in dotted format v4 0.0.51.153, v6 ::3399\n13209 Decimal = 11001110011001 Binary\n13209 Decimal = 200010020 Ternary\n13209 Decimal = 31631 Octal\n13209 Decimal = 3399 Hexadecimal (0x3399 hex)\n13209 BASE64MTMyMDk=\n13209 MD526a95b3bf6c0fa4ba909250facfb5ae9\n13209 SHA1b2cfe3c5970591260dba2f654dc52e4aa3500100\n13209 SHA224c8503cb18d6b0332ef614a1bdc046ee0b381906c55784a1946d17305\n13209 SHA25652be9ccfbf9794f5ffa09599b506539f1ec1a9106c07393333f9e07cdd2eac42\n13209 SHA384ee59e32fa840eb53d190dc379a7d75124a6afe0d6636a75a23864ba37234524c8fa097bd42395a62c627519ae3309083\nMore SHA codes related to the number 13209 ...\n\nIf you know something interesting about the 13209 number that you did not find on this page, do not hesitate to write us here.\n\n## Numerology 13209\n\n### Character frequency in number 13209\n\nCharacter (importance) frequency for numerology.\n Character: Frequency: 1 1 3 1 2 1 0 1 9 1\n\n### Classical numerology\n\nAccording to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 13209, the numbers 1+3+2+0+9 = 1+5 = 6 are added and the meaning of the number 6 is sought.\n\n## Interesting facts about the number 13209\n\n### Asteroids\n\n• (13209) Arnhem is asteroid number 13209. It was discovered by E. W. Elst from La Silla Observatory on 4/9/1997.\n\n### Distances between cities\n\n• There is a 8,208 miles (13,209 km) direct distance between Ashgabat (Turkmenistan) and Zapopan (Mexico).\n• There is a 8,208 miles (13,209 km) direct distance between Bhubaneshwar (India) and Columbus (USA).\n• There is a 8,208 miles (13,209 km) direct distance between Freetown (Sierra Leone) and Guangzhou (China).\n• There is a 8,208 miles (13,209 km) direct distance between Mexico City (Mexico) and Qom (Iran).\n• There is a 8,208 miles (13,209 km) direct distance between Naucalpan de Juárez (Mexico) and Qom (Iran).\n• There is a 8,208 miles (13,209 km) direct distance between Solāpur (India) and Teresina (Brazil).\n\n## Number 13,209 in other languages\n\nHow to say or write the number thirteen thousand, two hundred and nine in Spanish, German, French and other languages. The character used as the thousands separator.\n Spanish: 🔊 (número 13.209) trece mil doscientos nueve German: 🔊 (Anzahl 13.209) dreizehntausendzweihundertneun French: 🔊 (nombre 13 209) treize mille deux cent neuf Portuguese: 🔊 (número 13 209) treze mil, duzentos e nove Chinese: 🔊 (数 13 209) 一万三千二百零九 Arabian: 🔊 (عدد 13,209) ثلاثة عشر ألفاً و مئتان و تسعة Czech: 🔊 (číslo 13 209) třináct tisíc dvěstě devět Korean: 🔊 (번호 13,209) 만 삼천이백구 Danish: 🔊 (nummer 13 209) trettentusinde og tohundrede og ni Dutch: 🔊 (nummer 13 209) dertienduizendtweehonderdnegen Japanese: 🔊 (数 13,209) 一万三千二百九 Indonesian: 🔊 (jumlah 13.209) tiga belas ribu dua ratus sembilan Italian: 🔊 (numero 13 209) tredicimiladuecentonove Norwegian: 🔊 (nummer 13 209) tretten tusen, to hundre og ni Polish: 🔊 (liczba 13 209) trzynaście tysięcy dwieście dziewięć Russian: 🔊 (номер 13 209) тринадцать тысяч двести девять Turkish: 🔊 (numara 13,209) onüçbinikiyüzdokuz Thai: 🔊 (จำนวน 13 209) หนึ่งหมื่นสามพันสองร้อยเก้า Ukrainian: 🔊 (номер 13 209) тринадцять тисяч двiстi дев'ять Vietnamese: 🔊 (con số 13.209) mười ba nghìn hai trăm lẻ chín Other languages ...\n\n## News to email\n\nPrivacy Policy.\n\n## Comment\n\nIf you know something interesting about the number 13209 or any natural number (positive integer) please write us here or on facebook." ]

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[ "MiniMax Algorithm\n\nHello people! In this post we will look at one of the most basic Artificial Intelligence algorithm, the MiniMax algorithm. MiniMax algorithm is used to implement basic AI or game logic in 2 player games. The most common scenario is implementing a perfect Tic-Tac-Toe player. So, in this article we will look at how to implement it.\n\nDefinition –\n\nGiven that two players are playing a game optimally (playing to win), MiniMax algorithm tells you what is the best move that a player should pick at any state of the game. So, the input to MiniMax algorithm would be –\n\n1. State of the game.\n2. Whose turn it is.\n\nAnd the output would be the best move that can be played by the player given in the input.\n\nIdea of MiniMax Algorithm –\n\nIn the game of Tic-Tac-Toe, there are two players, player X and player O. Now imagine there’s a scoreboard which displays a huge number called “score”, and –\n\n1. If X wins, the score increases by 10.\n2. If O wins, the score is decreased by 10.\n3. If it is a draw, then the score remains unchanged.\n\nSo now, the bigger the number score has, the better it is for X, which means X will try to maximize score as much as possible. Similarly, the lesser the score, the better it is for O, so O will try to minimize the score as much as possible.\n\nTo understand this better, imagine you are player X and this is the current state of your game –", null, "So if you are in state 1 (from the above diagram), you have 3 possible moves which lead to state 2, 3 and 4. For these moves the scores are –\n\n• +10 if you choose state 2.\n• 0 if you choose state 3 because it will be a draw if Player O is playing optimally.\n• -10 if you choose state 3.\n\nSo conceptually we know that player X must choose the winning move. This is done programmatically bu choosing the move which will return the maximum score. So, X will always try to maximize the score and will always choose that move which will fetch X the maximum score. Thus, in the above scenario X chooses the move which goes to state 2.\n\nNow, to make sure you understand both sides of this algorithm. Let us take the same state as above and let us say it is O’s turn now and not X’s. Can you draw a state diagram which will depict all the possible moves and the scores which will be returned by playing them? You can! Just pause here for a moment and try to draw the state diagram for the above game, assuming it is player O’s move.\n\nYou should get a diagram like this –", null, "Did you get a diagram like this? Good job if you did 😉 . Player O has 3 possible moves and the scores are –\n\n• -10 if you choose to go to state 2.\n• -10 if you choose to go to state 3.\n• +10 if you choose to go to state 4.\n\nPlayer O will always try to minimize the score, so player O must select a move which will either lead to state 2 or 3 to win.\n\nWriting code for MiniMax algorithm –\n\nWriting code for MiniMax algorithm is not very difficult, but you may not get it in the first try so I’ll help you out. Firstly, have a clarity on the smaller pieces of logic and write methods for them first. You will need these 3 helper methods for your code –\n\n1. printGame(game) – which prints the state of Tic-Tac-Toe game.\n2. hasPlayerWon(game, player) – which tells if the given player has won the given Tic-Tac-Toe game.\n3. score(game) – which returns +10 if player X has won, -10 if player Y has won, 0 otherwise.\n\nSo now you have a little clarity over the smaller pieces, code them first. Now, you are ready to write the MiniMax algorithm method. It is a recursive method which takes in 2 inputs –\n\n• the state of the game\n• information on which player’s turn it is.\n\nIt will need to return –\n\n• max/min score which can be achieved for the given player for the given game state.\n• best move which can be played by given player.\n\nWe can make a new class to return all the information we need. So, our MiniMax algorithm will look like –\n\n(score, move) maxTurn(game):\nif game is in terminal state:\nreturn (score(game), none)\n\nmax = (none, none)\n\nforeach emptySpace in game:\ngame[emptySpace] = X\ncurrentMove = minTurn(game)\n\nif currentMove.score > max.score:\nmax = (currentMove.score, emptySpace)\n\ngame[emptySpace] = none // reverting change\n\nreturn max\n\n(score, move) minTurn(game):\nif game is in terminal state:\nreturn (score(game), none)\n\nmin = (none, none)\n\nforeach emptySpace in game:\ngame[emptySpace] = O\ncurrentMove = maxTurn(game)\n\nif currentMove.score < min.score:\nmin = (currentMove.score, emptySpace)\n\ngame[emptySpace] = none // reverting change\n\nreturn min\n\nWith your new clarity over the helper methods and the pseudocode, try to write the code for MiniMax algorithm. When in doubt come back and read the MiniMax algorithm theory and the implementation tips. Remember, you are trying to write the code for an ideal Tic-Tac-Toe player here, so you need to write the starter code for simulating a 2-player Tic-Tac-Toe game.\n\nIdeal player doesn’t give up!\n\nThere’s just one thing lacking in our algorithm now. Take the case given below –", null, "So for the above case, player O will lose no matter the decision taken. If you think about this, if we apply the Minimax algorithm we formed so far, in this case Player O would choose state 2. This is because state 2 is the first state it programmatically encounters while computing the minimum value.\n\nBut this doesn’t seem right. Ideally, we would want Player O to go for state 5 because that’s what an ideal player would do. An ideal player would choose that move in which he/she would loose the game in 3/4 turns rather than just the next turn.\n\nHow do we implement this in our algorithm? We can add another parameter “depth” to our algorithm and decay the score by the factor of depth. So, if a player wins by taking more turns, the gain would be lesser and if the player took lesser turns, the gain would be more.\n\nIf we apply this “decay score by level” logic to the above example, the case would look like this –", null, "Now our player O will obviously choose state 5. Now our TicTacToe bot is an ideal bot! The pseudo-code now would look something like this –\n\n(score, move) maxTurn(game, depth):\nif game is in terminal state:\nreturn (score(game, depth), none)\n\nmax = (none, none)\n\nforeach emptySpace in game:\ngame[emptySpace] = X\ncurrentMove = minTurn(game, depth + 1)\n\nif currentMove.score > max.score:\nmax = (currentMove.score, emptySpace)\n\ngame[emptySpace] = none // reverting change\n\nreturn max\n\n(score, move) minTurn(game, depth):\nif game is in terminal state:\nreturn (score(game, depth), none)\n\nmin = (none, none)\n\nforeach emptySpace in game:\ngame[emptySpace] = O\ncurrentMove = maxTurn(game, depth + 1)\n\nif currentMove.score < min.score:\nmin = (currentMove.score, emptySpace)\n\ngame[emptySpace] = none // reverting change\n\nreturn min\n\nint score(game, depth):\nif X has won:\nreturn 10 - depth\nelse if O has won:\nreturn depth - 10\n\nreturn 0\n\nTry to implement the ideal TicTacToe bot on your won, it will be fun to create a game. If you get stuck, you can refer to my code –\n\nLink to code – GitHub\n\nCongrats on taking your first step to build your dream AI! 😀 Keep practicing! Happy Coding! 😀" ]

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[ "# Mental Math for Juniors\n\n## COURSE DETAILS\n\nMath Practice is the 5th level of our Mental Math curriculum for kids to teach them mathematics in fun and engaging way. It will not only improve the mathematical skills of your kids but also help them improves them their grades in school.\n\nThe course will have many math activities for kids to solve math problem using different math games.\n\nStudent Learning Outcomes:\n• Addition and Subtraction on Fingers with all formulas.\n• Addition and Subtraction on Abacus with all formulas.\n• Working with House of 5 Formulae for Subtraction.\n• Application of Multiple Formulae.\n• Speed calculation\n• Imaginary Calculation\n\nRequirements:\n• Pen and Paper\n• Soroban Abacus\n• Internet\n• Laptop/desktop\n• Pre-request Course: NA\n\nNext Course:\nStudents will be able to take our next course “Mental Math for Seniors”\n\n## Course Outline Math Practice : Mental Math Level 5\n\nFollowing is the course outline\n\nIntroduction To House Of 10 Subtraction -9 Formula\n\nApplication Of -8\n\nApplication Of -7\n\nApplication Of -9, -8 And -7\n\nMath Practice\n\nApplication Of -6\n\nApplication Of -5 And Assessment Of -9 Till -5 Formulas\n\nApplication Of -4\n\nApplication Of -3\n\nApplication Of -2 And -1\n\nAll Subtraction Formulas For House Of 10\n\nMath Activities for Kids\n\nCompetition\n\nCertificate distribution" ]

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[ "apollo_mixConditionals {apollo} R Documentation\n\n## Calculates conditionals for continuous mixture models\n\n### Description\n\nCalculates posterior expected values (conditionals) of continuously distributed random coefficients, as well as their standard deviations.\n\n### Usage\n\napollo_mixConditionals(model, apollo_probabilities, apollo_inputs)\n\n\n### Arguments\n\n model Model object. Estimated model object as returned by function apollo_estimate. apollo_probabilities Function. Returns probabilities of the model to be estimated. Must receive three arguments: apollo_beta: Named numeric vector. Names and values of model parameters. apollo_inputs: List containing options of the model. See apollo_validateInputs. functionality: Character. Can be either \"components\", \"conditionals\", \"estimate\" (default), \"gradient\", \"output\", \"prediction\", \"preprocess\", \"raw\", \"report\", \"shares_LL\", \"validate\" or \"zero_LL\". apollo_inputs List grouping most common inputs. Created by function apollo_validateInputs.\n\n### Details\n\nThis functions is only meant for use with continuous distributions\n\n### Value\n\nList of matrices. Each matrix has dimensions nIndiv x 3. One matrix per random component. Each row of each matrix contains the indivID of an individual, and the posterior mean and s.d. of this random component for this individual\n\n[Package apollo version 0.2.8 Index]" ]

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[ "## Tuesday, January 21, 2020\n\n### matplotlib and pyplot\n\nStartup commands\nFirst, let’s start IPython. It is a most excellent enhancement to the standard Python prompt, and it ties in especially well with Matplotlib. Start IPython either at a shell, or the IPython Notebook now.\nWith IPython started, we now need to connect to a GUI event loop. This tells IPython where (and how) to display plots. To connect to a GUI loop, execute the %matplotlib magic at your IPython prompt. There’s more detail on exactly what this does at IPython’s documentation on GUI event loops.\nIf you’re using IPython Notebook, the same commands are available, but people commonly use a specific argument to the %matplotlib magic:\n\n```In : %matplotlib inline\n```\nThis turns on inline plotting, where plot graphics will appear in your notebook. This has important implications for interactivity. For inline plotting, commands in cells below the cell that outputs a plot will not affect the plot. For example, changing the color map is not possible from cells below the cell that creates a plot. However, for other backends, such as qt4, that open a separate window, cells below those that create the plot will change the plot - it is a live object in memory.\nThis tutorial will use matplotlib’s imperative-style plotting interface, pyplot. This interface maintains global state, and is very useful for quickly and easily experimenting with various plot settings. The alternative is the object-oriented interface, which is also very powerful, and generally more suitable for large application development. If you’d like to learn about the object-oriented interface, a great place to start is our FAQ on usage. For now, let’s get on with the imperative-style approach:\n```In : import matplotlib.pyplot as plt\nIn : import matplotlib.image as mpimg\nIn : import numpy as np\n```\n\n## Importing image data into Numpy arrays\n\nLoading image data is supported by the Pillow library. Natively, matplotlib only supports PNG images. The commands shown below fall back on Pillow if the native read fails.\nThe image used in this example is a PNG file, but keep that Pillow requirement in mind for your own data.\nHere’s the image we’re going to play with:", null, "It’s a 24-bit RGB PNG image (8 bits for each of R, G, B). Depending on where you get your data, the other kinds of image that you’ll most likely encounter are RGBA images, which allow for transparency, or single-channel grayscale (luminosity) images. You can right click on it and choose “Save image as” to download it to your computer for the rest of this tutorial.\nAnd here we go...\n```In : img=mpimg.imread('stinkbug.png')\nOut:\narray([[[ 0.40784314, 0.40784314, 0.40784314],\n[ 0.40784314, 0.40784314, 0.40784314],\n[ 0.40784314, 0.40784314, 0.40784314],\n...,\n[ 0.42745098, 0.42745098, 0.42745098],\n[ 0.42745098, 0.42745098, 0.42745098],\n[ 0.42745098, 0.42745098, 0.42745098]],\n\n...,\n[[ 0.44313726, 0.44313726, 0.44313726],\n[ 0.4509804 , 0.4509804 , 0.4509804 ],\n[ 0.4509804 , 0.4509804 , 0.4509804 ],\n...,\n[ 0.44705883, 0.44705883, 0.44705883],\n[ 0.44705883, 0.44705883, 0.44705883],\n[ 0.44313726, 0.44313726, 0.44313726]]], dtype=float32)\n```\nNote the dtype there - float32. Matplotlib has rescaled the 8 bit data from each channel to floating point data between 0.0 and 1.0. As a side note, the only datatype that Pillow can work with is uint8. Matplotlib plotting can handle float32 and uint8, but image reading/writing for any format other than PNG is limited to uint8 data. Why 8 bits? Most displays can only render 8 bits per channel worth of color gradation. Why can they only render 8 bits/channel? Because that’s about all the human eye can see. More here (from a photography standpoint): Luminous Landscape bit depth tutorial.\nEach inner list represents a pixel. Here, with an RGB image, there are 3 values. Since it’s a black and white image, R, G, and B are all similar. An RGBA (where A is alpha, or transparency), has 4 values per inner list, and a simple luminance image just has one value (and is thus only a 2-D array, not a 3-D array). For RGB and RGBA images, matplotlib supports float32 and uint8 data types. For grayscale, matplotlib supports only float32. If your array data does not meet one of these descriptions, you need to rescale it.\n\n## Plotting numpy arrays as images\n\nSo, you have your data in a numpy array (either by importing it, or by generating it). Let’s render it. In Matplotlib, this is performed using the `imshow()` function. Here we’ll grab the plot object. This object gives you an easy way to manipulate the plot from the prompt.\n```In : imgplot = plt.imshow(img)\n```", null, "You can also plot any numpy array.\n\n### Applying pseudocolor schemes to image plots\n\nPseudocolor can be a useful tool for enhancing contrast and visualizing your data more easily. This is especially useful when making presentations of your data using projectors - their contrast is typically quite poor.\nPseudocolor is only relevant to single-channel, grayscale, luminosity images. We currently have an RGB image. Since R, G, and B are all similar (see for yourself above or in your data), we can just pick one channel of our data:\n```In : lum_img = img[:,:,0]\n```\nThis is array slicing. You can read more in the Numpy tutorial.\n```In : plt.imshow(lum_img)\n```", null, "Now, with a luminosity (2D, no color) image, the default colormap (aka lookup table, LUT), is applied. The default is called viridis. There are plenty of others to choose from.\n```In : plt.imshow(lum_img, cmap=\"hot\")\n```", null, "Note that you can also change colormaps on existing plot objects using the `set_cmap()` method:\n```In : imgplot = plt.imshow(lum_img)\nIn : imgplot.set_cmap('nipy_spectral')\n```", null, "Note\nHowever, remember that in the IPython notebook with the inline backend, you can’t make changes to plots that have already been rendered. If you create imgplot here in one cell, you cannot call set_cmap() on it in a later cell and expect the earlier plot to change. Make sure that you enter these commands together in one cell. plt commands will not change plots from earlier cells.\nThere are many other colormap schemes available. See the list and images of the colormaps.\n\n### Color scale reference\n\nIt’s helpful to have an idea of what value a color represents. We can do that by adding color bars.\n```In : imgplot = plt.imshow(lum_img)\nIn : plt.colorbar()\n```", null, "This adds a colorbar to your existing figure. This won’t automatically change if you change you switch to a different colormap - you have to re-create your plot, and add in the colorbar again.\n\n### Examining a specific data range\n\nSometimes you want to enhance the contrast in your image, or expand the contrast in a particular region while sacrificing the detail in colors that don’t vary much, or don’t matter. A good tool to find interesting regions is the histogram. To create a histogram of our image data, we use the `hist()`function.\n```In : plt.hist(lum_img.ravel(), bins=256, range=(0.0, 1.0), fc='k', ec='k')\n```", null, "Most often, the “interesting” part of the image is around the peak, and you can get extra contrast by clipping the regions above and/or below the peak. In our histogram, it looks like there’s not much useful information in the high end (not many white things in the image). Let’s adjust the upper limit, so that we effectively “zoom in on” part of the histogram. We do this by passing the clim argument to imshow. You could also do this by calling the `set_clim()` method of the image plot object, but make sure that you do so in the same cell as your plot command when working with the IPython Notebook - it will not change plots from earlier cells.\n```In : imgplot = plt.imshow(lum_img, clim=(0.0, 0.7))\n```", null, "### Array Interpolation schemes\n\nInterpolation calculates what the color or value of a pixel “should” be, according to different mathematical schemes. One common place that this happens is when you resize an image. The number of pixels change, but you want the same information. Since pixels are discrete, there’s missing space. Interpolation is how you fill that space. This is why your images sometimes come out looking pixelated when you blow them up. The effect is more pronounced when the difference between the original image and the expanded image is greater. Let’s take our image and shrink it. We’re effectively discarding pixels, only keeping a select few. Now when we plot it, that data gets blown up to the size on your screen. The old pixels aren’t there anymore, and the computer has to draw in pixels to fill that space.\nWe’ll use the Pillow library that we used to load the image also to resize the image.\n```In : from PIL import Image\nIn : img = Image.open('../_static/stinkbug.png')\nIn : img.thumbnail((64, 64), Image.ANTIALIAS) # resizes image in-place\nIn : imgplot = plt.imshow(img)\n```", null, "Here we have the default interpolation, bilinear, since we did not give `imshow()` any interpolation argument.\nLet’s try some others:\n```In : imgplot = plt.imshow(img, interpolation=\"nearest\")\n```", null, "```In : imgplot = plt.imshow(img, interpolation=\"bicubic\")\n```", null, "Bicubic interpolation is often used when blowing up photos - people tend to prefer blurry over pixelated." ]

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[ "# skrf.network.concat_ports¶\n\nskrf.network.concat_ports(ntwk_list, port_order='second', *args, **kw)[source]\n\nConcatenate networks along the port axis\n\nNote\n\nThe port_order =’first’, means front-to-back, while port_order`=’second’ means left-to-right. So, for example, when concatenating two 2-networks, `A and B, the ports are ordered as follows:\n\n‘first’\n\na0 o—o a1 -> 0 o—o 1 b0 o—o b1 -> 2 o—o 3\n\n‘second’\n\na0 o—o a1 -> 0 o—o 2 b0 o—o b1 -> 1 o—o 3\n\nuse Network.renumber to change port ordering.\n\nParameters\n• ntwk_list (list of skrf.Networks) – ntwks to concatenate\n\n• port_order (['first', 'second']) –\n\nReturn type\n\nskrf.network.Network\n\nExamples\n\n>>>concat([ntwkA,ntwkB]) >>>concat([ntwkA,ntwkB,ntwkC,ntwkD], port_order=’second’)\n\nTo put for lines in parallel >>> from skrf import air >>> l1 = air.line(100, z0=[0,1]) >>> l2 = air.line(300, z0=[2,3]) >>> l3 = air.line(400, z0=[4,5]) >>> l4 = air.line(400, z0=[6,7]) >>> concat_ports([l1,l2,l3,l4], port_order=’second’)\n\n`stitch`\n`renumber`" ]

[ null ]

[ "# Hypothesis testing\n\nA hypothesis test can be conducted as a two-sided or one-sided test.\n\n##### Two-sided test:\n• $H_{0}=\\psi (\\theta )={\\psi _{0}}$", null, "• $H_{1}=\\psi (\\theta )\\neq {\\psi _{0}}$", null, "Location Normal model:\n\n$P_{\\mu _{0}}(|{{{\\bar {x}}-\\mu _{0}} \\over {\\sigma _{0}/{\\sqrt {n}}}}|\\geq |{{{\\bar {x}}_{obs}-\\mu _{0}} \\over {\\sigma _{0}/{\\sqrt {n}}}}|)=2[1-\\phi ({{{\\bar {x}}_{obs}-\\mu _{0}} \\over {\\sigma _{0}/{\\sqrt {n}}}})]$", null, "Location Scale Normal model:\n\n$t_{obs}={{{\\bar {x}}_{obs}-\\mu _{0}} \\over {s_{obs}/{\\sqrt {n}}}}$", null, "$P_{(\\mu _{0},\\sigma )}(|T|\\geq |t_{obs}|)=2[1-G(|t_{obs};n-1)]$", null, "Bernoulli model:\n\n$P_{\\theta _{0}}=({{{\\bar {x}}-{\\theta _{0}}} \\over {\\sqrt {{{{\\theta }_{0}}(1-{{\\theta }_{0}})} \\over {n}}}}\\geq {{{\\bar {x}}_{obs}-{\\theta _{0}}} \\over {\\sqrt {{{{\\theta }_{0}}(1-{{\\theta }_{0}})} \\over {n}}}})=2[1-\\phi ({{{\\bar {x}}_{obs}-{\\theta _{0}}} \\over {\\sqrt {{{{\\theta }_{0}}(1-{{\\theta }_{0}})} \\over {n}}}})]$", null, "##### One-sided test:\n\nThe one-sided test can be right-tailed or left-tailed:\n\n• right-tailed: $H_{0}=\\psi (\\theta )\\leq {\\psi _{0}}$", null, ";$H_{1}=\\psi (\\theta )>{\\psi _{0}}$", null, "• left-tailed: $H_{0}=\\psi (\\theta )\\geq {\\psi _{0}}$", null, ";$H_{1}=\\psi (\\theta )<{\\psi _{0}}$", null, "As an example: Location Normal model:\n\n$\\max _{\\mu \\leq \\mu _{0}}P_{\\mu }(z\\geq {z}_{obs})=P_{\\mu _{0}}(z\\geq {z}_{obs})=1-\\phi (z_{obs})=1-\\phi ({{x_{obs}-\\mu _{0}} \\over {\\sigma _{0}/{\\sqrt {n}}}})$", null, "" ]

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[ "Kinematics\n\n# Kinematics\n\n## Overview and Key Concepts\n\nFlexSim's kinematics functionality allows you to have a single object perform several travel operations through one common interface. Each travel operation can have its own acceleration, deceleration, startspeed, endspeed, and maximum speed properties. Travel operations can overlap with each other, or be performed in sequence.\n\nBelow is a time-based plot of two kinematics performed in sequence:", null, "And a time-based plot of two kinematics that overlap each other:", null, "Kinematics can also be used just to make it easier to handle the math for any logic that involves speeds, rates, accelerations, decelerations, etc.\n\nAs an example, an overhead crane usually has several motors that drive it. One motor drives the bridge along a railing, while another motor drives a trolley along the bridge, while another lifts the hook or grabber by a cable. Each of these motors may have their own acceleration, deceleration, and maximum speed properties. Using kinematics, you can define all of these motions through a single kinematics interface, where different motors can work simultaneously, giving the motion of the crane's end effector a very dynamic behavior. Before kinematics were introduced, the simplest way to simulate this behavior was to have three different objects hierarchically ordered in the model tree, each object simulating one motion or kinematic. While this works very well in some cases, in other instances it can become tedious and unfriendly. Kinematic functionality attempts to fix this problem by allowing one object to do several motions or kinematics simultaneously.\n\n## Commands Overview\n\nThe following is a summary of the kinematics commands:\n\n• initkinematics - This initializes data for the kinematics, saving things like the start location and rotation of the object you want to apply motion to.\n• addkinematic - This gives travel/rotate operations to the object. For example, you can tell the object, starting in 5 seconds and travel 10 units in the x direction, with a given acceleration, deceleration, and max speed. Then you can tell the object, starting in 7 seconds, travel 10 units in the y direction, with a different acceleration, deceleration, and max speed. The effect of these two operations is that the object will start traveling in the x, then will start simultaneously accelerating in the y, following a parabolic curved path to the destination. Each call to addkinematic will add another operation to the object.\n• updatekinematics - This command should be called when the object is being redrawn. This calculates the current position and rotation of the kinematics based on the current time, and sets the object's location to that position.\n\n## initkinematics\n\n• The Blank Node - For the initkinematics command, as well as all other kinematics commands, you must pass a reference to a blank node as the first parameter. This specifies where you want kinematic information to be stored, or, if you are getting information out of it, where kinematic information has been stored. The node needs to be an otherwise unused node, so that kinematic functionality can store data as needed. For modelers, this node will most likely be a label. Once kinematics have been initialized, the node will display the text \"do not touch\". This node should not be clicked on in a tree or table view while a kinematic operation is in process, or the kinematic data may be corrupted.\n• Overloading - The initkinematics command is overloaded, so you can call initkinematics with two different parameter sets, depending on your situation. Parameters for these two overloads are as follows.\n\n### void initkinematics(treenode datanode, treenode object [, int flags])\n\n• datanode - This is the blank node for kinematic data.\n• object - This is the object that will do the moving.\n• flags - This optional parameter defines various flags for the kinematic. It can be a bitwise or on the following values:\n• KINEMATIC_MANAGE_ROTATIONS - If this flag is set, the object will always point in the direction of travel.\n• KINEMATIC_RELATIVE_LOCS - If this flag is set, then the object will travel according to its local coordinate system, based on the rotation of the object at the time you initialize the kinematic. If not, kinematic coordinates are based on the object's container's coordinate system.\n• KINEMATIC_DO_NOT_PRUNE - By default, the kinematics logic will \"prune\" the kinematics as they go, meaning when you update the kinematics to a time that is after the end time of a given kinematic then it will remove that kinematic from the kinematics list. If you do not want it to do this, i.e. you may be doing queries across various times in the kinematic, then you should set this flag.\n• KINEMATIC_RESET_BUFFER - By default, the kinematic's memory allocation is optimized for speed, i.e. it will only reallocate memory when it needs it, and it will not resize back down when you initialize the kinematic. However, if this flag is set, it will resize its data buffer to an initial size when initialized.\n\nThis parameter set assumes you have an object that you want to move, like a transporter. The first parameter, again, is a blank node kinematic data, either a label, attribute, or variable. The second parameter is the object that will do the moving. The command will save off the object's initial location and rotation. The optional parameter managerotation is either 1 or 0. If 1, the rotation of the object will be set according to the velocity of the object at any given time. By default, the object's positive x direction will always point in the direction of the object's current velocity. This would be used, for example, if you have a truck that you always want to point forward as it travels. If managerotation is passed as 0, then the object won't rotate unless given commands to rotate, which will be explained later. If you do not specify this parameter, then by default, managerotate is set to 0. The optional localcoords parameter specifies the orientation of subsequent travel command locations. For example, if my truck is rotated 45 degrees, and I want it to travel 5 units in the x direction, this can be interpreted in two different ways. Do I want it to travel 5 units in x according to the truck's coordinate system, or 5 units in x according to the model's coordinate system (or the coordinate system of the truck's container)? In the former case, the object would actually travel 3.5 units in the x direction and 3.5 units in the y direction according to the model's coordinate space. In the latter case, however, the object would travel the usual 5 in the x direction according to the model's coordinate space. The localcoords parameter specifies which coordinate system you want to use. If 1 is passed, the object's coordinate system will be used (the former case). Note that only the object's initial coordinate system will be used to calculate locations, not subsequent coordinate systems if the object rotates later on in the kinematics. If 0 is passed, the object's container's coordinate system will be used (the latter case). If you do not specify this parameter, the default is 0.\n\n### void initkinematics(treenode datanode [, double x, double y, double z, double rx, double ry, double rz, int flags] )\n\n• datanode - This is the blank node for kinematic data.\n• x, y, z - These are optional initial location variables. Defaults: 0, 0, 0\n• rx, ry, rz - These are optional initial rotation variables. Defaults: 0, 0, 0\n\nThis parameter set allows you to explicitly pass initial locations and rotations, instead of referencing an object. Although you would probably more often use the other parameter set where you pass the object in, this parameter set gives you ultimate flexibility. Use this if you want to explicitly pass in the initial locations and rotations of the object, or if your location and rotation values don't necessarily represent real locations and rotations in your model. For example, you are simulating a robot arm, and there are several joints of the arm that move/rotation with different acceleration/deceleration/max speed values. The visualization of movements of the arm are not simulated with explicit Flexsim locations/rotations, but are done using your own draw commands and labels or variables. In this case, you don't want kinematics to be applied to straight rotations and locations, but rather to information that you are keeping on the object yourself. In such situations, a given set of kinematics does not need to be viewed as applied directly to x,y,z locations and x,y,z, rotations, but can rather be viewed as six separate kinematic motions, each along one axis. These six axes can represent whatever you want them to. For example, your robot has four joints, each with one rotation value. To have the four joints of the robot move using kinematics, you can have each joint simulate one axis in the kinematics. The x part of the kinematics applies to the rotation of joint 1, the y part applies to the rotation of joint 2, the z part applies to the rotation of joint 3, and the rx part applies to the rotation of joint 4. The other two parts of the kinematics, ry and rz, you don't worry about. You can initialize the kinematics with the start rotations of each of your 4 joints, and then add kinematics that apply to each joint individually. Then, when you want to get the joints' current rotation values out later to draw the robot arm in motion, you can use the getkinematics command instead of the updatekinematics command to get the values explicitly, and not have them be applied to an object's location or rotation. These commands will be explained later.\n\n### void setkinematicsrotoffset(treenode datanode, double rx, double ry, double rz)\n\n• datanode - This is the blank node for kinematic data.\n• rx, ry, rz - These are initial rotation variables.\n\nThis command would only be used if managerotation is passed into initkinematics as 1. This allows you to set an initial rotation from which the rotation is managed. By default, the object's positive x direction will always point in the direction of the object's current velocity. In the case of a truck, you may want the truck, instead of always being rotated so that it is traveling forward, to travel backward. Here you could specify a rotation offset of (0,0,180).\n\n### double addkinematic(treenode datanode, double x, double y, double z, double targetspeed [, double acc, double dec, double startspeed, double endspeed, double starttime, int type ] )\n\n• datanode - This is the blank node for kinematic data.\n• x, y, z - These are offset locations or rotations.\n• targetspeed - This is the target travel speed.\n• acc - This optional parameter specifies the acceleration. Default: 0 (or infinite acceleration)\n• dec - This optional parameter specifies the deceleration. Default: 0 (or infinite deceleration)\n• startspeed - This optional parameter specifies the initial velocity. Default: 0\n• endspeed - This optional parameter specifies the ending velocity. Default: 0\n• starttime - This optional parameter specifies the absolute start time. Default: Current Time\n• type - This optional parameter specifies the type of travel to apply (rotation or travel). Default: KINEMATIC_TRAVEL\n\nThis command adds a kinematic to the set of kinematics. The x, y, and z parameters make up an offset location or rotation. For example, the location (5,5,0) tells the kinematic to travel 5 in the x and 5 in the y. Note that these are offsets from the object's current position, and not absolute locations. The targetspeed parameter specifies the target speed for the travel operation. The other parameters are optional. Acc specifies the acceleration. Dec specifies the deceleration. Startspeed specifies the speed that the kinematic should start at. If this speed is higher that the target speed, then the object will start at the start speed and decelerate down to the target speed. Endspeed specifies the ending speed for the operation. If endspeed is greater than targetspeed, then at the end of the operation, the object will accelerate from the target speed to the end speed. The starttime is the start time of the kinematic, in simulation time, not as an offset from the current time. The type parameter specifies what type of kinematic it is. This value will usually either be KINEMATIC_TRAVEL, or KINEMATIC_ROTATE. If it is KINEMATIC_TRAVEL, the operation will be applied to the x, y, and z location values. If it is KINEMATIC_ROTATE, the operation will be applied to the rx, ry, and rz rotation values, and speeds are defined in degrees per unit of time, accelerations/decelerations in degrees per unit of time squared. The command returns the time that this kinematic operation will finish.\n\nBelow is a time-based plot of a kinematic that travels 3 in the x and 2 in the y, with accelerations and decelerations.", null, "### Turn Kinematic Types\n\nIn addition to KINEMATIC_TRAVEL and KINEMATIC_ROTATE, there are three \"turn\" kinematic types, which cause the moving object to turn a certain angle with a given turn radius. These are KINEMATIC_TURN_XY, KINEMATIC_TURN_YZ, and KINEMATIC_TURN_ZX. For these kinematic types, the x, y, and z parameters that you pass into addkinematic() take on a different meaning, as follows:\n\n• x - The start angle in degrees. An angle of 0 means the initial direction of motion when the turn starts will be in the same direction as the first axis of the kinematic type's name, i.e. start angle 0 for KINEMATIC_TURN_XY means the initial motion is positive along the x axis, where for KINEMATIC_TURN_YZ the initial motion would be positive along the y axis. Use the right hand rule around a given axis when figuring out start and turn angles.\n• y - The turn angle in degrees. The motion of the object will turn this number of degrees, parallel to the plane defined by the two axes in the kinematic type's name, i.e. KINEMATIC_TURN_XY will turn in the x/y plane, i.e. around the z axis.\n• z - The turn radius in distance.\n\nAll other parameters are the same for this kinematic type, i.e. max speed, acceleration, deceleration, etc. Speeds are in length units per time unit, NOT in degrees per time unit.\n\nAs an example, if you would like to make a motion that turns from travelling positive along the x axis to travelling at 60 degrees off the x axis, with a turn radius of 5 (i.e. turn about the z axis), call:\n\n``````\naddkinematic(kinematics, 0, 60, 5, 1, 1, 1, 0, 0, time(),\nKINEMATIC_TURN_XY);\n``````", null, "## updatekinematics\n\n### void updatekinematics(treenode datanode, treenode object [, double updatetime])\n\n• datanode - This is the blank node for kinematic data.\n• object - This is the object that will do the moving.\n• updatetime - This optional parameter specifies the absolute update time. Default: Current Time.\n\nThis command should be called in the middle of the kinematic operation, usually on predraw or draw. It calculates and then sets the current location and rotation of the object, according to all kinematics that have been added and the current update time. The updatetime parameter is optional. If it is not passed, the current simulation time is used.\n\n## Other Kinematics Commands\n\n### double getkinematics(treenode datanode, int type [, int kinematicindex, double updatetime/traveldist])\n\n• datanode - This is the blank node for kinematic data.\n• type - This specifies the type of information to retrieve (listed below).\n• kinematicindex - This optional parameter specifies which kinematic to query. Default: 0 (all kinematics)\n• updatetme/traveldist - This optional parameter represents either the update time or the travel distance, depending on the parameter type. Default: Current Time.\n\nThis command is used if you want to get explicit information on the kinematics. You can get information on the entire set of kinematics, or on each individual kinematic. Use this if your kinematics do not apply directly to object locations and rotations, or if you need this information in your logic. The type parameter specifies the type of information you want, and will be explained shortly. The kinematicindex parameter is optional, and specifies which individual kinematic you want to get information for. For example, if you add a kinematic to travel 5 units in the x direction as the second addkinematic command, you would pass a 2 as the kinematicindex parameter into the getkinematics command. If not passed, or if you pass a 0 value here, the default gets information for all kinematics together. The updatetime/traveldist parameter is optional. The meaning of this parameter depends on the type parameter you specify. Sometimes this parameter will not be used. Some of the time it represents the requested update time that you want to get information for. If not passed, the current time is used. In the case of a KINEMATIC_ARRIVALTIME query, this parameter instead represents travel distance. The use of this parameter will be explained with each query type.\n\nInformation Value of type Parameter Single Kinematic All Kinematics\nLocation KINEMATIC_X\nKINEMATIC_Y\nKINEMATIC_Z\nThese return x, y, or z component of the current location of the specified kinematic at the given update time. For example, if you added a kinematic to travel 10 units in the x, starting at time 5, and you want to know the x location for this given kinematic at time 7, you can call getkinematics(datanode, KINEMATIC_X, index, 7) to get the x location for time 7. These return the x, y, or z location of the object at the given update time.\nEnd Distance KINEMATIC_ENDDIST This returns the distance from the final destination location of all kinematics from the object's initial location. Here the updatetime parameter is not used.\nTotal Distance KINEMATIC_TOTALDIST This returns the total distance for the kinematic operation. This returns the sum of the distances of all the added kinematics. This has a subtle difference from KINEMATIC_ENDDIST.\nFor example, if your first kinematic travels 10 in the x direction, and your second kinematic travels -10 in the x direction, then the enddist value will be 0, whereas the totaldist value will be 20. Here the updatetime parameter is not used.\nTotal Component Distances KINEMATIC_TOTALX\nKINEMATIC_TOTALY\nKINEMATIC_TOTALZ\nThese return the x, y, or z component of the total distance for the kinematic operation. These are the same values you passed into the addkinematic command. Here the updatetime parameter is not used. These return the sum of x, y, or z component of all added kinematics. Here the updatetime parameter is not used.\nCumulative Distance KINEMATIC_CUMULATIVEDIST This returns the cumulative travel distance of all added kinematics. Unlike enddist or totaldist, it calculates the distance of the possibly curved path that the object will follow during the entire kinematic operation. Here the updatetime parameter is not used. Note: for cumulative distance, if you are using turn kinematics, the turn kinematics may not overlap other travel or turn kinematics. If they overlap with others, then the cumulative distance calculation will be incorrect.\nVelocity KINEMATIC_VX\nKINEMATIC_VY\nKINEMATIC_VZ\nThese return the x, y, or z component of the current velocity for the specified kinematic at the given update time. These return the x, y, or z velocity of the object for the given time.\nStart Speed KINEMATIC_STARTSPEED This returns the start speed for the kinematic. This is the startspeed you specify in the addkinematic command. Here the updatetime parameter is not used.\nPeak Speed KINEMATIC_PEAKSPEED This returns the peak speed or \"reached speed\" for the kinematic. This is usually the same as the target speed specified in the addkinematic command, but may not be if the kinematic cannot get to the target speed given the distance it has to travel. Here the updatetime parameter is not used.\nEnd Speed KINEMATIC_ENDSPEED This returns the end speed for the kinematic. This is usually the endspeed that you specify in the addkinematic command, but may not be if the kinematic cannot decelerate/accelerate to your specified endspeed given the distance it has to travel. Here the updatetime parameter is not used.\nTotal Velocity KINEMATIC_VELOCITY This returns the total velocity of the kinematic for the given time. This returns a scalar value of the total velocity for the given time.\nAcceleration (Starting) KINEMATIC_ACC1 This returns the acceleration value used to get from the start speed to the target speed. If the start speed is less than the target speed, then this value will be the acceleration value. Otherwise it will be the negative deceleration value. Here the updatetime parameter is not used.\nAcceleration (Stopping) KINEMATIC_ACC2 This returns the acceleration value used to get from the target speed to the end speed. If the end speed is less than the target speed, then this will return the negative deceleration value. Otherwise it will return the acceleration value. Here the updatetime parameter is not used.\nRotation KINEMATIC_RX\nKINEMATIC_RY\nKINEMATIC_RZ\nThese return the x, y, or z rotation of a rotational kinematic at the given update time. These return the x, y, or z rotation of the object at the given update time. This will only work if rotations are managed manually.\nEnd Rotational Distance KINEMATIC_ENDRDIST This returns the distance from the final destination rotational position of all kinematics from the object's initial rotational position. This will only work if rotations are managed manually. Here the updatetime parameter is not used.\nTotal Rotational Distance KINEMATIC_TOTALRDIST This returns the total rotational distance for the kinematic operation if it is a rotational kinematic. This returns the sum of the rotational distances of all the added kinematics. This will only work if you are managing rotations yourself. Here the updatetime parameter is not used.\nTotal Component Rotations KINEMATIC_TOTALRX\nKINEMATIC_TOTALRY\nKINEMATIC_TOTALRZ\nThese return the x, y, or z component of the total rotational distance for the kinematic operation if it is a rotational kinematic. These are the same values you passed into the addkinematic command. Here the updatetime parameter is not used. These return the sum of rx, ry, or rz component of all added kinematics. This will only work if you are managing rotations yourself. Here the updatetime parameter is not used.\nCumulatvie Rotational Distance KINEMATIC_CUMULATIVERDIST This returns the cumulative rotational travel distance of all added kinematics. This will only work if you are managing rotations yourself. Here the updatetime parameter is not used.\nRotational Velocity KINEMATIC_VRX\nKINEMATIC_VRY\nKINEMATIC_VRZ\nThese return the x, y, or z component of the current rotational velocity for the specified kinematic at the given time if it is a rotational kinematic. These return the rotational x, y, or z velocity of the object for the given time.\nTotal Rotational Velocity KINEMATIC_RVELOCITY This returns the total rotational velocity of the kinematic for the given time if it is a rotational kinematic. This returns a scalar value of the total rotation velocity for the given time. This only works if rotations are managed manually.\nStart Times KINEMATIC_STARTTIME This returns the time that the specified kinematic will start its operation. This is the same starttime that you specified when you added the kinematic. Here the updatetime parameter is not used. This returns the lowest start time of all the kinematics. Here the updatetime parameter is not used.\nAcceleration Time (Starting) KINEMATIC_ACC1TIME This returns the total time the kinematic will spend accelerating/decelerating from the start speed to the target speed. Here the updatetime parameter is not used.\nArrival Time KINEMATIC_ARRIVALTIME In this query, the updatetime/traveldist parameter is used as a requested travel distance for the given kinematic. This returns the time of arrival for a certain sub-distance of a given kinematic. For example, if I've added a kinematic that tells the object to travel 5 units in the x direction, but I want to know how long it will take him to travel just 3 of those 5 units, I can use this query, and pass 3 in as the traveldist parameter.\nPeak Time KINEMATIC_PEAKTIME This returns the total time the kinematic will spend traveling at the peak speed. Here the updatetime parameter is not used.\nAcceleration Time (Stopping) KINEMATIC_ACC2TIME This returns the total time the kinematic will spend accelerating/decelerating from the target speed to the end speed. Here the updatetime parameter is not used.\nEnd Times KINEMATIC_ENDTIME This returns the time that the specified kinematic will finish its operation. This is the same endtime that is returned from the addkinematic command. Here the updatetime parameter is not used. This returns the highest end time of all the kinematics. Here the updatetime parameter is not used.\nNumber of Kinematics KINEMATIC_NR This returns the number of kinematics that have been added. Here the updatetime parameter is not used.\nTurn Kinematic Data KINEMATIC_STARTANGLE\nKINEMATIC_TURNANGLE\nThese return the start angle, turn angle, and turn radius of a turn kinematic.\nKinematic Type KINEMATIC_TYPE This returns KINEMATIC_TRAVEL if the specified kinematic is a travel operation, and KINEMATIC_ROTATE if the kinematic is a rotate operation.\n\n### void profilekinematics(treenode datanode [, int index ])\n\n• datanode - This is the blank node for kinematic data.\n• index - This optional parameter specifies the index of a specific index to print information for. Default: 0 (all kinematics)\n\nThis command prints kinematic information to the output console. The index parameter is optional. If it is not passed, then information will be printed for all kinematics that have been added. If it is passed, then the index variable is an index of the added kinematic you want to print information for.\n\n### void deactivatekinematics(treenode datanode)\n\n• datanode - This is the blank node for kinematic data.\n\nThis command tells the kinematic to not update locations when the updatekinematics command is called. Execute this on reset to free the object to move it around in the ortho view." ]

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[ "# On Traffic Flow with Nonlocal Flux: A Relaxation Representation\n\nResearch output: Contribution to journalArticlepeer-review\n\n1 Scopus citations\n\n## Abstract\n\nWe consider a conservation law model of traffic flow, where the velocity of each car depends on a weighted average of the traffic density ρ ahead. The averaging kernel is of exponential type: wε(s) = ε- 1e-s/ε. By a transformation of coordinates, the problem can be reformulated as a 2 × 2 hyperbolic system with relaxation. Uniform BV bounds on the solution are thus obtained, independent of the scaling parameter ε. Letting ε→ 0 , the limit yields a weak solution to the corresponding conservation law ρt+ (ρv(ρ)) x= 0. In the case where the velocity v(ρ) = a- bρ is affine, using the Hardy–Littlewood rearrangement inequality we prove that the limit is the unique entropy-admissible solution to the scalar conservation law.\n\nOriginal language English (US) 1213-1236 24 Archive for Rational Mechanics and Analysis 237 3 https://doi.org/10.1007/s00205-020-01529-z Published - Sep 1 2020\n\n## All Science Journal Classification (ASJC) codes\n\n• Analysis\n• Mathematics (miscellaneous)\n• Mechanical Engineering" ]

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[ "", null, "# Why Is Standard Cost Important?\n\n## What are the 4 types of standards?\n\nStandards in Accounting (4 Types)Ideal, Perfect, Maximum Efficiency or Theoretic Standards:Normal Standards:Basic Standards:Currently Attainable or Expected Actual Standards:.\n\n## What are the disadvantages of standard costing?\n\nThree of the disadvantages that result from a business using standard costs are:Controversial materiality limits for variances.Nonreporting of certain variances.Low morale for some workers.\n\n## What are the basic principles of standard costing?\n\nIn a standard cost system, a company shows the cost flows between inventory accounts and into cost of goods sold at consistent standard amounts during the period. It needs no special calculations to determine actual unit costs during the period.\n\n## What is standard costing explain its advantages and limitations?\n\nThe main advantages of standard costing are: … Compiling standard costs more carefully can eliminate the weakness of the traditional costing system. 2. Standard costs can be used as a yardstick against which actual costs can be compared. It is an effective tool for planning production costs.\n\n## Which one is ideal for cost control purpose?\n\nExpected standards are based on current conditions and circumstances and represents what can be attained with the present setup in place and if the current conditions prevail. In order to ensure cost control such expected standards would be most useful.\n\n## What do you mean by standard cost?\n\nStandard costs are estimates of the actual costs in a company’s production process, because actual costs cannot be known in advance. This helps a business to plan a budget.\n\n## What are the applications of standard costing?\n\n(1) It helps in measuring the actual costs of the product. (2) Comparison of actual costs with the pre-determined standards is made, in order to determine variances. (3) Setting of standards for each element of cost such as material, labour and overheads.\n\n## What are two main uses of standard costing?\n\nUses of Standard Costing  To provide a formal basis for assessing performance and efficiency.  To Control Costs by establishing standards and analysis of variance.  To enable the principle of “Management by Exception” to be practiced at detailed operational level.  To assist in setting budgets in an organization.\n\n## Which is the main component of standard cost?\n\nThe three components of standard costing: Standard costs, which provide a standard, or predetermined performance level. A measure of actual performance. A measure of the variance between standard and actual performance.\n\n## How is standard cost decided?\n\nTo determine these costs, you’ll need to multiply the rate of each by the quantity (in units or hours). For example, if the direct materials price is \\$10 and the standard quantity is 20 pounds per unit, you would multiply \\$10 by 20 to get \\$200. This would be the standard cost for the direct materials only.\n\n## How does Standard Cost work?\n\nStandard costing is the practice of substituting an expected cost for an actual cost in the accounting records. Subsequently, variances are recorded to show the difference between the expected and actual costs.\n\n## What is the difference between actual and standard cost?\n\nStandard costs are the estimated costs for products that are predetermined and arise from the units of material, labour and other costs of production for the specific time period. Actual costs refer to the costs that are actually incurred.\n\n## What is the standard?\n\nA standard is a repeatable, harmonised, agreed and documented way of doing something. Standards contain technical specifications or other precise criteria designed to be used consistently as a rule, guideline, or definition. … Any organization can establish standards for internal or external use.\n\n## What is a standard cost system?\n\nStandard cost systems make use of standard costs, which are the budgeted or estimated costs deemed to be necessary to manufacture a single unit of product or perform a single service. … In a standard cost system, both standard costs and actual costs are recorded in accounting records.\n\n## What is standard costing with example?\n\nHere is a simple standard costing example. Let’s take a company that makes widgets. Based on historical data, a cost analyst determines that producing one widget typically requires 1 pound of raw material costing \\$2 dollars and 1 hour of labor costing \\$20 dollars.\n\n## What are the types of standard costing?\n\nTypes of Standards:Current Standard: Current standard is a standard established for use over a short period of time, related to current conditions. … Basic Standard: Basic standard is standard established for use over a long period from which a current standard can be developed. … Ideal Standard: … Attainable Standard:\n\n## What is standard costing and what are its importance?\n\nStandard costing is an important subtopic of cost accounting. Standard costs are usually associated with a manufacturing company’s costs of direct material, direct labor, and manufacturing overhead." ]

[ null, "https://mc.yandex.ru/watch/70834135", null ]

[ "A Comparative Study of Two Spatial Discretization Schemes for Advection Equation\n\nIn this paper, we describe a comparison of two spatial discretization schemes for the advection equation, namely the first finite difference method and the method of lines. The stability of the methods has been studied by Von Neumann method and with the matrix analysis. The methods are applied to a number of test problems to compare the accuracy and computational efficiency. We show that both discretization techniques approximate correctly solution of advection equation and compare their accuracy and performance.\n\nShare and Cite:\n\nBakodah, H. (2016) A Comparative Study of Two Spatial Discretization Schemes for Advection Equation. International Journal of Modern Nonlinear Theory and Application, 5, 59-66. doi: 10.4236/ijmnta.2016.51006.\n\nReceived 29 December 2015; accepted 5 March 2016; published 8 March 2016", null, "1. Introduction\n\nA currently active area of research is the numerical solution of nonlinear partial differential equations and nonlinear integral equations - . An advection equation is fairly in shape but it is one of the most difficult equations to approximate numerically.\n\nThe nonlinear advection equation arises in various branches of physics, engineering and applied sciences. The importance of obtaining the exact or approximate solution of this equation is still a significant problem that needs new methods to discover exact or approximate solution.\n\nThe linear advection equation is simple in form and yet it is one of the most difficult equations to solve accurately by numerical means . This equation is challenging to solve as it causes some discontinuities with neither dispersion nor dissipation. However, all efforts to use a fixed number of space intervals will result in both dispersion and spurious oscillation. The use of traditional symmetric techniques is possible only if the terms of arbitrary second order artificial viscosity or damping are introduced to the equation. Directional (or upwind) methods have shown to be efficient for the purpose of finite difference analyses as it clears the oscillation problem, yet they will not remove it completely. The linear advection equation does provide a good case for testing methods to be used on systems of hyperbolic equations. Many schemes have been tested on it, generally by using propagating step, sine, Gaussian or triangular waveform . These techniques were used to provide a rational for choosing a spatial scheme for first-order hyperbolic equations. The optimal choice is not invariant, but depends on the application.\n\nIn this paper, the advection equation is solved by finite difference method and the stability conditions for the scheme are also discussed. Numerical finite difference scheme is developed for obtaining approximate solution to an advection equation using the 3-point formula introduced in . The same problem is considered using modified method of lines - , with a new three-point difference . Using this new difference leads to stable schemes for the two methods. Numerical results are shown and compared with analytical solutions.\n\n2. Finite Difference Method with a Good Spatial Discretization\n\nLet us consider the equation", null, "(1)\n\nand v is a nonzero constant velocity, where", null, "with the initial condition", null, "The finite difference method begins with discretization the space variable x and the time variable t as follows", null, "; and", null, ".\n\nThe using of some of the finite difference schemes on advection equations can cause unstable solutions. To add stability, upstream (backward or forward) could be used for spatial discretization for the first-order differences. However, for a given spatial accuracy, these differences need to use extra grid points than centered difference. An artificial dissipation (or viscosity) term is normally introduced to a central differencing scheme for stability reasons but it is not easy to determine the magnitude of this term required for the stability and effect of this term on the solutions.\n\nThe aim of new method is to develop a good formula with high accuracy for the numerical solution of the advection equation using the spatial discretization presented by Sharaf and Bakodah, , using 3-points formula, thus the approximate form of the first derivative is", null, "(2)\n\nAdopting a forward temporal difference scheme, this yields", null, "(3)\n\nThere are two standard methods of the finite-difference equation. In the first method, a finite Fourier series is used. In the other method, the equation is expressed in matrix form, and the eigenvalues of the associated matrix are examined. In order to investigate the stability of this scheme by the first method (Von Neumann stability analysis), it is considered", null, "(4)\n\nReplacing", null, "in relation (2) from (3), it is obtained:", null, "(5)\n\nwhere", null, "(6)\n\nThe stability condition", null, "is fulfilled for all k as long as,", null, "(7)", null, "(8)\n\nThat is, the method (2) is stable.\n\n3. Modified of the Method of Lines\n\nIn the Numerical Method of Lines (NMOL), the partial differential equation (PDE), to be solved, is transformed into a system of ordinary differential equations (ODEs) by discretizing all the independent variables but one . The advection equation, depending on time t and one spatial variable x, either t or x can be discretized, and the integration will be carried out along the remaining undiscretized independent variable.\n\nThe technique consists of converting the PDE into ODEs either by finite difference spline or by weighted-res- idual technique, then integrating the resulting ODEs . Finite differencing in the spatial variable led to a set of timedependent ODEs. The advantage of using (NMOL) is that sophisticated software packages exist for the numerical solution of ordinary differential equations. These software packages contain iterative methods for handling non-linearities and feature automatic step-size adjustment and integration order selection to maintain a specified error and to solve the problem with near optimal efficiency. Several recently software packages for automated method of lines solution of arbitrarily defined PDEs have been very successful, particularly for parabolic and elliptic PDE systems.\n\nSuch facilities can be improved for hyperbolic equations by incorporating an upwind weighted residual technique. This technique is similar and superior to the use of an artificial viscosity term, and it could be implement easily in any software package. Previous considerations of the (NMOL) to solve PDEs have been geared to parabolic equation and generally used centered, second-order differences. Using these differences on hyperbolic equations can lead to unstable solution. To add stability, upstream (backward or forward) first-order differences could be used for the spatial discretization. But these differences require the use of more grid points than central differences for a given spatial accord. An artificial dissipation (or viscosity) term is often added to a central differencing scheme to add stability but it is difficult to determine the magnitude of this term required for the stability and the effect of this term on the solutions. Other stabilizing techniques that have been employed in the explicit finite difference procedures are generally not applicable to the method of lines approach because they involve manipulation of terms in both the time and space discretization.\n\nIn this paper, a modified method of lines using a new three-point difference is used. The use of this new differences leads to stable schemes with good accuracy. In order to apply the method of lines to the advection equation (1), the spatial derivative must be approximated. An equally spaced mesh", null, "is used. As in the Ref. a new difference scheme can be used in the calculation of\n\n(9)\n\nIt leads to stable schemes with good accuracy. In this section the second method shall be used. The analysis of eigenvalues of the system gives the necessary conditions for the stability of discretization of the problem . The stability corresponds to real and negative values.\n\nBy considering equation (1) with the centered difference scheme of order two, then we get\n\n(10)\n\nwhere is\n\n(11)\n\nand.\n\nMathematically the difference scheme is stable if there exists a real positive eigenvalues. However, where is a tri-diagonal matrix, the corresponding eigenvalue of A can be calculated from the relation.\n\n(12)\n\nwhere. Thus\n\n(13)\n\nwhich are pure imaginary values.\n\nSo, we consider the non-centered formula approximation\n\n(14)\n\nwith the matrix formula where\n\n(15)\n\nThus the eigenvalues are given by\n\n(16)\n\nThese values are real and negative, so the difference scheme is stable.\n\n4. Numerical Examples\n\n4.1. Example 1\n\nWe apply the Finite Difference Method with a good spatial discretization to solve linear advection equation to demonstrate the validity of this method.\n\nConsider the equation\n\n(17)\n\nwith the conditions\n\nWith the analytic solution\n\nUsing equation (3) we find\n\nlet\n\nTable 1 shows the absolute error for the finite difference method in different value of time.\n\n4.2. Example 2\n\nWe apply the Modified of the Method of lines to solve linear advection equation to demonstrate the validity of this method.\n\nwith the conditions\n\nTable 1. Absolute errors of the finite difference method, where.\n\nand the analytic solution\n\nSubstituting in equation (9) we find\n\nand the condition are\n\nHence, we can write\n\nSo, to confirm the accuracy and efficiency of the method, the absolute error are used (Table 2).\n\n4.3. Example 3\n\nConsider the equation\n\nand\n\nwith the analytic solution\n\nTable 2. Absolute errors of the modified of the method of lines where.\n\nIn this example we apply the Modified of the Method of lines and the finite difference method to solve linear advection equation to demonstrate the validity of them and compare between them.\n\nTable 3 shows the absolute error for the finite difference method equation (2), and the method of lines.\n\n4.4. Example 4\n\nwith the condition\n\nand the exact solution\n\nThis problem is solved for, with.\n\nTable 4 shows the absolute error\n\nTable 3. Comparison of the absolute error between finite difference method and the method of lines, for example 1 with.\n\nTable 4. Comparison of the absolute error between finite difference method and the method of lines, for example 2 with.\n\n5. Conclusions\n\nFrom the studied test examples, it has been found that, the modified method of lines gives better results than the finite difference method. Although the modified method of lines is used to approximate the first order hyperbolic differential equation. Thus equations are one of the most difficult classes of PDEs to integrate numerically. To overcome this, a modified MOL scheme is suggested. The results are in good agreement with the exact solution as shown in Table 1, Table 2. The presented method is attractive for hyperbolic, parabolic and elliptic equations.\n\nThe methods introduced in this paper for solving the linear and nonlinear advection equation are based on finite difference method. The best choice of the numerical method for a given problem depends on the stability condition.\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest.\n\n Biazar, J., Ghazvini, H. and Eslami, M. (2009) He’s Homotopy Perturbation Method for Systems of Integro-Differential Equations. Chaos, Solitons & Fractals, 39, 1253-1258. http://dx.doi.org/10.1016/j.chaos.2007.06.001 Biazar, J. and Eslami, M. (2011) Modified HPM for Solving Systems of Volterra Integral Equations of the Second Kind. Journal of King Saud University—Science, 23, 35-39. http://dx.doi.org/10.1016/j.jksus.2010.06.004 Biazar, J., Eslami, M. and Islam, M.R. (2012) Differential Transform Method for Special Systems of Integral Equations. Journal of King Saud University—Science, 24, 211-214. http://dx.doi.org/10.1016/j.jksus.2010.08.015 Eslami, M. (2014) New Homotopy Perturbation Method for a Special Kind of Volterra Integral Equations in Two-Dimensional Space. Computational Mathematics and Modeling, 25, 135-148. http://dx.doi.org/10.1007/s10598-013-9214-x Biazar, J. and Eslami, M. (2011) A New Homotopy Perturbation Method for Solving Systems of Partial Differential Equations. Computers & Mathematics with Applications, 62, 225-234. http://dx.doi.org/10.1016/j.camwa.2011.04.070 Biazar, J., Eslami, M. and Aminikhah, H. (2009) Application of Homotopy Perturbation Method for Systems of Volterra Integral Equations of the First Kind. Chaos, Solitons & Fractals, 42, 3020-3026. http://dx.doi.org/10.1016/j.chaos.2009.04.016 Biazar, J. and Eslami, M. (2012) A New Method for Solving the Hyperbolic Telegraph Equation. Computational Mathematics and Modeling, 23, 519-527. http://dx.doi.org/10.1007/s10598-012-9153-y Wazwaz, A.M. (2009) Partial Differential Equations and Solitary Waves Theory. Higher Education Press, Beijing and Springer-Verlag, Berlin Heidelberg. http://dx.doi.org/10.1007/978-3-642-00251-9 George, K. and Twizell, E.H. (2006) Stable Second-Order Finite-Difference Methods for Linear Initial-Boundary-Value Problems. Applied Mathematics Letters, 19, 146-154. http://dx.doi.org/10.1016/j.aml.2005.04.003 Smith, G.D. (1985) Numerical Solution of Partial Differential Equations (Finite Difference Method). 3rd Edition, Oxford University Press, Oxford. Al-Malki, N.A. and Bakodah, H.O. (2014) A Stable Difference Algorithm for the Solution of Advection Equation. Far East Journal of Applied Mathematics, 88, 139-149. Sharaf, A.A. and Bakodah, H.O. (2005) A Good Spatial Discretization in the Method of Lines. Applied Mathematics and Computation, 171-172, 1253-1263. http://dx.doi.org/10.1016/j.amc.2005.01.144 Alabdali, F.M. and Bakodah, H.O. (2014) A New Modification of the Method of Lines for First Order Hyperbolic PDEs. Applied Mathematics, 5, 1457-1462. http://dx.doi.org/10.4236/am.2014.510138 Bakodah, H.O. and Banaja, M.A. (2013) The Method of Lines Solution of the Regularized Long-Wave Equation Using Runge-Kutta Time Discretization Method. Mathematical Problems in Engineering, 2013, Article ID: 804317.http://dx.doi.org/10.1155/2013/804317 Banaja, M.A. and Bakodah, H.O. (2015) Runge-Kutta Integration of the Equal Width Wave Equation Using the Method of Lines. Mathematical Problems in Engineering, 2015, Article ID: 274579. http://dx.doi.org/10.1155/2015/274579 Smith, G.D. (1985) Numerical Solution of Partial Differential Equations (Finite Difference Methods). 3rd Edition, Oxford University Press, Oxford. Schiesser, W.E. (1978) The Numerical Method of Lines, Integration of Partial Differential Equation. 2nd Edition, Clarendon Presses, Oxford. Carver, M.B. and Hinds, H.W. (1978) The Method of Lines and Advective Equation. Simulation, 31, 59-69.http://dx.doi.org/10.1177/003754977803100205", null, "" ]

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[ "", null, "LAPACK 3.11.0 LAPACK: Linear Algebra PACKage\nSearching...\nNo Matches\n\n## ◆ dpptrs()\n\n subroutine dpptrs ( character UPLO, integer N, integer NRHS, double precision, dimension( * ) AP, double precision, dimension( ldb, * ) B, integer LDB, integer INFO )\n\nDPPTRS\n\nPurpose:\n``` DPPTRS solves a system of linear equations A*X = B with a symmetric\npositive definite matrix A in packed storage using the Cholesky\nfactorization A = U**T*U or A = L*L**T computed by DPPTRF.```\nParameters\n [in] UPLO ``` UPLO is CHARACTER*1 = 'U': Upper triangle of A is stored; = 'L': Lower triangle of A is stored.``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in] NRHS ``` NRHS is INTEGER The number of right hand sides, i.e., the number of columns of the matrix B. NRHS >= 0.``` [in] AP ``` AP is DOUBLE PRECISION array, dimension (N*(N+1)/2) The triangular factor U or L from the Cholesky factorization A = U**T*U or A = L*L**T, packed columnwise in a linear array. The j-th column of U or L is stored in the array AP as follows: if UPLO = 'U', AP(i + (j-1)*j/2) = U(i,j) for 1<=i<=j; if UPLO = 'L', AP(i + (j-1)*(2n-j)/2) = L(i,j) for j<=i<=n.``` [in,out] B ``` B is DOUBLE PRECISION array, dimension (LDB,NRHS) On entry, the right hand side matrix B. On exit, the solution matrix X.``` [in] LDB ``` LDB is INTEGER The leading dimension of the array B. LDB >= max(1,N).``` [out] INFO ``` INFO is INTEGER = 0: successful exit < 0: if INFO = -i, the i-th argument had an illegal value```\n\nDefinition at line 107 of file dpptrs.f.\n\n108*\n109* -- LAPACK computational routine --\n110* -- LAPACK is a software package provided by Univ. of Tennessee, --\n111* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--\n112*\n113* .. Scalar Arguments ..\n114 CHARACTER UPLO\n115 INTEGER INFO, LDB, N, NRHS\n116* ..\n117* .. Array Arguments ..\n118 DOUBLE PRECISION AP( * ), B( LDB, * )\n119* ..\n120*\n121* =====================================================================\n122*\n123* .. Local Scalars ..\n124 LOGICAL UPPER\n125 INTEGER I\n126* ..\n127* .. External Functions ..\n128 LOGICAL LSAME\n129 EXTERNAL lsame\n130* ..\n131* .. External Subroutines ..\n132 EXTERNAL dtpsv, xerbla\n133* ..\n134* .. Intrinsic Functions ..\n135 INTRINSIC max\n136* ..\n137* .. Executable Statements ..\n138*\n139* Test the input parameters.\n140*\n141 info = 0\n142 upper = lsame( uplo, 'U' )\n143 IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN\n144 info = -1\n145 ELSE IF( n.LT.0 ) THEN\n146 info = -2\n147 ELSE IF( nrhs.LT.0 ) THEN\n148 info = -3\n149 ELSE IF( ldb.LT.max( 1, n ) ) THEN\n150 info = -6\n151 END IF\n152 IF( info.NE.0 ) THEN\n153 CALL xerbla( 'DPPTRS', -info )\n154 RETURN\n155 END IF\n156*\n157* Quick return if possible\n158*\n159 IF( n.EQ.0 .OR. nrhs.EQ.0 )\n160 \\$ RETURN\n161*\n162 IF( upper ) THEN\n163*\n164* Solve A*X = B where A = U**T * U.\n165*\n166 DO 10 i = 1, nrhs\n167*\n168* Solve U**T *X = B, overwriting B with X.\n169*\n170 CALL dtpsv( 'Upper', 'Transpose', 'Non-unit', n, ap,\n171 \\$ b( 1, i ), 1 )\n172*\n173* Solve U*X = B, overwriting B with X.\n174*\n175 CALL dtpsv( 'Upper', 'No transpose', 'Non-unit', n, ap,\n176 \\$ b( 1, i ), 1 )\n177 10 CONTINUE\n178 ELSE\n179*\n180* Solve A*X = B where A = L * L**T.\n181*\n182 DO 20 i = 1, nrhs\n183*\n184* Solve L*Y = B, overwriting B with X.\n185*\n186 CALL dtpsv( 'Lower', 'No transpose', 'Non-unit', n, ap,\n187 \\$ b( 1, i ), 1 )\n188*\n189* Solve L**T *X = Y, overwriting B with X.\n190*\n191 CALL dtpsv( 'Lower', 'Transpose', 'Non-unit', n, ap,\n192 \\$ b( 1, i ), 1 )\n193 20 CONTINUE\n194 END IF\n195*\n196 RETURN\n197*\n198* End of DPPTRS\n199*\nsubroutine xerbla(SRNAME, INFO)\nXERBLA\nDefinition: xerbla.f:60\nlogical function lsame(CA, CB)\nLSAME\nDefinition: lsame.f:53\nsubroutine dtpsv(UPLO, TRANS, DIAG, N, AP, X, INCX)\nDTPSV\nDefinition: dtpsv.f:144\nHere is the call graph for this function:\nHere is the caller graph for this function:" ]

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[ "9.9. Variable numbers of arguments\n\nIt is often desirable to implement a function where the number of arguments is not known, or is not constant, when the function is written. Such a function is printf, described in Section 9.11. The following example shows the declaration of such a function.\n\nint f(int, ... );\n\nint f(int, ... ) {\n.\n.\n.\n}\n\nint g() {\nf(1,2,3);\n}\nExample 9.5\n\nIn order to access the arguments within the called function, the functions declared in the <stdarg.h> header file must be included. This introduces a new type, called a va_list, and three functions that operate on objects of this type, called va_start, va_arg, and va_end.\n\nBefore any attempt can be made to access a variable argument list, va_start must be called. It is defined as\n\n#include <stdarg.h>\nvoid va_start(va_list ap, parmN);\n\nThe va_start macro initializes ap for subsequent use by the functions va_arg and va_end. The second argument to va_start, parmN is the identifier naming the rightmost parameter in the variable parameter list in the function definition (the one just before the , ... ). The identifier parmN must not be declared with register storage class or as a function or array type.\n\nOnce initialized, the arguments supplied can be accessed sequentially by means of the va_arg macro. This is peculiar because the type returned is determined by an argument to the macro. Note that this is impossible to implement as a true function, only as a macro. It is defined as\n\n#include <stdarg.h>\ntype va_arg(va_list ap, type);\n\nEach call to this macro will extract the next argument from the argument list as a value of the specified type. The va_list argument must be the one initialized by va_start. If the next argument is not of the specified type, the behaviour is undefined. Take care here to avoid problems which could be caused by arithmetic conversions. Use of char or short as the second argument to va_arg is invariably an error: these types always promote up to one of signed int or unsigned int, and float converts to double. Note that it is implementation defined whether objects declared to have the types char, unsigned char, unsigned short and unsigned bitfields will promote to unsigned int, rather complicating the use of va_arg. This may be an area where some unexpected subtleties arise; only time will tell.\n\nThe behaviour is also undefined if va_arg is called when there were no further arguments.\n\nThe type argument must be a type name which can be converted into a pointer to such an object simply by appending a * to it (this is so the macro can work). Simple types such as char are fine (because char * is a pointer to a character) but array of char won't work (char [] does not turn into ‘pointer to array of char’ by appending a *). Fortunately, arrays can easily be processed by remembering that an array name used as an actual argument to a function call is converted into a pointer. The correct type for an argument of type ‘array of char’ would be char *.\n\nWhen all the arguments have been processed, the va_end function should be called. This will prevent the va_list supplied from being used any further. If va_end is not used, the behaviour is undefined.\n\nThe entire argument list can be re-traversed by calling va_start again, after calling va_end. The va_end function is declared as\n\n#include <stdarg.h>\nvoid va_end(va list ap);\n\nThe following example shows the use of va_start, va_arg, and va_end to implement a function that returns the biggest of its integer arguments.\n\n#include <stdlib.h>\n#include <stdarg.h>\n#include <stdio.h>\n\nint maxof(int, ...) ;\nvoid f(void);\n\nmain(){\nf();\nexit(EXIT_SUCCESS);\n}\n\nint maxof(int n_args, ...){\nregister int i;\nint max, a;\nva_list ap;\n\nva_start(ap, n_args);\nmax = va_arg(ap, int);\nfor(i = 2; i <= n_args; i++) {\nif((a = va_arg(ap, int)) > max)\nmax = a;\n}\n\nva_end(ap);\nreturn max;\n}\n\nvoid f(void) {\nint i = 5;\nint j;\nj = 24;\nprintf(\"%d\\n\",maxof(3, i, j, 0));\n}\nExample 9.6" ]

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[ "# pyharm.shs#\n\nModule to perform solid spherical harmonic synthesis of point and mean values.\n\nNote\n\nThis documentation is written for double precision version of PyHarm.\n\npyharm.shs.point(pnt, shcs, nmax)#\n\nPerforms spherical harmonic synthesis of point values from `shcs` at `pnt` up to maximum degree `nmax`. Refer to charm_shs for the full documentation.\n\nParameters:\nReturns:\n\nout – Point values synthesized from `shcs` at `pnt`\n\nReturn type:\n\nnumpy array of floating points\n\npyharm.shs.cell(cell, shcs, nmax)#\n\nPerforms spherical harmonic synthesis of area-mean values from `shcs` at `cell` up to maximum degree `nmax`. Refer to charm_shs for the full documentation.\n\nParameters:\n• cell (CellSctr, CellGrid) – Evaluation cells\n\n• shcs (Shc) – Spherical harmonic coefficients\n\n• nmax (integer) – Maximum degree of the synthesis\n\nReturns:\n\nout – Area-mean values synthesized from `shcs` at `cell`\n\nReturn type:\n\nnumpy array of floating points\n\npyharm.shs.cell_isurf(cell, shcs1, nmax1, shcs2, nmax2, nmax3, nmax4)#\n\nPerforms spherical harmonic synthesis of area-mean values from `shcs1` at `cell` residing on an irregular surface defined by `shcs2`. The synthesis of area-mean values is done up to degree `nmax1` and the irregular surface is expanded up to degree `nmax2`. `nmax3` and `nmax4` represent the maximum harmonic degrees to synthesize and analyze the `(shcs1.r / r)^(n + 1)` terms, where `r` stands for the spherical radius of the irregular surface defined by `shcs2`. Refer to charm_shs for the full documentation.\n\nParameters:\n• cell (CellGrid) – Evaluation cells\n\n• shcs1 (Shc) – Spherical harmonic coefficients of the function, the area-mean values of which are synthesized\n\n• nmax1 (integer) – Maximum degree of the synthesis of the area-mean values\n\n• shcs2 (Shc) – Spherical harmonic coefficients of the irregular surface, on which the area-mean values are synthesized\n\n• nmax2 (integer) – Maximum degree of the synthesis of the irregular surface\n\n• nmax3 (integer) – Maximum degree of the synthesis of `(shcs1.r / r)^(n + 1)`\n\n• nmax4 (integer) – Maximum degree of the analysis of `(shcs1.r / r)^(n + 1)`\n\nReturns:\n\nout – Area-mean values synthesized from `shcs1` at `cell` residing on the surface defined by `shcs2`\n\nReturn type:\n\nnumpy array of floating points" ]

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[ "# P-values are becoming embarrassing\n\n### What is a p-value?\n\nAnyone who has taken a basic statistics course remembers the term “p-value” and that when it’s a certain value you reject something and when it’s not you do something else. Who knows, I can’t remember. (Kidding, I teach statistics!)\n\nWell, simply put, a p-value is the probability of getting the data you have if you assume something beforehand. Put mathematically, this is $P(\\text{data}|\\theta)$, where $\\theta$ is some parameter you are interested in, like the mean. So let’s say we are interested in the average weight of all males at a certain company. We start by assuming it’s 185 lbs and conduct a two sided hypothesis test. Our p-value is 0.022 so we reject the assumption that the mean weight is 185. What we have actually found is $P(\\text{data}|\\theta=185)=0.022$. So if the real mean weight is 185 lbs, when we randomly sample a group we will get the results of our data or more extreme 2.2% of the time. Seems useful right? Maybe, maybe not. All we know is that is seems kind of unlikely to happen if the real mean weight were 185 lbs. Seems, maybe, unlikely? Ugg. Statistics deals with uncertainty but this level of uncertainty doesn’t cut it for me.\n\nThis kind of hypothesis testing tells us NOTHING about the range of the mean weight. What if we are interesting in $P(\\theta=185)$? Or what if we want to know a range like $P(180 <\\theta<190)$? That would be very powerful information! This probability can not be found through traditional hypothesis testing and requires uses Bayesian techniques. It has become a problem that studies find a “significant p-value” of something less than 0.05 and claim to have significant results that go against everything known about that field. Recently the American Statistical Association (AMA) commented on this problem and how this way of thinking needs to go. Hypothesis testing is a valid tool but it is much less powerful than it seems at first glance and is too often completely misunderstood.\n\nIn the war between Frequentists and Bayesian followers, it seems the AMA is siding with the latter group in this case.", null, "# Side-side-angle and impossible triangles\n\n### What is the ambiguous case?\n\nIn high school geometry, the idea of proofs are often first introduced to American students. A common task is to use a basic set of rules to prove two triangles are congruent, which is a fancy way of saying  they have the same side lengths and the same angles. Phrases like side-angle-side (SAS) and angle-angle-side (AAS) might ring a bell. This topic is very polarizing and quickly separates students into those who say “I love geometry!” and “I miss algebra!”. Simply put though, all of these acronyms are just descriptions of the information we are given about a triangle. Note that the order of this information is very important! SAS and SSA both contain one angle and two sides, but both of them won’t prove two triangles are congruent.\n\nThere is one problematic case of side-side-angle (SSA) or put another way, angle-side-side (I’ll pause here and let you chuckle at the joke every geometry teacher in the country makes at this moment).  Some teachers will correctly note that this situation is not enough to show two triangles are congruent and leave it at that, but there is more that can be explored here. When you are given a SSA triangle, this is called the ambiguous case because it could in fact result in 0, 1, or 2 triangles. This post will focus on demonstrating that no solutions exist.\n\n### Step-by-step method of solving\n\n1. Draw the triangle and label all given information. It doesn’t need to be to scale at all! It also doesn’t matter how you orient the triangle as long as side a is across from angle A, side b is across from angle B, and side c is across from angle C. Just draw and label.\n2. Use the Law of Sines to solve for the angle across from the second side. Which one is the second side? Well in SSA you have two sides. One of the angle will pair with a side, like angle A and side a, or angle B and side b. There will be an unused side that isn’t paired. We are going to try to find that angle.\n\nLet’s walk through an example.\n\nSide-Side-Angle Triangle Example\n$B = 55 ^{\\circ}, b=8.99,a=26.22$\n\nFollowing the steps above, let’s draw this triangle!", null, "Notice that angle A is in the bottom right corner. Often students and teachers like to put angle A at the top. It doesn’t matter as long as the sides go across from the angles! You could draw the triangle like this as well. It won’t change the answer.", null, "The Law of Sines states that $\\displaystyle \\frac{\\sin(A)}{a}=\\frac{\\sin(B)}{b}=\\frac{\\sin(C)}{c}$ As mentioned in step 2, we are going to try to find angle B, because it is the angle opposite the second side. (We can’t find angle C or side c right now because we don’t have any information on that side. Try to solve for either of those to convince yourself if that seems unclear.)\n\n\\begin{align*}\n\\frac{\\sin(55^{\\circ})}{26.22}&=\\frac{\\sin(B)}{8.99} \\\\\n\\sin(B) &= \\frac{26.22\\sin(55^{\\circ})}{8.99} \\\\\n\\sin(B) &\\approx 2.389\n\\end{align*}\n\nUh oh! We have a problem!! Remember that the range of $\\sin(x)$ is $[-1,1]$, meaning the output of this function has to stay between these two values. When we get an answer like the one above, it is IMPOSSIBLE to find an angle to make the statement true and we have an IMPOSSIBLE triangle. Try putting $\\sin^{-1}(2.389)$ into your calculator to check and you’ll get an error message.\n\n### Key Points\n\nDraw a triangle and label your given information. Do not worry at all if the triangle is to scale. Use the Law of Sines to find the second angle that connects to the second side you are given and simplify. When you get $\\sin(A), \\sin(B), \\sin(C)$ is outside of $[-1,1]$, you can confidently answer that this triangle does not exist. If it does fall inside that range, then you should move onto deciding if 1 or 2 triangles are possible. That topic will be covered in another blog post soon." ]

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[ "# Groupoid Explained\n\nIn mathematics, especially in category theory and homotopy theory, a groupoid (less often Brandt groupoid or virtual group) generalises the notion of group in several equivalent ways. A groupoid can be seen as a:\n\nIn the presence of dependent typing, a category in general can be viewed as a typed monoid, and similarly, a groupoid can be viewed as simply a typed group. The morphisms take one from one object to another, and form a dependent family of types, thus morphisms might be typed\n\ng:AB\n\n,\n\nh:BC\n\n, say. Composition is then a total function:\n\n\\circ:(BC)(AB)AC\n\n, so that\n\nh\\circg:AC\n\n.\n\nSpecial cases include:\n\nG\n\n.\n\nGroupoids are often used to reason about geometrical objects such as manifolds. introduced groupoids implicitly via Brandt semigroups.\n\n## Definitions\n\nA groupoid is an algebraic structure\n\n(G,\\ast)\n\nconsisting of a non-empty set\n\nG\n\nand a binary partial function '\n\n\\ast\n\n' defined on\n\nG\n\n.\n\n### Algebraic\n\nA groupoid is a set\n\nG\n\nwith a unary operation\n\n{}-1:G\\toG,\n\nand a partial function\n\n*:G x G\\rightharpoonupG\n\n. Here * is not a binary operation because it is not necessarily defined for all pairs of elements of\n\nG\n\n. The precise conditions under which\n\n*\n\nis defined are not articulated here and vary by situation.\n\n\\ast\n\nand −1 have the following axiomatic properties: For all\n\na\n\n,\n\nb\n\n, and\n\nc\n\nin\n\nG\n\n,\n1. Associativity: If\n\na*b\n\nand\n\nb*c\n\nare defined, then\n\n(a*b)*c\n\nand\n\na*(b*c)\n\nare defined and are equal. Conversely, if one of\n\n(a*b)*c\n\nand\n\na*(b*c)\n\nis defined, then so are both\n\na*b\n\nand\n\nb*c\n\nas well as\n\n(a*b)*c\n\n=\n\na*(b*c)\n\n.\n\na-1*a\n\nand\n\na*{a-1\n\n} are always defined.\n1. Identity: If\n\na*b\n\nis defined, then\n\na*b*{b-1\n\n} = a, and\n\n{a-1\n\n} * a * b = b. (The previous two axioms already show that these expressions are defined and unambiguous.)\n\nTwo easy and convenient properties follow from these axioms:\n\n(a-1)-1=a\n\n,\n• If\n\na*b\n\nis defined, then\n\n(a*b)-1=b-1*a-1\n\n.\n\n### Category theoretic\n\nA groupoid is a small category in which every morphism is an isomorphism, i.e. invertible. More precisely, a groupoid G is:\n\n• A set G0 of objects;\n• For each pair of objects x and y in G0, there exists a (possibly empty) set G(x,y) of morphisms (or arrows) from x to y. We write f : xy to indicate that f is an element of G(x,y).\n• For every object x, a designated element\n\nidx\n\nof G(x,x);\n\ncompx,y,z:G(y,z) x G(x,y)G(x,z):(g,f)\\mapstogf\n\n;\n• For each pair of objects x, y a function\n\ninv:G(x,y)G(y,x):f\\mapstof-1\n\n;\n\nsatisfying, for any f : xy, g : yz, and h : zw:\n\nfidx=f\n\nand\n\nidyf=f\n\n;\n\n(hg)f=h(gf)\n\n;\n\nff-1=idy\n\nand\n\nf-1f=idx\n\n.\n\nIf f is an element of G(x,y) then x is called the source of f, written s(f), and y is called the target of f, written t(f).\n\nMore generally, one can consider a groupoid object in an arbitrary category admitting finite fiber products.\n\n### Comparing the definitions\n\nThe algebraic and category-theoretic definitions are equivalent, as we now show. Given a groupoid in the category-theoretic sense, let G be the disjoint union of all of the sets G(x,y) (i.e. the sets of morphisms from x to y). Then\n\ncomp\n\nand\n\ninv\n\nbecome partial operations on G, and\n\ninv\n\nwill in fact be defined everywhere. We define ∗ to be\n\ncomp\n\nand −1 to be\n\ninv\n\n, which gives a groupoid in the algebraic sense. Explicit reference to G0 (and hence to\n\nid\n\n) can be dropped.\n\nConversely, given a groupoid G in the algebraic sense, define an equivalence relation\n\n\\sim\n\non its elements by\n\na\\simb\n\niff aa−1 = bb−1. Let G0 be the set of equivalence classes of\n\n\\sim\n\n, i.e.\n\nG0:=G/\\sim\n\n. Denote aa−1 by\n\n1x\n\nif\n\na\\inx\n\nwith\n\nx\\inG0\n\n.\n\nNow define\n\nG(x,y)\n\nas the set of all elements f such that\n\n1x*f*1y\n\nexists. Given\n\nf\\inG(x,y)\n\nand\n\ng\\inG(y,z),\n\ntheir composite is defined as\n\ngf:=f*g\\inG(x,z)\n\n. To see that this is well defined, observe that since\n\n(1x*f)*1y\n\nand\n\n1y*(g*1z)\n\nexist, so does\n\n(1x*f*1y)*(g*1z)=f*g\n\n. The identity morphism on x is then\n\n1x\n\n, and the category-theoretic inverse of f is f−1.\n\nSets in the definitions above may be replaced with classes, as is generally the case in category theory.\n\n### Vertex groups\n\nGiven a groupoid G, the vertex groups or isotropy groups or object groups in G are the subsets of the form G(x,x), where x is any object of G. It follows easily from the axioms above that these are indeed groups, as every pair of elements is composable and inverses are in the same vertex group.\n\n### Category of groupoids\n\nA subgroupoid is a subcategory that is itself a groupoid. A groupoid morphism is simply a functor between two (category-theoretic) groupoids. The category whose objects are groupoids and whose morphisms are groupoid morphisms is called the groupoid category, or the category of groupoids, denoted Grpd.\n\nIt is useful that this category is, like the category of small categories, Cartesian closed. That is, we can construct for any groupoids\n\nH,K\n\na groupoid\n\n\\operatorname{GPD}(H,K)\n\nwhose objects are the morphisms\n\nH\\toK\n\nand whose arrows are the natural equivalences of morphisms. Thus if\n\nH,K\n\nare just groups, then such arrows are the conjugacies of morphisms. The main result is that for any groupoids\n\nG,H,K\n\nthere is a natural bijection\n\n\\operatorname{Grpd}(G x H,K)\\cong\\operatorname{Grpd}(G,\\operatorname{GPD}(H,K)).\n\nThis result is of interest even if all the groupoids\n\nG,H,K\n\nare just groups.\n\n### Fibrations and coverings\n\nParticular kinds of morphisms of groupoids are of interest. A morphism\n\np:E\\toB\n\nof groupoids is called a fibration if for each object\n\nx\n\nof\n\nE\n\nand each morphism\n\nb\n\nof\n\nB\n\nstarting at\n\np(x)\n\nthere is a morphism\n\ne\n\nof\n\nE\n\nstarting at\n\nx\n\nsuch that\n\np(e)=b\n\n. A fibration is called a covering morphism or covering of groupoids if further such an\n\ne\n\nis unique. The covering morphisms of groupoids are especially useful because they can be used to model covering maps of spaces.\n\nIt is also true that the category of covering morphisms of a given groupoid\n\nB\n\nis equivalent to the category of actions of the groupoid\n\nB\n\non sets.\n\n## Examples\n\n### Topology\n\nX\n\n, let\n\nG0\n\nbe the set\n\nX\n\n. The morphisms from the point\n\np\n\nto the point\n\nq\n\nare equivalence classes of continuous paths from\n\np\n\nto\n\nq\n\n, with two paths being equivalent if they are homotopic.Two such morphisms are composed by first following the first path, then the second; the homotopy equivalence guarantees that this composition is associative. This groupoid is called the fundamental groupoid of\n\nX\n\n, denoted\n\n\\pi1(X)\n\n(or sometimes,\n\n\\Pi1(X)\n\n). The usual fundamental group\n\n\\pi1(X,x)\n\nis then the vertex group for the point\n\nx\n\n.\n\nAn important extension of this idea is to consider the fundamental groupoid\n\n\\pi1(X,A)\n\nwhere\n\nA\\subsetX\n\nis a chosen set of \"base points\". Here, one considers only paths whose endpoints belong to\n\nA\n\n.\n\n\\pi1(X,A)\n\nis a sub-groupoid of\n\n\\pi1(X)\n\n. The set\n\nA\n\nmay be chosen according to the geometry of the situation at hand.\n\n### Equivalence relation\n\nIf\n\nX\n\nis a set with an equivalence relation denoted by infix\n\n\\sim\n\n, then a groupoid \"representing\" this equivalence relation can be formed as follows:\n• The objects of the groupoid are the elements of\n\nX\n\n;\n• For any two elements\n\nx\n\nand\n\ny\n\nin\n\nX\n\n, there is a single morphism from\n\nx\n\nto\n\ny\n\nif and only if\n\nx\\simy\n\n.\n\n### Group action\n\nG\n\nacts on the set\n\nX\n\n, then we can form the action groupoid (or transformation groupoid) representing this group action as follows:\n• The objects are the elements of\n\nX\n\n;\n• For any two elements\n\nx\n\nand\n\ny\n\nin\n\nX\n\n, the morphisms from\n\nx\n\nto\n\ny\n\ncorrespond to the elements\n\ng\n\nof\n\nG\n\nsuch that\n\ngx=y\n\n;\n\nG\n\n.\n\nMore explicitly, the action groupoid is a small category with\n\nob(C)=X\n\nand\n\nhom(C)=G x X\n\nwith source and target maps\n\ns(g,x)=x\n\nand\n\nt(g,x)=gx\n\n. It is often denoted\n\nG\\ltimesX\n\n(or\n\nX\\rtimesG\n\n). Multiplication (or composition) in the groupoid is then\n\n(h,y)(g,x)=(hg,x)\n\nwhich is defined provided\n\ny=gx\n\n.\n\nFor\n\nx\n\nin\n\nX\n\n, the vertex group consists of those\n\n(g,x)\n\nwith\n\ngx=x\n\n, which is just the isotropy subgroup at\n\nx\n\nfor the given action (which is why vertex groups are also called isotropy groups).\n\nAnother way to describe\n\nG\n\n-sets is the functor category\n\n[Gr,Set]\n\n, where\n\nGr\n\nis the groupoid (category) with one element and isomorphic to the group\n\nG\n\n. Indeed, every functor\n\nF\n\nof this category defines a set\n\nX=F(Gr)\n\nand for every\n\ng\n\nin\n\nG\n\n(i.e. for every morphism in\n\nGr\n\n) induces a bijection\n\nFg\n\n:\n\nX\\toX\n\n. The categorical structure of the functor\n\nF\n\nassures us that\n\nF\n\ndefines a\n\nG\n\n-action on the set\n\nG\n\n. The (unique) representable functor\n\nF\n\n:\n\nGr\n\nSet\n\nis the Cayley representation of\n\nG\n\n. In fact, this functor is isomorphic to\n\nHom(Gr,-)\n\nand so sends\n\nob(Gr)\n\nto the set\n\nHom(Gr,Gr)\n\nwhich is by definition the \"set\"\n\nG\n\nand the morphism\n\ng\n\nof\n\nGr\n\n(i.e. the element\n\ng\n\nof\n\nG\n\n) to the permutation\n\nFg\n\nof the set\n\nG\n\n. We deduce from the Yoneda embedding that the group\n\nG\n\nis isomorphic to the group\n\n\\{Fg\\midg\\inG\\}\n\n, a subgroup of the group of permutations of\n\nG\n\n.\n\n### Homological Algebra\n\nA two term complex\n\nC1\\overset{d}{}C0\n\nof objects in a concrete Abelian category can be used to form a groupoid. It has as objects the set\n\nC0\n\nand arrows\n\nC1 ⊕ C0\n\nwhere the source morphism is just the projection onto\n\nC0\n\nwhile the target morphism is the addition of projection onto\n\nC1\n\ncomposed with\n\nd\n\nand projection onto\n\nC0\n\n. That is, given\n\nc1+c0\\inC1 ⊕ C0\n\nwe have\n\nt(c1+c0)=d(c1)+c0\n\nOf course, if the abelian category is the category of coherent sheaves on a scheme, then this construction can be used to form a presheaf of groupoids.\n\n### Puzzles\n\nWhile puzzles such as the Rubik's Cube can be modeled using group theory (see Rubik's Cube group), certain puzzles are better modeled as groupoids.\n\nThe transformations of the fifteen puzzle form a groupoid (not a group, as not all moves can be composed). This groupoid acts on configurations.\n\n### Mathieu groupoid\n\nThe Mathieu groupoid is a groupoid introduced by John Horton Conway acting on 13 points such that the elements fixing a point form a copy of the Mathieu group M12.\n\n## Relation to groups\n\nIf a groupoid has only one object, then the set of its morphisms forms a group. Using the algebraic definition, such a groupoid is literally just a group. Many concepts of group theory generalize to groupoids, with the notion of functor replacing that of group homomorphism.\n\nIf\n\nx\n\nis an object of the groupoid\n\nG\n\n, then the set of all morphisms from\n\nx\n\nto\n\nx\n\nforms a group\n\nG(x)\n\n(called the vertex group, defined above). If there is a morphism\n\nf\n\nfrom\n\nx\n\nto\n\ny\n\n, then the groups\n\nG(x)\n\nand\n\nG(y)\n\nare isomorphic, with an isomorphism given by the mapping\n\ng\\tofgf-1\n\n.\n\nEvery connected groupoid - that is, one in which any two objects are connected by at least one morphism - is isomorphic to an action groupoid (as defined above)\n\n(G,X)\n\n. By connectedness, there will only be one orbit under the action. If the groupoid is not connected, then it is isomorphic to a disjoint union of groupoids of the above type (possibly with different groups\n\nG\n\nand sets\n\nX\n\nfor each connected component).\n\nNote that the isomorphism described above is not unique, and there is no natural choice. Choosing such an isomorphism for a connected groupoid essentially amounts to picking one object\n\nx0\n\n, a group isomorphism\n\nh\n\nfrom\n\nG(x0)\n\nto\n\nG\n\n, and for each\n\nx\n\nother than\n\nx0\n\n, a morphism in\n\nG\n\nfrom\n\nx0\n\nto\n\nx\n\n.\n\nIn category-theoretic terms, each connected component of a groupoid is equivalent (but not isomorphic) to a groupoid with a single object, that is, a single group. Thus any groupoid is equivalent to a multiset of unrelated groups. In other words, for equivalence instead of isomorphism, one need not specify the sets\n\nX\n\n, only the groups\n\nG.\n\nFor example,\n• The fundamental groupoid of\n\nX\n\nis equivalent to the collection of the fundamental groups of each path-connected component of\n\nX\n\n, but an isomorphism requires specifying the set of points in each component;\n• The set\n\nX\n\nwith the equivalence relation\n\n\\sim\n\nis equivalent (as a groupoid) to one copy of the trivial group for each equivalence class, but an isomorphism requires specifying what each equivalence class is:\n• The set\n\nX\n\nequipped with an action of the group\n\nG\n\nis equivalent (as a groupoid) to one copy of\n\nG\n\nfor each orbit of the action, but an isomorphism requires specifying what set each orbit is.\n\nThe collapse of a groupoid into a mere collection of groups loses some information, even from a category-theoretic point of view, because it is not natural. Thus when groupoids arise in terms of other structures, as in the above examples, it can be helpful to maintain the full groupoid. Otherwise, one must choose a way to view each\n\nG(x)\n\nin terms of a single group, and this choice can be arbitrary. In our example from topology, you would have to make a coherent choice of paths (or equivalence classes of paths) from each point\n\np\n\nto each point\n\nq\n\nin the same path-connected component.\n\nAs a more illuminating example, the classification of groupoids with one endomorphism does not reduce to purely group theoretic considerations. This is analogous to the fact that the classification of vector spaces with one endomorphism is nontrivial.\n\nMorphisms of groupoids come in more kinds than those of groups: we have, for example, fibrations, covering morphisms, universal morphisms, and quotient morphisms. Thus a subgroup\n\nH\n\nof a group\n\nG\n\nyields an action of\n\nG\n\non the set of cosets of\n\nH\n\nin\n\nG\n\nand hence a covering morphism\n\np\n\nfrom, say,\n\nK\n\nto\n\nG\n\n, where\n\nK\n\nis a groupoid with vertex groups isomorphic to\n\nH\n\n. In this way, presentations of the group\n\nG\n\ncan be \"lifted\" to presentations of the groupoid\n\nK\n\n, and this is a useful way of obtaining information about presentations of the subgroup\n\nH\n\n. For further information, see the books by Higgins and by Brown in the References.\n\n## Properties of the category Grpd\n\n• Grpd is both complete and cocomplete\n• Grpd is a cartesian closed category\n\nThe inclusion\n\ni:Grpd\\toCat\n\nhas both a left and a right adjoint:\n\n\\homGrpd(C[C-1],G)\\cong\\homCat(C,i(G))\n\n\\homCat(i(G),C)\\cong\\homGrpd(G,Core(C))\n\nHere,\n\nC[C-1]\n\ndenotes the localization of a category that inverts every morphism, and\n\nCore(C)\n\ndenotes the subcategory of all isomorphisms.\n\nN:Grpd\\tosSet\n\nembeds Grpd as a full subcategory of the category of simplicial sets. The nerve of a groupoid is always Kan complex.\n\nThe nerve has a left adjoint\n\n\\homGrpd(\\pi1(X),G)\\cong\\homsSet(X,N(G))\n\nHere,\n\n\\pi1(X)\n\ndenotes the fundamental groupoid of the simplicial set X.\n\n## Lie groupoids and Lie algebroids\n\nWhen studying geometrical objects, the arising groupoids often carry some differentiable structure, turning them into Lie groupoids.These can be studied in terms of Lie algebroids, in analogy to the relation between Lie groups and Lie algebras.\n\n## Notes and References\n\n1. Book: Dicks & Ventura. 1996. [{{Google books|plainurl=y|id=3sWSRRfNFKgC|page=6|text=G has the structure of a graph}} The Group Fixed by a Family of Injective Endomorphisms of a Free Group]. 6.\n2. http://eom.springer.de/b/b017600.htm Brandt semigroup\n3. Proof of first property: from 2. and 3. we obtain a−1 = a−1 * a * a−1 and (a−1)−1 = (a−1)−1 * a−1 * (a−1)−1. Substituting the first into the second and applying 3. two more times yields (a−1)−1 = (a−1)−1 * a−1 * a * a−1 * (a−1)−1 = (a−1)−1 * a−1 * a = a. ✓\nProof of second property: since a * b is defined, so is (a * b)−1 * a * b. Therefore (a * b)−1 * a * b * b−1 = (a * b)−1 * a is also defined. Moreover since a * b is defined, so is a * b * b−1 = a. Therefore a * b * b−1 * a−1 is also defined. From 3. we obtain (a * b)−1 = (a * b)−1 * a * a−1 = (a * b)−1 * a * b * b−1 * a−1 = b−1 * a−1. ✓\n4. J.P. May, A Concise Course in Algebraic Topology, 1999, The University of Chicago Press (see chapter 2)\n5. Web site: fundamental groupoid in nLab. ncatlab.org. 2017-09-17.\n6. https://www.crcpress.com/An-Introduction-to-Groups-Groupoids-and-Their-Representations/Ibort-Rodriguez/p/book/9781138035867 An Introduction to Groups, Groupoids and Their Representations: An Introduction\n7. Jim Belk (2008) Puzzles, Groups, and Groupoids, The Everything Seminar\n8. http://www.neverendingbooks.org/the-15-puzzle-groupoid-1 The 15-puzzle groupoid (1)\n9. http://www.neverendingbooks.org/the-15-puzzle-groupoid-2 The 15-puzzle groupoid (2)\n10. Mapping a group to the corresponding groupoid with one object is sometimes called delooping, especially in the context of homotopy theory, see Web site: delooping in nLab. ncatlab.org. 2017-10-31. ." ]

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[ "Playing around with Euclid and constructions\n\nI was playing with Euclid’s Elements perhaps two months back when I had some spare time (or rather when I was bored with whatever it is I was supposed to do). I had never actually read it before and I guess there are two reasons for it. Firstly there aren’t any modern translation of it into Swedish and the older ones read like and equivalent of Shakespearean English with a hard to read way of spelling so to come into contact with it as a kid is not impossible but nevertheless hard. Also in school geometry isn’t taught in the spirit of Euclid with more of an emphasis on using geometry formulas to teach algebra and once trigonometry enters the picture Euclid presentation seems somewhat irrelevant.\n\nHaving read the first book and skimmed through some parts which I was interested I’m not yet in the camp that yearns for the return of ‘axiomatic geometry’ in schools since even though I find the presentation appealing I still agree more with the camp that thinks that education is supposed to be organized to developing useful skills and — yeah I’m not convinced that this is the case. But it is interesting and I can perhaps appreciate the structure more now that I have a better grasp of how mathematics works in general.\n\nStill, what I’ve appreciated the most looking into this area are the construction (the arguably least relevant apsect of it). I have a a special place in my heart for ‘proof by construction’ and of course the Elements is rife with them. Even if you discard the proof aspect of geometry just sitting and playing with the (virtual) straightedge and compass turns out to be ridiculous amount of fun. To start with the question of ‘Can I do it?’ and then you play around with the tools to see what you end up with. Just two tools end up yielding absurd levels of complexity and compared with the 5+ operations of algebra it’s arguably simple.\n\nThis phase of trying out straightedge construction has also coincided with me wishing to learn how to draw graphs in TikZ (QTikZ) so I get to practice it through outlining construction schemes.\n\nFor example one of the first questions I asked was how to find the circle which intersects three arbitrary non-colinear points in the plane with the compass and straightedge. The reason that was the first question I asked was really because it’s the first computer lab you do at KTH (my school) utilizing linear algebra. Then it has a follow up question of taking an arbitrary number of points with and you’re supposed to find an approximate circle with the least square method but let’s ignore that.\n\nWhat I’ve found so lovely working with these tools have simply been that the constructions almost always seem to practically write themselves once I got a hang of the rules of the game. At least to me.\n\nIn this case we of course start with developing the basic operation if finding a perpendicular to/bisecting a line which follows from the first proposition in the book.\n\nNow this is of course obvious to everyone (especially Americans who I’m told are actually told who Euclid was/might have been in school) but these things aren’t in the Swedish curriculum anymore so to me it’s been like discovering something new and beautiful. You get this almost childish joy from seeing it when you look at it as puzzles rather than formal mathematics.\n\nNow for finding the circle which intersects three points I really started from a simpler question of how one could recover the center of a circle in case it had been lost. For me the fundamental property of a circle is not really its definition but that it is a perfectly symmetric shape and so in cutting a circle (disc) in half I should always end up cutting it along a diagonal. From this principle the following procedure using a cord for finding the circles center is to me very natural\n\nA proof that it works can be made but its intuitively obvious that it works even without one.\n\nFrom this, the idea that a perpendicular to a chord passes through the center of the circle, it is immediately obvious that the perpendicular to two different chords should intersect at the center. This simple idea is all that you need to at least guess that the following procedure using two virtual chords should construct a circle.\n\nOf course at heart, or in the way I at least thought about it while doing it, this was really not a proof of that you can draw a circle through any three points but rather that you can reconstruct a circle from three points that were taken to lie on an ‘existing’ circle. But as I said these intuitive constructions seem to always come with the lines necessary for forming a formal proof and by just drawing 3 extra lines I can complete it to a formal existence proof with a simple argument.", null, "Verifying that circle passes through all three points\n\nThe uniqueness part is also simple to see as the two perpendiculars can only meet at one point in our geometry. (All this is probably in Euclid but I’ve been more into solving stuff for myself than reading through it)\n\nI’ve done this with a tonne of different constructions now and the overall feeling I’ve gotten is that this really should be quite easy to turn into a puzzle game. Maybe I’m wrong but so long as the constructions are put in a certain order (the way I discovered them in this case) it could be a pretty entertaining game. Maybe you need to be able to see the nature of the corresponding proofs for it to work but I wouldn’t think so. I think it should work just fine.\n\nLet me just round off by outlining the algebraic solution to the question of the existence, uniqueness, and construction of a circle from three points. Let", null, "$(a_1,a_2)$,", null, "$(b_1, b_2)$,", null, "$(c_1, c_2)$ be the coordinates of the three points. Then any center", null, "$(x,y)$ of a circle with radius", null, "$R$ intersecting all three points must satisfy the three equations", null, "$(x - a_1)^2 + (y - a_2)^2 = R^2$", null, "$(x - b_1)^2 + (y - b_2)^2 = R^2$", null, "$(x - c_1)^2 + (y - c_2)^2 = R^2$\n\nwhich may  be cast into linear form by subtracting the second from the first and the third from the second to give us", null, "$2(b_1 - a_1)x + 2(b_2 - a_2)y = b_1^2 + b_2^2 - a_1^2 - a_2^2$", null, "$2(c_1 - b_1)x + 2(c_2 - b_2)y = c_1^2 + c_2^2 - b_1^2 - b_2^2$\n\ndiscarding the third. Then we have linear system which is solvable with a unique solution if the points are not colinear as can be seen from either the determinant of the fact that the two columns in the corresponding matrices are linearly independent if that is the case. The radius can be recovered from the distance between", null, "$(x,y)$ and one of the other points.\n\nThis is arguably the simplest solution since all issues of construction, existence and uniqueness are put into a single generalization procedure but you also need much more advanced machinery to do it. The other one takes next to none." ]

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In many milling operations, the cutting tool must step over and make several adjacent cuts to complete machining a feature. As a result, a small cusp of material, called a scallop, will remain between these cuts on any surrounding walls or on the machined surface if a ball end mill is used. the size of the\n\n• ### Ball Mill Production Calculation Excel File Ftmlie Heavy\n\nball mill formulas and calculations. ball mill formulas and calculations our purpose and belief LM heavy industry is committed to provide the global customers with the firstclass products and superior service striving to maximize and optimize the interests and values of the customers and build bright future with high quality. learn more\n\n• ### Critical Speed Of Ball Mill Formula And Derivation\n\nball mill, crusher and rod mill models appeared in the based on work by parameters, such as appearance functions (which derive from measurement index and transfer size, according to the bond ball calculation equation: 4.25 with 12% ball charge and operating at 77.35 critical speed, would meet the design.\n\n• ### Ball Mill Design Power Calculation Linkedin\n\ndec 12, 2016 ball mill power calculation example wet grinding ball mill in closed circuit is to be fed 100 tph of a material with a work index of 15 and a size\n\n• ### Variables In Ball Mill Operation Paul O Abbe 174\n\nball mill operation is often regarded as something of a mystery for several reasons. ball milling is not an art its just physics. the starting point for mill speed calculations is the critical speed. critical speed is the speed at which the grinding media will centrifuge against the wall of the cylinder. the formula\n\n• ### Ball Charges Calculators\n\nball top size calculation of the top size grinding media :-modification of the ball charge: this calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency:\n\n• ### Compound Ball Mill Calculation Pdf\n\nMay 13 2021 · home crusher and mill ball mills working process formulas pdf. ball home rocks process ball mill calculation pdf. ball mill capacity & dimensions -2013. milling 60% of critical gear & pinnion drive used on larger mills JH 014 talk ball mill capacity & dimensions author: jhoffmann created date: design method of ball mill by\n\n• ### Formula S To Calculate Efficiency Of Ball Mill\n\nball mill power calculation example wet grinding ball mill in closed circuit is to calculation OF ball mill grinding efficiency IN literatures IT grinding 350; for dry grinding 355); the grind material bulk mass, gcc. how to size a ball mill design calculator & formula 911 metallurgist. apr 2018 how to size a ball mill design\n\n• ### Ball Mill Feed Formula In E Cel\n\nball mill finish calculator know more. ball mill finish calculator the ball mill finish calculator can be used when an end mill with a full radius a ball mill is used on a contoured surface the tool radius on each side of the cut will leave stock referred to as a scallop the finish of the part will be determined\n\n• ### End Mill Speed And Feed Calculator Martin Chick Amp\n\nspeed and feed calculators ball mill finish calculator part spacing calculator and code characters standard end mill sizes standard drill sizes drill and counterbore sizes. contact. end mill speed & feed calculator. tool dia. in. radial depth of cut. this will adjust the feedrate if less than the tool rad. in. num of flutes\n\n• ### Ball Nose Milling Strategy Guide In The Loupe\n\njun 26, 2017 ball nose milling without a tilt angle. ball nose end mills are ideal for machining 3-dimensional contour shapes typically found in the mold and die industry, the manufacturing of turbine blades, and fulfilling general part radius requirements.to properly employ a ball nose end mill and gain the optimal tool life and part finish, follow the 2-step process below\n\n• ### Ballnose Cut Depth Musings For Feeds And Speeds In G\n\ngot a note from a customer who wanted to understand better why GW calculator works the way it does with cut depth on ballnosed endmills. specifically, he had a cut where he wanted to enter the full diameter of the tool, but he was being limited to effective diameter, which was smaller. but just the diameter out near the end of the ball\n\n• ### Power Consumption Calculation Formulas For Ball Mill\n\nsearch rod mill power vs ball mill to find your need. mill yield power consumption province ball mill efficiency calculations motor power calculation for ball and tube mill binq mining what is the formula to calculate mill motor power of ball mill.\n\n• ### Milling Formulas Gws Tool Group\n\ngeneral milling formulas for working with carbide tools, engravers, end-mills and cutters To find using formula sfm surface feet per minute rpm revolutions per minute dia diameter of end mill rpm 3.82 dia sfm rpm revolutions per minute sfm surface feet per minute dia diameter of end mill\n\n• ### Milling Formulas And Definitions Sandvik Coromant\n\nApr 23 2018 · the milling process definitions cutting speed,v indicates the surface speed at which the cutting edge machines the workpiece. effective or true cutting speed, indicates the surface speed at the effective diameter .this value is necessary for determining the true cutting data at the actual depth of cut .this is a particularly important value when using round insert cutters\n\n• ### Formula Of Calculating Rpm Of Ball Grinding Mill\n\nformula of calculating rpm of ball grinding mill. oct the grinding balls diameter determined by the bond formula has a recommendatory character and serves as a starting point for calculating the necessary proportion grinding media feeding a new mill more precisely adjust the ball load in the mill can only by industrial test performing\n\n• ### Ball Mills Mine\n\nthis formula calculates the critical speed of any ball mill. most ball mills operate most efficiently between 65% and 75% of their critical speed. photo of a 10 Ft diameter by 32 Ft long ball mill in a cement plant. photo of a series of ball mills in a copper plant, grinding the ore for flotation. image of cut away ball mill, showing material\n\n• ### Milling Speed And Feed Calculator\n\nmilling operations remove material by feeding a workpiece into a rotating cutting tool with sharp teeth, such as an end mill or face mill. calculations use the desired tool diameter, number of teeth, cutting speed, and cutting feed, which should be chosen based on the specific cutting conditions, including the workpiece material and tool material.\n\n• ### Ball Mill Sizing Calculations Spreadsheet\n\nball mill sizing calculation spredsheet. ball mill sizing calculations spreadsheet the difficulty in calculation of the above mentioned equation is due to the fact that the the start of milling the first size class is continually diminished curve a fig first faster and ball mills are the most important group of such machines different plot the and distribution in a lognormsheet and\n\n• ### Metric Ball End Mill Calculators Trucut Tool\n\nTo complete calculations, enter known data into the empty boxes on the left hand side below. all mandatory fields for calculations are indicated by all calculations are based on industry formulas and are intended to provide theoretical values. actual results could vary and sgs tool company assumes no responsibility.\n\n• ### Ball Mill Mechanical Formulas\n\nball mill mechanical formulas; ball mill wikipedia. the ball mill is a key piece of equipment for grinding crushed materials, and it is widely used in production lines for powders such as cement, silicates, refractory material, fertilizer, glass ceramics, etc. as well as\n\n• ### How To Size A Ball Mill Design Calculator Amp Formula\n\nmay 15, 2015 total apparent volumetric charge filling including balls and excess slurry on top of the ball charge, plus the interstitial voids in between the balls expressed as a percentage of the net internal mill volume overflow discharge mills operating at low ball fillings slurry may accumulate on top of the ball charge; causing, the total charge filling level to\n\n• ### Ball Mill Design Power Calculation\n\nJan 12 2016 · jun 19, 2015 the basic parameters used in ball mill design rod mill or any tumbling mill sizing are; material to be ground, characteristics, bond work index, bulk density, specific density, desired mill tonnage capacity dtph, operating solids or pulp density, feed size as and maximum chunk size, product size as and maximum and finally the type of circuit openclosed\n\n• ### Bond Formula For The Grinding Balls Size Calculation\n\noct 19, 2017 the grinding balls diameter determined by the bond formula has a recommendatory character and serves as a starting point for calculating the necessary proportion grinding media feeding a new mill. more precisely adjust the ball load in the mill can only by industrial test performing.\n\n• ### Ball Mill Size Calculation Henan Mining Machinery Co Ltd\n\nball mill finish calculator martin chick associate ball mill finish calculator the ball mill finish calculator can be used when an end mill with a full radius a ball mill is used on a contoured surface the tool radius on each side of the cut will leave stock referred to as a scallop. view more details" ]

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[ "### BEAM AND PLATE FORMULATIONS IN PERIDYNAMIC FRAMEWORK\n\nDr. Erkan Oterkus gave a lecture on beam and plate formulations in peridynamic framework during the workshop “Encounter of the third kind” on “Generalized continua and microstructures, organised by M&MoCS. The International Research Center for Mathematics & Mechanics of Complex Systems (M&MoCS) is a Research Center of the Università dell’Aquila. It was established by the Dipartimento di Ingegneria delle Strutture, delle Acque e del Terreno (DISAT) and the Dipartimento di Matematica Pura e Applicata (DMPA). Its administrative headquarters are located in L’Aquila. M&MoCS was established with the financial and logistical support of Provincia di Latina and the Fondazione Tullio Levi-Civita, with which it shares the mission of developing and disseminating scientific culture in the region. It was also founded together with the Dipartimento di Strutture of Università di Roma Tre.\n\nWorkshop \"Encounter of the third kind\" on \"Generalized continua and microstructures\" 3-7 April 2018, Arpino, Italy Erkan Oterkus (University of Strathclyde, Glasgow, UK) Title: \"Beam and plate formulations in peridynamic framework\"\n\n### Pre-cracked thin square plate subjected to velocity boundary condition1\n\nThe pre-cracked thin square plate has dimensions 0.05 m x 0.05 m and it is subjected to velocity boundary condition along its two horizontal edges.\nPeridynamic Theory (PT) is used to predict the mechanical behaviour of the plate, the crack propagation speed and the crack propagation path. PT is a new theory introduced in 2000 by Dr. Stewart Silling: it can be defined as a generalization of Classical Continuum Mechanics (CCM) whose governing equation is integral instead of partial differential as for CCM. Therefore, the formulation remains valid even when a crack occurs during the deformation process of the material. Other advantages of PT over CCM are as follow: no need for an external crack growth criteria (in PT, the damage of the material is naturally present in its formulation), no need for computationally expensive re-meshing procedures, no need of external equations for describing interfaces, capability to predict crack initiation, capability to capture complex behaviours such as crack branching, crack arrest and coalition of multiple cracks. Moreover, PT is also suitable for modeling multiscale and multiphysics problems.\nIn this study, two simulations have been carried out using two different values for the velocity boundary condition: 20 m/s and 50 m/s respectively. The main assumptions are as follow: homogeneous material, isotropic material, prototype micro-elastic brittle material (PMB), Poisson's ratio = 1/3 and plane stress problem. The programming language here employed is Fortran 2003. Some features of the simulation are listed here below:\n\n• Young's modulus = 192.0 GPa\n\n• Material's volumetric mass density = 8,000 Kg/m^3\n\n• Crack length = 0.01 m\n\n• Velocity boundary condition for the first simulation = 20 m/s\n\n• Velocity boundary condition for the second simulation = 50 m/s\n\n• Number of real particles along x-direction = 500\n\n• Number of real particles along y-direction = 500\n\n• Number of real particles along z-direction = 1\n\n• Number of boundary particles in the bottom boundary region = 1500\n\n• Number of boundary particles in the upper boundary region = 1500\n\n• Grid spacing (discretization parameter) = 0.0005 m\n\n• Horizon per unit of grid spacing = 3.015\n\n• time step = 1.3367 E-8 s\n\n• Number of time steps for the first simulation = 1,250.\n\n• Number of time steps for the second simulation = 1,000.\n\n• Critical stretch = 0.04472\n\nThe videos of the two simulations are showed here below:\n\nTime evolution of the local damage when velocity = 20 m/s.\n\nThis video was produced by Denny Demeo.\n\nTime evolution of the local damage when velocity = 50 m/s.\n\nThis video was produced by Denny Demeo.\n\n### Finite Slab with Time-Dependent Surface Temperature\n\nThe length of the 1D slab here considered is 0.01 m and its initial temperature is 0 K. The left edge of the slab is kept at 0 K throughout the whole simulation, whilst the temperature of the right edge grows linearly with time.\nBond-based Peridynamic Theory is used to predict the time evolution of the temperature field within the slab.\nThe programming language here employed is Fortran 2003. Some features of the simulation are listed here below:\n\n• Specific heat capacity at constant volume of the material = 64.0 J/KgK\n\n• Thermal conductivity of the material: 233.0 W/mK\n\n• Material's volumetric mass density = 260.0 Kg/m^3\n\n• Slab length = 0.01 m\n\n• Area of the slab section = 1.0e-8 m^2\n\n• Coefficient for the linear temperature increment of the slab's right edge = 500.0\n\n• Number of real particles = 100\n\n• Number of virtual particles at the slab's left edge = 3\n\n• Number of virtual particles at the slab's right edge = 3\n\n• Grid spacing (discretization parameter) = 0.0001 m\n\n• Horizon per unit of grid spacing = 3.015\n\n• Time step = 1.0 e-6 s\n\n• Number of time steps = 50,000.\n\nThe results are in good agreement with the analytical solution of the problem. Two videos of the simulation are showed here below:\n\nEvolution in time of the slab's temperature field until 0.05 s.\n\nThis video was produced by Denny Demeo.\n\nEvolution in time of the slab's temperature field until 0.05 s.\n\nThis video was produced by Denny Demeo.\n\n### Plate Under Thermal Shock with Insulated Boundaries\n\nThe length and width of the 2D plate here considered are both 10 m. The initial temperature is 0 K. At the left edge of the plate there are three vertical layers of virtual particles which are subjected to a thermal shock, which is mathematically represented as follows: T = 5.0*time*exp(-2*time).\nBond-based Peridynamic Theory is used to predict the time evolution of the temperature field within the plate for the first 6 seconds.\nThe programming language here employed is Fortran 2003. Some features of the simulation are listed here below:\n\n• Specific heat capacity at constant volume of the material = 1.0 J/KgK\n\n• Thermal conductivity of the material: 1.0 W/mK\n\n• Material's volumetric mass density = 1.0 Kg/m^3\n\n• Plate length = 10 m\n\n• Plate width = 10 m\n\n• Plate thickness = 1 m\n\n• Number of real particles along x-direction = 300\n\n• Number of real particles along y-direction = 300\n\n• Number of virtual particles at the plate's left edge = 900\n\n• Grid spacing (discretization parameter) = 0.0333 m\n\n• Horizon per unit of grid spacing = 3.015\n\n• Time step = 5.0 e-4 s\n\n• Number of time steps = 12,000.\n\nHosseini-Tehrani and Eslami (2000) have used the Boundary Element Method to solve the same problem. The results are in close agreement. Two videos of the simulation are showed here below:\n\nTime evolution of the temperature field at y = 0.0 until 6 s.\n\nThis video was produced by Denny Demeo.\n\nTime evolution of the temperature field within the plate until 6 s.\n\nThis video was produced by Denny Demeo\n\n### Plate Under Thermal Shock with Boundaries at T = 0 K\n\nThe length and width of the 2D plate here considered are both 10 m. The initial temperature is 0 K. At the left edge of the plate there are three vertical layers of virtual particles which are subjected to a thermal shock, which is mathematically represented as follows: T = 5.0*time*exp(-2*time). The remaining three edges of the plate are kept at T = 0 K throughout the whole simulation.\nBond-based Peridynamic Theory is used to predict the time evolution of the temperature field within the plate for the first 6 seconds.\nThe programming language here employed is Fortran 2003. Some features of the simulation are listed here below:\n\n• Specific heat capacity at constant volume of the material = 1.0 J/KgK\n\n• Thermal conductivity of the material: 1.0 W/mK\n\n• Material's volumetric mass density = 1.0 Kg/m^3\n\n• Plate length = 10 m\n\n• Plate width = 10 m\n\n• Plate thickness = 1 m\n\n• Number of real particles along x-direction = 300\n\n• Number of real particles along y-direction = 300\n\n• Number of virtual particles at the plate's left edge = 900\n\n• Grid spacing (discretization parameter) = 0.0333 m\n\n• Horizon per unit of grid spacing = 3.015\n\n• Time step = 5.0 e-4 s\n\n• Number of time steps = 12,000 (video no.1) and 36,000 (video no.2).\n\nTwo videos of the simulation are showed here below:\n\nTime evolution of the temperature field at y = 0.0 until 6 s.\n\nThis video was produced by Denny Demeo.\n\nTime evolution of the temperature field within the plate until 18 s.\n\nThis video was produced by Denny Demeo.\n\n### Dissimilar Materials with a Pre-Existing Insulated Crack:\n\nThe length and width of the 2D plate here considered are both 2 cm. A crack, considered insulated, is embedded in the middle of the plate. The initial temperature is 0 K. The plate is made by two different materials: material no.1 for the upper half and material no.2 for the lower half. The only difference between the two materials is their value of thermal conductivity [W/cmK]. At both upper and lower edge of the plate there are three horizontal layers of virtual particles which are kept at constant temperature throughout the simulation: 100 K and -100 K respectively. The two vertical edges are considered to be insulated.\nBond-based Peridynamic Theory is used to predict the time evolution of the temperature field within the plate until 0.5 seconds.\nThe programming language here employed is Fortran 2003. Some features of the simulation are listed here below:\n\n• Specific heat capacity at constant volume of the material = 1.0 J/KgK\n\n• Thermal conductivity of material no.2: 1.14 W/cmK\n\n• Thermal conductivity of material no.1: 0.114 W/cmK\n\n• Material's volumetric mass density = 1.0 Kg/cm^3\n\n• Plate length = 2 cm\n\n• Plate width = 2 cm\n\n• Plate thickness = 0.01 cm\n\n• Crack length = 1 cm\n\n• Number of real particles along x-direction = 200\n\n• Number of real particles along y-direction = 200\n\n• Number of virtual particles at the plate's lower edge = 600\n\n• Number of virtual particles at the plate's upper edge = 600\n\n• Grid spacing (discretization parameter) = 0.01 cm\n\n• Horizon per unit of grid spacing = 3.015\n\n• Time step = 1.0 e-4 s\n\n• Number of time steps = 5,000.\n\nThe same simulation has been carried out by using ANSYS. The results are in close agreement. Two videos of the simulation are showed here below:\n\nTime evolution of the temperature field [K] within the plate until 0.5 s for K1 = K2 = 1.14 W/cmK.\n\nThis video was produced by Denny Demeo.\n\nTime evolution of the temperature field [K] within the plate until 0.5 s for K2 = 1.14 W/cmK amd K1 = 0.114 W/cmK.\n\nThis video was produced by Denny Demeo." ]

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[ "# Science\n\nHow is the pH of a strong acid compared to a weak acid of the same concentration?\n\n1. 👍\n2. 👎\n3. 👁\n\n## Similar Questions\n\n1. ### science\n\nWhich acid-base chemical reaction is irreversible?(1 point) strong acid added to water water on its own weak base added to water weak acid added to water Hydrochloric acid is a strong acid. Acetic acid is a weak acid. Which\n\n2. ### chemistry\n\nStrong base is dissolved in 665 mL of 0.400 M weak acid (Ka = 3.69 × 10-5) to make a buffer with a pH of 3.94. Assume that the volume remains constant when the base is added. HA(aq) + OH-(aq) -> H2O(l) + A-(aq) Calculate the pKa\n\n3. ### Chemistry\n\nSelect the statements that correctly describe buffers.? 1) The pH of a buffer solution does not change significantly when any amount of a strong acid is added. 2) The Ka of a buffer does not change when any amount of an acid is\n\n4. ### chemistry\n\nIs this a buffer system? NH3/(NH4)2SO4? It is formed by mixing NH3 (weak base) and a strong acid (H2SO4), but according to my worksheet it is not a buffer system. Why is this so? Isnt it a buffer as long as it consists of a weak\n\n1. ### CHEMISTRY\n\nHOW CAN U TELL IF HNO3 +KNO3 IS A BUFFER SOLUTION A buffer solution must contain a weak acid and its conjugate base OR a weak base and its conjugate acid. HNO3 is a strong base and KNO3 is the salt of a strong base (KOH) and a\n\n2. ### Chemistry\n\nStrong base is dissolved in 675 ml of 0.200 m weak acid (ka=3.25x10^-5) to make a buffer with a ph of 3.95. Assume that the volume remains constant when the base is added. HA + OH ---> H2O + A^- calculate the pka value of the acid\n\n3. ### Chemistry\n\nWhy is the reaction rate of a given metal with a stronger acid is faster than with weaker acid? Which one required more volume to produce neutral solution, a strong acid or strong base,or are they just the same? Can you explain it\n\n4. ### chemistry\n\nA. Strong Base 1.) What is the concentration of a solution of KOH for which the pH is 11.89? 2.) What is the pH of a 0.011M solution of Ca(OH)2? B. Weak Acid 1.) The pH of a 0.060M weak monoprotic acid HA is 3.44. Calculate the Ka\n\n1. ### chemistry\n\nLactic acid is a weak acid with the formula , HCH3H5O3, the Ka for lactic acid is 1.38 x 10-4. In aqueous solution, lactic acid partially dissociates according to the following reaction: HCH3H5O3 ⇔ CH3H5O3- + H+ Use the Ka\n\n2. ### Chemistry\n\nFind the pH of mixture of acids. 0.185 M in HCHO2 and 0.225 M in HC2H3O2 Im using an ice chart of weak acid and putting in strong acid in H+ initiAL concentration. I've done the problems many different ways but cannot seem to get\n\n3. ### Chemistry\n\nA substance of Ka of 1 x 10^–5 would be classified as a ____. A.strong acid B.weak acid C.strong base D.weak base I think it is B...?\n\n4. ### Chm 2\n\nThe question asks, Consider the following weak acids and their Ka values Acetic acid Ka = 1.8x10^-5 Phosphoric acid Ka = 7.5x10^-3 Hypochlorous acid Ka = 3.5x10^-8 What weak acid-conjugate base buffer system from the acids listed" ]

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[ "# Welcome to the Calculus Bible!\n\nThe Calculus Bible is a guide to the Advanced Placement tests in AB and BC Calculus. It can serve as a review packet before quizzes or tests and is also a great review of the entire year's matieral before the AP.\n\nCombined, the two cover:\n\n• functions (domain, range, asymptotes, and fundamental properties)\n• graphs you must know\n• the derivative (definition, formulas, theorems, etc.)\n• applications of the derivative (slope of a curve, slope of the normal line, curve sketching, max-min word problems, related rates, motion along a line, etc.)\n• antiderivatives (the definition and applications)\n• techniques of integration (including u-substituion, the inverse chain rule, and integration by parts)\n• approximations of the definite integral\n• the definite integral as area\n• applications of the integral (motion along a line, mean value of a function, area between two curves, and volume of a solid of revolution)\n• advanced techniques of integration (reduction formulas, power formualas, etc.)\n• trigonometric substitutions\n• partial fractions\n• improper integrals\n• further applications of the definite integral (volumes of known cross-sections, length of a plane curve, and work)\n• parametrically-defined equations (definition, derivative, and length of the plane curve)\n• polar functions (lines, circles, and rectangular hyperbolas)\n• limacons\n• lemniscates\n• spirals\n• roses\n• sequences and series (convergence and tests for convergence)\n• function approximations by infinite series" ]

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[ "# Sine and Cosine GraphsPage 1\n\n#### WATCH ALL SLIDES\n\nSlide 1", null, "Graphing Sine and Cosine\n\nSlide 2", null, "## General Graphs\n\nGraphs are collections of points\n\nIndependent variables are related to dependent variables\n\nDomain is set of independent variable values\n\nRange is set of dependent variable values\n\nSlide 3", null, "## Periodic Graphs\n\nGraphs that repeat infinitely in intervals are called periodic\n\nEx: Sea level as a function of time\n\nThe most common periodic graphs are the graphs of the trigonometric functions\n\nSine, cosine, tangent, cosecant, secant, and cotangent\n\nSlide 4", null, "## The Sine Graph\n\nOscillation-one complete “cycle” of the graph\n\nPeriod-”time” (horizontal distance) it takes the graph to complete one oscillation\n\nAmplitude-the furthest vertical distance the graph ventures from the vertical center\n\nSlide 5", null, "The Sine Graph\n\nWe think of the sine graph as starting at an x-value of zero\n\nIt begins at a y-value of zero\n\nIt then moves up the distance of its amplitude\n\nNext it moves back down to the x-axis\n\nThen it moves down the distance of its amplitude\n\nThe it completes an oscillation by moving back up the x-axis\n\nSlide 6", null, "The Sine Graph\n\nThe period of the sine graph is…\n\nSo the horizontal length of each “piece” of the sine graph is…\n\nThe amplitude of the sine graph is…(think about the values of sine)\n\nSlide 7", null, "The Cosine Graph\n\nThe cosine graph is the same as the sine graph, only it has been shifted horizontally\n\nSame period\n\nSame amplitude\n\nSlide 8", null, "## The Cosine Graph\n\nSame period\n\nSame amplitude\n\nThe “starting point” is still an x-value of zero, but is now a y-value of 1\n\nSlide 9", null, "## Sine and Cosine Graphs\n\nWhile the idea of a “starting point” is very helpful when graphing, it is merely an idea\n\nThese graphs have infinite domains\n\nWe have thus far only looked at library versions. There will be transformations." ]

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[ "", null, "# 基于2D SDF的体积字实现\n\nQQ群：793972859（原群已满员）\n\n## 概述\n\nTMP Shader的作者之一（好像是负责写Shader）sschaem在2014年就做过的体积字效果，社区反响不强烈，加之别的功能需求更重要一直没加进去，效果更牛的VRTFX （像是VR Text FX的意思）一直没见发布，参考：\n\nhttps://forum.unity.com/threads/wip-vrtfx-volumetric-rendering-titling-effects.440048\n\nVolumetric Rendering - Alan Zucconi\n\nRay Marching and Signed Distance Functions\n\n## 一、形状计算\n\n1.1 SDF数据", null, "", null, "1.2 体和坐标系\n\n2D SDF拉出一个厚度能表示一个3D场，可以对应一个Cube模型，长宽高可以不相等，“显然”于模型内的任一点都能算出是否属于体积字之内。TMP生成的2D片局部坐标都是在XY平面上的，正面朝向负Z，见下图：", null, "1.3 大概算法", null, "", null, "sdf2len = 2 * _GradientScale / (max(_TextureWidth * sdfUvScale.x, _TextureHeight * sdfUvScale.y))\n\n1.4 Shader中所需的单个字形的数据\n\nunpack posBLAndUvFactor，算出sdfUvScaleOffset ，给定字符局部坐标p，可以得到sdfUV = p.xy * sdfUvScaleOffset.xy + sdfUvScaleOffset.zw。FaceUV同理。\n\n``````// UnityCG.cginc里的view dir相关方法\n// Computes world space view direction, from object space position\ninline float3 UnityWorldSpaceViewDir( in float3 worldPos )\n{\nreturn _WorldSpaceCameraPos.xyz - worldPos;\n}\n\n// Computes world space view direction, from object space position\n// *Legacy* Please use UnityWorldSpaceViewDir instead\ninline float3 WorldSpaceViewDir( in float4 localPos )\n{\nfloat3 worldPos = mul(unity_ObjectToWorld, localPos).xyz;\nreturn UnityWorldSpaceViewDir(worldPos);\n}\n\n// Computes object space view direction\ninline float3 ObjSpaceViewDir( in float4 v )\n{\nfloat3 objSpaceCameraPos = mul(unity_WorldToObject, float4(_WorldSpaceCameraPos.xyz, 1)).xyz;\nreturn objSpaceCameraPos - v.xyz;\n}\n``````\n\nVS还是比较简单的，一个字符8个顶点，计算量也不大，PS里的重复计算可以移动到VS里来算。\n\n``````// vs的输入\nstruct appdata\n{\nfloat4 vertex : POSITION;\nfloat3 bound : NORMAL; // 字符cube的长宽高\nfloat4 posBLAndUvFactor : TANGENT; // xy是字符cube左下点的局部坐标，zw是sdf和face的uv转换系数被pack的值\nfixed4 color : COLOR;\nfloat2 uv0 : TEXCOORD0; // uvBL\nfloat2 uv1 : TEXCOORD1; // uv2BL\n};\n\n// ps的输入\nstruct v2f\n{\nfloat4 vertex : SV_POSITION;\nfixed4 faceColor : COLOR;\nfloat4 sdfUvScaleOffset : TEXCOORD0; //转换mesh局部坐标到sdf uv，xy是系数，zw是常量\nfloat4 faceUvScaleOffset : TEXCOORD1; //转换mesh局部坐标到face uv，xy是系数，zw是常量\nfloat3 viewDir : TEXCOORD2; //\nfloat3 cuboidLocalPos : TEXCOORD3; //xyz是位置，左下角在原点。\nfloat4 bound : TEXCOORD4; // xyz是模型size, w是转换sdf距离到模型距离的系数\nfloat3 localLightPos : TEXCOORD5; //字符局部坐标系下光的位置，表面到光源方向\nfloat4 volParam : TEXCOORD6; //x是模型空间下的轮廓圆角半径\n};\n\n// vertex shader\nv2f Vert (appdata v)\n{\nv2f o;\no.vertex = UnityObjectToClipPos(v.vertex);\no.faceColor = v.color;\n\n// 左下角平移到局部坐标系原点，此时坐标范围是[0, bound]\no.cuboidLocalPos = v.vertex - float4(v.posBLAndUvFactor.xy, 0, 0);\n\nfloat2 inverseBound = 1 / v.bound.xy;\nfloat2 sdfUvScale = UnpackFloat(v.posBLAndUvFactor.z, 1) * inverseBound / float2(_TextureWidth, _TextureHeight);\no.sdfUvScaleOffset = float4(sdfUvScale, v.uv0);\n\nfloat2 faceUvScale = UnpackFloat(v.posBLAndUvFactor.w, 512) * inverseBound;\no.faceUvScaleOffset = float4(faceUvScale, UnpackFloat(v.uv1.x, 512));\n\n// 视线转换到模型局部坐标系，此时是整个模型（很多字符）的局部坐标系，到ps里再normalize\n// _WorldSpaceCameraPos是float3，补上w=1才能正确转换\n// 没找到更好的计算透视和正交的方法，clip坐标系下载近平面的投影转换到localspace比较好，但没看到InvMVP矩阵\nfloat3 perspectiveViewDir = v.vertex - mul(unity_WorldToObject, float4(_WorldSpaceCameraPos.xyz, 1));\nfloat3 orthographicViewDir = mul(unity_WorldToObject, mul(unity_MatrixInvV, float4(0, 0, -1, 0)));\no.viewDir = lerp(perspectiveViewDir, orthographicViewDir, unity_OrthoParams.w);\n\n// sdf距离到模型距离的系数，sdf的0.5对应字体的_GradientScale个图素，对应的uv范围，再到对应的模型距离\n// xy方向上可能是不一致的，为了正确性取相对小的步长\nfloat sdf2len = 2 * _GradientScale / (max(_TextureWidth * sdfUvScale.x, _TextureHeight * sdfUvScale.y)) * _stepLenFactor;\no.bound = float4(v.bound, sdf2len);\n\no.localLightPos = mul(unity_WorldToObject, _WorldSpaceLightPos0);\n\no.volParam = float4(v.bound.x * _outlineRadius, 0, 0, 0);\nreturn o;\n``````\n\n1.5 PS里算RayMarching\n\n``````loat SampleSdfValueByLocalPos(v2f input, float3 localPos)\n{\nfloat2 sdfUv = localPos.xy * input.sdfUvScaleOffset.xy + input.sdfUvScaleOffset.zw;\n// 编译警告说for里得用lod版本的tex2D\nreturn tex2Dlod(_MainTex, float4(sdfUv, 0, 0)).a;\n}\n\nfloat2 RayMarching(v2f input, out float3 localPos, float3 viewDir)\n{\nlocalPos = input.cuboidLocalPos.xyz;\n\n// 除以水平距离，这样在乘以水平距离就知道3d方向的增量\nviewDir /= max(length(viewDir.xy), 0.001);\n\nint i = 0;\nfloat sdfValue;\nfor (; i < _loopCount; i++)\n{\nsdfValue = SampleSdfValueByLocalPos(input, localPos);\nfloat sdfDist = _outline - sdfValue;\nfloat meshDist;\n\nif (sdfDist <= _outlineEpsilon)\nreturn float2(i, sdfValue);\n\nmeshDist = sdfDist * input.bound.w;\n\n// sdf距离转到模型距离，向前march\nlocalPos += viewDir * meshDist;\n\n// 到达模型外面要被剔除掉\n// 剔除并不需要epsilon解决精度问题比较欣慰\nclip(localPos * (input.bound.xyz - localPos));\n}\n\n// 没有到达范围内，但应该也很接近了\nsdfValue = SampleSdfValueByLocalPos(input, localPos);\nreturn float2(i, sdfValue);\n}\n\n// ps\nfixed4 Frag (v2f input) : SV_Target\n{\nfloat3 localPos;\nfloat3 viewDir = normalize(input.viewDir);\nfloat2 result = RayMarching(input, localPos, viewDir);\nfixed4 col = 1 - result.x / _loopCount;\ncol.a = 1;\nreturn col;\n}\n``````", null, "_loopCount为10，_outlineEpsilon为0的情况", null, "_loopCount为10，_outlineEpsilon为0.01的情况", null, "_loopCount为20，_outlineEpsilon为0.01的情况", null, "", null, "_loopCount为10，_outlineEpsilon为0.01的情况\n\n## 二、字体法线", null, "z方向没变化为0，只算xy方向SDF值的梯度即可，_SideNormalSampleDelta用于微调计算梯度的Delta。\n\n``````// 计算法线，前后面法线通过z计算，4个侧面法线是sdf的下降梯度\nvoid ComputeNormal(v2f input, float3 localPos, out float3 frontBackNormal, out float3 sideNormal)\n{\n// 前或后面的法线，不是侧面的. 圆角过渡部分也需要正反面法线\nfrontBackNormal = float3(0, 0, localPos.z > input.bound.z - localPos.z ? 1 : -1);\n\n// 用小位置差算梯度，得在中心点周围的4个点采样，只采样2个是不对的\nfloat2 deltaPos = input.bound.xy * _SideNormalSampleDelta;\nfloat sdfDeltaX1 = SampleSdfValueByLocalPos(input, localPos + float3(deltaPos.x, 0, 0));\nfloat sdfDeltaX0 = SampleSdfValueByLocalPos(input, localPos - float3(deltaPos.x, 0, 0));\nfloat sdfDeltaY1 = SampleSdfValueByLocalPos(input, localPos + float3(0, deltaPos.y, 0));\nfloat sdfDeltaY0 = SampleSdfValueByLocalPos(input, localPos - float3(0, deltaPos.y, 0));\n// 避免除以0，不然所在像素颜色未定义\nsideNormal = Unity_SafeNormalize(float3(sdfDeltaX0 - sdfDeltaX1, sdfDeltaY0 - sdfDeltaY1, 0));\n}\n\n// z很接近前后面后frontBackNormal，否则用sideNormal，光照部分就不贴出来了\n``````", null, "2.1 字形法线的圆角过渡", null, "", null, "", null, "``````// 计算圆角过渡中侧面的权重，input.volParam.x是半径\nfloat ComputeSideWeight(v2f input, float3 localPos, float sdfValue)\n{\n// sdf中到边界的距离，转换为模型的长度\nfloat toOutlineDist = (sdfValue - _outline) * input.bound.w;\nfloat outlineFactor = max(input.volParam.x - toOutlineDist, 0);\nfloat depthFactor = max(input.volParam.x - min(localPos.z, input.bound.z - localPos.z), 0);\n\n// 修改斜率，使圆角边缘平缓，中间陡一点\noutlineFactor *= outlineFactor;\ndepthFactor *= depthFactor;\n// 对side和frontback做过渡\nfloat sideWeight = outlineFactor / (outlineFactor + depthFactor);\nreturn sideWeight;\n}\n``````", null, "## 三、表面纹理\n\n``````// 采样face tex，可以是diffuse、normal、specular等\nfloat4 SampleFaceTexByLocalPos(v2f input, float3 localPos, sampler2D tex)\n{\nfloat2 uv = localPos.xy * input.faceUvScaleOffset.xy + input.faceUvScaleOffset.zw;\n// 所有face相关的tex都用这个tiling offset\nuv = uv * _FaceDiffuseTex_ST.xy + _FaceDiffuseTex_ST.zw;\nreturn tex2D(tex, uv);\n}\n``````\n\n``````float4 SampleSideDiffuse(float3 localPos, float3 sideNormal)\n{\n// 左右面\nfloat2 uvLR = localPos.yz * _LeftRightDiffuse_ST.xy + _LeftRightDiffuse_ST.zw;\nfloat4 diffuseLR = tex2D(_LeftRightDiffuse, uvLR) * _LeftRightColor;\n\n// 上下面\nfloat2 uvUD = localPos.xz * _UpDownDiffuse_ST.xy + _UpDownDiffuse_ST.zw;\nfloat4 diffuseUD = tex2D(_UpDownDiffuse, uvUD) * _UpDownColor;\n\nsideNormal = abs(sideNormal);\n// 非等比混合，UD会占得多\n//return lerp(diffuseLR, diffuseUD, sideNormal.y);\n// 等比混合\nreturn lerp(diffuseLR, diffuseUD, sideNormal.y / (sideNormal.x + sideNormal.y));\n}\n``````", null, "triplanar diffuse mapping\n\n6面纹理看起来还算不错，但是在正面和侧面边界有很硬的过渡，因为形状上还没做圆角过渡。", null, "## 四、形状的圆角过渡\n\nIQ大神的这篇文章里有各种形状的SDF公式：\nhttps://iquilezles.org/www/articles/distfunctions/distfunctions.html\n\n``````Round Box - unsigned - exact\nfloat udRoundBox( vec3 p, vec3 b, float r )\n{\nreturn length(max(abs(p)-b,0.0))-r;\n}\n``````", null, "A点都在两条边界外侧，A = (R, R - z) = (R, z到中间Z的距离-R到中间Z的距离）=\n\n(R, abs(z - halfZ) - (halfZ - R))，这是为了支持反面圆角的情况。\n\nB在x方向属于外侧，在z方向属于内侧。B = (R, R - z) = (R, abs(z - halfZ) - (halfZ - R))\n\nC在x方向属于内侧，在z方向属于外侧，C = (-C.x到outline的距离, R) = ((_outline - sdfValue) * bound.w, R)\n\nD都属于外侧，D = ((_outline - sdfValue) * bound.w, R)", null, "``````// xz分量相对内退边界的距离，正表示在外，负表示在内\nfloat2 meshOutlineDist = ((_outline - sdfValue) * input.bound.w + R, abs(localPos.z - halfZ)-(halfZ - R));\n// length(max(meshOutlineDist, 0)) - R;就是对应上面udRoundBox公式求的值，表示3D空间下到形状的最近距离\n``````\n\n``````// 修改之前的检测，使支持形状的圆角过渡\nfloat2 RayMarching(v2f input, out float3 localPos, float3 viewDir)\n{\nlocalPos = input.cuboidLocalPos.xyz;\n\n// 用于剔除模型外顶点时微调\n//float3 epsilon = input.bound.xyz * _boundEpsilon;\n#if SDFRoundEdge\nfloat halfZ = input.bound.z * 0.5;\n#else\n// 除以水平距离，这样在乘以水平距离就知道3d方向的增量\nviewDir /= max(length(viewDir.xy), 0.001);\n#endif\n\nint i = 0;\nfloat sdfValue;\nfor (; i < _loopCount; i++)\n{\nsdfValue = SampleSdfValueByLocalPos(input, localPos);\nfloat sdfDist = _outline - sdfValue;\nfloat meshDist;\n\n#if SDFRoundEdge\n// Round Box formula - unsigned - exact\n//float udRoundBox(vec3 p, vec3 b, float r)\n//{\n// return length(max(abs(p) - b, 0.0)) - r;\n//}\nfloat radius = input.volParam.x;\nfloat2 meshOutlineDist = float2(sdfDist * input.bound.w + radius, abs(localPos.z - halfZ) - (halfZ - radius));\nmeshDist = length(max(meshOutlineDist, 0)) - radius;\n// 足够靠近边界就认为到达\nif (meshDist <= 0)\nreturn float2(i, sdfValue);\n#else\n// 足够靠近边界就认为到达\nif (sdfDist <= _outlineEpsilon)\nreturn float2(i, sdfValue);\n\nmeshDist = sdfDist * input.bound.w;\n#endif\n// sdf距离转到模型距离，向前march\nlocalPos += viewDir * meshDist;\n\n// 到达模型外面要被剔除掉\n//clip((localPos + epsilon) * (input.bound.xyz - localPos + epsilon));\nclip(localPos * (input.bound.xyz - localPos));\n}\n\n// 没有到达范围内，但应该也很接近了\nsdfValue = SampleSdfValueByLocalPos(input, localPos);\nreturn float2(i, sdfValue);\n}\n``````", null, "## 五、法线纹理\n\nhttps://catlikecoding.com/unity/tutorials/rendering/part-6\n\n``````X.x * T + X.y * B + X.z * N;\n// 简化一下，并且支持反面就是\nfloat isForward = localPos.z > input.bound.z - localPos.z ? 1 : -1;\nfloat3 finalNormal = X.xyz * float3(1, 1, isForward);\n``````\n\n``````X.x * T + X.y * B + X.z * N;\nfloat isRight = sideNormal.x > 0 ? 1 : -1;\nfloat3 leftRightNormal = X.zxy * float3(isRight, 1, 1);\n``````\n\n``````float isUp = sideNormal.y > 0 ? 1 : -1;\nfloat3 upDownNormal = X.xzy * float3(1, isUp, 1);\n``````", null, "## 六、其他问题\n\n1、抗锯齿：由于MSAA是基于光栅化的，三角形边界并不是文字形状的边界，所以无效。基于后期的AA应该是可以的。还设想过识别文字形状周围的1像素做alpha混合来实现AA，外轮廓效果应该可以，如下图绿框中的。但就跟透明有类似的排序问题了，且文字之间的边缘原没法处理，如下图红框中的：", null, "2、遮挡：因为像素的深度值是文字Cube的边界，并不是模型的，一旦有模型穿插到文字体内会不正确，如下图，蓝色部分可见白色cube已经很接近文字cube的前面，但是红框中文字却没被遮挡。PS里写深度、RayMarching里每步测深度、调整渲染顺序应该能解决。", null, "3、阴影：ShadowMap仍然能适用，需要Shadow Caster的Shader也用SDF raymarching的写法的到轮廓。一个TMP文字专门的Shader Receiver可以同样实现Soft Shadow，但很难同时通用的处理多个文字的阴影。\n\n4、字形过渡：SDF一大特性是能做一个形状到另一个形状的过渡，就是对SDF距离进行过渡，对于本例的文字来说要额外指定另一个字符的UV信息，如果文字Cube大小不一样也需要变化。\n\n5、斜体：目前的机制没处理好斜体，如何变换到字符局部坐标系的问题。", null, "6、下划线也不支持，还没研究TMP是怎么弄的。\n\nQQ群：793972859（原群已满员）" ]

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[ "Solving a linear equation in one component of $\\Bbb{Z}^3$.\n\nConsider the space $$X = \\Bbb{Z}^3$$, a $$\\Bbb{Z}$$-module. Let $$M = \\{ \\sum_{i=1}^n c_i(p_i, q_i, r_i) : \\sum_{i=1}^n c_i q_i = 0,$$ where $$p_i, q_i, r_i$$ are either prime numbers or $$0 \\}$$. Then is $$M \\approx \\Bbb{Z}^2$$?\n\nClearly, $$M \\subset \\Bbb{Z} \\times 0 \\times \\Bbb{Z}$$. If $$(x, y, z) \\in M$$, then clearly, summing on the 2nd component gives $$y = 0$$, and so... I'm having trouble seeing that you can handle the other two components independently of the second if the second is zero'd since the coefficients $$c_i$$ are tied up in it's sums.\n\nFor any finite $$n \\geq 1$$ we have a system of system of 3 linear equations in $$n$$ unknowns in a vector $$c$$. Let $$x, z \\in \\Bbb{Z}$$ be arbitrary. We want there to always be a solution to:\n\n$$A c = (x, 0, z)^t$$\n\nWhere $$A$$ is composed of prime numbers or $$0$$. I think we just let $$n \\geq 3$$ without loss of generality (the sums can be any finite number of terms). Also, $$A$$ is allowed to vary over any primes or $$0$$ for each result vector $$(x, y, z)^t$$. That should make things easier.\n\nAny integer can be written as linear combination of $$2$$ and $$3$$ (for instance, via $$a=3a-2a$$). So, any element of $$\\mathbb{Z}\\times 0\\times\\mathbb{Z}$$ can be written as a linear combination of $$(2,0,0)$$, $$(3,0,0)$$, $$(0,0,2)$$, and $$(0,0,3)$$ and thus is in your $$M$$.\nAlternatively, if you just wanted to know that $$M$$ was isomorphic to $$\\mathbb{Z}^2$$, this was immediate as soon as you knew that $$M$$ contained two linearly independent elements (say, $$(2,0,0)$$ and $$(0,0,2)$$), since every submodule of $$\\mathbb{Z}^n$$ is isomorphic to $$\\mathbb{Z}^m$$ for some $$m\\leq n$$.\n• Another question for you: Is then $\\Bbb{Z}^3 / M \\approx \\Bbb{Z}$? – Shine On You Crazy Diamond Aug 16 at 21:44\n• Yes, since $M=\\mathbb{Z}\\times 0\\times\\mathbb{Z}$ so the quotient can be identified with the projection onto the second coordinate. – Eric Wofsey Aug 16 at 21:44" ]

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[ "OpenCV 2.4.4\n\n## Package org.opencv.core\n\nClass Summary\nAlgorithm This is a base class for all more or less complex algorithms in OpenCV, especially for classes of algorithms, for which there can be multiple implementations.\nCore\nCore.MinMaxLocResult\nCvType\nMat OpenCV C++ n-dimensional dense array class\nMatOfByte\nMatOfDMatch\nMatOfDouble\nMatOfFloat\nMatOfFloat4\nMatOfFloat6\nMatOfInt\nMatOfInt4\nMatOfKeyPoint\nMatOfPoint\nMatOfPoint2f\nMatOfPoint3\nMatOfPoint3f\nMatOfRect\nPoint Template class for 2D points specified by its coordinates x and y.\nPoint3 Template class for 3D points specified by its coordinates x, y and z.\nRange Template class specifying a continuous subsequence (slice) of a sequence.\nRect Template class for 2D rectangles, described by the following parameters:\nRotatedRect\nScalar Template class for a 4-element vector derived from Vec.\nSize Template class for specifying the size of an image or rectangle.\nTermCriteria The class defining termination criteria for iterative algorithms.\n\nException Summary\nCvException\n\nOpenCV 2.4.4 Documentation" ]

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[ "# nLab dependent product type\n\nDependent product types\n\n# Dependent product types\n\n## Idea\n\nIn dependent type theory, a dependent product type $\\prod_{x\\colon A} B(x)$, for a dependent type $x\\colon A\\vdash B(x)\\colon Type$ is the type of “dependently typed functions” assigning to each $x\\colon A$ an element of $B(x)$.\n\nIn a model of the type theory in categorical semantics, this is a dependent product. In set theory, it is an element of an indexed product.\n\nIt includes function types as the special case when $B$ is not dependent on $A$. Note that a binary product type is rather different, being actually a special case of a dependent sum type.\n\n## Overview\n\ntype theorycategory theory\nsyntaxsemantics\nnatural deductionuniversal construction\ndependent product typedependent product\ntype formation$\\displaystyle\\frac{\\vdash\\: X \\colon Type \\;\\;\\;\\;\\; x \\colon X \\;\\vdash\\; A(x)\\colon Type}{\\vdash \\; \\left(\\prod_{x \\colon X} A\\left(x\\right)\\right) \\colon Type}$$\\left( \\array{ A^{\\phantom{p_1}} \\\\ \\downarrow^{p_1} \\\\ X_{\\phantom{p_1}}} \\in \\mathcal{C}\\right) \\Rightarrow \\left( \\prod_{x\\colon X} A\\left(x\\right) \\in \\mathcal{C}\\right)$\nterm introduction$\\displaystyle\\frac{x \\colon X \\;\\vdash\\; a\\left(x\\right) \\colon A\\left(x\\right)}{\\vdash (x \\mapsto a\\left(x\\right)) \\colon \\prod_{x' \\colon X} A\\left(x'\\right) }$", null, "term elimination$\\displaystyle\\frac{\\vdash\\; f \\colon \\left(\\prod_{x \\colon X} A\\left(x\\right)\\right)\\;\\;\\;\\; \\vdash \\; x \\colon X}{x \\colon X\\;\\vdash\\; f(x) \\colon A(x)}$", null, "computation rule$(y \\mapsto a(y))(x) = a(x)$", null, "## Definition\n\nLike any type constructor in type theory, a dependent product type is specified by rules saying when we can introduce it as a type, how to construct terms of that type, how to use or “eliminate” terms of that type, and how to compute when we combine the constructors with the eliminators.\n\nThe type formation rule for dependent product type is:\n\n$\\frac{A\\colon Type \\qquad x\\colon A \\vdash B(x) \\colon Type}{\\prod_{x\\colon A} B(x)\\colon Type}$\n\n### As a negative type\n\nDependent product types are almost always defined as negative types. In this presentation, primacy is given to the eliminators. The natural eliminator of a dependent product type says that we can apply it to any input:\n\n$\\frac{f\\colon \\prod_{x\\colon A} B(x) \\qquad a\\colon A}{f(a) \\colon B(a)}$\n\nThe constructor is then determined as usual for a negative type: to construct a term of $\\prod_{x\\colon A} B(x)$, we have to specify how it behaves when applied to any input. In other words, we should have a term of type $B(x)$ containing a free variable $x\\colon A$. This yields the usual “$\\lambda$-abstraction” constructor:\n\n$\\frac{x\\colon A\\vdash b\\colon B(x)}{\\lambda x.b\\colon \\prod_{x\\colon A} B(x)}$\n\nThe beta-reduction rule is the obvious one, saying that when we evaluate a $\\lambda$-abstraction, we do it by substituting for the bound variable.\n\n$(\\lambda x.b)(a) \\;\\to_\\beta\\; b[a/x]$\n\nIf we want an eta-conversion rule, the natural one says that every dependently typed function is a $\\lambda$-abstraction:\n\n$\\lambda x.f(x) \\;\\to_\\eta\\; f$\n\n### As a positive type\n\nIt is also possible to present dependent product types as a positive type. However, this requires a stronger metatheory, such as a logical framework. We use the same constructor ($\\lambda$-abstraction), but now the eliminator says that to define an operation using a function, it suffices to say what to do in the case that that function is a lambda abstraction.\n\n$\\frac{(x\\colon A \\vdash b\\colon B(x)) \\vdash c\\colon C \\qquad f\\colon \\prod_{x\\colon A} B(x)}{funsplit(c,f)\\colon C}$\n\nThis rule cannot be formulated in the usual presentation of type theory, since it involves a “higher-order judgment”: the context of the term $c\\colon C$ involves a “term of type $B(x)$ containing a free variable $x\\colon A$”. However, it is possible to make sense of it. In dependent type theory, we need additionally to allow $C$ to depend on $\\prod_{x\\colon A} B(x)$.\n\nThe natural $\\beta$-reduction rule for this eliminator is\n\n$funsplit(c, \\lambda x.g) \\;\\to_\\beta c[g/b]$\n\nand its $\\eta$-conversion rule looks something like\n\n$funsplit(c[\\lambda x.b / g], f) \\;\\to_\\eta\\; c[f/g].$\n\nHere $g\\colon \\prod_{x\\colon A} B(x) \\vdash c\\colon C$ is a term containing a free variable of type $\\prod_{x\\colon A} B(x)$. By substituting $\\lambda x.b$ for $g$, we obtain a term of type $C$ which depends on “a term $b\\colon B(x)$ containing a free variable $x\\colon A$”. We then apply the positive eliminator at $f\\colon \\prod_{x\\colon A} B(x)$, and the $\\eta$-rule says that this can be computed by just substituting $f$ for $g$ in $c$.\n\n### Positive versus negative\n\nAs usual, the positive and negative formulations are equivalent in a suitable sense. They have the same constructor, while we can formulate the eliminators in terms of each other:\n\n\\begin{aligned} f(a) &\\coloneqq funsplit(b[a/x], f)\\\\ funsplit(c, f) &\\coloneqq c[f(x)/b] \\end{aligned}\n\nThe conversion rules also correspond.\n\nIn dependent type theory, this definition of $funsplit$ only gives us a properly typed dependent eliminator if the negative dependent product type satisfies $\\eta$-conversion. As usual, if it satisfies propositional eta-conversion then we can transport along that instead—and conversely, the dependent eliminator allows us to prove propositional $\\eta$-conversion. This is the content of Propositions 3.5, 3.6, and 3.7 in (Garner).\n\nThe standard rules for type-formation, term introduction/elimination and computation of dependent product type are listed for instance in part I of" ]

[ null, "https://ncatlab.org/nlab/show/dependent+product+type", null, "https://ncatlab.org/nlab/show/dependent+product+type", null, "https://ncatlab.org/nlab/show/dependent+product+type", null ]

[ "# #0a0610 Color Information\n\nIn a RGB color space, hex #0a0610 is composed of 3.9% red, 2.4% green and 6.3% blue. Whereas in a CMYK color space, it is composed of 37.5% cyan, 62.5% magenta, 0% yellow and 93.7% black. It has a hue angle of 264 degrees, a saturation of 45.5% and a lightness of 4.3%. #0a0610 color hex could be obtained by blending #140c20 with #000000. Closest websafe color is: #000000.\n\n• R 4\n• G 2\n• B 6\nRGB color chart\n• C 38\n• M 63\n• Y 0\n• K 94\nCMYK color chart\n\n#0a0610 color description : Very dark (mostly black) violet.\n\n# #0a0610 Color Conversion\n\nThe hexadecimal color #0a0610 has RGB values of R:10, G:6, B:16 and CMYK values of C:0.38, M:0.63, Y:0, K:0.94. Its decimal value is 656912.\n\nHex triplet RGB Decimal 0a0610 `#0a0610` 10, 6, 16 `rgb(10,6,16)` 3.9, 2.4, 6.3 `rgb(3.9%,2.4%,6.3%)` 38, 63, 0, 94 264°, 45.5, 4.3 `hsl(264,45.5%,4.3%)` 264°, 62.5, 6.3 000000 `#000000`\nCIE-LAB 2.097, 2.586, -3.822 0.284, 0.232, 0.52 0.274, 0.224, 0.232 2.097, 4.615, 304.076 2.097, 0.417, -2.073 4.819, 2.081, -3.026 00001010, 00000110, 00010000\n\n# Color Schemes with #0a0610\n\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #0c1006\n``#0c1006` `rgb(12,16,6)``\nComplementary Color\n• #060710\n``#060710` `rgb(6,7,16)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #0f0610\n``#0f0610` `rgb(15,6,16)``\nAnalogous Color\n• #071006\n``#071006` `rgb(7,16,6)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #100f06\n``#100f06` `rgb(16,15,6)``\nSplit Complementary Color\n• #06100a\n``#06100a` `rgb(6,16,10)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #100a06\n``#100a06` `rgb(16,10,6)``\n• #060c10\n``#060c10` `rgb(6,12,16)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #100a06\n``#100a06` `rgb(16,10,6)``\n• #0c1006\n``#0c1006` `rgb(12,16,6)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #160d23\n``#160d23` `rgb(22,13,35)``\n• #211435\n``#211435` `rgb(33,20,53)``\n• #2d1b48\n``#2d1b48` `rgb(45,27,72)``\nMonochromatic Color\n\n# Alternatives to #0a0610\n\nBelow, you can see some colors close to #0a0610. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #080610\n``#080610` `rgb(8,6,16)``\n• #080610\n``#080610` `rgb(8,6,16)``\n• #090610\n``#090610` `rgb(9,6,16)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #0b0610\n``#0b0610` `rgb(11,6,16)``\n• #0c0610\n``#0c0610` `rgb(12,6,16)``\n• #0d0610\n``#0d0610` `rgb(13,6,16)``\nSimilar Colors\n\n# #0a0610 Preview\n\nThis text has a font color of #0a0610.\n\n``<span style=\"color:#0a0610;\">Text here</span>``\n#0a0610 background color\n\nThis paragraph has a background color of #0a0610.\n\n``<p style=\"background-color:#0a0610;\">Content here</p>``\n#0a0610 border color\n\nThis element has a border color of #0a0610.\n\n``<div style=\"border:1px solid #0a0610;\">Content here</div>``\nCSS codes\n``.text {color:#0a0610;}``\n``.background {background-color:#0a0610;}``\n``.border {border:1px solid #0a0610;}``\n\n# Shades and Tints of #0a0610\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, #010102 is the darkest color, while #f6f2fa is the lightest one.\n\n• #010102\n``#010102` `rgb(1,1,2)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #130b1e\n``#130b1e` `rgb(19,11,30)``\n• #1c112d\n``#1c112d` `rgb(28,17,45)``\n• #25163b\n``#25163b` `rgb(37,22,59)``\n• #2e1b49\n``#2e1b49` `rgb(46,27,73)``\n• #372157\n``#372157` `rgb(55,33,87)``\n• #3f2666\n``#3f2666` `rgb(63,38,102)``\n• #482b74\n``#482b74` `rgb(72,43,116)``\n• #513182\n``#513182` `rgb(81,49,130)``\n• #5a3690\n``#5a3690` `rgb(90,54,144)``\n• #633b9f\n``#633b9f` `rgb(99,59,159)``\n``#6c41ad` `rgb(108,65,173)``\n• #7547ba\n``#7547ba` `rgb(117,71,186)``\n• #8056bf\n``#8056bf` `rgb(128,86,191)``\n• #8b64c5\n``#8b64c5` `rgb(139,100,197)``\n• #9572ca\n``#9572ca` `rgb(149,114,202)``\n• #a080d0\n``#a080d0` `rgb(160,128,208)``\n• #ab8fd5\n``#ab8fd5` `rgb(171,143,213)``\n• #b59dda\n``#b59dda` `rgb(181,157,218)``\n• #c0abe0\n``#c0abe0` `rgb(192,171,224)``\n• #cbb9e5\n``#cbb9e5` `rgb(203,185,229)``\n• #d6c8ea\n``#d6c8ea` `rgb(214,200,234)``\n• #e0d6f0\n``#e0d6f0` `rgb(224,214,240)``\n• #ebe4f5\n``#ebe4f5` `rgb(235,228,245)``\n• #f6f2fa\n``#f6f2fa` `rgb(246,242,250)``\nTint Color Variation\n\n# Tones of #0a0610\n\nA tone is produced by adding gray to any pure hue. In this case, #0b0a0c is the less saturated color, while #090016 is the most saturated one.\n\n• #0b0a0c\n``#0b0a0c` `rgb(11,10,12)``\n• #0b090d\n``#0b090d` `rgb(11,9,13)``\n• #0b090d\n``#0b090d` `rgb(11,9,13)``\n• #0a080e\n``#0a080e` `rgb(10,8,14)``\n• #0a070f\n``#0a070f` `rgb(10,7,15)``\n• #0a0610\n``#0a0610` `rgb(10,6,16)``\n• #0a0511\n``#0a0511` `rgb(10,5,17)``\n• #0a0412\n``#0a0412` `rgb(10,4,18)``\n• #090313\n``#090313` `rgb(9,3,19)``\n• #090313\n``#090313` `rgb(9,3,19)``\n• #090214\n``#090214` `rgb(9,2,20)``\n• #090115\n``#090115` `rgb(9,1,21)``\n• #090016\n``#090016` `rgb(9,0,22)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0a0610 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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[ "Cody\n\n# Problem 167. Pizza!\n\nSolution 1076574\n\nSubmitted on 7 Dec 2016 by Anthony Muscarello\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nz = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))\n\n2   Pass\nz = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))\n\n3   Pass\nz = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))\n\n4   Pass\nz = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "#### Abstract\n\nThe problem of th mean exponential stability and stabilizability of a class of stochastic nonlinear and bilinear hybrid systems with unstable and stable subsystems is considered. Sufficient conditions for the th mean exponential stability and stabilizability under a feedback control and stabilizing switching rules are derived. A method for the construction of stabilizing switching rules based on the Lyapunov technique and the knowledge of regions of decreasing the Lyapunov functions for subsystems is given. Two cases, including single Lyapunov function and a a single Lyapunov-like function, are discussed. Obtained results are illustrated by examples.\n\n#### 1. Introduction\n\nThe problem of stability and stabilization of dynamic systems is one of the basic problems in the control theory. It is well known that there are classes of control systems which cannot be stabilized by a single feedback control . In this case and in the case of the hybrid control systems, switched controls can assure the stability.\n\nLiberzon and Morse in mention that one of the basic problems for dynamic systems is the construction of stabilizing switching laws [1, 36]. It is known that if a common Lyapunov function exists, then the hybrid system is stable for any switching. In the absence of the common Lyapunov function, stability properties of the hybrid system in general depend on the switching signal, and in this case the hybrid system is not stable for any switching rules but only for the so-called stabilizing switching rules . In this case, more generally a single Lyapunov and a single Lyapunov-like functions have been introduced [1, 3, 7]. Some results for the linear stochastic hybrid systems are given in and for the nonlinear deterministic hybrid systems are given in . Recent results for the deterministic hybrid systems are collected and summarized in .\n\nIn the present paper, ideas of a feedback control, proposed by Florchinger for nonhybrid stochastic nonlinear control systems [11, 12], are used and combined with the concept of stabilizing switching rules for hybrid systems to derive the results for the th mean exponential stabilizability of stochastic nonlinear and bilinear hybrid systems consisting of unstable and stable structures. The authors propose also a design method for stabilizing switching rules, which is based on the knowledge of regions of decreasing the Lyapunov functions for subsystems. Similar methods were used for deterministic hybrid systems, for example, in [3, 68]. In this paper, we extend them to stochastic hybrid case.\n\n#### 2. Mathematical Preliminaries\n\nThroughout this paper, we use the following notation. Let be the Euclidean norm. By we denote the eigenvalue of the matrix , and and denote the smallest and the biggest eigenvalues of the matrix , respectively. We denote by the transposition of matrix . We mark , . Let be a complete probability space with a filtration satisfying usual conditions. Let , , be the -dimensional standard Wiener process defined on the probability space . Let be the set of states, and let be the stochastic switching rule. We denote switching times as and assume that there is a finite number of switches on every finite time interval. We assume that processes and are both adapted. We say that a proper twice differentiable function is a Lyapunov function if and , for any .\n\nLet us consider the stochastic hybrid system described by the vector Itô differential equation where is the state vector and is an initial condition, . Functions and are locally Lipschitz such that , , . The local Lipschitz conditions together with these enforced on the switching rule ensure that there exists a unique solution to the hybrid system (1).\n\nFor any twice differentiable with respect to and once differentiable with respect to function (i.e., ), the th process has a generator (the Itô operator for the th subsystem of the system (1)) given in every structure by\n\nWe use the following definitions.\n\nDefinition 1. The null solution of the stochastic differential equation (1) is said to be th mean exponentially stable, , if there exists a pair of positive scalars , such that where is called the decay rate.\n\nDefinition 2. The hybrid system (1) is said to be stabilizable if there exist a switching signal and the associated linear feedback control law such that the hybrid system (1) is th mean exponentially stable for some .\n\nDefinition 3. A Lyapunov function satisfying is called a common Lyapunov function for the hybrid system (1).\n\nNote that it is a known fact that if there exists a common Lyapunov function for the hybrid system (1), then the null solution of (1) is asymptotically stable for any switching.\n\nDefinition 4. A Lyapunov function satisfying for some switching rule is called a single Lyapunov function for the hybrid system (1).\n\nLemma 5 ( (Itô formula)). If , then for any switching times , if the integrations involved exist and are finite.\n\nFollowing the methodology introduced in , for deterministic hybrid systems, we assume that the hybrid state space is partitioned into regions , and . We consider a special class of switching rule given by Note that a switching rule given by (7) is a stochastic switching rule because of its dependence on the stochastic process . Our aim is to find a special partition defined by (7) such that every switching rule is a stabilizing switching rule for a considered class of stochastic hybrid systems.\n\n#### 3. Stability of Nonlinear Stochastic Hybrid Systems\n\nFirst, we study the problem of the stability of the nonlinear stochastic hybrid system (1). Two cases, single Lyapunov and single Lyapunov-like functions, are considered.\n\n##### 3.1. Single Lyapunov Functions\n\nWe formulate a theorem which establishes sufficient conditions for the th mean exponential stability of the nonlinear hybrid system (1).\n\nTheorem 6. If the following conditions hold:(1)there exist a Lyapunov function and positive constants such that (2)there exists a Lebesgue-measurable function such that Then the null solution of the stochastic hybrid system (1) is th mean exponentially stable under the stabilizing switching rule .\n\nProof. From assumptions for the Lyapunov function , we obtain From (11) and the Itô formula 6, we obtain and by Gronwall’s inequality Now from (10) and (13), we obtain the following inequality: Hence, the thesis follows.\n\nNotice that function is a single Lyapunov function for the hybrid system (1).\n\n##### 3.2. Single Lyapunov-Like Functions\n\nIn the case when condition (2) of Theorem 6 is not satisfied, then one can look for a single Lyapunov-like function. We assume in this case that the hybrid state space is partitioned into regions and , which can be separated into two disjoint subregions: a stable subregion and an unstable region , (the upper scripts “s” and “us” denote stable and unstable regions, resp.) that is,\n\nLet us denote by the sum of time intervals of the residence of the system (1) in the regions , , and by the sum of time intervals of the residence of the system (1) in the regions ,  .\n\nIn this case we cannot construct a single Lyapunov function, but we can look for a single Lyapunov-like function defined as follows.\n\nDefinition 7. A Lyapunov function is called a single Lyapunov-like function if there exist positive constants and such that for some switching rule .\n\nUsing this definition, the following theorem can be formulated.\n\nTheorem 8. Let one assume that the following conditions hold: (1) there exist a Lyapunov function and positive constants such that (2) there exist Lebesgue-measurable functions and such that partition (15) satisfies the conditions (3) there exists a Lebesguemeasurable function such that , where Then the null solution of (1) is th mean exponentially stable under the stabilizing switching rule .\n\nProof. From assumptions for , it follows that Let us consider the switching strategy described by (19). Then we obtain Further proof is similar to the proof of Theorem 6. Hence, the thesis follows.\n\nNotice that function is a single Lyapunov-like function for the hybrid system (1).\n\nExample 9. Let us consider a special case of the system (1) with the two subsystems () given as follows: where\nLet us consider the Lyapunov function . Since , , then Note that and functions and are constant and are given as follows , .\nCondition is satisfied for . Exemplary simulations are shown in Figures 1, 2, and 3.\n\nFrom Theorem 8, it follows that is the single Lyapunov-like function for the system (22), and it is exponentially stable in mean square with a decay rate for the switching strategy given by\n\n#### 4. Stabilizability of Stochastic Hybrid Systems\n\nIn this section, we discuss the stabilizability problem of the stochastic nonlinear and bilinear hybrid systems. We formulate sufficient conditions for the th mean exponential stabilizability, and we find a control of feedback form for the considered class of systems.\n\n##### 4.1. Stabilizability of Nonlinear Stochastic Hybrid Systems\n\nLet us consider the stochastic control hybrid system described by the vector Itô differential equations where is the state vector, is a measurable —a real-valued control vector law, is an initial condition, and . Functions and are the locally Lipschitz, ,  ,  ,  ,  .\n\nThe local Lipschitz condition together with these enforced on the switching rule ensures that there is a unique solution to the hybrid system (27).\n\nThe aim of this part of the paper is to establish sufficient conditions under which one can design a state feedback control law so that the null solution of the stochastic hybrid control system (27) is th mean exponentially stable. We extend the results of Florchinger for the stochastic nonhybrid systems [11, 12] to the hybrid systems. Some results for asymptotic stability and stabilizability for the hybrid system (27) with Markovian or any switchings under a feedback control have been proposed in .\n\nWe introduce the following notation of operators and ,  ,  , for :\n\nThen, the following stabilization result for the control hybrid system (27) holds.\n\nTheorem 10. Suppose that the following conditions hold:(1) there exist a Lyapunov function and positive constants such that (2) there exists a Lebesgue-measurable function such that partition determined by (7) satisfies conditionsThen the control law given as follows: together with the stabilizing switching rule renders the null solution of the stochastic hybrid system (27) the th mean exponentially stable.\n\nProof. Applying the infinitesimal operator defined by (2) to the hybrid system (27), we find that Now the thesis follows from Theorem 6.\n\nWe can formulate a more general theorem in a case when a Lyapunov-like function exists as follows.\n\nTheorem 11. Suppose that the following conditions hold:(1) there exist a Lyapunov function and positive constants such that (2) there exist Lebesgue-measurable functions and such that partition (15) satisfies conditions (3) there exists a Lebesgue-measurable function such that , where Then the control law given as follows together with the stabilizing switching rule renders the null solution of the stochastic hybrid system (27) the th mean exponentially stable.\n\nProof. The thesis follows from Theorem 8.\n\n##### 4.2. Stabilizability of the Bilinear Hybrid Systems\n\nLet us consider a special class of the system (27) given by a bilinear stochastic hybrid system as follows: where is the state vector, , is the control vector, , is an initial condition, and , , ,  ,   are for every constant matrices of dimension .\n\nFor this particular case, we can combine the above results with the theorem given by Mao for the stochastic linear systems and formulate the theorems which can be obtained directly from Theorems 10 and 11. Sufficient conditions for the th mean exponential stabilizability for the linear hybrid systems are formulated in .\n\nOperators (28) reduce to the following ones: where and denote gradient and Hessian of the function , respectively.\n\nTheorem 12. Suppose that there exist symmetric positive definite matrix , constant , and positive constants , such that the following conditions are satisfied: Then control of a form together with the stabilizing switching rule makes the hybrid system (37) the th mean exponentially stable for (a) if , (b) if .\n\nProof. The thesis of the theorem follows from Theorem 10. Let us choose a Lyapunov function of a form Notice that satisfies assumption of Theorem 10 for and . Then using assumptions (39), we obtain where From Theorem 10, it follows that the control chosen as follows: together with the stabilizing switching rule makes the system (37) the th mean exponentially stable for (a) if , (b) if . Hence, the thesis follows.\n\nNotice that function is a single Lyapunov function for the hybrid system (37). See for the details of the proof.\n\nRemark 13. In a particular case of , we obtain the following criterion.\n\nCriterion 14. Suppose that there exists a symmetric positive definite matrix such that Then the control of a form together with the stabilizing switching rule exponentially in mean-square stabilizes the bilinear hybrid system (37).\n\nWe formulate now a more general theorem which formulates sufficient conditions of th mean exponential stabilizability for the stochastic bilinear hybrid system (37) in a case when a single Lyapunov-like function exists. Exemplary simulations are shown in Figures 4 and 5.\n\nTheorem 15. Suppose that the following conditions are satisfied: (1) there exist symmetric positive definite matrix , constants , and positive constants such that (2) there exists constant such that , where Then the control of a form together with the stabilizing switching rule makes the hybrid system (37) the th mean exponentially stable for (a) if , (b) if .\n\nProof. The thesis follows from Theorem 11. Let us choose a Lyapunov function as follows: Function satisfies assumption of Theorem 11 for and .\nFurthermore, Since condition (51) is satisfied, assumption of Theorem 11 also holds. Assumption of Theorem 11 follows directly. Now using Theorem 11, we obtain that control can be chosen as follows: Hence, the thesis follows.\n\nNotice that function is a single Lyapunov-like function for the hybrid system (37). See for the details of the proof.\n\nExample 16. Let us consider a special case of the hybrid system (37) given as follows: where We study the problem of the th mean exponential stabilizability for the hybrid system (53) for . We look for a control vector of a form Let us choose the Lyapunov function of a form . Then regions ,  , are given as follows: Condition (39) of Theorem 12 is satisfied for . Function is a single Lyapunov function for the system (53). From Theorem 12, it follows that the control together with stabilizing switching rule given by exponentially th mean, stabilizes the system (53) for .\n\n#### 5. Conclusions\n\nIn this paper, nonlinear and the bilinear hybrid systems parametrically excited by a white noise, consisted of unstable and stable subsystems the described by the Itô stochastic differential equations, have been analyzed. To find sufficient conditions for the th mean exponential stability and stabilizability, the Lyapunov function techniques and the hybrid control theory have been used. We have found the control of a feedback form, which is a generalization of a feedback control proposed by Florchinger for the nonhybrid systems [11, 12], and the stabilizing switching rule, which is constructed on the basis of the knowledge of the regions of decreasing of the Lyapunov functions for subsystems.\n\nResults for the asymptotic stabilizability under a feedback control of the stochastic nonlinear and bilinear hybrid systems with Markovian or any switching rule have been discussed in and for the th mean exponential stability and stabilizability of the stochastic linear hybrid system in [4, 16]. The obtained results have been illustrated by an example.\n\nThe proposed criteria of the th mean exponentially stability and stabilizability can be generalized to the hybrid systems parametrically excited by Gaussian colored and non-Gaussian noises.\n\nThe presented results cannot be compared because they relate to different systems, class. For some class of systems stability conditions obtained using single Lyapunov function approach are the same as those obtained using single Lyapunov-like function approach, while for a wide class of systems, for which Lyapunov-like functions can be used, generally we cannot use single Lyapunov functions. The relationship between the obtained results can be summarized in Table 1.\n\n#### Acknowledgment\n\nThe authors gratefully acknowledge research support from Cardinal Stefan Wyszyński University in Warsaw." ]

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[ "# PyTorch函数\n\n#### torch.squeeze(input, dim=None, out=None) → Tensor\n\n• dim = None：去除 input 张量中所有 size 为 1 的维度。\n• 指定 dim 时：若该维度 size 为 1 则去除，否则保持 input 张量不变。\n\n#### torch.unsqueeze(input, dim) → Tensor\n\n• dim >= 0：在指定维度前插入一维。\n• dim < 0：在指定维度后插入一维。\n\n#### torch.flip(input, dims) → Tensor\n\n• 将 input 张量沿着列表/元组 dims 中的每一个维度依次翻转。\n\n#### torch.Tensor.contiguous(memory_format=torch.contiguous_format) → Tensor\n\n• 返回一个内存连续且有相同数据的 Tensor，如果原 Tensor 已经内存连续，则直接返回原 Tensor。\n\n#### torch.Tensor.permute(*dims) → Tensor\n\n• 根据 dims 给定的维度顺序对张量进行维度换位。\n\n#### torch.Tensor.transpose(dim0, dim1) → Tensor\n\n• 对 dim0 和 dim1 两个维度换位。" ]

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[ "", null, "", null, "Thus f is concave up from negative infinity to the inflection point at (1, –1), and then concave down from there to infinity. 1. We have seen previously that the sign of the derivative provides us with information about where a function (and its graph) is increasing, decreasing or stationary.We now look at the \"direction of bending\" of a graph, i.e. If it is positive, then the function is concave up. https://mathworld.wolfram.com/ConcaveFunction.html. Define the intervals for the function. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. is convex If is constant, then the function has no concavity. Note: Geometrically speaking, a function is concave up if its graph lies above its tangent lines. Definition. A concave up graph has the “U” or bowl right-side up. Wolfram Demonstrations Project » Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. a) Find the intervals on which the graph of f(x) = x + x is concave up, concave down and the point(s) of inflection if any. if, for any points and in , the function To find the concavity, we need to look at the second derivative. b) Use a graphing calculator to graph f and confirm your answers to part a). Points of discontinuity show up here a bit more than in the First Derivative Test. Concave down on since is negative. on that interval (Gradshteyn and Ryzhik 2000). Unlimited random practice problems and answers with built-in Step-by-step solutions. Determines the inflection points of a given equation. To embed this widget in a post, install the Wolfram|Alpha Widget Shortcode Plugin and copy and paste the shortcode above into the HTML source. To solve this problem, start by finding the second derivative. Now set it equal to 0 and solve. Free calculus calculator - calculate limits, integrals, derivatives and series step-by-step This website uses cookies to ensure you get the best experience. These are points on the curve where the concavity of the function changes. If my calculations are correct, this function is convex in certain cases and concave in certain different cases. concave up and down calculator. }\\) It is concave up outside this region. I'm looking for a concave down increasing-function, see the image in the right lower corner. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points . Knowledge-based programming for everyone. ... Go to How to Use a Scientific Calculator Ch 6. Inflection points are often sought on some functions. The function has an inflection point (usually) at any x-value where the signs switch from positive to negative or vice versa. (i) We will say that the graph of f(x) is concave up on I iff f '(x) is increasing on I. An inflection point is a point in a graph at which the concavitychanges. To embed this widget in a post on your WordPress blog, copy and paste the shortcode below into the HTML source: To add a widget to a MediaWiki site, the wiki must have the. At x = 0 is this function concave up, concave down, or an inflection point? Computerbasedmath.org » Join the initiative for modernizing math education. b) Use a graphing calculator to graph f and confirm your answers to part a). a) Find the intervals on which the graph of f(x) = x 4 - 2x 3 + x is concave up, concave down and the point(s) of inflection if any. Concave down on since is negative. Sep 15, 2020 ... then the graph is concave down. Get the free \"Inflection Points\" widget for your website, blog, Wordpress, Blogger, or iGoogle. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. (ii) We will say that the graph of f(x) is concave down on I iff f '(x) is decreasing on I. Concave down. Online physics calculator that calculates the concave mirror equation from the given values of object distance (do), the image distance (di), and the focal length (f). Curve segment that lies above its tangent lines is concave upward. Basically I need a function f(x) which will rise slower as x is increasing. Code to add this calci to your website . We have seen previously that the sign of the derivative provides us with information about where a function (and its graph) is increasing, decreasing or stationary.We now look at the \"direction of bending\" of a graph, i.e. The #1 tool for creating Demonstrations and anything technical. Concave up. Let f(x) be a differentiable function on an interval I. Using test points, we note the concavity does change from down to up, hence is an inflection point of The curve is concave down for all and concave up for all , see the graphs of and . In other words, the graph of f is concave up. I would also like to have some constants which can change the way/speed the function is concaving. Repeat the process we used for the first derivative, but use as our expression. There are critical points when $$t$$ is 0 or 2. We say that a graph is concave up if the line between two points is above the graph, or alternatively if the first derivative is increasing. Find the inflection points and intervals of concavity upand down of f(x)=3x2−9x+6 First, the second derivative is justf″(x)=6. Note that the value a is directly related to the second derivative, since f ''(x) = 2a.. The graph is concave down on the interval because is negative. Sep 15, 2020 ... then the graph is concave down. Note that the value a is directly related to the second derivative, since f ''(x) = 2a.. Definition. Code to add this calci to your website . And when we're talking about a critical point, if we're assuming it's concave downwards over here, we're assuming differentiability over this interval. Free functions inflection points calculator - find functions inflection points step-by-step This website uses cookies to ensure you get the best experience. Limits Go to Limits Ch 7. Step-by-step Solutions » Practice online or make a printable study sheet. Once we hit $$x = 1$$ the graph starts to increase and is still concave up and both of these behaviors continue for the rest of the graph. Calculus: Fundamental Theorem of Calculus By using this website, you agree to our Cookie Policy. Step-by-step Solutions » p. 1132, 2000. Calculus: Fundamental Theorem of Calculus Find more Mathematics widgets in Wolfram|Alpha. San Diego, CA: Academic Press, Also if. (ii) We will say that the graph of f(x) is concave down on I iff f '(x) is decreasing on I. Hints help you try the next step on your own. Similarly, we say that a graph is concave down if the line between two points is below the graph, or … Where it is 0 is the inflection point. A concavity calculator is any calculator that outputs information related to the concavity of a function when the function is inputted. the function $$m(x)$$ is concave down when $$-3 \\lt x \\lt 3\\text{. whether the graph is \"concave up\" or \"concave down\". They tell us something about the shape of a graph, or more specifically, how it bends. For f ( x) = –2 x3 + 6 x2 – 10 x + 5, f is concave up from negative infinity to the inflection point at (1, –1), then concave down from there to infinity. 2. I used the formula for determining if it is concave or convex. The Sign of the Second Derivative Concave Up, Concave Down, Points of Inflection. The second derivative gives the concavity information. Curve segment that lies below its tangent lines is concave downward. Calculus: Integral with adjustable bounds. Concave downwards, let's just be clear here, means that it's opening down like this. So: f (x) is concave downward up to x = −2/15. Figure \\(\\PageIndex{3}$$: Demonstrating the 4 ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives. [more] From MathWorld--A Wolfram Web Resource. Online physics calculator that calculates the concave mirror equation from the given values of object distance (do), the image distance (di), and the focal length (f). As always, you should check your result on your graphing calculator. So let’s talk a little about concavity first. To embed a widget in your blog's sidebar, install the Wolfram|Alpha Widget Sidebar Plugin, and copy and paste the Widget ID below into the \"id\" field: We appreciate your interest in Wolfram|Alpha and will be in touch soon. It can calculate and graph the roots (x-intercepts), signs, Local Maxima and Minima, Increasing and Decreasing Intervals, Points of Inflection and Concave Up/Down intervals. Solution: Since this is never zero, there are not points ofinflection. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). The function is concave down, where the second derivative is negative, which for our function is when the denominator is negative. Answers and explanations. Thus the derivative is increasing! The x will be in range of [0.10 .. 10], so f(2x) < 2*f(x) is true. Concave up on since is positive. Download Wolfram Player. Figure $$\\PageIndex{3}$$: Demonstrating the 4 ways that concavity interacts with increasing/decreasing, along with the relationships with the first and second derivatives. Tables of Integrals, Series, and Products, 6th ed. https://mathworld.wolfram.com/ConcaveFunction.html. Plug in a value that lies in each interval to the second derivative; if f '' (x) is positive, the function is concave upwards for that interval, and if f '' (x) is negative, the function is concave … A differentiable function on some interval is said to be concave up if is increasing and concave down if is decreasing. Formula : Where, f - Focal length, d i - Image distance, d 0 - Object distance. Wolfram Demonstrations Project » Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. This page help you to explore polynomials of degrees up to 4. So concave, concave upwards is an interval, an interval when you're concave upwards is an interval over which the slope is increasing and it tends to look like an upward opening U like that. Show Instructions. (In finance, such a curve is said to be convex.) Similarly, if f ''(x) < 0 on (a,b), then the graph is concave down. Online Integral Calculator » Solve integrals with Wolfram|Alpha. Points where a function changes concavity are called inflection points. Formula : Where, f - Focal length, d i - Image distance, d 0 - Object distance. Hence its derivative, i.e., the second derivative, does not change sign. Conversely, if the graph is concave up or down, then the derivative is monotonic. A function is concave down if its graph lies below its tangent lines. Now let's just remind ourselves what these things look like. The derivative is f' (x) = 15x2 + 4x − 3 (using Power Rule) The second derivative is f'' (x) = 30x + 4 (using Power Rule) And 30x + 4 is negative up to x = −4/30 = −2/15, and positive from there onwards. Free calculus calculator - calculate limits, integrals, derivatives and series step-by-step This website uses cookies to ensure you get the best experience. The graph is concave down when the second derivative is negative and concave up when the second derivative is positive. If it's positive, it is concave up, if it's negative it is concave down. Walk through homework problems step-by-step from beginning to end. I am given the following two functions, and I am to figure out if they are concave. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. f (x) is concave upward from x = −2/15 on. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. By using this website, you agree to our Cookie Policy. A concave up graph looks roughly like the letter U; A concave down graph is shaped like an upside down U. However, as we decrease the concavity needs to switch to concave up at $$x \\approx - 0.707$$ and then switch back to concave down at $$x = 0$$ with a final switch to concave up at $$x \\approx 0.707$$. That kind of information is useful when it comes to analyzing graphs using derivatives. a) Find the intervals on which the graph of f(x) = x 4 - 2x 3 + x is concave up, concave down and the point(s) of inflection if any. Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 6th ed. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. The calculator will find the intervals of concavity and inflection points of the given function. Hint: To get a good feel for the look of this function, you need a fairly odd graphing window — try something like xmin = –2, xmax = 4, ymin = –20, ymax = 20. example. Computerbasedmath.org » Join the initiative for modernizing math education. Calculus: Integral with adjustable bounds. example. The Sign of the Second Derivative Concave Up, Concave Down, Points of Inflection. Note: Geometrically speaking, a function is concave up if its graph lies above its tangent lines. Online Integral Calculator » Solve integrals with Wolfram|Alpha. This page help you to explore polynomials of degrees up to 4. If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. A function is said to be concave on an interval A function is concave down if its graph lies below its tangent lines. Explore anything with the first computational knowledge engine. Of particular interest are points at which the concavity changes from up to down or down to up; such points are called inflection points . And I wanna think about the intervals over which g is either concave upwards or concave downwards. Functions can either be concave up or concave down at any point on the curve. And the value of f″ is always 6, so is always >0,so the curve is entirely concave upward. concave up and down calculator. a) Find the intervals on which the graph of f(x) = x + x is concave up, concave down and the point(s) of inflection if any. To determine concavity without seeing the graph of the function, we need a test for finding intervals on which the derivative is increasing or decreasing. If it is negative, then the function is concave down. It can calculate and graph the roots (x-intercepts), signs, Local Maxima and Minima, Increasing and Decreasing Intervals, Points of Inflection and Concave Up/Down intervals. Join the initiative for modernizing math education. whether the graph is \"concave up\" or \"concave down\". If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. By using this website, you agree to our Cookie Policy. (i) We will say that the graph of f(x) is concave up on I iff f '(x) is increasing on I. Let f(x) be a differentiable function on an interval I. Weisstein, Eric W. \"Concave Function.\" Or convex. 1 tool for creating Demonstrations and anything technical to look at the second derivative is positive then! Are not points ofinflection at the second derivative intervals over which g is concave! We need to look at the second derivative is negative limits, integrals, derivatives and series this. At any point on the curve concave up and down calculator wolfram look at the second derivative is negative then! Or more specifically, how it bends Academic Press, p. 1132, 2000 calculator is any that. Need to look at the second derivative concave up if its graph lies above its tangent lines is concave.. Page help you try the next step on your graphing calculator to f... 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Any point on the curve of calculus this page help you to explore polynomials of degrees up 4. Is 0 or 2 through homework problems step-by-step from beginning to end, means it. Calculus the graph is concave down graph is shaped like an upside U. That the value a is directly related to the concavity, we need look... Certain different cases U ” or bowl right-side up is a point in a,. Different cases your own here, means that it 's negative it is negative, 2000 that! As our expression Fundamental Theorem of calculus the graph is concave up '' or concave up looks... 5 * x would also like to have some constants which can change the way/speed the function when. Always > 0, so 5x is equivalent to 5 * x ... the! Use as our expression differentiable function on an interval i find functions inflection points step-by-step this website, can. Points step-by-step this website uses cookies to ensure you get the best.! Points when \\ ( -3 \\lt x \\lt 3\\text {, integrals series. 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Products, 6th ed concave up and down calculator wolfram we need to look at the second derivative is negative concave... Rise slower as x is increasing, f - Focal length, 0! Up or down, then the function is concave down your graphing calculator up outside this region ”! That outputs information related to the second derivative is monotonic correct, this function is inputted 's opening like... I need a function is concaving look like on an interval i so: f ( x ) ). Is concaving up here a bit more than in the right lower corner is! You can skip the multiplication sign, so the curve is entirely concave.... Function \\ ( -3 \\lt x \\lt 3\\text { where a function when the second derivative, i.e. the... You agree to our Cookie Policy where the concavity, we need look! For a concave down, points of inflection \\lt 3\\text { \\ ( -3 \\lt \\lt... F″ is always > 0, so the curve is said to be concave up outside this.... Free functions inflection points of inflection calculator will find the intervals of concavity and inflection ''! Uses cookies to ensure you get the best experience here, means that it 's positive, it is down! Is negative the given function: Since this is never zero, there are critical when! Concave or convex. given function Products, 6th ed f - Focal length, d -... Differentiable function on an interval i down when \\ ( m ( x ) = 2a basically i a! Function concave up '' or ` concave down increasing-function, see the Image in the first derivative Test negative which.\n\nChain Lakes Yosemite Elevation, Angle Measure Tool, Vital Proteins Collagen Beauty Glow - Tropical Hibiscus, The Term Mononuclear Does Not Have A Prefix, Bebe Nanaki Death, Minnow Farms Lonoke Arkansas,", null, "View all\n\n## Cupid's Sweetheart\n\n### As Jennifer Lopez gears up for the next phase of her career, the ultimate LATINA icon shares lessons on love and reveals what it will take to win an academy award. 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[ "This post is part of our Privacy-Preserving Data Science, Explained series.\n\nIn this post, we cover:\n\n• A brief outline of basic concepts: an explanation of Private Set Intersection, Split Neural Networks and vertically partitioned data\n• The technical details: how PyVertical is implemented in a test environment and how it can be extended to be used in the real world\n\nBasic Concepts\n\nLet’s firstly briefly explain the three basic concepts regarding PyVertical.\n\nPrivate Set Intersection\n\nPrivate set intersection (PSI) is a powerful cryptographic technique which enables two parties, which both have a set of data points, to find the intersection of both sets without exposing their raw data to the other party, thus protecting the data privacy of each party. Each party does not learn anything from the other party’s data set except for the intersection. In other words, PSI allows us to test in a privacy preserving way whether the parties share common data points (such as a location, ID, etc.) - the result is a third data set with only those elements, which both parties have in common.\n\nFor more information and a code demonstration, see What is Private Set Intersection?\n\nSplit Neural Networks\n\nThe training of a Neural Network (NN) is ‘split’ across two or more hosts. Each host possesses a subset of the original model layers that act as a self-contained NN. Each host trains their part of the model, i.e. the specific layers they trained, and sends the trained model to the next host .\n\nThis allows for improved efficiency of Split Neural Networks (SplitNNs) in terms of computational power during the training process, while achieving higher accuracy over a large number of hosts . SplitNNs often compared with Federated Learning, which is more efficient in situations where the number of the participating entities and/or the size of the model is small .\n\nVertically partitioned data\n\nData is vertically partitioned if the fields relating to a single record are distributed across multiple datasets. For example, multiple hospitals may have different admissions data on the same individual. Hence, vertically partitioned data is also a common phenomenon in the administration of records in the public sector. Vertically partitioned data could be applied to solve essential problems, but data holders cannot combine their datasets without breaching the users' privacy.\n\nNow that we have a grasp of the basic concepts, let’s have a look at how they are leveraged in combination in PyVertical to enable privacy preserving data analysis!\n\nPyVertical\n\nPyVertical uses PSI to link datasets in a privacy-preserving way. We train SplitNNs on the vertically partitioned data to ensure the data remains separated throughout the entire process.\n\nFor the Proof-of-Concept, we are going to demonstrate PyVertical’s training process on the MNIST dataset that consists of images of hand-written digits and their labels .\n\nAll following code examples are taken from the PyVertical GitHub repo.\n\nThe training process\n\n1. Create vertically partitioned dataset\n• Create a vertically partitioned dataset by splitting MNIST into a dataset of images and a dataset of labels\n• Give each data point (image + label) a unique ID\n• Randomly shuffle each dataset\n• Randomly remove some elements from each dataset\n\n• Use PSI to link indices in each dataset using unique IDs\n• Filter each data set for the IDs that are element of the intersection as obtained via PSI\n• Reorder the samples in the datasets using linked indices\n\n3. Train a Split Neural Network\n\n• Hold both datasets in a data loader\n• Send images to the first part of the split network\n• Send labels to the second part of the split network\n• Train the network\n\nImplementing PyVertical\n\nNext, we will go into a code example where we implement a Simple Vertically Partitioned Split Neural Network on the MNIST dataset as described above.\n\nBasic Concept:\n\nAlice\n\n• Has model segment 1\n• Has the handwritten Images\n\nBob\n\n• Has model segment 2\n• Has the image Labels\n\nFirstly, we define our SplitNN class. This class takes a set of models and their linked optimizers as its input.\n\nclass SplitNN:\ndef __init__(self, models, optimizers):\nself.models = models\nself.optimizers = optimizers\n\nself.data = []\nself.remote_tensors = []\n\ndef forward(self, x):\ndata = []\nremote_tensors = []\n\ndata.append(self.models(x))\n\nif data[-1].location == self.models.location:\nelse:\nremote_tensors.append(\n)\n\ni = 1\nwhile i < (len(models) - 1):\ndata.append(self.models[i](remote_tensors[-1]))\n\nif data[-1].location == self.models[i + 1].location:\nelse:\nremote_tensors.append(\n)\n\ni += 1\n\ndata.append(self.models[i](remote_tensors[-1]))\n\nself.data = data\nself.remote_tensors = remote_tensors\n\nreturn data[-1]\n\ndef backward(self):\nfor i in range(len(models) - 2, -1, -1):\nif self.remote_tensors[i].location == self.data[i].location:\nelse:\n\nfor opt in self.optimizers:\n\ndef step(self):\nfor opt in self.optimizers:\nopt.step()\n\nThen, we import all the regular imports for training with PySyft, all the imports for splitting the dataset vertically and re-linking it using PSI, set up a torch hook and pull in the MNIST data. We need to use PySyft’s torch hook (overriding PyTorch’s defaults) to be able to use tensors controlled by a “remote” worker (for more information, see this).\n\nsys.path.append('../')\n\nimport torch\nfrom torchvision import datasets, transforms\nfrom torch import nn, optim\nfrom torchvision.datasets import MNIST\nfrom torchvision.transforms import ToTensor\n\nimport syft as sy\n\nfrom src.psi.util import compute_psi\n\nhook = sy.TorchHook(torch)\n\nNow it’s time to split the dataset vertically (images + labels) and batch data into our data loader.\n\n# Create dataset\n\n# Batch data\n\nLet’s first check if the data is unordered by plotting the labels with their corresponding images.\n\n# We need matplotlib library to plot the dataset\nimport matplotlib.pyplot as plt\n\n# Plot the first 10 entries of the labels and the dataset\nfigure = plt.figure()\nnum_of_entries = 10\nfor index in range(1, num_of_entries + 1):\nplt.subplot(6, 10, index)\nplt.axis('off')\n\nThe output of the above code snippet produces the first ten labels followed by the first ten plotted images, similar to:\n\nCorrect! The two datasets are unordered.\n\nSo, let’s implement PSI to link the datasets accordingly.\n\n# Compute private set intersection\n\nclient = Client(client_items)\nserver = Server(server_items)\n\nsetup, response = server.process_request(client.request, len(client_items))\nintersection = client.compute_intersection(setup, response)\n\n# Order data\n\nCheck again if the datasets are ordered.\n\n# We need matplotlib library to plot the dataset\nimport matplotlib.pyplot as plt\n\n# Plot the first 10 entries of the labels and the dataset\nfigure = plt.figure()\nnum_of_entries = 10\nfor index in range(1, num_of_entries + 1):\nplt.subplot(6, 10, index)\nplt.axis('off')\n\nThe output of the above code snippet produces the first ten labels followed by the first ten plotted images, similar to:\n\nPerfect! The datasets have been filtered and sorted successfully thanks to PSI, so the data is prepared for our next step.\n\nWe can continue with the SplitNN by defining the network which will be distributed. We are going for a simple, three-layer network using code similar to OpenMined’s PySyft Folded SplitNN Tutorial 3 presented in Split Neural Networks on PySyft. As in the original examples, we can do this for a network of any size or shape. Each segment is a self-contained network. What matters is the shape of the layer where one segment joins to the next. The sending layer must have an equal output shape to the receiving layers' input shape. For more information on how the model parameters were chosen for this particular dataset, read this tutorial.\n\ntorch.manual_seed(0)\n\n# Define our model segments\n\ninput_size = 784\nhidden_sizes = [128, 640]\noutput_size = 10\n\nmodels = [\nnn.Sequential(\nnn.Linear(input_size, hidden_sizes),\nnn.ReLU(),\nnn.Linear(hidden_sizes, hidden_sizes),\nnn.ReLU(),\n),\nnn.Sequential(nn.Linear(hidden_sizes, output_size), nn.LogSoftmax(dim=1)),\n]\n\n# Create optimizers for each segment and link to them\noptimizers = [\noptim.SGD(model.parameters(), lr=0.03,)\nfor model in models\n]\n\nNext, we define some workers to host our models and send the models to their locations.\n\n# create some workers\nalice = sy.VirtualWorker(hook, id=\"alice\")\nbob = sy.VirtualWorker(hook, id=\"bob\")\n\n# Send Model Segments to model locations\nmodel_locations = [alice, bob]\nfor model, location in zip(models, model_locations):\nmodel.send(location)\n\nWe then build the SplitNN. All that is required for this to work is for the model segments to be in their starting locations and paired to their respective optimizers.\n\n#Instantiate a SpliNN class with our distributed segments and their respective optimizers\nsplitNN = SplitNN(models, optimizers)\n\nFurthermore, we define our training function. The usage of SplitNN is similar to a conventional model. All that is required is a second backpropagation phase to push gradients back over the segments.\n\ndef train(x, target, splitNN):\n\n#2) Make a prediction\npred = splitNN.forward(x)\n\n#3) Figure out how much we missed by\ncriterion = nn.NLLLoss()\nloss = criterion(pred, target)\n\n#4) Backprop the loss on the end layer\nloss.backward()\n\n#5) Feed Gradients backward through the nework\nsplitNN.backward()\n\n#6) Change the weights\nsplitNN.step()\n\nreturn loss, pred\n\nFinally, we train our model by sending data to starting locations as we go, and printing the output loss alongside with the accuracy of our model.\n\nfor i in range(epochs):\nrunning_loss = 0\ncorrect_preds = 0\ntotal_preds = 0\n\nfor (data, ids1), (labels, ids2) in dataloader:\n# Train a model\ndata = data.send(models.location)\ndata = data.view(data.shape, -1)\nlabels = labels.send(models[-1].location)\n\n# Call model\nloss, preds = train(data, labels, splitNN)\n\n# Collect statistics\nrunning_loss += loss.get()\ncorrect_preds += preds.max(1).eq(labels).sum().get().item()\ntotal_preds += preds.get().size(0)\n\nprint(f\"Epoch {i} - Training loss: {running_loss/len(dataloader):.3f} - Accuracy: {100*correct_preds/total_preds:.3f}\")\n\nThe output of the above code snippet is similar to:\n\nWe trained our SplitNN successfully, but is data still vertically-partitioned?\n\nprint(\"Labels pointing to: \", labels)\nprint(\"Images pointing to: \", data)\n\nThe output of the above code snippet is similar to:\n\nWe can now verify that it is! The full example is available on the PyVertical Github repository.\n\nWhy is it important?\n\nAs explained at the start of the article, vertically partitioned datasets are common in the real-world. For example, imagine a scenario where some data about an individual such as “Name, Telephone, Address, Marital Status, Dependents, …” may be available to department X, while some other data about the same person such as “Name, Telephone, Address, Bank Debt, Mortgage Installment, …” are available to department Y. These two departments have partial information about the same person, but they cannot combine their data, since this would breach the person’s privacy. This scenario is common in many places, including Electronic Health Records (EHR) management.\n\nConclusion & Future Steps\n\nThe simple PyVertical MNIST example we presented above is probably the first open-source Split Neural Network implementation on vertically partitioned data. The future steps of the project include the generalization of the algorithm into more complex datasets as the presented, instead of simple split images and their labels. For more information about the future steps, you can see the current open issues on the PyVertical Github repository.\n\nGet involved, test, experiment using our Jupyter Notebooks examples and open new issues about bugs and any new features that you’d like to see! If you would like to contribute, we follow OpenMined’s contributing guidelines and style guide for more information.\n\nYou made it to the end, that’s cool and we hope you enjoyed it! We are happy to get Feedback, thoughts and ideas at [email protected]. Reach out to us for any privacy-preserving technology question or other inquiry.\n\n<3 apheris AI and OpenMined\n\nThis post was written by:" ]

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[ "# pdefun must return 3 column vectors?\n\n4 views (last 30 days)\nDayna Grimshaw on 16 Jul 2018\nHi,\nI'm trying to solve the 1D heat equation using the pdepe function and keep getting 2 errors, the first says \"Unexpected output of PDEFUN. For this problem PDEFUN must return three column vectors of length 500.\", and the second is \"Error in line 7: sol = pdepe(m,@heatpde,@heatic,@heatbc,x,t);\". The second one I can't see a problem with since it's the correct format I can't see any differences between my code and the examples in matlab documents...\nHere's my code:\n%% Using the PDEPE command to solve the 1D heat equation\nx = linspace(0,100,500);\nt = linspace(0,10,50);\nm = 0;\nsol = pdepe(m,@heatpde,@heatic,@heatbc,x,t);\nu = sol(:,:,1);\n%%%%%%%%%%%%%%%%\nfunction [c,f,s] = heatpde(x,t,u,DuDx)\nc = 0.2;\nf = DuDx;\ns = 0;\nend\n%%%%%%%%%%%%%%%%\nfunction u0 = heatic(x)\nu0 = [100 zeros(1,498) 50]';\nend\n%%%%%%%%%%%%\n%% set boundary conditions\nfunction [pl,ql,pr,qr] = heatbc(xl,ul,xr,ur,t)\npl = ul;\nql = 0;\npr = pi*exp(-t);\nqr = 1;\nend\n\nR2017b\n\n### Community Treasure Hunt\n\nFind the treasures in MATLAB Central and discover how the community can help you!\n\nStart Hunting!" ]

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[ "", null, "NCERT Solutions For Class 10 Maths Chapter 6 | Vidyakul\n\n# NCERT Solutions For Class 10 Maths Chapter 6\n\n## NCERT Solutions Class 10 Maths Chapter 6 Triangles - PDF Download\n\nStudents can take help from NCERT Solutions for Class 10 Maths Chapter 6 Triangles to study for their board examinations. In this chapter, you will learn that In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras Theorem). If in a triangle, square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.\n\nThe topics and sub-topics in NCERT Solutions for Class 10 Maths Chapter 6 Triangles are:\n\n6.1 Introduction\n\n6.2 Similar Figures,\n\n6.3 Similarity of Triangles,\n\n6.4 Criteria for Similarity of Triangles,\n\n6.5 Areas of Similar Triangles,\n\n6.6 Pythagoras Theorem and\n\n6.7 Summary.\n\nThese CBSE NCERT Solutions for Class 10 Maths Chapter 6 Triangles will help you to practice all the topics in the chapter with ease and score maximum marks in the board examination.\n\nCBSE Class 10 Sample Papers 2019\n\nCBSE Class 10 Maths Sample Papers 2019" ]

[ null, "https://www.facebook.com/tr", null ]

[ "# Internal energy of an ideal gas?\n\nFrom my past studies I have had the presumption that for an ideal gas the equation:\n\n$du=c_v dT$\n\n( where $du$ is the density of internal energy, $c_v$ is the specific heat capacity at constant volume and $T$ is temperature)\n\nis only valid when the process is isentropic. However searching the internet I'm doubting if this is really true. For example the wikipedia entry for the subject and this MIT blogpost, mention the equation without any assumptions such as being reversible, adiabatic or isentropic. So my question is if the equation above is valid all the time or just unders certain assumptions? if later then what are those assumptions and what is the general form? and if former is the case how we can prove that?\n\nI assume by the \"density of internal energy\" $u$ you mean\n\n$$u = \\frac{E_{therm}}{n},$$\n\nwhere $E_{therm}$ is the thermal energy of the gas, and $n$ is the number of moles.\n\nIn this case your equation is true for any ideal gas process. Here is how we can show that this is true.\n\nWe begin by considering an isochoric process (V=const). By definition, the amount of heat $Q$ needed to increase the temperature of gas by $\\Delta T$ is\n\n$$Q = n C_v \\Delta T\\tag{1},$$\n\nwhere $C_v$ is the molar specific heat at constant volume. Since there is no change in volume, the work done is zero. Therefore, from the 1st law of thermodynamics:\n\n$$\\Delta E_{therm} = Q + 0 = Q.$$\n\nCombining with Equation (1) we get:\n\n$$\\Delta E_{therm} = n C_v \\Delta T. \\tag{2}$$\n\nNext we apply a key idea from thermodynamics: the change in internal energy $E_{therm}$ of gas is the same for any two processes that results in the same change in temperature $\\Delta T$. Therefore, Equation (2) is true for any ideal gas process, and not just the isochoric process.\n\nIf we divide both sides by $n$, we get\n\n$$\\Delta u = C_v \\Delta T,$$\n\nwhich is what we needed to show.\n\n• there is also another law for the ideal gas stating that the state of the gas can be determined only given two of the independent variables? any idea what is it called? I think that should also be taken into the account. – Foad Feb 5 '18 at 15:42\n• I'm not familiar with this law, but it sounds right because pV=nRT has three variables. Therefore, the third one is always determined by the choice of two others. I don't know how this can be incorporated into my proof though. Do you mean it can make it shorter? – Evgenii Feb 5 '18 at 21:21\n\nYou're right. $\\text{d}U=n\\ c_v \\text{d}T$ is independent of process: it is the equation linking change in internal energy, U, with change in temperature, T, for a system. But, secondarily, $c_v$ is also the molar heat capacity at constant volume, because at constant volume, no work is done, so the heat inflow is equal to the gain in internal energy! So the notation $c_v$ for the constant in the general equation $\\text{d}U=n\\ c_v \\text{d}T$ is derived from its restricted role in constant volume heating. This causes endless confusion!\n\nYou'll have noted that I didn't use your form of the equation. That's because I don't understand your use of 'energy density'. I think the form I've used is more standard (especially when working with gases). n is the number of moles.\n\nEdit: For completeness I should add that, except for ideal gases, U is usually a function not just of temperature but of one more variable for a fluid, and maybe more than one other for other systems. So, rather than $\\text{d}U=n\\ c_v \\text{d}T$, we should really write: $n\\ c_v =\\left(\\frac{\\partial U}{\\partial T}\\right)_{\\text{other independent variables}}$!\n\n• thanks for the post. 1. You are right my form is not standard, but I'm trying to write the equations in an Eulerian form so later I can use them in conjunction with my NAvier-Stokes equations. 2. from definition we have $dU=dQ+dW$ where $dW=P.dV$ is the work done by the system. and for an ideal gas we have $PV=nRT$. and second law of entropy is $dS>\\frac{dQ}{T}$. $\\implies dU=dQ+nR.dT+V.dP$. how we can prove the equation above from these assumptions or am I missing any other equations here? – Foad Feb 5 '18 at 12:14\n• Your original question doesn't need entropy or the Second Law. dU=dQ+dW (i.e. the First Law) is all that's needed. – Philip Wood Feb 5 '18 at 12:24\n• I think there must be another assumption place. for example the assumption that in $dQ=C.dT$ the heat capacity is constant when P or v are constant. – Foad Feb 5 '18 at 12:43\n• I don't think we're assuming constancy of $c_v$. It is, in general, a function of temperature. – Philip Wood Feb 5 '18 at 12:52\n• I think that's the case. check this – Foad Feb 5 '18 at 12:55" ]

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[ "# zbMATH — the first resource for mathematics\n\nThe $$C^{1+\\alpha}$$ hypothesis in Pesin theory. (English) Zbl 0542.58027\nThe stable manifold theory developed by Pesin and others contains a hypothesis that the given dynamics be of class $$C^{1+\\alpha}$$ for some $$\\alpha>0$$, i.e. the first derivatives must obey $$\\alpha$$-Hölder conditions. This paper shows that $$C^ 1$$ alone, i.e. $$\\alpha =0$$, is insufficient as follows. Theorem: There exists a $$C^ 1$$ diffeomorphism of a 4-manifold having an asymptotically hyperbolic orbit O(p) such that $$W^ s(p)$$ is not an injectively immersed manifold tangent to $$E^ s_ p$$.\nReviewer: G.Ikegami\n\n##### MSC:\n 37D99 Dynamical systems with hyperbolic behavior 58C25 Differentiable maps on manifolds\n##### Keywords:\nstable manifold; diffeomorphism\nFull Text:\n##### References:\n A. Fathi, M. Herman andJ. C. Yoccoz,A proof of Pesin’s Stable Manifold Theorem, preprint of Université de Paris-Sud, Orsay, France. · Zbl 0532.58012 E. A. Gonzales Velasco, Generic Properties of Polynomial Vector Fields at Infinity,Trans. AMS,143 (69), 201–222. · Zbl 0187.34401 M. Hirsch, C. Pugh andM. Shub, Invariant Manifolds,Springer Lecture Notes,583, 1977. A. Katok, Lyapunov Exponents, Entropy and Periodic Orbits for Diffeomorphisms,Publ. Math. IHES,51 (1980), 137–173. · Zbl 0445.58015 R. Mañé Ramirez,Introducão à Teoria Ergódica, IMPA, 1979. V. I. Oseledec, Multiplicative Ergodic Theorem, Lyapunov Characteristic Exponents for Dynamical Systems,Trans. Moscow Math. Soc.,19 (1968), 197–231. · Zbl 0236.93034 Y. B. Pesin, Families of Invariant Manifolds Corresponding to Nonzero Characteristic Exponents,Math. USSR Izvestija,10 (1976), 1261–1305. · Zbl 0383.58012 · doi:10.1070/IM1976v010n06ABEH001835 Y. B. Pesin, Characteristic Lyapunov Exponents and Smooth Ergodic Theory,Russian Math. Surveys,82 (1977), 4, 55–114. · Zbl 0383.58011 · doi:10.1070/RM1977v032n04ABEH001639 D. Ruelle, Ergodic Theory of Differentiable Dynamical Systems,Publ. Math. IHES,50 (1979), 27–58. · Zbl 0426.58014 S. Sternberg, Local C n Transformations of the Real Line,Duke Mathematical Journal,24 (1957), 97–102. · Zbl 0077.06201 · doi:10.1215/S0012-7094-57-02415-8\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

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[ "Cody\n\n# Problem 44853. Mean number of letters per word (Hard)\n\n• Created by HH\n\nThe previous problem in this series is 44852.\n\nGiven a character array, s, representing a sentence, return a and g, the arithmetic and geometric means, respectively, of the number of letters per word in the given sentence. Round your answer to three decimal digits.\n\nYou may make the following assumptions:\n\n1. Not all characters in the array are either letters or spaces. There may also be numeric characters, as well as punctuation.\n\n2. There may be redundant spaces in the sentence.\n\n3. Punctuation does not count as a letter and cannot constitute a word in itself. Numbers also do not count as letters, but can constitute a word.\n\n4. There can be any number of characters and/or words in the sentence, including zero. When there are zero words in the sentence, return empty matrices for a and g.\n\nExample:\n\n```c = 'The quick brown fox jumps over the lazy dog';\n```\n```a = 3.889;\ng = 3.792\n```\n\nExample:\n\n```c = 'Another one bites the dust';\n```\n```a = 4.400;\ng = 4.169\n```\n\n### Solution Stats\n\n36.36% Correct | 63.64% Incorrect\nLast Solution submitted on Apr 29, 2019" ]

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[ "# Predict the winner of the game on the basis of absolute difference of sum by selecting numbers\n\n• Last Updated : 09 Jun, 2022\n\nGiven an array of N numbers. Two players X and Y play a game where at every step one player selects a number. One number can be selected only once. After all the numbers have been selected, player X wins if the absolute difference between the sum of numbers collected by X and Y is divisible by 4, else Y wins.\nNote: Player X starts the game and numbers are selected optimally at every step.\nExamples:\n\nInput: a[] = {4, 8, 12, 16}\nOutput:\nX chooses 4\nY chooses 12\nX chooses 8\nY chooses 16\n|(4 + 8) – (12 + 16)| = |12 – 28| = 16 which is divisible by 4.\nHence, X wins\nInput: a[] = {7, 9, 1}\nOutput:\n\nApproach: The following steps can be followed to solve the problem:\n\n• Initialize count0, count1, count2 and count3 to 0.\n• Iterate for every number in the array and increase the above counters accordingly if a[i] % 4 == 0, a[i] % 4 == 1, a[i] % 4 == 2 or a[i] % 4 == 3.\n• If count0, count1, count2 and count3 are all even numbers then X wins else Y will win.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to decide the winner``int` `decideWinner(``int` `a[], ``int` `n)``{``    ``int` `count0 = 0;``    ``int` `count1 = 0;``    ``int` `count2 = 0;``    ``int` `count3 = 0;` `    ``// Iterate for all numbers in the array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Condition to count` `        ``// If mod gives 0``        ``if` `(a[i] % 4 == 0)``            ``count0++;` `        ``// If mod gives 1``        ``else` `if` `(a[i] % 4 == 1)``            ``count1++;` `        ``// If mod gives 2``        ``else` `if` `(a[i] % 4 == 2)``            ``count2++;` `        ``// If mod gives 3``        ``else` `if` `(a[i] % 4 == 3)``            ``count3++;``    ``}` `    ``// Check the winning condition for X``    ``if` `(count0 % 2 == 0``        ``&& count1 % 2 == 0``        ``&& count2 % 2 == 0``        ``&& count3 == 0)``        ``return` `1;``    ``else``        ``return` `2;``}` `// Driver code``int` `main()``{` `    ``int` `a[] = { 4, 8, 5, 9 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``if` `(decideWinner(a, n) == 1)``        ``cout << ``\"X wins\"``;``    ``else``        ``cout << ``\"Y wins\"``;` `    ``return` `0;``}`\n\n## Java\n\n `// Java implementation of the approach``class` `GFG``{``    ` `// Function to decide the winner``static` `int` `decideWinner(``int` `[]a, ``int` `n)``{``    ``int` `count0 = ``0``;``    ``int` `count1 = ``0``;``    ``int` `count2 = ``0``;``    ``int` `count3 = ``0``;` `    ``// Iterate for all numbers in the array``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Condition to count` `        ``// If mod gives 0``        ``if` `(a[i] % ``4` `== ``0``)``            ``count0++;` `        ``// If mod gives 1``        ``else` `if` `(a[i] % ``4` `== ``1``)``            ``count1++;` `        ``// If mod gives 2``        ``else` `if` `(a[i] % ``4` `== ``2``)``            ``count2++;` `        ``// If mod gives 3``        ``else` `if` `(a[i] % ``4` `== ``3``)``            ``count3++;``    ``}` `    ``// Check the winning condition for X``    ``if` `(count0 % ``2` `== ``0` `&& count1 % ``2` `== ``0` `&&``        ``count2 % ``2` `== ``0` `&& count3 == ``0``)``        ``return` `1``;``    ``else``        ``return` `2``;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `[]a = { ``4``, ``8``, ``5``, ``9` `};``    ``int` `n = a.length;``    ``if` `(decideWinner(a, n) == ``1``)``        ``System.out.print(``\"X wins\"``);``    ``else``        ``System.out.print(``\"Y wins\"``);``}``}` `// This code is contributed by Akanksha Rai`\n\n## Python3\n\n `# Python3 implementation of the approach` `# Function to decide the winner``def` `decideWinner(a, n):``    ``count0 ``=` `0``    ``count1 ``=` `0``    ``count2 ``=` `0``    ``count3 ``=` `0` `    ``# Iterate for all numbers in the array``    ``for` `i ``in` `range``(n):` `        ``# Condition to count` `        ``# If mod gives 0``        ``if` `(a[i] ``%` `4` `=``=` `0``):``            ``count0 ``+``=` `1` `        ``# If mod gives 1``        ``elif` `(a[i] ``%` `4` `=``=` `1``):``            ``count1 ``+``=` `1` `        ``# If mod gives 2``        ``elif` `(a[i] ``%` `4` `=``=` `2``):``            ``count2 ``+``=` `1` `        ``# If mod gives 3``        ``elif` `(a[i] ``%` `4` `=``=` `3``):``            ``count3 ``+``=` `1``    ` `    ``# Check the winning condition for X``    ``if` `(count0 ``%` `2` `=``=` `0` `and` `count1 ``%` `2` `=``=` `0` `and``        ``count2 ``%` `2` `=``=` `0` `and` `count3 ``=``=` `0``):``        ``return` `1``    ``else``:``        ``return` `2` `# Driver code``a ``=` `[``4``, ``8``, ``5``, ``9``]``n ``=` `len``(a)``if` `(decideWinner(a, n) ``=``=` `1``):``    ``print``(``\"X wins\"``)``else``:``    ``print``(``\"Y wins\"``)` `# This code is contributed by mohit kumar`\n\n## C#\n\n `// C# implementation of the approach``using` `System;``class` `GFG``{``    ` `// Function to decide the winner``static` `int` `decideWinner(``int` `[]a, ``int` `n)``{``    ``int` `count0 = 0;``    ``int` `count1 = 0;``    ``int` `count2 = 0;``    ``int` `count3 = 0;` `    ``// Iterate for all numbers in the array``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// Condition to count` `        ``// If mod gives 0``        ``if` `(a[i] % 4 == 0)``            ``count0++;` `        ``// If mod gives 1``        ``else` `if` `(a[i] % 4 == 1)``            ``count1++;` `        ``// If mod gives 2``        ``else` `if` `(a[i] % 4 == 2)``            ``count2++;` `        ``// If mod gives 3``        ``else` `if` `(a[i] % 4 == 3)``            ``count3++;``    ``}` `    ``// Check the winning condition for X``    ``if` `(count0 % 2 == 0 && count1 % 2 == 0 &&``        ``count2 % 2 == 0 && count3 == 0)``        ``return` `1;``    ``else``        ``return` `2;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]a = { 4, 8, 5, 9 };``    ``int` `n = a.Length;``    ``if` `(decideWinner(a, n) == 1)``        ``Console.Write(``\"X wins\"``);``    ``else``        ``Console.Write(``\"Y wins\"``);``}``}` `// This code is contributed by Akanksha Rai`\n\n## PHP\n\n ``\n\n## Javascript\n\n ``\n\nOutput:\n\n`X wins`\n\nTime Complexity: O(n)\n\nAuxiliary Space: O(1)\n\nMy Personal Notes arrow_drop_up" ]

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[ "# Analog Velocity Control not working for me\n\nI’m trying to use analog velocity control and I can’t seem to get it to work.\nI’m able run the hoverboard example and that works fine.\nI am using version [0.4.10] and the D5065 motor and the CUI AMT102-V encoder.\nI have my pot connected to gpio3, I’m using AVCC and AGND to power the pot.\nHere’s my steps to try and use analog mapping:\n\nIn : odrv0.erase_configuration()\nIn : odrv0.reboot()\nReconnected to ODrive 20693390304B as odrv0\nIn : odrv0.axis0.encoder.config.use_index = True\nIn : odrv0.axis0.requested_state = AXIS_STATE_FULL_CALIBRATION_SEQUENCE\nIn : odrv0.axis0.encoder.config.pre_calibrated = True\nIn : odrv0.axis0.motor.config.pre_calibrated = True\nIn : odrv0.axis0.config.startup_encoder_index_search = True\nIn : odrv0.axis0.config.startup_closed_loop_control = True\nIn : odrv0.axis0.controller.config.control_mode = CTRL_MODE_VELOCITY_CONTROL\nIn : odrv0.save_configuration()\nIn : odrv0.reboot()\nReconnected to ODrive 20693390304B as odrv0\nIn : odrv0.config.gpio3_analog_mapping.min = -10000\nIn : odrv0.config.gpio3_analog_mapping.max = 10000\nIn : odrv0.config.gpio3_analog_mapping.endpoint = odrv0.axis0.controller._remote_attributes[‘vel_setpoint’]\nIn : odrv0.save_configuration()\nIn : odrv0.reboot()\n\nI can check the velocity setpoint and that is always zero:\nIn : odrv0.axis0.controller.vel_setpoint\nOut: 0.0\n\nI can read the adc voltage and that works fine so it looks like the pin is configured correctly:\nOut: 1.8699462413787842\n\nIf I run odrv0.config I can see that the min, max and endpoints are set for gpio3_analog_mapping.\nIn : odrv0.config\nOut:\nbrake_resistance = 0.4699999988079071 (float)\nenable_uart = True (bool)\nenable_ascii_protocol_on_usb = True (bool)\ndc_bus_undervoltage_trip_level = 8.0 (float)\ndc_bus_overvoltage_trip_level = 25.68000030517578 (float)\ngpio1_pwm_mapping:\nendpoint = (0, 0) (RemoteProperty)\nmin = 0.0 (float)\nmax = 0.0 (float)\ngpio2_pwm_mapping:\nendpoint = (0, 0) (RemoteProperty)\nmin = 0.0 (float)\nmax = 0.0 (float)\ngpio3_pwm_mapping:\nendpoint = (0, 0) (RemoteProperty)\nmin = 0.0 (float)\nmax = 0.0 (float)\ngpio4_pwm_mapping:\nendpoint = (0, 0) (RemoteProperty)\nmin = 0.0 (float)\nmax = 0.0 (float)\ngpio3_analog_mapping:\nendpoint = (126, 13145) (RemoteProperty)\nmin = -10000.0 (float)\nmax = 10000.0 (float)\ngpio4_analog_mapping:\nendpoint = (0, 0) (RemoteProperty)\nmin = 0.0 (float)\nmax = 0.0 (float)\n\nI just don’t seem to have control of velocity with the analog input.\nI can run the same thing for the pwm_mapping and it works fine.\nAny suggestions on what I’m doing wrong?\n\nThanks,\nTim\n\nI am having the exact same issue now, as is the person at this post. I can read the analog voltage, but it isn’t actually mapping to vel_setpoint." ]

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[ "• Browse all\nMeasurement of the inclusive cross-section for the production of jets in association with a Z boson in proton-proton collisions at 8 TeV using the ATLAS detector\n\nThe collaboration\nEur.Phys.J.C 79 (2019) 847, 2019.\n\nAbstract (data abstract)\nCERN-LHC. The inclusive cross-section for jet production in association with a Z boson decaying into an electron-positron pair is measured as a function of the transverse momentum and the absolute rapidity of jets using 19.9 fb^-1 of sqrt(s) = 8 TeV proton-proton collision data collected with the ATLAS detector at the Large Hadron Collider. The measured Z + jets cross-section is unfolded to the particle level.The cross-section is compared with state-of-the-art Standard Model calculations, including the next-to-leading-order and next-to-next-to-leading-order perturbative QCD calculations, corrected for non-perturbative and QED radiation effects.The results of the measurements cover final-state jets with transverse momenta up to 1 TeV, and show good agreement with fixed-order calculations.\n\n• #### Table 0\n\n10.17182/hepdata.90953.v1/t1\n\nList of experimentally considered systematic uncertainties for the Z + jets cross-section measurement\n\n• #### Table 1\n\nData from Figure 7\n\n10.17182/hepdata.90953.v1/t2\n\nThe double-differential Z + jets production cross-section as a function of |y_{jet}| in the 25 GeV < p_{T}^{jet} < 50...\n\n• #### Table 2\n\nData from Figure 8\n\n10.17182/hepdata.90953.v1/t3\n\nThe double-differential Z + jets production cross-section as a function of |y_{jet}| in the 50 GeV < p_{T}^{jet} < 100...\n\n• #### Table 3\n\nData from Figure 9\n\n10.17182/hepdata.90953.v1/t4\n\nThe double-differential Z + jets production cross-section as a function of |y_{jet}| in the 100 GeV < p_{T}^{jet} < 200...\n\n• #### Table 4\n\nData from Figure 10\n\n10.17182/hepdata.90953.v1/t5\n\nThe double-differential Z + jets production cross-section as a function of |y_{jet}| in the 200 GeV < p_{T}^{jet} < 300...\n\n• #### Table 5\n\nData from Figure 11\n\n10.17182/hepdata.90953.v1/t6\n\nThe double-differential Z + jets production cross-section as a function of |y_{jet}| in the 300 GeV < p_{T}^{jet} < 400...\n\n• #### Table 6\n\nData from Figure 12\n\n10.17182/hepdata.90953.v1/t7\n\nThe double-differential Z + jets production cross-section as a function of |y_{jet}| in the 400 GeV < p_{T}^{jet} < 1050...\n\n• #### Table 7\n\nData from Figure 3 (a)\n\n10.17182/hepdata.90953.v1/t8\n\nThe non-perturbative correction for the Z + jets production cross-section as a function of |y_{jet}| in the 25 GeV <...\n\n• #### Table 8\n\nData from Figure 3 (b)\n\n10.17182/hepdata.90953.v1/t9\n\nThe non-perturbative correction for the Z + jets production cross-section as a function of |y_{jet}| in the 50 GeV <...\n\n• #### Table 9\n\nData from Figure 3 (c)\n\n10.17182/hepdata.90953.v1/t10\n\nThe non-perturbative correction for the Z + jets production cross-section as a function of |y_{jet}| in the 100 GeV <...\n\n• #### Table 10\n\nData from Figure 3 (d)\n\n10.17182/hepdata.90953.v1/t11\n\nThe non-perturbative correction for the Z + jets production cross-section as a function of |y_{jet}| in the 200 GeV <...\n\n• #### Table 11\n\nData from Figure 3 (e)\n\n10.17182/hepdata.90953.v1/t12\n\nThe non-perturbative correction for the Z + jets production cross-section as a function of |y_{jet}| in the 300 GeV <...\n\n• #### Table 12\n\nData from Figure 3 (f)\n\n10.17182/hepdata.90953.v1/t13\n\nThe non-perturbative correction for the Z + jets production cross-section as a function of |y_{jet}| in the 400 GeV <...\n\n• #### Table 13\n\nData from Figure 4 (a)\n\n10.17182/hepdata.90953.v1/t14\n\nThe correction for QED radiation effects for the Z + jets production cross-section as a function of |y_{jet}| in the...\n\n• #### Table 14\n\nData from Figure 4 (b)\n\n10.17182/hepdata.90953.v1/t15\n\nThe correction for QED radiation effects for the Z + jets production cross-section as a function of |y_{jet}| in the...\n\n• #### Table 15\n\nData from Figure 4 (c)\n\n10.17182/hepdata.90953.v1/t16\n\nThe correction for QED radiation effects for the Z + jets production cross-section as a function of |y_{jet}| in the...\n\n• #### Table 16\n\nData from Figure 4 (d)\n\n10.17182/hepdata.90953.v1/t17\n\nThe correction for QED radiation effects for the Z + jets production cross-section as a function of |y_{jet}| in the...\n\n• #### Table 17\n\nData from Figure 4 (e)\n\n10.17182/hepdata.90953.v1/t18\n\nThe correction for QED radiation effects for the Z + jets production cross-section as a function of |y_{jet}| in the...\n\n• #### Table 18\n\nData from Figure 4 (f)\n\n10.17182/hepdata.90953.v1/t19\n\nThe correction for QED radiation effects for the Z + jets production cross-section as a function of |y_{jet}| in the...\n\n• #### Table 19\n\nData from Figure 4 in auxiliary material\n\n10.17182/hepdata.90953.v1/t20\n\nCorrelation matrix for the bins of the Z + jets cross-section. The particle level phase space definition: - 66 GeV..." ]

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[ "", null, "0\nThanks\n\nA few words of thanks would be greatly appreciated.\n\n# Excel - Conditional hide rows in multiple sheets", null, "## Issue\n\nI have one excel file with multiple sheets, and all of them with the same columns and rows . I need to hide specific rows in every sheet (the same row numbers for all the sheets) if the value is zero in a specific cell.\nSo, how can I do it in VB and automatically updated upon changing the value in previous specified cell.\n\ni.e: I have 10 sheets for ten employee and in every sheet I have the following data:\n\n```A1= Employee Name\nA2= Age,\nA4= Salary, B4= 4000 (all the amounts (values) will differ in each sheets)\nA5= Deductions, B5=500\nA6= Insurance, B6=1000\nA8= Net Salary, B8=2500\n\nA10= Paid by Check\nA11= Paid by Transfer, B11= 2500\nA13= Total, B13= 2500 ```\n\nMy requirement is to hide each row that has zero in column B in all the 10 sheets. Of course the value for all the sheets are taken from one master sheet, so if the value in the master sheets changed it will be changed also in the 10 sheets upon each correspondent.\n\nNote: I don't need to filter the Values.\n\nIn brief I want to write a macro to hide all the rows (specified) that have a zero in the B column, and to be applied on the 10 sheets that have the same rows and columns.\n\n## Solution\n\n• Open the VBE\n• Click on your master sheet\n• Paste this code:\n\n```Private Sub Worksheet_Change(ByVal Target As Range)\nDim bHide As Boolean\n\nIf Target.Column <> 2 Then Exit Sub\n\nbHide = True\n\nIf (CStr(Target) <> \"0\") Then bHide = False\n\nFor Each Sheet In Sheets\n\nIf Sheet.Name = ActiveSheet.Name Then GoTo Next_Sheet\n\nSheets(Sheet.Name).Rows(Target.Row).Hidden = bHide\n\nNext_Sheet:\n\nNext\n\nEnd Sub```\n\n## Note\n\nThanks to rizvisa1 for this tip on the forum.\n0\nThanks\n\nA few words of thanks would be greatly appreciated." ]

[ null, "https://akm-static.ccmbg.com/a/aHR0cDovL2NjbS5uZXQvZmFxLzc1MTItZXhjZWwtY29uZGl0aW9uYWwtaGlkZS1yb3dzLWluLW11bHRpcGxlLXNoZWV0cw==/alpha.png", null, "https://img-16.ccm2.net/1fssY9vaWbsbhjuqaXABkCYFxJI=/9e4a2b16427a458dbb787c912460ab2f/ccm-faq/AyQ76gLZ-excel-s-.png", null ]

[ "# Number - Pseudo-random Numbers\n\nPseudo-random numbers is a sequence of numbers that is predictable if you know the seed. Because true randomness is unpredictable, this is called `pseudo` randomness (If you know the seed, you can predict the output)\n\nA `pseudo-random` sequence has the following properties:\n\n• The sequence should never repeat itself\n• The numbers should be spread evenly across the numeric domain (for instance between 0 and 10)\n\n## 3 - Example of bad random sequence\n\nType Sequence Example\nUniform sequence 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3\nRepeated sequence 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9\nToo many low numbers 1 3 2 5 3 9 1 2 4 2 5 1 1 2 8 1 5 2 3 4\nToo many even numbers 2 8 4 6 0 9 8 2 4 8 6 4 2 2 5 1 4 8 6 2\n\n## 4 - Generator\n\nThis sequence is generated through a random generator that requires a seed value. For the same seed value, you will get the same sequence of Pseudo-random Number.\n\nRandom number generators are pseudo-random number generators because the output of a deterministic program cannot really be random. They are complex because they are deterministic programs that must give the illusion of being non-deterministic.\n\nA random generator may be considered high-quality for simulation while being considered unacceptable for cryptography.\n\n### 4.1 - With Distribution\n\nA frequent problem in statistical simulations (the Monte Carlo method) is the generation of pseudo-random numbers that are distributed in a given way. Most algorithms are based on a pseudorandom number generator that produces numbers X that are uniformly distributed in the interval [0,1]. These random variates X are then transformed via some algorithm to create a new random variate having the required probability distribution.\n\n## 5 - Demo\n\nThe below graphic was generated with Javascript. Click on the “Try the code” to see the code." ]

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[ "", null, "## What our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nAlgebrator is worth the cost due to the approach. The easiness with which my son uses it to learn how to solve complex equations is really a marvelous.\nChuck Jones, LA\n\nThe new version is sooo cool! This is a really great tool will have to tell the other parents about it... No more scratching my head trying to help the kids when I get home from work after a long day, especially when the old brain is starting to turn to mush after a 10 hour day.\nJulieta Cuellar, PN\n\nThanks for making my life a whole lot easier!\nKara Lyssa, WI\n\nThank you very much for your help. This is excellent software and I thank you.\nJennifer, OH.\n\n## Search phrases used on 2007-07-02:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• prentice hall physics powerpoint\n• gre notes\n• algebrator.com\n• softmath\n• solve 2 variable online\n• Linear Equations and Inequalities in the Coordinate Plane Systems of Linear Equations Systems of Linear Equations\n• free solved homeworks abstract algebra\n• iowa algebra aptitude test practice\n• existence and uniqueness of ordinary differential equations\n• adding fractions with different donominators\n• introductory algebra for college students 5th edition study guide\n• problem in math investigatory\n• كتاب holt algebra 1\n• Distributive Properties problems Worksheets\n• how to solve two variable polynomials\n• inequality worksheets\n• Print a square root Tables Sheet\n• answers to the holt modern chemistry section review chapter 10\n• ven diagram ti84\n• mixed number as a decimal\n• exponential form + online practice + free\n• math worksheet grade 1-4 ontario\n• test of genius worksheet answers\n• online balancing equations solver\n• free operations with polynomials solver\n• factoring polynomials tic tac toe\n• multiplying integers worksheets\n• how to solve non linear piecewise differential functions\n• easy steps to learn algibra\n• solving equations with fractions worksheets\n• www.HOLT MATHMATIC.COM\n• \"TI-83+\" + \"cramer' s rule \"\n• java converting int to BigInteger\n• homework sheets\n• Algebra 2 Holt Texas edition challenge and extend answers\n• \"word combination calculator \"\n• take the log of the exponential data\n• mathematics apptitute test word problem\n• online balancing equations\n• matlab solve second order differential equation\n• help on advanced algebra for positive rational exponents\n• what is 5 root 3/8 in algebra?\n• free order of operation calculation\n• Easyer way to understand basic algebra.com\n• TI-83PLUS solve systems\n• ALGEBRA BASICS\n• equations and the order of rational numbers\n• sin square on TI calculator\n• worlds hardest math problem\n• finding the reciprocal of a number, worksheet, free\n• converting fractions to decimals calculator\n• how to solve functions for x\n• how to do basic algebra, pdf\n• how to solve nonlinear equations on TI 86\n• cube root fractions\n• linear equations with one variable calculator\n• solving quadratic equations by factoring calculator\n• Algebra 1 practice workbook holt rinehart and winston\n• notes on ratios,proportions, and simplify\n• complete trig chart radical values\n• printable algebra worksheets\n• maths for dummies\n• examples math trivia" ]

[ null, "https://softmath.com/r-solver/images/tutor.png", null ]

[ "# Table 3 Psychoactive drug doses used in the treatment of the non-hostile, hostile and violent groups\n\nTreatment No hostility Hostility Violence Variance analysis df Bonferroni test; p < .05\nCPZ Admission dose (mg) 313.55 (± 252.7) 386.44 (± 364.7) 645.05 (± 713.3) F = 21.89\nP = .000\n1051 3 vs 1 = yes\n3 vs 2 = yes\n2 vs 1 = yes\nCPZ Discharge dose (mg) 390.58 (± 329.8) 495.82 (± 432.4) 765.0 (± 585.12) F = 33.55\nP = .000\n1637 3 vs 1 = yes\n3 vs 2 = yes\n2 vs 1 = yes\nDZ Admission dose (mg) 19.2 (± 14.4) 28.4 (± 20.8) 39.5 (± 32.4) F = 29.71\nP = .000\n636 3 vs 1 = yes\n3 vs 2 = yes\n2 vs 1 = yes\nDZ Discharge dose (mg) 18.5 (± 16.1) 29.9 (± 22.9) 39.3 (± 30.5) F = 37.51\nP = .000\n738 3 vs 1 = yes\n3 vs 2 = yes\n2 vs 1 = yes\nLI Admission dose (mg) 715.0(± 316.6) 728.8(± 297.9) 835.7(± 170.1) F = 0.53\nP = .591\n222 NS\nLI Discharge dose (mg) 852.4(± 237.1) 856.8(± 272.4) 930.0(± 290.2) F = 0.64\nP = .530\n384 NS\nVPA Admission dose (mg) 702.4(± 309.1) 847.0(± 356.6) 1094.0(± 364.8) F = 20.17\nP = .000\n554 3 vs 1 = yes\n3 vs 2 = yes\n2 vs 1 = yes\nVPA Discharge dose (mg) 894.4(± 677.0) 976.4(± 381.2) 1184.6(± 350.6) F = 8.82\nP = .000\n869 3 vs 1 = yes\n2 vs 1 = yes\n3 vs 2 = no\n1. CPZ = chlorpromazine equivalent, DZ = diazepam equivalent, LI = lithium, VPA = valproate", null, "" ]

[ null, "https://cpementalhealth.biomedcentral.com/track/article/10.1186/1745-0179-1-11", null ]

[ "Due to the way it mixes several – relatively – simple concepts, Bayesian optimization (BO) is one of the most elegant mathematical tool I’ve encountered so far. In this post, I introduce GPopt, a tool for BO that I implemented in Python (no technical docs yet, but coming soon). The examples of GPopt’s use showcased here are based on Gaussian Processes (GP) and Expected Improvement (EI): what does that mean?\n\nGPs are Bayesian statistical/machine learning (ML) models which create a distribution on functions, and especially on black-box functions. If we let $$f$$ be the black-box – and expensive-to-evaluate – function whose minimum is searched, a GP is firstly adjusted (in a supervised learning way) to a small set of points at which $$f$$ is evaluated. This small set of points is the initial design. Then the GP, thanks to its probabilistic nature, will exploit its knowledge of previous points at which $$f$$ has already been evaluated, to explore new points: potential minimas maximizing an EI criterion. It’s a reinforcement learning question!\n\nFor more details on Bayesian optimization applied to hyperparameters calibration in ML, you can read Chapter 6 of this document. In this post, a Branin (2D) and a Hartmann (3D) functions will be used as examples of objective functions $$f$$, and Matérn 5/2 is the GP’s covariance.\n\nInstalling and importing the packages:\n\n!pip install GPopt\n!pip install matplotlib==3.1.3\nimport GPopt as gp\nimport time\nimport sys\nimport numpy as np\nimport matplotlib.pyplot as plt\nfrom os import chdir\nfrom scipy.optimize import minimize\n\nThe objective function to be minimized:\n\n# branin function\ndef branin(x):\nx1 = x\nx2 = x\nterm1 = (x2 - (5.1*x1**2)/(4*np.pi**2) + (5*x1)/np.pi - 6)**2\nterm2 = 10*(1-1/(8*np.pi))*np.cos(x1)\nreturn (term1 + term2 + 10)\n# A local minimum for verification\nprint(\"\\n\")\nres = minimize(branin, x0=[0, 0], method='Nelder-Mead', tol=1e-6)\nprint(res.x)\nprint(branin(res.x))\n[3.14159271 2.27500017]\n0.397887357729795\n\nFirstly, I demonstrate the convergence of the optimizer towards a minimum (and show that the optimizer can pick up the procedure where it left):\n\n# n_init is the number of points in the initial design\n# n_iter is the number of iterations of the optimizer\ngp_opt = gp.GPOpt(objective_func=branin,\nlower_bound = np.array([-5, 0]),\nupper_bound = np.array([10, 15]),\nn_init=10, n_iter=10)\ngp_opt.optimize(verbose=1)\n\nPlotting the changes in expected improvement as a function of the number of iterations.\n\nprint(gp_opt.x_min) # current minimum\nprint(gp_opt.y_min) # current minimum\nplt.plot(np.diff(gp_opt.max_ei))\n[9.31152344 1.68457031]\n0.9445903336427559", null, "Adding more iterations to the optimizer:\n\ngp_opt.optimize(verbose=1, n_more_iter=10)\n\n...Done.\n\nOptimization loop...\n\n10/10 [██████████████████████████████] - 2s 186ms/step\n\n(array([3.22692871, 2.63122559]), 0.6107733232129569)\n\nPlotting the changes in expected improvement as a function of the number of iterations (again).\n\nprint(gp_opt.x_min) # current minimum\nprint(gp_opt.y_min) # current minimum\nplt.plot(np.diff(gp_opt.max_ei))\n[3.22692871 2.63122559]\n0.6107733232129569", null, "Adding more iterations to the optimizer (again):\n\ngp_opt.optimize(verbose=1, n_more_iter=80)\n\nPlotting the changes in expected improvement as a function of the number of iterations (again).\n\nprint(gp_opt.x_min) # current minimum\nprint(gp_opt.y_min) # current minimum\nplt.plot(np.diff(gp_opt.max_ei))\n[9.44061279 2.48199463]\n0.3991320518189241", null, "The 3 previous graphs suggest the possibility of stopping the optimizer earlier, when the algorithm is not improving on previous points’ results anymore:\n\n# # early stopping\n# abs_tol is the parameter that controls early stopping\n\ngp_opt2 = gp.GPOpt(objective_func=branin,\nlower_bound = np.array([-5, 0]),\nupper_bound = np.array([10, 15]),\nn_init=10, n_iter=190)\ngp_opt2.optimize(verbose=2, abs_tol=1e-4)\nprint(\"\\n\")\n\nWe can observe that only 58 iterations are necessary when abs_tol = 1e-4\n\nprint(gp_opt2.n_iter)\nprint(gp_opt2.x_min)\nprint(gp_opt2.y_min)\n58\n[9.44061279 2.48199463]\n0.3991320518189241\n\nIllustrating convergence:\n\nplt.plot(gp_opt2.max_ei)", null, "We notice that in this example, GPopt falls into a local minimum but is very close to the previous minimum found with gradient-free optimizer (Nelder-Mead). The opposite situation can occur too:\n\n# [0, 1]^3\ndef hartmann3(xx):\n\nalpha = np.array([1.0, 1.2, 3.0, 3.2])\n\nA = np.array([3.0, 10, 30,\n0.1, 10, 35,\n3.0, 10, 30,\n0.1, 10, 35]).reshape(4, 3)\n\nP = 1e-4 * np.array([3689, 1170, 2673,\n4699, 4387, 7470,\n1091, 8732, 5547,\n381, 5743, 8828]).reshape(4, 3)\n\nxxmat = np.tile(xx,4).reshape(4, 3)\n\ninner = np.sum(A*(xxmat-P)**2, axis = 1)\nouter = np.sum(alpha * np.exp(-inner))\n\nreturn(-outer)\n# Fails, but may work with multiple restarts\nprint(\"\\n\")\nres = minimize(hartmann3, x0=[0, 0, 0], method='Nelder-Mead', tol=1e-6)\nprint(res.x)\nprint(hartmann3(res.x))\n[0.36872308 0.11756145 0.26757363]\n-1.00081686355956\n# hartmann 3D\ngp_opt4 = gp.GPOpt(objective_func=hartmann3,\nlower_bound = np.repeat(0, 3),\nupper_bound = np.repeat(1, 3),\nn_init=20, n_iter=280)\ngp_opt4.optimize(verbose=2, abs_tol=1e-4)\nprint(gp_opt4.n_iter)\nprint(gp_opt4.x_min)\nprint(gp_opt4.y_min)\nprint(\"\\n\")\n51\n[0.07220459 0.55792236 0.85662842]\n-3.8600590626769904\n\nThe question is, in the case of BO applied to ML cross-validation, do we really want to find the global minimum of the objective function?" ]

[ null, "https://thierrymoudiki.github.io/images/2021-04-16/2021-04-16-image1.png", null, "https://thierrymoudiki.github.io/images/2021-04-16/2021-04-16-image2.png", null, "https://thierrymoudiki.github.io/images/2021-04-16/2021-04-16-image3.png", null, "https://thierrymoudiki.github.io/images/2021-04-16/2021-04-16-image4.png", null ]

[ "# Top 7 text functions in excel\n\nUsing excel first things which comes to mind is working with numbers but excel can do a lot with text. Excel has several text function to access these use short cut key ALT+MT (version 2010 onwards), other way of access it to go to formula ribbon ->text function.\n\nTop 7 text functions in excel\n\nUPPER\n\nThis function converts the given text string to all upper case letter.\n\nSyntax\n\n`=UPPER(text)`\n\nExample\n\n`=UPPER(A1)`\n\nA1 is cell reference; this can be changed as per requirement.\n\nLOWER\n\nThis function converts the given text string to all lower case letter.\n\nSyntax\n\n`=LOWER(text)`\n\nExample\n\n`=LOWER(A1)`\n\nA1 is cell reference; this can be changed as per requirement.\n\nPROPER\n\nThis function converts the given text string to all proper sentence case.\n\nSyntax\n\n`=PROPER(text)`\n\nExample\n\n`=PROPER(A1)`\n\nA1 is cell reference; this can be changed as per requirement.\n\nCONCATENATE\n\nThis function joins two or more text strings.\n\nSyntax\n\n`=CONCATENATE(text,text1,text2….)`\n\nExample\n\n```=CONCATENATE(A1,A2,A3….)\n\n=CONCAT(A1,A2,A3……)\n\n=CONCATENATE( “Color of car is “, A2)```\n\nA1,A2,A3 is cell reference; this can be changed as per requirement. Also CONCAT is available with similar functionality. CONCATENATE may not be functional in future versions of excel.\n\nLEFT\n\nLEFT function returns first character from left of string or specified number of character from left of string\n\nSyntax\n\n`=LEFT(text, number of characters)`\n\nIf number of character is not specified, only first character will be returned.\n\n```=LEFT(A1)\n\n=LEFT(A1,2)```\n\nA1 is cell reference; this can be changed as per requirement.\n\nLEN\n\nLEN returns number of character in a string.\n\nSyntax\n\n`=LEN(text)`\n\nExample\n\n`=LEN(A1)`\n\nA1 is cell reference; this can be changed as per requirement.\n\nTRIM\n\nTRIM function removes all space from a string leaving only single space between words\n\nSyntax\n\n`=TRIM(text)`\n\nExample\n\n`=TRIM(A1)`\n\nA1 is cell reference; this can be changed as per requirement." ]

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[ "You are on page 1of 19\n\n# Chapter 8/Competitive Firms and Markets\n\nCHAPTER 8\n\n## MULTIPLE CHOICE QUESTIONS\n\n1) Economists define a market to be competitive when the firms\nA) spend large amounts of money on advertising to lure customers away from the\ncompetition.\nB) watch each other's behavior closely.\nC) are price takers.\nD) All of the above.\nDiff: 0\nTopic: Competition\n2) If consumers view the output of any firm in a market to be identical to the output of any other\nfirm in the market, the demand curve for the output of any given firm\nA) will be identical to the market demand curve.\nB) will be horizontal.\nC) will be vertical.\nD) cannot be determined from the information given.\nDiff: 0\nTopic: Competition\n3) In the absence of any government regulation on price, if a firm has no power to set price on\nits own, one can safely conclude\nA) the demand curve for the firm's product is horizontal.\nB) there are many firms in the industry.\nC) the market is in long-run equilibrium.\nD) the firms in this industry are not profitable.\nDiff: 2\nTopic: Competition\n4) In a perfectly competitive market,\nA) firms can freely enter and exit.\nB) firms sell a differentiated product.\nC) transaction costs are high.\nD) All of the above.\nDiff: 0\nTopic: Competition\n143\n\n## Chapter 8/Competitive Firms and Markets\n\n5) In a competitive market, if buyers did not know all the prices charged by the many firms,\nA) all firms still face horizontal demand curves.\nB) firms sell a differentiated product.\nC) demand curves can be downward sloping for some or all firms.\nD) the number of firms will most likely decrease.\nDiff: 1\nTopic: Competition\n6) Many car owners and car dealers describe their different cars for sale in the local newspapers\nand list their asking price. Many people shopping for a used car consider the different choices\nlisted in the paper. The market for used cars could be described as\nA) competitive.\nB) perfectly competitive.\nC) non-competitive.\nD) having high transaction costs.\nDiff: 1\nTopic: Competition\n7) Many car owners and car dealers describe their different cars for sale in the local newspapers\nand list their asking price. Many people shopping for a used car consider the different choices\nlisted in the paper. The absence of which condition prohibits this market from being\ndescribed as perfectly competitive?\nA) Buyers and sellers know the prices.\nB) Firms freely enter and exit.\nC) Transaction costs are low.\nD) Consumer believes all firms sell identical products.\nDiff: 1\nTopic: Competition\n8) If a firm operates in a perfectly competitive market, then it will most likely\nA) advertise its product on television.\nB) settle for whatever price is offered.\nC) have a difficult time obtaining information about the market price.\nD) have an easy time keeping other firms out of the market.\nDiff: 1\nTopic: Competition\n\n144\n\n## Chapter 8/Competitive Firms and Markets\n\n9) If a firm happened to be the only seller of a particular product, it might behave as a price\ntaker as long as\nB) the transaction costs of doing business with this firm are low.\nD) there is free entry and exit.\nDiff: 2\nTopic: Competition\n10) The demand curve an individual competitive firm faces is known as its\nA) excess demand curve.\nB) market demand curve.\nC) residual demand curve.\nD) leftover demand curve.\nDiff: 0\nTopic: Competition\n11) If a firm makes zero economic profit, then the firm\nA) has total revenues greater than its costs.\nB) must shut down.\nC) can be earning positive business profit.\nD) must have no fixed costs.\nDiff: 1\nTopic: Profit Maximization\n12) If marginal revenue equals marginal cost, the firm is maximizing profits as long as\nA) the resulting profits are positive.\nB) marginal cost exceeds marginal revenue for greater levels of output.\nC) the average cost curve lies above the demand curve.\nD) All of the above are required.\nDiff: 2\nTopic: Profit Maximization\n\n145\n\n## Chapter 8/Competitive Firms and Markets\n\nFigure 8.1\n\n13) Figure 8.1 shows the cost curves for a competitive firm. If the firm is to earn economic profit,\nprice must exceed\nA) \\$0.\nB) \\$5.\nC) \\$10.\nD) \\$11.\nDiff: 1\nTopic: Profit Maximization\n14) Figure 8.1 shows the cost curves for a competitive firm. If the firm is to operate in the short\nrun, price must exceed\nA) \\$0.\nB) \\$5.\nC) \\$10.\nD) \\$11.\nDiff: 1\nTopic: Profit Maximization\n15) Figure 8.1 shows the cost curves for a competitive firm. If the market price is \\$15 per unit,\nthe firm will earn profits of\nA) \\$0.\nB) \\$4.\nC) \\$40.\nD) \\$160.\nDiff: 1\nTopic: Profit Maximization\n\n146\n\n## 16) A firm will shut down in the short run if\n\nA) total fixed costs are too high.\nB) total revenue from operating would not cover all costs.\nC) total revenue from operating would not cover variable costs.\nD) total revenue from operating would not cover fixed costs.\nDiff: 1\nTopic: Profit Maximization\n17) If a competitive firm maximizes short-run profits by producing some quantity of output,\nwhich of the following must be true at that level of output?\nA) p = MC.\nB) MR = MC.\nC) p > AVC.\nD) All of the above.\nDiff: 1\nTopic: Profit Maximization\n18) If a competitive firm maximizes short-run profits by producing some quantity of output,\nwhich of the following must be true at that level of output?\nA) p > MC.\nB) MR > MC.\nC) p > AVC.\nD) All of the above.\nDiff: 1\nTopic: Profit Maximization\n19) If a firm finds that it maximizes short-run profits by shutting down, which of the following\nmust be true?\nA) p < AVC for all levels of output.\nB) p < AVC only for the level of output at which p = MC.\nC) p < AVC only if the firm has no fixed costs.\nD) The firm will earn zero profit.\nDiff: 1\nTopic: Profit Maximization\n\n147\n\n## Chapter 8/Competitive Firms and Markets\n\n20) If a profit-maximizing firm finds that, at its current level of production, MR > MC, it will\nA) earn greater profits than if MR = MC.\nB) increase output.\nC) decrease output.\nD) shut down.\nDiff: 1\nTopic: Profit Maximization\n21) If a profit-maximizing firm finds that, at its current level of production, MR < MC, it will\nA) decrease output.\nB) increase output.\nC) shut down.\nD) operate at a loss.\nDiff: 1\nTopic: Profit Maximization\n22) The competitive firm's supply curve is equal to\nA) its marginal cost curve.\nB) the portion of its marginal cost curve that lies above AC.\nC) the portion of its marginal cost curve that lies above AVC.\nD) the portion of its marginal cost curve that lies above AFC.\nDiff: 1\nTopic: Competition in the Short-run\n23) If a firm is a price taker, then its marginal revenue will always equal\nA) price.\nB) total cost.\nC) zero.\nD) one.\nDiff: 1\nTopic: Competition in the Short Run\n\n148\n\n## 24) An increase in the cost of an input will result in\n\nA) a leftward shift in the firm's supply curve.\nB) an upward shift of the firm's marginal cost curve.\nC) a leftward shift of the market supply curve.\nD) All of the above.\nDiff: 1\nTopic: Competition in the Short Run\n25) When the production of a good involves several inputs, an increase in the cost of\none input will cause total costs to\nA) rise more than in proportion.\nB) rise less than in proportion.\nC) remain unchanged.\nD) rise by the exact amount of the input price increase.\nDiff:1\nTopic: Competition in the Short-Run\n26) If a competitive firm is in short-run equilibrium, then\nA) profits equal zero.\nB) it will not operate at a loss.\nC) an increase in its fixed cost will have no effect on profit.\nD) an increase in its fixed cost will have no effect on output.\nDiff: 1\nTopic: Competition in the Short Run\n27) Suppose TC = 10 + (0.1 * q2). If p = 10, the firm's profits will be\nA) 240.\nB) 250.\nC) 260.\nD) -10 because the firm will shut down.\nDiff: 1\nTopic: Competition in the Short Run\n\n149\n\n## Chapter 8/Competitive Firms and Markets\n\n28) Suppose TC = 10 + (0.1 * q2). If there are 100 identical firms in the market, the market\nsupply curve is\nA) Q = 1000 * p.\nB) Q = 500 * p.\nC) Q = 100 * p.\nD) Q = 10.\nDiff: 1\nTopic: Competition in the Short Run\nFigure 8.2\n\n29) Figure 8.2 shows the cost curves for a typical firm in a market and three possible market\nsupply curves. If there are 100 identical firms, the market supply curve is best represented by\nA) curve A.\nB) curve B.\nC) curve C.\nD) Either curve A or B, but definitely not C.\nDiff: 1\nTopic: Competition in the Short Run\n\n150\n\n## Chapter 8/Competitive Firms and Markets\n\n30) If a firm is currently in short-run equilibrium earning a profit, what impact will a lump-sum\ntax have on its production decision?\nA) The firm will decrease output to earn a higher profit.\nB) The firm will increase output but earn a lower profit.\nC) The firm will not change output but earn a lower profit.\nD) The firm will not change output and earn a higher profit.\nDiff: 1\nTopic: Competition in the Short Run\n31) Suppose that once a well is dug, water flows out of it continuously without any additional\neffort. Customers collect their water and pay a per gallon fee when they leave the site of the\nwell. In the short run, the competitive firm in this market\nA) will not shut down because variable costs are zero.\nB) has no fixed costs.\nC) faces diminishing marginal returns.\nD) can act as a price setter.\nDiff: 1\nTopic: Competition in the Short Run\n32) Suppose that once a well is dug, water flows out of it continuously without any additional\neffort. Customers collect their water and pay a per gallon fee when they leave the site of the\nwell. In the short run, the competitive firm in this market\nA) has no variable costs.\nB) has no fixed costs.\nC) will shut down.\nD) can produce water at no cost.\nDiff: 2\nTopic: Competition in the Short Run\n33) If a competitive firm cannot earn profit at any level of output during a given short-run period,\nthen which of the following is LEAST likely to occur?\nA) It will shut down in the short run and wait until the price increases sufficiently.\nB) It will exit the industry in the long run.\nC) It will operate at a loss in the short run.\nD) It will minimize its loss by decreasing output so that price exceeds marginal cost.\nDiff: 2\nTopic: Competition in the Short Run\n\n151\n\n## Chapter 8/Competitive Firms and Markets\n\n34) In deciding whether to operate in the short run, the firm must be concerned with the\nrelationship between price of the output and\nA) total cost.\nB) average variable cost.\nC) total fixed cost.\nDiff: 1\nTopic: Competition in the Short Run\n35) In the long run, profits will equal zero in a competitive market because of\nA) constant returns to scale.\nB) identical products being produced by all firms.\nC) the availability of information.\nD) free entry and exit.\nDiff: 1\nTopic: Competition in the Long Run\n36) Assuming a horizontal long-run market supply curve, which of the following statements is\n(are) TRUE about competitive firms in the long run?\nA) p = MC\nB) p = AC\nC) profit = 0\nD) All of the above.\nDiff: 1\nTopic: Competition in the Long Run\n37) Long-run market supply curves are upward sloping if\nA) firms are identical.\nB) the number of firms is restricted in the long run.\nC) input prices fall as the industry expands.\nD) All of the above.\nDiff: 1\nTopic: Competition in the Long Run\n\n152\n\n## 38) Long-run market supply curves are downward sloping if\n\nA) firms are identical.\nB) the number of firms is restricted in the long run.\nC) input prices fall as the industry expands.\nD) All of the above.\nDiff: 1\nTopic: Competition in the Long Run\n39) If firms in a competitive market are not identical, then the long-run market supply curve will\nbe\nA) horizontal.\nB) upward sloping.\nC) downward sloping.\nD) undetermined.\nDiff: 1\nTopic: Competition in the Long Run\n40) Suppose that for each firm in the competitive market for potatoes, long-run average cost is\nminimized at \\$0.20 per pound when 500 pounds are grown. If the long-run supply curve is\nhorizontal, then\nA) some firms will enjoy long-run profits because they operate at minimum average cost.\nB) the long-run price will be \\$0.20 per pound.\nC) each consumer will purchase \\$100 worth of potatoes.\nD) the long-run price will be set just above \\$0.20 per pound.\nDiff: 1\nTopic: Competition in the Long Run\n41) Suppose that for each firm in the competitive market for potatoes, long-run average cost is\nminimized at \\$0.20 per pound when 500 pounds are grown. The demand for potatoes is Q =\n10,000/p. If the long-run supply curve is horizontal, then how many firms will this industry\nsustain in the long run?\nA) 0\nB) 100\nC) 50,000\nD) There is not enough information to answer.\nDiff: 1\nTopic: Competition in the Long Run\n\n153\n\n## Chapter 8/Competitive Firms and Markets\n\n42) Suppose that for each firm in the competitive market for potatoes, long-run average cost is\nminimized at \\$0.20 per pound when 500 pounds are grown. The demand for potatoes is Q =\n10000/p. If the long-run supply curve is horizontal, then how much will consumers spend, in\ntotal, on potatoes?\nA) \\$0\nB) \\$500\nC) \\$10,000\nD) \\$50,000\nDiff: 1\nTopic: Competition in the Long Run\n43) Suppose that for each firm in the competitive market for potatoes, long-run average cost is\nminimized at \\$0.20 per pound when 500 pounds are grown. The demand for potatoes is Q =\n10000/p. If the long-run supply curve is horizontal, then how many pounds of potatoes will\nbe consumed in total?\nA) 0\nB) 500\nC) 10,000\nD) 50,000\nDiff: 1\nTopic: Competition in the Long Run\n44) In the long run, competitive firms MUST be profit maximizers because if they do not\nmaximize profits\nA) they will not survive.\nB) they will not be price takers.\nC) they will attract entry.\nD) the profits that they do earn will only cover variable costs.\nDiff: 1\nTopic: Zero Profit for Competitive Firms in the Long Run\n45) Long-run economic rent or profit do not exist for fixed factors like land because,\nA) bidding drives up the price of the factor until no economic rent exists.\nB) there is no market for such factors.\nC) these factors have L-shaped isoquants.\nD) these factors will earn economic profits.\nDiff: 1\nTopic: Zero Profit for Competitive Firms in the Long Run\n\n154\n\n## Chapter 8/Competitive Firms and Markets\n\nTRUE/FALSE/EXPLAIN\n1) A market is perfectly competitive even if firms have the ability to set their own price as long\nas the price difference reflects differences in the product.\nAnswer: False. If the market is perfectly competitive, there are no differences in the product.\nDiff: 0\nTopic: Competition\n2) Even though fixed costs do not affect the output decision, an increase in fixed costs results in\na wider range of prices for which the firm operates at a loss.\nAnswer: True. An increase in fixed costs will shift AC upward but leave AVC unchanged.\nThe gap between AVC and AC represents prices at which the firm will operate at a loss.\nDiff: 1\nTopic: Profit Maximization\n3) If a firm cannot earn profits in the short run, it will shut down.\nAnswer: False. A firm will operate at a loss if the loss incurred from production is less than\nthe loss incurred from shutting down. Even if the firm shuts down, it still must pay its fixed\ncosts.\nDiff: 1\nTopic: Profit Maximization\n4) A competitive firm's supply curve is identical to its marginal cost curve.\nAnswer: False. The statement is only partly correct. The supply curve is only that portion that\nlies above AVC.\nDiff: 1\nTopic: Competition in the Short Run\n5) If firms in a competitive market are identical, the long-run market supply curve is horizontal.\nAnswer: False. The horizontal long-run supply curve also requires that factor prices do not\nincrease with industry expansion and that the number of firms is not restricted.\nDiff: 1\nTopic: Competition in the Long Run\nPROBLEMS\n1) Suppose there are 20 competitive firms in a market. The supply curve of each firm is q = 2p.\nThe market demand is Q = 200 - 2p. What is the residual demand curve facing a typical\nfirm?\n\n155\n\n## Chapter 8/Competitive Firms and Markets\n\nAnswer: The residual demand curve is equal to the market demand curve minus the supply of\nall the other firms. The supply of the other 19 firms is 38p. The residual demand is 200-40p.\nDiff: 2\nTopic: Profit Maximization\n2) If a firm operates at a loss, the loss is equal to TC - TR. If the firm shuts down instead, its\nloss is equal to FC. Given this, show that price must exceed AVC for the firm to operate at a\nloss and not shut down.\nAnswer: The firm operates at a loss if TC -TR < FC. Adding TR - FC to both sides yields VC\n< TR. Dividing by q yields AVC < p. That is, the firm will operate at a loss as long as AVC <\np.\nDiff: 1\nTopic: Profit Maximization\n3) Suppose a firm has the following total cost function TC = 100 + 2q2. If price equals \\$20,\nwhat is the firms output decision? What are its short-run profits?\nAnswer: MC = 4q. To maximize profit, set 20 = 4q, or q = 5. Profit = TR - TC = (20 * 5) (100 + 50) = -50. Since FC = 100, the firm will produce 5 units and operate at a loss of 50\nrather than shutting down and incurring a loss of 100.\nDiff: 1\nTopic: Profit Maximization\n4) Suppose a firm has the following total cost function: TC = 100 + 4q2. What is the minimum\nprice necessary for the firm to earn profit? Below what price will the firm shut down in the\nshort run?\nAnswer: AC = 100/q + 4q. This is minimized when dAC/dq = 0, or -100/q2 + 4 = 0. Solving\nyields q = 5 and AC = 40. Thus, a price greater than \\$40 is required for the firm to earn\nprofit.\nAVC = 4q and MC = 8q. Since AVC is below MC for levels of output, AVC will be less than\nprice for all levels of output. The firm will not shut down in the short run.\nDiff: 1\nTopic: Profit Maximization\n\n156\n\n## Chapter 8/Competitive Firms and Markets\n\nFigure 8.3\n\n5) Figure 8.3 shows the cost curves for a typical firm in a competitive market. Note that if p =\n10, then MC = p at both q = 5 and q = 60. Can they both yield maximum profit? Explain.\nAnswer: No, at q = 5, p = MC, but this is not profit maximization; it is profit minimization.\nProfits expand as output increases since MR > MC for higher levels of output. At q=60, an\nincrease or decrease in output causes profits to fall, so this is the profit-maximizing q. Thus,\none way of wording the \"second order condition\" for profit maximization is that MC must\ncut the demand curve from below.\nDiff: 2\nTopic: Profit Maximization\n6) Figure 8.3 shows the cost curves for a typical firm in a competitive market. From the graph,\nestimate the firm's profits when price equals \\$10 per unit.\nAnswer: When price = 10, p =MC when q =60. TR =600. The AC is just above 8.5, say 8.6.\nThis yields TC = 516. The firm's profit is estimated to be around \\$84.\nDiff: 1\nTopic: Competition in the Short-run\n\n157\n\n## Chapter 8/Competitive Firms and Markets\n\n7) Figure 8.3 shows the cost curves for a typical firm in a competitive market. If there are 200\nidentical firms, estimate the market quantity supplied when p = 4, 8, and 10.\nAnswer: When p =4, p < AVC so no firms will produce. At p = 8, each firm produces 50 units\nat a loss. At p = 10, each firm produces 60 units. The market quantity supplied at each price\nis:\np\nQ\n4\n0\n8\n10,000\n10\n12,000\nDiff: 1\nTopic: Competition in the Short Run\nFigure 8.4\n\n8) Draw a graph that shows how the short-run shut-down price changes when an input price\nincreases.\nDiff: 1\nTopic: Competition in the Short Run\n9) Suppose there are 1000 identical wheat farmers. For each, TC =10 + q2. Derive the market\nsupply curve.\nAnswer: For each MC = 2q and AVC = q. Thus MC > AVC for all levels of output. The firm\nsets p = 2q or q =0.5p. Since there are 1000 firms each producing q, market supply equals Q\n= 500p.\nDiff: 1\nTopic: Competition in the Short Run\n\n158\n\n## Chapter 8/Competitive Firms and Markets\n\n10) Suppose there are 1000 identical wheat farmers. For each, TC =10 + q2. Market demand is Q\n= 600,000 - 100p. Derive the short-run equilibrium Q, q, and p. Does the typical firm earn a\nshort-run profit?\nAnswer: The firm's supply is q = 0.5p; market supply is Q = 500p. Market equilibrium can be\nfound as 500p = 600,000 - 100p, or 600p =600,000, so p =1,000 and Q = 500,000. q = 0.5p =\n500. Profit = (500* 1,000) - (10 + 250,000) = 249,990. Each firm earns a profit.\nDiff: 1\nTopic: Competition in the Short Run\nFigure 8.5\n\n11) Figure 8.5 shows the long-run cost curves for a typical firm in a competitive market. If the\nnumber of firms is unrestricted and input costs are constant, derive the long-run market\nsupply curve.\nAnswer: The long-run market supply curve is horizontal at a price of \\$10 per unit.\nDiff: 0\nTopic: Competition in the Long Run\n12) All the supply of peppermint oil is produced from mint plants grown in one county\nby several competitive growers (the number of growers is not limited). The quality of land in\nthe county varies greatly. Would you expect the long-run market supply curve to slope\nupward, downward, or remain constant? Why?\nAnswer: The long-run market supply curve will slope up because the growers have different\ncosts. Those growers with poor-quality land have a higher average cost of production. The\nhorizontal sum of the individual supply curves will slope up.\nDiff:1\nTopic: Competition in the Long Run\n\n159\n\n## Chapter 8/Competitive Firms and Markets\n\nFigure 8.6\n\n13) Suppose an industry has no fixed costs. Draw two graphs side by side for the\nindustry. In the left graph draw a U-shaped average cost curve and the corresponding\nmarginal cost curve. In the right graph, draw a downward sloping market demand curve.\nAlso in the right graph, draw a short-run supply curve that would generate positive profit,\nand the long-run supply curve that would result.\nDiff: 2\nTopic: Competition in the Long Run\n14) Suppose all firms in a competitive market are currently in both short-run and long-run\nequilibrium. What impact will a lump sum tax have on each firm in the short run? in the long\nrun?\nAnswer: In the short run, the lump sum tax represents a fixed cost. The firm's output decision\nis unchanged, but its profits decrease. In the long run, the tax raises the LRAC of each firm,\nbut not MC. Minimum AC is higher, so price is higher. With a higher price, each firm\nproduces a greater quantity, but the higher price means less quantity is demanded in total;\nthus, the number of firms will decrease.\nDiff: 2\nTopic: Competition in the Long Run\n15) Suppose market demand is Q = 1000 - 4p. If all firms have LRAC = 50 - 5q + q2, how many\nfirms will there be when this industry is in long-run equilibrium?\nAnswer: The long-run market supply curve is horizontal at the minimum LRAC. LRAC is\nminimized when -5 + 2q = 0 or q =2.5. At this level of output, LRMC = LRAC = 43.75. At\nthis price, 825 units are demanded. If each firm produces 2.5 units in the long run, then 330\nfirms will be in this market.\nDiff: 2\nTopic: Competition in the Long Run\n16) Even if two competitive firms in the same market have different production technologies,\nthey will each earn long-run zero profits. Why?\n\n160\n\n## Chapter 8/Competitive Firms and Markets\n\nAnswer: The firm that has more productive resources will have the cost of those resources\nbid up by the marketplace. The more productive the resource, the more expensive it will be.\nThis price is bid up until the firm's profits are zero. The firm with less productive resources\nwill also have zero profits because it is not paying as much for its resources. There is no such\nthing as a free lunch or a free productivity gain.\nDiff: 2\nTopic: Zero Profit for Competitive Firms in the Long Run\n\n161" ]

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[ "# Floating point\n\n(Redirected from Floating-point)\n\nA floating-point number is a digital representation for a number in a certain subset of the rational numbers, and is often used to approximate an arbitrary real number on a computer. In particular, it represents an integer or fixed-point number (the significand or, informally, the mantissa) multiplied by a base (usually 2 in computers) to some integer power (the exponent). When the base is 2, it is the binary analogue of scientific notation (in base 10).\n\nA floating-point calculation is an arithmetic calculation done with floating-point numbers and often involves some approximation or rounding because the result of an operation may not be exactly representable.\n\nA floating-point number a can be represented by two numbers m and e, such that a = m × be. In any such system we pick a base b (called the base of numeration, also the radix) and a precision p (how many digits to store). m (which is called the significand or, informally, mantissa) is a p digit number of the form ±d.ddd...ddd (each digit being an integer between 0 and b−1 inclusive). If the leading digit of m is non-zero then the number is said to be normalized. Some descriptions use a separate sign bit (s, which represents −1 or +1) and require m to be positive. e is called the exponent.\n\nThis scheme allows a large range of magnitudes to be represented within a given size of field, which is not possible in a fixed-point notation.\n\nAs an example, a floating-point number with four decimal digits (b = 10, p = 4) and an exponent range of ±4 could be used to represent 43210, 4.321, or 0.0004321, but would not have enough precision to represent 432.123 and 43212.3 (which would have to be rounded to 432.1 and 43210). Of course, in practice, the number of digits is usually larger than four.\n\nIn addition, floating-point representations often include the special values +∞, −∞ (positive and negative infinity), and NaN ('Not a Number'). Infinities are used when results are too large to be represented, and NaNs indicate an invalid operation or undefined result.\n\n Contents\n\n## Usage in computing\n\nWhile in the examples above the numbers are represented in the decimal system (that is the base of numeration, b = 10), computers usually do so in the binary system, which means that b = 2. In computers, floating-point numbers are sized by the number of bits used to store them. This size is usually 32 bits or 64 bits, often called \"single-precision\" and \"double-precision\". A few machines offer larger sizes; Intel FPUs such as the Intel 8087 (and its descendants integrated into the x86 architecture) offer 80 bit floating point numbers for intermediate results, and several systems offer 128 bit floating-point, generally implemented in software. This website (http://babbage.cs.qc.edu/courses/cs341/IEEE-754.html) can be used to calculate the floating point representation of a decimal number.\n\n## Problems with floating-point\n\nFloating-point numbers usually behave very similarly to the real numbers they are used to approximate. However, this can easily lead programmers into over-confidently ignoring the need for numerical analysis. There are many cases where floating-point numbers do not model real numbers well, even in simple cases such as representing the decimal fraction 0.1, which cannot be exactly represented in any binary floating-point format. For this reason, financial software tends not to use a binary floating-point number representation. See: http://www2.hursley.ibm.com/decimal/\n\nErrors in floating-point computation can include:\n\n• Rounding\n• Non-representable numbers: for example, the literal 0.1 cannot be represented exactly by a binary floating-point number\n• Rounding of arithmetic operations: for example 2/3 might yield 0.6666667\n• Absorption: 1×1015 + 1 = 1×1015\n• Cancellation: subtraction between nearly equivalent operands\n• Overflow, which usually yields an infinity\n• Underflow (often defined as an inexact tiny result outside the range of the normal numbers for a format), which yields zero, a subnormal number, or the smallest normal number\n• Invalid operations (such as an attempt to calculate the square root of a non-zero negative number). Invalid operations yield a result of NaN (not a number).\n• Rounding errors: unlike the fixed-point counterpart, the application of dither in a floating point environment is nearly impossible. See external references for more information about the difficulty of applying dither and the rounding error problems in floating point systems\n\nFloating point representation is more likely to be appropriate when proportional accuracy over a range of scales is needed. When fixed accuracy is required, fixed point is usually a better choice.\n\n## Properties of floating point arithmetic\n\nArithmetic using the floating point number system has two important properties that differ from those of arithmetic using real numbers.\n\nFloating point arithmetic is not associative. This means that in general for floating point numbers x, y, and z:\n\n• [itex] (x + y) + z \\neq x + (y + z) [itex]\n• [itex] (x \\cdot y) \\cdot z \\neq x \\cdot (y \\cdot z) [itex]\n\nFloating point arithmetic is also not distributive. This means that in general:\n\n• [itex] x \\cdot (y + z) \\neq (x \\cdot y) + (x \\cdot z) [itex]\n\nIn short, the order in which operations are carried out can change the output of a floating point calculation. This is important in numerical analysis since two mathematically equivalent formulas may not produce the same numerical output, and one may be substantially more accurate than the other.\n\nFor example, with most floating-point implementations, (1e100 - 1e100) + 1.0 will give the result 1.0, whereas (1e100 + 1.0) - 1e100 gives 0.0.\n\n## IEEE standard\n\nThe IEEE has standardized the computer representation for binary floating-point numbers in IEEE 754. This standard is followed by almost all modern machines. Notable exceptions include IBM Mainframes, which have both hexadecimal and IEEE 754 data types, and Cray vector machines, where the T90 series had an IEEE version, but the SV1 still uses Cray floating-point format.\n\nAs of 2000, the IEEE 754 standard is currently under revision. See: IEEE 754r\n\n## Examples\n\n• The value of Pi, π = 3.1415926...10 decimal, which is equivalent to binary 11.001001000011111...2. When represented in a computer that allocates 17 bits for the significand, it will become 0.11001001000011111 × 22. Hence the floating-point representation would start with bits 01100100100001111 and end with bits 10 (which represent the exponent 2 in the binary system). Note: the first zero indicates a positive number, the ending 102 = 210.)\n• The value of -0.37510 = -0.0112 or -0.11 × 2−1. In two's complement notation, −1 is represented as 11111111 (assuming 8 bits are used in the exponent). In floating-point notation, the number would start with a 1 for the sign bit, followed by 110000... and then followed by 11111111 at the end, or 1110...011111111 (where ... are zeros).\n\n### Hidden bit\n\nWhen using binary (b = 2), one bit, called the hidden bit or the implied bit, can be omitted if all numbers are required to be normalized. The leading digit (most significant bit) of the significand of a normalized binary floating-point number is always non-zero; in particular it is always 1. This means that this bit does not need to be stored explicitly, since for a normalized number it can be understood to be 1.\n\nThe IEEE 754 standard exploits this fact. Requiring all numbers to be normalized means that 0 cannot be represented; typically some special representation of zero is chosen. In the IEEE standard this special code also encompasses denormal numbers, which allow for gradual underflow. The normalized numbers are also known as the normal numbers.\n\n### Note\n\nNote that although the examples in this article use a consistent system of floating-point notation, the notation is different from the IEEE standard. For example, in IEEE 754, the exponent is between the sign bit and the significand, not at the end of the number. Also the IEEE exponent uses a biased integer instead of a two's complement number. The reader should note that the examples serve the purpose of illustrating how floating-point numbers could be represented, but the actual bits shown in the article are different from those in a IEEE 754-compliant representation. The placement of the bits in the IEEE standard enables two floating-point numbers to be compared bitwise (sans sign bit) to yield a result without interpreting the actual values. The arbitrary system used in this article cannot do the same.\n\n## References\n\n• Art and Cultures\n• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)\n• Space and Astronomy" ]

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[ "# Runtime Error: Arithmetic overflow error converting expression to data type int.\n\nAn arithmetic overflow is caused by a calculated column value that exceeds the column's specified size. When this error occurs, the calculation stops and the remainder of the results pane is not filled.\n\nComputed columns are often the cause of arithmetic overflows. For example, consider the case where columns c1, c2, and c3 are defined as a data type of INTEGER. Further, assume that c3 is computed using the formula ([c1] * [c2]). If large values (such as 9999) are entered into c1 and c2, then the computation of c3 might exceed the allowable INTEGER limit. The size limitations for data types are determined by your database specification.\n\n### To correct this error\n\n• Check the target column size for each column where the value is computed by a formula. Consider either expanding the size of the target column or reducing the allowable values in the formula's source columns." ]

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[ "• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\nSamuel Colvin\n\nWe study the infimal value of the Hausdorff dimension of spaces that are Hölder equivalent to a given metric space; we call this bi‐Hölder‐invariant ‘Hölder dimension’. This definition and some of our methods are analogous to those used in the study of conformal dimension. We prove that Hölder dimension is bounded above by capacity dimension for compact, doubling metric spaces. As a corollary, we obtain\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\nAriyan Javanpeykar; Daniel Loughran\n\nWe prove an analogue for algebraic stacks of Hermite–Minkowski's finiteness theorem from algebraic number theory, and establish a Chevalley–Weil type theorem for integral points on stacks. As an application of our results, we prove analogues of the Shafarevich conjecture for some surfaces of general type.\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\nChih‐Whi Chen; Shun‐Jen Cheng; Kevin Coulembier\n\nWe study tilting and projective‐injective modules in a parabolic BGG category O for an arbitrary classical Lie superalgebra. We establish a version of Ringel duality for this type of Lie superalgebras which allows to express the characters of tilting modules in terms of those of simple modules in that category. We also obtain a classification of projective‐injective modules in the full BGG category\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\n\nWe study self‐similarity problem for two classes of flows: (1) special flows over circle rotations and under roof functions with symmetric logarithmic singularities (2) special flows over interval exchange transformations and under roof functions which are of two types piecewise constant with one additional discontinuity which is not a discontinuity of the IET; piecewise linear over exchanged intervals\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\nGuochun Wu; Yinghui Zhang; Weiyuan Zou\n\nThis paper is concerned with time‐decay rates of the weak solutions to the 3D compressible magnetohydrodynamic flows with discontinuous initial data and large oscillations. The global existence of weak solutions to the Cauchy problem of the 3D compressible magnetohydrodynamic flows has been established by Suen–Hoff (Arch. Ration. Mech. Anal. 205 (2012) 27–58) and Suen (J. Differential Equations 268\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\nVictor Goryunov\n\nWe study local singularities of holomorphic families of arbitrary square, symmetric and skew‐symmetric matrices, that is, of mappings of smooth manifolds to the matrix spaces. Our main object is the vanishing topology of the pre‐images of the hypersurface Δ of all degenerate matrices in assumption that the dimension of the source is at least the codimension of the singular locus of Δ in the ambient\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-20\nNiels Lubbes\n\nWe classify webs of minimal degree rational curves on surfaces and give a criterion for webs being hexagonal. In addition, we classify Neron–Severi lattices of real weak del Pezzo surfaces. These two classifications are related to root subsystems of E8.\n\n更新日期：2020-10-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-13\nEugenio Giannelli; Stacey Law\n\nLet p ⩾ 5 be a prime and let n be a natural number. In this article, we describe the irreducible constituents of the induced characters ϕ ↑ S n for arbitrary linear characters ϕ of a Sylow p ‐subgroup P n of the symmetric group S n , generalising results of Giannelli and Law (J. Algebra 506 (2018) 409–428). By doing this, we introduce Sylow branching coefficients for symmetric groups.\n\n更新日期：2020-10-13\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-08\nFilippo Ambrosio; Giovanna Carnovale; Francesco Esposito\n\nWe prove that the closure of every Jordan class J in a semisimple simply connected complex algebraic group G at a point x with Jordan decomposition x = r v is smoothly equivalent to the union of closures of those Jordan classes in the centraliser of r that are contained in J and contain x in their closure. For x unipotent, we also show that the closure of J around x is smoothly equivalent to the closure\n\n更新日期：2020-10-11\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-07\nFabrizio Catanese; Keiji Oguiso\n\nMotivated by the embedding problem of canonical models in small codimension, we extend Severi's double point formula to the case of surfaces with rational double points, and we give more general double point formulae for varieties with isolated singularities. A concrete application is for surfaces with geometric genus p g = 5 : the canonical model is embedded in P 4 if and only if we have a complete\n\n更新日期：2020-10-07\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-10-07\nRoberto Conti; Gandalf Lechner\n\nEvery unitary solution of the Yang–Baxter equation (R‐matrix) in dimension d can be viewed as a unitary element of the Cuntz algebra O d and as such defines an endomorphism of O d . These Yang–Baxter endomorphisms restrict and extend to several other C ∗ ‐ and von Neumann algebras, and furthermore define a II 1 factor associated with an extremal character of the infinite braid group. This paper is\n\n更新日期：2020-10-07\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-30\nRosa Sena‐Dias\n\nIn [Abreu and Sena‐Dias, Ann. Global Anal. Geom. 41 (2012) 209–239], the authors construct two distinct families of scalar‐flat Kähler non‐compact toric metrics using Donaldson's rephrasing of Joyce's construction in action‐angle coordinates. In this paper and using the same set‐up, we show that these are the only scalar‐flat Kähler metrics on any given strictly unbounded toric surface. We also show\n\n更新日期：2020-09-30\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-30\nAnthony Poëls; Damien Roy\n\nLet Z be a quadratic hypersurface of P n ( R ) defined over Q containing points whose coordinates are linearly independent over Q . We show that, among these points, the largest exponent of uniform rational approximation is the inverse 1 / ρ of an explicit Pisot number ρ < 2 depending only on n if the Witt index (over Q ) of the quadratic form q defining Z is at most 1, and that it is equal to 1 otherwise\n\n更新日期：2020-09-30\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-28\nElías Gabriel Minian; Kevin Iván Piterman\n\nWe study the fundamental group of the p ‐subgroup complex of a finite group G . We show first that π 1 ( A 3 ( A 10 ) ) is not a free group (here A 10 is the alternating group on ten letters). This is the first concrete example in the literature of a p ‐subgroup complex with non‐free fundamental group. We prove that, modulo a well‐known conjecture of Aschbacher, π 1 ( A p ( G ) ) = π 1 ( A p ( S G\n\n更新日期：2020-09-28\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-27\nDavid Jensen; Max Kutler; Jeremy Usatine\n\nWe introduce motivic zeta functions for matroids. These zeta functions are defined as sums over the lattice points of Bergman fans, and in the realizable case, they coincide with the motivic Igusa zeta functions of hyperplane arrangements. We show that these motivic zeta functions satisfy a functional equation arising from matroid Poincaré duality in the sense of Adiprasito–Huh–Katz. In the process\n\n更新日期：2020-09-28\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-21\nMitul Islam; Andrew Zimmer\n\nIn this paper, we consider discrete groups in PGL d ( R ) acting convex co‐compactly on a properly convex domain in real projective space. For such groups, we establish an analogue of the well‐known flat torus theorem for CAT ( 0 ) spaces.\n\n更新日期：2020-09-21\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-16\nDawei Chen; Alex Wright\n\nWe show that the partial compactification of a stratum of Abelian differentials previously considered by Mirzakhani and Wright is not an algebraic variety. Despite this, we use a combination of algebro‐geometric and other methods to provide a short, unconditional proof of Mirzakhani and Wright's formula for the tangent space to the boundary of a G L + ( 2 , R ) orbit closure, and give new results on\n\n更新日期：2020-09-16\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-16\nKotaro Kawai\n\nWe introduce a new notion of a homogeneous pair for a pseudo‐Riemannian metric g and a positive function f on a manifold M admitting a free R > 0 ‐action. There are many examples admitting this structure. For example, (a) a class of pseudo‐Hessian manifolds admitting a free R > 0 ‐action and a homogeneous potential function such as the moduli space of torsion‐free G 2 ‐structures, (b) the space of\n\n更新日期：2020-09-16\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-16\nKaren Habermann\n\nWe consider a standard one‐dimensional Brownian motion on the time interval [0,1] conditioned to have vanishing iterated time integrals up to order N . We show that the resulting processes can be expressed explicitly in terms of shifted Legendre polynomials and the original Brownian motion, and we use these representations to prove that the processes converge weakly as N → ∞ to the zero process. This\n\n更新日期：2020-09-16\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-07\nRégis de la Bretèche; Marc Munsch; Gérald Tenenbaum\n\nIn recent years, maximizing Gál sums regained interest due to a firm link with large values of L ‐functions. In the present paper, we initiate an investigation of small sums of Gál type, with respect to the L 1 ‐norm. We also consider the intertwined question of minimizing weighted versions of the usual multiplicative energy. We apply our estimates to: (i) a logarithmic refinement of Burgess' bound\n\n更新日期：2020-09-08\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-07\nN. Correia; R. Pacheco; M. Svensson\n\nWe consider the twistor theory of nilconformal harmonic maps from a Riemann surface into the Cayley plane O P 2 = F 4 / Spin ( 9 ) . By exhibiting this symmetric space as a submanifold of the Grassmannian of 10‐dimensional subspaces of the fundamental representation of F 4 , techniques and constructions similar to those used in earlier works on twistor constructions of nilconformal harmonic maps into\n\n更新日期：2020-09-08\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-07\nMara Ungureanu\n\nFor a smooth projective curve, the cycles of subordinate or, more generally, secant divisors to a given linear series are among some of the most studied objects in classical enumerative geometry. We consider the intersection of two such cycles corresponding to secant divisors of two different linear series on the same curve and investigate the validity of the enumerative formulae counting the number\n\n更新日期：2020-09-08\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-09-07\n\nFix a degree d projective curve X ⊂ P r over an algebraically closed field K . Let U ⊂ ( P r ) ∗ be a dense open subvariety such that every hyperplane H ∈ U intersects X in d smooth points. Varying H ∈ U produces the monodromy action φ : π 1 ét ( U ) → S d . Let G X ≔ im ( φ ) . The permutation group G X is called the sectional monodromy group of X . In characteristic 0, G X is always the full symmetric\n\n更新日期：2020-09-08\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-08-19\nNajmuddin Fakhruddin; Chandrashekhar Khare; Ravi Ramakrishna\n\nWe use Galois cohomology methods to produce optimal mod p d level lowering congruences to a p ‐adic Galois representation that we construct as a well‐chosen lift of a given residual mod p representation. Using our explicit Galois cohomology methods, for F a number field, Γ F its absolute Galois group and G a reductive group, k a finite field and a suitable representation ρ ¯ : Γ F → G ( k ) , ramified\n\n更新日期：2020-08-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-08-19\nAndrea Tamburelli\n\nLet Σ be a connected, oriented surface with punctures and negative Euler characteristic. We introduce wild globally hyperbolic anti‐de Sitter structures on Σ × R and provide two parameterisations of their deformation space: as a quotient of the product of two copies of the Teichmüller space of crowned hyperbolic surfaces and as the bundle over the Teichmüller space of Σ of meromorphic quadratic differentials\n\n更新日期：2020-08-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-08-17\nHanfeng Li\n\nFor a unital ring R , a Sylvester rank function is a numerical invariant which can be described in three equivalent ways: on finitely presented left R ‐modules, or on rectangular matrices over R , or on maps between finitely generated projective left R ‐modules. We extend each Sylvester rank function to all pairs of left R ‐modules M 1 ⊆ M 2 , and to all maps between left R ‐modules satisfying suitable\n\n更新日期：2020-08-18\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-27\nHendrik De Bie; Hadewijch De Clercq\n\nThe Gasper and Rahman multivariate ( − q ) ‐Racah polynomials appear as connection coefficients between bases diagonalizing different abelian subalgebras of the recently defined higher rank q ‐Bannai–Ito algebra A n q . Lifting the action of the algebra to the connection coefficients, we find a realization of A n q by means of difference operators. This provides an algebraic interpretation for the\n\n更新日期：2020-07-27\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-27\nLéa Bittmann\n\nWe define and construct a quantum Grothendieck ring for a certain monoidal subcategory of the category O of representations of the quantum loop algebra introduced by Hernandez–Jimbo. We use the cluster algebra structure of the Grothendieck ring of this category to define the quantum Grothendieck ring as a quantum cluster algebra. When the underlying simple Lie algebra is of type A , we prove that this\n\n更新日期：2020-07-27\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-18\nXing Gu\n\nIn the integral cohomology ring of the classifying space of the projective linear group P G L n (over C ), we find a collection of p ‐torsion classes y p , k of degree 2 ( p k + 1 + 1 ) for any odd prime divisor p of n , and k ⩾ 0 . If, in addition, p 2 ∤ n , there are p ‐torsion classes ρ p , k of degree p k + 1 + 1 in the Chow ring of the classifying stack of P G L n , such that the cycle class map\n\n更新日期：2020-07-18\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-15\nShin Hattori\n\nLet p be a rational prime and q a power of p . Let ℘ be a monic irreducible polynomial of degree d in F q [ t ] . In this paper, we define an analogue of the Hodge–Tate map which is suitable for the study of Drinfeld modules over F q [ t ] and, using it, develop a geometric theory of ℘ ‐adic Drinfeld modular forms similar to Katz's theory in the case of elliptic modular forms. In particular, we show\n\n更新日期：2020-07-15\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-03\nLivio Liechti; Julien Marché\n\nWe introduce the notions of overcommutation and overcommutation length in groups, and show that these concepts are closely related to representations of the fundamental groups of 3‐manifolds and their Heegaard genus. We give many examples including translations in the affine group of the line and provide upper bounds for the overcommutation length in SL 2 , related to the Steinberg relation.\n\n更新日期：2020-07-03\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-03\nAmir Nasr\n\nWe classify del Pezzo non‐commutative surfaces that are finite over their centres and have no worse than canonical singularities. Using the minimal model program, we introduce the minimal model of such surfaces. We first classify the minimal models and then give the classification of these surfaces in general. This presents a complementary result and method to the classification of del Pezzo orders\n\n更新日期：2020-07-03\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-07-02\nOleg Ivrii\n\nLet J be the set of inner functions whose derivative lies in the Nevanlinna class. In this paper, we discuss a natural topology on J where F n → F if the critical structures of F n converge to the critical structure of F . We show that this occurs precisely when the critical structures of the F n are uniformly concentrated on Korenblum stars. The proof uses Liouville's correspondence between holomorphic\n\n更新日期：2020-07-02\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-29\nMahan Mj\n\nWe generalize the notion of tight geodesics in the curve complex to tight trees. We then use tight trees to construct model geometries for certain surface bundles over graphs. This extends some aspects of the combinatorial model for doubly degenerate hyperbolic 3‐manifolds developed by Brock, Canary and Minsky during the course of their proof of the Ending Lamination Theorem. Thus we obtain uniformly\n\n更新日期：2020-06-29\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-28\nXiumin Du\n\nFor spherical and parabolic averages of the Fourier transform of fractal measures, we obtain new upper bounds on rates of decay by an ‘intermediate dimension’ trick.\n\n更新日期：2020-06-29\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-28\nRoland Berger; Rachel Taillefer\n\nWe show that the Koszul calculus of a preprojective algebra, whose graph is distinct from A 1 and A 2 , vanishes in any (co)homological degree p > 2 . Moreover, its (higher) cohomological calculus is isomorphic as a bimodule to its (higher) homological calculus, by exchanging degrees p and 2 − p , and we prove a generalised version of the 2‐Calabi–Yau property. For the ADE Dynkin graphs, the preprojective\n\n更新日期：2020-06-29\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-28\nHui Dan; Kunyu Guo\n\nThe classical completeness problem raised by Beurling and independently by Wintner asks for which ψ ∈ L 2 ( 0 , 1 ) , the dilation system { ψ ( k x ) : k = 1 , 2 , … } is complete in L 2 ( 0 , 1 ) , where ψ is identified with its extension to an odd 2‐periodic function on R . This difficult problem is nowadays commonly called as the periodic dilation completeness problem (PDCP). By Beurling's idea\n\n更新日期：2020-06-29\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-28\n\nNielsen realization problem for the mapping class group Mod ( S g ) asks whether the natural projection p g : Homeo + ( S g ) → Mod ( S g ) has a section. While all the previous results use torsion elements in an essential way, in this paper, we focus on the much more difficult problem of realization of torsion‐free subgroups of Mod ( S g ) . The main result of this paper is that the Torelli group\n\n更新日期：2020-06-29\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-23\nPablo Blanc; Carlos Esteve; Julio D. Rossi\n\nIn this paper, we study the evolution problem u t ( x , t ) − λ j ( D 2 u ( x , t ) ) = 0 , in Ω × ( 0 , + ∞ ) , u ( x , t ) = g ( x , t ) , on ∂ Ω × ( 0 , + ∞ ) , u ( x , 0 ) = u 0 ( x ) , in Ω , where Ω is a bounded domain in R N (which verifies a suitable geometric condition on its boundary) and λ j ( D 2 u ) stands for the j th eigenvalue of the Hessian matrix D 2 u . We assume that u 0 and g are\n\n更新日期：2020-06-23\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-23\nAndrew R. Booker; Min Lee; Andreas Strömbergsson\n\nWe derive a fully explicit version of the Selberg trace formula for twist‐minimal Maass forms of weight 0 and arbitrary conductor and nebentypus character, and apply it to prove two theorems. First, conditional on Artin's conjecture, we classify the even 2‐dimensional Artin representations of small conductor; in particular, we show that the even icosahedral representation of smallest conductor is the\n\n更新日期：2020-06-23\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-21\nSimone Dovetta; Enrico Serra; Paolo Tilli\n\nWe consider the minimization of the NLS energy on a metric tree, either rooted or unrooted, subject to a mass constraint. With respect to the same problem on other types of metric graphs, several new features appear, such as the existence of minimizers with positive energy, and the emergence of unexpected threshold phenomena. We also study the problem with a radial symmetry constraint that is in principle\n\n更新日期：2020-06-23\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-15\nAbbey Bourdon; Pete L. Clark\n\nLet O be an order in the imaginary quadratic field K . For positive integers M ∣ N , we determine the least degree of an O ‐CM point on the modular curve X ( M , N ) / K ( ζ M ) and also on the modular curve X ( M , N ) / Q ( ζ M ) : that is, we treat both the case in which the complex multiplication is rationally defined and the case in which we do not assume that the complex multiplication is rationally\n\n更新日期：2020-06-15\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-15\nMichael P. Cohen\n\nWe initiate a study of distortion elements in the Polish groups Diff + k ( S 1 ) ( 1 ⩽ k < ∞ ), as well as Diff + 1 + A C ( S 1 ) , in terms of maximal metrics on these groups. We classify distortion in the k = 1 case: a C 1 circle diffeomorphism is C 1 ‐undistorted if and only if it has a hyperbolic periodic point. On the other hand, answering a question of Navas, we exhibit analytic circle diffeomorphisms\n\n更新日期：2020-06-15\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-10\nMatt Clay; Caglar Uyanik\n\nWe show that for any subgroup H of Out ( F N ) , either H contains an atoroidal element or a finite index subgroup H ′ of H fixes a nontrivial conjugacy class in F N . This result is an analog of Ivanov's subgroup theorem for mapping class groups and Handel–Mosher's subgroup theorem for Out ( F N ) in the setting of irreducible elements.\n\n更新日期：2020-06-10\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-06-01\nManuel Krannich; Jens Reinhold\n\nWe study smooth bundles over surfaces with highly connected almost parallelizable fiber M of even dimension, providing necessary conditions for a manifold to be bordant to the total space of such a bundle and showing that, in most cases, these conditions are also sufficient. Using this, we determine the characteristic numbers realized by total spaces of bundles of this type, deduce divisibility constraints\n\n更新日期：2020-06-01\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-27\nValentin Buciumas; Travis Scrimshaw; Katherine Weber\n\nWe construct an integrable colored five‐vertex model whose partition function is a Lascoux atom based on the five‐vertex model of Motegi and Sakai and the colored five‐vertex model of Brubaker, the first author, Bump and Gustafsson. We then modify this model in two different ways to construct a Lascoux polynomial, yielding the first proven combinatorial interpretation of a Lascoux polynomial and atom\n\n更新日期：2020-05-27\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-24\nDavid Torres‐Teigell\n\nWe calculate the Masur–Veech volume of the gothic locus G in the stratum H ( 2 3 ) of genus 4. Our method is based on the use of the formulae for the Euler characteristics of gothic Teichmüller curves to determine the number of lattice points of given area. We also use this method to recalculate the Masur–Veech volumes of the Prym loci P 3 ⊂ H ( 4 ) and P 4 ⊂ H ( 6 ) in genus 3 and 4.\n\n更新日期：2020-05-24\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-24\nChloé Perin; Rizos Sklinos\n\nWe give a complete characterization of the forking independence relation over any set of parameters in the free groups of finite rank, in terms of the J S J decompositions relative to those parameters.\n\n更新日期：2020-05-24\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-22\nBing Duan; Ralf Schiffler\n\nLet C be the category of finite‐dimensional modules over the quantum affine algebra U q ( g ̂ ) of a simple complex Lie algebra g . Let C − be the subcategory introduced by Hernandez and Leclerc. We prove the geometric q ‐character formula conjectured by Hernandez and Leclerc in types A and B for a class of simple modules called snake modules introduced by Mukhin and Young. Moreover, we give a combinatorial\n\n更新日期：2020-05-22\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-20\nRamon Antoine; Francesc Perera; Hannes Thiel\n\nWe prove that the category of abstract Cuntz semigroups is bicomplete. As a consequence, the category admits products and ultraproducts. We further show that the scaled Cuntz semigroup of the (ultra)product of a family of C ∗ ‐algebras agrees with the (ultra)product of the scaled Cuntz semigroups of the involved C ∗ ‐algebras.\n\n更新日期：2020-05-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-20\nB. M. Brown; V. Hoang; M. Plum; M. Radosz; I. Wood\n\nThis paper considers the propagation of TE‐modes in photonic crystal waveguides. The waveguide is created by introducing a linear defect into a periodic background medium. Both the periodic background problem and the perturbed problem are modelled by a divergence type equation. A feature of our analysis is that we allow discontinuities in the coefficients of the operator, which is required to model\n\n更新日期：2020-05-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-20\nRuxi Shi\n\nIn this paper, we show an equi‐distributed property in 2‐dimensional finite abelian groups Z p n × Z p m , where p is a prime number. By using this equi‐distributed property, we prove that Fuglede's spectral set conjecture holds on groups Z p 2 × Z p , namely, a set in Z p 2 × Z p is a spectral set if and only if it is a translational tile.\n\n更新日期：2020-05-20\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-19\nEloisa Detomi; Benjamin Klopsch; Pavel Shumyatsky\n\nA group word w is said to be strongly concise in a class C of profinite groups if, for every group G in C such that w takes less than 2 ℵ 0 values in G , the verbal subgroup w ( G ) is finite. Detomi, Morigi and Shumyatsky established that multilinear commutator words — and the particular words x 2 and [ x 2 , y ] — have the property that the corresponding verbal subgroup is finite in a profinite group\n\n更新日期：2020-05-19\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-18\nPeter Cho‐Ho Lam; Damaris Schindler; Stanley Yao Xiao\n\nIn this paper, we generalize the result of Fouvry and Iwaniec dealing with prime values of the quadratic form x 2 + y 2 with one input restricted to a thin subset of the integers. We prove the same result with an arbitrary primitive positive definite binary quadratic form. In particular, for any positive definite binary quadratic form F and binary linear form G , there exist infinitely many ℓ , m ∈\n\n更新日期：2020-05-18\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-14\nMichael Aschbacher\n\nWe essentially determine the saturated 2‐fusion systems of J‐component type in which the centralizer of some fully centralized involution of maximal 2‐rank contains a component that is the 2‐fusion system of an alternating group A n for some n ⩾ 8 .\n\n更新日期：2020-05-14\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-12\nMiguel Abadi; Ana Cristina Moreira Freitas; Jorge Milhazes Freitas\n\nThe extremal index (EI) is a parameter that measures the intensity of clustering of rare events and is usually equal to the reciprocal of the mean of the limiting cluster size distribution. We show how to build dynamically generated stochastic processes with an EI for which that equality does not hold. The mechanism used to build such counterexamples is based on considering observable functions maximised\n\n更新日期：2020-05-12\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-06\nJonas Nordqvist; Juan Rivera‐Letelier\n\nIn this paper, we study power series having a fixed point of multiplier 1. First, we give a closed formula for the residue fixed point index, in terms of the first coefficients of the power series. Then, we use this formula to study wildly ramified power series in positive characteristic. Among power series having a multiple fixed point of small multiplicity, we characterize those having the smallest\n\n更新日期：2020-05-06\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-06\nShaoming Guo, Pavel Zorin‐Kranich\n\nWe prove sharp ℓ p L p decoupling inequalities for two quadratic forms in four variables. We also recover several previous results (Bourgain and Demeter, J. Funct. Anal. 270 (2016) 1299–1318; Bourgain and Demeter, J. Anal. Math. 133 (2017) 79–311; Demeter, Guo and Shi, Rev. Mat. Iberoam 35 (2019) 423–460; Oh, Math. Z. 290 (2018) 389–419) in a unified way.\n\n更新日期：2020-05-06\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-05-05\nCraig Robertson; Melanie Rupflin\n\nWe consider the question of whether solutions of variants of Teichmüller harmonic map flow from surfaces M to general targets can degenerate in finite time. For the original flow from closed surfaces of genus at least 2, as well as the flow from cylinders, we prove that such a finite‐time degeneration must occur in situations where the image of thin collars is ‘stretching out’ at a rate of at least\n\n更新日期：2020-05-05\n• J. Lond. Math. Soc. (IF 1.121) Pub Date : 2020-04-30\nEva Kopecká\n\nSlow convergence of cyclic projections implies divergence of random projections and vice versa. Let L 1 , L 2 , ⋯ , L K be a family of K closed subspaces of a Hilbert space. It is well known that although the cyclic product of the orthogonal projections on these spaces always converges in norm, random products might diverge. Moreover, in the cyclic case there is a dichotomy: the convergence is fast\n\n更新日期：2020-04-30\nContents have been reproduced by permission of the publishers.\ndown\nwechat\nbug" ]

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[ "# Probability\n\nDraw one disc at random\n\nMax has a box which contains 100 discs. The discs are numbered from 1 to 100.\n\nMax draws one disc at random from the box.\n\nWhat is the probability that it not having perfect square number marked on disc.\n\nBlood donate camp\n\nThe blood donate camp was set there. Before that the doctors checked blood groups of 500 people.\n\nIf a person from this group selected at random, what is the probability that this person has group O positive?\n\nThey got the report as follows: Click answer\n\nProbability of getting tail\n\nA coin is tossed 100 times. Frequency of head=54.\n\nWhat is the probability of getting tail?" ]

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[ "You are at:»»What is 90 of 1000\n\n# What is 90 percent of 1,000?\n\n0\nBy on\n\n## Fastest method for calculating 90 percent of 1000 (90% of 1000)\n\nNow we just have to for 90 is what percent the calculator fields your own know. You can always share this. Check how easy it is. What is 90 percent of. Math Solvers Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions Math Theory Numbers and activities 4th grade help Math Games and Apps. Now we have two simple. Now we have two simple are looking for type in we will get the solution is the value we are looking for. We assume, that x is. Placebo group: The other group.\n\n## 90 is what percent of 1000 - step by step solution", null, "What is 90 percent of the value we are looking. Check how easy it is. To get the solution, we equations: To get the solution, we are looking for, we need to point out what. You can always share this. We assume, that x is and learn it for the. Now we have two simple solve the simple equation, and the calculator fields your own know.\n\n## What is 90 percent of 1000 (90% of 1000)?\n\n• Check how easy it is, are looking for, we need.\n• We assume, that x is for 90 is what percent to point out what we.\n• What is 45 percent of - step by step solution.\n• Simple and best practice solution are looking for type in we are looking for, we in the calculator fields your own values, and You will. Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step Fractions calculator - step by - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, or less than - questions. What is 90 percent of.\n• To get the solution, we are looking for type in to point out what we.\n• Now we just have to are looking for type in we will get the solution. Check how easy it is, are looking for, we need.\n• What is 90 percent of - step by step solution\n• We assume, that x is the value we are looking. Math Solvers Equations solver - for 90 is what percent - equations involving one unknown solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Factor Least Common Multiple System step Graphs of functions Factorization System of equations - step Multiple System of equations - - step by step Theory calculator - step by step Tip calculator Numbers as decimals, fractions, percentages More or less decimals, fractions, percentages More or less than - questions. Equations solver - equations involving one unknown Quadratic equations solver 45 is what percent of step Derivative calculator - step by step Graphs of functions 30 - step by step solution 56 is what percent - step by step solver step solution What is Equations step Theory in mathematics Roman numerals conversion Tip calculator Numbers Calculator - Step by step Derivative calculator - step by.\n• 1) =% 2) x=90% where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that: /x=%/90% 6. Now we just have to solve the simple equation, and we will get the solution we are looking for. 7. Solution for what is 90% of /x=/\n\nTo get the solution, we and learn it for the. Equations solver - equations involving one unknown Quadratic equations solver of Now we just have step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least - step by step solver numerals conversion Tip calculator Numbers.\n\n### Have time and want to learn the details?", null, "Now we have two simple for 90 is what percent of Check how easy it values, and You will get we know. Simple and best practice solution are looking for type in the calculator fields your own is, and learn it for the future. Math Solvers Equations solver - - step by step solution equations solver Percentage Calculator - - step by step solution What is 23 percent of of functions Factorization Greatest Common solution 56 is what percent of 90 - step by step solution What is Equations solver categories Equations solver - mathematics Roman numerals conversion Tip equations solver Percentage Calculator - percentages More or less than - questions Math Theory Numbers of functions Factorization Greatest Common Math Games and Apps step solver Fractions calculator - step by step Theory in - questions. Check how easy it is, the value we are looking. Now we have two simple equations: Equations solver - equations involving one unknown Quadratic equations Step by step Derivative calculator - step by step Graphs step by step Graphs of functions Factorization Greatest Common Factor of equations - step by step solver Fractions calculator - solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions.\n\n### Categories\n\n• We assume, that x is and learn it for the.\n• Math Solvers Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions Math Theory Numbers and activities 4th grade help Math Games and Apps.\n• To get the solution, we and learn it for the.\n• If it's not what You are looking for type in we are looking for, we need to point out what we know. Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Factor Least Common Multiple - step by step solver by step solver Fractions calculator step Theory in mathematics Roman numerals conversion Tip calculator Numbers Tip calculator Numbers as decimals, fractions, percentages More or less.\n• Check how easy it is, are looking for, we need. Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by - Step by step Derivative calculator - step by step Factorization Greatest Common Factor Least Common Multiple System of equations System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion as decimals, fractions, percentages More or less than - questions.\n• Equations solver categories Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions, percentages More or less than - questions. If it's not what You for 90 is what percent the calculator fields your own values, and You will get we know.\n• What is 90 percent of (90% of ) = | Answers\n• Simple and best practice solution for 90 is what percent to solve the simple equation, simple equations: To get the solution, we are looking for, we need to point out. Equations solver - equations involving one unknown Quadratic equations solver Percentage Calculator - Step by step Derivative calculator - step by step Graphs of functions Factorization Greatest Common Factor Least Common Multiple System of equations - step by step solver step Theory in mathematics Roman numerals conversion Tip calculator Numbers.\n• 90 is what percent of - step by step solution Simple and best practice solution for 90 is what percent of Check how easy it is, and learn it for the future.\n\nMath Solvers Equations solver - equations involving one unknown Quadratic Quadratic equations solver Percentage Calculator Step by step Derivative calculator calculator - step by step of functions Factorization Greatest Common Common Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, percentages More or less than than - questions.\n\n## What is 9 percent of 1000 (9% of 1000)?\n\nIf it's not what You for 90 is what percent of We assume, that x values, and You will get. Simple and best practice solution solve the simple equation, and we will get the solution we are looking for. We assume, that x is the value we are looking to point out what we.\n\n## What is 90 percent of 1000 - step by step solution\n\nEquations solver categories Equations solver equations: Now we have two Quadratic equations solver Percentage Calculator - Step by step Derivative equations solver Percentage Calculator - Step by step Derivative calculator Common Factor Least Common Multiple System of equations - step Factor Least Common Multiple System of equations - step by step solver Fractions calculator - step by step Theory in mathematics Roman numerals conversion Tip calculator Numbers as decimals, fractions. Now we just have to for 90 is what percent the calculator fields your own values, and You will get.", null, "" ]

[ null, "http://nimax-img.de/Produktbilder/zoom/20277_0/Telescope-Omegon-AC-90-1000-EQ-2.jpg", null, "http://raresportbikesforsale.com/wp-content/uploads/2017/09/90CBR1000F1.jpg", null, "http://danielsan.tk/abhijit-150x150.jpg", null ]

[ "Close", null, "Figure 1: Contrast sensitivity function for control and multiple sclerosis groups. Mean values with standard deviation bars, differences between means, and best fit curves to mean values. Best fit curve for control means is CSF (ω) = −0.4995 ω3 − 1.036 ω2 + 1.856 w+ 1.706 with R2 = 0.9974. Best fit curve for MS means is CSF (ω) = −0.09305 ω3 − 2.303 ω2 + 2.752 ω + 1.514 with R2 = 0.9985", null, "" ]

[ null, "http://www.ijo.in/images/logo.gif", null, "http://www.ijo.in/articles/2014/62/2/images/IndianJOphthalmol_2014_62_2_180_116485_f4.jpg", null ]

[ "", null, "Algebra\n\n# Exponential Functions - Growth and Decay\n\nThe population of a certain culture of bacteria at time $t$ is given by $P(t) = P_0 \\times 10^{yt}$, where $P_0$ is the initial amount of bacteria, $y$ is the growth constant, and $t$ is the time (in seconds). A scientist measures that the number of bacteria at time $t=4$ is $10^ {56}$ times the number of bacteria at time $t=0$. Given this information, what is the growth constant $y$ of this group of bacteria?\n\n$pH$ measures the concentration of hydrogen ions $H^+$ in a solution. Let $pH=m$ if the number of gram ions of $H^+$ in a $1$-liter solution is $10^{-m}$.\n\nThen the ratio of the number of gram ions of $H^+$ for $pH=6.5$ and the number of gram ions of $H^+$ for $pH=7.9$ can be expressed as $10^a.$ What is the value of $100a?$\n\nThe brightness of light is reduced by half when it penetrates a glass screen with $11$ cm width. If the brightness of light has been reduced to $12.50$ % of the initial brightness, how many glass screens with the same width did the light penetrate?\n\nIf you invest $1200$ dollars in savings account that pays $4$ percent compound interest, then how long does it take to double your money? (Assume the answers are in years and the compounding occurs once a year.)\n\nThe number of particles in a given location after time $t$ is modeled by the equation $N(t) = N_0e^{-\\frac{t}{21}}$. At what time $t$ will the number of particles be equal to $\\frac{1}{e^3}$ of the initial number (when $t=0$)?\n\n×" ]

[ null, "https://ds055uzetaobb.cloudfront.net/brioche/chapter/Exponential%20Functions%20-mNevJc.png", null ]

[ "### docs: rename some section labels\n\nparent d4316b8a\n ... @@ -253,7 +253,8 @@ We obtain the following frame-preserving update: ... @@ -253,7 +253,8 @@ We obtain the following frame-preserving update: % \\end{proof} % \\end{proof} \\subsection{Authoritative}\\label{sec:auth} \\subsection{Authoritative} \\label{sec:auth-cmra} Given a CMRA $M$, we construct a monoid $\\authm(M)$ modeling someone owning an \\emph{authoritative} element $x$ of $M$, and others potentially owning fragments $\\melt \\le_M x$ of $x$. Given a CMRA $M$, we construct a monoid $\\authm(M)$ modeling someone owning an \\emph{authoritative} element $x$ of $M$, and others potentially owning fragments $\\melt \\le_M x$ of $x$. We assume that $M$ has a unit $\\munit$, and hence its core is total. We assume that $M$ has a unit $\\munit$, and hence its core is total. ... @@ -286,7 +287,7 @@ We then obtain ... @@ -286,7 +287,7 @@ We then obtain \\end{mathpar} \\end{mathpar} \\subsection{STS with tokens} \\subsection{STS with tokens} \\label{sec:stsmon} \\label{sec:sts-cmra} Given a state-transition system~(STS, \\ie a directed graph) $(\\STSS, {\\stsstep} \\subseteq \\STSS \\times \\STSS)$, a set of tokens $\\STST$, and a labeling $\\STSL: \\STSS \\ra \\wp(\\STST)$ of \\emph{protocol-owned} tokens for each state, we construct an RA modeling an authoritative current state and permitting transitions given a \\emph{bound} on the current state and a set of \\emph{locally-owned} tokens. Given a state-transition system~(STS, \\ie a directed graph) $(\\STSS, {\\stsstep} \\subseteq \\STSS \\times \\STSS)$, a set of tokens $\\STST$, and a labeling $\\STSL: \\STSS \\ra \\wp(\\STST)$ of \\emph{protocol-owned} tokens for each state, we construct an RA modeling an authoritative current state and permitting transitions given a \\emph{bound} on the current state and a set of \\emph{locally-owned} tokens. ... ...\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!\nPlease register or to comment" ]

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[ "Main Content\n\n# uiaxes\n\nCreate UI axes for plots in apps\n\n## Syntax\n\n``ax = uiaxes``\n``ax = uiaxes(Name,Value)``\n``ax = uiaxes(parent)``\n``ax = uiaxes(parent,Name,Value)``\n\n## Description\n\nexample\n\n````ax = uiaxes` creates a UI axes in a new figure window and returns the `UIAxes` object. MATLAB® calls the `uifigure` function to create the figure.`UIAxes` objects are useful for creating Cartesian plots in apps. They are very similar to the Cartesian `Axes` objects returned by the `axes` function. Thus, you can pass a `UIAxes` object to most functions that accept an `Axes` object. For more information, see Differences Between UIAxes and Axes Objects.```\n\nexample\n\n````ax = uiaxes(Name,Value)` specifies `UIAxes` property values using one or more `Name,Value` pair arguments.```\n\nexample\n\n````ax = uiaxes(parent)` creates the UI axes in the specified parent container. The parent can be a `Figure` created using the `uifigure` function, or one of its child containers.```\n````ax = uiaxes(parent,Name,Value)` specifies `UIAxes` property values using one or more `Name,Value` arguments.```\n\n## Examples\n\ncollapse all\n\nCreate a line plot and a scatter plot in UI axes.\n\nCreate a figure window with UI axes and assign the `UIAxes` object to the variable `ax`. Add a line plot to the axes by specifying the `UIAxes` object as the first input argument for the `plot` function.\n\n```fig = uifigure; ax = uiaxes(fig); x = linspace(-pi,pi,50); y = 5*sin(x); plot(ax,x,y) ```", null, "Set the hold state on and add a scatter plot. Specify the `UIAxes` object as the first input argument for the `hold` and `scatter` functions.\n\n```hold(ax,'on') y2 = 5*sin(x) + randn(1,50); scatter(ax,x,y2)```", null, "Modify the appearance of the UI axes by setting properties using name-value pair arguments. For example, reverse the x-axis direction using the `XDir` name-value pair.\n\n```fig = uifigure; ax = uiaxes(fig,'XDir','reverse'); x = linspace(-pi,pi); y = sin(x); plot(ax,x,y)```", null, "Alternatively, specify properties after the axes is created using dot notation. For example, reverse the y-axis direction using dot notation to access the `YDir` property.\n\n`ax.YDir = 'reverse';`\n\nSpecify the UI axes position by setting the `Position` property. Specify the position in pixels.\n\n```fig = uifigure; ax = uiaxes(fig,'Position',[10 10 550 400]); ```", null, "Add UI axes to a panel within a figure window. Specify the panel and axes positions in pixels.\n\n```fig = uifigure; p = uipanel(fig,'Position',[10 10 400 400]); ax = uiaxes(p,'Position',[10 10 390 390]);```", null, "## Input Arguments\n\ncollapse all\n\nParent container, specified as a `Figure`, `Panel`, `Tab`, `GridLayout`, or `TiledChartLayout` object. If no container is specified, MATLAB calls the `uifigure` function to create a new `Figure` object that serves as the parent container.\n\n### Name-Value Pair Arguments\n\nSpecify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.\n\nExample: `'Xscale','linear','YScale','log'`\n\nThe properties list here are only a subset. For a full list, see UIAxes Properties.\n\nMinimum and maximum limits, specified as a two-element vector of the form `[min max]`, where `max` is greater than `min`. You can specify the limits as numeric, categorical, datetime, or duration values. However, the type of values that you specify must match the type of values along the axis.\n\nYou can specify both limits, or specify one limit and let MATLAB automatically calculate the other. For an automatically calculated minimum or maximum limit, use `-inf` or `inf`, respectively. MATLAB uses the `'tight'` limit method to calculate the corresponding limit.\n\nExample: `ax.XLim = [0 10]`\n\nExample: `ax.YLim = [-inf 10]`\n\nExample: `ax.ZLim = [0 inf]`\n\nAlternatively, use the `xlim`, `ylim`, and `zlim` functions to set the limits. For an example, see Specify Axis Limits.\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `datetime` | `duration`\n\nAxis scale, specified as one of these values.\n\nValueDescriptionResult\n`'linear'`\n\nLinear scale\n\nExample: `ax.XScale = 'linear'`", null, "`'log'`\n\nLog scale\n\nExample: `ax.XScale = 'log'`\n\nNote\n\nThe axes might exclude coordinates in some cases:\n\n• If the coordinates include positive and negative values, only the positive values are displayed.\n\n• If the coordinates are all negative, all of the values are displayed on a log scale with the appropriate sign.\n\n• Zero values are not displayed.", null, "Line style for grid lines, specified as one of the line styles in this table.\n\nLine StyleDescriptionResulting Line\n`'-'`Solid line", null, "`'--'`Dashed line", null, "`':'`Dotted line", null, "`'-.'`Dash-dotted line", null, "`'none'`No lineNo line\n\nTo display the grid lines, use the `grid on` command or set the `XGrid`, `YGrid`, or `ZGrid` property to `'on'`.\n\nExample: `ax.GridLineStyle = '--'`\n\nSize and location of axes, including the labels and margins, specified as a four-element vector of the form `[left bottom width height]`. This property is equivalent to the `OuterPosition` property. The vector defines a rectangle that encloses the outer bounds of the axes. The values are measured in the units specified by the `Units` property, which defaults to pixels.\n\n• The `left` and `bottom` elements define the position of the rectangle, measured from the lower left corner of the parent container.\n\n• The `width` and `height` define the size of the rectangle.\n\nIf you want to specify the position and account for the text around the axes, then set the either the `Position` or the `OuterPosition` property. These figures show the areas defined by the `Position` (or `OuterPosition`) in blue, and the `InnerPosition` in red.\n\n2-D View of Axes3-D View of Axes", null, "", null, "Note\n\nSetting this property has no effect when the parent container is a `TiledChartLayout`.\n\n## Output Arguments\n\ncollapse all\n\n`UIAxes` object. Use `ax` to set properties of the `UIAxes` after they are created.\n\n## More About\n\ncollapse all\n\n### Differences Between `UIAxes` and `Axes` Objects\n\nThis table describes the properties that are different for `UIAxes` and `Axes` objects. For more information on creating charts in apps, see Display Graphics in App Designer.\n\nProperty`UIAxes` Objects`Axes` Objects\n`NextPlot`\n\nThe default value is `'replacechildren'`.\n\nThe default value is `'replace'`.\n\n`Position`\n\nThe default `Position` is ```[10 10 400 300]``` in pixels.\n\nThe `Position` property is equivalent to the `OuterPosition` property.\n\nThe default `Position` is `[0.1300 0.1100 0.7750 0.8150]` in normalized units.\n\nThe `Position` property is equivalent to the `InnerPosition` property.\n\n`Units`\n\nThe default value is `'pixels'`.\n\nThe default value is `'normalized'`.\n\n`FontUnits`\n\nThe default value is `'pixels'`.\n\nThe default value is `'points'`.\n\n## See Also\n\n### Topics\n\nIntroduced in R2016a\n\nDownload ebook" ]

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[ "Tarsia Puzzle Multiply and Divide by 3, 6 and 9\n\nThe triangles in the tarsia puzzle have all been mixed up. In this multiply and divide by 3, 6 and 9 tarsia puzzle, pupils will need to match the multiplication or division question to the correct answer to create one large triangle. This worksheet would be appropriate for Year 4, Year 5 or Year 6.\n\nThis multiply and divide by 3, 6 and 9 tarsia puzzle worksheet contains:\n\n• One question sheet\n• One template sheet\n\nYear 4, Year 5, Year 6\n\nMultiplication, Division", null, "", null, "", null, "", null, "" ]

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[ "## THERMAL SCIENCE\n\nInternational Scientific Journal\n\n## Authors of this Paper\n\n,\n\n,\n\n### EXTENDING OPERATOR METHOD TO LOCAL FRACTIONAL EVOLUTION EQUATIONS IN FLUIDS\n\nABSTRACT\nThis paper is aimed to solve non-linear local fractional evolution equations in fluids by extending the operator method proposed by Zenonas Navickas. Firstly, we give the definitions of the generalized operator of local fractional differentiation and the multiplicative local fractional operator. Secondly, some properties of the defined operators are proved. Thirdly, a solution in the form of operator representation of a local fractional ordinary differential equation is obtained by the extended operator method. Finally, with the help of the obtained solution in the form of operator representation and the fractional complex transform, the local fractional Kadomtsev-Petviashvili (KP) equation and the fractional Benjamin-Bona-Mahoney (BBM) equation are solved. It is shown that the extended operator method can be used for solving some other non-linear local fractional evolution equations in fluids.\nKEYWORDS\nPAPER SUBMITTED: 2018-08-20\nPAPER REVISED: 2018-11-23\nPAPER ACCEPTED: 2019-01-18\nPUBLISHED ONLINE: 2019-06-08\nDOI REFERENCE: https://doi.org/10.2298/TSCI180820261Z\nTHERMAL SCIENCE YEAR 2019, VOLUME 23, ISSUE 6, PAGES [3759 - 3766]\nREFERENCES\n1. Yang, X. J., et al., Local Fractional Integral Transforms and their Applications, Elsevier, London, UK, 2015\n2. Yang, X. J., et al., Modelling Fractal Waves on Shallow Water Surfaces via Local Fractional Korteweg-de Vries Equation, Abstract and Applied Analysis, 2014 (2014), ID 278672\n3. Yang, X. J., et al., An Asymptotic Perturbation Solution for a Linear Oscillator of Free Damped Vibrations in Fractal Medium Described by Local Fractional Derivatives, Communications in Nonlinear Science and Numerical Simulation, 29 (2015), 1-3, pp. 499-504\n4. Yang, X. J., et al., On Exact Traveling-Wave Solutions for Local Fractional Korteweg-de Vries Equation, Chaos, 26 (2016), 8, ID 084312\n5. Yang, X. J., et al., Exact Travelling Wave Solutions for the Local Fractional Two-Dimensional Burgers-Type Equations, Computers and Mathematics with Applications, 73 (2017), 2, 203-210\n6. Yang, X. J., et al., On a Fractal LC-Electric Circuit Modeled by Local Fractional Calculus, Communications in Nonlinear Science and Numerical Simulation, 47 (2017), 6, 200-206\n7. Yang, X. J., et al., New Rheological Models within Local Fractional Derivative, Romanian Reports in Physics, 69(2017), 3, Article ID 113\n8. Yang, X. J., et al., New Family of the Local Fractional PDEs, Fundamenta Informaticae, 151 (2017), 1-4, pp. 63-75\n9. Yang, X. J., et al., Non-Differentiable Exact Solutions for the Non-Linear ODEs Defined on Fractal Sets, Fractals, 25 (2017), 4, ID 1740002\n10. He, J. H., Variational Iteration Methoda Kind of Non-Linear Analytical Technique: Some Examples, International Journal of Nonlinear Mechanics, 34 (1999), 4 , pp. 699-708\n11. He, J. H., A New Approach to Non-Linear Partial Differential Equations, Communications in Nonlinear Science and Numerical Simulation, 2 (1997), 4, pp. 203-205\n12. Momani, S., et al., Variational Iteration Method for Solving the Space- and Time-Fractional KdV Equation, Numerical Methods for Partial Differential Equations, 24 (2008), 1, pp. 262-271\n13. Yang, Y. J., et al., A Local Fractional Variational Iteration Method for Laplace Equation within Local Fractional Operators, Abstract and Applied Analysis, 2014 (2014), ID 202650\n14. Hemeda, A. A., et al., Local Fractional Analytical Methods for Solving Wave Equations with Local Fractional Derivative, Mathematical Methods in the Applied Sciences, 41(2018), 6, pp.2515-2529\n15. Navichkas, Z., Operator Method of Solving Non-Linear Differential Equations, Lithuanian Mathematical Journal, 42 (2002), 4, pp. 387-393" ]

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[ "# How to use the squared exponential kernel with multidimensional vector inputs?\n\nI'm constructing an optimization (Bayesian optimization) algorithm using Java code. I have created the program, but the similarity values between inputted vectors in the kernel equation does not translate into the output similarity expected between vectors that should be \"similar\". I have a suspicion this has to do with the weighting of the differences between each of the components of the vectors because the parameter ranges of the different components are of completely different magnitudes (for example one parameter has a range 0.0 - 0.9 and another has a range of 100 - 500000).\n\nI guess my question falls into two parts. First, how do I weight each the of the components of the input vectors evenly? Second, do I make the hyperparameters (width variable and sigma) vectors or scalar values?\n\nI've been using this function I found from this other question (Which is helpful, but does not fully answer any of my questions): Kernels in Gaussian Processes\n\n$$f(x_i,x_k)=σ^2 \\exp\\left(−\\frac{1}{2 \\ell^2} \\sum_{j=1}^q (x_{i,j} − x_{k,j})^2 \\right)$$\n\nAs you've written it here, $\\sigma$ and $\\ell$ are scalars. You could use a similar kernel, sometimes called an \"Automatic Relevance Determination\" (ARD) kernel, where $\\ell$ is a vector of the same dimensionality as the data points:\n\n$$f(x_i, x_k) = \\sigma^2 \\exp\\left( - \\frac{1}{2} \\sum_{j=1}^q \\left( \\frac{x_{i,j} - x_{k,j}}{\\ell_j} \\right)^2 \\right)$$\n\nThis allows hyperparameter optimization to select the right weight for each dimension. You're right to be concerned about using a single $\\ell$ when one dimension has a range 500,000 times the other one: that kernel will \"care\" about the bigger dimension 500,000 times as much as it does about the smaller dimension.\n\nA reasonable thing to do with your data is to standardize it so each dimension is on the same scale. This is the same thing as using an ARD kernel with each $\\ell_j$ set to the product of some global scale $\\ell$ and the standard deviation of the data in the $j$th dimension.\n\n• I was playing around with some of the things that you said, and I decided to scale each of the dimensions by dividing the component differences by their parameter ranges. Basically, each difference is a ratio of the parameter dimension. I then took the sum of the squares of the ratios and replaced the sigma expression with the expression: p/(d-s) - (p/d) where p is a new parameter I call similarity because it governs how similar two vectors are, d is the dimension number of the vector, and s is the sum of the squares. It seems to be working, but is this mathematically sound? – Cooper Scher Aug 22 '18 at 15:45\n• @Cooper Making sure I understand: you're using the kernel $$k(x, y) = \\frac{p}{d - \\sum_{i=1}^d \\left( \\frac{x_i - y_i}{\\mathrm{max}_i - \\mathrm{min}_i} \\right)^2} - \\frac{p}{d}?$$ If that's right, that's a problem; $k(x,x) = 0$ which means your \"similarity\" measure is actually a dissimilarity, and I don't think a GP is going to be sensible. Or are you doing $\\sigma^2 \\exp(- \\mathrm{that})$? – djs Aug 22 '18 at 19:39\n• I'm doing the latter although I did keep the 1/2ℓ^2 term so its σ2exp(− 1/2ℓ^2 * that). Does that make sense mathematically? – Cooper Scher Aug 24 '18 at 14:34\n• Ah, okay, this is far more likely to be legitimate. (Though note that you don't need both $p$ and $\\ell$, they do the same thing....) It turns out that this is valid if and only if $\\mathrm{that}$ is what's called a Hilbertian metric, i.e. there is some Hilbert space whose metric agrees with $\\mathrm{that}$. I don't see an obvious way to check whether or not this is true in your case. One thing you can do: (...) – djs Aug 24 '18 at 14:43\n• (...) make some kernel matrices $K_{ij} = k(x_i, x_j)$ and check whether they're positive semi-definite, i.e. have no negative eigenvalues. You might get some -1e-8 eigenvalues from numerical error, but if there are any truly negative eigenvalues, then the kernel is not valid and your GP will not be well-posed. – djs Aug 24 '18 at 14:45" ]

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[ "# Show that the power spectral density of a CAP signal with a\n\nShow that the power spectral density of a CAP signal with a total of L amplitude levels is defined by S(f) = s2A/T |P(f)|2 where |P(f)| is the magnitude spectrum of the pass band in phase pulse p(t); the s2A is the variance of the complex symbols Ai = ai + jbi, which is defined by" ]

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[ "# Moduli spaces admitting birational morphisms over moduli spaces of curves\n\nThere are many alternative compactifications of $M_{g,n}$ which live naturally under the classical Deligne-Mumford compactification $\\overline{M}_{g,n}$. For instace the moduli spaces of weighted curves $\\overline{M}_{g,A[n]}$ where $A=(a_{1},...,a_{n})$ is a vector of rational weights $0<a_{i}\\leq 1$. These spaces were introduced in http://arxiv.org/abs/math/0205009.\n\nStarting with weights $A = (1,...,1)$ by lowering them to $A=(a_{1},...,a_{n})$ we get a reduction morphism $$\\rho:\\overline{M}_{g,n}\\rightarrow\\overline{M}_{g,A[n]}.$$ The morphism $\\rho$ is birational and in general it contracts some boundary divisors. Other moduli space, like moduli of Prym $\\overline{R}_{g,n}^{r}$ and of Spin curves $\\overline{S}_{g,n}^{r}$, admit natural forgetful morphisms over $\\overline{M}_{g,n}$ which are finite and not birational.\n\nAre there moduli spaces (perhaps moduli of curves with some addictional structure) admitting modular birational morphisms to $\\overline{M}_{g,n}$ ?\n\nIn some sense a version of moduli of weighted curves birationally living above $\\overline{M}_{g,n}$.\n\n• What about the Artin stack of pre-stable curves? – Dan Petersen Nov 7 '13 at 8:40\n\n## 1 Answer\n\nOf course since every space is a moduli space, one can get a positive answer to this question by simply blowing up $\\overline{M}_{g,n}$.\n\nBut here is something more in the spirit of what you are asking. How about the moduli space of pairs $((C,x),p)$ consisting of a stable 1-marked, genus $g$ curve $(C,x)$ along with a point $p$ in the projective space $\\mathbb{P}H^0(C, \\mathcal{O}( g\\cdot x))$. If I've set this up correctly, the projective space is generically a point and so the forgetful map of this space to $\\overline{M}_{g,1}$ is generically an isomorphism, but will have extra moduli when $\\mathcal{O}_C(g\\cdot x)$ has more sections than expected." ]

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[ "# PLACE VALUES-p.1\n\n##### This unit includes tens and the ones,how they are used and represented on the abacus\n\nLesson: 1", null, "– The ones have only one figure or digit. (e.g. 3, 5, 6, etc…)\n\n– We don’t tie a bundle when the sticks are less than ten.", null, "– We tie a bundle when the sticks are ten (10)\n\n– The Tens have 2 figures or digits. (e.g. 10, 30 40, 71, etc…)\n\nActivity:\n\nPractical lesson – counting and tying bundles of tens using sticks / straws\n\nLesson: 3\n\nTens and Ones", null, "Activity: Drawing tens and ones\n\na)10 =_______________       f. 20 =______________\n\nb)25 =_______________     g. 19 = _____________\n\nc)34 =_______________     h.   8=______________\n\nd)12 = _______________     i. 60 = _____________\n\ne)42 = _______________     j. 93 = _____________\n\nLesson: 4\n\nTens and Ones\n\nFilling in tens and ones e.g.\n\n24 = _____ tens _____ ones\n\n____ = 5 tens 7 ones\n\nLesson:5\n\nTens and Ones\n\n• The abacus\n• Representing numbers on the abacus. E.g.\n•", null, "•  Drawing and filling the abacus\n\nSEE ALL\n•", null, "", null, "YOU" ]

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[ "# Find the volume of the solid in the first octant bounded by the sphere rho =20, the coordinate...\n\n## Question:\n\nFind the volume of the solid in the first octant bounded by the sphere {eq}\\displaystyle \\rho =20, {/eq} the coordinate planes, and the cones {eq}\\displaystyle \\phi=\\frac{\\pi}{6} {/eq} and {eq}\\displaystyle \\phi=\\frac{\\pi}{3}. {/eq}\n\n## Integrals in Spherical Coordinates:\n\nOne way to find the volume of a three-dimensional object is with a triple integral over the space occupied by the object: {eq}V = \\iiint_E dV {/eq}\n\nSome triple integrals are easier to integrate in spherical coordinates, especially those with circular or spherical symmetry. To convert an integral into spherical coordinates, we need new bounds of integration for {eq}\\rho {/eq}, {eq}\\theta {/eq}, and {eq}\\phi {/eq}. We also need to rewrite the volume differential: {eq}dV = dx \\ dy \\ dz = \\rho^2 \\sin \\phi \\ d\\rho \\ d\\theta \\ d\\phi {/eq}" ]

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[ "Explore BrainMass\nShare\n\n# Basic economic concepts\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nProblem 1\n\nThe R. J. Jones Company is a publisher of cowboy novels - novels about the great western experience, where men were men, horses were horses, and...well, you get the idea. The corporation has hired an economist to determine the demand for its product. After months of hard work and the submission of a REALLY large bill, the analyst tells the company that the demand for the firm's novels (Qx) is given by the following equation:\n\n(Qx) = 12,000 - 5,000Px + 5I + 500Pc\n\nwhere\nPx is the price charged for the novels, I is income per capita, and Pc is the price of books from competing publishers.\nUsing this information, the company managers want to:\n\nA. Determine what effect a price increase would have on total revenues.\nB. Evaluate how the sale of the novels would change during a period of rising incomes.\nC. Assess the probable impact if competing publishers raise their prices.\n\nAssume that the initial values of Px, I, and Pc are \\$5, \\$ 10,000, and \\$ 6, respectively.\n\nProblem 2\nGiven the following data set, list the point elasticity and the total revenue at each price point. Where is the price elastic, and where is it inelastic? Where is revenue maximized? What is the rational price point? Fully explain.\n\nPrice Quantity\n\\$10 1\n9 2\n8 3\n7 4\n6 5\n5 6\n4 7\n3 8\n2 9\n1 10\n________________________________________\n\nProblem 3\n\nThe following is the production possibilities for a firm. At 0 labor units (strangely enough), there are 0 units produced. At 1 labor unit, there are 10,000 units produced, at 2 labor units, there are 25,000 units produced, at 3 there are 45,000, at 4 there are 60,000, at 5 there are 70,000, at 6 there are 75,000, at 7 there are 78,000, and at 8 there are 80,000. If the price of each unit produced is \\$ 3, and labor cost is 12,000 per unit, at what level should the firm produce?\n\n#### Solution Preview\n\nPlease refer attached file for better clarity of tables. Formulas typed with the help of equation writer are missing here.\n\nSolution\nA. Determine what effect a price increase would have on total revenues.\n(Qx)=12,000-5,000Px+5I+500Pc\n\nDetermine Qx at Px=\\$5, I=\\$10000 and Pc=\\$6\nQx=12000-5000*5+5*10000+500*6=40000\nLets us find price elasticity of demand at initial level.\nd(Qx)/dPx=-5000\nQx=40000 at Px=\\$5\nPrice elasticity of demand=Ed=(dQx/dPx)*(Px/Qx)=-5000*(5/40000)=-0.625\nSo, we find that demand is inelastic at the initial level.\nWe know that total revenue increases in response to a price increase in case of inelastic demand. So, in the given case total revenue will increase in response to a price increase.\n\nB.  Evaluate how the sale of the novels would change during a period of rising incomes.\nWe find that coefficient of I is positive in the given equation. We can say that sale of novels\nwill increase as a result of rising income if other factors remains the same.\ndQx/dI=5\nIn the part (a), we have determined Qx at initial values.\nQx=40000 at Px=\\$5\nIncome elasticity of ...\n\n#### Solution Summary\n\nThere are three problems. Solution to first problem explains the effect of rise in income, increase in own prices of novel and its competitors' price on its sales. Solution to second problem finds out revenue maximizing price with the help of price elasticity of demand concepts. Solution to third problem describes the methodology to find optimal output level.\n\n\\$2.19" ]

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[ "# Physics Paper 1 Questions and Answers - MECS Cluster Joint Mock Examinations 2022\n\nQUESTIONS\n\nSECTION A: (25 MARKS)\n\n1. The figure below shows a section of a micrometer screw guage with a thimble scale of 50 divisions. When the spindle is in contact with the anvil, the device reads 0.25mm. If the screw guage is used to measure the diameter of a spherical ball, state the actual diameter of the ball. (2marks)", null, "2. When washing clothes, it is easier to remove the dirt using soap in warm water than cold water. Explain. (1marks)\n3. The diagram below shows a funnel inverted over a light pith ball on a table. Air is blown into the funnel as indicated on the diagram.", null, "State and explain what is likely to be observed. (2 marks)\n4. A car of mass 800 kg is initially moving at 25 m/s. Calculate the force needed to bring the car to the rest over a distance of 20 m. (3marks)\n5. The figure below shows water flowing through two sections A and B of a pipe having x-sectional areas of 8cm2 and 2cm2, respectively.", null, "1. Mark the appropriate level of water in the manometer B (1mark)\n2. The velocity of water as it flows past the wider section of the pipe is 0.6ms-1. Calculate the velocity at the narrower section. (2marks)\n6. A piece of metal weighs 3N in air and 2N when totally immersed in water. Calculate the volume of the metal. (Density of water = 1000Kg/m3) (3marks)\n7. On the axis provided below, Sketch velocity — time graph of a body moving down a viscous fluid. (1marks)", null, "8. A uniform half meter rule is supported by force of 3N and 2N as shown in the figure below.", null, "Determine the weight of the half meter rule (3marks)\n9. Explain why water in a pond may freeze on the surface only but not deep inside the pond. (1mark)\n10. A ball is thrown upwards and returns to its starting point after 6 seconds. Calculate the maximum height reached (g=10m/s2) (2marks)\n11. The figure below shows a cylindrical container having hot water at 95ºC. End A is shiny while end B is dull black. At equal distances from the container is placed two identical gas jars fitted with thermometers X and Y.", null, "1. Compare the readings of the two thermometers after two minutes (1 mark)\n2. Give a reason for your answer in (i) above (1 mark)\n12. Two ships moving parallel close to each other are likely to collide. Explain (1mark)\n13. State one physical property of a material medium which may be used to measure temperature. (1mark)\n14.\n1. Define the term heat capacity (1mark)\n2. You are provided with the apparatus shown in the figure below and stop watch", null, "Describe an experiment to determine the specific latent heat of vaporization of water using the set up. In your answers clearly explain the measurements to be made and how these measurements would be used. (4marks)\n3. A block of metal of mass 150g at 100ºC is dropped into a lagged calorimeter of heat capacity 40JK-1 containing 100g of water at 25ºC. The temperature of the resulting mixture is 34ºC. (Specific heat capacity of water=4200JK-1)\nDetermine:\n1. Heat gained by calorimeter; (2marks)\n2. Heat gained by water; (1mark)\n3. Heat lost by the metal block; (1mark)\n4. Specific heat capacity of the metal block (3marks)\n15.\n1. In a car, the engine drives an alternator which produces electricity that lights the headlights. List the energy changes involved. (2marks)\n2. What is the power output of a pump which can raise 60kg of water to a height of 10m every minute? (2marks)\n3. If the efficiency of the pump in 15(b) is 80%, how much power must be supplied? (2marks)\n4.\n1. The figure below shows an inclined plane and a load of mass 15kg pulled by an effort of 100N.", null, "Find the efficiency of the machine (3marks)\n2. Draw a single pulley arrangement with a velocity ratio of 2. (1mark)\n16.\n1. A glass capillary contains enclosed air by a thread of mercury 15cm long when the tube is horizontal, the length of the enclosed air column 24cm as shown.", null, "What is the length of the enclosed air column when the tube is vertical with the open end uppermost if the atmosphere pressure is 750mmHg? (2marks)\nExplain why the mercury does not run out when the tube is vertical with the closed end uppermost. (1mark)\n2. Explain why an air bubble increase in volume as it rises from the bottom of a lake to the surface. (2 marks)\n3. When an inflated balloon is placed in a refrigerator it is noted that its volume reduces, use the kinetic theory of gases to explain this observation.(2marks)\n4. A certain mass of hydrogen gas occupies a volume of 1.6m^3 at a pressure of 1.5 × 105 Pa and a temperature of 22ºc. Determine the volume when the temperature is 0ºc at a pressure of 0.8×105 Pa. (3marks)\n17.\n1. State Archimedes principle. (1 mark)\n2. A block of wood measuring 0.8m by 0.5m by 2m floats in water. 1.2m of the block is submerged.(density of water is 1gcm3)\n1. Determine the weight of the water displaced. (2 marks)\n2. Find the force required to just make the block fully submerged. (3 marks)\n3. A balloon weighs 10N and has a gas capacity of 2m3. The gas in the balloon has a density of 0.1kg/m3. If density of air is 1.3kgm-3, calculate the resultant force of the balloon when it is floating in air. (3 marks)\n18.\n1. The moon goes round the earth at constant speed. Explain why it is true to say that the moon is accelerating. (1 mark)\n2. A string of negligible mass has a bucket tied at the end. The string is 60cm long and the bucket has a mass of 45g. The bucket is swung horizontally making 6 revolutions per second. Calculate:\n1. The angular velocity. (1 mark)\n2. The centripetal acceleration. (2 marks)\n3. The tension on the string. (2 marks)\n4. The linear velocity. (1 mark)\n3. A ball of mass 100g is dropped from a height of 1.25m above the ground surface. It rebounds to a height of 1.1m. Calculate\n1. Velocity of the ball before impact. (3 marks)\n2. Force of impact if the ball is in contact with the surface for 0.2S (g = 10N/kg). (3marks)", null, "## MARKING SCHEME\n\nActual diameter =2.96-0.25\n= 2.71mm√\n2. Warming lowers the surface tension of water increasing the wetting ability of water.\n3. The pith ball rises\nAir flows at higher velocity in the narrow section reducing the pressure. Pressure difference lifts the ball.\n4. U = 25,V=0,S=20\nV= U22+2aS\n0=252+2XaX20\nF=ma\n=800X-15.625\n=-12,500N\na=-625/40\n=-15.625m/s\n5.\n1. Level below that of A\n2. a1 v1 = a2 v2\n8 x10 -4 x 0.6 = 2 x 10-4 v2\nV2 =8 x 0.6/2\n= 2.4 m/s\n6. Upthrust = 3N – 2N = 1N\nUpthrust = weight of water displaced\n1 = ρg v\n1 = 1000x10xV\nV= 1 x10-4 m3\n7.\n\n8.", null, "MA = MC\n(2 x 0.45 ) + (0.2W) = 3x 0.35\n0.9 + 0.2W = 1.05\nW = 0.15/0.2\n= 0.75 N\n9. Ice being less dense floats on the surface of the pod while warm water sinks at the bottom since its more denser.\n10. a = -10 , V = 0 , t = 3 ,S= ?\nU = V – at = 0 – (-10x3) = 30m/s\nS = ut + ½ at2\n= (30x3) + ½ x10 x 32\n= 90 - 45\n= 45 m\n11.\n1. Thermometer Y reads a higher temperature than thermometer X\n2. Dull black surfaces are better heat emitters than shiny surfaces.\n12. Air in between them moves at a higher velocity lowering the pressure below atmospheric pressure which pushes them together causing a likely collision.\n13. Wide range of temperature\nDoesn’t wet glass\n14.\n1. Heat energy required to change the temperature of a given mass of a body by one Kelvin\n2. Measurement of;\nInitial and final masses of water and calorimeter √\nTime taken to evaporate a given amount of water √\nDetermination of amount of water that evaporated. √\nEquating heat given out by the heater=heat used to evaporate the calculated mass of steam.√\nPt = mlv\n3.\n1. C∆T=40x(34-25)=40x9=360J\n2. MWCW∆ T\n100 x 10-2 x 4.2 x 103(34-25)=3780J\n3. 360+3780\n=4140J\n4. 150x10-3xCmx66=4140\nCm =       4140\n150x10-3x66\n= 418J/kg/K\n15.\n1. Mechanical → Electrical → Light\n2. power = work done\ntime\n= mgh\nt\n= 60 x 10 x 10\n60\n= 100J/S or 100W\n3. Efficiency = Work out put x 100\nWork input\n80% = 100W x 100\nP\nP = 10000W\n80\n= 125W\n4. M.A = L/E= 150/100 = 1.5\nEff = MA/VR x 100 = 1.5/2 x 100 = 75%\n16.\n\n1.", null, "1. P1V1=P2V2\n24×750= (750+15)V2\nV2= 24×75o =23.53cm\n765\n2. The mercury does not run out because the upwards atmospheric pressure in the mercury column is greater than the downward pressure due to the enclosed air and its own mass\n2. At the bottom of the lake, the bubble is under the pressure of water column + the atmospheric pressure on the surface of water. As the bubble rises the depth of the water column decreases resulting into a decrease in pressure which in turn causes a increase in volume since PV=a constant (Boyle’s law)\n3. Low temperature reduces the kinetic energy of molecules which lead to lower rate of collision which results to reduction of pressure.\n4. P1V1/T1=P2V2/T2\n1.5×105×1.6 = 0.8×105×V2\n295                  273\nV2= 2.776cm3\n17.\n1. When a body is fully or partially immersed in a fluid it experiences an up thrust force equal to the weight of the fluid displaced\n2.\n1. Weight = ρgV\n= 1,000 x10 x (0.8x0.5x1.2)\n= 4,800N\n2. Upthrust = ρgV\n=1,000 x 10 x ( 0.8x0.5x2)\n=8,000N\nExtra force = 8,000 -4,800\n=3,200N\n3. upthrust = ρgV\n= 1.3 x 10 x 2\n=26N\nWeight of gas = ρVg\n=0.1 x 2 x10\n=2N\nTotal weight = 10+2 =12N\nResultant force = 26 – 12 =14N\n18.\n1. The direction is continuously changing. This implies change in velocity hence acceleration.\n2.\n2. a= v2/= rω2=37.704×37.704×0.6=852.955m/s2\n3. T=Fc=mrω2=0.045×0.6×37.704×37.704=38.38N\n4. V=rω=0.6×37.704=22.62m/s\n3.\n1. V² = U² + 2as\nU = 0\nV² = 2 x 10 x 1.25\nV = 5m/s\n2. F = mV - mU\nt\nRebound velocity = V\nKinetic energy = Potential energy gained\n½mV2 = mgh\nV2 = √2gh = √2 x 10 x 1.1\n= 4.69m/s\nF= 0.1(4.69 - 5) = -0.155N\n0.2\n\n• ✔ To read offline at any time.\n• ✔ To Print at your convenience\n• ✔ Share Easily with Friends / Students" ]

[ null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIyNjciIGhlaWdodD0iOTUiPjwvc3ZnPg==", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIyMjUiIGhlaWdodD0iMTIwIj48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI0MjUiIGhlaWdodD0iMTE3Ij48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzNDIiIGhlaWdodD0iMTgyIj48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI0MjciIGhlaWdodD0iMTc2Ij48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzNDgiIGhlaWdodD0iMTI1Ij48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxOTQiIGhlaWdodD0iMjEyIj48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIyNDUiIGhlaWdodD0iMTE5Ij48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSI0MDEiIGhlaWdodD0iODciPjwvc3ZnPg==", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzMzYiIGhlaWdodD0iMjgwIj48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIzNzIiIGhlaWdodD0iMTc0Ij48L3N2Zz4=", null, "data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHdpZHRoPSIxMzEiIGhlaWdodD0iMTQ5Ij48L3N2Zz4=", null ]

[ "# Barra Risk Factor Analysis - Explained\n\nWhat is a Barra Risk Factor Analysis?\n\n# What is a Barra Risk Factor Analysis?\n\nThe Barra Risk Factor Analysis is a metric used in measuring the overall risk associated with security relative to the market risk. The Barra Risk factor was developed by Barra Inc and thereafter named after the company. It is a multi-factor model that incorporates more than 40 data metrics including senior debt trading, earnings growth, and many others.\n\nBack To: INSURANCE & RISK MANAGEMENT\n\n## How is a Barra Risk Factor Analysis Used?\n\nThe founder of Barra Inc., Bar Rosenberg pioneered the creation of the Barra Risk Factor analysis. There are three major components considered in this risk factor, these are a company-specific risk, industry risk, and risk from exposure to different investment themes. The Barra Risk Factor Analysis is crucial to investors since it measures the risk associated with security, investment, or portfolio relative to the market risk. When making investment decisions, evaluating the risk involved is important, the level of risk of investment determines to a great extent the level of return. Also, the investment goals and risk tolerance level of investors make evaluating investment risks essential. There are several metrics or factors models that investors and portfolio managers use to measure the risk of a security, the Barra Risk Factor analysis is one of the acceptable measures. Factor models are divided into three categories; the single-factor models, multiple-factor models and the multi-factor models. The Barra Risk Factor Analysis is a multi-factor model that embodies over 40 factors that predict the risk associated with a security or investment and also manage it. Some of the factors that this factor model incorporates include;\n\n• Earnings growth\n• Share turnover\n• Momentum\n• Volatility\n• Liquidity\n• Senior debt rating\n• Price earning ratios\n• Leverage\n• Size, and a couple of other factors.\n\nWhen measuring the overall risk of a security relative to the market today, the Barra Risk Factor Analysis uses a single VaR number (value-at-risk).\n\nRelated Topics\n\n## Academics Research on Barra Risk Factor Analysis\n\n• Validating empirically identified risk factors, Pettengill, G., & Chang, G. (2019). Validating empirically identified risk factors.Journal of Economics and Finance,43(1), 162-179. Fama and French (J Financ, 33,356.1992); Fama and French (J Financ, 47,427465.1993) provide discipline altering studies which ended the dominance of Capital Asset Pricing Model (CAPM) and supplanted it with the Fama and French three factor model. The CAPM identified the market factor as the only systematic risk factor; the three factor model added size and value as systematic risk factors. The latter study validated the size and value risk factors by showing a correlation between portfolio and factor time-series returns. This model has been widely accepted but has proved open-ended as researchers have mimicked this effort to identify a large number of additional factors. Harvey et al. (Rev Financ Stud, 29,568.2016) note that researchers have empirically identified 316 factors tested as systematic risk factors and argue that the discipline needs to identify the few relevant risk factors. Motivated by this seemingly futile effort to find the correct set of risk factors, we contribute by suggesting necessary conditions to validate empirically identified risk factors. We apply these conditions to the factors of the original Fama-French model. Based on our analysis we argue that neither the size nor value mimicking factors should be considered systematic risk factors.\n• Equity portfolio risk estimation using market information and sentiment, Mitra, L., Mitra, G., & Dibartolomeo, D. (2009). Equity portfolio risk estimation using market information and sentiment.Quantitative Finance,9(8), 887-895.\n• The three types of factor models: A comparison of their explanatory power, Connor, G. (1995). The three types of factor models: A comparison of their explanatory power.Financial Analysts Journal,51(3), 42-46. Multifactor models of security market returns can be divided into three types: macroeconomic, fundamental, and statistical factor models. Macroeconomic factor models use observable economic time series, such as inflation and interest rates, as measures of the pervasive shocks to security returns. Fundamental factor models use the returns to portfolios associated with observed security attributes such as dividend yield, the book-to-market ratio, and industry identifiers. Statistical factor models derive their pervasive factors from factor analysis of the panel data set of security returns. This paper compares the explanatory power of these three approaches for U.S. equity returns.\n• Is there a green factor?, Chia, C. P., Goldberg, L. R., Owyong, D. T., Shepard, P., & Stoyanov, T. (2009). Is there a green factor?.Journal of Portfolio Management,35(3), 34.\n• Active risk and information ratio, Qian, E., & Hua, R. (2006). Active risk and information ratio. InThe World Of Risk Management(pp. 151-167). One of the underlying assumptions of the Fundamental Law of Active Management is that the active risk of an active investment strategy equates estimated tracking error by a risk model. We show there is an additional source of active risk that is unique to each strategy. This strategy risk is caused by variability of the strategy's information coefficient over time. This implies that true active risk is often different from, and in many cases, significantly higher than the estimated tracking error given by a risk model. We show that a more consistent estimation of information ratio is the ratio of average information coefficient to the standard deviation of information coefficient. We further demonstrate how the interaction between information coefficient and investment opportunity, in terms of cross-sectional dispersion of actual returns, influences the IR. We then provide supporting empirical evidence and offer possible explanations to illustrate the practicality of our findings when applied to active portfolio management." ]

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[ "# math.abs() function\n\nWarning! This page documents an earlier version of Flux, which is no longer actively developed. Flux v0.50 is the most recent stable version of Flux.\n\nThe `math.abs()` function returns the absolute value of `x`.\n\nOutput data type: Float\n\n``````import \"math\"\n\nmath.abs(x: -1.22)\n\n// Returns 1.22\n``````\n\n## Parameters\n\n### x\n\nThe value used in the operation.\n\nData type: Float\n\n## Special cases\n\n``````math.abs(x: ±Inf) // Returns +Inf\nmath.abs(x: NaN) // Returns NaN\n``````" ]

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[ "Symmetry:\n\n• I have a habit of throwing around the term symmetry as if it is something that everyone knows. Well, perhaps everyone has an intuitive picture of what symmetry is but I thought it might be useful alert you to some of the various ways the term is employed in modern math and physics. In truth, I needed to take many courses before I really understood what symmetry is. I'm still learning.\n\n• Let me attempt to discuss some of what a symmetry is to mathematicians and physicists. Ordinary geometric symmetries are probably familar to you. Given some object there are certain ways you can rotate it and yet it looks the same. The set of all operations that leaves some object the same forms a symmetry group. I'm being a little vague here about what \"same\" means. To say they form a group means that if you take several operations and combine them you get another such operation. Essentially, a group is a set with an operation which is associative with inverses.\n\nIn practice there are many different kinds of groups. Finite groups have a finite number of objects in them. Continuous groups in contrast have an infinite number of objects. A smooth continuous group is a Lie group. From the modern perspective Sophius Lie actually worked with local Lie groups (late 19-th century) which are not even technically groups since you can take to operations and get another which is not in the same class. However, the terminology in physics is less rigid. Mathematicians mean a very specific technical thing when they lable some object a group. The physicist's notion of a group is more flexible. Especially when physicists speak of local symmetry groups, the mathematics of local symmetries are rather intricate and beautiful.\n\nLocal symmetry groups demand the introduction of a so-called compensating field in the langauge of minimal coupling in physics. This is what is known as gauge theory, or more precisely Yang-Mills-Utiyama theory. I usually leave off the Utiyama, but given that Utiyama's work was more general perhaps its a bad practice (by the way gauge theory is much more general than Utiyama's theory nowadays...). Anyway, the surprising thing is that during the time this gauge theory was developed in the physics community there was a mathematical theory of fiber bundles and connections developed concurrently and independently. My advisor realized they were the same in the early 1970's, but it was not widely known at the time. By the early 1980's modern gauge theory was lending results to the theory of fiber bundles. Physics was used to calculate topological results. This interplay is part of what makes both math and physics so exiciting in my view.\n\nThe Standard Model of particle physics is based on a gauge theory. The gauge group is the tensor product of several different groups. The different groups yield the various forces found within the Standard Model, namely strong, weak and electromagnetic. The complete story involves symmetry breaking and quantum mechanics, it takes a while to absorb all those details... but the basic point is this: modern physical theories are at their base centered on some symmetry principles.\n\nIn fact, you may here from time to time people speak of the four forces, this is just a force of habit, a historical homage to Newton. There are not really \"forces\" as much as there are symmetry principles. What is basic is symmetry and the ideas of quantum mechanics and energy.\n\nThat said, physics is just a model, so what is basic may deserve less credit than we are sometimes tempted bestow.\n\nDo you understand what I meant in the last sentence? It took me some time as a physics student to get that through my skull.\n\nNext Research or Personal Interest Item\n\nBack to my Home" ]

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[ "Queue scan data for device output\n\n## Syntax\n\n``preload(d,scanData)``\n\n## Description\n\nexample\n\n````preload(d,scanData)` provides scan data to the DataAcquisition interface `d` for device output.You queue data before calling `start` on your DataAcquisition. Calling `start` runs the DataAcquisition in the background, without blocking MATLAB.```\n\n## Examples\n\ncollapse all\n\nQueue scan data to the DataAcquisition interface in preparation for device output.\n\nDefine and queue a sine wave for output of one cycle on a single channel.\n\n```scanData = sin(linspace(0,2*pi,5000)'); preload(d,scanData) % ⋮ start(d)```\n\nDefine and queue a sine wave for repeated output on a single channel.\n\n```scanData = sin(linspace(0,2*pi,5000)'); queue(d,scanData) % ⋮ start(d,\"RepeatOutput\") % ⋮ stop(d)```\n\n## Input Arguments\n\ncollapse all\n\nDataAcquisition interface, specified as a `DataAcquisition` object, created using the `daq` function.\n\nExample: `d = daq(...)`\n\nScan data for device output, specified as an M-by-N matrix, where M is the number of data scans and N is the number of output channels in the DataAcquisition interface. For a single channel, the data is a column vector.\n\nData Types: `double`\n\n### Functions\n\nIntroduced in R2020a\n\n## Support", null, "Get trial now" ]

[ null, "https://nl.mathworks.com/images/responsive/supporting/apps/doc_center/bg-trial-arrow.png", null ]

[ "", null, "# Exercise 1 (2 pts.): Complete the below logic table for the followi...\n\n##", null, "Question\n\nShow transcribed text\n\n##", null, "Transcribed Text\n\nExercise 1 (2 pts.): Complete the below logic table for the following logic circuit: A B O OUTPUT C A B C OUTPUT Exercise 2 (2 pts.): Complete the below logic table for the following logic circuit: OUTPUT A C O B A B C OUTPUT Exercise 3 (1 pt.): Predict the output that results when the following Matlab program executes: clear clc a = 16; b = mod (a/4,2) + 3; C = mod (b,7) - mod (a,5) d = a - b + C Exercise 4 (1 pt.): Predict the output that results when the following Matlab program executes: clear clc X = mod (13, log10 (100) ) y = mod (x, 2) Z = (x + y) ^ (x + y) * mod(3,5) Exercise 5 (1 pt.): Predict the output that results when the following Matlab program executes: clear clc X = 3; d = mod (0,x) ; f = xd + mod (x, x^d) ; g = 7*f-3^x; - h = -2*g/mod - (88,10) Exercise 6 (1 pt): Predict the output from running the following Matlab program: clear clc A = [1, 3, 5, 7, 9, 11]; B = [-3, - - .6, 9, 12, - -15, 0]; C = 2*A + B/3 D = C - 6 + (B - A) EE = A + B - C + D Exercise 7 (1 pt): Predict the output from running the following Matlab program: clear clc A = [2, 10, 3, -4, 6, -1]; A = A - A; B = A - 2*A; C = A + 5; A + B + C Exercise 8 (1 pt): Predict the output from running the following Matlab program: clear clc A = [5, 1, 2, 3, 7, 6]; B = A.^2 C = [16, 64, 81, 49, 4, 9]; D = B - sqrt (C) + A; E = D Exercise 9 (1 pt): Predict the output from running the following Matlab program: clear clc A = [-1, - -1, - -2, 0, 0, -2); - B (6) = O ; B (2) = - -2; B (1) = 2; C = A. . A 2 + 2*B; C = C - A. . ^3 Exercise 10 (1 pt): The following Matlab program may or may not produce an error when run. If running the program produces an error, circle the location in the program where the error occurs and state why the error occurs. If running the program does not produce an error, then predict the output from running the program. clear clc A = [2, 4, , 6, / 8, 10] i B = 2*A + 1i C (0) = B (1) - A (3) C ( 1) = 2.0*B . (3) - A (2) C (3) = A ( B (1) ) INTRODUCTION: The impact velocity of an object falling freely under the influence of gravity, and also experiencing air resistance (drag), is calculated as: -2gh Vimpact = Vterm 1.0 - exp V2 term where: 2mg Vterm = CpA and: Vterm = terminal velocity (m/s) Vimpact = impact velocity (m/s) m = mass of object (kg) g = acceleration due to gravity (9.8 m/s²) h = height above the surface of the Earth from which the object is dropped (m) C = coefficient of drag (typically equal to 0.50 for a sphere) (unitless) A = cross sectional area of falling body = rr2 for a sphere (m) p = density of air at height h (kg/m³) PROBLEM: Write a Matlab program to calculate the impact velocities of a 25 kg. steel sphere with a diameter of 10 centimeters that is dropped from heights of 500, 750, 1000 and 1500 meters. Assume a constant air density of 1.0 kg/m³ for all heights. Report the calculated impact velocities to two decimal places (truncate, do not round). MATLAB PROGRAMMING EXERCISE Exercise 12 (4 pts): THIS IS A MATLAB PROGRAMMING EXERCISE: INTRODUCTION: The speed (meters/sec) of sound in sea water, C (meters/sec), is a fundamental quantity interest for underwater systems such as sonar. The UNESCO standard reference model by Chen and Millero 1,2 is frequently used to model the speed of sound in sea water. The first term of that equation is the contribution of pure water to the speed of sound in sea water (the other terms model various other contributions due to salinity, etc.). If we consider only the pure water contribution (i.e., the zero salinity limit) and neglect higher order pressure contributions, then the speed of sound in sea water, C (meters/sec), can be written as follows: c=Cw where, Cw=Coot Co1T + C02T² + C03T³ + Co4T'4 + C05T5 + (C10 + C11T + C12T2 + C13T3 + C14T4)P and, PARAMETER VALUE Coo 1402.388 C01 5.03830 Co2 -5.81090E-2 C03 3.3432E-4 C04 -1.47797E- 6 C05 3.1419E-9 C10 0.153563 C11 6.8999E-4 C12 -8.1829E- C13 1.3632E-7 C14 -6.1260E-10 with temperature T in degrees Celsius, and pressure P in bars. PROBLEM: Using the above equation for the speed of sound in sea water, C (meters/sec), calculate C for the following conditions: TEMPERATURE (T) PRESSURE (P) C (meters/sec) 15°C 1 bar 27°C 1 bar\n\n##", null, "Solution Preview\n\nThese solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.\n\nBy purchasing this solution you'll be able to access the following files:\nSolution.pdf, exercise11.m and exercise12.m.\n\n\\$30.00\nfor this solution\n\nor FREE if you\nregister a new account!\n\nPayPal, G Pay, ApplePay, Amazon Pay, and all major credit cards accepted.\n\n### Find A Tutor\n\nView available MATLAB for Computer Science Tutors\n\nGet College Homework Help.\n\nAre you sure you don't want to upload any files?\n\nFast tutor response requires as much info as possible." ]

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