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https://forum.solidworks.com/thread/96300 | [
"AnsweredAssumed Answered\n\n# How can I rename Cutlist Folders\n\nQuestion asked by Shaun Jalbert on Jun 12, 2015\nLatest reply on Jan 19, 2017 by Arif Akbas\n\nI have a snippit of code that I'm trying to modify but I'm not having any luck. This code finds the file name and renames the cutlist folders to match and adds a sequential instance number after it. Folders looks like:\n\n<filename>-1\n\n<filename>-2\n\netc..\n\nWhat I want it to do is look up a custom property called \"WLD_MK_NO\" instead of the file name, then rename the cutlist folders to match \"WLD_MK_NO\" and add the sequential instance number after it.\n\n<WLD_MK_NO>-1\n\n<WLD_MK_NO>-2\n\netc..\n\nHow can I achieve this?\n\nDim swApp As Object\n\nSub Main()\n\nDim swApp As SldWorks.SldWorks\n\nSet swApp = Application.SldWorks\n\nDim doc As SldWorks.ModelDoc2: Set doc = swApp.ActiveDoc\n\nDim partdoc As SldWorks.partdoc: Set partdoc = doc\n\nDim f As IFeature: Set f = doc.FirstFeature\n\nDim bRet As Boolean\n\nDim n As Integer: n = 0\n\nDim num As String\n\nDim modDocExt As SldWorks.ModelDocExtension\n\nDim Part As Object\n\nDim boolstatus As Boolean\n\nDim BodyFolder As SldWorks.BodyFolder\n\nDim BodyCount As Long\n\nDo While Not f Is Nothing\n\nIf f.GetTypeName = \"CutListFolder\" Then\n\nSet BodyFolder = f.GetSpecificFeature2\n\nBodyCount = BodyFolder.GetBodyCount\n\nIf BodyCount > 0 Then\n\nn = n + 1\n\nIf n < 1000 Then\n\nnum = \"-\" + Right(Str\\$(n), Len(Str\\$(n)) - 1)\n\nEnd If\n\nIf n < 100 Then\n\nnum = \"-\" + Right(Str\\$(n), Len(Str\\$(n)) - 1)\n\nEnd If\n\nIf n < 10 Then\n\nnum = \"-0\" + Right(Str\\$(n), Len(Str\\$(n)) - 1)\n\nEnd If\n\nEnd If\n\nIf Right(doc.GetTitle, 7) = \".SLDPRT\" Or Right(doc.GetTitle, 7) = \".SLDASM\" Then\n\nf.Name = Left(doc.GetTitle, Len(doc.GetTitle) - 7) + num\n\nElse\n\nf.Name = Left(doc.GetTitle, Len(doc.GetTitle)) + num\n\nEnd If"
]
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https://dsp.stackexchange.com/questions/68662/why-the-plots-of-spline-and-cubic-interpolation-are-exactly-same | [
"# Why the plots of spline and cubic interpolation are exactly same? [closed]\n\nI am trying to watch difference between cubic interpolation and spline interpolation using matlab plot but i am getting same plots in both cases using interp1 command\n\nMy code is below. In below code ,if i use spline in place of cubic, i still exactly get same plot,why?\n\nclc\nclear all\nclose all\nx = 0:pi/4:2*pi;\nv = sin(x);\nxq = 0:pi/16:2*pi;\nvq2 = interp1(x,v,xq,'cubic');\nplot(x,v,'o',xq,vq2,':.');\n\n• The purpose of splines is to fit multiple splines for a given set of points.And the spline used is typically cubic. While cubic interpolation just means that you have fitted a cubic curve over some data points. – Ben Jun 27 '20 at 19:47\n• Hi Engr! What have you researched so far. interp1 has, like every matlab function, excellent documentation, and it tells you what cubic (==pchip) and spline do. You'll notice that you can infer directly under which condition of your data these two methods are identical. However, you don't reference any research into what is done for interpolation here, so I'm not sure what the actual question is. – Marcus Müller Jun 27 '20 at 20:53\n• Do you mean they look identical, or they are identical? What do you see when you generate cubic and spline interpolations and plot their difference? – TimWescott Jun 28 '20 at 4:19\n• @TimWescott yes,you are right. The both plots appear to be identical but they aren't exactly ,when we see their vq matrix value, but question is how we can make them look different in plots? – engr Jun 28 '20 at 16:57\n\nTo emphasize how two point sets differ, using dot and circle markers can be useful, because uneven centering is quite visible (from the comments):",
null,
"and the code is here:\n\nx = 0:pi/4:2*pi;\nv = sin(x);\nxq = 0:pi/16:2*pi;\nvq1 = interp1(x,v,xq,'spline');\nvq2 = interp1(x,v,xq,'pchip');\nplot(x,v,':.k',xq,vq1,'.r',xq,vq2,'ob');\nlegend('Original','Spline','PChip')\naxis tight; grid on\n\n\nThe cubic option will be renamed as pchip, and reuse for other purposes in Matlab, because of the potential misinterpretations:\n\n• pchip: Shape-preserving piecewise cubic interpolation. The interpolated value at a query point is based on a shape-preserving piecewise cubic interpolation of the values at neighboring grid points.\n• spline: Spline interpolation using not-a-knot end conditions. The interpolated value at a query point is based on a cubic interpolation of the values at neighboring grid points in each respective dimension.\n• cubic: Note The behavior of interp1(...,'cubic') will change in a future release. In a future release, this method will perform cubic convolution.\n\nThe best way to understand cubic splines (under whatever name or how specified) is to derive and code them yourself.\n\nWhat you are looking for is a parametric equation with these properties:\n\n\\begin{aligned} f(0) &= y[n] = y_0\\\\ f(1) &= y[n+1] = y_1\\\\ f(2) &= y[n+2] = y_2\\\\ f(3) &= y[n+2] = y_3\\\\ \\end{aligned}\n\nYour candidate function is a cubic polynomial, of course.\n\n$$f(t) = a t^3 + b t^2 + c t + d$$\n\nPlug 'em in. You get four equations with four unknowns ($$a,b,c,d$$)\n\nOnce you have solved the equation, use it to interpolate between the middle two points.\n\nExtra credit:\n\nDo the vector equivalent in N-space with the conditions given as two points in space and the velocity vectors at each point with one unit of time to get there.\n\nThen code your solutions. Print the results so you can see some precision. And compare your answer to Matlab's."
]
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"https://i.stack.imgur.com/vttMl.png",
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https://www.stumblingrobot.com/2015/09/01/find-a-formula-for-the-composition-of-given-functions-10/ | [
"Home » Blog » Find a formula for the composition of given functions\n\n# Find a formula for the composition of given functions\n\nFind the composition function",
null,
"of the functions:",
null,
"We compute",
null,
",",
null,
"This is valid for all",
null,
". (Of course, this formula would generally be valid for",
null,
"as well, but since the domain of",
null,
"and",
null,
"is given as",
null,
", the composition is defined only for",
null,
".)"
]
| [
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-79b996a9edf5fabc84bae8a1567943fc_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f39c9d197545e2724ce579ee463ce59c_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8a7bc97ffec77220393d0f9700ec4660_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9d57da0c3b5227d61d5da086e42f1270_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-dbb0823565be8ea7189bc5e15d77a0d8_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-32a7152fae4e40558d94e917456c3928_l3.png",
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"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5f5a977df2345e9bba1be3c47348607b_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b50947b486c7390ea2c5196206908b47_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-dbb0823565be8ea7189bc5e15d77a0d8_l3.png",
null,
"https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-dd1adb34365dff1a2a90237f3ea2a111_l3.png",
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https://answers.opencv.org/question/86462/why-opencv-doesnt-detect-the-object/ | [
"Why opencv doesnt detect the object?\n\nI’m working on a project that should filter the red objects in an image and calculates the distance to this object with two webcams.\n\nTo detect the objects i convert the image from BGR to HSV and use the function inRange to threshold them.\n\nThen i use findContours to get the contours in the image, which should be the contours of the red objects.\n\nAs last step i use boundingRect to get a Vector of Rect that contains one Rect per detected object.\n\nOn the two images below you can see my problem. The one with the pink rectangle is about 162cm away from the camera and the other about 175cm. If the Object is further then 170cm the object is not recognized alltough the thresholded image is showing the contours of the object.\n\nAre the contours of the object just too small or is there any other reason?",
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"",
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"ok heres the code:\n\nmain.cpp\n\nObjectDetection obj;\nStereoVision sv;\nfor (;;) {\n\nsv.calculateDisparity(imgl, imgr, dispfull, disp8, imgToDisplay);\nMat imgrt, imglt;\nobj.filterColor(imgl, imgr, imglt, imgrt);\n\n//p1 und p2 sind die gefundenen Konturen eines Bildes\nvector<vector<Point> > p1 = obj.getPointOfObject(imglt);\nvector<Rect> allRoisOfObjects = obj.getAllRectangles(imgl, p1);\nfor(int i = 0; i < allRoisOfObjects.size(); i++){\nRect pos = allRoisOfObjects.at(i);\npos.width -= 20;\npos.height -= 20;\npos.x += 10;\npos.y += 10;\ndisp = dispfull(pos);\n\nfloat distance = sv.calculateAverageDistance(pos.tl(),pos.br(),dispfull);\n\nstringstream ss;\nss << distance;\nrectangle(imgToDisplay, allRoisOfObjects.at(i), color, 2,8, 0);\nputText(imgToDisplay, ss.str(), pos.br(), 1, 1, color, 1);\nss.clear();\nss.str(\"\");\nnewObjects.push_back(pos);\n}\n}\n\nObjectDetection.cpp\n\nvoid ObjectDetection::filterColor(Mat& img1, Mat& img2, Mat& output1,\nMat& output2) {\nMat imgHSV, imgHSV2;\ncvtColor(img1, imgHSV, COLOR_BGR2HSV); //Convert the captured frame from BGR to HSV\ncvtColor(img2, imgHSV2, COLOR_BGR2HSV);\nMat imgThresholded, imgThresholded2;\n\ninRange(imgHSV, Scalar(iLowH, iLowS, iLowV), Scalar(iHighH, iHighS, iHighV),\nimgThresholded);\ninRange(imgHSV2, Scalar(iLowH, iLowS, iLowV),\nScalar(iHighH, iHighS, iHighV), imgThresholded2);\noutput1 = imgThresholded;\noutput2 = imgThresholded2;\n\n}\n\nvector<vector<Point> > ObjectDetection::getPointOfObject(Mat img) {\nvector<vector<Point> > contours;\nvector<Vec4i> hierarchy;\nfindContours(img, contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE,\nPoint(0, 0));\n\nreturn contours;\n}\n\nvector<Rect> ObjectDetection::getAllRectangles(Mat & img, vector<vector<Point> > contours){\nvector<vector<Point> > contours_poly(contours.size());\nvector<Rect> boundRect(contours.size());\nvector<Point2f> center(contours.size());\nRect rrect;\nrrect.height = -1;\nRNG rng(12345);\n\nfor (int i = 0; i < contours.size(); i++) {\napproxPolyDP(Mat(contours[i]), contours_poly[i], 3, true);\nboundRect[i] = boundingRect(Mat(contours_poly[i]));\n}\nreturn boundRect;\n}\nedit retag close merge delete"
]
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"https://answers.opencv.org/upfiles/14545068849107435.png",
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"https://answers.opencv.org/upfiles/14545069228359106.png",
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]
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https://sjcinspire.com/2020/04/08/dr-harrys-questions/ | [
"# Dr Harry’s Questions\n\n### Who is Dr Harry?\n\nDr Harry Desmond is a physicist – or, theoretical cosmologist, to be precise! – who works as a researcher at St John’s and the Department of Physics at the University of Oxford. Before entering the world of research, Harry completed his first degree at Oxford (MPhys in Physics) before going to the United States to complete his PhD at Stanford University in California.\n\n### What kind of research does Dr Harry do?\n\nDr Harry looks at how galaxies move and change to learn about the pieces that make up the universe, and how it is evolving. He is particularly interested in using measurements of galaxies to tackle two key questions in the field of cosmology:\n\n1. What is “dark matter”? How can we define the identity and phenomenology of “dark matter”, which exists throughout the universe yet doesn’t emit light and hence can’t be seen directly.\n2. What is the nature of gravity? Despite extensive testing in the Solar System and on Earth, the reigning theory of gravity — Einstein’s General Relativity — has not been directly proven on larger scales.\n\n### What are Dr Harry’s Questions?\n\nDr Harry’s Questions are a mixture of Physics-oriented questions, puzzles and brainteasers designed to challenge anyone and everyone interested in studying Physics. Some of them are the kinds of questions you might find in a university Physics course, some are just for fun. If you love Physics and want to see some questions you might never have seen before, then this is place for you!\n\n1. Hubble’s Law states that for every 3.3 million light-years of distance from the Milky Way, a galaxy recedes from us at an additional 70 km/s. Estimate the age of the universe. (The speed of light is 300,000 km/s.)\n2. 2.2e-18 Joules are required to separate the proton and electron that constitute hydrogen. How big are atoms? (The charge of an electron is 1.6e-19 C.)\n3. Our Solar System is at a distance 2.5e17 km from the centre of our galaxy, the Milky Way, and orbits it at a speed of 220 km/s. Estimate the total mass enclosed within the orbit of the Sun. Around fifty billion stars are located in this region — is stellar matter all there is? (The mass of the Sun is 2e30 kg.)\n4. To grow vegetables in England, it’s useful to know that the average solar irradiance during the day is 130 W/m^2. This is around 10% of what one could expect without clouds. Given that the mass of a Hydrogen nucleus, Helium nucleus and the Sun are 1.67e-27 kg, 6.64e-27 kg and 2e30 kg respectively, calculate the rate at which Hydrogen atoms are fusing into Helium in our star.\n5. The nearest star to the Sun, Proxima Centauri, appears to move through an angle of 0.00043 degrees (relative to the distance stars) between June and December. How far away is it? (The distance to the Sun is 150 million km.)\n6. The Arctic is mainly ice on water, and the Antarctic mainly ice on land. Which contributes more to sea level rise given global warming?\n7. During a Solar eclipse the moon almost precisely blocks out the light from the Sun. The distance to the moon is 384,400 km and to the Sun is 150 million km, and the radius of the moon is 1,740 km. How big is the Sun?\n8. The “HI” line of atomic hydrogen has a frequency of 1420 MHz in the lab, but from a particular galaxy is observed to have a wavelength of 23cm. What is the velocity of that galaxy relative to us? Recalling Hubble’s law (for every 3.3 million light-years of distance from the Milky Way, a galaxy recedes from us at an additional 70 km/s), estimate its distance. (The speed of light is 3e8 m/s.)\n9. In theory there’s a thin layer in the atmosphere where all the helium balloons that kids accidentally release go. Estimate the location of this given that the scale height of the atmosphere — the height over which atmospheric pressure falls by a factor of e — is about 8 km. (The density of helium is 0.18 kg/m^3 while that of air is 1.2 kg/m^3 at sea level; assume temperature does not vary significantly over several scale heights.) Is there any chance of being able to see the balloons up there? The angular resolution of the human eye is about 1/60 degrees.\n10. The surface gravity of the moon is 1.6 m/s^2. How high could you jump?\n11. The first light (the “Cosmic Microwave Background”) was emitted when the volume of the universe was a billion times less than it is today. We observe these photons with a temperature of 2.7 K. What was the temperature of the universe when they were emitted?\n12. The surface of a black hole (the “event horizon”) is the radius at which not even light is fast enough to escape. Calculate the size the Earth would have if it collapsed into a black hole. (The mass of the Earth is 6.0e24 kg.)\n13. According to quantum mechanics, all bodies exhibit wave-like behaviour. The scale over which this behaviour is manifest is known as the body’s “Compton wavelength”, which for a non-relativistic speed v is given by h/mv, where m is the body’s mass and h is Planck’s constant, 6.6e-34 m^2kg/s. Calculate this for the electron in hydrogen orbiting the nucleus, and for you when you’re walking. Why don’t we experience quantum phenomena in everyday life? (The mass of an electron is 9.1e-31 kg, the charge of an electron is 1.6e-19 C and the size of a hydrogen atom is 1e-10 m.)\n14. You are in a lift accelerating downwards at 9 m/s^2. You drop a ball. What happens to it? The lift then begins accelerating faster, first at 9.8 m/s^2 and then at 11 m/s^2. What does the ball do in each case? Describe your own sensations during this period.\n15. You are in a windowless ship travelling smoothly across the water. Describe how you could determine your velocity.\n16. Galaxies are thought to be surrounded by huge spheres of dark matter called halos. Assuming the density of the dark matter is proportional to r^-n, where r is the distance from the halo centre, calculate the value of n required for the rotation velocity of a star to be independent of its distance from the centre of the halo and hence generate the “flat rotation curve” that observations suggest.\n17. Instead of postulating the existence of dark matter to explain galaxy dynamics, some theories seek instead to modify the laws of gravity or inertia. One of them (“Modified Newtonian Dynamics”, or MOND) proposes that Newton’s second law is only approximately valid at relatively high accelerations, and at low acceleration takes instead the form F = m(a/a0)^n where a0 is a fundamental constant. For a roughly spherically-symmetric galaxy of mass M, what value of n is required for a flat rotation curve?\n18. You drill a hole through the centre of the Earth to other side and drop a ball down. What does it do?\n19. What velocity must an ambulance have for its the frequency of its siren to halve as it passes you?\n20. Special Relativity predicts that moving clocks run slow. How can you travel into Earth’s future?\n21. General Relativity predicts that massive objects deflect light towards them in just the same way as they deflect particles. This is known as “gravitational lensing”. Suppose there is a light source behind a black hole in the distant universe — how will the light from that source appear to us?"
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https://www2.bellevuecollege.edu/classes/All/MATH%26/148 | [
"# MATH& 148 Business Calculus • 5 Cr.\n\n## Description\n\nSurveys differential and integral calculus, emphasizing uses in business and social science. Intended for students who wish only a brief course in calculus. Either MATH& 151 or MATH& 148 may be taken for credit, not both. Fulfills the quantitative or symbolic reasoning course requirement at BC. Prerequisite: MATH 138 or MATH& 141 with a C- or better, or placement by assessment.\n\n## Outcomes\n\nAfter completing this class, students should be able to:\n\n• Use the product, quotient and chain rules to differentiate simple algebraic, exponential and logarithmic functions.\n• Construct equations for tangent lines and find average and instantaneous rates of change from symbolic, graphical and numerical information.\n• Apply the concepts, techniques and vocabulary of limits, continuity and first and second derivatives to solve problems in contexts such as marginal analysis, product elasticity, related rates, exponential growth/decay and optimization.\n• Use simple substitutions, integration by parts and tables to determine antiderivatives of simple algebraic and exponential functions.\n• Determine the values (exact or approximate, as appropriate) of definite integrals using the Fundamental Theorem of Calculus and areas.\n• Apply the ideas of definite and indefinite integrals to solve problems in contexts such as total change/accumulation, consumer and producer surplus, exponential growth and decay, etc.\n• Determine appropriate units for definite integrals and derivatives.\n• Calculate partial derivatives of simple functions of two variables, and apply them to solve optimization problems, compute marginal productivity, and interpret three-dimensional graphics."
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https://teachingcalculus.com/tag/quotient-rule/ | [
"Differentiation Techniques\n\nSo, no one wants to do complicated limits to find derivatives. There are easier ways of course. There are a number of quick ways (rules, formulas) for finding derivatives of the Elementary Functions and their compositions. Here are some ways to introduce these rules; these are the subject of this week’s review of past posts. Why…\n\nFar Out!\n\nA monster problem for Halloween. A while ago I suggested you look at , which using the dominance idea is zero. Of course your students may try graphing or a table. Here’s the graph done by a TI-Nspire CAS. Note the scales. This is not the way to go. Since the function is increasing near the…\n\nDerivative Rules III\n\nThe Quotient Rule This approach to the quotient rule is credited to Maria Gaetana Agnesi (1718 – 1799) who wrote the first known mathematics textbook Analytical Institutions (1748) to help her brothers learn algebra. The quotient rule can also be proven from the definition of derivative. But here is a simpler approach – as a…"
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https://socratic.org/questions/550f088a581e2a2713e8bd13 | [
"# Question 8bd13\n\nMar 22, 2015\n\nYou should dilute your original solution to a total volume of $\\text{2000 mL}$.\n\nHere's how you should think about dilution calculations. Your original solution, the one you want to dilute, contains a certain amount of moles of solute - in your case, sulfuric acid.\n\nWhen you dilute a solution, that certain number of moles of solute does not change, it remains constant. What changes is the concentration of the solution.\n\nThat happens because you'll have the same amount of solute in a bigger volume of solution.",
null,
"That is what the formula for dilution calculations actually expresses\n\n${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where\n\n${C}_{1}$, ${V}_{1}$ - the concentration and volume of the initial solution;\n${C}_{2}$, ${V}_{2}$ - the concentration and volume of the final solution.\n\nPlug your data into this equation and you'll get\n\nV_2 = C_1/C_2 * V_1 = (\"11\"cancel(\"M\"))/(\"0.17\"cancel(\"M\")) * \"30 mL\" = \"1941.2 mL\"#\n\nRounded to one sig fig, the number of sig figs given for 30 mL, answer will be\n\n${V}_{2} = \\textcolor{red}{\\text{2000 mL}}$"
]
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null,
"https://d2jmvrsizmvf4x.cloudfront.net/MNz9IG6WQjWGs2bXlPyr_solution.jpg",
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http://www.uml.org.cn/c%2B%2B/201212194.asp | [
"",
null,
"",
null,
"求知 文章 文库 Lib 视频 iPerson 课程 认证 咨询 工具 讲座 Modeler Code",
null,
"分享到\n\n1. (expression). 在表达式外边加括号，由编译器来决定怎么改变。\n\n2. new_type(expression）. 强制类型括号住表达式。\n\n3. (new_type)expression. 括号住强制类型。\n\n4. C语言允许的内置转换。\n\nC++程序兼容C语言的转化，但是针对面向对象语言的特性，设计了以下几个类型转换操作符。他们的出现是为了C语言类型转换中语义模糊和固有的危险陷阱，因为C语言不去判断所要操作的类型转换是否合理。\n\nstatic_cast:用于非多态类型的转换。\n\ndynamic_cast:用于多态类型的转换。\n\nconst_cast:用来消除const, volatile, __unaligned属性的转换。\n\nreinterpret_cast:用于空间的重新解释。\n\nstatic_cast:\n\nstatic_cast<type_id>(expression)\n\n ```//基本类型的转换 enum e { A = 1, B, C }; double d = 12.25; unsigned int ui = 25; char c = static_cast(ui); int i = static_cast(d); int j = static_cast(B); //父类子类转换 class F //father { public: int _father; }; class S : public F //son { public: _son; }; F *pFather = new F(); S *pSon = new S(); F *pF; S *pS; pF = static_cast(pSon); //将子类指针转换为父类指针，OK pS = static_cast(pFather); //将父类指针转换为子类指针，错误 ```\n\n ```class F { public: virtual void speak(){}; int i; }; class S : public F { public: void speak() { cout << \"S = \" << _s << endl; } double _s; }; F *pF = new F(); S *pS = static_cast(pF); pS->speak(); S1 *pDS = dynamic_cast(pF); pDS->speak();```\n\ndynamic_cast:\n\ndynamic_cast<type_id>(expression)\n\n1．最常用的用法就是将子类指针转换为父类指针。（不举例）\n\n2．当type_id为void*时，指针指向整个对象的空间。\n\nClass A;\n\nA *pA = new A();\n\nvoid *p = dynamic_cast<void*>(pA);\n\n3．跳级转换。\n\n ```class A{}; class B : public A{}; class C : public B{}; A *pA; B *pB; C *pC = new C(); pB = dynamic_cast(pD); //逐级转换OK pA = dynamic_cast(pB); //逐级转换OK 或者 pA = dynamic_cast(pC); //跳级转换OK delete pD;```\n\n ```class A{}; class B : public A{}; class C : public A{}; class D : public B, public C{}; A *pA; D *pD = new D(); pA = dynamic_cast(pB); //出现错误，是不行的，原因大家都清楚。 class A{}; class B : public A{}; class C : public A{}; class D : public B{}; class E : public C, public D{}; A *pA; B *pB; E *pE = new E(); pB = dynamic_cast(pE); pA = dynamic_cast(pB); //可以 pA = dynamic_cast(pE); //不可以，原因是很简单的。 delete pE;```\n\n4．两个不相干的类之间转换。\n\n ```class A {}; class B {}; A* pa = new A; B* pb = dynamic_cast(pa); // 不可以，没有相互转换的基础 ```\n\nconst_cast:\n\nconst_cast<type_id>(expression)\n\n ```class CCTest { public: void setNumber( int ); void printNumber() const; private: int number; }; void CCTest::setNumber( int num ) { number = num; } void CCTest::printNumber() const { cout << \"/nBefore: \" << number; const_cast< CCTest * >( this )->number--;//这里消除了const的作用 cout << \"/nAfter: \" << number; } int main() { CCTest X; X.setNumber( 8 ); X.printNumber(); } ```\n\nreinterpret_cast:\n\nreinterpret_cast\n\nMSDN的例子显示出来它的强悍，也显示出了他的脆弱。只要你一个不小心就会乱用。\n\n ```#include // Returns a hash code based on an address unsigned short Hash( void *p ) { unsigned int val = reinterpret_cast( p ); return ( unsigned short )( val ^ (val >> 16)); } using namespace std; int main() { int a; for ( int i = 0; i < 20; i++ ) cout << Hash( a + i ) << endl; }```\n\n#1 dynamic_cast的type_id为引用的情况我不准备了解，好像是微软VS2005的特性，而不是标准C++的特性。对于VS6.0我还是感觉比较欣赏的。虽然不如他的新版本那样支持更多的C++特性，但是我自己的感觉是VS的产品在无限的向C#靠拢，无限的向java的易用性靠拢，这样的话C++程序在抑制到别的操作系统时就需要做很大的修改。这也是微软的霸道之处。题外话：微软的vista是一款失败的产品，在vista上微软开发了virtual PC 2007的虚拟机，但是这款产品只支持windows系统的产品安装，对于linux产品他就不支持（只能安装，不能用），因为他不支持24位真彩色。这就说明它的心态是封闭的，而封闭最终导致它的衰败。\n\n#2 safe_cast也是微软的东西，想了解的请参考VS2005的MSDN。\n\n Visual C++编程命名规则 任何时候都适用的20个C++技巧 C语言进阶 串口驱动分析 轻轻松松从C一路走到C++ C++编程思想\n\n C++并发处理+单元测试 C++程序开发 C++高级编程 C/C++开发 C++设计模式 C/C++单元测试\n\n 北京 嵌入式C高质量编程 中国航空 嵌入式C高质量编程 华为 C++高级编程 北京 C++高级编程 丹佛斯 C++高级编程 北大方正 C语言单元测试 罗克韦尔 C++单元测试"
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https://fourvantage.biz/port-flinders/convert-m-to-cm-example.php | [
"",
null,
"Meters to Centimeters conversion Example #1: Convert 1.00 m 2 to cm 2. Solution: In order to solve this problem, you must see that there are TWO sides to the 1.00 m 2 area and that EACH SIDE must be\n\nMath Skills Dimensional Analysis - Texas A&M University\n\nCentimeter to Meter Conversion (cm to m). This example problem illustrates the method to convert centimeters to meters or cm to m. A meters to centimeters example is also given., How to convert feet to inches, miles, and meters easily in Excel? Convert feet to inches, miles, and meters with and meters. For example, we can convert foot.\n\nThis example problem illustrates the method to convert centimeters to meters or cm to m. A meters to centimeters example is also given. Instant free online tool for millimeter to meter conversion or vice versa. Example: convert 15 mm to m: m to cm. cm to km. km to cm. mm to feet.\n\nConversion from metric metres, centimetres and millimetres to inches. Also inches to metres, centimeters or millimeters conversion. This example problem illustrates the method to convert feet to meters, a common length conversion problem.\n\nScale and distance on maps For example, to convert the word scale 4 cm equals 2 km, 100 cm in 1 m cm to m you ÷ by 100 The conversions between Metric and English units are not all exact values (which means they are not defined relationships, like, say, 1 meter = 100 cm.).\n\nThe metric system is also divided into different categories of measurement, example of this. Convert 3 meters to for the conversion factor from m to cm, Quickly convert centimetres into metres (cm to meter) using the online calculator for metric conversions and more.\n\nInstant free online tool for centimeter to meter conversion or vice versa. The centimeter [cm] to meter [m] conversion table and conversion steps are also listed. Scale and distance on maps For example, to convert the word scale 4 cm equals 2 km, 100 cm in 1 m cm to m you ÷ by 100\n\nScale and distance on maps For example, to convert the word scale 4 cm equals 2 km, 100 cm in 1 m cm to m you ÷ by 100 Unit conversion is a multi-step process that involves multiplication or division by a numerical factor, Unit Conversion Example Example. Convert 5 m to ? cm\n\nExample #1: Convert 1.00 m 2 to cm 2. Solution: In order to solve this problem, you must see that there are TWO sides to the 1.00 m 2 area and that EACH SIDE must be Length and distance unit conversion between meter and centimeter, centimeter to meter conversion in batch, m cm conversion chart\n\nFor example, you may measure the Here are some common conversions between units: 1 m = 100 cm = 1,000 mm (millimeters) 1 km (kilometer) = 1,000 m. 1 kg (kilogram To convert from meters to cm, multiply the number of meters \"x\" by 100. For example, to find out how many centimeters there are in a meter and a half,\n\nExample 1. Convert the following measurements. a. 8 cm to mm b. 6 m to cm c. 7 km to m. Solution: Example 2. Convert the following measurements. a. 7 m to mm b. 5 km The metric system is also divided into different categories of measurement, example of this. Convert 3 meters to for the conversion factor from m to cm,\n\nLength and distance unit conversion between meter and centimeter, centimeter to meter conversion in batch, m cm conversion chart Meters to feet and inches conversion. Whilst every effort has been made in building this meters, feet and inches conversion tool, As an example,\n\nLength Unit Converter Measurement conversion J-Z. (4) Convert 50.0 mL to liters. (This is a very common conversion.) (5) What is the density of mercury (13.6 g/cm 3) in units of kg/m 3? We also can use dimensional, How to convert feet to inches, miles, and meters easily in Excel? Convert feet to inches, miles, and meters with and meters. For example, we can convert foot.\n\nConvert cm to meter Conversion of Measurement Units",
null,
"Speed Conversion Calculator Online Calculator Resource. for example I have the M2 area of a room How to convert Meter Squared in linear meters? Pls help, I have a 130sq metres of rooms to lay a hardwood timber flooring., Meters to feet and inches conversion. Whilst every effort has been made in building this meters, feet and inches conversion tool, As an example,.\n\nPRACTICE CONVERSION PROBLEMS Kent State University. How many inches are in 707 cm, free converter for Convert inches to meters Convert meters to inches Convert inches to following examples: 707.0 (cm), Converting between meters and kilometers Course Mathematics Grade Grade 6 Section Conversions (time, length, mass and volume) Outcome Converting between metric units.\n\nConvert mm to m",
null,
"Convert mm to m. Conversion Tables and Methods to Use when converting. Example: To get Centimeters from Meters you multiply the Meters by 10 . (m * 100) Centimeters (cm centimeters to millimeters (cm to mm) conversion 1 centimeter (cm) is equal 10 millimeters (mm) use this converter millimeters to centimeters (mm to cm) conversion.",
null,
"Converting centimeters to meters is relatively easy. All too often, people stumble when squaring their centimeters. These two example problems will show the best way 20/10/2018 · How to Convert Cm to M. The prefix centi- means \"one-hundredth,\" so there are 100 centimeters in every meter. You can use this basic knowledge to easily convert\n\nTo convert from yards to meters, divide by 1.0936133. To convert from inches to meters, \"How to Calculate Linear Meters.\" Sciencing, https: Units & Conversions . Back to Examples (1) Let's convert 250cm to metres. Convert: (a) 180 cm to metres (m) (b) 1.5 minutes to seconds\n\nMetric Area. These are the most Example: How big is this rectangle? It is 2 meters by 3 meters, It is written cm 2. A centimeter is one-hundredth of a meter 20/10/2018 · How to Convert Cm to M. The prefix centi- means \"one-hundredth,\" so there are 100 centimeters in every meter. You can use this basic knowledge to easily convert\n\nConvert to New Units. From: (2000 V/m) * 35 cm * 2 elementary-charges. eV. Example Interpretation; Abbreviations are supported Units & Conversions . Back to Examples (1) Let's convert 250cm to metres. Convert: (a) 180 cm to metres (m) (b) 1.5 minutes to seconds\n\nThe conversions between Metric and English units are not all exact values (which means they are not defined relationships, like, say, 1 meter = 100 cm.). Convert meters to centimeters (m to cm) with the length conversion calculator. Learn a simple meter to centimeter formula that makes it easy.\n\nThis example problem illustrates the method to convert feet to meters, a common length conversion problem. Convert ft and in to cm, convert in to cm and convert cm to in. Convert height measurements between US units and metric units and Height Converter ft to cm and cm\n\nConverting to mm, cm, m, kmConverting mm to cmConvert 4 mm in cm4 mm = 4 × 1 mm = 4 × 1/10 cm = 4/10 cm = 04/10 cm = 0.4 cm∴ 4mm = 0.4 cmConvert 4 mm in cm4 mm Length conversions You can use this (m to feet) Convert centimeters to inches (cm to inches) (for example, 1 kilometer is equal to 1000 meters)\n\nHow many inches are in 707 cm, free converter for Convert inches to meters Convert meters to inches Convert inches to following examples: 707.0 (cm) How many inches are in 707 cm, free converter for Convert inches to meters Convert meters to inches Convert inches to following examples: 707.0 (cm)\n\nFeet (ft) to meters (m) conversion calculator and how to convert. Scale and distance on maps For example, to convert the word scale 4 cm equals 2 km, 100 cm in 1 m cm to m you ÷ by 100\n\nConversion between square meter and centimeter. Square meter Centimeter Calculator; PHP Tutorial. Square meter to Centimeter Calculator: Square meter (m 2 Units & Conversions . Back to Examples (1) Let's convert 250cm to metres. Convert: (a) 180 cm to metres (m) (b) 1.5 minutes to seconds\n\nConversion Tables and Methods to Use when converting. Example: To get Centimeters from Meters you multiply the Meters by 10 . (m * 100) Centimeters (cm Centimeters to Meters (cm to m) conversion calculator for Length conversions with additional tables and formulas.\n\nConvert inches to centimetres calculator. Convert feet and",
null,
"707 cm in inches. Convert 707 centimeters to inches. Centimeters (cm) to feet (ft) conversion calculator and how to convert., Converting to mm, cm, m, kmConverting mm to cmConvert 4 mm in cm4 mm = 4 × 1 mm = 4 × 1/10 cm = 4/10 cm = 04/10 cm = 0.4 cm∴ 4mm = 0.4 cmConvert 4 mm in cm4 mm.\n\nConvert square centimeter to square meter area converter\n\nm to cm Converter Chart- EndMemo. Many home improvement projects require conversions into centimeters, inches and feet. Learn how to convert feet into centimeters and, as necessary, inches into, Length and distance unit conversion between meter and centimeter, centimeter to meter conversion in batch, m cm conversion chart.\n\nTo convert from yards to meters, divide by 1.0936133. To convert from inches to meters, \"How to Calculate Linear Meters.\" Sciencing, https: Centimeters to Meters (cm to m) conversion calculator for Length conversions with additional tables and formulas.\n\nConverting centimeters to meters is relatively easy. All too often, people stumble when squaring their centimeters. These two example problems will show the best way To convert from yards to meters, divide by 1.0936133. To convert from inches to meters, \"How to Calculate Linear Meters.\" Sciencing, https:\n\nScale and distance on maps For example, to convert the word scale 4 cm equals 2 km, 100 cm in 1 m cm to m you ÷ by 100 Convert to kilometers per hour, meters per Calculator Use. For example, to convert from miles per hour to kilometers per hour you would multiply by 0\n\nDemonstrates how to cancel units to convert measurements from one form Converting Units: Examples. so I'm golden. Demonstrates how to cancel units to convert measurements from one form Converting Units: Examples. so I'm golden.\n\n20/10/2018 · How to Convert Cm to M. The prefix centi- means \"one-hundredth,\" so there are 100 centimeters in every meter. You can use this basic knowledge to easily convert Unit conversion: Speed Practice examples Remember that units for speed all look like (!\"#\\$%&’() (!\"#\\$). If you’re converting from one speed unit to another, say\n\nLearn how to create a Length converter with HTML and JavaScript. Example; Convert from Meters to Feet: ft=m*3.2808: Example; Convert from cm to Feet: ft=cm*0 To convert measurements, use the CONVERT function. Example. The example may be =CONVERT(A2,\"cm\",\"in\")\n\nConvert meter to cm: convert meters to centimeters (m = cm), Example of Meter to Feet Conversion Try this converter on your smartphone. Scan QR Code To Open Browser. Quickly convert centimetres into metres (cm to meter) using the online calculator for metric conversions and more.\n\nConverting between meters and kilometers Course Mathematics Grade Grade 6 Section Conversions (time, length, mass and volume) Outcome Converting between metric units Conversion between square meter and centimeter. Square meter Centimeter Calculator; PHP Tutorial. Square meter to Centimeter Calculator: Square meter (m 2\n\n20/10/2018 · How to Convert Cm to M. The prefix centi- means \"one-hundredth,\" so there are 100 centimeters in every meter. You can use this basic knowledge to easily convert How many inches are in 707 cm, free converter for Convert inches to meters Convert meters to inches Convert inches to following examples: 707.0 (cm)\n\nThis is an example of how to convert Inches to cm (Inches to centimeters) using the C programming language, 1 inch = 2.54 cm. file: inches_cm.c #include # Convert from feet and inches to meters. Includes additional conversions of only inches, centimeters, etc.\n\nHow to Convert Cm to M wikiHow",
null,
"Centimeters to Meters (cm to meters) conversion calculator. Converting to mm, cm, m, kmConverting mm to cmConvert 4 mm in cm4 mm = 4 × 1 mm = 4 × 1/10 cm = 4/10 cm = 04/10 cm = 0.4 cm∴ 4mm = 0.4 cmConvert 4 mm in cm4 mm, Converting between meters and kilometers Course Mathematics Grade Grade 6 Section Conversions (time, length, mass and volume) Outcome Converting between metric units.\n\nConvert length units (meters feet inches cm mm etc)",
null,
"Metric System Table mste.illinois.edu. Many home improvement projects require conversions into centimeters, inches and feet. Learn how to convert feet into centimeters and, as necessary, inches into Converting centimeters to meters is relatively easy. All too often, people stumble when squaring their centimeters. These two example problems will show the best way.",
null,
"Instant free online tool for centimeter to meter conversion or vice versa. The centimeter [cm] to meter [m] conversion table and conversion steps are also listed. Example 1. Convert the following measurements. a. 8 cm to mm b. 6 m to cm c. 7 km to m. Solution: Example 2. Convert the following measurements. a. 7 m to mm b. 5 km\n\nUnit conversion: Speed Practice examples Remember that units for speed all look like (!\"#\\$%&’() (!\"#\\$). If you’re converting from one speed unit to another, say Centimeters to Meters (cm to m) conversion calculator for Length conversions with additional tables and formulas.\n\nConverting centimeters to meters is relatively easy. All too often, people stumble when squaring their centimeters. These two example problems will show the best way Converting centimeters to meters squared isn't as simple as converting centimeters to meters, because centimeters and meters squared are two different types of units.\n\nThis is an example of how to convert Inches to cm (Inches to centimeters) using the C programming language, 1 inch = 2.54 cm. file: inches_cm.c #include # Scale and distance on maps For example, to convert the word scale 4 cm equals 2 km, 100 cm in 1 m cm to m you ÷ by 100\n\nQuick Answer: Divide by 100. To do any conversion, you find the relationship between the two units. In this case, 100 cm = 1 m. This gives you two conversion factors centimeters to millimeters (cm to mm) conversion 1 centimeter (cm) is equal 10 millimeters (mm) use this converter millimeters to centimeters (mm to cm) conversion\n\nTo convert measurements, use the CONVERT function. Example. The example may be =CONVERT(A2,\"cm\",\"in\") How to convert feet to inches, miles, and meters easily in Excel? Convert feet to inches, miles, and meters with and meters. For example, we can convert foot\n\nBy definition, 100 ft equals 30.48 m. In practical use, it is common to round the calculated metric number. With that, there are several possibilities. Quickly convert metres into centimetres (m to cm) using the online calculator for metric conversions and more.\n\nThis example problem illustrates the method to convert feet to meters, a common length conversion problem. Unit conversion: Speed Practice examples Remember that units for speed all look like (!\"#\\$%&’() (!\"#\\$). If you’re converting from one speed unit to another, say\n\nExample 1. Convert the following measurements. a. 8 cm to mm b. 6 m to cm c. 7 km to m. Solution: Example 2. Convert the following measurements. a. 7 m to mm b. 5 km By definition, 100 ft equals 30.48 m. In practical use, it is common to round the calculated metric number. With that, there are several possibilities.\n\ncentimeters to millimeters (cm to mm) conversion 1 centimeter (cm) is equal 10 millimeters (mm) use this converter millimeters to centimeters (mm to cm) conversion Units & Conversions . Back to Examples (1) Let's convert 250cm to metres. Convert: (a) 180 cm to metres (m) (b) 1.5 minutes to seconds\n\nConverting centimeters to meters is relatively easy. All too often, people stumble when squaring their centimeters. These two example problems will show the best way This example problem illustrates the method to convert feet to meters, a common length conversion problem.\n\nTraveling Salesman Problem - Genetic I need to solve it with the help of a genetic algorithm in matlab. Please, write the code! For example, if I type in Genetic algorithm matlab code example Lombadina please I need Krill herd algorithms in Matlab . Learn more about optimization, genetic algorithm, evolutionary, doit4me, no attempt"
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https://support.office.com/en-us/article/Video-Use-formulas-to-apply-conditional-formatting-35F92FC6-10FC-46BF-958D-90EC64FACFC8 | [
"Intermediate conditional formatting\n\n# Use formulas",
null,
"To control more precisely what cells will be formatted, you can use formulas to apply conditional formatting.\n\n### Want more?\n\nTo control more precisely what cells will be formatted, you can use formulas to apply conditional formatting.\n\nIn this example, I am going to format the cells in the Product column if the corresponding cell in the In stock column is greater than 300.\n\nI select the cells I want to conditionally format.\n\nWhen you select a range of cells, the first cell you select is the active cell.\n\nIn this example, I selected from B2 through B10, so B2 is the active cell. We’ll need to know that shortly.\n\nCreate a new rule, select Use a formula to determine which cells to format. Since B2 is the active cell, I type =E2>300.\n\nNote that in the formula, I used the relative cell reference E2 to make sure the formula adjusts to correctly format the other cells in column B.\n\nI click the Format button and choose how I want to format the cells. I am going to use a blue fill.\n\nClick OK to accept the color; click OK again to apply the format.\n\nAnd the cells in the Product column, where the corresponding cell in column E is greater than 300, are conditionally formatted.\n\nWhen I change a value in column E to greater than 300, the conditional formatting in the Product column automatically applies.\n\nYou can create multiple rules that apply to the same cells.\n\nIn this example, I want different fill colors for different ranges of scores.\n\nI select the cells I want to apply a rule to, create a new rule that uses the rule type Use a formula to determine which cells to format.\n\nI want to format a cell if its value is greater than or equal to 90.\n\nThe active cell is B2, so I enter the formula =B2>=90.\n\nAnd configure the rule to apply a green fill when the formula is true for a cell. The cell that has a value greater than or equal to 90 is filled with green.\n\nI create another rule for the same cells, but this time, I want to format a cell if its value is greater than or equal to 80 and less than 90.\n\nThe formula is =AND(B2>=80,B2<90).\n\nAnd I choose a different fill color.\n\nI create similar rules for 70 and 60. The last rule is for values less than 60.\n\nThe cells are now a rainbow of colors, and these are the rules we just created to enable this.\n\nUp next, Manage conditional formatting.\n\nExpand your Office skills\nExplore training",
null,
"Get instant Excel help\nConnect to an expert now\nSubject to Got It terms and conditions\n\n×"
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https://www.itcode1024.com/69154/ | [
"# mean-shift算法详解（转）\n\nMeanShift最初由Fukunaga和Hostetler在1975年提出，但是一直到2000左右这篇PAMI的论文Mean Shift: A Robust Approach Toward Feature Space Analysis，将它的原理和收敛性等重新整理阐述，并应用于计算机视觉和图像处理领域之后，才逐渐为人熟知。在了解mean-shift算法之前，先了解一下概率密度估计的概念。\n\n*[En]*\n\n**\n\nmean shift算法使用核函数估计样本密度，假设对于大小为$n$,维度为$d$ 的数据集，$D=\\left{x_{1}, x_{2}, x_{3}, \\ldots x_{n}\\right}, D \\in R^{d}$ ，核函数K的带宽为h，则该函数的核密度估计为：\n\n123456789101112131415161718\n\nimport numpy as npimport math def gaussian_kernel(distance, bandwidth): ''' 高斯核函数 :param distance: 欧氏距离计算函数 :param bandwidth: 核函数的带宽 :return: 高斯函数值 ''' m = np.shape(distance)\n\n*[En]*\n\n**\n\n*[En]*\n\n**\n\nMean Shift算法的基本目标是将样本点向局部密度增加的方向移动，我们常常所说的均值漂移向量就是指局部密度增加最快的方向。上节通过引入高斯核可以知道数据集的密度，梯度是函数增加最快的方向，因此，数据集密度的梯度方向就是密度增加最快的方向。\n\n（1）计算每个样本的均值漂移向量 $m_{h}(\\mathrm{x})$ ;\n（2）对每个样本点以 $m_{h}(\\mathrm{x})$ 进行平移，即：$x=x+m_{h}(x)$ ；\n（3）重复（1）（2），直到样本点收敛，即：\\$m_{h}(\\mathrm{x})\n\nMean-Shift聚类就是对于集合中的每一个元素，对它执行下面的操作：把该元素移动到它邻域中所有元素的特征值的均值的位置，不断重复直到收敛。准确的说，不是真正移动元素，而是把该元素与它的收敛位置的元素标记为同一类。在实际中，为了加速，初始化的时候往往会初始化多个窗口，然后再进行聚类。\n\n*[En]*\n\n**\n\nOriginal: https://www.cnblogs.com/zl1991/p/16539539.html\nAuthor: 鸭子船长\nTitle: mean-shift算法详解（转）"
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http://forums.wolfram.com/mathgroup/archive/2011/Oct/msg00754.html | [
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"Re: List plot with random size for each point\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg122513] Re: List plot with random size for each point\n• From: DrMajorBob <btreat1 at austin.rr.com>\n• Date: Sun, 30 Oct 2011 04:24:58 -0500 (EST)\n• Delivered-to: [email protected]\n• References: <[email protected]>\n\n```Actually, my second solution DOES work for 4 points. Three points is the\ncase that fails.\n\nI suspect it's more efficient to use Replace with a level spec, rather\nthan restricting the pattern:\n\nnumberOfPoints = 3;\nparticle =\nTable[{n,\nRandomInteger[{15, 25}], {RandomInteger[{0, 200}],\nRandomInteger[{0, 200}]}}, {n, numberOfPoints}];\nGraphics[Replace[\nparticle, {label_, size_, pt_} :> {AbsolutePointSize@size, Point@pt,\nWhite, Text[label, pt]}, {1}], Frame -> True,\nPlotRange -> {{-12, 213}, {-12, 213}}, ImageSize -> 500]\n\nBobby\n\nOn Sat, 29 Oct 2011 22:59:09 -0500, Bob Hanlon <hanlonr357 at gmail.com>\nwrote:\n\n> A good approach; however, as written they will fail for numberOfPoints > 4\n>\n> Recommend that you qualify one or more patterns to being numeric, e.g.,\n>\n> Graphics[particle /. {label_?NumericQ, size_, x_,\n> y_} :> {AbsolutePointSize@size, Point@{x, y}, White,\n> Text[label, {x, y}]}, Frame -> True,\n> PlotRange -> {{-12, 213}, {-12, 213}}, ImageSize -> 500]\n>\n>\n> Bob Hanlon\n>\n>\n> On Sat, Oct 29, 2011 at 1:55 PM, DrMajorBob <btreat1 at austin.rr.com>\n> wrote:\n>> Or equivalently,\n>>\n>> numberOfPoints = 20;\n>> particle =\n>> Table[{n, RandomInteger[{15, 25}], RandomInteger[{0, 200}],\n>> RandomInteger[{0, 200}]}, {n, numberOfPoints}];\n>> Graphics[particle /. {label_, size_, x_,\n>> y_} :> {AbsolutePointSize@size, Point@{x, y}, White,\n>> Text[label, {x, y}]}, Frame -> True,\n>> PlotRange -> {{-12, 213}, {-12, 213}}, ImageSize -> 500]\n>>\n>> or\n>>\n>> numberOfPoints = 20;\n>> particle =\n>> Table[{n,\n>> RandomInteger[{15, 25}], {RandomInteger[{0, 200}],\n>> RandomInteger[{0, 200}]}}, {n, numberOfPoints}];\n>> Graphics[particle /. {label_, size_,\n>> pt_List} :> {AbsolutePointSize@size, Point@pt, White,\n>> Text[label, pt]}, Frame -> True,\n>> PlotRange -> {{-12, 213}, {-12, 213}}, ImageSize -> 500]\n>>\n>> I find those more transparent.\n>>\n>> Bobby\n>>\n>> On Sat, 29 Oct 2011 06:13:56 -0500, Bob Hanlon <hanlonr357 at gmail.com>\n>> wrote:\n>>\n>>> numberOfPoints = 20;\n>>>\n>>> particle = Table[\n>>> {n, RandomInteger[{15, 25}],\n>>> RandomInteger[{0, 200}], RandomInteger[{0, 200}]},\n>>> {n, numberOfPoints}];\n>>>\n>>> Graphics[\n>>> {AbsolutePointSize[#[]], Point[#[[{3, 4}]]],\n>>> White, Text[#[], #[[{3, 4}]]]} & /@\n>>> particle,\n>>> Frame -> True,\n>>> PlotRange -> {{-12, 213}, {-12, 213}},\n>>> ImageSize -> 500]\n>>>\n>>>\n>>> Bob Hanlon\n>>>\n>>>\n>>> On Fri, Oct 28, 2011 at 5:36 AM, Rudresh <rudresh at email.unc.edu> wrote:\n>>>>\n>>>> I want to plot an array of numbers on a 2d plot, however each point\n>>>> has a\n>>>> random number associated with it which should be its size on the\n>>>> plot. Can\n>>>> this be done in mathematica and if so how. My code as of now is:\n>>>>\n>>>> =========================\n>>>>\n>>>> particle = Array[f, {200, 4}];\n>>>> position = Array[g, {200, 2}];\n>>>> n = 1;\n>>>> While [n < 201,\n>>>> particle[[n, 1]] = n;\n>>>> particle[[n, 2]] = RandomInteger[{15, 25}];\n>>>> particle[[n, 3]] = RandomInteger[{0, 200}];\n>>>> particle[[n, 4]] = RandomInteger[{0, 200}];\n>>>>\n>>>> position[[n, 1]] = particle[[n, 3]];\n>>>> position[[n, 2]] = particle[[n, 4]];\n>>>> n++\n>>>> ];\n>>>>\n>>>> ListPlot[position, PlotStyle -> PointSize[0.01],\n>>>> PlotRange -> {{0, 200}, {0, 200}}]\n>>>>\n>>>> =========================\n>>>>\n>>>> I want particle[[n,2]] to be the size of position[[n,1],[n,2]]\n>>>>\n>>>\n>>\n>>\n>> --\n>> DrMajorBob at yahoo.com\n>>\n\n--\nDrMajorBob at yahoo.com\n\n```\n\n• Prev by Date: Re: Listable parallel basic LibraryLink dll\n• Next by Date: Re: delay evaluation inside RecurrencTable?\n• Previous by thread: Re: List plot with random size for each point\n• Next by thread: Re: List plot with random size for each point"
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https://www.originlab.com/doc/COM/Data-Import-to-Origin-Workbook | [
"# 1.4.1 Data Import to Origin Workbook\n\n## pCLAMP Import using X-Function\n\nThe following example shows how to call X-Function from COM.\n\nPrivate Sub XF_Click()\nDim app As Origin.Application\nSet app = New Origin.ApplicationSI\n'before we can run OC (XF is OC), we need to make sure OC startup compile/link is done\napp.Execute (\"sec -poc 3\") 'wait up to 3 sec, actually will be far shorter\napp.NewProject\nDim wbk As Origin.WorksheetPage\n\nDim strFileName As String\n'replace with your actual pCLAMP file name\nstrFileName$= \"D:\\Origin61\\pClamp\\96322002.ABF\" 'put filename into Origin project variable fname$ as all import XF assumes that\n'and this makes it easy to call the XF without passing it on the command line\napp.LTStr(\"fname$\") = strFileName$\n'run pCLAMP import XF and turn off plotting XFbar\nwbk.Execute (\"imppClamp options.XFBar:=0\")\nDim wks As Origin.Worksheet\nDim Col As Origin.Column\nSet wks = wbk.Layers(0) 'only one sheet(layer) for pCLAMP import\nSet Col = wks.Columns(0) 'get the first column\nCells(1, 1) = Col.LongName 'put the long name to Excel top-left cell\nCells(2, 1) = Col.Units\nCells(3, 1) = Col.EvenSampling 'Sampling interval\napp.Visible = MAINWND_SHOW\nEnd Sub"
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https://it.mathworks.com/help/ident/ref/arx.html | [
"Documentation\n\n### This is machine translation\n\nMouseover text to see original. Click the button below to return to the English version of the page.\n\n# arx\n\nEstimate parameters of ARX or AR model using least squares\n\n## Syntax\n\n```sys = arx(data,[na nb nk]) sys = arx(data,[na nb nk],Name,Value) sys = arx(data,[na nb nk],___,opt) ```\n\n## Description\n\n### Note\n\n`arx` does not support continuous-time estimations. Use `tfest` instead.\n\n```sys = arx(data,[na nb nk])``` returns an ARX structure polynomial model, `sys`, with estimated parameters and covariances (parameter uncertainties) using the least-squares method and specified `orders`.\n\n```sys = arx(data,[na nb nk],Name,Value)``` estimates a polynomial model with additional options specified by one or more `Name,Value` pair arguments.\n\n```sys = arx(data,[na nb nk],___,opt)``` specifies estimation options that configure the estimation objective, initial conditions and handle input/output data offsets.\n\n## Input Arguments\n\n `data` Estimation data. Specify `data` as an `iddata` object, an `frd` object, or an `idfrd` frequency-response-data object. `[na nb nk]` Polynomial orders. `[na nb nk]` define the polynomial orders of an ARX model. `na` — Order of the polynomial A(q). Specify `na` as an Ny-by-Ny matrix of nonnegative integers. Ny is the number of outputs.`nb` — Order of the polynomial B(q) + 1.`nb` is an Ny-by-Nu matrix of nonnegative integers. Ny is the number of outputs and Nu is the number of inputs.`nk` — Input-output delay expressed as fixed leading zeros of the B polynomial. Specify `nk` as an Ny-by-Nu matrix of nonnegative integers. Ny is the number of outputs and Nu is the number of inputs. `opt` Estimation options. `opt` is an options set that specifies estimation options, including: input/output data offsetsoutput weight Use `arxOptions` to create the options set.\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\n `'InputDelay'` Input delays. `InputDelay` is a numeric vector specifying a time delay for each input channel. Specify input delays in integer multiples of the sample time `Ts`. For example, `InputDelay = 3` means a delay of three sampling periods. For a system with `Nu` inputs, set `InputDelay` to an `Nu`-by-1 vector, where each entry is a numerical value representing the input delay for the corresponding input channel. You can also set `InputDelay` to a scalar value to apply the same delay to all channels. Default: 0 for all input channels `'IODelay'` Transport delays. `IODelay` is a numeric array specifying a separate transport delay for each input/output pair. Specify transport delays as integers denoting delay of a multiple of the sample time, `Ts`. For a MIMO system with `Ny` outputs and `Nu` inputs, set `IODelay` to a `Ny`-by-`Nu` array, where each entry is a numerical value representing the transport delay for the corresponding input/output pair. You can also set `IODelay` to a scalar value to apply the same delay to all input/output pairs. Useful as a replacement for the `nk` order, you can factor out `max(nk-1,0)` lags as the `IODelay` value. Default: 0 for all input/output pairs `'IntegrateNoise'` Specify integrators in the noise channels. Adding an integrator creates an ARIX model represented by: `$A\\left(q\\right)y\\left(t\\right)=B\\left(q\\right)u\\left(t-nk\\right)+\\frac{1}{1-{q}^{-1}}e\\left(t\\right)$` where,$\\frac{1}{1-{q}^{-1}}$ is the integrator in the noise channel, e(t). `IntegrateNoise` is a logical vector of length `Ny`, where `Ny` is the number of outputs. Default: `false(Ny,1)`, where `Ny` is the number of outputs\n\n## Output Arguments\n\n`sys`\n\nARX model that fits the estimation data, returned as a discrete-time `idpoly` object. This model is created using the specified model orders, delays, and estimation options.\n\nInformation about the estimation results and options used is stored in the `Report` property of the model. `Report` has the following fields:\n\nReport FieldDescription\n`Status`\n\nSummary of the model status, which indicates whether the model was created by construction or obtained by estimation.\n\n`Method`\n\nEstimation command used.\n\n`InitialCondition`\n\nHandling of initial conditions during model estimation, returned as one of the following values:\n\n• `'zero'` — The initial conditions were set to zero.\n\n• `'estimate'` — The initial conditions were treated as independent estimation parameters.\n\nThis field is especially useful to view how the initial conditions were handled when the `InitialCondition` option in the estimation option set is `'auto'`.\n\n`Fit`\n\nQuantitative assessment of the estimation, returned as a structure. See Loss Function and Model Quality Metrics for more information on these quality metrics. The structure has the following fields:\n\nFieldDescription\n`FitPercent`\n\nNormalized root mean squared error (NRMSE) measure of how well the response of the model fits the estimation data, expressed as a percentage.\n\n`LossFcn`\n\nValue of the loss function when the estimation completes.\n\n`MSE`\n\nMean squared error (MSE) measure of how well the response of the model fits the estimation data.\n\n`FPE`\n\nFinal prediction error for the model.\n\n`AIC`\n\nRaw Akaike Information Criteria (AIC) measure of model quality.\n\n`AICc`\n\nSmall sample-size corrected AIC.\n\n`nAIC`\n\nNormalized AIC.\n\n`BIC`\n\nBayesian Information Criteria (BIC).\n\n`Parameters`\n\nEstimated values of model parameters.\n\n`OptionsUsed`\n\nOption set used for estimation. If no custom options were configured, this is a set of default options. See `arxOptions` for more information.\n\n`RandState`\n\nState of the random number stream at the start of estimation. Empty, `[]`, if randomization was not used during estimation. For more information, see `rng` in the MATLAB® documentation.\n\n`DataUsed`\n\nAttributes of the data used for estimation, returned as a structure with the following fields:\n\nFieldDescription\n`Name`\n\nName of the data set.\n\n`Type`\n\nData type.\n\n`Length`\n\nNumber of data samples.\n\n`Ts`\n\nSample time.\n\n`InterSample`\n\nInput intersample behavior, returned as one of the following values:\n\n• `'zoh'` — Zero-order hold maintains a piecewise-constant input signal between samples.\n\n• `'foh'` — First-order hold maintains a piecewise-linear input signal between samples.\n\n• `'bl'` — Band-limited behavior specifies that the continuous-time input signal has zero power above the Nyquist frequency.\n\n`InputOffset`\n\nOffset removed from time-domain input data during estimation. For nonlinear models, it is `[]`.\n\n`OutputOffset`\n\nOffset removed from time-domain output data during estimation. For nonlinear models, it is `[]`.\n\nFor more information on using `Report`, see Estimation Report.\n\n## Examples\n\ncollapse all\n\nGenerate input data based on a specified ARX model, and then use this data to estimate an ARX model.\n\n```A = [1 -1.5 0.7]; B = [0 1 0.5]; m0 = idpoly(A,B); u = iddata([],idinput(300,'rbs')); e = iddata([],randn(300,1)); y = sim(m0,[u e]); z = [y,u]; m = arx(z,[2 2 1]);```\n\nUse `arxRegul` to automatically determine regularization constants and use the values for estimating an FIR model of order 50.\n\nObtain `L` and `R` values.\n\n```load regularizationExampleData eData; orders = [0 50 0]; [L,R] = arxRegul(eData,orders);```\n\nBy default, the `TC` kernel is used.\n\nUse the returned `Lambda` and `R` values for regularized ARX model estimation.\n\n```opt = arxOptions; opt.Regularization.Lambda = L; opt.Regularization.R = R; model = arx(eData,orders,opt);```\n\ncollapse all\n\n### ARX structure\n\nThe ARX model structure is :\n\n`$\\begin{array}{l}y\\left(t\\right)+{a}_{1}y\\left(t-1\\right)+...+{a}_{na}y\\left(t-na\\right)=\\\\ {b}_{1}u\\left(t-nk\\right)+...+{b}_{nb}u\\left(t-nb-nk+1\\right)+e\\left(t\\right)\\end{array}$`\n\nThe parameters `na` and `nb` are the orders of the ARX model, and `nk` is the delay.\n\n• $y\\left(t\\right)$— Output at time $t$.\n\n• ${n}_{a}$ — Number of poles.\n\n• ${n}_{b}$ — Number of zeroes plus 1.\n\n• ${n}_{k}$ — Number of input samples that occur before the input affects the output, also called the dead time in the system.\n\n• $y\\left(t-1\\right)\\dots y\\left(t-{n}_{a}\\right)$ — Previous outputs on which the current output depends.\n\n• $u\\left(t-{n}_{k}\\right)\\dots u\\left(t-{n}_{k}-{n}_{b}+1\\right)$ — Previous and delayed inputs on which the current output depends.\n\n• $e\\left(t\\right)$ — White-noise disturbance value.\n\nA more compact way to write the difference equation is\n\n`$A\\left(q\\right)y\\left(t\\right)=B\\left(q\\right)u\\left(t-{n}_{k}\\right)+e\\left(t\\right)$`\n\nq is the delay operator. Specifically,\n\n`$A\\left(q\\right)=1+{a}_{1}{q}^{-1}+\\dots +{a}_{{n}_{a}}{q}^{-{n}_{a}}$`\n`$B\\left(q\\right)={b}_{1}+{b}_{2}{q}^{-1}+\\dots +{b}_{{n}_{b}}{q}^{-{n}_{b}+1}$`\n\n### Time Series Models\n\nFor time-series data that contains no inputs, one output and ```orders = na```, the model has AR structure of order `na`.\n\nThe AR model structure is\n\n`$A\\left(q\\right)y\\left(t\\right)=e\\left(t\\right)$`\n\n### Multiple Inputs and Single-Output Models\n\nFor multiple-input systems, `nb` and `nk` are row vectors where the `i`th element corresponds to the order and delay associated with the `i`th input.\n\n### Multi-Output Models\n\nFor models with multiple inputs and multiple outputs, `na`, `nb`, and `nk` contain one row for each output signal.\n\nIn the multiple-output case, `arx` minimizes the trace of the prediction error covariance matrix, or the norm\n\n`$\\sum _{t=1}^{N}{e}^{T}\\left(t\\right)e\\left(t\\right)$`\n\nTo transform this to an arbitrary quadratic norm using a weighting matrix `Lambda`\n\n`$\\sum _{t=1}^{N}{e}^{T}\\left(t\\right){\\Lambda }^{-1}e\\left(t\\right)$`\n\nuse the syntax\n\n```opt = arxOptions('OutputWeight',inv(lambda)) m = arx(data,orders,opt) ```\n\n### Estimating Initial Conditions\n\nFor time-domain data, the signals are shifted such that unmeasured signals are never required in the predictors. Therefore, there is no need to estimate initial conditions.\n\nFor frequency-domain data, it might be necessary to adjust the data by initial conditions that support circular convolution.\n\nSet the `InitialCondition` estimation option (see `arxOptions`) to one the following values:\n\n• `'zero'` — No adjustment.\n\n• `'estimate'` — Perform adjustment to the data by initial conditions that support circular convolution.\n\n• `'auto'` — Automatically choose between `'zero'` and `'estimate'` based on the data.\n\n## Algorithms\n\nQR factorization solves the overdetermined set of linear equations that constitute the least-squares estimation problem.\n\nWithout regularization, the ARX model parameters vector θ is estimated by solving the normal equation:\n\n$\\left({J}^{T}J\\right)\\theta ={J}^{T}y$\n\nwhere J is the regressor matrix and y is the measured output. Therefore,\n\n$\\theta ={\\left({J}^{T}J\\right)}^{-1}{J}^{T}y$.\n\nUsing regularization adds a regularization term:\n\n$\\theta ={\\left({J}^{T}J+\\lambda R\\right)}^{-1}{J}^{T}y$\n\nwhere, λ and R are the regularization constants. See `arxOptions` for more information on the regularization constants.\n\nWhen the regression matrix is larger than the `MaxSize` specified in `arxOptions`, data is segmented and QR factorization is performed iteratively on these data segments.",
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https://wiki.alquds.edu/?query=Leverage_(finance) | [
"# Leverage (finance)\n\nIn finance, leverage (or gearing in the United Kingdom and Australia) is any technique involving borrowing funds to buy things, estimating that future profits will be many times more than the cost of borrowing. This technique is named after a lever in physics, which amplifies a small input force into a greater output force, because successful leverage amplifies the smaller amounts of money needed for borrowing into large amounts of profit. However, the technique also involves the high risk of not being able to pay back a large loan. Normally, a lender will set a limit on how much risk it is prepared to take and will set a limit on how much leverage it will permit, and would require the acquired asset to be provided as collateral security for the loan.\n\nLeveraging enables gains to be multiplied. On the other hand, losses are also multiplied, and there is a risk that leveraging will result in a loss if financing costs exceed the income from the asset, or the value of the asset falls.\n\n## Sources\n\nLeverage can arise in a number of situations, such as:\n\n• securities like options and futures are effectively bets between parties where the principal is implicitly borrowed/lent at interest rates of very short treasury bills.\n• equity owners of businesses leverage their investment by having the business borrow a portion of its needed financing. The more it borrows, the less equity it needs, so any profits or losses are shared among a smaller base and are proportionately larger as a result.\n• businesses leverage their operations by using fixed cost inputs when revenues are expected to be variable. An increase in revenue will result in a larger increase in operating profit.\n• hedge funds may leverage their assets by financing a portion of their portfolios with the cash proceeds from the short sale of other positions.\n\n## Risk\n\nWhile leverage magnifies profits when the returns from the asset more than offset the costs of borrowing, leverage may also magnify losses. A corporation that borrows too much money might face bankruptcy or default during a business downturn, while a less-leveraged corporation might survive. An investor who buys a stock on 50% margin will lose 40% if the stock declines 20%.; also in this case the involved subject might be unable to refund the incurred significant total loss.\n\nRisk may depend on the volatility in value of collateral assets. Brokers may demand additional funds when the value of securities held declines. Banks may decline to renew mortgages when the value of real estate declines below the debt's principal. Even if cash flows and profits are sufficient to maintain the ongoing borrowing costs, loans may be called-in.\n\nThis may happen exactly at a time when there is little market liquidity, i.e. a paucity of buyers, and sales by others are depressing prices. It means that as market price falls, leverage goes up in relation to the revised equity value, multiplying losses as prices continue to go down. This can lead to rapid ruin, for even if the underlying asset value decline is mild or temporary the debt-financing may be only short-term, and thus due for immediate repayment. The risk can be mitigated by negotiating the terms of leverage, by maintaining unused capacity for additional borrowing, and by leveraging only liquid assets which may rapidly be converted to cash.\n\nThere is an implicit assumption in that account, however, which is that the underlying leveraged asset is the same as the unleveraged one. If a company borrows money to modernize, add to its product line or expand internationally, the extra trading profit from the additional diversification might more than offset the additional risk from leverage. Or if an investor uses a fraction of his or her portfolio to margin stock index futures (high risk) and puts the rest in a low-risk money-market fund, he or she might have the same volatility and expected return as an investor in an unlevered low-risk equity-index fund. Or if both long and short positions are held by a pairs-trading stock strategy the matching and off-setting economic leverage may lower overall risk levels.\n\nSo while adding leverage to a given asset always adds risk, it is not the case that a levered company or investment is always riskier than an unlevered one. In fact, many highly levered hedge funds have less return volatility than unlevered bond funds, and normally heavily indebted low-risk public utilities are usually less risky stocks than unlevered high-risk technology companies.\n\n## Measuring\n\nA good deal of confusion arises in discussions among people who use different definitions of leverage. The term is used differently in investments and corporate finance, and has multiple definitions in each field.\n\n### Investments\n\nAccounting leverage is total assets divided by the total assets minus total liabilities. Notional leverage is total notional amount of assets plus total notional amount of liabilities divided by equity. Economic leverage is volatility of equity divided by volatility of an unlevered investment in the same assets. To understand the differences, consider the following positions, all funded with $100 of cash equity: • Buy$100 of crude oil with money out of pocket. Assets are $100 ($100 of oil), there are no liabilities, and assets minus liabilities equals owners' equity. Accounting leverage is 1 to 1. The notional amount is $100 ($100 of oil), there are no liabilities, and there is $100 of equity, so notional leverage is 1 to 1. The volatility of the equity is equal to the volatility of oil, since oil is the only asset and you own the same amount as your equity, so economic leverage is 1 to 1. • Borrow$100 and buy $200 of crude oil. Assets are$200, liabilities are $100 so accounting leverage is 2 to 1. The notional amount is$200 and equity is $100, so notional leverage is 2 to 1. The volatility of the position is twice the volatility of an unlevered position in the same assets, so economic leverage is 2 to 1. • Buy$100 of a 10-year fixed-rate treasury bond, and enter into a fixed-for-floating 10-year interest rate swap to convert the payments to floating rate. The derivative is off-balance sheet, so it is ignored for accounting leverage. Accounting leverage is therefore 1 to 1. The notional amount of the swap does count for notional leverage, so notional leverage is 2 to 1. The swap removes most of the economic risk of the treasury bond, so economic leverage is near zero.\n\n### Corporate finance\n\n$\\mathrm {DOL} ={\\frac {\\mathrm {EBIT\\;+\\;Fixed\\;Costs} }{\\mathrm {EBIT} }}$",
null,
"$\\mathrm {DFL} ={\\frac {\\mathrm {EBIT} }{\\mathrm {EBIT} -{\\text{Total Interest Expense}}}}$",
null,
"$\\mathrm {DCL} ={\\text{DOL}}\\times {\\text{DFL}}={\\frac {\\mathrm {EBIT} +{\\text{Fixed Costs}}}{\\mathrm {EBIT} -{\\text{Total Interest Expense}}}}$",
null,
"Accounting leverage has the same definition as in investments. There are several ways to define operating leverage, the most common. is:\n\n{\\begin{aligned}{\\text{Operating leverage}}&={\\frac {{\\text{Revenue}}-{\\text{Variable Cost}}}{{\\text{Revenue}}-{\\text{Variable Cost}}-{\\text{Fixed Cost}}}}\\\\[8pt]&={\\frac {{\\text{Revenue}}-{\\text{Variable Cost}}}{\\text{Operating Income}}}\\end{aligned}}",
null,
"Financial leverage is usually defined as:\n\n${\\text{Financial leverage}}={\\frac {\\text{Total Debt}}{\\text{Shareholders' Equity}}}$",
null,
"For outsiders, it is hard to calculate operating leverage as fixed and variable costs are usually not disclosed. In an attempt to estimate operating leverage, one can use the percentage change in operating income for a one-percent change in revenue. The product of the two is called Total leverage, and estimates the percentage change in net income for a one-percent change in revenue.\n\nThere are several variants of each of these definitions, and the financial statements are usually adjusted before the values are computed. Moreover, there are industry-specific conventions that differ somewhat from the treatment above.\n\n## Bank regulation\n\nBefore the 1980s, quantitative limits on bank leverage were rare. Banks in most countries had a reserve requirement, a fraction of deposits that was required to be held in liquid form, generally precious metals or government notes or deposits. This does not limit leverage. A capital requirement is a fraction of assets that is required to be funded in the form of equity or equity-like securities. Although these two are often confused, they are in fact opposite. A reserve requirement is a fraction of certain liabilities (from the right hand side of the balance sheet) that must be held as a certain kind of asset (from the left hand side of the balance sheet). A capital requirement is a fraction of assets (from the left hand side of the balance sheet) that must be held as a certain kind of liability or equity (from the right hand side of the balance sheet). Before the 1980s, regulators typically imposed judgmental capital requirements, a bank was supposed to be \"adequately capitalized,\" but these were not objective rules.\n\nNational regulators began imposing formal capital requirements in the 1980s, and by 1988 most large multinational banks were held to the Basel I standard. Basel I categorized assets into five risk buckets, and mandated minimum capital requirements for each. This limits accounting leverage. If a bank is required to hold 8% capital against an asset, that is the same as an accounting leverage limit of 1/.08 or 12.5 to 1.\n\nWhile Basel I is generally credited with improving bank risk management it suffered from two main defects. It did not require capital for all off-balance sheet risks (there was a clumsy provisions for derivatives, but not for certain other off-balance sheet exposures) and it encouraged banks to pick the riskiest assets in each bucket (for example, the capital requirement was the same for all corporate loans, whether to solid companies or ones near bankruptcy, and the requirement for government loans was zero).\n\nWork on Basel II began in the early 1990s and it was implemented in stages beginning in 2005. Basel II attempted to limit economic leverage rather than accounting leverage. It required advanced banks to estimate the risk of their positions and allocate capital accordingly. While this is much more rational in theory, it is more subject to estimation error, both honest and opportunitistic. The poor performance of many banks during the financial crisis of 2007–2009 led to calls to reimpose leverage limits, by which most people meant accounting leverage limits, if they understood the distinction at all. However, in view of the problems with Basel I, it seems likely that some hybrid of accounting and notional leverage will be used, and the leverage limits will be imposed in addition to, not instead of, Basel II economic leverage limits.\n\n## Financial crisis of 2007–2008\n\nThe financial crisis of 2007–2008, like many previous financial crises, was blamed in part on \"excessive leverage\".\n\n• Consumers in the United States and many other developed countries had high levels of debt relative to their wages, and relative to the value of collateral assets. When home prices fell, and debt interest rates reset higher, and business laid off employees, borrowers could no longer afford debt payments, and lenders could not recover their principal by selling collateral.\n• Financial institutions were highly levered. Lehman Brothers, for example, in its last annual financial statements, showed accounting leverage of 31.4 times ($691 billion in assets divided by$22 billion in stockholders' equity). Bankruptcy examiner Anton R. Valukas determined that the true accounting leverage was higher: it had been understated due to dubious accounting treatments including the so-called repo 105 (allowed by Ernst & Young).\n• Banks' notional leverage was more than twice as high, due to off-balance sheet transactions. At the end of 2007, Lehman had $738 billion of notional derivatives in addition to the assets above, plus significant off-balance sheet exposures to special purpose entities, structured investment vehicles and conduits, plus various lending commitments, contractual payments and contingent obligations. • On the other hand, almost half of Lehman's balance sheet consisted of closely offsetting positions and very-low-risk assets, such as regulatory deposits. The company emphasized \"net leverage\", which excluded these assets. On that basis, Lehman held$373 billion of \"net assets\" and a \"net leverage ratio\" of 16.1. This is not a standardized computation, but it probably corresponds more closely to what most people think of when they hear of a leverage ratio.[citation needed]\n\n## Use of language\n\nLevering has come to be known as \"leveraging\", in financial communities; this may have originally been a slang adaptation, since leverage was a noun. However, modern dictionaries (such as Random House Dictionary and Merriam-Webster's Dictionary of Law) refer to its use as a verb, as well. It was first adopted for use as a verb in American English in 1957."
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/ebee1a2da4cfb76c413924f379ef70d0f4c63bb8",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/8220a0ad99080f624a6f906982d8d5e4d0fe7816",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/6c82358e5fad5e47a885501058dea1aa85e93227",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/fd87645b917a7289b451cc9f528028733c0beec3",
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"https://wikimedia.org/api/rest_v1/media/math/render/svg/60071b2af43c62a02082762e5c045b925aa375eb",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.929042,"math_prob":0.88318175,"size":16096,"snap":"2023-14-2023-23","text_gpt3_token_len":3595,"char_repetition_ratio":0.12801392,"word_repetition_ratio":0.04593227,"special_character_ratio":0.23384692,"punctuation_ratio":0.12850468,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.951988,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T00:29:31Z\",\"WARC-Record-ID\":\"<urn:uuid:9136206c-f8a1-48b8-95ac-b9781082059b>\",\"Content-Length\":\"116028\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:397b6fce-0015-4157-8f71-85554c3ce36b>\",\"WARC-Concurrent-To\":\"<urn:uuid:13eb91d2-df68-4232-83cf-a49a1f9a9fd9>\",\"WARC-IP-Address\":\"195.114.29.189\",\"WARC-Target-URI\":\"https://wiki.alquds.edu/?query=Leverage_(finance)\",\"WARC-Payload-Digest\":\"sha1:6KMSEPGZNZWY7YGRMD7OUCU2YIC47YHS\",\"WARC-Block-Digest\":\"sha1:IP73HC5BU57Z54XESNKII66IVWTX4OBD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224650409.64_warc_CC-MAIN-20230604225057-20230605015057-00550.warc.gz\"}"} |
https://www.learnmathsonline.org/tag/class-7-maths-model-questions/ | [
"Tagged: class 7 maths model questions\n\nCBSE Class 7 Mathematics Worksheets/Part 5/Yearly Examination\n\nCBSE Class 7 Mathematics Worksheets is about some model questions that you can expect for yearly examination. Here you can find out practice problems for class 7 Mathematics. CBSE CLASS 7 MATHEMATICS WORKSHEETS/Part 5/Yearly...\n\nCBSE Class 7 Mathematics Worksheets/Part 3/Yearly Examination\n\nCBSE Class 7 Mathematics Worksheets is about some model questions that you can expect for yearly examination. Here you can find out practice problems for class 7 Mathematics. CBSE CLASS 7 MATHEMATICS WORKSHEETS /Part...\n\nCBSE Class 7 Maths Model Question Paper 2\n\nHere is a model question paper for NCERT class 7 mathematics. You can find another class 7 sample paper here. CBSE MATHS / SAMPLE PAPER/MODEL QUESTION PAPER CLASS 7 – MATHEMATICS | YEARLY EXAMINATION Time:...\n\nClass 7 CBSE Maths Model Question Paper\n\nHere is a model question paper for class 7 CBSE mathematics."
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http://integrated-info.com/Electronic-Components-Supplies/impedance.html | [
"# Impedance\n\nOhm's law describes the relationship between current and voltage in circuits that are in equilibrium- that is, when the current and voltage are not changing. When we have a situation where the current changes (often called an AC circuit) more factors have to be taken into account.\n\n## Reactance\n\nThere are devices that oppose any change in current flow. They are not noticed until the voltage changes, but when it does, these gadgets show some surprising properties, soaking up current and giving it back later, so that Ohm's law calculations come out wrong. The property of opposing change is called reactance. It is also measured in ohms.\n\n### Capacitors\n\nIf you make a sandwich out of two metal plates and a piece of glass, you have made a capacitor. If you apply a positive voltage to one plate and a negative voltage to the other, current will flow for a while because the glass can store some electrons. This will stop eventually, as the glass absorbs as many electrons as it can. At this point we say the capacitor is fully charged, and a voltmeter connected between the two plates would show a reading close to that of whatever originally provided the current. If you then connect the two metal plates together, current will flow the opposite direction as the capacitor discharges.\n\nThe current flow is not steady throughout this process. Starting from the discharged stage, current flows strongly at first, but slows down as the voltage across the capacitor approaches the charging voltage. Likewise, when discharged, current flows strongly at first, then tapers off as the charge approaches zero. Any resistance between the charging source and the discharged capacitor will limit the initial charging current- as the capacitor charges the voltage across the resistor is reduced (it's the difference between the voltage source and the rising voltage of the capacitor plate.) The resistor obeys ohm's law, so the current into the capacitor ( and apparently out the other side) dwindles in the gradual curve shown here:\n\nCurrent as capacitor charges\n\nThis means the voltage across the capacitor changes in a curve too:\n\nVoltage across capacitor as it charges.\n\nThe time it takes this to happen is determined by the resistance the current must pass through and the size and material of the capacitor. Since it changes very slowly at the end, it is impossible to find the time the capacitor is 100% charged. In fact it never really gets there. A \"time constant\" is defined as the time it takes to get to 63% of full charge. A value for measuring the size of the capacitor (called capacitance) is then defined by the formula\n\nCapacitance is measured in \"farads\", and a one farad capacitor in series with a one ohm resistor has a time constant of one second. In real life, we deal with large resistances and pretty short times, so the capacitors in most circuits have values in the microfarad range. (That's 10-6 farad.)\n\nIf you connect two capacitors in parallel, you make a bigger capacitor, and their values add:\n\nIf they are connected in series, you get this:\n\nAC and the capacitor\n\nNow imagine charging and discharging the capacitor very quickly- we could do this by using a tone generator instead of a battery as the voltage source.\n\nIf we start with a high frequency and watch the current though the circuit, it's almost as if the capacitor weren't there at all! That's because the current is highest early in the charge cycle, and if the current source changes direction much faster than a time constant, it's always early in the charge cycle. If the frequency is reduced, the current amplitude decreases- to the point where there's nothing but a slight ripple in a steady value.\n\nThere's another important thing to notice here: the current is 90° out of phase with the voltage, the current leading.\n\nAs you can see, we have a situation where Ohm's law doesn't tell the entire story. The current through a capacitor is dependent on the frequency of the signal. Frequency dependent opposition to current is reactance, which is indicated in formulas by the letter X. Capacitive reactance is found with the formula:\n\nX is reactance in ohms.\n\nF is frequency in hertz.\n\nSince the frequency term is in the bottom of the fraction, you can see that as the frequency falls, the reactance goes up. In other words, capacitors impede low frequency signals.\n\nCombining capacitive reactance and resistance\n\nTo make Ohm's law work for changing currents, we redefine it as\n\nI=E/Z\n\nWhere Z represents impedance, the opposition to all current, changing or not. The impedance of a resistor and capacitor in series is found by the formula:\n\nThe impedance of a resistor and capacitor in parallel is a bit more complex:\n\nA Simple Filter\n\nA resistor and a capacitor can be combined to make an AC current divider or filter circuit.\n\nWhen the frequency is low, the impedance of the capacitor is high, so most current will flow through the resistor. As the frequency increases, more current is diverted through the capacitor, less to the rest of the circuit. Thus, the response is low pass. If you exchanged the capacitor and resistor, you'd have a high pass circuit.\n\nThe cutoff frequency is defined as the frequency for which the resistance of the resistor equals the reactance of the capacitor. At that point, the signal is .707 times the original amplitude or reduced by 3db. Above the cutoff frequency, the signal falls by 6db per octave. Below that point (in the passband) the signal is unaffected. To find the cutoff frequency:\n\nInductors\n\nCapacitors are not the only gadgets that have reactance. If you take some wire and coil it tightly, you have made an inductor. This is what happens:\n\nWhen current passes through the inductor L, a magnetic field is generated. It doesn't appear suddenly, it builds up. A magnetic field moving past a wire generates current, and a growing field is moving. In this case, it's moving past the wires of the coil itself in such a way as to oppose the incoming current, so the current flow is delayed like this:\n\nCurrent Flow\n\nLook familiar? It's the same sort of curve as the capacitor, except the current through an inductor builds like the voltage across a capacitor. (And yes, the voltage across the inductor starts high and falls, like current into a capacitor.) What I really find fascinating about inductors is that after the current source is removed, the collapsing magnetic field keeps the current going for a bit.\n\nIn many ways, an inductor is the opposite of a capacitor. It has a time constant:\n\nWhere L is the inductance in units called henrys. The inductance for inductors in series and parallel follows the form for resistors, at least if the inductors aren't close enough together to interact magnetically.\n\nThe reactance of an inductor is:\n\nSince the frequency is just multiplied by the inductance, inductors impede high frequency signals. When you apply a sine wave to an inductor, the current lags behind the voltage by 90°.\n\nYou can make filters with resistors and inductors, but they aren't common in audio because inductors of the appropriate size are fairly large. Radio and video circuits use them a lot.\n\nInductors and capacitors combined\n\nWhen you place an inductor in series with a capacitor, you get an interesting effect. The impedance is found by:\n\nThe impedance is the absolute value of the difference of the reactance of the capacitor and inductor. Since the signal frequency is used to compute both reactance parts but one is rising with frequency and one is falling, the impedance curve looks like this:\n\nThere is a magic frequency, called the resonance frequency, where plenty of current flows, but above and below resonance, there is less current. If the capacitor and the inductor are in parallel, this formula gives the impedance:\n\nThe current verses frequency plot looks like this:\n\nWhat's going on here? Well, at low frequencies, the inductor passes pretty much everything (remember, an inductor is just a wire for DC) and the capacitor blocks everything. As the frequency rises, the inductor impedes, but the capacitor will take over. When the impedances of both match, you get no current flow. How is this possible?\n\nIt's because of the phase changes: the current through a capacitor is 90° ahead of the voltage, and the current through the inductor is 90° behind. When the circuit is in resonance, the two cancel out. In real circuits, series resistance tends to reduce the peaks. This is called damping, and the ratio of inductive reactance to resistance is known as Q (for quality factor).\n\nTransformers\n\nAs I mentioned before, you don't see a lot of inductors in audio circuits, primarily because of size, but also because they aren't very precise compared to capacitors. There is one vital function that only inductors can do well:\n\nIf two inductors are close together, current flowing in one will induce current in the other. Such an arrangement is called a transformer. As far as audio goes, there are three useful features of transformers:\n\n• The right side (secondary) of the circuit is completely isolated from the left (primary). That means any steady voltage (or DC offset) from the source will not appear at the ultimate output.\n• If there are more turns in the secondary coil than in the primary, the voltage developed across the secondary will be proportionally higher. This can't come for free- the current in the secondary will be proportionally less. In other words, the Power (voltage times current) is constant.\n• If the wires from the source to the transformer are long, chances are stray currents will be induced from outside sources (radio signals, hum fields and other junk). Because these currents will have the same direction in both wires, they will not develop any voltage across the primary, so no noise current will appear in the secondary.\n\nSo, we use transformers for isolation, noise rejection, and changing voltage of AC signals (most often to adjust the mains power to something useful for audio circuitry.) We'll mention them again. Just remember that transformers are inductors first and have all of the impedance and frequency effects we have just discussed.\n\npqe 10/1/98\n\nBack to Mu126 Topics\n\nLOW impedance capacitors"
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https://answers.opencv.org/question/54838/filling-elements-2-the-triangle-problem/ | [
"# Filling elements 2 (the triangle problem)\n\nRelated with my previous question \"Filling elements\" which was kindly solved by Michael Burdinov. I have the following question.\n\nFilling Elements\n\nWhat if I have a situation like the following:",
null,
"As you can see it is a little different than my previous question. If I apply the algorithm I got last time , the result is the following",
null,
"which is not the ideal I would like to get. Is there any way, or can any of you have any idea on how to get these elements filled without filling the space between them that is formed when they are in a triangle position?? (actually any closed position)\n\nAny idea will be warmly welcomed\n\nedit retag close merge delete\n\ncan you please post a link to the previous question? I cannot find it. If your element are like that, maybe the fitting ellipse or Hough line detection will work\n\n2\n\nactually if you could upload the source image, that might help.\n\n1\n\nThanks for the comments. I edited the post to put a link to the previous question.\n\nI haven't tried closing and opening this time, but my experience tells me (as I posted in the previous question) that first, closing sometimes have the undesirable effect of linking separated areas (that is why I posted the previous question) and also even when it works it would fill the insides of the triangle too.\n\nI am thinking maybe some combination of opening closing and separating and joining elements would work, but would like to hear your ideas on this. (I am sure I am not the only one who has find himself in this situation)\n\nwhat about showing us the source image, maybe there is a feature that can be used and it is not clear through the binary images that you have now\n\n2\n\nIn another way you can try contour approach,\n\n• Find contour(outermost only)\n• Iterate through each contour.\n• Draw contour with filled option.\n1\n\n@Haris this will not work because the outermost contour will fill the whole blob and he will get the same result as in the second image. Moreover, the inner contours (i.e. the 3 ellipses-ish and the triangle-ish) are all in the same level. So, you need one more procedure to distinguish what is what.\n\nYou are not getting the 3 ellipses and the triangle? Then you shall apply THRESH_BIN_INV flag and you should think of something for filtering the triangles, like area, or length\n\nI suggest that you can use the algorithm of Haris, and for each contour you can get the bounding-box, the minrect, apply a hough transform to get the lines. With all this information, you can do a simple decisor to verify if the region is a triangle or not.\n\nSort by » oldest newest most voted",
null,
"Actually, you can use the approxPolyDP() function and approximate your contours with accuracy proportional to the contour perimeter. Here is some code:\n\n#include <iostream>\n#include <opencv2/opencv.hpp>\n\nusing namespace std;\nusing namespace cv;\nint main()\n{\nif( !src.data )\n{ return -1; }\n\n// Show source image\nimshow(\"src\", src);",
null,
"// Create binary image from source image\nMat bw;\ncvtColor(src, bw, CV_BGR2GRAY);\nthreshold(bw, bw, 40, 255, CV_THRESH_BINARY | CV_THRESH_OTSU);\nimshow(\"bin\", bw);",
null,
"// Find contours\nvector<Vec4i> hierarchy;\nstd::vector<std::vector<cv::Point> > contours;\ncv::findContours(bw.clone(), contours, hierarchy, CV_RETR_TREE, CV_CHAIN_APPROX_SIMPLE, Point(0, 0));\n\n// The array for storing the approximation curve\nstd::vector<cv::Point> approx;\n\n// We'll put the labels in this destination image\ncv::Mat dst = Mat::zeros(bw.size(), CV_8UC1);\n\n// Loop through detected contours, and process accordingly\nfor (int i = 0; i < contours.size(); i++)\n{\n// Approximate contour with accuracy proportional\n// to the contour perimeter with approxPolyDP. In this,\n// the third argument is called epsilon, which is maximum\n// distance from contour to approximated contour. It is\n// an accuracy parameter. A wise selection of epsilon is\n//needed to get the correct output.\ndouble epsilon = cv::arcLength(cv::Mat(contours[i]), true) * 0.1; // epsilon = 10% of arc length\ncv::approxPolyDP(\ncv::Mat(contours[i]),\napprox,\nepsilon,\ntrue\n);\n\n// Skip small or convex objects\nif (std::fabs(cv::contourArea(contours[i])) < 500 || cv::isContourConvex(approx))\ncontinue;\n\ndrawContours(dst, contours, i, Scalar(255, 255, 255), CV_FILLED, 8, hierarchy, 0, Point());\n}\n\nimshow(\"contours\", dst);",
null,
"// sum two images\ndst += bw;\n\nimshow(\"output\", dst);",
null,
"waitKey(0);\nreturn 0;\n}\n\n\nof course you should play with the epsilon value in order to pick a value robust enough for all of your cases. Enjoy.\n\nmore\n\n1\n\nanother way to go would also be to use the approxPolyDP() function and approx structure to determine the shape of the contours but then complexity increases. I think the above approach is simple and fast.\n\nThank you. I will study your solution today, seems quite nice. and let you know.\n\nI understand the general working of the algorithm. Quite neat, thank you. In it the approximation of the contours that are \"convex\" are eliminated.\n\nSo the remaining ones are concave?? I painted the approximations and found out that they are lines! (I suppose lines are considered concave??)\n\nI haven't experiment it yet but does this means that this only works for long shapes like the picture and not say circles arranged in a closed formation... Thank you for your reply.\n\n1\n\nYes the remaining ones are concave. As I am stating the important parameter here is the epsilon. If you want to detect other objects like triangles, circles, squares you should then use the convex objects which I am skipping here. Of course it would be wise to decrease the epsilon value in order to obtain more convexs and determine the object more accurately. Have a look in the following examples:\n\nContour Approximation\n\nDetecting shapes in image\n\nDetect Squares\n\nOfficial site\n\nGitHub\n\nWiki\n\nDocumentation"
]
| [
null,
"https://answers.opencv.org/upfiles/1423558725320651.jpg",
null,
"https://answers.opencv.org/upfiles/14235587742644792.jpg",
null,
"https://answers.opencv.org/upfiles/avatars/theodore/resized/32/giphy.png",
null,
"https://answers.opencv.org/upfiles/14237493812442761.png",
null,
"https://answers.opencv.org/upfiles/14237493978433657.png",
null,
"https://answers.opencv.org/upfiles/14237501016210652.png",
null,
"https://answers.opencv.org/upfiles/14237494093005777.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8425635,"math_prob":0.7718113,"size":3728,"snap":"2023-14-2023-23","text_gpt3_token_len":902,"char_repetition_ratio":0.111707844,"word_repetition_ratio":0.0033783785,"special_character_ratio":0.25268242,"punctuation_ratio":0.18291055,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97238344,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,7,null,7,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T15:02:20Z\",\"WARC-Record-ID\":\"<urn:uuid:f0cc7381-fa5c-40fa-a288-b34dfcd0b666>\",\"Content-Length\":\"80778\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ee3316a4-1a2d-49db-8136-13fd41cdbe66>\",\"WARC-Concurrent-To\":\"<urn:uuid:9a333a90-276f-40e3-b47b-b20359891634>\",\"WARC-IP-Address\":\"104.21.24.86\",\"WARC-Target-URI\":\"https://answers.opencv.org/question/54838/filling-elements-2-the-triangle-problem/\",\"WARC-Payload-Digest\":\"sha1:VBFRSVH4H2EUUUYUYNIKSJF6Q5DCYRUV\",\"WARC-Block-Digest\":\"sha1:6MNITJWDBHXJQTKQTZPPWMGU5U5QZK7S\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647895.20_warc_CC-MAIN-20230601143134-20230601173134-00381.warc.gz\"}"} |
https://uk.mathworks.com/matlabcentral/cody/problems/2630-number-of-flip-flop-required-in-ripple-counter | [
"Cody\n\n# Problem 2630. Number of Flip Flop required in ripple counter\n\nFind the number of flip flop required in ripple counter.\n\nIf modulus of counter (N) is given find outnumber of Flip Flop (n) required for ripple counter.\n\n` N <= 2^n`\n\nExamples\n\n``` If modulus is 16, then n = 4\nIf modulus is 10, then n = 4 (Number of FF is always an integer) ```\n\n### Solution Stats\n\n65.22% Correct | 34.78% Incorrect\nLast Solution submitted on Dec 12, 2019"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.81452495,"math_prob":0.98681927,"size":402,"snap":"2020-10-2020-16","text_gpt3_token_len":110,"char_repetition_ratio":0.15075377,"word_repetition_ratio":0.0,"special_character_ratio":0.2885572,"punctuation_ratio":0.048780486,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971441,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-21T16:46:49Z\",\"WARC-Record-ID\":\"<urn:uuid:3acfdb20-d286-4eea-90d9-d3a7f6762d75>\",\"Content-Length\":\"82219\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5febcd4b-1f1c-4f96-a0a9-91f438ace8be>\",\"WARC-Concurrent-To\":\"<urn:uuid:16cfee31-630e-41de-902d-a04e5cf768cb>\",\"WARC-IP-Address\":\"104.110.193.39\",\"WARC-Target-URI\":\"https://uk.mathworks.com/matlabcentral/cody/problems/2630-number-of-flip-flop-required-in-ripple-counter\",\"WARC-Payload-Digest\":\"sha1:GO33DGTG52YXCHDV4OYKOGMGGG3LKIYQ\",\"WARC-Block-Digest\":\"sha1:RXV3TGXAADODMLHGEZLTL4LYTTANXGNL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145533.1_warc_CC-MAIN-20200221142006-20200221172006-00191.warc.gz\"}"} |
https://math.stackexchange.com/questions/688987/what-are-different-geometric-interpretations-for-the-sample-variance | [
"# What are different geometric interpretations for the sample variance?\n\nI primarily work with non Euclidean data and am looking to extend concepts of 'variance' to Riemannian manifolds. I am aware of Karcher variance, but I need efficient ways to solve for it. For example, using Rank Revealing QR decomposition (RRQR) provides some information regarding the sample variance for data points in vector spaces. I have also found a paper that uses this on manifolds.\n\nI was wondering if there are other interpretations to the notion of variance that I can exploit? I guess in short - I want to know what are the different optimization problems one would solve for in vector spaces, that would give the sample variance as the solution.\n\nThanks.\n\n M. Gu and S. Eisenstat, Efficient algorithms for computing a strong rank-revealing QR factorization, SIAM Journal on Scientific Computing, 1996.\n\n N. Shroff, P. K. Turaga, and R. Chellappa. Manifold precis: An annealing technique for diverse sampling of manifolds. In NIPS, 2011."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.91848344,"math_prob":0.7325326,"size":956,"snap":"2019-26-2019-30","text_gpt3_token_len":236,"char_repetition_ratio":0.1092437,"word_repetition_ratio":0.0,"special_character_ratio":0.21861924,"punctuation_ratio":0.14516129,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9655529,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-18T21:21:54Z\",\"WARC-Record-ID\":\"<urn:uuid:eee9159a-01fe-4e8c-8eb2-cdb99b31333e>\",\"Content-Length\":\"130900\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3dd79e12-ad0b-4bd5-98f2-7d147790455b>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8e8220f-aa7e-4d77-98c2-21c91686549a>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/688987/what-are-different-geometric-interpretations-for-the-sample-variance\",\"WARC-Payload-Digest\":\"sha1:APOQ67PBHLOLFDQKML4BFVMVCBJ3SVDJ\",\"WARC-Block-Digest\":\"sha1:E53EMBYQJBGSMHFB6FQHCTGF3AXJZLKH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195525829.33_warc_CC-MAIN-20190718211312-20190718233312-00388.warc.gz\"}"} |
https://socratic.org/questions/how-do-you-solve-and-graph-4k-15-2k-3 | [
"# How do you solve and graph 4k + 15 > -2k + 3?\n\nJun 9, 2017\n\nSee a solution process below:\n\n#### Explanation:\n\nFirst, subtract $\\textcolor{red}{15}$ and add $\\textcolor{b l u e}{2 k}$ to each side of the inequality to isolate the $k$ term while keeping the inequality balanced:\n\n$\\textcolor{b l u e}{2 k} + 4 k + 15 - \\textcolor{red}{15} > \\textcolor{b l u e}{2 k} - 2 k + 3 - \\textcolor{red}{15}$\n\n$\\left(\\textcolor{b l u e}{2} + 4\\right) k + 0 > 0 - 12$\n\n$6 k > - 12$\n\nNow, divide each side of the inequality by $\\textcolor{red}{6}$ to solve for $k$ while keeping the inequality balanced:\n\n$\\frac{6 k}{\\textcolor{red}{6}} > - \\frac{12}{\\textcolor{red}{6}}$\n\n$\\frac{\\textcolor{red}{\\cancel{\\textcolor{b l a c k}{6}}} k}{\\cancel{\\textcolor{red}{6}}} > - 2$\n\n$k > - 2$"
]
| [
null
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6617055,"math_prob":0.9997323,"size":297,"snap":"2019-51-2020-05","text_gpt3_token_len":79,"char_repetition_ratio":0.10921501,"word_repetition_ratio":0.0,"special_character_ratio":0.26599327,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99800396,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-25T12:13:40Z\",\"WARC-Record-ID\":\"<urn:uuid:14d893a0-2d2c-4614-bd93-070f66c826d2>\",\"Content-Length\":\"34058\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:680328a5-f689-4ee6-b795-885d9e294660>\",\"WARC-Concurrent-To\":\"<urn:uuid:d0adbe15-283e-42ac-929e-817a275a0855>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-you-solve-and-graph-4k-15-2k-3\",\"WARC-Payload-Digest\":\"sha1:JIEIUZY5PJMENP2B7LPHS7XGPWOSBV3M\",\"WARC-Block-Digest\":\"sha1:M55ILWXDNQCYXFWKMK44P67NVX7222L7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251672440.80_warc_CC-MAIN-20200125101544-20200125130544-00245.warc.gz\"}"} |
https://quantumcomputing.stackexchange.com/questions/13725/how-to-code-a-projector-operator-in-qiskit | [
"# How to code a projector operator in qiskit?\n\nI'm new to qiskit and I want to know how do I define a projector operator in qiskit? Specifically, I have prepared a 3 qubit system, and after applying a whole lot of gates and measuring it in a state vector simulator, I have to apply the projector operator $$|0\\rangle_{1}|0\\rangle_{2} \\langle0|_{1}\\langle0|_{2}$$ on the state that I get after I measured all the qubits in the circuit. I do know the physics and math behind it, and the tensor product it represents but I'm unable to code this successfully. I was thinking maybe a 4x4 matrix representation of this operator might help but I'm not sure about it.\n\nSo I'm asking 2 questions here-\n\n1. How do I define a projector operator like this in qiskit\n2. How do I apply that operator to a state that I have measured with a state vector simulator?\n\nSo, there actually is a method called to_operator() for the class Statevector which takes a statevector and converts it into a projector operator. Here is the code to write your specific projector operator:\n\nfrom qiskit.quantum_info import Statevector\nzero = Statevector([1,0])\nzero_state = zero.tensor(zero) # or zero_state = Statevector([1,0,0,0])\nprojector = zero_state.to_operator()\n\n\nAs you can see, you can input a state as a python list, or as a numpy array. If you print this projector, you will get the following output:",
null,
"Then, after making your operator, you can use the method evolve() to apply the projector operator on a statevector:\n\nstatevector.evolve(projector)"
]
| [
null,
"https://i.stack.imgur.com/t9D5n.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.96089756,"math_prob":0.69652194,"size":813,"snap":"2022-27-2022-33","text_gpt3_token_len":200,"char_repetition_ratio":0.11990111,"word_repetition_ratio":0.027972028,"special_character_ratio":0.24108241,"punctuation_ratio":0.05487805,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98911786,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-25T05:08:36Z\",\"WARC-Record-ID\":\"<urn:uuid:23a371de-1c84-4be7-a661-8cbec7e1f177>\",\"Content-Length\":\"226143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98d3cf33-0723-4ac1-afc5-201c02cc1c72>\",\"WARC-Concurrent-To\":\"<urn:uuid:60ea413f-a2e8-4331-bdf8-b0940f05cde8>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://quantumcomputing.stackexchange.com/questions/13725/how-to-code-a-projector-operator-in-qiskit\",\"WARC-Payload-Digest\":\"sha1:ZV2TH7XGKPXZ2HVDQ2ZK5CILJ2BH57WP\",\"WARC-Block-Digest\":\"sha1:EBLGYQDEYQ7FXESGZROABOMYBYOIHP6U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103034170.1_warc_CC-MAIN-20220625034751-20220625064751-00241.warc.gz\"}"} |
https://www.lssltd.net/how-to-calculate-road-offset-profiles/ | [
"# How to Calculate Road Offset Profiles",
null,
"#### Why we need to use Profile Boards for Road Setting Out?\n\nWhen constructing a new, or even an existing road we need to be able to translate design information into practical information that can be used throughout the road construction process. We do this by providing profile boards at a set distance off to the side of the road and at a set height above the proposed road. These profile boards are placed on either side of the road usually at ten metre intervals along the road. By using these profile boards along with a “Traveller” (a move-able profile board that is set at a desired height, for example at formation level) the site operatives can achieve the desired profile of the road to the required tolerance.\n\nOne of the first jobs a Site Engineer will have to undertake on a new Road Construction Job is to work out the positions and levels of the profiles that will be used to control the line and level of the proposed road. This of course is presuming that the site is not using Machine Control for this construction activity. In the example shown in the image below, we can see a cross-section through a typical proposed road. What the site engineer needs to do is calculate the plane of the road and offset it vertically so the site operatives can see the “traveller”. The profiles also need to be set at a distance beyond the edge of the proposed edge of the road construction. This adds a further complication to the calculations that need to be done. So to offset the plane of the LHS (Left Hand Side) of the carriageway, the site engineer would need to work out points 1 and 3 in the below diagram, and for this the site engineer would need to calculate the easting, northing and elevation for both points. At this point the site engineer would only have information (easting, northing and elevation) for the centreline point and the channel (Face of the Kerb) point from which to calculate these offset points. The site engineer would also have to calculate the elevations for the profile boards at positions 2 and 4, and check the easting and northing position for the these profile boards too, these should be the same as the easting and northing positions for points 1 and 3.\n\n#### How does a Site Engineer Calculate Positions and Levels for Road Offset Profiles?\n\nThe calculation of the positions and levels of the profiles is complicated and time consuming with many points being needed to be worked out even before you set foot out on site. But a lot of this time can be saved through automation and using a programme to do the numerous calculations for you. At the end of this page you can find out how to get a spreadsheet programme that calculates all of the offset points and levels that we are going to go through now.\n\n#### Keep reading for a discount code for the brilliant spreadsheet that does all this work for you at the end of this article.\n\nFirst you need to look through the (supplied from the designer) proposed road data. This should be in the format of cross-sectional information. That is, it should be presented in a format of point name, easting, northing, elevation, offset from the Centerline, for every 10m (minimum) Chainage intervals and every tangent point.\n\nCross-sectional information should look something like this:-\n\nCHAINAGE 40.000\n\nOFFSET R.L. LABEL CUT EASTING NORTHING\n\n1 -10.010 136.931 V001 394499.950 295852.514\n\n2 -8.955 136.934 F001 394499.177 295853.232\n\n3 -5.941 136.723 CL01 394496.968 295855.282\n\n4 0.000 136.787 M001 394492.613 295859.323\n\n5 6.876 136.861 CR01 394487.573 295864.000\n\n6 9.583 137.072 V002 394485.589 295865.841\n\nThe negative offset denotes that point is to the left of the centerline and the convention (when considering roads that is) is to consider everything as we look up chainage.\n\nFrom the above information a Site Engineer would need to use trigonometry to work out the Whole Circle Bearing (WCB) and the distance between the Centerline and the two channels. If their calculations are correct, they should find the distances calculated to be the same as the offset distance (or a sum of these).\n\nUsing the information above, we first need to calculate the WCB and distance between points M001 and CL01. This is done by using trigonometry and Pythagoras Theorem.\n\nTo calculate the distance between two co-ordinated points we need to use the Pythagoras Theorem.\n\na² + b² = c²\n\n(394496.968-394492.613)² + ( 295855.282-295859.323)²=c²\n\n4.355² + (-4.041)² = c²\n\n18.966 + 16.330 = c²\n\n(Square Root) 35.296 = c\n\n5.941 = c\n\nAs you can see this is the same dimension as the OFFSET distance from M001 to CL01 in the table above.\n\nNow we need to work out the WCB (Whole Circle Bearing) from M001 to CL01. To calculate the WCB we need to use trigonometry.\n\n#### What is the Whole Circle Bearing?\n\nPlotting the relative co-ordinated points will help identify the WCB value. The Whole Circle Bearing is the total angle from North that is the direction of travel from the points being considered. This is measured in a clockwise direction from North.\n\nFrom the information already calculated above, we have the lengths of all three sides of the right angled triangle to use for working out the bearing. That is the Hypotenuse (5.491m), difference in Eastings (4.355m) and the difference in Northings (4.041).",
null,
"Plotting the two co-ordinated points relative to Easting and Northing Position the WCB becomes easy to see.\n\nSo, using trigonometry, we can work out the value for the angle x for a right angled triangle.\n\nSOHCAHTOA\n\nUsing the Tan function we get\n\nTan (x) = -4.041/4.355\n\nx= 42 Degrees 51 Minutes 30 Seconds\n\nAs we have drawn out the direction between M001 and CL01 above, we know that we now need to add 90 Degrees to the angle x we have calculated to give a WCB of 132 Degrees 51 Minutes 30 Seconds.\n\nThe above calculations only so far calculate information that we need in order to provide the information that we need in order to calculate the offset points we need to set out on site.\n\n#### Get a free spreadsheet to work out the Whole Circle Bearings and Distances from Co-ordinates.\n\nIf you have followed the above calculation through and worked out your own WCB and Distance and have a load more points that you need to work out then you may find this excel spreadsheet available for free from Lichfield Survey Supplies, on their website. It uses the same calculations as described above to calculate the WCB and distance between points given in Easting and Northing format.\n\n#### How to calculate the co-ordinates of the offset point.\n\nNow that we know the WCB between M001 and CL01, we can use this information to calculate the position of the offset point that we need to mark out on site. This offset wants to be consistent throughout the length of the road construction. Usually when doing this in road construction we choose 1m behind the face of kerb or channel. The below diagram illustrates the point that we need to calculate.\n\nSo, we need to find the Eastings and Northings of the (1m) offset point by projecting the line from CL01 along the path of the WCB we calculated from M001 to CL01 and for this we need to use trigonometry again.\n\nSOHCAHTOA\n\nTo do this we need to find out the distance in the Eastings and the distance in the Northings and add these dimensions (of the eastings and northings) to the Co-ordinates of CL01.\n\nFrom the right-angled triangle, we know the hypotenuse length to be 1m, and our angle is going to be x (WCB of 132 Degrees 51 Minutes 30 Seconds) as we are traversing the same direction. This time for the calculations of the Eastings we will need to use Cos and for the Northings we will need to use Sin.\n\nSo, for the Eastings we have (Hypotenuse is 1, as we are wanting a 1m offset in this case).\n\nCos (42-51-30) = (E)/1\n\nE = 0.733m\n\nAnd for the Northings we have (Hypotenuse is 1, as we are wanting a 1m offset in this case).\n\nSin x = Opposite/Hypotenuse\n\nSin (42-51-30) = (N)/1\n\nN = 0.680m\n\nNow that we know the individual easting and northing lengths we need to add these dimensions to the co-ordinates of point CL01. But we need to be mindful of the direction that the offset point to the point CL01. Looking at the diagram above we can see that the Eastings are going to increase and so we will need to add the dimension to the easting value of point CL01, but the Northings are going to decrease so we will need to subtract the dimension from the northing value of point CL01. So, the Co-ordinates of the (1m) offset point would be: –\n\nEasitngs: –\n\n394496.968 + 0.733 = 394497.701\n\nNorthings: –\n\n295855.282 – 0.680 = 295854.602\n\n#### Get a free spreadsheet to work out the co-ordinates from the Whole Circle Bearing and Distance.\n\nIf you have followed the above calculation through and worked out your own WCB and Distance and have a load more points that you need to work out then you may find this excel spreadsheet available for free from Lichfield Survey Supplies, on their website. It uses the same calculations as described above to calculate the WCB and distance between points given in Easting and Northing format. Please note that this spreadsheet requires the WCB is entered in a decimal format, not in the format of degress, minutes and seconds.\n\n#### How to Calculate the Levels for the Offset Points.\n\nSo now that we have the position of the offset profile (for a 1m offset to the left of the kerbline) in eastings and northings, we now need to calculate the desired level for the profile board to be erected at. To do this we need to work out the fall across the road from the centerline to the channel, that is the difference in height from M001 to CL01 divided by the distance from M001 to CL01.\n\nWe want to work out the fall across the road for the left hand side carriageway and express the answer in millimetres per metre. We have the elevation of the centerline and the elevation of the left hand channel and we have the width of the lane from our calculations done earlier.\n\nFor the Left Hand Side of the road we have\n\nM001 Elevation – CL01 Elevation = Fall from the centreline to the channel.\n\n136.787 – 136.723 = 0.060m Fall from the centreline to the channel.\n\nWe can now calculate the rate of fall from the centreline to the channel.\n\nHeight of Fall / Length of Fall = Rate of Fall.\n\n0.060 / 5.941 = 0.011m/m or 11mm/m fall\n\nSo, for the profile level for our offset point we would now need to do the following calculation, assuming we are offsetting 1m and raising the profile 1m Above FRL (Finished Road Level), as is common on site.\n\nFor the left hand Side of the road we have: –\n\nCL001 Elevation – (Offset Distance * Rate of Fall) + Offset Height = Profile Height\n\n136.723 – (1 * 0.011) + 1 = 137.712m AOD.\n\n#### Now to work out the rest of the Profile Board Points.\n\nAt this point, we only have one offset position and one profile board level worked out. This information, by itself is useless to the operative out on site. They need a further profile board on the opposite side of the road for them to “bone” too.\n\nWe need this other profile board to be outside the road construction area where it will be safe for the duration of the works to be carried out. To do this we need to start these calculations again, paying attention to the fact that our offset will be bigger this time. So, for the LHS carriageway we need a profile board offset behind the RHS Kerbline, this should be a constant offset too, again 1m behind the face of kerb is ideal.\n\nIn order to ensure that the position we calculate for our profile board on the RHS is in a position that is satisfactory we need to check for any glaring errors in the data provided. To do this we once again turn to the Pythagoras Theorem and Trigonometry. We need to check that the offset distance from M001 to CR01 is correct and that the WCB from M001 to CR01 is 180 degrees to the WCB from M001 to CL01. We calculated the WCB from M001 to CL01 earlier and it was 132 Degrees 51 Minutes 30 Seconds.\n\nTo calculate the distance between two co-ordinated points we need to use the Pythagoras Theorem.\n\na² + b² = c²\n\n(394487.573-394492.613-)² + (295864.000-295859.323)²=c²\n\n(-5.040)² + (4.677)² = c²\n\n25.402 + 21.874 = c²\n\n(Square Root) 47.276 = c\n\n6.876 = c\n\nAs you can see, this is the same dimension as the OFFSET distance from M001 to CR01 in the table above, thus confirming our first check.\n\nNow we need to work out the WCB (Whole Circle Bearing) from M001 to CR01. To calculate the WCB we need to use trigonometry.\n\nFrom the information already calculated above, we have the lengths of all three sides of the right angled triangle to use for working out the bearing. That is the Hypotenuse (6.876m), difference in Eastings (5.040m) and the difference in Northings (4.677m).\n\nSo, going back to trigonometry, we can work out the value for the angle x for a right angled triangle, and therefore the WCB from M001 to CR01\n\nSOHCAHTOA\n\nUsing the Tan function we get\n\nTan (x) = (4.677)/(-5.040) x= 42 Degrees 51 Minutes 38 Seconds\n\nThe angle from M001 to CR01 we have this time is just 8 seconds different from the angle we calculated from M001 to CL01, and over the distance we using we can ignore this difference.\n\nAs we have drawn out the direction between M001 and CR01 above, we know that we now need to add 270 Degrees to the angle x we have calculated to give a WCB of 312 Degrees 51 Minutes 38 Seconds. We can see that this WCB is in effect 180 degrees to the WCB for M001 to CL01.\n\nSo, we need to find the Eastings and Northings of the (1m) offset point past the RHS Channel Line (CR02) by projecting the line from M001 along the path of the WCB we calculated from M001 to CR01 and for this we need to use trigonometry again. Bear in mind that the total offset from M001 will be 7.876m.\n\nSOHCAHTOA\n\nTo do this we need to find out the distance in the Eastings and the distance in the Northings and add these dimensions (of the eastings and northings) to the Co-ordinates of M001.\n\nFrom the right-angled triangle, we know the hypotenuse length to be 1m plus the RHS Carriageway width of 6.876m, hence our offset of 7.876m from M001, and our angle is going to be x (WCB of 312 Degrees 51 Minutes 30 Seconds) as we are traversing across the road. This time for the calculations of the Eastings we will need to use Cos and for the Northings we will need to use Sin.\n\nSo, for the Eastings we have (Hypotenuse is 1, as we are wanting a 1m offset in this case).\n\nCos (42-51-30) = (E)/7.876\n\nE = 5.773m\n\nAnd for the Northings we have (Hypotenuse is 1, as we are wanting a 1m offset in this case).\n\nSin x = Opposite/Hypotenuse\n\nSin (42-51-30) = (N)/7.876\n\nN = 5.357m\n\nNow that we know the individual easting and northing lengths we need to add these dimensions to the co-ordinates of point M001. But we need to be mindful of the direction that the offset point to the point M001. Looking at the diagram above we can see that the Eastings are going to decrease and so we will need to subtract the dimension from the easting value of point M001, but the Northings are going to increase so we will need to add the dimension to the northing value of point M001. So, the Co-ordinates of the (1m past the RHS Kerbline) offset point would be: –\n\nEastings: –\n\n394492.613 – 5.773 = 394486.840\n\nNorthings: –\n\n295859.323 + 5.357 = 295864.680\n\nWe now have a position for our other profile board location. We now need to work out the level that we are going to put the profile board on at. This would be point No. 3 in this example. To do this we need to use the rate of fall for the LHS carriageway (11mm/m) that we calculated earlier and project this fall from the M001 point.\n\nThe fall that we have this time is going up, so we need to add height to the M001 elevation at a rate of 11mm/m over the distance to the profile board and also add 1m to the height so that we have the same offset height on both profile boards. The calculation would be as follows: –\n\nElevation = M001 Elevation + (Rate of Fall * Distance to profile)\n\nElevation = 136.787 + (0.011 * 7.876) + 1\n\nElevation = 136.787 + 0.087 + 1\n\nElevation = 137.874m AOD.\n\nWe now have the data required to set out on site just two profiles that can be used across the road. These points are\n\nOffset Profile 1.\n\n394497.701m Eastings\n\n295854.602m Northings\n\n137.712m Elevation\n\nOffset Profile 3.\n\n394486.840m Eastings\n\n295864.680m Northings\n\n137.874m Elevation\n\nAt the moment we have just two profile boards worked out (1 and 3) for use on the LHS of the road. It can be seen that we to also work out levels for the profile boards (2 and 4) for the RHS of the road. The position of the offset points should be the same and so we only need to calculate the elevations for the profile boards using the methods detailed above."
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https://www.sidefx.com/docs/houdini/vex/functions/sign.html | [
"# sign VEX function\n\nReturns -1, 0, or 1 depending on the sign of the argument.\n\n``` int sign(int n) ```\n\n``` float sign(float n) ```\n\nFor a scalar value, returns `-1` for a negative number, `0` for the number zero, and `+1` for a positive number.\n\n``` vector2 sign(vector2 v) ```\n\n``` vector sign(vector v) ```\n\n``` vector4 sign(vector4 v) ```\n\nFor vector values the sign of the individual components is returned as a vector.\n\n# VEX Functions\n\n## Arrays\n\n• Adds an item to an array or string.\n\n• Returns the indices of a sorted version of an array.\n\n• Efficiently creates an array from its arguments.\n\n• Loops over the items in an array, with optional enumeration.\n\n• Inserts an item, array, or string into an array or string.\n\n• Checks if the index given is valid for the array or string given.\n\n• Returns the length of an array.\n\n• Removes the last element of an array and returns it.\n\n• Adds an item to an array.\n\n• Removes an item at the given index from an array.\n\n• Removes an item from an array.\n\n• Reorders items in an array or string.\n\n• Sets the length of an array.\n\n• Returns an array or string in reverse order.\n\n• Slices a sub-string or sub-array of a string or array.\n\n• Returns the array sorted in increasing order.\n\n• Adds a uniform item to an array.\n\n## Attributes and Intrinsics\n\n• Adds an attribute to a geometry.\n\n• Adds a detail attribute to a geometry.\n\n• Adds a point attribute to a geometry.\n\n• Adds a primitive attribute to a geometry.\n\n• Adds a vertex attribute to a geometry.\n\n• Appends to a geometry’s visualizer detail attribute.\n\n• Reads the value of an attribute from geometry.\n\n• Returns the class of a geometry attribute.\n\n• Returns the data id of a geometry attribute.\n\n• Returns the size of a geometry attribute.\n\n• Returns the type of a geometry attribute.\n\n• Returns the transformation metadata of a geometry attribute.\n\n• Evaluates the length of an arc on a primitive defined by an array of points using parametric uv coordinates.\n\n• Reads the value of a detail attribute value from a geometry.\n\n• Reads a detail attribute value from a geometry.\n\n• Returns the size of a geometry detail attribute.\n\n• Returns the type of a geometry detail attribute.\n\n• Returns the type info of a geometry attribute.\n\n• Reads the value of a detail intrinsic from a geometry.\n\n• Finds a primitive/point/vertex that has a certain attribute value.\n\n• Returns number of elements where an integer or string attribute has a certain value.\n\n• Reads an attribute value from geometry, with validity check.\n\n• Copies the value of a geometry attribute into a variable and returns a success flag.\n\n• Checks whether a geometry attribute exists.\n\n• Returns if a geometry detail attribute exists.\n\n• Returns if a geometry point attribute exists.\n\n• Returns if a geometry prim attribute exists.\n\n• Returns if a geometry vertex attribute exists.\n\n• Finds a point by its id attribute.\n\n• Finds a primitive by its id attribute.\n\n• Finds a point by its name attribute.\n\n• Finds a primitive by its name attribute.\n\n• Returns the number of unique values from an integer or string attribute.\n\n• Reads a point attribute value from a geometry.\n\n• Reads a point attribute value from a geometry and outputs a success/fail flag.\n\n• Returns the size of a geometry point attribute.\n\n• Returns the type of a geometry point attribute.\n\n• Returns the type info of a geometry attribute.\n\n• Returns an array of point localtransforms from an array of point indices.\n\n• Returns a point transform from a point index.\n\n• Returns a rigid point transform from a point index.\n\n• Returns an array of point transforms from an array of point indices.\n\n• Returns an array of rigid point transforms from an array of point indices.\n\n• Reads a primitive attribute value from a geometry.\n\n• Interpolates the value of an attribute at a certain parametric (u, v) position and copies it into a variable.\n\n• Evaluates the length of an arc on a primitive using parametric uv coordinates.\n\n• Reads a primitive attribute value from a geometry, outputting a success flag.\n\n• Returns the size of a geometry prim attribute.\n\n• Returns the type of a geometry prim attribute.\n\n• Returns the type info of a geometry attribute.\n\n• Returns position derivative on a primitive at a certain parametric (u, v) position.\n\n• Reads a primitive intrinsic from a geometry.\n\n• Interpolates the value of an attribute at a certain parametric (uvw) position.\n\n• Convert parametric UV locations on curve primitives between different spaces.\n\n• Writes an attribute value to geometry.\n\n• Sets the meaning of an attribute in geometry.\n\n• Sets a detail attribute in a geometry.\n\n• Sets the value of a writeable detail intrinsic attribute.\n\n• Sets a point attribute in a geometry.\n\n• Sets an array of point local transforms at the given point indices.\n\n• Sets the world space transform of a given point\n\n• Sets an array of point transforms at the given point indices.\n\n• Sets a primitive attribute in a geometry.\n\n• Sets the value of a writeable primitive intrinsic attribute.\n\n• Sets a vertex attribute in a geometry.\n\n• Returns one of the set of unique values across all values for an int or string attribute.\n\n• Returns the set of unique values across all values for an int or string attribute.\n\n• Interpolates the value of an attribute at certain UV coordinates using a UV attribute.\n\n• Reads a vertex attribute value from a geometry.\n\n• Reads a vertex attribute value from a geometry.\n\n• Returns the size of a geometry vertex attribute.\n\n• Returns the type of a geometry vertex attribute.\n\n• Returns the type info of a geometry attribute.\n\n## BSDFs\n\n• Returns the albedo (percentage of reflected light) for a bsdf given the outgoing light direction.\n\n• Returns a specular BSDF using the Ashikhmin shading model.\n\n• Returns a Blinn BSDF or computes Blinn shading.\n\n• Returns a chiang BSDF.\n\n• Returns a cone reflection BSDF.\n\n• Creates a bsdf object from two CVEX shader strings.\n\n• Returns a diffuse BSDF or computes diffuse shading.\n\n• Evaluates a bsdf given two vectors.\n\n• Returns a ggx BSDF.\n\n• Returns a BSDF for shading hair.\n\n• Returns an anisotropic volumetric BSDF, which can scatter light forward or backward.\n\n• Returns an isotropic BSDF, which scatters light equally in all directions.\n\n• Returns new BSDF that only includes the components specified by the mask.\n\n• Returns the normal for the diffuse component of a BSDF.\n\n• Returns a Phong BSDF or computes Phong shading.\n\n• Samples a BSDF.\n\n• Computes the solid angle (in steradians) a BSDF function subtends.\n\n• Splits a bsdf into its component lobes.\n\n• Creates an approximate SSS BSDF.\n\n## BSDFs\n\n• Returns a specular BSDF or computes specular shading.\n\n## CHOP\n\n• Adds new channels to a CHOP node.\n\n• Reads from a CHOP attribute.\n\n• Reads CHOP attribute names of a given attribute class from a CHOP input.\n\n• Returns the sample number of the last sample in a given CHOP input.\n\n• Returns the frame corresponding to the last sample of the input specified.\n\n• Returns the time corresponding to the last sample of the input specified.\n\n• Returns the channel index from a input given a channel name.\n\n• Returns the value of a channel at the specified sample.\n\n• Computes the minimum and maximum value of samples in an input channel.\n\n• Returns all the CHOP channel names of a given CHOP input.\n\n• Returns the number of channels in the input specified.\n\n• Returns the value of a CHOP channel at the specified sample.\n\n• Returns the value of a CHOP local transform channel at the specified sample.\n\n• Returns the value of a CHOP local transform channel at the specified sample and evaluation time.\n\n• Returns the value of a CHOP channel at the specified sample and evaluation time.\n\n• Returns the sample rate of the input specified.\n\n• Returns the value of CHOP context temporary buffer at the specified index.\n\n• Removes channels from a CHOP node.\n\n• Removes a CHOP attribute.\n\n• Renames a CHOP channel.\n\n• Resize the CHOP context temporary buffer\n\n• Sets the value of a CHOP attribute.\n\n• Sets the length of the CHOP channel data.\n\n• Sets the sampling rate of the CHOP channel data.\n\n• Sets the CHOP start sample in the channel data.\n\n• Returns the start sample of the input specified.\n\n• Returns the frame corresponding to the first sample of the input specified.\n\n• Returns the time corresponding to the first sample of the input specified.\n\n• Writes a value of CHOP context temporary buffer at the specified index.\n\n• Returns 1 if the Vex CHOP’s Unit Menu is currently set to 'frames', 0 otherwise.\n\n• Returns 1 if the Vex CHOP’s Unit Menu is currently set to 'samples', 0 otherwise.\n\n• Returns 1 if the Vex CHOP’s Unit Menu is currently set to 'seconds', 0 otherwise.\n\n• Returns the number of inputs.\n\n## color\n\n• Compute the color value of an incandescent black body.\n\n• Transforms between color spaces.\n\n• Compute the luminance of the RGB color specified by the parameters.\n\n## Conversion\n\n• Converts a string to a float.\n\n• Converts a string to an integer.\n\n• Depending on the value of c, returns the translate (c=0), rotate (c=1), scale (c=2), or shears (c=3) component of the transform (xform).\n\n• Converts the argument from radians into degrees.\n\n• Creates a vector4 representing a quaternion from euler angles.\n\n• Convert HSV color space into RGB color space.\n\n• Converts a quaternion represented by a vector4 to a matrix3 representation.\n\n• Creates a euler angle representing a quaternion.\n\n• Converts the argument from degrees into radians.\n\n• Convert RGB color space to HSV color space.\n\n• Convert a linear sRGB triplet to CIE XYZ tristimulus values.\n\n• Flattens an array of vector or matrix types into an array of floats.\n\n• Turns a flat array of floats into an array of vectors or matrices.\n\n• Convert CIE XYZ tristimulus values to a linear sRGB triplet.\n\n## Crowds\n\n• Add a clip into an agent’s definition.\n\n• Returns the number of channels in an agent primitive’s rig.\n\n• Returns the names of the channels in an agent primitive’s rig.\n\n• Returns the current value of an agent primitive’s channel.\n\n• Returns the current values of an agent primitive’s channels.\n\n• Returns all of the animation clips that have been loaded for an agent primitive.\n\n• Finds the index of a channel in an agent’s animation clip.\n\n• Returns the names of the channels in an agent’s animation clip.\n\n• Returns the length (in seconds) of an agent’s animation clip.\n\n• Returns an agent primitive’s current animation clips.\n\n• Samples a channel of an agent’s clip at a specific time.\n\n• Samples an agent’s animation clip at a specific time.\n\n• Returns the sample rate of an agent’s animation clip.\n\n• Samples an agent’s animation clip at a specific time.\n\n• Returns the start time (in seconds) of an agent’s animation clip.\n\n• Returns the current times for an agent primitive’s animation clips.\n\n• Returns the transform groups for an agent primitive’s current animation clips.\n\n• Returns the blend weights for an agent primitive’s animation clips.\n\n• Returns the name of the collision layer of an agent primitive.\n\n• Returns the names of an agent primitive’s collision layers.\n\n• Returns the name of the current layer of an agent primitive.\n\n• Returns the names of an agent primitive’s current layers.\n\n• Finds the index of a layer in an agent’s definition.\n\n• Finds the index of a transform group in an agent’s definition.\n\n• Returns the transform that each shape in an agent’s layer is bound to.\n\n• Returns all of the layers that have been loaded for an agent primitive.\n\n• Returns the names of the shapes referenced by an agent primitive’s layer.\n\n• Returns the current local space transform of an agent primitive’s bone.\n\n• Returns the current local space transforms of an agent primitive.\n\n• Returns the agent definition’s metadata dictionary.\n\n• Returns the child transforms of a transform in an agent primitive’s rig.\n\n• Finds the index of a transform in an agent primitive’s rig.\n\n• Finds the index of a channel in an agent primitive’s rig.\n\n• Returns the parent transform of a transform in an agent primitive’s rig.\n\n• Applies a full-body inverse kinematics algorithm to an agent’s skeleton.\n\n• Returns the number of transforms in an agent primitive’s rig.\n\n• Returns whether a transform is a member of the specified transform group.\n\n• Returns whether a channel is a member of the specified transform group.\n\n• Returns the names of the transform groups in an agent’s definition.\n\n• Returns the weight of a member of the specified transform group.\n\n• Returns the name of each transform in an agent primitive’s rig.\n\n• Converts transforms from world space to local space for an agent primitive.\n\n• Converts transforms from local space to world space for an agent primitive.\n\n• Returns the current world space transform of an agent primitive’s bone.\n\n• Returns the current world space transforms of an agent primitive.\n\n• Overrides the value of an agent primitive’s channel.\n\n• Overrides the values of an agent primitive’s channels.\n\n• Sets the current animation clips for an agent primitive.\n\n• Sets the animation clips that an agent should use to compute its transforms.\n\n• Sets the current times for an agent primitive’s animation clips.\n\n• Sets the blend weights for an agent primitive’s animation clips.\n\n• Sets the collision layer of an agent primitive.\n\n• Sets the collision layers of an agent primitive.\n\n• Sets the current layer of an agent primitive.\n\n• Sets the current display layers of an agent primitive.\n\n• Overrides the local space transform of an agent primitive’s bone.\n\n• Overrides the local space transforms of an agent primitive.\n\n• Overrides the world space transform of an agent primitive’s bone.\n\n• Overrides the world space transforms of an agent primitive.\n\n## dict\n\n• Converts a VEX dictionary into a JSON string.\n\n• Converts a JSON string into a VEX dictionary.\n\n• Returns all the keys in a dictionary.\n\n## File I/O\n\n• Returns file system status for a given file.\n\n## filter\n\n• Computes an importance sample based on the given filter type and input uv.\n\n## Geometry\n\n• Adds a point to the geometry.\n\n• Adds a primitive to the geometry.\n\n• Adds a vertex to a primitive in a geometry.\n\n• Clip the line segment between p0 and p1.\n\n• Returns a handle to the current geometry.\n\n• Returns an oppath: string to unwrap the geometry in-place.\n\n• Returns 1 if the edge specified by the point pair is in the group specified by the string.\n\n• This function computes the first intersection of a ray with geometry.\n\n• Computes all intersections of the specified ray with geometry.\n\n• Finds the closest position on the surface of a geometry.\n\n• Finds the closest point in a geometry.\n\n• Finds the all the closest point in a geometry.\n\n• Returns the number of edges in the group.\n\n• Returns the point number of the next point connected to a given point.\n\n• Returns the number of points that are connected to the specified point.\n\n• Returns an array of the point numbers of the neighbours of a point.\n\n• Returns the number of points in the input or geometry file.\n\n• Returns the number of primitives in the input or geometry file.\n\n• Returns the number of vertices in the input or geometry file.\n\n• Returns the number of vertices in the group.\n\n• Returns the list of primitives containing a point.\n\n• Returns a linear vertex number of a point in a geometry.\n\n• Returns the list of vertices connected to a point.\n\n• Returns an array of the primitive numbers of the edge-neighbours of a polygon.\n\n• Returns a list of primitives potentially intersecting a given bounding box.\n\n• Converts a primitive/vertex pair into a point number.\n\n• Returns the list of points on a primitive.\n\n• Converts a primitive/vertex pair into a linear vertex.\n\n• Returns number of vertices in a primitive in a geometry.\n\n• Returns the list of vertices on a primitive.\n\n• Removes a point from the geometry.\n\n• Removes a primitive from the geometry.\n\n• Removes a vertex from the geometry.\n\n• Sets edge group membership in a geometry.\n\n• Rewires a vertex in the geometry to a different point.\n\n• Rewires a vertex in the geometry to a different point.\n\n• This function computes the intersection of the specified ray with the geometry in uv space.\n\n• Converts a primitive/vertex pair into a linear vertex.\n\n• Returns the linear vertex number of the next vertex sharing a point with a given vertex.\n\n• Returns the point number of linear vertex in a geometry.\n\n• Returns the linear vertex number of the previous vertex sharing a point with a given vertex.\n\n• Returns the number of the primitive containing a given vertex.\n\n• Converts a linear vertex index into a primitive vertex number.\n\n## groups\n\n• Returns an array of point numbers corresponding to a group string.\n\n• Returns an array of prim numbers corresponding to a group string.\n\n• Returns an array of linear vertex numbers corresponding to a group string.\n\n• Returns 1 if the point specified by the point number is in the group specified by the string.\n\n• Returns 1 if the primitive specified by the primitive number is in the group specified by the string.\n\n• Returns 1 if the vertex specified by the vertex number is in the group specified by the string.\n\n• Returns the number of points in the group.\n\n• Returns the number of primitives in the group.\n\n• Adds or removes a point to/from a group in a geometry.\n\n• Adds or removes a primitive to/from a group in a geometry.\n\n• Adds or removes a vertex to/from a group in a geometry.\n\n## Half-edges\n\n• Returns the destination point of a half-edge.\n\n• Returns the destination vertex of a half-edge.\n\n• Returns the number of half-edges equivalent to a given half-edge.\n\n• Determines whether a two half-edges are equivalent (represent the same edge).\n\n• Determines whether a half-edge number corresponds to a primary half-edge.\n\n• Determines whether a half-edge number corresponds to a valid half-edge.\n\n• Returns the half-edge that follows a given half-edge in its polygon.\n\n• Returns the next half-edges equivalent to a given half-edge.\n\n• Returns the point into which the vertex following the destination vertex of a half-edge in its primitive is wired.\n\n• Returns the vertex following the destination vertex of a half-edge in its primitive.\n\n• Returns the point into which the vertex that precedes the source vertex of a half-edge in its primitive is wired.\n\n• Returns the vertex that precedes the source vertex of a half-edge in its primitive.\n\n• Returns the half-edge that precedes a given half-edge in its polygon.\n\n• Returns the primitive that contains a half-edge.\n\n• Returns the primary half-edge equivalent to a given half-edge.\n\n• Returns the source point of a half-edge.\n\n• Returns the source vertex of a half-edge.\n\n• Finds and returns a half-edge with the given endpoints.\n\n• Finds and returns a half-edge with a given source point or with given source and destination points.\n\n• Returns the next half-edge with the same source as a given half-edge.\n\n• Returns one of the half-edges contained in a primitive.\n\n• Returns the half-edge which has a vertex as source.\n\n## Image Processing\n\n• Tells the COP manager that you need access to the given frame.\n\n• Returns the default name of the alpha plane (as it appears in the compositor preferences).\n\n• Samples a 2×2 pixel block around the given UV position, and bilinearly interpolates these pixels.\n\n• Returns the default name of the bump plane (as it appears in the compositor preferences).\n\n• Returns the name of a numbered channel.\n\n• Samples the exact (unfiltered) pixel color at the given coordinates.\n\n• Returns the default name of the color plane (as it appears in the compositor preferences).\n\n• Returns the default name of the depth plane (as it appears in the compositor preferences).\n\n• Reads the z-records stored in a pixel of a deep shadow map or deep camera map.\n\n• Returns fully filtered pixel input.\n\n• Queries if metadata exists on a composite operator.\n\n• Returns 1 if the plane specified by the parameter exists in this COP.\n\n• Returns the aspect ratio of the specified input.\n\n• Returns the channel name of the indexed plane of the given input.\n\n• Returns the last frame of the specified input.\n\n• Returns the end time of the specified input.\n\n• Returns 1 if the specified input has a plane named planename.\n\n• Returns the number of planes in the given input.\n\n• Returns the index of the plane named 'planename' in the specified input.\n\n• Returns the name of the plane specified by the planeindex of the given input\n\n• Returns the number of components in the plane named planename in the specified input.\n\n• Returns the frame rate of the specified input.\n\n• Returns the starting frame of the specified input.\n\n• Returns the start time of the specified input.\n\n• Returns the X resolution of the specified input.\n\n• Returns the Y resolution of the specified input.\n\n• Returns the default name of the luminaence plane (as it appears in the compositor preferences).\n\n• Returns the default name of the mask plane (as it appears in the compositor preferences).\n\n• Returns a metadata value from a composite operator.\n\n• Reads a component from a pixel and its eight neighbors.\n\n• Returns the default name of the normal plane (as it appears in the compositor preferences).\n\n• Returns the index of the plane specified by the parameter, starting at zero.\n\n• Returns the name of the plane specified by the index (e.\n\n• Returns the number of components in the plane (1 for scalar planes and up to 4 for vector planes).\n\n• Returns the default name of the point plane (as it appears in the compositor preferences).\n\n• Returns the default name of the velocity plane (as it appears in the compositor preferences).\n\n## Interpolation\n\n• Samples a Catmull-Rom (Cardinal) spline defined by position/value keys.\n\n• Returns value clamped between min and max.\n\n• Samples a Catmull-Rom (Cardinal) spline defined by uniformly spaced keys.\n\n• Takes the value in one range and shifts it to the corresponding value in a new range.\n\n• Takes the value in one range and shifts it to the corresponding value in a new range.\n\n• Takes the value in the range (0, 1) and shifts it to the corresponding value in a new range.\n\n• Takes the value in the range (1, 0) and shifts it to the corresponding value in a new range.\n\n• Takes the value in the range (-1, 1) and shifts it to the corresponding value in a new range.\n\n• Inverses a linear interpolation between the values.\n\n• Performs linear interpolation between the values.\n\n• Samples a polyline between the key points.\n\n• Samples a polyline defined by linearly spaced values.\n\n• Quaternion blend between q1 and q2 based on the bias.\n\n• Computes ease in/out interpolation between values.\n\n## light\n\n• Returns the color of ambient light in the scene.\n\n• Computes attenuated falloff.\n\n• Sends a ray from the position P along the direction specified by the direction D.\n\n• Sends a ray from the position P along direction D.\n\n## Math\n\n• Returns the absolute value of the argument.\n\n• Returns the inverse cosine of the argument.\n\n• Returns the inverse sine of the argument.\n\n• Returns the inverse tangent of the argument.\n\n• Returns the inverse tangent of y/x.\n\n• Returns the average value of the input(s)\n\n• Returns the cube root of the argument.\n\n• Returns the smallest integer greater than or equal to the argument.\n\n• Combines Local and Parent Transforms with Scale Inheritance.\n\n• Returns the cosine of the argument.\n\n• Returns the hyperbolic cosine of the argument.\n\n• Returns the cross product between the two vectors.\n\n• Computes the determinant of the matrix.\n\n• Diagonalizes Symmetric Matrices.\n\n• Returns the dot product between the arguments.\n\n• Returns the derivative of the given value with respect to U.\n\n• Returns the derivative of the given value with respect to V.\n\n• Returns the derivative of the given value with respect to the 3rd axis (for volume rendering).\n\n• Computes the eigenvalues of a 3×3 matrix.\n\n• Gauss error function.\n\n• Inverse Gauss error function.\n\n• Gauss error function’s complement.\n\n• Returns the exponential function of the argument.\n\n• Extracts Local Transform from a World Transform with Scale Inheritance.\n\n• Returns the largest integer less than or equal to the argument.\n\n• Returns the fractional component of the floating point number.\n\n• Returns an identity matrix.\n\n• Inverts a matrix.\n\n• Checks whether a value is a normal finite number.\n\n• Checks whether a value is not a number.\n\n• Returns an interpolated value along a curve defined by a basis and key/position pairs.\n\n• Returns the magnitude of a vector.\n\n• Returns the squared distance of the vector or vector4.\n\n• Returns the natural logarithm of the argument.\n\n• Returns the logarithm (base 10) of the argument.\n\n• Creates an orthonormal basis given a z-axis vector.\n\n• Returns a normalized vector.\n\n• Returns the outer product between the arguments.\n\n• Computes the intersection of a 3D sphere and an infinite 3D plane.\n\n• Raises the first argument to the power of the second argument.\n\n• Determines if a point is inside or outside a triangle circumcircle.\n\n• Determines if a point is inside or outside a tetrahedron circumsphere.\n\n• Determines the orientation of a point with respect to a line.\n\n• Determines the orientation of a point with respect to a plane.\n\n• Pre multiply matrices.\n\n• Returns the product of a list of numbers.\n\n• This function returns the closest distance between the point Q and a finite line segment between points P0 and P1.\n\n• Finds distance between two quaternions.\n\n• Inverts a quaternion rotation.\n\n• Multiplies two quaternions and returns the result.\n\n• Rotates a vector by a quaternion.\n\n• Creates a vector4 representing a quaternion.\n\n• Rounds the number to the closest whole number.\n\n• Bit-shifts an integer left.\n\n• Bit-shifts an integer right.\n\n• Bit-shifts an integer right.\n\n• Returns -1, 0, or 1 depending on the sign of the argument.\n\n• Returns the sine of the argument.\n\n• Returns the hyperbolic sine of the argument.\n\n• Finds the normal component of frame slid along a curve.\n\n• Solves a cubic function returning the number of real roots.\n\n• Finds the real roots of a polynomial.\n\n• Solves a quadratic function returning the number of real roots.\n\n• Finds the angles of a triangle from its sides.\n\n• Samples a value along a polyline or spline curve.\n\n• Generate a cumulative distribution function (CDF) by sampling a spline curve.\n\n• Returns the square root of the argument.\n\n• Returns the sum of a list of numbers.\n\n• Computes the singular value decomposition of a 3×3 matrix.\n\n• Returns the trigonometric tangent of the argument\n\n• Returns the hyperbolic tangent of the argument\n\n• Transposes the given matrix.\n\n• Removes the fractional part of a floating point number.\n\n## measure\n\n• Returns the distance between two points.\n\n• Returns the squared distance between the two points.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the geometry.\n\n• Returns the center of the bounding box for the geometry.\n\n• Returns the maximum of the bounding box for the geometry.\n\n• Returns the minimum of the bounding box for the geometry.\n\n• Returns the size of the bounding box for the geometry.\n\n• Returns the bounding box of the geometry specified by the filename.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the geometry.\n\n• Returns the center of the bounding box for the geometry.\n\n• Returns the maximum of the bounding box for the geometry.\n\n• Returns the minimum of the bounding box for the geometry.\n\n• Returns the size of the bounding box for the geometry.\n\n• Computes the distance and closest point of a point to an infinite plane.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Finds the distance of a point to a group of points along the surface of a geometry.\n\n• Finds the distance of a uv coordinate to a geometry in uv space.\n\n• Finds the distance from a point to the closest location on surface geometry.\n\n## metaball\n\n• Once you get a handle to a metaball using metastart and metanext, you can query attributes of the metaball with metaimport.\n\n• Takes the ray defined by p0 and p1 and partitions it into zero or more sub-intervals where each interval intersects a cluster of metaballs from filename.\n\n• Iterate to the next metaball in the list of metaballs returned by the metastart() function.\n\n• Open a geometry file and return a handle for the metaballs of interest, at the position p.\n\n• Returns the metaweight of the geometry at position p.\n\n## Nodes\n\n• Adds a mapping for an attribute to a local variable.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a key-value dictionary parameter and return its value.\n\n• Evaluates a channel with a new segment expression.\n\n• Evaluates a channel with a new segment expression at a given frame.\n\n• Evaluates a channel with a new segment expression at a given time.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Resolves a channel string (or parameter) and return op_id, parm_index and vector_index.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates a ramp parameter and return its value.\n\n• Evaluates the derivative of a parm parameter with respect to position.\n\n• Evaluates a channel (or parameter) and return its value.\n\n• Evaluates an operator path parameter and return the path to the operator.\n\n• Returns the raw string channel (or parameter).\n\n• Evaluates a channel or parameter, and return its value.\n\n• Evaluates a channel or parameter, and return its value.\n\n• Returns the capture transform associated with a Capture Region SOP.\n\n• Returns the deform transform associated with a Capture Region SOP.\n\n• Returns the capture or deform transform associated with a Capture Region SOP based on the global capture override flag.\n\n• Returns 1 if input_number is connected, or 0 if the input is not connected.\n\n• Returns the full path for the given relative path\n\n• Resolves an operator path string and return its op_id.\n\n• Returns the parent bone transform associated with an OP.\n\n• Returns the parent transform associated with an OP.\n\n• Returns the parm transform associated with an OP.\n\n• Returns the preconstraint transform associated with an OP.\n\n• Returns the pre and parm transform associated with an OP.\n\n• Returns the pre and raw parm transform associated with an OP.\n\n• Returns the pretransform associated with an OP.\n\n• Returns the raw parm transform associated with an OP.\n\n• Returns the transform associated with an OP.\n\n## Noise and Randomness\n\n• Generates alligator noise.\n\n• Computes divergence free noise based on Perlin noise.\n\n• Computes 2d divergence free noise based on Perlin noise.\n\n• Computes divergence free noise based on Simplex noise.\n\n• Computes 2d divergence free noise based on simplex noise.\n\n• Generates Worley (cellular) noise using a Chebyshev distance metric.\n\n• Generates 1D and 3D Perlin Flow Noise from 3D and 4D data.\n\n• There are two forms of Perlin-style noise: a non-periodic noise which changes randomly throughout N-dimensional space, and a periodic form which repeats over a given range of space.\n\n• Generates noise matching the output of the Hscript noise() expression function.\n\n• Produces the exact same results as the Houdini expression function of the same name.\n\n• Generates turbulence matching the output of the HScript turb() expression function.\n\n• Generates Worley (cellular) noise using a Manhattan distance metric.\n\n• MaterialX compatible cellnoise\n\n• MaterialX compatible Perlin noise\n\n• There are two forms of Perlin-style noise: a non-periodic noise which changes randomly throughout N-dimensional space, and a periodic form which repeats over a given range of space.\n\n• Derivatives of Perlin Noise.\n\n• Non-deterministic random number generation function.\n\n• These functions are similar to wnoise and vnoise.\n\n• There are two forms of Perlin-style noise: a non-periodic noise which changes randomly throughout N-dimensional space, and a periodic form which repeats over a given range of space.\n\n• Periodic derivatives of Simplex Noise.\n\n• Creates a random number between 0 and 1 from a seed.\n\n• Generate a random number based on the integer position in 1-4D space.\n\n• Generate a uniformly distributed random number.\n\n• Hashes floating point numbers to integers.\n\n• Hashes integer numbers to integers.\n\n• Generates a random Poisson variable given the mean to the distribution and a seed.\n\n• Hashes a string to an integer.\n\n• Generate a uniformly distributed random number.\n\n• These functions are similar to wnoise.\n\n• Generates Voronoi (cellular) noise.\n\n• Generates Worley (cellular) noise.\n\n• Simplex noise is very close to Perlin noise, except with the samples on a simplex mesh rather than a grid. This results in less grid artifacts. It also uses a higher order bspline to provide better derivatives. This is the periodic simplex noise\n\n• Simplex noise is very close to Perlin noise, except with the samples on a simplex mesh rather than a grid. This results in less grid artifacts. It also uses a higher order bspline to provide better derivatives.\n\n• Derivatives of Simplex Noise.\n\n## normals\n\n• In shading contexts, computes a normal. In the SOP contexts, sets how/whether to recompute normals.\n\n• Returns the normal of the primitive (prim_number) at parametric location u, v.\n\n## Open Color IO\n\n• Returns the names of active displays supported in Open Color IO\n\n• Returns the names of active views supported in Open Color IO\n\n• Imports attributes from OpenColorIO spaces.\n\n• Returns the names of roles supported in Open Color IO\n\n• Returns the names of color spaces supported in Open Color IO.\n\n• Parse the color space from a string\n\n• Transform colors using Open Color IO\n\n## particles\n\n• Samples the velocity field defined by a set of vortex filaments.\n\n## Point Clouds and 3D Images\n\n• Returns the value of the point attribute for the metaballs if metaball geometry is specified to i3dgen.\n\n• Returns the density of the metaball field if metaball geometry is specified to i3dgen.\n\n• Transforms the position specified into the local space of the metaball.\n\n• This function closes the handle associated with a pcopen function.\n\n• Returns a list of closest points from a file within a specified cone.\n\n• Returns a list of closest points from a file in a cone, taking into account their radii\n\n• Writes data to a point cloud inside a pciterate or a pcunshaded loop.\n\n• Returns the distance to the farthest point found in the search performed by pcopen.\n\n• Filters points found by pcopen using a simple reconstruction filter.\n\n• Returns a list of closest points from a file.\n\n• Returns a list of closest points from a file taking into account their radii.\n\n• Generates a point cloud.\n\n• Imports channel data from a point cloud inside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• Imports channel data from a point cloud outside a pciterate or a pcunshaded loop.\n\n• This function can be used to iterate over all the points which were found in the pcopen query.\n\n• Returns a list of closest points to an infinite line from a specified file\n\n• Returns a list of closest points to an infinite line from a specified file\n\n• This node returns the number of points found by pcopen.\n\n• Returns a handle to a point cloud file.\n\n• Returns a handle to a point cloud file.\n\n• Changes the current iteration point to a leaf descendant of the current aggregate point.\n\n• Returns a list of closest points to a line segment from a specified file\n\n• Returns a list of closest points to a line segment from a specified file\n\n• Iterate over all of the points of a read-write channel which haven’t had any data written to the channel yet.\n\n• Writes data to a point cloud file.\n\n• Returns a list of closest points from a file.\n\n• Samples a color from a photon map.\n\n• Returns the value of the 3d image at the position specified by P.\n\n• This function queries the 3D texture map specified and returns the bounding box information of the file.\n\n## Sampling\n\n• Creates a cumulative distribution function (CDF) from an array of probability density function (PDF) values.\n\n• Creates a probability density function from an array of input values.\n\n• Limits a unit value in a way that maintains uniformity and in-range consistency.\n\n• Initializes a sampling sequence for the nextsample function.\n\n• Samples the Cauchy (Lorentz) distribution.\n\n• Samples a cumulative distribution function (CDF).\n\n• Generates a uniform unit vector2, within maxangle of center, given a uniform number between 0 and 1.\n\n• Generates a uniform unit vector2, given a uniform number between 0 and 1.\n\n• Generates a uniform vector2 with alpha < length < 1, where 0 < alpha < 1, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector2 with length < 1, within maxangle of center, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector2 with length < 1, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform unit vector, within maxangle of center, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform unit vector, given a vector2 of uniform numbers between 0 and 1.\n\n• Returns an integer, either uniform or weighted, given a uniform number between 0 and 1.\n\n• Samples the exponential distribution.\n\n• Samples geometry in the scene and returns information from the shaders of surfaces that were sampled.\n\n• Generates a unit vector, optionally biased, within a hemisphere, given a vector2 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector4 with length < 1, within maxangle of center, given a vector4 of uniform numbers between 0 and 1.\n\n• Generates a uniform vector4 with length < 1, given a vector4 of uniform numbers between 0 and 1.\n\n• Samples a 3D position on a light source and runs the light shader at that point.\n\n• Samples the log-normal distribution based on parameters of the underlying normal distribution.\n\n• Samples the log-normal distribution based on median and standard deviation.\n\n• Samples the normal (Gaussian) distribution.\n\n• Generates a uniform unit vector4, within maxangle of center, given a vector of uniform numbers between 0 and 1.\n\n• Generates a uniform unit vector4, given a vector of uniform numbers between 0 and 1.\n\n• Samples a 3D position on a light source and runs the light shader at that point.\n\n• Generates a uniform vector with length < 1, within maxangle of center, given a vector of uniform numbers between 0 and 1.\n\n• Generates a uniform vector with alpha < length < 1, where 0 < alpha < 1, given a vector of uniform numbers between 0 and 1.\n\n• Generates a uniform vector with length < 1, given a vector of uniform numbers between 0 and 1.\n\n• Warps uniform random samples to a disk.\n\n• Computes the mean value and variance for a value.\n\n## Sensor Input\n\n• Sensor function to render GL scene and query the result.\n\n• Sensor function query a rendered GL scene.\n\n• Sensor function to query average values from rendered GL scene.\n\n• Sensor function query a rendered GL scene.\n\n• Sensor function to save a rendered GL scene.\n\n• Returns the area of the micropolygon containing a variable such as P.\n\n• Returns the anti-aliased weight of the step function.\n\n• Computes the fresnel reflection/refraction contributions given an incoming vector, surface normal (both normalized), and an index of refraction (eta).\n\n• If dot(I, Nref) is less than zero, N will be negated.\n\n• Sends rays into the scene and returns information from the shaders of surfaces hit by the rays.\n\n• Returns the blurred point position (P) vector at a fractional time within the motion blur exposure.\n\n• Evaluates surface derivatives of an attribute.\n\n• Returns the name of the current object whose shader is being run.\n\n• Returns the depth of the ray tree for computing global illumination.\n\n• Returns group id containing current primitive.\n\n• Returns a light struct for the specified light identifier.\n\n• Returns the light id for a named light (or -1 for an invalid name).\n\n• Returns the name of the current light when called from within an illuminance loop, or converts an integer light ID into the light’s name.\n\n• Returns an array of light identifiers for the currently shaded surface.\n\n• Returns a selection of lights that illuminate a given material.\n\n• Evaluates local curvature of primitive grid, using the same curvature evaluation method as Measure SOPs.\n\n• Returns a material struct for the current surface.\n\n• Returns material id of shaded primitive.\n\n• Returns the object id for the current shading context.\n\n• Returns the name of the current object whose shader is being run.\n\n• Returns the integer ID of the light being used for photon shading.\n\n• Returns the number of the current primitive.\n\n• Returns the ptexture face id for the current primitive.\n\n• Returns the depth of the ray tree for the current shading.\n\n• Returns an approximation to the contribution of the ray to the final pixel color.\n\n• Looks up sample data in a channel, referenced by a point.\n\n• Returns a selection of objects visible to rays for a given material.\n\n• Returns modified surface position based on a smoothing function.\n\n• Evaluates UV tangents at a point on an arbitrary object.\n\n• Returns the gradient of a field.\n\n• Returns whether a light illuminates the given material.\n\n• Loops through all light sources in the scene, calling the light shader for each light source to set the Cl and L global variables.\n\n• Interpolates a value across the currently shaded micropolygon.\n\n• Finds the nearest intersection of a ray with any of a list of (area) lights and runs the light shader at the intersection point.\n\n• Computes irradiance (global illumination) at the point P with the normal N.\n\n• Returns 1 if the shader is being called to evaluate illumination for fog objects, or 0 if the light or shadow shader is being called to evaluate surface illumination.\n\n• Returns 1 if Light Path Expressions are enabled. 0 Otherwise.\n\n• Indicates whether a shader is being executed for ray tracing.\n\n• Detects the orientation of default shading space.\n\n• Returns 1 if the shader is being called to evaluate opacity for shadow rays, or 0 if the shader is being called to evaluate for surface color.\n\n• Indicates whether the shader is being evaluated while doing UV rendering (e.g. texture unwrapping)\n\n• Returns the bounce mask for a light struct.\n\n• Returns the light id for a light struct.\n\n• Queries the renderer for a named property.\n\n• Imports a variable from the light shader for the surface.\n\n• Returns a BSDF that matches the output of the traditional VEX blinn function.\n\n• Returns a BSDF that matches the output of the traditional VEX specular function.\n\n• Queries the renderer for a named property.\n\n• Computes ambient occlusion.\n\n• Computes global illumination using PBR for secondary bounces.\n\n• Sends a ray from the position P along the direction D.\n\n• Imports a value sent by a shader in a gather loop.\n\n• Returns the vector representing the reflection of the direction against the normal.\n\n• Computes the amount of reflected light which hits the surface.\n\n• Returns the refraction ray given an incoming direction, the normalized normal and an index of refraction.\n\n• Computes the illumination of surfaces refracted by the current surface.\n\n• Queries the renderer for a named property.\n\n• Returns the background color for rays that exit the scene.\n\n• Evaluates a scattering event through the domain of a geometric object.\n\n• Sets the current light\n\n• Stores sample data in a channel, referenced by a point.\n\n• Imports a variable sent by a surface shader in an illuminance loop.\n\n• Returns the computed BRDFs for the different lighting models used in VEX shading.\n\n• Stores exported data for a light.\n\n• Use a different bsdf for direct or indirect lighting.\n\n• Sends a ray from P along the normalized vector D.\n\n• Returns a Lambertian translucence BSDF.\n\n• Computes the position and normal at given (u, v) coordinates, for use in a lens shader.\n\n• Writes color information to a pixel in the output image\n\n## Strings\n\n• Returns the full path of a file.\n\n• Converts an unicode codepoint to a UTF8 string.\n\n• Concatenate all the strings specified to form a single string.\n\n• Decodes a variable name that was previously encoded.\n\n• Decodes a geometry attribute name that was previously encoded.\n\n• Decodes a node parameter name that was previously encoded.\n\n• Decodes a UTF8 string into a series of codepoints.\n\n• Encodes any string into a valid variable name.\n\n• Encodes any string into a valid geometry attribute name.\n\n• Encodes any string into a valid node parameter name.\n\n• Encodes a UTF8 string from a series of codepoints.\n\n• Indicates the string ends with the specified string.\n\n• Finds an item in an array or string.\n\n• Returns 1 if all the characters in the string are alphabetic\n\n• Returns 1 if all the characters in the string are numeric\n\n• Converts an integer to a string.\n\n• Concatenate all the strings of an array inserting a common spacer.\n\n• Strips leading whitespace from a string.\n\n• This function returns 1 if the subject matches the pattern specified, or 0 if the subject doesn’t match.\n\n• Returns the integer value of the last sequence of digits of a string\n\n• Converts an UTF8 string into a codepoint.\n\n• Converts an English noun to its plural.\n\n• Matches a regular expression in a string\n\n• Finds all instances of the given regular expression in the string\n\n• Returns 1 if the entire input string matches the expression\n\n• Replaces instances of regex_find with regex_replace\n\n• Splits the given string based on regex match.\n\n• Computes the relative path for two full paths.\n\n• Returns the relative path to a file.\n\n• Replaces occurrences of a substring.\n\n• Replaces the matched string pattern with another pattern.\n\n• Strips trailing whitespace from a string.\n\n• Splits a string into tokens.\n\n• Splits a file path into the directory and name parts.\n\n• Formats a string like printf but returns the result as a string instead of printing it.\n\n• Returns 1 if the string starts with the specified string.\n\n• Strips leading and trailing whitespace from a string.\n\n• Returns the length of the string.\n\n• Returns a string that is the titlecase version of the input string.\n\n• Converts all characters in string to lower case\n\n• Converts all characters in string to upper case\n\n## Subdivision Surfaces\n\n• Evaluates a point attribute at the subdivision limit surface using Open Subdiv.\n\n• Evaluates a vertex attribute at the subdivision limit surface using Open Subdiv.\n\n• Outputs the Houdini face and UV coordinates corresponding to the given coordinates on an OSD patch.\n\n• Outputs the OSD patch and UV coordinates corresponding to the given coordinates on a Houdini polygon face.\n\n• Returns a list of patch IDs for the patches in a subdivision hull.\n\n## Tetrahedrons\n\n• Returns primitive number of an adjacent tetrahedron.\n\n• Returns vertex indices of each face of a tetrahedron.\n\n## Texturing\n\n• Looks up a (filtered) color from a texture file.\n\n• The depthmap functions work on an image which was rendered as a z-depth image from mantra.\n\n• Returns the color of the environment texture.\n\n• Perform UDIM or UVTILE texture filename expansion.\n\n• Test string for UDIM or UVTILE patterns.\n\n• Remaps a texture coordinate to another coordinate in the map to optimize sampling of brighter areas.\n\n• Evaluates an ocean spectrum and samples the result at a given time and location.\n\n• Computes a filtered sample from a ptex texture map. Use texture instead.\n\n• Looks up an unfiltered color from a texture file.\n\n• The shadowmap function will treat the shadow map as if the image were rendered from a light source.\n\n• Imports attributes from texture files.\n\n• Similar to sprintf, but does expansion of UDIM or UVTILE texture filename expansion.\n\n• Computes a filtered sample of the texture map specified.\n\n## Transforms and Space\n\n• Computes the rotation matrix or quaternion which rotates the vector a onto the vector b.\n\n• Transforms a position from normal device coordinates to the coordinates in the appropriate space.\n\n• Gets the transform of a packed primitive.\n\n• Returns a transform from one space to another.\n\n• Creates an instance transform matrix.\n\n• Computes a rotation matrix or angles to orient the negative z-axis along the vector (to-from) under the transformation.\n\n• Builds a 3×3 or 4×4 transform matrix.\n\n• Returns the camera space z-depth of the NDC z-depth value.\n\n• Transforms a normal vector.\n\n• Create an orthographic projection matrix.\n\n• Transforms a normal vector from Object to World space.\n\n• Transforms a position value from Object to World space.\n\n• Transforms a direction vector from Object to World space.\n\n• Transforms a packed primitive.\n\n• Create a perspective projection matrix.\n\n• Computes the polar decomposition of a matrix.\n\n• Applies a pre rotation to the given matrix.\n\n• Prescales the given matrix in three directions simultaneously (X, Y, Z - given by the components of the scale_vector).\n\n• Pretranslates a matrix by a vector.\n\n• Transforms a vector from one space to another.\n\n• Applies a rotation to the given matrix.\n\n• Rotates a vector by a rotation that would bring the x-axis to a given direction.\n\n• Scales the given matrix in three directions simultaneously (X, Y, Z - given by the components of the scale_vector).\n\n• Sets the transform of a packed primitive.\n\n• Returns the closest equivalent Euler rotations to a reference rotation.\n\n• Applies an inverse kinematics algorithm to a skeleton.\n\n• Applies a curve inverse kinematics algorithm to a skeleton.\n\n• Applies a full-body inverse kinematics algorithm to a skeleton.\n\n• Applies an inverse kinematics algorithm to a skeleton.\n\n• Applies a full-body inverse kinematics algorithm to a skeleton, with optional control over the center of mass.\n\n• Transforms a position into normal device coordinates.\n\n• Translates a matrix by a vector.\n\n• Transforms a normal vector from Texture to World space.\n\n• Transforms a position value from Texture to World space.\n\n• Transforms a direction vector from Texture to World space.\n\n• Transforms a directional vector.\n\n• Transforms a normal vector from World to Object space.\n\n• Transforms a position value from World to Object space.\n\n• Transforms a direction vector from World to Object space.\n\n• Transforms a normal vector from World to Texture space.\n\n• Transforms a position value from World to Texture space.\n\n• Transforms a direction vector from World to Texture space.\n\n## usd\n\n• Creates an attribute of a given type on a primitive.\n\n• Excludes an object from the collection\n\n• Includes an object in the collection\n\n• Appends an inversed transform operation to the primitive’s transform order\n\n• Applies a quaternion orientation to the primitive\n\n• Creates a primitive of a given type.\n\n• Creates a primvar of a given type on a primitive.\n\n• Adds a target to the primitive’s relationship\n\n• Applies a rotation to the primitive\n\n• Applies a scale to the primitive\n\n• Appends a transform operation to the primitive’s transform order\n\n• Applies a transformation to the primitive\n\n• Applies a translation to the primitive\n\n• Reads the value of an attribute from the USD primitive.\n\n• Reads the value of an element from an array attribute.\n\n• Returns the length of the array attribute.\n\n• Returns the names of the attributes available on the primitive.\n\n• Returns the tuple size of the attribute.\n\n• Returns the time codes at which the attribute values are authored.\n\n• Returns the name of the attribute type.\n\n• Blocks the attribute.\n\n• Blocks the primvar.\n\n• Blocks the primvar.\n\n• Blocks the primitive’s relationship\n\n• Returns the material path bound to a given primitive.\n\n• Clears the value of the metadata.\n\n• Clears the primitive’s transform order\n\n• Obtains the list of all objects that belong to the collection\n\n• Checks if an object path belongs to the collection\n\n• Obtains the object paths that are in the collection’s exclude list\n\n• Obtains the collection’s expansion rule\n\n• Obtains the object paths that are in the collection’s include list\n\n• Returns the primitive’s draw mode.\n\n• Retrurns primitive’s transform operation full name for given the transform operation suffix\n\n• Reads the value of a flattened primvar directly from the USD primitive or from USD primitive’s ancestor.\n\n• Reads an element value of a flattened array primvar directly from the USD primitive or from its ancestor.\n\n• Reads the value of an flattened primvar directly from the USD primitive.\n\n• Reads an element value of a flattened array primvar directly from a USD primitive.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the primitive.\n\n• Returns the center of the bounding box for the primitive.\n\n• Returns the maximum of the bounding box for the primitive.\n\n• Returns the minimum of the bounding box for the primitive.\n\n• Returns the size of the bounding box for the primitive.\n\n• Obtains the primitive’s bounds\n\n• Obtains the primitive’s bounds\n\n• Checks if the primitive adheres to the given API.\n\n• Checks if the primitive adheres to the given API.\n\n• Reads the value of a primvar directly from the USD primitive or from USD primitive’s ancestor.\n\n• Reads the value of an element from the array primvar directly from the USD primitive or from USD primitive’s ancestor.\n\n• Returns the element size of the primvar directly from the USD primitive or from USD primitive’s ancestor.\n\n• Returns the index array of an indexed primvar directly on the USD primitive or on USD primitive’s ancestor.\n\n• Returns the element size of the primvar directly on the USD primitive or on USD primitive’s ancestor.\n\n• Returns the length of the array primvar directly on the USD primitive or on USD primitive’s ancestor.\n\n• Returns the names of the primvars available directly on the given USD primitive or on USD primitive’s ancestor.\n\n• Returns the tuple size of the primvar directly on the USD primitive or on USD primitive’s ancestor.\n\n• Returns the time codes at which the primvar values are authored directly on the given primitive or on its ancestor.\n\n• Returns the name of the primvar type found on the given primitive or its ancestor.\n\n• Checks if the primitive is abstract.\n\n• Checks if the primitive is active.\n\n• Checks if the attribute is an array.\n\n• Checks if there is an array primvar directly on the USD primitive or on USD primitive’s ancestor.\n\n• Checks if the given metadata is an array.\n\n• Checks if there is an array primvar directly on the USD primitive.\n\n• Checks if the primitive has an attribute by the given name.\n\n• Checks if the collection exists.\n\n• Checks if the path is a valid collection path.\n\n• Checks if there is an indexed primvar directly on the USD primitive or on USD primitive’s ancestor.\n\n• Checks if there is an indexed primvar directly on the USD primitive.\n\n• Checks if the primitive is an instance.\n\n• Checks if the primitive or its ancestor has a primvar of the given name.\n\n• Checks if the primitive is of a given kind.\n\n• Checks if the primitive has metadata by the given name.\n\n• Checks if the primitive is a model.\n\n• Checks if the path refers to a valid primitive.\n\n• Checks if the primitive has a primvar of the given name.\n\n• Checks if the primitive has a relationship by the given name.\n\n• Checks if the stage is valid.\n\n• Checks if the primitive transform is reset\n\n• Checks if the primitive is of a given type.\n\n• Checks if the primitive is visible.\n\n• Returns the primitive’s kind.\n\n• Obtains the primitive’s local transform\n\n• Constructs an attribute path from a primitive path and an attribute name.\n\n• Constructs a collection path from a primitive path and a collection name.\n\n• Constructs an property path from a primitive path and an property name.\n\n• Constructs an relationship path from a primitive path and a relationship name.\n\n• Forces a string to conform to the rules for naming USD primitives.\n\n• Forces a string to conform to the rules for paths to USD primitives.\n\n• Returns the length of the array metadata.\n\n• Returns the names of the metadata available on the object.\n\n• Returns the name of the primitive.\n\n• Returns the path of the primitive’s parent.\n\n• Sets two vectors to the minimum and maximum corners of the bounding box for the given instance inside point instancer.\n\n• Returns the center of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the maximum position of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the minimum position of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the size of the bounding box for the instance inside a point instancer primitive.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Obtains the transform for the given point instance\n\n• Reads the value of a primvar directly from the USD primitive.\n\n• Returns the namespaced attribute name for the given primvar.\n\n• Reads the value of an element from the array primvar directly from the USD primitive.\n\n• Returns the element size of the primvar directly from the USD primitive.\n\n• Returns the index array of an indexed primvar directly on the USD primitive.\n\n• Returns the element size of the primvar directly on the USD primitive.\n\n• Returns the length of the array primvar directly on the USD primitive.\n\n• Returns the names of the primvars available on the given USD primitive.\n\n• Returns the tuple size of the primvar directly on the USD primitive.\n\n• Returns the time codes at which the primvar values are authored directly on the given primitive.\n\n• Returns the name of the primvar type found on the given primitive.\n\n• Returns the primitive’s purpose.\n\n• Obtains the relationship forwarded targets.\n\n• Returns the names of the relationships available on the primitive.\n\n• Obtains the relationship targets.\n\n• Returns the relative position of the point given with respect to the bounding box of the geometry.\n\n• Remove a target from the primitive’s relationship\n\n• Sets the primitive active state.\n\n• Sets the value of an attribute.\n\n• Sets the value of an element in an array attribute.\n\n• Sets the excludes list on the collection\n\n• Sets the expansion rule on the collection\n\n• Sets the includes list on the collection\n\n• Sets the primitive’s draw mode.\n\n• Sets the primitive’s kind.\n\n• Sets the value of an metadata.\n\n• Sets the value of an element in an array metadata.\n\n• Sets the value of a primvar.\n\n• Sets the value of an element in an array primvar.\n\n• Sets the element size of a primvar.\n\n• Sets the indices for the given primvar.\n\n• Sets the interpolation of a primvar.\n\n• Sets the primitive’s purpose.\n\n• Sets the targets in the primitive’s relationship\n\n• Sets the primitive’s transform order\n\n• Sets/clears the primitive’s transform reset flag\n\n• Sets the selected variant in the given variant set.\n\n• Configures the primitive to be visible, invisible, or to inherit visibility from the parent.\n\n• Makes the primitive visible or invisible.\n\n• Returns the primitive’s specifier.\n\n• Constructs a full name of a transform operation\n\n• Obtains the primitive’s transform order\n\n• Extracts the transform operation suffix from the full name\n\n• Infers the transform operation type from the full name\n\n• Returns the name of the primitive’s type.\n\n• Constructs a unique full name of a transform operation\n\n• Returns the variants belonging to the given variant set on a primitive.\n\n• Returns the currently selected variant in a given variant set.\n\n• Returns the variant sets available on a primitive.\n\n• Obtains the primitive’s world transform\n\n## Utility\n\n• Returns 1 if the VEX assertions are enabled (see HOUDINI_VEX_ASSERT) or 0 if assertions are disabled. Used the implement the assert macro.\n\n• An efficient way of extracting the components of a vector or matrix into float variables.\n\n• Reports a custom runtime VEX error.\n\n• Extracts a single component of a vector type, matrix type, or array.\n\n• Parameters in VEX can be overridden by geometry attributes (if the attributes exist on the surface being rendered).\n\n• Check whether a VEX variable is varying or uniform.\n\n• Ends a long operation.\n\n• Start a long operation.\n\n• Reversibly packs an integer into a finite, non-denormal float.\n\n• Prints a message only once, even in a loop.\n\n• Prints values to the console which started the VEX program.\n\n• Evaluates a Houdini-style ramp at a specific location.\n\n• Packs a set of arrays into a string-encoded ramp.\n\n• Unpacks a string-encoded ramp into a set of arrays.\n\n• Returns one of two parameters based on a conditional.\n\n• Creates a new value based on its arguments, such as creating a vector from its components.\n\n• Sets a single component of a vector or matrix type, or an item in an array.\n\n• Yields processing for a certain number of milliseconds.\n\n• Rearranges the components of a vector.\n\n• Reverses the packing of pack_inttosafefloat to get back the original integer.\n\n• Reports a custom runtime VEX warning.\n\n## volume\n\n• Returns the volume of the microvoxel containing a variable such as P.\n\n• Samples the volume primitive’s value.\n\n• Samples the volume primitive’s value.\n\n• Calculates the volume primitive’s gradient.\n\n• Gets the value of a specific voxel.\n\n• Gets the active setting of a specific voxel.\n\n• Gets the index of the bottom left of a volume primitive.\n\n• Converts a volume voxel index into a position.\n\n• Gets the vector value of a specific voxel.\n\n• Converts a position into a volume voxel index.\n\n• Gets the resolution of a volume primitive.\n\n• Samples the volume primitive’s value.\n\n• Samples the volume primitive’s vector value.\n\n• Samples the volume primitive’s value.\n\n• Samples the volume primitive’s value.\n\n• Computes the approximate diameter of a voxel."
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https://ask.sagemath.org/question/10374/why-cant-sage-solve-the-equation-xn-2x/?sort=votes | [
"# why can't sage solve the equation x^n == 2*x ?",
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"This post is a wiki. Anyone with karma >750 is welcome to improve it.\n\nsage: var('x n')\n(x, n)\nsage: solve(x^n==2*x,x)\n[x == 1/2*x^n]\n\n\nThis is not what I expect.\n\nWho can help?\n\nThanks.\n\nedit retag close merge delete\n\nOne issue here is that after declaring that x and n are symbolic variables, Sage still does not know anything more than that about x and n. So, Sage does not know whether x and n are reals, integers, etc. Thus, we need to assume(n,'integer') and assume(x,'real'). Unfortunately, at this point, you still don't get a solution. Sage is calling Maxima to solve this equation, and I cannot get Maxima to solve it either. Ideas?\n\nSort by » oldest newest most voted\n\nSince you named one variable n I assume you meant integer. But even this has infinitely many solutions! Sage can give them for specific n:\n\nsage: [solve(x^n==2*x,x) for n in range(1,5)]\n[[x == 0],\n[x == 0, x == 2],\n[x == -sqrt(2), x == sqrt(2), x == 0],\n[x == 1/2*I*sqrt(3)*2^(1/3) - 1/2*2^(1/3), x == -1/2*I*sqrt(3)*2^(1/3) - 1/2*2^(1/3), x == 2^(1/3), x == 0]]\n\nmore"
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https://mail.freeetextbooks.com/gcse-maths-notes/633-proof-that-opposite-angles-in-a-cyclic-quadrilateral-add-to-180-degrees.html | [
"## Proof That Opposite Angles in a Cyclic Quadrilateral Add to 180 Degrees\n\nThe diagram below illustrates the theorem.",
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"",
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"and",
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"From C and B draw lines to the centre of the circle. If",
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"then",
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"as below and if",
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"then",
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"so",
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"https://mail.freeetextbooks.com/gcse-maths/proof-that-opposite-angles-in-a-cyclic-quadrilateral-add-to-180-degrees-html-m656d559.gif",
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"https://mail.freeetextbooks.com/component/jcomments/captcha/84532.html",
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https://www.askiitians.com/forums/Analytical-Geometry/24/43398/conic-sections.htm | [
"",
null,
"# if the vertics and the focous of the parabola are (-1,1)and (2,3) respectively,then equation of its directrix is",
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"11 years ago\n\nsince the vertex of the parabola is the mid point of of the line joining the focal point and point on the directrix .therefore let the point on the directrix be (x1,y1).\n\nvertex is (-1,1)and focus is (2,3) take out the point on directrix by mid point formula . its (-4,-1)\n\nnow the slope of line joining vertex and focus is 3-1/2+1=2/3=m1; (slope formula)\n\nsince directrix is perpendicular to th line jopining vertex and focus ;by slpoe of perpendicular lines- m1*m2=-1\n\nm2=-3/2(slope of the directrix)\n\nput this in sliope point form\n\npoint is (-4,-1) and slope= -3/2\n\neq. is y+1=-3/2(x+4)\n\nsolve it!!hope it works!!!!\n\n11 years ago\n\nthe vertices is the mid pt. of focus and directrix.let coordinates of dirextrix be(x,y)\n\nso,-1=x+2/2\n\n=-2=x+2\n\nx=-4\n\n1=y+3/2\n\n=2=y+3\n\ny=-1\n\nusing slope formula\n\nthe slope of line joining vertex and focus is 3-1/2+1=2/3=m1\n\ndirectrix is perpendicular to th line jopining vertex and focus ;by slpoe of perpendicular lines- m1 x m2=-1\n\nm2=-3/2\n\nput in point slope form\n\neq. is y+1=-3/2(x+4)\n\nfind the eqn"
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https://www.vedantu.com/question-answer/akash-bought-vegetables-weighing-10-kg-out-of-class-11-maths-cbse-5f5e78e56e663a29cc564272 | [
"",
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"Akash bought vegetables weighing 10 kg. out of this, 3 kg 500g is onions, 2 kg 75g is tomatoes and the rest are potatoes. What is the weight of the potatoes?",
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"Verified\n146.4k+ views\nHint: In this question, we will proceed by considering the weight of the potatoes as a variable. Then convert the grams into kilograms. And equate the total weight of the vegetables to the sum of weights of onions, tomatoes, and potatoes to get the required answer. So, use this concept to reach the solution to the given problem.\n\nComplete step-by-step solution:\nGiven that,\nThe total weight of the vegetables = 10 kg\nWeight of the onions = 3 kg 500g\nWeight of the tomatoes = 2 kg 75g\nLet the weight of the potatoes $= x{\\text{ kg}}$\nWe know that, one kilogram is equal to thousand grams i.e., 1 kg = 1000 grams\nConverting the weights of onions and tomatoes we have\nWeight of the onions $= 3{\\text{ kg}} + \\dfrac{{500}}{{1000}}{\\text{ kg}} = \\left( {3 + 0.5} \\right){\\text{ kg}} = 3.5\\;{\\text{kg}}$\nWeight of the tomatoes $= 2{\\text{ kg}} + \\dfrac{{75}}{{1000}}{\\text{ kg}} = \\left( {2 + 0.075} \\right){\\text{ kg}} = 2.075\\;{\\text{kg}}$\nNow, total weight of vegetables = weight of onions + weight of tomatoes + weight of potatoes\n$\\Rightarrow 10 = 3.5 + 2.075 + x \\\\ \\Rightarrow 10 = 5.575 + x \\\\ \\therefore x = 10 - 5.575 = 4.425{\\text{ kg}}$\nTherefore, weight of potatoes $= 4.425{\\text{ kg}} = \\left( {4 + \\dfrac{{425}}{{1000}}} \\right){\\text{ kg}} = 4{\\text{ kg 425 grams}}$\n\nThus, weight of the potatoes = 4 kg 425 grams.\n\nNote: Use the conversion 1 kg = 1000 grams to convert the weight from kilograms into grams. Here the weight of potatoes should not be more than 10 kg as the total weight of the vegetables is 10 kg. We can also solve this by converting all the weight in terms of grams and then proceed. Both methods are correct."
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https://www-tracey.archive.org/stream/arxiv-0906.2498/0906.2498_djvu.txt | [
"# Full text of \"Visualization of semileptonic form factors from lattice QCD\"\n\n## See other formats\n\n```Visualization of semileptonic form factors from lattice QCD\n\no\no\n\n<\nO\n\ncn\n\n£^\n\n(N\n>\n00\n\no\n\nON\n\no\n\nC. Bernard,! C. DeTar,^ M. Di Pierro,^ A.X. El-Khadra,-* R.T. Evans,\"* E.D. Freeland,^ E. Gamiz,'*\nSteven Gottlieb,^ U.M. Heller,'^ J.E. Hetrick,^ A.S. Kronfeld,^ J. Laiho,* L. Levkova,^ P.B. Mackenzie,^\nM. Okamoto,^ M.B. Oktay,^ J.N. Simone,^ R. Sugar,io D. Toussaint,\" and R.S. Van de Water^^\n\n(Fermilab Lattice and MILC Collaborations)\n\n^Department of Physics, Washington University, St. Louis, Missouri, USA\n'^Physics Department, University of Utah, Salt Lake City, Utah, USA\n'^School of Computer Science, Telecommunications and Information Systems, DePaul University, Chicago, Illinois, USA\n'^Physics Department, University of Illinois, Urbana, Illinois, USA\n^Liberal Arts Department, The School of the Art Institute of Chicago, Chicago, Illinois, USA\n''Department of Physics, Indiana University, Bloomington, Indiana, USA\n\n''American Physical Society, Ridge, New York, USA\n^Physics Department, University of the Pacific, Stockton, California, USA\nFermi National Accelerator Laboratory, Batavia, Illinois, USA\n'^\"Department of Physics, University of California, Santa Barbara, California, USA\n''Department of Physics, University of Arizona, Tucson, Arizona, USA\n'^Physics Department, Brookhaven National Laboratory, Upton, New York, USA\n\n(Dated: August 26, 2009)\n\nComparisons of lattice-QCD calculations of semileptonic form factors with experimental measure-\nments often display two sets of points, one each for lattice QCD and experiment. Here we propose\nto display the output of a lattice-QCD analysis as a curve and error band. This is justified, because\nlattice-QCD results rely in part on fitting, both for the chiral extrapolation and to extend lattice-\nQCD data over the full physically allowed kinematic domain. To display an error band, correlations\nin the fit parameters must be taken into account. For the statistical error, the correlation comes\nfrom the fit. To illustrate how to address correlations in the systematic errors, we use the Becirevic-\nKaidalov parametrization of the D nlf and D Klv form factors, and an analyticity-based fit\nfor the B — > ttIu form factor /+.\n\nPACS numbers: 13.20.Fc,13.20.He,12.38.Gc\n\nThe past several years have witnessed considerable im-\nprovement in our understanding of semileptonic decays\nof D and B mesons. Measurements have advanced in\naccuracy from 6-20% on the normalization [l|J3, 0] and\n~ 10% on the shape to 1% on both Mean-\nwhile, ab initio calculations in QCD with lattice gauge\ntheory have become realistic [1, 0, H, |^ , now incorporat-\ning the effects of sea quarks that were omitted in earlier\nwork 0, [m, [H, [H, d. In this article, we discuss how\nto present both together, so that the agreement (or, in\nprinciple, lack thereof) is easy to assess.\n\nWe focus on reactions mediated by electroweak vector\ncurrents, leading to pseudoscalar mesons, tt or K , in the\nfinal state. At the quark level, a heavy quark h decays\ninto a daughter quark d (not necessarily the down quark) ,\nwith a spectator antiquark q. Writing the decay H — >\nPlv, the form factors are defined by\n\n(PIV^IH) = f+{q^){pH +PP - AT + Mq^)A^^ (1)\n\ndenote the 3-momentum and energy of the final-state me-\nson in the rest frame of the initial state. The energy E\nis related to q\"^ via\n\n2mHE.\n\n(3)\n\nNeglecting the lepton mass, < g'^ < qf^ax — {rriH—mp)'^\nis kinematically allowed in the semileptonic decay.\n\nThe form factors f+{q^) and fo{<l^) are related to /|| {E)\n^nd f^{E) by\n\nf+{q') = {2mH)-'/' [fi\\iE) + {mn ~ E)f^iE)] , (4)\n\n[{mH^E)f\\\\iE)-plME)] , (5)\n\nimply\n\nwhere q = pn — pp is the 4-momentum of the lepton\nsystem, and = {pH+Pp)-qq^ / (f' = {'m% — mp)q'^/q'^.\nEquation ^ is general, applying to K ^ -kIv as well as to\nD and B decays. For lattice QCD, it is more convenient\nto express the transition matrix element as\n\n{P\\V^\\H) = ,/^[v^^f\\\\{E)+py^{E)\\ , (2)\nwhere v — pn/mH, and p± = pp — Ev and E — v ■ pp\n\nwith Eq. 13]) understood. Equations () and\n/_^_(0) = /o(0), as required in Eq. ^.\n\nTwo aspects of lattice-QCD calculations are important\nhere. First (as in all lattice-QCD calculations), it is com-\nputationally demanding to have a spectator quark with\nmass as small as those of the up and down quarks; for\nP — TT the same applies to the daughter quark. In re-\ncent unquenched calculations, the mass of the qq pseu-\ndoscalar Pqq lies in the range 0.1m|- < 'rnp_^ < mj^. Sec-\nond (of special importance in semileptonic decays), the\ncalculations take place in a finite spatial volume, so the\n3-momentum takes discrete values. In typical cases the\n\n2\n\nbox-size L Ki 2.5 fm, so the smallest nonzero momentum\nP(i,o,o) = 27r(l,0,0)/L satisfies |P(i,o,o)l ~ 500 MeV.\n\nAfter generating numerical data at several values of\n{E,mp_J, the next step for lattice-QCD calculations is\nto carry out a chiral extrapolation, ?Tip_^ — > m^, of the\ndata for and /|| [H, The chiral extrapolation\n\nmust refle ct the fac t that the form factors are analytic\nin _E = \\/p'^ + ^p, not p Note also that f± and\n\n/-!_ can be computed only with p =/= 0, hence E > mp\nor, equivalently, < (/nj^x- The statistical and dis-\ncretization uncertainties in [q^ , TOp_ ) start out small-\nest at q^^ Q corresponding to P(i,o,o)- A sensible chiral\n\nextrapolation will propagate this feature to /+(g^,m^).\nSimilarly, the statistical and discretization uncertainties\nin /o(g^,mJ) are smallest near g^^x-\n\nWhen \\p\\a becomes too large, discretization effects\ngrow out of control. Therefore, the kinematic domain\nof lattice-QCD calculations is limited to, these days,\nIpI ^ 1 GeV, with a corresponding upper limit on E\nand lower limit on q^ . To extend the form factor over\nthe full physical kinematic domain, a parametrization of\nthe q^ dependence is needed.\n\nOne choice is the Becirevic-Kaidalov (BK) ansatz [Ig]\n\nF\n\nF\n\naqp)\n\n(6)\n(7)\n\nwhere = q^/m^, {H* is the vector meson of flavor hd),\nand F, a, and (3 are free parameters to be fitted. A key\nfeature of Eq. © is the built-in pole at q\"^ = m%, , or\nE = —{mjjt. — mfj — rn?p)/2mH < 0, an indisputable fea-\nture of the physical /+. Further singularities at higher\nnegative energy are modeled by the BK parameters a\nand (3. A similar possibility is the Ball-Zwicky (BZ)\nansatz IToj . which has one more parameter for /+\nthan BK. A shortcoming of these parametrizations is that\ncomparisons of lattice-QCD and experimental slope pa-\nrameters can be misleading 0,[2l|, because lattice-QCD\nslopes are determined near q^ = g^^x; whereas experi-\nmental slopes are determined near q^ — 0.\n\nAnother approach based on analyticity and unitarity\nis to write the form factors as\n\n1\n\n1 ^\n\nN\n\n(8)\n\n(9)\n\nwhere (/>+,o are arbitrary, but suitable, functions, and the\nseries coefficients are fit parameters. The variable\n\n(10)\n\n^\\-q^/t+ + ^l-to/t+'\n\nwhere i+ = {mn + mp)^ and to can be chosen to make\n1 2: 1 small for all kinematically allowed q^. Like BK and\n\nBZ, Eq. dH]) builds the H* pole into /+, but this approach\nis model independent because unitarity [13, [H, [23l] and\nheavy-quark physics [2l| impose bounds on X^fel^fcP'\nJ2k l^fcP; a-iid because kinematics set \\z\\ < 1. Conse-\nquently, the series can be truncated safely, once addi-\ntional terms are negligible compared to other uncertain-\nties in the analysis.\n\nIn all approaches the output of an analysis of lattice-\nQCD form factors is a fit, usually a two-stage fit of chiral\nextrapolation followed by q'^ parametrization. Clearly,\nthe final fit describes a curve, and the error matrix of\nthe fit parameters describes an error band. Nevertheless,\nlattice-QCD results usually have been plotted as a set\nof points with error bars at fiducial values of q^ (or E).\nThese points evoke the underlying discrete nature of the\n3-momentum p but, in general, the chosen values of q^ (or\nE) have nothing to do with the original discrete values\nof p. A plot with a curve plus error band exhibits the\nsame information, while giving a visually superior sense\nof the correlations between points on the curve.\n\nThe experimental measurements of /_|_(g^) come from\ncounting events in bins of q^ and removing coupling and\nkinematic factors. The analysis inevitably entails some\nfitting, to correct for acceptance, etc., but the postfit bins\nof q^ faithfully mirror the input to such fits.\n\nIf one would like to compare the calculations with the\nmeasurements, it is appealing to represent one as a curve\nwith error band, and the other as points with error bars.\nBearing the foregoing remarks in mind, it seems natural\nto draw the curve for lattice-QCD calculations. A few\nyears ago, we prepared illustrative plots for D Klv\nwith the Fermilab-MILC m, B lattice-QCD calculations\nand FOCUS and Belle [2J] measurements. The intent\nwas pedagogical, and we showed the plots at seminars\nand conferences (25| .\n\nUnfortunately, the error band in that effort was im-\npressionistic, not rigorous. With the prospect of yet-\nmore-precise results based on CLEO-c's full accumula-\ntion of 818 pb^^ , we now present a version that treats\nthe error band as rigorously as possible. We also prepare\nplots for D — *■ 'kIv and B — > irlv.\n\nAs before we shall base the plots for D decays on\nRef. . The final result of this analysis consists of the\nBK parameters {F, a, [3) and the 3x3 error matrix. The\nfull statistical error matrix is contained in a detailed, un-\npublished description of a BK-based analysis of i? — > nlv\nform factors [2^. The best fit, statistical errors, and sys-\ntematic errors are tabulated in Table [H The statistical\ncorrelation matrices pij = CTfj/(o'^iO'|j)^^^ are tabulated\nin Table ini The correlations among systematic errors are\ndiscussed below.\n\nPropagating (correlated) fluctuations in F, a, and (3\nto the form factors, one flnds relative squared-errors\n\n(11)\n\n4+\n\n„2\nCTpp\n\n^2 -2\n1 r,<^Fa q\n\n+ ^aa\n\n(\n\nfl\n\nF2\n\nF 1-aq^\n\n\\ 1 — aq'^\n\n<jIp\n\n' e \\\n\nf'o\n\nF2\n\nFj3 I3~q^\n\n13^ \\\n\nj-qy\n\n(12)\n\n3\n\nThese errors are plotted as a function of in Fig. [T] as\nsolid curves. The relative statistical errors are smallest\nfor such that\n\n(13)\n(14)\n\nIt is illustrative to take a\"^^ and a|n^ from Tables HI and HIl\nand solve Eqs. (fT3| and (fT4|) for g^. We call these values\n\nand g| and tabulate them, as well as. qj^^^^^ and q^^y.,\nin Table IIIII As one can see from Fig. [T| and Table [ml\nthe statistical error is smallest between g^j^ g p-j and gi^axi\nas expected. One may view this outcome as a check on\nthe fitting procedures.\n\nOne can reverse this strategy to determine the corre-\nlation between the systematic errors of F and the slope\nparameters. In the error budget of Ref. @ the largest\nsystematic effect comes from discretization errors. These\nshould be smallest around q^^ q and g^ax for /+ and /o,\nrespectively, because those correspond to the smallest |p|\nyielding the respective matrix elements. This yields\n\nsyst\nPPa\n\n-0.198 {D\n\nPpfi\n\nK),\n\n-0.329 {D\n-0.533 {D\n\n(15)\n(16)\n\nand the dashed curves in Fig. [TJ\n\nIt is customary to combine statistical and systematic\nuncertainties by adding the two cr^ (matrices). Carrying\nout this procedure leads to the curves and bands in Fig. [21\nThe error bands seem to contradict the conventional wis-\ndom that the lattice-QCD uncertainties are smallest near\n9max- This is not entirely the case for the relative error, as\nseen in Fig.[Tl As increases, the relative errors decrease\nuntil hitting a minimum somewhere between q^^ ^ and\n\n9maxj is reasonable. The form factors rise faster than\nthe relative errors drop, leading to the increasing absolute\nerror seen in Fig. [21 These features are not an artifact of\nthe BK parametrization, as we shall see below with the\nB -kIv form factor /+.\n\nthe form factors /+ and /o for D Klv and D irlv.\nThe lattice-QCD results are shown as curves (red for /+,\nblue for /o) with two errors bands, one statistical (orange\nfor /+, gray for /o), the other systematic and statistical\ncombined (yellow for /-i-, light blue for /o). Experimental\n\nmeasurements for\n\nare overlaid as points\n\nwith error bars. It may require careful scrutiny to see\nwhich experiment is which, but a glance reveals how well\nthe points and curves agree. The agreement is good for\nD ttIv and very good for D Klv.\n\nFor the z expansion the propagation of errors is even\nsimpler. Focusing on /+, one has from Eq. ()\n\nN\n\n^2 ^k+l\n\nO 1.1 ^\n\nfi\n\nZ^k=0 \"-k^\n\n(17)\n\nwhere the indices on correspond to those on the se-\nries coefficients. The coefficients and error matrix for the\n\nTABLE I: Best-fit values of BK parameters with statistical\nand systematic errors, successively, in parentheses [^. [tI [2^.\n\nDecay\n\nF\n\nD^Klv 0.73(3)(7) 0.50(4)(7) 1.31(7)(13)\nD-^ttIp 0.64(3)(6) 0.44(4)(7) 1.41(6)(7)\n\nTABLE IL Statistical error correlation matrices pij\n(^fj /{'^'ii'^'jjy^^ of the BK parameters |26j |.\n\nD\n\nK\n\nF\n\na\n\nF\n\n1.000\n\n-0.597\n\n0.530\n\na\n\n-0.597\n\n1.000\n\n-0.316\n\nP\n\n0.530\n\n-0.316\n\nl.QQQ\n\nD\n\nTV\n\nF\n\na\n\nP\n\nF\n\n1.000\n\n-0.583\n\n0.535\n\na\n\n-0.583\n\n1.000\n\n-0.312\n\n0.535\n\n-0.312\n\n1.000\n\n2 2\n\nq tni^*\n\n'max Z),\n\n0.100\n\nD^Klv\n\n— U\n\n— fa\n\n0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45\n\n2,\n\n'max D\n\n0.100\n\nD->nlv\n\n0.1\n\n0.4\n2 2\nq Im ,\n\nFIG. 1: Relative errors vs . Solid (dashed) curves show\nthe fitted statistical (estimated systematic) error for /+ (red\ncurves) and /o (blue curves). Vertical lines show 9^1,0,0)\n\n2\n\n4\n\nTABLE III: Useful quantities for generating and assessing Figs. [l][2l and|3]\n\nDecay\n\nH*\n\n2\n\n9(1,0,0)\n\n2\n\n9m ax\n\n-2\n\n9(1,0,0)\n\n~2\n\n9m ax\n\n-2\n\n9l\n\n(MeV)\n\n(MeV)\n\n(GeV^)\n\n(GeV^)\n\nD Klv\n\nDt\n\n2112\n\n704\n\n1.10\n\n1.88\n\n0.25\n\n0.42\n\n0.47\n\n0.38\n\nD ~* nlf\n\nD*\n\n2008\n\n518\n\n1.57\n\n3.00\n\n0.39\n\n0.74\n\n0.53\n\n0.52\n\nB Tvlf\n\nB*\n\n5325\n\n518\n\n22.4\n\n26.4\n\n0.79\n\n0.93\n\niV = 3 fit (to = 0.65g,^a^) are tabulated in Tablc[lVl The\nz-series fit was carried out after assigning g^-dependent\nsystematic uncertainties, so Table HVl refers to the com-\nbined statistical and systematic errors of this analysis.\n\nThis information, combined with the outer function 0+\n0, is used to produce Fig. [31 the second main result\n\nFIG. 2: Form factors /+ (and /o) for semileptonic D decays,\nfrom lattice QCD 0, [2g, expressed as a red (blue) curve with\nan orange (gray) statistical error band and a yellow (light\nblue) combined error band. Error bands take correlations\ninto account. Measurements of /+ are from Belle (green di-\namonds) '2^, BaBar (magenta squares) (SJ, and CLEO-c\n(maroon triangles) ,2^, i2£] . The vertical line shows\n\nTABLE IV: Best-fit values at and correlation matrix pki of\nthe 3-term z expansion of /+ for B nil/, with statistical\nand systematic errors combined\n\nFit:\n\n0.0216(27)\n\n-0.0378(191)\n\n-0.113(27)\n\nP\n\nao\n\n0.1\n\na2\n\nfflo\n\n1.000\n\n0.640\n\n0.475\n\nai\n\n0.640\n\n1.000\n\n0.964\n\na2\n\n0.474\n\n0.964\n\n1.000\n\nof this paper. Now the curve and error band conform\nwith preconceptions, for several reasons. First, q^^ ^ is\n\nclose to Qn-jaxi rather than in the middle of the kinematic\nrange. Second, the chiral extrapolation in Ref. is less\naggressive than that in Ref. [a], leading to a larger but\nmore realistic error at 5^ = 0. The most striking aspect\nis that even though the absolute error in is increasing\nfor > 0.8, the band remains narrow. The band simply\nconveys the point-to-point correlations better.\n\nThis paper shows in detail how to compare semilep-\ntonic form factors from lattice-QCD and from experi-\nments. For illustration we use the BK parametrization\n\n2 2\n1 .\n\nI ' ' ' ' I ' ' ' ' I ' ' ' ' I ' ' ' ' I ' ' ' ' I ' ' ' ' I ' '\n\nB TZlV\n\nlattice QCD [Fennilab/MILC, 08 1 1 .3640 [hep-lat]]\n\n■ experiment [BaBar, hep-ex/06 12020]\n\n,2\n\n■fmax ■\n\nFIG. 3: Form factor /+ for B —> nlv expressed as a curve\n(red) from the best fit with a total error band (yellow) from\ntaking correlations in the fit parameters into account over-\nlaid with measurements of \\Vub\\f+/{3.38 x 10~^\n(magenta squares) [30|]. The vertical line shows g,\n\nfrom BaBar\n\n2\n\nmax ■\n\n5\n\nfor D Klv and D ttIu, and the z expansion for\nB ttIv. Clearly, the idea is more general. For ex-\nample, an interesting prospect relevant to semileptonic\nform factors is to inject 3-momenta smaller that P(i,o,o)\nusing \"twisted\" boundary conditions [3l|, [H, [s^l . That\nstrategy should improve the accuracy of parameters in\nthe chiral extrapolation and, hence, the BK, BZ, or z\nfits. The output of any fit could still be exhibited as\noutlined here, although one should bear in mind that su-\nperior visualization of a fitting procedure does not repair\nany shortcomings of the fit itself.\n\nAcknowledgments\n\nWe would like to thank Ian Shipsey for encouraging us\nto think carefully about the correlations in the systematic\nerrors. We would like to thank Laurenz Widhalm for pro-\nviding the Belle data in numerical form 2^] , and Shi psey\nfor the BaBar and CLEO D-decay data [13, IH, llj.\n\nComputations for this work were carried out in part on\nfacilities of the USQCD Collaboration, which are funded\nby the Office of Science of the United States Depart-\nment of Energy. This work was supported in part by\nthe U.S. Department of Energy under Grants No. DE-\nFC02-06ER41446 (CD., L.L., M.B.O.), No. DE-FG02-\n91ER40661 (S.G.), No. DE-FG02-91ER40677 (A.X.K.,\nR.T.E., E.G.), No. DE-FG02-91ER40628 (C.B., J.L.),\nand No. DE-FG02-04ER41298 (D.T.); by the Na-\ntional Science Foundation under Grants No. PHY-\n0555243, No. PHY-0757333, No. PHY-0703296 (C.D.,\nL.L., M.B.O.), No. PHY-0555235 (J.L.), and No. PHY-\n0757035 (R.S.); and by Universities Research Associates\n(R.T.E., E.G.). This manuscript has been coauthored\nby an employee of Brookhaven Science Associates, LLC,\nunder Contract No. DE-AC02-98CH10886 with the U.S.\nDepartment of Energy. Fermilab is operated by Fermi\nResearch AUiance, LLC, under Contract No. DE-AC02-\n07CH11359 with the U.S. Department of Energy.\n\n M. Ablikim et al. [BES Collaboration], Phys. Lett. B\n\n597, 39 (2004) [arXiv:hep-ex/0406028].\n G. S. Huang et al. [CLEO Collaboration], Phys. Rev.\n\nLett. 94, 011802 (2005) [arXiv:hep-ex/0407035].\n J. M. Link et al. [FOCUS Collaboration], Phys. Lett. B\n\n607, 51 (2005) [arXiv:hep-ex/0410068].\n J. M. Link et al. [FOCUS Collaboration], Phys. Lett. B\n\n607, 233 (2005) [arXiv:hep-ex/0410037].\n D. Besson et al. [CLEO Collaboration], Phys. Rev. D 80,\n\n032005 (2009) [arXiv:0906.2983 [hep-ex]].\n C. Aubin et al. [Fermilab Lattice, MILC, and HPQCD\nCollaborations], Phys. Rev. Lett. 94, 011601 (2005)\n[arXiv:hep-ph/0408306].\n M. Okamoto et al. [Fermilab Lattice, MILC, and HPQCD\nCollaborations], Nucl. Phys. Proc. Suppl. 140, 461\n(2005) [arXiv:hep-lat/0409116].\n E. Dalgic nee Gulez et al. [HPQCD Collaboration], Phys.\nRev. D 73, 074502 (2006) [arXiv:hep-lat/0601021]; 75,\n119906(E) (2007).\n J. A. Bailey et al, Phys. Rev. D 79, 054507 (2009)\n[arXiv:0811.3640 [hep-lat]].\n K. C. Bowler et al. [UKQCD Collaboration], Phys. Lett.\n\nB 486 (2000) 111 [arXiv:hep-lat/9911011].\n A. Abada et al, Nucl. Phys. B 619 (2001) 565\n\n[arXiv:hep-lat/0011065].\n A. X. El-Khadra et al, Phys. Rev. D 64, 014502 (2001).\n\n[arXiv:hep-ph/0101023].\n S. Aoki et al [JLQCD Collaboration], Phys. Rev. D 64\n\n(2001) 114505 [arXiv:hep-lat/0106024].\n J. Shigemitsu et al, Phys. Rev. D 66, 074506 (2002)\n\n[arXiv:hep-lat/0207011].\n D. Becirevic, S. Prelovsek, and J. Zupan, Phys. Rev. D\n\n67, 054010 (2003) [arXiv:hep-lat/0210048].\n C. Aubin and C. Bernard, Phys. Rev. D 76, 014002\n\n(2007) [arXiv:0704.0795 [hep-lat]].\n L. Lellouch, Nucl. Phys. B 479, 353 (1996) [arXiv:hep-\nph/9509358].\n\n D. Becirevic and A. B. Kaidalov, Phys. Lett. B 478, 417\n\n(2000) [arXiv:hep-ph/9904490].\n P. Ball and R. Zwicky, Phys. Rev. D 71, 014015 (2005)\n\n[arXiv:hep-ph/0406232].\n R. J. Hill, Phys. Rev. D 73, 014012 (2006) [arXiv:hep-\n\nph/0505129].\n\n T. Becher and R. J. Hill, Phys. Lett. B 633, 61 (2006)\n\n[arXiv:hep-ph/0509090].\n C. Bourrely, B. Machet, and E. de Rafael, Nucl. Phys. B\n\n189, 157 (1981).\n C. G. Boyd, B. Grinstein, and R. F. Lebed, Phys.\n\nRev. Lett. 74, 4603 (1995) [arXiv:hep-ph/9412324];\n\nC. G. Boyd and M. J. Savage, Phys. Rev. D 56, 303\n\n(1997) [arXiv:hep-ph/9702300].\n L. Widhalm et al [Belle Collaboration], Phys. Rev. Lett.\n\n97, 061804 (2006) [arXiv:hep-ex/0604049]; K. Abe et al\n[Belle Collaboration], arXiv:hep-ex/0510003.\n\n A. S. Kronfeld et al [Fermilab Lattice, MILC, and\nHPQCD Collaborations], PoS LAT2GG5, 206 (2006)\n[Int. J. Mod. Phys. A 21, 713 (2006)] [arXiv:hep-\nlat/0509169]; A. S. Kronfeld [Fermilab Lattice, MILC,\nand HPQCD Collaborations], J. Phys. Conf. Ser. 46, 147\n(2006) [arXiv:hep-lat/0607011].\n\n M. Okamoto [for the Fermilab Lattice and MILC Col-\nlaborations], unpublished (2005). A public report of this\nwork is contained in Ref. 0). This line of analysis was\nsuperseded by Ref. Q.\n\n B. Aubert et al [BaBar Collaboration], Phys. Rev. D 76,\n052005 (2007) [arXiv:0704.0020 [hep-ex]].\n\n D. Cronin-Hennessy et al. [CLEO Collaboration], Phys.\nRev. Lett. 100, 251802 (2008) [arXiv:0712.0998 [hep-ex]];\nS. Dobbs et al [CLEO Collaboration], Phys. Rev. D 77,\n112005 (2008) [arXiv:0712.1020 [hep-ex]].\n\n J. Y. Ge et al [CLEO Collaboration], Phys. Rev. D 79,\n052010 (2009) [arXiv:0810.3878 [hep-ex]].\n\n B. Aubert et al. [BaBar Collaboration], Phys. Rev. Lett.\n\n98, 091801 (2007) [arXiv:hep-ex/0612020].\n\n6\n\n P. F. Bodaquc, Phys. Lett. B 593, 82 (2004) [arXiv:nucl- J. M. Flynu, A. Jiittncr, and C. T. Sachrajda [UKQCD\n\nth/0402051]. Collaboration], Phys. Lett. B 632, 313 (2006) [arXiv:hep-\n\n C. T. Sachrajda and G. Villadoro, Phys. Lett. B 609, 73 lat/0506016].\n(2005) [arXiv:hep-lat/0411033].\n\n```",
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https://kr.mathworks.com/matlabcentral/cody/problems/2019-dimensions-of-a-rectangle/solutions/847733 | [
"Cody\n\n# Problem 2019. Dimensions of a rectangle\n\nSolution 847733\n\nSubmitted on 9 Mar 2016 by Carlos Zúñiga\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = 1 length_correct = 3/sqrt(10); width_correct = 1/sqrt(10); [width, length] = rectangle(x); tolerance = 1e-12; assert( abs(length-length_correct)<tolerance && abs(width-width_correct)<tolerance )\n\nx = 1 y = 0.3162\n\n2 Pass\nx = 2; width_correct = 2/sqrt(10); length_correct = 6/sqrt(10); [width, length] = rectangle(x); tolerance = 1e-12; assert(abs(length-length_correct)<tolerance && abs(width-width_correct)<tolerance)\n\ny = 0.6325"
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https://www.numberempire.com/11284 | [
"Home | Menu | Get Involved | Contact webmaster",
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"# Number 11284\n\neleven thousand two hundred eighty four\n\n### Properties of the number 11284\n\n Factorization 2 * 2 * 7 * 13 * 31 Divisors 1, 2, 4, 7, 13, 14, 26, 28, 31, 52, 62, 91, 124, 182, 217, 364, 403, 434, 806, 868, 1612, 2821, 5642, 11284 Count of divisors 24 Sum of divisors 25088 Previous integer 11283 Next integer 11285 Is prime? NO Previous prime 11279 Next prime 11287 11284th prime 119813 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 10110000010100 Octal 26024 Duodecimal 6644 Hexadecimal 2c14 Square 127328656 Square root 106.22617379912 Natural logarithm 9.3311410721219 Decimal logarithm 4.0524630774833 Sine -0.56531220772477 Cosine 0.82487702586346 Tangent -0.685329073304\nNumber 11284 is pronounced eleven thousand two hundred eighty four. Number 11284 is a composite number. Factors of 11284 are 2 * 2 * 7 * 13 * 31. Number 11284 has 24 divisors: 1, 2, 4, 7, 13, 14, 26, 28, 31, 52, 62, 91, 124, 182, 217, 364, 403, 434, 806, 868, 1612, 2821, 5642, 11284. Sum of the divisors is 25088. Number 11284 is not a Fibonacci number. It is not a Bell number. Number 11284 is not a Catalan number. Number 11284 is not a regular number (Hamming number). It is a not factorial of any number. Number 11284 is an abundant number and therefore is not a perfect number. Binary numeral for number 11284 is 10110000010100. Octal numeral is 26024. Duodecimal value is 6644. Hexadecimal representation is 2c14. Square of the number 11284 is 127328656. Square root of the number 11284 is 106.22617379912. Natural logarithm of 11284 is 9.3311410721219 Decimal logarithm of the number 11284 is 4.0524630774833 Sine of 11284 is -0.56531220772477. Cosine of the number 11284 is 0.82487702586346. Tangent of the number 11284 is -0.685329073304"
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https://aerospace.technion.ac.il/gvahim-members/nachum-heiman/ | [
"",
null,
"Nachum Eisen\n\[email protected]\n\nSpring-2013/2014\n\nTal Shima\nModelling of an inverted pendulum\n\nThe research dealt with modeling and simulating the motion of a rigid pendulum installed on a rotating horizontal bar. The coupled and non-linear equations of motion were developed. The equations of motion were linearized around both equilibrium points of the system. By a set of experiments demonstrated in the Lab, the numerical values of the different physical parameters of the experiment system were determined. A comparison between the motion of the pendulum and the bar as measured in the Lab to the motion simulated by a simulation, showed compatible results of both the non-linear and linear models of the system."
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https://numbermatics.com/n/11479/ | [
"# 11479\n\n## 11,479 is an odd composite number composed of two prime numbers multiplied together.\n\nWhat does the number 11479 look like?\n\nThis visualization shows the relationship between its 2 prime factors (large circles) and 4 divisors.\n\n11479 is an odd composite number. It is composed of two distinct prime numbers multiplied together. It has a total of four divisors.\n\n## Prime factorization of 11479:\n\n### 13 × 883\n\nSee below for interesting mathematical facts about the number 11479 from the Numbermatics database.\n\n### Names of 11479\n\n• Cardinal: 11479 can be written as Eleven thousand, four hundred seventy-nine.\n\n### Scientific notation\n\n• Scientific notation: 1.1479 × 104\n\n### Factors of 11479\n\n• Number of distinct prime factors ω(n): 2\n• Total number of prime factors Ω(n): 2\n• Sum of prime factors: 896\n\n### Divisors of 11479\n\n• Number of divisors d(n): 4\n• Complete list of divisors:\n• Sum of all divisors σ(n): 12376\n• Sum of proper divisors (its aliquot sum) s(n): 897\n• 11479 is a deficient number, because the sum of its proper divisors (897) is less than itself. Its deficiency is 10582\n\n### Bases of 11479\n\n• Binary: 101100110101112\n• Base-36: 8UV\n\n### Squares and roots of 11479\n\n• 11479 squared (114792) is 131767441\n• 11479 cubed (114793) is 1512558455239\n• The square root of 11479 is 107.1400952025\n• The cube root of 11479 is 22.5580394571\n\n### Scales and comparisons\n\nHow big is 11479?\n• 11,479 seconds is equal to 3 hours, 11 minutes, 19 seconds.\n• To count from 1 to 11,479 would take you about three hours.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 11479 cubic inches would be around 1.9 feet tall.\n\n### Recreational maths with 11479\n\n• 11479 backwards is 97411\n• The number of decimal digits it has is: 5\n• The sum of 11479's digits is 22\n• More coming soon!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.84351087,"math_prob":0.92815584,"size":3098,"snap":"2019-43-2019-47","text_gpt3_token_len":839,"char_repetition_ratio":0.123141564,"word_repetition_ratio":0.021276595,"special_character_ratio":0.32375726,"punctuation_ratio":0.15409836,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9954145,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T23:11:26Z\",\"WARC-Record-ID\":\"<urn:uuid:c17f0cfe-7fbd-4b8b-9fbf-2cc4c712a53a>\",\"Content-Length\":\"15540\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:85db45d3-2044-4646-b9b7-8f791e6d2661>\",\"WARC-Concurrent-To\":\"<urn:uuid:187678e6-b91e-45aa-94af-2abaed96b672>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/11479/\",\"WARC-Payload-Digest\":\"sha1:7IGLODTY6UHYMG46QU6ZLZX7OWPG5ALW\",\"WARC-Block-Digest\":\"sha1:6GTD3V2EWV65S47POEP4AQCK4VDHSZF2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665809.73_warc_CC-MAIN-20191112230002-20191113014002-00516.warc.gz\"}"} |
https://dividedby.org/998000-divided-by-36 | [
"Home » Divided by 36 » 998000 Divided by 36\n\n# 998000 Divided by 36\n\nWelcome to 998000 divided by 36, our post which explains the division of nine hundred ninety-eight thousand by thirty-six to you. 🙂\n\nThe number 998000 is called the numerator or dividend, and the number 36 is called the denominator or divisor.\n\nThe quotient (integer division) of of 998000 and 36, the ratio of 998000 and 36, as well as the fraction of 998000 and 36 all mean (almost) the same:\n\n998000 divided by 36, often written as 998000/36.\n\nRead on to find the result in various notations, along with its properties.\n\n## Calculator\n\nShow Steps\n27722.2222222222\n=\n27722 Remainder 8\nThe Long Division Steps are explained here. Read them now!\n\n## What is 998000 Divided by 36?\n\nWe provide you with the result of the division 998000 by 36 straightaway:\n\n998000 divided by 36 = 27722.2\n\nThe result of 998000/36 is a non-terminating, repeating decimal.\n\nThe repeating pattern above, 2, is called repetend, and denoted overlined with a vinculum.\n\nThis notation in parentheses is also common: 998000/36 = 27722.(2): However, in daily use it’s likely you come across the reptend indicated as ellipsis: 998000 / 36 = 27722.2… .\n• 998000 divided by 36 in decimal = 27722.2\n• 998000 divided by 36 in fraction = 998000/36\n• 998000 divided by 36 in percentage = 27722.2%\nNote that you may use our state-of-the-art calculator above to obtain the quotient of any two integers or whole numbers, including 998000 and 36, of course.\n\nRepetends, if any, are denoted in ().\n\nThe conversion is done automatically once the nominator, e.g. 998000, and the denominator, e.g. 36, have been inserted.\n\nTo start over overwrite the values of our calculator.\n\n## What is the Quotient and Remainder of 998000 Divided by 36?\n\nThe quotient and remainder of 998000 divided by 36 = 27722 R 8\n\nThe quotient (integer division) of 998000/36 equals 27722; the remainder (“left over”) is 8.\n\n998000 is the dividend, and 36 is the divisor.\n\nIn the next section of this post you can find the frequently asked questions in the context of nine hundred ninety-eight thousand over thirty-six, followed by the summary of our information.\n\n## Nine Hundred Ninety-Eight Thousand Divided by Thirty-Six\n\nYou already know what 998000 / 36 is, but you may also be interested in learning what other visitors have been searching for when coming to this page.\n\nThe FAQs include, for example:\n\n### Is 998000 divisible by 36?\n\nNo, the result of the division has decimal places.\n\n### What is the answer for 998000 divided by 36?\n\n27722 Remainder 8.\n\n### How many times does 36 go into 998000?\n\n36 goes into 998000 27722.2222222222 times.\n\n### What is the remainder of 998000 divided by 36?\n\nR = 8, the remainder R is the is the left-over after the whole number division.\n\n### What is 998000 divided by 36 as a decimal?\n\nThe result in decimal notation is 27722.2222222222.\n\nIf you have read our article up to this line, then we take it for granted that you can answer these FAQs and similar questions about the ratio of 998000 and 36.\n\nObserve that you may also locate many calculations such as 998000 ÷ 36 using the search form in the sidebar.\n\nThe result page lists all entries which are relevant to your query.\n\nGive the search box a go now, inserting, for instance, nine hundred ninety-eight thousand divided by thirty-six, or what’s 998000 over 36 in decimal, just to name a few potential search terms.\n\nFurther information, such as how to solve the division of nine hundred ninety-eight thousand by thirty-six, can be found in our article Divided by, along with links to further readings.\n\nSubmitting...\n{{message}}\n\n## Conclusion\n\nTo sum up, 998000/36 = 27722.(2). The indefinitely repeating sequence of this decimal is 2.\n\nAs division with remainder the result of 998000 ÷ 36 = 27722 R 8.\n\nYou may want to check out What is a Long Division?\n\nFor questions and comments about the division of 998000 by 36 fill in the comment form at the bottom, or get in touch by email using a meaningful subject line.\n\nIf our content has been helpful to you, then you might also be interested in the Remainder of 102000 Divided by 37.\n\nPlease push the sharing buttons to let your friends know about the quotient of 998000 and 36, and make sure to place a bookmark in your browser.\n\nEven better: install our website right now!\n\nThanks for visiting our article explaining the division of 998000 by 36.\n\nFor a list of our similar sites check out the sidebar of our home page.\n\nBefore you leave, what about taking our poll below?"
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https://bugs.call-cc.org/attachment/ticket/1293/srfi-69-testcases.patch | [
"# Ticket #1293: srfi-69-testcases.patch\n\nFile srfi-69-testcases.patch, 2.1 KB (added by sjamaan, 6 years ago)\n\nA patch for the srfi-69 test suite to detect these problems\n\n• ## tests/hash-table-tests.scm\n\n```diff --git a/tests/hash-table-tests.scm b/tests/hash-table-tests.scm\nindex 99960bd..61c849e 100644```\n a ;;;; hash-table-tests.scm (require-extension srfi-69 data-structures extras) (require-extension srfi-69 data-structures extras lolevel) (print \"SRFI 69 procedures\") (assert (eq? hash equal?-hash)) (recursive-hash-max-length dl) (print hsh1 \" \" hsh2) (assert (not (= hsh1 hsh2))) ) ) ) ;; Regression test for #1293, found by John Croisant (print \"HT - Retrieve compound objects after GC and mutate contents\") (set! ht (make-hash-table eq? hash-by-identity)) (let* ((lst (list 1 2 3)) (vec (vector 1 2 3)) (pvec (make-pointer-vector 5 (address->pointer 1))) (loc (make-locative (cons 1 2))) (str (string #\\x #\\y #\\z)) (objects (list lst vec pvec loc str)) (hashes (map hash-by-identity objects))) (for-each (lambda (obj) (hash-table-set! ht obj obj)) objects) ; (gc #t) (for-each (lambda (obj) (print \"Exists? \" obj) (assert (hash-table-exists? ht obj))) objects) (for-each (lambda (obj hash) (print \"Same hash? \" obj) (assert (= (hash-by-identity obj) hash))) objects hashes) (print \"After mutation\") (set-car! lst (string #\\l #\\s #\\t)) (vector-set! vec 0 (string #\\v #\\e #\\c)) (pointer-vector-set! pvec 0 (address->pointer 0)) (locative-set! loc (string #\\h #\\a #\\i)) (string-set! str 0 #\\f) ;; In each case, the object is the same: only contents have changed (for-each (lambda (obj) (print \"Exists? \" obj) (assert (hash-table-exists? ht obj))) objects) (for-each (lambda (obj hash) (print \"Same hash? \" obj) (assert (= (hash-by-identity obj) hash))) objects hashes) (gc #t) (for-each (lambda (obj) (print \"Exists? \" obj) (assert (hash-table-exists? ht obj))) objects) (for-each (lambda (obj hash) (print \"Same hash? \" obj) (assert (= (hash-by-identity obj) hash))) objects hashes))"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5937774,"math_prob":0.41987526,"size":1905,"snap":"2022-05-2022-21","text_gpt3_token_len":743,"char_repetition_ratio":0.19936876,"word_repetition_ratio":0.10684931,"special_character_ratio":0.5223097,"punctuation_ratio":0.08064516,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9537468,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T08:00:44Z\",\"WARC-Record-ID\":\"<urn:uuid:368240a7-aa03-4432-be81-a0a1533b63d4>\",\"Content-Length\":\"18266\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:077d9a1b-bec0-40f3-bd4c-29624445d425>\",\"WARC-Concurrent-To\":\"<urn:uuid:001e7b3f-a647-48fd-8a46-42bee3615005>\",\"WARC-IP-Address\":\"78.47.93.131\",\"WARC-Target-URI\":\"https://bugs.call-cc.org/attachment/ticket/1293/srfi-69-testcases.patch\",\"WARC-Payload-Digest\":\"sha1:IN534QBOATHCKYJYAHOYIQUSOIU4HJNJ\",\"WARC-Block-Digest\":\"sha1:FBSGHSQEMSHT4N22L5IPY4AY2HWZGGUH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662604495.84_warc_CC-MAIN-20220526065603-20220526095603-00312.warc.gz\"}"} |
https://blog.fips.fi/tomography/x-ray/total-variation-regularization-for-x-ray-tomography-experimental-data/ | [
"# Total variation regularization for X-ray tomography – Experimental Data\n\nSummary: Tomography means reconstructing the internal structure of a physical body using X-ray images of the body taken from different directions. Mathematically, the problem is to recover a non-negative function f(x) from a collection of line integrals of f. Here we show how to solve the tomography problem using Total Variation (TV) regularization on experimental data. TV is an edge-preserving reconstruction method for ill-posed inverse problems that favors sharp piecewise constant reconstructions. TV was first introduced by Rudin, Osher and Fatemi in 1992 .\n\nAuthors: Kristian Bredies, Tatiana Bubba and Samuli Siltanen.\n\nSoftware: Tested on Matlab 2018a.\nData: The open data set lotusroot.\n\nLiterature: L. Rudin, S. Osher, and E. Fatemi, Nonlinear total variation based noise removal algorithms, Physica D 60, 259–268 (1992).\n\nPlease read also the post Total variation regularization for X-ray tomography for an introduction on TV tested on simulated data. A summary of the experimental data can be found on arXiv.\n\nThe usual Total Variation (TV) seminorm is defined as",
null,
"TV is usually a suitable model for images with edges and have become a very popular tool in image processing and inverse problems due to peculiar features that cannot be realized with smooth regularizations .\n\nIn the following we compare the reconstructions obtained by minimizing the penalty functional:",
null,
"for different values of the regularization parameter α. Here, A is the measurement matrix, m the experimental data, f a discretization of the object to reconstruct, and ι≥0 the indicator function of the nonnegative orthant. The discretization of TV formulas is carried out by finite differences as described in details in .\n\nFor the solution of the minimization problem above, any kind of numerical algorithm for the solution of convex-concave saddle point problems can be used. The algorithm implemented in the dowloadable package above is a primal-dual ascent-descent method with primal extragradient, as described in the famous 2011 paper by Chambolle and Pock . The main bulk of the code is in tomo_tv.m. The script recon_tv.m can be used to set up the experiment: loading the data and setting up the values for different parameters (the regularization parameter and the maximum number of iterations).\n\nIn the following computational examples we use the 256×256 lotus root dataset obtainable from here (documentation in arXiv). This dataset uses 120 projection directions. We set the maximum number of iterations equal to 10000 and the regularization parameter α = 10−5 (see Figure 1).\n\nTrue object 1000 iterations 10000 iterations\n\nFigure 1: Reconstructions with α = 10−5 .\n\nAs always, the value for the regularization parameter is chosen to balance the data mismatch term and the prior information given by the TV penalty. We can play around with the value of α (still keeping the maximum number of iterations equal to 10000) to see how this affects the reconstructions (see Figures 2).\n\nTrue object 1000 iterations 10000 iterations\n\nTrue object 1000 iterations 10000 iterations\n\nFigure 2: Reconstructions with α = 10−4 (top) and α = 10−6 (bottom).\n\nUnfortunately, sometimes it can also happen that TV reconstructs undesired edges: this artifact is called staircasing effect. This is due to the fact that the model assumption for TV is that an image is piecewise constant up to a discontinuity set. However, natural images are often piecewise smooth due to shading, for instance. Stay tuned for the next post to see how this can be overcome.\n\nReferences\n\n L. Rudin, S. Osher and E. Fatemi, Nonlinear total variation based noise removal algorithms, Physica D 60, 259–268 (1992).\n\n M. Burger and S. Osher, A guide to the TV zoo (chapter 1 in M. Burger and S. Osher, Level-Set and PDE-based Reconstruction Methods, Springer, 2013).\n\n K. Bredies, Recovering piecewise smooth multichannel images by minimization of convex functionals with Total Generalized Variation penalty, Lecture Notes in Computer Science 8293, 44–77 (2014).\n\n A. Chambolle and T. Pock, A first-order primal-dual algorithm for convex problems with applications to imaging, Journal of Mathematical Imaging and Vision 40, 120–145 (2011)."
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"https://blog.fips.fi/wp-content/uploads/2018/12/Screen-Shot-2018-12-05-at-14.50.48-300x32.png",
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"https://blog.fips.fi/wp-content/uploads/2018/12/Screen-Shot-2018-12-05-at-14.36.31-300x53.png",
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https://www.tutorialspoint.com/day-of-the-year-in-python | [
"# Day of the Year in Python\n\nPythonServer Side ProgrammingProgramming\n\nSuppose, we have a date in the format “YYYY-MM-DD”. We have to return the day number of the year. So if the date is “2019-02-10”, then this is 41st day of the year.\n\nTo solve this, we will follow these steps −\n\n• Suppose D is an array of day count like [0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]\n• Convert the date into list of year, month and day\n• if the year is leap year then set date D = 29\n• Add up the day count up to the month mm – 1. and day count after that.\n\n## Example\n\nLet us see the following implementation to get better understanding −\n\nLive Demo\n\nclass Solution(object):\ndef dayOfYear(self, date):\ndays = [0,31,28,31,30,31,30,31,31,30,31,30,31]\nd = list(map(int,date.split(\"-\")))\nif d % 400 == 0:\ndays+=1\nelif d%4 == 0 and d%100!=0:\ndays+=1\nfor i in range(1,len(days)):\ndays[i]+=days[i-1]\nreturn days[d-1]+d\nob1 = Solution()\nprint(ob1.dayOfYear(\"2019-02-10\"))\n\n## Input\n\n\"2019-02-10\"\n\n## Output\n\n41"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.78024995,"math_prob":0.99152917,"size":2437,"snap":"2022-27-2022-33","text_gpt3_token_len":702,"char_repetition_ratio":0.16358405,"word_repetition_ratio":0.052287582,"special_character_ratio":0.31678292,"punctuation_ratio":0.10980392,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9681739,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-29T23:21:44Z\",\"WARC-Record-ID\":\"<urn:uuid:1a7111e7-27bb-4487-8bcf-a607ebb63ca3>\",\"Content-Length\":\"31356\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:66538505-1704-4648-82dc-86e31fb57ea3>\",\"WARC-Concurrent-To\":\"<urn:uuid:4ef438db-ec87-4a32-8be4-e96fa9193b38>\",\"WARC-IP-Address\":\"192.229.210.176\",\"WARC-Target-URI\":\"https://www.tutorialspoint.com/day-of-the-year-in-python\",\"WARC-Payload-Digest\":\"sha1:D6XEXSU4XL2ACJW6THPTL7CFMAPK6DJJ\",\"WARC-Block-Digest\":\"sha1:ECBJ4UEZDDBN6XCB77K3TTPAP7AGIFUB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103645173.39_warc_CC-MAIN-20220629211420-20220630001420-00315.warc.gz\"}"} |
https://www.diffen.com/difference/Billion_vs_Million | [
"A million is 106, or 1,000,000. A billion is one thousand million, or 1,000,000,000 (109). This is the common usage in English-speaking countries and is called the short scale. Countries in continental Europe and Latin America use the long scale where a billion is a million millions (1012).\n\nThe word billion originated from French word bi- (“two”) + -illion; i.e. a million million. It was first coined by Jehan Adam in 1475 as by-million and then rendered as byllion by Nicolas Chuquet in 1484.\n\nMillion originated from the Italian milione, from the Latin mille + the augmentative suffix -one.\n\nComparison chart\n\nBillion versus Million comparison chart",
null,
"BillionMillion\nPower of 10 10 to the 9th power (10^9) 10 to the 6th power (10^6)\nNumber 1,000,000,000 1,000,000\n\nMagnitude of the difference\n\nThe magnitude of difference between billion and million can be illustrated with this example of the time scale:\n\n• A million seconds is 12 days.\n• A billion seconds is 31 years.\n• A trillion seconds is 31,688 years.\n\nThe video further compares the three numbers\n\nOther large numbers\n\n• Million=1,000,000\n• Billion=1,000,000,000\n• Trillion=1,000,000,000,000\n• Quintillion=1,000,000,000,000,000,000\n• Sextillion=1,000,000,000,000,000,000,000,000.\n• Nonillion=1,000,000,000,000,000,000,000,000,000,000\n• CENTILLION=1 followed by 303 zeros"
]
| [
null,
"https://static.diffen.com/css/img/edit.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8651878,"math_prob":0.9873581,"size":1321,"snap":"2019-26-2019-30","text_gpt3_token_len":389,"char_repetition_ratio":0.24601367,"word_repetition_ratio":0.0,"special_character_ratio":0.3694171,"punctuation_ratio":0.20792079,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97114694,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T14:53:41Z\",\"WARC-Record-ID\":\"<urn:uuid:3dfe73ca-e02a-438e-bbbd-4943cde3cdbe>\",\"Content-Length\":\"35440\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:146dbd5e-2eb8-4054-8610-9d4e4d2b0ccb>\",\"WARC-Concurrent-To\":\"<urn:uuid:1c8fe1e2-f4eb-4525-b21c-db4d04c7bcad>\",\"WARC-IP-Address\":\"151.101.250.217\",\"WARC-Target-URI\":\"https://www.diffen.com/difference/Billion_vs_Million\",\"WARC-Payload-Digest\":\"sha1:6JXCLCR5CY7XFFSZKHEJFNR6TOUQMSDU\",\"WARC-Block-Digest\":\"sha1:ELAIC3BPLAK7RDGUDWXUMSU5DU2JEXWG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999000.76_warc_CC-MAIN-20190619143832-20190619165832-00179.warc.gz\"}"} |
https://scholars.direct/Articles/aerospace-engineering-and-mechanics/jaem-2-014.xml | [
"Journal of Aerospace Engineering and Mechanics J Aerosp Eng Mech 2578-6350 Scholars.Direct 561 Thomas L Berkley Way, Oakland, California 94612, USA Photon Mass Orlov SA The atomic structure is presented on the basis of the theory of vortex gravitation. The feasibility and calculation of the values of the density and mass of electromagnetic particles are proposed. A calculation is made, which proves that the photon must have mass. In the calculations, some physical characteristics of electromagnetic particles that are accepted by modern physics are refuted. REVIEW ARTICLE 2 1 OPEN ACCESS Photon Mass SA Orlov Petrozavodsk State University, Russia SA Orlov\nPetrozavodsk State University, Russia, Tel: +79535496127.\n04 June 2018 06 June 2018 Orlov SA 2018 Photon Mass J Aerosp Eng Mech 2018 Orlov SA, et al © This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.\n\nThe atomic structure is presented on the basis of the theory of vortex gravitation. The feasibility and calculation of the values of the density and mass of electromagnetic particles are proposed. A calculation is made, which proves that the photon must have mass. In the calculations, some physical characteristics of electromagnetic particles that are accepted by modern physics are refuted.\n\nTheory of vortex gravity, Cosmology and cosmogony, Celestial mechanics\n\nTheory of light\n\nThe nature of light has been studied by scientists since the Renaissance. According to some scientists, light had a wave nature. Others defended the corpuscular theory of the origin of light. To the founders of the wave theory, in the first place, should be attributed Rene Descartes. He represented light as a disturbance in the world substance . The founder of the corpuscular theory was Pierre Gassendi . The same point of view was followed by Isaac Newton . Later, the wave theory of light was investigated by Robert Hooke and Christian Huygens . Thomas Jung in the early 19th century, his experiments with diffraction received evidence for the recognition of the wave theory. In his opinion, different colors correspond to different wavelengths. In 1817 the wave theory of light was followed by Augustin Fresnel in 1817 . When considering the problem of thermal equilibrium of an absolutely black body, Max Planck formulated his idea of the emission of light by portions - light quanta, which were called photons. Experiments of Malus and Bio with polarization provided, as it seemed then, convincing evidence in favor of corpuscular theory and against the wave theory. In quantum mechanics, the idea of Dui de Brogliea about corpuscular-wave dualism was confirmed.\n\naL. de Broglie: a) Some important papers: C.R.A.S., 177, p. 517, p. 548, p. 630, 1923; Philosophical Mag., XLVII, p. 446, 1924; J. Phys., serie VI, t. VII, n° 11, p. 321, 1926; J. Phys., serie VI, t. VIII, n° 5, p. 225, 1927; J. Phys., 20, p. 963, 1959; C.R.A.S., 277, serie B, p. 71, 1973; Found. of Phys. (with G. Lochak, J.A. Beswick and J. Vassalo Pereira), 6, p. 3, 1976. b) Books (among 44 titles): - Recherches sur la theorie des quanta (Thesis, 1924): First published in: Annales de Physique, 10e serie, t. III, p. 22-128, 1925; german translation: Akademische Verlaggesellschaft, Leipzig, 1927; Reprinted by Masson, Paris, 1963. - Selected papers on wave mechanics (with Leon Brillouin), Blackie & Son, Glasgow, 1928. - Une tentative d'interpretation causale et non lineaire de la mecanique ondulatoire: la theorie de la double solution, Gauthier-Villars, Paris, 1956; English translation: Elsevier, Amsterdam, 1960. - La theorie de la mesure en mecanique ondulatoire (interpretation usuelle et interpretation causale), Gauthier-Villars, Paris, 1960. - Etude critique des bases de l-interpretation actuelle de la mecanique ondulatoire, Gauthier-Villars, Paris, 1963; English translation, Elsevier, Amsterdam, 1965. - La reinterpretation de la mecanique ondulatoire, Gauthier-Villars, Paris, 1971. - Contribution to: Foundations of quantum mechanics, Proc. Int. Schools Enrico Fermi, Varenna, 1971. - Les incertitudes d'Heisenberg et l'interpretation probabiliste de la mecanique ondulatoire, Gauthier-Villars, Paris, 1982 (with a long preface of G. Lochak on the evolution of the ideas of Louis de Broglie on the interpretation of wave-mechanics: Translated in English in the Special issue of Found, of Physics devoted to Louis de Broglie, 1982.\n\nIn modern science, a photon is represented as a massless particle.\n\nIn this article, it is proposed to consider the nature of electromagnetic particles based on the author's \"Theory of vortex gravity, cosmology and cosmogony\" . A photon is an extra small particle that has a mass. This mass can be determined according to the above theory.\n\nThe following section presents the basic principles of this theory.\n\nThe theory of vortex gravitation\n\nThe theory of vortex gravity, cosmology and cosmogony is based on the assumption that gravity, all celestial bodies and elementary particles are created by etheric vortices (torsions). The values of the bodies (the system of bodies) and the corresponding vortices can vary by an infinite amount. The largest etheric vortex that a person can observe is the universal whirlwind, the smallest - the atomic whirlwind.\n\nThe orbital velocities of the ether in each vortex decrease in the direction from the center to the periphery, according to the inverse square law. The change in orbital velocities, in accordance with the Bernoulli principle, causes an inversely proportional change (increase) in the pressure in the ether. The pressure gradient creates the forces of vortex gravity and pushes the substance (body) into the zones with the least pressure, that is, in the center of the torsion bar. This pattern operates in the same way in ethereal vortices of any size.\n\nThe ether is an excessively little dense gas that permeates all bodies (substances), except for superdense ones. Therefore, the ether can only push these superdense bodies, which include nucleons and particles of electromagnetic radiation. In the theory of vortex gravity , the Navier-Stokes equation for the motion of a viscous fluid (gas) was used to determine the pressure gradient in an ether vortex. ρ[∂∂t + vρ.grad]vρ = Fρ - grad P + ηΔvρ (1) Where - ether velocity vector, P- ether pressure, η- viscosity.\n\nIn cylindrical coordinates, taking into account the radial symmetry, vr = vz = 0, vφ= v(r), P = P the equation can be written in the form of a system\n\n⎧⎩⎨⎪⎪⎪⎪−v(r)2r = 1ρdPdrη.(∂2v(r)∂r2 + ∂v(r)r∂r - v(r)r2) = 0⎫⎭⎬⎪⎪⎪⎪ (2) After the transformations, an equation is obtained for determining the gravitational forces in the ether vortex: F = V × ρ × v2er (3) With the following dependence Ve ∼ 1r√ where V - the volume of nucleons in the body that is in the orbit of the torsion with a radius of - r. ρ = 8.85 × 10-12 κг/м3 - the density of the ether Ve - the speed of the ether in the orbit r r - the radius of the considered orbit of the ether vortex.\n\nIn the central part of any celestial, ethereal torsion, the centripetal acceleration of the ether reaches enormous values. Directly proportional to this acceleration corresponds to the pressure gradient in the ether or vortex gravity. Under the influence of a huge force of gravitation, the ether thickens and forms in the center of the torsion the super dense core of the celestial body. The density of the nucleus is so great that it cannot be pierced by ether. Ether, with its rotation near the nucleus, touches its fixed surface. In these zones there is turbulence of the ether, the formation of vortices and atomic micro-torsions. These torsions create atoms, as well as the forces of atomic gravity. In the central part of the atomic torsion, according to the same scheme, the ether is condensed into a superdense state and a fixed nucleus of the atom is formed. In the zone of contact of the ether with the core, the same turbulence of the ether and the appearance of torsion corpuscles occur. The size of these torsions is many orders of magnitude smaller than the atomic torsion. In the torsion corpuscule, also, the ether is consolidated and the formation of particles. In these particles, the density of the substance also reaches a value that the ether cannot permeate. Consequently, torsion corpuscles are transformed into material bodies that have masses. These bodies are various electromagnetic particles (electrons, photons or quarks, etc.). With increasing intrinsic mass, according to the law of conservation of the angular momentum of the rotation, electromagnetic particles inversely proportionally decrease their orbital velocity - from the velocity of the ether ve = 7.4 × 1017 to the speed of light vc = 3 × 108 m/s.\n\nThe strength of the substance is provided by interatomic bonds. In the author's scientific work it was determined that interatomic bonds are created only by vortex gravitation in an atomic torsion. Vortex gravity is created by the centripetal acceleration of the ether in this torsion. In , this acceleration was calculated on the surface of the atomic nucleus as: v2ern = (7.4 × 1017)210−14 (4) Where ve = 7.4 × 1017 м/c - ether velocity on the surface of the atomic nucleus. r = 10-14 m - the radius of the nucleus.\n\nWith this acceleration, on the basis of equation (3), the pressure gradient in the atomic, ether torsion creates a force of vortex gravity equal to:\n\nFg = 4.7 × 1038 × V (5)\n\nThis force of atomic attraction corresponds to the known force of interatomic attraction of atoms and ensures the strength of the substance.\n\nIn the orbital circular circulation of electromagnetic particles around the nucleus of the atom, reactive centrifugal forces act on them, which are equal to:\n\nFc = m × v2cr (6) In a circular orbital motion, the reactive forces (6) should be equal to the forces of attraction (5), that is:\n\nFg = Fc = V × ρe × v2er = m × v2cr from where: v2ev2c = mV × ρ, где mv = ρh− density of the corpuscle-photon or electron, then\n\nv2ev2c = ρhρe (7)\n\nFrom the equation (7), we determine the density of the light photon-corpuscle- ρn, taking into account the fact that the speed of the newly formed photon is equal to the speed of light vc = 3 × 108 m/s. We transform equation (7) and determine the photon density ρn, ρh = v2e × ρev2c = (7.4 × 1017)2 × 8.85 × 10−12(3 × 108)2 = 5.4 × 107 kg/m3.\n\nThe calculated density of an electromagnetic particle has objective evidence of its existence under the condition that the interatomic attraction is created by the atomic vortex rotation of the ether.\n\nTo determine the mass of a photon, you need to know its volume. To do this, we will be guided by the research of Stephen Weinberg . According to his statement, at least 20 billion (2 × 1010) photons per nucleon.\n\nThe nucleon volume is equal to Vn = 10-45 m3\n\nThen the photon volume must be equal to Vh = 10−452 × 1010 = 0.5 × 10−56 m3.\n\nAnd the photon mass will be: Mh = Vh × ρh = 0.5 × 10−56 × 5.4 × 107 = 2.5 × 10−49 kg.\n\nOn the basis of equations 5 and 6 it follows that the photon can detach from the atom and move along a spiral, in accordance with the trajectory of Hohmann , for the following reasons:\n\n1. Under the influence of its own gravity, the photon will be condensed more than the calculated values and then the centrifugal forces will exceed the gravitational ones.\n\n2. The power of atomic gravity will decrease its value. In particular, this can be due to the heating of matter or atom. The pressure in the air at the center of the atomic torsion should thus increase, in accordance with the law of Charles : T ~ P\n\nIn this case, the pressure gradient should decrease, which will cause a decrease in the vortex atomic gravity. Then the centrifugal forces will prevail over the gravitational attraction.\n\nIn this paper, all calculations are made from the assumption that electromagnetic particles arise at the lowest orbit of the atomic torsion. It is possible that torsion corpuscles can appear on other orbits, or in atomic torsion, having other dynamic characteristics. In this case, the results presented can be subject to adjustment, but the physico-mathematical principle of calculations will correspond to this article.\n\nThe radius of an electron in modern physics is assumed to be Rel = 10-22 м .\n\nConsequently, the volume of an electron must be of the order of 10-66 m3. Its mass, with a higher calculated density in the order of 108, should be equal to the order of 10-58 kg.\n\nThe electron mass, in accordance with the fundamental constants, is assumed to be 10-39 kg. Then, with an electron volume in the order of 10-66 m3, the electron density must be ρel = mv = 10−3910−66 = 1027 m/kg\n\nWith such a density, the orbital revolution of the electron along atomic orbits would be impossible. That is, on the basis of equation (5), the electron must be torn from the atom tangentially to the orbit much earlier than it will gain such a density. Consequently, the electron density should be slightly different from the above photon density.\n\nAs is known with increasing temperature of the conductor-metal, its electrical conductivity decreases. This phenomenon is explained by the decrease in the volume of the atomic torsion and its gravity, which causes instability of the electrons in their orbits. Such violations in the torsion coupling of electrons prevent the electrical conductivity of the heated metal.\n\nReferences René Descartes (1998) The Treatise on Light. In: Stephen Gaukroger, Descartes: The world and other writings. Cambridge University Press, UK, 3-75.https://www.cambridge.org/core/books/descartes-the-world-and-other-writings/treatise-on-light/531EA674F9FBFEFACF65F7C75D8AC814 Pierre Gassendi (2009) Stanford encyclopedia of philosophyhttps://plato.stanford.edu/entries/gassendi/ Isaac Newton (2007) Stanford encyclopedia of philosophy.https://plato.stanford.edu/entries/newton/ Robert Hooke (1757) The history of the royal society. London, 10-15. C Huygens (1912) Treatise on light. Macmillan, London. Young Thomas (2016) A cambridge alumni database. University of Cambridge, UK. A Fresnel (1818) Fresnel's prize memoir on the diffraction of light. Crew, 81-144. Planck M (1901) Über das gesetz der energieverteilung in normalspektrum. Ann Physik 4: 553-563. Kahr Bart, Claborn Kacey (2008) The lives of Malus and his bicentennial law. Chemphyschem: A European Journal of Chemical Physics and Physical Chemistry 9: 43-58. S Orlov (1969) Foundation of vortex gravitation, cosmology and cosmogony. Global Journal of Science Frontier Research.https://journalofscience.org/index.php/GJSFR/article/view/361 VA Atsurovskiy (1990) General ether-dynamics. Energoatomizdat, Moscow, Russia, 278.http://rusnauka.narod.ru/lib/phisic/acukov/3/about.html http://www.tsijournals.com/articles/gravitational-properties-of-atom-13432.html Steven Weinberg (1977) The First three minutes: A modern view of the origin of the universe. Walter Hohmann (1925) Die erreichbarkeit der himmelskörper. Verlag Oldenbourg, München, Germany. Castka Joseph F, Metcalfe H Clark, Davis Raymond E, et al. (2002) Modern Chemistry. Dehmelt Hans (1988) A single atomic particle forever floating at rest in free space: New value for electron radius. Physica Scripta.http://iopscience.iop.org/article/10.1088/0031-8949/1988/T22/016/meta"
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https://nxlog.co/question/1654/convert-log-date | [
"2\nresponses\n\nHello,\n\nI'm trying to convert a date in NXlog from 06/15/16 to 2016-06-15 because NXlog is not able to parse the date (DEBUG couldn't parse date: 06/14/16).\n\nI created a regular expression (\\$Date =~ s/(\\d+)\\/(\\d+)\\/(\\d+)/20\\$3-\\$2-\\$1/;) in my module to convert the date. See the module below\n\n Exec if \\$raw_event =~ /^[0-9][0-9],/ \\ { \\ ParseDHCP->parse_csv(); \\ if \\$raw_event =~ /^00/ \\$IDdef = \"The log was started.\"; \\ if \\$raw_event =~ /^01/ \\$IDdef = \"The log was stopped.\"; \\ if \\$raw_event =~ /^02/ \\$IDdef = \"The log was temporarily paused due to low disk space.\"; \\ if \\$raw_event =~ /^10/ \\$IDdef = \"A new IP address was leased to a client.\"; \\ if \\$raw_event =~ /^11/ \\$IDdef = \"A lease was renewed by a client.\"; \\ if \\$raw_event =~ /^12/ \\$IDdef = \"A lease was released by a client.\"; \\ if \\$raw_event =~ /^13/ \\$IDdef = \"An IP address was found to be in use on the network.\"; \\ if \\$raw_event =~ /^14/ \\$IDdef = \"A lease request could not be satisfied because the scope's address pool was exhausted.\"; \\ if \\$raw_event =~ /^15/ \\$IDdef = \"A lease was denied.\"; \\ if \\$raw_event =~ /^16/ \\$IDdef = \"A lease was deleted.\"; \\ if \\$raw_event =~ /^17/ \\$IDdef = \"A lease was expired and DNS records for an expired leases have not been deleted.\"; \\ if \\$raw_event =~ /^18/ \\$IDdef = \"A lease was expired and DNS records were deleted.\"; \\ if \\$raw_event =~ /^20/ \\$IDdef = \"A BOOTP address was leased to a client.\"; \\ if \\$raw_event =~ /^21/ \\$IDdef = \"A dynamic BOOTP address was leased to a client.\"; \\ if \\$raw_event =~ /^22/ \\$IDdef = \"A BOOTP request could not be satisfied because the scope's address pool for BOOTP was exhausted.\"; \\ if \\$raw_event =~ /^23/ \\$IDdef = \"A BOOTP IP address was deleted after checking to see it was not in use.\"; \\ if \\$raw_event =~ /^24/ \\$IDdef = \"IP address cleanup operation has began.\"; \\ if \\$raw_event =~ /^25/ \\$IDdef = \"IP address cleanup statistics.\"; \\ if \\$raw_event =~ /^30/ \\$IDdef = \"DNS update request to the named DNS server.\"; \\ if \\$raw_event =~ /^31/ \\$IDdef = \"DNS update failed.\"; \\ if \\$raw_event =~ /^32/ \\$IDdef = \"DNS update successful.\"; \\ if \\$raw_event =~ /^33/ \\$IDdef = \"Packet dropped due to NAP policy.\"; \\ if \\$raw_event =~ /^34/ \\$IDdef = \"DNS update request failed.as the DNS update request queue limit exceeded.\"; \\ if \\$raw_event =~ /^35/ \\$IDdef = \"DNS update request failed.\"; \\ if \\$raw_event =~ /^36/ \\$IDdef = \"Packet dropped because the server is in failover standby role or the hash of the client ID does not match.\"; \\ if \\$raw_event =~ /^[5-9][0-9]/ \\$IDdef = \"Codes above 50 are used for Rogue Server Detection information.\"; \\ if \\$raw_event =~ /^.+,.+,.+,.+,.+,.+,.+,.+,0,/ \\$QResultDef = \"NoQuarantine\"; \\ if \\$raw_event =~ /^.+,.+,.+,.+,.+,.+,.+,.+,1,/ \\$QResultDef = \"Quarantine\"; \\ if \\$raw_event =~ /^.+,.+,.+,.+,.+,.+,.+,.+,2,/ \\$QResultDef = \"Drop Packet\"; \\ if \\$raw_event =~ /^.+,.+,.+,.+,.+,.+,.+,.+,3,/ \\$QResultDef = \"Probation\"; \\ if \\$raw_event =~ /^.+,.+,.+,.+,.+,.+,.+,.+,6,/ \\$QResultDef = \"No Quarantine Information ProbationTime:Year-Month-Day Hour:Minute:Second:MilliSecond.\"; \\ \\$host = hostname_fqdn(); \\ \\$Date =~ s/(\\d+)\\/(\\d+)\\/(\\d+)/20\\$3-\\$2-\\$1/; \\ \\$EventTime = parsedate(\\$Date + \" \" + \\$Time); \\ \\$SourceName = \"DHCPEvents\"; \\ \\$Message = to_json(); \\ } \\ else \\ drop();\n\nHowever it returns 2016-06-15 17:37:29 INFO EventTime: 20\\$3-\\$2-\\$1\n\nAskedJune 15, 2016 - 5:40pm\n\nIt's not possible to use captured value references inside the regexp substitution so you'll need something like this:\n\n`if \\$Date =~ /(\\d+)\\/(\\d+)\\/(\\d+)/ { \\$Date = '20' + \\$3 + '-' + \\$2 + '-' + \\$1; };`\n\nThere is also strptime().\n\n•",
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https://s7097.gridserver.com/cjiutt/5330f5-topology-book-with-solutions-pdf | [
"MATH-305. BruceH Edwards. Many of the solutions to the proofs are just hints but still, this is better than nothing:) A Topology Book with Solutions Bookmark File PDF Topology Textbooks Answers Topology Textbooks Answers This is likewise one of the factors by Zuckerman, An introduction to number theory ( Wiley Eastern. Although this book is subject to a. This is pages Hint. Semester III. The book contains as an appendix some selected solutions to exercises to 9. Printed Version: The book was published by Cambridge University Press in 2002 in both paperback and hardback editions, but only the paperback version is currently available (ISBN 0-521-79540-0). Some solutions have figures, which are done directly in LaTeX using the TikZ and PGFPLOTS packages. This is why we offer the books compilations in this website. Start by pressing the button below! L. Lamport, “Latex: A Documentation Preparation System User's Guide and Ref- erence Manual” .... G. F. 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All books are in clear copy here, and all files are secure so don't worry about it. 197. ~~ Book Introductory Topology Exercises And Solutions ~~ Uploaded By Dr. Seuss, introductory topology exercises and solutions mortad mohammed hichem year 2017 edition second edition publisher world scientific language english pages 356 375 isbn 10 9813148020 isbn 13 9789813148024 file pdf 647 mb preview send to kindle or Solutions Manual pdf free topology without tears solutions manual manual pdf pdf file Page 1/16. Essentials of ... Topology. solutions manual to apostol book introduction to mathematical analysis-point set topology-chapter 3. There are many domains in the broad field of topology, of which the following ... [DM] “Introduction to Hilbert Spaces with Applications” by L. Debnath and P. Mikusinski. Thanks for your help; hope I'm posting in right forum. General topology is the branch of topology dealing with the basic set-theoretic definitions and constructions used in topology. Vector and Tensor Analysis. Download [Book] Topology Munkres Solutions book pdf free download link or read online here in PDF. Further Complex Variable Theory & General Topology Solutions to Problem Sheet 4 Jos e A. Canizo~ March 2013 Unless otherwise speci ed, the symbols X, Y and Zrepresent topological spaces in the following exercises. To get an idea you can look at the Table of Contents and the Preface.. General topology is the subject of the first one. As is common, the problems that have seemed to be most difficult to Printed Version: The book was published by Cambridge University Press in 2002 in both paperback and hardback editions, but only the paperback version is currently available (ISBN 0-521-79540-0). In the second edition, some significant changes have been made, other than the additional exercises. Text Book :- Ivan Nivam & H.S. Then f is bounded. Joseph J. R0tman. Solutions Manual with solutions to the exercises in the book in terms of a PDF. Mechanics. Rather than enjoying a fine book taking into consideration a cup of coffee in the afternoon, otherwise they juggled in the manner of some harmful virus inside their computer. The python directory contains some quick and dirty Sections: 2.1 to ... Review : General remarks on solutions of differential equations, Families of curves, Orthogonal trajectories. So, this weekend. MATH-304. Munkres - Topology - Chapter 4 Solutions … Exercise 4.1. All books are in clear copy here, and all files are secure so don't worry about it. All solutions of problems are put in the end of the book. Hall, S. Paulo Apipe. and Topology. Then is strictly finer than and , where the latter two topologies are not comparable. In the process of problem solving, it is possible to distinguish several more or less ... a set of machine instructions. The solution is complete. 11. Readers of this book may wish to communicate with each other regarding difficulties, solutions to exercises, comments on this book, and further reading. Exercise 4.1. José A. Ca˜nizo. 9. The Munkres text gave a brief introduction to homotopy and the TOPOLOGY: NOTES AND PROBLEMS Some Topology Problems and Solutions - Free download as PDF File (.pdf), Text File (.txt) or read online for free. So, A is bounded above. Perhaps not as easy for a beginner as the preceding book. This site is like a library, you could find million book here by using search box in the header. chapter 02: algebra of sets. MT. Most exercises are given with detailed solutions. A Survey ... TOPOLOGY. SIM. Terry Lawson: Topology: A Geometric Approach. The converse of the previous theorem is not always true. Test 9. Topology problems and solutions pdf - secondary objective of many point set topology courses is to is to build the students' proaching and solving mathematical problems, and the file delightfulart.org Solutions to Problem Sheet 4. Truth be told, this is more of an advanced analysis book than a Topology book,... Topology First Course by James Munkres - AbeBooks j.r.munkres topology a first course pdf However, to make sense of this, we must first give the abstract Once the foundations of Topology … This book provides an introduction to Number Theory from a point of view that is more geometric than is usual for the subject, inspired by the idea that pictures are often a great aid to understanding. Also, two balls with the same radius and different centers may also be equal. Key words and phrases. Mathematics 490 – Introduction to Topology Winter 2007 What is this? MATH- 306. (Mc-Graw Hill). It is mainly intended for undergraduate students. MT. It does not have any exercises and is very tersely written, so it is not a substitute for a standard text like Munkres, but as a beginner I liked this book because it gave me … cations, 2003. To make this easier I have created a Facebook Group called “Topology Without Tears Readers”. Stuck on a topology question that's not in your textbook? of maths -csir,cmi,isi,gate,nbhm (How to qualify CSIR NET easily)Topology - Bruno 514. This book also proposes a number of exercises marked with , which I per-sonally believe to be harder than the typical exam question. Topology (from Greek topos [place/location] and logos [discourse/ reason/logic]) can be viewed as the study of continuous functions, also .... rational coefficients, and a topological condition is satisfied, then there are only finitely many rational solutions. This book was written to be a readable introduction to algebraic topology with rather broad coverage of the subject. [Si] “Introduction to Topology and Modern Analysis” by G.F. Simmons. 09. [NS] “Linear Operator Theory in ... have previously taken this or an equivalent course, from a solutions manual, or from the web will treated as a serious offense. Prove that two discrete spaces are homeomorphic iff they have the same cardinality. 203. Prerequisite: Nil. Download File PDF Topology Munkres Solutions Topology Munkres Solutions Recognizing the pretension ways to acquire this book topology munkres solutions is additionally useful. or scientific purpose. 8. Introduction To Topology. Chegg's step-by-step topology guided textbook solutions will help you learn and understand how to solve topology textbook problems and be better prepared for class. Differential geometry and topology are essential tools for many theoretical physicists, particularly in the study of condensed matter physics, gravity, and particle physics. A downloadable textbook in algebraic topology. 1. Thus A is dense in R equipped with this topology and hence so are the two given sets. of maths -csir,cmi,isi,gate,nbhm (How to qualify CSIR NET easily)Topology - Bruno I aim in this book to provide a thorough grounding in general topology… Free Download Introduction To Topology And Modern ... April 25th, 2018 - Introduction To Topology And Modern Analysis By G F Simmons Book This pdf file has Introduction To Topology And Modern 2017 Solution Slsa Powercraft Manual 7th' Munkres - Topology - Chapter 4 Solutions Section 30 Problem 30.1. Hint. Does anyone know solution book of those? Solutions Manual with solutions to the exercises in the book in terms of a PDF. This content was uploaded by our users and we assume good faith they have the permission to share this book. Topology is an important and interesting area of mathematics, the study of which will not only introduce you to new concepts and theorems but also put into context old ones like continuous ... itself, that it is not necessary to provide answers to exercises – indeed it is probably undesirable. However, charges for profit beyond reasonable printing costs are prohibited. Solution of equations involving absolute values. Topology problems and solutions pdf - secondary objective of many point set topology courses is to is to build the students' proaching and solving mathematical problems, and the file delightfulart.org Solutions to Problem Sheet 4. Topology james munkres solutions manual - Google Документи Topology Solutions Manual 0, Problem 4. The title of the book, Topology of Numbers, is intended to express this visual slant, where we are using the term “Topology\" with its This is why we offer the books compilations in this website. University of Chicago Press, 1999. A prerequisite for the course is an introductory course in real analysis. Show that [a, b] is compact using total boundedness. Read online Munkres - Topology - Chapter 3 Solutions book pdf free download link book now. Topology. This exercise suggests a way to show that a quotient space is homeomorphic to some other space. The study of topology and its spaces is an important aspect of mathematics,topological spaces like other mathematical spaces have axioms that must be satisfied for a topological space to hold. VectK vectors spaces over a given field K linear transformations. Elementary Linerar Algebra. 2. The book offers a good introduction to topology through solved exercises. Exercise 5. and Topology. Ordinary Differential Equations (MA4009):. Company ... Introduction. Download Topology 2nd Edition By James Munkres Solutions book pdf free download link or read online here in PDF. A Concise Course in Algebraic Topology. If m 1 >m 2 then consider open sets fm 1 + (n 1)(m 1 + m 2 + 1)g and fm 2 + (n 1)(m 1 + m 2 + 1)g. The following observation justi es the terminology basis: Proposition 4.6. Harold Simmons. 10. Parent Topic: Topology Munkres (2000) Topology with Solutions Below are links to answers and solutions for exercises in the Munkres (2000) Topology, Second Edition . G.F. Simmons, Introduction to Topology and Modern Analysis, McGraw-Hill, New York,1963. The main solutions manual is solutions.tex. Simmons Topology Modern Analysis Solutions PDF Download. You have remained in right site to begin getting this info. KEL. CMA. Mathematics Semester VI. University Courses ( Exactly Three). Hence f is discontinuous on the whole of R. Theorem See Exercise 2. George F. Simmons Introduction to topology and modern analysis. 519.5. Search for the Group, andthenfromtherejointheGroup. 9. 10. 3 cr. Universities were conducted at various institutions. Read online Munkres - Topology - Chapter 4 Solutions book pdf free download link book now. Functional Analysis. All books are in clear copy here, and all files are secure so don't worry about it. [\\$70] — Includes basics on smooth manifolds, and even some point-set topology. Linear Algebra. Download File PDF Solution Manual Munkres Topology Services are book distributors in the UK and worldwide and we are one of the most experienced book distribution companies in Europe, We offer a fast, flexible and effective book distribution service stretching across the UK & Continental Europe to Scandinavia, the Baltics and Eastern Europe. To apostol book Introduction to Topology and Modern Analysis them as due on but! Library, you could find million book here by using search box in the.... Solutions book pdf free download link book now `` Topology without tears solutions Manual Manual pdf File... Analysis ; McGraw -Hill, N.Y. ( 1963 ) general remarks on solutions of equations. Of R. Mahmo in Topology and Topology ( the latter is the subject... may 3,.... Exam question vectors spaces over a given... Simmons: Introduction to Topology for free five.... Solution of problems are put in the header in this website the subject of the equations. Manifolds, and our terms of a pdf to every single problem and t-equivalence Completeness. So do n't worry about it teachers from affiliated colleges and inviting experts from.. Analysis ; McGraw -Hill, N.Y. ( 1963 ) book in terms of a.. Have figures, which must be closed in Topology virtually any Topology problem, often in as little 2., two balls with the participation of the first one, then show that a quotient space is to. Where the latter is the... itself in the process of problem solving, is... Than the typical exam question pxf on X offer the books compilations in website! Also, two balls with the discrete Topology that it carries non-trivial “ (..., Use of a set endowed with the practical solution of problems on puters. 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[ Si ] “ Introduction to Topology and hence so are the two given sets... Sep 1 2010! This exercise suggests a way to show that the finite product of separable spaces is.! Online here in pdf ( CS ) helicity charges ” as 2 hours, the X... Topological non-trivial solutions in field theory associate only with non-linear equations of motion is separable [ 18! Expanded version of a pdf in clear copy here, and all files are secure so do n't about... Book Topology Munkres solutions Chapter 9, but end happening in harmful downloads differential,. Xis a nite-countable T 1 space Si ] “ Introduction to Topology Modern. A nite-countable T 1 space a right action from a given field K transformations... … Introduction to set topology-chapter 3 a given... Simmons: Introduction to Topology solved! Coverage of the teachers from affiliated colleges and inviting experts from topology book with solutions pdf changes have been made other! Exercises in the book `` Topology without tears solutions Manual with solutions to virtually any Topology,. F. Simmons Introduction to algebraic Topology with rather broad coverage of the subject of subject. Than the typical exam question solutions this is a collection of neighborhoods of xsuch that neighborhood. A topological property topological spaces in the Post Graduate programmes as well with. Finite product of separable spaces is separable 30 problem 30.1 specified, solutions... It actually has solutions to the book offers a good Introduction to.. Topology problems com- puters Analysis ” by G.F. Simmons, topology book with solutions pdf to Topology and Analysis! By Sydney A. Morris Wiley Eastern of action for introducing the CBCSS in the that... This website over 200 exercises and solutions available expanded version of a set endowed with participation! ; hope I 'm posting in right forum it actually has solutions every! And check out the link Part ( a ) Suppose Xis a nite-countable T space! This info 0, problem 4 introductory topics of point-set and algebraic Topology experts can provide answers solutions... Munkres SolutionsTopology by James Munkres, 2nd Edition solutions Manual with solutions this is why offer! Little as 2 hours Readers ” question that 's not in your textbook in! Content was uploaded by our users and we assume good faith they the. Beginner as the product Topology ) easy for a course given to year... On com- puters and solutions available book and it actually has solutions the. The growth of abstract algebra – Introduction to Topology through solved exercises approximately 400 exercises of varying difficulty viewpoint! A Topology question that 's not in your textbook and t-equivalence O. Completeness is a topological property problem. Topology James Munkres solutions book pdf free download link book now solutions.! Read free Topology without tears Readers ” a series of five chapters exercise 2 for profit beyond reasonable costs! Of curves, Orthogonal trajectories... an Introduction to Topology and Modern Analysis ” by G.F.,! Case pronlems sets are dense in X, Y and Z represent topological spaces continuous open.. Fxgbe a one-point set in X X and Y respectively mathematical analysis-point set topology-chapter.. Are an expanded version of a set written for a beginner as the preceding book book ] Munkres... The... itself in the book offers a good Introduction to Topology Winter 2007 semester of equations. Not in your textbook notes are an expanded version of a set with. Review: general remarks on solutions of differential equations, Families of curves, Orthogonal trajectories version..... topological in! Of differential equations, Use of a pdf Michigan in the fact it. K linear transformations of abstract algebra solution: Part ( a ) Suppose Xis a nite-countable T 1 space and! Actually has solutions to exercises from the calculation of the previous theorem is not always true K linear.... Your help ; hope I 'm posting in right site to begin this. The course is an introductory course topology book with solutions pdf real Analysis to prepare a comprehensive of! Dense in X from affiliated colleges and inviting experts from other exams, solutions, handouts etc! At any point of R. Mahmo be closed can provide answers and solutions available spaces open. Product of separable spaces is separable, Families of curves, Orthogonal trajectories “. Exam question G.F.Simmons: Introduction to number theory ( Wiley Eastern so n't. Like a library, you could find million book here by using search box in header. Covered in a series of five chapters discrete Topology through solved exercises “ Chern-Simons CS. Is discontinuous on the whole of R. theorem See exercise 2 are an expanded version of set! 2011... Introduction online for free topology book with solutions pdf: general remarks on solutions of problems com-! Depository with free Fifty challenging problems in Probability with solutions to exercises from the offers! Introductory topics of point-set and algebraic Topology ] “ Introduction to Topology and Modern Analysis ( the latter topologies!\nToilet Paper Millionaire, Log Cabin With Hot Tub And Open Fire Scotland, M92 Folding Brace, 1979 Mazda 626 Coupe For Sale, Culpeper County Circuit Court Case Information, Culpeper County Circuit Court Case Information, Code Compliance Inspection, Summer Public Health Scholars Program, Questions On Community Helpers For Grade 2,"
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https://arizona.pure.elsevier.com/en/publications/on-the-number-of-regular-vertices-of-the-union-of-jordan-regions | [
"# On the number of regular vertices of the union of Jordan regions\n\nB. Aronov, Alon Efrat, D. Halperin, M. Sharir\n\nResearch output: Contribution to journalArticle\n\n7 Citations (Scopus)\n\n### Abstract\n\nLet C be a collection of n Jordan regions in the plane in general position, such that each pair of their boundaries intersect in at most s points, where s is a constant. If the boundaries of two sets in C cross exactly twice, then their intersection points are called regular vertices of the arrangement A(C). Let R(C) denote the set of regular vertices on the boundary of the union of C. We present several bounds on |R(C)|, depending on the type of the sets of C. (i) If each set of C is convex, then |R(C)| = O(n1.5+ε) for any ε > 0.1 (ii) If no further assumptions are made on the sets of C, then we show that there is a positive integer r that depends only on s such that |R(C)| = O(n2-1/r). (iii) If C consists of two collections C1 and C2 where C1 is a collection of m convex pseudo-disks in the plane (closed Jordan regions with the property that the boundaries of any two of them intersect at most twice), and C2 is a collection of polygons with a total of n sides, then |R(C)| = O(m2/3n2/3 + m + n), and this bound is tight in the worst case.\n\nOriginal language English (US) 203-220 18 Discrete and Computational Geometry 25 2 Published - 2001 Yes\n\nUnion\nIntersect\nPolygon\nArrangement\nIntersection\nDenote\nClosed\nInteger\n\n### ASJC Scopus subject areas\n\n• Theoretical Computer Science\n• Computational Theory and Mathematics\n• Discrete Mathematics and Combinatorics\n• Geometry and Topology\n\n### Cite this\n\nOn the number of regular vertices of the union of Jordan regions. / Aronov, B.; Efrat, Alon; Halperin, D.; Sharir, M.\n\nIn: Discrete and Computational Geometry, Vol. 25, No. 2, 2001, p. 203-220.\n\nResearch output: Contribution to journalArticle\n\nAronov, B, Efrat, A, Halperin, D & Sharir, M 2001, 'On the number of regular vertices of the union of Jordan regions', Discrete and Computational Geometry, vol. 25, no. 2, pp. 203-220.\nAronov, B. ; Efrat, Alon ; Halperin, D. ; Sharir, M. / On the number of regular vertices of the union of Jordan regions. In: Discrete and Computational Geometry. 2001 ; Vol. 25, No. 2. pp. 203-220.\n@article{469cb780513f440484c6e9ab9a278c80,\ntitle = \"On the number of regular vertices of the union of Jordan regions\",\nabstract = \"Let C be a collection of n Jordan regions in the plane in general position, such that each pair of their boundaries intersect in at most s points, where s is a constant. If the boundaries of two sets in C cross exactly twice, then their intersection points are called regular vertices of the arrangement A(C). Let R(C) denote the set of regular vertices on the boundary of the union of C. We present several bounds on |R(C)|, depending on the type of the sets of C. (i) If each set of C is convex, then |R(C)| = O(n1.5+ε) for any ε > 0.1 (ii) If no further assumptions are made on the sets of C, then we show that there is a positive integer r that depends only on s such that |R(C)| = O(n2-1/r). (iii) If C consists of two collections C1 and C2 where C1 is a collection of m convex pseudo-disks in the plane (closed Jordan regions with the property that the boundaries of any two of them intersect at most twice), and C2 is a collection of polygons with a total of n sides, then |R(C)| = O(m2/3n2/3 + m + n), and this bound is tight in the worst case.\",\nauthor = \"B. Aronov and Alon Efrat and D. Halperin and M. Sharir\",\nyear = \"2001\",\nlanguage = \"English (US)\",\nvolume = \"25\",\npages = \"203--220\",\njournal = \"Discrete and Computational Geometry\",\nissn = \"0179-5376\",\npublisher = \"Springer New York\",\nnumber = \"2\",\n\n}\n\nTY - JOUR\n\nT1 - On the number of regular vertices of the union of Jordan regions\n\nAU - Aronov, B.\n\nAU - Efrat, Alon\n\nAU - Halperin, D.\n\nAU - Sharir, M.\n\nPY - 2001\n\nY1 - 2001\n\nN2 - Let C be a collection of n Jordan regions in the plane in general position, such that each pair of their boundaries intersect in at most s points, where s is a constant. If the boundaries of two sets in C cross exactly twice, then their intersection points are called regular vertices of the arrangement A(C). Let R(C) denote the set of regular vertices on the boundary of the union of C. We present several bounds on |R(C)|, depending on the type of the sets of C. (i) If each set of C is convex, then |R(C)| = O(n1.5+ε) for any ε > 0.1 (ii) If no further assumptions are made on the sets of C, then we show that there is a positive integer r that depends only on s such that |R(C)| = O(n2-1/r). (iii) If C consists of two collections C1 and C2 where C1 is a collection of m convex pseudo-disks in the plane (closed Jordan regions with the property that the boundaries of any two of them intersect at most twice), and C2 is a collection of polygons with a total of n sides, then |R(C)| = O(m2/3n2/3 + m + n), and this bound is tight in the worst case.\n\nAB - Let C be a collection of n Jordan regions in the plane in general position, such that each pair of their boundaries intersect in at most s points, where s is a constant. If the boundaries of two sets in C cross exactly twice, then their intersection points are called regular vertices of the arrangement A(C). Let R(C) denote the set of regular vertices on the boundary of the union of C. We present several bounds on |R(C)|, depending on the type of the sets of C. (i) If each set of C is convex, then |R(C)| = O(n1.5+ε) for any ε > 0.1 (ii) If no further assumptions are made on the sets of C, then we show that there is a positive integer r that depends only on s such that |R(C)| = O(n2-1/r). (iii) If C consists of two collections C1 and C2 where C1 is a collection of m convex pseudo-disks in the plane (closed Jordan regions with the property that the boundaries of any two of them intersect at most twice), and C2 is a collection of polygons with a total of n sides, then |R(C)| = O(m2/3n2/3 + m + n), and this bound is tight in the worst case.\n\nUR - http://www.scopus.com/inward/record.url?scp=0035584446&partnerID=8YFLogxK\n\nUR - http://www.scopus.com/inward/citedby.url?scp=0035584446&partnerID=8YFLogxK\n\nM3 - Article\n\nAN - SCOPUS:0035584446\n\nVL - 25\n\nSP - 203\n\nEP - 220\n\nJO - Discrete and Computational Geometry\n\nJF - Discrete and Computational Geometry\n\nSN - 0179-5376\n\nIS - 2\n\nER -"
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https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-with-the-following-percentages-75-69 | [
"# What is the molecular formula of a compound with the following percentages: 75.69% C, 8.80% H, and 15.51% O?\n\nJan 28, 2016\n\nC ${C}_{7} {H}_{9} O$\n\n#### Explanation:\n\nI might be mistaken, but here's what I did.\nAssume you have 100 g of the compund. Then you take these percentages as normal weight, meaning, you can use formula for mole number. You count it for every atom and then compare the results.\n\nSo:\n$n = \\frac{m}{\\text{M}}$\nFor C it's 6,3, for H 8,8 and for O it's 0,96\n\n6,3:8,8:0,96 /*100\n630:880:96 /:96\n7:9:1"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.85453105,"math_prob":0.9711826,"size":599,"snap":"2021-43-2021-49","text_gpt3_token_len":186,"char_repetition_ratio":0.090756305,"word_repetition_ratio":0.0,"special_character_ratio":0.33555928,"punctuation_ratio":0.2012987,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946143,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-03T22:57:16Z\",\"WARC-Record-ID\":\"<urn:uuid:1e9be8ff-da12-44d3-8fa6-d13a426b85da>\",\"Content-Length\":\"33119\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ae491213-63bf-455c-bad2-d26a867b0857>\",\"WARC-Concurrent-To\":\"<urn:uuid:e3edf88d-d364-44cc-a0eb-02e28842fbd1>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/what-is-the-molecular-formula-of-a-compound-with-the-following-percentages-75-69\",\"WARC-Payload-Digest\":\"sha1:2ZA6ROLNZ2XGR5HUU7CWJ274PHC7EJO5\",\"WARC-Block-Digest\":\"sha1:7FRSWYHARDKAM4KS44CRO62BDN7L6OLL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362919.65_warc_CC-MAIN-20211203212721-20211204002721-00620.warc.gz\"}"} |
https://www.uni-ulm.de/mawi/finmath/courses/winter-20162017/seminar-rough-paths.html | [
"# Seminar: Rough Paths\n\n## Rough Paths\n\n Lecturer: Robert Stelzer, Karsten Urban, Thai Nguyen Type: Master (all mathematical programmes including Finance) Registration: To register for the seminar, please write an E-Mail to Eva Nacca until July 13th, 2016, 11:30 a.m. In the e-mail please give your name, matriculation number, your course of studies and subjects you have taken in the area of Financial Mathematics, Numerical Mathematics or Probability. The number of participants is limited to 15 students. Content: In this seminar we shall together work on understanding the modern rough path integration theory (for stochastic integrals) and its application in rough (partial) differential equations based on a textbook coauthored by a fields medaillist. The books back cover text:\"Lyons’ rough path analysis has provided new insights in the analysis of stochastic differential equations and stochastic partial differential equations, such as the KPZ equation. This textbook presents the first thorough and easily accessible introduction to rough path analysis.When applied to stochastic systems, rough path analysis provides a means to construct a pathwise solution theory which, in many respects, behaves much like the theory of deterministic differential equations and provides a clean break between analytical and probabilistic arguments. It provides a toolbox allowing to recover many classical results without using specific probabilistic properties such as predictability or the martingale property. The study of stochastic PDEs has recently led to a significant extension – the theory of regularity structures – and the last parts of this book are devoted to a gentle introduction.Most of this course is written as an essentially self-contained textbook, with an emphasis on ideas and short arguments, rather than pushing for the strongest possible statements. A typical reader will have been exposed to upper undergraduate analysis courses and has some interest in stochastic analysis. For a large part of the text, little more than Itô integration against Brownian motion is required as background.\" (quoted from http://www.springer.com/de/book/9783319083315 ) First meeting (assignment of topics): July 13th, 2016, 15:05 (sharp!), Heho 22, E18 Prerequisites: An Introduction to Probability and Statistics, Stochastik 2 Master in Finance: An Introduction to Measure Theoretic Probability, Stochastics 2 Time and Venue: to be determined Literature: Friz, Peter K. and Hairer, Martin, A Course On Rough Paths: With An Introduction to Regularity Structures, Springer, Cham, 2014"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8846716,"math_prob":0.7621374,"size":2628,"snap":"2019-35-2019-39","text_gpt3_token_len":552,"char_repetition_ratio":0.10632622,"word_repetition_ratio":0.0,"special_character_ratio":0.20319635,"punctuation_ratio":0.1308204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95645285,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-19T22:56:31Z\",\"WARC-Record-ID\":\"<urn:uuid:6fd10340-f4c2-42c8-98e3-361f5c62bd59>\",\"Content-Length\":\"39300\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3fb9fd9c-3e98-447a-9968-3627628cb9c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:a254e4a1-cd42-48fa-b30b-9df63cfc125d>\",\"WARC-IP-Address\":\"134.60.1.22\",\"WARC-Target-URI\":\"https://www.uni-ulm.de/mawi/finmath/courses/winter-20162017/seminar-rough-paths.html\",\"WARC-Payload-Digest\":\"sha1:O4YJXCI523R462NAMFN2RE2E6WRC3BBQ\",\"WARC-Block-Digest\":\"sha1:TL4UOM6JEACGMY2MW6KRFTKFWTCJSOE4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315132.71_warc_CC-MAIN-20190819221806-20190820003806-00421.warc.gz\"}"} |
https://gamedev.stackexchange.com/questions/26362/how-to-do-reflective-collisions-with-particles-hitting-background-tiles | [
"# How to do reflective collisions with particles hitting background tiles?\n\nIn my 2d pixel old-school platformer, I'm looking for methods for bouncing particles off of background tiles. Particles aren't affected by gravity and collisions are \"reflective\". By that I mean a particle hitting the side of a square tile at 45 degrees should bounce off at 45 degrees as well.\n\nWe can assume that tiles will always be perfectly square. No slopes or anything.\n\nWhat are efficient methods and algorithms to do this? I'd be implementing this on a Sega Genesis.\n\n## 2 Answers\n\nWhen testing a particle to see if it's hit a block, you'll usually know if it hits a side or top or bottom.\n\nIf you express the velocity of the particle as a delta_x and delta_y, you can invert an individual axis when hitting a wall.\n\nFor instance, if you hit a vertical surface, invert the x-axis, and the y-axis if it was a horizontal surface.\n\n• A refinement on this is to not invert the axis but rather reset the sign appropriately (x = abs(x) or x = -abs(x)). Inverting means that if a particle gets in a bad position inside an object it will jitter and not escape, which is a very visible glitch. – Kevin Reid Apr 28 '12 at 13:09\n\nIn a perfectly reflective collision, the angle of reflection is equal to the angle of incidence reflected over the surface normal.\n\nI'm not sure if I'm understanding your question in its entirety, but assuming you have access to the surface normal at the point of contact and the particles' movement vectors, it should be as simple as reflecting one vector over the other according to this formula:\n\nVr = Vi - 2(N dot Vi) * N\n\nWhere N is the normal vector, Vr is the reflected vector and Vi is the incident vector. Hope that helps!\n\n• Thanks for your answer. Unfortunately, it's not exactly what I'm looking for. In my case I'm working with a very low power CPU (Sega Genesis 68000 running at 7mhz) and doing all those dot products and multiplications and normal calculations are going to be bog slow. I'll try to clear up my question to better explain what I need. – DJCouchyCouch Mar 29 '12 at 13:21"
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https://www.notjustnumbers.co.uk/2014/01/guest-post-calculating-median-and-mode.html | [
"## Tuesday, 21 January 2014\n\n### Guest Post - Calculating Median and Mode in Excel\n\nThis week you've got a break from me with a very useful guest post from Alan Murray of Computergaga.\n\nBefore we get into that though, just a quick reminder that the early bird discount on Mynda Treacy's excellent Excel Dashboards course expires on Thursday, so if you don't want to miss out, click over there now.\n\nOK, on with Alan's tips for when you don't just want the average AVERAGE!\n\n## Calculating Median and Mode in Excel\n\nWhen calculating averages in Excel, this normally involves using the AVERAGE function. The AVERAGE function in Excel calculates the mean of a set of numbers.\n\nThere are two other types of averages known as median and mode. Let’s take a look at calculating median and mode in Excel.\n\n### Calculating the Median\n\nThe median is the middle number in a sorted list of numbers. If you were to calculate the median with a pen and paper you would have to;\n1. Write down the list of numbers in numerical order.\n2. If there were an odd number of results, then the median would be the middle number.\n3. If there were an even number of results, then the median would be the mean of the two central numbers.\nFortunately Excel can handle all of this for us with the MEDIAN function. The following formula can be written to calculate the median from the list of exam scores.\n\n=MEDIAN(B4:B13)\n\nThe list of exam scores is not sorted in numerical order, but that is no problem for Excel. The median has been found by finding the mean of 55 and 65 because there are 10 scores which is an even number of results.\n\n39, 48, 51, 52, 5565, 75, 77, 77,90\n(55 + 65) / 2 = 60\n\n### Finding the Mode\n\nThe mode, or modal number, is the number that occurs most frequently in a list. There can be more than one mode.\n\nIn Excel, the MODE function can be used on a list to return the number that occurs most often. In this example the formula would look like below.\n\n=MODE(B4:B13)\n\nThe mode in this example can easily be seen to be 77 as it is the only one to appear more than once. If there is more than one modal number, this formula will only return the first one.\n\n### Returning Multiple Modes\n\nTo return multiple modes the MODE.MULT function can be used. This function was released with Excel 2010 so is not available on versions previous to that.\n\nThis function is an array function, so to run the function you should press Ctrl + Shift + Enter instead of only Enter. It will also appear in the Formula Bar wrapped within curly braces.\n\nYou will need to select a range of cells to apply the function to, or copy the formula and run it again.\n\nIn the example below there are 3 modal values. The following MODE.MULT function has been used in cells D5:D7 to return the results.\n\nWhen there are no more modal values the function will return the #N/A error.\n\n### Create a Frequency Distribution Table\n\nResults are often broken down into groups, or classes. This frequency distribution table gives us a good view of the spread of our data and also identifies the class with the most occurrences, known as the modal class.\n\nThe table below shows the grades of 34 pupils in range C4:C37.\n\nThe COUNTIFS function has been used to calculate these frequencies. The formula below was entered into cell F4 and then copied to the other cells of the table, with the criteria altered to match the class.\n\nThe formula counts the number of scores that are greater than or equals to 31, but less than 40. This is then repeated for each class.\n\nThe COUNTIFS function was released with Excel 2007. If you are using a version previous to this, the SUMPRODUCT function could be used to count based on multiple conditions.\n.",
null,
"About the author: Alan Murray is an IT Trainer and the founder of Computergaga. Author of the popular Computergaga Blog providing the latest Excel, Word, PowerPoint and Project tips and techniques.\n\nIf you enjoyed this post, go to the top of the blog, where you can subscribe for regular updates and get two freebies \"The 5 Excel features that you NEED to know\" and \"30 Chants for Better Charts\".\n\n#### 3 comments:\n\n1.",
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"great post.\nSometimes we dont even know the basics of excel or you can say microsoft office.\nAppreciate your work\n\n1.",
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"Thanks for the feedback. Pleased you found it useful.\n\n2.",
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"hi, great !\nmay i have a copy via my email:[email protected].\nregards."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9205891,"math_prob":0.9240318,"size":4172,"snap":"2021-04-2021-17","text_gpt3_token_len":947,"char_repetition_ratio":0.123320535,"word_repetition_ratio":0.021276595,"special_character_ratio":0.22411314,"punctuation_ratio":0.10600707,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99005073,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,8,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-11T11:05:06Z\",\"WARC-Record-ID\":\"<urn:uuid:a1d20b62-7f81-4ba2-a1e6-e0143eb240a0>\",\"Content-Length\":\"123918\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ea9d2844-cbf2-460d-b5e8-84f86dfe3e6b>\",\"WARC-Concurrent-To\":\"<urn:uuid:e1263510-dc2b-46b4-9d16-0ac92e70ad67>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://www.notjustnumbers.co.uk/2014/01/guest-post-calculating-median-and-mode.html\",\"WARC-Payload-Digest\":\"sha1:JG4F2XOEYX7C2IGVN7WMDDYL5HNGQYV6\",\"WARC-Block-Digest\":\"sha1:5X42577C676KJ3YIISMV4GMICE7S4ODG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038061820.19_warc_CC-MAIN-20210411085610-20210411115610-00460.warc.gz\"}"} |
https://es.scribd.com/document/300728903/A-Simple-Model-of-Price-Formation | [
"Está en la página 1de 4\n\n# A simple model of price formation\n\nK. SznajdWeron , R. Weron\n\n## Institute of Theoretical Physics, University of Wroclaw,\n\npl. Maxa Borna 9, 50-204 Wroclaw, Poland\nHugo Steinhaus Center, Wroclaw University of Technology,\nWyspianskiego 27, 50-370 Wroclaw, Poland\n\n## arXiv:cond-mat/0101001 v2 7 Nov 2001\n\nA simple Ising spin model which can describe the mechanism of price formation in nancial markets\nis proposed. In contrast to other agent-based models, the inuence does not ow inward from the\nsurrounding neighbors to the center site, but spreads outward from the center to the neighbors. The\nmodel thus describes the spread of opinions among traders. It is shown via standard Monte Carlo\nsimulations that very simple rules lead to dynamics that duplicate those of asset prices.\nPACS numbers: 05.45.Tp, 05.50.+q, 87.23.Ge, 89.90.+n\n\n## generalized hyperbolic [13,14], and some reject any single\n\ndistribution [15,16].\nInstead of looking at the central part of the distribution, an alternative way is to look at the tails. Most\ntypes of distributions can be classified into three categories: 1 thin-tailed for which all moments exist and\nwhose density function decays exponentially in the tails;\n2 fat-tailed whose density function decays in a powerlaw fashion; 3 bounded which have no tails.\nVirtually all quantitative analysts suggest that asset\nreturns fall into the second category. If we plot returns\nagainst time we can notice many more outlying (away\nfrom the mean) observations than for white noise. This\nphenomenon can be seen even better on normal probability plots, where the cumulative distribution function\n(CDF) is drawn on the scale of the cumulative Gaussian distribution function. Normal distributions have the\nform of a straight line in this representation, which is approximately the case for the distribution of weekly or\nmonthly returns. However, distributions of daily and\nhigher-frequency returns are distinctly fat-tailed .\nThis can be easily seen in the top panels of Fig. 1, where\ndaily returns of the DJIA index for the period Jan. 2nd,\n1990 Dec. 30th, 1999, are presented.\nClustering and dependence. Despite the wishes of\nmany researchers asset returns cannot be modeled adequately by series of iid (independent and identically\ndistributed) realizations of a random variable described\nby a certain fat-tailed distribution. This is caused by\nthe fact that financial time series depend on the evolution of a large number of strongly interacting systems and belong to the class of complex evolving systems. As a result, if we plot returns against time\nwe can observe the non-stationarity (heteroscedasticity) of the process in the form of clusters, i.e. periods during which the volatility (measured by standard\ndeviation or the equivalent l1 norm ) of the process is much higher than usual, see the top-left panel\nof Fig. 1. Thus it is natural to expect dependence\nin asset returns. Fortunately, there are many methods to quantify dependence. The direct method consists in plotting the autocorrelation function: acf (r, k) =\nPN\nPN\n)(rtk r)/ t=1 (rt r)2 , where N is the\nt=k+1 (rt r\n\n## The Ising spin system is one of the most frequently\n\nused models of statistical mechanics. Its simplicity (binary variables) makes it appealing to researchers from\nother branches of science including biology , sociology\n and economy . It is rather obvious that Isingtype models cannot explain origins of very complicated\nphenomena observed in complex systems. However, it\nis believed that these kind of models can describe some\nuniversal behavior.\nRecently, an Ising spin model which can describe the\nmechanism of making a decision in a closed community\nwas proposed . In spite of simple rules the model exhibited complicated dynamics in one and two dimensions. In contrast to usual majority rules , in\nthis model the influence was spreading outward from the\ncenter. This idea seemed appealing and we adapted it\nto model financial markets. We introduced new dynamic\nrules describing the behavior of two types of market players: trend followers and fundamentalists. The obtained\nresults were astonishing the properties of simulated\nprice trajectories duplicated those of analyzed historic\ndata sets. Three simple rules led to a fat-tailed distribution of returns, long-term dependence in volatility and\nno dependence in returns themselves.\nWe strongly believe that this simple and parameter free\nmodel is a good first approximation of a number of real\nfinancial markets. But before we introduce our model we\nreview some of the stylized facts about price formation\nin the financial markets.\nStylized facts. Adequate analysis of financial\ndata relies on an explicit definition of the variables under study. Among others these include the price and the\nchange of price. In our studies the price xt is the daily\nclosing price for a given asset. The change of price rt\nat time t is defined as rt = log xt+1 log xt . In fact,\nthis is the change of the logarithmic price and is often\nreferred to as return. The change of price, rather than\nthe price itself, is the variable of interest for traders (and\nresearchers as well).\nFat-tailed distribution of returns. The variety of\nopinions about the distributions of asset returns and their\ngenerating processes is wide. Some authors claim the\ndistributions to be close to Paretian stable , some to\n1\n\nPN\nsample length and r = N1 t=1 rt , for different time lags\nk. For most financial data autocorrelation of returns dies\nout (or more precisely: falls into the confidence interval\nof Gaussian random walk) after at most a few days and\nlong-term autocorrelations are found only for squared or\nabsolute value of returns , see bottom panels of\nFig. 1. Recall that for Brownian motion the classical\nmodel of price fluctuations [19,21] autocorrelations of\nrt , rt2 and |rt | are not significant for lags greater or equal\nto one.\n\n## fluctuation, respectively) and plotted against box size on\n\na double-logarithmic paper. Linear regression yields the\nHurst exponent H, whose value can lie in one of the three\nregimes: 1 H > 0.5 persistent time series (strict long\nmemory), 2 H = 0.5 random walk or a short-memory\nprocess, 3 H < 0.5 anti-persistent (or mean-reverting)\ntime series. Unfortunately, no asymptotic distribution\ntheory has been derived for the R/S and DFA statistics\nso far. However, it is possible to estimate confidence intervals based on Monte Carlo simulations .\nThe model. Recently a simple model for opinion\nevolution in a closed community was proposed [9,10]. In\nthis model (called USDF) the community is represented\nby a horizontal chain of Ising spins which are either up or\ndown. A pair of parallel neighbors forces its two neighbors to have the same orientation (in random sequential\nupdating), while for an antiparallel pair, the left neighbor takes the orientation of the right part of the pair,\nand the right neighbor follows the left part of the pair.\nIn contrast to usual majority rules , in the USDF\nmodel the influence does not flow inward from the surrounding neighbors to the center site, but spreads outward from the center to the neighbors. The model thus\nto three steady states: two ferromagnetic (all spins up or\nall spins down) and one antiferromagnetic (an up-spin is\nfollowed by a down-spin, which is again followed by an\nup-spin, etc.).\nIn this paper we modify the model to simulate price\nformation in a financial market. The spins are interpreted\nas market participants attitude. An up-spin (Si = 1)\nwhereas a down-spin (Si = 1) represents a trader who\nis bearish and places sell orders. In our model the first\ndynamic rule of the USDF model remains unchanged,\ni.e. if Si Si+1 = 1 then Si1 and Si+2 take the direction\nof the pair (i,i+1). This can be justified by the fact\nthat a lot of market participants are trend followers and\nplace their orders on the basis of a local gurus opinion.\nHowever, the second dynamic rule of the USDF model\nhas to be changed to incorporate the fact that the absence\nof a local guru (two neighboring spins are in different\ndirections) causes market participants to act randomly\nrather than make the opposite decision to his neighbor:\nif Si Si+1 = 1 then Si1 and Si+2 take one of the two\ndirections at random.\nSuch a model has two stable states (both ferromagnetic), which is not very realistic for a financial market.\nFortunately, trend followers are not the only participants\nof the market . There are also fundamentalists players that know much more about the system and have a\nstrategy (or perhaps we should call them rationalists).\nTo make things simple, in our model we introduce one\nfundamentalist. He knows exactly what is the current\ndifference between demand and supply in the whole system. If supply is greater than demand he places buy\norders, if lower sell orders.\nIt is not clear a priori how to define the price in a\n\n0.08\n0.999\n0.99\n\nReturns\n0.04\n\nCDF\n\n0.90\n0.50\n\n0.10\n0.04\n0.08\n\n0.01\n0.001\n500\n\nDays\n\n0.05\n\n0.05\n\n1\nACF\n\nACF\n\n0.8\n\n0.8\n\n0.6\n\n0.6\n\n0.4\n\n0.4\n\n0.2\n\n0.2\n\n0\n0.2\n0\n\n0\nReturns\n\n0\n10\n\n20\n30\n40\nTime lag (days)\n\n50\n\n0.2\n0\n\n10\n\n20\n30\n40\nTime lag (days)\n\n50\n\n## FIG. 1. Daily returns of the Dow Jones Industrial Average\n\nindex during the last decade (top left), normal probability plot\nof DJIA returns (top right), lagged autocorrelation function\nof DJIA daily returns (bottom left), and lagged autocorrelation function of absolute value of DJIA daily returns (bottom\nright). Dashed horizontal lines represent the 95% condence\ninterval of a Gaussian random walk.\n\n## Another way to examine the dependence structure is\n\nthe power spectrum analysis, also known as the frequency domain analysis. One of the most often used\ntechniques was proposed by Geweke and Porter-Hudak\n (GPH) and is based on observations of the slope of\nthe spectral density function of a fractionally integrated\nseries around the angular frequency = 0. A simple\nlinear regression of the periodogram In (a sample analogue of the spectral density) at low Fourier frequencies\nk : log{In (k )} = a dlog{4 sin2 (k /2)} + k yields the\ndifferencing parameter d = H 0.5 through the rela The GPH estimate of the Hurst exponent\ntion d = d.\nH has well known asymptotic properties and allows for\nconstruction of confidence intervals [22,23].\nYet another method is the Hurst R/S analysis [24,25]\nor its younger sister the Detrended Fluctuation Analysis (DFA) . Both methods are based on a similar\nalgorithm, which begins with dividing the time series\n(of returns) into boxes of equal length and normalizing\nthe data in each box by subtracting the sample mean\n(R/S) or a linear trend (DFA). Next some sort of volatility statistics is calculated (rescaled range or mean square\n\n## The returns rt are obtained from the simulated price\n\ncurve xt (see Fig. 2) after it is shifted (incremented by\none) to make it positive. Alternatively we could have defined the up-spin to be equal to two and the down-spin\nto zero. However, this would have made the calculations more difficult and the description of the model less\nappealing.\nIn Figure 3 we present daily returns and normal probability plots for the simulated (left panels) and USD/DEM\nexchange rate (right panels) time series. Without prior\nknowledge as to the magnitude of historical returns it is\nimpossible to judge which process is real and which is a\nfraud. The same is true for the simulated and the DJIA\nreturns of Fig. 1.\n\n## market. The only obvious requirement is, that the price\n\nshould go up, when there is more demand than supply,\nand vice versa. For simplicity, we define the price xt in\nour model as the normalized differencePbetween demand\nN\nand supply (magnetization): xt = N1 i=1 Si (t), where\nN is the system size. Note that xt [1, 1], so |xt | can be\ntreated as probability. Now we can formulate the third\nrule of our model: the fundamentalist will buy (i.e. take\nvalue 1) at time t with probability |xt | if xt < 0 and sell\n(i.e. take value 1) with probability |xt | if xt > 0.\nThe third rule means that if the system will be close\nto the stable state all up, the fundamentalist will place\nsell orders with probability close to one (in the limiting\nstate exactly with probability one) and start to reverse\nthe system. So the price will start to fall. On the contrary, when r will be close to 1, the fundamentalist will\nplace buy orders (take the value 1) and the price will\nstart to grow. This means that ferromagnetic states will\nnot be stable states anymore.\nResults. To investigate our model we perform a\nstandard Monte Carlo simulation with random updating.\nWe consider a chain of N Ising spins with free boundary\nconditions. We were usually taking N = 1000, but we\nhave done simulations for N = 10000 as well. We start\nfrom a totally random initial state, i.e. to each site of\nthe chain we assign an arrow with a randomly chosen\ndirection: up or down (Ising spin).\n\n0.1\n\n0.04\nReturns\n\nReturns\n\n0.05\n\n0.02\n\n0.05\n\n0.02\n\n0.1\n\n0.999\n0.99\n\n500\n\n1000 1500\nDays\n\n2000\n\n0.04\n\n0.999\n0.99\n\nCDF\n\n0.90\n\n0.90\n\n0.50\n\n0.50\n\n0.75\n\n0.10\n\n0.10\n\n0.9\n\n0.7\n\n0.01\n0.001\n\n0.01\n0.001\n\n0.8\n\n0.65\n\n0.7\n\n0.6\n\n0.6\n500\n\n1000 1500\nDays\n\nUSD/DEM\n2000\n\n0.5\n\n500\n\n1000 1500\nDays\n\n0\nReturns\n\n0.05\n\n1000 1500\nDays\n\n2000\n\nCDF\n\n0.02\n\n0\nReturns\n\n0.02\n\n## FIG. 3. Returns of the simulated price process xt and\n\ndaily returns of the USD/DEM exchange rate during the last\ndecade, respectively (top panels). Normal probability plots of\nrt and USD/DEM returns, respectively, clearly showing the\nfat tails of price returns distributions (bottom panels).\n\n0.55\nSimulation\n\n0.5\n\n0.05\n\n500\n\n2000\n\n## FIG. 2. A typical path of the simulated price process xt\n\nand the USD/DEM exchange rate, respectively.\n\n## In Figure 4 we present the lagged autocorrelation study\n\nfor the simulated (left panels) and USD/DEM exchange\nrate (right panels) time series. Again the properties of\nthe simulated price process duplicate those of historical\ndata. The lagged autocorrelation of returns falls into\nthe confidence interval of Gaussian random walk immediately, like for the dollar-mark exchange rate. The same\nplot for the DJIA returns shows a bit more dependence.\nThis can be seen also in Table 1, where values and\nsignificance of the R/S, DFA and GPH statistics for the\nsimulated and four historical data sets are presented. In\nall but one case (DFA for DJIA) the Hurst exponents\nof daily returns are insignificantly different from those\nof white noise. On the other hand, the Hurst exponents\nof the absolute value of daily returns are persistent in\nall cases, with the results being significant even at the\ntwo-sided 99% level. Moreover, the estimates of H from\nthe simulation are indistinguishable from those of real\nmarket data.\n\nIn our simulations each Monte Carlo step (MCS) represents one trading hour; eight steps constitute one trading day. We typically simulate 20000 MCSs, which corresponds to 2500 trading days or roughly 10 years. We\nchose such a period of time, because in this paper we compare the empirical results with historical data sets (two\nFX rates and two stock indices) of about the same size:\n2245 daily quotations of the dollar-mark (USD/DEM) exchange rate for the period Aug. 9th, 1990 Aug. 20th,\n1999 (see Fig. 2 and Table 1), 2809 daily quotations\nof the yen-dollar (JPY/USD) exchange rate for the period Jan. 2nd, 1990 Feb. 28th, 2001 (see Table 1),\n2527 daily quotations of the Dow Jones Industrial Average (DJIA) index for the period Jan. 2nd, 1990 Dec.\n30th, 1999 (see Fig. 1), and 1561 daily quotations of the\nWIG20 Warsaw Stock Exchange index (based on 20 blue\nchip stocks from the Polish capital market) for the period\nJan. 2nd, 1995 Mar. 30th, 2001 (see Table 1).\n\n1\nACF\n0.8\n\n0.6\n\n0.6\n\n0.4\n\n0.4\n\n0.2\n\n0.2\n\n0\n0.2\n0\n\n0\n10\n\n20\n30\n40\nTime lag (days)\n\n50\n\n0.2\n0\n\n10\n\n20\n30\n40\nTime lag (days)\n\n50\n\n1\nACF\n\nACF\n\n0.8\n\n0.8\n\n0.6\n\n0.6\n\n0.4\n\n0.4\n\n0.2\n\n0.2\n\n0\n0.2\n0\n\n## in Finance, Wiley, 2000.\n\n R.C. Blattberg, N. Gonedes, J. Business 47 (1974) 244.\n E. Eberlein, U. Keller, Bernoulli 1 (1995) 281.\n J.R. Calderon-Rossel, M. Ben-Horim, J. Intern. Business\nStudies 13 (1982) 99.\n R. Mantegna, H.E. Stanley, Phys. Rev. Lett. 73 (1994)\n2946.\n U.A. M\nuller, M.M. Dacorogna, R.B. Olsen, O.V. Pictet,\nM. Schwarz, C. Morgenegg, J. Banking & Finance 14\n(1990) 1189.\n D.M. Guillaume, M.M. Dacorogna, R.R. Dave, U.A.\nM\nuller, R.B. Olsen, O.V. Pictet, Finance Stochast. 1\n(1997) 95.\n A. Weron, R. Weron, Financial Engineering: Derivatives\nPricing, Computer Simulations, Market Statistics, (in\nPolish), WNT, Warsaw, 1998.\n R. Weron, Physica A 285 (2000) 127.\n L. Bachelier, Theorie de la speculation, Annales Scientiques de lEcole Normale Superieure III-17 (1900) 21.\n J. Geweke, S. Porter-Hudak, J. Time Series Analysis 4\n(1983) 221.\n R. Weron, cond-mat/0103510 (2001).\n H.E. Hurst, Trans. Am. Soc. Civil Engineers 116 (1951)\n770.\n B.B. Mandelbrot, J.R. Wallis, Water Resources Res. 5\n(1969) 967.\n C.-K. Peng, S.V. Buldyrev, S. Havlin, M. Simons, H.E.\nStanley, A.L. Goldberger, Phys. Rev. E 49 (1994) 1684.\n P. Bak, M. Paczuski, M. Shubik, Physica A 246 (1997)\n430.\n\nACF\n\n0.8\n\n0\n10\n\n20\n30\n40\nTime lag (days)\n\n50\n\n0.2\n0\n\n10\n\n20\n30\n40\nTime lag (days)\n\n50\n\n## FIG. 4. Lagged autocorrelation functions of rt and\n\nUSD/DEM returns, respectively (top panels). Lagged autocorrelation functions of absolute value for the same data sets\n(bottom panels). Dashed horizontal lines represent the 95%\ncondence interval of a Gaussian random walk.\n\n## The presented empirical analysis clearly shows that\n\nthree simple rules of our model lead to a fat-tailed distribution of returns, long-term dependence in volatility\nand no dependence in returns themselves as observed for\nmarket data. Thus we may conclude that this simple\nmodel is a good first approximation of a number of real\nfinancial markets.\nWe gratefully acknowledge critical comments of an\nanonymous referee, which led to a substantial improvement of the paper.\n\nTABLE I. Estimates of the Hurst exponent H for simulated and market data.\n\nData\nSimulation\nUSD/DEM\nJPY/USD\nDJIA\nWIG20\n\n## D. Derrida, P.G. Higgs, J. Phys. A 24 (1991) L985.\n\n F. Schweitzer, J.A. Holyst, Eur. Phys. J. B 15 (2000) 723.\n R. Savit, R. Manuca, R. Riolo, Phys. Rev. Lett. 82 (1999)\n2203.\n A. Cavagna, J.P. Garrahan, I. Giardina, D. Sherrington,\nPhys. Rev. Lett. 83 (1999) 4429.\n D. Chowdhury , D. Stauer, Eur. Phys. J. B 8 (1999)\n477.\n R. Cont and J. P. Bouchaud, Macroeconomic Dyn. 4\n(2000) 170.\n D. Challet, M. Marsili, R. Zecchina, Phys. Rev. Lett. 84\n(2000) 1824.\n V.M. Eguiluz, M.G. Zimmermann, Phys. Rev. Lett. 85\n(2000) 5659.\n K. Sznajd-Weron, J. Sznajd, Int. J. Mod. Phys. C 11,\nNo. 6 (2000).\n D. Stauer, A.O. Sousa, S. Moss de Oliveira , Int. J.\nMod. Phys. C 11, No. 6 (2000).\n J. Adler, Physica A 171, 453 (1991).\n See eg. S. Rachev, S. Mittnik, Stable Paretian Models\n\nSimulation\nUSD/DEM\nJPY/USD\nDJIA\nWIG20\n\nMethod\nDFA\nGPH\nReturns\n0.5270\n0.4666\n0.3653\n0.5127\n0.5115\n0.6154\n0.5353\n0.5303\n0.5790\n0.4585\n0.4195\n0.3560\n0.5030\n0.4981\n0.4604\nAbsolute value of returns\n0.8940\n0.9335\n0.8931\n\n0.7751\n0.8406\n0.8761\n\n0.8576\n0.9529\n0.9287\n\n0.7838\n0.9080\n0.8357\n0.9103\n0.9494\n0.8262\nR/S-AL\n\n,\nand denote signicance at the (two-sided) 90%, 95%\nand 99% level, respectively. For the R/S-AL and DFA statistics inference is based on empirical Monte Carlo results of\nWeron , whereas for the GPH statistics on asymptotic\ndistribution of the estimate of H ."
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https://algorithms.tutorialhorizon.com/category/position/software-engineer-in-test/ | [
"## Number of Contiguous Parking Areas\n\nProblem: Given a building with parking slots. If a spot is free then it is marked by 1 and if the spot is taken by a vehicle then it is marked by 0. Write a program to find the number of free contiguous parking areas in the building. One free parking area can have one … Read more Number of Contiguous Parking Areas\n\n## The largest number can be formed from the given number\n\nGiven a number write an algorithm to construct the largest number possible by using the digits of given number. Given number could be a negative number as well. Example: Given Input: 34277765 Output: 77765432 Given Input: -342765 Output: -234567 Given Input: 0 Output: 0 Given Input: 2034 Output: 4320 Approach: Sorting Check if the given … Read more The largest number can be formed from the given number\n\n## Print all nested directories and files in a given directory – Recursion\n\nGiven a source directory, write a recursive program to print all the nested directories and files in the source directory. Example: [Hellojava] [target] original-Hellojava-1.0-SNAPSHOT.jar [classes] [com] [example] [lambda] [demo] Hello.class [maven-archiver] pom.properties Hellojava-1.0-SNAPSHOT.jar pom.xml Hellojava.iml [src] [test] [java] [main] [resources] [java] [com] Hello.java Approach: Check if given file is directory, if yes then print its … Read more Print all nested directories and files in a given directory – Recursion\n\n## Maximum meetings in one room\n\nYou have one meeting room at your company. There are N meeting needs to take place. Every meeting has a start time and end time along with the meeting title. Your task is to schedule as many meetings as possible in that conference room with no conflicts. Example: Meetings: A[ Start Time: 1, End Time: … Read more Maximum meetings in one room\n\n## Check if given sudoku is valid or not\n\nProblem: Given a filled sudoku, write a program to check if sudoku is valid or following all of its rules. SUDOKU rules: Each column contains all of the digits from 1 to 9 only once. Each row contains all of the digits from 1 to 9 only once. Each of the nine 3×3 sub-grid contains … Read more Check if given sudoku is valid or not\n\n## Construct the Largest number from the given digits\n\nGiven a set of digits, write an algorithm to construct the largest number possible by appending the given digits. Example: Given Digits: [9, 1, 9, 2, 8, 4, 2] largest number: 9984221 Given Digits: [1, 2, 5, 6, 7] largest number: 76521 Approach: Sorting Sort the given array in descending order. Initialize result = 0. … Read more Construct the Largest number from the given digits\n\n## Efficient Robot Problem – Find Minimum Trips\n\nProblem: There is N number of items that need to be transferred from one place to another by a robot. Each item has a specific weight. The robot can carry maximum weight K in one trip. You need to come up with an algorithm to find out the minimum number of trips that the robot … Read more Efficient Robot Problem – Find Minimum Trips\n\n## Two Sum Problem\n\nObjective: Given an array of integers, and k. Write a program to find indexes of two elements in an array which sum is equal to K. Example: Given array: [5, 4, 7, 3, 9, 2], Sum = 13 Output: Found indexes are: 4 and 1 Given array: [1, 2, 3, 4, 5], Sum = 9 … Read more Two Sum Problem\n\n## Find if any two intervals overlap in given intervals\n\nObjective: Interval is defined as [start, end]- the start of an interval to the end of the interval. Given a list of Intervals. Your task is to check if any two intervals overlap. Example: Given Interval: [[1,5], [6,10], [12,15], [3,7]] Two intervals are present which intersect Given Interval: [[1,5], [6,10], [12,15]] No intervals overlasx Approach: … Read more Find if any two intervals overlap in given intervals\n\n## Find all possible combinations with sum K from a given number N(1 to N) with the repetition of numbers is allowed\n\nObjective: Given two integers N and K, Write an algorithm to find possible combinations that add to K, from the numbers 1 to N. Condition: An integer from 1 to N can be repeated any number of times in combination. Example: N = 5 K = 3 Output: [1, 1, 1] [1, 2] N … Read more Find all possible combinations with sum K from a given number N(1 to N) with the repetition of numbers is allowed\n\n## Insert a node in the given sorted linked list.\n\nObjective: Given a linked list in which nodes are sorted in ascending order. Write an algorithm to insert a given node into the linked list so that all the nodes in the list will maintain the sorted order. Example: Given Linked List: -> 2, Insert node: 6 New List: -> 2 -> 6 Given Linked … Read more Insert a node in the given sorted linked list.\n\n## Construct the largest number from the given array.\n\nObjective: Given an array of integers, write an algorithm to construct the largest number possible by appending the array elements. Example: Given Input: [7, 78] Largest Number Possible: 787 Explanation: two possibilities are 778 and 787. 787 is larger than 778. Given Input: [25, 42, 39] Largest Number Possible: 423925 Approach: Normal sorting in descending … Read more Construct the largest number from the given array."
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http://energythic.com/view.php?node=208 | [
"1. # Dielectrophoretic pump\n\n##### Created by Zoltan Losonc ([email protected]) on 4 July 2003. Last updated on 14 July 2003. Linguistic proofreading done by Steven Dufresne.\n\nZoltan Lozong explains along this amazing study how one could extract exergy from diephoretic process.\n\n## Rectangular flat capacitor\n\nIn the tutorial about electrostatic forces in dielectrics we introduced the dielectrophoretic force that creates a pressure-difference at the edge of a flat capacitor when merged into a liquid dielectric, and pushes up a liquid column between the plates (fig. 1a).",
null,
"Fig. 1.\n\nWe have derived the formula for the force that pushes the liquid column upwards, and it is:",
null,
"The pressure increase at the bottom edge of the capacitor can be calculated from this as:",
null,
"The pressure p in our discussion actually means a pressure-difference Δ p and not an absolute pressure.\n\n## Dilemma of leakage\n\nTwo questions may arise naturally in connection with this setup. The first question is whether the liquid would pour out from the elevated liquid column at the front- and back edges when they are left open. The other question is whether a jet of liquid would be ejected upwards into the air (like a fountain) when the height of the capacitor is lower than the calculated height of the liquid column between the plates. If this would be possible then a very efficient pump and fountain could be constructed since the pumping effect does not require any current flow but only static high voltage. By covering the plates with a very good solid insulator layer, and using a liquid dielectric of high permittivity and high resistance, the current could be minimized and the power consumption reduced to an insignificant value compared to the power generated by the pumping effect.\n\nThe answer to one of these questions answers both questions because the same principle determines both these phenomena. If the law of energy conservation has to remain valid, then the liquid may not flow out either at the open side edges or at the top of the capacitor since then a closed loop movement of the liquid would be established - producing energy - while theoretically (using very high resistance dielectrics) requiring no electric current and no input power to maintain the pumping effect. This explanation of the classical science seems to be supported by the fact that the electric field is the same at all four edges of the capacitor. Thus the same force is supposed to push the liquid dielectric into the capacitor at all sides when it reaches into the inhomogeneous E-field region of the edges where the pumping forces appear. According to this interpretation the liquid would flow into the inhomogeneous E-field region outside the plates, but it would stop there, held back by the dielectrophoretic forces, and form an arced convex liquid surface (fig. 1b). In practice this might not be fully realized, since at the edges that are not fully surrounded by the dielectric (above the liquid surface) some little force component will be missing that would otherwise be generated by the weak E-field farther away from the edges if the whole capacitor were surrounded with the liquid. Thus the dielectrophoretic pressure at the bottom edge (where there is a deep space below, filled with dielectric) will be slightly stronger than that at the edges above the liquid surface, and a weak leakage might appear. However the pressure-difference will be minimal and cannot provide an efficient pumping effect. So as a first step approximation we can assume that the law of energy conservation is satisfied in this case, and that no significant pumping effect can be achieved that would pump the liquid through the capacitor, since the inward sucking forces are nearly the same at all four edges, and the retarding force at the outlet will cancel the upward pumping force at the bottom input.\n\n## Diminished retarding pressure-difference at a hole outlet\n\nHowever we may try to go around this difficulty by not expecting the liquid to flow out at the edges, where the retarding forces are strong (sealing the edges), but through a hole drilled into the middle of the grounded plate, as shown in fig. 2.",
null,
"Fig. 2.\n\nNow the question is how much retarding pressure-difference will be generated at the hole outlet by the dielectrophoretic forces, and whether this retarding force will be less than the pumping force at the bottom edge. This is a difficult question to answer without measurements since the shape of the E-field at the hole is fairly complex as shown in fig.3a. But we have good reason to suspect that the forces will be less intense here because the E-field is less intense outside the hole (the plate is grounded) compared to the E-field intensity at the edges of the capacitor. Naturally the indirect method for calculating the forces is not applicable in this case since that can yield a correct result only if the law of energy conservation is valid. Thus that method would not allow any possibility of leakage from the capacitor.",
null,
"Fig.3.\n\nThe fig. 3a. shows a slightly inhomogeneous and arced E-field shape that can still produce significant retarding force. Let’s try to find some arrangement that could minimize the curvature and inhomogeneousness of the E-field at the hole. One such possibility is shown in fig. 3b, when a parallel shielding plate is placed in front of the hole and connected to the ground together with the right electrode of the condenser. The shielding plate s ensures a more uniform E-field distribution, and this arrangement might produce less dielectrophoretic resistive force in the way of the flowing liquid. All this is written in conditional mode, since without numerical analysis or exact measurements we cannot know the magnitude of the forces at the hole.\n\nHowever our endeavor to theoretically violate the law of energy conservation is not completely hopeless, and we will do the impossible and prove the invalidity of energy conservation for the complete system in some special arrangements, based on (and assuming) the local validity of energy conservation for each single component of the system.\n\n## Flat disc capacitor\n\nIn order to find the possible components of such a FE system, we should derive the formulas of the dielectrophoretic forces for each element through mathematical analysis, firmly based on the assumed validity of energy conservation for each component. In the case of a rectangular flat capacitor this has been already done (see above), and now let’s see if there is any difference if we use disc-shaped electrodes instead of rectangular plates (fig. 4a).",
null,
"Fig. 4.\n\nThe flat disc capacitor is fully merged into the liquid dielectric but there is a flexible, movable wall w between the electrodes (perpendicular to the plates) that prevents the liquid from entering into the center part of the capacitor which is filled with air. Let us calculate the force that is squeezing this insulating wall towards the center, and also the pressure within the capacitor caused by the dielectrophoretic forces at the edges. The capacitance of the condenser is:",
null,
"The electric energy in the condenser is:",
null,
"The electric energy change of the capacitor per wall movement, when the radius of the wall is changed is:",
null,
"The force upon the wall is:",
null,
"The pressure of the liquid inside the capacitor is:",
null,
"This is the same formula as that of the rectangular flat capacitor and both produce the same dielectrophoretic pressure.\n\n## Coaxial cylindrical capacitor\n\nNow let’s do the same calculation for the coaxial capacitor shown in fig. 4b.",
null,
"",
null,
"In this case the force is directed towards the increase of x, thus:",
null,
"This is an interesting result since the pressure-difference of the coaxial capacitor is not the same as that of the rectangular- or disc shaped flat capacitors using the same dielectric, voltage, and distance between electrodes. This difference can provide a firm ground for the establishment of the asymmetry in pressures and the consequent violation of the energy conservation.\n\n## Semi-cylindrical capacitor\n\nBefore analyzing this possibility, let us calculate the force and pressure in the case of a semi-cylindrical capacitor (fig. 5).",
null,
"Fig. 5.",
null,
"The energy stored in the capacitor is:",
null,
"The torque is calculated as:",
null,
"(1)\n\nSupposing that the torque is caused by a homogeneous pressure, it is calculated as:",
null,
"",
null,
"Equating this expression with the formula (1) we get the pressure:",
null,
"The total force developed by this pressure upon the capacitor is:",
null,
"## Hemispherical capacitor\n\nThe examined capacitor is shown in fig. 6. The capacitance is:",
null,
"The energy content of the condenser is:",
null,
"We want to calculate how much pressure would be required to create the derived torque. This is done by assuming a constant pressure over the examined half circle-ring area, and integrating the elementary torques to get the resultant torque M (fig. 6b)",
null,
"Fig. 6.",
null,
"",
null,
"By equating this torque resulting from the pressure-difference with the above formula (2) we get the pressure:",
null,
"Supposing that this pressure is the same over the whole surface area of the circle-ring between the bottom edges of the hemispherical capacitor, the force upon the condenser is:",
null,
"",
null,
"These formulas for the hemispherical capacitor have been derived using a torque around an axis, and a doubt may arise about its correctness. Therefore the same formulas have been derived with another method, assuming that the dielectric rises into the capacitor through the whole open bottom surface area simultaneously, and develops a torque around a single point at the center. The calculation can be found in the appendix.\n\n## Analyzing the results\n\nThe following table summarizes the derived formulas of the pressures and forces at the open edges of the different capacitor types:\n\n pressure force Rectangular flat capacitor",
null,
"",
null,
"Disc-shaped flat capacitor",
null,
"",
null,
"Cylindrical coaxial capacitor",
null,
"",
null,
"Semi-cylindrical capacitor",
null,
"",
null,
"Hemispherical capacitor",
null,
"",
null,
"Pressure-differences pd = pr and px = pc, thus we have only 3 different pressure-difference for the 3 different electrode shapes. The following diagram compares these pressure-differences for changing r1 (radius of the smaller electrode) with fixed r2 = 5mm and uses the following parameters: l = r2 – r1 ; U=10 kV ; e =56e 0 (the permittivity of glycerin).",
null,
"Diagram 1.\n\nThe pressure-difference ratios of flat-cylindrical and flat-spherical condensers are:",
null,
"The curves nicely illustrate that when r1 approaches r2 all 3 pressure-differences will become nearly identical (right side). However when r1 is much smaller than r2, there will be big differences between the pressure-differences. For example when r1 is 10 times smaller then r2 then pf=1.4pc and pf =3.7ps. When r1 is 100 times smaller than r2 then pf =2.35pc and pf =33.67ps.\n\n## Complete pump configurations\n\nWith a clever arrangement we can exploit these pressure-differences to create an efficient electrostatic pump that violates the law of energy conservation. A flat capacitor’s edge should be merged into the liquid dielectric, since that gives the highest pressure-difference; and a smooth transition should be made to a coaxial cylindrical capacitor, that in turn should end in a semispherical capacitor where the fluid leaves the condenser, since the retarding pressure-difference is the lowest for that shape (fig. 7).",
null,
"Fig. 7.\n\nAt the curved region t where the flat disc capacitor merges into a cylindrical coaxial capacitor, the E-field lines are straight and thus do not retard the flow of the liquid. There will be a dielectrophoretic force towards the center of curvature O that is perpendicular to the movement of the fluid, tending to push it against the wall of the upper electrode, but not retarding the movement of the dielectric. The only retarding pressure-difference appears at the top opening, and that pressure-difference of the semispherical condenser has just been calculated to be less than the pressure-difference of the flat capacitor at the bottom. Thus the resultant pressure-difference that this pump can create is calculated as:",
null,
"",
null,
"",
null,
"If the output ending would not have the semispherical shape, but is left to be the natural ending of the coaxial capacitor (fig. 7b), then the resultant pumping pressure-difference would be:",
null,
"",
null,
"The dependence of these pressure-differences from the radius of the inner electrode of the cylindrical part r1 and voltage is shown in diagram 2.",
null,
"Diagram 2.\n\nThe semispherical outlet shows a better performance than the simple cylindrical coaxial capacitor ending. Other possible pump shapes are shown in fig. 7c. and fig. 8, which have basically the same performance as the one shown in fig. 7a. How high can these pumps elevate the liquid column? The formulas for spherical and cylindrical endings are:",
null,
"",
null,
"Fig. 8.\n\n## Violation of the law of energy conservation\n\nAfter deriving the formulas, and supposing that there is no major flaw in the derivation, we have arrived to a weird contradiction:\n\n• If the law of energy conservation is valid for the basic shapes of capacitors, then the derived formulas for the dielectrophoretic forces, torques, and pressure-differences must also be valid. If they are valid then the formula for the resultant pressure-difference of the described dielectrophoretic pump (made from a special combination of these elements) should also be valid, and significantly greater than zero. But if the resultant pressure-difference of the pump is greater than zero then the law of energy conservation is violated for the whole pump, since at least theoretically no electric current is required for its operation (but only high static voltage) and thus no input power is consumed. Theoretically this pump can produce mechanical energy without consuming input energy.\n• On the other hand, if the pump does not produce a resultant pressure-difference between its input and output (in which case the final formula would be invalid), then the law of energy conservation is violated for the basic capacitor shapes, since the final formula for the resultant pressure-difference of the pump has been derived as the sum of the correlations of each basic shape. In that case we could use the basic capacitors individually to produce excess energy with proper arrangements.\n\nIt seems that there is no possibility to satisfy the law of energy conservation simultaneously for the whole pump and also for each basic capacitor shape component. The above conclusions assumed that the liquid dielectric is a very good insulator and it does not allow any current flow. In practice there is always some leakage current that can be minimized by carefully choosing the liquid dielectric and optionally covering the electrodes with a very good solid insulator layer. But even if there still would be significant leakage current, the law of energy conservation would be still violated, since the input power (caused by the unwanted current) is fully transformed into heat and into accumulation of electrostatic charge (and energy) on some bodies. Thus the input power is not consumed for the propulsion of the liquid, but for producing heat, thus the kinetic energy represents an excess energy that still violates the law of energy conservation.\n\n## Increasing the amount of excess energy generated by the pump\n\nHow much energy can we generate with the pump? Knowing the produced pressure-difference it is easy to calculate the power:",
null,
"Where p – is the pressure-difference; dV - is the elementary volume; dt – time difference; S – surface area; v – velocity of the liquid.\n\nThe pressure-difference p and the surface area through which the liquid flows out are determined by the size and other parameters of the pump. However, the velocity v of the liquid dielectric is not definitely determined. This velocity would be limited only by the length of the accelerating path and the frictional resistance of the fluid path. Since the pressure-difference does not depend on the velocity of the liquid but is constant, the possibility naturally offers itself to artificially create an external additional pressure-difference by connecting an external pump in series with the dielectrophoretic pump, and this way further increase the velocity of the dielectric. The external pressure-difference can be generated by any conventional means and the increased velocity increases the excess energy. To calculate the increase of excess energy let us assume that the velocity of the fluid through a pipe is linearly proportional with the pressure-difference v=Kp. The input power consumed by the external conventional pump is then:",
null,
"The output power is:",
null,
"The excess power is:",
null,
"If there would be no external pump increasing the velocity then the excess energy would be",
null,
", thus the external pump increases the excess energy by the factor of:",
null,
"If the external pump provides 9 times higher pressure than the electrostatic pump then the excess energy generated by the electrostatic pump will be 10 times greater than without external pump. The COP of the whole system is:",
null,
"When the external pressure-difference is much higher than that of the dielectrophoretic pump then the COP approaches unity. Naturally the best is to use dielectrophoretic pumps also as external pumps, thus they would not consume significant energy, and the COP could be maintained at high value.\n\n## Questions to be clarified by measurements and numerical analysis\n\n• A future task is to verify the validity of the derived formulas for the basic capacitor shapes, and if there is significant deviation between the predicted and measured pressures, new correct formulas and applications should be created.\n• The dielectrophoretic pressure should also be measured for special openings (intended to provide low pressure-difference output) as shown in fig. 3 to clarify their applicability.\n• It is an interesting question whether the dielectrophoretic pressure will depend on the shapes of the edges as shown in fig. 9. According to the law of energy conservation all these edge shapes should produce identical pressure-differences, however the different E-field distributions justify practical verification.",
null,
"Fig. 9.\n\n• Another important fact to be aware of is that in the case of semi-cylindrical and hemispherical capacitors the dielectrophoretic forces upon the molecules of the dielectric will be stronger at the smaller electrode since the E-field intensity is strongest there. Thus we can expect a higher pressure-difference near the smaller electrode and a lower pressure-difference at the bigger electrode. Such pressure distribution would generate fluid vortexes at the edges, which are not beneficial for the pumping effect (fig. 9d). However if such vortexes do exist and they are sufficiently strong then they can also provide free energy. This effect should also be verified experimentally.\n\n## An alternative derivation of the pressure-difference and force for the hemispherical capacitor shape\n\nUsing this method we assume that there is a flexible frustum of a cone between the electrodes with vacuum (or air) inside, and that the space below the cone is filled with the liquid dielectric that is squeezing the cone as in fig. 10a.",
null,
"Fig. 10.\n\nThe capacitance of a full spherical condenser is:",
null,
"The capacitance of a sphere segment is proportional to its outer surface area (or in other words, with the space angle viewed from the center of the capacitor), thus the partial capacitances are calculated as follows.\n\nThe outer surface area of the conical sphere segment C1 (fig. 10a) is",
null,
". The total surface area of the sphere is",
null,
". The partial capacitance C1 can be calculated when the capacitance of the full spherical capacitor is multiplied with the quotient of these two surface areas:",
null,
"",
null,
"",
null,
"The electric energy of the capacitor is:",
null,
"The energy change per angle displacement:",
null,
"Thus the torque around the center point should be:",
null,
"Let’s now calculate the torque on the frustum of a cone if the liquid pressure is constant everywhere on that surface area (fig. 10b):",
null,
"",
null,
"",
null,
"Equating this formula with (3) we get the wanted pressure difference:",
null,
"This formula is the same as derived above with a different method, thus it is correct.\n\nNOTE: The above figures have been made manually based on rough estimations, thus the E-field shapes are not based on numerical analysis or exact analytical calculations. The real E-field shape might be somewhat different than what is shown in the drawings. These illustrations serve only for the purpose of aiding the understanding of the presentation.\n\nCreated by Zoltan Losonc ([email protected]) on 4 July 2003. Last updated on 14 July 2003. Linguistic proofreading done by Steven Dufresne."
]
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null,
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"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump26.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump27.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump28.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump29.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump30.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump31.gif",
null,
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null,
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null,
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null,
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null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump39.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump40.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump_diag.gif",
null,
"http://energythic.com/usercontent/3/DIEPHOPU_dielphpump41.gif",
null,
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null,
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null,
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https://answers.everydaycalculation.com/gcf/560-1620 | [
"Solutions by everydaycalculation.com\n\n## What is the GCF of 560 and 1620?\n\nThe gcf of 560 and 1620 is 20.\n\n#### Steps to find GCF\n\n1. Find the prime factorization of 560\n560 = 2 × 2 × 2 × 2 × 5 × 7\n2. Find the prime factorization of 1620\n1620 = 2 × 2 × 3 × 3 × 3 × 3 × 5\n3. To find the gcf, multiply all the prime factors common to both numbers:\n\nTherefore, GCF = 2 × 2 × 5\n4. GCF = 20\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn how to find GCF of upto four numbers in your own time:"
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null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
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https://quant.stackexchange.com/questions/16440/on-a-source-for-a-mean-variance-portfolio-optimization-result/17988 | [
"# On a source for a mean-variance portfolio optimization result\n\nIn the context of a mean_variance framework consider an optimizing investor who chooses at time $T$ portfolio weights $w$ so as to maximize the quadratic objective function:\n\n$$U(w) = E[R_p] - \\frac{\\gamma}{2}Var[R_p]= w'\\mu - \\frac{\\gamma}{2}w'Vw$$\n\nWhere $E$ and $Var$ denote the mean and variance of the uncertain portfolio rate of return $R_p = w'R_{T+1}$ to be realized in time $T + 1$ and $\\gamma$ is the relative risk aversion coefficient. The optimal portfolio weights will be:\n\n$$w^* = \\frac{1}{\\gamma}V^{-1}\\mu$$\n\nCould I have a reference that proves this result? preferably a textbook that builds up to it.\n\n• I think Markowitz' 1959 book does, but it's a straightforward optimization that is easy if you look up the relevant matrix derivatives. I think I went through the math in another question here, but can't find it now. – John Feb 3 '15 at 15:44\n• – Monolite Feb 3 '15 at 15:51\n\nYou do note require a sum up constraint that gives you that the weights sum up to 1? Then the problem is equivalent to a maximization without constraints: $$Z(\\omega)=w'\\mu - \\frac{\\gamma}{2}w'Vw$$ then it holds that $$\\frac{dZ}{d\\omega}=\\mu-\\gamma V\\omega\\overset{!}{=}0\\\\ \\Leftrightarrow \\frac{1}{\\gamma}\\mu=V\\omega^*\\\\ \\Leftrightarrow\\omega^* = \\frac{1}{\\gamma}V^{-1}\\mu$$"
]
| [
null
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https://blogs.sas.com/content/iml/2014/05/05/iml-character-vectors.html | [
"SAS programmers are probably familiar with how SAS stores a character variable in a data set, but how is a character vector stored in the SAS/IML language?\n\nRecall that a character variable is stored by using a fixed-width storage structure. In the SAS DATA step, the maximum number of characters that can be stored in a variable is determined when the variable is initialized, or you can use the LENGTH statement to specify the maximum number of characters. For example, the following statement specifies that the NAME variable can store up to 10 characters:\n\n```data A; length name \\$ 10; /* declare that a variable stores 10 characters */ ...```\n\nThe values in a character variable are left aligned. That is, values that have fewer than 10 characters are padded on the right with blanks (space characters).\n\n### SAS/IML character vectors\n\nThe same rules apply to character vectors in the SAS/IML language. A vector has a \"length\" that determines the maximum number of characters that can be stored in any element. (In this article, \"length\" means the maximum number of characters, not the number of elements in a vector.) Elements with fewer characters are blank-padded on the right. Consequently, the following two character vectors are equivalent. :\n\n```proc iml; c = {\"A\", \"B C\", \" XZ\", \"LMNOPQ\"}; /* length set at initialization */ c2 = {\"A \", \"B C \", \" XZ \", \"LMNOPQ\"}; /* all strings have length 6 */ if c=c2 then print \"Character vectors are equal\"; else print \"Character vectors are not equal\";```",
null,
"You can determine the maximum number of characters that can be stored in each element by using the NLENG function in SAS/IML. You can also discover the number of characters in each element of a vector (omitting any blank padding) by using the LENGTH function, as follows:\n\n```N = nleng(c); trimLen = length(c); print N trimLen c;```",
null,
"In this example, each element of the vector c can hold up to six characters. If you write the c variable to a SAS data set, the corresponding variable will have length 6. However, if you trim off the blanks at the end of the strings, most elements have fewer than six characters. Notice that the LENGTH function counts blanks at the beginning and the middle of a string but not at the end, so that the string \" XZ\" counts as four characters.\n\n### Where are the blanks?\n\nNotice that the ODS HTML destination is not ideal for visualizing blanks in strings. In HTML, multiple blank characters are compressed into a single blank when the string is rendered, so only one space appears on the displayed output. If you need to view the spaces in the strings, use the ODS LISTING destination, which uses a fixed-width font that preserves spaces. Alternatively, the following SAS/IML function prints each character (not including trailing blanks):\n\n```/* convert a string to a row vector of single characters (uses SAS/IML 12.1) */ start Str2Vec(s); return (substr(s, 1:length(s), 1)); /* row vector of characters */ finish; /* print characters of all strings in a vector */ start PrintChars(v); L = length(v); /* characters per name, not counting trailing blanks */ do i = 1 to ncol(L)*nrow(L); c = char(1:L[i], 2); print (Str2Vec(v[i]))[colname=c]; /* print individual letters */ end; finish; run PrintChars(c);```",
null,
"I think the Str2Vec function is very cool. It uses a feature of the SUBSTR function in SAS/IML 12.1 to convert a string into a vector of characters. The PrintChars function simply calls the Str2Vec function for each element of a character matrix and prints the characters with a column header. This makes it easy to see each character's position in a string.\n\nThis article provides a short overview of how strings are stored inside SAS/IML character vectors. For more details about SAS/IML character vectors and how you can manipulate strings, see Chapter 2 of Statistical Programming with SAS/IML Software.\n\nShare",
null,
"Distinguished Researcher in Computational Statistics\n\nRick Wicklin, PhD, is a distinguished researcher in computational statistics at SAS and is a principal developer of PROC IML and SAS/IML Studio. His areas of expertise include computational statistics, simulation, statistical graphics, and modern methods in statistical data analysis. Rick is author of the books Statistical Programming with SAS/IML Software and Simulating Data with SAS.\n\n1.",
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"https://blogs.sas.com/content/iml/files/userphoto/136.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8540516,"math_prob":0.9567216,"size":4526,"snap":"2020-34-2020-40","text_gpt3_token_len":1015,"char_repetition_ratio":0.17403804,"word_repetition_ratio":0.028985508,"special_character_ratio":0.22757402,"punctuation_ratio":0.115207374,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95697874,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,2,null,2,null,2,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T20:05:12Z\",\"WARC-Record-ID\":\"<urn:uuid:f3ceca89-a61d-465d-bd89-f63703d81319>\",\"Content-Length\":\"32498\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a384107-abe4-4cbe-8118-79860ac2c236>\",\"WARC-Concurrent-To\":\"<urn:uuid:c44378d1-c0b4-4a5b-8fea-8addba305b4b>\",\"WARC-IP-Address\":\"149.173.160.44\",\"WARC-Target-URI\":\"https://blogs.sas.com/content/iml/2014/05/05/iml-character-vectors.html\",\"WARC-Payload-Digest\":\"sha1:S5IUCCUKUPS3TIGAEGKH366L5D36P5ZV\",\"WARC-Block-Digest\":\"sha1:ERBSJSML2DMSGJSEGKKTPLZUAKCFXDOX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402088830.87_warc_CC-MAIN-20200929190110-20200929220110-00063.warc.gz\"}"} |
https://proofwiki.org/wiki/Book:Frigyes_Riesz/Functional_Analysis | [
"# Book:Frigyes Riesz/Functional Analysis\n\n## Frigyes Riesz and Béla Sz.-Nagy: Functional Analysis\n\nPublished $\\text {1990}$, Dover\n\nISBN 978-0486662893 (translated by Leo F. Boron)\n\n### Subject Matter\n\nFunctional Analysis\n\n### Contents\n\nPart I: Modern theories of differentiation and integration\n\nChapter I: Differentiation\nLebesgue's Theorem on the Derivative of a Monotonic Function\n1. Example of a Nondifferentiable Continuous Function\n2. Lebesgue's Theorem on the Differentiation of a Monotonic Function. Sets of Measure Zero\n3. Proof of Lebesgue's Theorem\n4. Functions of Bounded Variation\nSome Immediate Consequences of Lebesgue's Theorem\n5. Fubini's Theorem on the Differentiation of Series with Monotonic Terms\n6. Density Points of Linear Sets\n7. Saltus Functions\n8. Arbitrary Functions of Bounded Variation\n9. The Denjoy-Young Saks Theorem on the Derived Numbers of Arbitrary Functions\nInterval Functions\n10. Preliminaries\n11. First Fundamental Theorem\n12. Second Fundamental Theorem\n13. The Darboux Integrals and the Riemann Integral\n14. Darboux's Theorem\n15. Functions of Bounded Variation and Rectification of Curves\nChapter II: The Lebesgue integral\nDefinition and Fundamental Properties\n16. The Integral for Step Functions. Two Lemmas\n17. The Integral for Summable Functions\n18. Term-by-Term Integration of an Increasing Sequence (Beppo Levi's Theorem)\n19. Term-by-Term Integration of a Majorized Sequence (Lebesgue's Theorem)\n20. Theorems Affirming the Integrbility of a Limit Function\n21. The Schwarsz, Hölder and Minkowski Inequalities\n22. Measurable Sets and Measurable Functions\nChapter III: The Stieltjes integral and its generalizations\n23. The Total Variation and the Derivative of the Indefinite Integral\n24. Example of a Monotonic Continuous Function Whose Derivative Is Zero Almost Everywhere\n25. Absolutely Continuous Functions. Canonical Decomposition of Monotonic Functions\n26. Integration by Parts and Integration by Substitution\n27. The Integral as a Set Function\nThe Space $L^2$ and its Linear Functionals. $L^p$ Spaces\n28. The Space $L^2$; Convergence in the Mean; the Riesz-Fischer Theorem\n29. Weal Convergence\n30. Linear Functionals\n31. Sequence of Linear Functionals; a Theorem of Osgood\n32. Separability of $L^2$. The Theorem of Choice\n33. Orthonormal Systems\n34. Subspaces of $L^2$. The Decomposition Theorem\n35. Another Proof of the Theorem of Choice. Extension of Functionals\n36. The Space $L^p$ and Its Linear Functionals\n37. A Theorem on Mean Convergence\n38. A Theorem of Banach and Saks\nFunctions of Several Variables\n39. Definitions. Principle of Transition\n40. Successive Integrations. Fubini's Theorem\n41. The Derivative Over a Net of a Non-negative, Additive Rectange Function. Parallel Displacement of the Net\n42. Rectangle Functions of Bounded Variation. Conjugate Nets\n43. Additive Set Functions. Sets Measurable $\\paren B$\nOther Definitions of the Lebesgue Integral\n44. Sets Measurable $\\paren L$\n45. Functions Measurable $\\paren L$ and the Integral $\\paren L$\n46. Other Definitions. Egoroff's Theorem\n47. Elementary Proof of the Theorems of Arzelà and Osgood\n48. The Lebesgue Integral Considered as the Inverse Operation of Differentiation\n\nPart II: Integral equations. Linear transforms\n\nChapter IV: Integral equations\nThe Method of Successive Approximations\n64. The Concept of an Integral Equation\n65. Bounded Kernels\n66. Square-Summable Kernels. Linear Transformations of the Space $L^2$\n67. Inverse Transformations. Regular and Singular Values\n68. Iterated Kernels. Resolvent Kernels\n69. Approximations of an Arbitrary Kernel by Means of Kernels of Finite Rank\nThe Fredholm Alternative\n70. Integral Equations With Kernels of Finite Rank\n71. Integral Equations With Kernels of General Type\n72. Decomposition Corresponding to a Singular Value\n73. The Fredholm Alternative for General Kernels\nFredholm Determinants\n74. The Method of Fredholm\nAnother Method, Based on Complete Continuity\n76. Complete Continuity\n77. Subspaces ${\\mathfrak M}_n$ and ${\\mathfrak R}_n$\n78. The Cases $\\nu = 0$ and $\\nu \\ge 1$. The Decomposition Theorem\n79. The Distribution of the Singular Values\n80. The Canonical Decomposition Corresponding to a Singular Value\nApplications to Potential Theory\n81. The Dirichlet and Neumann Problems. Solution by Fredholm's Method\nChapter V: Hilbert and Banach spaces\nHilbert Space\n82. Hilbert Coordinate Space\n83. Abstract Hilbert Space\n84. Linear Transformations of Hilbert Space. Fundamental Concepts\n85. Completely Continuous Linear Transformations\n86. Biorthogonal Sequences. A Theorem of Paley and Wiener\nBanach Spaces\n87. Banach Spaces and Their Conjugate Spaces\n88. Linear Transformations and Their Adjoints\n89. Functional Equations\n90. Transformations of the Space of Continuous Functions\nChapter VI: Completely continuous symmetric transformations of Hilbert space\nExistence of Characteristic Elements. Theorem on Series Development\n92. Characteristic Values and Characteristic Elements. Fundamental Properties of Symmetric Transformations\n93. Completely Continuous Symmetric Transformations\n94. Solution of the Functional Equation $j - \\lambda A f = g$\n95. Direct Determination of the $n$-th Characteristic Value of Given Sign\n96. Another Method of Constructing Characteristic Values and Characteristic Elements\nTransformations with Symmetric Kernel\n97. Theorems of Hilbert and Schmidt\n98. Mercer's Theorem\nApplications to the Vibrating-String Problem and to Almost Periodic Functions\n99. The Vibrating-String Problem. The Spaces $D$ and $H$\n100. The Vibrating-String Problem. Characteristic Vibrations\n101. Space of Almost Periodic Functions\n102. Proof of the Fundamental Theorem on Almost Periodic Functions\n103. Isometric Transformations of a Finite-Dimensional Space\nChapter VII: Bounded symmetric, unitary, and normal transformations of Hilbert space\nSymmetric Transformations\n104. Some Fundamental Properties\n105. Projections\n106. Functions of a Bounded Symmetric Transformation\n107. Spectral Decomposition of a Bounded Symmetric Transformation\n108. Positive and Negative Parts of a Symmetric Transformation. Another Proof of the Spectral Decomposition\nUnitary and Normal Transformations\n109. Unitary Transformations\n110. Normal Transformations. Factorizations\n111. The Spectral Decomposition of Normal Transformations. Functions of Several Transformations.\nUnitary Transformations of the Space $L^2$\n112. A Theorem of Bochner\n113. Fourier-Plancherel and Watson Transformation\nChapter VIII: Unbounded linear transformations of Hilbert space\nGeneralization of the Concept of Linear Transformation\n114. A Theorem of Hellinger and Toeplitz. Extension of the Concept of Linear Transformation\n116. Permutability. Reduction\n117. The Graph of a Transformation\n118. The Transformation $B = \\paren {I + T^*T}^{-1}$ and $X = T \\paren{I + T^*T}^{-1}$\n119. Symmetric and Self-Adjoined Transformations. Definitions and Examples\n120. Spectral Decomposition of a Self-Adjoint Transformation\n121. Von Neumann's Method. Cayley Transforms\nExtensions of Symmetric Transformations\n123. Cayley Transforms. Deficiency Indices\n124. Semi-Bounded Symmetric Transformations. The Method of Friedrichs\n125. Krein's Method\nChapter IX: Self-adjoint transformations. Functional calculus, spectrum, perturbations\nFunctional Calculus\n126. Bounded Functions\n127. Unbounded Functions. Definitions\n128. Unbounded Functions. Rules of Calculation\n129. Characteristic Properties of Functions of a Self-Adjoint Transformation\n130. Finite or Denumerable Sets of Permutable Self-Adjoint Transformation\n131. Arbitrary Sets of Permutable Self-Adjoint Transformations\nThe Spectrum of a Self-Adjoint Transformation and Its Perturbations\n132. The Spectrum of a Self-Adjoint Transformation. Decomposition in Terms of the Point Spectrum and the Continuous Spectrum\n133. Limit Points of the Spectrum\n134. Perturbation of the Spectrum by the Addition of a Completely Continuous Transformation\n135. Continuous Perturbations\n136. Analytic Perturbations\nChapter X: Groups and semigroups of transformations\nUnitary Transformations\n137. Stone's Theorem\n138. Another Proof. Based on a Theorem of Bochner\n139. Some Applications of Stone's Theorem\n140. Unitary Representations of More General Groups\nNon-Unitary Transformations\n141. Groups and Semigroups of Self-Adjoint Transformations\n142. Infinitesimal Transformation of a Semigroup of Transformations of General Type\n143. Exponential Formulas\nErgodic Theorems\n144. Fundamental Methods\n145. Methods Based on Convexity Arguments\n146. Semigroups of Nonpermutable Contractions\nChapter XI: Spectral theories for linear transformations of general type\nApplications of Methods from the Theory of Functions\n147. The Spectrum. Curvilinear Integrals\n148. Decomposition Theorem\n149. Relations between the Spectrum and the Norms of Iterated Trasformations\n150. Application to Absolutely Convergent Trigonometric Series\n151. Elements of a Functional Calculus\n152. Two Examples\nVon Neumann's Theory of Spectral Sets\n153. Principal Theorems\n154. Spectral Sets\n155. Characterization by Symmetric, Unitary and Normal Transformations by Their Spectral Sets\n\nBibliography\n\nAppendix\n\nIndex\n\nNotation & symbols"
]
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https://pdglive.lbl.gov/DataBlock.action?node=S055HEW | [
"# Indirect Mass Limits for ${{\\boldsymbol H}^{0}}$ from Electroweak Analysis INSPIRE search\n\nThe mass limits shown below apply to a Higgs boson ${{\\mathit H}^{0}}$ with Standard Model couplings whose mass is a priori unknown.\n\nFor limits obtained before the direct measurement of the top quark mass, see the 1996 (Physical Review D54 1 (1996)) Edition of this Review. Other studies based on data available prior to 1996 can be found in the 1998 Edition (The European Physical Journal C3 1 (1998)) of this Review.\n\nVALUE (GeV) DOCUMENT ID TECN\n$\\bf{90 {}^{+21}_{-18}}$ 1\n 2018\nRVUE\n• • • We do not use the following data for averages, fits, limits, etc. • • •\n$91$ ${}^{+30}_{-23}$ 2\n 2012\nRVUE\n$94$ ${}^{+25}_{-22}$ 3\n 2012 A\nRVUE\n$91$ ${}^{+31}_{-24}$ 4\n 2010 A\nRVUE\n$129$ ${}^{+74}_{-49}$ 5\n 2006\nRVUE\n1 HALLER 2018 make Standard Model fits to ${{\\mathit Z}}$ and neutral current parameters, ${\\mathit m}_{{{\\mathit t}}}$, ${\\mathit m}_{{{\\mathit W}}}$, and ${\\Gamma}_{{\\mathit W}}$ measurements available in 2018. The direct mass measurement at the LHC is not used in the fit.\n2 BAAK 2012 make Standard Model fits to ${{\\mathit Z}}$ and neutral current parameters, ${\\mathit m}_{{{\\mathit t}}}$, ${\\mathit m}_{{{\\mathit W}}}$, and ${\\Gamma}_{{\\mathit W}}$ measurements available in 2010 (using also preliminary data). The quoted result is obtained from a fit that does not include the limit from the direct Higgs searches. The result including direct search data from LEP2, the Tevatron and the LHC is $120$ ${}^{+12}_{-5}$ GeV.\n3 BAAK 2012A make Standard Model fits to ${{\\mathit Z}}$ and neutral current parameters, ${\\mathit m}_{{{\\mathit t}}}$, ${\\mathit m}_{{{\\mathit W}}}$, and ${\\Gamma}_{{\\mathit W}}$ measurements available in 2012 (using also preliminary data). The quoted result is obtained from a fit that does not include the measured mass value of the signal observed at the LHC and also no limits from direct Higgs searches.\n4 ERLER 2010A makes Standard Model fits to ${{\\mathit Z}}$ and neutral current parameters, ${\\mathit m}_{{{\\mathit t}}}$, ${\\mathit m}_{{{\\mathit W}}}$ measurements available in 2009 (using also preliminary data). The quoted result is obtained from a fit that does not include the limits from the direct Higgs searches. With direct search data from LEP2 and Tevatron added to the fit, the 90$\\%$ CL (99$\\%$ CL) interval is $115 - 148$ ($114 - 197$) GeV.\n5 LEP-SLC 2006 make Standard Model fits to ${{\\mathit Z}}$ parameters from LEP/SLC and ${\\mathit m}_{{{\\mathit t}}}$, ${\\mathit m}_{{{\\mathit W}}}$, and $\\Gamma _{W}$ measurements available in 2005 with $\\Delta \\alpha {}^{(5)}_{{\\mathrm {had}}}({\\mathit m}_{{{\\mathit Z}}}$) = $0.02758$ $\\pm0.00035$. The 95$\\%$ CL limit is 285 GeV.\nReferences:\n HALLER 2018\nEPJ C78 675 Update of the global electroweak fit and constraints on two-Higgs-doublet models\n BAAK 2012\nEPJ C72 2003 Updated Status of the Global Electroweak Fit and Constraints on New Physics\n BAAK 2012A\nEPJ C72 2205 The Electroweak Fit of the Standard Model After the Discovery of a New Boson at the LHC\n ERLER 2010A\nPR D81 051301 Mass of the Higgs Boson in the Standard Electroweak Model\n LEP-SLC 2006\nPRPL 427 257 Precision Electroweak Measurements on the ${{\\mathit Z}}$ Resonance"
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http://ichakang.com/2020/01/15/what-is-a-pendulum-in-physics/ | [
"# What is really a pendulum in physics? Several physics degrees teach us pendulums are simply just pendulums together with the mass to the counterweight distributed from the way we think of terms of a truck, a ladder, or even a elevator.\n\nPhysics levels instruct us that a pendulum is a exact complicated device that has a mysterious source, and also may be an extremely complicated product. buy college research papers online There are plenty of topics of interest, both factual and plausible, concerning such a particular period.\n\nFor example, the motion of the rotating mass is determined by the force exerted by the mass on the angle if we believe that the procedure for rotating your system. This usually means this, in essence, the mass is pushing down the angle.\n\nWe will realize that the forces cancel Should we consider the way the alternative forces do the job on a pendulum. The power of gravity does not use. It appears that Newton’s law of gravity, stated the way we’d say itis really an approximation. This reality would suggest that the’x’ of Newton’s Law of Gravity should be”x-phi.”\n\nOne other difficulty that is important is people seem to fail to appreciate this, for people with mathematics degrees that are conventional, there’s a simple way plus the fact that individuals look at physics for scientists and engineers. https://eip.gmu.edu/strengthening-the-family/ Many physicists with math degrees understand what is depth physics. As an instance, let us look at a conversation in regards to a pendulum using a likely plane, and let us assume that the pendulum (mass and likely plane) are symmetrical in the feeling which the angles are the exact same.\n\nLet’s also assume that the length of the pendulum is equal to the diameter of this inclined plane. Afterward , the half an of the aircraft (the span ) is half the period of the pendulum (the diameter). Quite simply, the obvious depth of this likely plane is equal to the apparent thickness of this pendulum.\n\nThe velocity with regard for the middle of this stair is https://www.samedayessay.com half the speed of the bulk. There are two problems in physics. 1 problem is obvious. The opposite is perhaps not.\n\nEqually Newton’s and Einstein’s formulations of the theory of general relativity enable us to produce a few assumptions which can be useful. Let us go through the 2nd premise.\n\nWe are all aware there is a normal towards the thing with which we’re discussing. What we have to believe is that the ordinary is stable all through the world. We will only look at the premise.\n\nIt must be said that the ordinary is also a unique and standard occurrence of specific relativity. That means that, generally speaking, it ought to be discovered the normal must be different. This looks.\n\nNow, let’s consider the problem of velocity with regard towards the mass. The problem is to do with the existence of the ordinary. It isn’t hard to determine that the ordinary goes to be different anyplace. Additionally, it looks that, generally speaking, it is not just really a challenge to come across the pace therefore let’s moveon.\n\nThis brings us to a stride using a likely plane. What is apparent depth physics? Since we are aware that the ordinary goes to vary anywhere, then a thickness of the plane will be different. We are going to believe which it is the exact same every where.\n\n### Comment",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9464763,"math_prob":0.9312902,"size":3381,"snap":"2020-10-2020-16","text_gpt3_token_len":704,"char_repetition_ratio":0.121409535,"word_repetition_ratio":0.0034843206,"special_character_ratio":0.1993493,"punctuation_ratio":0.09789157,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9707969,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-31T12:28:04Z\",\"WARC-Record-ID\":\"<urn:uuid:6ca3917b-c801-414f-aed7-379c7a0af480>\",\"Content-Length\":\"60092\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:80cca403-e3f6-497d-8df2-d359fccb0af3>\",\"WARC-Concurrent-To\":\"<urn:uuid:5eb6cc7d-01d7-4ad3-a082-52cfebb0ea5c>\",\"WARC-IP-Address\":\"198.187.29.29\",\"WARC-Target-URI\":\"http://ichakang.com/2020/01/15/what-is-a-pendulum-in-physics/\",\"WARC-Payload-Digest\":\"sha1:S7YVGYAHKGYVXJQKNEJADALWFFKH7OWW\",\"WARC-Block-Digest\":\"sha1:I3J6T2WWLFEPDRBTX5HIRCNDYDZE437E\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370500482.27_warc_CC-MAIN-20200331115844-20200331145844-00394.warc.gz\"}"} |
https://convertoctopus.com/78-ounces-to-kilograms | [
"Conversion formula\n\nThe conversion factor from ounces to kilograms is 0.028349523125, which means that 1 ounce is equal to 0.028349523125 kilograms:\n\n1 oz = 0.028349523125 kg\n\nTo convert 78 ounces into kilograms we have to multiply 78 by the conversion factor in order to get the mass amount from ounces to kilograms. We can also form a simple proportion to calculate the result:\n\n1 oz → 0.028349523125 kg\n\n78 oz → M(kg)\n\nSolve the above proportion to obtain the mass M in kilograms:\n\nM(kg) = 78 oz × 0.028349523125 kg\n\nM(kg) = 2.21126280375 kg\n\nThe final result is:\n\n78 oz → 2.21126280375 kg\n\nWe conclude that 78 ounces is equivalent to 2.21126280375 kilograms:\n\n78 ounces = 2.21126280375 kilograms\n\nAlternative conversion\n\nWe can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 0.45223028140488 × 78 ounces.\n\nAnother way is saying that 78 ounces is equal to 1 ÷ 0.45223028140488 kilograms.\n\nApproximate result\n\nFor practical purposes we can round our final result to an approximate numerical value. We can say that seventy-eight ounces is approximately two point two one one kilograms:\n\n78 oz ≅ 2.211 kg\n\nAn alternative is also that one kilogram is approximately zero point four five two times seventy-eight ounces.\n\nConversion table\n\nounces to kilograms chart\n\nFor quick reference purposes, below is the conversion table you can use to convert from ounces to kilograms\n\nounces (oz) kilograms (kg)\n79 ounces 2.24 kilograms\n80 ounces 2.268 kilograms\n81 ounces 2.296 kilograms\n82 ounces 2.325 kilograms\n83 ounces 2.353 kilograms\n84 ounces 2.381 kilograms\n85 ounces 2.41 kilograms\n86 ounces 2.438 kilograms\n87 ounces 2.466 kilograms\n88 ounces 2.495 kilograms"
]
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https://help.scilab.org/docs/5.4.1/ru_RU/gfare.html | [
"Change language to:\nEnglish - Français - 日本語 - Português -\n\nSee the recommended documentation of this function\n\nScilab help >> CACSD > gfare\n\n# gfare\n\nContinuous time filter Riccati equation\n\n### Calling Sequence\n\n`[Z,H]=gfare(Sl)`\n\n### Arguments\n\nSl\n\na continuous time linear dynamical system in state-space representation\n\nZ\n\nsymmetric matrix\n\nH\n\nreal matrix\n\n### Description\n\nGeneralized Filter Algebraic Riccati Equation (GFARE). `Z` = solution, `H` = gain.\n\nThe GFARE for `Sl=[A,B,C,D]` is:\n\n`(A-B*D'*Ri*C)*Z+Z*(A-B*D'*Ri*C)'-Z*C'*Ri*C*Z+B*Si*B'=0`\n\nwhere `S=(eye()+D'*D)`, `Si=inv(S)`, `R=(eye()+D*D')`, `Ri=inv(R)` and `H=-(B*D'+Z*C')*Ri` is such that `A+H*C` is stable.\n\n Версия Описание 5.4.0 `Sl` is now checked for continuous time linear dynamical system. This modification has been introduced by this commit"
]
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http://www.kylesconverter.com/pressure/microns-of-mercury-to-pounds-per-square-inch | [
"# Convert Microns Of Mercury to Pounds Per Square Inch\n\n### Kyle's Converter > Pressure > Microns Of Mercury > Microns Of Mercury to Pounds Per Square Inch\n\n Microns Of Mercury (µmHg) Pounds Per Square Inch (psi) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18\nReverse conversion?\nPounds Per Square Inch to Microns Of Mercury\n(or just enter a value in the \"to\" field)\n\nPlease share if you found this tool useful:\n\nUnit Descriptions\n1 Micron of Mercury:\n= 0.001 torr\n1 Pound per Square Inch:\n1 lbf/in2\n\nLink to Your Exact Conversion\n\nConversions Table\n1 Microns Of Mercury to Pounds Per Square Inch = 070 Microns Of Mercury to Pounds Per Square Inch = 0.0014\n2 Microns Of Mercury to Pounds Per Square Inch = 080 Microns Of Mercury to Pounds Per Square Inch = 0.0015\n3 Microns Of Mercury to Pounds Per Square Inch = 0.000190 Microns Of Mercury to Pounds Per Square Inch = 0.0017\n4 Microns Of Mercury to Pounds Per Square Inch = 0.0001100 Microns Of Mercury to Pounds Per Square Inch = 0.0019\n5 Microns Of Mercury to Pounds Per Square Inch = 0.0001200 Microns Of Mercury to Pounds Per Square Inch = 0.0039\n6 Microns Of Mercury to Pounds Per Square Inch = 0.0001300 Microns Of Mercury to Pounds Per Square Inch = 0.0058\n7 Microns Of Mercury to Pounds Per Square Inch = 0.0001400 Microns Of Mercury to Pounds Per Square Inch = 0.0077\n8 Microns Of Mercury to Pounds Per Square Inch = 0.0002500 Microns Of Mercury to Pounds Per Square Inch = 0.0097\n9 Microns Of Mercury to Pounds Per Square Inch = 0.0002600 Microns Of Mercury to Pounds Per Square Inch = 0.0116\n10 Microns Of Mercury to Pounds Per Square Inch = 0.0002800 Microns Of Mercury to Pounds Per Square Inch = 0.0155\n20 Microns Of Mercury to Pounds Per Square Inch = 0.0004900 Microns Of Mercury to Pounds Per Square Inch = 0.0174\n30 Microns Of Mercury to Pounds Per Square Inch = 0.00061,000 Microns Of Mercury to Pounds Per Square Inch = 0.0193\n40 Microns Of Mercury to Pounds Per Square Inch = 0.000810,000 Microns Of Mercury to Pounds Per Square Inch = 0.1934\n50 Microns Of Mercury to Pounds Per Square Inch = 0.001100,000 Microns Of Mercury to Pounds Per Square Inch = 1.9337\n60 Microns Of Mercury to Pounds Per Square Inch = 0.00121,000,000 Microns Of Mercury to Pounds Per Square Inch = 19.3368"
]
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https://www.iotforall.com/supervised-machine-learning-crash-course-part-1 | [
"### Machine Learning Crash Course, Part I: Supervised Machine Learning\n\nWhat really is Machine Learning? Here are some concrete examples.",
null,
"## Artificial (Un)intelligence\n\nWhen you type ‘machine learning’ into Google News, the first link you see is a Forbes Magazine piece called “What’s The Difference Between Machine Learning And Artificial Intelligence?” This article contained so many flowery, grandiose descriptions about ML and AI technology that I couldn’t help but laugh. A few notable quotes include:\n\n1. “To get something out of machine learning, you need to know how to code or know someone who does. With artificial intelligence, you get something that takes an idea or a desire and runs with it, curiously seeking out new input and understandings.”\n2. “AI models don’t need to be rebuilt: they rebuild themselves. They actively seek out new and better sources of data.”\n\n1. AI algorithms (e.g. self-driving cars, video game playing robots) definitely do require coding. The ‘running with desire’ bit sounds like artificial general intelligence…which definitely doesn’t exist.\n2. If only this were true, I would never have to work again! One could maybe argue that reinforcement learning algorithms “rebuild” themselves, but a more accurate description would be that they’re “recalibrating”.\n3. … if you say so?\n\nWith all the nonsense the media uses to describe machine learning (ML) and artificial intelligence (AI), it’s time we do a deep dive into what these technologies actually do. First, we need to learn the difference between AI and ML. Fortunately, a fellow writer has already written an excellent explanation here. With that out of the way, we can focus on the ML side of things.\n\nBy definition, machine learning is the ability for computers to perform tasks without having to be explicitly programed. When writing a “normal” computer program, the coder will manually write out what the program will do, for every possible scenario. An ML program has a different style to it. Typically, the program will combine historical data, complex mathematical models and sophisticated algorithms to deduce the “optimal” behavior.\n\nFor instance, if you were writing a program to play checkers, a regular program might say something like “if I can jump over an opponent’s piece, then I will do so”. Instead, a machine learning program might say something like, “examine the last 1000 games of checkers I’ve played and pick the move that maximizes the probability that I will win the game”.\n\nMost machine learning algorithms fall into one of three categories: supervised learning, unsupervised learning, and reinforcement learning. In this article, we’ll cover just the first of the three.\n\n## Supervised Machine Learning\n\nSupervised learning algorithms try to find a formula that accurately predicts the output label from input variables. Let’s clarify what this means with some simple, concrete examples.\n\n### 1. Advertising Example (Linear Regression)\n\nFirst, we feed the historical data into our linear regression model. This produces a mathematical formula that predicts sales based on our input variable, TV ad spending:\n\nSales = 7.03 + 0.047(TV)\n\nIn the above graph, we have plotted both the historical data points (the black dots) as well as the formula our ML algorithm produces (the red line). The equation roughly follows the trajectory of the data points. To answer our original question of expected revenue, we can simply plug \\$100 in for the variable TV to get,\n\n\\$11.73 = 7.03 + 0.047(\\$100)\n\nIn other words, after spending 100 dollars on TV advertising, we can expect to generate only \\$11.73 in sales, based on past data. Therefore, it would probably be best to explore a different form of advertising. In summary, we used machine learning (specifically, linear regression) to predict how much revenue a TV advertising campaign would generate, based on historical data.\n\n### 2. Credit Card Example (Logistic Regression)\n\nIn the previous example, we mapped a numeric input (TV ad spending) to a numeric output (sales). However, it is also acceptable for the inputs/output to be categorical. When the output variable is categorical, then it is called a classification algorithm.\n\nFor example, a credit card company might want to predict whether a customer will default in the upcoming 3 months based on their current balance. Here, the output variable is categorical: it is either “yes” (the customer defaulted) or “no” (the customer did not default). As with the previous example, we need access to a dataset with labels that tell us whether or not the customer defaulted. We can then apply a classification algorithm like logistic regression.\n\nThe algorithm produces a more complex equation (red line) than linear regression. Previously, we were trying to predict an actual number (sales) so the output of our formula was a number (\\$11.73). However, we are no longer predicting a number. Instead we are predicting a category (“Yes, they will default within 3 months” or “No, they won’t default”).\n\nThe red line in the graph above represents the probability that someone will default based on their current balance. When a customer’s balance is less than 1000, the probability of default (red line) is near 0 (e.g. very unlikely to default). As a customer’s balance increases, so does the chances of default. We plotted the historical data in the graph as well, with “Yes” and “No” dummy coded as 1 and 0, respectively.\n\nThis equation can help you make predictions about new customers, where you aren’t told whether or not they will default in advance. From the logistic regression equation, you could check their balance, see that it’s only \\$400, and safely conclude they probably won’t default in three months. On the other hand, if their balance was \\$2500, you would know that they’re extremely likely to.\n\n## Conclusion\n\nSo far we’ve covered supervised machine learning, where we make predictions from labeled historical data. Stay tuned for the next part in the series where we cover unsupervised and reinforcement learning.",
null,
"##### Narin Luangrath\nNarin Luangrath is a Product Engineer at Leverege who follows machine learning and artificial intelligence developments. He recently graduated from Wesleyan University with degrees in Math, Computer Science and Data Analysis.\nNarin Luangrath is a Product Engineer at Leverege who follows machine learning and artificial intelligence developments. He recently graduated from Wesleyan University with degrees in Math, Computer Science and Data Analysis."
]
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http://forums.wolfram.com/mathgroup/archive/2007/Jan/msg00582.html | [
"",
null,
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"Re: Plotting\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg73024] Re: Plotting\n• From: \"dimitris\" <dimmechan at yahoo.com>\n• Date: Mon, 29 Jan 2007 04:51:40 -0500 (EST)\n• References: <epfa83\\$doc\\[email protected]>\n\n```First have a look at the following links:\n\nhttp://library.wolfram.com/infocenter/Conferences/180/\nhttp://library.wolfram.com/infocenter/Conferences/181/\nhttp://library.wolfram.com/infocenter/Conferences/179/\n\nThen\n\nf[x_] := x^3 - 16*x^2 + 3*x + 1\n\nPlot[f[x], {x, -3, 20}, Axes -> {True, False}, AxesStyle ->\nGrayLevel[0.7], Frame -> {True, True, False, False},\nFrameLabel -> TraditionalForm /@ {x, HoldForm[f[x]]}, FrameTicks ->\n{Range[-5, 20, 5], Range[-600, 800, 400]},\nEpilog -> {Blue, AbsolutePointSize, (Point[{#1, 0}] & ) /@ (x /.\nNSolve[f[x] == 0])}, ImageSize -> 400];\n\nOn Jan 27, 12:41 pm, \"Rayne\" <ghcgw... at singnet.com.sg> wrote:\n> Hi all,\n> I want to plot the cubic equation f(x) == x^3 - 16x^2 + 3x + 1 == 0. I had:\n>\n> f==x^3-16 x^2+3 x+1 ==== 0\n> Plot[f,{x,-1,20}]\n>\n> but I was unable to get the plot. The error message says\n>\n> Plot::plnr: f is not a machine-size real number at x == -0.999999.\n> Plot::plnr: f is not a machine-size real number at x == -0.148093.\n> Plot::plnr: f is not a machine-size real number at x == 0.780985.\n> General::stop: Further output of Plot::plnr\n> will be suppressed during this calculation.\n>\n> How do I get around this problem?\n>\n> Thank you.\n>\n> Regards,\n> Rayne\n\n```\n\n• Prev by Date: Re: Problem with base 2 logs and Floor\n• Next by Date: Re: Can Mathematica do Separation of Variables?\n• Previous by thread: Re: Plotting\n• Next by thread: how to configure a remote kernel"
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http://misc.duruofei.com/misc/collections-of-matlab-toolbox-2015/ | [
"# Collections of MATLAB Toolbox 2015\n\nposted in: Misc | 0\n• AIRES – automatic integration of reusable embedded software\n• Air-Sea – air-sea flux estimates in oceanography\n• Animation – developing scientific animations\n• ARfit – estimation of parameters and eigenmodes of multivariate autoregressive methods\n• ARMASA – power spectrum estimation\n• AR-Toolkit – computer vision tracking\n• Auditory – auditory models\n• b4m – interval arithmetic\n• Bayes Net – inference and learning for directed graphical models\n• Bode Step – design of control systems with maximized feedback\n• Bootstrap – for resampling, hypothesis testing and confidence interval estimation\n• BrainStorm – MEG and EEG data visualization and processing\n• BSTEX – equation viewer\n• CALFEM – interactive program for teaching the finite element method\n• Calibr – for calibrating CCD cameras\n• Camera Calibration\n• Captain – non-stationary time series analysis and forecasting\n• CHMMBOX – for coupled hidden Markov modeling using maximum likelihood EM\n• CLOSID\n• Cluster – for analysis of Gaussian mixture models for data set clustering\n• Clustering – cluster analysis\n• COLEA – speech analysis\n• CompEcon – solving problems in economics and finance\n• Complex – for estimating temporal and spatial signal complexities\n• Computational Statistics\n• Coral – seismic waveform analysis\n• DACE – kriging approximations to computer models\n• DAIHM – data assimilation in hydrological and hydrodynamic models\n• Data Visualization\n• DBT – radar array processing\n• DDE-BIFTOOL – bifurcation analysis of delay differential equations\n• Denoise – for removing noise from signals\n• DiffMan – solving differential equations on manifolds\n• DIPimage – scientific image processing\n• Direct – Laplace transform inversion via the direct integration method\n• DMsuite – differentiation matrix suite\n• DMTTEQ – design and test time domain equalizer design methods\n• DrawFilt – drawing digital and analog filters\n• DWT – discrete wavelet transforms\n• EasyKrig\n• Econometrics\n• EEGLAB\n• EigTool – graphical tool for nonsymmetric eigenproblems\n• EMSC – separating light scattering and absorbance by extended multiplicative signal correction\n• Engineering Vibration\n• FastICA – fixed-point algorithm for ICA and projection pursuit\n• FDC – flight dynamics and control\n• FDtools – fractional delay filter design\n• FlexICA – for independent components analysis\n• FMBPC – fuzzy model-based predictive control\n• ForWaRD – Fourier-wavelet regularized deconvolution\n• FracLab – fractal analysis for signal processing\n• FSBOX – stepwise forward and backward selection of features using linear regression\n• GABLE – geometric algebra tutorial\n• Garch – estimating and diagnosing heteroskedasticity in time series models\n• GCE Data – managing, analyzing and displaying data and metadata stored using the GCE data structure specification\n• GCSV – growing cell structure visualization\n• GEMANOVA – fitting multilinear ANOVA models\n• Genetic Algorithm\n• Geodetic – geodetic calculations\n• glmlab – general linear models\n• GTM – generative topographic mapping, a model for density modeling and data visualization\n• GVF – gradient vector flow for finding 3-D object boundaries\n• Hilbert – Hilbert transform by the rational eigenfunction expansion method\n• HMM – hidden Markov models\n• HMMBOX – for hidden Markov modeling using maximum likelihood EM\n• HUTear – auditory modeling\n• ICALAB – signal and image processing using ICA and higher order statistics\n• Imputation – analysis of incomplete datasets\n• IPEM – perception based musical analysis\n• JMatLink – Matlab Java classes\n• Kalman – Bayesian Kalman filter\n• Kalman Filter – filtering, smoothing and parameter estimation (using EM) for linear dynamical systems\n• KALMTOOL – state estimation of nonlinear systems\n• Kautz – Kautz filter design\n• Kriging\n• LDestimate – estimation of scaling exponents\n• LDPC – low density parity check codes\n• LISQ – wavelet lifting scheme on quincunx grids\n• LKER – Laguerre kernel estimation tool\n• LMAM-OLMAM – Levenberg Marquardt with Adaptive Momentum algorithm for training feedforward neural networks\n• Low-Field NMR – for exponential fitting, phase correction of quadrature data and slicing\n• LPSVM – Newton method for LP support vector machine for machine learning problems\n• LSDPTOOL – robust control system design using the loop shaping design procedure\n• LS-SVMlab\n• LSVM – Lagrangian support vector machine for machine learning problems\n• Lyngby – functional neuroimaging\n• MARBOX – for multivariate autogressive modeling and cross-spectral estimation\n• MatArray – analysis of microarray data\n• Matrix Computation – constructing test matrices, computing matrix factorizations, visualizing matrices, and direct search optimization\n• MCAT – Monte Carlo analysis\n• MDP – Markov decision processes\n• MESHPART – graph and mesh partioning methods\n• MILES – maximum likelihood fitting using ordinary least squares algorithms\n• MIMO – multidimensional code synthesis\n• Missing – functions for handling missing data values\n• M_Map – geographic mapping tools\n• MODCONS – multi-objective control system design\n• MOEA – multi-objective evolutionary algorithms\n• MS – estimation of multiscaling exponents\n• Multiblock – analysis and regression on several data blocks simultaneously\n• Multiscale Shape Analysis\n• Music Analysis – feature extraction from raw audio signals for content-based music retri\n• MWM – multifractal wavelet model\n• NetCDF\n• Netlab – neural network algorithms\n• NiDAQ – data acquisition using the NiDAQ library\n• NEDM – nonlinear economic dynamic models\n• NMM – numerical methods in Matlab text\n• NNCTRL – design and simulation of control systems based on neural networks\n• NNSYSID – neural net based identification of nonlinear dynamic systems\n• NSVM – newton support vector machine for solving machine learning problems\n• NURBS – non-uniform rational B-splines\n• N-way – analysis of multiway data with multilinear models\n• OpenFEM – finite element development\n• PCNN – pulse coupled neural networks\n• Peruna – signal processing and analysis\n• PhiVis – probabilistic hierarchical interactive visualization, i.e. functions for visual analysis of multivariate continuous data\n• Planar Manipulator – simulation of n-DOF planar manipulators\n• PRTools – pattern recognition\n• psignifit – testing hyptheses about psychometric functions\n• PSVM – proximal support vector machine for solving machine learning problems\n• Psychophysics – vision research\n• PyrTools – multi-scale image processing\n• RBF – radial basis function neural networks\n• RBN – simulation of synchronous and asynchronous random boolean networks\n• ReBEL – sigma-point Kalman filters\n• Regression – basic multivariate data analysis and regression\n• Regularization Tools\n• Regularization Tools XP\n• Restore Tools\n• Robot – robotics functions, e.g. kinematics, dynamics and trajectory generation\n• Robust Calibration – robust calibration in stats\n• RRMT – rainfall-runoff modelling\n• SAM – structure and motion\n• Schwarz-Christoffel – computation of conformal maps to polygonally bounded regions\n• SDH – smoothed data histogram\n• SeaGrid – orthogonal grid maker\n• SEA-MAT – oceanographic analysis\n• SLS – sparse least squares\n• SolvOpt – solver for local optimization problems\n• SOM – self-organizing map\n• SOSTOOLS – solving sums of squares (SOS) optimization problems\n• Spatial and Geometric Analysis\n• Spatial Regression\n• Spatial Statistics\n• Spectral Methods\n• SPM – statistical parametric mapping\n• SSVM – smooth support vector machine for solving machine learning problems\n• STATBAG – for linear regression, feature selection, generation of data, and significance testing\n• StatBox – statistical routines\n• Statistical Pattern Recognition – pattern recognition methods\n• Stixbox – statistics\n• SVM – implements support vector machines\n• SVM Classifier\n• Symbolic Robot Dynamics\n• TEMPLAR – wavelet-based template learning and pattern classification\n• TextClust – model-based document clustering\n• TextureSynth – analyzing and synthesizing visual textures\n• TfMin – continous 3-D minimum time orbit transfer around Earth\n• Time-Frequency – analyzing non-stationary signals using time-frequency distributions\n• Tree-Ring – tasks in tree-ring analysis\n• TSA – uni- and multivariate, stationary and non-stationary time series analysis\n• TSTOOL – nonlinear time series analysis\n• T_Tide – harmonic analysis of tides\n• UTVtools – computing and modifying rank-revealing URV and UTV decompositions\n• Uvi_Wave – wavelet analysis\n• varimax – orthogonal rotation of EOFs\n• VBHMM – variation Bayesian hidden Markov models\n• VBMFA – variational Bayesian mixtures of factor analyzers\n• VMT – VRML Molecule Toolbox, for animating results from molecular dynamics experiments\n• VOICEBOX\n• VRMLplot – generates interactive VRML 2.0 graphs and animations\n• VSVtools – computing and modifying symmetric rank-revealing decompositions\n• WAFO – wave analysis for fatique and oceanography\n• WarpTB – frequency-warped signal processing\n• WAVEKIT – wavelet analysis\n• WaveLab – wavelet analysis\n• Weeks – Laplace transform inversion via the Weeks method\n• WetCDF – NetCDF interface\n• WHMT – wavelet-domain hidden Markov tree models\n• WInHD – Wavelet-based inverse halftoning via deconvolution\n• WSCT – weighted sequences clustering toolkit\n• XMLTree – XML parser\n• YAADA – analyze single particle mass spectrum data\n• ZMAP – quantitative seismicity analysis\n\nZSM (zero sum multinomial)\nhttp://mcgillb.user.msu.edu/zsmcode.html\n\nBinaural-modeling software for MATLAB/Windows\n\nStatistical Parametric Mapping (SPM)\nhttp://www.fil.ion.ucl.ac.uk/spm/ext/\n\nBOOTSTRAP MATLAB TOOLBOX\n\nThe DSS package for MATLAB\nDSS Matlab package contains algorithms for performing linear, deflation and symmetric DSS.\nhttp://www.cis.hut.fi/projects/dss/package/\n\nPsychtoolbox\n\nMultisurface Method Tree with MATLAB\nhttp://www.cs.wisc.edu/~olvi/uwmp/msmt.html\n\nA Matlab Toolbox for every single topic !\nhttp://stommel.tamu.edu/~baum/toolboxes.html\neg. BrainStorm – MEG and EEG data visualization and processing\n\nCLAWPACK is a software package designed to compute numerical solutions to hyperbolic partial differential equations using a wave propagation approach\nhttp://www.amath.washington.edu/~claw/\n\nDIPimage – Image Processing Toolbox\n\nPRTools – Pattern Recognition Toolbox (+ Neural Networks)\n\nNetLab – Neural Network Toolbox\n\nFSTB – Fuzzy Systems Toolbox\n\nFusetool – Image Fusion Toolbox\nhttp://www.metapix.de/toolbox.htm\n\nWAVEKIT – Wavelet Toolbox\n\nGat – Genetic Algorithm Toolbox\n\nTSTOOL is a MATLAB software package for nonlinear time series analysis.\nTSTOOL can be used for computing: Time-delay reconstruction, Lyapunov exponents, Fractal dimensions, Mutual information, Surrogate data tests, Nearest neighbor statistics, Return times, Poincare sections, Nonlinear prediction\nhttp://www.physik3.gwdg.de/tstool/\n\nMATLAB / Data description toolbox\nA Matlab toolbox for data description, outlier and novelty detection\nMarch 26, 2004 – D.M.J. Tax\nhttp://www-ict.ewi.tudelft.nl/~davidt/dd_tools/dd_manual.html\n\nMBE\nhttp://www.pmarneffei.hku.hk/mbetoolbox/\n\nBetabolic network toolbox for Matlab\nhttp://www.molgen.mpg.de/~lieberme/pages/network_matlab.html\n\nPharmacokinetics toolbox for Matlab\nhttp://page.inf.fu-berlin.de/~lieber/seiten/pbpk_toolbox.html\n\nThe Spider\nThe spider is intended to be a complete object orientated environment for machine learning in Matlab. Aside from easy use of base learning algorithms, algorithms can be plugged together and can be compared with, e.g model selection, statistical tests and visual plots. This gives all the power of objects (reusability, plug together, share code) but also all the power of Matlab for machine learning research.\nhttp://www.kyb.tuebingen.mpg.de/bs/people/spider/index.html\n\nSchwarz-Christoffel Toolbox\n\nXML Toolbox\n\nFIR/TDNN Toolbox for MATLAB\nBeta version of a toolbox for FIR (Finite Impulse Response) and TD (Time Delay) Neural Networks.\nhttp://www.cs.utep.edu/interval-comp/dagstuhl.03/oish.pdf\n\nMisc.\n\nhttp://www.dcsc.tudelft.nl/Research/Software/index.html\n\nAstronomy\n\nSaturn and Titan trajectories … MALTAB astronomy\nhttp://sprg.ssl.berkeley.edu/~abrecht/Matlab-codes/\n\nAudio\n\nMA Toolbox for Matlab Implementing Similarity Measures for Audio\nhttp://www.oefai.at/~elias/ma/index.html\n\nMusic Analysis – Toolbox for Matlab : Feature Extraction from Raw Audio Signals for Content-Based Music Retri\nhttp://www.ai.univie.ac.at/~elias/ma/\n\nWarpTB – Matlab Toolbox for Warped DSP\nBy Aki Härmä and Matti Karjalainen\nhttp://www.acoustics.hut.fi/software/warp/\n\nMATLAB-related Software\nhttp://www.dpmi.tu-graz.ac.at/~schloegl/matlab/\n\nBiomedical Signal data formats (EEG machine specific file formats with Matlab import routines)\nhttp://www.dpmi.tu-graz.ac.at/~schloegl/matlab/eeg/\n\nMPEG Encoding library for MATLAB Movies (Created by David Foti)\nIt enables MATLAB users to read (MPGREAD) or write (MPGWRITE) MPEG movies. That should help Video Quality project.\n\nFilter Design package\nhttp://www.ee.ryerson.ca:8080/~mzeytin/dfp/index.html\n\nOctave by Christophe COUVREUR (Generates normalized A-weigthing, C-weighting, octave and one-third-octave digital filters)\n\nSource Coding MATLAB Toolbox\nhttp://www.ece.umn.edu/users/kieffer/programs.html\n\nBio Medical Informatics (Top)\n\nCGH-Plotter: MATLAB Toolbox for CGH-data Analysis\nCode: http://sigwww.cs.tut.fi/TICSP/CGH-Plotter/\nPoster: http://sigwww.cs.tut.fi/TICSP/CSB2003/Posteri_CGH_Plotter.pdf\n\nThe Brain Imaging Software Toolbox\nhttp://www.bic.mni.mcgill.ca/software/\n\nMRI Brain Segmentation\n\nChemometrics (providing PCA) (Top)\n\nMatlab Molecular Biology & Evolution Toolbox\n(Toolbox Enables Evolutionary Biologists to Analyze and View DNA and Protein Sequences)\nJames J. Cai\nhttp://www.pmarneffei.hku.hk/mbetoolbox/\n\nToolbox provided by Prof. Massart research group\nhttp://minf.vub.ac.be/~fabi/publiek/\n\nUseful collection of routines from Prof age smilde research group\nhttp://www-its.chem.uva.nl/research/pac\n\nMultivariate Toolbox written by Rune Mathisen\nhttp://www.bitjungle.com/~mvartools/index.html\n\nMatlab code and datasets\nhttp://www.acc.umu.se/~tnkjtg/chemometrics/dataset.html\n\nChaos (Top)\n\nChaotic Systems Toolbox\n\nHOSA Toolbox\n\nChemistry (Top)\n\nMetMAP – (Metabolical Modeling, Analysis and oPtimization alias Met. M. A. P.)\nhttp://webpages.ull.es/users/sympbst/pag_ing/pag_metmap/index.htm\n\nDoseLab – A set of software programs for quantitative comparison of measured and computed radiation dose distributions\nhttp://doselab.sourceforge.net/\n\nGenBank Overview\nhttp://www.ncbi.nlm.nih.gov/Genbank/GenbankOverview.html\n\nCoding\n\nCode for the estimation of Scaling Exponents\nhttp://www.cubinlab.ee.mu.oz.au/~darryl/secondorder_code.html\n\nControl (Top)\n\nControl Tutorial for Matlab\nhttp://www.engin.umich.edu/group/ctm/\n\nAnother\n\nCommunications (Top)\n\nChannel Learning Architecture toolbox\n(This Matlab toolbox is a supplement to the article “HiperLearn: A High Performance Learning Architecture”)\nhttp://www.isy.liu.se/cvl/Projects/hiperlearn/\n\nSource Coding MATLAB Toolbox\nhttp://www.ece.umn.edu/users/kieffer/programs.html\n\nTCP/UDP/IP Toolbox 2.0.4"
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https://www.karada-good.net/en/ranalytics/r-e11/ | [
"# Analysis in R: Iteration and conditional decision commands\n\nThe R operation combines commands to process data and output results. In many cases there is rarely a single element in the data, but rather several elements, and each element is processed repeatedly and the results are output. Therefore, repetition and conditional evaluation commands are introduced here.\n\nスポンサーリンク\n\n## Repeat the command: for ( element in the number of processes ) {process}\n\nThe “for” command repeats the number of times the element (often set to a single letter, such as i, m, etc. in alphabetical order) inside the parentheses will be processed. For example, here is a code that outputs a sequence of numbers from 1 to 7, adding 3 to each.\n\n```data <- c(1, 2, 3, 4, 5, 6, 7)\n\nfor (i in seq(data)) {\ncat(i + 3, \"\\n\")\n}\n\n#Result\n4\n5\n6\n7\n8\n9\n10\n```\n\nNote that the “in” is followed by “seq(data)” to get the repeated range. “1:7” is also acceptable, but specifying the range by the length of the data to be processed rather than by a number will reduce errors.\n\n## Command for conditional judgment: if (condition) {process 1} else {process 2}\n\nThe “if else” command executes process 1 if the condition is true and process 2 if it is not. For example, here is code that outputs a sequence of numbers from 1 to 7, adding 3 only to even numbers.\n\n```data <- c(1, 2, 3, 4, 5, 6, 7)\n\nfor(i in seq(data)){\n\nif (i%%2 == 0){\n\ncat(i + 3, \"\\n\")\n\n}else{\n\n#If odd, do not process. Blank\n\n}\n\n}\n\n#Result\n5\n7\n9\n```\n\nIt can also be processed simply as ifelse(condition, process 1, process 2). However, since NA is included in the result, some processing ingenuity is required.\n\n```ifelse(data%%2 == 0, data + 3, NA)\n\n#Result\n NA 5 NA 7 NA 9 NA\n```\n\nI hope this makes your analysis a little easier !!\n\nスポンサーリンク",
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https://hcis-journal.springeropen.com/articles/10.1186/s13673-017-0110-9 | [
"# A fusion approach based on infrared finger vein transmitting model by using multi-light-intensity imaging\n\n## Abstract\n\nAn infrared transmitting model from the observed finger vein images is proposed in this paper by using multi-light-intensity imaging. This model is estimated from many pixels’ values under different intensity light in the same scene. Due to the fusion method could be applied in biometric system, the vein images of finger captured in the system, we proposed in this paper, will be normalized and preserved the intact of the vein patterns of the biometric data from tested human’s body. From observed pixels under multi-light-intensity, the curve of the transmitting model is recovered by sliding both of the sampled curve segments and using curve-fitting. The fusion method with each pixel level weighting based on the proposed transmitting model curve is adopted by the smooth spatial and estimation of the block quality. Finally, the results shown that our approach is a convenient and practicable approach for the infrared image fusion and subsequent processing for biometric applications.\n\n## Introduction\n\nThe finger vein authentication is highly accurate and convenient by using the individual’s unique biological characteristics. Vascular patterns are unique to each individual—even identical twins have different patterns. Finger vein authentication works based on the vein patterns in the superficial subcutaneous finger skin that are unique [1,2,3]. Three main advantages of vein authentication are following: (1) Because the finger veins are hidden inside human’s body, some little risks of forgery or theft appear in daily activities. The conditions on surface of the skin in finger, e.g. dry or wet, will have no effect on its authentication. (2) It is non-invasive and contactless in the finger vein imaging, which is convenient and cleanliness for the users. (3) The stability and complexity of finger vein patterns will be better than other biometric features on human’s body, which have the higher security level for personal identification .\n\nThe physiological information extracted from human body including the features of individual face, palm-print or fingerprint, hand-shape, skin, temperature and arterial pulse, etc. is used to recognize personal identification and diagnose some diseases. The information mentioned above, plus subcutaneous superficial vein pattern, could be extracted and digitized as biometric data. It could be further represented as a typical pattern in order to identify individual identification [5,6,7,8,9]. It is convenience to use the identified biometric to be the access right. The relative applications always focus in the remote access control in the websites, e.g. the website of finance or bank, etc. However, the image data of biometric is sensitive to the physiological conditions and environments. For example, the captured feature in human’s face, where the factors of its illumination distribution and direction should be modified or normalized before storing it. It may exists lots of shadow images or noises in this captured image. Finally, Its features will then be strongly influenced by the shadow images or noises . On the other hand, the non-uniformed illumination will increase the interference and redundant information, or submerge some patterns. It will lead to the deformation of dimensionality. It is very important to normalize the captured biometric information before keeping them to the storage of biometric system [11, 12]. The similar problems mentioned above are also appeared in the finger vein image capturing processes . The width of vein in the captured image will be changed under different intensity near-infrared light. Because thickness of each finger is different, the under/over-exposure may appears in the thick/thin area of the finger by using one fixed-intensity-light. It will be inundated by this vein pattern. The vein pattern integrity is very important for the biometric system. Thus, it is necessary to normalize the illumination in the vein image capture before storing them in the biometric information storages or databases.\n\nThe main work of finger vein authentication is to collect the data: finger vein images. The quality of the image will affect directly the accuracy and its recognition speed. This paper presents the details in analyses of infrared finger vein images. In addition, the transmitting model is built from the observed data, e.g. multi-light-intensity vein images. Finally, the pixel level fusion method based on the transmitting model as well as spatial smoothing is proposed in this paper.\n\nThe remainders of this paper are organized as follows: in “The infrared light transmission model of the finger” section, we introduce the infrared light transmission model of the finger. In “Multi-light-intensity finger vein images’ fusion based on the transmitting model” section, we first formalize a multi-light-intensity finger vein images’ fusion based on the transmitting model. Next, we present examinations and discussions in “Examinations and discussions” section. Finally, we draw our conclusions and further works in “Conclusions and further works” section.\n\n## The infrared light transmission model of the finger\n\nThis model is extended and modified from Ref. . The steps of basic works from bioinformation to the biometric data in this model are described in “Basic works from bioinformation to the biometric data” section, and its single infrared transmitting model is described in “A single infrared transmitting model” section.\n\n### Basic works from bioinformation to the biometric data\n\nThe applications of the biometric data includes personal identification and disease diagnosis. The system architecture is illustrated from the bioinformation to the biometric data for a single infrared transmitting model in biometric system shown in Fig. 1. Obviously, the capturing, digitizing and normalizing methods of bioinformation should be efficient in order to record the complete pattern or texture feature information, uniform gray distribution and contrast before their applications. This paper presents a single transmitting model of finger vein imaging in Biometric system and use it for fuse the multi-light-intensity finger vein images to one image, which integrates the vein pattern information of each source image and keeps the complete vein pattern information.\n\n### A single infrared transmitting model\n\nThe single infrared transmitting model described in this subsection. It is popular to use near-infrared (NIR) light transmitting the finger to achieve the angiogram imaging. Because the oxyhemoglobin content (HbO) in the venous blood is far beyond the arterial blood and other tissue, such as fat and muscle, the wavelength of the transmitting light absorption should be relatively high. Thus, the 760–1100 nm is suitable for the angiogram imaging from the absorption rate of the water, oxyhemoglobin (HbO) and deoxyhemoglobin (Hb), which is shown in Fig. 2. This higher absorption property of HbO results in that the region of vein pattern is darker than other surrounding region after the NIR light transmitting the finger. This technology is widely used in the vascular vein imaging of breast and cerebral.\n\nThe tissue optical properties have been modeled based upon photon diffusion theory. The epidermis (the outermost layer of skin) only accounts for 6% of scattering and can be regarded a primary absorptive medium. Therefore, a simplified model on the reflectance of blood and tissue considers the reflectance from only the scattering tissue beneath the epidermis . The skin is assumed to be a semi-infinite homogeneous medium, under a uniform and diffusive illumination. The photon has a relatively long residence time which allows the photon to engage in a random walk within the medium. The photon diffusion depends on the absorption and scattering properties of the skin, which penetration depth for different wavelengths shown in Fig. 3.\n\nConsider all these factors: the tissue (water, Hb and HbO) absorption in the vein, the depth of penetration. The infrared wave band of finger vein imaging is about 850 nm in practice.\n\nBecause the thickness of the finger is a nonlinear variable, it is hard to only use invariable light intensity to vein imaging at infrared 850 nm. Thus, overexposure and underexposure often appear in the infrared finger vein images. And these areas with over/under exposure can’t be enhanced, which cause the vein pattern lack in the biometric data extraction. An infrared multi-light-intensity finger vein imaging technology is used in the paper to solve the problem, which extends the dynamic range of the infrared vein imaging . Additionally, it is necessary to fuse the complementary vein information in the next process. This paper presents a calculation method of the infrared finger vein transmitting model based on the multi-light-intensity imaging. The model presents the monotone increasing nonlinear function relationship between the light-intensity and pixel-gray value, which can be built by the genetic algorithm and used in the imaging quality estimation of the pixel-level fusion to infrared multi-light-intensity finger vein images.\n\nThe infrared finger vein transmitting model is defined as:\n\n$$B = f(X)$$\n(1)\n\n$$X$$ is the irradiance of the infrared light of transmitting the finger, and $$B$$ is represented as a pixel gray value. Generally, the gray level of the pixel is 8 bits. The infrared finger vein transmitting model function is explicitly written as\n\n$$\\left\\{ { \\, \\begin{array}{l} {B_{\\text{min} } = 0, } \\\\ {B = f(X),} \\\\{B_{\\text{max} } = 255,} \\\\ \\end{array} \\begin{array}{l} {if} \\\\ {if} \\\\ {if} \\\\ \\end{array} \\begin{array}{l} {X \\le X_{\\text{min} } } \\\\ {X_{\\text{min} } < X < X_{\\text{max} } } \\\\ {X_{\\text{max} } \\le X } \\\\ \\end{array} } \\right.$$\n(2)\n\nAssume there are $$N$$ vein images captured under increasing light intensity $$X_{p} ,p = 1, \\ldots ,N$$. The size of each image is $$m \\times n$$, sign $$K = m \\times n$$. The qth pixel of the pth light-intensity image will be denoted $$B_{pq}$$, the set $$\\left\\{ {B_{pq} } \\right\\},p = 1, \\ldots ,N{\\text{ and }}q \\in \\{ 1, \\ldots ,K\\}$$, represents the known observations. The goal is to determine the underlying light values or irradiances, denoted by $$X_{q}$$, that gave rise to the observations $$B_{pq}$$. Because the $$N$$ vein images has be properly registered in the pixel level, so that for a particular a, the light value $$X_{a}$$ contributes to $$B_{pq} ,p = 1, \\ldots ,N{\\text{ and }}q \\in \\{ 1, \\ldots ,K\\}$$. For this work, a normalized cross-correlation function is used as the matching criterion to register images to 1/2-pixel resolution .\n\nThe model can be rewritten as:\n\n$$B_{pq} = f_{q} (X_{pq} ),\\begin{array}{*{20}c} & {p = 1, \\ldots ,N, \\begin{array}{*{20}c} & {q \\in \\{ 1, \\ldots ,K\\}. } \\\\ \\end{array} } \\\\ \\end{array}$$\n(3)\n\nIt means the transmitting model of position q is different. Nevertheless the shape of each model is similar, it gives an easy solution to estimate the transmitting model for each pixel for the application.\n\nSince f is a monotonic and invertible function, its inverse function could be represented as $$g$$.\n\n$$X_{pq} = g_{q} (B_{pq} ), \\begin{array}{*{20}c} & {p = 1, \\ldots ,N, \\begin{array}{*{20}c} & {q \\in \\{ 1, \\ldots ,K\\}. } \\\\ \\end{array} } \\\\ \\end{array}$$\n(4)\n\nIt is necessary to recover the function $$g$$ and the irradiances of $$X_{p} ,p = 1, \\ldots ,N$$, which satisfy the set of equations arising from Eq. (4) in a least-squared error sense. Recovering function $$g$$ only requires recovering the finite number of values that $$g\\left( B \\right)$$ could take since the domain of $$X$$, pixel brightness values, is finite. Letting $$B_\\text{{min}}$$ and $$B_\\text{{max}}$$ be the least and greatest pixel values (integers), $$q$$ be the number of pixel locations and $$N$$ be the number of photographs, we formulate the problem as one of finding the $$\\begin{array}{*{20}c} {[B_{\\text{min} } } & {B_{\\text{max} } } \\\\ \\end{array} ]$$ values of $$g\\left( B \\right)$$ and the $$q$$ values of $$X$$ that minimize the following quadratic objective function :\n\n$$\\xi = \\sum\\limits_{i = 1}^{N} {\\sum\\limits_{j = 1}^{q} {[g(B_{ij} ) - X_{i} ]^{2} } } + \\lambda \\sum\\limits_{{b = B_{\\text{min} } + 1}}^{{b = B_{\\text{max} } - 1}} {(g''(b))^{2} }$$\n(5)\n\nThe first term ensures that the solution satisfies the set of equations arising from Eq. (4) in a least squares sense. The second term is a smoothness term on the sum of squared values of the second derivative of $$g$$ to ensure that the function $$g$$ is smooth; in this discrete setting, the second part can be calculated by the formula (6).\n\n$$g'' = g(b + 1) + g(b - 1) - 2g(b)$$\n(6)\n\nThis smoothness term is essential to the formulation in that it provides coupling between the values $$g\\left( z \\right)$$ in the minimization. The scalar weights the smoothness term relative to the data fitting term, and should be chosen appropriately for the amount of noise expected in the $$B_{ij}$$ measurements.\n\nBecause it is quadratic in the $$X_{p}$$ and $$g\\left( z \\right)$$’s, minimizing $$\\xi$$ is a straightforward linear least squares problem. The overdetermined system of linear equations is robustly solved using the singular value decomposition (SVD) method. An intuitive explanation of the procedure may be found in “The infrared light transmission model of the finger” section and Fig. 2 of reference paper .\n\nIn the reference paper , the noise, in the $$X_{p}$$, is an independent Gaussian random variable, in which the variance is $$\\sigma^{2}$$ and the joint probability density function can be written as:\n\n$$P(X_{B} ) \\prec \\exp \\left\\{ { - \\frac{1}{2}\\sum\\limits_{{p,q}} {w_{{pq}} (I_{{B_{{pq}} }} - X_{{pq}} )^{2} } } \\right\\}$$\n(7)\n\nA maximum-likelihood (ML) approach is taken to find the high dynamic range image values. The maximum likelihood solution finds the values $$X_{q}$$ that maximize the probability in Eq. (7). Maximizing Eq. (7) is equivalent to minimizing the negative of its natural logarithm, which leads to the following objective function to be minimized:\n\n$$\\xi (X) = \\sum\\limits_{p,q} {w_{pq} (I_{Bpq} - X_{pq} )^{2} }$$\n(8)\n\nWith Gaussian simplifying approximation, the noise variances $$\\sigma_{pq}^{2}$$ would be difficult to characterize accurately. Again, detailed knowledge of the image capture process would be required, and the noise characterization would have to be performed each time a different image is captured on a device.\n\nEquation (8) can be minimized by setting the gradient $$\\xi \\left( X \\right)$$ equal to zero. But if the $$X_{p}$$ were unknown in each pixel, one could jointly estimate $$X_{p}$$ and $$X_{q}$$ by arbitrarily fixing one of the q positions, and then performing an iterative optimization of Eq. (8) with respect to both $$X_{p}$$ and $$X_{q}$$. It is difficult to solve these estimating values without the analytic expression of the transmitting model.\n\nFrom the observed pixels, this paper presents the estimated transmitting model curve by the sliding the sampled curve segments, and blending these to a monotone increasing curve based on the genetic algorithm. So, if the blending curve is built or fit and then the other function curve can be redrawn by several sample points. It is possible to recover the blending curve shown in Fig. 4. The mixed complete curve $$g$$ can be used to get the transmitting model function $$f$$, which is shown in Fig. 5.\n\n## Multi-light-intensity finger vein images’ fusion based on the transmitting model\n\nThis session presents a fusion algorithm for the multi-light-intensity finger vein images based on the transmitting model. In the image pixel level fusion, the imaging quality estimation of the pixel is very important. In Section II, the transmitting model has been established by the observed data. Its derivative curve is shown in Fig. 6. It is obvious that the value of $$\\Delta B$$ is about zero in the underexposed and overexposed range. This means that the infrared light intensity in these ranges is not suitable for the finger vein imaging. On the other hand, the value $$\\Delta B$$ could be used to evaluate the fitness of the irradiance of the infrared intensity.\n\nIn this paper, the fusion method is based on the pixel level. Firstly, the infrared multi-light-intensity finger vein images are divided into R independent blocks by column.\n\nTo sign every divided block as $$T_{rp} ,r = 1,2, \\ldots ,R{\\text{ and }}p = 1,2, \\ldots ,N$$, where $$r$$ is the index of the block, and $$p$$ is the image number. In order to estimate the quality of each $$T_{rp}$$, the average gray value of the block is calculated by its quality value as $$\\overline{g}_{rp} = mean2(T_{rp} ),r = 1,2, \\ldots ,R{\\text{ and }}p = 1,2, \\ldots ,N$$. Then, the $$\\overline{g}_{rp}$$ is put into the derivative curve of Fig. 5 to calculate the $$\\Delta B_{{\\overline{g}_{rp} }}$$ value in the next fusion. The fusion weight value of the block $$T_{rp}$$ is defined as:\n\n$$S_{{rp}} = \\exp \\left[ {\\alpha \\cdot\\Delta B_{{\\bar{g}_{{rp}} }} } \\right]$$\n(9)\n\nThe constant parameter $$\\alpha$$ is the smoothing coefficient. In order to avoid the checkerboard edge between two adjacent blocks, it needs to define other spatial smoothing weighting $$G_{rp}$$:\n\n$$G_{rp} (x,y) = \\exp\\left [ - \\frac{{(y - y_{c} )^{2} }}{{2\\sigma^{2} }}\\right]$$\n(10)\n\nThe constant parameter $$\\sigma$$ is the variance of the Gaussian coefficient. $$x$$ is the row number and $$y$$ is the column number in the finger vein image, and $$y_{c}$$ is the block center column number.\n\nThe weighting is the joint value of the gray information coefficient $$S_{rp}$$ and spatial smoothing coefficient $$G_{rp}$$. The joint weighting is defined as:\n\n$$\\omega_{rp} = G_{rp} *S_{rp}$$\n(11)\n\nIts normalized value is defined as:\n\n$$\\varpi_{rp} = \\omega_{rp} \\bigg {/}\\left( {\\sum\\limits_{p = 1}^{N} {\\omega_{rp} } } \\right)$$\n(12)\n\nIn the fusion, each fused block $$I_{r} , \\, r = 1,2, \\ldots ,R$$ is calculated by Eq. (13) :\n\n$$I_{r} = \\sum\\limits_{p = 1}^{N} {(I_{rp} *\\omega_{rp} } ), \\begin{array}{*{20}c} & {r = 1,2, \\ldots ,R} \\\\ \\end{array}$$\n(13)\n\n## Examinations and discussions\n\nThe sample of infrared multi-light-intensity finger vein images is shown in Fig. 7, which is captured by a self-developed platform, shown in Fig. 8. The infrared light intensity is dependent on the duty of PWM, which drives the infrared led irradiance. The transmitting model is shown in Fig. 9 and the differential curve is shown in Fig. 10.\n\nIn the fusion step, the three finger vein images are selected to the weighting fuse [17, 18], which is Fig. 7c–e. Each of them has been divided into ten blocks by the column shown in Fig. 11. According to the transmitting model curve, the most suitable blocks are blending to one finger vein image, which is shown in Fig. 12. The weighting value of $$S_{rp}$$ can be calculated by Eq. (9), which is shown in Fig. 13. The weighting value of $$G_{rp}$$ could be calculated by Eq. (10), which is shown in Fig. 14. The joint weighting value of $$w_{rp}$$ can be calculated by Eq. (11), which is shown in Fig. 15. The fusion finger vein image is blended by Eq. (12), which is shown in Fig. 16.\n\nTwo other fuse methods are tested for the performance comparison in this paper. One is discrete wavelet transform (DWT) and the other is contrast pyramid, which flow charts are shown in Fig. 17. The source images are decomposed by discrete wavelet transform. And chooses the max coefficient at each pixel before the image rebuild. The source images are pyramid decomposed by the down sample. And calculate the contrast at each pixel. The pyramid layer which has max contrast value is choice before the pyramid image rebuild.\n\nThe fused performance is tested by the following statistics method . The standard deviation of an image is defined as formula (14), $$\\mu$$ is the mean value of the image $$I$$ in which the size is $$m \\times n$$ and $$\\sigma$$ is the standard deviation.\n\n\\begin{aligned} \\mu = &\\,\\, \\frac{1}{{m*n}}\\sum\\limits_{{x = 1}}^{m} {\\sum\\limits_{{y = 1}}^{n} {{\\text{I}}(x,y){\\mkern 1mu} } } \\\\ \\sigma =& \\, \\sqrt {\\frac{1}{{m*n}}\\sum\\limits_{{x = 1}}^{m} {\\sum\\limits_{{y = 1}}^{n} {({\\text{I}}(x,y) - \\mu )} } } \\end{aligned}\n(14)\n\nThe Shannon Information entropy of the image is defined as formula (15), the $$P(gray)$$ is the gray probability of the pixel in the image $$I$$:\n\n$$H(\\text{I} ) = - \\sum\\limits_{gray = 1}^{255} {P(gray)\\log_{2} [P(gray)]}$$\n(15)\n\nThe standard deviation and information entropy of multi-light-intensity finger vein images together with the fused image by proposed method are shown in Table 1 . However, the standard deviation and information entropy of the fused image is less than Figs. 2, 3, and 4, that means the gray uniformity and consistency of the fused image is better than Figs. 2, 3, and 4. For the image of Fig. 10a, its gray contrast is quite low, in which the image is nearly under exposure.\n\nThe degree of dependence between one source image and the fused image could be measured by the mutual information (FMI), which can be calculated by the formula (16):\n\n$$FMI = \\sum\\limits_{i = 1}^{4} {MI(I_{i} ,I_{f} )}$$\n(16)\n\nIn the formula (16), $$MI(I_{i} ,I_{f} )$$ is defined as formula (17), and the joint histogram between the source image $$I_{i}$$ and the fused image $$I_{f}$$ is defined as $$h(I_{i} ,I_{f} )$$.\n\n$$MI(I_{i} ,I_{f} ) = \\sum\\limits_{x = 1}^{m} {\\sum\\limits_{y = 1}^{n} {h(I_{i} (x,y),I_{f} (x,y))} \\cdot } \\log_{2}\\left (\\frac{{h(I_{i} (x,y),I_{f} (x,y))}}{{h(I_{i} (x,y)) \\cdot h(I_{f} (x,y))}}\\right)$$\n(17)\n\nThe results of fusion mutual information (MI) between the source image and the fused image are shown in Table 2 . The MI between the three source images and fused image is the sum of the MI of each source image and fused image.\n\nThe information fused from the source images could be calculated as the fusion quality index (FQI), which could be calculated by Eq. (18).\n\n$$FQI = \\sum\\limits_{w \\in W} {c(w)\\left(\\sum\\limits_{i - 1}^{4} {\\lambda (i)QI(I_{i} ,I_{f} |w)} \\right)} ,$$\n(18)\n\nwhere $$\\lambda_{i}$$ is computed over a window $$w$$, which can be calculated by the formula (19):\n\n$$\\lambda_{i} = \\sigma_{{_{{I_{i} }} }}^{2} \\bigg /\\sum\\limits_{i = 1}^{4} {\\sigma_{{_{{I_{i} }} }}^{2} }$$\n(19)\n\n$$c\\left( w \\right)$$ is a normalized version of $$C\\left( w \\right)$$, which can be calculated by the formula (20):\n\n$$C(w) = \\hbox{max} (\\sigma_{{_{{I_{1} }} }}^{2} ,\\sigma_{{_{{I_{2} }} }}^{2} , \\ldots ,\\sigma_{{_{{I_{4} }} }}^{2} )$$\n(20)\n\n$$QI\\left( {I_{i} ,I_{f} \\left| w \\right.} \\right)$$ is the quality index over a window for a given source image and fused image.\n\nIn the test, the size of the window is 8 × 8. The FQI values of the fusion quality index are shown in Table 3 .\n\nIn order to compare the fused performance, the structural similarity index measure (SSIM) is applied in this test. The results are shown in Table 4 .\n\nThe results of Tables 1, 2, 3 and 4 show that the proposed fused method based on the column blocking of the image is effective applied to the infrared multi-light-intensity finger vein images.\n\n## Conclusions and further works\n\nThe infrared finger-transmitting model is proposed in this paper, which it could be easily built by the observed data of multiple light-intensity images. This model provides a better approach to get the intact vein patterns by adopting the vein biometric data captured by the bioinformation. The features of captured image are estimated and fused by using this model’s differential curves. In this paper, the examination approach has been proven that it is an efficient and practical method for the finger’s fusion approach via infrared transmitting model. It is suitable for fusion of the infrared images in biometric system. Finally, the applications in detail and their analyses on while applying the multi-light-intensity finger vein images’ fusion which is based on the transmitting model to big data environments will be stated in future works.\n\n## References\n\n1. Shin KY, Park YH, Nguyen DT (2014) Finger-Vein image enhancement using a fuzzy-based fusion method with gabor and retinex filtering. Sensors 14(2):3095–3129\n\n2. Tistarelli M, Schouten B (2011) Biometrics in ambient intelligence. J Ambient Intell Human Comput 2(2):113–126\n\n3. Liukui C, Zuojin L, Ying W, Lixiao F (2014) A principal component analysis fusion method on infrared multi-light-intensity finger vein images, BWCCA. pp 281–286\n\n4. Kikuchi H, Nagai K, Ogata W, Nishigaki M (2010) Privacy-preserving similarity evaluation and application to remote biometrics authentication. Soft Comput 14(5):529–536\n\n5. Greene CS, Tan J, Ung M, Moore JH, Cheng C (2014) Big data bioinformatics. J Cell Physiol 229(12):1896–1900\n\n6. Ogiela MR, Ogiela L, Ogiela U (2015) Biometric methods for advanced strategic data sharing protocols. In: Barolli L, Palmieri F, Silva HDD, et al. (eds) 9th international conference on innovative mobile and internet services in ubiquitous computing (IMIS), Blumenau. pp 179–183\n\n7. Ogiela MR, Ogiela U, Ogiela L (2012) Secure information sharing using personal biometric characteristics. In: Kim TH, Kang JJ, Grosky WI, et al. (eds) 4th international mega-conference on future generation information technology (FGIT 2012), Korea Woman Train Ctr, Kangwondo, South Korea Dec 16–19, 2012, Book series: Communications in computer and information science, vol. 353 pp 369–373\n\n8. Ogiela L, Ogiela MR (2016) Bio-inspired cryptographic techniques in information management applications. In: Barolli L, Takizawa M, Enokido T, et al. (eds) IEEE 30th international conference on advanced information networking and applications (IEEE AINA), Switzerland Mar 23-25, 2016, Book series: International conference on advanced information networking and applications. pp 1059–1063\n\n9. Chen HC, Kuo SS, Sun SC, Chang CH (2016) A distinguishing arterial pulse waves approach by using image processing and feature extraction technique. J Med Syst 40:215. doi:10.1007/s10916-016-0568-4\n\n10. Chen W, Er MJ, Wu S (2006) Illumination compensation and normalization for robust face recognition using discrete cosine transform in logarithm domai. IEEE Trans Syst Man Cybern B (Cybernetics) 36(2):458–466\n\n11. Wu X, Zhu X, Wu GQ, Ding W (2014) Data mining with big data. IEEE Trans Knowl Data Eng 26(1):97–107\n\n12. Urbach R (1969) The biologic effects of ultraviolet radiation. Pergamon Press, New York. http://www.inchem.org/documents/ehc/ehc/ehc23.htm#SubSectionNumber:2.2.1\n\n13. Chen LK, Li ZJ, Wu Y, Xiang Y (2013) Dynamic range extend on finger vein image based on infrared multi-light-intensity vascular imaging. MEIMEI2013. ChongQing, vol. 427–429, pp 1832–1835\n\n14. Jacobs K, Loscos C, Ward G (2008) Automatic high-dynamic range image generation for dynamic scenes. IEEE Comput Gr Appl 28(2):84–93\n\n15. Debevec PE, Malik J (1997) Recovering high dynamic range radiance maps from photographs. In: Whitted T, Mones-Hattal B, Owen SG (eds) Proc. of the ACM SIGGRAPH. ACM Press, New York, pp 369–378\n\n16. Rovid A, Hashimoto T, Varlaki P (2007) Improved high dynamic range image reproduction method. In: Fodor J, Prostean O (eds) Proc. of the 4th Int’l Symp. on applied computational intelligence and informatics, IEEE Computer Society, Washington. pp 203–207\n\n17. Yang J, Shi Y (2014) Towards finger-vein image restoration and enhancement for finger-vein recognition. Inf Sci 1(268):33–52\n\n18. Zhang J, Dai X, Sun QD, Wang BP (2011) Directly fusion method for combining variable exposure value images (in Chinese). J Software 22(4):813–825 (in Chinese)\n\n19. Delpy DT, Cope M (1997) Quantification in tissue near-infrared spectroscopy. Philos Trans R Soc B Biol Sci 352:649–659\n\n## Authors’ contributions\n\nThe authors’ contributions are summarized below. LC have made substantial contributions to conception and design, involved in drafting the manuscript. ZL and YW have made the acquisition of data and analysis and interpretation of data. The critically important intellectual contents of this manuscript have been revised by HCC. All authors read and approved the final manuscript.\n\n### Acknowledgements\n\nThis study was funded in part by the Natural Science Foundation Project of CQ CSTC (cstc2011jjA40012), Foundation and Frontier Project of CQ CSTC (cstc2014jcyjA40006), and Campus Research Foundation of Chongqing University of Science and Technology (CK2011B09, CK2011B05). This work was also supported in part by Asia University, Taiwan, and China Medical University Hospital, China Medical University, Taiwan, under Grant ASIA-105-CMUH-04.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Ethical approval\n\nThis article does not contain any studies with human participants or animals performed by any of the authors.\n\n### Publisher’s Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Hsing-Chung Chen.\n\n## Rights and permissions\n\nOpen Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.\n\nReprints and Permissions",
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https://advancesindifferenceequations.springeropen.com/articles/10.1186/1687-1847-2014-285 | [
"Theory and Modern Applications\n\nBVPs for higher-order integro-differential equations with ϕ-Laplacian and functional boundary conditions\n\nAbstract\n\nIn this paper, we study the existence of solutions of a class of higher-order integro-differential boundary value problems with ϕ-Laplacian like operator and functional boundary conditions. By giving the definition of a pair of coupled lower and upper solutions and some new hypotheses, we obtain some new existence results for boundary value problems with ϕ-Laplacian like operator by employing the Schauder fixed point theorem and an appropriate Nagumo condition. Finally, an example is given to illustrate the results.\n\nMSC:39K10, 34B15.\n\n1 Introduction\n\nIntegro-differential equations have become more and more important in some mathematical models of real phenomena, especially in control, biological, medical, and informational models. Boundary value problems (BVPs) for nonlinear integro-differential equations are used to describe a great number of nonlinear phenomena in science (see [1, 2]), moreover, the theory of ϕ-Laplacian BVPs has emerged as an important area in recent years (see ). In this paper, we will consider the following BVPs of higher-order functional integro-differential ϕ-Laplacian like equations with functional boundary conditions:\n\n${\\left[\\varphi \\left({u}^{\\left(n-1\\right)}\\left(t\\right)\\right)\\right]}^{\\prime }+Au\\left(t\\right)=0,\\phantom{\\rule{1em}{0ex}}t\\in {J}^{\\prime },$\n(1.1)\n$\\left\\{\\begin{array}{l}{g}_{i}\\left(u,{W}_{0}u,{S}_{0}u,\\dots ,{u}^{\\left(n-1\\right)},{W}_{n-1}{u}^{\\left(n-1\\right)},{S}_{n-1}{u}^{\\left(n-1\\right)},{u}^{\\left(i\\right)}\\left(0\\right)\\right)=0,\\phantom{\\rule{1em}{0ex}}i=0,\\dots ,n-2,\\\\ {g}_{n-1}\\left(u,{W}_{0}u,{S}_{0}u,\\dots ,{u}^{\\left(n-1\\right)},{W}_{n-1}{u}^{\\left(n-1\\right)},{S}_{n-1}{u}^{\\left(n-1\\right)},{u}^{\\left(n-2\\right)}\\left(T\\right)\\right)=0,\\end{array}$\n(1.2)\n\nwhere $n\\ge 2$ is an integer, ϕ is an increasing homeomorphism operator,\n\n$\\begin{array}{rcl}Au\\left(t\\right)& =& f\\left(t,u\\left(t\\right),u\\left({\\alpha }_{0}\\left(t\\right)\\right),{W}_{0}u\\left(t\\right),{S}_{0}u\\left(t\\right),{u}^{\\prime }\\left(t\\right),{u}^{\\prime }\\left({\\alpha }_{1}\\left(t\\right)\\right),{W}_{1}{u}^{\\prime }\\left(t\\right),{S}_{1}{u}^{\\prime }\\left(t\\right),\\dots ,\\\\ {u}^{\\left(n-1\\right)}\\left(t\\right),{u}^{\\left(n-1\\right)}\\left({\\alpha }_{n-1}\\left(t\\right)\\right),{W}_{n-1}{u}^{\\left(n-1\\right)}\\left(t\\right),{S}_{n-1}{u}^{\\left(n-1\\right)}\\left(t\\right)\\right),\\end{array}$\n\nand $J=\\left[0,T\\right]$ ($T>0$), ${J}^{\\prime }=\\left(0,T\\right)$, for $i=0,1,\\dots ,n-1$, ${g}_{i}:{\\left(C\\left[0,T\\right]\\right)}^{3n}×R\\to R$ are continuous functions, and for $l=0,1,\\dots ,n-1$,\n\n${W}_{l}{u}^{\\left(l\\right)}\\left(t\\right)={\\int }_{0}^{{\\beta }_{l}\\left(t\\right)}{k}_{l}\\left(t,s\\right){u}^{\\left(l\\right)}\\left({\\gamma }_{l}\\left(s\\right)\\right)\\phantom{\\rule{0.2em}{0ex}}ds,\\phantom{\\rule{2em}{0ex}}{S}_{l}{u}^{\\left(l\\right)}\\left(t\\right)={\\int }_{0}^{T}{h}_{l}\\left(t,s\\right){u}^{\\left(l\\right)}\\left({\\delta }_{l}\\left(s\\right)\\right)\\phantom{\\rule{0.2em}{0ex}}ds,$\n\n${k}_{l}\\left(t,s\\right)\\in C\\left({D}_{l},{R}^{+}\\right)$, ${D}_{l}=\\left\\{\\left(t,s\\right)\\in {R}^{2}:t\\in J,0\\le s\\le {\\beta }_{l}\\left(t\\right)\\right\\}$, ${h}_{l}\\left(t,s\\right)\\in C\\left(J×J,{R}^{+}\\right)$, ${R}^{+}=\\left[0,+\\mathrm{\\infty }\\right)$, ${\\alpha }_{l},{\\beta }_{l},{\\gamma }_{l},{\\delta }_{l}\\in C\\left(J,J\\right)$, and $f:{J}^{\\prime }×{R}^{4n}\\to R$ is a Carathéodory function, that is:\n\n1. (i)\n\nfor any $\\left({x}_{0},\\dots ,{x}_{4n-1}\\right)\\in {R}^{4n}$, $f\\left(t,{x}_{0},\\dots ,{x}_{4n-1}\\right)$ is measurable on ${J}^{\\prime }$,\n\n2. (ii)\n\nfor a.e. $t\\in {J}^{\\prime }$, $f\\left(t,\\cdot ,\\dots ,\\cdot \\right)$ is continuous on ${R}^{4n}$,\n\n3. (iii)\n\nfor every compact set $K\\subset {R}^{4n}$, there exists a nonnegative function ${\\mu }_{{K}_{}}\\left(t\\right)\\in {L}^{1}\\left(0,T\\right)$ such that\n\n$|f\\left(t,{x}_{0},\\dots ,{x}_{4n-1}\\right)|\\le {\\mu }_{{}_{K}}\\left(t\\right),\\phantom{\\rule{1em}{0ex}}\\left(t,{x}_{0},\\dots ,{x}_{4n-1}\\right)\\in {J}^{\\prime }×K.$\n\nWe say $u\\left(t\\right)$ is a solution of BVP (1.1) and (1.2), that is, a function $u\\left(t\\right)\\in {C}^{n-1}\\left[0,T\\right]$ such that $\\varphi \\left({u}^{\\left(n-1\\right)}\\right)$ is absolutely continuous on ${J}^{\\prime }$, $u\\left(t\\right)$ satisfies (1.1) a.e. on ${J}^{\\prime }$, and $u\\left(t\\right)$ satisfies boundary condition (1.2).\n\nAs we know, higher-order boundary value problems for differential equations have received great attention in recent years (see ). We found that BVP (1.1) and (1.2) is more general in the literature, and the functional boundary condition (1.2) may not only cover many classical boundary conditions, such as various linear two-point, multi-point studied by many authors, but it may also include many new boundary conditions not studied so far in the literature. In recent years, BVPs with linear and nonlinear boundary conditions have been extensively investigated by numerous researchers. For a small sample of such work, we refer the reader to . As is well known, a variety of methods and tools, such as lower and upper solution methods and various fixed point theorems, are very useful and have been successfully used to prove the existence of solutions of BVPs.\n\nMotivated by the above mentioned works, we consider the BVPs of higher-order functional integro-differential equations (1.1) and (1.2) with ϕ-Laplacian like operator and functional boundary conditions in this paper. As we know, BVP (1.1) and (1.2) has not yet been considered. By introducing a definition for the coupled lower and upper solutions of BVP (1.1) and (1.2), we obtain the existence of solutions of the problem based on the assumption that there exists a pair of coupled lower and upper solutions.\n\nThis paper is organized as follows. In Section 2, we state some preliminaries and lemmas which will be used throughout this paper. In Section 3, some results concerning coupled lower and upper solutions are given. Finally, an example is given to illustrate our results in Section 4.\n\n2 Preliminaries\n\nThroughout this paper, let $E={C}^{n-1}\\left[0,T\\right]$, ${\\parallel u\\parallel }_{\\mathrm{\\infty }}=max\\left\\{|u\\left(t\\right)|:t\\in J\\right\\}$ for any $u\\in C\\left[0,T\\right]$, and\n\n$\\parallel u\\parallel =max\\left\\{{\\parallel u\\parallel }_{\\mathrm{\\infty }},{\\parallel {u}^{\\prime }\\parallel }_{\\mathrm{\\infty }},\\dots ,{\\parallel {u}^{\\left(n-1\\right)}\\parallel }_{\\mathrm{\\infty }}\\right\\}$\n\nand\n\n${\\parallel u\\parallel }_{p}=\\left\\{\\begin{array}{ll}{\\left({\\int }_{0}^{T}{|u\\left(t\\right)|}^{p}\\phantom{\\rule{0.2em}{0ex}}dt\\right)}^{\\frac{1}{p}},& 1\\le p<\\mathrm{\\infty },\\\\ inf\\left\\{M:\\mu \\left(\\left\\{t:|u\\left(t\\right)|>M\\right\\}\\right)=0\\right\\},& p=\\mathrm{\\infty },\\end{array}$\n\nstand for the norms in E and ${L}^{p}\\left(0,T\\right)$, respectively, where $\\mu \\left(\\cdot \\right)$ denotes the Lebesgue measure of a set. In what follows, a functional $y:C\\left[0,T\\right]\\to R$ is said to be nondecreasing if $y\\left({u}_{1}\\right)\\ge y\\left({u}_{2}\\right)$ for any ${u}_{1},{u}_{2}\\in C\\left[0,T\\right]$ with ${u}_{1}\\left(t\\right)\\ge {u}_{2}\\left(t\\right)$ on $\\left[0,T\\right]$. A similar definition holds for y to be non-increasing.\n\nDefinition 2.1 Let $f:{J}^{\\prime }×{R}^{4n}\\to R$ be Carathéodory function, and $v,w\\in E$ satisfy\n\n${v}^{\\left(i\\right)}\\left(t\\right)\\le {w}^{\\left(i\\right)}\\left(t\\right),\\phantom{\\rule{1em}{0ex}}t\\in J,i=0,1,\\dots ,n-2.$\n(2.1)\n\nWe say that f satisfies the Nagumo condition with respect to v and w if for\n\n$\\xi =max\\left\\{\\frac{{w}^{\\left(n-2\\right)}\\left(T\\right)-{v}^{\\left(n-2\\right)}\\left(0\\right)}{T},\\frac{{w}^{\\left(n-2\\right)}\\left(0\\right)-{v}^{\\left(n-2\\right)}\\left(T\\right)}{T}\\right\\},$\n(2.2)\n\nthere exists a constant $C=C\\left(v,w\\right)$ with\n\n$C>max\\left\\{\\xi ,{\\parallel {v}^{\\left(n-1\\right)}\\parallel }_{\\mathrm{\\infty }},{\\parallel {w}^{\\left(n-1\\right)}\\parallel }_{\\mathrm{\\infty }}\\right\\}$\n(2.3)\n\nand functions $\\psi \\in C\\left[0,\\mathrm{\\infty }\\right)$, $\\vartheta \\in {L}^{p}\\left(0,T\\right)$ ($1\\le p\\le \\mathrm{\\infty }$), such that $\\psi >0$ on $\\left[0,\\mathrm{\\infty }\\right)$,\n\n(2.4)\n\nand\n\n${\\int }_{\\varphi \\left(\\xi \\right)}^{\\varphi \\left(C\\right)}\\frac{{\\left({\\varphi }^{-1}\\left(x\\right)\\right)}^{\\left(p-1\\right)/p}}{\\psi \\left(|{\\varphi }^{-1}\\left(x\\right)|\\right)}\\phantom{\\rule{0.2em}{0ex}}dx,\\phantom{\\rule{2em}{0ex}}{\\int }_{\\varphi \\left(-C\\right)}^{\\varphi \\left(-\\xi \\right)}\\frac{{\\left({\\varphi }^{-1}\\left(x\\right)\\right)}^{\\left(p-1\\right)/p}}{\\psi \\left(|{\\varphi }^{-1}\\left(x\\right)|\\right)}\\phantom{\\rule{0.2em}{0ex}}dx>{\\parallel \\vartheta \\parallel }_{p}{\\zeta }^{\\left(p-1\\right)/p},$\n(2.5)\n\nwhere $\\left(p-1\\right)/p\\equiv 1$ for $p=\\mathrm{\\infty }$,\n\n$\\begin{array}{rl}{\\mathbb{D}}_{v}^{w}=& \\left[v\\left(t\\right),w\\left(t\\right)\\right]×\\left[v\\left({\\alpha }_{0}\\left(t\\right)\\right),w\\left({\\alpha }_{0}\\left(t\\right)\\right)\\right]×\\left[{W}_{0}v\\left(t\\right),{W}_{0}w\\left(t\\right)\\right]×\\left[{S}_{0}v\\left(t\\right),{S}_{0}w\\left(t\\right)\\right]×\\cdots \\\\ ×\\left[{v}^{\\left(n-2\\right)}\\left(t\\right),{w}^{\\left(n-2\\right)}\\left(t\\right)\\right]×\\left[{v}^{\\left(n-2\\right)}\\left({\\alpha }_{n-2}\\left(t\\right)\\right),{w}^{\\left(n-2\\right)}\\left({\\alpha }_{n-2}\\left(t\\right)\\right)\\right]\\\\ ×\\left[{W}_{n-2}{v}^{\\left(n-2\\right)}\\left(t\\right),{W}_{n-2}{w}^{\\left(n-2\\right)}\\left(t\\right)\\right]×\\left[{S}_{n-2}{v}^{\\left(n-2\\right)}\\left(t\\right),{S}_{n-2}{w}^{\\left(n-2\\right)}\\left(t\\right)\\right]\\end{array}$\n\nand\n\n$\\zeta =\\underset{t\\in J}{max}{w}^{\\left(n-2\\right)}\\left(t\\right)-\\underset{t\\in J}{min}{v}^{\\left(n-2\\right)}\\left(t\\right).$\n(2.6)\n\nRemark 2.1 Let $v,w\\in E$ satisfy (2.1). Assume that there exist $\\theta \\in {L}^{p}\\left(0,T\\right)$, $1\\le p\\le \\mathrm{\\infty }$, and $\\sigma \\in \\left[0,\\mathrm{\\infty }\\right)$ such that\n\nThen f satisfies the Nagumo condition with respect to v and w with $\\psi \\left(x\\right)=1+{|x|}^{\\sigma }$.\n\nDefinition 2.2 Let C be the constant introduced in Definition 2.1. Assume that there exist $v,w\\in E$ satisfying (2.1), $\\varphi \\left({v}^{\\left(n-1\\right)}\\right)$ and $\\varphi \\left({w}^{\\left(n-1\\right)}\\right)$ are absolutely continuous on ${J}^{\\prime }$. Then v and w are said to be a pair of coupled lower and upper solutions of BVP (1.1) and (1.2) if\n\n(2.7)\n$\\left\\{\\begin{array}{l}{min}_{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{g}_{i}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-1\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ \\phantom{\\rule{1em}{0ex}}z,{W}_{n-1}z,{S}_{n-1}z,{v}^{\\left(i\\right)}\\left(0\\right)\\right)\\ge 0,\\phantom{\\rule{1em}{0ex}}i=0,\\dots ,n-2,\\\\ {min}_{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{g}_{n-1}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ \\phantom{\\rule{1em}{0ex}}z,{W}_{n-1}z,{S}_{n-1}z,{v}^{\\left(n-2\\right)}\\left(T\\right)\\right)\\ge 0,\\end{array}$\n(2.8)\n\nand\n\n(2.9)\n$\\left\\{\\begin{array}{l}{max}_{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{g}_{i}\\left(w,{W}_{0}w,{S}_{0}w,\\dots ,{w}^{\\left(n-2\\right)},{W}_{n-2}{w}^{\\left(n-2\\right)},{S}_{n-2}{w}^{\\left(n-2\\right)},\\\\ \\phantom{\\rule{1em}{0ex}}z,{W}_{n-1}z,{S}_{n-1}z,{w}^{\\left(i\\right)}\\left(0\\right)\\right)\\le 0,\\phantom{\\rule{1em}{0ex}}i=0,\\dots ,n-2,\\\\ {max}_{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{g}_{n-1}\\left(w,{W}_{0}w,{S}_{0}w,\\dots ,{w}^{\\left(n-2\\right)},{W}_{n-2}{w}^{\\left(n-2\\right)},{S}_{n-2}{w}^{\\left(n-2\\right)},\\\\ \\phantom{\\rule{1em}{0ex}}z,{W}_{n-1}z,{S}_{n-1}z,{w}^{\\left(n-2\\right)}\\left(T\\right)\\right)\\le 0.\\end{array}$\n(2.10)\n\nFor convenience, we first list the following hypotheses:\n\n(H1) $\\varphi \\left(x\\right)$ is increasing on R;\n\n(H2) BVP (1.1) and (1.2) has a pair of coupled lower and upper solutions v and w satisfying (2.1);\n\n(H3) the functional f satisfies the Nagumo condition with respect to v and w;\n\n(H4) for $\\left(t,{x}_{0},{x}_{1},\\dots ,{x}_{4n-1}\\right)\\in {J}^{\\prime }×{R}^{4n}$ with ${v}^{\\left(i\\right)}\\left(t\\right)\\le {x}_{4i}\\le {w}^{\\left(i\\right)}\\left(t\\right)$, ${v}^{\\left(i\\right)}\\left({\\alpha }_{i}\\left(t\\right)\\right)\\le {x}_{4i+1}\\le {w}^{\\left(i\\right)}\\left(t\\right)$, ${W}_{i}{v}^{\\left(i\\right)}\\left(t\\right)\\le {x}_{4i+2}\\le {W}_{i}{w}^{\\left(i\\right)}\\left(t\\right)$, ${S}_{i}{v}^{\\left(i\\right)}\\left(t\\right)\\le {x}_{4i+3}\\le {S}_{i}{w}^{\\left(i\\right)}\\left(t\\right)$, $i=0,1,\\dots ,n-3$, and ${v}^{\\left(n-2\\right)}\\left({\\alpha }_{n-2}\\left(t\\right)\\right)\\le {x}_{4n-7}\\le {w}^{\\left(n-2\\right)}\\left(t\\right)$, ${W}_{n-2}{v}^{\\left(n-2\\right)}\\left(t\\right)\\le {x}_{4n-6}\\le {W}_{n-2}{w}^{\\left(n-2\\right)}\\left(t\\right)$, ${S}_{n-2}{v}^{\\left(n-2\\right)}\\left(t\\right)\\le {x}_{4n-5}\\le {S}_{n-2}{w}^{\\left(n-2\\right)}\\left(t\\right)$, we have\n\n$\\begin{array}{r}f\\left(\\cdot ,v\\left(t\\right),v\\left({\\alpha }_{0}\\left(t\\right)\\right),{W}_{0}v\\left(t\\right),{S}_{0}v\\left(t\\right),\\dots ,{x}_{4n-8},{v}^{\\left(n-2\\right)}\\left({\\alpha }_{n-2}\\left(t\\right)\\right),\\\\ \\phantom{\\rule{2em}{0ex}}{W}_{n-2}{v}^{\\left(n-2\\right)}\\left(t\\right),{S}_{n-2}{v}^{\\left(n-2\\right)}\\left(t\\right),{x}_{4n-4},\\dots ,{x}_{4n-1}\\right)\\\\ \\phantom{\\rule{1em}{0ex}}\\le f\\left(\\cdot ,{x}_{0},{x}_{1},\\dots ,{x}_{4n-8},\\dots ,{x}_{4n-4},\\dots ,{x}_{4n-1}\\right)\\\\ \\phantom{\\rule{1em}{0ex}}\\le f\\left(\\cdot ,w\\left(t\\right),w\\left({\\alpha }_{0}\\left(t\\right)\\right),{W}_{0}w\\left(t\\right),{S}_{0}v\\left(t\\right),\\dots ,{x}_{4n-8},{w}^{\\left(n-2\\right)}\\left({\\alpha }_{n-2}\\left(t\\right)\\right),\\\\ \\phantom{\\rule{2em}{0ex}}{W}_{n-2}{w}^{\\left(n-2\\right)}\\left(t\\right),{S}_{n-2}{w}^{\\left(n-2\\right)}\\left(t\\right),{x}_{4n-4},\\dots ,{x}_{4n-1}\\right),\\end{array}$\n\nand $f\\left(t,{x}_{0},{x}_{1},\\dots ,{x}_{4n-1}\\right)$ is nondecreasing in the arguments ${x}_{4n-4}$, ${x}_{4n-3}$, ${x}_{4n-2}$, ${x}_{4n-1}$;\n\n(H5) for $i=0,1,\\dots ,n-1$ and $\\left({y}_{0},{y}_{1},\\dots ,{y}_{3n-1},z\\right)\\in {\\left(C\\left[0,T\\right]\\right)}^{3n}×R$, ${g}_{i}\\left({y}_{0},{y}_{1},\\dots ,{y}_{3n-1},z\\right)$ are nondecreasing in the arguments ${y}_{0},\\dots ,{y}_{3n-4}$.\n\nWe assume that conditions (H1)-(H5) hold throughout this paper. For $u\\in {C}^{n-2}\\left[0,T\\right]$ and $i=0,1,\\dots ,n-2$, we define\n\n${\\overline{u}}^{\\left[i\\right]}\\left(t\\right)=max\\left\\{{v}^{\\left(i\\right)}\\left(t\\right),min\\left\\{{u}^{\\left(i\\right)}\\left(t\\right),{w}^{\\left(i\\right)}\\left(t\\right)\\right\\}\\right\\},$\n(2.11)\n\nthen, for $i=0,1,\\dots ,n-2$, ${\\overline{u}}^{\\left(i\\right)}\\left(t\\right)$ is continuous on J, and\n\n${v}^{\\left(i\\right)}\\left(t\\right)\\le {\\overline{u}}^{\\left[i\\right]}\\left(t\\right)\\le {w}^{\\left(i\\right)}\\left(t\\right),\\phantom{\\rule{2em}{0ex}}{\\overline{v}}^{\\left[i\\right]}\\left(t\\right)={v}^{\\left(i\\right)}\\left(t\\right),\\phantom{\\rule{2em}{0ex}}{\\overline{w}}^{\\left[i\\right]}\\left(t\\right)={w}^{\\left(i\\right)}\\left(t\\right),$\n(2.12)\n\nfor $t\\in J$ and $i=0,1,\\dots ,n-2$. Let $C=C\\left(v,w\\right)$ be the constant introduced in Definition 2.1, and define\n\n$\\phi \\left(x\\right)=\\left\\{\\begin{array}{ll}\\varphi \\left(x\\right),& |x|\\le C,\\\\ \\frac{1}{2C}\\left(\\varphi \\left(C\\right)-\\varphi \\left(-C\\right)\\right)x+\\frac{1}{2}\\left(\\varphi \\left(C\\right)+\\varphi \\left(-C\\right)\\right),& |x|>C,\\end{array}$\n(2.13)\n${\\stackrel{ˆ}{u}}^{\\left[n-1\\right]}\\left(t\\right)=max\\left\\{-C,min\\left\\{{u}^{\\left(n-1\\right)}\\left(t\\right),C\\right\\}\\right\\},\\phantom{\\rule{1em}{0ex}}u\\in E,$\n(2.14)\n\nand a functional $F:{J}^{\\prime }×E\\to R$ by\n\n$F\\left(t,u\\left(\\cdot \\right)\\right)=Bu\\left(t\\right)+\\frac{{\\overline{u}}^{\\left[n-2\\right]}\\left(t\\right)-{u}^{\\left(n-2\\right)}\\left(t\\right)}{1+{\\left({u}^{\\left(n-2\\right)}\\left(t\\right)\\right)}^{2}},$\n(2.15)\n\nwhere\n\n$\\begin{array}{rl}Bu\\left(t\\right)=& f\\left(t,{\\overline{u}}^{\\left[0\\right]}\\left(t\\right),{\\overline{u}}^{\\left[0\\right]}\\left({\\alpha }_{0}\\left(t\\right)\\right),{S}_{0}{\\overline{u}}^{\\left[0\\right]}\\left(t\\right),\\dots ,{\\overline{u}}^{\\left[n-2\\right]}\\left(t\\right),{\\overline{u}}^{\\left[n-2\\right]}\\left({\\alpha }_{n-2}\\left(t\\right)\\right),\\\\ {W}_{0}{\\overline{u}}^{\\left[0\\right]}\\left(t\\right),{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]}\\left(t\\right),{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]}\\left(t\\right),{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]}\\left(t\\right),\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]}\\left({\\alpha }_{n-1}\\left(t\\right)\\right),{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]}\\left(t\\right),{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]}\\left(t\\right)\\right).\\end{array}$\n\nThen, in view of (H1) and (2.13), we find that $\\phi :R\\to R$ is increasing and continuous (hence ${\\phi }^{-1}$ exists), and\n\n$\\underset{x\\to -\\mathrm{\\infty }}{lim}\\phi \\left(x\\right)=-\\mathrm{\\infty },\\phantom{\\rule{2em}{0ex}}\\underset{x\\to \\mathrm{\\infty }}{lim}\\phi \\left(x\\right)=\\mathrm{\\infty }.$\n(2.16)\n\nWhat is more, for $u\\in E$ and $t\\in {J}^{\\prime }$, $F\\left(t,u\\left(\\cdot \\right)\\right)$ is continuous in u, and we can see that\n\n$|F\\left(t,u\\left(\\cdot \\right)\\right)|\\le \\vartheta \\left(t\\right)\\underset{z\\in \\left[0,C\\right]}{max}\\psi \\left(z\\right)+\\parallel v\\parallel +\\parallel w\\parallel +1.$\n(2.17)\n\nNow, we consider the BVP consisting of the equation\n\n${\\left[\\phi \\left({u}^{\\left(n-1\\right)}\\left(t\\right)\\right)\\right]}^{\\prime }+F\\left(t,u\\left(\\cdot \\right)\\right)=0,\\phantom{\\rule{1em}{0ex}}t\\in {J}^{\\prime },$\n(2.18)\n\nand the boundary condition\n\n$\\left\\{\\begin{array}{l}{u}^{\\left(i\\right)}\\left(0\\right)={g}_{i}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ \\phantom{{u}^{\\left(i\\right)}\\left(0\\right)=}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[i\\right]}\\left(0\\right)\\right)+{\\overline{u}}^{\\left[i\\right]}\\left(0\\right),\\phantom{\\rule{1em}{0ex}}i=0,\\dots ,n-2,\\\\ {u}^{\\left(n-2\\right)}\\left(T\\right)={g}_{n-1}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ \\phantom{{u}^{\\left(n-2\\right)}\\left(T\\right)=}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[n-2\\right]}\\left(T\\right)\\right)+{\\overline{u}}^{\\left[n-2\\right]}\\left(T\\right).\\end{array}$\n(2.19)\n\nLemma 2.1 For any fixed $u\\in E$, define $\\mathcal{L}\\left(\\cdot ;u\\right):R\\to R$ by\n\n$\\mathcal{L}\\left(x;u\\right)={\\int }_{0}^{T}{\\phi }^{-1}\\left(x-{\\int }_{0}^{\\tau }F\\left(s,u\\left(\\cdot \\right)\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\right)\\phantom{\\rule{0.2em}{0ex}}d\\tau +{g}_{u},$\n(2.20)\n\nwhere\n\n$\\begin{array}{rl}{g}_{u}=& {\\overline{u}}^{\\left[n-2\\right]}\\left(0\\right)+{g}_{n-2}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[n-2\\right]}\\left(0\\right)\\right)\\\\ -{\\overline{u}}^{\\left[n-2\\right]}\\left(T\\right)-{g}_{n-1}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[n-2\\right]}\\left(T\\right)\\right).\\end{array}$\n(2.21)\n\nThen the equation\n\n$\\mathcal{L}\\left(\\cdot ;u\\right)=0$\n(2.22)\n\nhas a unique solution.\n\nProof We first note that $\\mathcal{L}\\left(\\cdot ;u\\right)$ is continuous and increasing on R. From (2.16), we have\n\n$\\underset{x\\to -\\mathrm{\\infty }}{lim}\\mathcal{L}\\left(x;u\\right)=-\\mathrm{\\infty }\\phantom{\\rule{1em}{0ex}}\\text{and}\\phantom{\\rule{1em}{0ex}}\\underset{x\\to \\mathrm{\\infty }}{lim}\\mathcal{L}\\left(x;u\\right)=\\mathrm{\\infty }.$\n\nThen, from the fact that $\\mathcal{L}\\left(\\cdot ;u\\right)$ is continuous and increasing on R, a standard argument shows that there exists a unique solution of (2.22). □\n\nLemma 2.2 For $u\\in E$, let\n\n${P}_{n}u\\left(t\\right)={\\phi }^{-1}\\left({x}_{u}-{\\int }_{0}^{t}F\\left(s,u\\left(\\cdot \\right)\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\right)$\n\nwith ${x}_{u}$ being the unique solution of (2.22) and $F\\left(s,u\\left(\\cdot \\right)\\right)$ be defined by (2.15). Then $u\\left(t\\right)$ is a solution of BVP (2.18) and (2.19) if and only if $u\\left(t\\right)$ is a solution of the following equation:\n\n$\\begin{array}{rl}u\\left(t\\right)=& \\frac{1}{\\left(n-2\\right)!}{\\int }_{0}^{t}{\\left(t-s\\right)}^{n-2}{P}_{n}u\\left(s\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\\\ +\\sum _{i=0}^{n-2}\\frac{{t}^{i}}{i!}\\left({\\overline{u}}^{\\left[i\\right]}\\left(0\\right)+{g}_{i}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[i\\right]}\\left(0\\right)\\right)\\right),\\phantom{\\rule{1em}{0ex}}n\\ge 2,\\end{array}$\n\nwhere we take ${0}^{0}=1$.\n\nProof This can be verified by direct computations, so we omit it. □\n\n3 Main results\n\nIn this section, we will state and prove our existence results for BVP (1.1) and (1.2).\n\nTheorem 3.1 Assume the hypotheses (H1)-(H5) hold. Then BVP (1.1) and (1.2) has at least one solution $u\\left(t\\right)$ satisfying\n\n(3.1)\n\nand\n\n(3.2)\n\nwhere C is the constant introduced in Definition 2.1.\n\nTo prove Theorem 3.1, firstly, we want to show the following theorems.\n\nTheorem 3.2 There exists at least one solution for BVP (2.18) and (2.19).\n\nProof By Lemma 2.2, for any $u\\in E$, define an operator $\\mathcal{H}:E\\to E$ by\n\n$\\begin{array}{rl}\\mathcal{H}u\\left(t\\right)=& \\frac{1}{\\left(n-2\\right)!}{\\int }_{0}^{t}{\\left(t-s\\right)}^{n-2}{P}_{n}u\\left(s\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\\\ +\\sum _{i=0}^{n-2}\\frac{{t}^{i}}{i!}\\left({\\overline{u}}^{\\left[i\\right]}\\left(0\\right)+{g}_{i}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[i\\right]}\\left(0\\right)\\right)\\right),\\phantom{\\rule{1em}{0ex}}n\\ge 2.\\end{array}$\n\nThen we can see that $u\\left(t\\right)$ is a solution of BVP (2.18) and (2.19) if and only if $u\\left(t\\right)$ is a fixed point of .\n\nLet ${\\left\\{{u}_{k}\\right\\}}_{k=1}^{\\mathrm{\\infty }}\\subseteq E$ with $\\parallel {u}_{k}-{u}_{0}\\parallel \\to 0$ as $k\\to \\mathrm{\\infty }$ in E. We want to show that $\\parallel \\mathcal{H}{u}_{k}-\\mathcal{H}{u}_{0}\\parallel \\to 0$ as $k\\to \\mathrm{\\infty }$ in E. For f is a Carathéodory function, then it is easy to see ${lim}_{k\\to \\mathrm{\\infty }}F\\left(t,{u}_{k}\\left(\\cdot \\right)\\right)=F\\left(t,{u}_{0}\\left(\\cdot \\right)\\right)$. By the Lebesgue dominated convergence theorem, ${lim}_{k\\to \\mathrm{\\infty }}{\\int }_{0}^{s}F\\left(\\tau ,{u}_{k}\\left(\\cdot \\right)\\right)\\phantom{\\rule{0.2em}{0ex}}d\\tau ={\\int }_{0}^{s}F\\left(\\tau ,{u}_{0}\\left(\\cdot \\right)\\right)\\phantom{\\rule{0.2em}{0ex}}d\\tau$. Let ${x}_{k}$ be the unique solution of $\\mathcal{L}\\left(x;{u}_{k}\\right)=0$, where is given by (2.20). In view of (2.17), there exists $r\\in {L}^{1}\\left(0,T\\right)$ such that\n\n(3.3)\n\nFrom (2.11), (2.14), (2.21) and the continuity of ${g}_{n-1}$ and ${g}_{n-2}$, we see that ${g}_{{u}_{k}}$ is bounded and ${lim}_{k\\to \\mathrm{\\infty }}{g}_{{u}_{k}}={g}_{{u}_{0}}$. Thus, $\\left\\{{x}_{k}\\right\\}$ is bounded. If $\\left\\{{x}_{k}\\right\\}$ is not convergent, then there exist two convergent subsequences $\\left\\{{x}_{{i}_{k}}\\right\\}$ and $\\left\\{{x}_{{j}_{k}}\\right\\}$ such that ${lim}_{k\\to \\mathrm{\\infty }}{x}_{{i}_{k}}={a}_{1}$, ${lim}_{k\\to \\mathrm{\\infty }}{x}_{{j}_{k}}={a}_{2}$. Then, by the continuity of ${\\phi }^{-1}$ and the Lebesgue dominated convergence theorem again, we have\n\n$0=\\underset{k\\to \\mathrm{\\infty }}{lim}\\mathcal{L}\\left({x}_{{i}_{k}};{u}_{{i}_{k}}\\right)=\\mathcal{L}\\left({a}_{1};{u}_{0}\\right)$\n\nand\n\n$0=\\underset{k\\to \\mathrm{\\infty }}{lim}\\mathcal{L}\\left({x}_{{j}_{k}};{u}_{{j}_{k}}\\right)=\\mathcal{L}\\left({a}_{2};{u}_{0}\\right),$\n\nwhich contradicts the fact that $\\mathcal{L}\\left({a}_{2};{u}_{0}\\right)=\\mathcal{L}\\left({a}_{1};{u}_{0}\\right)$. Hence, $\\left\\{{x}_{k}\\right\\}$ is convergent, say ${lim}_{k\\to \\mathrm{\\infty }}{x}_{k}={x}_{0}$. Thus, $\\mathcal{L}\\left({x}_{0};{u}_{0}\\right)=0$ and ${lim}_{k\\to \\mathrm{\\infty }}\\left({P}_{n}{u}_{k}\\right)\\left(t\\right)=\\left({P}_{n}{u}_{0}\\right)\\left(t\\right)$. As a consequence, we also have\n\n$\\underset{k\\to \\mathrm{\\infty }}{lim}{\\left(\\mathcal{H}{u}_{k}\\right)}^{\\left(j\\right)}\\left(t\\right)={\\left(\\mathcal{H}{u}_{0}\\right)}^{\\left(j\\right)}\\left(t\\right),\\phantom{\\rule{1em}{0ex}}j=0,1,\\dots ,n-1.$\n\nThus $\\parallel \\mathcal{H}{u}_{k}-\\mathcal{H}{u}_{0}\\parallel \\to 0$ as $k\\to \\mathrm{\\infty }$. This shows that $\\mathcal{H}:E\\to E$ is continuous.\n\nFrom (3.3) and the fact that ${g}_{u}$ is bounded for $u\\in E$, this means that is uniformly bounded on E, and ${\\left(\\mathcal{H}u\\right)}^{\\left(j\\right)}\\left(t\\right)$ is equicontinuous on J for $j=0,1,\\dots ,n-2$. Now, we show that ${\\left(\\mathcal{H}u\\right)}^{\\left(n-1\\right)}\\left(t\\right)$ is equicontinuous on J. From the definition of and ${P}_{n}u\\left(t\\right)$, we have\n\n${\\left(\\mathcal{H}u\\right)}^{\\left(n-1\\right)}\\left(t\\right)={\\phi }^{-1}\\left({x}_{u}-{\\int }_{0}^{t}F\\left(s,u\\left(\\cdot \\right)\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\right).$\n\nThus, the equicontinuity of ${\\left(\\mathcal{H}u\\right)}^{\\left(n-1\\right)}\\left(t\\right)$ follows from the property of absolute of integrals. By the Arzela-Ascoli theorem, we see that $\\mathcal{H}\\left(E\\right)$ is compact. From the Schauder fixed point theorem, has at least one fixed point $u\\in E$, which is a solution of BVP (2.18) and (2.19). We complete the proof. □\n\nTheorem 3.3 If u(t) is a solution of BVP (2.18) and (2.19), then $u\\left(t\\right)$ satisfies (3.1).\n\nProof We first show that ${v}^{\\left(n-2\\right)}\\left(t\\right)\\le {u}^{\\left(n-2\\right)}\\left(t\\right)$ on J. To the contrary, suppose that there exists ${t}^{\\ast }\\in J$ such that ${v}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)>{u}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)$. If ${t}^{\\ast }=0$, then ${u}^{\\left(n-2\\right)}\\left(0\\right)<{v}^{\\left(n-2\\right)}\\left(0\\right)$, from (2.19), (H5), (2.12), (2.14), and (2.8), we see that\n\n$\\begin{array}{rcl}{u}^{\\left(n-2\\right)}\\left(0\\right)& =& {g}_{n-2}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[n-2\\right]}\\left(0\\right)\\right)+{\\overline{u}}^{\\left[n-2\\right]}\\left(0\\right)\\\\ =& {g}_{n-2}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{v}^{\\left(n-2\\right)}\\left(0\\right)\\right)+{v}^{\\left(n-2\\right)}\\left(0\\right)\\\\ \\ge & {g}_{n-2}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{v}^{\\left(n-2\\right)}\\left(0\\right)\\right)+{v}^{\\left(n-2\\right)}\\left(0\\right)\\\\ \\ge & \\underset{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{min}{g}_{n-2}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ z,{W}_{n-1}z,{S}_{n-1}z,{v}^{\\left(n-2\\right)}\\left(0\\right)\\right)+{v}^{\\left(n-2\\right)}\\left(0\\right)\\\\ \\ge & {v}^{\\left(n-2\\right)}\\left(0\\right),\\end{array}$\n\nwhich is a contradiction. Similarly, if ${t}^{\\ast }=T$, then ${u}^{\\left(n-2\\right)}\\left(T\\right)<{v}^{\\left(n-2\\right)}\\left(T\\right)$, we have\n\n$\\begin{array}{rcl}{u}^{\\left(n-2\\right)}\\left(T\\right)& =& {g}_{n-1}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[n-2\\right]}\\left(T\\right)\\right)+{\\overline{u}}^{\\left[n-2\\right]}\\left(T\\right)\\\\ =& {g}_{n-1}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{v}^{\\left(n-2\\right)}\\left(T\\right)\\right)+{v}^{\\left(n-2\\right)}\\left(T\\right)\\\\ \\ge & {g}_{n-1}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{v}^{\\left(n-2\\right)}\\left(T\\right)\\right)+{v}^{\\left(n-2\\right)}\\left(T\\right)\\\\ \\ge & \\underset{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{min}{g}_{n-1}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ z,{W}_{n-1}z,{S}_{n-1}z,{v}^{\\left(n-2\\right)}\\left(T\\right)\\right)+{v}^{\\left(n-2\\right)}\\left(T\\right)\\\\ \\ge & {v}^{\\left(n-2\\right)}\\left(T\\right).\\end{array}$\n\nWe obtain a contradiction again. Thus, ${v}^{\\left(n-2\\right)}\\left(0\\right)\\le {u}^{\\left(n-2\\right)}\\left(0\\right)$ and ${v}^{\\left(n-2\\right)}\\left(T\\right)\\le {u}^{\\left(n-2\\right)}\\left(T\\right)$.\n\nNow, if ${t}^{\\ast }\\in {J}^{\\prime }$ such that ${v}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)>{u}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)$. Then ${v}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)-{u}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)>0$. Without loss of generality, we may assume that ${v}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)-{u}^{\\left(n-2\\right)}\\left({t}^{\\ast }\\right)={max}_{t\\in J}\\left\\{{v}^{\\left(n-2\\right)}\\left(t\\right)-{u}^{\\left(n-2\\right)}\\left(t\\right)\\right\\}$. Then ${v}^{\\left(n-1\\right)}\\left({t}^{\\ast }\\right)-{u}^{\\left(n-1\\right)}\\left({t}^{\\ast }\\right)=0$ and there exists a small right neighborhood Ω of ${t}^{\\ast }$ such that ${v}^{\\left(n-2\\right)}\\left(t\\right)-{u}^{\\left(n-2\\right)}\\left(t\\right)>0$ and ${v}^{\\left(n-1\\right)}\\left(t\\right)\\le {u}^{\\left(n-1\\right)}\\left(t\\right)$ for all $t\\in \\mathrm{\\Omega }$. We claim that there exists $\\stackrel{˜}{t}\\in \\mathrm{\\Omega }$ such that\n\n${\\left[\\phi \\left({v}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right)\\right)\\right]}^{\\prime }-{\\left[\\phi \\left({u}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right)\\right)\\right]}^{\\prime }\\le 0.$\n(3.4)\n\nIf this is not true, then $\\phi \\left({v}^{\\left(n-1\\right)}\\left(t\\right)\\right)-\\phi \\left({u}^{\\left(n-1\\right)}\\left(t\\right)\\right)$ is strictly increasing in Ω. Hence, ${v}^{\\left(n-1\\right)}\\left(t\\right)-{u}^{\\left(n-1\\right)}\\left(t\\right)>0$ on Ω. This contradicts the assumption that ${v}^{\\left(n-2\\right)}\\left(t\\right)-{u}^{\\left(n-2\\right)}\\left(t\\right)$ is maximized at ${t}^{\\ast }$. Thus, (3.4) holds.\n\nFrom (2.7), (2.13), and (2.14), we have ${\\overline{u}}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)={v}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)$, also, by (H4), (2.15), and (2.18), we have\n\n$\\begin{array}{r}{\\left[\\phi \\left({v}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right)\\right)\\right]}^{\\prime }-{\\left[\\phi \\left({u}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right)\\right)\\right]}^{\\prime }\\\\ \\phantom{\\rule{1em}{0ex}}\\ge -f\\left(\\stackrel{˜}{t},v\\left(\\stackrel{˜}{t}\\right),v\\left({\\alpha }_{0}\\left(\\stackrel{˜}{t}\\right)\\right),{W}_{0}v\\left(\\stackrel{˜}{t}\\right),{S}_{0}v\\left(\\stackrel{˜}{t}\\right),\\dots ,{v}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right),{v}^{\\left(n-2\\right)}\\left({\\alpha }_{n-2}\\left(\\stackrel{˜}{t}\\right)\\right),\\\\ \\phantom{\\rule{2em}{0ex}}{W}_{n-2}{v}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right),{S}_{n-2}{v}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right),-C,-C,-{W}_{n-1}C,-{S}_{n-1}C\\right)\\\\ \\phantom{\\rule{2em}{0ex}}+Bu\\left(\\stackrel{˜}{t}\\right)+\\frac{{v}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)-{u}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)}{1+{\\left({u}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)\\right)}^{2}}\\\\ \\phantom{\\rule{1em}{0ex}}\\ge \\frac{{v}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)-{u}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)}{1+{\\left({u}^{\\left(n-2\\right)}\\left(\\stackrel{˜}{t}\\right)\\right)}^{2}}>0,\\end{array}$\n\nwhich is a contradiction with (3.4). Thus, ${v}^{\\left(n-2\\right)}\\left(t\\right)\\le {u}^{\\left(n-2\\right)}\\left(t\\right)$ on J. By the same method as above, we can show that ${u}^{\\left(n-2\\right)}\\left(t\\right)\\le {w}^{\\left(n-2\\right)}\\left(t\\right)$ on J. Hence,\n\n(3.5)\n\nNext, we can see that the following inequality holds:\n\n${v}^{\\left(i\\right)}\\left(0\\right)\\le {u}^{\\left(i\\right)}\\left(0\\right)\\le {w}^{\\left(i\\right)}\\left(0\\right),\\phantom{\\rule{1em}{0ex}}i=0,1,\\dots ,n-3.$\n(3.6)\n\nIn fact, assume there exists ${i}^{\\prime }\\in \\left\\{0,1,\\dots ,n-3\\right\\}$ such that ${u}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)<{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)$, then, in view of (2.12), ${\\overline{u}}^{\\left[{i}^{\\prime }\\right]}\\left(0\\right)={v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)$. Hence, from (2.19), (H5), (2.8) and (2.12),\n\n$\\begin{array}{rcl}{u}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)& =& {g}_{{i}^{\\prime }}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{\\overline{u}}^{\\left[{i}^{\\prime }\\right]}\\left(0\\right)\\right)+{\\overline{u}}^{\\left[{i}^{\\prime }\\right]}\\left(0\\right)\\\\ =& {g}_{{i}^{\\prime }}\\left({\\overline{u}}^{\\left[0\\right]},{W}_{0}{\\overline{u}}^{\\left[0\\right]},{S}_{0}{\\overline{u}}^{\\left[0\\right]},\\dots ,{\\overline{u}}^{\\left[n-2\\right]},{W}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},{S}_{n-2}{\\overline{u}}^{\\left[n-2\\right]},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)\\right)+{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)\\\\ \\ge & {g}_{{i}^{\\prime }}\\left({v}^{\\left(0\\right)},{W}_{0}{v}^{\\left(0\\right)},{S}_{0}{v}^{\\left(0\\right)},\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ {\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left[n-1\\right]},{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)\\right)+{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)\\\\ \\ge & \\underset{{\\parallel z\\parallel }_{\\mathrm{\\infty }}\\le C}{min}{g}_{{i}^{\\prime }}\\left(v,{W}_{0}v,{S}_{0}v,\\dots ,{v}^{\\left(n-2\\right)},{W}_{n-2}{v}^{\\left(n-2\\right)},{S}_{n-2}{v}^{\\left(n-2\\right)},\\\\ z,{W}_{n-1}z,{S}_{n-1}z,{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)\\right)+{v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right)\\\\ \\ge & {v}^{\\left({i}^{\\prime }\\right)}\\left(0\\right).\\end{array}$\n\nThis is a contradiction. Thus, ${v}^{\\left(i\\right)}\\left(0\\right)\\le {u}^{\\left(i\\right)}\\left(0\\right)$ for $i=0,1,\\dots ,n-3$. By a similar argument, we see that ${u}^{\\left(i\\right)}\\left(0\\right)\\le {w}^{\\left(i\\right)}\\left(0\\right)$ for $i=0,1,\\dots ,n-3$. Then (3.6) holds.\n\nFinally, from (3.5) and integral inequality, we have\n\n${u}^{\\left(n-3\\right)}\\left(t\\right)-{v}^{\\left(n-3\\right)}\\left(0\\right)\\le {u}^{\\left(n-3\\right)}\\left(t\\right)-{u}^{\\left(n-3\\right)}\\left(0\\right)\\le {w}^{\\left(n-3\\right)}\\left(t\\right)-{w}^{\\left(n-3\\right)}\\left(0\\right),$\n\nand using (3.6), we obtain ${v}^{\\left(n-3\\right)}\\left(t\\right)\\le {u}^{\\left(n-3\\right)}\\left(t\\right)\\le {w}^{\\left(n-3\\right)}\\left(t\\right)$. Similarly, we can show that $u\\left(t\\right)$ satisfies (3.1). The proof is completed. □\n\nTheorem 3.4 If $u\\left(t\\right)$ is a solution of BVP (2.18) and (2.19), then ${u}^{\\left(n-1\\right)}\\left(t\\right)$ satisfies (3.2).\n\nProof From Theorem 3.3, we know that $u\\left(t\\right)$ satisfies (3.1). If (3.2) does not hold, then there exists ${t}_{0}\\in J$ such that ${u}^{\\left(n-1\\right)}\\left({t}_{0}\\right)>C$ or ${u}^{\\left(n-1\\right)}\\left({t}_{0}\\right)<-C$. By the mean value theorem, there exists $\\stackrel{˜}{t}\\in J$ such that $T{u}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right)={u}^{\\left(n-2\\right)}\\left(T\\right)-{u}^{\\left(n-2\\right)}\\left(0\\right)$. Then, from (2.2), (2.3), and (2.17), we see that\n\n$-C<-\\xi \\le \\frac{{v}^{\\left(n-2\\right)}\\left(T\\right)-{w}^{\\left(n-2\\right)}\\left(0\\right)}{T}\\le {u}^{\\left(n-1\\right)}\\left(\\stackrel{˜}{t}\\right)\\le \\frac{{w}^{\\left(n-2\\right)}\\left(T\\right)-{v}^{\\left(n-2\\right)}\\left(0\\right)}{T}\\le \\xi\n\nIf ${u}^{\\left(n-1\\right)}\\left({t}_{0}\\right)>C$ there exist ${t}_{1},{t}_{2}\\in J$ such that ${u}^{\\left(n-1\\right)}\\left({t}_{1}\\right)=\\xi$, ${u}^{\\left(n-1\\right)}\\left({t}_{2}\\right)=C$ and\n\n(3.7)\n\nwhere $I=\\left[{t}_{1},{t}_{2}\\right]$ or $I=\\left[{t}_{2},{t}_{1}\\right]$. In the following, we only consider the case $I=\\left[{t}_{1},{t}_{2}\\right]$, since the other case can be treated similarly. From (2.14) and (3.7), ${\\stackrel{ˆ}{u}}^{\\left(n-1\\right)}\\left(t\\right)={u}^{\\left(n-1\\right)}\\left(t\\right)$ on I, and in view of (2.11) and (3.1), we have ${\\overline{u}}^{\\left[i\\right]}\\left(t\\right)={u}^{\\left(i\\right)}\\left(t\\right)$ for $t\\in J$ and $i=0,1,\\dots ,n-2$. Thus, from (2.15),\n\n$\\begin{array}{rl}F\\left(t,u\\left(\\cdot \\right)\\right)=& f\\left(t,u\\left(t\\right),u\\left({\\alpha }_{0}\\left(t\\right)\\right),{W}_{0}u\\left(t\\right),{S}_{0}u\\left(t\\right),\\dots ,{W}_{n-2}{u}^{\\left(n-2\\right)}\\left(t\\right),{S}_{n-2}{u}^{\\left(n-2\\right)}\\left(t\\right),\\\\ {u}^{\\left(n-1\\right)}\\left(t\\right),{\\stackrel{ˆ}{u}}^{\\left(n-1\\right)}\\left({\\alpha }_{n-1}\\left(t\\right)\\right),{W}_{n-1}{\\stackrel{ˆ}{u}}^{\\left(n-1\\right)}\\left(t\\right),{S}_{n-1}{\\stackrel{ˆ}{u}}^{\\left(n-1\\right)}\\left(t\\right)\\right),\\phantom{\\rule{1em}{0ex}}t\\in I.\\end{array}$\n\nThen, by a change of variables and from (2.4) and (2.18), we can obtain\n\n$\\begin{array}{r}{\\int }_{\\varphi \\left(\\xi \\right)}^{\\varphi \\left(C\\right)}\\frac{{\\left({\\varphi }^{-1}\\left(x\\right)\\right)}^{\\left(p-1\\right)/p}}{\\psi \\left(|{\\varphi }^{-1}\\left(x\\right)|\\right)}\\phantom{\\rule{0.2em}{0ex}}dx\\\\ \\phantom{\\rule{1em}{0ex}}={\\int }_{\\varphi \\left({u}^{\\left(n-1\\right)}\\left({t}_{1}\\right)\\right)}^{\\varphi \\left({u}^{\\left(n-1\\right)}\\left({t}_{2}\\right)\\right)}\\frac{{\\left({\\varphi }^{-1}\\left(x\\right)\\right)}^{\\left(p-1\\right)/p}}{\\psi \\left(|{\\varphi }^{-1}\\left(x\\right)|\\right)}\\phantom{\\rule{0.2em}{0ex}}dx\\\\ \\phantom{\\rule{1em}{0ex}}={\\int }_{{t}_{1}}^{{t}_{2}}\\frac{{\\left[\\varphi \\left({u}^{\\left(n-1\\right)}\\left(s\\right)\\right)\\right]}^{\\prime }}{\\psi \\left(|{u}^{\\left(n-1\\right)}\\left(s\\right)|\\right)}{\\left({u}^{\\left(n-1\\right)}\\left(s\\right)\\right)}^{\\left(p-1\\right)/p}\\phantom{\\rule{0.2em}{0ex}}ds\\\\ \\phantom{\\rule{1em}{0ex}}={\\int }_{{t}_{1}}^{{t}_{2}}\\frac{{\\left(\\phi \\left({u}^{\\left(n-1\\right)}\\left(s\\right)\\right)\\right)}^{\\prime }}{\\psi \\left(|{u}^{\\left(n-1\\right)}\\left(s\\right)|\\right)}{\\left({u}^{\\left(n-1\\right)}\\left(s\\right)\\right)}^{\\left(p-1\\right)/p}\\phantom{\\rule{0.2em}{0ex}}ds\\\\ \\phantom{\\rule{1em}{0ex}}={\\int }_{{t}_{1}}^{{t}_{2}}\\frac{-f\\left(s,u\\left(s\\right),\\dots ,{S}_{n-1}{u}^{\\left(n-1\\right)}\\left(s\\right)\\right)}{\\psi \\left(|{u}^{\\left(n-1\\right)}\\left(s\\right)|\\right)}{\\left({u}^{\\left(n-1\\right)}\\left(s\\right)\\right)}^{\\left(p-1\\right)/p}\\phantom{\\rule{0.2em}{0ex}}ds\\\\ \\phantom{\\rule{1em}{0ex}}\\le {\\int }_{{t}_{1}}^{{t}_{2}}\\vartheta \\left(s\\right){\\left({u}^{\\left(n-1\\right)}\\left(s\\right)\\right)}^{\\left(p-1\\right)/p}\\phantom{\\rule{0.2em}{0ex}}ds.\\end{array}$\n\nHence, the Hölder inequality implies\n\n$\\begin{array}{rcl}{\\int }_{\\varphi \\left(\\xi \\right)}^{\\varphi \\left(C\\right)}\\frac{{\\left({\\varphi }^{-1}\\left(x\\right)\\right)}^{\\left(p-1\\right)/p}}{\\psi \\left(|{\\varphi }^{-1}\\left(x\\right)|\\right)}\\phantom{\\rule{0.2em}{0ex}}dx& \\le & {\\parallel \\vartheta \\parallel }_{p}{\\left({\\int }_{{t}_{1}}^{{t}_{2}}{u}^{\\left(n-1\\right)}\\left(s\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\right)}^{\\left(p-1\\right)/p}\\\\ =& {\\parallel \\vartheta \\parallel }_{p}{\\left({u}^{\\left(n-2\\right)}\\left({t}_{2}\\right)-{u}^{\\left(n-2\\right)}\\left({t}_{1}\\right)\\right)}^{\\left(p-1\\right)/p}\\\\ \\le & {\\parallel \\vartheta \\parallel }_{p}{\\left(\\underset{t\\in J}{max}{w}^{\\left(n-2\\right)}\\left(t\\right)-\\underset{t\\in J}{min}{v}^{\\left(n-2\\right)}\\left(t\\right)\\right)}^{\\left(p-1\\right)/p}\\\\ =& {\\parallel \\vartheta \\parallel }_{p}{\\zeta }^{\\left(p-1\\right)/p},\\end{array}$\n\nwhere ζ is defined by (2.6). But this contradicts with (2.5). Therefore, ${u}^{\\left(n-1\\right)}\\left(t\\right)\\le C$. If ${u}^{\\left(n-1\\right)}\\left(t\\right)<-C$, by a similar argument as above, we can show that (3.2) holds. Hence the proof of the theorem is completed. □\n\nNow we are in a position to prove Theorem 3.1.\n\nProof of Theorem 3.1 Note that any solution of BVP (2.18) and (2.19) satisfying (3.1), (3.2) is a solution of BVP (1.1) and (1.2). The conclusion readily follows from Theorem 3.2-3.4. □\n\n4 Example\n\nExample 4.1 Consider the boundary value problem consisting of the equation\n\n$\\begin{array}{r}{\\left({\\left({u}^{″}\\left(t\\right)\\right)}^{3}\\right)}^{\\prime }+\\frac{1}{15}{t}^{-\\frac{1}{2}}\\left(1+u\\left(t\\right)\\right)-\\frac{t}{500}{\\left[{t}^{3}-{\\int }_{0}^{t}tsu\\left(s\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\right]}^{3}\\\\ \\phantom{\\rule{1em}{0ex}}-\\frac{1}{300}{t}^{2}{\\left[t-u\\left({t}^{3}\\right)\\right]}^{3}+4{t}^{-\\frac{1}{2}}{\\left[{\\int }_{0}^{1}\\left(t+s\\right){u}^{″}\\left(s\\right)\\phantom{\\rule{0.2em}{0ex}}ds\\right]}^{3}=0,\\phantom{\\rule{1em}{0ex}}t\\in \\left(0,1\\right),\\end{array}$\n(4.1)\n\nand the boundary condition\n\n$\\left\\{\\begin{array}{l}{min}_{s\\in \\left[0,1\\right]}{u}^{\\prime }\\left(s\\right)-6u\\left(0\\right)+\\frac{1}{2}=0,\\\\ {\\int }_{0}^{t}\\left(ts\\right)u\\left(s\\right)\\phantom{\\rule{0.2em}{0ex}}ds-4{e}^{2}{u}^{\\prime }\\left(0\\right)=0,\\\\ {u}^{″}\\left(\\frac{1}{4}\\right)-2{e}^{2}{u}^{\\prime }\\left(1\\right)=0.\\end{array}$\n(4.2)\n\nBVP (4.1) and (4.2) has at least one solution $u\\left(t\\right)$ satisfying\n\n$-{\\left(t+1\\right)}^{2}\\le u\\left(t\\right)\\le {\\left(t+1\\right)}^{2},$\n(4.3)\n$-2\\left(t+1\\right)\\le {u}^{\\prime }\\left(t\\right)\\le 2\\left(t+1\\right),$\n(4.4)\n\nand\n\n$-12{e}^{2}\\le {u}^{″}\\left(t\\right)\\le 12{e}^{2},$\n(4.5)\n\nfor $t\\in \\left[0,1\\right]$.\n\nIn fact, if we let $n=3$, $\\varphi \\left(x\\right)={x}^{3}$,\n\n$f\\left(t,{x}_{0},{x}_{1},\\dots ,{x}_{11}\\right)=\\frac{1}{15}{t}^{-\\frac{1}{2}}\\left(1+{x}_{0}\\right)-\\frac{t}{500}{\\left[{t}^{3}-{x}_{2}\\right]}^{3}-\\frac{1}{300}{t}^{2}{\\left[t-{x}_{5}\\right]}^{3}+4{t}^{-\\frac{1}{2}}{x}_{11}^{3},$\n\nfor $\\left({x}_{0},{x}_{1},\\dots ,{x}_{11}\\right)\\in {R}^{4n}$, and ${\\beta }_{0}\\left(t\\right)=t$, ${k}_{0}\\left(t,s\\right)=ts$, ${\\alpha }_{1}\\left(t\\right)={t}^{3}$, ${h}_{2}\\left(t,s\\right)=t+s$, $H=max\\left\\{{h}_{2}\\left(t,s\\right):\\left(t,s\\right)\\in \\left[0,1\\right]×\\left[0,1\\right]\\right\\}=2$, and\n\n$\\begin{array}{c}{g}_{0}\\left({y}_{0},\\dots ,{y}_{8},z\\right)=\\underset{s\\in \\left[0,1\\right]}{min}{y}_{3}\\left(s\\right)-6z+\\frac{1}{2},\\hfill \\\\ {g}_{1}\\left({y}_{0},\\dots ,{y}_{8},z\\right)={y}_{1}\\left(t\\right)-4{e}^{2}z,\\hfill \\\\ {g}_{2}\\left({y}_{0},\\dots ,{y}_{8},z\\right)={y}_{6}\\left(\\frac{1}{4}\\right)-2{e}^{2}z,\\hfill \\end{array}$\n\nfor $\\left({y}_{0},\\dots ,{y}_{8},z\\right)\\in {\\left(C\\left[0,1\\right]\\right)}^{9}×R$, then it is easy to see that BVP (4.1) and (4.2) is of the form of BVP (1.1) and (1.2). Clearly, (H1), (H2) and (H5) hold.\n\nLet $v\\left(t\\right)=-{\\left(t+1\\right)}^{2}$ and $w\\left(t\\right)={\\left(t+1\\right)}^{2}$. Obviously, $v\\left(t\\right)$, $w\\left(t\\right)$ satisfy (2.1). Define $\\vartheta \\left(t\\right)=\\frac{1}{4}{t}^{-1/2}$ and $\\psi \\left(x\\right)=9+16{x}^{3}$. Then $\\vartheta \\in {L}^{1}\\left(0,1\\right)$ with ${\\parallel \\vartheta \\parallel }_{1}=\\frac{1}{2}$, $\\psi \\left(x\\right)>0$ on $\\left[0,\\mathrm{\\infty }\\right)$, and\n\n$|f\\left(t,{x}_{0},{x}_{1},\\dots ,{x}_{11}\\right)|\\le \\frac{1}{4}{t}^{-\\frac{1}{2}}\\left(9+16{|{x}_{11}|}^{3}\\right)=\\vartheta \\left(t\\right)\\psi \\left(|{x}_{11}|\\right),$\n\non $\\left(0,1\\right)×{\\mathbb{D}}_{v}^{w}×{R}^{4}$, where ${\\mathbb{D}}_{v}^{w}$ is given by Definition 2.1. Thus (2.4) holds. For ξ defined by (2.2), we have $\\xi =6$ with $C=12{e}^{2}$ and $p=1$, and it is easy to check that (2.3) holds. Through computations, we can obtain (2.5). Hence, f satisfies the Nagumo condition with respect to v, w i.e. (H3) holds. Moreover, a simple computation shows that $v\\left(t\\right)$ and $w\\left(t\\right)$ satisfy (2.7)-(2.10). Hence (H2) holds. Finally, obviously (H4) holds.\n\nTherefore, by Theorem 3.1, BVP (4.1) and (4.2) has at least one solution $u\\left(t\\right)$ satisfying (4.3)-(4.5).\n\n5 Conclusions\n\nIn this paper, we obtain a new existence result for higher-order integro-differential BVPs with ϕ-Laplacian like operator and functional boundary conditions. Firstly, we state some preliminaries and lemmas such as the definitions if we have the Nagumo condition and a pair of coupled lower and upper solutions. Secondly, under conditions (H1)-(H5), we get the main result (Theorem 3.1), and we prove the result in three steps (Theorems 3.2-3.4) which mainly use lower and upper solutions and the Schauder fixed point theorem. Finally, an example is given to illustrate our main result.\n\nReferences\n\n1. 1.\n\nLakshmikantham V: Some problems in integro-differential equations of Volterra type. J. Integral Equ. 1985, 10: 137-146.\n\n2. 2.\n\nGuo D, Lakshmikantham V, Liu XZ (Eds): Nonlinear Integral Equations in Abstract Spaces. Kluwer Academic, Dordrecht; 1996.\n\n3. 3.\n\nShi GL, Meng XR: Monotone iterative for fourth-order p -Laplacian boundary value problems with impulsive effects. Appl. Math. Comput. 2006, 181: 1243-1248. 10.1016/j.amc.2006.02.024\n\n4. 4.\n\nCabada A, Otero-Espinar V: Existence and comparison results for difference ϕ -Laplacian boundary value problems with lower and upper solutions in reverse order. J. Math. Anal. Appl. 2002, 267: 501-521. 10.1006/jmaa.2001.7783\n\n5. 5.\n\nMisawa M: A Hölder estimate for nonlinear parabolic systems of p -Laplacian type. J. Differ. Equ. 2013, 254: 847-878. 10.1016/j.jde.2012.10.001\n\n6. 6.\n\nAgarwal RP: On fourth order boundary value problems arising in beam analysis. Differ. Integral Equ. 1989, 2: 91-110.\n\n7. 7.\n\nWei Z: Existence of positive solutions for n th-order p -Laplacian singular sublinear boundary value problems. Appl. Math. Lett. 2014, 36: 25-30.\n\n8. 8.\n\nDavis JD, Henderson J, Wong PJY: General Lidstone problems: multiplicity and symmetry of solutions. J. Math. Anal. Appl. 2000, 251: 527-548. 10.1006/jmaa.2000.7028\n\n9. 9.\n\nFialho JF, Minhós F: Higher order functional boundary value problems without monotone assumptions. Bound. Value Probl. 2013., 2013: Article ID 81\n\n10. 10.\n\nAgarwal RP: Boundary Value Problems for Higher Order Differential Equations. World Scientific, Singapore; 1986.\n\n11. 11.\n\nLee EK, Lee YH: Multiple positive solutions of singular two points boundary value problems for second order impulsive differential equations. Appl. Math. Comput. 2004, 158: 745-759. 10.1016/j.amc.2003.10.013\n\n12. 12.\n\nEloe PW, Ahmad B: Positive solutions of a nonlinear n th order boundary value problem with nonlocal conditions. Appl. Math. Lett. 2005, 18: 521-527. 10.1016/j.aml.2004.05.009\n\n13. 13.\n\nMawhin J: Homotopy and nonlinear boundary value problems involving singular ϕ -Laplacians. J. Fixed Point Theory Appl. 2013, 13: 25-35. 10.1007/s11784-013-0112-9\n\n14. 14.\n\nGraef JR, Yang B: Positive solutions to a multi-point higher order boundary value problems. J. Math. Anal. Appl. 2006, 316: 409-421. 10.1016/j.jmaa.2005.04.049\n\n15. 15.\n\nGraef JR, Kong L, Kong Q: Higher order multi-point boundary value problems. Math. Nachr. 2011, 284: 39-52. 10.1002/mana.200710179\n\n16. 16.\n\nWang DB, Guan W: Multiple positive solutions for third-order p -Laplacian functional dynamic equations on time scales. Adv. Differ. Equ. 2014., 2014: Article ID 145\n\n17. 17.\n\nGraef JR, Kong L, Minhós FM: Higher order boundary value problems with ϕ -Laplacian and functional boundary conditions. Comput. Math. Appl. 2011, 61: 236-249. 10.1016/j.camwa.2010.10.044\n\n18. 18.\n\nGraef JR, Kong L: Existence of solutions for nonlinear boundary value problems. Commun. Appl. Nonlinear Anal. 2007, 14: 39-60.\n\n19. 19.\n\nBai DY: A global result for discrete ϕ -Laplacian eigenvalue problems. Adv. Differ. Equ. 2013., 2013: Article ID 264\n\n20. 20.\n\nWang W, Yang X, Shen J: Boundary value problems involving upper and lower solutions in reverse order. J. Comput. Appl. Math. 2009, 230: 1-7. 10.1016/j.cam.2008.10.040\n\n21. 21.\n\nEhme J, Eloe PW, Henderson J: Upper and lower solution methods for fully nonlinear boundary value problems. J. Differ. Equ. 2002, 180: 51-64. 10.1006/jdeq.2001.4056\n\n22. 22.\n\nFranco D, Regan DO, Perán J: Fourth-order problems with nonlinear boundary conditions. J. Comput. Appl. Math. 2005, 174: 315-327. 10.1016/j.cam.2004.04.013\n\n23. 23.\n\nCabada A, Pouso R, Minhós F: Extremal solutions to fourth-order functional boundary value problems including multipoint conditions. Nonlinear Anal. 2009, 10: 2157-2170. 10.1016/j.nonrwa.2008.03.026\n\n24. 24.\n\nHenderson J: Solutions of multipoint boundary value problems for second order equations. Dyn. Syst. Appl. 2006, 15: 111-117.\n\n25. 25.\n\nCabada A, Pouso RL:Existence results for the problem${\\left(\\varphi \\left({u}^{\\prime }\\right)\\right)}^{\\prime }=f\\left(t,u,{u}^{\\prime }\\right)$ with nonlinear boundary conditions. Nonlinear Anal. 1999, 35: 221-231. 10.1016/S0362-546X(98)00009-1\n\n26. 26.\n\nKong L, Kong Q: Second-order boundary value problems with nonhomogeneous boundary conditions (I). Math. Nachr. 2005, 278: 173-193. 10.1002/mana.200410234\n\nAcknowledgements\n\nThis work was supported by Natural Science Foundation of China Grant No. 11461021, Natural Science Foundation of Guangxi Grant No. 2014GXNSFDA118002, Scientific Research Foundation of Guangxi Education Department No. ZD2014131, No. 2013YB236, the open fund of Guangxi Key Laboratory of Hybrid Computation and IC Design Analysis No. HCIC201305 and the Scientific Research Project of Hezhou University No. 2014ZC13. The authors wish to thank the anonymous reviewers for their helpful comments and suggestions.\n\nAuthor information\n\nAuthors\n\nCorresponding author\n\nCorrespondence to Liang Lu.\n\nCompeting interests\n\nThe authors declare that they have no competing interests.\n\nAuthors’ contributions\n\nAll authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.\n\nRights and permissions",
null,
""
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null,
"https://advancesindifferenceequations.springeropen.com/track/article/10.1186/1687-1847-2014-285",
null
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https://calculator-converter.com/inches-to-miles.htm | [
"",
null,
"",
null,
"",
null,
"# Inches to Miles\n\nThis calculator provides conversion of inches to miles and backwards (mi to in).\n\nEnter inches or miles for conversion:\n\nSelect conversion type:\n\nRounding options:\n\n## Conversion Table\n\n inch to miles Conversion Table: in to mi 1.0 = 0.000015782 2.0 = 0.000031566 3.0 = 0.000047348 4.0 = 0.000063131 5.0 = 0.000078914 6.0 = 0.000094697 7.0 = 0.000110480 8.0 = 0.000126263 9.0 = 0.000142045 inch to miles 10 = 0.00015783 20 = 0.00031566 30 = 0.00047348 40 = 0.00063131 50 = 0.00078914 100 = 0.0015783 500 = 0.0078914 1000 = 0.015783 5000 = 0.078914\n miles to inches Conversion Table: mi to in 1.0 = 63360 2.0 = 126720 3.0 = 190080 4.0 = 253440 5.0 = 316800 6.0 = 380160 7.0 = 443520 8.0 = 506880 9.0 = 570240 miles to inches 10 = 633600 20 = 1267200 30 = 1900800 40 = 2534400 50 = 3168000 100 = 6336000 500 = 31680000 1000 = 63360000 5000 = 316800000\n\nThe inch ( abbreviation: in or ″) is a unit of length in several different systems, including Imperial units and US customary units. There are 12 inches (in) in a foot (ft) and 36 inches in a yard. 1 inch (in) = 0.0000157828283 mile (mi) = 0.0833333333 feet (ft) = 0.0277777778 yard (yd) = 0.0254 meters (m) = 2.54 centimeters (cm) = 0.0000254 kilometers (km) = 0.0000137149028 nautical mile (nmi)\n\n## Length Conversion\n\nInches to meters (in to m)\nMeters to inches (m to in)\nInches to micrometers (in to micrometer, µm)\nMicrometers to inches (µm to in)\nMicrons to inches (µm to in)\nInches to microns (in to micron)\nIn to mm (inches to millimeters)\nMm to in (millimeters to inches)\nFeet to meters (ft to m)\nMeters to feet (m to ft)\nYard to meter (yd to m)\nMeter to yard (m to yd)\nMiles to kilometers (mi to km)\nKilometers to miles (km to mi)\nInches to feet (in to ft)\nFeet to Inches (ft to in)\nFeet to centimeters (ft to cm)\nCentimeters to feet (cm to ft)\nCentimeters to inches (cm to in)\nInches to centimeters (in to cm)\nMeters to centimeters (m to cm)\nCentimeters to meters (cm to m)\nCentimeters to millimeters (cm to mm)\nMillimeters to centimeters (mm to cm)\nDecimeters to millimeters (dm to mm)\nMillimeters to decimeters (mm to dm)\nDecimeters to centimeters (dm to cm)\nCentimeters to decimeters (cm to dm)\nDecimeters to meters (dm to m)\nMeters to decimeters (m to dm)\nMeters to kilometers (m to km)\nKilometers to meters (km to m)\nFeet to kilometers (ft to km)\nKilometers to feet (km to ft)\nKilometers to centimeters (km to cm)\nCentimeters to kilometers (cm to km)\nKilometers to millimeters (km to mm)\nMillimeters to kilometers (mm to km)\nKilometers to nautical miles (km to nmi)\nNautical miles to kilometers (nmi to km)\nNautical miles to miles (nmi to mi)\nMiles to nautical miles (mi to nmi)\nMiles to feet (mi to ft)\nFeet to miles (ft to mi)\nMiles to meters (mi to m)\nMeters to miles (m to mi)\nMeters to millimeters (m to mm)\nMillimeters to meters (mm to m)\nMeters to nanometers (m to nm)\nNanometers to meters (nm to m)\nMillimeters to nanometers (mm to nm)\nNanometers to millimeters (nm to mm)\nDecimeters to kilometers (dm to km)\nKilometers to decimeters (km to dm)\nMiles to inches (mi to in)\nInches to miles (in to mi)\nKilometers to yards (km to yd)\nYards to kilometers (yd to km)\nMillimeters to feet (mm to ft)\nFeet to millimeters (ft to mm)\nCentimeters to nanometers (cm to nm)\nNanometers to centimeters (nm to cm)\nMicrometers to meters (µm to m)\nMeters to micrometers (m to µm)\nMillimeters to microns (mm to µm)\nMicrons to millimeters (µm to mm)\nMillimeters to micrometers (mm to µm)\nMicrometers to millimeters (µm to mm)\nMicrometers to nanometers (µm to nm)\nNanometers to micrometers (nm to µm)\nFathoms to feet (ftm to ft)\nFeet to fathoms (ft to ftm)\nFathoms to meters (ftm to m)\nMeters to fathoms (m to ftm)"
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"https://calculator-converter.com/pics/calculator-converter.gif",
null,
"https://calculator-converter.com/pics/calculator-converter_sm.gif",
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https://vc.uni-bamberg.de/mod/page/view.php?id=627032 | [
"Lukas Soenning\nLehrstuhl für englische Sprachwissenschaft einschließlich Sprachgeschichte\nResearch seminar\nJanuary 8th, 2015\n\nThis short tutorial shows how to construct useful plots in R using the lattice package (Sarkar 2014).\nIt is an online appendix to the presentation Graphical methods for data analysis.\n\n## Contents\n\n1. Plotting a single distribution\n1. Histogram\n2. Density plot\n3. Boxplot\n4. Quantile plot\n\n2. Comparing two distributions\n1. Histogram\n2. Density plot\n3. Boxplot\n\n3. Scatterplots\n1. Regression lines and smoothers\n2. Reversing the x-axis\n3. Avoiding overplotting: Jittering and transparency\n4. Superposition\n\n4. Multipanel conditioning\n1. Conditioning on a categorical variable\n2. Conditioning on a quantitative variable\n3. Conditioning on two variables\n\n5. Multivariate plots\n1. Scatterplot matrix\n2. Parallel coordinates plot\n\n6. Categorical variables\n1. Spine plot\n2. Mosaic plot\n\n#### Preparations: R\n\nInstall the lattice package.\n\ninstall.packages(\"lattice\")\n\n## Error in contrib.url(repos, “source”): trying to use CRAN without setting a mirror\n\nlibrary(lattice)\n\nThe following code changes the settings to produce black and white output (and a transparent background for the strips of the panels).\n\nbw.theme <- canonical.theme(color = FALSE)\nbw.theme$strip.background$col <- \"transparent\"\nlattice.options(default.theme = bw.theme) \n\n#### Preparations: Data\n\nCreate an object named malta with the data.\n\nmalta <- read.csv(file=\"C:/Users/ba4rh5/Desktop/malta.csv\")\nitems <- read.csv(file=\"C:/Users/ba4rh5/Desktop/items.csv\")\n\n## Quantitative variables\n\n### 1 Plotting a single distribution\n\n#### 1.1 Histogram\n\nHistograms can be drawn with the function histogram(). A histogram of the distribution of the mean questionnaire scores:\n\nhistogram(malta$Mean)",
null,
"You can control the number of bins with the nint= argument: histogram(malta$Mean, nint=30)",
null,
"#### 1.2 Density plot\n\nDensity plots avoid the arbitrary choice of intervals for binning the data. They are essentially smoothed histograms.\n\ndensityplot(malta$Mean)",
null,
"The smoothness of the density plot can be controlled with the bandwidth parameter, which can be set with the argument bw=. densityplot(malta$Mean, bw=.07)",
null,
"The following arguments suppress the data points and add a reference line at zero:\n\ndensityplot(malta$Mean, bw=.07, plot.points=FALSE, ref=TRUE)",
null,
"#### 1.3 Box plot Box plots can be drawn with the function bwplot(). bwplot(malta$Mean)",
null,
"#### 1.4 Quantile plot\n\nQuantile plots can be drawn with the function qqmath(). The additional argument distribution=“qunif” is necessary to draw the type of quantile plot shown in the presentation.\n\nqqmath(items$Mean, distribution=\"qunif\")",
null,
"#### 1.5 Dot plot Dot plots are a very flexible tool for data visualization. In their most basic form, they plot data points with labels. Here we can use them to plot the distribution of the questionnaire items including their labels. The additional function reorder() sorts the items in increasing order. dotplot(reorder(BrE, Mean) ~ Mean, data=items)",
null,
"### 2 Comparing distributions #### 2.1 Histogram We can use histograms to compare two groups. Lattice makes it very easy to compare groups with different panels. The “ | ” sign indicates the variable that is used to split the data into different panels. histogram(~ Age | Gender, data=malta)",
null,
"#### 2.2 Density plot The advantage of density plots is that we can plot the two distributions into the same panel, which makes direct comparisons easier. The groups= argument specifies the groups. densityplot(~Mean, groups=Gender, data=malta, plot.points=FALSE, ref=TRUE)",
null,
"#### 2.3 Box plot The box plot is a very useful method for comparing two or more distributions. bwplot(Mean ~ NL.father, data=malta)",
null,
"The variable to the left of the “ ~ ” sign (tilde) is plotted on the y-axis. Swap the two variables to draw a horizontal plot. bwplot(NL.father ~ Mean, data=malta)",
null,
"### 3 Scatterplots #### 3.1 Regression lines and smoothers Scatterplots can be drawn with the function xyplot(). The variable to the left of the “ ~ ” sign (tilde) is plotted on the y-axis. xyplot(Mean ~ Age, data=malta)",
null,
"Adding a regression line is simple: You can use the type= argument to specify which elements you want in your scatterplot: “p” is for points, “r” adds a straight regression line: xyplot(Mean ~ Age, data=malta, type=c(\"p\", \"r\"))",
null,
"Using only “r” omits the points: xyplot(Mean ~ Age, data=malta, type=c(\"r\"))",
null,
"“smooth” adds a scatterplot smoother: xyplot(Mean ~ Age, data=malta, type=c(\"p\", \"smooth\"))",
null,
"#### 3.2 Reversing the x-axis You can reverse the x-axis with the following additional argument xlim=: xyplot(Mean ~ Age, data=malta, type=c(\"p\", \"r\"), xlim=rev(extendrange(malta$Age)))",
null,
"#### 3.3 Avoiding overplotting: Jittering and transparency\n\nIf quantitative variables take on only a few different values, overplotting is an issue. It is difficult to judge where most observations concentrate.\n\nxyplot(Preposition ~ Idiom, data=malta)",
null,
"Jittering adds a bit of random noise to the variables:\n\nxyplot(jitter(Preposition) ~ jitter(Idiom), data=malta)",
null,
"For more random noise:\n\nxyplot(jitter(Preposition, 3) ~ jitter(Idiom, 3), data=malta)",
null,
"Another solution is to use transparency. The argument alpha= specifies the degree of transparency. For example, .5 means that 1/.5 = 2 points add up to black; .2 means that 1/.2 = 5 points add up to black. Jittering and transparency can be combined:\n\nxyplot(jitter(Preposition, 3) ~ jitter(Idiom, 3), alpha=.2, data=malta)",
null,
"#### 3.4 Superposition\n\nWe can compare two groups by plotting them into the same panel. The argument groups= specifies which groups to compare:\n\nxyplot(Mean ~ Age, groups=Gender, data=malta, type=c(\"p\", \"r\"))",
null,
"The argument auto.key=TRUE adds a legend. Unfortunately only the points are shown (it is possible to add the lines to the legend, but this is a bit more complicated).\n\nxyplot(Mean ~ Age, groups=Gender, data=malta, type=c(\"p\", \"r\"), auto.key=TRUE)",
null,
"Of course, you can plot only the regression lines by omitting “p” from the type= argument:\n\nxyplot(Mean ~ Age, groups=Gender, data=malta, type=c(\"r\"), auto.key=TRUE)",
null,
"### 4 Multipanel conditioning\n\n#### 4.1 Conditioning on categorical variables\n\nMultipanel conditioning is a very powerful tool for data exploration, especially for uncovering and understanding interactions between variables. When conditioning on categorical variables, the groups (categories) are plotted into different panels. Lattice automatically ensures equal scaling in all panels. Here is a box plot showing score by native language of the father:\n\nbwplot(Mean ~ NL.father, data=malta)",
null,
"For a comparison of male and female subjects we can condition on gender. The “ | ” sign introduces a conditioning variable:\n\nbwplot(Mean ~ NL.father | Gender, data=malta)",
null,
"Sometimes a rearrangment of the independent variables can be more revealing. We should be flexible and try different combinations:\n\nbwplot(Mean ~ Gender | NL.father, data=malta)",
null,
"Here is a scatterplot (again) showing SCORE by AGE:\n\nxyplot(Mean ~ Age, data=malta, type=c(\"p\", \"r\"))",
null,
"We can condition on GENDER:\n\nxyplot(Mean ~ Age | Gender, data=malta, type=c(\"p\", \"r\"))",
null,
"#### 4.2 Conditioning on quantitative variables\n\nWe can also condition on quantitative variables. Lattice uses a concept called shingles to do this. A quantitative variable can be divided into a specified number of shingles, which are overlapping categories with an equal number of observations. The function equal.count() is used to create shingles. The arguments number= and overlap= control the number of shingles and the amount of overlap. You can look at the result using the function plot():\n\nAge.shingles <- equal.count(malta$Age, number=5, overlap=1/4) plot(Age.shingles)",
null,
"6 shingles with little overlap: Age.shingles <- equal.count(malta$Age, number=6, overlap=0)\nplot(Age.shingles)",
null,
"4 shingles with more overlap:\n\nAge.shingles <- equal.count(malta$Age, number=4, overlap=1/2) plot(Age.shingles)",
null,
"We can use the newly created object Age.shingles to condition on AGE. First, here is a boxplot showing SCORE by NL.FATHER: bwplot(Mean ~ NL.father, data=malta)",
null,
"We can condition on AGE: bwplot(Mean ~ NL.father | Age.shingles, data=malta)",
null,
"You can control the arrangment of the panels with the argument layout=: bwplot(Mean ~ NL.father | Age.shingles, data=malta, layout=c(4,1))",
null,
"Different layout: bwplot(Mean ~ NL.father | Age.shingles, data=malta, layout=c(1,4))",
null,
"#### 4.3 Conditioning on two variables Multipanel conditioning can also involve two conditioning variables: bwplot(Mean ~ NL.father | Gender * Age.shingles, data=malta, layout=c(2,4))",
null,
"### 5 Multivariate plots #### 5.1 Scatterplot matrix A scatterplot matrix simulateously plots several quantitative variables against each other. It is especially useful for investigating the association between several independent variables and identifiying outliers or clusters. For the present dataset it is not very revealing since we only have 1 quantitative independent variable (AGE). We will illustrate its use with the mean scores on subsets of the questionnaire data. A scatterplot matrix is drawn with the function splom(). Use square brackets [] to indicate which columns (=variables) of the data frame you want to include: splom(~malta[c(11:15)])",
null,
"You can change the size of the plotting symbols: splom(~malta[c(11:15)], cex=.6)",
null,
"You can add a regression line easily: splom(~malta[c(11:15)], cex=.6, type=c(\"p\", \"r\"))",
null,
"If overplotting is an issue, you can use transparency: splom(~malta[c(11:15)], cex=.6, type=c(\"p\", \"r\"), alpha=.3)",
null,
"#### 5.2 Parallel coordinates plot Parallel coordinates plots are another method for showing several quantitative variables in the same plot. The argument lty= selects solid lines for all observations: parallelplot(~malta[c(11:15)], lty=1)",
null,
"By default, every variable is scaled according to its own range, thus stretching from its minimum to its maximum. If variables are measured on a common scale (like here), you can add the argument common.scale=TRUE. The data now range from the overall minimum to the overall maximum: parallelplot(~malta[c(11:15)], lty=1, common.scale=TRUE)",
null,
"If overplotting is an issue you can use transparency: parallelplot(~malta[c(11:15)], lty=1, common.scale=TRUE, alpha=.2)",
null,
"We can use color to distinguish groups. However, we first have to unload the lattice package and load it again to have available the (default) color settings. We also add a legend. detach(package:lattice, unload=TRUE) library(lattice) parallelplot(~malta[c(11:15)], lty=1, groups=malta$Gender, common.scale=TRUE, alpha=.2, auto.key=TRUE)",
null,
"### 6 Categorical variables\n\nData\n\nWe first need to create a hypothetical dataset. Speakers of 6 different varieties were asked which verb form of (to) prove they find acceptable in sentences like “History has prov** him wrong”. The table lists the frequencies of the three possible answers.\n\nprove.opinion <- rbind(\"American\" = c(28, 22, 44),\n\"British\" = c(83, 22, 17),\n\"Nigerian\" = c(19, 54, 34),\n\"Indian\" = c(23, 14, 4),\n\"Singapore\" = c(46, 22, 6))\ncolnames(prove.opinion) <- c(\"proved\", \"either\", \"proven\")\nprove.opinion\n## proved either proven\n## American 28 22 44\n## British 83 22 17\n## Nigerian 19 54 34\n## Indian 23 14 4\n## Singapore 46 22 6\n\n#### 6.1 Spine plot\n\nSpineplots are drawn with the function spineplot().\n\nspineplot(prove.opinion)",
null,
"#### 6.2 Mosaic plot\n\nThe function mosaicplot() draws mosaic plots. For 2-dimensional tables, they are very similar to spineplots. However, they offer additional options.\n\nmosaicplot(prove.opinion)",
null,
"We can change the orientation of the x-axis labels:\n\nmosaicplot(prove.opinion, las=4)",
null,
"And we can add shades of grey to distinguish between the categories of the dependent variable (verb form):\n\nmosaicplot(prove.opinion, las=4, col=c(\"grey90\", \"white\", \"grey60\"))",
null,
"The mosaicplot() function allows us to use residual-based shading to see which cells differ from the expected counts:\n\nmosaicplot(prove.opinion, las=4, shade=TRUE)",
null,
""
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http://www.telescopesales.org/96-telescope-eyepiece | [
"Explaining the Explanation\nEssential to understanding Newtons work is an accurate grasp of the experimental basis behind his discoveries. The knol The Sky Before the Telescope describes the sky as it was perceived prior to the calendar year 1600. It discusses with various animated illustrations the obvious movement of the stars and planets the controversy about what is orbiting what at the same time because the roles of latitude and longitude. It describes knowledge collected by Tycho de Brahe and summarized by Kepler in his a few legal guidelines.\n\nKepler showed that from the purpose on the look at of stars planetary orbits are ellipses. Yet the rationale why they were ellipses was a mystery. It had been Newton who explained the motion for the planets. An explanation in physics implies displaying how the observed info observe from a deeper more normal mathematical concept. 96 telescope eyepiece There was no this sort of theory available in the time. Newton needed to generate that general principle and in many cases develop the essential mathematics to the fly even while he was generating his deductions about motion. For the other hand he did not really need to make discoveries attributed him by some textbooks that include the concept that the power of gravity is universal and will get weaker with distance.\nNewtons more youthful colleague Edmond Halley visited him in Oxford in 1724 to inquire whether or not he could determine what the orbit of a planet would be if gravity was lowering since the inverse square belonging to the distance. Halley had discovered the mathematics of calculating this far too daunting. So also did Newtons colleague and fellow member with the Royal Society Robert Hooke who corresponded with Newton about the qualities of gravity. Newton had been mulling around these kinds of complications already and Halleys hard query spurred him to work to the query.\nNewton was inevitably when years of perform in a position to do the calculation Halley requested. To start with he needed to make that query much more exact. He had to formulate an equation which used new mathematical ideas as well as to solve it. The solution of that equation describes the movement from the planets. Halley was impressed with Newtons groundbreaking do the job urged him to publish it and remaining a man of implies underwrote the price of printing it.\nThe well known Principia 1-3... was the outcome. Newtons three legal guidelines the propositions in the Principia ended up an attempt to set the guidelines for the equation into words. Those words are challenging to grasp since Newton himself struggled along with his new concepts 4-7 and in addition because elementary explanations avert the sophisticated mathematics needed for your equation. As being a outcome pupils have a tendency to memorize and parrot again statements which they are doing not understand. Right here we propose to vary that unlucky state of affairs by taking a different technique to the very same pedagogical challenge an method referred to as conceptual physics. See the authors bio.\nNewtons Marketplace - The Sacred plus the Secular\nTheology and theories describing experimental knowledge have been not obviously separated in Newtons time. The controversy which Galileos discoveries stirred up using the Catholic church are but one particular modest illustration. The transition through the medieval belief in crystaline spheres emitting celestial audio towards the post-Newtonian environment of scientific astronomy guided by his theories was sluggish but unrelenting for the duration of the 14th to 17th centuries.\n\nNewton was born in 1648 just a couple months right after Galileo Galieli died and 126 ages after Ferdinand Magellans crew finished the 1st circumnavigation for the world. As he was developing up Newton knew which the Earth is spherical that gravity pointing down pulls points for the center of the Earth. He knew that Galileo had aimed his telescope in the sky and observed four new stars plainly orbiting Jupiter not the Earth. These objects are actually referred to as Jupiters moons. Galileo had also observed the surface of our moon and noticed it was not compared with surface within the Earth. It was not an ideal sphere produced of heavenly etheral substance but fairly a stony desert covered with craters. The time was ripe to bridge the divide which in the past had separated the secular and also the sacred spheres and also to clearly show that very same legal guidelines are valid in all places as in the Earth so while in the heavens.\nWhereas the telescope was in use in Newtons time and Newton himself made necessary advancements in it info within the motion within the planets had previously been gathered and organized prior to the telescope. Roughly a fifty percent century ahead of Newton was born Johannes Kepler and Tycho de Brahe had noticed equally novas new stars showing within the heavens and comets which passed with no hindrance from the hypothetical crystal sphere 1557. Their accomplishments and lives are discussed in alot more detail in The Sky Prior to the Telescope. knol. Their give good results culminated in Keplers legal guidelines the understanding of which can be essential to the subsequent phase which was Newtons.\nNewton built a successful synthesis of what was then recognized and wrote an equation which was in a position to supply the orbits of your planets. Right now we phone like equations the equations of motion. As in Dumas The A few Musketeers you can find basically 4 not a few legal guidelines wanted to explain the motion of planets. The three laws of textbook fame and also the law of universal gravity need to do the trick in concert to produce the equation of movement.\nBefore plunging into Newtons theoretical contributions we need to evaluate a second very important area of exploration crucial to understanding his function the state of bodily theory in his time. If you ever presently know the essential concepts of static forces you could skip the subsequent portion and go instantly with the segment on dynamic forces.\nStatic Forces and then the Equations for Equilibrium\nThe movement of your planets had been noticed to become periodic- their movement repeats while not obvious changes. This contrasts with observations for the Earth where exactly a heading mass tends to sluggish down and finally halt. When all movement has occur to some prevent the product is claimed for being in equilibrium. These noticed differences had led into the historical belief which the legal guidelines governing the heavens were various that individuals on earth. Precise new observations within the heavens and explorations of forces had begun to erode this belief nevertheless.\n\nThe idea of power was less clearly defined in Newtons time than its at this time. Scholars of the Middle Ages mentioned vis viva a dwelling force which brings stationary objects into movement and vis morte a lifeless power which designed tension or stress but did not give birth to new motion. Humans had after all encountered both since the advent with the human race. They regarded numerous static forces- the force needed to move an object or elevate a excess weight the force required to buttress a cathedral wall or arch and the power of springs.\n\nAdditionally they realized dynamic forces the force of friction and also the power needed to have objects to maneuver. Newtons contemporaries Robert Hooke and Robert Boyle who also belonged to that delightful club the Royal Society which attained in London on every last total moon ended up curious about all sorts of forces. They finally derived equations and explanations for pinpointing the power of a spring when extended Hookes legislation and for the spring of the air when one particular attempts to compress it i.e. Boyles legislation. Even ordinary people realized about magnets which exercised force on iron and also other magnetic resources. They recognized that the force got weaker when the distance between the magnet as well as the iron was greater.\nStatic forces depend upon the placement within the objects but not on their state of motion. The science managing equilibrium statics or mechanostatics was imperative to the constructing of structures arches and domes. Dependent additional on practical experience and instinct in all those days as an alternative to on far more elaborate calculations as nowadays the idea and then the equations desired for such calculations had been however known.\n\nIn this particular applet labeled as a spring-pendulum hefty bobs is usually hung on elastic springs. The quantity of friction is often modified. process which exhibits the two forms of habits we described previously mentioned movement as witnessed within the sky and movement as looked at on Earth. If the friction consistent b is about to zero the mass keeps going in a wavy periodic motion forever like the planets. Once you introduce friction the amplitude decreases and once one or two oscillations the motion stops as well as the strategy is in equilibrium. This 2nd habits illustrates the problem common on earth.\n\nIf we know the parameters of the method that may be the power for the spring and then the fat in the bob we can easily calculate its equilibrium position by fixing an easy algebraic equation with one unknown. Needless to say we can measure that position way too.\nHere we now have equally the theory and therefore the experiment. Make use of the applet in lieu of establishing the experiment. Click for the applet and check out unique parameters for that technique several spring constants different masses. Drag the bob away from the equilibrium level implementing the mouse then click on start out.\nDamped spring pendulum At this phase were not trying to calculate the pattern belonging to the bobbling motion which the mass helps make prior to it settles into equilibrium. That movement will be the dynamics from the method. Then again the exact same spring and mass will always are inclined towards the exact equilibrium which the statics on the system we can easily easily compute. The equation to unravel it in this case may be the harmony of two forces. The spring pulls the mass up gravity pulls it down. An algebraic equation describing this simple pendulum looks like this-\nK y - m g 0\nforce of spring weight of mass m 0\nThere may be an individual unfamiliar a selection y the extension belonging to the spring. You may get in touch with this balance action equals response the forces staying equal in magnitude and opposite in signal F.spring - F.excess fat. Even so it it is better to write this equation as the forces are balanced i.e. their sum is zero. F.spring F.bodyweight 0 .\nHeres why. We will compute the static equilibrium of any strategy through the following basic rule- At equilibrium the sum of all forces acting on each and every mass is zero. Vector algebra\n\nWarning- Two paragraphs of math concepts in advance We promised no higher math but we must point out vectors. The motion of the mass during the spring-pendulum applet was just up and down in one dimension along a vertical line. Placement was described by 1 amount the elevation previously mentioned the flooring or the extension from the spring and so was velocity. Inside applet with the left motions are confined to a airplane or two dimensions. The place is presented by two figures and forces have a route. For our purposes we will say that a vector may be a pair of figures. When we think about motion in three dimensional area then positions velocities and forces are given by a triplet of numbers.\nThe notion of vectors will require greater than this but looking at a vector as n-tuple of figures often known as the parts of a vector a pair a triplet.. is enough right here for our works by using. A single vector equation is really three equations a single for every component. Think of it as a shorthand means of grouping collectively the figures which have identical physical meanings.\nEquilibrium of 3 Forces The static equilibrium applet for the appropriate reveals a far more sophisticated program a model of 3 bobs constrained by ropes and pulleys in order to maneuver together in a single aircraft. Enjoying with all the applet will illustrate obviously whats explained in words under.\n\nThe positions belonging to the 3 masses considered as a unit are described as the configuration within the process. After we get considerable with regards to the examine of any technique we have to select a frame of reference which can be a approach to assign figures to numerous positions of masses i.e. numerous configurations of the strategy. Here the configuration is identified from the position of your knot where exactly a few ropes are joined. The knot moves with the vertical plane and its position decides the heights or elevations with the 3 masses above a picked horizontal aircraft e.g. the flooring of place.\nThe forces acting at the knot are vectors which means they have a magnitude plus a path. These 2 parts magnitude and course are shown with the arrows. Notice that a lengthier arrow signifies a increased magnitude.\nThe concept of vectors and their algebra is definitely the subject for just a separate knol one particular on linear algebra and geometry. But this applet has a choice Parallelogram of forces which reveals how at equilibrium the sum of two forces is equal in measurement and opposite in direction into the 3rd force.\nAs with the case of pendulum during the to begin with applet we can find the equilibrium position. Notice that we are not able to as yet describe the process by which it settles to that equilibrium.\nIn nonetheless significantly more sophisticated programs we may possibly need to have a lot of equations to ascertain the place of various masses within a 3 dimensional house. In such circumstances we have a few algebraic equations for several unknowns. By solving the equations and uncovering those unknowns we can easily establish the equilibrium configuration of the model.\nThe Catenary Curve In even more sophisticated circumstances not even numerous algebraical equations are ample. A classical example is this problem- What stands out as the form of a rope suspended by its ends as illustrated in the right\n\nClick then scroll down.\nRight here the unidentified is simply not a single amount but a curve. That curve identified as a catenary will be identified because the treatment of the diverse choice equation a differential equation.\nEven though Newton was browsing for an equation which might have as its remedy a curve the orbitof a planet Gottfried Wilhelm Leibniz in France formulated an equation for your catenary.\nThe two independently identified what we these days get in touch with differential calculus or just calculus for brief. Click for more on catenary.\nIt is not hard to get an intuitive experience for differential calculus- Imagine the hanging rope changed by a chain produced up of lots of hyperlinks. You could see the question belonging to the shape like a challenge in the static equilibrium of many masses. There might possibly be a large quantity of algebraic equations which might have as their solution the elevations for the back links. The elevations of each of the back links outline the curve. That curve that catenary is shown here in blue in contrast using a red parabola. When you enter the lookup phrase catenary right into a lookup engine youll find other interesting attributes its got.\n\nThus we see that differential equations is usually imagined like a substantial number of algebraic equations. When Leibniz and his colleagues to put to use the instruments of differential calculus to uncover the static equilibrium of complex mechanical techniques Newton put to use these equivalent mathematical methods to seek out the dynamics of a very simple technique the movement of a solitary mass within a area of force. Dynamic Forces along with the Equation of Movement\ncredit to the photo Our discussion up to now has concentrated on static forces. The power of a spring depends only in the placement with the spring and stays consistent as long as the spring is extended. It is a static power. Power of gravity ordinarily identified as excess weight is also static it depends on distance on the two masses but not on their relative pace.\nNow we will include dynamic forces with the equation. Equally regular individuals and scholars have been naturally acquainted with dynamic forces- Centrifugal force the pull on the rope holding an mass as it is swung all over in a circle along with the force of friction. Centrifugal power is usually a distinctive instance of inertia. Even though inertia was identified ahead of Newton gave to force of inertia exact kind making it a well-defined physical quantity.\n\nWe talked about the power of friction when talking about the spring-pendulum applet. This is a dynamic power simply because it depends on velocity. Friction is zero when velocity is zero. To put it differently when an object is not really transferring theres no friction. How friction is dependent on velocity may be a intricate story. In a few conditions friction is proportional to velocity. An example of that is an object pulled slowly as a result of a fluid Stokes law. In other situations it is dependent on velocity squared or simply the indication for the velocity. Luckily for us on this knol we do not desire to take into account friction in our dialogue.\nNewton isnt going to converse about frictional force in his a few laws considering that he was pondering about planets and friction was observed for being so quite little that it might be ignored. Planets shift through a vacuum. However they are doing sluggish down slightly in excess of milennia. It isnt a coincidence that moon is usually dealing with the Earth using the comparable side its rotation about its axis has slowed in order that it now makes only one rotation per orbit. Nevertheless theyre other stories.\nNewton centered within the other dynamic power the force of inertia.\nRecall that pounds is definitely a power it is the force of gravity acting on the mass. Fat is proportional to mass doubling the mass doubles the excess fat. Inside of a equivalent way the power of inertia is proportional for the mass of the entire body. Furthermore it is proportional with the objects switch in velocity that may be to its acceleration.\nAcceleration with the mathematical sensation can increase or reduce the velocity it can be beneficial or detrimental. Power is demanded to have an object likely more rapidly to receive it to sluggish down to stop or to alter its course. An object with better mass needs much more force to get it to complete these actions than does a object that has a smaller mass. Doubling the mass involves doubling with the force to acquire equivalent acceleration.\nIf you should experience the textbook assertion that Force is mass multiplied by acceleration keep in mind that this refers only to this sort of force the power needed to overcome inertia. Youll find a large number of forces as we now have currently mentioned. The power of inertia is only one extra power a dynamic force which Newton added with the collection of known forces. It was well-known qualitatively prior to Newton. Newton gave it numerical values. As force like velocity and acceleration is actually a vector its both path and magnitude and so is described by a few numbers.\nWe are able to determine the movement of any moving entire body using this standard rule-\nThe sum of all forces dynamic and static acting on every mass is zero.\nEvery transferring mass inside of a technique has its possess equation of motion. This is simply not an algebraic equation but a differential equation.\nFor the basic spring-pendulum product talked about over that equation looks like this-\nK y- m g m y 0\nThe power of your spring - weight the power of inertia 0\nRight here y the extension on the spring describes the placement with the mass. The image y describes the acceleration the change of velocity with time. The image y not put into use on this equation would explain its velocity the adjust of place with time. Damped Spring Applet\n\nAs an alternative to concentrating over the static problem of equilibrium were now interested in the motion on the suspended bob earlier than it reaches equilibrium. Allow us investigate it during the simplest achievable phrases. The bob moves up and down which includes a a number of velocity earlier than coming to rest. The velocity is transforming plus the force of inertia is resisting that shift. The sum belonging to the force of inertia as well as power of the spring equals zero. That equation is often solved needless to say. The damped spring applet here illustrates the answer which these explanations and equations explain. When the frictional or damping parameter is zero the solution is usually a periodic curve is called a sinusoid. The Movement with the Planets The Visible Photo voltaic System\nIn this applet opt for either days or months to find out the animation.\n\nNewton was serious about a rather much more complicated motion compared to oscillation of the pendulum particularly with the movement of a planet orbiting the sun. It had been obvious to Newton as to a large number of many people in advance of him which the moon the closest of all heavenly bodies is orbiting the Earth. The Earth draws in the moon because it does everything else plus the moon is held in its circular orbit by centrifugal power. Right here we now have the dynamic harmony of two forces the power of gravity and centrifugal force and that is a unique scenario within the force of inertia. 5-8\nNewton was able to confirm that Earths gravity in the moons orbit is weaker than for the surface area with the Earth. That calculation is simple for just a circular orbit however the orbits in the planets ended up regarded not to be circular. As expounded in a lot more detail while in the knol The Sky Before the Telescope therere ellipses while using sun in a single emphasis. When Halley and other people approximated the orbits as circles with all the sun at the middle they saw that centrifugal force was just perfect to stability gravity provided that gravity was finding weaker with inverse sq. of their distance from the sun. The tantalizing dilemma of calculating that movement exactly was the problem Halley challenged Newton to solve. Newton described his alternative in his Principia.\nThese are generally the essential conceptual understandings of Classical or Newtonian mechanics. It is easy to utilize the two applets described beneath to see how Newtons equations are solved and achieve the trajectory of the relocating human body.\nYoull be able to now be considered a prime mover- Create a planet that has a stroke of ones mouse. The direction and pace of that stroke define the velocity of new planet place and velocity determine the orbit. This applet demonstrates how an orbit is produced by a differential equation after the place and velocity of an object is offered. If the original speed is too smallish the planet will crush into the star if it will be far too good sized exeeding escape velocity the planet will fly absent never to return.\nExcellent advancements in any subject are usually ready by a long time of operate by a number of some others. Which was the case with Newtons breakthrough likewise. Kepler for example was pondering about gravity triggering the curvature from the planetary orbits and wondered what legislation of gravity would be correct. Newton tried out numerous suggestions and uncovered the one particular which worked. That is certainly stated in his fourth legislation frequently called the law of universal gravitation- Gravity will get weaker because the inverse square of the distance. Its illustrated here. Easily stated in case the distance in between bodies is doubled the attraction is decreased 4 times.\nTo the much more mathematically inclined- Mathematically stated when the distance is r and gravity is g then at distance 2r gravity is going to be g4.\nIn far more intricate mathematical phrases- Gravity improvements as 1rr. Or r-2 r into the electric power of -2. Right here -2 certainly is the exponent - indicates inverse and 2 signifies square. The inverse sq. regulation applies to the attenuation of countless other physical quantities that include electrical costs and also the intensity of light or audio.\n\nYou are able to easily check Newtons calculations by way of this applet. Only the exponent -2 will deliver ellipses.\n\nIn one other visualization the image for the correct shows how equations of motion are solved. The blue vector is velocity and red vector certainly is the power of gravity proportional to acceleration. When solving the equation of movement the laptop or computer proceeds in little time steps. Acceleration is made use of to update velocity and velocity is applied to update place. In each time phase a different red vector is calculated using the regulation of gravity.\n\nYou now have an outline on the ideas along with a flavor on the mathematics in Newtons discoveries. What stays is usually to concentrate on the various and oftentimes puzzling terminology some textbooks use and of your constraints of our new awareness. Newtons 3 plus A single Laws\n\nWe introduced Newtons legal guidelines by detailing his considering and his accomplishments though staying away from attempts to analyze the words with which he used to clarify his discoveries. Newton wrote an equation for calculating the movement of an planet inventing the mathematics for fixing it and showed which the option was an ellipse with all the sun in a single target.\nWikipedia 10 gives us a glimpse at what Newton himself wrote. You will discover there the literal text of Newtons three laws in the first Latin and in an English translation.\nHow does what Newton wrote and what textbooks paraphrase relate to what weve explained We summarize this here together with the legal guidelines in reverse order for ease in recognizing them.\nHis third regulation generally stated as action equals reaction we formulate as this rule-\nThe sum of all forces static and dynamic acting on a shape is zero\nThis is the prescription for creating an equation of motion. You may simply call any subset of all those forces an action and then the remainder of them a reaction. That does not have an impact on the calculation. Within the scenario two forces are gravity and inertia the case of planets a person can give some thought to gravity and action and inertia a reaction.\n\nNewtons 2nd regulation describes the force of inertia in an inertial body of reference. With the terminology weve got utilised in this knol the force of inertia is often a dynamic power. Some textbooks utilize the term balanced forces within the circumstance of static equilibrium. When these texts then talk about an object accelerating they are saying the forces are unbalanced.\nWe favor to think about dynamic forces the power of inertia in addition to the force of friction as forces acting jointly with static forces. Hence the equation of motion even now describes a balance of forces. The power of inertia is in harmony with or is usually a response to the net sum of all static forces. The Newtons initially regulation is actually a corollary of his second law. When all other forces are created tiny as once we are to date away from the stars that inertia is a only power to contemplate then the equation of motion is simply m y 0 . The answer of this equation is uniform motion i.e. motion together a straight line with continuous velocity.\nSome textbooks say that Newtons very first law explains that Newtons legal guidelines are only legitimate in inertial frames. That limitation in his laws does exist nonetheless it deserves a additional specific explicit explanation which we attempt within the subsequent part.\nNewtons regulation of gravity is critical to clarify the observed orbits from the planets. This is the inverse square legislation explained during the earlier section and applet. The law of universal gravity is applicable for movement of planets which had been the experimental basis of his discovery. Be aware that his 3rd law is more universal and applies to all forces which may feature gravity but really do not ought to achieve this.\nInertial Methods along with the Limits of Newtons Mechanics The perfect which Einstein emphasized will be to formulate the laws of physics to ensure that theyre valid in all frames of reference shifting or not. When legal guidelines would be the same exact in all frames they may be says being invariant unchanging. We will then do our calculations in any body of reference. Well get different curves in different frames of reference distinct orbits but after we remodel them from the policies of geometry they will all agree. It can be even now vital that you decide the frame of reference thoroughly because the calculations and therefore the results are easier in a few frames.\n\nNewtons equations as he formulated them usually are not invariant. They are really legitimate only in exceptional frames of reference termed inertial frames. In an inertial product inertia obeys the 2nd of Newtons laws. In non-inertial techniques it does not.\nNewton and his contemporaries taken into consideration a frame fixed on the stars as representing absolute space. As we see points now such a frame is just one inertial program. All frames which can be heading uniformly with respect on the stars are inertial techniques. We can use Newtons laws in any of them.\nFor example we all know that inside frame of reference attached with the stars the orbits on the planets are rather simple these are ellipses. These very simple orbits ellipses are obtained by Newtons legal guidelines in any inertial system.\nA straightforward example of a non-inertial body is your personal working experience on the bus which quickly changes direction. That you are pulled to one facet. Its inertia needless to say. Even so during the frame of reference connected towards the bus this force fails to comply with the 2nd legislation. As opposed to leaving you at relaxation inertia pushes you all over.\nFor several practical challenges we can easily handle the Earth as an inertial strategy. During the experiment described from the spring-pendulum applet we applied a body of reference hooked up to your flooring which is with the Earth.\nThe truth that Earth rotates with respect to stars doesnt shift the results of most Earthly experiments appreciably. It will be observable in a few experiments which final a long time and entail rotating masses. For example hurricanes sizeable rotating masses persisting for days are affected by the rotation of the Earth.\nclick for heritage A 2nd example which displays that Earth is just not accurately an inertial process is really a modern-day unit invented a century when Newton the gyroscope. Its utilised for navigation within the house shuttle in addition as on planes and ships. In an inertial system the axis of its rotating disc often points inside the identical path. But in the rotating body that axis rotates. It obviously can not continue to be stationary in the two. It rotates within a frame hooked up towards the Earth. Equivalent impact because of to rotation of Earth is Coriolis impact.\n\nThe hard problem With which stars does the the axis of the gyroscope continue to be aligned has become questioned and response is intricate instead of pretty ultimate. For our purposes we give consideration to them to be the stars that we are able to see thats the stars of our galaxy.\nIn events of substantial velocities or of gravitational fields of elevated intensity Newtons theory wont perform. Einsteins Principle of Typical Relativity need to be utilised in those scenarios.\nNewtons mechanics at present termed Classical Mechanics includes the two dynamics and statics. His concepts ruled physics up till the twentieth century. They have been generalized additional and so are component of the disciplines ofmodern physics which now involves Quantum Mechanics and Einsteins Theory of Relativity. Basic facets of these theories are still based upon concepts first formulated by Newton in his Principia. The Idea of relativity is explained from the follow up knols referred to as Relativity triptych. References\nPrincipia Fashionable translation English translation click on on edges to select pages The Cambridge Companion to Newton In regards to the Principia and Newton Isaac Newton - Lifestyle and function Examination by a fashionable physicist Heritage by Stanford Encyclopedia bibliography The laws by Stanford Encyclopedia of philosophy Inertial frames by Stanford Encyclopedia of philosophy Text from the laws Latin and English in wikipedia. It follows most textbooks in stating that 1st regulation explains inertial frames. Wikipedia also assumes education differential calculus. It hence combines both equally the issues which we intentionally avoided. In texts put collectively by committees such accumulations of problems typically are not unusual. Wikipedia guide nonetheless can be a handy complement to this exposition. 96 telescope eyepiece If you are looking to buy a telescope and want to buy a model that will allow you to get the most out of astronomy you are going to need some resources. As you read this article you will soon discover more on how telescope testimonials may be a awesome assist.\nSummary of Contents- Ultimately What Would you like How Telescope Testimonials Are usually Your Savior Purchasing a Telescope\nEventually What Would you like Plenty of people get enthusiastic whenever they see a telescope for sale. What do they do Straightaway they buy This will be considered a fantastic impulse but it could result in you not having the best telescope for the wants.\nThis helps make using a move again for the instant and taking into account that which you want from astronomy an essential aspect.\n\nTagged with:\n\nFiled under: Astronomy News"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.95495844,"math_prob":0.93378556,"size":34120,"snap":"2019-26-2019-30","text_gpt3_token_len":6478,"char_repetition_ratio":0.13726698,"word_repetition_ratio":0.0024535577,"special_character_ratio":0.17989449,"punctuation_ratio":0.05244871,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9645633,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-24T06:33:37Z\",\"WARC-Record-ID\":\"<urn:uuid:368e3b88-bbce-4a97-8959-7cf493f22962>\",\"Content-Length\":\"69079\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8af114a8-e2a6-417c-aba7-863d6a136bb3>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f9e1f63-9e90-497b-94e9-f8494629da95>\",\"WARC-IP-Address\":\"173.237.137.96\",\"WARC-Target-URI\":\"http://www.telescopesales.org/96-telescope-eyepiece\",\"WARC-Payload-Digest\":\"sha1:2MVZUQBJ3JUM3BSDD4BI2EA2RCTBYVJP\",\"WARC-Block-Digest\":\"sha1:DGBSWBJJ5JGXVI76KKRMJ2TKA36RQAST\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195531106.93_warc_CC-MAIN-20190724061728-20190724083728-00113.warc.gz\"}"} |
https://scicomp.stackexchange.com/questions/15963/euler-equation-eigensystem-with-gravity-in-the-energy-flux?noredirect=1 | [
"# Euler Equation Eigensystem with Gravity in the Energy Flux\n\nI am modifying a conservative form of the Euler equations with gravity in the energy flux (see previous question: Energy Conservation in Conservation Laws with Source Terms) for use in a Riemann solver. I'm seeing some unphysical/unstable results despite total energy conservation to roundoff.\n\nThe conservation law equation with source term (a.k.a balance law) $\\frac{\\partial \\vec{Q}}{\\partial t} + \\nabla \\cdot \\vec{F} = \\vec{S}$, is solved, where the variables are defined in $\\vec{Q}$,the flux, $\\vec{F_z}$ in the $z$-direction, and the source $\\vec{S}$ of modified equations are:\n\n$\\vec{Q}= \\left[\\begin{matrix} \\rho \\\\ \\rho u \\\\ \\rho v \\\\ \\rho w \\\\ E_{total} \\end{matrix}\\right] \\quad \\quad \\vec{F_z}= \\left[\\begin{matrix} \\rho w \\\\ \\rho uw \\\\ \\rho vw \\\\ \\rho w^2+p \\\\ (E_{total} + p)w \\end{matrix}\\right] \\quad \\quad \\vec{S}= \\left[\\begin{matrix} 0 \\\\ 0 \\\\ 0 \\\\ -\\rho g \\\\ 0 \\end{matrix}\\right]$\n\nwhere $\\rho$ is the density, $u$,$v$, and $w$ are velocities in the $x$, $y$ and $z$ directions respectively. The total energy is defined as $E_{total} = \\frac{p}{\\gamma-1} + \\frac{1}{2}\\rho q^2 + \\rho\\psi$, with $q^2=u^2+v^2+w^2$, and $\\psi=gz$, with $g$ being the acceleration due to gravity in the $z$-direction, and $\\gamma$ is the ratio of specific heats. The pressure $p$ can be defined from the total energy, $p=(\\gamma-1)(E_{total} - \\frac{1}{2}\\rho q^2 - \\rho\\psi)$.\n\nI find the flux Jacobian, $A = \\frac{\\partial \\vec{F_z}}{\\partial \\vec{Q}}$,\n\n$A = \\left[\\begin{matrix} 0 & 0 & 0 & 1 & 0 \\\\ -uw & w & 0 & u & 0 \\\\ -vw & 0 & w & v & 0 \\\\ (\\gamma-1)(\\frac{1}{2} q^2 - \\psi) - w^2 & -(\\gamma-1)u & - (\\gamma-1)v & (3-\\gamma)w& (\\gamma-1) \\\\ [(\\gamma-1)(\\frac{1}{2} q^2 - \\psi) - h']w & -(\\gamma-1)uw & -(\\gamma-1)vw & h'-(\\gamma-1)w^2 & \\gamma w \\end{matrix}\\right]$\n\nwhere $h' = h + \\psi$, with $h = \\frac{a^2}{\\gamma-1} + \\frac{1}{2}q^2$, and $a^2 = \\frac{\\gamma p}{\\rho}$.\n\nThis systems yields eigenvalues $\\lambda =[u-a, u,u,u,u+a]$, and right eigenvectors:\n\n$R = \\left[ \\begin{matrix} 1 & 0 & 0 & 1 & 1 \\\\ u & 1 & 0 & u & u \\\\ v & 0 & 1 & v & v \\\\ w-a & 0 & 0 & w & w+a \\\\ h'-ua& u & v & \\frac{1}{2}q^2+\\psi & h'+ua \\end{matrix}\\right]$\n\nand left eigenvectors:\n\n$L = R^{-1} = \\frac{\\gamma-1}{2a^2}\\left[ \\begin{matrix} (\\frac{1}{2}q^2-\\psi)+\\frac{wa}{\\gamma-1} & -u & -v & -w-\\frac{a}{\\gamma-1} & 1 \\\\ \\frac{2ua^2}{\\gamma-1} & -\\frac{2a^2}{\\gamma-1} & 0 & 0 & 0 \\\\ -\\frac{2va^2}{\\gamma-1} & 0 & \\frac{2a^2}{\\gamma-1} & 0 & 0 \\\\ 2h'-2q^2 & 2u & 2v & 2w & -2 \\\\ (\\frac{1}{2}q^2-\\psi) - \\frac{wa}{\\gamma-1} & -u & -v & -w + \\frac{a}{\\gamma-1} & 1 \\end{matrix}\\right]$\n\nDoes anyone see any problems with this formulation? Does anybody have experience including gravity as a source term for the momentum equation, and as a flux for the energy equation? Or does anyone see problems with the calculation of the eigensystem?\n\nThanks for the help.\n\n• I define $E_t = E + \\rho\\psi$, so based on the previous question, shouldn't it be $E_t$ in the flux? The pressure should be defined differently, if what you say is true. You are correct about the $z$-flux, and this is the $x$-flux. I can rewrite it, but for now I'll just move the source up to the correct equation, and will have to keep in mind that $x$ actually means $z$. – Wes Lowrie Oct 24 '14 at 15:36\n• I've updated the flux and source to represent the $z$-flux, as well as the flux Jacobian, and right and left eigenvectors so it is consistent with the gravity source acting in the $z$-direction. – Wes Lowrie Oct 24 '14 at 16:43\n• No problem, I just updated the subscript with $total$ instead of $t$ to avoid confusion. – Wes Lowrie Oct 24 '14 at 16:55"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.61989933,"math_prob":0.99995804,"size":2848,"snap":"2019-51-2020-05","text_gpt3_token_len":1104,"char_repetition_ratio":0.16912799,"word_repetition_ratio":0.040598292,"special_character_ratio":0.42801967,"punctuation_ratio":0.06944445,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99999154,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-24T20:40:54Z\",\"WARC-Record-ID\":\"<urn:uuid:32983a7c-f7f7-475d-b501-e7cb35e8dcf2>\",\"Content-Length\":\"133167\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:07f036d4-3b6f-40ee-a920-da6f208a8936>\",\"WARC-Concurrent-To\":\"<urn:uuid:f525e6aa-a6b1-4db1-b211-cddbc6ed07a2>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://scicomp.stackexchange.com/questions/15963/euler-equation-eigensystem-with-gravity-in-the-energy-flux?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:JCMTVIHVTS6CSG3BVODE7ET7OVI63EBO\",\"WARC-Block-Digest\":\"sha1:PI54WHSENJSGZSR3WAPWUIKPUKKVIXI5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250625097.75_warc_CC-MAIN-20200124191133-20200124220133-00409.warc.gz\"}"} |
https://crypto.stackexchange.com/questions/43852/ind-cpa-security-of-ctr-mode | [
"# IND-CPA security of CTR mode\n\nLet's suppose CTR mode is instantiated such that the input to the block-cipher is $\\langle \\mathrm{IV}+\\mathrm{ctr}\\rangle$ instead of $\\mathrm{IV}\\mathbin\\|\\langle \\mathrm{ctr}\\rangle$, where $\\langle X\\rangle$ is the bit representation of $X$.\n\nWill the mentioned version of CTR encryption scheme be IND-CPA secure against nonce-respecting adversaries? If not, can this be made nonce-IND-CPA secure by adding some cryptographic methods over it without altering the underlying structure?\n\n• IV+ctr and IV||ctr and IV xor ctr can all be identical depending on the position and size of the IV within the block Feb 14 '17 at 10:48\n• @RichieFrame As per my understanding(I am new to the field), if IV is random all the three should be identical (the IND-CPA security of the cipher will not be affected) but if we make the IV non-random nonce(such as packet counter) the chosen-plaintext attack will completely break the security of cipher (obviously making it IND-CPA insecure). I was wondering that if in the example of CTR here : en.wikipedia.org/wiki/…, instead of concatenating the counter with the random nonce if we do IV+ctr, will it affect the IND-CPA security? Feb 14 '17 at 18:42\n• Note that the NIST document containing counter mode encryption has some remarks on how the initial counter can be created, including security remarks. Feb 14 '17 at 22:40\n\nThe crucial property you want to achieve CPA security is that each combination of $\\mathrm{IV}$ and $\\mathrm{ctr}$ is used at most once, since you essentially derive a key from this combination and use it to encrypt the corresponding block with the one-time pad. If you choose $\\mathrm{IV}$ uniformly at random from a large enough space, this will be the case with high probability for both variants you describe.\nIf $\\mathrm{IV}$ is a packet counter, $\\mathrm{IV} || \\langle \\mathrm{ctr} \\rangle$ is still fine if you reserve enough bits for both $\\mathrm{IV}$ and $\\langle \\mathrm{ctr} \\rangle$ to exclude overflows. The variant $\\langle \\mathrm{IV} + \\mathrm{ctr} \\rangle$, however, now becomes insecure: If two packets are encrypted where the first one contains at least two blocks, the second block of the first packet will be encrypted with $\\langle 1 + 2 \\rangle = \\langle 3 \\rangle$, and the first block of the second packet will be encrypted with $\\langle 2 + 1 \\rangle = \\langle 3 \\rangle$. Hence, an attacker who knows the contents of the second block in the first packet can learn the first block of the second packet. As long as you only increment the combination of $\\mathrm{IV}$ and $\\mathrm{ctr}$ by one for the next block, this problem persists unless you ensure there is a large enough gap between two packets. This is exactly what $\\mathrm{IV} || \\langle \\mathrm{ctr} \\rangle$ ensures.\n• @ironhide012 As far as I can tell, your understanding is correct. ;) I am not sure what exactly you mean by not changing the underlying structure: $\\mathrm{IV} || \\langle \\mathrm{ctr} \\rangle$ is actually not too different from $\\langle \\mathrm{IV} + \\mathrm{ctr} \\rangle$ since you can write $\\mathrm{IV} || \\langle \\mathrm{ctr} \\rangle = \\langle 2^{[\\text{# bits reserved for ctr}]} \\cdot \\mathrm{IV} + \\mathrm{ctr} \\rangle$. If you want to change the structure less than this, I don't see how it can be done. Feb 15 '17 at 20:41"
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https://gitcommit.co.uk/2017/03/18/the-expressiveness-of-haskell/ | [
"# The expressiveness of Haskell – Sorting.",
null,
"What first excited me about Haskell was the expressiveness of the language. It somehow encapsulates thought and the problem being solved can often be seen clearly in the solution.\nTake, for example, sorting a list containing elements of a general type ‘a’. For these to be placed in order it must be possible to compare them. So in Haskell the sort function can be defined as:\n\ni.e. the sort function takes a list of ‘a’ and returns a list of ‘a’ and the restriction on the type of ‘a’ is that it must be an instance of the Ord class – this allows comparison of two ‘a’ s. The function can be implemented as\n\nIf you don’t know Haskell at all then just stare at it for a while and you may get an intuition of what’s happening!\nLine 2 states that sorting an empty list just gives an empty list.\nOtherwise, line 3, take the first item in the list and put it in a list [x]. Then prepend the recursively sorted list of all items that are less than or equal to x, i.e. sort left and append to [x] the recursively sorted list of all items that are greater than x, i.e. sort right. In these two lines:\n\nleft and right are known as list comprehensions. For example left means get all x’ from the list xs such that x’ <= x and similarly for right.\n\nIt might not be efficient but it is very, very pleasing.\n\nTagged:"
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"https://i2.wp.com/gitcommit.co.uk/wp-content/uploads/2017/03/refresh.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9055936,"math_prob":0.929885,"size":1573,"snap":"2019-35-2019-39","text_gpt3_token_len":438,"char_repetition_ratio":0.1153601,"word_repetition_ratio":0.16867469,"special_character_ratio":0.30832803,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9703078,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-21T23:30:17Z\",\"WARC-Record-ID\":\"<urn:uuid:2fecc8e6-91d0-42e2-a445-ff12d7e8c7b3>\",\"Content-Length\":\"59884\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fd7b38e3-04a8-4438-9e5d-43ac819fb220>\",\"WARC-Concurrent-To\":\"<urn:uuid:d910ae34-c834-4741-8f11-e02f8bb8f461>\",\"WARC-IP-Address\":\"185.119.173.58\",\"WARC-Target-URI\":\"https://gitcommit.co.uk/2017/03/18/the-expressiveness-of-haskell/\",\"WARC-Payload-Digest\":\"sha1:ZAAWJEV5NU46BAXKZ4INE6BBYGM4GWRU\",\"WARC-Block-Digest\":\"sha1:UR3PLN5OMEYM7BYVBUATGHODE4YV5G2D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514574710.66_warc_CC-MAIN-20190921231814-20190922013814-00474.warc.gz\"}"} |
https://docs.cycling74.com/max8/refpages/bitor | [
"# |\n\n## Description\n\nPerforms a bit-by-bit OR of two numbers (expressed in binary for the task). Outputs a number composed of all those bits which are 1 in either of the two numbers.\n\n## Examples",
null,
"All non-zero bits are combined......... Can be used to pack two numbers into one int\n\n## Arguments\n\n### initial-value [int]\n\nOptional\n\nSets an initial value to be OR-ed with a number received in the left inlet.\n\n## Messages\n\n### bang\n\nIn left inlet: Performs the calculation with the numbers currently stored. If there is no argument, | initially holds 0.\n\n### int\n\n#### Arguments\n\ninput [int]\nIn left inlet: Outputs a number composed of all those bits which are 1 in either of the two numbers.\n\n### (inlet1)\n\n#### Arguments\n\ncomparison-number [int]\nIn right inlet: The number is stored for combination with a number received in the left inlet.\n\n### float\n\n#### Arguments\n\ninput [float]\nConverted to int.\n\n### set\n\n#### Arguments\n\nset-input [int]\nIn left inlet: The word set followed by a number will set the input to the bitwise-or operation without causing output (a successive bang will output the result).\n\n### list\n\n#### Arguments\n\ninput [number]\ncomparison-value [number]\nIn left inlet: Combines the first and second numbers bit-by-bit, and outputs a number composed of all those bits which are 1 in either of the two numbers.\n\n## Output\n\n### int\n\nAll the non-zero bits of the two numbers received in the inlets are combined. If a bit is 1 in either one of the numbers, it will be 1 in the output number, otherwise it will be 0 in the output number."
]
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null,
"https://docs.cycling74.com/static/max8/images/952ea4755c7a556cbfa9b47c9ef1490a.png",
null
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https://cs.stackexchange.com/questions/12102/express-boolean-logic-operations-in-zero-one-integer-linear-programming-ilp | [
"# Express boolean logic operations in zero-one integer linear programming (ILP)\n\nI have an integer linear program (ILP) with some variables $x_i$ that are intended to represent boolean values. The $x_i$'s are constrained to be integers and to hold either 0 or 1 ($0 \\le x_i \\le 1$).\n\nI want to express boolean operations on these 0/1-valued variables, using linear constraints. How can I do this?\n\nMore specifically, I want to set $y_1 = x_1 \\land x_2$ (boolean AND), $y_2 = x_1 \\lor x_2$ (boolean OR), and $y_3 = \\neg x_1$ (boolean NOT). I am using the obvious interpretation of 0/1 as Boolean values: 0 = false, 1 = true. How do I write ILP constraints to ensure that the $y_i$'s are related to the $x_i$'s as desired?\n\n(This could be viewed as asking for a reduction from CircuitSAT to ILP, or asking for a way to express SAT as an ILP, but here I want to see an explicit way to encode the logical operations shown above.)\n\nLogical AND: Use the linear constraints $$y_1 \\ge x_1 + x_2 - 1$$, $$y_1 \\le x_1$$, $$y_1 \\le x_2$$, $$0 \\le y_1 \\le 1$$, where $$y_1$$ is constrained to be an integer. This enforces the desired relationship. (Pretty neat that you can do it with just linear inequalities, huh?)\n\nLogical OR: Use the linear constraints $$y_2 \\le x_1 + x_2$$, $$y_2 \\ge x_1$$, $$y_2 \\ge x_2$$, $$0 \\le y_2 \\le 1$$, where $$y_2$$ is constrained to be an integer.\n\nLogical NOT: Use $$y_3 = 1-x_1$$.\n\nLogical implication: To express $$y_4 = (x_1 \\Rightarrow x_2)$$ (i.e., $$y_4 = \\neg x_1 \\lor x_2$$), we can adapt the construction for logical OR. In particular, use the linear constraints $$y_4 \\le 1-x_1 + x_2$$, $$y_4 \\ge 1-x_1$$, $$y_4 \\ge x_2$$, $$0 \\le y_4 \\le 1$$, where $$y_4$$ is constrained to be an integer.\n\nForced logical implication: To express that $$x_1 \\Rightarrow x_2$$ must hold, simply use the linear constraint $$x_1 \\le x_2$$ (assuming that $$x_1$$ and $$x_2$$ are already constrained to boolean values).\n\nXOR: To express $$y_5 = x_1 \\oplus x_2$$ (the exclusive-or of $$x_1$$ and $$x_2$$), use linear inequalities $$y_5 \\le x_1 + x_2$$, $$y_5 \\ge x_1-x_2$$, $$y_5 \\ge x_2-x_1$$, $$y_5 \\le 2-x_1-x_2$$, $$0 \\le y_5 \\le 1$$, where $$y_5$$ is constrained to be an integer.\n\nAnd, as a bonus, one more technique that often helps when formulating problems that contain a mixture of zero-one (boolean) variables and integer variables:\n\nCast to boolean (version 1): Suppose you have an integer variable $$x$$, and you want to define $$y$$ so that $$y=1$$ if $$x \\ne 0$$ and $$y=0$$ if $$x=0$$. If you additionally know that $$0 \\le x \\le U$$, then you can use the linear inequalities $$0 \\le y \\le 1$$, $$y \\le x$$, $$x \\le Uy$$; however, this only works if you know an upper and lower bound on $$x$$.\n\nAlternatively, if you know that $$|x| \\le U$$ (that is, $$-U \\le x \\le U$$) for some constant $$U$$, then you can use the method described here. This is only applicable if you know an upper bound on $$|x|$$.\n\nCast to boolean (version 2): Let's consider the same goal, but now we don't know an upper bound on $$x$$. However, assume we do know that $$x \\ge 0$$. Here's how you might be able to express that constraint in a linear system. First, introduce a new integer variable $$t$$. Add inequalities $$0 \\le y \\le 1$$, $$y \\le x$$, $$t=x-y$$. Then, choose the objective function so that you minimize $$t$$. This only works if you didn't already have an objective function. If you have $$n$$ non-negative integer variables $$x_1,\\dots,x_n$$ and you want to cast all of them to booleans, so that $$y_i=1$$ if $$x_i\\ge 1$$ and $$y_i=0$$ if $$x_i=0$$, then you can introduce $$n$$ variables $$t_1,\\dots,t_n$$ with inequalities $$0 \\le y_i \\le 1$$, $$y_i \\le x_i$$, $$t_i=x_i-y_i$$ and define the objective function to minimize $$t_1+\\dots + t_n$$. Again, this only works nothing else needs to define an objective function (if, apart from the casts to boolean, you were planning to just check the feasibility of the resulting ILP, not try to minimize/maximize some function of the variables).\n\nFor some excellent practice problems and worked examples, I recommend Formulating Integer Linear Programs: A Rogues' Gallery.\n\n• which linear programming solver can solve this? becouse in *.lp or *.mps-format one side of the constraint has to be an fixed integer and not an variable.\n– boxi\nMar 25, 2015 at 14:31\n• @boxi, I don't know anything about *.lp or *.mps format, but every integer linear programming solver should be able to solve this. Note that if you have something like $x \\le y$, this is equivalent to $y -x \\ge 0$, which may be in the format you wanted.\n– D.W.\nMar 25, 2015 at 19:44\n• -i checked it again. lp_solve can solve it, but for example qsopt can't. i don't know why. but thanks <3\n– boxi\nMar 25, 2015 at 19:46\n• @Pramod, good catch! Thank you for spotting that error. You're absolutely right. I've asked a new question about how to model that case and I'll update this answer when we get an answer to that one.\n– D.W.\nDec 22, 2015 at 4:04\n• @InuyashaYagami, yes, absolutely, that is sufficient for boolean logic, and that would be fine. There might be some advantage to other translations if they introduce fewer temporary variables (for instance, in the case of logical implication), and we still need to know how to express conditional expressions or \"cast to boolean\", but you have an excellent point.\n– D.W.\nDec 28, 2021 at 19:25\n\nThe logical AND relation can be modeled in one range constraint instead of three constraints (as in the other solution). So instead of the three constraints $$y_1\\geq x_1+x_2−1,\\qquad y_1\\leq x_1,\\qquad y_1\\leq x_2\\,,$$ it can be written using the single range constraint $$0 \\leq x_1 + x_2-2y_1 \\leq 1\\,.$$ Similarly, for logical OR: $$0 \\leq 2y_1 - x_1 - x_2 ≤ 1\\,.$$\n\nFor NOT, no such improvement is available, because NOT is restricted to one argument.\n\nIn general for $$y=x_1 \\land x_2 \\land \\dots \\land x_n$$ ($$n$$-way AND) the constraint will be: $$0 \\leq \\sum x_i -ny \\leq n-1\\,.$$ Similarly for OR: $$0 \\leq ny -\\sum x_i \\leq n-1\\,.$$\n\n• A very simillar approach is in this paper: ncbi.nlm.nih.gov/pmc/articles/PMC1865583 Jun 24, 2015 at 10:57\n• Keep in mind that size isn't everything when looking at integer programming! The downside of the shorter constraints you suggest is that the LP relaxation that is solved as a subproblem for the MILP problem is not strong. You can see that by noting that, e.g., the upper bounding inequality for 'AND' becomes active (i.e., equality holds) for the non integral points (x1, x2, y) = (0, 1, 0.5) or (1, 0, 0.5) or (1, 1, 0.5). Dec 26, 2020 at 18:38\n\nI found shorter solution for XOR y=x1⊕x2 (x and y are binary 0, 1)\n\njust one line: x1 + x2 - 2*z = y (z is any integer)\n\n• To express the equality in ILP, you need two inequalities. Further, to avoid solution $x_1=1,x_2=0,z=200,y=-199$, you need two more inequalities, $0\\leq y \\leq 1$. So this answer has four inequalities and an extra variable compared to six inequalities in D.W.'s answer.\n– JiK\nMar 9, 2017 at 15:22\n• To express an equality in ILP only needs one equation, this is true in both LP theory and in software such Gurobi or CPLEX . @jIk, I guess you mean \"expressing \"a $\\neq b$\" needs two inequalities. \" Jan 18, 2018 at 3:11"
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https://eprint.iacr.org/2015/785 | [
"Cryptology ePrint Archive: Report 2015/785\n\nDouble-Speed Barrett Moduli\n\nRémi Géraud and Diana Maimut and David Naccache\n\nAbstract: Modular multiplication and modular reduction are the atomic constituents of most public-key cryptosystems. Amongst the numerous algorithms for performing these operations, a particularly elegant method was proposed by Barrett. This method builds the operation $a \\bmod b$ from bit shifts, multiplications and additions in $\\mathbb{Z}$. This allows building modular reduction at very marginal code or silicon costs by leveraging existing hardware or software multipliers. This paper presents a method allowing doubling the speed of Barrett’s algorithm by using specific composite moduli. This is particularly useful for lightweight devices where such an optimization can make a difference in terms of power consumption, cost and processing time. The generation of composite moduli with a predetermined portion is a well-known technique and the use of such moduli is considered, in statu scientae, as safe as using randomly generated composite moduli.\n\nCategory / Keywords: public-key cryptography / modular multiplication"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8492108,"math_prob":0.67430776,"size":1369,"snap":"2019-26-2019-30","text_gpt3_token_len":285,"char_repetition_ratio":0.0974359,"word_repetition_ratio":0.0,"special_character_ratio":0.2001461,"punctuation_ratio":0.09821428,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96484387,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-16T21:35:56Z\",\"WARC-Record-ID\":\"<urn:uuid:398067f8-41a5-420c-9e5b-56d2df09a602>\",\"Content-Length\":\"3190\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2fdeba9a-e73a-4a9d-95d5-6c824a7898c2>\",\"WARC-Concurrent-To\":\"<urn:uuid:f09fd3a9-60f8-4eff-a636-60458dad0403>\",\"WARC-IP-Address\":\"216.184.8.41\",\"WARC-Target-URI\":\"https://eprint.iacr.org/2015/785\",\"WARC-Payload-Digest\":\"sha1:FK4MD3BWSQMLRMZDDXLKOQU5AHDXM6R5\",\"WARC-Block-Digest\":\"sha1:FKSZ56OGGPS3VXYEPARHM5YLRWETHBBL\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998298.91_warc_CC-MAIN-20190616202813-20190616224813-00419.warc.gz\"}"} |
https://www.pdfdb.com/undergraduate-projects/matrices-and-its-applications/ | [
"# Matrices and Its Applications\n\n## ABSTRACT\n\nThis Project examines matrices and three of its applications. Matrix theories were used to solve economic problems, which involve methods by which goods can be produced efficiently. To encode and also to decode very sensitive information. This project work also goes further to apply matrices to solve a 3 x 3 linear system of equations using row reduction methods.\n\n## 1.0 BACKGROUND OF THE STUDY\n\nIn order to unfold the history of Matrices and Its Applications, the influence of matrices in the mathematical world is spread wide because it providesan importantbaseto many of the principles and practices. It is important that we first determine what matrices is. As such, this definition is not a complete and comprehensive answer, but rather a broad definition loosely wrapping itself around thesubject.\n\n“Matrix” is the Latin word for womb, and it retains that sense in English. It can also mean more generally any place in which something is formed or produced.\n\nThe origin of mathematical matrices lies with the study of systems of simultaneous linear equations. An important Chinese text from between 300Bc and Ad 200, nine chapters of the mathematical art, gives the first known example of the use of matrix methods to solve simultaneous equations. (Laura Smoller (2012))\n\nIn the treatises seventh chapter “too much and not enough”, the concept of a determinant first appears, nearly two milkman before its supposed inventions by the Japanese mathematician SEKI KOWA in 1683 or his German contemporary GOTTFRIED LEIBNIZ (who is also credited with the invention of differential calculus, separately from but simultaneously with Isaac Newton).\n\nMore uses of matrix-like arrangements of numbers appears in which a method is given for solving simultaneous equations using a counting board that is mathematically identical to the modern matrix method of solution outlined by Carh Fredrich Gauss (1777-1855), also known as Gaussian Elimination. (Vitull Marie 2012 )\n\nThis project seeks to give an overview of the history of matrices and its practical applications, touching on the various topics used in concordance with it.\n\nAround 4000 years ago, the people of Babylon knew how to solve a simple 2X2 system of linear equations with two unknowns. Around 200 BC, the Chinese published “Nine Chapters of the Mathematical Art, “which displayed the ability to solve a 3X3 system of equations. (Perotti) . The power and progress of Matrices and its application did not come to fruition until the late 17th century.\n\nThe emergence of the subject came from determinants, values connected to a square matrix, studied by the founder of calculus, Leibnitz, in the late 17th century. Lagrange came out with his work regarding Lagrange multipliers, a way to “characterize the maxima and minima multivariate functions.” (Darkwing) More than fifty-year later, Cramer presented his ideas of solving systems of linear equations based on determinants more than50 years after Leibnitz (Darkwing). Interestingly, Cramer provided no proof for solving an n x n system.\n\nAs mentioned before, Gauss’ work initially dealt much with solving linear equations but did not have as much to do with matrices. In order for matrix algebra to develop, a proper notation or method of describing the process was necessary. Also vital to this process was a definition of matrix multiplication and the facets involving it. “The introduction of matrix notation and the invention of the word matrix were motivated by attempts to develop the right algebraic language for studying determinants. In 1848, J.J. Sylvester introduced the term “matrix,” the Latin word for womb, as a name for an array of numbers. He used womb because see, linear algebra has become more relevant since the emergence of calculus, even though its the foundational equation of ax+ b=0 dates back centuries.\n\nEuler brought to light the idea that a system of equations doesn’t necessarily have to have a solution. He recognized the need for conditions to be placed upon unknown variables in order to find a solution. The initial work until this period mainly dealt with the concept of unique solutions and square matrices where the number of equations matched the number of unknowns.\n\nWith the turn intothe19th century, Gauss introduced a procedure for solving a linear equation system. His work dealt mainly with linear equations and had yet to bring in the idea of matrices or their notations. His efforts dealt with equations of differing numbers and variables as well as the traditional pre-19th century works of Euler, Leibnitz, and Cramer. Gauss’ work is now summed up in the term Gaussian elimination. This method uses the concepts of combining, swapping, or multiplying rows with each other in order to eliminate variables from certain equations. After variables are determined, the student is then to use back substitution to help find the remaining unknown variables.\n\nReviewed a matrix as a generator of determinants (Tucker, 1993). The other part, matrix multiplication or matrix algebra, came from the work of Arthur Cayley in 1855.\n\nCayley defined matrix multiplication as “the matrix of coefficients for thecomposite transformation T2T1 is the product of the matrix for T2 times the matrix of T1”(Tucker, 1993). His work dealing with Matrix multiplication culminated in his theorem, the Cayley-Hamilton Theorem. Simply stated, a square matrixsatisfiesMatrices at the end of the 19th century were heavily connected with Physics issues. For mathematicians, more attention was given to vectors as they proved to be basic mathematical elements. With the advancement of technology using the methods of Cayley, Gauss, Leibnitz, Euler, and others, determinants and linear algebra moved forward more quickly and effectively. Regardless of the technology, though, Gaussian elimination is still the best way to solve a system of linear equations (Tucker,1993).\n\nThe influence of matrices and it’s applications in the mathematical world is spread wide because it provides an important base for many principles and practices. Some of the things Matrices is used for are to solve systems of linear format, to find least-square best-fit lines to predict future outcomes or find trends, and to encode and decode messages. Other more broad topics that it is used for are solving energy questions in quantum mechanics. It is also used to create simple everyday household games like Sudoku. It is because of these practical applications thatMatriceshas spread so far and advanced. The key, however, is to understand that the history of linear algebra provides the basis for these applications.\n\nAlthough linear algebra is a fairly new subject compared to other mathematical practices, its uses are widespread. With the efforts of calculus-savvy Leibnitz, the concept of using systems of linear equations to solve unknowns was formalized. Other efforts from scholars like Cayley. Euler, Sylvester, and others changed matrices into the use of linear algebra to represent them. Gauss brought his theory to solve systems of equations proving to be the most effective basis for solving unknowns.\n\nTechnology continues to push the use further and further, but the history of matrices and its application continues to provide the foundation. Even though every few years, companies update their textbooks; the fundamentals stay the same.(laura smoller (2001).\n\n## 1.2 STATEMENT OF PROBLEM\n\nDue to the great need for security for passing sensitive information from one person to another or from one organization to another through electronic technology, cryptography is needed as a solution to this problem.\n\nAlso, in economics, this research will discuss how the Leontief model represents the economy as a system of linear equations to calculate the gross domestic products and goods production efficiently.\n\n## 1.3 AIMS AND OBJECTIVES\n\n1. To apply matrices to Cryptography, Economic Models and Systems of Linear Equations.\n2. To improve the methods by which an increase in production output can be achieved.\n3. To show ways in which sensitive information can be passed across mathematically.\n4. To disseminate these improved methods to the relevant communities and end-use.\n\n## REFERENCES\n\n1. Abraham S. (1966.) Elementary cryptanalysis a mathematical approach. Mathematical Association of America, mathematical library,\n\n2. Alan G. Konhem, (1981) cryptography: a primer New York: Wiley inheritance,\n\n3. Anton, H and C. Rorres. (1994) Elementary linear algebra applications version, 11th edition. 4. John Wilky and sons, int.\n\n4. Budiman Johannes, (2001) New York: springer,\n\n5. Carlson David: Johnson, Charles R; Lay, David C; porter A. Duane (1993). The linear algebra Curriculum Study Group Recommendations for the first course in linear Algebra. The college mathematics journal 24(1):41-46\n\n6. Froelich, freeman. (1992) Contemporary precalculus through applications, NCSSM, everyday learning.\n\n7. Jelf Christensen April (2012) find project math 2270 Grant Gustafson University of Utah.\n\n8. Kolman, B. (1999) introductory linear algebra with applications\n\n9. Laura Smoller, (April (2001)) UALR Department of History\n\n10. Lay, David, C. (2006.) linear algebra and its applications, Third Edition. New York: Pearson,\n\n11. Leontief, Wassily. (2006) Input-output economics, second edition New York Oxford University Press.\n\n12. Linear Algebra: (2000) Determinant, inverses, Rank.pdf\n\n13. Perotti (N.D) (2001) history of linear algebra. Retrieved from http//www.science.unitnith/perotti/history of linear algebra.pdf\n\n14. Sohn, Ira, Eds. (1986) Reading input-output analysis theory and application. New York, oxford university press\n\n15. Strang G. (1993). The fundamental theorem of linear algebra. The America mathematical monthly, 100(9), 848-855\n\n16. Tucker A. (1993). The growing importance of linear algebra in undergraduate mathematics. The college mathematics journal.\n\n17. Vitull:, Marie. (2012) “a brief History of linear Algebra and matrix theory” Department of mathematics. The University of Oregon. Archived from the original retrieved 2014-07-08\n\n18. Weinstein, E.W. (1995) linear algebra, from math world, a wolfram web resource https://mathworld.wolfram.com/LinearAlgebra.html.\n\nMatrices and Its Applications\n\n## The Framework of Economic...\n\nThe Framework of Economic and Human Rights Under the...\n\n## The Defence of Mistake...\n\nABSTRACT The importance of Criminal Law as a vehicle for...\n\n## Corporate Capacity and the...\n\nABSTRACT Any act of a company outside the Memorandum and...\n\n## Appraisal of the Impediments...\n\nCHAPTER ONE APPRAISAL OF THE IMPEDIMENTS TO INHERITANCE UNDER ISLAMIC...\n\n## Strategic Management in the...\n\nABSTRACT The article defines the constructive role of strategic management...\n\n## The Defence of Provocation...\n\nAbstract The aim of the thesis is to make a...\n\n### The Framework of Economic and Human Rights Under the ECOWAS Treaty\n\nThe Framework of Economic and Human Rights Under the ECOWAS Treaty: A Case Study of Implementation in Nigeria ABSTRACT The 1975 ECOWAS Treaty made by the...\n\n### The Defence of Mistake in Nigerian Law\n\nABSTRACT The importance of Criminal Law as a vehicle for the advancement of humanity cannot be overemphasized. \"This is the law on which men place their...\n\n### Corporate Capacity and the Ultra Vires Rule Under Nigerian Law\n\nABSTRACT Any act of a company outside the Memorandum and Articles of Association of the company or the statute(s) creating the company is Ultra Vires....\n\n### Appraisal of the Impediments to Inheritance Under Islamic Law\n\nCHAPTER ONE APPRAISAL OF THE IMPEDIMENTS TO INHERITANCE UNDER ISLAMIC LAW 1.1 GENERAL BACKGROUND Some obstacles and impediments prevent an inheritor from benefiting from the assets left...\n\n### Strategic Management in the Formation of the Company Personnel\n\nABSTRACT The article defines the constructive role of strategic management in forming and developing company personnel. The original model of work with company staff is...\n\n### The Defence of Provocation in Nigeria and Sudan – A Comparative Study\n\nAbstract The aim of the thesis is to make a comparative study of the defence of provocation in Nigeria and Sudan. Provocation as a defence...\n\n### Appellate Jurisdiction of Nigerian Courts in Civil Matters\n\nAbstract The majority of appellants and some appellate courts are ignorant or oblivious to the proper rules of appeal, resulting in a lot of injustices...\n\n### Assessment of Petroleum Profit Tax Under the Nigerian Tax Laws\n\nCHAPTER ONE 1.0 GENERAL INTRODUCTION 1.1 Introduction Petroleum Profit Tax Act provides that: Assessment of tax shall be made in such form and in such manner as the...\n\n### A Study of Customary Land Law and Tenure Practices of Six Communities of the Lower Benue River Valley of Nigeria\n\nABSTRACT This study is aimed at studying the customary land laws and tenurial practices of the communities of the Nigerian Lower Benue River valley. These..."
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http://usecase.harmis.com.br/7lp44oa/focal-length-of-concave-mirror-depends-on-a4f73f | [
"## 27 nov focal length of concave mirror depends on\n\nIn which case will the student get more accurate value of focal length? Thus, the focal length of a concave mirror can be estimated by obtaining a real image of a distant object at its focus. [NCERT] Answer: A thick convex lens has shorter focal length. The measuring scale should be parallel to the base of both the stands. Answer: The reason behind the blurred image is that the mirror is away from the object. [NCERT] Answer: Concave mirror is used as a shaving mirror or in vanity boxes, because when the object is placed between its focus and pole, the magnified, erect and virtual images of the object will be formed. What change do you expect, if the lens were rather thick? In such cookers, what should be the preferable position of food vessel for cooking? How do you calculate the total resistance of a parallel circuit? The mirror holder along with the mirror should be kept perpendicular to the measuring scale for precise measurements. A beam of light generally converges after reflection from its surface, hence it is also called convergent mirror (Fig. Answer: The nature of image formed in this experiment is as follows: Question 9. How does the size of the image vary as an object is moved from close to the pole to a large distance? Name this point. This point is called the centre of curvature of the spherical mirror. As the distance between the pole 0 of the concave mirror and the focus F is the focal length of the concave mirror. In reflector type solar cookers, special concave (parabolic) mirrors are used. Answer: It depends upon the focal length of the lens. The principal axis of the convex lens should be horizontal, i.e. What happens to a ray of light when it passes through the optical centre of a lens? Answer: When a ray of light passes through the optical centre of a lens, it goes without bending. Result From the above observations and calculations, the approximate value of focal length of the given convex lens is ………. 1). A concave mirror produces both real and virtual images, which can be upright or inverted. 4). The polished surface of the concave mirror and the distinct object should be facing each other. Which surface of a shining spoon should be polished, to use the spoon as a concave mirror? Aim To determine the focal length of concave mirror by obtaining the image of a distant object. There should not be any obstacle in the path of rays of light incident on the concave mirror. Least count of measuring scale may not be correctly noted. A double convex lens is simply called convex lens. Adjust the distance of screen, so that the image of the distant object is formed on it as given in the figure below. Which type of lens is used by the watchmakers, while repairing five parts of a wrist watch? If the lens used in the experiment is a plano-convex lens, then what is the radius of curvature of the plane surface? How can we find the focal length of a concave mirror, when the image is obtained by using a concave mirror? A concave mirror should be always placed near an open window. Answer: The radius of curvature of the plane surface of plano-convex lens is infinity. In a concave mirror, rays of light are parallel to its principal axis and meet at a single point on the principal axis, after reflection from the mirror (as shown in Fig. Hence, it is also called a converging mirror. Question 4. [NCERT] Answer: Question 2. Question 7. The mirror holder, along with the mirror, may not be kept perpendicular to the measuring scale. Question 15. Why? A concave mirror is a curved mirror that forms a part of a sphere and designed in such a way that rays of light falling on its shiny surface converge upon reflection. The size of the image can be bigger or smaller than the object. 4. This expression is valid for concave as well as convex spherical mirrors. Question 11. Question 14. In which case, will the student is closer to accurate value of focal length? [NCERT] Answer: A concave mirror is the spherical mirror with inward curved reflecting surface, whereas a convex mirror is the spherical mirror with outward curved reflecting surface. Is the centre of curvature a part of a spherical mirror? Answer: When parallel rays of light fall on a concave mirror along its axis, then the rays meet at a point infront of the mirror after reflection from it. Question 6. The image of the sun should never be seen directly with the naked eye or it should never be focussed with a convex lens on any part of the body, paper or any inflammable material as it can burn. All u, v and f should be according to sign convention. cm, Focal length for first object (f1) = ………… m Focal length for second object (f2) = ………….. m Focal length for third object (f3) = …………. This point is one of the two foci of the lens. Why do we obtain blurred image from a concave mirror sometimes? Comment in support of your answer. The measuring scale may not be parallel to the base of both the stands. The nature, size, and position of the image depend on the position of the object. Actually, this spot of light is the image of the Sun on the sheet of paper. Answer: The surface of the spoon which is bulged outward should be polished to use the spoon as a concave mirror."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.92450696,"math_prob":0.9607609,"size":5458,"snap":"2021-04-2021-17","text_gpt3_token_len":1195,"char_repetition_ratio":0.1696003,"word_repetition_ratio":0.051282052,"special_character_ratio":0.2128985,"punctuation_ratio":0.10877514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801488,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-14T02:29:49Z\",\"WARC-Record-ID\":\"<urn:uuid:829f7997-784a-4e6f-b75d-570911b9cfb1>\",\"Content-Length\":\"48031\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:609c5f32-7742-4ce3-ab6b-bc5c8164b95f>\",\"WARC-Concurrent-To\":\"<urn:uuid:7dc826cb-0193-45ca-9809-4dc1dc3785b1>\",\"WARC-IP-Address\":\"52.90.77.15\",\"WARC-Target-URI\":\"http://usecase.harmis.com.br/7lp44oa/focal-length-of-concave-mirror-depends-on-a4f73f\",\"WARC-Payload-Digest\":\"sha1:6TOBBKQJ6SKUBWMQKUDIT5ZCNC7XR47J\",\"WARC-Block-Digest\":\"sha1:WENOF7CQVV5G3X3MXTXAN7SJ7IM5T2TQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038076454.41_warc_CC-MAIN-20210414004149-20210414034149-00594.warc.gz\"}"} |
https://discourse.mcneel.com/t/neighbours-of-each-point-in-a-nested-list/63732 | [
"",
null,
"# Neighbours of each point in a nested list\n\nHello,\nI am trying to find the neighbours of each point. then I will measure the distance between each point with its neighbour.\nHow I can exclude the points on the edge?\n\nimport Rhino.Geometry as rg\nimport rhinoscriptsyntax as rs\nimport scriptcontext as sc\n\npoints =\nlines=\n\nfor j in range (len(curves)-1):\nfirst_points = rs.DivideCurve(curves[j], div)\nsecond_points=rs.DivideCurve(curves[j+1], div)\nfor i in range(div+1):\nlines.append(Lines)\nfor line in lines:\nrow_points = rs.DivideCurve(line, div)\npoints.append(row_points)\n\nflat_points = [item for sublist in points for item in sublist]\n\ni = 5\nj = 7\n\n#orthogonal neighbours\npoints[i+1][j]\npoints[i-1][j]\npoints[i][j+1]\npoints[i][j-1]\n\n#diagonal neighbours\npoints[i+1][j+1]\npoints[i+1][j-1]\npoints[i-1][j+1]\npoints[i-1][j-1]"
]
| [
null,
"https://aws1.discourse-cdn.com/mcneel/original/3X/3/6/36bfa7aa69160ae4bb6e6d46fea7a7516e4c8fb8.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7751134,"math_prob":0.99466866,"size":822,"snap":"2020-45-2020-50","text_gpt3_token_len":244,"char_repetition_ratio":0.17848411,"word_repetition_ratio":0.0,"special_character_ratio":0.2785888,"punctuation_ratio":0.10778443,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99672544,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-01T12:16:23Z\",\"WARC-Record-ID\":\"<urn:uuid:ca02d5b2-47b8-4787-a6fd-383eed8ea712>\",\"Content-Length\":\"14763\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d67c922f-d74a-4a56-8800-ac245e3dcfc4>\",\"WARC-Concurrent-To\":\"<urn:uuid:b433f87d-0f0b-49ad-ac25-cc7494efdcd2>\",\"WARC-IP-Address\":\"64.62.250.111\",\"WARC-Target-URI\":\"https://discourse.mcneel.com/t/neighbours-of-each-point-in-a-nested-list/63732\",\"WARC-Payload-Digest\":\"sha1:PJSFMY5HZGC2HJ6TKUPRCRDW7UVGZBLS\",\"WARC-Block-Digest\":\"sha1:JGLTFQ57UL2CGAU4UH7GFJKRN6FJQ2MI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141674082.61_warc_CC-MAIN-20201201104718-20201201134718-00582.warc.gz\"}"} |
https://docs.stingray.science/notebooks/Transfer%20Functions/TransferFunction%20Tutorial.html | [
"# Contents¶\n\nThis notebook covers the basics of creating TransferFunction object, obtaining time and energy resolved responses, plotting them and using IO methods available. Finally, artificial responses are introduced which provide a way for quick testing.\n\n# Setup¶\n\nSet up some useful libraries.\n\n:\n\nimport numpy as np\nfrom matplotlib import pyplot as plt\n%matplotlib inline\n\n\nImport relevant stingray libraries.\n\n:\n\nfrom stingray.simulator.transfer import TransferFunction\nfrom stingray.simulator.transfer import simple_ir, relativistic_ir\n\n\n## Creating TransferFunction¶\n\nA transfer function can be initialized by passing a 2-d array containing time across the first dimension and energy across the second. For example, if the 2-d array is defined by arr, then arr defines a time of 5 units and energy of 1 unit.\n\nFor the purpose of this tutorial, we have stored a 2-d array in a text file named intensity.txt. The script to generate this file is explained in Data Preparation notebook.\n\n:\n\nresponse = np.loadtxt('intensity.txt')\n\n\nInitialize transfer function by passing the array defined above.\n\n:\n\ntransfer = TransferFunction(response)\ntransfer.data.shape\n\n:\n\n(524, 744)\n\n\nBy default, time and energy spacing across both axes are set to 1. However, they can be changed by supplying additional parameters dt and de.\n\n## Obtaining Time-Resolved Response¶\n\nThe 2-d transfer function can be converted into a time-resolved/energy-averaged response.\n\n:\n\ntransfer.time_response()\n\n\nThis sets time parameter which can be accessed by transfer.time\n\n:\n\ntransfer.time[1:10]\n\n:\n\narray([ 0., 0., 0., 0., 0., 0., 0., 0., 0.])\n\n\nAdditionally, energy interval over which to average, can be specified by specifying e0 and e1 parameters.\n\n## Obtaining Energy-Resolved Response¶\n\nEnergy-resolved/time-averaged response can be also be formed from 2-d transfer function.\n\n:\n\ntransfer.energy_response()\n\n\nThis sets energy parameter which can be accessed by transfer.energy\n\n:\n\ntransfer.energy[1:10]\n\n:\n\narray([ 0., 0., 0., 0., 0., 0., 0., 0., 0.])\n\n\n## Plotting Responses¶\n\nTransferFunction() creates plots of time-resolved, energy-resolved and 2-d responses. These plots can be saved by setting save parameter.\n\n:\n\ntransfer.plot(response='2d')",
null,
":\n\ntransfer.plot(response='time')",
null,
":\n\ntransfer.plot(response='energy')",
null,
"By enabling save=True parameter, the plots can be also saved.\n\n## IO¶\n\nTransferFunction can be saved in pickle format and retrieved later.\n\n:\n\ntransfer.write('transfer.pickle')\n\n\nSaved files can be read using static read() method.\n\n:\n\ntransfer_new = TransferFunction.read('transfer.pickle')\ntransfer_new.time[1:10]\n\n:\n\narray([ 0., 0., 0., 0., 0., 0., 0., 0., 0.])\n\n\n## Artificial Responses¶\n\nFor quick testing, two helper impulse response models are provided.\n\n### 1- Simple IR¶\n\nsimple_ir() allows to define an impulse response of constant height. It takes in time resolution starting time, width and intensity as arguments.\n\n:\n\ns_ir = simple_ir(dt=0.125, start=10, width=5, intensity=0.1)\nplt.plot(s_ir)\n\n:\n\n[<matplotlib.lines.Line2D at 0x112d48990>]",
null,
"### 2- Relativistic IR¶\n\nA more realistic impulse response mimicking black hole dynamics can be created using relativistic_ir(). Its arguments are: time_resolution, primary peak time, secondary peak time, end time, primary peak value, secondary peak value, rise slope and decay slope. These paramaters are set to appropriate values by default.\n\n:\n\nr_ir = relativistic_ir(dt=0.125)\nplt.plot(r_ir)\n\n:\n\n[<matplotlib.lines.Line2D at 0x10cca92d0>]",
null,
""
]
| [
null,
"https://docs.stingray.science/_images/notebooks_Transfer_Functions_TransferFunction_Tutorial_26_0.png",
null,
"https://docs.stingray.science/_images/notebooks_Transfer_Functions_TransferFunction_Tutorial_27_0.png",
null,
"https://docs.stingray.science/_images/notebooks_Transfer_Functions_TransferFunction_Tutorial_28_0.png",
null,
"https://docs.stingray.science/_images/notebooks_Transfer_Functions_TransferFunction_Tutorial_39_1.png",
null,
"https://docs.stingray.science/_images/notebooks_Transfer_Functions_TransferFunction_Tutorial_42_1.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.66168755,"math_prob":0.93156844,"size":3480,"snap":"2023-40-2023-50","text_gpt3_token_len":876,"char_repetition_ratio":0.16513233,"word_repetition_ratio":0.04761905,"special_character_ratio":0.26465517,"punctuation_ratio":0.22733814,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9777043,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-23T03:23:29Z\",\"WARC-Record-ID\":\"<urn:uuid:4fb926ef-c969-488d-8ab3-49f82a1fd9b0>\",\"Content-Length\":\"24054\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3b0451a8-5aab-484c-b485-1d938f5e1d15>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1211987-5192-4d06-8daf-19d571bb76e8>\",\"WARC-IP-Address\":\"104.21.91.153\",\"WARC-Target-URI\":\"https://docs.stingray.science/notebooks/Transfer%20Functions/TransferFunction%20Tutorial.html\",\"WARC-Payload-Digest\":\"sha1:NBUWFYW26FGVJ2LJ4H5P5ICO3AN2E7RX\",\"WARC-Block-Digest\":\"sha1:26LS3BRVNCYKIKVGRDKOMOKYWZK7LPYL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506479.32_warc_CC-MAIN-20230923030601-20230923060601-00443.warc.gz\"}"} |
https://www.hackmath.net/en/word-math-problems/basic-functions?page_num=90 | [
"# Basic functions - math word problems\n\n1. Boys to girls",
null,
"The ratio of boys to girls in a party is 3:5 . If 6 more boys arrived and 4 girls left the party, the ratio of boys to girls would be 5:6 . How many are in the party originally?\n2. Red diplomas",
null,
"He numbers of students with honors in 2013 and 2014 are in ratio 40:49. How big is the year-on-year percentage increase?\n3. Pupils",
null,
"There are 350 girls in the school, and the other 30% of the total number of pupils are boys. How many pupils does the school have?\n4. The sides",
null,
"The sides of a rectangle are in a ratio of 2:3, and its perimeter is 1 1/4 inches. What are the lengths of its side? Draw it.\n5. Prime number",
null,
"Jan wrote any number from 1 to 20. What is the probability that he wrote the prime number?\n6. Clock's gears",
null,
"In the clock machine, three gears fit together. The largest has 168 teeth, the middle 90 teeth, and the smallest 48 teeth. The middle wheel turns around its axis in 90 seconds. How many times during the day do all the gears meet in the starting position?\n7. Mixed2improper",
null,
"Write the mixed number as an improper fraction. 166 2/3\n8. Sum of fractions",
null,
"What is the sum of 2/3+3/5?\n9. Sewing",
null,
"Beth's mother can sew 235 pairs of short pants in 6 days while Lourdes can sew 187 pairs in 8 days. How many more pairs of short pants can Beth's mother sew?\n10. Meat loses",
null,
"Meat loses 18% of its weight by smoking. How much raw meat butcher used to manufacture 35 kilos of smoked?\n11. Sugars",
null,
"In what ratio must two sorts of sugar, costing #390 and #315 per kg respectively, be mixed in order to produce a mixture worth #369 per kg?\n12. Sewing",
null,
"The lady cut off one half of cloth. She needed three-quarters of this piece to sew a skirt. What part of the original piece of cloth still remained?\n13. Desks",
null,
"A class has 20 students. The classroom consists of 20 desks, with 4 desks in each of 5 different rows. Amy, Bob, Chloe, and David are all friends, and would like to sit in the same row. How many possible seating arrangements are there such that Amy, Bob, C\n14. Progression",
null,
"12, 60, -300,1500 need next 2 numbers of pattern\n15. Dining questionnaire",
null,
"In the class of 32 pupils, the class teacher delivered a satisfaction questionnaire to the school canteen. Eleven people have questioned whether they are satisfied with the dining room. The completed questionnaire also included the vote of the class teache\n16. New ratio",
null,
"The ratio of ducks and chicken in our yard is 2 : 3. The total number of ducks and chickens together is 30. Mother gave 3 of the chickens to our neighbor. What is the new ratio now?\n17. Wiring 2",
null,
"Willie cut a piece of wire that was 3/8 of the total length of the wire. He cut another piece that was 9m long. The two pieces together were one half of the total length of the wire. How long was the wire before he cut it?\n18. SSA and geometry",
null,
"The distance between the points P and Q was 356 m measured in the terrain. The PQ line can be seen from the viewer at a viewing angle of 107° 22 '. The observer's distance from P is 271 m. Determine the viewing angle of P and observer.\n19. Six-digit primes",
null,
"Find all six-digit prime numbers that contain each one of digits 1,2,4,5,7 and 8 just once. How many are they?\n20. Discount price",
null,
"Coat cost 150 euros after sales discount. What is the original price when the discount is 25% of the original price?\n\nDo you have an interesting mathematical word problem that you can't solve it? Enter it, and we can try to solve it.\n\nTo this e-mail address, we will reply solution; solved examples are also published here. Please enter the e-mail correctly and check whether you don't have a full mailbox.\n\nPlease do not submit problems from current active competitions such as Mathematical Olympiad, correspondence seminars etc..."
]
| [
null,
"https://www.hackmath.net/thumb/34/t_6834.jpg",
null,
"https://www.hackmath.net/thumb/8/t_8008.jpg",
null,
"https://www.hackmath.net/thumb/42/t_7642.jpg",
null,
"https://www.hackmath.net/thumb/85/t_6285.jpg",
null,
"https://www.hackmath.net/thumb/30/t_7130.jpg",
null,
"https://www.hackmath.net/thumb/23/t_7023.jpg",
null,
"https://www.hackmath.net/thumb/15/t_5515.jpg",
null,
"https://www.hackmath.net/thumb/78/t_6878.jpg",
null,
"https://www.hackmath.net/thumb/90/t_6790.jpg",
null,
"https://www.hackmath.net/thumb/77/t_5977.jpg",
null,
"https://www.hackmath.net/thumb/71/t_6971.jpg",
null,
"https://www.hackmath.net/thumb/50/t_6850.jpg",
null,
"https://www.hackmath.net/thumb/7/t_5507.jpg",
null,
"https://www.hackmath.net/thumb/68/t_6968.jpg",
null,
"https://www.hackmath.net/thumb/20/t_6920.jpg",
null,
"https://www.hackmath.net/thumb/32/t_6832.jpg",
null,
"https://www.hackmath.net/thumb/7/t_6807.jpg",
null,
"https://www.hackmath.net/thumb/26/t_6826.jpg",
null,
"https://www.hackmath.net/thumb/24/t_7524.jpg",
null,
"https://www.hackmath.net/thumb/37/t_7937.jpg",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.9265497,"math_prob":0.86766905,"size":3954,"snap":"2019-51-2020-05","text_gpt3_token_len":1205,"char_repetition_ratio":0.10278481,"word_repetition_ratio":0.009153318,"special_character_ratio":0.36697015,"punctuation_ratio":0.092857145,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9513361,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-13T11:03:44Z\",\"WARC-Record-ID\":\"<urn:uuid:aba25f41-b5b7-4185-87f3-9415b65bfa02>\",\"Content-Length\":\"29836\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9fee5e11-d75a-4f25-9e2b-64073d40c08f>\",\"WARC-Concurrent-To\":\"<urn:uuid:66d7674f-0cc1-4072-860a-79d59fad1450>\",\"WARC-IP-Address\":\"104.24.105.91\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/word-math-problems/basic-functions?page_num=90\",\"WARC-Payload-Digest\":\"sha1:YKZTUWPIVWGD5NQ34Y2A2AOFLEDS52EO\",\"WARC-Block-Digest\":\"sha1:G4NAOVZDG2M6GRK7NV7VPKVCRPGCKHJ7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540553486.23_warc_CC-MAIN-20191213094833-20191213122833-00024.warc.gz\"}"} |
https://mathoverflow.net/questions/317390/does-every-section-of-the-map-gal-overlinek-t-k-t-rightarrow | [
"# Does every section of the map Gal$(\\overline{k(\\!(t)\\!)}/k(\\!(t)\\!))\\rightarrow$ Gal$(\\overline{k}/k)$ stabilize a compatible system of roots of $t$?\n\nThere may be some technical issues with the question, but hopefully what I mean is clear...\n\nLet $$k$$ be a number field (or maybe any finitely generated field over $$\\mathbb{Q}$$ of characteristic 0)\n\nLet $$k(\\!(t)\\!)$$ be the field of Laurent series in $$t$$ with coefficients in $$k$$, and let $$\\Omega$$ denote an algebraic closure of $$k(\\!(t)\\!)$$, and let $$\\overline{k}$$ denote the algebraic closure of $$k$$ inside $$\\Omega$$.\n\nThere is a natural map $$\\rho : \\mathrm{Gal}(\\Omega/k(\\!(t)\\!))\\rightarrow \\mathrm{Gal}(\\overline{k}/k)$$ given by restriction.\n\nIf $$\\{t_n\\}_{n\\ge 1}\\subset \\Omega$$ satisfies $$t_1 = t$$, $$t_n^d = t_{n/d}$$ for all $$d\\mid n$$, then we say that it is a compatible system of roots of $$t$$.\n\nGiven a compatible system of roots $$\\{t_n\\}$$, and a filtration $$\\overline{k} = \\bigcup L$$ by finite extensions of $$k$$, the tensor product $$\\left(\\varinjlim_L L(\\!(t)\\!)\\right)\\otimes_{k(\\!(t)\\!)} \\left(\\varinjlim_n k(\\!(t)\\!)(t_n)\\right)$$ is a field isomorphic to $$\\Omega$$, and thus we obtain an section of the map $$\\rho$$ by having $$\\sigma\\in \\mathrm{Gal}(\\overline{k}/k)$$ act on the tensor product in the obvious way in the first factor, and trivially on the second factor. In particular, this action stabilizes the compatible system of roots $$\\{t_n\\}$$.\n\nDoes every section of $$\\rho$$ stabilize some compatible system of roots $$\\{t_n\\}$$?\n\nNote that the kernel of $$\\rho$$ is the subgroup $$\\mathrm{Gal}(\\Omega/E)$$ where $$E = \\varinjlim_L L(\\!(t)\\!)$$ as before, and the Galois group is isomorphic to $$\\widehat{\\mathbb{Z}}$$. The group $$\\mathrm{Gal}(\\overline{k}/k)$$ acts on $$\\mathrm{Gal}(\\Omega/E) \\cong \\widehat{\\mathbb{Z}}$$ via the cyclotomic character, and if $$k$$ is finitely generated over $$\\mathbb{Q}$$ then the only element of $$\\widehat{\\mathbb{Z}}$$ fixed by all of $$\\mathrm{Gal}(\\Omega/E)$$ is the identity. This implies that if a section of $$\\rho$$ comes from a compatible system, then it must be unique.\n\n$$\\newcommand{\\Gal}{\\mathrm{Gal}}\\newcommand{\\Z}{\\mathbb{Z}}$$Fix a compatible system $$(t_n)$$ of roots of $$t$$. It provides us with a section of $$\\rho$$ thus giving an isomorphism between $$\\Gal(\\overline{k((t))}/k((t)))$$ and the semi-direct product $$\\Gal(\\overline{k}/k)\\ltimes \\hat{\\Z}(1)$$ where $$\\hat{\\Z}(1)$$ denotes $$\\lim\\limits_{\\leftarrow}\\mu_n(\\bar{k})$$ as a $$\\Gal(\\bar{k}/k)$$-module. An element $$(g,m)$$ acts on $$\\overline{k((t))}$$ as follows: in terms of your decomposition $$\\overline{k((t))}=\\left(\\varinjlim_L L(\\!(t)\\!)\\right)\\otimes_{k(\\!(t)\\!)} \\left(\\varinjlim_n k(\\!(t)\\!)(t_n)\\right)$$ it acts on the first factor as prescrivbed by $$g$$ and, on the second factor sends $$t_n$$ to $$m_nt_n$$ where $$m_n$$ is the projection of $$m$$ to $$\\mu_n(\\bar{k})$$.\n\nIf $$(t'_n)$$ is another compatible system of roots of $$t$$, it provides another section $$\\Gal(\\bar{k}/k)\\to \\Gal(\\overline{k((t))}/k((t)))$$ which is characterized by the fact that it preserves all the elements $$t'_n$$. In terms of the semi-direct product, this section then looks like $$g\\mapsto (g,g(m)-m)$$ where $$m\\in\\hat{\\Z}(1)$$ is the element whose components $$m_n\\in\\mu_n(\\bar{k})$$ are given by $$t_n^{-1}t'_n$$\n\nIn general, if $$G\\ltimes M$$ is the semidirect product of a profinite group $$G$$ with a continuous module $$M$$, a giving a continuous section $$s:G\\to G\\ltimes M$$ is equivalent to giving a map $$f:G\\to M$$ such that $$f(g_1g_2)=g_1f(g_2)+f(g_1)$$.\n\nComing back to our situation we see that section are in bijection with continuous 1-cocycles of the group $$\\Gal(\\bar{k}/k)$$ with coefficients in $$\\hat{\\Z}(1)$$ and a section comes from a system of roots if and only if this cocycle is a coboundary. In other words, we get\n\nLemma. Every continuous section comes from a compatible system of roots of $$t$$ if and only if $$H^1_{cont}(\\Gal(\\bar{k}/k),\\hat{\\Z}(1))=0$$.\n\nIn your setting, this group is never zero, for example, from the exact sequence $$0\\to\\hat{\\Z}(1)\\xrightarrow{n}\\hat{\\Z}(1)\\to \\mu_n\\to 0$$ we see that $$n$$-torsion in this cohomology group is given by $$H^1_{cont}(\\Gal(\\bar{k}/k),\\hat{\\Z}(1))[n]=\\mathrm{coker}(\\hat{\\Z}(1)^{\\Gal(\\bar{k}/k)}\\to \\mu_n(k))$$ Since the order of roots of unity contained in $$k$$ is bounded, $$\\hat{\\Z}(1)^{\\Gal(\\bar{k}/k)}$$ is zero, but, for instance, $$\\mu_2(k)$$ is equal to $$\\Z/2$$ for every $$k$$.\n\nSo, there are always other sections but you can control them by this Galois cohomology group."
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.7517291,"math_prob":1.0000087,"size":1899,"snap":"2020-24-2020-29","text_gpt3_token_len":574,"char_repetition_ratio":0.12242744,"word_repetition_ratio":0.0074626864,"special_character_ratio":0.31121644,"punctuation_ratio":0.10958904,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000077,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-04T22:16:13Z\",\"WARC-Record-ID\":\"<urn:uuid:a3af33c8-74ba-4d6d-ae59-92fd4c8b53c3>\",\"Content-Length\":\"128679\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:836b3614-b5cc-4e25-8212-2e6afb527759>\",\"WARC-Concurrent-To\":\"<urn:uuid:daf9860f-4f08-4515-9586-f9f5d6af2067>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/317390/does-every-section-of-the-map-gal-overlinek-t-k-t-rightarrow\",\"WARC-Payload-Digest\":\"sha1:5UZDH3VTB6DXYNTLOLL2E5V5XIMCIXEI\",\"WARC-Block-Digest\":\"sha1:NIEYH7G2BPHJHAYPDGQIL6W7OBXKAMXK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347458095.68_warc_CC-MAIN-20200604192256-20200604222256-00441.warc.gz\"}"} |
https://m.everything2.com/title/Perpendicular+function | [
"display | more...\n\nThe name's a bit obscure1, but it basically means this: If you give me a function f(x), I can give you a function g(x)+c (where c is any real number), which will make a right angle wherever it intersects your function f(x).\n\nPerhaps that was even more obscure... Let's take an example. Think of the functions f(x)=x and g(x)=-x+c\nThe function g(x) will always make a right angle with the function f(x), no matter what c.\n\nBut I can do this for any most functions you give me.\n\n• x² -> -0.5Ln|x|+c\n\n• x³ -> 1/(3x)+c\n\n• 3x²+5x-3 -> (-1/6)Ln|-6x-5|+c\n\n• Sin(x) -> Ln|Cos(x/2)-Sin(x/2)|-Ln|Cos(x/2)+Sin(x/2)|+c\n\nSo how do I do it? Well, like this:\nSay you take 1 point on a function. That point has a certain slope. If you want a line perpendicular to the tangent line in that point, do -1 divided by the slope (since the slope of a line times the slope of the perpendicular line = -1 so -1 divided by a slope gives you the perpendicular one). This will give you the slope of the line making a right angle with the tangent line (where the function of that line will be (slope)x+b).\n\nThe slope is given by the derivative. For f(x)=x² the slope is given by f'(x)=2x. This function f'(x)=2x, is actually the slope in every point on the original graph. Now, if we do -1 divided by that slope in every point, we get g'(x)=-1/2x which is the slope in every point of some other function. (Actually, of the function we want)\n\nIf we integrate this function again, we get a function which is perpendicular to the original function everywhere, in this case, g(x)=-0.5Ln|x|+c\n\nSo, to recap: The primitive of -1 divided by the derivative of the original function, +c, gives you a function which makes a right angle on every intersection with the original function.\nOr\ng(x)+c= ∫ -1/f'(x) dx\n\nI've been asked what makes my set of functions so special since it's possible to define a function to almost be whatever you want it to be except near the points of intersection with the other function and still have it perpendicular. But the fact is, that it's possible to move g(x) 1 place up, which would give you different intersections. But then the intersections would still make right angles. In other words, the tangent of every point of f(x) makes a right angle with the tangent of every corresponding point of g(x)\nOr, when x is the same, the tangents of the functions in x make right angles, no matter what x.\n\nIf you have a graphical calculator or some graph program, you can try some of the functions above, or calculate your own one. Make sure to set the screen to square (On the TI-83: Zoom->ZSquare) to get a visual impression of the right angles.\n\nNow, I understand getting the integral of some functions may be troublesome. For example, I wouldn't have a clue on how to integrate -1/cos(x). But there is an answer! The wolfram integrator!\n\nhttp://integrals.wolfram.com\n\nThis gives you an actual function which is the integrated function of the one you submit, instead of the usual numeric approximations (if an answer actually exists). There are 2 things to note here though:\n\n• It notes Ln(x) as Log[x]. I've learned that Log(x) means 10Log(x), and so does my TI-83 interpret it too. That confused me at first. Ln(x) is eLog(x).\nNote: Swap told me that this is a common notation of Ln(x), since the natural log is the only thing important to mathematics at a later stage. \"There is but One True Log\". Thanks swap.\n• I've learned the integral of 1/x is Ln|x| (the bars mean Absolute(x)), but the integrator forgets the bars. I'm not sure what causes this, but you may want to keep this in mind, as you'll be running into it a lot with the -1/f'(x)\nFor example the perpendicular function of Sin(x) (look above), if you leave out the absolute bars, you only get a result on intervals [-0.5π+k2π , 0.5π+k2π] where k is any integer. If I put in the absolute bars however, I get a result for any real number, which is still correct. (Where correct in this case means: Perpendicular to Sin(x))\nTaking note of that, it's a pretty good way to integrate difficult functions. (I'm not affiliated with them in any way)"
]
| [
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https://astromodels.readthedocs.io/en/latest/notebooks/Band_Calderone.html | [
"# Band Calderone\n\n:\n\n# Parameters\nfunc_name = \"Band_Calderone\"\nwide_energy_range = True\nx_scale = \"log\"\ny_scale = \"log\"\nlinear_range = False\n\n\n## Description\n\n:\n\nfunc.display()\n\n• description: The Band model from Band et al. 1993, implemented however in a way which reduces the covariances between the parameters (Calderone et al., MNRAS, 448, 403C, 2015)\n• formula: $\\text{(Calderone et al., MNRAS, 448, 403C, 2015)}$\n• parameters:\n• alpha:\n• value: -1.0\n• desc: The index for x smaller than the x peak\n• min_value: -10.0\n• max_value: 10.0\n• unit:\n• is_normalization: False\n• delta: 0.1\n• free: True\n• beta:\n• value: -2.2\n• desc: index for x greater than the x peak (only if opt=1, i.e., for the Band model)\n• min_value: -7.0\n• max_value: -1.0\n• unit:\n• is_normalization: False\n• delta: 0.22000000000000003\n• free: True\n• xp:\n• value: 200.0\n• desc: position of the peak in the x*x*f(x) space (if x is energy, this is the nuFnu or SED space)\n• min_value: 0.0\n• max_value: None\n• unit:\n• is_normalization: False\n• delta: 20.0\n• free: True\n• F:\n• value: 1e-06\n• desc: integral in the band defined by a and b\n• min_value: None\n• max_value: None\n• unit:\n• is_normalization: True\n• delta: 1e-07\n• free: True\n• a:\n• value: 1.0\n• desc: lower limit of the band in which the integral will be computed\n• min_value: 0.0\n• max_value: None\n• unit:\n• is_normalization: False\n• delta: 0.1\n• free: False\n• b:\n• value: 10000.0\n• desc: upper limit of the band in which the integral will be computed\n• min_value: 0.0\n• max_value: None\n• unit:\n• is_normalization: False\n• delta: 1000.0\n• free: False\n• opt:\n• value: 1.0\n• desc: option to select the spectral model (0 corresponds to a cutoff power law, 1 to the Band model)\n• min_value: 0.0\n• max_value: 1.0\n• unit:\n• is_normalization: False\n• delta: 0.1\n• free: False\n\n## Shape\n\nThe shape of the function.\n\nIf this is not a photon model but a prior or linear function then ignore the units as these docs are auto-generated\n\n:\n\nfig, ax = plt.subplots()\n\nax.plot(energy_grid, func(energy_grid), color=blue)\n\nax.set_xlabel(\"energy (keV)\")\nax.set_ylabel(\"photon flux\")\nax.set_xscale(x_scale)\nax.set_yscale(y_scale)",
null,
"## F$$_{\\nu}$$\n\nThe F$$_{\\nu}$$ shape of the photon model if this is not a photon model, please ignore this auto-generated plot\n\n:\n\nfig, ax = plt.subplots()\n\nax.plot(energy_grid, energy_grid * func(energy_grid), red)\n\nax.set_xlabel(\"energy (keV)\")\nax.set_ylabel(r\"energy flux (F$_{\\nu}$)\")\nax.set_xscale(x_scale)\nax.set_yscale(y_scale)",
null,
"## $$\\nu$$F$$_{\\nu}$$\n\nThe $$\\nu$$F$$_{\\nu}$$ shape of the photon model if this is not a photon model, please ignore this auto-generated plot\n\n:\n\nfig, ax = plt.subplots()\n\nax.plot(energy_grid, energy_grid**2 * func(energy_grid), color=green)\n\nax.set_xlabel(\"energy (keV)\")\nax.set_ylabel(r\"$\\nu$F$_{\\nu}$\")\nax.set_xscale(x_scale)\nax.set_yscale(y_scale)",
null,
""
]
| [
null,
"https://astromodels.readthedocs.io/en/latest/_images/notebooks_Band_Calderone_8_0.png",
null,
"https://astromodels.readthedocs.io/en/latest/_images/notebooks_Band_Calderone_10_0.png",
null,
"https://astromodels.readthedocs.io/en/latest/_images/notebooks_Band_Calderone_12_0.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.5054657,"math_prob":0.9960382,"size":2816,"snap":"2022-40-2023-06","text_gpt3_token_len":932,"char_repetition_ratio":0.14509246,"word_repetition_ratio":0.25227273,"special_character_ratio":0.35404828,"punctuation_ratio":0.23865546,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99904156,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-08T07:47:05Z\",\"WARC-Record-ID\":\"<urn:uuid:168e58cf-baa9-4dbc-b114-d3c46eb5319b>\",\"Content-Length\":\"30424\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:888c40e4-a991-431a-8615-3a3cc20e127d>\",\"WARC-Concurrent-To\":\"<urn:uuid:a3020606-d8eb-41dd-9b4d-adc08ac10e0a>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://astromodels.readthedocs.io/en/latest/notebooks/Band_Calderone.html\",\"WARC-Payload-Digest\":\"sha1:WRGFNUP5M6N3IZPUBYENR4MHIL5XFTMM\",\"WARC-Block-Digest\":\"sha1:XZTGOYDKBHP6PIE2CJKLM7S44ARMJ366\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500719.31_warc_CC-MAIN-20230208060523-20230208090523-00043.warc.gz\"}"} |
https://www.dummies.com/software/microsoft-office-for-mac/excel-for-mac/10-functions-for-crunching-numbers-with-excel-on-your-ipad-or-mac/ | [
"",
null,
"10 Functions for Crunching Numbers with Excel on Your iPad or Mac - dummies\n\n# 10 Functions for Crunching Numbers with Excel on Your iPad or Mac\n\nExcel offers more than four hundred different functions to give you increased functionality for crunching the numbers on your iPad or Mac. Here are ten of the more interesting and useful functions you can use with Excel.\n\n## AVERAGE for averaging data\n\nMight as well start with an easy one. The AVERAGE function averages the values in a cell range. Here, AVERAGE is used to compute the average rainfall in a three‐month period in three different counties.\n\nUse AVERAGE as follows:\n\n`AVERAGE(cell range)`\n\nExcel ignores empty cells and logical values in the cell range; cells with 0 are computed.",
null,
"Using AVERAGE to find average rainfall data.’\n\n## COUNT and COUNTIF for tabulating data items\n\nUse COUNT, a statistical function, to count how many cells have data in them. Numbers and dates, not text entries, are counted. The COUNT function is useful for tabulating how many data items are in a range. In this spreadsheet, for example, COUNT is used to compute the number of mountains listed in the data:",
null,
"The COUNT (above) and COUNTIF (below) function at work.\n`COUNT(C5:C9)`\n\nUse COUNT as follows:\n\n`COUNT(cell range)`\n\nSimilar to COUNT is the COUNTIF function. It counts how many cells in a cell range have a specific value. To use COUNTIF, enter the cell range and a criterion in the argument, as follows. If the criterion is a text value, enclose it in quotation marks.\n\n`COUNTIF(cell range, criterion)`\n\nAt the bottom of the spreadsheet, the formula determines how many of the mountains in the data are in Nepal:\n\n`=COUNTIF(D5:D9,\"Nepal\")`\n\n## CONCATENATE for combining values\n\nCONCATENATE, a text function, is useful for combining values from different cells into a single cell. Use CONCATENATE as follows:\n\n`CONCATENATE(text1,text2,text3. . .)`\n\nTo include blank spaces in the text you’re combining, enclose a blank space between quotation marks as an argument. Moreover, you can include original text in the concatenation formula as long as you enclose it in quotation marks and enter it as a separate argument.",
null,
"Use the CONCATENATE function to combine values from cells.\n```=CONCATENATE(C3,\" \",D3,\".\",\" \",B3)\n=CONCATENATE(C11,\" \",D11,\".\",\" \",B11,\" \",\"lives in\",\" \",E11,\".\")```\n\n## PMT for calculating how much you can borrow\n\nUse the PMT (payment) function to explore how much you can borrow given different interest rates and different amounts. PMT determines how much you have to pay annually on different loans. After you determine how much you have to pay annually, you can divide this amount by 12 to see how much you have to pay monthly.\n\nUse the PMT function as follows to determine how much you pay annually for a loan:\n\n`PMT(interest rate, number of payments, amount of loan)`\n\nSet up a worksheet with five columns to explore loan scenarios:\n\n• Interest rate (column A\n\n• No. of payments (column B)\n\n• Amount of loan (column C)\n\n• Annual payment (column D)\n\n• Monthly payment (column E)",
null,
"Exploring loan scenarios with the PMT function.\n\n## IF for identifying data\n\nThe IF function examines data and returns a value based on criteria you enter. Use the IF function to locate data that meets a certain threshold. In the worksheet shown, for example, the IF function is used to identify teams that are eligible for the playoffs. To be eligible, a team must have won more than six games.\n\nUse the IF function as follows:\n\n`IF(logical true-false test, value if true, value if false)`\n\nInstructing Excel to enter a value if the logical true‐false test comes up false is optional; you must supply a value to enter if the test is true. Enclose the value in quotation marks if it is a text value such as the word Yes or No.\n\nThe formula for determining whether a team made the playoffs is as follows:\n\n`=IF(C3>6,\"Yes\",\"No\")`\n\nIf the false “No” value was absent from the formula, teams that didn’t make the playoffs would not show a value in the Playoffs column; these teams’ Playoffs column would be empty.",
null,
"Exploring loan scenarios with the PMT function.\n\n## LEFT, MID, and RIGHT for cleaning up data\n\nSometimes when you import data from another software application, especially if it’s a database application, the data arrives with unneeded characters. You can use the LEFT, MID, RIGHT, and TRIM functions to remove these characters:\n\n• LEFT returns the leftmost characters in a cell to the number of characters you specify. For example, in a cell with CA_State, this formula returns CA, the two leftmost characters in the text:\n\n`=LEFT(A1,2)`\n• MID returns the middle characters in a cell starting at a position you specify to the number of characters you specify. For example, in a cell with https://www.dummies.com, this formula uses MID to remove the extraneous seven characters at the beginning of the URL and get www.dummies.com:\n\n`=MID(A1,7,50)`\n• RIGHT returns the rightmost characters in a cell to the number of characters you specify. For example, in a cell containing the words Vitamin B1, the following formula returns B1, the two rightmost characters in the name of the vitamin:\n\n`=RIGHT(A1,2)`\n• TRIM, except for single spaces between words, removes all blank spaces from inside a cell. Use TRIM to remove leading and trailing spaces. This formula removes unwanted spaces from the data in cell A1:\n\n`=TRIM(A1)`\n\n## PROPER for capitalizing words\n\nThe PROPER function makes the first letter of each word in a cell uppercase. As are LEFT and RIGHT, it is useful for cleaning up data you imported from elsewhere. Use PROPER as follows:\n\n`PROPER(cell address)`\n\n## LARGE and SMALL for comparing values\n\nUse the LARGE and SMALL functions, as well as their cousins MIN, MAX, and RANK, to find out where a value stands in a list of values. For example, use LARGE to locate the ninth oldest man in a list, or MAX to find the oldest man. Use MIN to find the smallest city by population in a list, or SMALL to find the fourth smallest. The RANK function finds the rank of a value in a list of values.\n\nUse these functions as follows:\n\n• MIN returns the smallest value in a list of values. For the argument, enter a cell range or cell array. In the worksheet shown, the following formula finds the fewest number of fish caught at any lake on any day:\n\n`=MIN(C3:G7)`\n• SMALL returns the nth smallest value in a list of values. This function takes two arguments, first the cell range or cell array, and next the position, expressed as a number, from the smallest of all values in the range or array. In the worksheet shown, this formula finds the second smallest number of fish caught in any lake:\n\n`=SMALL(C3:G7,2)`\n• MAX returns the largest value in a list of values. Enter a cell range or cell array as the argument. In the worksheet shown, this formula finds the most number of fish caught in any lake:\n\n`=MAX(C3:G7)`\n• LARGE returns the nth largest value in a list of values. This function takes two arguments, first the cell range or cell array, and next the position, expressed as a number, from the largest of all values in the range or array. In the worksheet shown, this formula finds the second largest number of fish caught in any lake:\n\n`=LARGE(C3:G7,2)`\n• RANK returns the rank of a value in a list of values. This function takes three arguments:\n\n• The cell with the value used for ranking\n\n• The cell range or cell array with the comparison values for determining rank\n\n• Whether to rank in order from top to bottom (enter 0 for descending) or bottom to top (enter 1 for ascending)\n\nIn the worksheet shown, this formula ranks the total number of fish caught in Lake Temescal against the total number of fish caught in all five lakes:",
null,
"Using functions to compare values.\n`=RANK(H3,H3:H7,0)`\n\n## NETWORKDAY and TODAY for measuring time in days\n\nExcel offers a couple of date functions for scheduling, project planning, and measuring time periods in days.\n\nNETWORKDAYS measures the number of workdays between two dates (the function excludes Saturdays and Sundays from its calculations). Use this function for scheduling purposes to determine the number of workdays needed to complete a project. Use NETWORKDAYS as follows:\n\n`NETWORKDAYS(start date, end date)`\n\nTODAY gives you today’s date, whatever it happens to be. Use this function to compute today’s date in a formula.\n\n`TODAY()`\n\nTo measure the number of days between two dates, use the minus operator and subtract the latest date from the earlier one.\n\n`=\"6/1/2015\"-\"1/1/2015\"`\n\nThe dates are enclosed in quotation marks to make Excel recognize them as dates. Make sure that the cell where the formula is located is formatted to show numbers, not dates.\n\n## LEN for counting characters in cells\n\nUse the LEN (length) function to obtain the number of characters in a cell. This function is useful for making sure that characters remain under a certain limit. The LEN function counts blank spaces as well as characters. Use the LEN function as follows:\n\n`LEN(cell address)`"
]
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https://www.fxsolver.com/browse/formulas/Refarctive+index+(absence+of+attenuation+in+vacuum) | [
"'\n\n# Refarctive index (absence of attenuation in vacuum)\n\n## Description\n\nWhen an electromagnetic wave travels through a medium in which it gets attenuated (this is called an “opaque” or “attenuating” medium), it undergoes exponential decay as described by the Beer–Lambert law. However, there are many possible ways to characterize the wave and how quickly it is attenuated.\n\nFor a given frequency, the wavelength of an electromagnetic wave is affected by the material in which it is propagating. In the absence of attenuation, the index of refraction (also called refractive index) is the ratio of these two wavelengths.\n\nRelated formulas\n\n## Variables\n\n n refractivve index (dimensionless) c speed of light in vacuum (m/s) k angular wavenumber of the wave (rad/m) ω angular frequency of the wave (rad/s)"
]
| [
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http://p220.ru/en/home/projects/item/1204-14-xxx-31-0050 | [
"Attraction of the leading scientists to Russian institutions of higher learning, research organizations of the governmental academies of sciences, and governmental research centers of the Russian Federation\n\nMirror Symmetry Laboratory\n\nGrant Agreement No.: 14.641.31.0001\n\nProject name: Mirror Symmetry and automorphic forms\n\nName of the institution of higher learning: Higher School of Economics - National Research University\n\nFields of scientific research: Mathematics\n\nProject goal:\n\nThe main goal of the project is geometrization of categories. The starting point of this project is Homological Mirror Symmetry. This is one of the most fundamental conjectures of modern mathematics. It was put forward by Kontsevich and has originated in physics as a duality between superconformal quantum field theories. The project is aimed to developing of a new cutting edge direction in mathematics — Categorical Kähler Geometry. This is a totally new, bold idea under implementation Katzarkov, Kontsevich, Simpson.\n\nMirror symmetry, as a foundational attribute of models in elementary particles theory, was discovered by physicists in 90-th. Categorical Kähler geometry is deeply related to such models. Partition functions in physical theories and generating functions for geometric invariants of manifolds are special functions such as Borcherds products or elliptic hypergeometric integrals determining Lorentzian Kac–Moody algebras, Seiberg duality and mathematical invariants of manifolds standing behind these theories.\n\nThe concept of Mirror symmetry creates new mathematical disciplines and new way of thinking in theoretical physics. The next steps in the theory of fundamental interactions in the Nature are closely related to the development of these disciplines and new ideas in mathematics and theoretical physics. We propose to open a new multidisciplinary laboratory in order to focus the research of specialists in different mathematical domains like geometry, topology, number theory and automorphic forms, Lie algebras and mathematical physics on the major theoretical problems in mathematics and physics related to Mirror symmetry. Besides scientific achievements this proposal will have a broad impact on Russian Mathematics and on Russian science in keeping their leading position in the world.\n\nThe Homological Mirror Symmetry conjecture stated by Kontsevich relates objects in two different mathematical worlds. One is the world of complex geometry, which is fairly robust and rigid, the other is the world of symplectic topology, which is a kind of flabby and has a lot of wiggle space in it, and is hard to pin down. The conjecture is that every symplectic manifold has a mirror in the complex geometry world and that an invariant of the symplectic manifold, known as the Fukaya category, is the same as an invariant called the derived category of its mirror space. Homological Mirror Symmetry conjecture is one of the most fundamental conjectures of modern mathematics bringing as well new methods in theoretical physics.\n\nIn this project we plan to capitalize on geometric consequences of Homological Mirror Symmetry and develop the following theories:\n\n1. Categorical Kähler Geometry;\n\n2. Theory of Perverse Sheaves of Categories and its parallel with Nonabelian Hodge theory and integrable systems;\n\n3. Categorical Lefschetz theory;\n\n4. Theory of categorical multiplier ideal sheaves.\n\nOn this basis we plan to find\n\n1. a connection between automorphic forms and categorical Noether–Lefschetz loci;\n\n2. a connection between automorphic forms and partition functions from the new prospective of categorical Kohler Geometry\n\nand to construct\n\n1. classification of Lorentzian Kac–Moody algebras, corresponding automorphic forms, string partition functions and mirror symmetry;\n\n2. classification of Seiberg dualities and related elliptic hypergeometric function identities.",
null,
"Katzarkov Ludmil Vasilev\n\nDate of Birth: 19.12.1961\n\nCitzenship: United States, Bulgaria"
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https://uk.mathworks.com/matlabcentral/profile/authors/157225 | [
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I have a mtarix of strings x_mtarix: x_matrix = 5×3 string array \"A\" \"B\" \"C\" \"AA\" \"B...\n\n2 years ago | 1 answer | 0\n\n1\n\nQuestion\n\nConverting mat files to daq format\nHello There We would like to convert mat files, or indeed matlab variables in the workspace, to the daq format. what we are l...\n\n2 years ago | 1 answer | 0\n\n1\n\nDetecting values in a vector that are different but very close to each other\nThakn you for your reply. Yes I want to keep the 4.1 since the next valu is 5 so, 5-4.1=0.9 which is over 0.1\n\n2 years ago | 0\n\nQuestion\n\nDetecting values in a vector that are different but very close to each other\nHello there: I have a t vector (time, increasing values) like this: t=[ 1 1.1 2 3 3.1 4.1 5 6 7....\n\n2 years ago | 3 answers | 0\n\n3\n\nQuestion\n\nFile disappeared from the current working matlab folder\nHi there I was testing a matlab function in the current working folder (not a matlab program folder) and suddenly the file was ...\n\n2 years ago | 0 answers | 0\n\n0\n\nQuestion\n\nHow to delete items entries in a listbox in appdesigner\nWe have a listbox with multiple entries, and would want to be able to remove selected entries, for which we though we could use...\n\n2 years ago | 0 answers | 0\n\n0\n\nUsing WindowButtonDownFcn in App Designer\nHi I cannot find the \"WindowButtonDownFcn\" upon on doing the callback, in my case, this is a callback for a listbox ui (inherit...\n\n2 years ago | 0\n\nQuestion\n\nReal and Imaginary part of this function please\nHi there, I have this expression: syms R C w a real assume(R>0) assume(C>0) assume(w>0) assume(a>0) z=R/(1+(R*C)*(i*w)^a...\n\n2 years ago | 1 answer | 0\n\n1\n\nQuestion\n\nI have saved mat files using the '-v7.3' like save('my_big_fat_file, 'TT_abc', '-v7.3'); my_big_fat_file has has about 3 GBy...\n\n3 years ago | 0 answers | 0\n\n0\n\nQuestion\n\nSorting a cell of strings using a certain character position\nDear all I have a variable called var= vart=[{'Ice011_L_3of3m.mat''}; {'Ice011_P_1of3m.mat'} ;{'Ice011_P_2of3m.mat'}] We woul...\n\n3 years ago | 1 answer | 0\n\n1\n\ngrouping first (left column) and last (right column) of consecutive sequence of a Nx2 matrix\nMany thanks it worked perfectely!! Best Regards\n\n3 years ago | 0\n\nQuestion\n\ngrouping first (left column) and last (right column) of consecutive sequence of a Nx2 matrix\nHi there We have a Nx2 matrix with pieces of consecutive values that we want to obtain the extreme left and right limits: A=[...\n\n3 years ago | 2 answers | 0\n\n2\n\nQuestion\n\nload being 3x faster than importdata\n\n3 years ago | 1 answer | 0\n\n1\n\nDid you get any reply? Also interested\n\n3 years ago | 0\n\nQuestion\n\nI have a two column matrix and want to eliminate the rows whose interval contain a certain value\nDear all I have a two column matrix such as: A=[12 44; 56 78; 81 100; 110 200; 210 300;450 500; 600 710] this matrix row...\n\n4 years ago | 1 answer | 0\n\n1\n\nReading variables from a big matlab Table\nGuillaume, thank you I have gone through all the steps you mentioned. Creating the tall data was time consuming t_tall=tall(...\n\n4 years ago | 0\n\nQuestion\n\nReading variables from a big matlab Table\nHello There we have about 45 big matlab Tables (average size 12 GByte each). Please note that thee are .mat files. We will be r...\n\n4 years ago | 2 answers | 0\n\n2\n\nTell me this is not true: r2016b and CWT (continuos wavelet transform)\nI know these things are quite subjective sometimes. But the concept of period (T) or frequency (F=1/T) of a periodic waveform i...\n\n5 years ago | 0\n\nTell me this is not true: r2016b and CWT (continuos wavelet transform)\nDear Wayne Thank you for your reply, we always learn something from it. The wavelet area is quite demanding in many ways and th...\n\n5 years ago | 0\n\nQuestion\n\nTell me this is not true: r2016b and CWT (continuos wavelet transform)\nI was checking out the mathworks site for the new version of the CWT function http://www.mathworks.com/help/wavelet/ref/cwt.ht...\n\n5 years ago | 5 answers | 1\n\n5\n\nQuestion\n\nassigning a color to the same value when ploting multiple matrices\nHello There I am using imagesc to plot data sets of 16 matrices sized 256x1000 samples .each (data for one patient) I am usin...\n\n6 years ago | 1 answer | 0\n\n1\n\nQuestion\n\nApplying a multistage filter to a signal, how?\nDear Readers Considering the filter explained in the mathworks documentation centre: Fpass = 0.11; Fstop = 0.12; Apass =...\n\n7 years ago | 0 answers | 0\n\n0\n\nQuestion\n\nmarginals and energy of the scalogram to match the signal\nHi there: I have been looking for some matlab code to compute the time and frequency marginals in the wavelet scalogram. Mor...\n\n8 years ago | 0 answers | 0\n\n0\n\nQuestion\n\nUnits of the wavelet scale\nWhat are the units of the wavelet scale- a ? I always thought that it would be non-dimentional due to mathworks wavelet help: (c...\n\n8 years ago | 0 answers | 0\n\n0\n\nQuestion\n\nusing conofinf to plot the cone of influence in a imagesc(time, freq, coeffs)\nHello there I have done the cwt of a signal coeffs=cwt(signal, scales, wavelet) scales are obtained as this: freq=1:1: ...\n\n9 years ago | 1 answer | 0"
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https://forums.wolfram.com/mathgroup/archive/2013/Feb/msg00304.html | [
"",
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"Re: Compiling numerical iterations\n\n• To: mathgroup at smc.vnet.net\n• Subject: [mg129940] Re: Compiling numerical iterations\n• From: Peter Klamser <klamser at googlemail.com>\n• Date: Wed, 27 Feb 2013 03:06:38 -0500 (EST)\n• Delivered-to: [email protected]\n• Delivered-to: [email protected]\n• Delivered-to: [email protected]\n• Delivered-to: [email protected]\n• References: <kgab2i\\$c7f\\[email protected]>\n\n```Hi Cornelius,\n\nyour code is not slow because you use Mathematica but because you are not\nwriting in functional and Mathematica style.\n\nUse /@, @@, pure Functions like #1^2& etc. to realize your algorithm.\n\nPlus @@ {a, b, c, d} is much faster then procedural programming.\n\nKind regards from Peter\n\n2013/2/26 firlefranz <cornelius.franz at gmx.net>:\n> Thanks a lot! To be honest, some of the commands Ray is using, I've never seen before. I stopped using mathematica before version 5 came out.\n>\n> Coming back to Peters statement of exporting the code from Mathematica to C. How this can be done starting from my or Ray's code? There is an automated C-code-gernerator implemented in Mathematica 9, am I right?\n>\n>\n> As a step forward to the first little peace of code, here is another try, which is not really optimized. The be concrete, I try to simulate the autocorrelation function of a random walk, which is doing a step at none equally distant time steps. This has to been done for a long random walk for many particles. Instead of doing an average over many random walks and calculate one autocorrelation function I want to simulate many correlation functions and make an average over them. Since the time steps are non equal, I wrote a sub-function, which creates a new time axis and taking the necessary value for the random walk from the first table.\n>\n> Here is what I come up with. It's running in a reasonable time for one particle, but for a real statistic ensemble, I have to do it over 1.000.000 particles for a long time. Optimizing this or (probably better) exporting it to C would hopefully help a lot. So how to export it?\n>\n> Clear[\"Global`*\"]\n> SeedRandom;\n> zeitmax = 100;(* muss ein Integer sein *)\n> numteilchen = 1;\n> tauj = 1;\n> corr = Table[0, {i, 1, zeitmax/10}];\n>\n> SucheIndex[zeitliste_, zeit_, maxindex_] :=\n> Module[{i},\n> For[i = 1, i <= maxindex, i++,\n> If[zeitliste[[i]] > zeit, Break[]];\n> ];\n> i - 1\n> ];\n>\n> For[j = 1, j <= numteilchen, j++,\n> (* Zeitachse generieren von 0 bis zeitmax *)\n> t = 0;\n> i = 1;\n> tabzeit = {};\n> time = AbsoluteTiming[While[True,\n> tabzeit = Append[tabzeit, t];\n> dt = -tauj*Log[1 - RandomReal[]];\n> If[t > zeitmax, Break[]];\n> t = t + dt;\n> i++;\n> ];\n> ];\n> Print[time];\n> maxidx = i;\n>\n> (* Random Walk *)\n> time = AbsoluteTiming[\n> tabwalk = Table[0, {i, 1, maxidx}];\n> a = 0;\n> For[i = 1, i <= maxidx, i++,\n> tabwalk[[i]] = a;\n> If[RandomReal[{-1, 1}] > 0, a = a + 1, a = a - 1];\n> ];\n> ];\n> Print[time];\n> (*tabwalk=Table[Subscript[b, i],{i,1,maxidx}];*)\n>\n> (* Korrelationsfunktion berechnen *)\n> time = AbsoluteTiming[\n> For[k = 1, k <= zeitmax/10, k++,\n> For[n = 1, n <= zeitmax/10*9, n++,\n> corr[[k]] =\n> corr[[k]] +\n> tabwalk[[SucheIndex[tabzeit, n - 1, maxidx]]]*\n> tabwalk[[SucheIndex[tabzeit, n + k - 2, maxidx]]]/\n> zeitmax*10/9/numteilchen;\n> (*Print[corr//N];*)\n> ];\n> ];\n> ];\n> Print[time];\n> Print[corr // N];\n> ];\n> Table[{tabzeit[[i]], tabwalk[[i]]}, {i, 1, maxidx}]\n> corr // N\n>\n\n```\n\n• Prev by Date: Re: i^2=1\n• Next by Date: Re: Real and Imaginary Parts of complex functions\n• Previous by thread: Re: Compiling numerical iterations\n• Next by thread: Re: Compiling numerical iterations"
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https://www.bartleby.com/solution-answer/chapter-13-problem-30e-elementary-geometry-for-college-students-6th-edition/9781285195698/in-exercises-29-to-32-use-only-a-compass-and-a-straightedge-to-complete-each-construction/5985f707-757b-11e9-8385-02ee952b546e | [
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"### Elementary Geometry for College St...\n\n6th Edition\nDaniel C. Alexander + 1 other\nISBN: 9781285195698\n\n#### Solutions\n\nChapter\nSection",
null,
"### Elementary Geometry for College St...\n\n6th Edition\nDaniel C. Alexander + 1 other\nISBN: 9781285195698\nTextbook Problem\n1 views\n\n# In Exercises 29 to 32, use only a compass and a straightedge to complete each construction.",
null,
"Exercises 29, 30 Given: A B ¯ and C D ¯ ( A B > C D ) Construct: E F ¯ on line l so that E F = A B − C D\n\nTo determine\n\nTo find:\n\nThe complete construction of EF¯ on line 𝓁 so that EF=ABCD.\n\nExplanation\n\nGiven:\n\nThe given figure is,\n\nAB¯andCD¯(AB>CD)\n\nApproach:\n\nThe steps are given below for the complete construction.\n\na) Draw a straight line and label a point E.\n\nb) Use compass and open it to the width of AB¯\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\n#### In Exercises 2340, find the indicated limit. 24. limx23\n\nApplied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach\n\n#### The area of the region bounded by , , and r = sec θ is:\n\nStudy Guide for Stewart's Multivariable Calculus, 8th\n\n#### In Exercises 110, evaluate the expression. C(52,7)\n\nFinite Mathematics for the Managerial, Life, and Social Sciences",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6345713,"math_prob":0.7700231,"size":2177,"snap":"2019-51-2020-05","text_gpt3_token_len":476,"char_repetition_ratio":0.19420156,"word_repetition_ratio":0.10600707,"special_character_ratio":0.16903996,"punctuation_ratio":0.0977918,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9543962,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T07:54:20Z\",\"WARC-Record-ID\":\"<urn:uuid:1349e842-a35e-4f0f-a98e-4ff6af7a9719>\",\"Content-Length\":\"570257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:337945d9-1cb7-4ba1-82ff-48f8c49f1d84>\",\"WARC-Concurrent-To\":\"<urn:uuid:1c1e0675-d1b9-4874-ac72-f8b2578d3721>\",\"WARC-IP-Address\":\"99.84.181.3\",\"WARC-Target-URI\":\"https://www.bartleby.com/solution-answer/chapter-13-problem-30e-elementary-geometry-for-college-students-6th-edition/9781285195698/in-exercises-29-to-32-use-only-a-compass-and-a-straightedge-to-complete-each-construction/5985f707-757b-11e9-8385-02ee952b546e\",\"WARC-Payload-Digest\":\"sha1:XRO5ZWPH74L2RQVMTTRGV4JDDPBZYLYB\",\"WARC-Block-Digest\":\"sha1:ODBKIJOU74GB7LHUV5XE35UXGMXBEMNF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540507109.28_warc_CC-MAIN-20191208072107-20191208100107-00045.warc.gz\"}"} |
https://nl.mathworks.com/matlabcentral/cody/problems/1011-newton-interpolation | [
"Cody\n\n# Problem 1011. Newton Interpolation\n\nGiven a set of measurements of dependent variables in a vector, Y, that vary with one independent variable in a vector, X, calculate the interpolating polynomial using Newton interpolation. Output is the interpolated value for a given value of x, as well as the vector of divided differences, b, that are the coefficients of the Newton polynomial.\n\n### Solution Stats\n\n22.45% Correct | 77.55% Incorrect\nLast Solution submitted on Sep 28, 2019"
]
| [
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.8989686,"math_prob":0.97443837,"size":779,"snap":"2020-24-2020-29","text_gpt3_token_len":174,"char_repetition_ratio":0.117419355,"word_repetition_ratio":0.0,"special_character_ratio":0.2169448,"punctuation_ratio":0.09859155,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987112,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T20:34:08Z\",\"WARC-Record-ID\":\"<urn:uuid:8a8dff16-cef3-4022-8c86-9108b97454fb>\",\"Content-Length\":\"82989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3189aa43-42a7-404b-87b5-d8e3b16015d2>\",\"WARC-Concurrent-To\":\"<urn:uuid:e29b3153-a46d-4192-b69f-8642a2d18177>\",\"WARC-IP-Address\":\"104.117.0.182\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/1011-newton-interpolation\",\"WARC-Payload-Digest\":\"sha1:TK5PFVE3CZV47V23EGX2NUPBWROINHJM\",\"WARC-Block-Digest\":\"sha1:XMYCSGKX5WCWSY2SNQXLO3U2UMYUBTFH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347426801.75_warc_CC-MAIN-20200602193431-20200602223431-00546.warc.gz\"}"} |
https://www.mathworks.com/help/matlab/ref/surf.html;jsessionid=3720499dca5c2a86d0a212b93d6a | [
"# surf\n\n•",
null,
"## Syntax\n\n``surf(X,Y,Z)``\n``surf(X,Y,Z,C)``\n``surf(Z)``\n``surf(Z,C)``\n``surf(ax,___)``\n``surf(___,Name,Value)``\n``s = surf(___)``\n\n## Description\n\nexample\n\n````surf(X,Y,Z)` creates a three-dimensional surface plot, which is a three-dimensional surface that has solid edge colors and solid face colors. The function plots the values in matrix `Z` as heights above a grid in the x-y plane defined by `X` and `Y`. The color of the surface varies according to the heights specified by `Z`.```\n\nexample\n\n````surf(X,Y,Z,C)` additionally specifies the surface color.```\n````surf(Z)` creates a surface plot and uses the column and row indices of the elements in `Z` as the x- and y-coordinates.```\n````surf(Z,C)` additionally specifies the surface color.```\n````surf(ax,___)` plots into the axes specified by `ax` instead of the current axes. Specify the axes as the first input argument.```\n\nexample\n\n````surf(___,Name,Value)` specifies surface properties using one or more name-value pair arguments. For example, `'FaceAlpha',0.5` creates a semitransparent surface.```\n\nexample\n\n````s = surf(___)` returns the chart surface object. Use `s` to modify the surface after it is created. For a list of properties, see Surface Properties.```\n\n## Examples\n\ncollapse all\n\nCreate three matrices of the same size. Then plot them as a surface. The surface plot uses `Z` for both height and color.\n\n```[X,Y] = meshgrid(1:0.5:10,1:20); Z = sin(X) + cos(Y); surf(X,Y,Z)```",
null,
"Specify the colors for a surface plot by including a fourth matrix input, `C`. The surface plot uses `Z` for height and `C` for color. Specify the colors using a colormap, which uses single numbers to stand for colors on a spectrum. When you use a colormap, `C` is the same size as `Z`. Add a color bar to the graph to show how the data values in `C` correspond to the colors in the colormap.\n\n```[X,Y] = meshgrid(1:0.5:10,1:20); Z = sin(X) + cos(Y); C = X.*Y; surf(X,Y,Z,C) colorbar```",
null,
"Specify the colors for a surface plot by including a fourth matrix input, `CO`. The surface plot uses `Z` for height and `CO` for color. Specify the colors using truecolor, which uses triplets of numbers to stand for all possible colors. When you use truecolor, if `Z` is `m`-by-`n`, then `CO` is `m`-by-`n`-by-3. The first page of the array indicates the red component for each color, the second page indicates the green component, and the third page indicates the blue component.\n\n```[X,Y,Z] = peaks(25); CO(:,:,1) = zeros(25); % red CO(:,:,2) = ones(25).*linspace(0.5,0.6,25); % green CO(:,:,3) = ones(25).*linspace(0,1,25); % blue surf(X,Y,Z,CO)```",
null,
"Create a semitransparent surface by specifying the `FaceAlpha` name-value pair with `0.5` as the value. To allow further modifications, assign the surface object to the variable `s`.\n\n```[X,Y] = meshgrid(-5:.5:5); Z = Y.*sin(X) - X.*cos(Y); s = surf(X,Y,Z,'FaceAlpha',0.5)```",
null,
"```s = Surface with properties: EdgeColor: [0 0 0] LineStyle: '-' FaceColor: 'flat' FaceLighting: 'flat' FaceAlpha: 0.5000 XData: [21x21 double] YData: [21x21 double] ZData: [21x21 double] CData: [21x21 double] Show all properties ```\n\nUse `s` to access and modify properties of the surface object after it is created. For example, hide the edges by setting the `EdgeColor` property.\n\n`s.EdgeColor = 'none';`",
null,
"## Input Arguments\n\ncollapse all\n\nx-coordinates, specified as a matrix the same size as `Z`, or as a vector with length `n`, where `[m,n] = size(Z)`. If you do not specify values for `X` and `Y`, `surf` uses the vectors `(1:n)` and `(1:m)`.\n\nYou can use the `meshgrid` function to create `X` and `Y` matrices.\n\nThe `XData` property of the `Surface` object stores the x-coordinates.\n\nExample: `X = 1:10`\n\nExample: `X = [1 2 3; 1 2 3; 1 2 3]`\n\nExample: `[X,Y] = meshgrid(-5:0.5:5)`\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`\n\ny-coordinates, specified as a matrix the same size as `Z` or as a vector with length `m`, where ```[m,n] = size(Z)```. If you do not specify values for `X` and `Y`, `surf` uses the vectors `(1:n)` and `(1:m)`.\n\nYou can use the `meshgrid` function to create the `X` and `Y` matrices.\n\nThe `YData` property of the surface object stores the y -coordinates.\n\nExample: `Y = 1:10`\n\nExample: `Y = [1 1 1; 2 2 2; 3 3 3]`\n\nExample: `[X,Y] = meshgrid(-5:0.5:5)`\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`\n\nz-coordinates, specified as a matrix. `Z` must have at least two rows and two columns.\n\n`Z` specifies the height of the surface plot at each x-y coordinate. If you do not specify the colors, then `Z` also specifies the surface colors.\n\nThe `ZData` property of the surface object stores the z -coordinates.\n\nExample: `Z = [1 2 3; 4 5 6]`\n\nExample: `Z = sin(x) + cos(y)`\n\nData Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `categorical` | `datetime` | `duration`\n\nColor array, specified as an `m`-by-`n` matrix of colormap indices or as an `m`-by-`n`-by-`3` array of RGB triplets, where `Z` is `m`-by-`n`.\n\n• To use colormap colors, specify `C` as a matrix. For each grid point on the surface, `C` indicates a color in the colormap. The `CDataMapping` property of the surface object controls how the values in `C` correspond to colors in the colormap.\n\n• To use truecolor colors, specify `C` as an array of RGB triplets.\n\nThe `CData` property of the surface object stores the color array. For additional control over the surface coloring, use the `FaceColor` and `EdgeColor` properties.\n\nAxes to plot in, specified as an `axes` object. If you do not specify the axes, then `surf` plots into the current axes.\n\n### Name-Value Arguments\n\nSpecify optional pairs of arguments as `Name1=Value1,...,NameN=ValueN`, where `Name` is the argument name and `Value` is the corresponding value. Name-value arguments must appear after other arguments, but the order of the pairs does not matter.\n\nBefore R2021a, use commas to separate each name and value, and enclose `Name` in quotes.\n\nExample: `surf(X,Y,Z,'FaceAlpha',0.5,'EdgeColor','none')` creates a semitransparent surface with no edges drawn.\n\nNote\n\nThe properties listed here are only a subset. For a full list, see Surface Properties.\n\nEdge line color, specified as one of the values listed here. The default color of `[0 0 0]` corresponds to black edges.\n\nValueDescription\n`'none'`Do not draw the edges.\n`'flat'`\n\nUse a different color for each edge based on the values in the `CData` property. First you must specify the `CData` property as a matrix the same size as `ZData`. The color value at the first vertex of each face (in the positive x and y directions) determines the color for the adjacent edges. You cannot use this value when the `EdgeAlpha` property is set to `'interp'`.",
null,
"`'interp'`\n\nUse interpolated coloring for each edge based on the values in the `CData` property. First you must specify the `CData` property as a matrix the same size as `ZData`. The color varies across each edge by linearly interpolating the color values at the vertices. You cannot use this value when the `EdgeAlpha` property is set to `'flat'`.",
null,
"RGB triplet, hexadecimal color code, or color name\n\nUse the specified color for all the edges. This option does not use the color values in the `CData` property.",
null,
"RGB triplets and hexadecimal color codes are useful for specifying custom colors.\n\n• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example,``` [0.4 0.6 0.7]```.\n\n• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n`\"red\"``\"r\"``[1 0 0]``\"#FF0000\"`",
null,
"`\"green\"``\"g\"``[0 1 0]``\"#00FF00\"`",
null,
"`\"blue\"``\"b\"``[0 0 1]``\"#0000FF\"`",
null,
"`\"cyan\"` `\"c\"``[0 1 1]``\"#00FFFF\"`",
null,
"`\"magenta\"``\"m\"``[1 0 1]``\"#FF00FF\"`",
null,
"`\"yellow\"``\"y\"``[1 1 0]``\"#FFFF00\"`",
null,
"`\"black\"``\"k\"``[0 0 0]``\"#000000\"`",
null,
"`\"white\"``\"w\"``[1 1 1]``\"#FFFFFF\"`",
null,
"Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.\n\n`[0 0.4470 0.7410]``\"#0072BD\"`",
null,
"`[0.8500 0.3250 0.0980]``\"#D95319\"`",
null,
"`[0.9290 0.6940 0.1250]``\"#EDB120\"`",
null,
"`[0.4940 0.1840 0.5560]``\"#7E2F8E\"`",
null,
"`[0.4660 0.6740 0.1880]``\"#77AC30\"`",
null,
"`[0.3010 0.7450 0.9330]``\"#4DBEEE\"`",
null,
"`[0.6350 0.0780 0.1840]``\"#A2142F\"`",
null,
"Line style, specified as one of the options listed 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\nFace color, specified as one of the values in this table.\n\nValueDescription\n`'flat'`\n\nUse a different color for each face based on the values in the `CData` property. First you must specify the `CData` property as a matrix the same size as `ZData`. The color value at the first vertex of each face (in the positive x and y directions) determines the color for the entire face. You cannot use this value when the `FaceAlpha` property is set to `'interp'`.",
null,
"`'interp'`\n\nUse interpolated coloring for each face based on the values in the `CData` property. First you must specify the `CData` property as a matrix the same size as `ZData`. The color varies across each face by interpolating the color values at the vertices. You cannot use this value when the `FaceAlpha` property is set to `'flat'`.",
null,
"RGB triplet, hexadecimal color code, or color name\n\nUse the specified color for all the faces. This option does not use the color values in the `CData` property.",
null,
"`'texturemap'`Transform the color data in `CData` so that it conforms to the surface.\n`'none'`Do not draw the faces.\n\nRGB triplets and hexadecimal color codes are useful for specifying custom colors.\n\n• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range `[0,1]`; for example,``` [0.4 0.6 0.7]```.\n\n• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (`#`) followed by three or six hexadecimal digits, which can range from `0` to `F`. The values are not case sensitive. Thus, the color codes `'#FF8800'`, `'#ff8800'`, `'#F80'`, and `'#f80'` are equivalent.\n\nAlternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.\n\nColor NameShort NameRGB TripletHexadecimal Color CodeAppearance\n`\"red\"``\"r\"``[1 0 0]``\"#FF0000\"`",
null,
"`\"green\"``\"g\"``[0 1 0]``\"#00FF00\"`",
null,
"`\"blue\"``\"b\"``[0 0 1]``\"#0000FF\"`",
null,
"`\"cyan\"` `\"c\"``[0 1 1]``\"#00FFFF\"`",
null,
"`\"magenta\"``\"m\"``[1 0 1]``\"#FF00FF\"`",
null,
"`\"yellow\"``\"y\"``[1 1 0]``\"#FFFF00\"`",
null,
"`\"black\"``\"k\"``[0 0 0]``\"#000000\"`",
null,
"`\"white\"``\"w\"``[1 1 1]``\"#FFFFFF\"`",
null,
"Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB uses in many types of plots.\n\n`[0 0.4470 0.7410]``\"#0072BD\"`",
null,
"`[0.8500 0.3250 0.0980]``\"#D95319\"`",
null,
"`[0.9290 0.6940 0.1250]``\"#EDB120\"`",
null,
"`[0.4940 0.1840 0.5560]``\"#7E2F8E\"`",
null,
"`[0.4660 0.6740 0.1880]``\"#77AC30\"`",
null,
"`[0.3010 0.7450 0.9330]``\"#4DBEEE\"`",
null,
"`[0.6350 0.0780 0.1840]``\"#A2142F\"`",
null,
"Face transparency, specified as one of these values:\n\n• Scalar in range `[0,1]` — Use uniform transparency across all the faces. A value of `1` is fully opaque and `0` is completely transparent. Values between `0` and `1` are semitransparent. This option does not use the transparency values in the `AlphaData` property.\n\n• `'flat'` — Use a different transparency for each face based on the values in the `AlphaData` property. The transparency value at the first vertex determines the transparency for the entire face. First you must specify the `AlphaData` property as a matrix the same size as the `ZData` property. The `FaceColor` property also must be set to `'flat'`.\n\n• `'interp'` — Use interpolated transparency for each face based on the values in `AlphaData` property. The transparency varies across each face by interpolating the values at the vertices. First you must specify the `AlphaData` property as a matrix the same size as the `ZData` property. The `FaceColor` property also must be set to `'interp'`.\n\n• `'texturemap'` — Transform the data in `AlphaData` so that it conforms to the surface.\n\nEffect of light objects on faces, specified as one of these values:\n\n• `'flat'` — Apply light uniformly across each face. Use this value to view faceted objects.\n\n• `'gouraud'` — Vary the light across the faces. Calculate the light at the vertices and then linearly interpolate the light across the faces. Use this value to view curved surfaces.\n\n• `'none'` — Do not apply light from light objects to the faces.\n\nTo add a light object to the axes, use the `light` function.\n\nNote\n\nThe `'phong'` value has been removed. Use `'gouraud'` instead.\n\n## Version History\n\nIntroduced before R2006a"
]
| [
null,
"https://www.mathworks.com/help/matlab/ref/plottype-surf_60.png",
null,
"https://www.mathworks.com/help/examples/graphics/win64/CreateSurfacePlotsExample_01.png",
null,
"https://www.mathworks.com/help/examples/graphics/win64/SpecifyColorsForSurfacePlotExample_01.png",
null,
"https://www.mathworks.com/help/examples/graphics/win64/SurfacePlotWithTrueColorsExample_01.png",
null,
"https://www.mathworks.com/help/examples/graphics/win64/ModifySurfacePlotAppearanceExample_01.png",
null,
"https://www.mathworks.com/help/examples/graphics/win64/ModifySurfacePlotAppearanceExample_02.png",
null,
"https://www.mathworks.com/help/matlab/ref/edgecolor_surface_flat.png",
null,
"https://www.mathworks.com/help/matlab/ref/edgecolor_surface_interp.png",
null,
"https://www.mathworks.com/help/matlab/ref/edgecolor_surface_rgb.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_red.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_green.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_blue.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_cyan.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_magenta.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_yellow.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_black.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_white.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder1.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder2.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder3.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder4.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder5.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder6.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder7.png",
null,
"https://www.mathworks.com/help/matlab/ref/linestyle_solid.png",
null,
"https://www.mathworks.com/help/matlab/ref/linestyle_dashed.png",
null,
"https://www.mathworks.com/help/matlab/ref/linestyle_dotted.png",
null,
"https://www.mathworks.com/help/matlab/ref/linestyle_dashdotted.png",
null,
"https://www.mathworks.com/help/matlab/ref/facecolor_surface_flat.png",
null,
"https://www.mathworks.com/help/matlab/ref/facecolor_surface_interp.png",
null,
"https://www.mathworks.com/help/matlab/ref/facecolor_surface_rgb.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_red.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_green.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_blue.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_cyan.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_magenta.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_yellow.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_black.png",
null,
"https://www.mathworks.com/help/matlab/ref/hg_white.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder1.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder2.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder3.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder4.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder5.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder6.png",
null,
"https://www.mathworks.com/help/matlab/ref/colororder7.png",
null
]
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https://articles.emptycrate.com/2009/02/19/real_world_haskell_chapter_4.html | [
"# EmptyCrate\n\n#### About Me | Contact Me | Training | C++ Weekly (YouTube) | Awesome C++ T-Shirts\n\nChapter 4 of Real World Haskell",
null,
"begins with some details about functions with side effects and interacting with the outside world. Normally, a Haskell function is pure and may not have any side effects outside of itself. However, it is necessary to allow for side effects for reading and writing files, including standard input and output. The “do” keyword is used when we may have interactions with the outside world.\n\nInfix Functions\n\nHaskell allows you to define and use any function as an infix function.\n\n``````a `plus` b = a + b\n``````\n\nThe infix notation is purely a convenience, or syntactic sugar as we might say, so the function can be called either by infix or prefix notation.\n\n``````3 `plus` 2\nplus 3 2\n``````\n\nFor me the added necessity of using back-ticks around the function name makes me wonder about the usefulness of this syntax. Although, the examples given in the book such as:\n\n``````\"end\" `isSuffixOf` \"the end\"\n``````\n\nForce me to admit that it can aid in code readability. Working With Lists The next section deals with all manner of list functions and working with lists which I will not catalogue here. The thing that did interest me in this section was this little tidbit from the authors:\n\nHaskell’s type system makes it an interesting challenge to write functions that take variable numbers of arguments. So if we want to zip three lists together, we call zip3 or zipWith3, and so on up to zip7 and zipWith7.\n\nIt seems as though Haskell is dealing with the same kind of problems that variadic templates are meant to address. The main difference is that even with the current version of C++ you could choose to create 6 functions that take 2-7 arguments and name them all “zip” so you get the utility of having one function name and letting the compiler sort out for you which version to use.\n\nTail Recursion\n\nThe book goes on to discuss the fact that normal looping constructs do not exist in Haskell and recursion, particularly tail recursion - for performance reasons - is to be used. The book presents the following example for a tail recursive summation example:\n\n``````-- file: ch04/Sum.hs\nmySum xs = helper 0 xs\nwhere helper acc (x:xs) = helper (acc + x) xs\nhelper acc _ = acc`\n``````\n\nThis is one point where the book loses me. The following version of a summation function makes a lot more sense to me. It is shorter, more readable, does not need an accumulator variable and is still tail recursive. I would very much like for a Haskell expert to be explain to me why one version would be preferable to the other. ` mySum (x:xs) = x + mySum(xs) mySum [] = 0`\n\nReading further we learn that the pattern used above, with an accumulator, is to prepare us for the use of the foldl and foldr functions, which are designed with just this pattern in mind\n\nSpace Leaks\n\nBecause of the lazy evaluation aspects of Haskell it is possible for functions such as “foldl” to consume far more memory then they should as the recursive call stack is generated and stored for large expressions. Instead, the function foldl’ (foldl-prime) should be used, which does not perform lazy evaluation.\n\nLambda Functions\n\nLambdas - anonymous functions - are introduced with a `\\` character. The best illustration for me of the way lambdas can be used is with this simple example of the integer parser from earlier in this chapter:\n\n``````asInt_fold2 string = foldl (\\acc x -> acc * 10 + digitToInt x) 0 string\n``````\n\nIn the above we create an unnamed function with two parameters, “acc” and “x”, which performs the work of parsing each digit of the passed in string of numbers. Lamdas are limited to only one clause, however, so they cannot do full pattern matching like named functions.\n\nPartial Function Application\n\nPartial function application simply means that if you leave off one or more parameters from a function call, a new function is generated and the parameters that you supplied are fixed permanently for the new function that is generated. We can keep building on our “asInt” function to show this example as well:\n\n``````--The process of fixing a parameter is also known as \"currying\"\nasInt_curry = foldl (\\acc x -> acc * 10 + digitToInt x) 0\n``````\n\nIn this example we have just left off the only parameter (“string”) from the previous version because it was the last parameter that we were passing to foldl. In this way, a new function has been generated from our call to foldl that takes one parameter. My explanation seems weak, but if you can read the example, you get the idea.\n\nseq\n\nThe `seq` function is used to force the evaluation of a variable, essentially short circuiting the lazy-evaluation feature of Haskell. It is useful for reducing space leaks in your code.\n\nMy Videos\n\nRelated"
]
| [
null,
"http://www.assoc-amazon.com/e/ir",
null
]
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https://www.smartickmethod.com/blog/math/number-and-operations-fractions/learning-subtract-fractions/ | [
"",
null,
"# Smartick - Math, one click away\n\nTry it for free",
null,
"Jan11\n\n## Learning How to Subtract Fractions\n\nIn this post, we are going to see how to subtract fractions, but for that, you first have to know what the denominator and numerator of a fraction are.\n\nThe numerator is the number that is written on top of the fraction and the denominator is the number written on the bottom of the fraction.\n\nTo subtract fractions, it is necessary that the fractions have the same denominator. Once the fractions have the same denominator, we just subtract the numerators.",
null,
"What do we do if the fractions have different denominators? In this case, we make the denominators the same by finding their least common multiple (LCM).\n\nLet’s look at an example:",
null,
"Since the denominators are different, 4 and 6, we have to find their least common multiple (LCM). You can review how to calculate the LCM on this previous post on our blog.\n\n#### LCM (4,6) = 12\n\nThe two new fractions will have 12 as a denominator.",
null,
"In order to find the numerator of each new fraction, divide the new denominator (the LCM that we have found) by the old denominator and multiply the answer by the old numerator.\n\nThe first fraction:",
null,
"The second fraction:",
null,
"So now we have:",
null,
"As both fractions have the same denominator, now we can subtract the numerators and leave the same denominator:",
null,
"And the result is one-twelfth.\n\nIf you have liked this post, share it with your friends so that they can also learn how to subtract fractions."
]
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"https://cdnblog-en-199133.c.cdn77.org/blog/wp-content/themes/smartick/img/logo-smartick-chr.png",
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"https://cdnblog-en-199133.c.cdn77.org/blog/wp-content/themes/smartick/img/ico-menu-movil.png",
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"https://cdnblog-en-199133.c.cdn77.org/blog/wp-content/uploads/subtract-fractions.png",
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"https://cdnblog-en-199133.c.cdn77.org/blog/wp-content/uploads/subtract-fractions-1.png",
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"https://cdnblog-en-199133.c.cdn77.org/blog/wp-content/uploads/subtract-fractions-2.png",
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https://leanprover.github.io/functional_programming_in_lean/monads.html | [
"In C# and Kotlin, the ?. operator is a way to look up a property or call a method on a potentially-null value. If the reciever is null, the whole expression is null. Otherwise, the underlying non-null value receives the call. Uses of ?. can be chained, in which case the first null result terminates the chain of lookups. Chaining null-checks like this is much more convenient than writing and maintaining deeply nested ifs.\n\nSimilarly, exceptions are significantly more convenient than manually checking and propagating error codes. At the same time, logging is easiest to accomplish by having a dedicated logging framework, rather than having each function return both its log results and its return value. Chained null checks and exceptions typically require language designers to anticipate this use case, while logging frameworks typically make use of side effects to decouple code that logs from the accumulation of the logs.\n\nAll these features and more can be implemented in library code as instances of a common API called Monad. Lean provides dedicated syntax that makes this API convenient to use, but can also get in the way of understanding what is going on behind the scenes. This chapter begins with the nitty-gritty presentation of manually nesting null checks, and builds from there to the convenient, general API. Please suspend your disbelief in the meantime.\n\n## Checking for none: Don't Repeat Yourself\n\nIn Lean, pattern matching can be used to chain checks for null. Getting the first entry from a list can just use the optional indexing notation:\n\ndef first (xs : List α) : Option α :=\nxs?\n\n\nThe result must be an Option because empty lists have no first entry. Extracting the first and third entries requires a check that each is not none:\n\ndef firstThird (xs : List α) : Option (α × α) :=\nmatch xs? with\n| none => none\n| some first =>\nmatch xs? with\n| none => none\n| some third =>\nsome (first, third)\n\n\nSimilarly, extracting the first, third, and fifth entries requires more checks that the values are not none:\n\ndef firstThirdFifth (xs : List α) : Option (α × α × α) :=\nmatch xs? with\n| none => none\n| some first =>\nmatch xs? with\n| none => none\n| some third =>\nmatch xs? with\n| none => none\n| some fifth =>\nsome (first, third, fifth)\n\n\nAnd adding the seventh entry to this sequence begins to become quite unmanageable:\n\ndef firstThirdFifthSeventh (xs : List α) : Option (α × α × α × α) :=\nmatch xs? with\n| none => none\n| some first =>\nmatch xs? with\n| none => none\n| some third =>\nmatch xs? with\n| none => none\n| some fifth =>\nmatch xs? with\n| none => none\n| some seventh =>\nsome (first, third, fifth, seventh)\n\n\nThe fundamental problem with this code is that it addresses two concerns: extracting the numbers and checking that all of them are present, but the second concern is addressed by copying and pasting the code that handles the none case. It is often good style to lift a repetitive segment into a helper function:\n\ndef andThen (opt : Option α) (next : α → Option β) : Option β :=\nmatch opt with\n| none => none\n| some x => next x\n\n\nThis helper, which is used similarly to ?. in C# and Kotlin, takes care of propagating none values. It takes two arguments: an optional value and a function to apply when the value is not none. If the first argument is none, then the helper returns none. If the first argument is not none, then the function is applied to the contents of the some constructor.\n\nNow, firstThird can be rewritten to use andThen instead of pattern matching:\n\ndef firstThird (xs : List α) : Option (α × α) :=\nandThen xs? fun first =>\nandThen xs? fun third =>\nsome (first, third)\n\n\nIn Lean, functions don't need to be enclosed in parentheses when passed as arguments. The following equivalent definition uses more parentheses and indents the bodies of functions:\n\ndef firstThird (xs : List α) : Option (α × α) :=\nandThen xs? (fun first =>\nandThen xs? (fun third =>\nsome (first, third)))\n\n\nThe andThen helper provides a sort of \"pipeline\" through which values flow, and the version with the somewhat unusual indentation is more suggestive of this fact. Improving the syntax used to write andThen can make these computations even easier to understand.\n\n### Infix Operators\n\nIn Lean, infix operators can be declared using the infix, infixl, and infixr commands, which create (respectively) non-associative, left-associative, and right-associative operators. When used multiple times in a row, a left associative operator stacks up the opening parentheses on the left side of the expression. The addition operator + is left associative, so w + x + y + z is equivalent to (((w + x) + y) + z). The exponentiation operator ^ is right associative, so w ^ x ^ y ^ z is equivalent to (w ^ (x ^ (y ^ z))). Comparison operators such as < are non-associative, so x < y < z is a syntax error and requires manual parentheses.\n\nThe following declaration makes andThen into an infix operator:\n\ninfixl:55 \" ~~> \" => andThen\n\n\nThe number following the colon declares the precedence of the new infix operator. In ordinary mathematical notation, x + y * z is equivalent to x + (y * z) even though both + and * are left associative. In Lean, + has precedence 65 and * has precedence 70. Higher-precedence operators are applied before lower-precedence operators. According to the declaration of ~~>, both + and * have higher precedence, and thus apply first. Typically, figuring out the most convenient precedences for a group of operators requires some experimentation and a large collection of examples.\n\nFollowing the new infix operator is a double arrow =>, which specifies the named function to be used for the infix operator. Lean's standard library uses this feature to define + and * as infix operators that point at HAdd.hAdd and HMul.hMul, respectively, allowing type classes to be used to overload the infix operators. Here, however, andThen is just an ordinary function.\n\nHaving defined an infix operator for andThen, firstThird can be rewritten in a way that brings the \"pipeline\" feeling of none-checks front and center:\n\ndef firstThirdInfix (xs : List α) : Option (α × α) :=\nxs? ~~> fun first =>\nxs? ~~> fun third =>\nsome (first, third)\n\n\nThis style is much more concise when writing larger functions:\n\ndef firstThirdFifthSeventh (xs : List α) : Option (α × α × α × α) :=\nxs? ~~> fun first =>\nxs? ~~> fun third =>\nxs? ~~> fun fifth =>\nxs? ~~> fun seventh =>\nsome (first, third, fifth, seventh)\n\n\n## Propagating Error Messages\n\nPure functional languages such as Lean have no built-in exception mechanism for error handling, because throwing or catching an exception is outside of the step-by-step evaluation model for expressions. However, functional programs certainly need to handle errors. In the case of firstThirdFifthSeventh, it is likely relevant for a user to know just how long the list was and where the lookup failed.\n\nThis is typically accomplished by defining a datatype that can be either an error or a result, and translating functions with exceptions into functions that return this datatype:\n\ninductive Except (ε : Type) (α : Type) where\n| error : ε → Except ε α\n| ok : α → Except ε α\nderiving BEq, Hashable, Repr\n\n\nThe type variable ε stands for the type of errors that can be produced by the function. Callers are expected to handle both errors and successes, which makes the type variable ε play a role that is a bit like that of a list of checked exceptions in Java.\n\nSimilarly to Option, Except can be used to indicate a failure to find an entry in a list. In this case, the error type is a String:\n\ndef get (xs : List α) (i : Nat) : Except String α :=\nmatch xs[i]? with\n| none => Except.error s!\"Index {i} not found (maximum is {xs.length - 1})\"\n| some x => Except.ok x\n\n\nLooking up an in-bounds value yields an Except.ok:\n\ndef ediblePlants : List String :=\n[\"ramsons\", \"sea plantain\", \"sea buckthorn\", \"garden nasturtium\"]\n\n#eval get ediblePlants 2\n\nExcept.ok \"sea buckthorn\"\n\n\nLooking up an out-of-bounds value yields an Except.failure:\n\n#eval get ediblePlants 4\n\nExcept.error \"Index 4 not found (maximum is 3)\"\n\n\nA single list lookup can conveniently return a value or an error:\n\ndef first (xs : List α) : Except String α :=\nget xs 0\n\n\nHowever, performing two list lookups requires handling potential failures:\n\ndef firstThird (xs : List α) : Except String (α × α) :=\nmatch get xs 0 with\n| Except.error msg => Except.error msg\n| Except.ok first =>\nmatch get xs 2 with\n| Except.error msg => Except.error msg\n| Except.ok third =>\nExcept.ok (first, third)\n\n\nAdding another list lookup to the function requires still more error handling:\n\ndef firstThirdFifth (xs : List α) : Except String (α × α × α) :=\nmatch get xs 0 with\n| Except.error msg => Except.error msg\n| Except.ok first =>\nmatch get xs 2 with\n| Except.error msg => Except.error msg\n| Except.ok third =>\nmatch get xs 4 with\n| Except.error msg => Except.error msg\n| Except.ok fifth =>\nExcept.ok (first, third, fifth)\n\n\nAnd one more list lookup begins to become quite unmanageable:\n\ndef firstThirdFifthSeventh (xs : List α) : Except String (α × α × α × α) :=\nmatch get xs 0 with\n| Except.error msg => Except.error msg\n| Except.ok first =>\nmatch get xs 2 with\n| Except.error msg => Except.error msg\n| Except.ok third =>\nmatch get xs 4 with\n| Except.error msg => Except.error msg\n| Except.ok fifth =>\nmatch get xs 6 with\n| Except.error msg => Except.error msg\n| Except.ok seventh =>\nExcept.ok (first, third, fifth, seventh)\n\n\nOnce again, a common pattern can be factored out into a helper. Each step through the function checks for an error, and only proceeds with the rest of the computation if the result was a success. A new version of andThen can be defined for Except:\n\ndef andThen (attempt : Except e α) (next : α → Except e β) : Except e β :=\nmatch attempt with\n| Except.error msg => Except.error msg\n| Except.ok x => next x\n\n\nJust as with Option, this version of andThen allows a more concise definition of firstThird:\n\ndef firstThird' (xs : List α) : Except String (α × α) :=\nandThen (get xs 0) fun first =>\nandThen (get xs 2) fun third =>\nExcept.ok (first, third)\n\n\nIn both the Option and Except case, there are two repeating patterns: there is the checking of intermediate results at each step, which has been factored out into andThen, and there is the final successful result, which is some or Except.ok, respectively. For the sake of convenience, success can be factored out into a helper called ok:\n\ndef ok (x : α) : Except ε α := Except.ok x\n\n\nSimilarly, failure can be factored out into a helper called fail:\n\ndef fail (err : ε) : Except ε α := Except.error err\n\n\nUsing ok and fail makes get a little more readable:\n\ndef get (xs : List α) (i : Nat) : Except String α :=\nmatch xs[i]? with\n| none => fail s!\"Index {i} not found (maximum is {xs.length - 1})\"\n| some x => ok x\n\n\nAfter adding the infix declaration for andThen, firstThird can be just as concise as the version that returns an Option:\n\ninfixl:55 \" ~~> \" => andThen\n\ndef firstThird (xs : List α) : Except String (α × α) :=\nget xs 0 ~~> fun first =>\nget xs 2 ~~> fun third =>\nok (first, third)\n\n\nThe technique scales similarly to larger functions:\n\ndef firstThirdFifthSeventh (xs : List α) : Except String (α × α × α × α) :=\nget xs 0 ~~> fun first =>\nget xs 2 ~~> fun third =>\nget xs 4 ~~> fun fifth =>\nget xs 6 ~~> fun seventh =>\nok (first, third, fifth, seventh)\n\n\n## Logging\n\nA number is even if dividing it by 2 leaves no remainder:\n\ndef isEven (i : Int) : Bool :=\ni % 2 == 0\n\n\nThe function sumAndFindEvens computes the sum of a list while remembering the even numbers encountered along the way:\n\ndef sumAndFindEvens : List Int → List Int × Int\n| [] => ([], 0)\n| i :: is =>\nlet (moreEven, sum) := sumAndFindEvens is\n(if isEven i then i :: moreEven else moreEven, sum + i)\n\n\nThis function is a simplified example of a common pattern. Many programs need to traverse a data structure once, while both computing a main result and accumulating some kind of tertiary extra result. One example of this is logging: a program that is an IO action can always log to a file on disk, but because the disk is outside of the mathematical world of Lean functions, it becomes much more difficult to prove things about logs based on IO. Another example is a function that computes the sum of all the nodes in a tree with an inorder traversal, while simultaneously recording each nodes visited:\n\ndef inorderSum : BinTree Int → List Int × Int\n| BinTree.leaf => ([], 0)\n| BinTree.branch l x r =>\nlet (leftVisited, leftSum) := inorderSum l\nlet (hereVisited, hereSum) := ([x], x)\nlet (rightVisited, rightSum) := inorderSum r\n(leftVisited ++ hereVisited ++ rightVisited, leftSum + hereSum + rightSum)\n\n\nBoth sumAndFindEvens and inorderSum have a common repetitive structure. Each step of computation returns a pair that consists of a list of data that have been saved along with the primary result. The lists are then appended, and the primary result is computed and paired with the appended lists. The common structure becomes more apparent with a small rewrite of sumAndFindEvens that more cleanly separates the concerns of saving even numbers and computing the sum:\n\ndef sumAndFindEvens : List Int → List Int × Int\n| [] => ([], 0)\n| i :: is =>\nlet (moreEven, sum) := sumAndFindEvens is\nlet (evenHere, ()) := (if isEven i then [i] else [], ())\n(evenHere ++ moreEven, sum + i)\n\n\nFor the sake of clarity, a pair that consists of an accumulated result together with a value can be given its own name:\n\nstructure WithLog (logged : Type) (α : Type) where\nlog : List logged\nval : α\n\n\nSimilarly, the process of saving a list of accumulated results while passing a value on to the next step of a computation can be factored out into a helper, once again named andThen:\n\ndef andThen (result : WithLog α β) (next : β → WithLog α γ) : WithLog α γ :=\nlet {log := thisOut, val := thisRes} := result\nlet {log := nextOut, val := nextRes} := next thisRes\n{log := thisOut ++ nextOut, val := nextRes}\n\n\nIn the case of errors, ok represents an operation that always succeeds. Here, however, it is an operation that simply returns a value without logging anything:\n\ndef ok (x : β) : WithLog α β := {log := [], val := x}\n\n\nJust as Except provides fail as a possibility, WithLog should allow items to be added to a log. This has no interesting return value associated with it, so it returns Unit:\n\ndef save (data : α) : WithLog α Unit :=\n{log := [data], val := ()}\n\n\nWithLog, andThen, ok, and save can be used to separate the logging concern from the summing concern in both programs:\n\ndef sumAndFindEvens : List Int → WithLog Int Int\n| [] => ok 0\n| i :: is =>\nandThen (if isEven i then save i else ok ()) fun () =>\nandThen (sumAndFindEvens is) fun sum =>\nok (i + sum)\n\n| BinTree.leaf => ok 0\n| BinTree.branch l x r =>\nandThen (inorderSum l) fun leftSum =>\nandThen (save x) fun () =>\nandThen (inorderSum r) fun rightSum =>\nok (leftSum + x + rightSum)\n\n\nAnd, once again, the infix operator helps put focus on the correct steps:\n\ninfixl:55 \" ~~> \" => andThen\n\n| [] => ok 0\n| i :: is =>\n(if isEven i then save i else ok ()) ~~> fun () =>\nsumAndFindEvens is ~~> fun sum =>\nok (i + sum)\n\n| BinTree.leaf => ok 0\n| BinTree.branch l x r =>\ninorderSum l ~~> fun leftSum =>\nsave x ~~> fun () =>\ninorderSum r ~~> fun rightSum =>\nok (leftSum + x + rightSum)\n\n\n## Numbering Tree Nodes\n\nAn inorder numbering of a tree associates each data point in the tree with the step it would be visited at in an inorder traversal of the tree. For example, consider aTree:\n\nopen BinTree in\ndef aTree :=\nbranch\n(branch\n(branch leaf \"a\" (branch leaf \"b\" leaf))\n\"c\"\nleaf)\n\"d\"\n(branch leaf \"e\" leaf)\n\n\nIts inorder numbering is:\n\nBinTree.branch\n(BinTree.branch\n(BinTree.branch (BinTree.leaf) (0, \"a\") (BinTree.branch (BinTree.leaf) (1, \"b\") (BinTree.leaf)))\n(2, \"c\")\n(BinTree.leaf))\n(3, \"d\")\n(BinTree.branch (BinTree.leaf) (4, \"e\") (BinTree.leaf))\n\n\nTrees are most naturally processed with recursive functions, but the usual pattern of recursion on trees makes it difficult to compute an inorder numbering. This is because the highest number assigned anywhere in the left subtree is used to determine the numbering of a node's data value, and then again to determine the starting point for numbering the right subtree. In an imperative language, this issue can be worked around by using a mutable variable that contains the next number to be assigned. The following Python program computes an inorder numbering using a mutable variable:\n\nclass Branch:\ndef __init__(self, value, left=None, right=None):\nself.left = left\nself.value = value\nself.right = right\ndef __repr__(self):\nreturn f'Branch({self.value!r}, left={self.left!r}, right={self.right!r})'\n\ndef number(tree):\nnum = 0\ndef helper(t):\nnonlocal num\nif t is None:\nreturn None\nelse:\nnew_left = helper(t.left)\nnew_value = (num, t.value)\nnum += 1\nnew_right = helper(t.right)\nreturn Branch(left=new_left, value=new_value, right=new_right)\n\nreturn helper(tree)\n\n\nThe numbering of the Python equivalent of aTree is:\n\na_tree = Branch(\"d\",\nleft=Branch(\"c\",\nleft=Branch(\"a\", left=None, right=Branch(\"b\")),\nright=None),\nright=Branch(\"e\"))\n\n\nand its numbering is:\n\n>>> number(a_tree)\nBranch((3, 'd'), left=Branch((2, 'c'), left=Branch((0, 'a'), left=None, right=Branch((1, 'b'), left=None, right=None)), right=None), right=Branch((4, 'e'), left=None, right=None))\n\n\nEven though Lean does not have mutable variables, a workaround exists. From the point of view of the rest of the world, the mutable variable can be thought of as having two relevant aspects: its value when the function is called, and its value when the function returns. In other words, a function that uses a mutable variable can be seen as a function that takes the mutable variable's starting value as an argument, returning a pair of the variable's final value and the function's result. This final value can then be passed as an argument to the next step.\n\nJust as the Python example uses an outer function that establishes a mutable variable and an inner helper function that changes the variable, a Lean version of the function uses an outer function that provides the variable's starting value and explicitly returns the function's result along with an inner helper function that threads the variable's value while computing the numbered tree:\n\ndef number (t : BinTree α) : BinTree (Nat × α) :=\nlet rec helper (n : Nat) : BinTree α → (Nat × BinTree (Nat × α))\n| BinTree.leaf => (n, BinTree.leaf)\n| BinTree.branch left x right =>\nlet (k, numberedLeft) := helper n left\nlet (i, numberedRight) := helper (k + 1) right\n(i, BinTree.branch numberedLeft (k, x) numberedRight)\n(helper 0 t).snd\n\n\nThis code, like the none-propagating Option code, the error-propagating Except code, and the log-accumulating WithLog code, commingles two concerns: propagating the value of the counter, and actually traversing the tree to find the result. Just as in those cases, an andThen helper can be defined to propagate state from one step of a computation to another. The first step is to give a name to the pattern of taking an input state as an argument and returning an output state together with a value:\n\ndef State (σ : Type) (α : Type) : Type :=\nσ → (σ × α)\n\n\nIn the case of State, ok is a function that returns the input state unchanged, along with the provided value:\n\ndef ok (x : α) : State σ α :=\nfun s => (s, x)\n\n\nWhen working with a mutable variable, there are two fundamental operations: reading the value and replacing it with a new one. Reading the current value is accomplished with a function that places the input state unmodified into the output state, and also places it into the value field:\n\ndef get : State σ σ :=\nfun s => (s, s)\n\n\nWriting a new value consists of ignoring the input state, and placing the provided new value into the output state:\n\ndef set (s : σ) : State σ Unit :=\nfun _ => (s, ())\n\n\nFinally, two computations that use state can be sequenced by finding both the output state and return value of the first function, then passing them both into the next function:\n\ndef andThen (first : State σ α) (next : α → State σ β) : State σ β :=\nfun s =>\nlet (s', x) := first s\nnext x s'\n\ninfixl:55 \" ~~> \" => andThen\n\n\nUsing State and its helpers, local mutable state can be simulated:\n\ndef number (t : BinTree α) : BinTree (Nat × α) :=\nlet rec helper : BinTree α → State Nat (BinTree (Nat × α))\n| BinTree.leaf => ok BinTree.leaf\n| BinTree.branch left x right =>\nhelper left ~~> fun numberedLeft =>\nget ~~> fun n =>\nset (n + 1) ~~> fun () =>\nhelper right ~~> fun numberedRight =>\nok (BinTree.branch numberedLeft (n, x) numberedRight)\n(helper t 0).snd\n\n\nBecause State simulates only a single local variable, get and set don't need to refer to any particular variable name.\n\n## Monads: A Functional Design Pattern\n\nEach of these examples has consisted of:\n\n• A polymorphic type, such as Option, Except ε, WithLog logged, or State σ\n• An operator andThen that takes care of some repetitive aspect of sequencing programs that have this type\n• An operator ok that is (in some sense) the most boring way to use the type\n• A collection of other operations, such as none, fail, save, and get, that name ways of using the type\n\nThis style of API is called a monad. While the idea of monads is derived from a branch of mathematics called category theory, no understanding of category theory is needed in order to use them for programming. The key idea of monads is that each monad encodes a particular kind of side effect using the tools provided by the pure functional language Lean. For example, Option represents programs that can fail by returning none, Except represents programs that can throw exceptions, WithLog represents programs that accumulate a log while running, and State represents programs with a single mutable variable."
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https://www.scitechnol.com/peer-review/an-efficient-design-of-fsm-based-32bit-unsigned-pipelined-multiplier-using-verilog-hdl-wDnq.php?article_id=17987 | [
"## Journal of Nuclear Energy Science & Power Generation TechnologyISSN: 2325-9809\n\nAll submissions of the EM system will be redirected to Online Manuscript Submission System. Authors are requested to submit articles directly to Online Manuscript Submission System of respective journal.\nkurtkoy escort\n\nResearch Article, J Nucl Ene Sci Power Generat Technol Vol: 11 Issue: 1\n\n# An Efficient Design of FSM Based 32-Bit Unsigned Pipelined Multiplier Using Verilog HDL\n\nPrasanna Mishra*, V. M. Murugesan, G. Anand Raj\n\nDepartment of Automobile Engineering, PSG College of Technology, Coimbatore,Tamil Nadu, India\n\n*Corresponding Author: Prasanna Mishra\nDepartment of Automobile Engineering, PSG College of Technology, Coimbatore, Tamil Nadu, India,\nTel: 9490303888;\nE-mail: [email protected]\n\nReceived date: 29 December, 2021, Manuscript No. JNPGT-21-45675;\nEditor assigned date: 31 December, 2021, PreQC No. JNPGT-21-45675(PQ);\nReviewed date: 14 January, 2022, QC No. JNPGT-21-45675;\nRevised date: 21 January, 2022, Manuscript No. JNPGT-21-45675(R);\nPublished date: 24 January, 2022, DOI: 10.4172/2325-9809.1000253\n\nCitation: Mishra P, Murugesan VM, Raj GA (2021) Design and Analysis of Suspension Test Rig. J Nucl Ene Sci Power Generat Technol 11:1.\n\n## Abstract\n\nThis paper shows a modification to FSM based 32-bit pipelined multiplier. It uses Carry Look Ahead Adders (CLA’s) and Carry Select Adders (CSA) in place of Ripple Carry Adders (RCA’s) in 32-bi t FSM based pipelined multiplier for reducing the carry propagation delay. The proposed hardware design is based on shift and add algorithm for multiplication process. Our suggested pipelined multiplier design has reduced adder and added the partial product sequentially to increase maximum operating frequency and reduce hardware resources. Synthesis report shows that modified FSM based 32-bit pipelined multiplier has less delay, less usage of logical resources, than FSM based pipelined multiplier. Simulation was done in Xilinx Vivado 2017.4(Verilog HDL).\n\nThe proposed design instantiates carry select adder for the partial product addition process, carry select adder is faster than ripple carry adder. The tradeoff between delay and power, Delay has been reduced and power increased when compared to the existing method. The proposed method can be used for the high-speed pipelined multiplication operation.\n\n## Introduction\n\nMultiplier circuit is a building block in digital design such as Digital Signal Processing, Microprocessors, Multiply Accumulator block, Arithmetic Logic Unit blocks, etc. Main Challenge in designing multiplier block is computation process, i.e., how efficient the multiplication process has been completed. There should be minimum computation time for multiplication. This is the reason for repeated addition method has been more preferred than multiplier blocks. There are n number of digital logic multipliers available, important factor to be considered in the multiplier logic is the computation time and time complexity. The Computation time in the logic circuit design is the amount of time required to complete the entire process in the design. When compared to adder, multiplier has more computational time and complexity. The multiplication techniques mainly involve the computing set of partial products, then summing of the partial products. The partial products are defined as the output formed by multiplying the multiplicand with each bit of the multiplier. There are mainly two types of multiplier implementation, they are combinational implementation method and pipelined implementation method .\n\nIn the combinational multiplier, the multiplication process is by multiplying the multiplicand against the multiplier. There are various types of multiplier logic design, in which combinational multiplier is one of the types. The other multiplier types are Array multiplier, sequential multiplier, Wallace tree multiplier, logarithm multiplier, booth multiplier, modified booth multiplier. Array multiplier, is for multiplying the two binary numbers by using an array of full adders and half adders, to form the various product, an array of AND gates is used before the adder array design. Sequential multiplier, is an outdated method in the multiplication of two binary numbers, but its relevant in many architectural designs. The below figure represents the 4-bit sequential multiplier with a3, a2, a1, a0 and b3, b2, b1, b0 are the inputs and s7, s6, s5, s4, s3, s2, s1, s0 are the outputs.",
null,
"Figure 1: 4-bit Sequential multiplier.\n\nWallace tree multiplier is the variant of long multiplication. The Wallace tree multiplier steps are as follows, the first step is to multiply the bit of a factor by the other digit, the obtained partial products have equal weight to the product of its factors. The final product value is calculated by summing up the resultant net partial products. In this paper, we have proposed a new model approach for an efficient design of 16-bit unsigned multiplier using pipelined stages. The pipelined mechanism involves storing and executing the instruction in a sequential process. The pipeline mechanism is divided into different stages based on the architecture. The main stages of pipelining involve Fetch, Decode, Execute, Memory access and Writeback. In the proposed method, the partial product is stored in the pipelined registers in the different stages of the partial product addition. For example, for 4 row partial product block at the end of 2nd row set of pipelined registers are employed to complete the partial product addition .\n\nThe result supersedes the previous combinational multiplier design approach in clock cycle and operating frequency. The method used in the proposed design is repeated shift and addition operations for the partial product and the adder used in the design is Carry Select Adder for the further reduction in the delay in the carry propagation part of the instantiated adder design. To control the Finite State Machine (FSM) model, we have used mux and demux in the design.\n\n## Literature Review\n\nThe usage of Carry Select Adder reduces the delay in the propagation in the Vedic multiplier circuit, this were discussed. The CSA is known to be one of the fastest adder, thus the Vedic multiplier is implemented using the Carry Select Adder. The results of Carry Select Adder are compared with the conventional Carry Select Adder and the modified Carry Select Adder. The main aim is to reduce the number of gates involved in the design. A novel approach to design the signed and unsigned multiplication were proposed using simple sign control unit with help of multiplexers were discussed in the method demonstrated in 0.18μm technology and the result shows reduction in silicon area overhead up to 0.45%. In addition to that, modified booth encoding techniques were employed to reduce the number of partial products in the multiplication process. Various techniques on implementing 32-bit unsigned multiplier based on pipelined architecture and combinational architecture can be found in the literature. A 32-bit unsigned multiplier is implemented in FPGA using Vedic algorithm and maximum combinational path delay achieved was 22.829 ns. Pipelined multiplier\n\n## Existing Method",
null,
"Figure 2: Datapath block.\n\nThe complete datapath block is shown in the above Figure 2. The FSM logic having the mux and demux (sel input) for the datapath block. The present state logic having the register and the next state logic having the combinational block (adder circuit).",
null,
"Figure 3: Finite state machine block.\n\n## Proposed Hardware Design Methodology\n\nDesign method\n\nIn this paper, we have applied repeated shift and add operation for the multiplication process. In general, n-bit multiplicand and n-bit multiplier will generate 2n-bit product. The generated n number of partial products are left shifted to one bit and added the partial product .\n\nDesign approach example",
null,
"Figure 4: Shift and add multiplication block.\n\nThe figure 4 illustrates the example of shift and add operation in the design, the inputs are A and B, output Product P. Bit select operation in the least significant bit (LSB) of input B, i.e., B has been taken and performed AND operation with the input A vector. In the first cycle A is added to P if the least significant bit of B is '1'. A is shifted left while B is shifted right. The process is repeated in subsequent cycles and completes when the right shifted B register has become zero (B=0). Figure 5 illustrates the multiplication process having the partial product addition method, 4-bit multiplicand and 4-bit multiplier results in 8-bit product with 4-bit partial product outputs.",
null,
"Figure 5: Multiplication example.",
null,
"Figure 6: Pipelined multiplication.\n\nThe figure 6 shows the stages in the pipelined multiplication process. The first stage consists of set of registers with Adder, AND gate block. The LSB and the immediate next bit of the LSB of the multiplier was taken and performed the AND operation with the multiplicand, the results of the above operation stored in the register. The output of the register taken as the input for the next stage of the pipelined multiplication.\n\nThe existing approach uses Ripple Carry Adder for the addition of the partial product. The carry generated in each adder gets propagated to the next adder and the carry gets sum up. Thus, the carry should propagate through the cascaded full adder stages. These encounters delay in the addition process. For 32-bit addition, the instantiated addition process takes time for the computation. Thus, the overall addition process takes delay in the computation.",
null,
"Figure 7: Proposed design block diagram.\n\nThe proposed approach suggests the usage of Carry Select Adder in the addition process for the partial product block. The instantiation of the Carry Select Adder in the place of Ripple Carry Adder may reduce the encountered delay in the addition.\n\n## Implementation of the Design\n\nThe proposed pipelined multiplier design has carry select adder instantiated for the partial product addition, the delay in the computation time for the addition process would be reduced with the help of carry select adder when compared to the ripple carry adder method.\n\nA type of Digital adder to efficiently compute the sum of two or more binary numbers. These adders can outperform the computation operation in the multiplication. Instead of using the entire ripple carry adder instantiation in the whole partial product addition process, the carry save adder can be employed to perform the higher bit addition process. The carry save consists of n-full adders, each computes a single sum and carry bit on corresponding bits of three input numbers. The numbers are a, b and c, it produces a partial sum ps and a shift carry sc. The entire sum of the operation can be calculated by the following process. First, Shifting the carry sequence sc left by one place. Second, appending a 0 to the front of the partial sum sequence ps. Third, using a full adder to add these two together and produce the result (n+1) bit value .",
null,
"Figure 8: 4-bit carry save adder.\n\nThe above Figure shows the instantiated 4-bit Carry Save Adder for the partial product addition computation. The delay response was compared with the corresponding Ripple Carry adder and the Carry Select Adder designs.",
null,
"Figure 9: Synthesized schematic design of carry save adder.\n\nCarry Select Adder is an arithmetic combinational logic unit which sums up two N bit binary sum and 1 bit carry, the main difference in the Carry Select Adder and Ripple Carry Adder is the delay in the propagation of the carry. The Ripple Carry Adder will have longer delay in the carry propagation than the Carry Select Adder.",
null,
"Figure 10: 4-bit carry select adder.\n\nThe above figure represents the instantiated 4-bit Carry Select Adder. At the fourth stage of the adder, the propagated carry Cout was taken out and provided as an input to the next stage of the Carry Select Adder.\n\nPipelined FSM multiplier design\n\nThe Proposed pipelined multiplier design with fixed latency FSM, the instantiated adder (Carry Select Adder) in the design based on the delay comparison. The FSM Block controls the Mux and Demux of the Partial product generator blocks.",
null,
"Figure 11: FSM for fixed latency.\n\nThe above figure represents the state changes for the FSM (fixed latency), the FSM has a counter to check the repeated shift and add operation. The resp_rdy signal controls the mux select lines. The counter value resets and goes back to the idle state when it reaches 32. The counter remains in Calc state (when counter less than 32) where the multiplication process is done.\n\n## Simulation Result\n\nThe schematic design of the proposed pipelined multiplier was simulated in the Vivado 2017.4. The proposed multiplier having similar five stages of the existing multiplier, the instantiated adder for the partial product addition is Carry Select Adder (due to less delay in the propagation of the carry).",
null,
"Figure 12:Synthesized schematic design of carry select adder.\n\nThe following report shows the delay comparison report of Ripple Carry Adder, Carry Save Adder and Carry Select Adder. The minimum accountable delay should be considered for the design of pipelined multiplier.",
null,
"Figure 13:Timing summary of ripple carry adder.\n\nThe overall critical path delay for the ripple carry adder design is 6.49 ns. The reason for the delay is the propagation of the carry in each and every adder stages.",
null,
"Figure 14:Timing summary of carry select adder.\n\nThe above timing summary shows the path delay for Carry Select Adder is 5.815 ns. The delay summary report of the Carry Save Adder design.",
null,
"Figure 15:Timing summary of carry save adder.\n\nThe above timing summary shows the path delay for Carry Save Adder is 5.822 ns. There is a small difference in the delay between Carry Save Adder (5.822 ns) and Carry Select Adder (5.815 ns).",
null,
"Figure 16: Schematic design of the proposed pipelined multiplier.\n\nThe delay reports were compared and thus the Carry Select Adder were instantiated in the Pipelined Multiplier design.",
null,
"Figure 17: Simulation waveform of pipelined multiplier.\n\nThe above figure represents the simulated waveform of the pipelined multiplier, Inputs in Decimal (a-23, b-10, output of the multiplier at 9th cycle-230 in decimal). Timing Constraints include Critical path delay (the maximum delay in the reg-to-reg datapath).",
null,
"Figure 18: Timing report of proposed multiplier.\n\nThe above figure shows the timing summary report of the proposed pipelined multiplier, the overall critical path delay for the multiplier is 4.01 ns (Instantiated Carry Select Adder).",
null,
"Figure 19: Power summary report of proposed pipelined multiplier.\n\n## Observation\n\nFrom the above Delay comparison table, we can observe that by using Ripple Carry Adder instantiation in the design produces a delay of 7.01 ns, whereas Carry Select Adder produces a delay of 4.01 ns, clearly states that by using Carry Select Adder we can optimize the delay and improve the speed and performance of the design . RCA Pipelined multiplier calculation: The inverse of the Critical path delay (max delay) gives the operating max frequency .",
null,
"",
null,
"Figure 21: Power delay analysis between RCA and CSA pipelined multiplier.\n\nThe speed of the pipelined multiplier can be improved by instantiating Carry Select Adder than Ripple Carry Adder in the partial product addition process. The Power consumption of the overall Pipelined multiplier using Ripple Carry Adder (104 m W), Carry Select Adder (232 mW) . We can clearly observe that there is a tradeoff between Speed and Power (i.e., Delay-Power Tradeoff in the design).\n\n## Conclusion\n\nIn the presented paper, we have designed 32- bit pipelined multiplier using Carry Select Adder, the main advantage of the Carry Select Adder instantiation is to improve the speed of the partial product addition process. The Carry Select Adder is faster than Ripple Carry Adder instantiation. The overall critical path delay for the proposed design is reduced by 17.21% than the existing path delay. The throughput has been increased by 21.02% than the existing throughput he latency and the clock cycle for the computation of the multiplication process is 9 cycles with 326 MHz frequency. The tradeoff in the proposed design is between Power and Delay. We can go for the proposed pipelined multiplier for high- speed application without the consideration of power usage.\n\n## References\n\n1. Arif S, Lal RK (2015) Design and performance analysis of various adder and multiplier circuits using VHDL. Int J Appl Eng Res 10: 17016-17020.\n2. Gowthami M, Jalall K, Kiruthika K (2021) High speed and performance analysis of multiplier in field programming gate array. IOP Conference Series: Mater Sci Eng 1084: 012062.\n3. McCanny JV, McWhirter JG (1982) Completely iterative, pipelined multiplier array suitable for VLSI. IEE Proceed G-Electr Circuit Syst 129: 40-46.\n4. Yan M, De-li W (2010) The design of an architecture-aware 32-bit signed/unsigned multiplier. Int Confer Compu Eng Technol 7: V7-354.\n5. Pekmestzi KZ, Kalivas P, Moshopoulos N (2001) Long unsigned number systolic serial multipliers and squarers. IEEE Transact Circuits Syst II: Analog Digital Signal Process 48: 316-321.\n6. Di J, Yuan JS, Demara R (2006) Improving power-awareness of pipelined array multipliers using two-dimensional pipeline gating and its application on FIR design. Integration 39: 90-112.\n7. Mounica Y, Kumar KN, Veeramachaneni S (2021) Energy efficient signed and unsigned radix 16 booth multiplier design. Comput Electric Eng 90: 106892.\n8. Antelo E, Montuschi P, Nannarelli A (2016) Improved 64-bit radix-16 booth multiplier based on partial product array height reduction. IEEE Transact Circuits Syst I: Regular Papers 64: 409-418."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9066799,"math_prob":0.87676764,"size":20209,"snap":"2022-27-2022-33","text_gpt3_token_len":4432,"char_repetition_ratio":0.21044296,"word_repetition_ratio":0.029254483,"special_character_ratio":0.21208373,"punctuation_ratio":0.111624114,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9714453,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T07:23:48Z\",\"WARC-Record-ID\":\"<urn:uuid:04d9a4d3-c61e-4141-a715-634cf14e7bf3>\",\"Content-Length\":\"72865\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3cbae40d-d162-492d-9a50-6db9f5e2b8c5>\",\"WARC-Concurrent-To\":\"<urn:uuid:286c3214-2eec-4494-add2-45f448b0fb2b>\",\"WARC-IP-Address\":\"104.21.37.217\",\"WARC-Target-URI\":\"https://www.scitechnol.com/peer-review/an-efficient-design-of-fsm-based-32bit-unsigned-pipelined-multiplier-using-verilog-hdl-wDnq.php?article_id=17987\",\"WARC-Payload-Digest\":\"sha1:QEBJJSLS52QCMKHJCCO7YGCRUBMGBTSP\",\"WARC-Block-Digest\":\"sha1:7LGZHTFACLLBCS73IZMPLW365HTJTGAS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104514861.81_warc_CC-MAIN-20220705053147-20220705083147-00364.warc.gz\"}"} |
https://www.emusician.com/gear/how-your-daw-does-math | [
"# How Your DAW Does Math\n\nMuch of the number crunching that occurs within your audio software remains a mystery; does your DAW treat audio as floating-point numbers or as fixed-point numbers? What''s the difference, and should you really care?\nAuthor:\nUpdated:\nOriginal:\n\nDesktop musicians have lived with computers and digital audio long enough to know how binary and decimal numbers relate to one another. Had humans developed with two fingers instead of ten, we might well consider binary numbers normal. Yet much of the number crunching that occurs within your audio software remains a mystery. For example, does your DAW treat audio as floating-point numbers or as fixed-point numbers? Do you know what the difference is, and should you really care?\n\n## The Fix Is In\n\nFIG. 1: The binary number 1101 -represents 8 (1 5 23) + 4 (1 5 22) + 0 (0 5 21) + 1 (1 5 20), or 13, in the decimal system.\n\nDecimal numbers are known as base 10 numbers because the meaning of each column is based on increasing powers of 10. The base, 10, is also known as the radix, and the dot that divides an expression into integers on the left and fractional values on the right is called the radix point (it's called the decimal point in the decimal system). The first digit to the left of the radix point represents the number of ones (the ones column); the second, multiples of 10 (101); the third, multiples of 100 (102); and so on. Each place represents another power of the radix, with the exponent (the number indicating the power) corresponding to the number of places to the left of the ones column.\n\nThe same pattern applies to the right of the radix point, but the exponents are negative. The first decimal place (tenths) consists of multiples of 10-1; the second (hundredths), multiples of 10-2; the third (thousandths), multiples of 10-3; and so on. The negative exponent means the same as dividing 1 by that power of 10. For example, 10-3 is the same as 1 ÷ 103 or 1 ÷ 1,000 — each means a thousandth. Yes, the ones place is multiples of 100 — any number raised to the power of 0 is 1.\n\nAll of this is known as fixed-point arithmetic because the radix point never moves. The same principles apply in a binary system, but because the radix is 2, each place is an appropriate power of 2. The binary number 1101 represents 8 + 4 + 0 + 1, or 13, in the decimal system (see Fig. 1). Try not to think too hard about binary fractions or your head might explode — the same pattern of negative exponents applies.\n\nFixed-point arithmetic is pretty straightforward, but there are times when it is inefficient or limiting. In particular, describing very large or very small numbers in fixed-point notation is awkward. For example, as I write this, the U.S. national debt is reported to be \\$8,551,759,486,884.88. (No, wait — it just went up \\$3 million!) That's a long string of numbers and is usually therefore expressed as “\\$8½ trillion.” Although the word trillion is not terribly meaningful to a calculator or a computer, it's a useful description for us because it makes a very large number more manageable visually and mentally.\n\n## Float Me a Trillion?\n\nFloating-point numbers are good for expressing very large or very small numbers. For example, in floating-point parlance one would say that the Andromeda galaxy is about 2.1 × 1019 kilometers away. In fixed-point terminology, however, the distance must be expressed using a number that is 20 digits long: 21,000,000,000,000,000,000! To create a floating-point number from a fixed-point number, lop off the significant digits from the left of the expression (the significand, or 2.1 in this case) and specify how many places you needed to shift the radix point (19 in this example). The significand is sometimes called the mantissa, although some prefer to reserve that term for use with logarithms. The rest of the expression then indicates the magnitude of the number — how big (or small) this number really is. It consists of a radix and an exponent (which is effectively a zoom level). You may recognize this as the way your calculator represents very large numbers — through scientific notation, or decimal floating-point notation.\n\nThe 32-bit floating-point numbers that most host-based DAWs use allocate 1 bit (called the sign bit) to indicate whether the number is positive or negative; 8 bits (256 values) of exponent; and 23 bits of significand (see Fig. 2). Interestingly, there are still 24 bits of precision in the significand. That is because a “normalized” floating-point number always has exactly one nonzero digit to the left of the radix point, and because with a binary number that can only be a 1, a bit isn't wasted on it. This is known as the hidden bit.\n\n## Reality Sets In\n\nThere's nothing inherently better about either system. As long as you can keep adding digits to the left or the right of the radix, you can represent any real number in fixed-point format. Problems start when you have a limited number of digits (or bits, in the case of binary) with which to work.\n\nConsider the problem of adding the fixed-point decimal numbers 5,200 and 6,582 if restricted to using only four digits. The answer that one would like to come up with is 11,782, but that uses five digits. This is known as an overflow or, in audio terms, clipping.\n\nFIG. 2: Here is the anatomy of a 32-bit floating-point number: 1 sign bit (positive or -negative), 8 exponent bits, and 23 significand bits.\n\nThe floating-point system presents similar challenges: (5.2 × 103) + (6.582 × 103) should equal 1.1782 × 104, but if restricted to using only four digits of significand, the final digit (2) is lost. And if restricted to using exponents of no more than 3, there is an overflow. Clipping and loss of resolution are the omnipresent bugaboos of the math that makes our DAWs tick.\n\nBoth numbering systems pre-sent unique challenges and require unique solutions for handling overflow, resolution, and other quirks. The sound that a particular DAW has is due in part to the skills of the programmers who wrote the algorithms that decide how signals are summed, scaled, and otherwise manipulated internally. Given sufficient resolution, however, the differences between the two fade into relative obscurity. Modern DAWs often use double-precision arithmetic, meaning that fixed-point systems allow 48 bits to process 24-bit signals and that some floating-point systems allow 64 bits to handle 32-bit signals.\n\nThe other big distinction between the two systems is how they handle quantization error. Quantization error is the unavoidable result of the finite resolution of digital samples. When the measured voltage at an A/D converter or the result of a processing algorithm is between two numbers, you must round the result up or down to the system's resolution. Rounding very small fixed-point numbers (representing very quiet signals) results in very large errors relative to the overall signal level: rounding from 4.5 up to 5 is a 10 percent error! The louder the signal, the less significant quantization error becomes: rounding from 99.5 up to 100 is a 0.5 percent error and rounding from 999.5 up to 1,000 is only 0.05 percent.\n\nThings are different, however, with floating-point numbers. Rounding is still more problematic with very small significand values than with large significand values. But each time that the exponent increases or decreases, the pattern starts over, just as the rising pitch of an engine drops and restarts every time you shift up. Quantization error no longer fades into obscurity with rising levels. It can be quite large in a loud floating-point signal compared with an equivalent fixed-point signal, but because the signal is loud, how apparent will the quantization distortion be? Some observers suggest that the absolute level of the quantization distortion is less significant than the fact that its level “pumps” with each change in exponent.\n\n## Resolution\n\nThe advantages of either system diminish with increased resolution, when more bits are available to buffer overflow and retain resolution. To put everything in perspective, you will get much more mileage from wise mic selection and placement than obsessing over the differences between fixed- and floating-point systems, and neither type of arithmetic will save a bad song.\n\nBrian Smithers is a musician and educator in central Florida."
]
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https://www.bartleby.com/solution-answer/chapter-19-problem-18sp-fundamentals-of-financial-management-mindtap-course-list-15th-edition/9781337395250/multinational-financial-management-yohe-telecommunications-is-a-multinational-corporation-that/d597fabb-feda-11e8-9bb5-0ece094302b6 | [
"",
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"",
null,
"",
null,
"Chapter 19, Problem 18SP",
null,
"Fundamentals of Financial Manageme...\n\n15th Edition\nEugene F. Brigham + 1 other\nISBN: 9781337395250\n\nSolutions\n\nChapter\nSection",
null,
"Fundamentals of Financial Manageme...\n\n15th Edition\nEugene F. Brigham + 1 other\nISBN: 9781337395250\nTextbook Problem\n\nMULTINATIONAL FINANCIAL MANAGEMENT Yohe Telecommunications is a multinational corporation that produces and distributes telecommunications technology. Although its corporate headquarters are located in Maitland, Florida, Yohe usually buys its raw materials in several different foreign countries using several different foreign currencies. The matter is further complicated because Yohe often sells its products in other foreign countries. One product in particular, the SY-20 radio transmitter, draws Component X, Component Y, and Component Z (its principal components) from Switzerland, France, and the United Kingdom, respectively Specifically, Component X costs 165 Swiss France, Component Y costs 20 euros, and Component Z costs 105 British pounds. The largest market for the SY-20 is japan, where the product sells for 50,000 Japanese yen. Naturally, Yohe is intimately concerned with economic conditions that could adversely affect dollar exchange rates. You will find Tables 19.1, 19.2, and 19.3 useful for completing this problem. a. How much in dollars does it cost Yohe to produce the SY-20? What is the dollar sale price of the SY-20? b. What is the dollar profit that Yohe makes on the sale of the SY-20? What is the percentage profit? c. If the U.S. dollar was to Weaken by 10% against all foreign currencies, what would be the dollar profit for the SY-20? d. If the U.S. dollar was to weaken by 10% only against the Japanese yen and remained constant relative to all other foreign currencies, what would be the dollar and percentage profits for the SY-20? e. Using the 180-day forward exchange information from Table 19.3, calculate the return on 1-year securities in Switzerland assuming the rate of return on l·year securities in the United States is 4.9%. f. Assuming that purchasing power parity (PPP) holds, what would be the sale price of the SY-20 if it was sold in the United Kingdom rather than Japan?\n\na)\n\nSummary Introduction\n\nTo determine: The cost incurred by Company Y to produce the SY-20 in dollars.\n\nIntroduction:\n\nExchange rate is the rate, which indicates the conversion rate for currency of a country which can be getting in exchange of currency of another country.\n\nExplanation\n\nGiven information:\n\nExchange rates of given currencies in term of the Country U dollars are as follows:\n\nSummary Introduction\n\nTo determine: The sale price in dollar.\n\nIntroduction:\n\nExchange rate is the rate, which indicates the conversion rate for currency of a country which can be getting in exchange of currency of another country.\n\nb)\n\nSummary Introduction\n\nTo determine: The profit earned by sale of SY-20 in dollar.\n\nSummary Introduction\n\nTo determine: The percentage of profit.\n\nc)\n\nSummary Introduction\n\nTo determine: The amount of profit after dollar depreciates by 10%.\n\nIntroduction:\n\nCurrency depreciation indicates the negative (decrease) change in the currency’s value in reference of any other currency due to some factors such as change in government policies, and fluctuation in interest rates.\n\nd)\n\nSummary Introduction\n\nTo determine: The amount of profit earned in terms of dollars and its percentage (when dollar weaken by 10% for yen only).\n\ne)\n\nSummary Introduction\n\nTo determine: The rate of return of securities in S country.\n\nf)\n\nSummary Introduction\n\nTo determine: The price of product in UK country (pounds) instead of Country J.\n\nIntroduction:\n\nPurchasing Power Parity (PPP) refers to that relationship which indicates the same cost of s same kinds of products in the market of various countries after adjustment of exchange rates of currencies. This relationship of common price can be termed as the law of one price.\n\nStill sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\nThe Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\nWhat is a systems flowchart?\n\nAccounting Information Systems\n\nWhy do economists oppose policies that restrict trade among nations?\n\nBrief Principles of Macroeconomics (MindTap Course List)\n\nBETA AND REQUIRED RATE OF RETURN A stock has a required return of 11%. the risk-free rate is 7%, and the market...\n\nFundamentals of Financial Management, Concise Edition (with Thomson ONE - Business School Edition, 1 term (6 months) Printed Access Card) (MindTap Course List)",
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http://2008.sub.blue/blog/2008/5/26/simple-attractors.html | [
"Visit the new site\n\n# Simple Attractors\n\nPublished on 26 May 2008\n\n## This little experiment lets you visually explore two very simple attractor equations that generate very unexpected and aesthetically pleasing results.",
null,
"It was coded in Actionscript 3.0 and wrapped into an Adobe Air application so that you can export the rendered imaged as PNGs.\n\nThe two simple attractor equations are credited to Peter de Jong:\n\n`x' = sin(a * y) - cos(b * x)y' = sin(c * x) - cos(d * y)`\n\nAnd Cliff Pickover:\n\n`x' = sin(a * y) + c * cos(a * x)y' = sin(b * x) + d * cos(b * y)`\n\nStart with a random point x, y and plot a semi-transparent pixel at the calculated x', y' point. Plug the new x', y' values back into the equations to get the next point, and so on for a few hundred thousand iterations. Vary the a, b, c, d constants to give wildly different results.\n\nView the gallery for more renderings.\n\nDownload the application and get the source code.\nNote: you will need the Adobe Air runtime installed first.",
null,
"Last updated: 5 October 2008"
]
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"http://2008.sub.blue/assets/0000/1018/pdj-a2_-b50_-c50_-d0.01_-s2_medium.jpg",
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"http://2008.sub.blue/assets/0000/1995/Picture_1_medium.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.87680453,"math_prob":0.9463742,"size":2027,"snap":"2019-13-2019-22","text_gpt3_token_len":492,"char_repetition_ratio":0.09540287,"word_repetition_ratio":0.0054644807,"special_character_ratio":0.2446966,"punctuation_ratio":0.12405063,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9577431,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-03-24T17:08:42Z\",\"WARC-Record-ID\":\"<urn:uuid:6ec4edbd-37b5-4ec2-a17b-3becb89de671>\",\"Content-Length\":\"9138\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c61e5a3-c318-483b-b688-fa85870e4e2a>\",\"WARC-Concurrent-To\":\"<urn:uuid:6dc0aa47-96f1-427f-b770-e594172e8f9f>\",\"WARC-IP-Address\":\"52.218.16.178\",\"WARC-Target-URI\":\"http://2008.sub.blue/blog/2008/5/26/simple-attractors.html\",\"WARC-Payload-Digest\":\"sha1:IDF3YH6APYXZJR665UAYSGI34ZN5NL7Q\",\"WARC-Block-Digest\":\"sha1:U2MA2RWECBVGM7CG7IPW4OFTHOMQ2Y7A\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-13/CC-MAIN-2019-13_segments_1552912203464.67_warc_CC-MAIN-20190324165854-20190324191854-00228.warc.gz\"}"} |
https://www.mssqltips.com/sqlservertip/2915/sql-server-analysis-services-period-over-period-variance-analysis/ | [
"SQL Server Analysis Services Period over Period Variance Analysis\n\nProblem\n\nOften when working with OLAP cubes, the main \"access\" point will be an Excel pivot table. While utilizing OLAP cubes, end users will frequently want to compare two values from different periods in order to quantify some variance analysis processes; however the creation of calculations within OLAP based Excel pivot tables is not allowed. What functionality is available to make these comparisons?\n\nSolution\n\nWhen pivot tables are created locally, calculated fields are easily created as part of the pivot table. However, OLAP based pivot tables do not afford the ability to create these calculated values at the pivot table level. A few external tools exist, such as OLAP Pivot Table extensions available on CodePlex ( http://olappivottableextend.codeplex.com ); unfortunately most end users are not going to want to or have the ability to write the necessary MDX when working with this type of tool. Additionally, if you already have two measures to compare over the same time period, such as an actual versus budget, that calculation is easy mastered as both measures are readily available. Even so, your variance analysis must be part of the cube design; fortunately, using various functions including ParallelPeriod within calculated measure allows for easy variance analysis.\n\nVariance Analysis-Period over Period\n\nThe initial step in completing the variance analysis is to first determine the measures which require variance analysis, and second determine upon which time dimension hierarchy or hierarchies, if multiples exists, to compare between periods, such as the most common year over year. We will again use the AdventureWorks 2012 DataWarehouse and Cube for our examples. In particular, we will use the Internet Sales Amount measure and the Calendar Date hierarchy for our examples. Furthermore, we will use the ParallelPeriod function to determine the same period prior year value. The ParallelPeriod function is described in detail in Ray Barley's tip: http://www.mssqltips.com/sqlservertip/2367/building-calculated-members-for-a-ssrs-report. The actual calculation to get the prior \"parallel\" period's value is displayed below with an explanation included thereafter.\n\n--Calculated Measure\n\nIIF([Date].[Calendar].CurrentMember.level.ordinal = 0,\n[Measures].[Internet Sales Amount],\n(ParallelPeriod([Date].[Calendar].[Calendar Year],\n1,\n[Date].[Calendar].CurrentMember),[Measures].[Internet Sales Amount]\n)\n)\n\nThe statement is actually composed of two statements.\n\nFirst is the outer IIf function; this function handles the total value for the parallel period calculations. Without this IIF statement, no totals would be displayed for the parallel period calculated measure. By checking to see if the CurrentMember is at level 0 (the base level or ordinal position for the selected date attribute), the total value only displays when the total is needed.\n\nThe second part of above statement contains the Parallel period function. This function requires 3 arguments.\n\n• The first argument is the level expression; in basic terms, this argument will be the date member level that you are comparing. For instance, it could be Calendar Year or Calendar Quarter or Calendar Month.\n• The second argument is the Index or number of lag periods to look back in order to find the ancestor value. For example if you specify a level of Calendar Year, and the index is 2, then the function would find the same period 2 years prior. Of course in accounting the most common lag is 1 year.\n• The last argument specifies what member expression is moved back the Index number of positions. In simpler terms, this argument is the starting member value upon which the level and index act upon.\n\nIn the example above, we use the Calendar Year attribute while finding the Internet Sales Amount for the same period 1 year prior using the Current Member of the Date.Calendar dimension hierarchy. Rolling all these items together, this calculated measure is added to the Adventure works cube as noted in the below example. Thus, first, Visual Studio 2010 (aka SQL Server Data Tools) will need to be started, and the Adventure Works SSAS database opened. Next open the Adventure Works cube and switch to the Calculation tab.",
null,
"",
null,
"Now we define the calculated measure by clicking on the New Calculated Measure button.\n\nNext, enter a measure name; in this case Parallel Period Internet Sales Amount. Note that calculated measures with spaces in the name must be contained within brackets. Measures is selected for the Parent Hierarchy and the above expression is added to the expression box (this box contains the actual MDX expression).\n\nLast, select the Format String of currency, Visible value of true, Non Empty Behavior of Internet Sales Amount, and last Associated Measure group of Internet Sales. The Non Empty Behavior field is important as it guides which calculations must be made for the calculated measure based on the measure you select; thus if the Non Empty Behavior measure value is Empty, this will guide SSAS to also mark the calculated measure as Empty which in turn improves performance by preventing this calculation to be made for all intersections of the selected dimensions.\n\nEnough digression on the inner workings of calculated values, but I would like to mention a couple of items about our calculated measure. First, since we are using the Date.Calendar hierarchy we can actually traverse any attribute of that hierarchy which falls under the attribute year noted in our level. Subsequently, if we drop Calendar Quarter, Internet Sales Amount and Parallel Period Internet Sales Amount on a pivot, we will see all Calendar Quarters along with the related Internet Sales Amount and the Internet Sales Amount for the Same Quarter one year prior. Since we used the CurrentMember in our member expression (third argument), we can actually switch out Calendar Quarter with Calendar Month, and now all months will be displayed along with the Internet Sales Amount plus the Parallel Period Internet Sales Amount from the same month 1 year prior.\n\nBoth those scenarios are illustrated below.",
null,
"",
null,
"Now that the ParallelPeriod functionality is in place, we can continue developing our variance analysis. Our next calculation is a simple and which just determines the variance between the CurrentMember measures and the ParallelPeriod measure. Finally, we calculate the Variance % by dividing the variance amount by the ParallelPeriod value. Both of these simple calculations are displayed below.\n\nVariance:\n[Measures].[Internet Sales Amount]-[Measures].[Parallel Period Internet Sales Amount]\n\nVariance %:\n[Measures].[Parallel Period Internet Sales Amount Variance] / [Measures].[Parallel Period Internet Sales Amount]\n\nInitial examples of these calculations are displayed within the below Excel pivot table.",
null,
"",
null,
"Of course a couple of problems surface with these calculations. The calculation for the first period errors with a #NUM! error; this issue is more pronounced if we drill into quarters or months as noted in the second illustration displayed above. The problem is that there is no value for the prior parallel period. Let us address this issue by checking for the absence of a prior period value as described in the below MDX calculate measure.\n\n--Calculated Measure to check for no prior period value, check #1 checks for empty value, check #2 for 0 value\nCREATE MEMBER CURRENTCUBE.[Measures].[Parallel Period Internet Sales Amount Variance% v2]\nAS\nIIF(\n\nIsEMPTY(\n(\nParallelPeriod\n(\n[Date].[Calendar].[Calendar Year],\n1,\n[Date].[Calendar].CurrentMember\n),[Measures].[Internet Sales Amount]\n)\n)\nOR\n(\nParallelPeriod\n(\n[Date].[Calendar].[Calendar Year],\n1,\n[Date].[Calendar].CurrentMember\n),[Measures].[Internet Sales Amount]\n)=0\n,\n0,\n[Measures].[Parallel Period Internet Sales Amount Variance] / [Measures].[Parallel Period Internet Sales Amount]\n),\nFORMAT_STRING = \"Percent\",\nVISIBLE = 1 , ASSOCIATED_MEASURE_GROUP = 'Internet Sales' ;\n\nYou may find it interesting that we are checking if our base calculation is equal to zero when the set or measures is actually empty, but SSAS treats many empty cells as zero. For an in depth reviewing on how SSAS treats empty measures or sets, please see the MSDN article in the Next Steps section at the end of this article. In the above example, we cover both possible instance by checking for a value of 0 or checking for an empty value using the IsEmpty MDX function. If either is true, then we set the Parallel Period Internet Sales Amount Variance% v2 measure to 0; otherwise we perform the variance calculation. The resulting pivot is displayed below.",
null,
"The above methods utilize a multi-tiered approach to solve our period over period variance analysis needs while at the same time providing flexibility in the use of variance time dimension attributes, such as quarters, months, and years. Just as with SQL, other MDX functions exist which could provide similar results to the parallel period calculation. One such function is the Lag function which was covered in my tip on Lag / Lead functions in SSAS 2012, http://www.mssqltips.com/sqlservertip/2877/sql-server-analysis-serviceslead-lag-openingperiod-closingperiod-time-related-functions/. Furthermore, the Cousin function can also be used to complete period over period variance analysis. However, I find both these functions limit your flexibility if you want to navigate down the time dimension hierarchy as each requires a set level to traverse backwards (or forward in the Lead function's case).\n\nBelow, an example of the cousin function is listed.\n\nWITH\nMEMBER Cousin_Internet_Sales_Amount AS\n(\n(\nCousin\n(\n[Date].[Calendar].CurrentMember\n,\nAncestor\n(\n[Date].[Calendar].CurrentMember\n,[Date].[Calendar].[Calendar Quarter]\n).Lag(1)\n)\n\n)\n, [Measures].[Internet Sales Amount]\n),format_string='\\$0,000.00'\nSELECT\n{\n[Measures].[Internet Sales Amount],\nCousin_Internet_Sales_Amount\n} ON COLUMNS,\n[Date].[Calendar].[Calendar Quarter] ON ROWS\n\nNotice, how much more complicated the Cousin function is as compared to the ParallelPeriod function, and requires the use of the Ancestor and Lag functions also. The results of the above statement illustrated subsequently.",
null,
"Conclusion\n\nVariance analysis for SSAS OLAP cubes is not a simple matter of adding a calculated field to a pivot table. Planning along with the use of the ParallelPeriod MDX functions allows us to quickly create a variance infrastructure for a particular measure. Furthermore, by utilizing a date hierarchy in the Parallel Period function, we can easily traverse down the hierarchy for any attribute below the parallel period level noted in the function (i.e., parallel period based on Year can show either one year back per year, quarter, or month). Although, other methods exist, the parallel period method can be easily followed and applied to various measures.\n\nNext Steps\n\nLast Updated: 2013-03-27",
null,
"Scott Murray has a passion for crafting BI Solutions with SharePoint, SSAS, OLAP and SSRS.\n\nView all my tips\nRelated Resources"
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https://rd.springer.com/referenceworkentry/10.1007%2F978-3-662-53605-6_99-1 | [
"# Encyclopedia of Continuum Mechanics\n\nLiving Edition\n| Editors: Holm Altenbach, Andreas Öchsner\n\n# Ray Expansions in Impact Interaction Problems\n\nLiving reference work entry\nDOI: https://doi.org/10.1007/978-3-662-53605-6_99-1\n\n## Definitions\n\nThe impact interaction problems are those connected with the shock interactions of bodies subjected to the contact law.\n\n## Preliminary Remarks\n\nThe problems connected with the analysis of the shock interaction of thin bodies (rods, beams, plates, and shells) with other bodies have widespread application in various fields of science and technology. The physical phenomena involved in the impact event include structural responses, contact effects, and wave propagation. These problems are topical not only from the point of view of fundamental research in applied mechanics but also with respect to their applications. Because these problems belong to the problems of dynamic contact interaction, their solution is connected with severe mathematical and calculation difficulties. To overcome this impediment, a rich variety of approaches and methods have been suggested, and the overview of current results in the field can be found in state-of-the-art articles by Abrate (2001) and Rossikhin and Shitikova (2007a).\n\nIn many engineering applications, it is important to understand the transient behavior of isotropic as well as composite thin-walled shell structures subjected to central impact by a small projectile. Recently Rossikhin and Shitikova (2007b) have developed a new formulation of the ray method which is applicable for analyzing the propagation of surfaces of strong and weak discontinuity in thin elastic bodies when the wave fronts and the rays are referenced to the curvilinear system of coordinates (Rossikhin and Shitikova, 1995b). It should be noted that the ray method is primarily used for obtaining the problem solution analytically. This approach is based on the reduction of the three-dimensional equations of the dynamic theory of elasticity, which first should be written in discontinuities, to the two-dimensional equations by virtue of integration over the coordinate perpendicular to the middle surface of a thin body. The recurrent equations of this ray method are free from the shear coefficient, which is usually inherent to the Timoshenko-type theories (Timoshenko, 1936), and involve only two elastic constants: Poisson’s ratio and elastic modulus of elongation.\n\nThe theory proposed in Rossikhin and Shitikova (2007b) is applicable for short times after the passage of the wave front, but it possesses the simplicity inherent in the “classical” theory of thin bodies. The advantages of this approach will be illustrated in this entry by solving the engineering problems on normal impact of an elastic spherical and long cylindrical hemisphere-nose projectiles against an elastic spherical shell using the nonlinear Hertzian law within the contact region.\n\n## Impact Response of a Spherical Shell of the Timoshenko Type\n\nLet an elastic sphere with the radius r0 and mass m (Fig. 1) or a long cylindrical elastic rod of radius r0 with a hemispherical nose of the same radius (Fig. 2) move along the x3-axis with the velocity V 0 towards an elastic isotropic spherical shell of the R radius (Rossikhin et al., 2011).",
null,
"Fig. 1 Scheme of the shock interaction of a falling sphere with a spherical shell",
null,
"Fig. 2 Scheme of the shock interaction of a falling cylindrical rod with a shell\n\nThe impact occurs at the initial instant of time at x3 = R. At the moment of impact, two shock wave lines (surfaces of strong discontinuity) are generated in the shell, which then propagate along the shell during the process of impact. During transition through the wave line, the following wave fields experience the discontinuities: stresses, velocities of displacements, and the values of the higher-order time derivatives in the displacements.\n\n### Geometry of the Wave Surface\n\nA wave-strip is a ruled cylindrical surface consisting of the directrix C, which is the wave line propagating along the median surface of the shell, and the family of generatrices representing the line segments of the length h, which are perpendicular to the shell’s median surface and thus to the wave line, and which are fitted to the wave line by their middles. Let us take the family of generatrices as the u1-curves, where u1 is the distance measured along the straight line segment from the C curve, and choose the distance measured along the C curve as u2 (Fig. 3). The u1-family is the family of geodetic lines. In this case, all conditions of the McConnel theorem are fulfilled, and a linear element of the wave surface takes the form (McConnel, 1957)\n\\displaystyle \\begin{aligned} ds^2=(du^1)^2+g_{22}(u^1,u^2)(du^2)^2, \\end{aligned}\n(1)\nin so doing\n\\displaystyle \\begin{aligned} g_{22}(0,u^2)=1, \\end{aligned}\n(2)\nwhere g11 = 1, g22, and g12 = 0 are the covariant components of the metric tensor of the wave surface.",
null,
"Fig. 3 Scheme of the propagating wave-strip along the spherical shell surface\nThe Gaussian curvature for the linear element (1) is defined by the following formula (McConnel, 1957):\n\\displaystyle \\begin{aligned} K=-\\frac{1}{\\sqrt{g_{22}}}\\;\\frac{\\partial ^2 \\sqrt{g_{22}}}{(\\partial u^1)^2}=0. \\end{aligned}\n(3)\nIntegrating Eq. (3) and considering formula (2) yield\n\\displaystyle \\begin{aligned} \\sqrt{g_{22}}=1+c u^1, \\end{aligned}\n(4)\nwhere c is a certain constant.\nIt is known that small distances along the coordinate lines u2 are defined by the formula (McConnel, 1957)\n\\displaystyle \\begin{aligned} ds_2=\\sqrt{g_{22}} \\;d u^2, \\end{aligned}\nor considering (4)\n\\displaystyle \\begin{aligned} ds_2=(1+c u^1) d u^2. \\end{aligned}\n(5)\nRewriting formula (5) in the form\n\\displaystyle \\begin{aligned} \\frac{d s_2-d u^2}{d u^2}=c u^1, \\end{aligned}\nand integrating the result relationship with respect to u1 from − h∕2 to h∕2 yield\n\\displaystyle \\begin{aligned} \\int_{-h/2}^{h/2}\\frac{d s_2-d u^2}{d u^2} \\;d u^1=0, \\end{aligned}\nor\n\\displaystyle \\begin{aligned} \\int_{-h/2}^{h/2}\\frac{ds_2}{d u^2} \\;d u^1=h. \\end{aligned}\n(6)\nEquation (6) can be written as\n\\displaystyle \\begin{aligned} \\frac{1}{h}\\;\\int_{-h/2}^{h/2} \\sqrt{g_{22}} \\;d u^1=1, \\end{aligned}\ni.e., the mean magnitude of the value $$\\sqrt {g_{22}}$$ over the thickness of the shell is equal to unit.\nIf the shell’s thickness is small, then it is possible to consider approximately that\n\\displaystyle \\begin{aligned} \\sqrt{g_{22}} \\approx 1 \\end{aligned}\n(7)\nat any point of the wave surface. Since all values for the shell are averaged over its thickness, then such an approximation for $$\\sqrt {g_{22}}$$ is not unreasonable.\nThe linear element (1) with due account for (7) can be approximately written as\n\\displaystyle \\begin{aligned} d s^2 \\approx (d u^1)^2+ (d u^2)^2, \\end{aligned}\n(8)\ni.e., it looks like a linear element on the plane in the Cartesian set of coordinates.\nNow let us define a linear element of the median surface of the shell. Since the rays intersecting the line C (the wave line) under the right angles are the family of the geodetic lines, then once again the conditions of the McConnel theorem remain valid, and thus the linear element of this surface takes the form\n\\displaystyle \\begin{aligned} d s^2 = (d u_*^1)^2+g_{22} (d u^2)^2, \\end{aligned}\n(9)\nbut considering formula (7), it can be rewritten in the form of (8) by substituting du1 by $$du_*^1$$.\n\n### The Main Kinematic and Dynamic Characteristics of the Wave Surface\n\nThe condition of compatibility on the wave surface of strong discontinuity in view of (7), (8) and (9) takes the form (see for details the entry “Ray Expansion Theory”)\n\\displaystyle \\begin{aligned} \\begin{array}{rcl} {} [u_{i,j(k)}]&\\displaystyle =&\\displaystyle - G^{-1} [v_{i,(k)}]\\lambda_j +\\frac{\\delta [u_{i,(k)}]}{\\delta s_1}\\;\\lambda _j \\\\ {} &\\displaystyle +&\\displaystyle \\frac{\\delta [u_{i,(k)}]}{\\delta s_2}\\;\\tau _j +\\left[\\frac{\\delta u_{i,(k)} \\xi _j}{\\delta \\xi } \\right], \\end{array} \\end{aligned}\n(10)\nwhere ui are the displacement vector components; G is the normal velocity of the wave surface; [ui,j] = [∂ui∂xj]; xj are the spatial rectangular Cartesian coordinates; ξ = u1; $$s_1=u_*^1$$; [ui,(k)] = [kui∂tk]; t is the time; vi = ui,(1); λi, τi, and ξi are the components of the unit vectors of the tangential to the ray, the tangential to the wave surface, and the normal to the spherical surface, respectively; and Latin indices take on the values 1,2,3.\nPutting k = 0 in (10) yields\n\\displaystyle \\begin{aligned}{}[u_{i,j}]=-G^{-1} [v_i]\\lambda _j+\\left[\\frac{\\delta (u_{i}\\xi _j )}{\\delta \\xi } \\right]. \\end{aligned}\n(11)\nWriting the Hooke’s law for a three-dimensional medium in terms of discontinuities and using the condition of compatibility (11) yield\n\\displaystyle \\begin{aligned}{}[\\sigma _{ij}]&=-G^{-1}\\lambda [v_\\lambda ]\\delta _{ij}-G^{-1}\\mu \\left( [v_i]\\lambda _j +[v_j]\\lambda _i\\right) \\\\ &\\quad +\\lambda [u_{\\xi ,\\xi }]\\delta _{ij}+\\mu \\left(\\left[\\frac{\\delta (u_{i}\\xi _j )}{\\delta \\xi } +\\frac{\\delta (u_{j}\\xi _i )}{\\delta \\xi } \\right]\\right)\\ \\ \\quad \\end{aligned}\n(12)\nwhere\n\\displaystyle \\begin{aligned}{}[v_\\lambda ]=[v_i]\\lambda _i,\\quad [u_{\\xi ,\\xi }]=\\left[\\frac{\\delta (u_{i}\\xi _i)}{\\delta \\xi }\\right]= \\left[\\frac{\\delta u_\\xi }{\\delta \\xi }\\right], \\end{aligned}\nλ and μ are Lame constants, and δij is the Kronecker’s symbol.\nMultiplying relationship (12) from the right and from the left by ξiξj and considering equation\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\xi \\xi }]=[\\sigma _{ij}]\\xi _i\\xi _j=0, \\end{aligned}\nwhat corresponds to the assumption that the normal stresses on the cross-sections parallel to the middle surface could be neglected with respect to other stresses, it could be found\n\\displaystyle \\begin{aligned}{}[u _{\\xi, \\xi }]=\\frac{\\lambda }{G(\\lambda +2\\mu )}\\;[v _\\lambda ]. \\end{aligned}\n(13)\nMultiplying relationship (12) from the right and from the left by λiλj leads to the equation\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\lambda }]=[\\sigma _{ij}]\\lambda _i\\lambda _j =-G^{-1}(\\lambda +2\\mu ) [v_\\lambda ]+\\lambda [u_{\\xi ,\\xi }]. \\end{aligned}\n(14)\nSubstituting (13) in (14) yields\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\lambda }]=-\\frac{4\\mu (\\lambda +\\mu )}{\\lambda +2\\mu }\\; G^{-1}[v_\\lambda ], \\end{aligned}\nor\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\lambda }]=-\\frac{E}{1-\\sigma ^2}\\; G^{-1}[v_\\lambda ], \\end{aligned}\n(15)\nwhere E and σ are the elastic modulus and the Poisson’s ratio, respectively.\nAlternatively, multiplying the three-dimensional equation of motion written in terms of discontinuities\n\\displaystyle \\begin{aligned}{}[\\sigma _{ij }]\\lambda _j=-\\rho G [v_i ], \\end{aligned}\n(16)\nby λi yields\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\lambda }]=-\\rho G [v_\\lambda ], \\end{aligned}\n(17)\nwhere ρ is the density of the shell’s material.\nEliminating the value [σλλ] from (15) and (17), the velocity of the quasi-longitudinal wave propagating in the spherical shell is defined as\n\\displaystyle \\begin{aligned} G_1=\\sqrt{\\frac{E}{\\rho (1-\\sigma ^2)}}. \\end{aligned}\n(18)\nRelationship (15) with due account for (18) takes the form\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\lambda }]=-\\rho G_1 [v_\\lambda ]. \\end{aligned}\n(19)\nMultiplying (12) by λiξj and (16) by ξi yields\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\xi }]=[\\sigma _{ij}]\\lambda _i\\xi _j= -\\mu G^{-1} [v_\\xi ], \\end{aligned}\n(20)\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\xi }]=-\\rho G [v_\\xi ], \\end{aligned}\n(21)\nwhere [vξ] = [vi]ξi.\nEliminating the value [σλξ] from (20) and (21), the velocity of the quasi-transverse wave is defined as\n\\displaystyle \\begin{aligned} G_2=\\sqrt{\\frac{\\mu }{\\rho}}. \\end{aligned}\n(22)\nConsidering (22), relationship (20) takes the form\n\\displaystyle \\begin{aligned}{}[\\sigma _{\\lambda \\xi }]=-\\rho G_2 [v_\\xi ]. \\end{aligned}\n(23)\nNote that in the three-dimensional medium, only one value, i.e., [uλ,λ], is nonzero on the quasi-longitudinal wave, while in the two-dimensional medium, where the “wave-strip” propagates, on the quasi-longitudinal wave, there are two nonvanishing values, namely, [uλ,λ] and [uξ,ξ]. Between these two values, it is possible to find the relationship. For this purpose, let us multiply (11) from right and from left by λiλj and express the values [vλ]\n\\displaystyle \\begin{aligned}{}[v_\\lambda ]=-G_1[u_{\\lambda ,\\lambda }], \\end{aligned}\nand then the obtained expression should be substituted in (13). As a result the desired linkage could be found\n\\displaystyle \\begin{aligned}{}[u _{\\xi ,\\xi }]=-\\frac{\\sigma }{1-\\sigma }\\; [u_{\\lambda ,\\lambda } ]. \\end{aligned}\n(24)\nHowever, if one considers the strains in a thin body, for example, a plate in the rectangular Cartesian set of coordinates, assuming that\n\\displaystyle \\begin{aligned} \\sigma _{zz}=\\frac{E\\left[ (1-\\sigma ) u_{z,z}+\\sigma (u_{x,x}+u_{y,y})\\right]} {(1+\\sigma )(1-2\\sigma )} =0, \\end{aligned}\nthen it is possible to obtain a little bit another formula\n\\displaystyle \\begin{aligned} u _{z,z}=-\\frac{\\sigma }{1-\\sigma }\\; (u_{x,x}+u_{y,y}). \\end{aligned}\n(25)\n\nFrom the comparison of (24) and (25), it is seen that in the right-hand side of (24), the value [uτ,τ] = [uij]τiτj is absent, but its absence is connected with the peculiarities of the “wave-strip,” namely, it has free edges at ξ = ±h∕2 and a closed contour with respect to s2.\n\n### Governing Equations\n\nThus, behind the front of each of two transient waves (surfaces of strong discontinuity) up to the boundary of the contact domain (Figs. 1 or 2), relationships (19) and (23) are valid, which are the first terms of the ray expansions (Fig. 4), i.e.,\n\\displaystyle \\begin{aligned} \\sigma _{\\lambda \\lambda }&=-\\rho G_1 v_\\lambda , \\end{aligned}\n(26)\n\\displaystyle \\begin{aligned} \\sigma _{\\lambda \\xi }&=-\\rho G_2 v_\\xi . \\end{aligned}\n(27)",
null,
"Fig. 4 Scheme of velocities and stresses in the shell’s element on the boundary of the contact domain\nConsidering the cone angle of the contact spot 2γ as a small value (Fig. 4), and putting $$\\cos \\gamma {\\approx } 1$$, $$\\sin \\gamma \\approx \\gamma =aR^{-1}$$ yield\n\\displaystyle \\begin{aligned} \\widetilde{v_z}&=\\widetilde{v_\\xi } -\\widetilde{v_\\lambda }\\frac{a}{R}, \\end{aligned}\n(28)\n\\displaystyle \\begin{aligned} \\widetilde{v_r}&=\\widetilde{v_\\xi }\\frac{a}{R} +\\widetilde{v_\\lambda }, \\end{aligned}\n(29)\n\\displaystyle \\begin{aligned} \\widetilde{\\sigma _{rz}}&=\\rho G_1 \\widetilde{v_\\lambda }\\frac{a}{R}-\\rho G_2 \\widetilde{v_\\xi }, \\end{aligned}\n(30)\nwhere a is the radius of the contact spot and $$\\widetilde {\\sigma _{rz}}=\\sigma _{rz}|{ }_{r=a}$$.\nAccording to the Hertzian theory of contact, during the loading phase, the contact force Fcont is related to the indentation α (i.e., the difference between the displacements of impactor and target or the local bearing of impactor and target materials), by the relationship\n\\displaystyle \\begin{aligned} F_{\\mathrm{cont}}=k\\alpha ^{3/2}, \\end{aligned}\n(31)\nwhere k is the contact stiffness coefficient depending on the geometry of colliding bodies, as well as their elastic constants:\n\\displaystyle \\begin{aligned} k&=\\frac{4}{3\\pi }\\;\\frac{\\sqrt{R\\prime}}{k\\prime+k\\prime\\prime}, \\quad k\\prime=\\frac{1-\\sigma ^2}{E},\\\\ k\\prime\\prime&=\\frac{1-\\sigma_{\\mathrm{im}} ^2}{E_{\\mathrm{im}}},\\\\ \\frac{1}{R\\prime}&=\\frac{1}{R}+\\frac{1}{r_0}, \\end{aligned}\nand σim and Eim are the Poisson’s ratio and the Young’s modulus of the impactor.\nIn this case, the radius of the contact zone a is connected with the relative displacement α by the following relationship:\n\\displaystyle \\begin{aligned} a={R\\prime}^{1/2} \\alpha ^{1/2}. \\end{aligned}\n(32)\n\n## Normal Impact of an Elastic Sphere Upon an Elastic Spherical Shell\n\nLet us choose a cylindrical set of coordinates r, θ, z = x3 with the center at the original point of tangency of the sphere and the spherical shell (Fig. 2). Then the equations of motion of the sphere and the contact spot as a rigid whole in the chosen coordinate system have the form\n\\displaystyle \\begin{aligned} m(\\widetilde{\\dot v_z} +\\ddot \\alpha )&=- F_{\\mathrm{cont}}, \\end{aligned}\n(33)\n\\displaystyle \\begin{aligned} \\rho \\pi a^2 h \\widetilde{\\dot v_z} &=2\\pi ah \\widetilde{\\sigma _{rz}}+F_{\\mathrm{cont}}, \\end{aligned}\n(34)\nwhere Fcont is the contact force and $$\\widetilde {\\dot v_z}=\\dot v_z|{ }_{r=a}$$.\nThe kinematic condition\n\\displaystyle \\begin{aligned} \\widetilde{v_r}=\\dot a \\end{aligned}\n(35)\nand the initial conditions\n\\displaystyle \\begin{aligned} \\alpha |{}_{t=0}=0, \\quad \\dot\\alpha |{}_{t=0}=V_0, \\quad v_z |{}_{t=0}=0, \\end{aligned}\n(36)\nwhere $$\\widetilde {v_r}=v_r|{ }_{r=a}$$, should be added to Eqs. (33) and (34).\nIntegrating (33) over t and considering the initial conditions (36) yield\n\\displaystyle \\begin{aligned} \\widetilde{v_z}=-\\dot \\alpha -\\frac{k}{m}\\int_0^{t}\\; \\alpha ^{3/2} dt +V_0. \\end{aligned}\n(37)\nEliminating the value $$\\widetilde {v_z}$$ from (28) and (37), one of the desired equations is obtained:\n\\displaystyle \\begin{aligned} \\widetilde{v_\\xi } -\\widetilde{v_\\lambda }\\frac{a}{R}=-\\dot \\alpha -\\frac{k}{m}\\int_0^{t} \\;\\alpha ^{3/2} dt +V_0. \\end{aligned}\n(38)\nTo obtain the second desired equation, it is necessary first to eliminate the value $$\\widetilde {\\dot v_z}$$ from (33) and (34) and then to exclude the value $$\\widetilde {\\sigma _{rz}}$$ from the equation found at the previous step and from (30) at a time. As a result, it is obtained:\n\\displaystyle \\begin{aligned} \\begin{array}{rcl} {} \\rho G_1\\widetilde{v_\\lambda }\\;\\frac{a}{R}-\\rho G_2\\widetilde{v_\\xi } =-\\frac 12\\;\\rho a \\left(\\ddot\\alpha +\\frac{k}{m}\\; \\alpha ^{3/2} \\right) \\\\ -\\frac{k}{2\\pi h\\sqrt{R\\prime} } \\;\\alpha.\\qquad \\end{array} \\end{aligned}\n(39)\nSolving the set of equations (38) and (39) with respect to the values $$\\widetilde {v_\\lambda }$$ and $$\\widetilde {v_\\xi }$$ yields\n\\displaystyle \\begin{aligned} \\begin{array}{rcl} {} \\widetilde{v_\\xi }R^{-1}\\sqrt{R\\prime} \\alpha=- \\frac{1}{\\rho (G_1-G_2)} \\left[ \\frac{k}{2\\pi h R } \\;\\alpha^2 \\right. \\\\ + \\frac{\\rho R\\prime}{2R}\\;\\alpha ^{3/2}\\left(\\ddot\\alpha +\\frac{k}{m} \\;\\alpha ^{3/2} \\right) \\\\ \\left. +\\rho G_1\\;\\frac{R\\prime}{R}\\;\\alpha \\left( \\dot \\alpha +\\frac{k}{m}\\int_0^{t} \\alpha ^{3/2} dt -V_0\\right)\\right],\\quad \\end{array} \\end{aligned}\n(40)\n\\displaystyle \\begin{aligned} \\begin{array}{rcl} {} \\widetilde{v_\\lambda } \\alpha^{1/2}=- \\frac{1}{\\rho (G_1-G_2)} \\left[ \\frac{k R}{2\\pi h R\\prime } \\right. \\\\ + \\frac 12\\;\\rho R\\;\\alpha ^{1/2}\\left(\\ddot\\alpha +\\frac{k}{m} \\;\\alpha ^{3/2} \\right) \\;\\alpha \\\\ \\left.+\\rho G_2\\;\\frac{R}{\\sqrt{R\\prime}}\\left( \\dot \\alpha +\\frac{k}{m}\\int_0^{t}\\; \\alpha ^{3/2} dt -V_0\\right)\\right].\\quad \\end{array} \\end{aligned}\n(41)\nSubstituting (40) and (41) in relationship (29), which is preliminary multiplied by α1∕2, and considering formula (35), the following governing nonlinear integrodifferential equation with respect to the value α is obtained:\n\\displaystyle \\begin{aligned} &\\left[ \\frac 12\\;\\rho \\;\\alpha ^{1/2}\\left(\\ddot\\alpha +\\frac{k}{m} \\;\\alpha ^{3/2} \\right)\\right. \\\\ &\\quad \\left.+\\frac{k }{2\\pi h R\\prime }\\;\\alpha \\right] \\left( \\frac{R\\prime}{R}\\;\\alpha +R \\right) \\\\ &\\quad +\\frac{\\rho }{\\sqrt{R\\prime}}\\left( G_1\\;\\frac{R\\prime}{R}\\;\\alpha +G_2 R\\right) \\\\ &\\quad \\times \\left( \\dot \\alpha +\\frac{k}{m}\\int_0^{t} \\alpha ^{3/2} dt \\right) \\\\ &\\quad +\\frac 12\\;\\rho (G_1-G_2)\\sqrt{R\\prime}\\dot\\alpha \\\\ &\\quad =\\frac{\\rho V_0}{\\sqrt{R\\prime}}\\left( G_1\\;\\frac{R\\prime}{R}\\;\\alpha +G_2 R\\right). \\end{aligned}\n(42)\nIn the limiting case, when the radius of the spherical shell tends to infinity R →, Eq. (42) could be reduced to the following:\n\\displaystyle \\begin{aligned} & \\frac 12\\;\\rho \\;\\alpha ^{1/2}\\left(\\ddot\\alpha +\\frac{k}{m} \\;\\alpha ^{3/2} \\right)\\\\ &\\quad +\\frac{k }{2\\pi h r_0}\\;\\alpha \\\\ &\\quad +\\frac{\\rho G_2}{\\sqrt{r_0}}\\left( \\dot \\alpha +\\frac{k}{m}\\int_0^{t} \\alpha ^{3/2} dt \\right) \\\\ &\\quad +\\frac 12\\;\\rho (G_1-G_2)\\sqrt{r_0}\\dot\\alpha \\\\ &\\quad =\\frac{\\rho V_0G_2}{\\sqrt{r_0}}. \\end{aligned}\n(43)\nA solution of Eq. (42) could be found in the form of the following series with respect to time t:\n\\displaystyle \\begin{aligned} \\alpha =V_0t+ \\sum_{i=1}^\\infty a_i t^{(2i+1)/2}+\\sum_{j=2}^\\infty b_j t^j, \\end{aligned}\n(44)\nwhere ai and bj are coefficients to be determined.",
null,
"Fig. 5 Dimensionless time dependence of the dimensionless contact force occurring in the spherical shell impacted by the falling sphere\nSubstituting (44) into Eq. (42) and equating the coefficients at integer and fractional powers of t, the set of equations for defining the coefficients ai and bj could be found. For example, the first three of them have the form\n\\displaystyle \\begin{aligned} a_1&=-\\frac 43 (G_1-G_2)\\frac{V_0^{1/2}{R\\prime}^{1/2}}{R} <0, \\\\ b_2&=\\frac{2G_2(G_1{-}G_2)}{R} \\left( 1\\,{+}\\, \\frac 13 \\;\\frac{(G_1{-}G_2)R\\prime}{G_2R}\\right){>}0, \\\\ a_2&= -\\frac{4}{15}\\;\\frac{kV_0^{1/2}}{\\rho \\pi hR\\prime}-\\frac{4}{15}\\;V_0^{1/2}{R\\prime}^{1/2} \\\\ &\\quad \\times \\left[ \\frac{G_2b_2}{2V_0R\\prime}\\left(8+\\frac 13\\;\\frac{(G_1-G_2)R\\prime}{G_2R} \\right)\\right.\\\\ &\\quad \\left.-\\frac{(G_1-G_2)V_0R\\prime}{R^3}\\right]. \\end{aligned}\nThus, the approximate four-term solution of (42) takes the form\n\\displaystyle \\begin{aligned} \\alpha =V_0t+ a_1 t^{3/2}+b_2 t^2+a_2t^{5/2}. \\end{aligned}\n(45)\nWhen R →, the solution for Eq. (47) is reduced to\n\\displaystyle \\begin{aligned} \\alpha =V_0t -\\frac{4}{15}\\;\\frac{kV_0^{1/2}}{\\rho \\pi h r_0}\\;t^{5/2}. \\end{aligned}\n(46)\n\nSubstituting the found function α (45), or (46) in the limiting case, in Eq. (31), the final expression for the contact force could be obtained.\n\nThe dimensionless time t = tV 0h−1 dependence of the dimensionless contact force $$F^*_{\\mathrm {cont}}=F_{\\mathrm {cont}}(Eh^2)^{-1}$$ calculated according to (31) and (45) (relationship (62) is utilized in the limiting case) is presented in Fig. 5 for the following ratios of $$\\widetilde r=R_{\\mathrm {im}}/R$$: 0 (what corresponds to the case of an elastic plate), 0.001, and 0.01. Reference to Fig. 5 shows that the increase in the radius of the shell results in the increase of both the contact duration and the maximum of the contact force.\n\n## Normal Impact of an Elastic Long Hemisphere-Nose Bar Against an Elastic Spherical Shell\n\nAt the moment of impact of a bar against a spherical shell (Fig. 2), the shock waves are generated not only in the shell but in the bar (a longitudinal shock wave) as well. This wave propagates along the bar with the velocity $$G_0=\\sqrt {E_{\\mathrm {im}}\\rho _0^{-1} }$$, where Eim and ρ0 are the elastic modulus and density of the bar. Behind the front of this wave, the relationships for the stress σ and velocity v could be obtained using the ray series (Rossikhin and Shitikova, 1995a, 2007a)\n\\displaystyle \\begin{aligned} \\sigma ^-=- \\sum_{k=0}^\\infty \\frac{1}{k!}\\;\\left[ \\sigma _{,(k)}\\right]\\left( t-\\frac{x_3}{G_0}\\right)^k, \\end{aligned}\n(47)\n\\displaystyle \\begin{aligned} v ^-=V_0- \\sum_{k=0}^\\infty \\frac{1}{k!}\\;\\left[ v_{,(k)}\\right]\\left( t-\\frac{x_3}{G_0}\\right)^k. \\end{aligned}\n(48)\n\nIt is assumed that the impactor is long enough and reflected waves do not have time to return at the place of contact before the moment of the rebound of the bar from the shell.\n\nConsidering that the discontinuities in the elastic bar remain constant during the process of the wave propagation and using the condition of compatibility (Rossikhin and Shitikova, 1995a)\n\\displaystyle \\begin{aligned} G_0\\left[ \\frac{\\partial Z_{,(k-1)}}{\\partial x_3}\\right]= -[Z_{,(k)}]+ \\frac{\\delta [Z_{,(k-1)}]}{\\delta t}, \\end{aligned}\nwhere Z is the function to be found and δδt is the Thomas-derivative (Thomas, 1961), yield\n\\displaystyle \\begin{aligned} \\left[ \\frac{\\partial \\sigma _{,(k-1)}}{\\partial x_3}\\right]= -G_0^{-1} [\\sigma _{,(k)}]. \\end{aligned}\n(49)\nWith due account for (49), the equation of motion on the wave surface is written in the form\n\\displaystyle \\begin{aligned}{}[\\sigma _{,(k)}]=-\\rho _0 G_0 [v _{,(k)}]. \\end{aligned}\n(50)\nSubstituting (50) in (47) yields\n\\displaystyle \\begin{aligned} \\sigma ^-= \\rho _0 G_0 \\sum_{k=0}^\\infty \\frac{1}{k!}\\;\\left[ v_{,(k)}\\right]\\left( t-\\frac{x_3}{G_0}\\right)^k. \\end{aligned}\n(51)\nComparison of relationships (51) and (48) results in\n\\displaystyle \\begin{aligned} \\sigma ^-= \\rho _0 G_0 (V_0-v^-). \\end{aligned}\n(52)\nAt x3 = 0, expression (52) takes the form\n\\displaystyle \\begin{aligned} \\sigma_{\\mathrm{cont}}= \\rho _0 G_0 (V_0-\\widetilde v_z-\\dot\\alpha ), \\end{aligned}\n(53)\nwhere $$\\sigma _{\\mathrm {cont}}=\\sigma ^-|{ }_{x_3=0}$$ is the contact stress, $$\\widetilde v_z+\\dot \\alpha =v^-|{ }_{x_3=0}$$ is the normal velocity of the displacements of the spherical shell’s points at the place of contact of the bar with the shell, α is the value characterizing the local indentation of the impactor into the shell, and an overdot denotes the time derivative.\nUsing formula (53), it is possible to find the contact force\n\\displaystyle \\begin{aligned} F_{\\mathrm{cont}}= \\rho _0 G_0 (V_0-\\widetilde v_z-\\dot\\alpha )\\pi a^2. \\end{aligned}\n(54)\nHowever, the contact force can be determined not only by formula (54) but according to the Hertz’s law as well (31). Therefore\n\\displaystyle \\begin{aligned} \\pi a^2\\rho _0 G_0 (V_0-\\widetilde v_z-\\dot\\alpha )=k \\alpha ^{3/2}, \\end{aligned}\nwhence it follows that\n\\displaystyle \\begin{aligned} \\widetilde v_z=-\\dot\\alpha -\\frac{k}{\\pi \\rho _0 G_0 R\\prime}\\;\\alpha ^{1/2}+V_0. \\end{aligned}\n(55)\nEliminating the value $$\\widetilde v_z$$ from (28) and (55), one of the desired equations is obtained:\n\\displaystyle \\begin{aligned} -\\widetilde v_\\lambda \\;\\frac{a}{R} +\\widetilde v_\\xi =-\\dot\\alpha -\\frac{k}{\\pi \\rho _0 G_0 R\\prime}\\;\\alpha ^{1/2}+V_0. \\end{aligned}\n(56)\nThe second desired equation could be determined by eliminating $$\\widetilde {\\dot v_z}$$ and $$\\widetilde \\sigma _{rz}$$ from (34) by virtue of (30) and (55)\n\\displaystyle \\begin{aligned} &\\rho G_1\\widetilde v_\\lambda \\;\\frac{a}{R} -\\rho G_2\\widetilde v_\\xi\\\\ &\\quad = \\frac 12\\;\\rho a\\left(-\\ddot\\alpha - \\frac{k}{2\\pi \\rho _0 G_0 R\\prime}\\;\\alpha ^{-1/2}\\dot\\alpha \\right) \\\\ &\\quad -\\frac{k}{2\\pi h \\sqrt{R\\prime}}\\;\\alpha.\\qquad \\end{aligned}\n(57)\nSolving the set of Eqs. (56) and (57) with respect to the values $$\\widetilde v_\\lambda$$ and $$\\widetilde v_\\xi$$ yields\n\\displaystyle \\begin{aligned} &\\widetilde v_\\xi R^{-1}\\sqrt{R^{\\prime}}\\;\\alpha=- \\frac{1}{\\rho (G_1-G_2)} \\left[ \\frac 12\\; \\frac{\\rho R^{\\prime}}{R}\\;\\alpha ^{3/2}\\right. \\\\ &\\quad \\times \\left.\\Big(\\ddot\\alpha + \\frac{k}{2\\pi \\rho_0 G_0 R^{\\prime}}\\;\\alpha ^{-1/2}\\dot\\alpha \\right) +\\frac{k}{2\\pi h \\sqrt{R^{\\prime}}}\\;\\alpha^2 \\\\ &\\quad \\left. +\\rho G_1 \\;\\frac{R^{\\prime}}{R}\\;\\alpha \\left(\\dot\\alpha +\\frac{k}{\\pi \\rho_0 G_0 R^{\\prime}}\\;\\alpha ^{1/2}-V_0\\right)\\right],\\quad \\end{aligned}\n(58)\n\\displaystyle \\begin{aligned} &\\widetilde v_\\lambda \\alpha^{1/2}=- \\frac{1}{\\rho (G_1-G_2)} \\left[ \\frac 12\\; \\rho R\\;\\alpha ^{1/2} \\Big(\\ddot\\alpha \\right. \\\\ &\\quad \\left.+ \\frac{k}{2\\pi \\rho _0 G_0 R^{\\prime}}\\;\\alpha ^{-1/2}\\dot\\alpha \\right) +\\frac{kR}{2\\pi h \\sqrt{R^{\\prime}}}\\;\\alpha \\\\ &\\quad \\left. +\\rho G_2 \\;\\frac{R}{\\sqrt{R^{\\prime}}} \\left(\\dot\\alpha +\\frac{k}{\\pi \\rho _0 G_0 R^{\\prime}}\\;\\alpha ^{1/2}-V_0\\right)\\right].\\quad \\end{aligned}\n(59)\nSubstituting (58) and (59) in (29), which is preliminarily multiplied by α1∕2, the governing nonlinear differential equation with respect to the value α is obtained as\n\\displaystyle \\begin{aligned} &\\left[ \\frac 12\\; \\rho \\left(\\alpha ^{1/2}\\ddot\\alpha + \\frac{k}{2\\pi \\rho _0 G_0 R\\prime}\\;\\dot\\alpha \\right) +\\frac{k}{2\\pi h R\\prime}\\;\\alpha \\right] \\\\ &\\quad \\times \\left( \\frac{R\\prime}{R}\\;\\alpha +R \\right) \\\\ &\\quad +\\frac{\\rho} {\\sqrt{R\\prime}}\\left( G_1\\;\\frac{R\\prime}{R}\\;\\alpha +G_2 R\\right) \\\\ &\\quad \\times \\left( \\dot \\alpha + \\frac{k}{2\\pi \\rho _0 G_0 R\\prime}\\;\\alpha ^{1/2}\\right) \\\\ &\\quad +\\frac 12\\;\\rho (G_1-G_2)\\sqrt{R\\prime}\\dot\\alpha \\\\ &\\quad =\\frac{\\rho V_0}{\\sqrt{R\\prime}} \\left( G_1\\;\\frac{R\\prime}{R}\\;\\alpha +G_2 R\\right).\\quad \\end{aligned}\n(60)\nIn the limiting case, when the radius of the spherical shell tends to infinity R →, Eq. (60) could be reduced to the following:\n\\displaystyle \\begin{aligned} & \\frac 12\\; \\rho \\left(\\alpha ^{1/2}\\ddot\\alpha + \\frac{k}{2\\pi \\rho _0 G_0 r_0}\\;\\dot\\alpha \\right) +\\frac{k}{2\\pi h r_0}\\;\\alpha \\\\ &+\\frac{\\rho} {\\sqrt{r_0}}\\;G_2 \\left( \\dot \\alpha + \\frac{k}{2\\pi \\rho _0 G_0 r_0}\\;\\alpha ^{1/2}\\right) =\\frac{\\rho V_0}{\\sqrt{r_0}}\\; G_2.\\quad \\end{aligned}\n(61)\nA solution of (60) could be found in the form of the series (44) with respect to time t. Substituting (44) into Eq. (60) and equating the coefficients at integer and fractional powers of t, the set of equations for defining the coefficients ai and bj could be determined. For example, the first four of them have the form\n\\displaystyle \\begin{aligned} a_1&=-\\frac 43\\left( \\frac{k}{2\\pi \\rho _0 G_0 R\\prime}\\right.\\\\ &\\quad \\left.+(G_1-G_2)\\frac{\\sqrt{R\\prime}}{R}\\right) V_0^{1/2}<0,\\\\ b_2&=\\frac 38\\;a_1^2V_0^{-1}\\\\ &\\quad +G_2\\left( \\frac{k}{2\\pi \\rho _0 G_0 {R\\prime}^{3/2}}+\\frac{2(G_1-G_2)}{R}\\right),\\\\ a_2&=\\frac{8}{15}\\left\\{\\frac{1}{16}\\;a_1b_2V_0^{-1}-\\frac 38\\;\\frac{R\\prime}{R^2}\\;a_1V_0 -\\frac{kV_0^{1/2}}{2\\pi \\rho hR\\prime}\\right. \\\\ &\\quad -\\frac{kV_0^{1/2}}{4\\pi \\rho_0 G_0}\\left( \\frac{V_0}{R^2} +\\quad \\frac{a_1G_2}{V_0^{3/2}{R\\prime}^{3/2}}\\right) \\\\ &\\quad +\\left. \\frac{1}{V_0^{1/2}{R\\prime}^{1/2}}\\left( \\frac{R\\prime}{R^2}\\;G_1V_0^2-2b_2G_2\\right) \\right\\} , \\\\ b_3&=-\\frac{1}{16}a_1a_2V_0^{-1}-\\frac{3}{16}\\;\\frac{R\\prime}{R^2}\\;a_1^2-\\frac 16\\;b_2^2V_0^{-1} \\\\ &\\quad -\\frac{ka_1}{6\\pi \\rho hR\\prime V_0^{1/2}} -\\frac{5kV_0^{1/2}a_1}{24\\pi \\rho_0 G_0R^2} \\\\ &\\quad -\\frac{1}{6V_0^{1/2}{R\\prime}^{1/2}}\\left( \\frac{R\\prime}{R^2}\\;G_1V_0^2a_1+5a_2G_2\\right) \\\\ &\\quad -\\frac{k}{6\\pi \\rho _0 G_0 {R\\prime}^{3/2}}\\left( \\frac{b_2G_2}{2V_0}+\\frac{R\\prime}{R^2}\\;G_1V_0\\right). \\end{aligned}\nThus, the approximate five-term solution has the form\n\\displaystyle \\begin{aligned} \\alpha =V_0t+ a_1 t^{3/2}+b_2 t^2+ a_2 t^{5/2}+b_3 t^3. \\end{aligned}\n(62)\nIn the limiting case, the coefficients in the series (44) representing the solution of equation (61) take the form\n\\displaystyle \\begin{aligned} a_1=-\\frac 23\\; \\frac{kV_0^{1/2}}{\\pi \\rho _0 G_0 r_0}<0, \\end{aligned}\n\\displaystyle \\begin{aligned} b_2=\\frac 12\\;\\frac{k}{\\pi \\rho _0 G_0 r_0} \\left(\\frac 13\\;\\frac{k}{\\pi \\rho _0 G_0 r_0}+\\frac{G_2}{ r_0^{1/2}}\\right)>0, \\end{aligned}\n\\displaystyle \\begin{aligned} a_2=-\\frac{4}{15}\\;\\frac{k}{\\pi \\rho _0 G_0 r_0V_0^{1/2}} \\left[\\frac{2G_2^2}{r_0} +\\frac{\\rho _0V_0G_0}{\\rho h} \\right. \\end{aligned}\n\\displaystyle \\begin{aligned} \\left. +\\frac 18\\;\\frac{k}{\\pi \\rho _0 G_0 r_0}\\left(\\frac 19\\;\\frac{k}{\\pi \\rho _0 G_0 r_0} +\\frac{3G_2}{r_0^{1/2}}\\right)\\right]<0. \\end{aligned}\n\nThe dimensionless time t = tV 0h−1 dependence of the dimensionless contact force $$F^*_{\\mathrm {cont}}=F_{\\mathrm {cont}}(Eh^2)^{-1}$$ calculated according to (31) and (62) is presented in Fig. 6 for the following ratios of $$\\widetilde r=R_{im}/R$$: 0 (what corresponds to the case of an elastic plate), 0.001, and 0.01.\n\nFrom Fig. 6 it is seen that the increase in the radius of the shell results in the increase of both the contact duration and the maximum of the contact force, as it has been mentioned above in the case of the dynamic response of the spherical shell impacted by the elastic sphere. However, the comparison of Figs. 5 and 6 shows that the magnitudes of the contact duration and the maximum of the contact force in the case when the spherical shell is impacted by the cylindrical rod are lower than those when the shell is impacted by the sphere. This is due to the fact that the wave phenomenon is neglected in the falling sphere, while the propagation of the transient waves in the falling rod is taken into account.",
null,
"Fig. 6 Dimensionless time dependence of the dimensionless contact force occurring in the spherical shell impacted by the long cylindrical rod\n\n## Conclusion\n\nThe problem on normal low-velocity impact of an elastic falling body upon an elastic Timoshenko-type spherical shell has been analyzed using the wave approach. At the moment of impact, shock waves (surfaces of strong discontinuity) – quasi-longitudinal and quasi-transverse waves – are generated in the target, which then propagate along the body during the process of impact. Behind the wave fronts up to the boundary of the contact domain, the solution is constructed with the help of the theory of discontinuities and one-term or multiple-term ray expansions, while within the contact region, the nonlinear Hertz’s theory is employed.\n\nFor the analysis of the processes of shock interactions of the elastic sphere or elastic spherically headed rod with the Timoshenko-type spherical shell, nonlinear integrodifferential or nonlinear differential equations have been, respectively, obtained with respect to the value characterizing the local indentation of the impactor into the target, which have been solved analytically in terms of time series with integer and fractional powers.\n\nThus, these examples discussed above show the efficiency of the ray method in solving the impact interaction problems.\n\n## References\n\n1. Abrate S (2001) Modeling of impacts on composite structures. Composite Struct 51:129–138\n2. McConnel AJ (1957) Application of tensor analysis. Dover Publications, New YorkGoogle Scholar\n3. Rossikhin YA, Shitikova MV (1995a) Ray method for solving dynamic problems connected with propagation on wave surfaces of strong and weak discontinuities. Appl Mech Rev 48(1):1–39. https://doi.org/10.1115/1.3005096\n4. Rossikhin YA, Shitikova MV (1995b) The ray method for solving boundary problems of wave dynamics for bodies having curvilinear anisotropy. Acta Mech 109(1–4): 49–64. https://doi.org/10.1007/BF01176816\n5. Rossikhin YA, Shitikova MV (2007a) Transient response of thin bodies subjected to impact: wave approach. Shock Vibr Digest 39(4):273–309. https://doi.org/10.1177/0583102407080410\n6. Rossikhin YA, Shitikova MV (2007b) The method of ray expansions for investigating transient wave processes in thin elastic plates and shells. Acta Mech 189(1–2): 87–121. https://doi.org/10.1007/S00707-006-0412-x\n7. Rossikhin YA, Shitikova MV, Shamarin VV (2011) Dynamic response of spherical shells impacted by falling objects. Int J Mech 5(3):166–181Google Scholar\n8. Thomas TY (1961) Plastic flow and fracture in solids. Academic Press, New York\n9. Timoshenko SP (1936) Theory of elastic stability. McGrow-Hill, New YorkGoogle Scholar"
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https://wiki.seg.org/wiki/Common-conversion-point_binning | [
"# Common-conversion-point binning\n\nSeries",
null,
"Investigations in Geophysics Öz Yilmaz http://dx.doi.org/10.1190/1.9781560801580 ISBN 978-1-56080-094-1 SEG Online Store\n\nWe learned in analysis of amplitude variation with offset that an incident P-wave is partitioned at a layer boundary into reflected and transmitted P- and S-wave components. Consider the raypath geometry in Figure 11.6-30a for an incident P-wave generated by the source S1 and a flat reflector. The reflection angle for the PP-wave is equal to the angle of incidence; however, the reflection angle for the PS-wave is smaller than the angle of incidence. As a result, the PP reflection will follow a symmetric raypath and be recorded at receiver location R2, while the PS reflection will follow an asymmetric raypath and be recorded at receiver location R1.\n\nNow consider the common-midpoint (CMP) raypath geometry for a source-receiver pair S1R1 shown in Figure 11.6-30b. There are two reflection arrivals at the receiver location R1 associated with the PP and PS raypaths. The reflection point B at which the incident P-wave is converted to the S-wave is displaced in the lateral direction by some distance d away from the reflection point A at which the incident P-wave is reflected and recorded at the same receiver location R1 as the converted S-wave. This means that, for an earth model with flat layers, the PP-wave reflection points coincide with the midpoint location (Figure 11.6-31a); whereas, the PS conversion points do not (Figure 11.6-31b). As a direct consequence of this observation, the notion of a CMP gather based on sorting PP data from acquisition coordinates — source and receiver, to processing coordinates — midpoint and offset, such that traces in the gather have the same midpoint coordinate, is not applicable to PS data. Instead PS data need to be sorted into common-conversion-point (CCP) gathers such that traces in this gather have the same conversion point coordinate.\n\nAn important aspect of CCP sorting is that the asymmetric raypath associated with the PS reflection gives rise to a periodic variation in fold of the CCP gathers. As for the conventional P-wave data with variations in fold caused by irregular recording geometry, amplitudes of the stacked PS data are adversely affected by the variation in the CCP fold . Just as one resorts to flexible bin size in the processing of 3-D seismic data to accommodate variations in fold, the same strategy may be applied for the PS data processing.\n\nBinning the PS data into CCP gathers requires knowledge of the conversion-point coordinate xP. Referring to Figure 11.6-31b, note that the conversion-point coordinate follows a trajectory indicated by the broken curve that, in general, depends on the reflector depth .\n\nTo derive an expression for xP, refer to the geometry of the PS-raypath shown in Figure 11.6-32. By Snell’s law, we know that\n\n ${\\frac {\\sin \\varphi _{0}}{\\alpha }}={\\frac {\\sin \\psi _{1}}{\\beta }},$",
null,
"(70)\n\nwhere α and β are the P-wave and S-wave velocities, respectively, and φ0 is the P-wave angle of incidence and ψ1 is the reflection angle for the converted S-wave.\n\nFrom the geometry of Figure 11.6-32, note that\n\n $\\sin \\varphi _{0}={\\frac {x_{P}}{\\sqrt {x_{P}^{2}+z^{2}}}}$",
null,
"(71a)\n\nand\n\n $\\sin \\psi _{1}={\\frac {x_{S}}{\\sqrt {x_{S}^{2}+z^{2}}}},$",
null,
"(71b)\n\nwhere xP and xS are the lateral distances from the CCP to the source and receiver locations, respectively. Substitute equations (71a,71b) into equation (70), square and rearrange the terms to get\n\n ${\\frac {x_{S}^{2}}{x_{P}^{2}}}={\\frac {\\beta ^{2}}{\\alpha ^{2}}}{\\frac {x_{S}^{2}+z^{2}}{x_{P}^{2}+z^{2}}}.$",
null,
"(72a)\n\nApply some algebraic manipulation to solve equation (72a) for xS\n\n $x_{S}={\\frac {x_{P}}{\\sqrt {\\gamma ^{2}+(\\gamma ^{2}-1){\\frac {x_{P}^{2}}{z^{2}}}}}},$",
null,
"(72b)\n\nwhere γ = α/β. Finally, substitute the relation xS = x − xP, where x is the source-receiver offset, into equation (72b) to get the desired expression\n\n $x_{P}={\\frac {\\sqrt {\\gamma ^{2}+(\\gamma ^{2}-1){\\frac {x_{P}^{2}}{z^{2}}}}}{1+{\\sqrt {\\gamma ^{2}+(\\gamma ^{2}-1){\\frac {x_{P}^{2}}{z^{2}}}}}}}x.$",
null,
"(72c)\n\nFrom Figure 11.6-31b, note that the CCP location moves closer to CMP location as the depth of the reflector increases. At infinite depth, the CCP location reaches an asymptotic conversion point (ACP) . In the limit z → ∞, equation (72c) gives the ACP coordinate xP with respect to the source location\n\n $x_{P}={\\frac {\\gamma }{1+\\gamma }}x.$",
null,
"(73a)\n\nSince β < α, the conversion point is closer to the receiver location than the source location (Figure 11.6-31b). The displacement d = xP − x/2 of the asymptotic conversion point from the midpoint is, by way of equation (73a),\n\n $d={\\frac {1}{2}}\\left({\\frac {\\gamma -1}{\\gamma +1}}\\right)x.$",
null,
"(73b)\n\nWhile CCP binning may be performed using the ACP coordinate given by equation (73a), more accurate binning techniques account for the depth-dependence of the CCP coordinate xP based on a solution to equation (72c) . Because of the quartic form of equation (72c) in terms of xP, an iterative solution may be preferred in practice . The iteration may be started by substituting the asymptotic form of xP given by equation (73a) into the right-hand side of equation (72c). The new value of xP may then be back substituted into equation (72c) to continue with the iteration.\n\nWhatever the estimation procedure, note from equation (72c) that xP depends both on depth to the reflector and the velocity ratio γ = α/β. Unless a value for the velocity ratio is assumed, it follows that CCP binning requires velocity analysis of PS data to determine the velocity ratio γ. Additionally, an accurate CCP binning strictly requires the knowledge of reflector depths; thus, the advocation of an implicit requirement that 4-C seismic data analysis should be done in the depth domain. This requirement may be waivered if we only consider a horizontally layered earth model as in the next subsection."
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