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https://de.maplesoft.com/support/help/maple/view.aspx?path=DEtools%2FXchange | [
"",
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"Xchange - Maple Help\n\nDEtools\n\n Xchange\n change variables in an ODE point symmetry generator",
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"Calling Sequence Xchange(tr, X, y(x), [t, u(t)])",
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"Parameters\n\n tr - set of transformation equations of the form $\\left\\{x=..,y\\left(x\\right)=..\\right\\}$ from the old variables on the left hand side to the new variables on the right hand side X - list of coefficients of a point symmetry generator (infinitesimals) as in [xi(x, y), eta(x, y)] y(x) - 'dependent variable' of the problem (it can be any indeterminate function of one variable) [t, u(t)] - (optional) new independent and dependent variables (required when they cannot be inferred from the transformation)",
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"Description\n\n • The Xchange command receives a change of variables transformation set and a pair of infinitesimals $\\left[\\mathrm{\\xi },\\mathrm{\\eta }\\right]$, representing the coefficients of a point symmetry generator of an ODE, and the dependent variable y(x), and returns the symmetry in the new variables. The change of variables is performed by making calls to dchange, and hence the same extra arguments accepted by dchange are accepted by Xchange as well. This change of variables takes into account that $\\left[\\mathrm{\\xi }\\left(x,y\\right),\\mathrm{\\eta }\\left(x,y\\right)\\right]$ are the coefficients of a differential operator $\\mathrm{expr}$. Thus, if the new variables are $\\left\\{t,u\\left(t\\right)\\right\\}$, then the returned list is constituted by the coefficients of the differential operator $\\mathrm{expr}$, where $\\mathrm{ξ2}\\left(t,u\\left(t\\right)\\right)$ and $\\mathrm{η2}\\left(t,u\\left(t\\right)\\right)$ also incorporate any factor that appears when changing variables in $\\frac{d}{\\mathrm{dx}}$ and $\\frac{d}{\\mathrm{dy}}$.\n • This function is part of the DEtools package, and so it can be used in the form Xchange(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[Xchange](..).",
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"Examples\n\n > $\\mathrm{with}\\left(\\mathrm{DEtools}\\right):$\n\nNote that the infinitesimals $\\left[\\mathrm{\\xi },\\mathrm{\\eta }\\right]$ can be expressed in terms of $y\\left(x\\right)$, but also directly in terms of $y$:\n\n > $X≔\\left[-y,x\\right]$\n ${X}{≔}\\left[{-}{y}{,}{x}\\right]$ (1)\n > $\\mathrm{tr}≔\\left\\{x=u\\left(t\\right)\\mathrm{cos}\\left(t\\right),y\\left(x\\right)=u\\left(t\\right)\\mathrm{sin}\\left(t\\right)\\right\\}$\n ${\\mathrm{tr}}{≔}\\left\\{{x}{=}{u}{}\\left({t}\\right){}{\\mathrm{cos}}{}\\left({t}\\right){,}{y}{}\\left({x}\\right){=}{u}{}\\left({t}\\right){}{\\mathrm{sin}}{}\\left({t}\\right)\\right\\}$ (2)\n > $\\mathrm{Xchange}\\left(\\mathrm{tr},X,y\\left(x\\right)\\right)$\n $\\left[{1}{,}{0}\\right]$ (3)\n > $X≔\\left[\\frac{1}{x+y},1\\right]$\n ${X}{≔}\\left[\\frac{{1}}{{x}{+}{y}}{,}{1}\\right]$ (4)\n > $\\mathrm{tr}≔\\left\\{x=-\\mathrm{LambertW}\\left(-r\\mathrm{exp}\\left(-s\\left(r\\right)-1\\right)\\right)-s\\left(r\\right)-1,y\\left(x\\right)=s\\left(r\\right)\\right\\}$\n ${\\mathrm{tr}}{≔}\\left\\{{x}{=}{-}{\\mathrm{LambertW}}{}\\left({-}{r}{}{{ⅇ}}^{{-}{s}{}\\left({r}\\right){-}{1}}\\right){-}{s}{}\\left({r}\\right){-}{1}{,}{y}{}\\left({x}\\right){=}{s}{}\\left({r}\\right)\\right\\}$ (5)\n > $\\mathrm{Xchange}\\left(\\mathrm{tr},X,y\\left(x\\right)\\right)$\n $\\left[{0}{,}{1}\\right]$ (6)\n\nThis example illustrates a more general change of variables where the new variables are $\\left[v,u\\left(v\\right)\\right]$:\n\n > $X≔\\left[x,y\\right]$\n ${X}{≔}\\left[{x}{,}{y}\\right]$ (7)\n > $\\mathrm{tr}≔\\left\\{x=\\mathrm{\\phi }\\left(u,v\\left(u\\right)\\right),y\\left(x\\right)=\\mathrm{\\psi }\\left(u,v\\left(u\\right)\\right)\\right\\}$\n ${\\mathrm{tr}}{≔}\\left\\{{x}{=}{\\mathrm{\\phi }}{}\\left({u}{,}{v}{}\\left({u}\\right)\\right){,}{y}{}\\left({x}\\right){=}{\\mathrm{\\psi }}{}\\left({u}{,}{v}{}\\left({u}\\right)\\right)\\right\\}$ (8)\n > $\\mathrm{Xchange}\\left(\\mathrm{tr},X,y\\left(x\\right),\\left[u,v\\left(u\\right)\\right]\\right)$\n $\\left[\\frac{{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right){}\\left(\\frac{{\\partial }}{{\\partial }{v}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right)\\right){-}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right){}\\left(\\frac{{\\partial }}{{\\partial }{v}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right)\\right)}{\\left(\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right)\\right){}\\left(\\frac{{\\partial }}{{\\partial }{v}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right)\\right){-}\\left(\\frac{{\\partial }}{{\\partial }{v}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right)\\right){}\\left(\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right)\\right)}{,}{-}\\frac{{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right){}\\left(\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right)\\right){-}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right){}\\left(\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right)\\right)}{\\left(\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right)\\right){}\\left(\\frac{{\\partial }}{{\\partial }{v}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right)\\right){-}\\left(\\frac{{\\partial }}{{\\partial }{v}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\phi }}{}\\left({u}{,}{v}\\right)\\right){}\\left(\\frac{{\\partial }}{{\\partial }{u}}\\phantom{\\rule[-0.0ex]{0.4em}{0.0ex}}{\\mathrm{\\psi }}{}\\left({u}{,}{v}\\right)\\right)}\\right]$ (9)"
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https://alexmennen.com/index.php/2016/03/ | [
"## Ordered algebraic geometry\n\nEdit: Shortly after posting this, I found where the machinery I develop here was discussed in the literature. Real Algebraic Geometry by Bochnak, Coste, and Roy covers at least most of this material. I may eventually edit this to clean it up and adopt more standard notation, but don't hold your breath.\n\n### Introduction\n\nIn algebraic geometry, an affine algebraic set is a subset of $\\mathbb{C}^{n}$ which is the set of solutions to some finite set of polynomials. Since all ideals of $\\mathbb{C}\\left[x_{1},...,x_{n}\\right]$ are finitely generated, this is equivalent to saying that an affine algebraic set is a subset of $\\mathbb{C}^{n}$ which is the set of solutions to some arbitrary set of polynomials.\n\nIn semialgebraic geometry, a closed semialgebraic set is a subset of $\\mathbb{R}^{n}$ of the form $\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in F\\right\\}$ for some finite set of polynomials $F\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$. Unlike in the case of affine algebraic sets, if $F\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is an arbitrary set of polynomials, $\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in F\\right\\}$ is not necessarily a closed semialgebraic set. As a result of this, the collection of closed semialgebraic sets are not the closed sets of a topology on $\\mathbb{R}^{n}$. In the topology on $\\mathbb{R}^{n}$ generated by closed semialgebraic sets being closed, the closed sets are the sets of the form $\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in F\\right\\}$ for arbitrary $F\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$. Semialgebraic geometry usually restricts itself to the study of semialgebraic sets, but here I wish to consider all the closed sets of this topology. Notice that closed semialgebraic sets are also closed in the standard topology, so the standard topology is a refinement of this one. Notice also that the open ball $B_{r}\\left(\\bar{p}\\right)$ of radius $r$ centered at $\\bar{p}$ is the complement of the closed semialgebraic set $\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid\\left|\\bar{x}-\\bar{p}\\right|^{2}-r^{2}\\geq0\\right\\}$, and these open balls are a basis for the standard topology, so this topology is a refinement of the standard one. Thus, the topology I have defined is exactly the standard topology on $\\mathbb{R}^{n}$.\n\nIn algebra, instead of referring to a set of polynomials, it is often nicer to talk about the ideal generated by that set instead. What is the analog of an ideal in ordered algebra? It's this thing:\n\nDefinition: If $A$ is a partially ordered commutative ring, a cone $C$ in $A$ is a subsemiring of $A$ which contains all positive elements, and such that $C\\cap-C$ is an ideal of $A$. By \"subsemiring\", I mean a subset that contains $0$ and $1$, and is closed under addition and multiplication (but not necessarily negation). If $F\\subseteq A$, the cone generated by $F$, denoted $\\left\\langle F\\right\\rangle$, is the smallest cone containing $F$. Given a cone $C$, the ideal $C\\cap-C$ will be called the interior ideal of $C$, and denoted $C^{\\circ}$.\n\n$\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is partially ordered by $f\\geq g\\iff f\\left(\\bar{x}\\right)\\geq g\\left(\\bar{x}\\right)\\,\\forall\\bar{x}\\in\\mathbb{R}^{n}$. If $F\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a set of polynomials and $\\bar{x}\\in\\mathbb{R}^{n}$, then $f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in F\\iff f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in\\left\\langle F\\right\\rangle$. Thus I can consider closed sets to be defined by cones. We now have a Galois connection between cones of $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ and subsets of $\\mathbb{R}^{n}$, given by, for a cone $C$, its positive-set is $P_{\\mathbb{R}}\\left(C\\right):=\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in C\\right\\}$ (I'm calling it the \"positive-set\" even though it is where the polynomials are all non-negative, because \"non-negative-set\" is kind of a mouthful), and for $X\\subseteq\\mathbb{R}^{n}$, its cone is $C_{\\mathbb{R}}\\left(X\\right):=\\left\\{ f\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid f\\left(\\bar{x}\\right)\\geq0\\,\\forall\\bar{x}\\in X\\right\\}$$P_{\\mathbb{R}}\\circ C_{\\mathbb{R}}$ is closure in the standard topology on $\\mathbb{R}^{n}$ (the analog in algebraic geometry is closure in the Zariski topology on $\\mathbb{C}^{n}$). A closed set $X$ is semialgebraic if and only if it is the positive-set of a finitely-generated cone.\n\n### Quotients by cones, and coordinate rings\n\nAn affine algebraic set $V$ is associated with its coordinate ring $\\mathbb{C}\\left[V\\right]:=\\mathbb{C}\\left[x_{1},...,x_{n}\\right]/I\\left(V\\right)$. We can do something analogous for closed subsets of $\\mathbb{R}^{n}$.\n\nDefinition: If $A$ is a partially ordered commutative ring and $C\\subseteq A$ is a cone, $A/C$ is the ring $A/C^{\\circ}$, equipped with the partial order given by $f+C^{\\circ}\\geq g+C^{\\circ}$ if and only if $f-g\\in C$, for $f,g\\in A$.\n\nDefinition: If $X\\subseteq\\mathbb{R}^{n}$ is closed, the coordinate ring of $X$ is $\\mathbb{R}\\left[X\\right]:=\\mathbb{R}\\left[x_{1},...,x_{n}\\right]/C\\left(X\\right)$. This is the ring of functions $X\\rightarrow\\mathbb{R}$ that are restrictions of polynomials, ordered by $f\\geq g$ if and only if $f\\left(\\bar{x}\\right)\\geq g\\left(\\bar{x}\\right)\\,\\forall\\bar{x}\\in X$. For arbitrary $X\\subseteq\\mathbb{R}^{n}$, the ring of regular functions on $X$, denoted $\\mathcal{O}\\left(X\\right)$, consists of functions on $X$ that are locally ratios of polynomials, again ordered by $f\\geq g$ if and only if $f\\left(\\bar{x}\\right)\\geq g\\left(\\bar{x}\\right)\\,\\forall\\bar{x}\\in X$. Assigning its ring of regular functions to each open subset of $X$ endows $X$ with a sheaf of partially ordered commutative rings.\n\nFor closed $X\\subseteq\\mathbb{R}^{n}$, $\\mathbb{R}\\left[X\\right]\\subseteq\\mathcal{O}\\left(X\\right)$, and this inclusion is generally proper, both because it is possible to divide by polynomials that do not have roots in $X$, and because $X$ may be disconnected, making it possible to have functions given by different polynomials on different connected components.\n\n### Positivstellensätze\n\nWhat is $C_{\\mathbb{R}}\\circ P_{\\mathbb{R}}$? The Nullstellensatz says that its analog in algebraic geometry is the radical of an ideal. As such, we could say that the radical of a cone $C$, denoted $\\text{Rad}_{\\mathbb{R}}\\left(C\\right)$, is $C_{\\mathbb{R}}\\left(P_{\\mathbb{R}}\\left(C\\right)\\right)$, and that a cone $C$ is radical if $C=\\text{Rad}_{\\mathbb{R}}\\left(C\\right)$. In algebraic geometry, the Nullstellensatz shows that a notion of radical ideal defined without reference to algebraic sets in fact characterizes the ideals which are closed in the corresponding Galois connection. It would be nice to have a description of the radical of a cone that does not refer to the Galois connection. There is a semialgebraic analog of the Nullstellensatz, but it does not quite characterize radical cones.\n\nPositivstellensatz 1: If $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a finitely-generated cone and $p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a polynomial, then $p\\left(\\bar{x}\\right)>0\\,\\forall\\bar{x}\\in P_{\\mathbb{R}}\\left(C\\right)$ if and only if $\\exists f\\in C$ such that $pf-1\\in C$.\n\nThere are two ways in which this is unsatisfactory: first, it applies only to finitely-generated cones, and second, it tells us exactly which polynomials are strictly positive everywhere on a closed semialgebraic set, whereas we want to know which polynomials are non-negative everywhere on a set.\n\nThe second problem is easier to handle: a polynomial $p$ is non-negative everywhere on a set $S$ if and only if there is a decreasing sequence of polynomials $\\left(p_{i}\\mid i\\in\\mathbb{N}\\right)$ converging to $p$ such that each $p_{i}$ is strictly positive everywhere on $S$. Thus, to find $\\text{Rad}_{\\mathbb{R}}\\left(C\\right)$, it is enough to first find all the polynomials that are strictly positive everywhere on $P_{\\mathbb{R}}\\left(C\\right)$, and then take the closure under lower limits. Thus we have a characterization of radicals of finitely-generated cones.\n\nPositivstellensatz 2: If $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a finitely-generated cone, $\\text{Rad}_{\\mathbb{R}}\\left(C\\right)$ is the closure of $\\left\\{ p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid\\exists f\\in C\\, pf-1\\in C\\right\\}$, where the closure of a subset $X\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is defined to be the set of all polynomials in $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ which are infima of chains contained in $X$.\n\nThis still doesn't even tell us what's going on for cones which are not finitely-generated. However, we can generalize the Positivstellensatz to some other cones.\n\nPositivstellensatz 3: Let $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ be a cone containing a finitely-generated subcone $D\\subseteq C$ such that $P_{\\mathbb{R}}\\left(D\\right)$ is compact. If $p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a polynomial, then $p\\left(\\bar{x}\\right)>0\\,\\forall\\bar{x}\\in P_{\\mathbb{R}}\\left(C\\right)$ if and only if $\\exists f\\in C$ such that $pf-1\\in C$. As before, it follows that $\\text{Rad}_{\\mathbb{R}}\\left(C\\right)$ is the closure of $\\left\\{ p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid\\exists f\\in C\\, pf-1\\in C\\right\\}$.\n\nproof: For a given $p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$$\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid p\\left(\\bar{x}\\right)\\leq0\\right\\} \\cap P_{\\mathbb{R}}\\left(C\\right)=\\left\\{ \\bar{x}\\in\\mathbb{R}^{n}\\mid p\\left(\\bar{x}\\right)\\leq0\\right\\} \\cap\\bigcap\\left\\{ P_{\\mathbb{R}}\\left(\\left\\langle f\\right\\rangle \\right)\\mid f\\in C\\right\\}$, an intersection of closed sets contained in the compact set $P_{\\mathbb{R}}\\left(D\\right)$, which is thus empty if and only if some finite subcollection of them has empty intersection within $P_{\\mathbb{R}}\\left(D\\right)$. Thus if $p$ is strictly positive everywhere on $P_{\\mathbb{R}}\\left(C\\right)$, then there is some finitely generated subcone $E\\subseteq C$ such that $p$ is strictly positive everywhere on $P_{\\mathbb{R}}\\left(E\\right)\\cap P_{\\mathbb{R}}\\left(D\\right)=P_{\\mathbb{R}}\\left(\\left\\langle E\\cup D\\right\\rangle \\right)$, and $\\left\\langle E\\cup D\\right\\rangle$ is finitely-generated, so by Positivstellensatz 1, there is $f\\in\\left\\langle E\\cup D\\right\\rangle \\subseteq C$ such that $pf-1\\in\\left\\langle E\\cup D\\right\\rangle \\subseteq C$$\\square$\n\nFor cones that are not finitely-generated and do not contain any finitely-generated subcones with compact positive-sets, the Positivstellensatz will usually fail. Thus, it seems likely that if there is a satisfactory general definition of radical for cones in arbitrary partially ordered commutative rings that agrees with this one in $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$, then there is also an abstract notion of \"having a compact positive-set\" for such cones, even though they don't even have positive-sets associated with them.\n\n### Beyond $\\mathbb{R}^{n}$\n\nAn example of cone for which the Positivstellensatz fails is $C_{\\infty}:=\\left\\{ f\\in\\mathbb{R}\\left[x\\right]\\mid\\exists x\\in\\mathbb{R}\\,\\forall y\\geq x\\, f\\left(y\\right)\\geq0\\right\\}$, the cone of polynomials that are non-negative on sufficiently large inputs (equivalently, the cone of polynomials that are either $0$ or have positive leading coefficient). $P_{\\mathbb{R}}\\left(C\\right)=\\emptyset$, and $-1$ is strictly positive on $\\emptyset$, but for $f\\in C_{\\infty}$$-f-1\\notin C_{\\infty}$.\n\nHowever, it doesn't really look $C_{\\infty}$ is trying to point to the empty set; instead, $C_{\\infty}$ is trying to describe the set of all infinitely large reals, which only looks like the empty set because there are no infinitely large reals. Similar phenomena can occur even for cones that do contain finitely-generated subcones with compact positive-sets. For example, let $C_{\\varepsilon}:=\\left\\{ f\\in\\mathbb{R}\\left[x\\right]\\mid\\exists x>0\\,\\forall y\\in\\left[0,x\\right]\\, f\\left(y\\right)\\geq0\\right\\}$$P_{\\mathbb{R}}\\left(C_{\\varepsilon}\\right)=\\left\\{ 0\\right\\}$, but $C_{\\varepsilon}$ is trying to point out the set containing $0$ and all positive infinitesimals. Since $\\mathbb{R}$ has no infinitesimals, this looks like $\\left\\{ 0\\right\\}$.\n\nTo formalize this intuition, we can change the Galois connection. We could say that for a cone $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$$P_{\\text{*}\\mathbb{R}}\\left(C\\right):=\\left\\{ \\bar{x}\\in\\left(\\text{*}\\mathbb{R}\\right)^{n}\\mid f\\left(\\bar{x}\\right)\\geq0\\,\\forall f\\in C\\right\\}$, where $\\text{*}\\mathbb{R}$ is the field of hyperreals. All you really need to know about $\\text{*}\\mathbb{R}$ is that it is a big ordered field extension of $\\mathbb{R}$. $P_{\\text{*}\\mathbb{R}}\\left(C_{\\infty}\\right)$ is the set of hyperreals that are bigger than any real number, and $P_{\\text{*}\\mathbb{R}}\\left(C_{\\varepsilon}\\right)$ is the set of hyperreals that are non-negative and smaller than any positive real. The cone of a subset $X\\subseteq\\left(\\text{*}\\mathbb{R}\\right)^{n}$, denoted $C_{\\text{*}\\mathbb{R}}$$\\left(X\\right)$ will be defined as before, still consisting only of polynomials with real coefficients. This defines a topology on $\\left(\\text{*}\\mathbb{R}\\right)^{n}$ by saying that the closed sets are the fixed points of $P_{\\text{*}\\mathbb{R}}\\circ C_{\\text{*}\\mathbb{R}}$. This topology is not $T_{0}$ because, for example, there are many hyperreals that are larger than all reals, and they cannot be distinguished by polynomials with real coefficients. There is no use keeping track of the difference between points that are in the same closed sets. If you have a topology that is not $T_{0}$, you can make it $T_{0}$ by identifying any pair of points that have the same closure. If we do this to $\\left(\\text{*}\\mathbb{R}\\right)^{n}$ , we get what I'm calling ordered affine $n$-space over $\\mathbb{R}$.\n\nDefinition: An $n$-type over $\\mathbb{R}$ is a set $\\Phi$ of inequalities, consisting of, for each polynomial $f\\in\\mathbb{R}\\left[x_{1},..,x_{n}\\right]$, one of the inequalities $f\\left(\\bar{x}\\right)\\geq0$ or $f\\left(\\bar{x}\\right)<0$, such that there is some totally ordered field extension $\\mathcal{R}\\supseteq\\mathbb{R}$ and $\\bar{x}\\in\\mathcal{R}^{n}$ such that all inequalities in $\\Phi$ are true about $\\bar{x}$. $\\Phi$ is called the type of $\\bar{x}$. Ordered affine $n$-space over $\\mathbb{R}$, denoted $\\mathbb{OA}_{\\mathbb{R}}^{n}$ is the set of $n$-types over $\\mathbb{R}$.\n\nCompactness Theorem: Let $\\Phi$ be a set of inequalities consisting of, for each polynomial $f\\in\\mathbb{R}\\left[x_{1},..,x_{n}\\right]$, one of the inequalities $f\\left(\\bar{x}\\right)\\geq0$ or $f\\left(\\bar{x}\\right)<0$. Then $\\Phi$ is an $n$-type if and only if for any finite subset $\\Delta\\subseteq\\Phi$, there is $\\bar{x}\\in\\mathbb{R}$ such that all inequalities in $\\Delta$ are true about $\\bar{x}$.\n\nproof: Follows from the compactness theorem of first-order logic and the fact that ordered field extensions of $\\mathbb{R}$ embed into elementary extensions of $\\mathbb{R}$. The theorem is not obvious if you do not know what those mean. $\\square$\n\nAn $n$-type represents an $n$-tuple of elements of an ordered field extension of $\\mathbb{R}$, up to the equivalence relation that identifies two such tuples that relate to $\\mathbb{R}$ by polynomials in the same way. One way that a tuple of elements of an extension of $\\mathbb{R}$ can relate to elements of $\\mathbb{R}$ is to equal a tuple of elements of $\\mathbb{R}$, so there is a natural inclusion $\\mathbb{R}^{n}\\subseteq\\mathbb{OA}_{\\mathbb{R}}^{n}$ that associates an $n$-tuple of reals with the set of polynomial inequalities that are true at that $n$-tuple.\n\nA tuple of polynomials $\\left(f_{1},...,f_{m}\\right)\\in\\left(\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\right)^{m}$ describes a function $f:\\mathbb{R}^{n}\\rightarrow\\mathbb{R}^{m}$, which extends naturally to a function $f:\\mathbb{OA}_{\\mathbb{R}}^{n}\\rightarrow\\mathbb{OA}_{\\mathbb{R}}^{m}$ by $f\\left(\\Phi\\right)$ is the type of $\\left(f_{1}\\left(\\bar{x}\\right),...,f_{m}\\left(\\bar{x}\\right)\\right)$, where $\\bar{x}$ is an $n$-tuple of elements of type $\\Phi$ in an extension of $\\mathbb{R}$. In particular, a polynomial $f\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ extends to a function $f:\\mathbb{OA}_{\\mathbb{R}}^{n}\\rightarrow\\mathbb{OA}_{\\mathbb{R}}^{1}$, and $\\mathbb{OA}_{\\mathbb{R}}^{1}$ is totally ordered by $\\Phi\\geq\\Psi$ if and only if $x\\geq y$, where $x$ and $y$ are elements of type $\\Phi$ and $\\Psi$, respectively, in an extension of $\\mathbb{R}$$f\\left(\\Phi\\right)\\geq0$ if and only if $\\text{\"}f\\left(\\bar{x}\\right)\\geq0\\text{\"}\\in\\Phi$, so we can talk about inequalities satisfied by types in place of talking about inequalities contained in types.\n\nI will now change the Galois connection that we are talking about yet again (last time, I promise). It will now be a Galois connection between the set of cones in $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ and the set of subsets of $\\mathbb{OA}_{\\mathbb{R}}^{n}$. For a cone $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$, $P\\left(C\\right):=\\left\\{ \\Phi\\in\\mathbb{OA}_{\\mathbb{R}}^{n}\\mid f\\left(\\Phi\\right)\\geq0\\,\\forall f\\in C\\right\\}$. For a set $X\\subseteq\\mathbb{OA}_{\\mathbb{R}}^{n}$, $C\\left(X\\right):=\\left\\{ f\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid f\\left(\\Phi\\right)\\geq0\\,\\forall\\Phi\\in X\\right\\}$. Again, this defines a topology on $\\mathbb{OA}_{\\mathbb{R}}^{n}$ by saying that fixed points of $P\\circ C$ are closed. $\\mathbb{OA}_{\\mathbb{R}}^{n}$ is $T_{0}$; in fact, it is the $T_{0}$ topological space obtained from $\\left(\\text{*}\\mathbb{R}\\right)^{n}$ by identifying points with the same closure as mentioned earlier. $\\mathbb{OA}_{\\mathbb{R}}^{n}$ is also compact, as can be seen from the compactness theorem. $\\mathbb{OA}_{\\mathbb{R}}^{n}$ is not $T_{1}$ (unless $n=0$). Note that model theorists have their own topology on $\\mathbb{OA}_{\\mathbb{R}}^{n}$, which is distinct from the one I use here, and is a refinement of it.\n\nThe new Galois connection is compatible with the old one via the inclusion $\\mathbb{R}^{n}\\subseteq\\mathbb{OA}_{\\mathbb{R}}^{n}$, in the sense that if $X\\subseteq\\mathbb{R}^{n}$, then $C_{\\mathbb{R}}\\left(X\\right)=C\\left(X\\right)$ (where we identify $X$ with its image in $\\mathbb{OA}_{\\mathbb{R}}^{n}$), and for a cone $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$$P_{\\mathbb{R}}=P\\left(C\\right)\\cap\\mathbb{R}^{n}$.\n\nLike our intermediate Galois connection $\\left(P_{\\text{*}\\mathbb{R}},C_{\\text{*}\\mathbb{R}}\\right)$, our final Galois connection $\\left(P,C\\right)$ succeeds in distinguishing $P\\left(C_{\\infty}\\right)$ and $P\\left(C_{\\varepsilon}\\right)$ from $\\emptyset$ and $\\left\\{ 0\\right\\}$, respectively, in the desirable manner. $P\\left(C_{\\infty}\\right)$ consists of the type of numbers larger than any real, and $P\\left(C_{\\varepsilon}\\right)$ consists of the types of $0$ and of positive numbers smaller than any positive real.\n\nJust like for subsets of $\\mathbb{R}^{n}$, a closed subset $X\\subseteq\\mathbb{OA}_{\\mathbb{R}}^{n}$ has a coordinate ring $\\mathbb{R}\\left[X\\right]:=\\mathbb{R}\\left[x_{1},...,x_{n}\\right]/C\\left(X\\right)$, and an arbitrary $X\\subseteq\\mathbb{OA}_{\\mathbb{R}}^{n}$ has a ring of regular functions $\\mathcal{O}\\left(X\\right)$ consisting of functions on $X$ that are locally ratios of polynomials, ordered by $f\\geq0$ if and only if $\\forall\\Phi\\in X$, where $f=\\frac{p}{q}$ is a representation of $f$ as a ratio of polynomials in a neighborhood of $\\Phi$, either $p\\left(\\Phi\\right)\\geq0$ and $q\\left(\\Phi\\right)>0$, or $p\\left(\\Phi\\right)\\leq0$ and $q\\left(\\Phi\\right)<0$, and $f\\geq g$ if and only if $f-g\\geq0$. As before, $\\mathbb{R}\\left[X\\right]\\subseteq\\mathcal{O}\\left(X\\right)$ for closed $X\\subseteq\\mathbb{OA}_{\\mathbb{R}}^{n}$.\n\n$\\mathbb{OA}_{\\mathbb{R}}^{n}$ is analogous to $\\mathbb{A}_{\\mathbb{C}}^{n}$ from algebraic geometry because if, in the above definitions, you replace \"$\\geq$\" and \"$<$\" with \"$=$\" and \"$\\neq$\", replace totally ordered field extensions with field extensions, and replace cones with ideals, then you recover a description of $\\mathbb{A}_{\\mathbb{C}}^{n}$, in the sense of $\\text{Spec}\\left(\\mathbb{C}\\left[x_{1},...,x_{n}\\right]\\right)$.\n\nWhat about an analog of projective space? Since we're paying attention to order, we should look at spheres, not real projective space. The $n$-sphere over $\\mathbb{R}$, denoted $\\mathbb{S}_{\\mathbb{R}}^{n}$, can be described as the locus of $\\left|\\bar{x}\\right|^{2}=1$ in $\\mathbb{OA}_{\\mathbb{R}}^{n}$.\n\nFor any totally ordered field $k$, we can define $\\mathbb{OA}_{k}^{n}$ similarly to $\\mathbb{OA}_{\\mathbb{R}}^{n}$, as the space of $n$-types over $k$, defined as above, replacing $\\mathbb{R}$ with $k$ (although a model theorist would no longer call it the space of $n$-types over $k$). The compactness theorem is not true for arbitrary $k$, but its corollary that $\\mathbb{OA}_{k}^{n}$ is compact still is true.\n\n### Visualizing $\\mathbb{OA}_{\\mathbb{R}}^{n}$ and $\\mathbb{S}_{\\mathbb{R}}^{n}$\n\n$\\mathbb{S}_{\\mathbb{R}}^{n}$ should be thought of as the $n$-sphere with infinitesimals in all directions around each point. Specifically, $\\mathbb{S}_{\\mathbb{R}}^{0}$ is just $\\mathbb{S}^{0}$, a pair of points. The closed points of $\\mathbb{S}_{\\mathbb{R}}^{n+1}$ are the points of $\\mathbb{S}^{n+1}$, and for each closed point $p$, there is an $n$-sphere of infinitesimals around $p$, meaning a copy of $\\mathbb{S}_{\\mathbb{R}}^{n}$, each point of which has $p$ in its closure.\n\n$\\mathbb{OA}_{\\mathbb{R}}^{n}$ should be thought of as $n$-space with infinitesimals in all directions around each point, and infinities in all directions. Specifically, $\\mathbb{OA}_{\\mathbb{R}}^{n}$ contains $\\mathbb{R}^{n}$, and for each point $p\\in\\mathbb{R}^{n}$, there is an $n-1$-sphere of infinitesimals around $p$, and there is also a copy of $\\mathbb{S}_{\\mathbb{R}}^{n-1}$ around the whole thing, the closed points of which are limits of rays in $\\mathbb{R}^{n}$.\n\n$\\mathbb{OA}_{\\mathbb{R}}^{n}$ and $\\mathbb{S}_{\\mathbb{R}}^{n}$ relate to each other the same way that $\\mathbb{R}^{n}$ and $\\mathbb{S}^{n}$ do. If you remove a closed point from $\\mathbb{S}_{\\mathbb{R}}^{n}$, you get $\\mathbb{OA}_{\\mathbb{R}}^{n}$, where the sphere of infinitesimals around the removed closed point becomes the sphere of infinities of $\\mathbb{OA}_{\\mathbb{R}}^{n}$.\n\nMore generally, if $k$ is a totally ordered field, let $k^{r}$ be its real closure. $\\mathbb{OA}_{k}^{n}$ consists of the Cauchy completion of $\\left(k^{r}\\right)^{n}$ (as a metric space with distances valued in $k^{r}$), and for each point $p\\in\\left(k^{r}\\right)^{n}$ (though not for points that are limits of Cauchy sequences that do not converge in $\\left(k^{r}\\right)^{n}$), an $n-1$-sphere $\\mathbb{S}_{k}^{n-1}$ of infinitesimals around $p$, and an $n-1$-sphere $\\mathbb{S}_{k}^{n-1}$ around the whole thing, where $\\mathbb{S}_{k}^{n}$ is the locus of $\\left|\\bar{x}\\right|^{2}=1$ in $\\mathbb{OA}_{k}^{n}$. $\\mathbb{OA}$ does not distinguish between fields with the same real closure.\n\n### More Positivstellensätze\n\nThis Galois connection gives us a new notion of what it means for a cone to be radical, which is distinct from the old one and is better, so I will define $\\text{Rad}\\left(C\\right)$ to be $C\\left(P\\left(C\\right)\\right)$. A cone $C$ will be called radical if $C=\\text{Rad}\\left(C\\right)$. Again, it would be nice to be able to characterize radical cones without referring to the Galois connection. And this time, I can do it. Note that since $\\mathbb{OA}_{\\mathbb{R}}^{n}$ is compact, the proof of Positivstellensatz 3 shows that in our new context, the Positivstellensatz holds for all cones, since even the subcone generated by $\\emptyset$ has a compact positive-set.\n\nPositivstellensatz 4: If $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a cone and $p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a polynomial, then $p\\left(\\Phi\\right)>0\\,\\forall\\Phi\\in P\\left(C\\right)$ if and only if $\\exists f\\in C$ such that $pf-1\\in C$.\n\nHowever, we can no longer add in lower limits of sequences of polynomials. For example, $-x+\\varepsilon\\in C_{\\varepsilon}$ for all real $\\varepsilon>0$, but $-x\\notin C_{\\varepsilon}$, even though $C_{\\varepsilon}$ is radical. This happens because, where $\\Sigma$ is the type of positive infinitesimals, $-\\Sigma+\\varepsilon>0$ for real $\\varepsilon>0$, but $-\\Sigma<0$. However, we can add in lower limits of sequences contained in finitely-generated subcones, and this is all we need to add, so this characterizes radical cones.\n\nPositivstellensatz 5: If $C\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is a cone, $\\text{Rad}\\left(C\\right)$ is the union over all finitely-generated subcones $D\\subseteq C$ of the closure of $\\left\\{ p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid\\exists f\\in D\\, pf-1\\in D\\right\\}$ (again the closure of a subset $X\\subseteq\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ is defined to be the set of all polynomials in $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ which are infima of chains contained in $X$).\n\nProof: Suppose $D\\subseteq C$ is a subcone generated by a finite set $\\left\\{ f_{1},...,f_{m}\\right\\}$, and $q$ is the infimum of a chain $\\left\\{ q_{\\alpha}\\right\\} _{\\alpha\\in A}\\subseteq\\left\\{ p\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid\\exists f\\in D\\, pf-1\\in D\\right\\}$. For any $\\bar{x}\\in\\mathbb{R}^{n}$, if $f_{i}\\left(\\bar{x}\\right)\\geq0$ for each $i$, then $q_{\\alpha}\\left(\\bar{x}\\right)>0$ for each $\\alpha$, and hence $q\\left(\\bar{x}\\right)\\geq0$. That is, the finite set of inequalities $\\left\\{ f_{i}\\left(\\bar{x}\\right)\\geq0\\mid1\\leq i\\leq m\\right\\} \\cup\\left\\{ q\\left(\\bar{x}\\right)<0\\right\\}$ does not hold anywhere in $\\mathbb{R}^{n}$. By the compactness theorem, there are no $n$-types satisfying all those inequalities. Given $\\Phi\\in P\\left(C\\right)$$f_{i}\\left(\\Phi\\right)\\geq0$, so $q\\left(\\Phi\\right)\\nless0$; that is, $q\\left(\\Phi\\right)\\geq0$.\n\nConversely, suppose $q\\in\\text{Rad}\\left(C\\right)$. Then by the compactness theorem, there are some $f_{1},...,f_{m}\\in C$ such that $q\\in\\text{Rad}\\left(\\left\\langle f_{1},...,f_{m}\\right\\rangle \\right)$. Then $\\forall\\varepsilon>0$, $q+\\varepsilon$ is strictly positive on $P\\left(\\left\\langle f_{1},...,f_{m}\\right\\rangle \\right)$, and hence by Positivstellensatz 4, $\\exists f\\in\\left\\langle f_{1},...,f_{m}\\right\\rangle$ such that $pf-1\\in\\left\\langle f_{1},...,f_{m}\\right\\rangle$. That is, $\\left\\{ q+\\varepsilon\\mid\\varepsilon>0\\right\\}$ is a chain contained in $\\left\\langle f_{1},...,f_{m}\\right\\rangle$, a finitely-generated subcone of $C$, whose infimum is $q$. $\\square$\n\n### Ordered commutative algebra\n\nEven though they are technically not isomorphic, $\\mathbb{C}^{n}$ and $\\text{Spec}\\left(\\mathbb{C}\\left[x_{1},...,x_{n}\\right]\\right)$ are closely related, and can often be used interchangeably. Of the two, $\\text{Spec}\\left(\\mathbb{C}\\left[x_{1},...,x_{n}\\right]\\right)$ is of a form that can be more easily generalized to more abstruse situations in algebraic geometry, which may indicate that it is the better thing to talk about, whereas $\\mathbb{C}^{n}$ is merely the simpler thing that is easier to think about and just as good in practice in many contexts. In contrast, $\\mathbb{R}^{n}$ and $\\mathbb{OA}_{\\mathbb{R}}^{n}$ are different in important ways. The situation in algebraic geometry provides further reason to pay more attention to $\\mathbb{OA}_{\\mathbb{R}}^{n}$ than to $\\mathbb{R}^{n}$.\n\nThe next thing to look for would be an analog of the spectrum of a ring for a partially ordered commutative ring (I will henceforth abbreviate \"partially ordered commutative ring\" as \"ordered ring\" in order to cut down on the profusion of adjectives) in a way that makes use of the order, and gives us $\\mathbb{OA}_{\\mathbb{R}}^{n}$ when applied to $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$. I will call it the order spectrum of an ordered ring $A$, denoted $\\text{OrdSpec}\\left(A\\right)$. Then of course $\\mathbb{OA}_{A}^{n}$ can be defined as $\\text{OrdSpec}\\left(A\\left[x_{1},...,x_{n}\\right]\\right)$$\\text{OrdSpec}\\left(A\\right)$ should be, of course, the set of prime cones. But what even is a prime cone?\n\nDefinition: A cone $\\mathfrak{p}\\subseteq A$ is prime if $A/\\mathfrak{p}$ is a totally ordered integral domain.\n\nDefinition: $\\text{OrdSpec}\\left(A\\right)$ is the set of prime cones in $A$, equipped with the topology whose closed sets are the sets of prime cones containing a given cone.\n\nAn $n$-type $\\Phi\\in\\mathbb{OA}_{\\mathbb{R}}^{n}$ can be seen as a cone, by identifying it with $\\left\\{ f\\in\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\mid f\\left(\\Phi\\right)\\geq0\\right\\}$, aka $C\\left(\\left\\{ \\Phi\\right\\} \\right)$. Under this identification, $\\mathbb{OA}_{\\mathbb{R}}^{n}=\\text{OrdSpec}\\left(\\mathbb{R}\\left[x_{1},...,x_{n}\\right]\\right)$, as desired. The prime cones in $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$ are also the radical cones $C$ such that $P\\left(C\\right)$ is irreducible. Notice that irreducible subsets of $\\mathbb{OA}_{\\mathbb{R}}^{n}$ are much smaller than irreducible subsets of $\\mathbb{A}_{\\mathbb{C}}^{n}$; in particular, none of them contain more than one element of $\\mathbb{R}^{n}$.\n\nThere is also a natural notion of maximal cone.\n\nDefinition: A cone $\\mathfrak{m}\\subseteq A$ is maximal if $\\mathfrak{m}\\neq A$ and there are no strictly intermediate cones between $\\mathfrak{m}$ and $A$. Equivalently, if $\\mathfrak{m}$ is prime and closed in $\\text{OrdSpec}\\left(A\\right)$.\n\nMaximal ideals of $\\mathbb{C}\\left[x_{1},...,x_{n}\\right]$ correspond to elements of $\\mathbb{C}^{n}$. And the cones of elements of $\\mathbb{R}^{n}$ are maximal cones in $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$, but unlike in the complex case, these are not all the maximal cones, since there are closed points in $\\mathbb{OA}_{\\mathbb{R}}^{n}$ outside of $\\mathbb{R}^{n}$. For example, $C_{\\infty}$ is a maximal cone, and the type of numbers greater than all reals is closed. To characterize the cones of elements of $\\mathbb{R}^{n}$, we need something slightly different.\n\nDefinition: A cone $\\mathfrak{m}\\subseteq A$ is ideally maximal if $A/\\mathfrak{m}$ is a totally ordered field. Equivalently, if $\\mathfrak{m}$ is maximal and $\\mathfrak{m}^{\\circ}$ is a maximal ideal.\n\nElements of $\\mathbb{R}^{n}$ correspond to ideally maximal cones of $\\mathbb{R}\\left[x_{1},...,x_{n}\\right]$.\n\n$\\text{OrdSpec}$ also allows us to define the radical of a cone in an arbitrary partially ordered commutative ring.\n\nDefinition: For a cone $C\\subseteq A$, $\\text{Rad}\\left(C\\right)$ is the intersection of all prime cones containing $C$. $C$ is radical if $C=\\text{Rad}\\left(C\\right)$.\n\nConjecture: $\\text{Rad}\\left(C\\right)$ is the union over all finitely-generated subcones $C\\subseteq D$ of the closure of $\\left\\{ p\\in A\\mid\\exists f\\in D\\, pf-1\\in D\\right\\}$ (as before, the closure of a subset $X\\subseteq A$ is defined to be the set of all elements of $A$ which are infima of chains contained in $X$).\n\n### Order schemes\n\nDefinition: An ordered ringed space is a topological space equipped with a sheaf of ordered rings. An ordered ring is local if it has a unique ideally maximal cone, and a locally ordered ringed space is an ordered ringed space whose stalks are local.\n\n$\\text{OrdSpec}\\left(A\\right)$ can be equipped with a sheaf of ordered rings $\\mathcal{O}_{A}$, making it a locally ordered ringed space.\n\nDefinition: For a prime cone $\\mathfrak{p}\\subseteq A$, the localization of $A$ at $\\mathfrak{p}$, denoted $A_{\\mathfrak{p}}$, is the ring $A_{\\mathfrak{p}^{\\circ}}$ equipped with an ordering that makes it a local ordered ring. This will be the stalk at $\\mathfrak{p}$ of $\\mathcal{O}_{A}$. A fraction $\\frac{a}{b}\\in A_{\\mathfrak{p}}$ ($b\\notin\\mathfrak{p}^{\\circ}$) is also an element of $A_{\\mathfrak{q}}$ for any prime cone $\\mathfrak{q}\\subseteq A$ whose interior ideal does not contain $b$. This is an open neighborhood of $\\mathfrak{p}$ (its complement is the set of prime cones containing $\\left\\langle b,-b\\right\\rangle$). There is a natural map $A_{\\mathfrak{p}}\\rightarrow\\text{Frac}\\left(A/\\mathfrak{p}\\right)$ given by $\\frac{a}{b}\\mapsto\\frac{a+\\mathfrak{p}^{\\circ}}{b+\\mathfrak{p}^{\\circ}}$, and the total order on $A/\\mathfrak{p}$ extends uniquely to a total order on the fraction field, so for $a,b\\in A_{\\mathfrak{p}}$, we can say that $a\\geq b$ at $\\mathfrak{p}$ if this is true of their images in $\\text{Frac}\\left(A/\\mathfrak{p}\\right)$. We can then say that $a\\geq b$ near $\\mathfrak{p}$ if $a\\geq b$ at every point in some neighborhood of $\\mathfrak{p}$, which defines the ordering on $A_{\\mathfrak{p}}$.\n\nDefinition: For open $U\\subseteq\\text{OrdSpec}\\left(A\\right)$, $\\mathcal{O}_{A}\\left(U\\right)$ consists of elements of $\\prod_{\\mathfrak{p}\\in U}A_{\\mathfrak{p}}$ that are locally ratios of elements of $A$. $\\mathcal{O}_{A}\\left(U\\right)$ is ordered by $a\\geq b$ if and only if $\\forall\\mathfrak{p}\\in\\text{OrdSpec}\\left(A\\right)$ $a\\geq b$ near $\\mathfrak{p}$ (equivalently, if $\\forall\\mathfrak{p}\\in\\text{OrdSpec}\\left(A\\right)$ $a\\geq b$ at $\\mathfrak{p}$).\n\n$A\\subseteq\\mathcal{O}_{A}\\left(\\text{OrdSpec}\\left(A\\right)\\right)$, and this inclusion can be proper. Conjecture: $\\text{OrdSpec}\\left(\\mathcal{O}_{A}\\left(U\\right)\\right)\\cong U$ as locally ordered ringed spaces for open $U\\subseteq\\text{OrdSpec}\\left(A\\right)$. This conjecture says that it makes sense to talk about whether or not a locally ordered ringed space looks locally like an order spectrum near a given point. Thus, if this conjecture is false, it would make the following definition look highly suspect.\n\nDefinition: An order scheme is a topological space $X$ equipped with a sheaf of ordered commutative rings $\\mathcal{O}_{X}$ such that for some open cover of $X$, the restrictions of $\\mathcal{O}_{X}$ to the open sets in the cover are all isomorphic to order spectra of ordered commutative rings.\n\nI don't have any uses in mind for order schemes, but then again, I don't know what ordinary schemes are for either and they are apparently useful, and order schemes seem like a natural analog of them."
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https://brainmass.com/computer-science/c/few-class-c-subnetting-questions-186719 | [
"Explore BrainMass\n\n# Few Class C subnetting questions\n\nNot what you're looking for? Search our solutions OR ask your own Custom question.\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nAnswer the following questions for each of the configuration in the table below.\n\n1. How many subnets?\n2. How many hosts per subnet?\n3. What are the valid subnets?\n4. What are the valid hosts per subnet?\n\nCONFIGURATION A 192.168.10.0 255.255.255.192\nCONFIGURATION B 220.110.16.0 255.255.255.224\nCONFIGURATION C 219.220.250.0 255.255.255.248\n\nhttps://brainmass.com/computer-science/c/few-class-c-subnetting-questions-186719\n\n#### Solution Preview\n\nAll the given configurations belong to Class C network, so subnetting involves only the last octet. This answer looks at the Class C specific subnetting computations in different ways.\n\nConfiguration A:\n\nFirst, we find out how many bits of host part are used for subnetting in this case.\n\n192 = (128 + 64 + 0 + 0 + 0 + 0 + 0 + 0) OR 11000000 in binary.\n\nNumber of bits used for denoting subnetwork (Ns) = 2\nNumber of bits for denoting hosts in subnetwork (Nh) = 8 - 2 = 6.\n\n1. No. of subnets = 2^Ns - 2 = 2^2 - 2 = 2\n\n2. No. of hosts per subnet = 2^Nh - 2 = 2^6 - 2 = 62\n\n3. To find the first subnet, we compute (256 - subnetmask in fourth octet), and to find further subnets we add this computed value to previous subnet id and continue till we get to the above mentioned subnetmask value.\n\nFirst subnet id = 256 - 192 = 64\nSecond subnet id = 64 + 64 = 128\n\nSo, the valid subnets are 192.168.10.64 and 192.168.10.128 .\n\n4. Since the bits in host id can not be all zeros or all ones. We compute the valid host ids in each of the above subnets as following.\n\nSubnet ...\n\n#### Solution Summary\n\nSolution not only gives detailed explanation along with computations, but also illustrates different ways of subnetting computations.\n\n\\$2.49"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8442882,"math_prob":0.886675,"size":1860,"snap":"2021-31-2021-39","text_gpt3_token_len":526,"char_repetition_ratio":0.14331897,"word_repetition_ratio":0.02173913,"special_character_ratio":0.3236559,"punctuation_ratio":0.16458853,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T13:19:21Z\",\"WARC-Record-ID\":\"<urn:uuid:a9fbe629-3e5d-4b7e-881e-1e97813b857f>\",\"Content-Length\":\"332022\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:13404d39-a01f-4fc0-9acc-8d8034eca2eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:394e3363-5c93-4dfb-973e-9d82c24c8044>\",\"WARC-IP-Address\":\"172.67.75.38\",\"WARC-Target-URI\":\"https://brainmass.com/computer-science/c/few-class-c-subnetting-questions-186719\",\"WARC-Payload-Digest\":\"sha1:Y3J6EIQEQT6QN5P33HPBUPI5FSVVWF53\",\"WARC-Block-Digest\":\"sha1:O2OYHM35PHTANLERAHCIHBX6CGVZTNXE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154214.36_warc_CC-MAIN-20210801123745-20210801153745-00535.warc.gz\"}"} |
https://socratic.org/questions/what-is-the-derivative-of-e-3x-2 | [
"# What is the derivative of -e^(3x^2)?\n\n##### 1 Answer\nApr 19, 2018\n\n$\\frac{\\mathrm{dy}}{\\mathrm{dx}} = - 6 x {e}^{3 {x}^{2}}$\n\n#### Explanation:\n\nBy using chain rule for the function of function concept\n\ny=-e^(3x)^2)\n\n$y = - {e}^{t}$\n\n$\\frac{\\mathrm{dy}}{\\mathrm{dx}} = - {e}^{t} \\frac{\\mathrm{dt}}{\\mathrm{dx}}$\n\n$t = 3 {x}^{2}$\n\n$t = 3 u$\n\n$u = {x}^{2}$\n\n$\\frac{\\mathrm{du}}{\\mathrm{dx}} = 2 x$\n\n(dt)/(dx)=3)du)/(dx)\n\n3(du/(dx)=3xx2x\n\n$\\frac{\\mathrm{dt}}{\\mathrm{dx}} = 3 \\times 2 x$\n\n$3 \\times 2 x = 6 x$\n\n$\\frac{\\mathrm{dt}}{\\mathrm{dx}} = 6 x$\n\n$\\frac{\\mathrm{dy}}{\\mathrm{dx}} = - {e}^{t} \\frac{\\mathrm{dt}}{\\mathrm{dx}}$\n\n$\\frac{\\mathrm{dy}}{\\mathrm{dx}} = - {e}^{3 {x}^{2}} \\left(6 x\\right)$\n\n$\\frac{\\mathrm{dy}}{\\mathrm{dx}} = - 6 x {e}^{3 {x}^{2}}$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.60022837,"math_prob":1.00001,"size":259,"snap":"2021-21-2021-25","text_gpt3_token_len":64,"char_repetition_ratio":0.12156863,"word_repetition_ratio":0.0,"special_character_ratio":0.25096524,"punctuation_ratio":0.06382979,"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\":\"2021-06-13T21:32:01Z\",\"WARC-Record-ID\":\"<urn:uuid:f838dda3-6ea0-4c90-9ce3-11c350e28c03>\",\"Content-Length\":\"33143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5bf1c5e-795d-43b7-a46b-d0f12f5288a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:693f0141-6478-46fd-a98b-242d98024d3b>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/what-is-the-derivative-of-e-3x-2\",\"WARC-Payload-Digest\":\"sha1:QWUQGJEAPN5XJ2STVCDKR74KAYMYEN42\",\"WARC-Block-Digest\":\"sha1:GMRJ2AKJN2DALC4JTFX5CBAAI3QIXCUG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487610841.7_warc_CC-MAIN-20210613192529-20210613222529-00139.warc.gz\"}"} |
https://listserv.uni-heidelberg.de/cgi-bin/wa?A3=0705&L=LATEX-L&E=7bit&P=2221&B=--&T=text%2Fplain;%20charset=ISO-8859-1&header=1 | [
"On 5/5/07, Andreas Matthias <[log in to unmask]> wrote:\n> 1) I suppose there's a bug in \\prop_map_function_aux:NNn.\n\nYes, thanks. Will fix in source as soon as I get access from my laptop\n(tonight I hope).\n\n> 2) I have no idea how to use \\prop_map_inline:NN. Can someone\n> show me? There seems to be a missing \\prop_map_function:N.\n\nHmm, the code does look odd. I wonder why it wasn't written in a more\nstraightforward manner. Here's a combined test+fix document.\n\n\\documentclass{article}\n\n\\usepackage{l3prop,l3quark}\n\\CodeStart\n\n% also missing:\n\\def:Npn \\prop_map_function:Nc {\\exp_args:NNc \\prop_map_function:NN}\n% fix\n\\def:Npn \\prop_map_function_aux:NNn #1#2#3{\n\\if_meaning:NN \\q_nil #2\n\\exp_after:NN \\prop_map_break:w\n\\fi:\n#1#2{#3}\n\\prop_map_function_aux:NNn #1\n}\n% fix\n\\def_long:Npn \\prop_map_inline:Nn #1#2 {\n\\num_incr:N \\l_prop_inline_level_num\n\\def_long:cpn {prop_map_inline_ \\num_use:N \\l_prop_inline_level_num :n}\n##1##2{#2}\n\\prop_map_function:Nc #1\n{prop_map_inline_ \\num_use:N \\l_prop_inline_level_num :n}\n\\num_decr:N \\l_prop_inline_level_num\n}\n\n\\prop_new:N \\l_testa_plist\n\\prop_put:NNn \\l_testa_plist \\keya {info1}\n\\prop_put:NNn \\l_testa_plist \\keyb {info2}\n\\prop_put:NNn \\l_testa_plist \\keyc {info3}\n\n\\prop_map_inline:Nn \\l_testa_plist{\n\\io_put_term:x{\\token_to_string:N #1\\space =\\space #2}\n}\n\n\\stop\n\n--\nMorten"
] | [
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https://numbermatics.com/n/6648538482576/ | [
"# 6648538482576\n\n## 6,648,538,482,576 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 6648538482576 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 135 divisors.\n\n6648538482576 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of one hundred thirty-five divisors.\n\n## Prime factorization of 6648538482576:\n\n### 24 × 32 × 1512 × 14232\n\n(2 × 2 × 2 × 2 × 3 × 3 × 151 × 151 × 1423 × 1423)\n\nSee below for interesting mathematical facts about the number 6648538482576 from the Numbermatics database.\n\n### Names of 6648538482576\n\n• Cardinal: 6648538482576 can be written as Six trillion, six hundred forty-eight billion, five hundred thirty-eight million, four hundred eighty-two thousand, five hundred seventy-six.\n\n### Scientific notation\n\n• Scientific notation: 6.648538482576 × 1012\n\n### Factors of 6648538482576\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 10\n• Sum of prime factors: 1579\n\n### Divisors of 6648538482576\n\n• Number of divisors d(n): 135\n• Complete list of divisors:\n• Sum of all divisors σ(n): 18743884804827\n• Sum of proper divisors (its aliquot sum) s(n): 12095346322251\n• 6648538482576 is an abundant number, because the sum of its proper divisors (12095346322251) is greater than itself. Its abundance is 5446807839675\n\n### Bases of 6648538482576\n\n• Binary: 11000001011111110111100011001000111100100002\n• Base-36: 2CUAO5HG0\n\n### Squares and roots of 6648538482576\n\n• 6648538482576 squared (66485384825762) is 44203063954293980655595776\n• 6648538482576 cubed (66485384825763) is 293885771747891584367365438230275198976\n• 6648538482576 is a perfect square number. Its square root is 2578476\n• The cube root of 6648538482576 is 18803.6458171681\n\n### Scales and comparisons\n\nHow big is 6648538482576?\n• 6,648,538,482,576 seconds is equal to 211,402 years, 49 weeks, 5 days, 21 hours, 9 minutes, 36 seconds.\n• To count from 1 to 6,648,538,482,576 would take you about five hundred twenty-eight thousand, five hundred seven years!\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 6648538482576 cubic inches would be around 1567 feet tall.\n\n### Recreational maths with 6648538482576\n\n• 6648538482576 backwards is 6752848358466\n• 6648538482576 is a Harshad number.\n• The number of decimal digits it has is: 13\n• The sum of 6648538482576's digits is 72\n• More coming soon!"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6864196,"math_prob":0.9528132,"size":4964,"snap":"2021-31-2021-39","text_gpt3_token_len":1587,"char_repetition_ratio":0.14758064,"word_repetition_ratio":0.052333806,"special_character_ratio":0.52699435,"punctuation_ratio":0.13801453,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903613,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-27T08:55:46Z\",\"WARC-Record-ID\":\"<urn:uuid:cb322c8f-775f-4076-a2ea-1e3c340a6e83>\",\"Content-Length\":\"28666\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c45e2bbe-e3f9-47d8-bc57-2ab71eb2a217>\",\"WARC-Concurrent-To\":\"<urn:uuid:27cf0011-5082-47a7-900b-b9aac68b314c>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/6648538482576/\",\"WARC-Payload-Digest\":\"sha1:Z5P3NBFUPVSOBNMC5TLHOKH72L3O6ZNG\",\"WARC-Block-Digest\":\"sha1:ZONHASAZGXCJ5DBILAZ2HDUUYD2K5WFV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153223.30_warc_CC-MAIN-20210727072531-20210727102531-00062.warc.gz\"}"} |
http://ilarioaltobelli.it/parent-functions-algebra-2.html | [
"# Parent Functions Algebra 2\n\nThis topic covers: - Evaluating functions - Domain & range of functions - Graphical features of functions - Average rate of change of functions - Function combination and composition - Function transformations (shift, reflect, stretch) - Piecewise functions - Inverse functions - Two-variable functions. Algebra 2 -27 - Functions, Equations, and Graphs SECTION 2. You may use your graphing calculator to compare & sketch. Algebra, functions, & patterns cover 20-30% of the GED Math test. From parent functions in algebra to value, we have got all the pieces included. x is not a function of y, because the input y = 3 has multiple outputs: x = 1 and x = 2. Merely said, the answers to parent function packet algebra 2 is universally compatible following any devices to read. 2 worksheet Name _____ Parent Functions and Transformations Per ____ Date _____ Graph the parent function (black), the transformation (another color), and the domain and range for the. As discussed in the previous section, quadratic functions have y = x 2 as their parent function. g (x) = 2x³ – 5 Part A State the…. They are simple, yet powerful in their ability to model real world situations. The quadratic formula and its uses. Jul 10, 2017 - Explore Marla Barkman's board \"parent functions\" on Pinterest. 2X-2 _ 1 2 d. = -2(x - 1/2) 2 + 7/2. By the end of. Introduction to Parent Functions. Can you identify different functions and their translations? 1. IF Interpreting Functions Understand the concept of a function and use function notation. The graphs of these functions are drawn on the next page. f (x) = −2x+1 f (x) = - 2 x + 1 The parent function is the simplest form of the type of function given. Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Last Name *. Square Root Function f(x) = x 7. A set of basic functions used as building blocks for more complicated functions. This function can be transformed in the same way as other functions. logs 2 (Oft) 7. For each function, graph the indicated change and write a function statement. (Do not use any of the examples from #1 -12!). We could alternatively write these functions as (x +2)2,(x 2)2, (2x) 2,(x 2) ,and(x). Algebra 1 Honors – Ch 3. If a parabola opens downward, it has a highest point. Each poster gives the parent function, graph, and the following characteristics: domain, range, x-intercept, y-intercept, maximum number of roots, end behavior, increasing interval (s), decreasing interval (s) and asymptotes. 4 Exercises - Page 203 28 including work step by step written by community members like you. Pre-AP Algebra 2 Function Transformations ©a x2b0U1\\8s mKEuatXa` DSgoxfYtvwAarr[eG FLCLaCt. Should you seek advice with math and in particular with parent function of algebra 2 or dividing rational expressions come pay a visit to us at Solve-variable. Algebra 2: Unit 4 Instructional Focus –– Modeling with Functions Topic Instructional Foci le In this unit students synthesize and generalize what they have learned about a variety of function families. Similar graph for all odd roots. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\\left(x\\right)={b}^{x}\\\\$ without loss of shape. BUSH ALGEBRA 2. Mathematics: Parent functions \"give birth\" to child functions. Planning a Parent Workshop: Toolkit for Parent Engagement Maine-Endwell Educator Named 2020 New York State Teacher of the Year Released 2019 3-8 ELA and Mathematics State Test Questions. horizontal stretching of functions common core algebra 2 homework answer key, Some students in my introductory algebra class struggle with basic concepts. A point of special interest is the point (0,0) on the parent function. Linear; translation down 6. In mathematics, you see certain graphs over and over again. Graphs of Common Parent Functions; Get full access to over 1,300 online videos and slideshows from multiple courses ranging from Algebra 1 to Calculus. Graph is a horizontal line. Horizontal Translation. Quadratic; translation left 5. Algebra 2 Rational Functions Test Answers algebra 2 rational functions test review sheet. Graphing parent functions with transformations by Dan Tating - September 10, 2012. Algebra 3/MAT 150 Adv. In order to translate any function to the right or left, place an addition or subtraction \"inside\" of the Parent function. Common Parent Functions Linear Function: f(x) = x Domain: All real numbers. d: Perform operations on functions, including function composition, and determine domain and range for each of the given functions. 2-7 Parent Functions & Transformations •identify and use parent functions •describe transformations of functions •F. What is a parent function in Algebra 2? A parent function is the simplest function of a family of functions. U2 C6 Parent Functions. Algebra 2 – Unit 4: Working with Polynomials; Algebra 2 – Unit 5: Polynomial Equations & Graphs; Algebra 2 – Unit 6: Rational Functions; Algebra 2 – Unit 7: Exponential and Logarithmic Functions; Algebra 2 – Unit 8: Inequalities; Algebra 2 – Unit 9: Sequences and Series; Algebra 2 – Unit 10: Cumulative Review; Algebra 2 – Unit. and has specific properties and key points to assist in graphing. Browse other questions tagged algebra-precalculus functions transformation absolute-value or ask your own question. Parent Function: y = x. quadratic. f(x) = -2/+3 Parent Function: List all transformations: Parent Function: List all transformations: 7. (Do not use any of the examples from #1 -12!). 00 Parent Functions Wall Posters $3. The parent function of a parabola is where are the vertex. The functions in Items 2 and 3 are examples of horizontal translations. For example, if we begin by graphing the parent function $$f(x)=2^x$$, we can then graph the two reflections alongside it. s of Symmetry: Plug back in to find the vertex! EXAMPLE: VERTEX FORM: Axis of Symmetry. Inverse of a Function 79 4. In addition to watching the pre-recorded lessons or viewing the online slides, you may alsopurchase the PowerPoint (PPT) or Keynote file for this lesson for$3. 3: LINEAR FUNCTIONS AND SLOPE-INTERCEPT FORM MACC. }y 5 1 3 Ï} x is a vertical shrink of y 5 Ï} x. Introduction to Parent Functions Identify the parent function for from its equation. Range of a Function, and Onto Functions 78 4. Featured on Meta State of the Stack Q1 2021 Blog Post. Be sure to graph the squaring function using a dashed curve because it will be used as a guide and is not the answer. See answer below Given: y = sqrt(x) - 2 The parent function is g(x) = sqrt(x) The domain is limited because of the radical: x >= 0 The -2 outside of the radical determines the vertical shift, 2 units down. The functions in Items 2 and 3 are examples of horizontal translations. = -2(x - 1/2) 2 + 7/2. Anchor Points for Parent Function Graphs f(x) = x2 Anchor Points: (-1, 1), (0, 0), (1, 1), (-2, 4), (2, 4) D = { x| x ∈ R } or (-∞, ∞) R = { x| x ∈ R , x ≥0. Worksheet 3 Graphing exponential functions g(x) =- Hour Identify each transformation from the parent function of Tell if the function is a decay or growth function. Teacher Resources Based on extensive research, SmartGraphs: Algebra is designed for students studying Algebra 1 or Algebra 2. 1) x y-8-6-4-22468-8-6-4-2 2 4 6 8 2) x y-8-6. Algebra 3/MAT 150 Adv. But you can see where this could get really hard, right?. A set of basic functions used as building blocks for more complicated functions. 1 vertex; 1 line of symmetry; The highest degree (the greatest exponent) of the function is 2; The graph is a parabola; Parent and Offspring. See back of book. That was a great start with my students in Algebra 2, as we were able to use it right away starting with linears. By the end of. d: Perform operations on functions, including function composition, and determine domain and range for each of the given functions. 2 WS: Evaluate Parent Functions Name _____ Pre-AP Algebra II Date _____ Per _____ Assessment: Match the function (write the number of the graph on the blank) with. Transformations of exponential graphs behave similarly to those of other functions. IF Interpreting Functions Understand the concept of a function and use function notation. and has specific properties and key points to assist in graphing. Study Guide and Intervention Parent Functions and Transformations Name Characteristics Parent Function. Identified parent functions and transformations. Our focus on in-depth instruction is also ideal for homeschool parents looking to offer their child the equivalent of a $30,000-a-year private school math education for a tiny fraction of the cost. Algebra 1 Parent Functions and Transformations. The word logarithm, abbreviated log, is introduced to satisfy this need. 1-5 Guided Notes SE - Parent Functions and Transformations. absolute value B. Functions in the same family are transformations of their parent function. Parabola Parent Function - MathBitsNotebook (A2 - CCSS Math) A parent function is the simplest function of a family of functions. Parent Functions 1. ) • Polar graphs (r=cos2θ) • Parametric. Given f (x) = –x2 + 5x + 2, find the expression, in terms of f , for a leftward shift of five units. g(x) = x 2 – 6 Parent: _____. share to google. Before exploring the family of related functions, let’s clarify some of the. Scaling means shrinking or magnifying the function. 2 Transformations of Linear & Absolute Value Functions 1. 2 Represent Functions and Relations A2. Learn about the properties and graphs of linear parent functions quadratic parent functions absolute value parent functions and reciprocal parent functions. This exercise differs from the previous one in that I not only have to do the operations with the functions, but I also have to evaluate at a particular x-value. Tell students that the parent function, y = x 2, can be modified to create other functions by adding or subtracting values from x 2, or by multiplying x 2 by a coefficient, or both. One to one functions: is a function from to. This is a set of nine parent function posters to display in your classroom. [Algebra 2] What is this parent function. Parent Function Posters for Algebra 2This is a set of nine parent function posters to display in your classroom. Linear Function (Identity) f(x) = x 3. Then describe the transformations. Algebra 2 Team Tutorials – If I am not available, try one of the other great Algebra 2 teachers. This kind of circumstance is so stressful and through the help of some. Constant Linear Quadratic Absolute Value Cubic Square Root. THEOREM 2. Function Family Fun. The foldable is a great guided practice, the interactive notebook is a great way for students to collaborate and create and manipulate, the practice sheet can IA-2. f(x) = 2x 2. Chapter 2 43 Glencoe Algebra 2 2-7 Parent Graphs The parent graph, which is the graph of the parent function, is the simplest of the graphs in a family. 2 Represent Functions and Relations A2. This is the graph that is transformed to create other members in a family of graphs. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Vertical Expansions and Compressions. Bookmark File PDF Algebra 2 Parent Function Project With Graphing Parent Functions Project - Algebra 2 2018-19 Create a “Family Album” for the 7 function families listed below. Algebra 1 Honors – Ch 3. The other leg (the region in Quadrant 1) points upwards. An absolute value function has an x-intercept. Merely said, the answers to parent function packet algebra 2 is universally compatible with any devices to read. g(x) = 3 9. Learn how to shift graphs up, down, left, and right by looking at their equations. Unit 2 – Linear Functions & Systems Unit 3 – Intro to Parent Functions and Transformations Unit 4 – Solving Quadratics and Complex Numbers Unit 5 – Polynomial Functions Unit 6 – Radical Functions Unit 7 – Exponential and Logarithmic Functions Unit 8 – Rational Functions Unit 9 – Conic Sections Unit 10 – Sequences and Series. We can use a handy trick to find out whether this data set matches a linear or quadratic function. PARENT FUNCTIONS Author: Pete Falzone Created Date: 8/2/2001 12:36:31 AM. The Vertex Formula. Lesson 2-6 NAME DATE PERIOD PDF Pass Chapter 2 37 Glencoe Algebra 2 Exercises Graph each function. translation parent function 1–8. What is the line of reflection? A. Square Root Function f(x) = x 7. y= constant f(x) = constant. For the family of quadratic functions, y = ax2 + bx + c, the simplest function. Algebra 2 TAP. 1) 2 x>10 3. b: Use transformation to draw the graph of a relation and determine a relation that fits a graph. When we multiply the parent function f (x) = b x f (x) = b x by −1, −1, we get a reflection about the x-axis. View ALG2 HW Parent Functions. The graph below represents the function f (x) = - x2. IF Interpreting Functions Understand the concept of a function and use function notation. Vertical Expansions and Compressions. Come to Algebra-net. 𝑓𝑥 𝑥2 v 𝑓 𝑥 2 v 𝑥 3 𝑥 u2. Provide at least 2 examples of a function Provide 1 example of a non-function 3. h(x) = −2 ⋅ f(x) Multiply the output by −2. (link on my webpage) An explanation of what a function is. Name the parent function. Parent Function: Domain: X. Transform rational functions by changing parameters. ex: f(x) = (1/2 x — 4)2 becomes (1/2 (x — 8))2 is stretched horizontally by a factor of 2 5. 2 - Parent Functions 2 4 6 8 ©v f2D0c1f4Y QK]untXao uStomfMtHwxavrEev XLaLBCd. The other leg (the region in Quadrant 1) points upwards. Monday, 4/20/15- 7. Quadratic; translation 2 units up 8. Analyze functions using different representations. Range of a Function, and Onto Functions 78 4. Use the video above instead of going on a field trip. Chapter 1: Linear Functions 1. Then describe which transformation of the parent function it represents. This Custom Polygraph is designed to spark vocabulary-rich conversations about graphs of parent functions. A scientist notes the bacteria count in a petrie dish is 50. Functions are equal if they have the same domain and rule of correspondence. The vertex is the point (1/2, 7/2). 1 5 Parent Functions And Transformations April 18th, 2019 - Function Of Time Is Given By F T At2 V 0 T X 0 Where A Is The Acceleration V 0 Is The Initial Velocity And X0 Is The Initial Position Of The Object Describe The Transformation S Of The Parent 2 / 7. One leg of this graph (the region in Quadrant 3) points downwards. PDF Lessons 9. Parabola Parent Function - MathBitsNotebook (A2 - CCSS Math) A parent function is the simplest function of a family of functions. Parent functions included: linear, absolute value, quadratic, cubic, square root, cube root, reciprocal, exponential, and logarithmic. General Form for Describing the TRANSOFMRATIONS for a function f(x): F(x) ( a•f(x – h. This lowest or highest point is the vertex of the parabola. com contains practical advice on Glencoe Algebra 2 Answer Key, synthetic division and equations in two variables and other math topics. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Algebra 2: Parent Functions Name _____ Parent function: A parent function is the simplest function in a family of function – the simplest form of the given function. 3 2 hx x= 11. Planning a Parent Workshop: Toolkit for Parent Engagement Maine-Endwell Educator Named 2020 New York State Teacher of the Year Released 2019 3-8 ELA and Mathematics State Test Questions. h(x) = - x2 + 1 10. Unit 2 – Linear Functions & Systems Unit 3 – Intro to Parent Functions and Transformations Unit 4 – Solving Quadratics and Complex Numbers Unit 5 – Polynomial Functions Unit 6 – Radical Functions Unit 7 – Exponential and Logarithmic Functions Unit 8 – Rational Functions Unit 9 – Conic Sections Unit 10 – Sequences and Series. What are parent functions? The parent functions are a base of functions you should be able to recognize the graph of given the function and the other way around. Algebra 2 - Parent Functions. The equation is y= 5(2)^x+1 +3 The parent function here is y= 2^x The parent function is just where the graph y= 5(2)^x+1 +3 originates from. Function Family Fun. Finally, they learned how to write equations given the graph with dilations. After reviewing the Warm Up, I hand students a list of Graphing Vocabulary, and a Graphic Organizer for students to take notes on the Parent Functions. 1{1 Functions 78 4. 1 Linear and Quadratic Functions; 2. Reflection through the x-axis. They are simple, yet powerful in their ability to model real world situations. ( )2 4 9 hx x=− +5 14. Then describe the transformation. function for the graph at the right. Examples of doing distributive property in algebra workshhet that i can print; how to use my casio in algebra; free geometry basic transformation worksheets; 7th grade pre algebra test; algebra 2 online book; how changes in one or two dimensional of an object affect perimeter, area, surface area, and volume. All notes and homework should be done in student binder. You must understand that there's different parent functions. Algebra, functions, & patterns cover 20-30% of the GED Math test. (Limited to linear and quadratic functions only. 3 Polynomial Functions of Higher degree; 2. 3 5 f xx= 10. In cases where you have to have assistance with algebra and in particular with essay on parent functions in algebra 2 or math come pay a visit to us at Emaths. When in the function , a horizontal shrinking of the graph of will occur. If n is an odd whole number (1,3,5, etc) the graph will generally start in the third quadrant and end in the first quadrant. Algebra 2 Parent Functions Worksheet Answers - It is tiring when your kids request you in assisting these algebra residence operates, and you also are not able to do that home works, or you may not find out about them where you have not. To shift the graph side to side, I need to add or subtract inside the argument of the function (that is, inside the parentheses). translation parent function 1–8. In order to translate any function to the right or left, place an addition or subtraction \"inside\" of the Parent function. Transformations Of Parent Functions. Lesson #1: Parent Functions and Transformations: Vertical & Horizontal Shifts and Reflections September 13, 2019 September 13, 2019 mshallmaa Leave a comment Homework – Due Monday, September 16. Big Idea To develop an understanding for the transformations of parent functions through repeated reasoning (MP8) and the structure (MP7) of the equation. The vertex of the parent function y = x 2 lies on the origin. Common Parent Functions Linear Function: f(x) = x Domain: All real numbers. The Concept of a Function 72 4. Teacher Resources Based on extensive research, SmartGraphs: Algebra is designed for students studying Algebra 1 or Algebra 2. 9A The student is expected to use the parent function to investigate, describe, and predict the effects of parameter changes on the graphs of square root functions and describe limitations on the domains and ranges. TYPES OF TRANSLATIONS. See back of book. Solution for State the parent function of g (x), and describe how the graph of g (x) is related to to its parent function. All for only$14. 9) x y 10) x y 11) x y 12) x y Identify the parent function f (x) and write an equation for the function given. Parent Organizations; Paw Print; This website is for all Unit 5 students taking Algebra 1. Algebra 2 Parent Function Toolkit 5 Name: Sine Parent Equation: yx sin Description of the Locator: y-intercept (h, a+k) General Equation: y a b x h k sin ( ) a = amplitude b = angular frequency pb 2S h = horizontal shift k = vertical shift Properties: starts in the middle and ends in the middle Domain: , f f all real numbers f fx. Algebra 2: Parent Functions Name _____ Parent function: A parent function is the simplest function in a family of function – the simplest form of the given function. Pascal's Triangle. Gradesheet – This is to help the students keep track of their grades and daily homework checks. Algebra 1 - Unit 5 Vocabulary: Domain Range Translation Function Linear Function Quadratic Function Absolute Value Function Transformation Relation Parent Function Maximum Minimum Increasing Decreasing x-intercept y-intercept Function Notation F. Home Functions Logarithms Exponents Compounding Interest Probability Parent Funtions. In a quadratic function, the variable is always squared. This graph is known as the \"Parent Function\" for parabolas, or quadratic functions. For the following functions, identify the parent function and the transformations, then sketch the graph. Step Functions 83 4. The quadratic formula and its uses. What is a parent function and how do you do them? My teacher is no help and i'm stuck not knowing how to do my homework. The other leg (the region in Quadrant 1) points upwards. The restricted (natural) domain of the parent function is also of special interest as it is the non-negative real numbers, ie the interval. The \"Parent\" Graph: The simplest parabola is y = x 2, whose graph is shown at the right. g x 3x Quadratic Cubic Linear Identify the parent function for each graph. They are simple, yet powerful in their ability to model real world situations. Glenn’s convinced me to use the “transformation form” (AKA h,k form) last year in Geometry. These worksheets are written so that you do not have to be a mathematician to help your child. > Algebra 2A: Unit 2 - Introduction to Functions. Learn about algebra 2 parent functions with free interactive flashcards. YES! Now is the time to redefine your true self using Slader’s Algebra 2: A Common Core Curriculum answers. Common Parent Functions Linear Function: f(x) = x Domain: All real numbers. Quadratic 2. Algebra 2: Unit 1. Partial Fractions. This Page 11/28. Given f (x) = –x2 + 5x + 2, find the expression, in terms of f , for a leftward shift of five units. Quadratic Function f(x) = x 2 5. Graphing quadratic functions Quadratic functions are functions in which the 2nd power, or square, is the highest to which […]. Parent Functions and Transformations Worksheet, Word Docs, & PowerPoints. Start studying Parent Functions Algebra 2. When we multiply the input by $$−1$$,we get a reflection about the y-axis. Parent Functions DRAFT. Describe the transformation from the parent graph. 1) x y-8-6-4-22468-8-6-4-2 2 4 6 8 2) x y-8-6. the graph of the function changes, compared to the parent function. Algebra 2 Team Tutorials – If I am not available, try one of the other great Algebra 2 teachers. Algebra, Functions, and Data Analysis 2014 AFDA Vocabulary Cards Page 2 Parent Functions Linear, Quadratic. , 46) but is having difficulty with more advanced exponent problems. Solution for State the parent function of g (x), and describe how the graph of g (x) is related to to its parent function. The parent function f(x) = x2 has its vertex at the origin. 0 mathematicsvisionproject. Transformations of exponential graphs behave similarly to those of other functions. ti 89 help videos; Pre algebra final. Algebra 1 - Unit 5 Vocabulary: Domain Range Translation Function Linear Function Quadratic Function Absolute Value Function Transformation Relation Parent Function Maximum Minimum Increasing Decreasing x-intercept y-intercept Function Notation F. answers to parent function packet algebra 2 is available in our digital library an online access to it is set as public so you can get it instantly. Algebra 2 TAP. The vertex is the point (1/2, 7/2). 00 Algebra 2 End of Year Review Escape Room Activity $8. Title: HONORS Algebra 2: Intro Activity to Graphing Absolute Value Functions Author: lausd_user. Transformations of exponential graphs behave similarly to those of other functions. 2) Algebra II. Here you will learn to identify primary function families by their equations and graphs. Let us start with a function, in this case it is f(x) = x 2, but it could be anything: f(x) = x 2. The graph of the function g(x) = is a reflection of the parent function f(x) =. f(x) = -x - 11 + 2 6b. What effect do the variables a and k have on the. ) Click to print the worksheet 2. Identify the domain and range. 7 Contrast to Parent Function: 2-1 0 1 2. The purpose of this lesson is to introduce students to some of the Parent Functions, and the vocabulary used to explain graphs. This is the graph that is transformed to create other members in a family of graphs. share to facebook share to twitter Questions. A quiz over the main 8 parent functions. Identified parent functions and transformations. September 22, 2019 0 Comment. Parent Function: y = x. 3: LINEAR FUNCTIONS AND SLOPE-INTERCEPT FORM MACC. Algebra 2 provides further development of the concept of functions, systems of functions, and more advanced nonlinear functions. Name _ Date _ Algebra 2, Parent Function Worksheet # 1- 6 Give the name of the parent function. But you can see where this could get really hard, right?. Algebra, functions, & patterns cover 20-30% of the GED Math test. Name the parent function. The restricted (natural) domain of the parent function is also of special interest as it is the non-negative real numbers, ie the interval. ( f ∘ g) ( x) = f [ g ( x)] = f [ 2 x − 1] ( f ∘ g) ( x) = f [ g ( x)] = f [ 2 x − 1] Now, notice that since we’ve got a formula for g ( x) g ( x) we went ahead and plugged that in first. Algebra 1 Parent Functions and Transformations. g(x) 5 x 3 2 4 Parent function: Transformation: 6. This equation is rewritten as y = log 2 x. Exponential functions increase faster than either linear or quadratic functions, while square root functions increase slower than either linear or quadratic functions. y x= +2 2 y x= −2 2 The graph of the function shifts up 2. Identified parent functions and transformations. Algebra Il: Unit 4 Group Competency Part 2 Name. Jul 10, 2017 - Explore Marla Barkman's board \"parent functions\" on Pinterest. 9A The student is expected to use the parent function to investigate, describe, and predict the effects of parameter changes on the graphs of square root functions and describe limitations on the domains and ranges. Specifically, we use th. 0 mathematicsvisionproject. Parent Letter; Unit 2 - Exponents; Parent Letter; Unit 3 - Linear Equations; Expanding Space Station; Parent Letter; Unit 4 - Functions Relations; Parent Letter; Unit 5 - Slippery Slope; Walk the Graph; Parent Letter; Unit 6 - Parallel Lines & Congruence; Parent Letter; Unit 7 - Systems of Equations and Inequalities; Cara’s Candles & DVD Club. A scientist notes the bacteria count in a petrie dish is 50. In cases where you have to have assistance with algebra and in particular with essay on parent functions in algebra 2 or math come pay a visit to us at Emaths. absolute value B. Algebra ‒ Parent Functions and Transformations THE PARENT FUNCTION GRAPHS AND TRANSFORMATIONS!. All notes and homework should be done in student binder. Algebra 1 - Unit 5 Vocabulary: Domain Range Translation Function Linear Function Quadratic Function Absolute Value Function Transformation Relation Parent Function Maximum Minimum Increasing Decreasing x-intercept y-intercept Function Notation F. Lesson 2-6 NAME DATE PERIOD PDF Pass Chapter 2 37 Glencoe Algebra 2 Exercises Graph each function. Then describe the transformations. represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change. Can you identify different functions and their translations? 1. Example: y = 2x The blue line shows the graph of the stretch of the parent function, y = 2x, by a scale factor of 2. y x= +2 2 y x= −2 2 The graph of the function shifts up 2. Study Guide and Intervention Parent Functions and Transformations Name Characteristics Parent Function. Get full access to over 1,300 online videos and slideshows from multiple courses ranging from Algebra 1 to Calculus. A shrink is also referred to as a widening of the graph of the function closer to the x-axis. 4 Combining Functions; 1. Functions in the same family are transformations of their parent function. 1 5 Parent Functions And Transformations April 18th, 2019 - Function Of Time Is Given By F T At2 V 0 T X 0 Where A Is The Acceleration V 0 Is The Initial Velocity And X0 Is The Initial Position Of The Object Describe The Transformation S Of The Parent 2 / 7. When the parent function f (x) = x2 has an a -value that is less than 0, the graph reflects across the x -axis before it is transformed. We can use a handy trick to find out whether this data set matches a linear or quadratic function. Write “none” for transformations that do not exist. h(x) = −2 ⋅ f(x) Multiply the output by −2. Let's look at the effect of the addition or subtraction. It's never been late to practice some questions before the test. 1{1 Functions 78 4. PARENT FUNCTIONS Pre -Ap Algebra 2 Complete your Parent Function Packet!!!! • There are two slides per Parent Function. We can use a handy trick to find out whether this data set matches a linear or quadratic function. 1 5 Parent Functions And Transformations April 18th, 2019 - Function Of Time Is Given By F T At2 V 0 T X 0 Where A Is The Acceleration V 0 Is The Initial Velocity And X0 Is The Initial Position Of The Object Describe The Transformation S Of The Parent 2 / 7. 1 3 fx x=− + −2. g(x) 5 x 3 2 4 Parent function: Transformation: 6. Algebra 1 Module 1 Unit 2 Scatterplots and Lines of Best Fit Student Edition; Algebra 1 Module 2 Unit 3 Solving Equations & Inequalities Student Edition; Algebra 1 Module 2 Unit 4 Plus Honors Matrices Student Edition; Algebra 1 Module 2 Unit 4 Systems of Equations & Inequalities Student Ed; Algebra 1 Module 3 Functions; Algebra 1 Module 4. Additition resources: PDF 1, PDF 2, PDF 3, PDF 4. Provide at least 2 examples of a function Provide 1 example of a non-function 3. Our practice tests include the answers to all math problems. [Algebra 2] What is this parent function. The graph of the function shifts down 2. 1 vertex; 1 line of symmetry; The highest degree (the greatest exponent) of the function is 2; The graph is a parabola; Parent and Offspring. 2X-2 _ 1 2 d. Last Name *. 03/17/21 3-4: Parent Functions 2 What is a parent function? The parent function is the simplest function with the defining characteristics of the family. PAP-Algebra II Parent Functions/Transformations Test 1 Name:_____ II. y = x +k y = a x: y = a. Algebra 3/MAT 150. When we multiply the parent function f (x) = b x f (x) = b x by −1, −1, we get a reflection about the x-axis. Algebra 1 Parent Functions and Transformations. and has specific properties and key points to assist in graphing. If x = 2 y, then y = (the power on base 2) to equal x. Introduction: Identifying and graphing parent functions and their translations has been a primary focus of. a) F(x)= x2-6x+1 b) F(x)= x 3 2 c) F(x)= 1 4 3 x d) F(x)= |x-6|+2 Sketch with graphing calculator and identify the requested point. This is the graph that is transformed to create other members in a family of graphs. The concept of parent function is less clear for polynomials of higher power because of the extra turning points, but for the family of n-degree polynomial functions for any given n, the parent function is sometimes taken as x n, or, to simplify further, x 2 when n is even and x 3 for odd n. Algebra 2 Pre-AP Overview 2020 - 2021 This document is designed provide parents/guardians/community an overview of the curriculum taught in the FBISD classroom. What effect do the variables a and k have on the. THEOREM 2. Algebra 2 - Parent Functions DRAFT. Unit 3 - Linear Functions. Homework Help. Jul 10, 2017 - Explore Marla Barkman's board \"parent functions\" on Pinterest. Duration : 25 min Activity 5. Learn about algebra 2 parent functions with free interactive flashcards. High School Math. Y Range: Class Notes special. horizontal stretching of functions common core algebra 2 homework answer key, Some students in my introductory algebra class struggle with basic concepts. 2X-2 _ 1 2 d. Square root; translation 4 units left 2. This document supports families in understanding the learning goals for the course, and how students will demonstrate what they know and are able to do. On this page you can read or download gina wilson all things algebra 2015 parent functions and transformations in PDF format. Jun 29, 2015 - Parent Functions and Transformations (Algebra 2 Curriculum - Unit 3) DISTANCE LEARNING UPDATE: This unit now contains a Google document with: (1) Links to instructional videos. Algebra 1 - Unit 5 Vocabulary: Domain Range Translation Function Linear Function Quadratic Function Absolute Value Function Transformation Relation Parent Function Maximum Minimum Increasing Decreasing x-intercept y-intercept Function Notation F. exponential. 2 Power Functions; 2. The restricted (natural) domain of the parent function is also of special interest as it is the non-negative real numbers, ie the interval. Use the video above instead of going on a field trip. Algebra, functions, & patterns cover 20-30% of the GED Math test. To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or else I can find the values of the functions at x = 2 and then work from there. Browse other questions tagged algebra-precalculus functions transformation absolute-value or ask your own question. To shift the graph side to side, I need to add or subtract inside the argument of the function (that is, inside the parentheses). Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Similar graph for all odd powers. image/svg+xml. SAT functions have the dubious honor of being one of the trickiest topics on the SAT math section. Y Quadratic; Linear; Cubic; reflection translation left translation down across the y-axis Graph the data from the table. Those being Linear (y=x), Absolute Value (y=IxI), Quadratic (y=x (power of 2)), Radical (y= (square root of x)), Cubic (y=x (power of 3)), Cube Root (y=3 (square root of x)), Exponential (y=b*), Log (y=logb (x. Algebra 2 - Parent Functions DRAFT. pdf from MATH 2410 at Fremont High School. 2: Tesing Cause and Effect with Experiments and Conclusions From Observational Studies PDF Lesson 9. Algebra 2 Parent Functions Worksheet Answers - It is tiring when your kids request you in assisting these algebra residence operates, and you also are not able to do that home works, or you may not find out about them where you have not. Observe that this function increases when x is positive and decreases while x is negative. This document supports families in understanding the learning goals for the course, and how students will demonstrate what they know and are able to do. We are talking about transformations of exponential functions here. com/ Lesson 5. In a quadratic function, the variable is always squared. Teacher Resources Based on extensive research, SmartGraphs: Algebra is designed for students studying Algebra 1 or Algebra 2. This will be a major grade project and should demonstrate thorough and thoughtful work. Similar graph for all odd roots. x +k : Math 2 Unit 6 Lesson 1 Absolute Value Functions Page 5. High School Math. For example, if we begin by graphing the parent function $$f(x)=2^x$$, we can then graph the two reflections alongside it. Algebra 2 TAP. Y Quadratic; Linear; Cubic; reflection translation left translation down across the y-axis Graph the data from the table. 1 - Algebra 2 - parent functions and transformations. translation parent function 1–8. 2: Parent Functions and Transformations Name: Algebra II, Unit 1: Functions Block: In this course, you will learn shortcuts that allow you to sketch many different types of graphs quickly and accurately. How many zeros of the function are there in this graph? 6. Algebra and Trigonometry 10th Edition answers to Chapter 2 - 2. The table shows the linear and quadratic parent functions. In cases where you have to have assistance with algebra and in particular with essay on parent functions in algebra 2 or math come pay a visit to us at Emaths. (link on my webpage) An explanation of what a function is. Learn how to shift graphs up, down, left, and right by looking at their equations. Jump to navigation Jump to search. 9th - University grade. The Parent Functions are numbered in the bottom right corner of each slide. To do this, we simply add a constant term to the function. The 'Parent' Graph: The simplest parabola is y = x2, whose graph is shown at the right. Parent Functions And Transformations. Range of a Function, and Onto Functions 78 4. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function $f\\left(x\\right)={b}^{x}\\\\$ without loss of shape. From Wikiversity. exponential. Introduction to Parent Functions. This is illustrated below in the graph. Transformers Parts 2-5 - Algebra 2 Parent Functions For Students 10th - 12th Standards. the previous day. Scribd is the world's largest social reading and publishing site. Choose from 500 different sets of flashcards about algebra 2 parent functions on Quizlet. Learn about the properties and graphs of linear parent functions quadratic parent functions absolute value parent functions and reciprocal parent functions. parent function - the most basic function in a family; strip away all coefficients, negative signs, constants, etc you get the parent. Parent Graph: the graph of the parent function - it's the ; Function. pdf), Text File (. The parent graph, which is the graph of the parent function, is the simplest of the graphs in a family. 2 units up of the graph of the parent linear function. Changing variable names does not change the function. Tip #1: To Connect The Math Vocabulary To Prior Learning Okay, so you've probably heard this before, but as the teacher, we must connect new information to. IF Interpreting Functions Understand the concept of a function and use function notation. The graphs of these functions are drawn on the next page. Take a look. Transform rational functions by changing parameters. Range of a Function, and Onto Functions 78 4. When we multiply the parent function f (x) = b x f (x) = b x by −1, −1, we get a reflection about the x-axis. Graph is a horizontal line. Linear; translation down 6. Algebra 1 - Unit 5 Vocabulary: Domain Range Translation Function Linear Function Quadratic Function Absolute Value Function Transformation Relation Parent Function Maximum Minimum Increasing Decreasing x-intercept y-intercept Function Notation F. Functions are equal if they have the same domain and rule of correspondence. add 2 to each y-coordinate. Identified parent functions and transformations. Range of a Function, and Onto Functions 78 4. Here are a few of the types of functions we’ll learn about this year, and what their basic equations are, what their basic graphs look like, and what their domain and range are. log, c 4 16 9. Please watch through fi. y x= +2 2 y x= −2 2 The graph of the function shifts up 2. Algebra 1 Unit 7 – Quadratic Functions Monday Tuesday Wednesday Thursday Friday Mar 2 A Day 3 B Day 4 A Day 5 B Day 6 A Day Quadratic Parent Function Characteristics − Find the AOS, vertex, roots/zeros/ x-intercepts/solutions. Transform rational functions by changing parameters. For example, the function A = s² giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a. 5 Inverses; 1. b: Use transformation to draw the graph of a relation and determine a relation that fits a graph. Describe what happened to the parent a. shifted up two units b. Day 5/6 - Parent Graphs and Transformations - Notes Day 6 - Parent. PARENT FUNCTIONS Author: Pete Falzone Created Date: 8/2/2001 12:36:31 AM. If you need a review on polynomials in general, feel free to go to Tutorial 6: Polynomials. Vertical Expansions and Compressions. Click on the lesson that. Discover (and save!) your own Pins on Pinterest. 0 mathematicsvisionproject. 2X-2 _ 1 2 d. View ALG2 HW Parent Functions. Parent Functions And Transformation - Displaying top 8 worksheets found for this concept. A point of special interest is the point (0,0) on the parent function. notebook 2 September 27, 2013 Sep 276:48 AM Vertex Axis of Symmetry Parent Function Sep 276:50 AM ex. 3 5 f xx= 10. 1-5 Assignment - Parent Functions and Transformations. Parent Functions In Algebra 2, we are going to learn about a lot of different types of functions. Plus each one comes with an answer key. Restrictions on Domain Most of the functions we have studied in Algebra I are defined for all real numbers. Get full access to over 1,300 online videos and slideshows from multiple courses ranging from Algebra 1 to Calculus. Basically, the graph of a polynomial function is a smooth continuous curve. The module culminates with the fundamental theorem of algebra as the ultimate result in factoring. f(x) = - 2 # 7- 12 Draw the basic graph for the parent function. Then write a function g that represents the translation of h. If you need a review on functions, feel free to go to Tutorial 30: Introduction to Functions. the graph of (x) = 2x, reflected across the x axis. 1) x y-8-6-4-22468-8-6-4-2 2 4 6 8 2) x y-8-6. To graph the function, plot the points on the graph with x-values 2, QUADRATIC FUNCTIONS PARENT FUNCTION: Creates a U-shape curve called a The is a vertical X line that divides the parabola into two equal parts. Similar graph for all even roots. IF Interpreting Functions Understand the concept of a function and use function notation. Jun 29, 2015 - Parent Functions and Transformations (Algebra 2 Curriculum - Unit 3) DISTANCE LEARNING UPDATE: This unit now contains a Google document with: (1) Links to instructional videos. In these examples, the k value is what is changing. If a parabola opens downward, it has a highest point. x is not a function of y, because the input y = 3 has multiple outputs: x = 1 and x = 2. That was a great start with my students in Algebra 2, as we were able to use it right away starting with linears. < Algebra II. Before exploring the family of related functions, let’s clarify some of the. The graph of the function shifts down 2. In Topic C, students solve rational and radical equations, identifying extraneous solutions, then modeling and solving equations in situations where rational and radical functions are necessary. Functions are equal if they have the same domain and rule of correspondence. Algebra 2 Team Tutorials – If I am not available, try one of the other great Algebra 2 teachers. Given no information regarding the specific equation of. The parent function is the simplest function with the defining characteristics of the family. The turning point is called the STANDARD FORM: Ax. A point of special interest is the point (0,0) on the parent function. It results in a graph of the same shape and size but possibly in a different position. Each poster gives the parent function, graph, and the following characteristics: domain, range, x-intercept, y-intercept, maximum number of roots, end behavior, increasing interval (s), decreasing interval (s) and asymptotes. Algebra 2 provides further development of the concept of functions, systems of functions, and more advanced nonlinear functions. By the end of. Unit 3 will include the following subtopics: Exponential Functions including Growth and Decay (compound interest) Direct and Inverse Variation ( Note: Please see Unit 2 for the above topics) Parent. Algebra and Trigonometry 10th Edition answers to Chapter 2 - 2. 2 Shift and Stretch A Solidify Understanding Task In 4. The math involved in the calculation is easy as long as you are careful in every step of … How to Tell if a Function is Even, Odd or Neither Read More ». Linear Absolute Value Quadratic Square Root Exponential Logarithmic. = -2(x - 1/2) 2 + 7/2. Algebra Il: Unit 4 Group Competency Part 2 Name. Yay Math in Studio returns, with the help of baby daughter, to share some knowledge about parent functions and their transformations. ) • Polar graphs (r=cos2θ) • Parametric. The parent graph of this family of graphs is shown below. Then graph the function. Graphing parent functions with transformations by Dan Tating - September 10, 2012. of this form is y = x2. Parent Function: y = x. They are simple, yet powerful in their ability to model real world situations. 1 Parent Functions & Transformations 1. owvc parent transformations intercept asymptote Simplify each logarithm. Common Parent Functions Linear Function: f(x) = x Domain: All real numbers. Parent functions included: linear, absolute value, quadratic, cubic, square root, cube root, reciprocal, exponential, and logarithmic. General Form for Describing the TRANSOFMRATIONS for a function f(x): F(x) ( a•f(x – h. 7 Statistics and Statistical Graphs. That was a great start with my students in Algebra 2, as we were able to use it right away starting with linears. On this page you can read or download gina wilson all things algebra 2015 parent functions and transformations in PDF format. Algebra 2 Syllabus – General information about Algebra 2 as well as materials needed for the class. Gina Wilson All Things Algebra Test. translation parent function 1–8. Mathematics: Parent functions \"give birth\" to child functions. g(x) = x 2 – 1 4. IF Interpreting Functions Understand the concept of a function and use function notation. The Parent and Student Study Guide Workbook includes: •A 1-page worksheet for every lesson in the. Grieser Page 2 3) Sketch the graphs using transformations. Parent Functions and Transformations. f(x) = -2/+3 Parent Function: List all transformations: Parent Function: List all transformations: 7. Graph is a horizontal line. Gina Wilson All Things Algebra Test. Unit# 4 Algebraic Expressions and Algebraic Formulas Exercise 4. Analyze functions using different representations. 1-0 0, describe how. Duration : 40 min Activity 5. Click on the lesson that. The phrase \"y is a function of x\" means that the value of y depends upon the value of. Transformers Parts 2-5 - Algebra 2 Parent Functions For Students 10th - 12th Standards. Identified parent functions and transformations. Functions in the same family are transformations of their parent function. Similar graph for all odd powers. High School Math. Relations The first section deals with the difference between relations and functions. Home Functions Logarithms Exponents Compounding Interest Probability Parent Funtions. 1 Algebra 2 Project With Free PDF ebook Download: Algebra 2 Project With Download or Read Online ebook algebra 2 parent function project with graphing in PDF Format From The Best User Guide Database 4th 9 Weeks Pre-AP Algebra II Project. From Wikiversity. Observe that this function increases when x is positive and decreases while x is negative. We add and subtract 1/4, because (-1/2) 2 = 1/4, and -1 is the coefficient of x. 00 Algebra 1 Review Flip Book$4. 4 Finding Real Zeros of Polynomials of Higher. Parent Functions. Unit 2: Functions & Their Graphs Date: Homework 6: Parent Functions & Transformations ** This is a 2-page document ** Directions: Given each function, identify both the parent function and the transformations from the parent function. 4 years ago by. Each graph in a family of graphs has similar characteristics. • Function notation provides an efficient way to define and communicate functions. Lesson 2-6 NAME DATE PERIOD PDF Pass Chapter 2 37 Glencoe Algebra 2 Exercises Graph each function."
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https://www.colorhexa.com/ebcb4f | [
"# #ebcb4f Color Information\n\nIn a RGB color space, hex #ebcb4f is composed of 92.2% red, 79.6% green and 31% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 13.6% magenta, 66.4% yellow and 7.8% black. It has a hue angle of 47.7 degrees, a saturation of 79.6% and a lightness of 61.6%. #ebcb4f color hex could be obtained by blending #ffff9e with #d79700. Closest websafe color is: #ffcc66.\n\n• R 92\n• G 80\n• B 31\nRGB color chart\n• C 0\n• M 14\n• Y 66\n• K 8\nCMYK color chart\n\n#ebcb4f color description : Soft yellow.\n\n# #ebcb4f Color Conversion\n\nThe hexadecimal color #ebcb4f has RGB values of R:235, G:203, B:79 and CMYK values of C:0, M:0.14, Y:0.66, K:0.08. Its decimal value is 15453007.\n\nHex triplet RGB Decimal ebcb4f `#ebcb4f` 235, 203, 79 `rgb(235,203,79)` 92.2, 79.6, 31 `rgb(92.2%,79.6%,31%)` 0, 14, 66, 8 47.7°, 79.6, 61.6 `hsl(47.7,79.6%,61.6%)` 47.7°, 66.4, 92.2 ffcc66 `#ffcc66`\nCIE-LAB 82.347, -2.191, 63.683 57.029, 60.941, 16.156 0.425, 0.454, 60.941 82.347, 63.72, 91.971 82.347, 27.714, 74.49 78.064, -6.213, 42.375 11101011, 11001011, 01001111\n\n# Color Schemes with #ebcb4f\n\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #4f6feb\n``#4f6feb` `rgb(79,111,235)``\nComplementary Color\n• #eb7d4f\n``#eb7d4f` `rgb(235,125,79)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #bdeb4f\n``#bdeb4f` `rgb(189,235,79)``\nAnalogous Color\n• #7d4feb\n``#7d4feb` `rgb(125,79,235)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #4fbdeb\n``#4fbdeb` `rgb(79,189,235)``\nSplit Complementary Color\n• #cb4feb\n``#cb4feb` `rgb(203,79,235)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #4febcb\n``#4febcb` `rgb(79,235,203)``\n• #eb4f6f\n``#eb4f6f` `rgb(235,79,111)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #4febcb\n``#4febcb` `rgb(79,235,203)``\n• #4f6feb\n``#4f6feb` `rgb(79,111,235)``\n• #d5ae18\n``#d5ae18` `rgb(213,174,24)``\n• #e6bd21\n``#e6bd21` `rgb(230,189,33)``\n• #e8c438\n``#e8c438` `rgb(232,196,56)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #eed266\n``#eed266` `rgb(238,210,102)``\n• #f0d97d\n``#f0d97d` `rgb(240,217,125)``\n• #f3df94\n``#f3df94` `rgb(243,223,148)``\nMonochromatic Color\n\n# Alternatives to #ebcb4f\n\nBelow, you can see some colors close to #ebcb4f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #eba44f\n``#eba44f` `rgb(235,164,79)``\n• #ebb14f\n``#ebb14f` `rgb(235,177,79)``\n• #ebbe4f\n``#ebbe4f` `rgb(235,190,79)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #ebd84f\n``#ebd84f` `rgb(235,216,79)``\n• #ebe54f\n``#ebe54f` `rgb(235,229,79)``\n• #e4eb4f\n``#e4eb4f` `rgb(228,235,79)``\nSimilar Colors\n\n# #ebcb4f Preview\n\nThis text has a font color of #ebcb4f.\n\n``<span style=\"color:#ebcb4f;\">Text here</span>``\n#ebcb4f background color\n\nThis paragraph has a background color of #ebcb4f.\n\n``<p style=\"background-color:#ebcb4f;\">Content here</p>``\n#ebcb4f border color\n\nThis element has a border color of #ebcb4f.\n\n``<div style=\"border:1px solid #ebcb4f;\">Content here</div>``\nCSS codes\n``.text {color:#ebcb4f;}``\n``.background {background-color:#ebcb4f;}``\n``.border {border:1px solid #ebcb4f;}``\n\n# Shades and Tints of #ebcb4f\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000000 is the darkest color, while #fdfaee is the lightest one.\n\n• #000000\n``#000000` `rgb(0,0,0)``\n• #120f02\n``#120f02` `rgb(18,15,2)``\n• #231d04\n``#231d04` `rgb(35,29,4)``\n• #352b06\n``#352b06` `rgb(53,43,6)``\n• #473a08\n``#473a08` `rgb(71,58,8)``\n• #58480a\n``#58480a` `rgb(88,72,10)``\n• #6a570c\n``#6a570c` `rgb(106,87,12)``\n• #7b650e\n``#7b650e` `rgb(123,101,14)``\n• #8d7310\n``#8d7310` `rgb(141,115,16)``\n• #9f8212\n``#9f8212` `rgb(159,130,18)``\n• #b09014\n``#b09014` `rgb(176,144,20)``\n• #c29f16\n``#c29f16` `rgb(194,159,22)``\n``#d4ad18` `rgb(212,173,24)``\n• #e5bb1a\n``#e5bb1a` `rgb(229,187,26)``\n• #e7c12c\n``#e7c12c` `rgb(231,193,44)``\n• #e9c63d\n``#e9c63d` `rgb(233,198,61)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #edd061\n``#edd061` `rgb(237,208,97)``\n• #efd572\n``#efd572` `rgb(239,213,114)``\n• #f1db84\n``#f1db84` `rgb(241,219,132)``\n• #f3e095\n``#f3e095` `rgb(243,224,149)``\n• #f5e5a7\n``#f5e5a7` `rgb(245,229,167)``\n• #f7eab9\n``#f7eab9` `rgb(247,234,185)``\n• #f9efca\n``#f9efca` `rgb(249,239,202)``\n• #fbf5dc\n``#fbf5dc` `rgb(251,245,220)``\n• #fdfaee\n``#fdfaee` `rgb(253,250,238)``\nTint Color Variation\n\n# Tones of #ebcb4f\n\nA tone is produced by adding gray to any pure hue. In this case, #a09f9a is the less saturated color, while #fad440 is the most saturated one.\n\n• #a09f9a\n``#a09f9a` `rgb(160,159,154)``\n• #a7a393\n``#a7a393` `rgb(167,163,147)``\n• #afa78b\n``#afa78b` `rgb(175,167,139)``\n• #b6ac84\n``#b6ac84` `rgb(182,172,132)``\n• #beb07c\n``#beb07c` `rgb(190,176,124)``\n• #c5b575\n``#c5b575` `rgb(197,181,117)``\n• #cdb96d\n``#cdb96d` `rgb(205,185,109)``\n• #d4be66\n``#d4be66` `rgb(212,190,102)``\n• #dcc25e\n``#dcc25e` `rgb(220,194,94)``\n• #e3c757\n``#e3c757` `rgb(227,199,87)``\n• #ebcb4f\n``#ebcb4f` `rgb(235,203,79)``\n• #f3cf47\n``#f3cf47` `rgb(243,207,71)``\n``#fad440` `rgb(250,212,64)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #ebcb4f is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://stats.stackexchange.com/questions/tagged/skellam | [
"# Questions tagged [skellam]\n\nThe Skellam distribution is a discrete distribution that describes the difference between two independent Poisson distributions with possibly different parameters.\n\n19 questions\nFilter by\nSorted by\nTagged with\n363 views\n\n### What probability distribution would be suitable for modelling scores in a basketball match?\n\nIn football (not American football but what Americans call soccer), it is pretty clear: we may consider the difference of two independent Poisson variables. In basketball, in theory, we could increase ...\n554 views\n\n### What is the expectation of the absolute value of the Skellam distribution?\n\nIn particular, for a Skellam distribution obtained by substracting two iid Poisson Processes. Thank you!\n293 views\n\n184 views\n\n### Estimator of Bessel function?\n\nI am trying to estimate the parameters of the modified Bessel function of the first kind for integer order case. $I_n(wt) = \\sum\\limits_{m=0}^\\infty \\frac{1}{m!(m+n)!}(\\frac{wt}{2})^{2m+n}$ In ...\n1k views\n\n### Is the absolute value of the difference between two Poisson distributions a Poisson distribution?\n\nWhat is the distribution of the absolute value of the Skellam distribution?\n218 views\n\n### How to compare rates of occurence in consecutive time series count data?\n\nMy data consists of occurrences of words in time windows. E.g.: Day; Word; Frequency 1; \"dog\"; 45 1; \"cat\"; 2 ... 2; \"dog\"; 90 2; \"cat\"; 4 ... I would like to ...\n3k views\n\n### The difference between discrete and continuous variables\n\nIs the number of hydrogen bonds or the number of rings in a molecule a discrete or a continuous variable ? Can I say that the number of rings: 1, 3, 4, $\\dots$ is a continuous variable because in ...\n401 views\n\n### Poisson processes\n\nI have two realizations of a poisson stochastic process, they are over the same space with rate $\\lambda_{1}$ and $\\lambda_{2}$. What is the probability that N elements in both sequences are the same, ...\n7k views\n\n### How to calculate cumulative Poisson probabilities without adding each one if no. of outcomes is large and hence, adding (without excel) is difficult?\n\nOne can calculate the probability of a correct score of a football match by Poisson(Number of Team1 goals, Mean average Team1 goals) x Poisson(Number of Team2 goals, Mean average Team2 goals) So for ...\nI want to estimate the parameter $\\mu$ using the difference between two Poisson distributions with the same parameter, i.e. a Skellam distribution with $\\mu_1=\\mu_2 = \\mu$. I can calculate the ..."
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https://number.rocks/add/3367/plus/18 | [
"What is the sum of 3367 plus 18 and How much is 3367 + 18 percent?\n\nAddition of 3367 and 18 is 3385 and 3367 plus 18 percent is 3973.06\n\n3367 + 18\n=\n3385\n3367 + 18 percent\n=\n3973.06\n\nStep by Step Calculation for what is 3367 plus 18%\n\n= 3367 + 18% of 3367\n\n= 3367 + 18 * 3367 / 100\n\n= 3367 + 18 * 3367/100\n\n= 3367 + 30303/50\n\n= 63569/16\n\n63569/16 or fraction as a decimal is 3973.06"
] | [
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https://westcottcourses.com/subsite-publishing/algebra1-standards.html | [
"### California State Standard for Algebra l\n\nSymbolic reasoning and calculations with symbols are central in algebra. Through the study of algebra, a student develops an understanding of the symbolic language of mathematics and the sciences. In addition, algebraic skills and concepts are developed and used in a wide variety of problem-solving situations.\n\nCalifornia Standards for Algebra l Omega Math's Alignment to California Standards\n1.0 Students identify and use the arithmetic properties of subsets of integers and rational, irrational, and real numbers, including closure properties for the four basic arithmetic operations where applicable: Algebra l - e1.1\n1.1 Students use properties of numbers to demonstrate whether assertions are true or false. Algebra l - e1.1, e1.5, e1.7, e3.1, e3.2, e4.1, e5.5, e6.5, e7.2, e7.3, e7.4, e8.5, e10.3\n2.0 Students understand and use such operations as taking the opposite, finding the reciprocal, taking a root, and raising to a fractional power. They understand and use the rules of exponents. Algebra l - e1.4, chapter 8, e2.1, e2.2, e6.4\n3.0 Students solve equations and inequalities involving absolute values. Algebra l - Chapter 3, e1.7, e6.5\n4.0 Students simplify expressions before solving linear equations and inequalities in one variable, such as 3(2x-5) + 4(x-2) = 12. Algebra l - e3.2\n5.0 Students solve multistep problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step. Algebra l - Chapter 10, e3.5\n6.0 Students graph a linear equation and compute the x-and y-intercepts (e.g., graph 2x + 6y = 4). They are also able to sketch the region defined by linear inequality (e.g., they sketch the region defined by 2x + 6y < 4). Algebra l - e6.2, e6.3, e6.5\n7.0 Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula. Algebra l - e6.4\n8.0 Students understand the concepts of parallel lines and perpendicular lines and how those slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point. Algebra l - e6.2, e6.4\n9.0 Students solve a system of two linear equations in two variables algebraically and are able to interpret the answer graphically. Students are able to solve a system of two linear inequalities in two variables and to sketch the solution sets. Algebra 1 - e7.4\n10.0 Students add, subtract, multiply, and divide monomials and polynomials. Students solve multistep problems, including word problems, by using these techniques. Algebra l - e2.4, e2.5, e2.7, e3.5, e4.6, e6.4, e7.3, e8.6, e10.1, e10.2, e10.3\n11.0 Students apply basic factoring techniques to second-and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. Algebra l - Chapter 4\n12.0 Students simplify fractions with polynomials in the numerator and denominator by factoring both and reducing them to the lowest terms. Algebra l - e5.4\n13.0 Students add, subtract, multiply, and divide rational expressions and functions. Students solve both computationally and conceptually challenging problems by using these techniques. Algebra l - Chapter 5\n14.0 Students solve a quadratic equation by factoring or completing the square. Algebra l - e4.6, e9.2\n15.0 Students apply algebraic techniques to solve rate problems, work problems, and percent mixture problems. Algebra l - Chapter 10\n16.0 Students understand the concepts of a relation and a function, determine whether a given relation defines a function, and give pertinent information about given relations and functions. Algebra ll - Chapter 10\n17.0 Students determine the domain of independent variables and the range of dependent variables defined by a graph, a set of ordered pairs, or a symbolic expression. Algebra ll - i10.2\n18.0 Students determine whether a relation defined by a graph, a set of ordered pairs, or a symbolic expression is a function and justify the conclusion. Algebra ll - i10.1\n19.0 Students know the quadratic formula and are familiar with its proof by completing the square. Algebra l - e9.3\n20.0 Students use the quadratic formula to find the roots of a second-degree polynomial and to solve quadratic equations. Algebra l - e9.3\n21.0 Students graph quadratic functions and know that their roots are the x-intercepts. Algebra ll - i8.5\n22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points. Algebra l-e9.3\n23.0 Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity. Algebra l - e4.6\n24.0 Students use and know simple aspects of a logical argument. Geometry - g2.1\n24.1 Students explain the difference between inductive and deductive reasoning and identify and provide examples of each. N/A\n24.2 Students identify the hypothesis and conclusion in logical deduction. 24.3 Students use counterexamples to show that an assertion is false and recognize that a single counterexample is sufficient to refute an assertion. Geometry - g2.1\n25.0 Students use properties of the number system to judge the validity of results, to justify each step of a procedure, and to prove or disprove statements: Algebra l - e1.1\n25.1 Students use properties of numbers to construct simple, valid arguments (direct and indirect) for, or formulate counterexamples to, claimed assertions. Geometry - g2.1, g2.2, g2.3\n25.2 Students judge the validity of an argument according to whether the properties of the real number system and the order of operations have been applied correctly at each step. Algebra l - e1.1, e1.5, e1.6, e1.7, e3.1, e3.2, e3.4, e5.5, e6.5, e7.2, e7.3, e7.4, e8.2, e8.5, e10.3\n25.3 Given a specific algebraic statement involving linear, quadratic, or absolute value expressions or equations or inequalities, students determine whether the statement is true sometimes, always, or never. Algebra 1 - e3.2"
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https://www.arxiv-vanity.com/papers/0809.1501/ | [
"# Quantum Semi-Markov Processes\n\nHeinz-Peter Breuer Physikalisches Institut, Universität Freiburg, Hermann-Herder-Strasse 3, D-79104 Freiburg, Germany Hanse-Wissenschaftskolleg, Institute for Advanced Study, D-27753 Delmenhorst, Germany Bassano Vacchini Dipartimento di Fisica dell’Università di Milano and INFN Sezione di Milano, Via Celoria 16, I-20133 Milano, Italy\nMarch 19, 2023\n###### Abstract\n\nWe construct a large class of non-Markovian master equations that describe the dynamics of open quantum systems featuring strong memory effects, which relies on a quantum generalization of the concept of classical semi-Markov processes. General conditions for the complete positivity of the corresponding quantum dynamical maps are formulated. The resulting non-Markovian quantum processes allow the treatment of a variety of physical systems, as is illustrated by means of various examples and applications, including quantum optical systems and models of quantum transport.\n\n###### pacs:\n03.65.Yz,42.50.Lc,02.50.Ga,03.65.Ta\n\nThe analysis of the time evolution of open systems plays a central role in many applications of modern quantum theory, including quantum information science, quantum transport theory, quantum thermodynamics, and quantum process tomography and control (see e.g. general ). The state of an open quantum system that is coupled to the degrees of freedom of its surroundings is represented by a time-dependent density matrix . In the Markovian regime the dynamics is governed by a master equation of the relatively simple form\n\n ddtρ(t)=Lρ(t), (1)\n\n Lρ=−i[H,ρ]+∑α[AαρA†α−12{A†αAα,ρ}]. (2)\n\nThe Hamiltonian describes the coherent part of the time evolution and the are certain operators representing the various decay modes. The solution of Eq. (1) can be written in terms of a linear map that transforms the initial state into the state at time . The physical interpretation of this map requires that it preserves the trace and the positivity of the density matrix . According to general physical principles must be a completely positive (CP) map KRAUS ; TheWork . Hence, represents a CP dynamical semigroup known as quantum Markov process, whose generator has been proven GORINI-LINDBLAD to be of the form (2).\n\nThe quantum dynamics given by Eq. (2) has a clearcut connection to a classical Markov process for the case in which one has a closed system of equations for the populations in a fixed orthonormal basis of the open system’s Hilbert space, typically the energy eigenbasis. In fact, in this case one recovers the Pauli master equation,\n\n ddtPn(t)=∑m[ΓnmPm(t)−ΓmnPn(t)], (3)\n\nwhich describes a classical Markovian jump process with transition rates , justifying the notion of a quantum Markov process.\n\nThe most important physical assumption which underlies the master equation (1) is the validity of the Markov approximation of short environmental correlation times. If this approximation is violated non-Markovian dynamics emerges which is characterized by pronounced memory effects, finite revival times and non-exponential relaxation and decoherence. These effects can result from long-range correlation functions, from correlations and entanglement in the initial state, as well as from the neglection of extra degrees of freedom affecting the dynamics BUDINI ; nmgroup . As a consequence the theoretical treatment of non-Markovian quantum dynamics is generally extremely demanding. A widely used non-Markovian generalization of Eq. (1) is given by the integrodifferential equation\n\n ddtρ(t)=∫t0dτK(τ)ρ(t−τ). (4)\n\nIn this equation one takes into account quantum memory effects through the introduction of the memory kernel which means that the rate of change of the state at time depends on the states at previous times . Equations of the form (4) arise, for instance, by employing the standard Nakajima-Zwanzig projection operator technique NAKAJIMA-ZWANZIG . Obviously, the Markovian master equation (1) is obtained if the memory kernel is taken to be proportional to a -function, .\n\nIn order to be physically acceptable the superoperator appearing in Eq. (4) must grant the CP of the resulting quantum dynamical map . This is a very stringent requirement and, in fact, the general structural characterization of physically admissible memory kernels is an unsolved problem of central importance in the field of non-Markovian quantum dynamics BUDINI ; Daffer-Lidar . It has been realized recently that even the most simple and natural choices for the memory kernel can lead to unphysical results BARNETT ; BUDINI . To improve this situation we will construct a class of non-Markovian quantum master equations that arises naturally as a quantum mechanical generalization of classical semi-Markov processes Feller . The approach proposed here leads to important physical insights guiding the phenomenological determination of the memory kernel, and, at the same time, enables a compact formulation of sufficient conditions that guarantee the existence and the CP of the quantum dynamical map. Moreover, for a specific class of processes one can formulate CP conditions which are not only sufficient but also necessary.\n\nWe consider memory kernels with the general structure\n\n K(τ)ρ = −i[H(τ),ρ]−12∑α{A†α(τ)Aα(τ),ρ} (5) +∑αAα(τ)ρA†α(τ),\n\nthat is to say of the form given by Eq. (2) apart from the time dependence of the considered operators. As previously done in the Markovian case let us consider the situation in which the populations obey a closed system of equations of motion, which then takes the form\n\n ddtPn(t) = ∫t0dτ∑m[Wnm(τ)Pm(t−τ) (6) −Wmn(τ)Pn(t−τ)],\n\nwhere . This is the master equation for a general type of classical non-Markovian processes known as semi-Markov processes Feller . Thus, whenever the populations obey closed equations, Eq. (2) yields the classical Markovian master equation (3), while Eq. (4) with the kernel (5) leads under the same conditions to the generalized master equation (6) for a classical semi-Markov process. This justifies on the same footing as before the name quantum semi-Markov process.\n\nTo clarify the physical content of Eq. (6) let us consider as an example the situation in which the kernel functions factorize as , where and . The corresponding process can then be interpreted as describing a particle moving on a lattice with sites labelled by , where the are the probabilities for jumps from site to site . Jumps out of a given site take place after a certain waiting time that follows the waiting time distribution . The characteristic feature of semi-Markov processes is the fact that, by contrast to the Markovian case, need not be an exponential function, but can be any probability distribution, thus giving rise to memory effects. These waiting time distributions are uniquely determined by the functions according to the relation GILLESPIE\n\n fn(t)=∫t0dτkn(τ)gn(t−τ)≡(kn∗gn)(t), (7)\n\nwhere the function\n\n gn(t)=1−∫t0dτfn(τ) (8)\n\ndenotes the probability not to have left site by time , the so-called survival probability, and is the usual convolution product. The generalized master equation (6) therefore provides a physically acceptable time evolution for the populations , granting in particular their positivity, provided the functions allow an interpretation in terms of waiting time distributions GILLESPIE ; Esposito2008a .\n\nHowever, these classical conditions are clearly not enough to ensure the existence of a well-defined dynamics in the quantum case, and a general characterization at the quantum level can hardly be achieved. Therefore our next goal is the formulation of sufficient conditions that guarantee the CP of the dynamical map corresponding to the non-Markovian master equation defined by Eqs. (4) and (5), no longer assuming that closed equations for the populations exist. This map is defined by\n\n ddtV(t)=∫t0dτK(τ)V(t−τ), (9)\n\ntogether with the initial condition , with the identity map. We now employ ideas recently formulated in Ref. KOSSAKOWSKI , decomposing the memory kernel as , where is the CP map defined by\n\n B(τ)ρ=∑αAα(τ)ρA†α(τ), (10)\n\nand is given by the first line of (5). We further introduce the map as the solution of the equation\n\n ddtV0(t)=∫t0dτC(τ)V0(t−τ), (11)\n\nwith the initial condition . Considering the Laplace transforms of Eqs. (9) and (11) one obtains a resolvent-like identity for the dynamical map leading in the time domain to the equation\n\n V(t)=V0(t)+(V0∗B∗V)(t). (12)\n\nRegarding formally the superoperator as a perturbation and iterating Eq. (12) one finds that the full dynamical map can be represented as a series,\n\n V(t) = V0(t)+(V0∗B∗V0)(t) (13) +(V0∗B∗V0∗B∗V0)(t)+….\n\nDue to the fact that the set of CP maps is closed under addition and convolution, we can conclude from Eq. (13) that is CP if is CP. To bring this condition into an explicit form let us assume that the Hermitian operators and are diagonal in the time-independent orthonormal basis , that is and\n\n ∑αA†α(τ)Aα(τ)=∑nkn(τ)|n⟩⟨n|. (14)\n\nThen we can solve Eq. (11) to obtain\n\n V0(t)ρ(0)=∑nmgnm(t)|n⟩⟨n|ρ(0)|m⟩⟨m|, (15)\n\nwhere the functions are the solutions of\n\n ˙gnm(t)=−∫t0dτ[zn(τ)+z∗m(τ)]gnm(t−τ), (16)\n\ncorresponding to the initial conditions , and . To prove Eq. (15) one first shows that . Using this relation one easily demonstrates that the expression (15) indeed represents the desired solution of Eq. (11). It is important to notice that the functions do actually coincide with the survival probabilities introduced by Eq. (8).\n\nEmploying the Kraus representation KRAUS we see that the map given by Eq. (15) is CP if and only if the matrix with elements is positive,\n\n G(t)=(gnm(t))≥0. (17)\n\nHence, we arrive at a sufficient condition for CP: The quantum dynamical map corresponding to the non-Markovian master equation (4) with the memory kernel (5) is CP if the condition (17) is fulfilled. A necessary condition for (17) to hold is the positivity of the diagonal elements of , which are given by the survival probabilities . This necessary condition in turn implies the positivity of the functions according to Eq. (7), which can then be interpreted as true waiting time distributions. The positivity of the matrix therefore represents a natural quantum generalization of the classical conditions.\n\nWe illustrate the result (17) with the help of several examples, which all fall into the class of quantum semi-Markov processes introduced by means of Eqs. (4) and (5). A prototypical system showing strong non-Markovian behavior is a two-level atom interacting with a damped field mode described by the memory kernel\n\n K(τ)ρ = −iε(τ)[σ+σ−,ρ] (18) +k(τ)[σ−ρσ+−12{σ+σ−,ρ}].\n\nExcited and ground state are denoted by and , respectively, and are the corresponding raising and lowering operators. The index thus takes on the two values . For a positive function the memory kernel (18) is of the form specified above with , , , and . Hence, the matrix takes the form\n\n G(t)=(g++(t)g+−(t)g∗+−(t)1), (19)\n\nwith and determined by Eq. (16). Thus we see that the condition (17) for CP is equivalent to . The master equation corresponding to the memory kernel (18) can be solved analytically. One then finds that for this case the condition (17) is not only sufficient but also necessary for CP.\n\nA further very instructive example involving an infinite dimensional Hilbert space is the model of a quantum oscillator with non-Markovian damping studied in Ref. BARNETT . The memory kernel for this model reads\n\n K(τ)ρ=k(τ)[aρa†−12{a†a,ρ}], (20)\n\nwhere and , are the raising and lowering operators of the oscillator. This kernel is again of the form (5) with a single Lindblad operator . Here, the basis states are the number states of the oscillator, and . Solving Eq. (16) by means of a Laplace transformation, we find\n\n gnn(t)=e−γt/2[cosh(dnt/2)+γdnsinh(dnt/2)],\n\nwhere . For the necessary condition to hold must be real. This shows that condition (17) is certainly violated if . Because can be arbitrary large we conclude that condition (17) is never fulfilled. The interesting aspect of this example is the fact that the non-Markovian master equation indeed violates not only CP but also positivity. This fact has been demonstrated in BARNETT and clearly shows again the relevance of our CP conditions.\n\nMany further physical systems lead to a generalized master equation of the form introduced here if one applies the Nakajima-Zwanzig projection operator technique, such as the tight-binding quantum diffusion model studied in GASPARD , and the quantum transport model introduced in DIFBAL , which leads to a memory kernel of the form\n\n K(τ)ρ=k(τ)[12TρT†+12T†ρT−ρ]. (21)\n\nThis kernel describes the motion of an excitation in a modular system consisting of weakly coupled subunits labelled by the index , where represents the corresponding translation operator. The model features strong non-Markovian behavior and a transition from diffusive to ballistic quantum transport. The memory kernel is obviously of the form introduced above. The Hamiltonian contribution vanishes, , and all kernel functions are equal to each other, , which corresponds to the special case treated in Refs. KOSSAKOWSKI ; BUDINI with a loss term proportional to the identity operator. Equation (16) shows that also all matrix elements of are equal, , and, hence, condition (17) reduces to the condition . Clearly this condition leads to important restrictions on the form of the kernel function which is determined by the correlation function of the microscopic model.\n\nAs our final example we discuss memory kernels of the following general structure,\n\n K(τ)ρ = −i[H(τ),ρ]−12∑nkn(τ){|n⟩⟨n|,ρ}, (22) +∑nmπnmkm(τ)|n⟩⟨m|ρ|m⟩⟨n|\n\nof which (18) provides an example. For this memory kernel the coherences of the density matrix, i. e. the off-diagonal elements , , are simply given by . On the other hand, the diagonals of the density matrix, i. e. the populations obey a closed transport equation as in (6). It is remarkable that in this case one can go one step further to derive a condition for the CP which is not only sufficient but also necessary. To this end one writes the quantum dynamical map corresponding to the non-Markovian quantum master equation (4) with the memory kernel (22) in terms of the functions and of the conditional transition probabilities obeying the classical master equation (6). The quantity represents the probability that the particle is at site at time given that it started at site at time . With the help of the resulting expression for the map we then find the following result. Given a classical semi-Markov process, the quantum dynamical map is CP if and only if the condition\n\n ~G(t)=(~gnm(t))≥0 (23)\n\nis satisfied. Here, the off-diagonal elements of the matrix coincide with those of , while the diagonals of are given by the conditional transition probabilities, . Note that the probabilities are in fact in general greater than the corresponding survival probabilities , since the system can be in state at time both because it has not left it and because it has come back to the initial state. Eq. (23) thus provides a complete characterization of the CP of the class of quantum semi-Markov processes given by (22).\n\nBuilding on an analogy with classical semi-Markov processes we have constructed a large class of non-Markovian master equations with memory kernel and formulated sufficient conditions for the CP of the resulting quantum dynamical map. The latter impose strong restrictions on the structure of physically acceptable non-Markovian quantum master equations, which are particularly useful in phenomenological approaches. For a specific class of quantum semi-Markov processes necessary and sufficient conditions for CP have also been formulated. Important further developments of the theory should include the case of temporarily negative kernel functions and effects from correlations and entanglement in the initial state.\n\n###### Acknowledgements.\nOne of us (HPB) gratefully acknowledges a Fellowship of the Hanse-Wissenschaftskolleg, Delmenhorst.\n\n## References\n\n• (1) J. Gemmer, M. Michel, and G. Mahler, Quantum Thermodynamics, Lecture Notes in Physics, vol 657 (Springer, Berlin, 2004); S. Mancini, V. I. Man’ko, and H. M. Wiseman (eds.), J. Opt. B: Quantum Semiclass. Opt. 7 (2005); R. Alicki, D. A. Lidar, and P. Zanardi Phys. Rev. A 73, 052311 (2006); H. P. Breuer, J. Gemmer, M. Michel, and U. Schollwöck (eds.), Eur. Phys. J. Special Topics 151 (2007).\n• (2) V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, J. Math. Phys. 17, 821 (1976); G. Lindblad, Commun. Math. Phys. 48, 119 (1976).\n• (3) E. C. G. Sudarshan, P. M. Mathews, and J. Rau, Phys. Rev. 121, 920 (1961); K. Kraus, States, Effects, and Operations, Lecture Notes in Physics, vol 190 (Springer, Berlin, 1983).\n• (4) H. P. Breuer and F. Petruccione, The Theory of Open Quantum Systems (OUP, Oxford, 2007).\n• (5) A. A. Budini, Phys. Rev. A 69, 042107 (2004).\n• (6) H. P. Breuer, J. Gemmer, and M. Michel, Phys. Rev. E 73, 016139 (2006); H. P. Breuer, Phys. Rev. A 75, 022103 (2007); A. A. Budini, Phys. Rev. E 72, 056106 (2005); Phys. Rev. A 74, 053815 (2006); A. A. Budini and H. Schomerus, J. Phys. A: Math. Gen. 38, 9251 (2005); B. Vacchini, Phys. Rev. A 78, 022112 (2008).\n• (7) S. Nakajima, Progr. Theor. Phys. 20, 948 (1958); R. Zwanzig, J. Chem. Phys. 33, 1338 (1960).\n• (8) S. Daffer et al., Phys. Rev. A 70, 010304 (R) (2004); A. Shabani and D. A. Lidar, Phys. Rev. A 71, 020101 (R) (2005).\n• (9) S. M. Barnett and S. Stenholm, Phys. Rev. A 64, 033808 (2001).\n• (10) W. Feller, An Introduction to Probability Theory and its Applications. Vol. II, John Wiley & Sons Inc., New York, 1971; Proc. Natl. Acad. Sci. U.S.A. 51, 653 (1964); B. D. Hughes, Random Walks and Random Environments (OUP, New York, 1995).\n• (11) D. T. Gillespie, Phys. Lett. A 64, 22 (1977).\n• (12) M. Esposito and K. Lindenberg, Phys. Rev. E 77, 051119 (2008).\n• (13) A. Kossakowski and R. Rebolledo, eprint arXiv:0801.1057.\n• (14) M. Esposito and P. Gaspard, Phys. Rev. B 71, 214302 (2005).\n• (15) R. Steinigeweg, H. P. Breuer, and J. Gemmer, Phys. Rev. Lett. 99, 150601 (2007)."
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https://stackoverflow.com/questions/2461667/centering-strings-with-printf | [
"# Centering strings with printf()\n\nBy default, `printf()` seems to align strings to the right.\n\n``````printf(\"%10s %20s %20s\\n\", \"col1\", \"col2\", \"col3\");\n/* col1 col2 col3 */\n``````\n\nI can also align text to the left like this:\n\n``````printf(\"%-10s %-20s %-20s\", \"col1\", \"col2\", \"col3\");\n``````\n\nIs there a quick way to center text? Or do I have to write a function that turns a string like `test` into `(space)(space)test(space)(space)` if the text width for that column is 8?\n\nprintf by itself can't do the trick, but you could play with the \"indirect\" width, which specifies the width by reading it from an argument. Lets' try this (ok, not perfect)\n\n``````void f(char *s)\n{\nprintf(\"---%*s%*s---\\n\",10+strlen(s)/2,s,10-strlen(s)/2,\"\");\n}\nint main(int argc, char **argv)\n{\nf(\"uno\");\nf(\"quattro\");\nreturn 0;\n}\n``````\n• I tested this in some C++ code I wrote to create table, and it was truncating valid text, and I had to figure out why. I've provided an alternative answer based on how I resolved it. Nov 25, 2018 at 15:57\n• Correct, indeed I warned that the solution is \"not perfect\", due to hard coded constants it contains. My goal was just to give a hint, your solution completes the idea. Nov 26, 2018 at 18:21\n• The `*` expects an `int`, `strlen()` will return `size_t`. This function causes UB when `SIZE_MAX>INT_MAX`. Possible solution: Change it to `(int) ((10+strlen(s)/2)&INT_MAX)` Aug 13, 2021 at 20:36\n• Correct. In most cases casting strlen directly to int is enough. It's unlikely that a string that was meant to be centered exceeds INT_MAX in length. Aug 23, 2021 at 7:24\n\n@GiuseppeGuerrini's was helpful, by suggesting how to use print format specifiers and dividing the whitespace. Unfortunately, it can truncate text.\n\nThe following solves the problem of truncation (assuming the field specified is actually large enough to hold the text).\n\n``````void centerText(char *text, int fieldWidth) {\nint padlen = (fieldWidth - strlen(text)) / 2;\n}\n``````\n• note that the resulting total width will be different depending on if `fieldWidth - strlen(text)` is even or odd Feb 21, 2019 at 16:38\n• added a modified version that always prints `fieldWidth` characters (e.g. for use in tables etc.) Dec 4, 2019 at 14:29\n• The behavior is implementations defined when the result of `(fieldWidth - strlen(text)) / 2;` is out of the range of an `int`. Which can be the case when `SIZE_MAX>INT_MAX` and `strlen(text)>fieldWidth`, causing `(fieldWidth - strlen(text)) == SIZE_MAX-(strlen(text)-fieldWidth-1)`. Aug 13, 2021 at 20:43\n• @12431234123412341234123 Reminds me of the old adage \"Just because you can do something, doesn't mean you should.\" Aug 16 at 0:28\n\nThere is no `printf()` format specifier to centre text.\n\nYou will need to write your own function or locate a library which provides the functionality that you're looking for.\n\n• Your opportunity to go the extra 9 yards to provide a solution, eh? Nov 25, 2018 at 15:14\n\nYou may try write own function for this problem.\n\n``````/**\n* Returns a sting \"str\" centered in string of a length width \"new_length\".\n* Padding is done using the specified fill character \"placeholder\".\n*/\nchar *\nstr_center(char str[], unsigned int new_length, char placeholder)\n{\nsize_t str_length = strlen(str);\n\n// if a new length is less or equal length of the original string, returns the original string\nif (new_length <= str_length)\nreturn str;\n\nchar *buffer;\nunsigned int i, total_rest_length;\n\nbuffer = malloc(sizeof(char) * new_length);\n\n// length of a wrapper of the original string\ntotal_rest_length = new_length - str_length;\n\n// write a prefix to buffer\ni = 0;\nwhile (i < (total_rest_length / 2)) {\nbuffer[i] = placeholder;\n++i;\n}\nbuffer[i + 1] = '\\0';\n\n// write the original string\nstrcat(buffer, str);\n\n// write a postfix to the buffer\ni += str_length;\nwhile (i < new_length) {\nbuffer[i] = placeholder;\n++i;\n}\nbuffer[i + 1] = '\\0';\n\nreturn buffer;\n}\n``````\n\nResults:\n\n``````puts(str_center(\"A\", 0, '-')); // A\nputs(str_center(\"A\", 1, '-')); // A\nputs(str_center(\"A\", 10, '-')); // ----A-----\nputs(str_center(\"text\", 10, '*')); // ***text***\nputs(str_center(\"The C programming language\", 26, '!')); // The C programming language\nputs(str_center(\"The C programming language\", 27, '!')); // The C programming language!\nputs(str_center(\"The C programming language\", 28, '!')); // !The C programming language!\nputs(str_center(\"The C programming language\", 29, '!')); // !The C programming language!!\nputs(str_center(\"The C programming language\", 30, '!')); // !!The C programming language!!\nputs(str_center(\"The C programming language\", 31, '!')); // !!The C programming language!!!\n``````\n• Cool that it is c/stdlib only implementation for the standpoint of educating about C but otherwise at least replace the prefix/postfix loops with `memcpy()` or `bcopy()` Nov 27, 2018 at 13:55\n• And replace malloc with alloca.. ewww Apr 22, 2019 at 15:33\n• alloca wouldn't work in this instance, as the string being returned is in that memory - you don't want it be free'd at the point of the 'str_center' returns! But as Martin points out, as written this leaks memory. Apr 19, 2021 at 10:06\n\nThere are two solutions, the first is similar to the above, by placing macros in printf, and the second is a custom macro, which calculates the length of the formatted string in advance through snprintf, and then calls the printf function to output.\n\n``````#include <stdio.h>\n#include <string.h>\n\n#define LEFT(str, w) \\\n({int m = w + strlen(str); m % 2 ? (m + 1) / 2 : m / 2;})\n\n#define RIGHT(str, w) \\\n({ int m = w - strlen(str); m % 2 ? (m - 1) / 2 : m / 2; })\n\n#define STR_CENTER(str, width) \\\nLEFT(str, width), str, RIGHT(str, width), \"\"\n\n#define PRINTF_CENTER(width, start, fmt, end, ...) ({ \\\nint n = snprintf(NULL, 0, fmt, __VA_ARGS__); \\\nint m = width - n; \\\nint left = m % 2 ? (m + 1) / 2 : m / 2; \\\nint right = m % 2 ? (m - 1) / 2 : m / 2; \\\nprintf(start \"%*s\" fmt \"%*s\" end, left, \"\", \\\n__VA_ARGS__, right, \"\"); \\\n})\n\n#define MYFORMAT_CENTER(width, fmt, ...) \\\nPRINTF_CENTER(40, \"[\", fmt , \"]\\n\", __VA_ARGS__)\n\nint main(int argc, char const *argv[])\n{\nprintf(\"%*s%*s\\n\\n\", STR_CENTER(\"--- Hello World ---\", 40));\nprintf(\"[%*s%*s]\\n\", STR_CENTER(\"I am okay today\", 40));\n\nMYFORMAT_CENTER(40, \"%d, e is %f\", 1, 2.71828);\nMYFORMAT_CENTER(40, \"%d, pi is %f\", 2, 3.1415926);\nMYFORMAT_CENTER(40, \"%s %d.\", \"This is such a long string that it exceeds the given size:\", 40);\n\nreturn 0;\n}\n``````\n\nOutput:\n\n`````` --- Hello World ---\n\n[ I am okay today ]\n[ 1, e is 2.718280 ]\n[ 2, pi is 3.141593 ]\n[ This is such a long string that it exceeds the given size: 40. ]\n``````\n\nIll drop my 2 cents after dealing with similar issue of trying to center a table headers in a row with printf.\n\nThe following macros will need to be printed before/after the text and will align regardless of the length of the text itself. Notice that if we have odd length strings, we will not align as should(because the normal devision will result in missing space). Therefor a round up is needed, and I think this is the elegant way to solve that issue:\n\n``````#define CALC_CENTER_POSITION_PREV(WIDTH, STR) (((WIDTH + ((int)strlen(STR))) % 2) \\\n? ((WIDTH + ((int)strlen(STR)) + 1)/2) : ((WIDTH + ((int)strlen(STR)))/2))\n#define CALC_CENTER_POSITION_POST(WIDTH, STR) (((WIDTH - ((int)strlen(STR))) % 2) \\\n? ((WIDTH - ((int)strlen(STR)) - 1)/2) : ((WIDTH - ((int)strlen(STR)))/2))\n``````\n\nUsage example:\n\n``````printf(\"%*s%*s\" , CALC_CENTER_POSITION_PREV(MY_COLUMN_WIDTH, \"Header\")\n``````\n\nYes, you will either have to write your own function that returns \" test \" etc, e.g.\n\n``````printf(\"%s %s %s\", center(\"col1\", 10), center(\"col2\", 20), center(\"col3\", 20));\n``````\n\nOr you have a center_print function, something like the following:\n\n``````void center_print(const char *s, int width)\n{\nint length = strlen(s);\nint i;\nfor (i=0; i<=(width-length)/2; i++) {\nfputs(\" \", stdout);\n}\nfputs(s, stdout);\ni += length;\nfor (; i<=width; i++) {\nfputs(\" \", stdout);\n}\n}\n``````\n• The first suggestion: How can this be impl'd without leaking memory? Mar 31, 2017 at 13:57\n• If you preallocate some buffers based on some criteria that does not seem unreasonable (like for instance no more than 20 arguments will be centred for one printf, and none of the centred results will be longer than 200 bytes), you could let the center function just rotate buffers on each invocation. Apr 2, 2017 at 15:18\n\nA more compact version of PADYMKO's function above (which still leaks memory):\n\n``````char *str_center(char str[], unsigned int new_length, char placeholder)\n{\nsize_t str_length = strlen(str);\nchar *buffer;\n/*------------------------------------------------------------------\n* If a new length is less or equal length of the original string,\n* returns the original string\n*------------------------------------------------------------------*/\nif (new_length <= str_length)\n{\nreturn(str);\n}\nbuffer = malloc(sizeof(char) * (new_length + 1));\nmemset(buffer, placeholder, new_length);\nbuffer[new_length] = '\\0';\nbcopy(str, buffer + (( new_length - str_length) / 2), str_length);\nreturn(buffer);\n}\n``````\n\nThis sets the whole of newly allocated buffer to the padding character, null terminates that, and then drops the string to be centred into the middle of the buffer - no loops, or keeping track of where to copy to..\n\nIf you want to be able to use a `printf()` format string for that and you accept to be limited to the GNU clib, you can extend `printf()` with your own conversion specifier for centering a string with. Add the conversion specifier with `register_printf_function()`. See here for the documentation: https://www.gnu.org/software/libc/manual/html_node/Customizing-Printf.html The other answers already provide you with a solution on how to manually print a string in the center, which you still need when using your own conversion specifier.\n\nYou can use either of the following two options:\n\n``````char name[] = \"Name1\";\n\n//Option One\nprintf(\"%*s\", 40+strlen(name)/2, name, 40-strlen(name)/2, \"\");\nputs(\"\");//skip one line\n``````\n``````//Option two\nprintf(\"%*s\", 40+strlen(\"Name2\")/2, \"Name2\", 40-strlen(\"Name2\")/2, \"\");\n``````\n\nThe output is:\n\nName1(center)\nName2(center)\n\n• First of all, this looks like a ripoff of Giuseppe's answer, and, secondly, you have more printf() arguments than you have format specifiers to accommodate them. Did you even test this? Nov 25, 2018 at 15:13"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5075467,"math_prob":0.9026396,"size":1433,"snap":"2023-40-2023-50","text_gpt3_token_len":480,"char_repetition_ratio":0.11336599,"word_repetition_ratio":0.19485295,"special_character_ratio":0.448709,"punctuation_ratio":0.27439025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9632326,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T14:23:00Z\",\"WARC-Record-ID\":\"<urn:uuid:f9d77788-39ec-4afb-8b43-66d0b5152db0>\",\"Content-Length\":\"281371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8bc52ebf-ab12-4a8d-8366-0bbfaf8d1471>\",\"WARC-Concurrent-To\":\"<urn:uuid:5964cefc-d8e7-4577-83f3-933bae3c599c>\",\"WARC-IP-Address\":\"104.18.23.201\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/2461667/centering-strings-with-printf\",\"WARC-Payload-Digest\":\"sha1:F62WA2KMSZN33OV7NVTLRC7K7YSVJUAF\",\"WARC-Block-Digest\":\"sha1:YGOYEEOF3Z5T3NWAAO7KKNLEGJUFHKJ3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510676.40_warc_CC-MAIN-20230930113949-20230930143949-00424.warc.gz\"}"} |
https://www.numbersaplenty.com/232909 | [
"Search a number\nBaseRepresentation\nbin111000110111001101\n3102211111021\n4320313031\n524423114\n64554141\n71660015\noct706715\n9384437\n10232909\n11149a96\n12b2951\n1382021\n1460c45\n1549024\nhex38dcd\n\n232909 has 4 divisors (see below), whose sum is σ = 234016. Its totient is φ = 231804.\n\nThe previous prime is 232907. The next prime is 232919. The reversal of 232909 is 909232.\n\nIt is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also a brilliant number, because the two primes have the same length.\n\nIt is a cyclic number.\n\nIt is not a de Polignac number, because 232909 - 21 = 232907 is a prime.\n\nIt is an alternating number because its digits alternate between even and odd.\n\nIt is a Duffinian number.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (232901) by changing a digit.\n\nIt is a pernicious number, because its binary representation contains a prime number (11) of ones.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 129 + ... + 694.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (58504).\n\n2232909 is an apocalyptic number.\n\nIt is an amenable number.\n\n232909 is a deficient number, since it is larger than the sum of its proper divisors (1107).\n\n232909 is an equidigital number, since it uses as much as digits as its factorization.\n\n232909 is an odious number, because the sum of its binary digits is odd.\n\nThe sum of its prime factors is 1106.\n\nThe product of its (nonzero) digits is 972, while the sum is 25.\n\nThe square root of 232909 is about 482.6064649380. The cubic root of 232909 is about 61.5264829650.\n\nThe spelling of 232909 in words is \"two hundred thirty-two thousand, nine hundred nine\".\n\nDivisors: 1 283 823 232909"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9253408,"math_prob":0.9935881,"size":1719,"snap":"2022-40-2023-06","text_gpt3_token_len":463,"char_repetition_ratio":0.19183673,"word_repetition_ratio":0.0063492064,"special_character_ratio":0.32693428,"punctuation_ratio":0.1401099,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99638176,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T14:06:20Z\",\"WARC-Record-ID\":\"<urn:uuid:75d7c18e-b3c3-44c0-9ede-a475eadac951>\",\"Content-Length\":\"9226\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d35f2cb-3a71-4d55-939b-db205e028f75>\",\"WARC-Concurrent-To\":\"<urn:uuid:6b0341f8-4764-4399-b27c-dc51f77b9e70>\",\"WARC-IP-Address\":\"89.46.108.74\",\"WARC-Target-URI\":\"https://www.numbersaplenty.com/232909\",\"WARC-Payload-Digest\":\"sha1:4O2SI6T7NGDN2KP7NL7SONQ2XEVLEVJW\",\"WARC-Block-Digest\":\"sha1:K47IBNUI64X24WXTFJIECJC3MLKS7IHC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499634.11_warc_CC-MAIN-20230128121809-20230128151809-00021.warc.gz\"}"} |
https://forum.uipath.com/t/filter-relative-comparison/564085 | [
"# Filter relative comparison\n\nHello all,\n\nI use filter to comparison but I want relative comparison ± 0.05. Thank all\nAmount column: 10602.38\nTong Tien column: 10602.39\n\[email protected]\n\n\"[Amount] >= \" + (10602.38 - 0.05).ToString + \" And [Amount] <= \" + (10602.39 + 0.05).ToString\n\nHope this works\n\n1 Like\n\nHi,\n\nUnfortunately, FilterDataTable activity doesn’t support such calculation.\nCan you try as the following?\n\n``````arrDr = dt.AsEnumerable.Where(Function(r) Math.Abs(CDbl(r(\"Amount\"))-CDbl(r(\"Tong Tien\")))<=0.05).ToArray()\n``````\n\nNote: arrDr is DataRow array type.\n\nRegards,\n\n1 Like\n\nHi @Yoichi\n\nBecause i need to compare 2sheet. How can I use your code ?\nAssign: Column ‘Tổng Tiền’ does not belong to table DataTable.\n\n“Tổng Tiền” column in dt2\n“AMOUNT” column in DTfilter\n\nHi,\n\nDo you want to compare same row index data in both datatable?\n\nIf so,can you try the following expression. it’s unnecesary to use ForEachRow activity.\n\n``````dtFilter = dt.AsEnumerable.Where(Function(r,i) Math.Abs(CDbl(r(\"Amount\"))-CDbl(dt2.Rows(i)(\"Tong Tien\")))<=0.05).ToArray()\n``````\n\nRegards,\n\nhi, when I assign have some issues, and the condition for true? Thank you\n\nHi,\n\nSorry, It’s my mistake. Can you try to use arrDR (DataRow array type) as the following? (Because if there is no row in filtered datatable, it will throws error)\n\nRegards,"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5808557,"math_prob":0.6772615,"size":1120,"snap":"2023-40-2023-50","text_gpt3_token_len":336,"char_repetition_ratio":0.12007169,"word_repetition_ratio":0.24675325,"special_character_ratio":0.31339285,"punctuation_ratio":0.20588236,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9883428,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T17:10:44Z\",\"WARC-Record-ID\":\"<urn:uuid:2b66b875-30c6-4eb9-a6c9-a0030b296ee5>\",\"Content-Length\":\"63319\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:58f65765-b611-4e7c-aaac-07232a2b5876>\",\"WARC-Concurrent-To\":\"<urn:uuid:e7af8940-9a91-4535-9d81-9d2560105672>\",\"WARC-IP-Address\":\"184.105.99.75\",\"WARC-Target-URI\":\"https://forum.uipath.com/t/filter-relative-comparison/564085\",\"WARC-Payload-Digest\":\"sha1:P2FW64XOWZYYDUHGEU7OQ3LD2DHZOZIP\",\"WARC-Block-Digest\":\"sha1:CL5R4DS6XLAZTSGE75LOV74GQD6YVRGN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100229.44_warc_CC-MAIN-20231130161920-20231130191920-00123.warc.gz\"}"} |
https://studyrankersonline.com/85511/prism-of-glass-is-shown-in-figure-ray-incident-normally-on-one-face-is-totally-reflected | [
"# A prism of glass is shown in figure A ray incident normally on one face is totally reflected.\n\n592 views\nin Science\n\nA prism of glass is shown in figure A, ray incident normally on one face is totally reflected. If θ is 45°, the index of refraction of glass is _______ .\n\n(A) < √2 (B) > √2 (C) = 2 (D) None of these",
null,
"by Expert (37.9k points)\n\nThe correct option is (B) > √2.\n\nExplanation:\n\nFor total internal reflection i > θc ......(1)\n\nwhen θc is critical angle\n\nsin θc = (1/η) i.e. η = (1 / sin θc)\n\ni = 45°\n\nhence 45° > sin –1(1/η) From (1)\n\nsin 45° > (1/η)\n\n(1 / √2) > (1/η) hence η > √2"
] | [
null,
"https://www.studyrankersonline.com/",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.77055854,"math_prob":0.9813811,"size":439,"snap":"2021-43-2021-49","text_gpt3_token_len":178,"char_repetition_ratio":0.10344828,"word_repetition_ratio":0.0,"special_character_ratio":0.44646925,"punctuation_ratio":0.13461539,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97577935,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T22:17:21Z\",\"WARC-Record-ID\":\"<urn:uuid:537ba703-465e-4cad-b34c-a4973309ac04>\",\"Content-Length\":\"77870\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab4acd87-06d7-417c-962b-a01d50dc4ca8>\",\"WARC-Concurrent-To\":\"<urn:uuid:47f4fc61-af37-4c8a-a773-8e913b3f59b7>\",\"WARC-IP-Address\":\"139.59.83.9\",\"WARC-Target-URI\":\"https://studyrankersonline.com/85511/prism-of-glass-is-shown-in-figure-ray-incident-normally-on-one-face-is-totally-reflected\",\"WARC-Payload-Digest\":\"sha1:OO5JYDFCB6PIAY4T6VAWPDTMZOQ4DI3H\",\"WARC-Block-Digest\":\"sha1:ST4SV7IT6NQQCVOH7GRGQDWC7U2A7MEK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585281.35_warc_CC-MAIN-20211019202148-20211019232148-00333.warc.gz\"}"} |
https://naumathstat.github.io/seminars/acgtSpring2020/ | [
"# ACGT Seminar\n\nThe Algebra, Combinatorics, Geometry, and Topology (ACGT) Seminar meets on Tuesdays at 1:00-1:50PM in Room 221 of the Adel Mathematics Building.\n\n# Schedule Spring 2020\n\nNote that talks are listed in reverse chronological order.\n\nNo meeting March 3 or March 10.\n\n### Hypergraphic polymatroids and Dilworth truncation\n\nDate: January 28, February 4, February 11, February 18, February 25\n\nSpeaker: Michael Falk (NAU)\n\nAbstract: Matroids capture the combinatorial structure of vector or line configurations (or, equivalently, linear subspaces). Graphs (allowing loops and multiple edges) naturally give rise to line configurations and matroids, which reflect the graph structure to a large degree. For instance, planarity of a graph is detected by realizability (over the field of two elements) of the dual matroid. Whitney’s 2-isomorphism theorem states that graphs with isomorphic matroids are related by splitting, joining, or twisting.\n\nPolymatroids capture the combinatorial structure of configurations of subspaces, and hypergraphs naturally give rise to such. The purpose of these talks is to examine the generalization of Whitney’s Theorem to hypergraphs, due to Vertigan and Whittle, which involves a very interesting operation on subspace arrangements or polymatroids known as Dilworth truncation.\n\n### Binary Linear Codes from Graphs\n\nDate: January 14, January 21\n\nSpeaker: Sudipta Mallik (NAU)\n\nAbstract: A binary linear code of length $n$ is a subspace of $\\mathbb{F}_2^n$. I will start with a brief introduction to binary linear codes. Then I will present a construction of binary linear codes from graphs, in particular, by the generator matrix $[I_n\\mid A]$ where A is the adjacency matrix of a graph on n vertices. Finally a graph theoretic interpretation of the minimum distance of such codes will be given."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92014754,"math_prob":0.8586755,"size":1697,"snap":"2022-27-2022-33","text_gpt3_token_len":391,"char_repetition_ratio":0.10277614,"word_repetition_ratio":0.007905139,"special_character_ratio":0.20860341,"punctuation_ratio":0.13029316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96542716,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T00:24:04Z\",\"WARC-Record-ID\":\"<urn:uuid:1e02d77e-c0bb-4ccc-b56e-ec6e3dd18df8>\",\"Content-Length\":\"5898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:591ae764-3eb9-40d2-a5fa-ebef507bf78f>\",\"WARC-Concurrent-To\":\"<urn:uuid:c2c0fdf3-13c2-45be-9e97-d6806cc42bfa>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://naumathstat.github.io/seminars/acgtSpring2020/\",\"WARC-Payload-Digest\":\"sha1:XUXHOKP6PQLRVI27733SIX6QVH53PSKS\",\"WARC-Block-Digest\":\"sha1:N6KZI3YMC6ET5SX7U2FSSWA7L7ADBNXV\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104506762.79_warc_CC-MAIN-20220704232527-20220705022527-00264.warc.gz\"}"} |
https://www.crio.do/projects/python-preprocessor-cli/ | [
"#### Objective\n\nYou will implement a Command Line Interface (CLI) which will preprocess your dataset and save your time.\n\n#### Project Context\n\nMachine Learning is a subset of the larger field of artificial intelligence (AI) that focuses on teaching computers how to learn without the need to be programmed for specific tasks. In fact, the key idea behind ML is that it is possible to create algorithms that learn from and make predictions on the data.\n\nExamples of Machine Learning are present everywhere including the spam filter that flags messages in your email, the recommendation engine Netflix uses to suggest content you might like, and the self-driving cars being developed by Google and other companies.\n\nBut before applying Machine Learning on any dataset, you need to convert it in such a way that the algorithms could understand the dataset. These steps are preprocessing steps.\n\nData preprocessing is an integral step in Machine Learning as the quality of data and the useful information that can be derived from it directly affects the ability of your model to learn; therefore, it is extremely crucial that you preprocess your data before feeding it into your model.\n\nOne more advantage of preprocessing is that it is considered time consuming for many machine learning developers. This simple CLI tool will save your time so that you can utilize it in applying different machine learning algorithms.\n\nYou will apply the following preprocessing steps:\n\n• Handling NULL values\n• Encoding Categorical Data\n• Feature Scaling\n\n#### Product Architecture\n\nThe Product Architecture consists of 6 parts which are as follows:",
null,
"#### High-level approach\n\nThis project consists of the following milestones:\n\n• Input the Dataset\n• Data Description\n• Handling NULL Values\n• Encoding Categorical Data\n• Feature Scaling\n\nPandas and scikit learn will be used throughout the project to perform the preprocessig steps.\n\nThe desired end result of this project is like this:\n\n#### Objective\n\nYou will implement a Command Line Interface (CLI) which will preprocess your dataset and save your time.\n\n#### Project Context\n\nMachine Learning is a subset of the larger field of artificial intelligence (AI) that focuses on teaching computers how to learn without the need to be programmed for specific tasks. In fact, the key idea behind ML is that it is possible to create algorithms that learn from and make predictions on the data.\n\nExamples of Machine Learning are present everywhere including the spam filter that flags messages in your email, the recommendation engine Netflix uses to suggest content you might like, and the self-driving cars being developed by Google and other companies.\n\nBut before applying Machine Learning on any dataset, you need to convert it in such a way that the algorithms could understand the dataset. These steps are preprocessing steps.\n\nData preprocessing is an integral step in Machine Learning as the quality of data and the useful information that can be derived from it directly affects the ability of your model to learn; therefore, it is extremely crucial that you preprocess your data before feeding it into your model.\n\nOne more advantage of preprocessing is that it is considered time consuming for many machine learning developers. This simple CLI tool will save your time so that you can utilize it in applying different machine learning algorithms.\n\nYou will apply the following preprocessing steps:\n\n• Handling NULL values\n• Encoding Categorical Data\n• Feature Scaling\n\n#### Product Architecture\n\nThe Product Architecture consists of 6 parts which are as follows:",
null,
"#### High-level approach\n\nThis project consists of the following milestones:\n\n• Input the Dataset\n• Data Description\n• Handling NULL Values\n• Encoding Categorical Data\n• Feature Scaling\n\nPandas and scikit learn will be used throughout the project to perform the preprocessig steps.\n\nThe desired end result of this project is like this:\n\n#### Input the Dataset\n\nThere are several types of Machine Learning such as Supervised learning, Unsupervised learning etc. Here, you are writing python scripts to make preprocessed dataset for performing supervised learning.\n\nSupervised learning consists of mapping input data (independent variables) to known targets (dependent variable), which humans have provided. Predicting house prices is a good example.\n\nImportant: For simultaneously testing out our application you will be performing the preprocessing on a very popular ML dataset - Titanic survival Dataset. You need to download train.csv dataset from the mentioned website.\n\nThe idea of this milestone is that you are correctly taking the input of the dataset.\n\n#### Requirements\n\n• This is a python application and the project structure is shown below.",
null,
"• Make a main class for the Project where all below functions will be defined -\n• A function that only takes a ‘.csv’ input from the command line.\n• A function that chooses a target (dependent) variable and removes it from the dataset, so that you can start preprocessing on all the independent variables.\n• Each task will have a separate class and will be called by their respective object.\n• Handle all the exceptions in the input.",
null,
"### References\n\n#### Expected Outcome\n\nAt the end of this milestone, the project should work like this.\n\n#### Data Description\n\nNow that you are done with the initial step, you can implement various components of your project.\n\nIn this milestone, you need to implement the functionality that will enable users to describe the dataset’s properties like mean, max, standard deviation etc.\n\nThe idea of this milestone is that you can correctly show some basic statistical details (mean, standard deviation, percentiles, total number of values, maximum, minimum), datatype of columns of the dataset using methods provided by pandas library.\n\n#### Requirements\n\n• Make a separate class for Data Description where all below functions will be defined -\n• A function that shows properties (mean, standard deviation, percentiles, total number of values, maximum, minimum) of each numeric column. This function should also show the datatypes along with the null value count of each column.\n• A function that shows the property of any specific column.\n• A numeric column should show properties like mean, standard deviation, percentiles, total number of values, maximum and minimum.\n• A string column should show properties like total number of values and number of distinct values.\n• A function that takes a number of rows ‘n’ as input and prints the dataset.\n• Handle all the exceptions in the input.",
null,
"### References\n\n#### Expected Outcome\n\nAt the end of this milestone, the project should work like this.\n\n#### Handling NULL Values\n\nThe next step of data preprocessing is to handle missing data in the datasets. If your dataset contains some missing data, then it may create a huge problem for your machine learning model. Hence it is necessary to handle missing values present in the dataset.\n\nThe handling of missing values is also called Data Imputation.\n\nThe idea of this milestone is to remove all the NULL values from the dataset.\n\n#### Requirements\n\n• Make a separate class for Data Imputation where all below functions will be defined -\n• A function that removes the specified column.\n• Three respective functions that fill the null values in a column with mean, median and mode.\n• Handle all the exceptions in the input.",
null,
"### References\n\n#### Expected Outcome\n\nAt the end of this milestone, the project should work like this.\n\n#### Encoding Categorical Data\n\nCategorical data is data which has some categories. Machine learning models completely works on mathematics and numbers, but if your dataset would have a categorical variable, then it may create trouble while building the model. So, it is necessary to encode these categorical variables into numbers.\n\nThe idea of this milestone is to map all the categorical columns into numbers.\n\n#### Requirements\n\n• Make a separate class for Encoding Categorical data where all below functions will be defined -\n• A function that shows all the Categorical columns present in the dataset\n• A function that performs one hot encoding for a categorical column.\n• Handle all the exceptions in the input.",
null,
"### References\n\n#### Expected Outcome\n\nAt the end of this milestone, the project should work like this.\n\n#### Feature Scaling\n\nFeature scaling is a method used to normalize the range of independent variables or columns of data. It is done to handle highly varying magnitudes among different columns.\n\nIf feature scaling is not done, then a machine learning algorithm tends to weigh greater values, higher and consider smaller values as the lower values, regardless of the unit of the values. To avoid this, feature scaling is done.\n\nThere are 2 main ways of doing feature scaling:\n\n• Normalization\n• Standardization\n\nThe idea of this milestone is that you are able to correctly scale the dataset.\n\n#### Requirements\n\n• Make a separate class for Feature Scaling where all below functions will be defined -\n• A function that performs normalization of any specific column or group of columns.\n• A function that performs standardization of any specific column or group of columns.\n• Handle all the exceptions in the input.",
null,
"### References\n\n#### Expected Outcome\n\nAt the end of this milestone, the project should work like this.\n\nAs all the preprocessing is done, you can implement the functionality to download the preprocessed dataset.\n\nThe idea of this milestone is that you can correctly download the dataset in the correct file format.\n\n#### Requirements",
null,
""
] | [
null,
"https://storage.googleapis.com/crio-content-container-assets/ME_ME_PROJECT_PREP_CLI_MODULE_ME_PROJECT_PREP_CLI_MODULE_PREP_CLI_product-architecture.png",
null,
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null,
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null,
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null,
"https://storage.googleapis.com/crio-content-container-assets/ME_ME_PROJECT_PREP_CLI_MODULE_ME_PROJECT_PREP_CLI_MODULE_PREP_CLI_step2-description.png",
null,
"https://storage.googleapis.com/crio-content-container-assets/ME_ME_PROJECT_PREP_CLI_MODULE_ME_PROJECT_PREP_CLI_MODULE_PREP_CLI_step3-imputation.png",
null,
"https://storage.googleapis.com/crio-content-container-assets/ME_ME_PROJECT_PREP_CLI_MODULE_ME_PROJECT_PREP_CLI_MODULE_PREP_CLI_step4-encoding_categorical.png",
null,
"https://storage.googleapis.com/crio-content-container-assets/ME_ME_PROJECT_PREP_CLI_MODULE_ME_PROJECT_PREP_CLI_MODULE_PREP_CLI_step5-feature_scaling.png",
null,
"https://storage.googleapis.com/crio-content-container-assets/ME_ME_PROJECT_PREP_CLI_MODULE_ME_PROJECT_PREP_CLI_MODULE_PREP_CLI_step6-download.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.914605,"math_prob":0.7936518,"size":6991,"snap":"2023-40-2023-50","text_gpt3_token_len":1288,"char_repetition_ratio":0.1509947,"word_repetition_ratio":0.61613476,"special_character_ratio":0.18166214,"punctuation_ratio":0.09272582,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9587907,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,8,null,8,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T08:12:38Z\",\"WARC-Record-ID\":\"<urn:uuid:4852a44e-bd12-452f-895c-241de5b4852f>\",\"Content-Length\":\"124000\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42711804-5099-447b-8015-db3189fc07c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:95ad2c6d-4971-4399-9ad3-aa5fe46bf119>\",\"WARC-IP-Address\":\"52.85.151.114\",\"WARC-Target-URI\":\"https://www.crio.do/projects/python-preprocessor-cli/\",\"WARC-Payload-Digest\":\"sha1:3CE7KEIYMG2QOKKKRZRVNDFDKNWYZUSL\",\"WARC-Block-Digest\":\"sha1:EXA6JUETBM2TRYEFJQICSIDZKFEUM7WO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100489.16_warc_CC-MAIN-20231203062445-20231203092445-00819.warc.gz\"}"} |
http://kali.kaysen.me/kalitools/powerfuzzer_zh.html | [
"## Powerfuzzer包装说明\n\nPowerfuzzer是基于从众多安全资源和网站上聚集了许多可供其他开放源代码模糊器和信息的高度自动化,完全可定制的web模糊器(HTTP协议的应用程序的fuzzer)。它被设计成用户友好的,现代化的,有效的工作。\n\n• 跨站点脚本(XSS)\n• 注射剂(SQL中,LDAP,代码,命令和XPATH)\n• CRLF\n• HTTP 500状态(通常是指示是否有误/安全漏洞(包括图表)缓冲区溢出)\n\nPowerfuzzer首页 | 卡利Powerfuzzer回购\n\n• 作者:马辛·科兹洛夫斯基\n• 许可:GPLv3的\n\nWeb应用程序漏洞扫描。\n\n### Powerfuzzer用法示例\n\n`[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */:~# powerfuzzer`"
] | [
null
] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.8746025,"math_prob":0.88898563,"size":460,"snap":"2020-10-2020-16","text_gpt3_token_len":334,"char_repetition_ratio":0.18421052,"word_repetition_ratio":0.0,"special_character_ratio":0.17608696,"punctuation_ratio":0.08163265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9537122,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-03T05:26:55Z\",\"WARC-Record-ID\":\"<urn:uuid:ae2e1d21-c342-4d8d-a5b5-a8b96a191aad>\",\"Content-Length\":\"3306\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dc27a0b3-42f1-4e73-af0e-e2bcec9da71e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e3b0b172-743e-42c0-81c9-87dcbad07029>\",\"WARC-IP-Address\":\"101.200.135.227\",\"WARC-Target-URI\":\"http://kali.kaysen.me/kalitools/powerfuzzer_zh.html\",\"WARC-Payload-Digest\":\"sha1:53RFWRIBTHU3OCQXB2CZ6BLPZONLGKWZ\",\"WARC-Block-Digest\":\"sha1:CDSJXF6H4D6C5ZXP6X2WH6BTIQUIMLA2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370510287.30_warc_CC-MAIN-20200403030659-20200403060659-00106.warc.gz\"}"} |
https://socratic.org/questions/a-line-segment-with-endpoints-at-1-2-and-6-7-is-rotated-clockwise-by-pi-2-what-a | [
"# A line segment with endpoints at (1 , -2 ) and (6, -7 ) is rotated clockwise by pi/2. What are the new endpoints of the line segment?\n\nFeb 22, 2018\n\nNew coordinates of the end points are\n\nA’((-2),(-1)) and B’((-7),(-6))\n\n#### Explanation:\n\n$A \\left(1 , - 2\\right) , B \\left(6 , - 7\\right)$. Rotted clockwise by $\\frac{\\pi}{2}$",
null,
"A((1),(-2)) => A’((-2),(-1)) IV to III quadrant.\n\nB((6),(-7)) => B’((-7),(-6)) IV to III quadrant."
] | [
null,
"https://useruploads.socratic.org/oZYbQRzaTjCwH9zctC06_ECD0D721-0026-4B6C-8948-AA966ED22DEA.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7582403,"math_prob":0.99960166,"size":340,"snap":"2023-40-2023-50","text_gpt3_token_len":86,"char_repetition_ratio":0.116071425,"word_repetition_ratio":0.0,"special_character_ratio":0.27941176,"punctuation_ratio":0.092307694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9573386,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T09:46:49Z\",\"WARC-Record-ID\":\"<urn:uuid:f05a7043-a11a-49a3-8e7c-06ea8d2d3020>\",\"Content-Length\":\"33772\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:38fbfebf-d5bb-4ab9-ba43-dee5b592853a>\",\"WARC-Concurrent-To\":\"<urn:uuid:97ddfbe2-4343-4e8a-af0f-5a153ca217b9>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/a-line-segment-with-endpoints-at-1-2-and-6-7-is-rotated-clockwise-by-pi-2-what-a\",\"WARC-Payload-Digest\":\"sha1:COSITLJA75VYWRLK6ZIJA6CAOL6QSUGN\",\"WARC-Block-Digest\":\"sha1:HOKWVXVKUX6DVEDDMUNIKVX3UIEKJIO6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233508959.20_warc_CC-MAIN-20230925083430-20230925113430-00869.warc.gz\"}"} |
https://socratic.org/questions/if-you-roll-two-dice-what-is-the-probability-of-rolling-a-number-less-than-5-and | [
"# If you roll two dice, what is the probability of rolling a number less than 5 and another odd number?\n\nNov 6, 2015\n\nBecause we see the word AND we will have to multiply, but see the \"but\" later on.\n\n#### Explanation:\n\nThere are 6 possible outcomes:\n\n$P \\left(< 5\\right) = \\frac{4}{6} = \\frac{2}{3}$\n\n$P \\left(o \\mathrm{dd}\\right) = \\frac{3}{6} = \\frac{1}{2}$\n\nMultiplying we get $\\frac{2}{3} \\cdot \\frac{1}{2} = \\frac{2}{6} = \\frac{1}{3}$\n\nBUT!\nThis only counts for the first die giving the under 5, and the second die giving the odd. The other way around has the same probability.\nIt's either..or , so we add .\n$\\frac{1}{3} + \\frac{1}{3} = \\frac{2}{3}$"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7985205,"math_prob":0.99995816,"size":567,"snap":"2022-27-2022-33","text_gpt3_token_len":137,"char_repetition_ratio":0.09058615,"word_repetition_ratio":0.0,"special_character_ratio":0.23809524,"punctuation_ratio":0.12605043,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999794,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T16:05:35Z\",\"WARC-Record-ID\":\"<urn:uuid:ca7ba46e-5ee7-4fc2-beb6-30a1f5ba6117>\",\"Content-Length\":\"33755\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e57dc174-c6fc-4d99-b265-33a0b8e0b73a>\",\"WARC-Concurrent-To\":\"<urn:uuid:366de50c-f162-4c45-98d6-5042ff230291>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/if-you-roll-two-dice-what-is-the-probability-of-rolling-a-number-less-than-5-and\",\"WARC-Payload-Digest\":\"sha1:334BQC5G6GSQBKUM4CDFFN6JEXHS4G3J\",\"WARC-Block-Digest\":\"sha1:R3FHLC2GFOUQKWOB6V4SBC3NBGDKDNS2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104585887.84_warc_CC-MAIN-20220705144321-20220705174321-00159.warc.gz\"}"} |
http://mechdesigner.support/md-qst13_3-rocker-rotating-rocker.htm | [
" Getting Started Tutorials - MechDesigner > Tutorial 13: Forces: Introduction > Step 13.3: Rotating Rocker\n\n# Step 13.3: A Rocker\n\n### A Rotating Rocker – a Crank\n\n We connect a Linear-Motion FB to rotate the Rocker with a constant angular-velocity. This Step helps to understand: 1.Centripetal Acceleration and Force from the circular motion of the Center-of-Mass 2.Superposition of Gravitational Force and Centripetal Force 3.Why the Moment of the Centripetal-Force equals zero. STEP 1: Delete the Spring FB or Open the Spring FB dialog-box and make the Spring-Rate = 0 N/mm STEP 2: Add a Linear-Motion FB STEP 3: Connect a wire from the output-connector of the Linear-Motion FB to the input-connector of the Motion-Dimension FB. The Machine Speed Setting is 60RPM, or 1 Cycle/second = 2π radians/second. The mechanism below shows the forces that act at the Pin-Joint - the rotation axis of the Rocker.",
null,
"Addition of Vertical Forces acting on the Rocker (Point 2) : (↑+ve). ∑FV=0 : R2V(N) - 1(kg)*9.807(m/s/s) = 0; R2V = 9.807N (upwards) Addition of Horizontal Forces acting on the Rocker (Point 2) : (→ +ve). ∑FV=0 : +R2H(N) + 1(kg)*0.1(m)*(2π)2(1/s/s) = 0; R2H = -3.948N (to the left) Moments are: •An Anti-clockwise Moment that acts on the Rocker is 0.98Nm - as before. •The Centripetal Force acts through the center of rotation. It does not apply a moment to the Arm. Hence, the Moment is the same. The image to the left is the same as the image above. I have added to the image the Horizontal and Vertical Force Vectors that ACT-ON the Rocker The two components are: 1.Reaction to the Gravitational Force - it acts on the Rocker. Equal to 9.81N 2.Centripetal Force acts on the Rocker to the left. Equal to 3.948N The forces are perpendicular(⊥) when the Rocker is horizontal. Hence, we can use Pythagoras, to give 10.571N Ftotal = √(Fg2 + Fc2) (Total Force = SQRT(SQR(Gravitation Force) + SQR(Centripetal Force)) = SQRT((m.g)2 + (m.r.ω2)2) Total Force = √((1*g)2 + (1*0.1*(2π)2)2) = √(9.812 + 3.9482) = 10.571N",
null,
""
] | [
null,
"http://mechdesigner.support/gst-rocker-forces-10.png",
null,
"http://mechdesigner.support/gst-rocker-forces-11.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7709826,"math_prob":0.99272776,"size":1941,"snap":"2022-27-2022-33","text_gpt3_token_len":623,"char_repetition_ratio":0.15229736,"word_repetition_ratio":0.025,"special_character_ratio":0.30602783,"punctuation_ratio":0.13625304,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986768,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-01T04:12:18Z\",\"WARC-Record-ID\":\"<urn:uuid:914a318e-a5f7-4850-8e7f-ccb46e6cdb6e>\",\"Content-Length\":\"18855\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3061195d-afb3-4a67-8ba2-c2cd38a3b676>\",\"WARC-Concurrent-To\":\"<urn:uuid:7d026eac-f0b3-49fe-ad99-56db1002d68c>\",\"WARC-IP-Address\":\"212.67.221.235\",\"WARC-Target-URI\":\"http://mechdesigner.support/md-qst13_3-rocker-rotating-rocker.htm\",\"WARC-Payload-Digest\":\"sha1:3Q5RZG54FWCSCHDUV6OFVI374X7RAEQL\",\"WARC-Block-Digest\":\"sha1:SDHUBTD4XAHU7RXUCHWN3MQS5JT67OM7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103920118.49_warc_CC-MAIN-20220701034437-20220701064437-00645.warc.gz\"}"} |
https://developers.google.com/earth-engine/guides/resample | [
"# Resampling and Reducing Resolution\n\nAs noted in the Projections doc, Earth Engine performs nearest neighbor resampling by default during reprojection. You can change this behavior with the `resample()` or `reduceResolution()` methods. Specifically, when one of these methods is applied to an input image, any required reprojection of the input will be done using the indicated resampling or aggregation method.\n\n## Resampling\n\n`resample()` causes the indicated resampling method (`'bilinear'` or `'bicubic'`) to be used at the next reprojection. Since inputs are requested in the output projection, an implicit reprojection may happen before any other operation on the input. For this reason, call `resample()` directly on the input image. Consider the following simple example:\n\n```// Load a Landsat image over San Francisco, California, UAS.\nvar landsat = ee.Image('LANDSAT/LC08/C01/T1_TOA/LC08_044034_20160323');\n\n// Set display and visualization parameters.\nMap.setCenter(-122.37383, 37.6193, 15);\nvar visParams = {bands: ['B4', 'B3', 'B2'], max: 0.3};\n\n// Display the Landsat image using the default nearest neighbor resampling.\n// when reprojecting to Mercator for the Code Editor map.\n\n// Force the next reprojection on this image to use bicubic resampling.\nvar resampled = landsat.resample('bicubic');\n\n// Display the Landsat image using bicubic resampling.\n```\n\nNote that the `'bicubic'` resampling results in the output pixels appearing smooth relative to the original image (Figure 1).\n\nThe order of operations for this code sample is diagrammed in Figure 2. Specifically, the implicit reprojection to the maps mercator projection takes place with the resampling method specified on the input image.",
null,
"Figure 2. Flow chart of operations when resample() is called on the input image prior to display in the Code Editor. Curved lines indicate the flow of information to the reprojection: specifically, the output projection, scale and resampling method to use.\n\n## Reduce Resolution\n\nSuppose that instead of resampling during reprojection, your goal is to aggregate pixels to larger pixels in a different projection. This is useful when comparing image datasets at different scales, for example 30-meter pixels from a Landsat-based product to coarse pixels (higher scale) from a MODIS-based product. You can control this aggregation process with the `reduceResolution()` method. As with `resample()`, call `reduceResolution()` on the input, in order to affect the next reprojection of the image. The following example uses `reduceResolution()` to compare forest cover data at 30-meters resolution to a vegetation index at 500-meters resolution:\n\n```// Load a MODIS EVI image.\nvar modis = ee.Image(ee.ImageCollection('MODIS/006/MOD13A1').first())\n.select('EVI');\n\n// Display the EVI image near La Honda, California.\nMap.setCenter(-122.3616, 37.5331, 12);\nMap.addLayer(modis, {min: 2000, max: 5000}, 'MODIS EVI');\n\n// Get information about the MODIS projection.\nvar modisProjection = modis.projection();\nprint('MODIS projection:', modisProjection);\n\n// Load and display forest cover data at 30 meters resolution.\nvar forest = ee.Image('UMD/hansen/global_forest_change_2015')\n.select('treecover2000');\nMap.addLayer(forest, {max: 80}, 'forest cover 30 m');\n\n// Get the forest cover data at MODIS scale and projection.\nvar forestMean = forest\n// Force the next reprojection to aggregate instead of resampling.\n.reduceResolution({\nreducer: ee.Reducer.mean(),\nmaxPixels: 1024\n})\n// Request the data at the scale and projection of the MODIS image.\n.reproject({\ncrs: modisProjection\n});\n\n// Display the aggregated, reprojected forest cover data.\nMap.addLayer(forestMean, {max: 80}, 'forest cover at MODIS scale');\n```\n\nIn this example, note that the output projection is explicitly set with `reproject()`. During the reprojection to the MODIS sinusoidal projection, rather than resampling, the smaller pixels are aggregated with the specified reducer (`ee.Reducer.mean()` in the example). This sequence of operations is illustrated in Figure 3. Although this example uses `reproject()` to help visualize the effect of `reduceResolution()`, most scripts don't need to explicitly reproject; see the warning here.",
null,
"Figure 3. Flow chart of operations when reduceResolution() is called on an input image prior to reproject(). Curved lines indicate the flow of information to the reprojection: specifically, the output projection, scale and pixel aggregation method to use.\n\nNote that a second reprojection occurs (implicitly) to display the data on the Code Editor map. Visually inspect the results and observe the correspondence between the pixels from the MODIS layer and the forest cover data reprojected to MODIS scale and projection. In general, you should rarely need to explicitly `reproject()` in Earth Engine.\n\n### Pixel weights for `ReduceResolution`\n\nThe weights of pixels used during the `reduceResolution()` aggregation process are based on the overlap between the smaller pixels being aggregated and the larger pixels specified by the output projection. This is illustrated in Figure 4.",
null,
"Figure 4. Input pixels (black) and output pixel (blue) for reduceResolution().\n\nThe default behavior is that input pixel weights are computed as the fraction of the output pixel area covered by the input pixel. In the diagram, the output pixel has area a, the weight of the input pixel with intersection area b is computed as b/a and the weight of the input pixel with intersection area c is computed as c/a. This behavior can result in unexpected results when using a reducer other than the mean reducer. For example, to compute forested area per pixel, use the mean reducer to compute the fraction of a pixel covered, then multiply by area (instead of computing areas in the smaller pixels then adding them up with the sum reducer):\n\n```// Compute forest area per MODIS pixel.\nvar forestArea = forest.gt(0)\n// Force the next reprojection to aggregate instead of resampling.\n.reduceResolution({\nreducer: ee.Reducer.mean(),\nmaxPixels: 1024\n})\n// The reduce resolution returns the fraction of the MODIS pixel\n// that's covered by 30 meter forest pixels. Convert to area\n// after the reduceResolution() call.\n.multiply(ee.Image.pixelArea())\n// Request the data at the scale and projection of the MODIS image.\n.reproject({\ncrs: modisProjection\n});\nMap.addLayer(forestArea, {max: 500 * 500}, 'forested area at MODIS scale');\n```"
] | [
null,
"https://developers.google.com/earth-engine/images/Resample.png",
null,
"https://developers.google.com/earth-engine/images/ReduceResolution.png",
null,
"https://developers.google.com/earth-engine/images/ReduceResolution_weights.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.76309055,"math_prob":0.85707366,"size":5804,"snap":"2020-34-2020-40","text_gpt3_token_len":1341,"char_repetition_ratio":0.16896552,"word_repetition_ratio":0.08043218,"special_character_ratio":0.23104756,"punctuation_ratio":0.15804878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96205556,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,7,null,7,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T20:13:20Z\",\"WARC-Record-ID\":\"<urn:uuid:2764b8e0-52f6-4225-b669-67eeb357a310>\",\"Content-Length\":\"86098\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:73caa1fb-7420-4c5b-a250-368474448699>\",\"WARC-Concurrent-To\":\"<urn:uuid:fca8d959-f8d2-487f-8815-f5b9aa34075c>\",\"WARC-IP-Address\":\"172.217.13.238\",\"WARC-Target-URI\":\"https://developers.google.com/earth-engine/guides/resample\",\"WARC-Payload-Digest\":\"sha1:ZVK2ZGZBFUF2YKQBVN6BO66V6M7CE3RE\",\"WARC-Block-Digest\":\"sha1:CZ7RKFUPLZX5D6WXSUCEM6SPSKXU6WS4\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401604940.65_warc_CC-MAIN-20200928171446-20200928201446-00313.warc.gz\"}"} |
https://www.reference.com/science/molar-mass-kci03-3a3f2b6015d6328b | [
"# What Is the Molar Mass of KCI03?\n\nThe molar mass of KCI03 is 431.82 grams per mole. KCI03 a compound containing one atom of potassium, one atom of chlorine and three atoms of iodine. The molar mass of the similarly written compound KClO3, which is the common chemical potassium chlorate, is 122.54 grams per mole.\n\nThe molar mass of either compound can be calculated by adding together the atomic mass of each atom of each element comprising the compound. For example, a compound containing one atom of potassium and one atom of chlorine includes at least the mass of 39.09 grams per mole for a potassium atom and 35.45 grams per mole for a chlorine atom.\n\nSimilar Articles"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87479424,"math_prob":0.9764706,"size":620,"snap":"2019-35-2019-39","text_gpt3_token_len":142,"char_repetition_ratio":0.16720779,"word_repetition_ratio":0.057142857,"special_character_ratio":0.23225807,"punctuation_ratio":0.103174604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99507266,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T11:32:50Z\",\"WARC-Record-ID\":\"<urn:uuid:b9d17177-1e65-401f-a7c3-4af3a5062470>\",\"Content-Length\":\"171073\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2c104b28-4dd3-40e9-ac70-8caa429fdac2>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c1991ee-8359-4f1f-8f27-3fb67cc17ab3>\",\"WARC-IP-Address\":\"151.101.250.114\",\"WARC-Target-URI\":\"https://www.reference.com/science/molar-mass-kci03-3a3f2b6015d6328b\",\"WARC-Payload-Digest\":\"sha1:GNIMKGNXIGHGXIWKUHUAZR4DETLRKRVT\",\"WARC-Block-Digest\":\"sha1:A23YCTEOU2EZTZWBAHMJCIC45MMRQVPA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317113.27_warc_CC-MAIN-20190822110215-20190822132215-00295.warc.gz\"}"} |
http://wellness.bcbsla.com/Conditions/Heart/90,P01598 | [
"# Heart Disease\n\nHealth Library Explorer\n\n# Determining Body Mass Index for Teens\n\n## What is body mass index?\n\nDetermining how much your teen should weigh is not a simple matter of looking at an insurance height-weight chart. It includes considering the amount of bone, muscle, and fat in his or her body. The amount of fat is the critical measurement.\n\nA good indicator of how much fat your teen carries is the body mass index (BMI). Although it's not a perfect measure, it gives a fairly accurate assessment of how much of your teen's body is composed of fat.\n\nThe formulas below apply to adults only. For children and teens ages 2 to 19 years, the BMI varies by age and sex. An additional step must be done after the BMI has been determined using one of the formulas below. The BMI-for-age percentile is determined by comparing your teen's weight to that of other teens of the same age and sex.\n\nIn other words, by plotting your child's BMI value into the CDC's BMI-for-age growth chart, you can determine if your child is underweight, within normal range, overweight, or obese. Or you can easily calculate it from CDC's online tool\n\n## To calculate an adult's BMI using the English formula\n\nFor example, a person who weighs 165 pounds and is 5 feet 4 inches tall has a BMI of 28.165 lbs x 703=28(64 inches) x (64 inches)\n\n## To calculate an adult's BMI using the metric formula\n\nA BMI between 25 to 29.9 is considered overweight. Anything over 30 is considered obese. Normal BMI is between 18.5 and 24.9. BMI cut-offs for people of South Asian descent are different.\n\nBMI can be calculated using pounds and inches.BMI=Weight in Pounds x 703(Height in Inches) x (Height in Inches)\n\nMultiply your weight in pounds by 703.\n\nBMI can be calculated using kilograms and meters.BMI=Weight in Kilograms(Height in Meters) x (Height in Meters)\n\nDivided by your height in meters."
] | [
null
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http://mathandmultimedia.com/tag/problem-solving-in-mathematics/ | [
"## Introduction to Math Word Problem Solving\n\nAs I have mentioned in the previous post, I will be starting a Math Word Problem Solving Series for elementary school and high school students. If you can recall, I have already started this series before, but it was discontinued for a while.\n\nIn this series, we are going to learn some of the basics of solving word problems and learn some strategies and tricks to make problem solving easier. This series will include tutorials on how to solve number problems, age problems, number digit problems, distance-rate-time problems, base-rate-percentage problems, mixture problems, etc.\n\nProblem solving ability is developed over time. If you want to be good at it, then you have to practice solving a lot of problems. Reading this series will probably help you a little, but solving the problems yourself will make you a better problem solver. » Read more"
] | [
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https://projecteuclid.org/journals/electronic-journal-of-probability/volume-21/issue-none/Optimal-binomial-Poisson-and-normal-left-tail-domination-for-sums/10.1214/16-EJP4474.full | [
"Translator Disclaimer\n2016 Optimal binomial, Poisson, and normal left-tail domination for sums of nonnegative random variables\nIosif Pinelis\nElectron. J. Probab. 21: 1-19 (2016). DOI: 10.1214/16-EJP4474\n\n## Abstract\n\nExact upper bounds on the generalized moments $\\operatorname{\\mathsf {E}} f(S_n)$ of sums $S_n$ of independent nonnegative random variables $X_i$ for certain classes $\\mathcal{F}$ of nonincreasing functions $f$ are given in terms of (the sums of) the first two moments of the $X_i$’s. These bounds are of the form $\\operatorname{\\mathsf {E}} f(\\eta )$, where the random variable $\\eta$ is either binomial or Poisson depending on whether $n$ is fixed or not. The classes $\\mathcal{F}$ contain, and are much wider than, the class of all decreasing exponential functions. As corollaries of these results, optimal in a certain sense upper bounds on the left-tail probabilities $\\operatorname{\\mathsf {P}} (S_n\\le x)$ are presented, for any real $x$. In fact, more general settings than the ones described above are considered. Exact upper bounds on the exponential moments $\\operatorname{\\mathsf {E}} \\exp \\{hS_n\\}$ for $h<0$, as well as the corresponding exponential bounds on the left-tail probabilities, were previously obtained by Pinelis and Utev. It is shown that the new bounds on the tails are substantially better.\n\n## Citation\n\nIosif Pinelis. \"Optimal binomial, Poisson, and normal left-tail domination for sums of nonnegative random variables.\" Electron. J. Probab. 21 1 - 19, 2016. https://doi.org/10.1214/16-EJP4474\n\n## Information\n\nReceived: 11 August 2015; Accepted: 25 February 2016; Published: 2016\nFirst available in Project Euclid: 11 March 2016\n\nzbMATH: 1338.60061\nMathSciNet: MR3485362\nDigital Object Identifier: 10.1214/16-EJP4474\n\nSubjects:\nPrimary: 60E15\nSecondary: 60G42, 60G48\n\nJOURNAL ARTICLE\n19 PAGES",
null,
"SHARE\nVol.21 • 2016",
null,
""
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https://www.encyclopediaofmath.org/index.php/Weil_algebra | [
"# Weil algebra\n\nMotivated by algebraic geometry, A. Weil [a3] suggested the treatment of infinitesimal objects as homomorphisms from algebras of smooth functions",
null,
"into some real finite-dimensional commutative algebra",
null,
"with unit. The points in",
null,
"correspond to the choice",
null,
", while the algebra",
null,
",",
null,
", of dual numbers (also called Study numbers) leads to the tangent vectors at points in",
null,
"(viewed as derivations on functions). At the same time, Ch. Ehresmann established similar objects, jets (cf. also Jet), in the realm of differential geometry, cf. [a1].\n\nSince",
null,
"is formally real (i.e.",
null,
"is invertible for all",
null,
"), the values of the homomorphisms in",
null,
"are in formally real subalgebras. Now, for each finite-dimensional real commutative unital algebra",
null,
"which is formally real, there is a decomposition of the unit",
null,
"into all minimal idempotent elements. Thus,",
null,
", where",
null,
", and",
null,
"are nilpotent ideals in",
null,
". A real unital finite-dimensional commutative algebra",
null,
"is called a Weil algebra if it is of the form",
null,
"where",
null,
"is the ideal of all nilpotent elements in",
null,
". The smallest",
null,
"with the property",
null,
"is called the depth, or order, of",
null,
".\n\nIn other words, one may also characterize the Weil algebras as the formally real and local (i.e. the ring structure is local, cf. also Local ring) finite-dimensional commutative real unital algebras. See [a2], 35.1, for details.\n\nAs a consequence of the Nakayama lemma, the Weil algebras can be also characterized as the local finite-dimensional quotients of the algebras of real polynomials",
null,
". Consequently, the Weil algebras",
null,
"correspond to choices of ideals",
null,
"in",
null,
"of finite codimension. The algebra of Study numbers",
null,
"is given by",
null,
", for example. Equivalently, one may consider the algebras of formal power series or the algebras of germs of smooth functions at the origin",
null,
"(cf. also Germ) instead of the polynomials.\n\nThe width of a Weil algebra",
null,
"is defined as the dimension of the vector space",
null,
". If",
null,
"is an ideal of finite codimension in",
null,
",",
null,
", then the width of",
null,
"equals",
null,
". For example, the Weil algebra",
null,
"has width",
null,
"and order",
null,
", and it coincides with the algebra",
null,
"of",
null,
"-jets of smooth functions at the origin in",
null,
". Moreover, each Weil algebra of width",
null,
"and order",
null,
"is a quotient of",
null,
".\n\nTensor products of Weil algebras are Weil algebras again. For instance,",
null,
".\n\nThe infinitesimal objects of type",
null,
"attached to points in",
null,
"are simply",
null,
". All smooth functions",
null,
"extend to",
null,
"by the evaluation of the Taylor series (cf. also Whitney extension theorem)",
null,
"where",
null,
",",
null,
",",
null,
"are multi-indices,",
null,
". Applying this formula to all components of a mapping",
null,
", one obtains an assignment functorial in both",
null,
"and",
null,
". Of course, this definition extends to a functor on all locally defined smooth mappings",
null,
"and so each Weil algebra gives rise to a Weil functor",
null,
". (See Weil bundle for more details.)\n\nThe automorphism group",
null,
"of a Weil algebra is a Lie subgroup (cf. also Lie group) in",
null,
"and its Lie algebra coincides with the space of all derivations (cf. also Derivation in a ring) on",
null,
",",
null,
", i.e. all mappings",
null,
"satisfying",
null,
", cf. [a2], 42.9."
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null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.83798206,"math_prob":0.9744947,"size":3643,"snap":"2020-10-2020-16","text_gpt3_token_len":926,"char_repetition_ratio":0.1415224,"word_repetition_ratio":0.010169491,"special_character_ratio":0.24430415,"punctuation_ratio":0.1764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9962325,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,2,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-03-31T16:51:55Z\",\"WARC-Record-ID\":\"<urn:uuid:4acb0111-a97c-4d9b-85fe-273c5cf9ea4e>\",\"Content-Length\":\"25862\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a000154e-c3d3-4c65-832b-920cd68fb807>\",\"WARC-Concurrent-To\":\"<urn:uuid:98ed99c9-5fba-45dd-8a79-e42d6940e822>\",\"WARC-IP-Address\":\"80.242.138.72\",\"WARC-Target-URI\":\"https://www.encyclopediaofmath.org/index.php/Weil_algebra\",\"WARC-Payload-Digest\":\"sha1:4U7LGUMOT2IDGYF6ZAIRIMEIRIXXBM46\",\"WARC-Block-Digest\":\"sha1:LOJUURCRKH2E6O7D6TURRKCEVQDKJMYU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370502513.35_warc_CC-MAIN-20200331150854-20200331180854-00432.warc.gz\"}"} |
https://slideplayer.com/slide/6381895/ | [
"",
null,
"Graphing in Science. Types of Charts Most scientific graphs are made as line graphs. However, occasionally bar graphs, pie charts, or scatter plots.\n\nPresentation on theme: \"Graphing in Science. Types of Charts Most scientific graphs are made as line graphs. However, occasionally bar graphs, pie charts, or scatter plots.\"— Presentation transcript:\n\nGraphing in Science\n\nTypes of Charts Most scientific graphs are made as line graphs. However, occasionally bar graphs, pie charts, or scatter plots are used.\n\nWhy do we use graphs? All types of graphs and charts make trends in data easier to see. It is a way for scientists to analyze their data. It can also be used to predict data that is not measured on the graph.\n\nSteps to constructing a line graph Identify your variables Independent variable goes on the x-axis Dependent variable goes on the y-axis Determine the variable range Subtract the lowest data value from the highest data value for each variable Determine the scale of the graph This is the numerical value for each square that best fits the range of each variable Use most of the graphing space available.\n\nSteps to constructing a line graph (continued) Number and Label each axis Do not forget to include units!! Plot the data points. Draw the graph Connect the dots Best Fit Title the graph Titles should clearly show what the graph is about If you have more than one set of data, you need to include a key to identify the different lines\n\nExample Using the data table, answer the questions below: What is the independent variable? pH of the water What is the dependent variable? Number of tadpoles What goes on the x-axis? What is the range? pH of water (2.5) What goes on the y-axis? What is the range? Number of tadpoles (65) pH of water# of tadpoles 8.045 7.569 7.078 6.588 6.043 5.523\n\nExample pH of water# of tadpoles 8.045 7.569 7.078 6.588 6.043 5.523\n\nDownload ppt \"Graphing in Science. Types of Charts Most scientific graphs are made as line graphs. However, occasionally bar graphs, pie charts, or scatter plots.\"\n\nSimilar presentations"
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null,
"https://slideplayer.com/static/blue_design/img/slide-loader4.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.87451947,"math_prob":0.9070901,"size":1718,"snap":"2022-05-2022-21","text_gpt3_token_len":411,"char_repetition_ratio":0.13710618,"word_repetition_ratio":0.06020067,"special_character_ratio":0.24388824,"punctuation_ratio":0.11111111,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9852575,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T17:40:02Z\",\"WARC-Record-ID\":\"<urn:uuid:9f82a5ab-b35e-4f8f-b909-03e38d5837fd>\",\"Content-Length\":\"148752\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dca8cd9f-56e2-49d3-8a92-5a772e57a7d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:1efc3d9c-5e67-406a-892c-360afa2b4f90>\",\"WARC-IP-Address\":\"138.201.58.10\",\"WARC-Target-URI\":\"https://slideplayer.com/slide/6381895/\",\"WARC-Payload-Digest\":\"sha1:XF23X2SIDLE5FJOIGPD6ESQJ6BNFVHJH\",\"WARC-Block-Digest\":\"sha1:5WMAPSKUPORZT7QRTJX5SE3TPJRXKLYE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306301.52_warc_CC-MAIN-20220128152530-20220128182530-00057.warc.gz\"}"} |
https://newpathworksheets.com/math/grade-8/linear-equations/virgin-islands-common-core-standards | [
"## ◂Math Worksheets and Study Guides Eighth Grade. Linear equations\n\n### The resources above correspond to the standards listed below:\n\n#### Virgin Islands Common Core Standards\n\nVI.CC.EE.8. Expressions and Equations\nUnderstand the connections between proportional relationships, lines, and linear equations.\nEE.8.5. Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.\nEE.8.6. Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.\nVI.CC.F.8. Functions\nDefine, evaluate, and compare functions.\nF.8.3. Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s^2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.\nUse functions to model relationships between quantities.\nF.8.4. Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8998012,"math_prob":0.9979954,"size":1648,"snap":"2020-10-2020-16","text_gpt3_token_len":358,"char_repetition_ratio":0.14902677,"word_repetition_ratio":0.055555556,"special_character_ratio":0.21723302,"punctuation_ratio":0.14749263,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993462,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-02T03:38:36Z\",\"WARC-Record-ID\":\"<urn:uuid:4c99a900-2da5-42d1-bee0-40a31bde7b86>\",\"Content-Length\":\"30664\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:191d9078-b772-44e4-b470-af655ddbd15d>\",\"WARC-Concurrent-To\":\"<urn:uuid:e14259d9-a65a-431c-914a-75a728c39df0>\",\"WARC-IP-Address\":\"13.58.13.153\",\"WARC-Target-URI\":\"https://newpathworksheets.com/math/grade-8/linear-equations/virgin-islands-common-core-standards\",\"WARC-Payload-Digest\":\"sha1:YU5PS74EOAOUFPFSHGCD457EXOZXYLQE\",\"WARC-Block-Digest\":\"sha1:A72P64BTBWJ7SCLAR4WRR6VLNU7U46ZY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370506580.20_warc_CC-MAIN-20200402014600-20200402044600-00281.warc.gz\"}"} |
http://sigmapedia.com/includes/term.cfm?word_id=2179&lang=eng | [
"",
null,
"# Normality Plot\n\nGo Back\n\nDefinition\n\nThis is the probability plot produced by statistical software when a normality test is performed. Perfectly normal data would form a straight line on the probability plot. Typically, a normality test such as the Anderson-Darling Normality Test statistic is shown on the graph to aid in its interpretation.\n\nApplication\n\nThe image shows normal probability plots for three different datasets. The one on the left comes from an Uniform distribution, the one in the center comes from a Normal distribution, and the one on the right comes from an Exponential distribution.\n\nAlthough there are more sophisticated ways of telling whether or not a sample comes from a normal population, a common informal test is the ‘Fat Pencil Test’. Take the normal probability plot and lay an imaginary fat pencil across the plotted data. If the pencil covers all the data points, the data are most likely from a normal distribution.",
null,
""
] | [
null,
"https://media.moresteam.com/main/pics/sigmapedia.png",
null,
"http://sigmapedia.com/sigmapedia/pics/norm_2.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9006504,"math_prob":0.9419583,"size":920,"snap":"2020-24-2020-29","text_gpt3_token_len":175,"char_repetition_ratio":0.14737992,"word_repetition_ratio":0.0,"special_character_ratio":0.17934783,"punctuation_ratio":0.07926829,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99067926,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-05T14:32:52Z\",\"WARC-Record-ID\":\"<urn:uuid:dfab41cd-5caa-4e8c-a240-f9f472f4148d>\",\"Content-Length\":\"19043\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ff214f2-b922-40b1-ba8d-65f3a04be992>\",\"WARC-Concurrent-To\":\"<urn:uuid:4c7e9368-b1a1-400e-919e-be11b28014fd>\",\"WARC-IP-Address\":\"52.176.89.191\",\"WARC-Target-URI\":\"http://sigmapedia.com/includes/term.cfm?word_id=2179&lang=eng\",\"WARC-Payload-Digest\":\"sha1:XQS67B7WNUNZYKYO2FRQTYEIRCEOWX6K\",\"WARC-Block-Digest\":\"sha1:53TNGMTVVP6CK4GU7DH2YGECEL7BXBEN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590348502097.77_warc_CC-MAIN-20200605143036-20200605173036-00051.warc.gz\"}"} |
https://www.physicsforums.com/threads/billiard-balls-elastic-collision.311193/ | [
"# Billiard Balls-Elastic Collision\n\nBilliard Balls--Elastic Collision\n\nI'm going through my old homework problems to prepare for the AP exam, and I got this right back in December, but now I have no idea how to solve it\n\n## Homework Statement\n\nA cue ball traveling at 7.0 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30° with its original direction of travel. Find:\na) the angle between the velocity vectors of the two balls after the collision\nb) the speed of each ball after the collision\n\nKE = 1/2mv2\np = mv\n\n## The Attempt at a Solution\n\nThe collision is elastic, so both kinetic energy and momentum are conserved, so\n\n1/2mv02 = 1/2mv12 + 1/2 mv22\nv02 = v12 + v22\n\nand\n\nmv0 = mv1 + mv2\nv0 = v1 + v2\n\nThe x component of the cue ball's final velocity is v1cos30\nThe x component of the target ball's final velocity is v2cosΘ\n\nthis gives me two equations, and three unknowns:\n\n49 = (v1cos30)2 + (v2cosΘ)2\n\n7 = v1cos30 + v2cosΘ\n\nAm I missing something that will tell me the angle of the velocity of the second ball? Thanks\n\n## The Attempt at a Solution\n\nRelated Introductory Physics Homework Help News on Phys.org\n\nFor KE, you use the total magnitude of the velocity and don't break it up into x components. Also, the third equation you might be looking for is the y-component of momentum. Three equations and three unknowns is solvable.\n\nOkay, so then my equations should be:\n\n49 = v12 + v22\n7 = v1cos30 + v2cosΘ\n\nand, since i feel like the y components of momentum ought to cancel out for some reason,\n\n0 = v1sin30 - v2sinΘ\n\nThe algebra is getting pretty thorny...\n\nThe y-components of the momentum are correct, since there is no initial y-momentum. The equations do look correct, although it is pretty late and I might be forgetting something, but lets hope not. Now you need to wade through some messy algebra, and definitely check your final answers when you are done.\n\nAndrew Mason\nHomework Helper\n\nI\nThe collision is elastic, so both kinetic energy and momentum are conserved, so\n\n1/2mv02 = 1/2mv12 + 1/2 mv22\nv02 = v12 + v22\nOk.\n\nand\n\nmv0 = mv1 + mv2\nv0 = v1 + v2\nThis is correct only if these are vectors.\n\n$$\\vec v_0 = \\vec v_1 + \\vec v_2$$\n\nDraw the vector triangle made by these velocity vectors.\n\nSince\n\n$$v_0^2 = v_1^2 + v_2^2$$\n\nwhat kind of triangle is it?\n\nAM\n\nOk.\n\nThis is correct only if these are vectors.\n\n$$\\vec v_0 = \\vec v_1 + \\vec v_2$$\n\nDraw the vector triangle made by these velocity vectors.\n\nSince\n\n$$v_0^2 = v_1^2 + v_2^2$$\n\nwhat kind of triangle is it?\n\nAM\n\ni think the answer you wanted me to get for the triangle question is that it's a right triangle, probably because I'm supposed to recognize the Pythagorean theorem? Didn't see that, maybe next time.\n\nAnyway, I slogged through the algebra and got the correct answer. I'll remember to look for that in the future though.\n\nThanks!\n\nAndrew Mason\nHomework Helper\n\ni think the answer you wanted me to get for the triangle question is that it's a right triangle, probably because I'm supposed to recognize the Pythagorean theorem? Didn't see that, maybe next time.\n\nAnyway, I slogged through the algebra and got the correct answer. I'll remember to look for that in the future though.\n\nThanks!\nIn an elastic collision between two objects of equal mass at an oblique angle, the angle between the directions after the collision is always 90 degrees. That makes it very easy to solve this problem.\n\nYou can observe this all the time in curling and billiards (except that in billiards the cue ball spin can dramatically change how the cue ball moves after collision).\n\nAM"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9012801,"math_prob":0.9860921,"size":2110,"snap":"2020-10-2020-16","text_gpt3_token_len":621,"char_repetition_ratio":0.11728395,"word_repetition_ratio":0.8856448,"special_character_ratio":0.28625593,"punctuation_ratio":0.056872036,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988631,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-26T05:05:31Z\",\"WARC-Record-ID\":\"<urn:uuid:c3eb5453-b7f4-40c0-ab56-c2e3ec0b3e5c>\",\"Content-Length\":\"88664\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7de71a1f-73e7-4209-b22e-2dfc0403a950>\",\"WARC-Concurrent-To\":\"<urn:uuid:e92e778f-bb8b-4e6c-a801-24c8054c3dd7>\",\"WARC-IP-Address\":\"23.111.143.85\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/billiard-balls-elastic-collision.311193/\",\"WARC-Payload-Digest\":\"sha1:YDRCRHC6ZTKUQEL4IQUS45JXPUF57HXW\",\"WARC-Block-Digest\":\"sha1:YYXUFKTOYOMI3M2A5HJLXH3VTYPWC5TF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875146186.62_warc_CC-MAIN-20200226023658-20200226053658-00046.warc.gz\"}"} |
https://www.tutorialfor.com/questions-324080.htm | [
"Home>\n\n### python - meaning of dlib face detection function arguments\n\nReference site\n\nI have a face detection&stamp program with reference to the above site, but I had one question.\nI didn't understand it even after checking it, so I asked a question.\n\nI don't understand the meaning of the number in the second argument of dlib.get_frontal_face_detector ().\nCould you tell me?\n\n(Hereafter detector.py in the site)\n\n``````#coding: utf-8\nimport json\nimport math\nimport os\nimport cv2\nimport dlib\n\ndef main ():\n## input path and read image\nimg_path = \"../data/img/\" + os.listdir (\"../ data/img /\") \nimg_filename = os.path.basename (img_path)\nimg_ext = img_filename.split (\".\") [-1]\nimg = cv2.imread (img_path, cv2.IMREAD_COLOR)\nif img is None:\nprint (\"Could not read input image\")\nexit ()\n\n## instance detector/predictor\ndetector = dlib.get_frontal_face_detector ()\npredictor = dlib.shape_predictor (\"predictor/shape_predictor_68_face_landmarks.dat\")\n\n## detect faces\n## I don't understand the meaning of the second argument below! !! !! !!\nfaces = detector (img, 1)\nif len (faces) == 0:\nprint (\"Could not detect any faces\")\nexit ()\n\n## compute face properties\nface_properties = {}\nfor i, f in enumerate (faces):\n## get 68 points\npoints = predictor (img, f) .parts ()\n## Left/Right temples\nt_L = points \nt_R = points \n## calc properties\nface_width = math.sqrt ((t_L.x --t_R.x) ** 2 + (t_L.y --t_R.y) ** 2)\nface_center = [(t_L.x + t_R.x)/2, (t_L.y + t_L.y)/2]\ndegree_acw = -1 * math.degrees (math.atan2 ((t_R.y --t_L.y), (t_R.x --t_L.x)))\n## add result\nface_properties [i] = {\n\"stamp\": \"01.png\",\n\"center\": face_center,\n\"width\": face_width,\n\"angle\": degree_acw\n}\n## save\nwith open (\"../ data/json /\" + img_filename.replace (img_ext, \"json\"), \"w\") as f:\njson.dump (face_properties, f, ensure_ascii = False, indent = 4, separators = (',',':'))\n\nif __name__ == \"__main__\":\nmain ()``````\n• Answer # 1\n\nIt is a parameter called upsample_num_times.\nIt seems that the original image is upsampled that number of times before it is applied to the detector.\nhttp://dlib.net/python/index.html\n\nThe higher the number, the smaller the area that can be detected, but the larger the processing time and memory required.\nThe value of the default parameter is 0, and the smallest area detected at that time is 80x80. It seems that 1 is 40x40 and 2 is 20x20.\nhttps://qiita.com/nonbiri15/items/9561c8194ba0b2041bd0 # About the minimum detection size of face detection with dlib"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7421677,"math_prob":0.92407346,"size":4641,"snap":"2021-21-2021-25","text_gpt3_token_len":1175,"char_repetition_ratio":0.13219754,"word_repetition_ratio":0.03307888,"special_character_ratio":0.2783883,"punctuation_ratio":0.12244898,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98589295,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-12T20:11:51Z\",\"WARC-Record-ID\":\"<urn:uuid:3f6106e4-264e-4a75-8489-39a28d898751>\",\"Content-Length\":\"15777\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:340a3a58-b372-435f-9e2b-1aa551c83c65>\",\"WARC-Concurrent-To\":\"<urn:uuid:1207ef14-7cd8-4199-9883-d696d2378d99>\",\"WARC-IP-Address\":\"172.67.222.192\",\"WARC-Target-URI\":\"https://www.tutorialfor.com/questions-324080.htm\",\"WARC-Payload-Digest\":\"sha1:B2E4ZDQEZTDIJUBUXZ3MRJ3JNVZY5AU7\",\"WARC-Block-Digest\":\"sha1:A4YBFBK7N5WROHEMOM6TM5BBSOPFH434\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487586390.4_warc_CC-MAIN-20210612193058-20210612223058-00190.warc.gz\"}"} |
https://www.homeandlearn.co.uk/php/php6p8.html | [
"Home and Learn: PHP Programming Course\n\n# The PHP count function\n\nThe count( ) function is useful when you want to return how many elements are in your array. You can then use this in a for loop. Here's an example we used earlier, only this time with the count function:\n\n\\$seasons = array(\"Autumn\", \"Winter\", \"Spring\", \"Summer\");\n\n\\$array_count = count(\\$seasons);\n\nfor (\\$key_Number = 0; \\$key_Number < \\$array_count; \\$key_Number++) {\n\nprint \\$seasons[\\$key_Number];\n\n}\n\nTo get how many elements are in the array, we used this:\n\n\\$array_count = count(\\$seasons);\n\nSo you type the word count and then the round brackets. In between the round brackets, you type the name of your array. The function then counts how many elements are in the array, which we then assign to a variable called \\$array_count. You can then use this value as the end condition in you loop:\n\nfor (\\$key_Number = 0; \\$key_Number < \\$array_count; \\$key_Number++)\n\nHere, we're saying, \"keep looping round as long as the value in \\$key_Number is less than the value in \\$array_count.\n\nTo round off this section on arrays, there are some script for you to try out in the next part.\n\nBack to the PHP Contents Page\n\nEmail us: enquiry at homeandlearn.co.uk"
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https://nforum.ncatlab.org/discussion/15134/a-noneditable-entry-double-derivation/ | [
"# Start a new discussion\n\n## Not signed in\n\nWant to take part in these discussions? Sign in if you have an account, or apply for one below\n\n## Site Tag Cloud\n\nVanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.\n\n• CommentRowNumber1.\n• CommentAuthorzskoda\n• CommentTime7 days ago\n\nI tried to edit the entry double derivation and even if I write a single letter or none (!) it sends back an error\n\n500 Internal Server Error\n\nYour edit was blocked by spam filtering\n\nPlease report this on the nForum in the nLab Technical Matters category. Please give as precise details as you can as to what triggered the error.\n\nHere is the error:\n\nNo such file or directory @ rb_sysopen - page_content/submitted_edits/nlab/double_derivation\n\netc. If I try to click on “discuss” button it does NOT open the latest changes page but just links to front page of $n$Forum, not linked to a page.\n\n• CommentRowNumber2.\n• CommentAuthorzskoda\n• CommentTime7 days ago\nHere is the edited page which I did not succeed to send:\n\nGiven a _commutative_ ring $k$ and an associative $k$-algebra $A$ over $k$, the tensor product $A\\otimes_k A$ is equipped with two bimodule structures, \"outer\" and \"inner\". For the outer structure $a\\cdot_o(b\\otimes c)\\cdot_o d = a b\\otimes c d$ and for the inner $a\\cdot_i(b\\otimes c)\\cdot_i d = b d\\otimes a c$. The two bimodule structures mutually commute. A $k$-linear map $\\alpha\\in Hom_k(A,A\\otimes A)$ is called a __double derivation__ if it is also a map of $A$-bimodules with respect to the _outer_ bimodule structure ($\\alpha\\in A Mod A({}_A A_A,{}_A A\\otimes_k A_A)$); thus the $k$-module $Der(A,A\\otimes A)$ of all double derivations becomes an $A$-bimodule with respect to the _inner_ $A$-bimodule structure.\n\nThe tensor algebra $T_A Der(A,A\\otimes A)$ of the $A$-bimodule $Der(A,A\\otimes A)$ (which is the free monoid on $Der(A,A\\otimes A)$ in the monoidal category of $A$-bimodules) is a step in the definition of the deformed preprojective algebras of Bill Crawley-Boevey. A theorem of Van den Bergh says that for any associative $A$ the tensor algebra $T_A Der(A,A\\otimes A)$ has a canonical double Poisson bracket.\n\n* Michel Van den Bergh, _Double Poisson algebras_, Trans. Amer. Math. Soc. __360__ (2008) 5711--5769, [arXiv:math.AG/0410528](https://arXiv.org/abs/math/0410528)\n* Anne Pichereau, Geert Van de Weyer, _Double Poisson cohomology of path algebras of quivers_, J. Alg. __319__, 5 (2008) 2166--2208 [doi](https://doi.org/10.1016/j.jalgebra.2007.09.021)\n* Jorge A. Guccione, Juan J. Guccione, _A characterization of quiver algebras based on double derivations_, [arXiv:0807.1148](https://arxiv.org/abs/0807.1148)\n\ncategory: algebra\n• CommentRowNumber3.\n• CommentAuthorzskoda\n• CommentTime7 days ago\nCopying the content of double derivation to another pagename does not work: spam message appears instead of creating a page, including at my personal web.\n\nI created William Crawley-Boevey with redirect Bill Crawley-Boevey wanted at double derivation which still shows that Bill Crawley-Boevey does not exist.\n• CommentRowNumber4.\n• CommentAuthorUrs\n• CommentTime6 days ago\n\nHave been busy all day. Am forwarding this to the technical team now…\n\n1. The problem was that \"gucci\" was a pattern in the spam filter. I've removed it. You should now be able to submit your edit.\n• CommentRowNumber6.\n• CommentAuthorzskoda\n• CommentTime5 days ago\n\nUnfortunately, it still seem to persist if I leave “Guccione” so I temporarily wrote G uccione with a space. Then it accepts.\n\n• V. Ginzburg, T. Schedler, Differential operators and BV structures in noncommutative geometry, Sel. Math. New Ser. 16, 673–730 (2010) doi\n• CommentRowNumber7.\n• CommentAuthorUrs\n• CommentTime5 days ago\n\nOne can work around this problem by escaping some html characters. For example:\n\n Guccione\n\n\ngives\n\nGuccione\n\n2. Unfortunately, it still seem to persist if I leave “Guccione”\n\nRight, I forgot to actually update the deployment. Should be fixed now.\n\n• CommentRowNumber9.\n• CommentAuthorzskoda\n• CommentTime5 days ago\n\n7 thanks for the tip"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89591026,"math_prob":0.6313333,"size":1265,"snap":"2022-40-2023-06","text_gpt3_token_len":307,"char_repetition_ratio":0.07930214,"word_repetition_ratio":0.06306306,"special_character_ratio":0.2229249,"punctuation_ratio":0.09881423,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96456826,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-26T00:03:15Z\",\"WARC-Record-ID\":\"<urn:uuid:f370b4d1-e9fb-493f-8e95-5e0ed0dcefa4>\",\"Content-Length\":\"55486\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f7d95606-3acf-41fe-a55a-0eca52a3c882>\",\"WARC-Concurrent-To\":\"<urn:uuid:4be271b0-eedd-4507-988e-ae9e8033ebd1>\",\"WARC-IP-Address\":\"172.67.177.13\",\"WARC-Target-URI\":\"https://nforum.ncatlab.org/discussion/15134/a-noneditable-entry-double-derivation/\",\"WARC-Payload-Digest\":\"sha1:EBWWKL5JJR5WMMHM3564T3HP3EHCLVI3\",\"WARC-Block-Digest\":\"sha1:RLY3RAX5L364FB5ZBYZ264PPZHMBCASS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334620.49_warc_CC-MAIN-20220925225000-20220926015000-00424.warc.gz\"}"} |
https://mruanova.medium.com/softmax-function-dfe8f223208e | [
"In mathematics, the softmax function, also known as softargmax or normalized exponential function, is a function that takes as input a vector z of K real numbers, and normalizes it into a probability distribution consisting of K probabilities proportional to the exponentials of the input numbers.\n\nThat is, prior to applying softmax, some vector components could be negative, or greater than one; and might not sum to 1; but after applying softmax, each component will be in the interval (0,1), and the components will add up to 1, so that they can be interpreted as probabilities.\n\nFurthermore, the larger input components will correspond to larger probabilities.\n\nSoftmax is often used in neural networks, to map the non-normalized output of a network to a probability distribution over predicted output classes.\n\nWritten by"
] | [
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https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1186/s13662-019-2362-3 | [
"Theory and Modern Applications\n\n# Some new inequalities for generalized fractional conformable integral operators\n\n## Abstract\n\nThe present paper aims to establish certain new classes of integral inequalities for a class of n ($$n\\in \\mathbb{N}$$) positive continuous and decreasing functions by utilizing the generalized fractional conformable integral operators (FCIO) recently defined by Khan and Khan. From these results, we also derive several particular cases.\n\n## 1 Introduction\n\nFractional calculus earned more recognition due to its applications in diverse domains. Recent research focuses on developing a large number of the fractional integral operators (FIO) and their applications in multiple disciplines of sciences (see [13, 14, 20, 26]). In , Liu et al. introduced interesting integral inequalities for continuous functions on $$[a,b]$$. Later on, Dahmani generalized the work of involving the Riemann–Liouville fractional integral operators. In , Dahmani and Tabharit introduced weighted Grüss type inequalities involving fractional integral operators. Dahmani established some new inequalities for fractional integrals. Polya–Szego and Chebyshev type inequalities involving the Riemann–Liouville fractional integral operators are found in . Nisar et al. established some inequalities involving extended gamma and the Kummar confluent hypergeometric k-functions. In , Set et al. established generalized Grüss type inequalities for k-fractional integrals and applications. Certain Gronwall inequalities associated with Riemann–Liouville k and Hadamard k-fractional derivatives and their applications are found in the work of Nisar et al. . The $$(k,s)$$-fractional integrals and their applications are found in . Rahman et al. presented certain inequalities via $$(k,\\rho )$$-fractional integral operators. In , the authors introduced the idea of fractional conformable derivative operators with a shortcoming that the new derivative operator does not tend to the original function when the order $$\\rho \\rightarrow 0$$. In , the author studied certain various properties of the fractional conformable derivative operators and raised the problem of how to use conformable derivative operators to generate more general types of nonlocal fractional derivative operators, after that the method was demonstrated in .\n\nThe generalized FCIO defined in is given by\n\n\\begin{aligned} {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r^{+}}^{\\beta }f(x)= \\frac{1}{\\varGamma ( \\beta )} \\int _{r}^{x} \\biggl(\\frac{x^{\\alpha +\\mu }-\\tau ^{\\alpha + \\mu }}{\\alpha +\\mu } \\biggr)^{\\beta -1}\\frac{f(\\tau )}{ \\tau ^{1-\\alpha -\\mu }}\\,d\\tau ,\\quad x>r, \\end{aligned}\n(1)\n\nand\n\n\\begin{aligned} {}_{\\alpha }^{\\mu }\\mathfrak{I}_{s^{-}}^{\\beta }f(x)= \\frac{1}{\\varGamma ( \\beta )} \\int _{x}^{s} \\biggl(\\frac{\\tau ^{\\alpha +\\mu }-x^{\\alpha + \\mu }}{\\alpha +\\mu } \\biggr)^{\\beta -1}\\frac{f(\\tau )}{ \\tau ^{1-\\alpha -\\mu }}\\,d\\tau ,\\quad x< s, \\end{aligned}\n(2)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\Re (\\beta )>0$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, and Γ is the gamma function .\n\n### Remark 1\n\n(i) If we set $$\\mu =0$$ in (1) and (2), then we have the following Riemann–Liouville (R-L) type FCIO:\n\n\\begin{aligned} {}_{\\alpha }\\mathfrak{I}_{r^{+}}^{\\beta }f(x)= \\frac{1}{\\varGamma (\\beta )} \\int _{r}^{x} \\biggl(\\frac{x^{\\alpha }-\\tau ^{\\alpha }}{\\alpha } \\biggr) ^{\\beta -1}\\frac{f(\\tau )}{\\tau ^{1-\\alpha }}\\,d\\tau ,\\quad x>r, \\end{aligned}\n(3)\n\nand\n\n\\begin{aligned} {}_{\\alpha }\\mathfrak{I}_{s^{-}}^{\\beta }f(x)= \\frac{1}{\\varGamma (\\beta )} \\int _{x}^{s} \\biggl(\\frac{\\tau ^{\\alpha }-x^{\\alpha }}{\\alpha } \\biggr) ^{\\beta -1}\\frac{f(\\tau )}{\\tau ^{1-\\alpha }}\\,d\\tau ,\\quad x< s, \\end{aligned}\n(4)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\Re (\\beta )>0$$, $$\\alpha \\in (0,1]$$.\n\n(ii) If $$\\alpha =1$$ in 3 and 4, then we obtain the following R-L FIO:\n\n\\begin{aligned} \\mathfrak{I}_{r^{+}}^{\\beta }f(x)=\\frac{1}{\\varGamma (\\beta )} \\int _{r} ^{x} (x-\\tau )^{\\beta -1}f(\\tau )\\,d \\tau ,\\quad x>r, \\end{aligned}\n(5)\n\nand\n\n\\begin{aligned} \\mathfrak{I}_{s^{-}}^{\\beta }f(x)=\\frac{1}{\\varGamma (\\beta )} \\int _{x} ^{s} (\\tau -x )^{\\beta -1}f(\\tau )\\,d \\tau ,\\quad x< s, \\end{aligned}\n(6)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\Re (\\beta )>0$$.\n\nRecently, the researchers [21, 24] established inequalities of Grüss type and Čebyšev type by utilizing fractional conformable integral operators. Rahman et al. established certain Chebyshev type inequalities involving fractional conformable integral operators. In , the authors introduced the Minkowski inequalities via generalized proportional fractional integral operators. Some new inequalities involving fractional conformable integrals are found in the work of Nisar et al. . Adjabi et al. presented generalized fractional integral operators and Gronwall type inequalities with applications. In , Abdeljawad established a Lyapunov type inequality for fractional operators with nonsingular Mittag-Leffler kernel. Abdeljawad et al. introduced Lyapunov type inequalities for mixed nonlinear forced differential equations within conformable derivatives. Fractional operators with exponential kernels and a Lyapunov type inequality are found in . Abdeljawad et al. presented a generalized Lyapunov type inequality in the frame of conformable derivatives.\n\nOur aim in this paper is to generalize the inequalities obtained earlier by [8, 15] by employing the left generalized fractional conformable integral operator (1).\n\n## 2 Main results\n\nIn this section, we employ the left generalized FCIO to establish the generalization of some classical inequalities.\n\n### Theorem 1\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ be n positive continuous and decreasing functions on the interval $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3, \\ldots ,n\\}$$. Then, for generalized fractional conformable integral (1), we have\n\n\\begin{aligned} \\frac{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [\\prod_{i \\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]}{{}_{\\alpha } ^{\\mu }\\mathfrak{I}_{r}^{\\beta } [\\prod_{i=1}^{n}g_{i}^{\\gamma_{i}}(x) ]}\\geq \\frac{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{ \\beta } [(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]}{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{ \\beta } [(x-r)^{\\vartheta }\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]}, \\end{aligned}\n(7)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, and $$\\Re (\\beta )>0$$.\n\n### Proof\n\nSince $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ are n positive continuous and decreasing functions on the interval $$[r,s]$$. Therefore, we have\n\n\\begin{aligned} \\bigl((\\rho -r)^{\\vartheta }-(t-r)^{\\vartheta } \\bigr) \\bigl(g_{p} ^{\\sigma -\\gamma _{p}}(t)-g_{p}^{\\sigma -\\gamma _{p}}(\\rho ) \\bigr) \\geq 0, \\end{aligned}\n(8)\n\nwhere $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$, $$t, \\rho \\in [r,x]$$ and for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$.\n\nDefine a function\n\n\\begin{aligned} {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta }(x,\\rho ,t) &=\\frac{1}{ \\varGamma (\\beta )} \\biggl(\\frac{x^{\\alpha +\\mu }-t^{\\alpha +\\mu }}{ \\alpha +\\mu } \\biggr)^{\\beta -1} \\\\ &\\quad {}\\times \\frac{\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(t)}{t^{1-\\alpha - \\mu }} \\bigl((\\rho -r)^{\\vartheta }-(t-r)^{\\vartheta } \\bigr) \\bigl(g_{p}^{\\sigma -\\gamma _{p}}(t)-g_{p}^{\\sigma -\\gamma _{p}}( \\rho ) \\bigr). \\end{aligned}\n(9)\n\nWe observe that the above function satisfies all the assumptions stated in Theorem 1, and hence the function $${}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta }(x,\\rho ,t)$$ is positive for all $$t\\in (r,x)$$ ($$x>r$$). Integrating both sides of (9) with respect to t over $$(r,x)$$, we have\n\n\\begin{aligned} 0&\\leq \\int _{r}^{x}{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta }(x, \\rho ,t)\\,dt \\\\ &=\\frac{1}{\\varGamma (\\beta )} \\int _{r}^{x} \\biggl(\\frac{x^{ \\alpha +\\mu }-t^{\\alpha +\\mu }}{\\alpha +\\mu } \\biggr)^{\\beta -1} \\\\ &\\quad {}\\times \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(t) \\bigl((\\rho -r)^{\\vartheta }-(t-r)^{\\vartheta } \\bigr) \\bigl(g_{p}^{\\sigma -\\gamma _{p}}(t)-g_{p}^{\\sigma -\\gamma _{p}}( \\rho ) \\bigr)\\frac{dt}{t^{1-\\alpha -\\mu }} \\\\ &= \\Biggl[(\\rho -r)^{\\vartheta }{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r} ^{\\beta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]+g_{p}^{\\sigma -\\gamma _{p}}(\\rho ){}_{\\alpha }^{\\mu } \\mathfrak{I}_{r} ^{\\beta } \\Biggl[(x-r)^{\\vartheta }\\prod _{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}- (\\rho -r)^{\\vartheta }g_{p}^{\\sigma -\\gamma _{p}}(\\rho ) \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\prod_{i=1 }^{n}g_{i}^{ \\gamma _{i}}(x) \\Biggr]-{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[(x-r)^{\\vartheta }\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr]. \\end{aligned}\n(10)\n\nMultiplying (10) by $$\\frac{1}{\\varGamma (\\beta )} (\\frac{x ^{\\alpha +\\mu }-\\rho ^{\\alpha +\\mu }}{\\alpha +\\mu } )^{\\beta -1}\\frac{ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(\\rho )}{\\rho ^{1-\\alpha -\\mu }}$$ and integrating the resultant identity with respect to ρ over $$(r,x)$$, we have\n\n\\begin{aligned} 0&\\leq \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\prod_{i \\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha } ^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[(x-r)^{\\vartheta }\\prod_{i=1} ^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}-{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r} ^{\\beta } \\Biggl[(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma_{i}}g_{p}^{\\sigma }(x) \\Biggr] \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\prod_{i=1 }^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr], \\end{aligned}\n\nwhich completes the desired inequality (7). □\n\n### Corollary 1\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ be n positive continuous and decreasing on $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for R-L type fractional conformable integral (3), we have\n\n\\begin{aligned} \\frac{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]}{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]} \\geq \\frac{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]}{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [(x-r)^{\\vartheta }\\prod_{i=1} ^{n}g_{i}^{\\gamma _{i}}(x) ]}, \\end{aligned}\n(11)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, and $$\\Re (\\beta )>0$$.\n\n### Corollary 2\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ be n positive continuous and decreasing on $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for R-L fractional integral (5), we have\n\n\\begin{aligned} \\frac{\\mathfrak{I}_{r}^{\\beta } [\\prod_{i\\neq p}^{n}g_{i}^{\\gamma_{i}}g_{p}^{\\sigma }(x) ]}{\\mathfrak{I}_{r}^{\\beta } [\\prod_{i=1} ^{n}g_{i}^{\\gamma _{i}}(x) ]}\\geq \\frac{\\mathfrak{I}_{r}^{\\beta } [(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p} ^{\\sigma }(x) ]}{\\mathfrak{I}_{r}^{\\beta } [(x-r)^{\\vartheta }\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]}, \\end{aligned}\n(12)\n\nwhere $$\\beta \\in \\mathbb{C}$$ and $$\\Re (\\beta )>0$$.\n\n### Remark 2\n\nThe inequality in Theorem 1 will reverse if $$(g_{i})_{i=1,2,3, \\ldots ,n}$$ are increasing on the interval $$[r,s]$$. If we let $$\\alpha =1$$, $$\\mu =0$$, then Theorem 1 will lead to Theorem 3.1 . Moreover, setting $$\\mu =0$$, $$\\alpha =\\beta =n=1$$, $$x=s$$, then Theorem 1 reduces to the well-known Theorem 3 .\n\n### Theorem 2\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ be n positive continuous and decreasing on $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for generalized fractional conformable integral (1), we have\n\n\\begin{aligned} &\\frac{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta } { \\prod_{i=1}^{n}} g_{i}^{\\gamma _{i}}(x) ] +{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta } { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta }{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}(x) ]{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+ {}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta } [(x-r)^{\\vartheta }{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}(x) ] {}_{\\alpha }^{ \\mu }\\mathfrak{I}_{r}^{\\lambda } [{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]} \\\\ &\\quad \\geq 1, \\end{aligned}\n(13)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, $$\\Re (\\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Proof\n\nFirstly, multiplying both sides of equation (10) by $$\\frac{1}{\\varGamma (\\lambda )} (\\frac{x^{\\alpha +\\mu }- \\rho ^{\\alpha +\\mu }}{\\alpha +\\mu } )^{\\lambda -1}\\frac{\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(\\rho )}{\\rho ^{1-\\alpha -\\mu }}$$ and integrating the resultant identity with respect to ρ over $$(r,x)$$, we have\n\n\\begin{aligned} 0&\\leq \\int _{r}^{x} \\int _{r}^{x}\\frac{1}{\\varGamma (\\lambda )} \\biggl( \\frac{x ^{\\alpha +\\mu }-\\rho ^{\\alpha +\\mu }}{\\alpha +\\mu } \\biggr)^{\\lambda -1}\\frac{ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(\\rho )}{\\rho ^{1-\\alpha -\\mu }} {}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta }(x,\\rho ,t)\\,dt\\,d\\rho \\\\ &= {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[\\prod_{i\\neq p} ^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } \\Biggl[(x-r)^{\\vartheta }\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}+{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{ \\lambda } \\Biggl[\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr] {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[(x-r)^{\\vartheta } \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}- {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\Biggl[ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}-{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{ \\lambda } \\Biggl[(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma_{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r} ^{\\beta } \\Biggl[ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr]. \\end{aligned}\n(14)\n\nHence, dividing both sides of (14) by\n\n$${}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[(x-r)^{\\vartheta } \\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{ \\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\Biggl[ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr]+{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{ \\lambda } \\Biggl[(x-r)^{\\vartheta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma_{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r} ^{\\beta } \\Biggl[ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr],$$\n\nwe get the desired proof. □\n\n### Corollary 3\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ be n positive continuous and decreasing on $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for R-L type fractional conformable integral (3), we have\n\n\\begin{aligned} &\\frac{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta }{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ] +{}_{\\alpha } \\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta }{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{{}_{\\alpha } \\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta }{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}(x) ]{}_{\\alpha } \\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+ {}_{\\alpha } \\mathfrak{I}_{r}^{\\beta } [(x-r)^{\\vartheta }{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}(x) ] {}_{\\alpha } \\mathfrak{I}_{r}^{\\lambda } [{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]} \\\\ &\\quad \\geq 1, \\end{aligned}\n(15)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\Re ( \\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Corollary 4\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ be n positive continuous and decreasing on $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for R-L fractional integral (5), we have\n\n\\begin{aligned} &\\frac{\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] \\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta }{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ] +\\mathfrak{I}_{r}^{ \\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] \\mathfrak{I}_{r}^{\\lambda } [(x-r)^{\\vartheta }{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{\\mathfrak{I}_{r}^{ \\lambda } [(x-r)^{\\vartheta }{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}(x) ]\\mathfrak{I}_{r}^{ \\beta } [{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{p}}(x) ]+ \\mathfrak{I}_{r}^{ \\beta } [(x-r)^{\\vartheta }{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}(x) ] \\mathfrak{I}_{r} ^{\\lambda } [{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]} \\\\ &\\quad \\geq 1, \\end{aligned}\n(16)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\Re (\\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Remark 3\n\nApplying Theorem 2 for $$\\beta =\\lambda$$, we get Theorem 1. Again, the inequality will reverse if $$(g_{i})_{i=1,2,3, \\ldots ,n}$$ are increasing functions on the interval $$[r,s]$$. If we let $$\\alpha =1$$, $$\\mu =0$$, then Theorem 1 will lead to Theorem 3.4 . Moreover, setting $$\\mu =0$$, $$\\alpha =\\beta = \\lambda =n=1$$, $$x=s$$, then again Theorem 1 reduces to the well-known Theorem 3 .\n\n### Theorem 3\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous on the interval $$[r,s]$$ such that h is increasing and $$(g_{i})_{i=1,2,3, \\ldots ,n}$$ are decreasing functions on $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3, \\ldots ,n\\}$$. Then, for generalized fractional conformable integral (1), we have\n\n\\begin{aligned} \\frac{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [\\prod_{i \\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha } ^{\\mu }\\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x) \\prod_{i=1} ^{n}g_{i}^{\\gamma _{i}}(x) ]}{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x)\\prod_{i\\neq p}^{n}g_{i}^{\\gamma_{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r} ^{\\beta } [ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]}\\geq 1, \\end{aligned}\n(17)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, and $$\\Re (\\beta )>0$$.\n\n### Proof\n\nUnder the conditions stated in Theorem 3, we can write\n\n\\begin{aligned} \\bigl(h^{\\vartheta }(\\rho )-h^{\\vartheta }(t) \\bigr) \\bigl(g_{p} ^{\\sigma -\\gamma _{p}}(t)-g_{p}^{\\sigma -\\gamma _{p}}(\\rho ) \\bigr) \\geq 0, \\end{aligned}\n(18)\n\nwhere $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$, $$t, \\rho \\in [r,x]$$ and for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$.\n\nDefine a function\n\n\\begin{aligned} {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta }(x,\\rho ,t) &=\\frac{1}{ \\varGamma (\\beta )} \\biggl(\\frac{x^{\\alpha +\\mu }-t^{\\alpha +\\mu }}{ \\alpha +\\mu } \\biggr)^{\\beta -1} \\\\ &\\quad {}\\times \\frac{\\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(t)}{t^{1-\\alpha - \\mu }} \\bigl(h^{\\vartheta }(\\rho )-h^{\\vartheta }(t) \\bigr) \\bigl(g_{p}^{\\sigma -\\gamma _{p}}(t)-g_{p}^{\\sigma -\\gamma _{p}}( \\rho ) \\bigr). \\end{aligned}\n(19)\n\nWe observe that the above function satisfies all the assumptions stated in Theorem 3, and hence the function $${}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta }(x,\\rho ,t)$$ is positive for all $$t\\in (r,x)$$ ($$x>r$$). Therefore, integrating both sides of (19) with respect to t over $$(r,x)$$, we have\n\n\\begin{aligned} 0&\\leq \\int _{r}^{x}{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta }(x, \\rho ,t)\\,dt \\\\ &=\\frac{1}{\\varGamma (\\beta )} \\int _{r}^{x} \\biggl(\\frac{x^{ \\alpha +\\mu }-t^{\\alpha +\\mu }}{\\alpha +\\mu } \\biggr)^{\\beta -1} \\\\ &\\quad {}\\times \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(t) \\bigl(h^{\\vartheta }( \\rho )-h^{\\vartheta }(t) \\bigr) \\bigl(g_{p}^{\\sigma -\\gamma _{p}}(t)-g_{p}^{\\sigma -\\gamma _{p}}( \\rho ) \\bigr)\\frac{dt}{t^{1-\\alpha -\\mu }} \\\\ &= \\Biggl[h(\\rho )^{\\vartheta }{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r} ^{\\beta }\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]+g_{p}^{\\sigma -\\gamma _{p}}(\\rho ){}_{\\alpha }^{\\mu } \\mathfrak{I}_{r} ^{\\beta } \\Biggl[h^{\\vartheta }(x) \\prod _{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}- h^{\\vartheta }(\\rho ) g_{p}^{\\sigma -\\gamma _{p}}(\\rho ) \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\prod_{i=1 }^{n}g_{i}^{ \\gamma _{i}}(x) \\Biggr]-{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[h^{\\vartheta }(x) \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr]. \\end{aligned}\n(20)\n\nMultiplying (20) by $$\\frac{1}{\\varGamma (\\beta )} (\\frac{x ^{\\alpha +\\mu }-\\rho ^{\\alpha +\\mu }}{\\alpha +\\mu } )^{\\beta -1}\\frac{ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(\\rho )}{\\rho ^{1-\\alpha -\\mu }}$$ and integrating the resultant identity with respect to ρ over $$(r,x)$$, we have\n\n\\begin{aligned} 0&\\leq \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\prod_{i \\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha } ^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[h^{\\vartheta }(x) \\prod_{i=1} ^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}-{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r} ^{\\beta } \\Biggl[h^{\\vartheta }(x)\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr] \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r} ^{\\beta }\\prod_{i=1 }^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr], \\end{aligned}\n\nwhich completes the desired inequality (17) of Theorem 3. □\n\n### Corollary 5\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous on $$[r,s]$$ such that h is increasing and $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ are decreasing functions on the interval $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3, \\ldots ,n\\}$$. Then, for R-L type fractional conformable integral (3), we have\n\n\\begin{aligned} \\frac{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x) \\prod_{i=1}^{n}g_{i}^{\\gamma_{i}}(x) ]}{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [h^{ \\vartheta }(x)\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] {}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [ \\prod_{i=1}^{n}g_{i}^{ \\gamma _{i}}(x) ]}\\geq 1, \\end{aligned}\n(21)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, and $$\\Re (\\beta )>0$$.\n\n### Corollary 6\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous on $$[r,s]$$ such that h is increasing and $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ are decreasing functions on the interval $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3, \\ldots ,n\\}$$. Then, for R-L fractional integral (5), we have\n\n\\begin{aligned} \\frac{\\mathfrak{I}_{r}^{\\beta } [\\prod_{i\\neq p}^{n}g_{i}^{\\gamma_{i}}g_{p}^{\\sigma }(x) ]\\mathfrak{I}_{r}^{\\beta } [h^{ \\vartheta }(x) \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]}{ \\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x)\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]\\mathfrak{I}_{r}^{\\beta } [ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]}\\geq 1, \\end{aligned}\n(22)\n\nwhere $$\\beta \\in \\mathbb{C}$$ and $$\\Re (\\beta )>0$$.\n\n### Theorem 4\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous functions on $$[r,s]$$ such that h is increasing and $$(g_{i})_{i=1,2,3, \\ldots ,n}$$ are decreasing functions on the interval $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for generalized fractional conformable integral (1), we have\n\n\\begin{aligned} &\\frac{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } [h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } [ { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]} \\\\ &\\quad \\geq 1, \\end{aligned}\n(23)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, $$\\Re (\\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Proof\n\nMultiplying (20) by $$\\frac{1}{\\varGamma (\\lambda )} (\\frac{x ^{\\alpha +\\mu }-\\rho ^{\\alpha +\\mu }}{\\alpha +\\mu } )^{\\lambda -1}\\frac{ \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(\\rho )}{\\rho ^{1-\\alpha -\\mu }}$$ then integrating with respect to ρ over $$(r,x)$$, we have\n\n\\begin{aligned} 0&\\leq \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\prod_{i \\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr]{}_{\\alpha } ^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\Biggl[h^{\\vartheta }(x) \\prod_{i=1} ^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}+ \\Biggl[{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } \\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p} ^{\\sigma }(x) \\Biggr]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[h^{\\vartheta }(x) \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {}- {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[h^{\\vartheta }(x) \\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) \\Biggr] \\Biggl[{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\prod_{i=1 }^{n}g_{i}^{ \\gamma _{i}}(x) \\Biggr] \\\\ &\\quad {} -{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\Biggl[h^{\\vartheta }(x)\\prod_{i\\neq p}^{n}g_{i}^{\\gamma _{i}}g_{p} ^{\\sigma }(x) \\Biggr] \\Biggl[{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{ \\beta }\\prod_{i=1 }^{n}g_{i}^{\\gamma _{i}}(x) \\Biggr]. \\end{aligned}\n\nAfter simplification, we get the desired result. □\n\n### Remark 4\n\nApplying Theorem 4 for $$\\beta =\\lambda$$, we get Theorem 3.\n\n### Corollary 7\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous on $$[r,s]$$ such that h is increasing and $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ are decreasing functions on the interval $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3, \\ldots ,n\\}$$. Then, for R-L type fractional conformable integral (3), we have\n\n\\begin{aligned} &\\frac{{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\lambda } [h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+{}_{\\alpha } \\mathfrak{I}_{r}^{\\lambda } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{{}_{\\alpha } \\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\lambda } [ { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+{}_{\\alpha } \\mathfrak{I}_{r}^{\\lambda } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }\\mathfrak{I}_{r}^{\\beta } [ { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]} \\\\ &\\quad \\geq 1, \\end{aligned}\n(24)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\Re ( \\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Corollary 8\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous on $$[r,s]$$ such that h is increasing and $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ are decreasing on the interval $$[r,s]$$. Let $$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$ for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$. Then, for R-L fractional integral (5), we have\n\n\\begin{aligned} &\\frac{\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] \\mathfrak{I}_{r}^{\\lambda } [h^{\\vartheta }(x) \\prod_{i=1}^{n}g_{i}^{\\gamma _{i}}(x) ]+\\mathfrak{I}_{r}^{\\lambda } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] \\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{\\mathfrak{I}_{r}^{ \\beta } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] \\mathfrak{I}_{r}^{\\lambda } [ { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]+\\mathfrak{I}_{r}^{ \\lambda } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ] \\mathfrak{I}_{r}^{\\beta } [ { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]} \\\\ &\\quad \\geq 1, \\end{aligned}\n(25)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\Re (\\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Theorem 5\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous on $$[r,s]$$, and let for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$,\n\n$$\\bigl(g_{p}^{\\vartheta }(t)h^{\\vartheta }(\\rho )-g_{p}^{\\vartheta }( \\rho )h^{\\vartheta }(t) \\bigr) \\bigl(g_{p}^{\\sigma -\\gamma _{p}}(t) -g_{p}^{\\sigma -\\gamma _{p}}(\\rho ) \\bigr)\\geq 0,$$\n\n$$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$, then we have\n\n\\begin{aligned} \\frac{{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma +\\vartheta }(x) ] {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}{{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta } [h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma }(x) ]{}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } [g_{p}^{\\vartheta } { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) ]}\\geq 1, \\end{aligned}\n(26)\n\nwhere $$\\beta \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, and $$\\Re (\\beta )>0$$.\n\n### Proof\n\nThe proof of Theorem 5 is similar to the proof of Theorem 3 if we replace $$h^{\\vartheta }(\\rho )-h^{\\vartheta }(t)$$ by $$(g_{p}^{\\vartheta }(t)h^{\\vartheta }(\\rho )-g_{p}^{\\vartheta }( \\rho )h^{\\vartheta }(t) )$$. □\n\n### Theorem 6\n\nLet $$(g_{i})_{i=1,2,3,\\ldots ,n}$$ and h be positive continuous functions on $$[r,s]$$, and let for any fixed $$p\\in \\{1,2,3,\\ldots ,n\\}$$,\n\n$$\\bigl(g_{p}^{\\vartheta }(t)h^{\\vartheta }(\\rho )-g_{p}^{\\vartheta }( \\rho )h^{\\vartheta }(t) \\bigr) \\bigl(g_{p}^{\\sigma -\\gamma _{p}}(t) -g_{p}^{\\sigma -\\gamma _{p}}(\\rho ) \\bigr)\\geq 0,$$\n\n$$r< x\\leq s$$, $$\\vartheta >0$$, $$\\sigma \\geq \\gamma _{p}>0$$, then we have\n\n\\begin{aligned} &\\Biggl({}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma +\\vartheta }(x) \\Biggr] {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\Biggl[h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\qquad {}+{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } \\Biggl[{ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma +\\vartheta }(x) \\Biggr] {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[h^{\\vartheta }(x) { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) \\Biggr]\\Biggr) \\\\ &\\qquad {}\\Big/\\Biggl({}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\beta } \\Biggl[h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma +\\vartheta }(x) \\Biggr] {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\lambda } \\Biggl[g_{p}^{\\vartheta }{ \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) \\Biggr] \\\\ &\\qquad {}+{}_{\\alpha }^{\\mu } \\mathfrak{I}_{r}^{\\lambda } \\Biggl[h^{\\vartheta }(x){ \\prod_{i\\neq p}^{n}}g_{i}^{\\gamma _{i}}g_{p}^{\\sigma +\\vartheta }(x) \\Biggr] {}_{\\alpha }^{\\mu }\\mathfrak{I}_{r}^{\\beta } \\Biggl[g_{p}^{\\vartheta } { \\prod_{i=1}^{n}}g_{i}^{\\gamma _{i}}(x) \\Biggr]\\Biggr) \\\\ &\\quad \\geq 1, \\end{aligned}\n(27)\n\nwhere $$\\beta ,\\lambda \\in \\mathbb{C}$$, $$\\alpha \\in (0,1]$$, $$\\mu \\in \\mathbb{R}$$, $$\\alpha +\\mu \\neq 0$$, $$\\Re (\\beta )>0$$, and $$\\Re (\\lambda )>0$$.\n\n### Proof\n\nThe proof of Theorem 6 runs parallel as to the proof of Theorem 4 if we replace $$h^{\\vartheta }(\\rho )-h^{\\vartheta }(t)$$ by $$(g_{p}^{\\vartheta }(t)h^{\\vartheta }(\\rho )-g_{p}^{\\vartheta }( \\rho )h^{\\vartheta }(t) )$$. □\n\n### Remark 5\n\nIn a similar way, we can get the inequalities for the generalized right FCIO (2) and special cases for integrals (4) and (6).\n\n## References\n\n1. Abdeljawad, T.: On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015). https://doi.org/10.1016/j.cam.2014.10.016\n\n2. Abdeljawad, T.: A Lyapunov type inequality for fractional operators with nonsingular Mittag-Leffler kernel. J. Inequal. Appl. 2017, 130 (2017). https://doi.org/10.1186/s13660-017-1400-5\n\n3. Abdeljawad, T.: Fractional operators with exponential kernels and a Lyapunov type inequality. Adv. Differ. Equ. 2017, 313 (2017). https://doi.org/10.1186/s13662-017-1285-0\n\n4. Abdeljawad, T., Agarwal, R.P., Alzabut, J., Jarad, F., ÖZbekler, A.: Lyapunov-type inequalities for mixed non-linear forced differential equations within conformable derivatives. J. Inequal. Appl. 2018, 143 (2018). https://doi.org/10.1186/s13660-018-1731-x\n\n5. Abdeljawad, T., Alzabut, J., Jarad, F.: A generalized Lyapunov-type inequality in the frame of conformable derivatives. A generalized Lyapunov-type inequality in the frame of conformable derivatives. Adv. Differ. Equ. 2017, 321 (2017). https://doi.org/10.1186/s13662-017-1383-z\n\n6. Adjabi, Y., Jarad, F., Abdeljawad, T.: On generalized fractional operators and a Gronwall type inequality with applications. Filomat 31(17), 5457–5473 (2017). https://doi.org/10.2298/FIL1717457A\n\n7. Dahmani, Z.: New inequalities in fractional integrals. Int. J. Nonlinear Sci. 9(4), 493–497 (2010)\n\n8. Dahmani, Z.: New classes of integral inequalities of fractional order. Matematiche 69(1), 237–247 (2014). https://doi.org/10.4418/2014.69.1.18\n\n9. Dahmani, Z., Tabharit, L.: On weighted Gruss type inequalities via fractional integration. J. Adv. Res. Pure Math. 2, 31–38 (2010)\n\n10. Jarad, F., Ugurlu, E., Abdeljawad, T., Baleanu, D.: On a new class of fractional operators. Adv. Differ. Equ. 2017, 247 (2017)\n\n11. Khalil, R., Horani, M.A., Yousef, A., Sababheh, M.: A new definition of fractional derivative. J. Comput. Appl. Math. 264(65), 65–70 (2014)\n\n12. Khan, T.U., Khan, M.A.: Generalized conformable fractional integral operators. J. Comput. Appl. Math. 346, 378–389 (2019). https://doi.org/10.1016/j.cam.2018.07.018\n\n13. Kilbas, A.A.: Hadamard-type fractional calculus. J. Korean Math. Soc. 38, 1191–1204 (2001)\n\n14. Kiryakova, V.: Generalized Fractional Calculus and Applications. Pitman Res. Notes Math. Ser., vol. 301. Longman, New York (1994)\n\n15. Liu, W., Ngǒ, Q.A., Huy, V.N.: Several interesting integral inequalities. J. Math. Inequal. 3(2), 201–212 (2009)\n\n16. Nisar, K.S., Qi, F., Rahman, G., Mubeen, S., Arshad, M.: Some inequalities involving the extended gamma function and the Kummer confluent hypergeometric k-function. J. Inequal. Appl. 2018, 135 (2018)\n\n17. Nisar, K.S., Rahman, G., Choi, V., Mubeen, S., Arshad, M.: Certain Gronwall type inequalities associated with Riemann–Liouville k- and Hadamard k-fractional derivatives and their applications. East Asian Math. J. 34(3), 249–263 (2018)\n\n18. Nisar, K.S., Tassaddiq, A., Rahman, G., Khan, A.: Some inequalities via fractional conformable integral operators. J. Inequal. Appl. 2019, 217 (2019)\n\n19. Ntouyas, K.S., Agarwal, P., Tariboon, J.: On Polya–Szego and Chebyshev types inequalities involving the Riemann–Liouville fractional integral operators. J. Math. Inequal. 10(2), 491–504 (2016)\n\n20. Podlubny, I.: Fractional Differential Equations. Academic Press, London (1999)\n\n21. Qi, F., Rahman, G., Hussain, S.M., Du, W.S., Nisar, K.S.: Some inequalities of Čebyšev type for conformable k-fractional integral operators. Symmetry 10, 614 (2018). https://doi.org/10.3390/sym10110614\n\n22. Rahman, G., Khan, A., Abdeljawad, T., Nisar, K.S.: The Minkowski inequalities via generalized proportional fractional integral operators. Adv. Differ. Equ. 2019, 287 (2019). https://doi.org/10.1186/s13662-019-2229-7\n\n23. Rahman, G., Nisar, K.S., Mubeen, S., Choi, J.: Certain inequalities involving the $$(k,\\rho )$$-fractional integral operator. Far East J. Math. Sci.: FJMS 103(11), 1879–1888 (2018)\n\n24. Rahman, G., Nisar, K.S., Qi, F.: Some new inequalities of the Gruss type for conformable fractional integrals. AIMS Mathematics 3(4), 575–583 (2018)\n\n25. Rahman, G., Ullah, Z., Khan, A., Set, E., Nisar, K.S.: Certain Chebyshev type inequalities involving fractional conformable integral operators. Mathematics 7, 364 (2019). https://doi.org/10.3390/math7040364\n\n26. Samko, S.G., Kilbas, A.A., Marichev, O.I.: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, Reading (1993)\n\n27. Sarikaya, M.Z., Dahmani, Z., Kiris, M.E., Ahmad, F.: $$(k, s)$$-Riemann–Liouville fractional integral and applications. Hacet. J. Math. Stat. 45(1), 77–89 (2016)\n\n28. Set, E., Tomar, M., Sarikaya, M.Z.: On generalized Grüss type inequalities for k-fractional integrals. Appl. Math. Comput. 269, 29–34 (2015)\n\n29. Srivastava, H.M., Choi, J.: Zeta and q-Zeta Functions and Associated Series and Integrals. Elsevier, Amsterdam (2012)\n\n### Acknowledgements\n\nThe authors G. Rahman and A. Khan thanks to the Higher Education Commission of Pakistan for the support under the Start-Up Research Grant Project.\n\nNot applicable.\n\nNot applicable.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nThe authors have contributed equally to this manuscript. They read and approved the final manuscript.\n\n### Corresponding author\n\nCorrespondence to Kottakkaran Sooppy Nisar.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.",
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"https://advancesincontinuousanddiscretemodels.springeropen.com/track/article/10.1186/s13662-019-2362-3",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.67107934,"math_prob":1.0000006,"size":20230,"snap":"2023-40-2023-50","text_gpt3_token_len":6763,"char_repetition_ratio":0.16103035,"word_repetition_ratio":0.4166368,"special_character_ratio":0.36307463,"punctuation_ratio":0.22624227,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000095,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-29T17:26:55Z\",\"WARC-Record-ID\":\"<urn:uuid:ea5af691-f49a-40ca-8c9b-d8546d6a69b3>\",\"Content-Length\":\"354589\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:92942dcc-45e2-44e2-b1c1-0624487f9526>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a8942d9-ce7a-4e13-a27f-b4d64d149739>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1186/s13662-019-2362-3\",\"WARC-Payload-Digest\":\"sha1:LF7DR4E2LZULB4MPOTZR3Y32QMDFDQWG\",\"WARC-Block-Digest\":\"sha1:WRMQN6ACVWIAUFAEFZUFPY4WHNAJVQAZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510520.98_warc_CC-MAIN-20230929154432-20230929184432-00407.warc.gz\"}"} |
https://codebun.com/category/problem-solving/page/13/ | [
"Category: Problem Solving\n\nWrite a java program to find Sum of Squares of Even Digits.\n\nWrite a program to read a number, calculate the sum of squares of even digits (values) present in the given number. Input and Output Format: Input consists of a positive integer n. Output is a single integer . Refer sample output for formatting specifications. Sample Input 1: 56895 Sample Output 1: 100 Sum of Squares\n\nWrite a java program check Number Validation.\n\nWrite a program to read a string of 10 digit number, check whether the string contains a 10 digit number in the format XXX-XXX-XXXX where ‘X’ is a digit(Number Validation). Input and Output Format: Input consists of a string. The output is a string specifying the given string is valid or not. Refer sample output for\n\nWrite a java program to check sum of odd digits.\n\nWrite a program to read a number, calculate the sum of odd digits (values) present in the given number. Check Sum of Odd Digits in java Input and Output Format: Input consists of a positive integer n. Refer sample output for formatting specifications. Sample Input 1: 56895 Sample Output 1: Sum of odd digits is"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.697768,"math_prob":0.99293154,"size":948,"snap":"2019-26-2019-30","text_gpt3_token_len":211,"char_repetition_ratio":0.14724576,"word_repetition_ratio":0.3580247,"special_character_ratio":0.2278481,"punctuation_ratio":0.10638298,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971382,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T18:05:57Z\",\"WARC-Record-ID\":\"<urn:uuid:1196fdde-468b-40e5-ab41-1d14140d6aa4>\",\"Content-Length\":\"42192\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:31dde19d-537f-4f08-9ecc-a07c8a94dee4>\",\"WARC-Concurrent-To\":\"<urn:uuid:d39d5ac0-b945-4955-a093-287a004c3940>\",\"WARC-IP-Address\":\"198.54.115.144\",\"WARC-Target-URI\":\"https://codebun.com/category/problem-solving/page/13/\",\"WARC-Payload-Digest\":\"sha1:KKN5THSY6L5WBODRS2AITDBF4JYFTWAP\",\"WARC-Block-Digest\":\"sha1:JPQKSSZ3QIGVGHRM4LJXIUNAUSLJXXQP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999003.64_warc_CC-MAIN-20190619163847-20190619185847-00145.warc.gz\"}"} |
https://www.mbatious.com/topic/1022/quant-boosters-sibanand-pattnaik-set-2/?page=2%E2%8C%A9=en-US | [
"Quant Boosters - Sibanand Pattnaik - Set 2\n\n• 2310 = 2 * 3 * 5 * 7 * 11\nThese 5 primes to be distributed into 3 places.\nTotal : 3^5 ways , but this is ordered ( arranged) and we need unordered.\nWe know that any triplet with distinct elements (a,b,c) is arranged in 3! ways, so all the distinct triplets in 3^5 = 243 are arranged in 3! Ways.\nExcept for one triplet (1, 1, 2310) which is arranged in 3!/2= 3 ways\nSo remove this from total and we are left with only distinct, is divide by 3! ,\nSo (3^5 - 3)/3! , and just add that one case of 1, 1, 2310\nSo (3^5-3)/3! + 1 = 41\n\n• Q11) How many pairs of positive integers (m, n) satisfy 1/m + 4/n = 1/12. Where n is an odd integer less than 60.\n\n• (m -12) * (n-48) = 12 * 48 = 2^6 * 3^2\nnow for \"n\" to be odd , (n-48) has to be odd\nso n - 48 can be either 1 or 3 or 3^2 (because we need ODD)\n(m -12) * (n - 48) = 2^6 * 3^2 * 1\nOR (m -12) * (n - 48) = 2^6 * 3 * 3\nor (m -12) * (n - 48) = 2^6 * 3^2\nwhen n - 48 = 1, n = 49 (acceptable)\nwhen n - 48 = 3, n = 51 (acceptable)\nwhen n - 48 = 3^2, n = 57 (acceptable)\nonly these 3 values are possible for n < 60 and odd\n\n• Q12) Find number of whole number solutions of a + b + c + d = 20 where a, b, c and d ≤ 8\n\n• Cofficient of x^20 in (1 + x + x^2 + x^3 + ... + x^8)^4\n= coff of x^20 in (1 - x^9)^4 * (1 - x)^-4\n= 23c3 - 4 * 14c3 + 4c2 * 5c3\n= 1771 - 1456 + 60\n= 375\n\nFor more detailed explanation of the concept, refer the video.\n\n• Q13) If k is a factor of (n+5) and (6n+54) where n is a natural number, how many possible values of k are there?\na) 6\nb) 7\nc) 8\nd) 9\n\n• As given k must be a factor of 6n+30 and is also a factor of 6n+54.\nSo it must be a factor of (6n+54) – (6n+30) = 24 = 2^3 x 3\nTherefore k can take 8 values\nHence (c)\n\n• Q14) The sum of all the factors of 480 including 1 and the number itself is 1512. What is the sum of the reciprocals of all the factors?\na) 3.15\nb) 2.85\nc) 2.55\nd) 3.45\n\n• Sum of reciprocals = (sum of factors)/480\n1512/480= 3.15\n\n• Q15) Find the last two digits of 89^102\na) 23\nb) 21\nc) 01\nd) 69\n\n• (89^2)^51\n89^2 same as 11^2 -> 21^51 = 21\n\n• Q16) The highest power of 5 in N! is 34. Find the highest power of 7 in N!.\na) 23\nb) 26\nc) 22\nd) Cannot be determined\n\n• Highest power of 5 in 125!= 31.\nHighest power of 5 in 140!=34\nBut highest power of 5 in 141!, 142!, 143! And 144! Is also 34.\nSo N can be 140 to 144, any of these values\nHowever the highest power of 7 in any of these values is same i.e. 22\n\n• Q17) A! + B! = N^2, Where A, B and N are whole numbers. How many possible combinations are there?\na) 1\nb) 2\nc) 3\nd) 4\n\n• 0! + 4! = 5^2\n1! + 4! = 5^2\n0! + 5! = 11^2\n1! + 5! = 11^2\nHence 4 combinations are possible\n\n• Q18) x, y and z are natural numbers such that x > y > z > 1 and PRODUCT= 3003. Let A and B be the maximum and minimum values of x + y + z. Find A-B.\na) 152\nb) 45\nc) 107\nd) None of the above\n\n• 3003= 3 x 7 x 11 x 13\nA= 143+7+3= 153\nB= 21 + 11 + 13= 45\nA-B= 108\nHence (d)\n\n• Q19) How many natural numbers less than 4444 are \"not co primes\" to 247?\na) 585\nb) 341\nc) 244\nd) 557\n\n• 247= 13 x 19\nSo all multiples of 13 and 19 has to be excluded till 4444\nMultiples of 13 : 341\nMultiples of 19: 233\nMultiple of both 13 and 19 i.e. of 247: 17\nTherefore number of numbers less than 4444 which are co-prime to 247: 341+233-17= 557\n\n• Q20) N , a natural number has 2 prime factors. N^3 can have which of the following number of factors?\na) 18\nb) 90\nc) 70\nd) 84\n\n61\n\n61\n\n61\n\n64\n\n58\n\n61\n\n61\n\n62"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7658342,"math_prob":0.99922,"size":3380,"snap":"2019-43-2019-47","text_gpt3_token_len":1421,"char_repetition_ratio":0.097748816,"word_repetition_ratio":0.039653037,"special_character_ratio":0.48994082,"punctuation_ratio":0.1056338,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996685,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-23T10:09:45Z\",\"WARC-Record-ID\":\"<urn:uuid:256829e6-0813-4586-b8ee-079e41cd0619>\",\"Content-Length\":\"184838\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:03bfe22d-296c-4214-a396-35d0e45ce28b>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3b11995-4f5d-4e73-9fb4-dc72d7680b93>\",\"WARC-IP-Address\":\"104.28.9.51\",\"WARC-Target-URI\":\"https://www.mbatious.com/topic/1022/quant-boosters-sibanand-pattnaik-set-2/?page=2%E2%8C%A9=en-US\",\"WARC-Payload-Digest\":\"sha1:OMAFMA3SBF54OJ4UG76F5P5GOR6MZOI6\",\"WARC-Block-Digest\":\"sha1:2XW3GW7TVMN5AEQ5ZSN5572NUBRBNUN2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987833089.90_warc_CC-MAIN-20191023094558-20191023122058-00122.warc.gz\"}"} |
https://studyres.com/doc/4781246/midterm-exam-2003-question-1-discuss-two-of-the-following--a | [
"# Download Midterm Exam 2003 Question 1 Discuss two of the following: a\n\nSurvey\n\n* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project\n\nDocument related concepts\n\nModern Monetary Theory wikipedia , lookup\n\nExchange rate wikipedia , lookup\n\nPensions crisis wikipedia , lookup\n\nRagnar Nurkse's balanced growth theory wikipedia , lookup\n\nMonetary policy wikipedia , lookup\n\nFear of floating wikipedia , lookup\n\nBalance of trade wikipedia , lookup\n\nEconomic calculation problem wikipedia , lookup\n\nBalance of payments wikipedia , lookup\n\nAustrian business cycle theory wikipedia , lookup\n\nGross fixed capital formation wikipedia , lookup\n\nInterest rate wikipedia , lookup\n\nFiscal multiplier wikipedia , lookup\n\nTranscript\n```Midterm Exam 2003\nQuestion 1\nDiscuss two of the following:\na)\nPaasche and Laspeyeres Indices\nb)\nc)\nThe money multiplier\nQuestion 2\nConsider the following numerical version of the IS-LM model:\nC = 296+0.6YD\nYD = Y-T\nY = tY\nt = 0.4\nI = 200+0.2Y-1000i\nG=200\n(M/P)d = 0.5Y-2500i\n(M/P)s = 500\na)\nb)\nc)\nc)\ne)\ng)\nh)\ni)\nFind the equation for the IS curve\nFind an equation for the LM curve\nSolve for equilibrium output (Y) and the equilibrium interest rate (i).\nSolve for the equilibrium levels of consumption (C) and investment (I).\nNow assume that the government decides to increase the tax rate (t) to 60%,\ncalculate what happens to the equilibrium values of Y, i, C and I.\nUsing diagrams show graphically the impact of this policy on output and the\ninterest rate. Explain in words how this policy affects output, interest rates,\ninvestment and consumption.\nSetting all variables back to their original levels, calculate the new equilibrium\nvalues for Y, i, C and I if the Central Bank increased the money supply from 500\nto 600.\nUsing diagrams, show graphically the impact of this policy on output and the\ninterest rate. Explain in words how this policy affects output, interest rates,\ninvestment and consumption.\nQuestion 3\nThe data below corresponds to the Austrian economy in 2002. The numbers are given at\ncurrent prices and in billions of Euros.\nGDP at market prices\nGross operating surplus and mixed income (Profit)\nTaxes less subsidies on production\nGross investment expenditure\nGovernment consumption\nExports of goods and services\nImports of goods and services\nBalance of primary income to the rest of the world\nBalance of current transfers to the rest of the world\nConsumption of fixed capital (Depreciation)\n218.4\n80.0\n26.5\n48.9\n40.7\n115.2\n110.4\n2.0\n1.7\n32.2\nCompute the following:\n(i)\nwages and salaries\n(ii)\nconsumption of households (including non-profit institutions serving households)\n(iii) current account balance\n(iv)\ncapital account balance\n(v)\ngross national income\n(vi)\nnet national income\n(vii) net national disposable income\n(viii) net domestic product\n(ix)\nnet domestic product to factor costs and net national product to factor costs\n```"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.82133424,"math_prob":0.93662566,"size":2798,"snap":"2023-14-2023-23","text_gpt3_token_len":641,"char_repetition_ratio":0.14030065,"word_repetition_ratio":0.111856826,"special_character_ratio":0.231594,"punctuation_ratio":0.1003937,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959095,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-01T15:09:43Z\",\"WARC-Record-ID\":\"<urn:uuid:d3e97cf2-0c65-4e27-bfee-b9a731b265fa>\",\"Content-Length\":\"33889\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9427d7e-7817-403b-957e-9abe5be3709c>\",\"WARC-Concurrent-To\":\"<urn:uuid:dd6b1c75-b4a8-4468-a2fe-65bbf18fc80b>\",\"WARC-IP-Address\":\"172.67.151.140\",\"WARC-Target-URI\":\"https://studyres.com/doc/4781246/midterm-exam-2003-question-1-discuss-two-of-the-following--a\",\"WARC-Payload-Digest\":\"sha1:6UN7NJHUADTO5UEILEXT67Z4W5WCNN5R\",\"WARC-Block-Digest\":\"sha1:FDLD4M36ZPFTXP2HVS4LFT2B6BHFX2ZY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296950030.57_warc_CC-MAIN-20230401125552-20230401155552-00075.warc.gz\"}"} |
https://kilthub.cmu.edu/articles/journal%20contribution/High-Dimensional_Graphical_Model_Selection_Using_sub_1_sub_-Regularized_Logistic_Regression/6606098/1 | [
"file.pdf (259.54 kB)\n0/0\n\n## High-Dimensional Graphical Model Selection Using ℓ1-Regularized Logistic Regression\n\njournal contribution\nposted on 01.07.2000 by Pradeep Ravikumar, Martin J. Wainwright, John Lafferty\nWe consider the problem of estimating the graph structure associated with a discrete Markov random field. We describe a method based on ℓ1-regularized logistic regression, in which the neighborhood of any given node is estimated by performing logistic regression subject to an ℓ1-constraint. Our framework applies to the high-dimensional setting, in which both the number of nodes p and maximum neighborhood sizes d are allowed to grow as a function of the number of observations n. Our main results provide sufficient conditions on the triple (n, p, d) for the method to succeed in consistently estimating the neighborhood of every node in the graph simultaneously. Under certain assumptions on the population Fisher information matrix, we prove that consistent neighborhood selection can be obtained for sample sizes n = Ω(d3 log p), with the error decaying as O(exp(−Cn/d3)) for some constant C. If these same assumptions are imposed directly on the sample matrices, we show that n= Ω(d2 log p) samples are sufficient.\n\n01/07/2000\n\n### Exports\n\nfigshare. credit for all your research."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.88839906,"math_prob":0.9451159,"size":1329,"snap":"2020-34-2020-40","text_gpt3_token_len":283,"char_repetition_ratio":0.0890566,"word_repetition_ratio":0.0,"special_character_ratio":0.18961625,"punctuation_ratio":0.08658009,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9605072,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T09:13:53Z\",\"WARC-Record-ID\":\"<urn:uuid:2cdbf859-e989-435f-9348-9385560b1121>\",\"Content-Length\":\"100259\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:89bcd145-4b91-4ed8-af70-2f0cd0b8ab85>\",\"WARC-Concurrent-To\":\"<urn:uuid:afb4c716-24a0-476b-989a-833772773866>\",\"WARC-IP-Address\":\"34.247.3.232\",\"WARC-Target-URI\":\"https://kilthub.cmu.edu/articles/journal%20contribution/High-Dimensional_Graphical_Model_Selection_Using_sub_1_sub_-Regularized_Logistic_Regression/6606098/1\",\"WARC-Payload-Digest\":\"sha1:73P6JMUKDCTKYXW2PT74SCV7ORDOH55S\",\"WARC-Block-Digest\":\"sha1:P5OAZRNQPR424CWP4OKBEDOOVHUMGRYI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402124756.81_warc_CC-MAIN-20201001062039-20201001092039-00709.warc.gz\"}"} |
https://ounces-to-grams.appspot.com/pl/1549-uncja-na-gram.html | [
"Ounces To Grams\n\n# 1549 oz to g1549 Ounce to Grams\n\noz\n=\ng\n\n## How to convert 1549 ounce to grams?\n\n 1549 oz * 28.349523125 g = 43913.4113206 g 1 oz\nA common question is How many ounce in 1549 gram? And the answer is 54.6393670599 oz in 1549 g. Likewise the question how many gram in 1549 ounce has the answer of 43913.4113206 g in 1549 oz.\n\n## How much are 1549 ounces in grams?\n\n1549 ounces equal 43913.4113206 grams (1549oz = 43913.4113206g). Converting 1549 oz to g is easy. Simply use our calculator above, or apply the formula to change the length 1549 oz to g.\n\n## Convert 1549 oz to common mass\n\nUnitMass\nMicrogram43913411320.6 µg\nMilligram43913411.3206 mg\nGram43913.4113206 g\nOunce1549.0 oz\nPound96.8125 lbs\nKilogram43.9134113206 kg\nStone6.9151785714 st\nUS ton0.04840625 ton\nTonne0.0439134113 t\nImperial ton0.0432198661 Long tons\n\n## What is 1549 ounces in g?\n\nTo convert 1549 oz to g multiply the mass in ounces by 28.349523125. The 1549 oz in g formula is [g] = 1549 * 28.349523125. Thus, for 1549 ounces in gram we get 43913.4113206 g.\n\n## 1549 Ounce Conversion Table",
null,
"## Alternative spelling\n\n1549 oz in Grams, 1549 oz to Gram, 1549 Ounces to g, 1549 oz to g, 1549 Ounce to Grams, 1549 Ounce in Grams, 1549 Ounce to Gram, 1549 Ounce in Gram, 1549 Ounce in g,"
] | [
null,
"https://ounces-to-grams.appspot.com/image/1549.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7691788,"math_prob":0.82944036,"size":892,"snap":"2023-40-2023-50","text_gpt3_token_len":314,"char_repetition_ratio":0.2117117,"word_repetition_ratio":0.0,"special_character_ratio":0.47309417,"punctuation_ratio":0.15384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9746438,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-04T21:00:50Z\",\"WARC-Record-ID\":\"<urn:uuid:69de1a97-3add-4777-b25d-2e08efcaf8e0>\",\"Content-Length\":\"28438\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0f09de09-d663-4225-ac85-dc511909763f>\",\"WARC-Concurrent-To\":\"<urn:uuid:b0cd2bd7-a42b-48b7-a630-4b23d558c03d>\",\"WARC-IP-Address\":\"172.253.122.153\",\"WARC-Target-URI\":\"https://ounces-to-grams.appspot.com/pl/1549-uncja-na-gram.html\",\"WARC-Payload-Digest\":\"sha1:UNT2DISAH3BVWB7B3XVHJN3N6CXP24UF\",\"WARC-Block-Digest\":\"sha1:C5AVU7EM7D3HQU77EEXAK5FH2USFUSPC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100534.18_warc_CC-MAIN-20231204182901-20231204212901-00310.warc.gz\"}"} |
https://perl5.git.perl.org/perl5.git/blob/bdce6f371e87eca951cb070d597587ec0dc9c25f:/dist/Math-BigInt-FastCalc/lib/Math/BigInt/FastCalc.pm | [
"This is a live mirror of the Perl 5 development currently hosted at https://github.com/perl/perl5\n1 package Math::BigInt::FastCalc;\n3 use 5.006;\n4 use strict;\n5 use warnings;\n7 use Math::BigInt::Calc 1.997;\n9 use vars '\\$VERSION';\n11 \\$VERSION = '0.31';\n13 ##############################################################################\n14 # global constants, flags and accessory\n16 # announce that we are compatible with MBI v1.83 and up\n17 sub api_version () { 2; }\n19 # use Calc to override the methods that we do not provide in XS\n21 for my \\$method (qw/\n22 str num\n24 rsft lsft\n25 mod modpow modinv\n26 gcd\n27 pow root sqrt log_int fac nok\n28 digit check\n29 from_hex from_bin from_oct as_hex as_bin as_oct\n30 zeros base_len\n31 xor or and\n32 alen 1ex\n33 /)\n34 {\n35 no strict 'refs';\n36 *{'Math::BigInt::FastCalc::_' . \\$method} = \\&{'Math::BigInt::Calc::_' . \\$method};\n37 }\n42 ##############################################################################\n43 ##############################################################################\n45 1;\n46 __END__\n47 =pod\n51 Math::BigInt::FastCalc - Math::BigInt::Calc with some XS for more speed\n55 Provides support for big integer calculations. Not intended to be used by\n56 other modules. Other modules which sport the same functions can also be used\n57 to support Math::BigInt, like L<Math::BigInt::GMP> or L<Math::BigInt::Pari>.\n61 In order to allow for multiple big integer libraries, Math::BigInt was\n62 rewritten to use library modules for core math routines. Any module which\n63 follows the same API as this can be used instead by using the following:\n65 use Math::BigInt lib => 'libname';\n67 'libname' is either the long name ('Math::BigInt::Pari'), or only the short\n68 version like 'Pari'. To use this library:\n70 use Math::BigInt lib => 'FastCalc';\n72 Note that from L<Math::BigInt> v1.76 onwards, FastCalc will be loaded\n73 automatically, if possible.\n77 FastCalc works exactly like Calc, in stores the numbers in decimal form,\n78 chopped into parts.\n82 The following functions are now implemented in FastCalc.xs:\n84 _is_odd _is_even _is_one _is_zero\n85 _is_two _is_ten\n86 _zero _one _two _ten\n87 _acmp _len\n88 _inc _dec\n89 __strip_zeros _copy"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5522286,"math_prob":0.842744,"size":2373,"snap":"2022-05-2022-21","text_gpt3_token_len":624,"char_repetition_ratio":0.23596454,"word_repetition_ratio":0.0,"special_character_ratio":0.3771597,"punctuation_ratio":0.2515213,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9753192,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T10:13:12Z\",\"WARC-Record-ID\":\"<urn:uuid:abb22f40-32f9-4131-8fac-e01389095a3b>\",\"Content-Length\":\"32180\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:11a990c9-a2d3-476c-a04f-0c1eceda7783>\",\"WARC-Concurrent-To\":\"<urn:uuid:c57569a3-4e5c-48fd-9d87-30a942ae95b4>\",\"WARC-IP-Address\":\"188.166.18.220\",\"WARC-Target-URI\":\"https://perl5.git.perl.org/perl5.git/blob/bdce6f371e87eca951cb070d597587ec0dc9c25f:/dist/Math-BigInt-FastCalc/lib/Math/BigInt/FastCalc.pm\",\"WARC-Payload-Digest\":\"sha1:6K5NYJDZ2DZED44TO4YOMTUHG5PAMXKS\",\"WARC-Block-Digest\":\"sha1:GEG2WYNN3IA6SB65ZQYATR7B7LL5SBRZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662526009.35_warc_CC-MAIN-20220519074217-20220519104217-00015.warc.gz\"}"} |
http://news.datascience.org.ua/2019/02/10/time-series-in-python%E2%80%8A-%E2%80%8Aexponential-smoothing-and-arima-processes/ | [
"# Time Series in Python — Exponential Smoothing and ARIMA processes\n\nTime Series in Python — Exponential Smoothing and ARIMA processesBenjamin EtienneBlockedUnblockFollowFollowingFeb 9TL;DR: In this article you’ll learn the basics steps to performing time-series analysis and concepts like trend, stationarity, moving averages, etc.\n\nYou’ll also explore exponential smoothing methods, and learn how to fit an ARIMA model on non-stationary data.\n\nDefinitionA time series is a data sequence ordered (or indexed) by time.\n\nIt is discrete, and the the interval between each point is constant.\n\nProperties and types of seriesTrend : A long-term increase or decrease in the data.\n\nThis can be seen as a slope (is doesn’t have to be linear) roughly going through the data.\n\nSeasonality : A time series is said to be seasonal when it is affected by seasonal factors (hour of day, week, month, year, etc.\n\n).\n\nSeasonality can be observed with nice cyclical patterns of fixed frequency.\n\nCyclicity : A cycle occurs when the data exhibits rises and falls that are not of a fixed frequency.\n\nThese fluctuations are usually due to economic conditions, and are often related to the “business cycle”.\n\nThe duration of these fluctuations is usually at least 2 years.\n\nLeft: Series with a trend.\n\nRight: Series exhibiting seasonal behaviorLeft: Cyclical series.\n\nRight: Random walkThe golden rule: StationarityBefore going any further into our analysis, our series has to be made stationary.\n\nStationarity is the property of exhibiting constant statistical properties (mean, variance, autocorrelation, etc.\n\n).\n\nIf the mean of a time-series increases over time, then it’s not stationary.\n\nTransforms used for stationarizing data:De-trending : We remove the underlying trend in the series.\n\nThis can be done in several ways, depending on the nature of data :- Indexed data: data measured in currencies are linked to a price index or related to inflation.\n\nDividing the series by this index (ie deflating) element-wise is therefore the solution to de-trend the data.\n\n– Non-indexed data: is it necessary to estimate if the trend is constant, linear or exponential.\n\nThe first two cases are easy, for the last one it is necessary to estimate a growth rate (inflation or deflation) and apply the same method as for indexed data.\n\nDifferencing : Seasonal or cyclical patterns can be removed by substracting periodical values.\n\nIf the data is 12-month seasonal, substracting the series with a 12-lag difference series will give a “flatter” seriesLogging : in the case where the compound rate in the trend is not due to a price index (ie the series is not measured in a currency), logging can help linearize a series with an exponential trend (recall that log(exp(x)) = x).\n\nIt does not remove an eventual trend whatsoever, unlike deflation.\n\nChecking StationarityPlotting rolling statisticsPlotting rolling means and variances is a first good way to visually inspect our series.\n\nIf the rolling statistics exhibit a clear trend (upwards or downwards) and show varying variance (increasing or decreasing amplitude), then you might conclude that the series is very likely not to be stationary.\n\nAugmented Dickey-Fuller TestThis test is used to assess whether or not a time-series is stationary.\n\nWithout getting into too much details about hypothesis testing, you should know that this test will give a result called a “test-statistic”, based on which you can say, with different levels (or percentage) of confidence, if the time-series is stationary or not.\n\nKPSSThe KPSS (Kwiatkowski-Phillips-Schmidt-Shin) test tests for the null hypothesis that the series is trend stationary.\n\nIn other words, if the test statistic is above the X% confidence threshold, this means we reject this hypothesis and that the series is not trend-stationary with X% confidence.\n\nA test statistic lower than the threshold will lead us to accept this hypothesis and conclude that the series is trend-stationary.\n\nAutocorrelation plots (ACF & PACF)An autocorrelation (ACF) plot represents the autocorrelation of the series with lags of itself.\n\nA partial autocorrelation (PACF) plot represents the amount of correlation between a series and a lag of itself that is not explained by correlations at all lower-order lags.\n\nIdeally, we want no correlation between the series and lags of itself.\n\nGraphically speaking, we would like all the spikes to fall in the blue region.\n\nAs we can see, there are several spikes above the blue region, meaning there are correlations at lags 1, 2, 3 and 4.\n\nChoosing a modelExponential smoothings methods are appropriate for non-stationary data (ie data with a trend and seasonal data).\n\nARIMA models should be used on stationary data only.\n\nOne should therefore remove the trend of the data (via deflating or logging), and then look at the differenced series.\n\nNon-Seasonal smoothing or trend-fittingSimple Exponential Smoothing & Linear Exponential SmoothingWhen to use?> Few data points> Irregular dataWinter’s Seasonal SmoothingThe problem with Holt’s Linear trend method is that the trend is constant in the future, increasing or decreasing indefinitely.\n\nFor long forecast horizons, this can be problematic.\n\nThe damped trend method is therefore a method which add a dampening parameter so that the trend converges to a constant value in the future (it flattens the trend).\n\nThe parameter ????.is replaced by ????????When to use?>Data has a trend and is seasonal>Use the multiplicative version, unless the data has been logged before.\n\nIn this case, use the additive versionARIMAARIMA models (which include ARMA, AR and MA models) are a general class of models to forecast stationary time series.\n\nARIMA models are made of three parts:A weighted sum of lagged values of the series (Auto-regressive (AR) part)A weighted sum of lagged forecasted errors of the series (Moving-average (MA) part)A difference of the time series (Integrated (I) part)An ARIMA model is often noted ARIMA(p, d, q) where p represents the order of the AR part, d the order of differencing (“I” part), and q the order of the MA term.\n\n1) Choosing the differencing orderThe first step of fitting an ARIMA model is to determine the differencing order to stationarize the series.\n\nTo do that, we look at the ACF and PACF plots, and keep in mind these two rules:“ — Rule 1 : If the series has positive autocorrelations out to a high number of lags, then it probably needs a higher order of differencing.\n\n— Rule 2 : If the lag-1 autocorrelation is zero or negative, or the autocorrelations are all small and patternless, then the series does not need a higher order of differencing.\n\nIf the lag-1 autocorrelation is -0.\n\n5 or more negative, the series may be overdifferenced.\n\n” (Robert Nau, Statistical Forecasting)We start by logging the data as the raw data exhibits an exponential trend:The logged series seems to be more flat, but is it stationary ?.Let’s compute a KPSS test to check this :The test statistic is above the critical values, we reject the null hypothesis, our series is not trend stationary.\n\nHowever, in order to keep this article short, we will continue as if it were.\n\n(— To have a trend-stationary series, we could for instance think of linearly regressing our logged series, and divide our series by the coefficient of the regression.\n\n— )Let’s now look at the ACF plots for the logged-first-difference data:The “ACF — Logged data” chart shows non-stationary data, characterized by the slow linear decay in the spikes (cf rule 1 above).\n\nAdding a 1st order difference gives a single negative spike at lag value 1.\n\nAccording to rule#2, we don’t need to differentiate the series any further.\n\nLet’s check our results by comparing a (0, 0, 0) and a (0, 1, 0) ARIMA model:The Akaike Information Criterion (AIC) is lower with the ARIMA(0,1,0), meaning this model is performing better than the ARIMA(0,0,0).\n\nLet’s have a look at the residuals and check their variance:Which residuals do you think look better ?2) Choosing the MA orderNow we know we have to include a 1st order difference in our model, we need to choose the Moving-Average order.\n\nThis is done by looking at the differenced series (because we just saw that the first-order difference series was stationary).\n\nAgain, we look at our ACF and PACF plots, with this rule in mind:“ If the lag-1 autocorrelation of the differenced series ACF is negative, and/or there is a sharp cutoff, then choose a MA order of 1”.\n\nQuestion : why choosing an MA order of 1, and not higher ?.Well it’s because we will proceed step by step.\n\nIf we observe autocorrelations at higher lags, and by looking at the autocorrelation residuals of our (1,1,0) model, we still observe these spikes, we can increase our MA order, though is usually not recommended to go beyond 2!Notice how the AIC has dropped again, and how the residuals variance decreased.\n\nThat’s a sign our (1,1,0) ARIMA is performing better than the (0,1,0) model !3) Choosing the AR orderNow you might wonder: Do we have to add an AR term at all ?.The answer is no.\n\nIn fact, you should add an AR term in the case where :“ If the lag-1 autocorrelation of the differenced series PACF is negative, and/or there is a sharp cutoff, then choose a AR order of 1”.\n\nIn our case, we observe negative lag-1 autocorrelations in both the ACF and the PACF.\n\nNotice that the PACF of the differenced series shows two negative spikes at lags 1 and 2, meaning we could in theory push our AR order up to 2.\n\nHere is what we get by fitting a (1,1,1) ARIMA:Adding an AR term has lowered the AIC and the variance went from 0.\n\n155 to 0.\n\n145, we’re doing good !.Should we add another AR term and pick a (2,1,1) ARIMA?.Let’s have a look at the ACF & PACF of the ARIMA(1,1,1) residuals:There are no significant autocorrelation lag values in the plots, our model doesn’t seem to need any additional AR terms.\n\nYou might point out the spike at lag 12: this could be a sign of seasonality.\n\nWe will look at it in the next part with SARIMA and SARIMAX.\n\nPlotting predictionsLuckily, statsmodels has a nice API which allows us to plot predictions from ARIMA processes.\n\nI highly recommend you put your data in a DataFrame with a DateTimeIndex, because the plot_predict() method really likes dates…In this case, we will choose 12 forecasting steps, and will set the dynamic keyword to True: this will force the algorithm to use its own forecasted values as the lagged values for future forecasts:Let’s have a look on a larger prediction window (34 forecasting steps):That’s all folks ! In the next article, I will discuss Regression models.\n\nHope you enjoyed it :)Additional lectures:I highly recommended the two below if you are interested in time-series forecasting and analysis:Forecasting, Principles and Practice (Hyndman & Athanasopoulos)Statistical Forecasting (Robert Nau)You can also check this article on Neural Turing Machines and this article on Attention-based Neural Networks !.. More details"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9060471,"math_prob":0.8853572,"size":10821,"snap":"2019-35-2019-39","text_gpt3_token_len":2484,"char_repetition_ratio":0.14283073,"word_repetition_ratio":0.01915493,"special_character_ratio":0.21291932,"punctuation_ratio":0.11907021,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9769339,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T17:46:11Z\",\"WARC-Record-ID\":\"<urn:uuid:cfcef87f-f324-46ed-8b25-d4e14696f462>\",\"Content-Length\":\"40745\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d1d3ba6e-1d4e-489c-81c5-f1f4570be84a>\",\"WARC-Concurrent-To\":\"<urn:uuid:71524ee8-cd73-43ff-a9b3-267162029ede>\",\"WARC-IP-Address\":\"78.47.26.210\",\"WARC-Target-URI\":\"http://news.datascience.org.ua/2019/02/10/time-series-in-python%E2%80%8A-%E2%80%8Aexponential-smoothing-and-arima-processes/\",\"WARC-Payload-Digest\":\"sha1:CUNXHZ4M7QXVFIA57EUZMU6343HUN7YL\",\"WARC-Block-Digest\":\"sha1:HORKWIWSNL2GXVDFI5RLGSEYTTA4EMRF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317339.12_warc_CC-MAIN-20190822172901-20190822194901-00182.warc.gz\"}"} |
https://socratic.org/questions/how-do-solve-2x-x-2-3-algebraically | [
"# How do solve (2x)/(x-2)<=3 algebraically?\n\nAug 6, 2018\n\nMultiply both sides by $x - 2$\n\n$2 x \\le 3 \\left(x - 2\\right)$\n\nDistribute the 3\n\n$2 x \\le 3 x - 6$\n\nMove $x$ terms to one side\n\n$- x \\le - 6$\n\nDivide out negative one from both sides\n\n$x \\ge 6$\n\n*When dividing or multiplying by a negative, you have to switch the greater or less than sign.\n\nAug 7, 2018\n\n$x \\in \\left(- 00 , 2\\right) \\cup \\left[6 , \\infty\\right)$\n\n#### Explanation:\n\n$\\text{subtract 3 from both sides}$\n\n$\\frac{2 x}{x - 2} - 3 \\le 0$\n\n$\\text{combine the left side as a single fraction}$\n\n$\\frac{2 x}{x - 2} - \\frac{3 \\left(x - 2\\right)}{x - 2} \\le 0$\n\n$\\frac{6 - x}{x - 2} \\le 0$\n\n$\\text{find the critical values of numerator/denominator}$\n\n$6 - x = 0 \\Rightarrow x = 6 \\leftarrow \\textcolor{b l u e}{\\text{is a zero}}$\n\n$x - 2 = 0 \\Rightarrow x = 2$\n\n$\\text{these values divide the domain into 3 intervals}$\n\n$\\left(- \\infty , 2\\right) \\cup \\left(2 , 6\\right] \\cup \\left[6 , \\infty\\right)$\n\n$\\text{select a value for x as a \"color(red)\"test point in each interval}$\n\n$x = 1 \\to \\frac{6 - 1}{1 - 2} = - 5 < 0 \\leftarrow \\textcolor{b l u e}{\\text{valid}}$\n\n$x = 3 \\to \\frac{6 - 3}{3 - 2} = 3 > 0 \\leftarrow \\textcolor{b l u e}{\\text{not valid}}$\n\n$x = 10 \\to \\frac{6 - 10}{10 - 2} = - \\frac{1}{2} < 0 \\leftarrow \\textcolor{b l u e}{\\text{valid}}$\n\n$x \\in \\left(- \\infty , 2\\right) \\cup \\left[6 , \\infty\\right)$\ngraph{(2x)/(x-2)-3 [-10, 10, -5, 5]}"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5203426,"math_prob":0.9999993,"size":901,"snap":"2020-34-2020-40","text_gpt3_token_len":350,"char_repetition_ratio":0.12374582,"word_repetition_ratio":0.0,"special_character_ratio":0.39289677,"punctuation_ratio":0.06282722,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999075,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-12T10:23:36Z\",\"WARC-Record-ID\":\"<urn:uuid:fec45d6d-bae3-4c6b-a9fc-cb9efd5fb65e>\",\"Content-Length\":\"34965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6fe30eff-61ac-4af1-8003-d16d6dae8f7a>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe2f0444-bb4c-4c17-bca5-00f468c89a71>\",\"WARC-IP-Address\":\"216.239.34.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/how-do-solve-2x-x-2-3-algebraically\",\"WARC-Payload-Digest\":\"sha1:KJUILLPFLSPZBJ2KM5E5CLEPLO5CPPEJ\",\"WARC-Block-Digest\":\"sha1:3UECKLEM7DQ3IQMZ4NCIKADDC3WN7ORN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738888.13_warc_CC-MAIN-20200812083025-20200812113025-00337.warc.gz\"}"} |
https://math.paperswithcode.com/paper/explicit-formulas-for-heat-kernels-on-diamond | [
"# Explicit formulas for heat kernels on diamond fractals\n\n19 Jul 2018 · Ruiz Patricia Alonso ·\n\nThis paper provides explicit pointwise formulas for the heat kernel on compact metric measure spaces that belong to a $(\\mathbb{N}\\times\\mathbb{N})$-parameter family of fractals which are regarded as projective limits of metric measure graphs and do not satisfy the volume doubling property. The formulas are applied to obtain uniform continuity estimates of the heat kernel and to derive an expression of the fundamental solution of the free Schr\\\"odinger equation... The results also open up the possibility to approach infinite dimensional spaces based on this model. read more\n\nPDF Abstract\n\n## Code Add Remove Mark official\n\nNo code implementations yet. Submit your code now\n\nProbability"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.84054476,"math_prob":0.9334382,"size":648,"snap":"2021-43-2021-49","text_gpt3_token_len":125,"char_repetition_ratio":0.10403727,"word_repetition_ratio":0.0,"special_character_ratio":0.17592593,"punctuation_ratio":0.046728972,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98186046,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-05T16:47:20Z\",\"WARC-Record-ID\":\"<urn:uuid:7318d287-a038-45df-804c-6eb29d0b7cf8>\",\"Content-Length\":\"76020\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ed527c6a-6b72-4805-ba60-be9cd7a17252>\",\"WARC-Concurrent-To\":\"<urn:uuid:9b890f38-f79f-4183-b760-d87c15e3d622>\",\"WARC-IP-Address\":\"104.26.13.155\",\"WARC-Target-URI\":\"https://math.paperswithcode.com/paper/explicit-formulas-for-heat-kernels-on-diamond\",\"WARC-Payload-Digest\":\"sha1:7YDMNRC4CW5P32NM52Q26QWFHICY5SRQ\",\"WARC-Block-Digest\":\"sha1:QCSQETPUHEMOWPUQZVZSL7UCTHKLSN7P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363215.8_warc_CC-MAIN-20211205160950-20211205190950-00096.warc.gz\"}"} |
https://vertx.io/docs/4.1.8/apidocs/io/vertx/pgclient/data/package-summary.html | [
"# Package io.vertx.pgclient.data\n\n• Class Summary\nClass Description\nBox\nRectangular box data type in Postgres represented by pairs of `Point`s that are opposite corners of the box.\nCircle\nCircle data type in Postgres represented by a center `Point` and radius.\nInet\nInterval\nPostgres Interval is date and time based such as 120 years 3 months 332 days 20 hours 20 minutes 20.999999 seconds\nLine\nLine data type in Postgres represented by the linear equation Ax + By + C = 0, where A and B are not both zero.\nLineSegment\nFinite line segment data type in Postgres represented by pairs of `Point`s that are the endpoints of the segment.\nPath\nPath data type in Postgres represented by lists of connected points.\nPoint\nA Postgresql point.\nPolygon\nPolygon data type in Postgres represented by lists of points (the vertexes of the polygon)."
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https://tools.carboncollective.co/compound-interest/32704-at-33-percent-in-20-years/ | [
"# What is the compound interest on $32704 at 33% over 20 years? If you want to invest$32,704 over 20 years, and you expect it will earn 33.00% in annual interest, your investment will have grown to become $9,809,202.35. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$32,704 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 20 years, but it could be monthly, weekly, daily, or even continuously compounding.\n\nThe formula for calculating compound interest is:\n\n$$A = P(1 + \\dfrac{r}{n})^{nt}$$\n\n• A is the amount of money after the compounding periods\n• P is the principal amount\n• r is the annual interest rate\n• n is the number of compounding periods per year\n• t is the number of years\n\nWe can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.\n\nFor this formula, we need to convert the rate, 33.00% into a decimal, which would be 0.33.\n\n$$A = 32704(1 + \\dfrac{ 0.33 }{1})^{ 20}$$\n\nAs you can see, we are ignoring the n when calculating this to the power of 20 because our example is for annual compounding, or one period per year, so 20 × 1 = 20.\n\n## How the compound interest on $32,704 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 20 years it is compounding: Start Balance Interest End Balance 1$32,704.00 $10,792.32$43,496.32\n2 $43,496.32$14,353.79 $57,850.11 3$57,850.11 $19,090.53$76,940.64\n4 $76,940.64$25,390.41 $102,331.05 5$102,331.05 $33,769.25$136,100.30\n6 $136,100.30$44,913.10 $181,013.40 7$181,013.40 $59,734.42$240,747.82\n8 $240,747.82$79,446.78 $320,194.60 9$320,194.60 $105,664.22$425,858.82\n10 $425,858.82$140,533.41 $566,392.23 11$566,392.23 $186,909.43$753,301.66\n12 $753,301.66$248,589.55 $1,001,891.21 13$1,001,891.21 $330,624.10$1,332,515.31\n14 $1,332,515.31$439,730.05 $1,772,245.36 15$1,772,245.36 $584,840.97$2,357,086.33\n16 $2,357,086.33$777,838.49 $3,134,924.81 17$3,134,924.81 $1,034,525.19$4,169,450.00\n18 $4,169,450.00$1,375,918.50 $5,545,368.51 19$5,545,368.51 $1,829,971.61$7,375,340.11\n20 $7,375,340.11$2,433,862.24 $9,809,202.35 We can also display this data on a chart to show you how the compounding increases with each compounding period. As you can see if you view the compounding chart for$32,704 at 33.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.\n\n## How long would it take to double $32,704 at 33% interest? Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of$32,704 assuming an interest rate of 33.00%.\n\nWe can calculate this very approximately using the Rule of 72.\n\nThe formula for this is very simple:\n\n$$Years = \\dfrac{72}{Interest\\: Rate}$$\n\nBy dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:\n\n$$Years = \\dfrac{72}{ 33 } = 2.18$$\n\nUsing this, we know that any amount we invest at 33.00% would double itself in approximately 2.18 years. So $32,704 would be worth$65,408 in ~2.18 years.\n\nWe can also calculate the exact length of time it will take to double an amount at 33.00% using a slightly more complex formula:\n\n$$Years = \\dfrac{log(2)}{log(1 + 0.33)} = 2.43\\; years$$\n\nHere, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.\n\nAs you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.\n\nHopefully, this article has helped you to understand the compound interest you might achieve from investing \\$32,704 at 33.00% over a 20 year investment period."
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9175824,"math_prob":0.99920815,"size":4230,"snap":"2022-40-2023-06","text_gpt3_token_len":1356,"char_repetition_ratio":0.13771889,"word_repetition_ratio":0.01490313,"special_character_ratio":0.42434987,"punctuation_ratio":0.20702754,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998753,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-27T03:54:07Z\",\"WARC-Record-ID\":\"<urn:uuid:ee2a8cc5-e829-4759-a0f4-6d6bffccd91c>\",\"Content-Length\":\"28212\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:461a60f9-5f11-4638-bf87-3f6f887c231b>\",\"WARC-Concurrent-To\":\"<urn:uuid:83853c5f-0549-4125-ac08-19260a233121>\",\"WARC-IP-Address\":\"138.197.3.89\",\"WARC-Target-URI\":\"https://tools.carboncollective.co/compound-interest/32704-at-33-percent-in-20-years/\",\"WARC-Payload-Digest\":\"sha1:UQREEACCY6JLFJFXADIH6E4QY7QV6MEH\",\"WARC-Block-Digest\":\"sha1:SSGEPFJYZXCSNELLDSGA63GDENMJLATT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334987.39_warc_CC-MAIN-20220927033539-20220927063539-00738.warc.gz\"}"} |
https://www.forecastsolutions.co.uk/price-sensitivity-analysis.htm | [
"Price Sensitivity Analysis using Demand Curves and Pricing Models",
null,
"",
null,
"Welcome to the Forecast Solutions website",
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"",
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"Demand Curves and Pricing Models for Price Sensitivity Analysis\n\nMeasurement of the effect of price changes on product demand is fundamental to the formulation of a pricing strategy. Price sensitivity analysis fulfills that need by measuring the effect of price on demand by defining the demand curve. If a pricing model can be formulated by analysis of historical demand, then a marketing strategy can be defined with much lower risk than would be the case with live experimentation. Because the effect of a company's prices depends also on competitor prices, it is often helpful to include in the price analysis a causal factor calculated as an index of price relative to total market or key competitor prices.\n\nPricing Models and Demand Curves\n\nA pricing model is the mathematical formula that is used to descibe price sensitivity in the form of a demand curve. A wide variety of mathematical models can be used. If weekly or monthly historical sales and price data is available over a good period of time, usually two years or more, this can be used in the price analysis. The historical data may only cover a relatively small range of prices and one should be careful not to extrapolate results in the form of a pricing model or demand curve much beyond the range of historical experience. So if the historical price of a product has varied between 1.35 and 1.60 it would be dangerous to use a fitted curve to predict the demand that would result from extremely different prices of 0.50 or 3.50.\n\nStraight line (linear) relationships often give a good starting point in the price analysis and have the benefit that perfect mathematical solutions are available using the linear regression method that is widely available in software. A number of other types of demand curve can be transformed to linear form using logarithmic mathematics, so are also relatively easy to use. These include poynomials, the exponential curve and the power curve. Other forms of curve may require the use of nonlinear regression using optimisation techniques. These require the use of specialist statistical software and may not always come up with the same solution across different software.\n\nStraight Line: Sales = a + bp\n\na= constant. b = coefficient of price (this will be -ve), p=price\n\nExponential curve: Sales = aebp\n\nwhere e = mathematical constant (approx. 2.7183)\n\na = constant, b = coefficient, p = price\n\nPolynomial (2nd order): Sales = a + bp + cp2\n\na = constant, b = coefficient, c = coefficient, p = price\n\nPower curve: Sales = apb\n\na = constant, b = coefficient (elasticity), p = price\n\nThe power curve is somewhat special in that it exhibits a constant price elasticity of b (see below).\n\nPrice Elasticity\n\nPrice elasticity is one way of describing price sensitivity and the effect of price change. Price elasticity is defined as the % change in sales likely to take place as a result of a 1% change in price. As increased price results in a reduction in demand the price elasticity of demand will always be a negative figure. Unit price elasticity refers to the specific situation where a 1% change in price causes exactly a 1% decrease in sales.\n\nIn most price models, including simple linear relationships, the price elasticity will vary depending on the particular point of reference on the demand curve. So the elasticity from a point where the price is 8.00 may vary from the elasticity when the price is 9.00. The elasticity at particular price may then be referred to as the 'point' price elasticity.\n\nCause and Effect Analysis\n\nIn cause and effect analysis, or causal analysis, the aim is to quantify the effect of factors which are suspected of causing shifts in sales volume or market share. Price sensitivity analysis in the form of pricing models and demand curves is one special example. Other causal factors such as unseasonal weather or economic indices may also play a part. So a full cause and effect analysis incorporating the effects of other factors as well as pricing may lead to a fuller understanding and a better platform for the formulation of pricing strategy and other business strategy\n\nWhen a causal relationship has been identified and quantified it can immediately help to explain variations that have been experienced in historical demand. That is in itself is very helpful, but to make full use of a pricing model or other causal model in forecasting it is necessary to forecast the future values of the causal factor itself before one can calculate the demand forecast. If the causal factor is a leading indicator and/or some well known index such as GDP or RPI for which other organisations publish forecasts, the task may be easier.\n\nNeed for expert help\n\nSpecialist software is invaluable in carrying out price sensitivity analysis and other cause and effect analysis. Care is needed to avoid confusion of the results with natural seasonality or inherent trends in market size or share. Forecast Solutions can expertly carry out the work using specialist statistical software, analysing the effect of price change and the influence of unseasonal weather, economic indices, sales force calling or other potential causal factors. The results can then be taken into account in defining pricing strategy, other business strategy and the demand forecasting process."
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"https://www.forecastsolutions.co.uk/images/merge3.jpg",
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"https://www.forecastsolutions.co.uk/images/welcome.jpg",
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"https://www.forecastsolutions.co.uk/images/services.jpg",
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"https://www.forecastsolutions.co.uk/images/email-contact.jpg",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.92506576,"math_prob":0.967631,"size":5413,"snap":"2019-43-2019-47","text_gpt3_token_len":1063,"char_repetition_ratio":0.13440561,"word_repetition_ratio":0.014348786,"special_character_ratio":0.19527064,"punctuation_ratio":0.07180021,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98397124,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-16T10:30:14Z\",\"WARC-Record-ID\":\"<urn:uuid:b8f003da-c4e1-4160-bfa1-504e5118dc2b>\",\"Content-Length\":\"13689\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ffe83fc-6a57-4053-bd42-c8319bdf6e33>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cfd5203-fb95-42be-a441-e737f157659d>\",\"WARC-IP-Address\":\"85.233.160.141\",\"WARC-Target-URI\":\"https://www.forecastsolutions.co.uk/price-sensitivity-analysis.htm\",\"WARC-Payload-Digest\":\"sha1:PWLJG3RR7FVROY3GTNA5ZHFWV4QSRKN5\",\"WARC-Block-Digest\":\"sha1:S56AE272P5KMQGB4XOJIYL7AL6NV373F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986666959.47_warc_CC-MAIN-20191016090425-20191016113925-00462.warc.gz\"}"} |
https://www.shaalaa.com/question-bank-solutions/solve-the-following-quadratic-equation-x2-4ix-4-0-solution-of-a-quadratic-equation-in-complex-number-system_163298 | [
"# Solve the following quadratic equation: x2 + 4ix – 4 = 0 - Mathematics and Statistics\n\nSum\n\nSolve the following quadratic equation:\n\nx2 + 4ix – 4 = 0\n\n#### Solution\n\nGiven equation is x2 + 4ix – 4 = 0\nComparing with ax2 + bx + c = 0, we get\na = 1, b = 4i, c = – 4\nDiscriminant = b2 – 4ac\n= (4i)2 – 4 x 1 x – 4\n= 16i2 + 16\n= – 16 + 16 ...[∵ i2 = – 1]\n= 0\nSo, the given equation has equal roots.\nThese roots are given by\n\nx = (-\"b\" ± sqrt(\"b\"^2 - 4\"ac\"))/(2\"a\")\n\n= (-4\"i\" +- sqrt(0))/(2(1)\n\n= (-4\"i\")/2\n\n∴ x = – 2i\n∴ the roots of the given equation are – 2i and – 2i.\n\nConcept: Solution of a Quadratic Equation in Complex Number System\nIs there an error in this question or solution?\n\n#### APPEARS IN\n\nBalbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board\nChapter 3 Complex Numbers\nExercise 3.2 | Q 3. (iii) | Page 40"
] | [
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https://scienceon.kisti.re.kr/srch/selectPORSrchArticle.do?cn=NPAP07422127 | [
"$\\require{mediawiki-texvc}$\n• 검색어에 아래의 연산자를 사용하시면 더 정확한 검색결과를 얻을 수 있습니다.\n• 검색연산자\n검색연산자 기능 검색시 예\n() 우선순위가 가장 높은 연산자 예1) (나노 (기계 | machine))\n공백 두 개의 검색어(식)을 모두 포함하고 있는 문서 검색 예1) (나노 기계)\n예2) 나노 장영실\n| 두 개의 검색어(식) 중 하나 이상 포함하고 있는 문서 검색 예1) (줄기세포 | 면역)\n예2) 줄기세포 | 장영실\n! NOT 이후에 있는 검색어가 포함된 문서는 제외 예1) (황금 !백금)\n예2) !image\n* 검색어의 *란에 0개 이상의 임의의 문자가 포함된 문서 검색 예) semi*\n\"\" 따옴표 내의 구문과 완전히 일치하는 문서만 검색 예) \"Transform and Quantization\"\n\n# 과학기술 지식인프라 ScienceON",
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"안녕하세요!\nScienceON 챗봇입니다.\n궁금한 것은 저에게 물어봐주세요.\n\n## A Study on the System Principle of PID Module Implementation\n\n###### 초록\n\nThe derivative equation measured of a MV=Kp8{(EVn-EVn-1)+Ki/S/1* EVn+(Kd/s)*(2PVn-1-{PVn-PVn-1)}(원문이미지참조) is used on the machine apparatus of industrial field, but this par doesn\\t able to educate now, because we didn\\t have the implementation device of PID module, so the principle implementation system of the PID Module is manufactured and developed. Through this system, the implementation system of PID Module is practiced with that the SV and the set of P, I, D is set on the derivative equation measured of PID. A things to be known of this experiment result is flow. 1)PID module is known that had to be used with the module of A/D and D/A. 2) In process of PV is approached to the SV to follow Kp, Ti and Td to cause a constant of set value on the MVp=Kp*EV, Mv=Ki/1 EVdt, MVd+tDBT/D EV(원문이미지참조) the variable rate of E and Kp, Td, Ti in that table 1 is analysed, is same as flow. ①If Kp is high, PV is near fast to the SV, but Kp is small, PV is near slowly to the SV. ②If Ki is shot, PV is close fast to the SV, but Ti is high, PV is close slowly to the SV ③If Td is high, the variable rate of E press hardly when because it doesn\\t increase, but Td is small, the variable rate of E press not hardly, upper with 1),2), PID module is supposed that be able to do the A/S and an implementation of that apparatus, and getting a success of aim that an engineer want, on control of temperature, tension, velocity, amount of flow, power of wind end so on, to get the principle of automatic implementation in industrial field with cooperation of A/D and D/A module.\n\n###### Abstract\n\nThe derivative equation measured of a ${\\Delta}MV=Kp*{(EVn-EVn-1)+\\frac{1}{Ki/S}* EVn+(Kd/S)*(2PVn-1-PVn-PVn-1)}$ is used on the machine apparatus of industrial field, but this par doesn\\t able to educate now, because we didn\\t have the implementation device of PID module, so the principle implementation system of the PID Module is manufactured and developed. Through this system, the implementation system of PID Module is practiced with that the SV and the set of P, I, D is set on the derivative equation measured of PID. A things to be known of this experiment result is flow. 1)PID module is known that had to be used with the module of A/D and D/A. 2) In process of PV is approached to the SV to follow Kp, Ti and Td to cause a constant of set value on the $MVp=Kp*EV, MV=\\frac{1}{Ki}{\\int}EVdt, MVd=Td\\frac{d}{dt}EV$, the variable rate of E and Kp, Td, Ti in that table 1 is analysed, is same as flow. (1)If Kp is high, PV is near fast to the SV, but Kp is small, PV is near slowly to the SV. (2)If Ki is shot, PV is close fast to the SV, but Ti is high, PV is close slowly to the SV (3)If Td is high, the variable rate of E press hardly when because it doesn\\t increase, but Td is small, the variable rate of E press not hardly, upper with 1), 2), PID module is supposed that be able to do the A/S and an implementation of that apparatus, and getting a success of aim that an engineer want, on control of temperature, tension, velocity, amount of flow, power of wind end so on, to get the principle of automatic implementation in industrial field with cooperation of A/D and D/A module.\n\n#### 참고문헌 (0)\n\n1. 이 논문의 참고문헌 없음\n\n#### 이 논문을 인용한 문헌 (0)\n\n1. 이 논문을 인용한 문헌 없음\n\n### 원문보기\n\n원문 PDF 다운로드\n\n• ScienceON :\n\n원문 URL 링크\n\n• 원문 URL 링크 정보가 존재하지 않습니다.\n\n원문 PDF 파일 및 링크정보가 존재하지 않을 경우 KISTI DDS 시스템에서 제공하는 원문복사서비스를 사용할 수 있습니다. (원문복사서비스 안내 바로 가기)\n\n상세조회 0건 원문조회 0건\n\nDOI 인용 스타일"
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https://answers.everydaycalculation.com/add-fractions/8-5-plus-14-6 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Add 8/5 and 14/6\n\n1st number: 1 3/5, 2nd number: 2 2/6\n\n8/5 + 14/6 is 59/15.\n\n#### Steps for adding fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 5 and 6 is 30\n\nNext, find the equivalent fraction of both fractional numbers with denominator 30\n2. For the 1st fraction, since 5 × 6 = 30,\n8/5 = 8 × 6/5 × 6 = 48/30\n3. Likewise, for the 2nd fraction, since 6 × 5 = 30,\n14/6 = 14 × 5/6 × 5 = 70/30\n4. Add the two like fractions:\n48/30 + 70/30 = 48 + 70/30 = 118/30\n5. 118/30 simplified gives 59/15\n6. So, 8/5 + 14/6 = 59/15\nIn mixed form: 314/15\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Add Fractions Calculator\n\n+\n\n© everydaycalculation.com"
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.56187797,"math_prob":0.99752426,"size":407,"snap":"2021-21-2021-25","text_gpt3_token_len":203,"char_repetition_ratio":0.22580644,"word_repetition_ratio":0.0,"special_character_ratio":0.55773956,"punctuation_ratio":0.07017544,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971857,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-19T07:07:50Z\",\"WARC-Record-ID\":\"<urn:uuid:b5d991a3-1a2a-4f50-9c07-0755a8f247d1>\",\"Content-Length\":\"8821\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c9df40ce-3ae2-415d-979c-31ceef7d9ebe>\",\"WARC-Concurrent-To\":\"<urn:uuid:363bd76a-0a61-4f0d-a808-99ac7e444fe7>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/8-5-plus-14-6\",\"WARC-Payload-Digest\":\"sha1:HRXQCG5KRBM76GWZ6XL3PAVEYFTDKDDA\",\"WARC-Block-Digest\":\"sha1:EUONIW74OPQARHLFWSS2EXUPYB5ZYC4M\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487643703.56_warc_CC-MAIN-20210619051239-20210619081239-00491.warc.gz\"}"} |
http://lib.mexmat.ru/books/108570 | [
"Электронная библиотека Попечительского советамеханико-математического факультета Московского государственного университета\n Главная Ex Libris Книги Журналы Статьи Серии Каталог Wanted Загрузка ХудЛит Справка Поиск по индексам Поиск Форум",
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"Авторизация",
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"Поиск по указателям",
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"",
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"",
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"",
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"Noble B. — Methods based on the Wiener-Hopf technique for the solution of PDEs",
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"Читать книгубесплатно\n\nСкачать книгу с нашего сайта нельзя\n\nОбсудите книгу на научном форуме",
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"Нашли опечатку?\nВыделите ее мышкой и нажмите Ctrl+Enter\n\nНазвание: Methods based on the Wiener-Hopf technique for the solution of PDEs\n\nАвтор: Noble B.\n\nЯзык:",
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"Рубрика: Математика/\n\nСтатус предметного указателя: Готов указатель с номерами страниц\n\ned2k: ed2k stats\n\nГод издания: 1958\n\nКоличество страниц: 255\n\nДобавлена в каталог: 18.11.2012\n\nОперации: Положить на полку | Скопировать ссылку для форума | Скопировать ID",
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"Предметный указатель\n Abelian theorems 36 46 57 Absorbent boundaries (see Resistive boundaries) Aerofoil, quarter-infinite 95 Aerofoil, two-dimensional 173 Alblas, J. B. 96 236 Ament, W. J. 47 Analytic continuation 7ff 38ff Antenna, biconical 218 Antenna, cylindrical rod 213 Approximate factorization 160ff Approximate methods 178ff 236 generalized) Approximate methods for integral equations 23ff 233ff Approximate methods for strips and slits 203ff Approximate methods for thick plates 187ff Approximate methods, meaning of “approximate” 180 Approximate methods, miscellaneous formulations 181ff Asymptotic expansions for contour integrals 33ff 46 Asymptotic expansions for transforms 36 Asymptotic expansions in approximate solutions 200ff (see also Far-field) Baiun, L. L. 122 Baker, B. B. and Copson, E. T. 70 206 Baldwin, G. L. and Heins, A. E. 125 Bazer, J. and Karp, S. N. 73 128 153 166 167 Biharmonic equation (see also Elastic medium) Biharmonic equation, approximate factorization for 162 Biharmonic equation, stresses in strip 138 Boundary conditions, specified by arbitrary functions 7 7ff Boundary layer on flat plate 96 97 Bouwkamp, C. J. 75 179 218 Branch Cuts 8ff Branch cuts, special cases 9 19 38 39 Branch points and poles (see Poles and branch points) Carlson, J. F. and Heins, A. E. 103 133 Carrier, G. F. and Munk, W. H. 94 Carslaw, H. S. 72 Chester, W. 20 132 Churchill, R. V. 27 Circular waveguide (see Duct cylindrical) Clemmow, P. C. 36 88 93 94 152 Cone, Laplace’s equation 165 Cone, wave equation 218 Contour integrals for sum and product decompositions 18ff Contour integrals special type 31ff. Convection of heat from flat plate 94 Conventions x 10 28 86 Copson, E. T. vii 5 6 70 95 231 Copson’s method 231 Cosine transform 147 212 Cylindrical pipe (see Duct cylindrical) Davison, B. 168 Dielectric slab 211 213 Diffraction (see Disc Duct Halfplanes Strips Diffraction, methods other than the Wiener-Hopf for 125 179 Dirac delta function 43 Disc, Laplace’s equation 166 231 Disc, low frequency 213ff 231 Disc, wave equation, high frequency 219 Divergent integrals 64 Doetsch, G. 27 36 Dual integral equations for boundary conditions specified arbitrarily 77ff Dual integral equations for three-part problems 231 Dual integral equations formulation in terms of 4 58 150 Dual integral equations general comments 48 77 221 236 Dual integral equations, exact solution of general cases 77ff 222ff Dual integral equations, half-plane problem solved by 58ff Dual integral equations, involving Bessel functions viii 221 Dual integral equations, multiplying factor method for 58ff 77ff 222ff Dual integral equations, reduction to complex variable equation 150 151 221 Duct, cylindrical, finite length 207 Duct, infinite length, resistive wall 134 Duct, propagated and attenuated modes 110 Duct, radiation from 11 Duct, radiation from coaxial 133 154 Duct, semi-infinite tube in 121 Duct, semi-infinite, electromagnetic waves 132 Duct, two-dimensional, finite length 207 Duct, two-dimensional, infinite length with semi-infinite obstacle, dielectric slab 211 Duct, two-dimensional, infinite length with step 196 207ff Duct, two-dimensional, infinite length with strip across 122ff 212 Duct, two-dimensional, propagated and attenuated modes 98ff Duct, two-dimensional, resistive wall 133 Duct, two-dimensional, semi-infinite (see Half-planes two Duct, two-dimensional, set of posts 174 Duct, two-dimensional, solution via simultaneous linear algebraic equations 175 Duct, two-dimensional, strip of finite thickness parallel to wall 212 Duct, two-dimensional, strip parallel to wall 118ff Duct, wave incident on 116ff Edge conditions 51 74ff Edge conditions, references 75 Elastic medium (see also Biharmonic equation) Elastic medium, diffraction by half-plane in 96 Elastic medium, stresses in wedge with mixed boundary condition 173 Electromagnetic waves, boundary conditions 134 Electromagnetic waves, diffraction by half-plane 95 Electromagnetic waves, diffraction by semi-infinite cylinder 132 Electromagnetic waves, edge conditions 76 Electromagnetic waves, terminology 85 92 End correction, cylindrical tube 115 End correction, thick semi-infinite strip 196 Erdelyi, A. 36 46 Erdelyi, A. et al. 148 176 213 Factorization (see Product factorization) Far-field, examples of calculation of 105 110 115 192 206 Fel’d, A. N. 174 Fock, V. 168 Fourier transforms 2 Fourier transforms, basic example 53ff Fox, E. N. 232 Fox’s integral identity 232 Fredholm integral equations, first kind (see Wiener-Hopf integral equations) Fredholm integral equations, second kind 221 Fresnel functions 33 45 73 213 Function-theoretic methods 153 Gamma functions 40 Gamma functions, factorization by means of 41 167 171 215 Greene, T. R. and Heins, A. E. 94 Green’s function 6 Half-planes with flange 183 Half-planes, algebraic equations 175 Half-planes, dielectric material 213 Half-planes, infinite set of staggered 133 Half-planes, single infinitely thin, basic (Sommerfeld) diffraction problem 48ff Half-planes, single infinitely thin, diffraction of electromagnetic waves by 95 Half-planes, single infinitely thin, finite resistivity 83 Half-planes, single infinitely thin, generalizations of basic problem 5 Half-planes, single of finite thickness 181 187ff Half-planes, three parallel 154 Half-planes, two parallel, radiation from 105ff Half-planes, two parallel, staggered 154 Half-planes, wave falling on 100ff 125 126 Hankel functions, asymptotic behaviour 45 Hankel functions, integral representation 32 Harrington, R. F. 88 Heins, A. E. 94 103 127 131 133 134 157 171 Heins, A. E. and Carlson, J. F. 133 Heins, A. E. and Feshbach, H. 92 93 134 Hilbert problem 49 14 Hilbert problem, equation application 169 Hopf, E. 168 Imperfect conductivity (see Resistive boundaries) Indentation of contour 30 44 86 Indentation of contour for transform in time 44 Infinite product 15 40 103 129 130 Infinite product or gamma functions 41 Infinite product, asymptotic behaviour of 104 128 129 Infinite product, computation of 127 128 Institute of Mathematical Sciences, New York, reports 237 Integral equation method, general remarks 49 71 76 236 Integral equations (see Fredholm Volterra Wiener-Hopf) Integral functions 6 178 197 Jets 139 Jones, D. S. vii, viii 48 52 75 95 97 127 130 131 167 173 187 190 195 196 199 200 207 231 Jones’s method for approximate solutions 17 8ff 236 Jones’s method for boundary conditions specified arbitrarily 79ff Jones’s method, detailed application to a special example 52ff Jones’s method, general remarks 48 71 76 98 Jones’s method, routine applications 100ff Jost, Res 219 Karp, S. N. and Russek, A. 206 Karp, S. N. and Williams, W. E. 125 Klunker, E. B. and Harder, K. C. 139 Koiter, W. T. 75 139 161 162 173 Kourganoff, V. 133 168 Laplace transform 10 21 52 86 Laplace transform, complete equivalence with Fourier transform 22 26 Laplace-type equations 135 (see also Biharmonic equation) Laplace’s equation as limit of wave equation 135ff Laplace’s equation in jet problem 140 Laplace’s equation in polar coordinates 164ff 177 Latter, R. 236 Lebedev — Kontorovich transform 148 213ff Lebesgue integration, reason for not using 23 Leitner, A. and WELLS,C.P. 213 216 Lennox, S. C. 139 Levine, H. 236 242 Levine, H. and Sch winger, J. 20 42 115 130 132 Levine, H. and Wu, T. T. 236 Lewin, L. 131 179 242 Lewis, J. A. and Carrier, G. F. 97 Linfoot, E. H. and Shepherd, W. M. 175 Liouville’s theorem 6 38 57 69 103 114 Magnus, W. 175 Magnus, W. and Oberhettinger, F. 175 Marcuvitz, N. 122 133 179 Mark, C. 168 Marshak, R. E. 168 Maue, A. W. 75 96 Meier, J. A. and Leitner, A. 218 Meixner, J. 75 Meromorphic functions 39 Miles, J. W. 95 Milne’s integral equation 167 Milne’s integral, equation 167 Milne’s integral, reduction of more general types to Fredholm equations of second kind 232 233ff Milne’s integral, reduction to repeated Volterra equations 230 Milne’s integral, solution by the Wiener — Hopf technique 67ff 167ff 227 Mixed boundary value problem 1 Mixed boundary value problem, solution of a specific example 77ff Mixed boundary value problem, three-part 183 203ff 231ff Mixed boundary value problem, two-part, general theory 147 Morse, P. M. and Fesiibaoh, H. 02 115 131 134 179 Multiplying factor method 58ff 78ff 22 Muskhelishvili, N. I. 142 144 146 158 Notation 54 Notation in certain approximate methods 182 197 Numerical evaluation of infinite, products 127 128 Numerical tables viii 122 131 Oberhettinger, F. 36 Oseen’s equations 97 Pack, D. C. viii 139 Pai, S. I. 139 Papadopoulos, V. M. 134 196 Pearson, J. D. 132 Perpendicular boundaries 178 181 187ff 207ff Placzek, G. 168 Placzek, G. and Seidel, W. 168 Plemelj formulae 47 145 Pol, B. van dee and Bremmer, H. 27 36 Polar coordinates, Laplace’s equation 164ff 177 Polar coordinates, steady-state wave equation 148 214ff Polar coordinates, wedge of elastic material 173 Poles and branch points 47 133 207 Product factorization 14ff Product factorization, approximate 160ff Product factorization, general theorems 15 41 Product factorization, involving gamma functions 41 167 171 215 Quantum mechanics, reference to problem in 219 Quarter-plane aerofoil 95 Radiation condition at infinity 2 31 Radiation-type boundary conditions (see Resistive boundaries) Radio-wave propagation over landsea interface 93 94 Reflection coefficient in duct 110 114 120 124 Regularity, of functions defined by integrals 1 Resistive boundaries, half-plane 83ff Resistive boundaries, notation 134 Resistive boundaries, strip across duct 212 Resistive boundaries, strip parallel to walls of duct 120 Rod, solid cylindrical, finite length 213 Rod, solid cylindrical, semi-infinite 196 Scattering cross-section of disc 218 Schwinger, J. vii Semi-infinite plane (see Half-planes) Senior, T. B. A. 83 85 92 93 134 Separation of variables 1 147 149 Simultaneous linear algebraic equations from simultaneous Wiener-Hopf equations 155 Simultaneous linear algebraic equations in approximate solutions 185 190ff 208 217 Simultaneous linear algebraic equations in duct problem 175 Simultaneous linear algebraic equations in half-plane problem 176 Simultaneous linear algebraic equations, solution of special sets 173 174ff Simultaneous Wiener-Hopf equations in electromagnetic cylindrical duct problem 132 Simultaneous Wiener-Hopf equations in electromagnetic half-plane problem 95 Simultaneous Wiener-Hopf equations in two-dimensional duct problems 121 154 Simultaneous Wiener-Hopf equations, general theory 153ff Slits (see Strips and slits) Sneddon, I. N. viii 2 96 168 231 236 Sommerfeld, half-plane problem 48ff Sommerfeld, radiation condition 31 Sound waves (see Duct Half-planes etc.) Source distribution 44 52 87 Sparenberg, J. A. 169 171 Steady-state wave equation 28 Steepest descent 34 Strips and slits of finite thickness and finite length 213 Strips and slits of finite thickness, induct 212 Strips and slits of finite thickness, semi-infinite length 181 187ff Strips and slits of finite width, Laplace’s equation 177 Strips and slits of finite width, low frequency 218 Strips and slits of finite width, wave equation, high frequency 183 203ff 219 Sum decomposition in approximate solutions 179 186 Sum decomposition, applications of 80 179 186 198 228 Sum decomposition, function with poles 39 Sum decomposition, general formulae for concrete cases 18 Sum decomposition, general theorem 13 Superposition, method of 81 87 Tables (see Numerical tables) Tides in permeable rock 94 Titchmarsh, E. C. 2 5 8 11 39 40 41 42 168 171 Titchmarsh’s method 171 Titchmarsh’s method in wedge problems 219 Titchmarsh’s method or evaluation of integrals 33 Titchmarsh’s method or sum decompositions 47 Titchmarsh’s method, applied to diffraction by half-plane 176 Transient problems 28 97 Tranter, C. J. 2 Uniqueness 72 Uniqueness and edge conditions 51 74 Uniqueness, equations 69 168 Vajnshtejn, L. A. 20 115 125 130 131 132 153 Vajnshtejn-Karp-Clemmow method 49 153 213 Verification of solution 72 via the Hilbert, problem 169 via the Hilbert, special equations 171 177 via the Hilbert, uniqueness of solution 69 168 Viscous medium 96 97 139 Volterra integral equations 229 230 233 Water waves 94 134 Watson, G. N. 36 46 Wave equation 27ff waveguide (see Duct two-dimensional Duct cylindrical Half-planes two Wedges of elastic material 173\n1 2",
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https://sportsbizusa.com/20-decimals-on-number-line-worksheets/decimals-on-number-line-worksheets-decimal-worksheets/ | [
"# Decimals On Number Line Worksheets Decimal Worksheets\n\nHome20 Decimals On Number Line Worksheets ➟ Decimals On Number Line Worksheets Decimal Worksheets\n\nDecimals On Number Line Worksheets Decimal Worksheets one of Worksheet From Home - ideas, to explore this Decimals On Number Line Worksheets Decimal Worksheets idea you can browse by and . We hope your happy with this Decimals On Number Line Worksheets Decimal Worksheets idea. You can download and please share this Decimals On Number Line Worksheets Decimal Worksheets ideas to your friends and family via your social media account. Back to 20 Decimals On Number Line Worksheets\n\n## 20 Multiples Of 4 Worksheet\n\nHundred Chart with Multiples of 4 multiples grade 4 worksheet, multiples year 4 worksheet, multiples of 4 and 8 worksheet, factors and multiples worksheet for grade 4 with answers pdf, factors and multiples worksheet for grade 4 with answers, via: math-drills.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their […]\n\n## 20 Simple Addition Worksheets with Pictures\n\nFree Simple Addition Worksheets Math Free simple addition pictures worksheets, simple picture addition worksheets for kindergarten, simple math addition worksheets with pictures, simple addition and subtraction worksheets with pictures, simple addition worksheets with pictures for kindergarten, via: pinterest.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The […]\n\n## 20 Number Line Inequalities Worksheet\n\nGraph Basic Inequalities on Number Lines A representing inequalities on a number line worksheet tes, graphing inequalities on a number line worksheet 6th grade, solving inequalities number line worksheet, graphing inequalities on a number line worksheet kuta, writing inequalities from a number line worksheet answers, via: math-drills.com Numbering Worksheets for Kids. Kids are usually introduced […]"
] | [
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https://www.bakemag.com/articles/724-king-arthur-flour-features-professional-baker-s-reference | [
"King Arthur Flour features the Professional Baker’s Reference at http://www.kingarthurflour.com/professional/bakers-reference.html\n\nFind tips, techniques, and handy references for commercial bakers here on topics such as dough temperatures:\n\ndough temperatures\n\nOne of the most important skills a baker must learn is the ability to accurately control dough temperature. The benefits are clear and immediate: more consistency in fermentation, in bread flavor, and more predictability in the overall production schedule.\n\nIf a dough is coming off the machine at 70°F one day and 80°F the next, we cannot expect to have uniformity in the results.\n\nBread yeasts attain an optimum growth temperature in the mid-90°F range. One might think, therefore, that in order to favor yeast development, we should aim for bread dough in the mid-90s. This high temperature, however, would be at the expense of flavor development through the production of organic acids, which requires considerably lower temperatures. In general, wheat-based doughs should be between 75°F and 78°F at the end of mix time. Some exceptions to this guideline are naturally leavened breads that will ferment in a retarder, which could have a slightly lower dough temperature; and sourdough breads made with a high proportion of rye flour, which benefit from a dough temperature in the low to mid 80s.\n\nThe calculation of Desired Dough Temperature involves taking into consideration three or four factors (occasionally but rarely more than this). These factors are the variables over which we have no control when we enter the bakeshop and prepare to mix the dough:\n\n1. air temperature\n2. temperature of the flour\n3. the \"friction factor\" of our mixer\n4. temperature of the preferment, if any\n\nAfter figuring these, we can simply and quickly establish the correct water temperature (the only variable over which we have control). At the end of this discussion we will talk about friction factor in more detail.\n\nLet’s assume that we want a desired dough temperature of 76°F. We will multiply this figure by 3 if we are mixing a straight dough, and by 4 if using a preferment. The result will be the total temperature factor. Once we have determined this factor, we subtract from it the sum of the known temperatures, and the result will be the required water temperature. Here are two examples:\n\n calculating water temperature straight dough with preferment Desired Dough Temperature (DDT) 76°F 75°F Multiplication Factor x 3 x 4 Total Temperature Factor (TTF) 228 300 Minus Flour Temperature – 72° – 68° Minus Room Temperature – 74° – 68° Minus Preferment Temperature – 71° Minus Friction Factor – 24° – 26° Water Temperature = 58° = 67°\n\nfriction factor\n\nAs a dough spins, heat is generated by the friction caused by the action of the dough hook and bowl on the dough. Some factors that affect the amount of friction generated during mixing are the type of mixer being used (for example, spiral, oblique, or planetary), the length of mix time, mixing speeds used, and the quantity of dough that is in the machine. Ascertaining the friction factor for your mixer can be difficult indeed. It is the most difficult variable to quantify in the computation of desired dough temperature.\n\nTo determine the friction factor, first a trial dough is made. We take the temperature of the dough after mixing and use the results to calculate the friction factor. It is important to note the length of time the dough mixes, and on what speeds. If we alter these times and speeds, we will generate different amounts of friction. Here are two examples:\n\n calculating friction factor straight dough with preferment Actual Dough Temperature (after mixing) 78°F 80°F Multiplication Factor x 3 x 4 Total Temperature Factor (TTF) 234 320 Minus Flour Temperature – 71° – 68° Minus Room Temperature – 73° – 68° Minus Water Temperature – 66° – 85° Minus Preferment Temperature – 71° Friction Factor = 24° = 28°"
] | [
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https://bbs.huaweicloud.com/blogs/273142 | [
"# 论文解读系列十一:图神经网络应用于半结构化文档的命名实体识别和关系提取\n\n【摘要】 随着用于传递和记录业务信息的管理文档的广泛使用,能够鲁棒且高效地从这些文档中自动提取和理解内容的方法成为一个迫切的需求。本次解读的文章提出利用图神经网络来解决半结构化文档中的实体识别(NER)和关系提取问题。",
null,
"# 摘要:\n\n随着用于传递和记录业务信息的管理文档的广泛使用,能够鲁棒且高效地从这些文档中自动提取和理解内容的方法成为一个迫切的需求。此外,基于图的表达方法对不同文档模版的变化具有灵活的适应性,从而使得图表达方式与这些管理文档的半结构化特性非常契合。正因为图神经网络(GNN)能够很好地学习出文档中数据元素间的关系,所以本次解读的文章提出利用图神经网络来解决半结构化文档中的实体识别(NER)和关系提取问题。经实验验证该文章提出的方法在单词分组、实体分类、关系预测三个任务上取得了SOTA结果,同时在FUNSD(表单理解)和IEHHR(手写婚姻档案理解)两个完全不同类别的数据集上取得的实验结果进一步验证了本次解读文章所提出的方法的泛化性。\n\n# 1. 方法\n\nGNN被广泛应用于NER和表格提取等任务中,本次解读的文章在此基础上提出将GNN应用于提取key-value对的任务中,不仅对文档图片中的实体进行分类,而且还会对实体间的关系进行预测。\n给定一个输入文档,模型需要完成的任务包括:(a)单词分组:检测文档实体,即将相同语义的单词进行分组;(b)实体分类:将检测到的实体分为预设的类别;(c)关系预测:发现实体间配对关系。\n\n## (1)图的构造\n\n本次解读的文章提出构造两张图来表示文档,并在此基础上训练三个不同的模型来解决对应的任务:单词分组 $f_1$、实体分类 $f_2$、关系预测 $f_3$。如图1所示,文档会被表示为由OCR结果构造的图 $G_1=(V_1,E_1)$,其中 $V_1$是由OCR结果中每个单词组成的节点集合;对每个单词文本框左上角间的距离进行 $k$-近邻(取 $k=10$)来生成边 $E_1$,对各边计算分数 $s=f_1 (G_1)$,筛选出大于阈值 $\\tau$(FUNSD设为0.65, IEHHR设为0.9)的边就可以得到单词分组的结果。",
null,
"",
null,
"如图2所示,在 $G_1$的基础上得到实体(即各单词分组)后,由每个实体构造得到图 $G_2=(V_2,E_2)$,其中 $V_2$表示由 $G_1$筛选得到的实体集合, $E_2$是由各实体节点间全连接得到的边集合。由 $c=f_2 (G_2)$得到实体分类结果;由 $s=f_3 (G_3)$得到关系预测结果。\n\n## (2)图的计算\n\n本次解读文章中的 $f_1$, $f_2$, $f_3$ $L$个GAT层(graph attention network)作为模型骨干结构并经过训练优化得到。\n给定 $G=(V,E)$。每个节点 $v_i$的初始化表达由 $h_i^0=x_i,y_i,w_i,h_i,w_{embed}]$拼接得到,其中 $x_i,y_i,w_i,h_i$是单词文本框的左上角横纵坐标和文本框宽高, $w_{embed}$为单词的词向量。根据GAT,每一对节点间计算出其注意力系数:\n\n$\\alpha_{ij}=\\frac{exp(LeakyRelu(V[Wh_i ||Wh_j]))}{∑_{k \\in N_{v_i}}exp(LeakyRelu(V[Wh_i ||Wh_k]))}$\n\n$h_i^{l+1}=g(h_i )=||_{k=1}^K \\sigma(∑_{j \\in N_i}α_{ij}^k W^k h_j^l )$\n\n$c_i=σ(Wh_i^L)$\n\n$s_ij=\\sigma(W(|h_i^L-h_j^L |))$\n\n$CE(y' )=-(y∙log(y')+(1-y)∙log(1-y' ))$\n\n# 2. 实验结果",
null,
"从FUNSD实验结果表明,本次解读文章提出的方法与LayoutLM相比较还有优化空间,原因可能在于FUNSD的数据量较小。从IEHHR实验结果表明,该方法在表单识别的其他领域即手写记录理解上也具有一定的效果,体现了其泛化性。\n【版权声明】本文为华为云社区用户原创内容,转载时必须标注文章的来源(华为云社区),文章链接,文章作者等基本信息,否则作者和本社区有权追究责任。如果您发现本社区中有涉嫌抄袭的内容,欢迎发送邮件至:[email protected]进行举报,并提供相关证据,一经查实,本社区将立刻删除涉嫌侵权内容。\n• 点赞\n• 收藏\n• 关注作者\n\n0/1000\n\n### 设置昵称\n\n*长度不超过10个汉字或20个英文字符,设置后3个月内不可修改。\n\n*长度不超过10个汉字或20个英文字符,设置后3个月内不可修改。"
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null,
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null,
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https://www.drdawnwright.com/tag/t-test/ | [
"# T-test\n\n## Single-sample t-test & Confidence Interval – Excel & StatCrunch\n\nWhen we do not know the population standard deviation sigma, σ, and the sample size, n, is less than 30, we use the t-test to evaluate a claim. Consider the following problem: A scientist thinks the mean waste recycled by adults in the US is now more than one pound per person per day. In …\n\n## Recognizing problem types: Hypothesis tests\n\nRead the problem looking for keywords and values: What type of variable is the focus of the problem? Is it quantitative, e.g. a mean, or categorical, e.g. a proportion or percent? How many variables are of concern? Is(are) the population(s) standard deviation, sigma, given? Are sample variances equal or are you instructed to assume they …\n\n## Setting up Hypothesis Tests\n\nThe Null and Alternative The most common problem I noticed on this assignment was caused by failing to properly identify the appropriate null and alternative hypotheses. In part, this is due to the Evans text’s somewhat confusing explanation of how to do this – the “burden of proof” approach. There is a much simpler approach …"
] | [
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https://jasonhemann.github.io/monads/ | [
"# Maybe? Definitely.\n\nHere is a use of Scheme’s findf operator.\n\n``````> (findf number? '(1 #f 2 dog))\n1\n``````\n\nFindf takes a predicate and a list, and returns the first element of the list for which that predicate holds true, returning #f in the case that the predicate holds for no member of the list.\n\nSo if we change 1 to one, then we will now get a different answer.\n\n``````> (findf number? '(one #f 2 dog))\n2\n``````\n\nAnd if we change 2 to two, we get #f, as now no elements of the list are numbers.\n\n``````> (findf number? '(one #f two dog))\n#f\n``````\n\n``````> (findf boolean? '(one #f two dog))\n#f\n``````\n\nWhoa ho ho. What’ve we got /here/?\n\nfindf returns the first element of the list for which that predicate holds. #f is a boolean, so that’s what’s returned. Lo, the confusion that can result.\n\n``````> (findf boolean? '(one two dog))\n#f\n``````\n\nLet’s look at the issue a bit more abstractly. The meat of the problem is that the value we’re using to denote an “error” (“failure”, “exception”, what have you) is /in/ the set of things that we could possibly return as /actual/ answers. Some languages use -1 as an error condition instead. Obviously, that’s no good as an error value here, either.\n\nThere’s not a safe value we can pick, because for whatever value we pick, it might be someone wants to hunt for precisely that value in their list. For instance, we might try to patch the problem above, saying to ourselves “Well, #f is useful. Comes up all the time. I’ll write my own findf, and pick a different thing.” Heck, we’ll choose this-obscure-symbol-means-failure-Qx29t7.\n\nAnd that works great until some dingbat comes along and does the following\n\n``````> (my-findf (lambda (y) (eqv? 'this-obscure-symbol-means-failure-Qx29t7 y))\n'(a b c this-obscure-symbol-means-failure-Qx29t7))\nthis-obscure-symbol-means-failure-Qx29t7\n``````\n\nBut as it turns out, that might not be such a dingbat thing to do, after all. All the time, we write code that generates code. Someone who’s generating Scheme code might very well have reason to hunt for that symbol in a list.\n\nAnyhow. What’s clear is that the above suggestion is just sort of a patch. These sorts of things come up all the time. We report #f, -1, Error 404, etc., etc. It seems like there ought to be a unified approach to handling this sort of problem, and one in which we don’t have to kludge together some sort of an error code for each distinct case.\n\nThe maybe monad gives us the power to do both of those. It provides a generic way to handle and propogate failure, and, by virtue of being general, it means there’s no ad-hoc solution to the value that denotes an error.\n\nWe’ve got return and maybe, those new relatives of eta and * that we introduced. And we’ve also got fail, which is our way of signaling a failure.\n\n``````(define return-maybe\n(lambda (a) `(Just ,a)))\n\n(define bind-maybe\n(lambda (ma f)\n(cond\n[(eq? (car ma) 'Just) (f (cadr ma))]\n[(eq? (car ma) 'Nothing) '(Nothing)])))\n\n(define fail\n(lambda ()\n'(Nothing)))\n``````\n\nreturn is just another name for eta (or unit). More on that in a sec, but a quick reminder.\n\nThere’s lots of monads, and they all have the same type signatures for the return and bind, and they all have to obey those monadic laws. We’re gonna walk though, at a type level, what return and bind /are/, how those things relate to eta and *, why their essence and behavior makes sense, and then look at how the actual implementations of return-maybe and bind-maybe make sense.\n\nWhew. That’s a tall order. Well, let’s get to it.\n\nAs we were. return is just another name for eta.\n\n``````return : alpha -> Malpha\n``````\n\nThe job of return/eta is to take a “pure” value – that is, a value from the regular land of computation, and wrap it up in the monad (burrito) (or like Just above.)\n\nFor our maybe monad, there are two classes of things that are monadic values.\n\n1 : (Just ), where is any old thing you choose. (Must it be a pure value?) 2 : (Nothing)\n\nThis second one is the value provided by invoking fail. Notice that because all the pure values are wrapped up in a list with Just at the front, then this (Nothing) doesn’t have the same problem as we had before with -1 or #f.\n\nIf I want to return #f out of my findf-maybe (which we’ll write in a second), it’ll come out as (Just #f). If I want to signal something’s not in there, it’ll come out as (Nothing). (Of course, that could be #f).\n\nAnd now, let’s consider that same dingbat from before, trying to be cute again.\n\n``````> (findf-maybe (lambda (v) (equal v `(Nothing))) '(-1 #f (Nothing) 404))\n(Just (Nothing))\n``````\n\nAh-HA! Lookie there. Unlike before with Scheme’s findf, now we can tell when the failure value is exactly that thing looked up in the list, vs. when it’s actually returned as a failure value. We’ve got the problem licked. Well, as soon as we figure out bind, relate it to *, explain why bind-maybe makes sense as a bind, and use this to implement findf-maybe.\n\n• was typed as follows.\n\n• : (alpha -> Mbeta) -> (Malpha -> Mbeta)\n\nThat is, it takes a function and returns a function. A few months ago that in and of itself might’ve been mind-blowing, but it’s pretty old-hat now. BTW, if you need a mundane example, integral is the sort of thing that takes a function and returns a function.\n\nIn the biz, we always associate curried functions to the left. If I have something that looks like\n\n``````f : int -> (bool -> (string -> (bool -> (string -> int))))\n``````\n\nthose parens start to get a bit annoying there.\n\nSo we instead write it as\n\n``````f : int -> bool -> string -> bool -> string -> int\n``````\n\nbut we all know what that means. So you’ll more likely see * typed like\n\n• : (alpha -> Mbeta) -> Malpha -> Mbeta\n\nThat is, it’s gonna be a function that takes an f, then it takes an ma.\n\n``````(define *\n(lambda (f)\n(lambda (ma) ...)))\n``````\n\nAnd does … well, who knows what the /particular/ implementation does at this point, but it does something like calling the function on that argument.\n\nA quick digression, seemingly apropos of nothing. Let’s define an operator curried-app.\n\n``````(define curried-app\n(lambda (f)\n(lambda (a)\n(f a))))\n``````\n\nNothing terribly fancy here.\n\nBUT–imagine a world in which that’s the only way you apply functions. You don’t get to call them directly on their arguments. Instead, when you wanna apply a function, you pass it to curried-app, to which you then pass the argument upon which you wanna call that function.\n\nThere’s not much use to doing such a thing when we do our regular run-of-the-mill Scheme programming. But when we’re working with monads, it’s a necessity. By the way, did you notice the type of curried-app?\n\n``````curried-app : (alpha -> beta) -> alpha -> beta\n``````\n\nRemarkably, strikingly, /suspiciously/ similar to the type of *. * does what curried-app does, except it works over monads. When using monads, we’ve given up the necessity of doing our own function applications to arguments. It’s passe. Well, not really that, so much as we’re modeling effects, and we want to make sure the effects happen the right way. * (well, really it’ll be bind) is going to make sure that happens, ‘cause it and it alone is allowed to /do/ the application of functions to monadic arguments. So if * (bind) is right, then the whole process is right. We’ve deputized * (bind) as the function-apper. And because * (bind) has to obey the Monadic Laws if it’s to be a genuine, bonified, monad. Since we’re using pre-certified monads, we’re certain that they’re up to snuff, so if we just use’em correctly * (bind) will make all the applications work properly.\n\nAlright, I’m done writing bind in parens. Let’s describe bind, so that we can just get to using it. A couple of things.\n\n1. Writing code with nested calls to * starts to get uuuugly.\n``````((* (lambda (x)\n((* (lambda (y)\n((* (lambda (z) ...))\narg3)))\narg2)))\narg1)\n``````\n\nThe flow of control is going to the right. Programs written this way end up building these giant rightward dagger shapes.\n\n1. By doing all of our monadic code with calls to a function-apper mechanism, it makes sense to think of that part of the program as doing work in units of function-apping. “Do a *, which then demands another *, which …”. Why bother to curry *? Nothing’s gonna happen until we give it both parts, and it’s not really of use to do’em one at a time.\n\nIt’d be nice if my code read linearly, down the page, rather than having to follow it out in that triangle pattern, and if I treated these applications of function to argument as the real unit of work.\n\nBind does both of those.\n\nThe type of bind is\n\n``````bind : (Malpha x (alpha -> Mbeta)) -> Mbeta\n``````\n\nThat is, it takes two arguments at once, the argument and the function, and it takes’em in that order. Notice, now, if we have some stuff that’s nested, then the work that we do next, that’s part of the function body, will be underneath the stuff above. We can read what’s going to happen in order down the page.\n\n``````(define bind-maybe\n(lambda (ma f)\n(cond\n[(eq? (car ma) 'Just) (f (cadr ma))]\n[(eq? (car ma) 'Nothing) '(Nothing)])))\n``````\n\nSo here’s our definition of bind-maybe, again. We can see it’s doing at least part of what we wanted it to do. It takes two arguments, in the right order. We know that ma is one of the things in my maybe monad. Which means it’s either a (Just ) or (Nothing). In the first case, it's a real and actual value. In the second, there was an error/failure/exception/badness up the road. If there was an error, we can't compute on it, and we wanna propagate that forward. So if we had an error, we return an error.\n\nIf instead we had a (Just ), a non-error value, well we wanted to do that application. f takes a pure value. We see that, here. Look, it's taking the cadr of ma. bind, as we theorized, knows how to take the ma, unwrap the burrito and get ahold of the a, and call f on it. Whatever comes out of bind needs to still be in the monad. Pretend this bind is the last thing we do before we end the computation. Well, I want to make sure I can tell the difference between returning (Just (Nothing)) and (Nothing), like that yahoo tried to pull earlier.\n\nf, the function we pass in, the one we’re gonna build, needs to be of alpha -> Mbeta. That way, the thing we hand back is still safely nestled in the monad. That’s the reason behind the “monads are space-suits” metaphor. You can’t open a space-suit willy-nilly. You’ve got to do it nestled safely inside a space-capsule. If you do it outside, or go outside not wearing one, you’ll die in the cold, cold, blackness of space. Alone. That’s what happens to you if you try to get at the pure inside the monad when you aren’t supposed to.\n\nA word about those alphas and betas there. They don’t /have/ to be distinct types. They can be. But they don’t have to be. I’m gonna put some !s around the parts that you really wanna pay attention to about the type of bind.\n\n``````bind : (!Malpha! x (!alpha! -> Mbeta)) -> Mbeta\n``````\n\nThe point is that the thing f takes as an argument is exacty the same type as the first argument to bind, except the first argument to bind has it all wrapped up in the monad (burrito).\n\nSo that’s the gist of what maybe is. Let’s real quick write findf-maybe and, ‘cause findf-maybe doesn’t show off everything we’re gonna wanna do, we’ll write a somewhat contrived example that’ll show off bind-maybe, too.\n\n``````(define findf-maybe\n(lambda (p ls)\n(cond\n((null? ls) (fail))\n...)))\n``````\n\nSo far, so good, this makes sense. We’ve been promising that we’ll return (Nothing) when we can’t findf the thing. Whelp, that’ll do it.\n\n``````(define findf-maybe\n(lambda (p ls)\n(cond\n((null? ls) (fail))\n(else (let ((a (car ls)))\n(cond\n((p a) ...)\n(else ...)))))))\n``````\n\nSo here, we’re just setting up and being a bit fancy. I happen to know for a fact that we’re going to need a couple of times, potentially, so I’m let-binding it above. There’s the remaining two cases: either (p a), or not. We’ll handle’em one at a time.\n\n``````(define findf-maybe\n(lambda (p ls)\n(cond\n((null? ls) (fail))\n(else (let ((a (car ls)))\n(cond\n((p a) (return-maybe a))\n(else ...)))))))\n``````\n\nWe wrap the thing up in the monad. findf-maybe is written to have return-type Mbeta. Which makes sense, according to our spec. We wanted it to return the stuff wrapped up, again, so we could tell the difference. That’s why we put it through return. Final case, the recursive call. Well, findf-maybe spits back something of Mbeta, which is what we’re after, so nothing special to do here, just the plain-ol’ regular recursion.\n\n``````(define findf-maybe\n(lambda (p ls)\n(cond\n((null? ls) (fail))\n(else (let ((a (car ls)))\n(cond\n((p a) (return-maybe a))\n(else (findf-maybe p (cdr ls)))))))))\n``````\n\nVoilà. Go forth and findf to your heart’s content. I promised another example, this time with bind-maybe. We’re gonna try to write an operator that’ll take a list of numbers and return the product of their reciprocals. Why? You caught me, no good reason. But let’s do it anyway, for sport. Or something.\n\nNow, the thing about reciprocals is that zero doesn’t have one. Division by zero is undefined. So when we try to do this, we’re gonna have to be on the lookout for a zero, and if we hit one, we come back out with (Nothing).\n\n``````(define prod-recip-maybe\n(lambda (ls)\n...))\n``````\n\nWe’re just gonna blast right through to the new and interesting bit.\n\n``````(define prod-recip-maybe\n(lambda (ls)\n(cond\n((null? ls) (return-maybe 1))\n(else\n(let ((a (car ls)))\n(cond\n((zero? a) (fail))\n(else ...)))))))\n``````\n\nSo we’ve got a non-zero number in the car. Let’s do the recursion, get that value, and then use it to return the value we’re really after. I’m doing the recursion, which means I’m coming back with a value in the monad. And then I wanna do something with the value of that recursive call, to get the value of the whole expression. This is a job for bind-maybe. If this was being done non-monadically, and we were using curried-app, again, it’d look like\n\n``````((curried-app (lambda (rec) (* a rec)))\n(prod-recip-maybe (cdr ls)))\n``````\n\nBut we’re not, we’re 1. in the monad, and 2. really /done/ with that curried-app thing. Since we’re in the monad, bind-maybe’s gonna do the job of pulling the pure out of the monadic value. And we’re gonna make sure that that function returns a monadic value, ‘cause the next fella up the line is expecting a value in the monad.\n\n``````(define prod-recip-maybe\n(lambda (ls)\n(cond\n((null? ls) (return-maybe 1))\n(else\n(let ((a (car ls)))\n(cond\n((zero? a) (fail))\n(else\n(bind-maybe\n(prod-recip-maybe (cdr ls))\n(lambda (rec)\n(return-maybe (* (/ 1 a) rec)))))))))))\n``````\n\nWell done. Buut, that is looking a little long, and is slightly obscuring. That do syntax we introduced would certainly clean some of this up.\n\n``````(define-syntax do\n(syntax-rules (<-)\n((_ bind e) e)\n((_ bind (v <- e) e* e** ...)\n(bind e (lambda (v) (do bind e* e** ...))))\n((_ bind e e* e** ...)\n(bind e (lambda (_) (do bind e* e** ...))))))\n``````\n\nIt’s your standard, run-of-the-mill syntax-rules macro. We take which bind function it is, and a whole bunch of expressions, and expand out into nested calls to that bind in the right fashion. For our case, this yields\n\n``````(define prod-recip-maybe\n(lambda (ls)\n(cond\n((null? ls) (return-maybe 1))\n(else\n(let ((a (car ls)))\n(cond\n((zero? a) (fail))\n(else\n(do bind-maybe\n(rec <- (prod-recip-maybe (cdr ls)))\n(return-maybe (* (/ 1 a) rec))))))))))\n``````\n\nAnd /there/ we go.\n\nNow, Go Eat A Burrito! Maybe While Wearing A Space-Suit!\n\nUpdated:"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8691756,"math_prob":0.9257842,"size":14276,"snap":"2022-40-2023-06","text_gpt3_token_len":3796,"char_repetition_ratio":0.12282792,"word_repetition_ratio":0.06523438,"special_character_ratio":0.27507704,"punctuation_ratio":0.1268198,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95980775,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-06T05:51:05Z\",\"WARC-Record-ID\":\"<urn:uuid:98ec0cdd-b3df-4039-8c66-af27773c93de>\",\"Content-Length\":\"35890\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d876b03a-f1b3-4627-a103-07d2bc73e19c>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd2f0732-1b2a-4ee9-a7e2-59da1aa5c259>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://jasonhemann.github.io/monads/\",\"WARC-Payload-Digest\":\"sha1:HVPVICMVCNRAYHYDFHDRNNJIUR6N7UIV\",\"WARC-Block-Digest\":\"sha1:DP5NX4FT6AJ7OLW42LC3NLIPC6I6Q2EI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500304.90_warc_CC-MAIN-20230206051215-20230206081215-00043.warc.gz\"}"} |
http://essay.helpstudents.xyz/page/Math-trivia-questions-4th-grade.html | [
"# Math trivia questions 4th grade\n\nTrivia questions are really healthy for minds and when it comes to math trivia questions than it really sharpens your mind and activates your responses. Here we are going to present you math trivia question and answers: The Perimeter Of A Circle Is Also Known As What? Pi Can Be Correctly Written As A Fraction.",
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"This quiz is full of questions you would find in a typical fourth-grade curriculum, and while you might think you’ve got this on lock, we are fairly certain some of these are going to be tougher than you can handle. For most of us, it has been quite a few years since we graced an elementary school classroom, and while we like to think that.",
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"Also, it is a mark gaining subject. Therefore, we provide you 5th Grade Math Trivia Questions and Answers which will help you to understand the basic concepts of the subject easily. 5th Grade MathTrivia Questions and Answers 5th Grade Trivia Questions. 61. Which letters are usually used in Roman numerical system?",
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"Visit this site for our 5th Grade Trivia Quiz Questions and Answers. Our printable 5th Grade Trivia Quiz Questions are suitable for children age 8-10 or elementary school age and the whole family. Free Kids 5th Grade Trivia Questions.",
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"Math Interactive Online Quizzes for Fourth (4th) Grade. On this page you will find interactive math quizzes for 4th grade in flash swf format. We have math quizzes that cover topics such as: Algebra, Patterns, Addition, Subtraction, Decimals, Geometry, Fractions, Probability, Venn Diagrams, Time and more. These quizzes offer a chance at teacher.",
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"JumpStart's 'Tiger Trivia' is a fun science worksheet for 4th grade. Help your 4th graders learn all about the majestic big cat. Download now!",
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"Math: trivia questions, facts and quizzes. Trivia Quizzes, Games, and Facts.\n\n## Interactive 4th Grade Math Quiz - YouTube.",
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"Math Trivia is information or math facts about a particular concept. In this page we have presented you some of the math facts. Answers are given at the end. Kids, parents and teachers can make use of these free printable math trivia or math facts worksheets. Math facts worksheets include information about numbers such as counting numbers.",
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"This quiz is designed for a 4th grade student. You will be using your problem solving skills. It will focus mostly on fractions and money. You will need to decide when to use multiplication, division, addition, or subtraction. Good Luck! Group: Math Math Quizzes: Topic: Fractions and Money.",
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"Grade 4 Math Questions With Answers. A set of grade 4 math questions on operations on numbers, converting units, algebraic expressions, evaluation of algebraic expressions, problems are presented along with their answers at the bottom of the page.",
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"Looking for some Math Trivia and Math Fun? Whether your are looking for math fun facts or trivia to spice up your lessons or simply to lighten up math work, here is a useful collection for you. Browse our extensive collection of math trivia and math fun stuff - math games, math tricks, math jokes, riddles, funny quotes, brain teasers, puzzles.",
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"Keep trying Jr. Scientist! Question 1 of 20. Who is credited with the discovery of Electricity? Benjamin Franklin. Sir Isaac Newton. Albert Einstein. Benjamin Franklin was the first scientist to discover electricity by tying a key to a kite. The correct answer is Benjamin Franklin. Benjamin Franklin was the first scientist to discover.",
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"Print your 4th grade math test before you start. Try to answer all the questions. Which fraction is not equivalent to the other two? Jetser has a business that is not doing well. On Friday, he earned 468 dollars. On Saturday, he earned 459. On Tuesday, he earned 432 dollars. How much money did he earned on Thursday?",
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"Visit this site for our 5th Grade Trivia Quiz Questions and Answers. Our printable 5th Grade Trivia Quiz Answers are suitable for children age 8-10 or elementary school age and the whole family. Free Kids 5th Grade Trivia Answers.\n\n## Tiger Trivia - Free Science Worksheet for 4th Grade.\n\nPlay this game to review Fun. Which of the following is not a name of Ninja Turtle.Can You Pass This Quiz of 4th Grade Science Questions? Emily DiNuzzo Elementary school science is probably where you learned a lot of what you know about how the natural world works.Everyday math 4th grade, Worksheets, Quizzes For Children, Fourth grade math activities for children, math games, quizzes, worksheets and more. Get your fourth graders engaged with our math fun games.\n\nThe following are some examples of 4th Grade Math Word Problems that use mixed operations (addition, subtraction, multiplication and division). These problems are solved with the help of block diagrams or bar models (Singapore Math) or tape diagrams (Common Core). There were 42 mangoes in each crate. 12 such crates of mangoes were delivered to.Science quizzes about Roman numerals, prime numbers, Fibonacci numbers, polygons, fractions and square roots. Have fun with these and get a grade for your trivia knowledge."
] | [
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https://www.daniweb.com/programming/web-development/threads/351935/problem-in-displaying-images | [
"",
null,
"Hello people,\n\nI have a problem in displaying my images in the dimensions I wish: I used the following code:\n\n``echo'<td><a href=\"portsdetails.php?idImg='.\\$idphoto.'\"><img src=\"'.\\$lien.'\" width=\"10\" height=\"10\" border=\"0\"/></a></td>';``\n\n## All 6 Replies\n\nI don't see your issue. Perhaps you could paste the whole html or link to the image?",
null,
"I don't see your issue. Perhaps you could paste the whole html or link to the image?\n\nThank you, but I want to resize my image (which name is here: \\$lien) when displaying it in the dimensions I want. When I put : width=\"100\" height=\"50\" it does not take effect.\n\n<?php\n\n\\$imagepath=\"phpimages/dog.jpg\";\n\n\\$image=imagecreatefromjpeg(\\$imagepath);\n\nimagejpeg(\\$image);\n\n?>\n\nUse above code that is working",
null,
"html should resize. It doesn't actually resize the filesize, just the display, so it will load just as slowly as if it were displaying the actual size. Can't see why\n\n``echo \"<a href=\\\"...\\\"><img src=\\\"\\$lien\\\" width=\\\"10\\\" height=\\\"10\\\" border=\\\"0\\\" /></a>\";``\n\ndoesn't work for you.\n\n``````<?php\n\n\\$p = \\$_GET['p'];\n\\$i = 1;\nwhile (false !== (\\$file = readdir(\\$handle))) {\nif (\\$file != \".\" && \\$file != \"..\") {\n\\$img[\\$i] = \\$file;\nif (\\$p == \\$img[\\$i]) {\n\\$ci = \\$i;\n}\n\\$i++;\n}\n}\nclosedir(\\$handle);\n\\$ti = \\$i - 1;\n\\$pi = \\$ci - 1;\nif (\\$p == \"\") {\n\\$ni = \\$ci + 2;\n} else {\n\\$ni = \\$ci + 1;\n}\n\\$prevNext = \"\";\nif (\\$pi > 0) {\n\\$piFile = \\$img[\\$pi];\n\\$prevNext .= \"<a href=\\\"\" . \\$_SERVER['PHP_SELF'] . \"?p=\" . \\$piFile . \"\\\" title=\\\"show previous image\\\"><strong>Previous</strong></a>\";\n} else {\n\\$prevNext .= \"<strong>Previous</strong>\";\n}\n\\$prevNext .= \" | \";\nif (\\$ni <= \\$ti) {\n\\$niFile = \\$img[\\$ni];\n\\$prevNext .= \"<a href=\\\"\" . \\$_SERVER['PHP_SELF'] . \"?p=\" . \\$niFile . \"\\\" title=\\\"show next image\\\"><strong>Next</strong></a>\";\n} else\n{\n\\$prevNext .= \"<strong>Next</strong>\";\n}\n\nif (\\$p == \"\")\n{\n\\$p = \\$img;\n}\n}\n?>\n\n<div class=\"hRule\"></div>\n\n<tr align=\"center\">\n<td class=\"nextPrevious\"><?php echo \\$prevNext; ?></td>\n</tr>\n<tr align=\"center\">\n<td> <img src=\"uploads/<?php echo \\$p; ?>\" border=\"2\"width=\"400\" height=\"400\"></td>\n</tr>\n</table>``````\n\ntry this..",
null,
"html should resize. It doesn't actually resize the filesize, just the display, so it will load just as slowly as if it were displaying the actual size. Can't see why\n\n``echo \"<a href=\\\"...\\\"><img src=\\\"\\$lien\\\" width=\\\"10\\\" height=\\\"10\\\" border=\\\"0\\\" /></a>\";``\n\ndoesn't work for you.\n\nThank you very much it works thanks to your code:\nI replaced echo'....';\nby echo \"....\"; as you wrote it and it really works very fine !!!\n\nThank you very much again Ardav !"
] | [
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"https://static.daniweb.com/connect/images/anonymous.png",
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"https://static.daniweb.com/connect/images/anonymous.png",
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"https://static.daniweb.com/connect/images/anonymous.png",
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"https://static.daniweb.com/connect/images/anonymous.png",
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https://www.indiabix.com/electronics/rl-circuits/134003 | [
"# Electronics - RL Circuits\n\n### Exercise :: RL Circuits - General Questions\n\n11.",
null,
"Which of the following statements is true if R = 100",
null,
"and XL = 100",
null,
"in the circuit in the given circuit?\n\n A. Each component drops 5 V. B. The impedance equals 200",
null,
". C. The power factor equals 1. D. The phase angle equals 45°.\n\nExplanation:\n\nNo answer description available for this question. Let us discuss.\n\n12.\n\nWhich of the following statements is true about a lag network?\n\n A. An increase in frequency causes an increase in phase lag. B. An increase in frequency causes an increase in the magnitude of the output voltage. C. A decrease in frequency causes an increase in phase lag. D. A decrease in frequency causes a decrease in the magnitude of the output voltage."
] | [
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"https://www.indiabix.com/_files/images/basic-electronics/basics/mcq12_1001_1.gif",
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"https://www.indiabix.com/_files/images/basic-electronics/basics/omega.gif",
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"https://www.indiabix.com/_files/images/basic-electronics/basics/omega.gif",
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"https://www.indiabix.com/_files/images/basic-electronics/basics/omega.gif",
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.7127636,"math_prob":0.9364107,"size":520,"snap":"2020-34-2020-40","text_gpt3_token_len":122,"char_repetition_ratio":0.14147286,"word_repetition_ratio":0.26666668,"special_character_ratio":0.25384617,"punctuation_ratio":0.14893617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9828584,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-07T09:00:18Z\",\"WARC-Record-ID\":\"<urn:uuid:25eb9902-a7f7-4c29-9699-3256a4162fed>\",\"Content-Length\":\"22380\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28bbc278-1810-4287-89ad-33f8cc3ea1e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:92790429-f16e-4b06-a6be-9e0567799df7>\",\"WARC-IP-Address\":\"35.244.11.196\",\"WARC-Target-URI\":\"https://www.indiabix.com/electronics/rl-circuits/134003\",\"WARC-Payload-Digest\":\"sha1:XUHKFGGAGPP4QZGBVEB26WLCKFFA7VM6\",\"WARC-Block-Digest\":\"sha1:TQ6ERIRAMX7HDO4MKZNDHBLYLQZ242IA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439737172.50_warc_CC-MAIN-20200807083754-20200807113754-00510.warc.gz\"}"} |
https://www.geeksforgeeks.org/find-minimum-number-of-steps-to-reach-the-end-of-string/?ref=rp | [
"# Find minimum number of steps to reach the end of String\n\n• Difficulty Level : Medium\n• Last Updated : 11 Jun, 2021\n\nGiven a binary string str of length N and an integer K, the task is to find the minimum number of steps required to move from str to str[N – 1] with the following moves:\n\n1. From an index i, the only valid moves are i + 1, i + 2 and i + K\n2. An index i can only be visited if str[i] = ‘1’\n\nExamples:\n\nInput: str = “101000011”, K = 5\nOutput:\nstr -> str -> str -> str\nInput: str = “1100000100111”, K = 6\nOutput: -1\nThere is no possible path.\nInput: str = “10101010101111010101”, K = 4\nOutput:\n\nApproach: The idea is to use dynamic programming to solve the problem.\n\n• It is given that for any index i, it is possible to move to an index i+1, i+2 or i+K.\n• One of the three possibilities will give the required result that is the minimum number of steps to reach the end.\n• Therefore, the dp[] array is created and is filled in a bottom-up manner.\n\nBelow is the implementation of the above approach:\n\n## C++\n\n `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum number``// of steps to reach the end``int` `minSteps(string str, ``int` `n, ``int` `k)``{` ` ``// If the end can't be reached`` ``if` `(str[n - 1] == ``'0'``)`` ``return` `-1;` ` ``// Already at the end`` ``if` `(n == 1)`` ``return` `0;` ` ``// If the length is 2 or 3 then the end`` ``// can be reached in a single step`` ``if` `(n < 4)`` ``return` `1;` ` ``// For the other cases, solve the problem`` ``// using dynamic programming`` ``int` `dp[n];` ` ``// It requires no move from the`` ``// end to reach the end`` ``dp[n - 1] = 0;` ` ``// From the 2nd last and the 3rd last`` ``// index, only a single move is required`` ``dp[n - 2] = 1;`` ``dp[n - 3] = 1;` ` ``// Update the answer for every index`` ``for` `(``int` `i = n - 4; i >= 0; i--) {` ` ``// If the current index is not reachable`` ``if` `(str[i] == ``'0'``)`` ``continue``;` ` ``// To store the minimum steps required`` ``// from the current index`` ``int` `steps = INT_MAX;` ` ``// If it is a valid move then update`` ``// the minimum steps required`` ``if` `(i + k < n && str[i + k] == ``'1'``)`` ``steps = min(steps, dp[i + k]);` ` ``if` `(str[i + 1] == ``'1'``)`` ``steps = min(steps, dp[i + 1]);` ` ``if` `(str[i + 2] == ``'1'``)`` ``steps = min(steps, dp[i + 2]);` ` ``// Update the minimum steps required starting`` ``// from the current index`` ``dp[i] = (steps == INT_MAX) ? steps : 1 + steps;`` ``}` ` ``// Cannot reach the end starting from str`` ``if` `(dp == INT_MAX)`` ``return` `-1;` ` ``// Return the minimum steps required`` ``return` `dp;``}` `// Driver code``int` `main()``{`` ``string str = ``\"101000011\"``;`` ``int` `n = str.length();`` ``int` `k = 5;` ` ``cout << minSteps(str, n, k);` ` ``return` `0;``}`\n\n## Java\n\n `// Java implementation of the approach``class` `GFG``{`` ` ` ``final` `static` `int` `INT_MAX = Integer.MAX_VALUE ;`` ` ` ``// Function to return the minimum number`` ``// of steps to reach the end`` ``static` `int` `minSteps(String str, ``int` `n, ``int` `k)`` ``{`` ` ` ``// If the end can't be reached`` ``if` `(str.charAt(n - ``1``) == ``'0'``)`` ``return` `-``1``;`` ` ` ``// Already at the end`` ``if` `(n == ``1``)`` ``return` `0``;`` ` ` ``// If the length is 2 or 3 then the end`` ``// can be reached in a single step`` ``if` `(n < ``4``)`` ``return` `1``;`` ` ` ``// For the other cases, solve the problem`` ``// using dynamic programming`` ``int` `dp[] = ``new` `int``[n];`` ` ` ``// It requires no move from the`` ``// end to reach the end`` ``dp[n - ``1``] = ``0``;`` ` ` ``// From the 2nd last and the 3rd last`` ``// index, only a single move is required`` ``dp[n - ``2``] = ``1``;`` ``dp[n - ``3``] = ``1``;`` ` ` ``// Update the answer for every index`` ``for` `(``int` `i = n - ``4``; i >= ``0``; i--)`` ``{`` ` ` ``// If the current index is not reachable`` ``if` `(str.charAt(i) == ``'0'``)`` ``continue``;`` ` ` ``// To store the minimum steps required`` ``// from the current index`` ``int` `steps =INT_MAX ;`` ` ` ``// If it is a valiINT_MAXd move then update`` ``// the minimum steps required`` ``if` `(i + k < n && str.charAt(i + k) == ``'1'``)`` ``steps = Math.min(steps, dp[i + k]);`` ` ` ``if` `(str.charAt(i + ``1``) == ``'1'``)`` ``steps = Math.min(steps, dp[i + ``1``]);`` ` ` ``if` `(str.charAt(i + ``2``) == ``'1'``)`` ``steps = Math.min(steps, dp[i + ``2``]);`` ` ` ``// Update the minimum steps required starting`` ``// from the current index`` ``dp[i] = (steps == INT_MAX) ? steps : ``1` `+ steps;`` ``}`` ` ` ``// Cannot reach the end starting from str`` ``if` `(dp[``0``] == INT_MAX)`` ``return` `-``1``;`` ` ` ``// Return the minimum steps required`` ``return` `dp[``0``];`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``String str = ``\"101000011\"``;`` ``int` `n = str.length();`` ``int` `k = ``5``;`` ` ` ``System.out.println(minSteps(str, n, k));`` ``}``}` `// This code is contributed by AnkitRai01`\n\n## Python3\n\n `# Python3 implementation of the approach``import` `sys` `INT_MAX ``=` `sys.maxsize;` `# Function to return the minimum number``# of steps to reach the end``def` `minSteps(string , n, k) :`` ` ` ``# If the end can't be reached`` ``if` `(string[n ``-` `1``] ``=``=` `'0'``) :`` ``return` `-``1``;` ` ``# Already at the end`` ``if` `(n ``=``=` `1``) :`` ``return` `0``;` ` ``# If the length is 2 or 3 then the end`` ``# can be reached in a single step`` ``if` `(n < ``4``) :`` ``return` `1``;` ` ``# For the other cases, solve the problem`` ``# using dynamic programming`` ``dp ``=` `[``0``] ``*` `n;` ` ``# It requires no move from the`` ``# end to reach the end`` ``dp[n ``-` `1``] ``=` `0``;` ` ``# From the 2nd last and the 3rd last`` ``# index, only a single move is required`` ``dp[n ``-` `2``] ``=` `1``;`` ``dp[n ``-` `3``] ``=` `1``;` ` ``# Update the answer for every index`` ``for` `i ``in` `range``(n ``-` `4``, ``-``1``, ``-``1``) :`` ` ` ``# If the current index is not reachable`` ``if` `(string[i] ``=``=` `'0'``) :`` ``continue``;` ` ``# To store the minimum steps required`` ``# from the current index`` ``steps ``=` `INT_MAX;` ` ``# If it is a valid move then update`` ``# the minimum steps required`` ``if` `(i ``+` `k < n ``and` `string[i ``+` `k] ``=``=` `'1'``) :`` ``steps ``=` `min``(steps, dp[i ``+` `k]);` ` ``if` `(string[i ``+` `1``] ``=``=` `'1'``) :`` ``steps ``=` `min``(steps, dp[i ``+` `1``]);` ` ``if` `(string[i ``+` `2``] ``=``=` `'1'``) :`` ``steps ``=` `min``(steps, dp[i ``+` `2``]);` ` ``# Update the minimum steps required starting`` ``# from the current index`` ``dp[i] ``=` `steps ``if` `(steps ``=``=` `INT_MAX) ``else` `(``1` `+` `steps);`` ` ` ``# Cannot reach the end starting from str`` ``if` `(dp[``0``] ``=``=` `INT_MAX) :`` ``return` `-``1``;` ` ``# Return the minimum steps required`` ``return` `dp[``0``];` `# Driver code``if` `__name__ ``=``=` `\"__main__\"` `:` ` ``string ``=` `\"101000011\"``;`` ``n ``=` `len``(string);`` ``k ``=` `5``;` ` ``print``(minSteps(string, n, k));` `# This code is contributed by AnkitRai01`\n\n## C#\n\n `// C# implementation of the approach``using` `System;` `class` `GFG``{`` ` ` ``static` `int` `INT_MAX = ``int``.MaxValue ;`` ` ` ``// Function to return the minimum number`` ``// of steps to reach the end`` ``static` `int` `minSteps(``string` `str, ``int` `n, ``int` `k)`` ``{`` ` ` ``// If the end can't be reached`` ``if` `(str[n - 1] == ``'0'``)`` ``return` `-1;`` ` ` ``// Already at the end`` ``if` `(n == 1)`` ``return` `0;`` ` ` ``// If the length is 2 or 3 then the end`` ``// can be reached in a single step`` ``if` `(n < 4)`` ``return` `1;`` ` ` ``// For the other cases, solve the problem`` ``// using dynamic programming`` ``int` `[]dp = ``new` `int``[n];`` ` ` ``// It requires no move from the`` ``// end to reach the end`` ``dp[n - 1] = 0;`` ` ` ``// From the 2nd last and the 3rd last`` ``// index, only a single move is required`` ``dp[n - 2] = 1;`` ``dp[n - 3] = 1;`` ` ` ``// Update the answer for every index`` ``for` `(``int` `i = n - 4; i >= 0; i--)`` ``{`` ` ` ``// If the current index is not reachable`` ``if` `(str[i] == ``'0'``)`` ``continue``;`` ` ` ``// To store the minimum steps required`` ``// from the current index`` ``int` `steps = INT_MAX ;`` ` ` ``// If it is a valiINT_MAXd move then update`` ``// the minimum steps required`` ``if` `(i + k < n && str[i + k] == ``'1'``)`` ``steps = Math.Min(steps, dp[i + k]);`` ` ` ``if` `(str[i + 1] == ``'1'``)`` ``steps = Math.Min(steps, dp[i + 1]);`` ` ` ``if` `(str[i + 2] == ``'1'``)`` ``steps = Math.Min(steps, dp[i + 2]);`` ` ` ``// Update the minimum steps required starting`` ``// from the current index`` ``dp[i] = (steps == INT_MAX) ? steps : 1 + steps;`` ``}`` ` ` ``// Cannot reach the end starting from str`` ``if` `(dp == INT_MAX)`` ``return` `-1;`` ` ` ``// Return the minimum steps required`` ``return` `dp;`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `Main()`` ``{`` ``string` `str = ``\"101000011\"``;`` ``int` `n = str.Length;`` ``int` `k = 5;`` ` ` ``Console.WriteLine(minSteps(str, n, k));`` ``}``}` `// This code is contributed by AnkitRai01`\n\n## Javascript\n\n ``\nOutput:\n`3`\n\nTime Complexity: O(N) where N is the length of the string.\n\nMy Personal Notes arrow_drop_up"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.57670397,"math_prob":0.98357576,"size":9098,"snap":"2022-05-2022-21","text_gpt3_token_len":2973,"char_repetition_ratio":0.16835275,"word_repetition_ratio":0.5863474,"special_character_ratio":0.38689822,"punctuation_ratio":0.12365011,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998883,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T19:00:28Z\",\"WARC-Record-ID\":\"<urn:uuid:41e1c39d-363c-4bd0-8b60-26b6af64722b>\",\"Content-Length\":\"202027\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0122e5ff-7b6f-4128-922c-4bf83bf13a2a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1b76947-66f6-459a-8843-e406315c4903>\",\"WARC-IP-Address\":\"23.12.144.230\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/find-minimum-number-of-steps-to-reach-the-end-of-string/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:2DQKOCSTKLMB55UI42XE5CJLTHEGIJUX\",\"WARC-Block-Digest\":\"sha1:5FVD76NDFMVIGEO27CNN2D5YVDPPD5XL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662675072.99_warc_CC-MAIN-20220527174336-20220527204336-00677.warc.gz\"}"} |
http://www.softmath.com/math-book-answers/multiplying-fractions/combinations-maths-year-8.html | [
"English | Español\n\nTry our Free Online Math Solver!",
null,
"Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nWhat our customers say...\n\nThousands of users are using our software to conquer their algebra homework. Here are some of their experiences:\n\nI use this great program for my Algebra lessons, maximize the program and use it as a blackboard. Students just love Algebrator presentations\nDebra Ratto, CO\n\nI was so proud when my son decided to take algebra honors, but I was disheartened when I realized that I could not help him with his homework. I had not taken algebra since high school, and simply did not remember how to complete some of the projects. Algebrator allowed us to go through each step together. Thank you for making a program that allows me to help my son!\nJenni Coburn, IN\n\nI've been using your system, and it breezed through every problem that couldn't be solved by PAT. I'm really impressed with the user friendly setup, and capabilities of your system. Thanks again!\nRomero Reynolds, AK.\n\nThe simple interface of Algebrator makes it easy for my son to get right down to solving math problems. Thanks for offering such a useful program.\nR.B., New Mexico\n\nSearch phrases used on 2010-09-26:\n\nStudents struggling with all kinds of algebra problems find out that our software is a life-saver. Here are the search phrases that today's searchers used to find our site. Can you find yours among them?\n\n• multipy algebra equations\n• algebra graphing linear equations\n• free ti-84 emulator\n• algebra calculator online free\n• math software programs\n• 3\n• algebraic equations\n• math help algebra\n• equation calculator online\n• simplifying rational expressions\n• how do you graph linear inequalities\n• graphing quadratic equations\n• free algebra solver step by step\n• how to simplify algebra\n• Answers Math Problems\n• matrix calculator\n• help solve rational expression in lowest term\n• simplifying equations\n• how do you solve L+M=2x\n• solving quadratic equation -x-2x+24=0 by graphing\n• Linear Inequality Prentice Hall Gold Algebra\n• systems of equations steps\n• what is the solution to this equation x+4(x+5)=40\n• intermediate algebra problem solver for free\n• solve ln .75 rad 7\n• holt algebra 1 worksheet answers\n• linear equations\n• computation of rational numbers\n• my calculator\n• free rationalize the denominator calculator\n• solve algebra 2 matrices for free\n• Course 2 Georgia Homework and Practice Workbook, Holt Mathematics answer\n• solve for y\n• polynomial solver long division\n• Algebra with Pizzazz Answers\n• how do you solve using a matrix method\n• algebra calculators\n• X^3/(.110)(.140)^3 sovle for x\n• solving linear equations\n• what is the answer for the Integrated Algebra Regents june 2011\n• help me solve algebra\n• algebra answer generator\n• solving quadratic equations\n• matrices solver\n• synthetic division solver\n• algebrator 2011\n• linear equations games\n• activities on teaching integers\n• algebra software\n• transposition of formula worksheet\n• algebra calculator\n• Dividing Rational Expressions Calculator\n• 8th pre algebra help\n• free college algebra tutoring calculators\n• solve x (x^2+9)^-1/2=2/5\n• factoring equations\n• solving 10th grade algebra problems\n• 3rd degree algebra solver\n• Algebra Calculator\n• How is doing operations-adding, subtracting, multiplying, and dividing-with rational expressions similar to or different from doing operations with fractions?\n• find the value of y when x= 4,9\n• how to solve quadratic equations\n• algebra\n• what is the algebraic expression for 3,6.11,18,27,38\n• algebra step by step calculator\n• roots and radicals\n• free printable math adding negatives\n• free least common denominator solver\n• ALGEBRATOR\n• algebra help\n• 3x+y=4 5x-7 y=11 solve by substitution\n• utube algebra.com\n• +multiplying negative numbers worksheet\n• sample contemporary math problems\n• least common denominator calculator"
] | [
null,
"http://www.softmath.com/images/video-pages/solver-top.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.874311,"math_prob":0.9772249,"size":4241,"snap":"2019-43-2019-47","text_gpt3_token_len":1018,"char_repetition_ratio":0.17441586,"word_repetition_ratio":0.0029069767,"special_character_ratio":0.2230606,"punctuation_ratio":0.06861314,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99913985,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T09:29:54Z\",\"WARC-Record-ID\":\"<urn:uuid:9d35f1f8-b3fc-47a3-9d13-d04412e9c19c>\",\"Content-Length\":\"89244\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:786b7f42-f0db-4443-b103-060d7a118b34>\",\"WARC-Concurrent-To\":\"<urn:uuid:28940da4-0d08-4534-b546-3d8be9fba129>\",\"WARC-IP-Address\":\"52.43.142.96\",\"WARC-Target-URI\":\"http://www.softmath.com/math-book-answers/multiplying-fractions/combinations-maths-year-8.html\",\"WARC-Payload-Digest\":\"sha1:LQSE4HFKOFKGUNEM662JIIJG7JJIJHBA\",\"WARC-Block-Digest\":\"sha1:LCZYAFIIFMGWKUN33WIHDPJM72ZSUI7J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986692723.54_warc_CC-MAIN-20191019090937-20191019114437-00414.warc.gz\"}"} |
https://www.lexico.com/en-es/translate/electron_volt | [
"# Traducción de electron volt en español:\n\n## electronvoltio, n.\n\n### nombre\n\n• 1\n\nelectronvoltio masculino\n• Its decay produces gamma rays with energies of a trillion electronvolts.\n• At today's accelerators, for example the CERN LEP accelerator, the value has been measured as 1/128 rather than 1/137 at energies corresponding to approximately 100 billion electronvolts.\n• It means that the energy of the source is of the order of a teraelectronvolt, that is, a trillion electronvolts.\n• Then if these electrons are attracted to a second plate B with a voltage (relative to A) of +1 volt, each electron gains one electron volt.\n• One electron volt is roughly half the energy carried by a single photon of red light."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.73429316,"math_prob":0.9796829,"size":715,"snap":"2021-31-2021-39","text_gpt3_token_len":164,"char_repetition_ratio":0.19268636,"word_repetition_ratio":0.0,"special_character_ratio":0.21538462,"punctuation_ratio":0.1015625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811907,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-17T01:56:07Z\",\"WARC-Record-ID\":\"<urn:uuid:127f16a1-3b1f-418c-9e89-7440774a4445>\",\"Content-Length\":\"72074\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:345f4b20-aef6-4c93-ba2c-bd0ee88616d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:4f9248f2-d77b-4921-9269-b16a0b4f13c0>\",\"WARC-IP-Address\":\"151.101.202.133\",\"WARC-Target-URI\":\"https://www.lexico.com/en-es/translate/electron_volt\",\"WARC-Payload-Digest\":\"sha1:TRP7UAUT5PXKJ4QP6KUWUP7EO3PWELDK\",\"WARC-Block-Digest\":\"sha1:YRILRKFUM7RHTHACR427TGNXYZR5GBZC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780053918.46_warc_CC-MAIN-20210916234514-20210917024514-00065.warc.gz\"}"} |
https://dsp.stackexchange.com/questions/55887/normalized-cross-correlation-in-frequency-domain | [
"# Normalized cross-correlation in frequency domain\n\nI never worked with signal processing and never really used Fourier transforms before, still I am working on a project consisting on taking the output of an accelerometer to detect some movement features between touching surfaces. Using a frame of the output, I compute the FFT. After analyzing the data for multiple experiments, it appeared to make sense to use gaussians to isolate bands of frequencies and use that to classify the features.\n\nThat does not work well, however, if I change the sensor I am using, the sampling rate or frame size. Thus, I want to calculate the normalized cross-correlation. Problem is: I could not find an algorithm to calculate the NCC in the frequency domain and I did not figure out how to transform the translated gaussians to the time domain.\n\nFor instance, I have a frequency spectrum from -F/2 to F/2 and L samples. I converted that to L/2+1 frequency bins from 0 to F/2 and defined a band using a gaussian with standard deviation s and centered on c.\nTo get the cross-correlation between the spectrum and the gaussian, I simply do the sum of the element-wise multiplication of them. Now, if I want that normalized, what can I do? Is there a way to normalize it right from the frequency domain?\n\n• the fourier transform of a gaussian is a gaussian. if your fft is large, using the continuous fourier result is typically reasonable. – user28715 Mar 9 '19 at 16:45\n\n$$F[k]$$ is the (discrete) Fourier transform of my sample, $$G[k]$$ is the gaussian curve in the frequency domain. The cross-correlation is $$CC = \\sum^{L-1}_{k=0}F[k]{\\cdot}G[k]$$ Then, to get the normalized cross-correlation we first normalize the vectors $$F$$ and $$G$$, as would be done to any vector: $$\\hat{v} = \\frac{v}{|v|} = \\frac{v}{\\sqrt{\\sum^{L-1}_{k=0}v^2[k]}}$$ Thus, the normalized cross-correlation should be the dot-product of the normalized vectors $$\\hat{F}$$ and $$\\hat{G}$$: $$NCC = \\frac{\\sum^{L-1}_{k=0}F[k]{\\cdot}G[k]}{\\sqrt{\\sum^{L-1}_{k=0}F^2[k]\\cdot\\sum^{L-1}_{k=0}G^2[k]}}$$"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9314582,"math_prob":0.996353,"size":1232,"snap":"2021-04-2021-17","text_gpt3_token_len":259,"char_repetition_ratio":0.10749186,"word_repetition_ratio":0.0,"special_character_ratio":0.20535715,"punctuation_ratio":0.09311741,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999405,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-19T06:06:19Z\",\"WARC-Record-ID\":\"<urn:uuid:fc089450-779a-44aa-85d9-558cc8588c3a>\",\"Content-Length\":\"152134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:18964b5b-f342-4d4b-92eb-f3bcce0bf222>\",\"WARC-Concurrent-To\":\"<urn:uuid:bc3bae67-961a-4a06-89c9-b60a06147899>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/55887/normalized-cross-correlation-in-frequency-domain\",\"WARC-Payload-Digest\":\"sha1:JWH6WZ4IUGCR4PJVBHUEXVAU7RRM36DU\",\"WARC-Block-Digest\":\"sha1:OUADYJUPL366CI5WIQ544A3BGGT4E6QJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703517966.39_warc_CC-MAIN-20210119042046-20210119072046-00023.warc.gz\"}"} |
https://www.codespeedy.com/cutting-a-rod-problem-in-cpp/ | [
"# Cutting a Rod problem in C++\n\nIn this tutorial, we will learn about cutting a rod problem in C++. We will also see the use of dynamic programming to solve the cutting of the rod problem.\n\nWe will also see examples to understand the concept in a better way.\n\n## Cutting Rod Problem using Dynamic Programming in C++\n\nIn cutting rod problem,\n\nWe have given a rod of length n and an array of prices of the length of pieces whose size is smaller than n. We need to determine the maximum price to cut the rod.\n\nLet,s see the example,\n\n```length of rod is given 4.\nlength of pieces---1 2 3 4\nprices of pieces---2 5 7 8\n\nmaximum prices will be 10.\nsize of pieces will be 2,2.```\n\nIn the above example, we can understand the cutting rod problem.\n\nTo solve the problem we take the temporary array to obtain the maximum price.\n\n`int dp[n+1]`\n\nLet,s see the pseudocode,\n\n```int max_value use to store the maximum cost.\ndp[n] will be the solution.\ninitialise the dp=0;\ni=1;\nwhile i<n+1 do\nint max_value=INT_MIN.\nj=0.\nwhile j<i do\nmax_value = max(max_value, price[j] + dp[i-j-1]);\ndp[i] = max_value;\ndp[n]is maximum price to cut a rod```\n\nLet,s see the value of an array dp for the above example,\n\n```length-0 1 2 3 4\ncost---0 2 5 7 10```\n\nSo, for each entry, we need to apply step,\n\n```max_value = max(max_value, price[j] + dp[i-j-1]);\ni.e if we select the length j so we need to take the maximum value of\nmax_value and cost of j length + maximum cost of i-j length```\n\nNow, let’s see the code.\n\n### Code implementation in C++ | Cutting a Rod\n\n```#include<iostream>\n#include<bits/stdc++.h>\nusing namespace std;\nint main(){\nint price[] = {2,5,7,8};\nint n = sizeof(price)/sizeof(price);\nint dp[n+1]; //store the maximum price of length i\ndp = 0;\nfor (int i = 1; i<=n; i++)\n{\nint max_val = INT_MIN;\nfor (int j = 0; j < i; j++)\nmax_val = max(max_val, price[j] + dp[i-j-1]);\ndp[i] = max_val;\n}\ncout<<\"Maximum price of rod is:\"<<\" \";\ncout<<dp[n]<<endl;\n}```\n\nOutput\n\n`Maximum price of rod is: 10`\n\nYou may also learn,\n\nActivity Selection Problem using Greedy method in C++\n\nProgram to find all distinct solutions to N-Queens problem in Java"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.70309085,"math_prob":0.9963933,"size":2078,"snap":"2020-34-2020-40","text_gpt3_token_len":589,"char_repetition_ratio":0.15621987,"word_repetition_ratio":0.016,"special_character_ratio":0.30654475,"punctuation_ratio":0.12343096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99944574,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-30T06:57:16Z\",\"WARC-Record-ID\":\"<urn:uuid:aeb4b70f-cf0d-48be-9bd1-10b5b15851a1>\",\"Content-Length\":\"32371\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c8e318aa-091a-4145-ae42-aa39d0040ca8>\",\"WARC-Concurrent-To\":\"<urn:uuid:854f44b6-5173-49a1-bcf2-1017277a1a49>\",\"WARC-IP-Address\":\"194.1.147.77\",\"WARC-Target-URI\":\"https://www.codespeedy.com/cutting-a-rod-problem-in-cpp/\",\"WARC-Payload-Digest\":\"sha1:AUM43YD3HHD3SRWGXPF5ZJR2R5HYRSTR\",\"WARC-Block-Digest\":\"sha1:MAOB2RN5LNDYQJ4Z425LZPGHABSMFGHX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402118004.92_warc_CC-MAIN-20200930044533-20200930074533-00577.warc.gz\"}"} |
https://math.stackexchange.com/questions/1810629/let-a-be-a-square-matrix-of-order-3-with-integer-entries-such-that-deta-1/1810647 | [
"# Let A be a square matrix of order $3$ with integer entries such that $\\det(A)=1$.\n\n• What is the maximum possible number of entries of A that are even? What if A is a matrix of order n?\n\n• What is the maximum possible number of entries of A that are PRIMES? What if A is a matrix of order n?\n\nThis question was asked during my class test and I was unable to do this.\n\nPlease help me in doing this.\n\n• What set are the entries from? – joriki Jun 3 '16 at 5:49\n• @joriki:Integers – Styles Aug 28 '16 at 14:39\n\n## 3 Answers\n\nFor the first question, you can show that $I_3$ is an example of such a matrix with $6$ even entries. If there are $7$ or more event entries, then necessarily there is a row with only even entries, calculate your determinant by expanding this line, it is a sum of even terms, thus the determinant is even and can't be $1$.\nThus $6$ is the maximum possible number of entries of A that are even and such as $\\det(A)=1$.\n\nThe generalisation for matrices of order $n$, for the first question is $n^2-n$, you can prove this with same reasoning that in the case $3 \\times 3$ : $I_n$ has $n^2-n$ even entries and its determinant is $1$. And if your matrix has more that $n^2-n$ even entries then there is necessary a whole row with even entries and you can expand your determinant from this row and thus it will be even and so different of $1$.\n\n• :But,its determinant is p,we've to consider those matrices whose determinant is 1 – Styles Jun 3 '16 at 6:12\n• Try $\\begin{matrix} 1 & 2 & 2 \\\\ 2& 1&2 \\\\ 0 &2 &1 \\end{matrix}$. – Hetebrij Jun 3 '16 at 6:22\n\nFor first part answer is $6$. You cab take $A$ as $$A=\\begin{bmatrix} 1&0&0\\\\b&a&ac-1\\\\d&1&c\\end{bmatrix}$$ where $a,b,c,d$ are any even number. Now consider $$e_3=6,$$ $$e_4=e_3+2(4-1)$$ $$e_n=e_{n-1}+2(n-1)$$ where $e_n$ is no. of even entries when $A$ is of order $n\\ge3$.\n\nFor second part take $A$ as $$A=\\begin{bmatrix} 1&0&0\\\\a&p_1&p_2\\\\b&p_3&p_4\\end{bmatrix}$$ where $a,b$ are any prime number and $p_1p_4-p_3p_2=1$ it is possible to find such primes e.g take $p_1=3,p_4=5,p_3=2,p_2=7$. Now consider $$p_3=6,$$ $$e_4=e_3+(4-1)$$ $$e_n=e_{n-1}+(n-1)$$ where $p_n$ is no. of prime entries when $A$ is of order $n\\ge3$.\n\n• For the first part, $\\frac{1}{ab}$ is not an integer. – Hetebrij Jun 3 '16 at 6:29\n• @Sry:How can we prove these recursion formulas. – Styles Jun 3 '16 at 7:00\n• Say we are going for matrix of order $4$, it will be of same format i.e first row has $a_{11}=1$ entry fixed so rest three are left for even choices again first column has 3 entries left for even choices. Apart from first row and first column, leftover matrix is of $3\\times3$ whose even choice we know. Similarly if we go for $5\\times5$ matrix First row and column has $4$ choices each for even entries and leftover $4\\times4$ matrix has $e_4$ choices we know already. So.... – Sry Jun 3 '16 at 7:34\n• So, what is the maximum possible number of entries of a $3 \\times 3$ matrix $A$ that are primes? – SARTHAK GUPTA Dec 6 '19 at 18:40\n\nTo a matrix of order 3, maximum possibility of even entries is 6 ( take identity matrix of order 3) but as you put even entries on more than 6 places then you can find a row consisting of even entries along which you can take determinant.We can also generalize this for order n, may be possibility for maximum even entries is n^{2}-n( which is Identity matrix)\n\n• see identity matrix is always a case – Sunny Jun 3 '16 at 6:10\n• There is no conviction in the argument – Shailesh Jun 3 '16 at 6:18"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.70578957,"math_prob":0.9998447,"size":622,"snap":"2021-04-2021-17","text_gpt3_token_len":294,"char_repetition_ratio":0.1262136,"word_repetition_ratio":0.077922076,"special_character_ratio":0.47588426,"punctuation_ratio":0.10691824,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99981374,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T09:10:24Z\",\"WARC-Record-ID\":\"<urn:uuid:3fdeae65-6162-4247-be2c-13e94eaa27da>\",\"Content-Length\":\"190588\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:def00242-995c-4c66-be04-3f81f0ae9e30>\",\"WARC-Concurrent-To\":\"<urn:uuid:9ffb13c4-f6e1-490e-84ea-b7286c239e55>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1810629/let-a-be-a-square-matrix-of-order-3-with-integer-entries-such-that-deta-1/1810647\",\"WARC-Payload-Digest\":\"sha1:N37OX7ES3V3P4PSFKDKY7RNRAJKDEDXN\",\"WARC-Block-Digest\":\"sha1:MXJYS7YLIVVIEVSFDUKLVT5TJ2OW4OHR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038084601.32_warc_CC-MAIN-20210415065312-20210415095312-00237.warc.gz\"}"} |
https://www.analyticsvidhya.com/blog/2021/06/part-17-step-by-step-guide-to-master-nlp-topic-modelling-using-plsa/ | [
"This article was published as a part of the Data Science Blogathon\n\n## Introduction\n\nThis article is part of an ongoing blog series on Natural Language Processing (NLP). In the previous article, we discussed a Topic modelling technique named Latent Semantic Analysis (LSA), but we observed that there are some disadvantages of LSA, so to overcome those problems, we come up with the concept of pLSA, which stands for Probabilistic Latent Semantic Analysis.\n\nSo, In this article, we will deep dive into the concepts of pLSA, which is a technique used to model information under a probabilistic framework, and also discuss the mathematics behind the different parametrization of this technique in a detailed manner.\n\nThis is part-17 of the blog series on the Step by Step Guide to Natural Language Processing.\n\n1. Familiar with variables involved in pLSA\n\n2. What is pLSA?\n\n3. Latent Variable Model for pLSA\n\n4. Matrix Factorization model for pLSA\n\n## Familiar with Variables involved in pLSA\n\nWe have to understand the following three sets of variables while studying pLSA:\n\n• Documents\n• Words\n• Topics\n\nLet’s discuss each of them one by one in a bit detailed manner:\n\n### Documents\n\n`Representation: D={d1,d2,d3,…dN}`\n\nwhere N represents the number of documents present in the corpus.\n\ndi denotes ith document in the set D.\n\nHere we can call a document also as a sentence since these two words are used interchangeably.\n\n### Words\n\n`Representation: W={w1,w2,…wM}`\n\nwhere M represents the size of our vocabulary or dictionary size.\n\nwi denotes ith word in the vocabulary W.\n\nHere we treat the set W as a bag of words implies that there is no particular order followed in the assignment of the index i.\n\n### Topics\n\n`Representation: Z={z1,z2,…zk} `\n\nThese are also called Latent or hidden variables.\n\nThe k value of the parameter is specified by the user.\n\n## What is pLSA?\n\nRecap the basic assumption of topic modelling algorithms:\n\n• Each document consists of a mixture of topics, and\n• Each topic consists of a collection of words.\n\npLSA stands for Probabilistic Latent Semantic Analysis, uses a probabilistic method instead of Singular Value Decomposition, which we used in LSA to tackle the problem.\n\nThe main goal is to find a probabilistic model with latent or hidden topics that can generate the data which we observe in our document-term matrix. In mathematical terms, we want a model P(D, W) such that for any document d and word w in the corpus, P(d,w) corresponds to that entry in the document-term matrix.\n\nSo, pLSA is an advancement to LSA. It is a statistical technique for the analysis of two-mode and co-occurrence data.\n\n## Latent Variable Model for pLSA\n\nHere we describe the two parametrizations for pLSA:\n\n### Parametrization -1\n\nIn this parametrization, we sample a document first then based on the document we sample a topic, and based on the topic we sample a word, which means d and w are conditionally independent given a hidden topic ‘z’.\n\nThe pictorial representation of this parametrization is as follows:\n\nAs discussed earlier, the topics are hidden variables. The only things we see are the words and the set of documents. So, In this framework, we have to find the relation between the hidden variables and the observed variables.\n\nAs we discussed the assumptions of the topic model, pLSA adds a probabilistic spin to these assumptions in the following way:\n\n• Given a document d, a topic z is present in that selected document with probability P(z|d)\n• Given a topic z, word w is drawn from the topic z with probability P(w|z)\n\nHere we associate z with (d,w) and described a generative process where we select a document, then a topic, and finally a word from that topic. Formally,\n\n1. We select a document from the corpus with a probability P(d)\n\n2. For every word in the selected document dn, and word wi\n\n• Select a topic zi from a conditional distribution with a probability P(z|dn).\n• Select a word with a probability P(w|zi)\n\nBefore diving into the mathematical equations, let’s discuss the two main assumptions this model makes.\n\n### Assumption-1(Bag of Words)\n\nAs we discussed while learning the text vectorization techniques that the word ordering in the vocabulary doesn’t matter. In simple words, the joint variable (d,w) is sampled independently.",
null,
"### Assumption-2(Conditional Independence)\n\nIt is one of the key assumptions that we make while formulating the theory is that the words and the documents are conditionally independent. Focus on the word conditionally. This implies\n\n`P(w,d|z) = P(w|z)*P(d|z) `\n\nThe model under the above-stated discussion can be specified in the following manner:",
null,
"Now,",
null,
"",
null,
"By using the assumption of conditional independence, we have:",
null,
"Now with the help of Bayes Rule, we get:",
null,
"Now, as we know that we have to determine the P(D) directly from our corpus. Therefore, we reduce the above expression to the following expression with the help of basic rules of conditional probability. Therefore, the joint probability of seeing a given document and word together is:",
null,
"### What does the expression on the right side of the above equation represent?\n\nThe right-hand side of the above equation tells us that how likely it is to observe some document and then based upon the distribution of topics in that document, how likely it is to find a certain word within that document. This is the exact interpretation of that component in the equation.\n\n## What are the Parameters of this model?\n\nThe two main parameters in the model are as follows:\n\nP(w|z): There is (M-1)*K of them. How? for every z we have M words.\n\nThe question is why we subtract 1 from the total number of words since the sum of these M probabilities should be 1, so we lose one degree of freedom, that’s why we have written (M-1) instead of M.\n\nP(z|d): There are (K-1)*N parameters to determine.\n\nBoth these parameters are modeled as multinomial distributions and can be trained using the expectation-maximization algorithm.\n\n## Short Recap of Expectation-Maximization Algorithm\n\nEM is a method of finding the likeliest parameter estimates for a model which depends on unobserved, latent variables (in our case, the latent variables are topics).\n\nEM algorithm has the following two steps:\n\n• Step-1: This step is known as the expectation (E) step, where posterior probabilities are computed for the latent variables,\n• Step-2: This step is known as the maximization (M) step, where parameters are updated according to the likelihood function.\n\n### Homework Problem\n\nHere in the above section, we not discussed the objective function for the above parametrization for the EM algorithm. As your homework, you have to find out what is the objective function and log-likelihood function for the above model?\n\nNote: You can take references from the paper, which I have given in the last section of the article.\n\n### Parametrization -2\n\nIn this parametrization, we are starting with the topic with P(z), and then independently generating the document with P(d|z) and the word with P(w|z).\n\nYou can see the differences between this parametrization from the following diagram:",
null,
"Interestingly, in this parametrization, P(D, W) can be equivalently parameterized using a different set of 3 parameters:",
null,
"We can also look at the equivalency of the model as a generative process. This parametrization is more interesting than the first one since we can see a direct parallel between our pLSA model and our LSA model:",
null,
"Now, a question comes to mind:\n\n### What do the different probabilities in this parametrization represent?\n\nP(Z): The probability of our topic corresponds to the diagonal matrix of our singular topic probabilities,\n\nP(D|Z): The probability of our document given the topic corresponds to our document-topic matrix U, and\n\nP(W|Z): The probability of our word given the topic corresponds to our term-topic matrix V.\n\nSo what does that tell us?\n\nAlthough it looks quite different and tackles the problem in a very different manner, and pLSA just adds a probabilistic treatment of topics and words on the top of LSA. Therefore, it is a far more flexible model but still faces the following issues.\n\n• Since we have no parameters to model the probability P(D), so we don’t know how to assign probabilities to new documents.\n• The number of parameters involved in the pLSA grows linearly with the number of documents we have, so it is prone to overfitting.\n\nIn general, when people are looking for a topic model beyond the baseline performance LSA gives, they try LDA, which is the most common type of topic model, and LDA is the extension of pLSA to overcome these issues.\n\n1. Which of the following are the instances of stemming according to Porter Stemmer?\n\n• programer -> program\n• programing -> program\n• programmers -> program\n• programmably -> program\n\n2. According to Porter Stemmer, “python” cannot be the base form for which of the following word?\n\n• pythoned\n• pythoning\n• pythonly\n• pythoner\n\n3. While preprocessing the text for POS tagging which of the following techniques will affect the POS tags?\n\n• Case Folding\n• Lemmatization\n• Both\n• None of the above\n\n4. Stemming refers to the removal of suffices by a simple rule-based approach. Which of the following options demonstrates the stemming of words?\n\n• was, am, is, are -> be\n• helped, helps -> help\n• troubled, troubling, trouble -> trouble\n• friend, friendship, friends, friendships -> friend\n\n## Matrix Factorization model for pLSA\n\nThe matrix Factorization Model is an alternative way to represent pLSA.\n\nConsider a document-word matrix of shape × M, where N represents the number of documents and M represents the dictionary size. The elements of the matrix represent the counts of the occurrences of a word in a document. The element ((j, i) ) in a matrix becomes one if a word wi occurs once in the document dj.\n\nThe matrix formed above is a sparse matrix since most of the elements are 0.\n\nFor Example, Let’s have a document of 10 words and a dictionary having 1000 words. Then, 990 elements of the row will have the value 0. Such a matrix is called a Sparse Matrix.\n\nMatrix Factorization breaks this matrix (let’s call it A) into lower dimension matrices with the help of Singular Value Decomposition.",
null,
"",
null,
"The shapes of the matrices L, U, and R are N × K, K × K, and K × M respectively.\n\nMatrix U is a diagonal matrix with diagonal values equals to the square root of the eigenvalues of AAT. For any given k, you select the first k rows of L, the first k elements of and the first k columns of R. And k represents the number of topics we want.\n\nRemember this model is not very different from the Latent Variable Model.",
null,
"### How to interpret the above three matrices related to probabilities?\n\nL matrix – This matrix contains the document probabilities P(d|z)\n\nU matrix This is a diagonal matrix that contains the prior probabilities of the topics P(z)\n\nR matrix – This matrix corresponds to the word probability P(w|z)\n\nSo if you do the multiplication of all the above described three matrices, then you actually do what the below equation says —",
null,
"Note that the elements of all these three matrices cannot be negative as they represent probabilities. Hence, to decompose the A matrix, we can take the help of the Non-Negative Matrix Factorization, which we completed in the previous part of this Blog Series.\n\n1. It models word-document co-occurrences as a mixture of conditionally independent multinomial distributions.\n\n2. It is considered as a mixture model instead of a clustering model.\n\n3. The results of pLSA have a clear probabilistic interpretation.\n\n4. It also allows for model combination.\n\n1. Potentially higher computational complexity.\n\n2. EM algorithm gives local maximum instead of Global Maximum.\n\n3. It is prone to overfitting.\n\n4. It is not a well-defined generative model for new documents.\n\n## Other Blog Posts by Me\n\nYou can also check my previous blog posts.\n\nPrevious Data Science Blog posts.\n\nHere is my Linkedin profile in case you want to connect with me. I’ll be happy to be connected with you.\n\n## Email\n\nFor any queries, you can mail me on Gmail.\n\n## End Notes",
null,
""
] | [
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"https://miro.medium.com/max/412/1*TaVn_YvvKd9sprJohLt2FA.jpeg",
null,
"https://miro.medium.com/max/306/1*sjYatEoQW9O1-j7QzGlwSw.jpeg",
null,
"https://miro.medium.com/max/328/1*HKmRyClbqX7tN0Z8E-js-w.jpeg",
null,
"https://miro.medium.com/max/302/1*wrlH47rrBxwEzsmHiPbqaA.jpeg",
null,
"https://miro.medium.com/max/390/1*Xw8OIr5bGCRgVUrPkMan5A.jpeg",
null,
"https://miro.medium.com/max/460/1*hTenLukkUKrqX1NMDpdzyA.jpeg",
null,
"https://lh6.googleusercontent.com/nx26UuzpyiLzWOf09om_UU5L78HFN1lUcwDPVbQzTtWkhJ0oc5sTvGY1uMHK0d8lY1qo1CPeueFgfgQfX4ra233z-WSw8xq8TOSQxrNs1TgiaaBIQxc5viS9NgtDZPMaddfE4B7I",
null,
"https://lh4.googleusercontent.com/gd2xuEr-3xx4qKqrnq-IBBX0rSW5_YorYEaumO8l4E1YYTHk3q2M1fsFPBjUOJ87cSmzr1VSc5l9NyRoFHgp41Bql5M1Cnhuvgg1mkltfll5plLHaIrPJ8zuW7w1lf38XimC8SgZ",
null,
"https://lh5.googleusercontent.com/BJ3wXpWOok0wRfxDt8Qe_DIALzUrmZrw3rXuVI468Lmiz_9U9So6yiXC8u6xPtr28EU13muJZQsGwSxEdQBfSIHIGuAwkxRuLHJi7yjc87aYXOv82A7gPOVYw-azvLJMWHSV6mLq",
null,
"https://lh4.googleusercontent.com/eyI_IQzWYDXNQuloJsJFclBffXyOr2a8BjB3renMfbZcq73pUPnWyk3yfUWO_-JruDZpItKz1jCadxwdilpY19X0lhZBYoUlAZrPshgxkY10PGq7J0VCVraCf8Yx76VIOtrvIa-v",
null,
"https://miro.medium.com/max/196/1*ldwEwRk-aJKqZ_caaO9VBg.jpeg",
null,
"https://miro.medium.com/max/956/1*9TCz0n3e-qADLlmJxVJ5aA.jpeg",
null,
"https://lh4.googleusercontent.com/i487DWwRXtVCbnGAw3XC5ZwN_zGEmoHVufc9Tx-J387kLbb4ESydtyBZeElSfab4glWGJzSqVBIA0p5IgTA4rqxi1PMfVgOP__RpLuBSn_dfvBBOHiPAoGkX_wTNhkvtxhiolcZ8",
null,
"https://miro.medium.com/max/390/1*Xw8OIr5bGCRgVUrPkMan5A.jpeg",
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"https://av-identity.s3.amazonaws.com/users/user/ZcU4ALTFT96MVCzfiGuhsQ.jpeg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8864314,"math_prob":0.935778,"size":12852,"snap":"2022-27-2022-33","text_gpt3_token_len":2847,"char_repetition_ratio":0.14679328,"word_repetition_ratio":0.014753343,"special_character_ratio":0.20915033,"punctuation_ratio":0.103559874,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9971984,"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],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,6,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,6,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-07T16:59:38Z\",\"WARC-Record-ID\":\"<urn:uuid:8802a26a-0e5d-4506-9d46-beb8f140ad21>\",\"Content-Length\":\"152203\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:08525fdd-8dc6-432d-9cec-4d00eb7c8ff1>\",\"WARC-Concurrent-To\":\"<urn:uuid:f83009f2-96f0-4c58-8328-cd253f5031be>\",\"WARC-IP-Address\":\"172.67.38.119\",\"WARC-Target-URI\":\"https://www.analyticsvidhya.com/blog/2021/06/part-17-step-by-step-guide-to-master-nlp-topic-modelling-using-plsa/\",\"WARC-Payload-Digest\":\"sha1:IQR5FD2CAAMDK3FTBKUDYOMK43I5RGDQ\",\"WARC-Block-Digest\":\"sha1:XDG5Z6CDWEAIRKWXOGDUY73C7VCLIHYZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104495692.77_warc_CC-MAIN-20220707154329-20220707184329-00178.warc.gz\"}"} |
http://mizar.uwb.edu.pl/version/current/html/proofs/chord/109_3_1 | [
"let G be _Graph; :: thesis: ( G is complete implies G is chordal )\nassume A21: G is complete ; :: thesis: G is chordal\nnow :: thesis: for W being Walk of G st W .length() > 3 & W is Cycle-like holds\nW is chordal\nreconsider t7 = (2 * 3) + 1 as odd Nat ;\nreconsider t3 = (2 * 1) + 1 as odd Nat ;\nlet W be Walk of G; :: thesis: ( W .length() > 3 & W is Cycle-like implies W is chordal )\nassume that\nA22: W .length() > 3 and\nA23: W is Cycle-like ; :: thesis: W is chordal\nW .length() >= 3 + 1 by ;\nthen 2 * () >= 2 * 4 by XREAL_1:64;\nthen (2 * ()) + 1 >= 8 + 1 by XREAL_1:7;\nthen A24: len W >= 9 by GLIB_001:112;\nthen reconsider W3 = W . t3 as Vertex of G by ;\nA25: ( not t3 = 3 or not t7 = len W ) by A24;\nreconsider W7 = W . t7 as Vertex of G by ;\nt7 <= len W by ;\nthen W3 <> W7 by ;\nthen W3,W7 are_adjacent by A21;\nthen A26: ex e being object st e Joins W3,W7,G ;\nA27: t3 + 2 < t7 ;\nt7 <= len W by ;\nhence W is chordal by A23, A26, A27, A25, Th84; :: thesis: verum\nend;\nhence G is chordal ; :: thesis: verum"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.711719,"math_prob":0.9990168,"size":1065,"snap":"2019-43-2019-47","text_gpt3_token_len":481,"char_repetition_ratio":0.1470311,"word_repetition_ratio":0.08097166,"special_character_ratio":0.4741784,"punctuation_ratio":0.21908127,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992986,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-20T11:05:39Z\",\"WARC-Record-ID\":\"<urn:uuid:427f6f68-9576-403c-be10-ad784aa3e5ea>\",\"Content-Length\":\"13019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9ba0e45b-511d-40bf-b016-d29553b64213>\",\"WARC-Concurrent-To\":\"<urn:uuid:926d1311-9c63-428e-a8ce-9885b498cbcc>\",\"WARC-IP-Address\":\"212.33.73.131\",\"WARC-Target-URI\":\"http://mizar.uwb.edu.pl/version/current/html/proofs/chord/109_3_1\",\"WARC-Payload-Digest\":\"sha1:LUMJHGGFKGGXOVHH72DQSVJQRSDZSMZ6\",\"WARC-Block-Digest\":\"sha1:RNBPIT55RCEDYZMIH2MMTRV5SGJXOJXE\",\"WARC-Identified-Payload-Type\":\"application/xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986707990.49_warc_CC-MAIN-20191020105426-20191020132926-00439.warc.gz\"}"} |
https://metacpan.org/pod/Crypt::PRNG::ChaCha20 | [
"# NAME\n\nCrypt::PRNG::ChaCha20 - Cryptographically secure PRNG based on ChaCha20 (stream cipher) algorithm\n\n# SYNOPSIS\n\n`````` ### Functional interface:\nuse Crypt::PRNG::ChaCha20 qw(random_bytes random_bytes_hex random_bytes_b64 random_string random_string_from rand irand);\n\n\\$octets = random_bytes(45);\n\\$hex_string = random_bytes_hex(45);\n\\$base64_string = random_bytes_b64(45);\n\\$base64url_string = random_bytes_b64u(45);\n\\$alphanumeric_string = random_string(30);\n\\$string = random_string_from('ACGT', 64);\n\\$floating_point_number_0_to_1 = rand;\n\\$floating_point_number_0_to_88 = rand(88);\n\\$unsigned_32bit_int = irand;\n\n### OO interface:\nuse Crypt::PRNG::ChaCha20;\n\n\\$prng = Crypt::PRNG::ChaCha20->new;\n#or\n\\$prng = Crypt::PRNG::ChaCha20->new(\"some data used for seeding PRNG\");\n\n\\$octets = \\$prng->bytes(45);\n\\$hex_string = \\$prng->bytes_hex(45);\n\\$base64_string = \\$prng->bytes_b64(45);\n\\$base64url_string = \\$prng->bytes_b64u(45);\n\\$alphanumeric_string = \\$prng->string(30);\n\\$string = \\$prng->string_from('ACGT', 64);\n\\$floating_point_number_0_to_1 = rand;\n\\$floating_point_number_0_to_88 = rand(88);\n\\$unsigned_32bit_int = irand;``````\n\n# DESCRIPTION\n\nProvides an interface to the ChaCha20 based pseudo random number generator\n\nAll methods and functions are the same as for Crypt::PRNG."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6226654,"math_prob":0.97993654,"size":2067,"snap":"2023-40-2023-50","text_gpt3_token_len":632,"char_repetition_ratio":0.26902568,"word_repetition_ratio":0.054298643,"special_character_ratio":0.3091437,"punctuation_ratio":0.29539296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97949016,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T13:42:55Z\",\"WARC-Record-ID\":\"<urn:uuid:592f83f7-679d-4ed6-81cc-81605257d0e6>\",\"Content-Length\":\"80583\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41d90b1b-a0c0-413d-a205-a8d0734ed253>\",\"WARC-Concurrent-To\":\"<urn:uuid:427d3810-506c-4c18-bdf1-67f51a6d3c95>\",\"WARC-IP-Address\":\"151.101.2.217\",\"WARC-Target-URI\":\"https://metacpan.org/pod/Crypt::PRNG::ChaCha20\",\"WARC-Payload-Digest\":\"sha1:FTBLKYKOIJSKXW7XUMFSXPHGKTBPHRB6\",\"WARC-Block-Digest\":\"sha1:6XIPLDN5ITN552C2EFWRF7QKD2DZD2NW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233508977.50_warc_CC-MAIN-20230925115505-20230925145505-00323.warc.gz\"}"} |
https://answers.everydaycalculation.com/simplify-fraction/900-526 | [
"Solutions by everydaycalculation.com\n\n## Reduce 900/526 to lowest terms\n\nThe simplest form of 900/526 is 450/263.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 900 and 526 is 2\n2. Divide both the numerator and denominator by the GCD\n900 ÷ 2/526 ÷ 2\n3. Reduced fraction: 450/263\nTherefore, 900/526 simplified to lowest terms is 450/263.\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
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https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-020-01854-7/tables/1 | [
"Proposed $${N_t}^2+N_{RF}^2(R_{\\mathrm{smpl}}-1)$$ $$N_{RF}^2$$\nTaylor expansion $${N_t}^2R_{\\mathrm{smpl}}$$ $${N_t}^2$$",
null,
""
] | [
null,
"https://jwcn-eurasipjournals.springeropen.com/track/article/10.1186/s13638-020-01854-7",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.554197,"math_prob":1.00001,"size":309,"snap":"2022-27-2022-33","text_gpt3_token_len":115,"char_repetition_ratio":0.13442624,"word_repetition_ratio":0.0,"special_character_ratio":0.3398058,"punctuation_ratio":0.019607844,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999958,"pos_list":[0,1,2],"im_url_duplicate_count":[null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-14T00:25:59Z\",\"WARC-Record-ID\":\"<urn:uuid:571d2d4f-ed22-4531-bbf5-2531d1ad4ac3>\",\"Content-Length\":\"149623\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:05f86c12-2a66-4ea1-9716-c94895e0d264>\",\"WARC-Concurrent-To\":\"<urn:uuid:0bb7a0c0-03ce-43d0-8c13-b0b9b0657213>\",\"WARC-IP-Address\":\"146.75.32.95\",\"WARC-Target-URI\":\"https://jwcn-eurasipjournals.springeropen.com/articles/10.1186/s13638-020-01854-7/tables/1\",\"WARC-Payload-Digest\":\"sha1:XXGYN7AXMU3LUPXJBULLUVAM6ULLLHKA\",\"WARC-Block-Digest\":\"sha1:V6WZOUMGZM3DZGFDTBKV23RDB3POCWCW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571989.67_warc_CC-MAIN-20220813232744-20220814022744-00612.warc.gz\"}"} |
https://answers.everydaycalculation.com/percent-of-what-number/49-255 | [
"Solutions by everydaycalculation.com\n\n## 49 percent of what number is 255?\n\n255 is 49% of 520.41\n\n#### Steps to solve \"255 is 49 percent of what number?\"\n\n1. We have, 49% × x = 255\n2. or, 49/100 × x = 255\n3. Multiplying both sides by 100 and dividing both sides by 49,\nwe have x = 255 × 100/49\n4. x = 520.41\n\nIf you are using a calculator, simply enter 255×100÷49, which will give you the answer.\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn how to work with percentages in your own time:"
] | [
null,
"https://answers.everydaycalculation.com/mathstep-app-icon.png",
null
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https://toppoolleague.nl/Dec/28_heating-vacuum-water-boiler-machine---boiler.html | [
"heating vacuum water boiler machine - Boiler\n\n•",
null,
"• What Size Boiler Do I Need For My House? | U.S. Boiler …\n\n•",
null,
"• Boiler Size Calculator | What Boiler Size do I Need | Aquatek\n\nTo effectively calculate the size of the boiler you'll need, it's important to look at the following criteria: The type of boiler you want. This could be a system or regular boilers that store your hot water in a cylinder and thus making it better for those properties with multiple bathrooms.Learn More\n\n•",
null,
"• How do you calculate gpm on a boiler? - FindAnyAnswer.com\n\nMay 31, 2020 · Example: What Size Boiler Do I Need for 2,000 sq ft? So all you have to do is take the square footage and multiply it by 30. If your home is 2,000 square feet than you will need about 60,000 BTUs. Finding the right size of the boiler for your home is so very important.Learn More\n\n•",
null,
"• Water Handbook - Boiler Blowdown Control | SUEZ\n\nIn higher-pressure boilers, a soluble, inert material may be added to the boiler water as a tracer to determine the percentage of blowdown. The formula for calculating blowdown percentage using chloride and its derivation are shown in Table 13-1. Table 13-1. Algebraic proof of blowdown formula. The size of the lines and control valves Learn More\n\n•",
null,
"• What Size Boiler Do I Need? [Definitive Guide] | Boiler Guide\n\nCombi boiler sizing guide. The central heating outputs of combi boilers can be split into three categories to help you find a suitable model depending on the number of radiators in your property. Number of Radiators. Recommended Combi Boiler Size (CH output) Up to 10. 24-27 kW. 10-15.Learn More\n\n•",
null,
"• Boiler Size Calculator | What KW Boiler Do I Need\n\nApr 12, 2021 · How to calculate what size system or regular boiler you need. Regular boilers are also called heat only boilers, featuring a boiler, hot water tank and a cold water tank. The hot water tank is usually found in an airing cupboard and the cold water tank …Learn More\n\n•",
null,
"• Boiler Size Calculator | What Boiler Size do I Need | Aquatek\n\nTo effectively calculate the size of the boiler you'll need, it's important to look at the following criteria: The type of boiler you want. This could be a system or regular boilers that store your hot water in a cylinder and thus making it better for those properties …Learn More\n\n•",
null,
"• How to calculate steam boiler capacity?-ZBG Boiler\n\n•",
null,
"• HOW TO CALCULATE BOILER CAPACITY - YouTube\n\nMar 31, 2020 · Boiler load - the capacity of a steam boiler - is often rated as boiler horse powers, lbs of steam delivered per hour, or BTU.Lbs Steam delivered per HourLarLearn More\n\n•",
null,
"• Eco Mathi - Boiler supervisor - BEE JOO INDUSTRIES PTE LTD\n\nDistributed Control System Operator di PLTU Jateng 2 Adipala. Central Java, Indonesia. Ebenezer Mills. Ebenezer Mills. Mechanical Engineering Technician at TAQA. Ghana. herlina abdullah. herlina abdullah. HR senior Manager at PT Indonesia Power.Learn More\n\n•",
null,
"• Proper sizing of Boiler Feedwater System\n\nWe also know that one-gallon of water weighs 8.37 lbs. To calculate the storage tank needed use the following formula: BHP X 34.5 ÷ 8.337 lbs ÷ 60 min. X 10 = minimum useable capacity in gallons. For example, if you have a 500 HP boiler the calculation will be …Learn More\n\n•",
null,
"• Boiler Cheat Sheet - Novatherm\n\ncalculate the boiler generated steam rate without considering that the boiler does not work at atmospheric pressure. ex. 500 hp x 34.5 lb / hp hr = 17,250 lb / hr produced by the boiler. This is what we call the nominal boiler capacity. In reality, the operating pressure andLearn More\n\n•",
null,
"• Proper sizing of Boiler Feedwater System\n\nWe also know that one-gallon of water weighs 8.37 lbs. To calculate the storage tank needed use the following formula: BHP X 34.5 ÷ 8.337 lbs ÷ 60 min. X 10 = minimum useable capacity in gallons. For example, if you have a 500 HP boiler the calculation will be …Learn More\n\n•",
null,
"• Sizing a Steam Boiler | U.S. Boiler Company\n\nDec 05, 2017 · An important note on boiler sizing – all the supply piping in the basement should be insulated with a minimum of 1\" of pipe insulation. A better choice would be 1-1/2\" or 2\" of pipe insulation. An un-insulated steam main is equal to a radiator and must be calculated and added to the above calculation. When operating a steam boiler Learn More\n\n•",
null,
"• Water Handbook - Boiler Blowdown Control | SUEZ\n\nIn higher-pressure boilers, a soluble, inert material may be added to the boiler water as a tracer to determine the percentage of blowdown. The formula for calculating blowdown percentage using chloride and its derivation are shown in Table 13-1. Table 13-1. Algebraic proof of blowdown formula. The size of the lines and control valves Learn More\n\n•",
null,
"• Boiler Sizes Explained | 24 30 35 40kW Output | Ideal Heating\n\nWith modern boilers being so efficient, it's now much easier to calculate what size boiler you need than in years gone by. How Many Radiators Can a 12kW Boiler Run? Here at Ideal, you can find 12kW boilers in our heat only range, and 15kW in our heat only and system models.Learn More\n\n•",
null,
"• How to calculate steam boiler capacity?-ZBG Boiler\n\nWhat Size Boiler Do I Need For My House - U.S. Boiler Learn More\n\n•",
null,
"• Steam Calculators: Boiler Calculator\n\nMar 17, 2015 · Steam, Boiler, and Blowdown Pressure are the same. Combustion Efficiency is the % of fuel energy that is directly added to the feedwater and not otherwise lost or used. Blowdown Rate is the % of incoming feedwater mass flow rate that leaves the boiler as a saturated liquid at boiler pressure.Learn More\n\n•",
null,
"• Power plant and calculations: Boiler calculations for\n\nFeb 12, 2021 · 4-A Coal fired boiler having total heating surface area 5200 M2 produces 18 kg of steam per square meter per hour of heating surface, then calculate the Boiler capacity in TPH. Boiler Capacity = (Heating g surface area X Steam generation per square meter) Boiler Capacity = 5200 X 18 = 93600 kg/hr =93600 / 1000 = 93.6 TPHLearn More\n\n•",
null,
"• Water Volume Calculations in Systems\n\nof the boilers must be used in the formula (1400 kW 3+ 1400 kW) * 0.014 = 39.2 m3 39.2 m - %20 = 31.36 m3 estimated water volume in the system. NOTE: If there is more than one boiler in the system and it is uncertain whether the other boilers are back ups or not, a different method should be used to calculate the water volume.Learn More"
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https://literateprograms.org/jacobi_symbol__erlang_.html | [
"# Jacobi Symbol (Erlang)\n\nOther implementations: Erlang | Haskell | Python\n\nThe Jacobi symbol satisfies the six properties listed below, and these can be used to compute the Jacobi symbol in polynomial time.\n\n1. ```<<property 1>>=\njacobi(A, N) when A rem 2 =:= 0 ->\njacobi(2, N) * jacobi(A div 2 + A rem 2, N);\n```\n2. For ,.\n```<<property 2>>=\njacobi(A, N) when A >= N ->\njacobi(A rem N, N);\n```\n3. For odd coprimes a and n, .\n```<<property 3>>=\njacobi(A, N) when A rem 4 =:= 3, N rem 4 =:= 3 ->\n-jacobi(N, A);\njacobi(A, N) ->\njacobi(N, A).\n```\n4. .\n```<<property 4>>=\njacobi(1, _N) ->\n1;\n```\n5. .\n```<<property 5>>=\njacobi(2, N) ->\ncase (N rem 8) of\n1 -> 1;\n3 -> -1;\n5 -> -1;\n7 -> 1\nend;\n```\n6. .\n```<<property 6>>=\njacobi(0, _N) ->\n0;\n```\n\nFrom these properties we can come up with a polynomial time algorithm.\n\nThe function always complete as property 1 can be used to remove all even factors, and these will end at property 4. The odd factors will be reduced by property 2 if and if a < n property 3 will swap a and n and property 2 will again reduce a until it matches property 4 or property 6.\n\n```<<jacobi symbol>>=\nproperty 4\nproperty 5\nproperty 6\nproperty 1\nproperty 2\nproperty 3\n```\n\nThis is an escript, it can be run from the command line on unix type computers.\n\n```<<jacobi>>=\n#!/usr/bin/env escript\nmain([A, P]) ->\nio:fwrite(\"~p~n\", [xjacobi(string:to_integer(A), string:to_integer(P))]).\nxjacobi({A, _}, {B, _}) ->\njacobi(A, B).\njacobi symbol\n```"
] | [
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https://www.geeksforgeeks.org/class-8-rd-sharma-solutions-chapter-11-time-and-work-exercise-11-1-set-2/?ref=rp | [
"Related Articles\nClass 8 RD Sharma Solutions – Chapter 11 Time And Work – Exercise 11.1 | Set 2\n• Last Updated : 07 Apr, 2021\n\n### Question 15. A and B can do a piece of work in 6 days and 4 days respectively. A started the work, worked at it for 2 days and then was joined by B. Find the total time taken to complete the work.\n\nSolution:\n\nGiven:\n\nA can do a piece of work in = 6 days\n\nA’s 1 day work = 1/6\n\nB can do a piece of work in = 4 days\n\nIn 2 days the work completed by A = 2 × 1/6 = 1/3 part of work\n\nRemaining work = 1 – 1/3\n\n= (3-1)/3\n\n= 2/3\n\nA and B can finish the remaining work in = (2/3)/(1/6 + 1/4)\n\n= (2/3)/(5/12)\n\n= (2×12)/(3×5)\n\n= 8/5 days\n\nTherefore,\n\nTotal time taken to complete the work by A and B = 2 + 8/5\n\n= (10+8)/5 = 18/5\n\n= 3 days.\n5\n\n### Question 16. 6 men can complete the electric fitting in a building in 7 days. How many days will it take if 21 men do the job?\n\nSolution:\n\nGiven:\n\n6 men can complete the job in = 7 days\n\n1 man can complete the job in = 6 × 7 = 42 days\n\nTherefore,\n\n21 men can complete the job in = 42/21 = 2 days\n\n### Question 17. 8 men can do a piece of work in 9 days. In how many days will 6 men do it?\n\nSolution:\n\nGiven:\n\n8 men can do a piece of work = 9 days\n\n1 man can complete a piece of work in = 8 × 9 = 72 days\n\nTherefore,\n\n6 men can complete the work in = 72/6 = 12 days\n\n### Question 18. Reema weaves 35 baskets in 25 days. In how many days will she weave 55 baskets?\n\nSolution:\n\nGiven:\n\nReema weaves 35 baskets in = 25 days\n\nTime taken by Reema to weave 1 basket = 25/35\n\nTherefore,\n\nTime taken by Reema to weave 55 baskets = 25/35 × 55\n\n= 5/7 × 55\n\n= 275/7\n\n= 39 2 days.\n7\n\n### Question 19. Neha types 75 pages in 14 hours. How many pages will she type in 20 hours?\n\nSolution:\n\nGiven:\n\nNumber of pages Neha can type in 14 hours = 75 pages\n\nNumber of pages Neha can type in 1 hour = 75/14\n\nTherefore,\n\nNumber of pages Neha can type in 20 hours = 20 × 75/14\n\n= 750/7\n\n= 107 pages.\n7\n\n### Question 20. If 12 boys earn Rs 840 in 7 days, what will 15 boys earn in 6 days?\n\nSolution:\n\nGiven:\n\nEarning of 12 boys in 7 days = Rs 840\n\nEarning of 12 boys in 1 day = Rs 840/7\n\n= Rs 120\n\nEarning of 1 boy in 1 day = 120/12 = Rs 10\n\nEarning of 1 boy in 6 days = 10 × 6 = Rs 60\n\nTherefore\n\nEarning of 15 boys in 6 days = 60 × 15 = Rs 900\n\n### Question 21. If 25 men earn Rs 1000 in 10 days, how much will 15 men earn in 15 days?\n\nSolution:\n\nGiven:\n\nIn 10 days 25 men can earn = Rs.1000\n\nIn 1 day 25 men can earn = 1000/10 = Rs 100\n\nIn 1 day 1 man can earn = 100/25 = Rs 4\n\nIn 15 days 1 man can earn = 15 × 4 = Rs 60\n\nTherefore,\n\nIn 15 days 15 men can earn = 60 × 15 = Rs 900\n\n### Question 22. Working 8 hours a day, Ashu can copy a book in 18 days. How many hours a day should he work so as to finish the work in 12 days?\n\nSolution:\n\nGiven:\n\nWorking 8 hours a day, Ashu can complete a work in = 18 days\n\nWorking 1 hour a day, Ashu can complete the work in = 18 × 8 = 144 days\n\nTherefore,\n\nNumber of hours Ashu should work to complete the work in 12 days = 144/12 = 12 hours/day\n\n### Question 23. If 9 girls can prepare 135 garlands in 3 hours, how many girls are needed to prepare 270 garlands in 1 hour?\n\nSolution:\n\nGiven:\n\nIn 3 hours, 9 Girls can prepare = 135 garlands\n\nIn 1 hour, 9 girls can prepare = 135/3 = 45 garlands\n\nIn 1 hour, 1 girl can prepare = 45/9 = 5 garlands\n\nTherefore,\n\nIn 1 hour, Number of girls required to prepare 270 garlands = 270/5 = 54 girls\n\n### Question 24. A cistern can be filled by one tap in 8 hours, and by another in 4 hours. How long will it take to fill the cistern if both taps are opened together?\n\nSolution:\n\nGiven:\n\nA cistern can be filled by one tap in = 8 hours\n\nA cistern filled by one tap in 1 hour = 1/8\n\nAnother cistern can be filled in = 4 hours\n\nCistern filled by another tap in 1 hour = 1/4\n\nTotal cistern filled in 1 hour = 1/4 + 1/8\n\n= (2+1)/8\n\n= 3/8\n\nTherefore,\n\nCistern can be filled when both the taps are opened together in = 8/3 = 223 hours\n\n### Question 25. Two taps A and B can fill an overhead tank in 10 hours and 15 hours respectively. Both the taps are opened for 4 hours and then B is turned off. How much time will A take to fill the remaining tank?\n\nSolution:\n\nGiven:\n\nTap A can fill the tank in = 10 hours\n\nTap A can fill the tank in 1 hour = 1/10\n\nTap B can fill the tank in = 15 hours\n\nTap B can fill the tank in 1 hour = 1/15\n\nBoth taps together can fill the tank in 1 hour = 1/10 + 1/15\n\n= (3+2)/30 = 5/30\n\n= 1/6\n\nBoth taps together can fill the tank in 4 hours = 4 × 1/6 = 2/3\n\nRemaining tank to be filled = 1 – 2/3\n\n= (3-2)/3\n\n= 1/3\n\nTherefore,\n\nTime taken by A to fill the remaining tank = (1/3)/(1/10)\n\n= 10/3\n\n= 3 hours.\n3\n\n### Question 26. A pipe can fill a cistern in 10 hours. Due to a leak in the bottom it is filled in 12 hours. When the cistern is full, in how much time will it be emptied by the leak?\n\nSolution:\n\nGiven:\n\nWhen there is no leakage A pipe can fill the cistern in = 10 hours\n\nIn 1 hour without leakage A pipe can fill the cistern in = 1/10 hours\n\nWhen there is leakage cistern gets filled in = 12 hours\n\nIn 1 hour, when there is leakage cistern gets filled in = 1/12 hours\n\nIn 1 hour, due to leakage cistern gets filled to = 1/10 – 1/12\n\n= (12-10)/120\n\n= 2/120\n\n= 1/60 part\n\nTherefore,\n\nDue to leakage the cistern gets emptied in = 1/(1/60) = 60 hours.\n\n### Question 27. A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?\n\nSolution:\n\nGiven:\n\nInlet A can fill the cistern in = 12 hours\n\nInlet A can fill the cistern in 1 hour = 1/12\n\nInlet B can fill the cistern in = 15 hours\n\nInlet B can fill the cistern in 1 hour = 1/15\n\nOutlet pipe can empty the cistern in = 10 hours\n\nOutlet pipe can empty the cistern in 1 hour = 1/10\n\nSo we have, (1/12 + 1/15) – 1/10\n\n= (9/60) – 1/10\n\n= (9-6)/60\n\n= 3/60\n\n= 1/20 part\n\nTherefore,\n\nWhen all 3 pipes are opened together to empty the cistern, time taken to fill the cistern completly = 1/(1/20) = 20 hours\n\n### Question 28. A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both the tap and the pipe are opened together?\n\nSolution:\n\nGiven:\n\nInlet tap can fill a cistern in = 4 hours\n\nInlet tap can fill a cistern in 1 hour = 1/4\n\nOutlet tap can empty the cistern in = 6 hours\n\nOutlet tap can empty the cistern in 1 hour = 1/6\n\nWork done by both pipe in 1 hour = (1/4 – 1/6)\n\n= (3-2)/12\n\n= 1/12\n\nTherefore,\n\nWhen both tap and pipe are opened together the cistern can be filled in = 1/(1/12) = 12 hours.\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.\n\nMy Personal Notes arrow_drop_up"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8723015,"math_prob":0.9609109,"size":6567,"snap":"2021-04-2021-17","text_gpt3_token_len":2245,"char_repetition_ratio":0.17720555,"word_repetition_ratio":0.16274238,"special_character_ratio":0.36911833,"punctuation_ratio":0.074489124,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9969558,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-14T03:11:01Z\",\"WARC-Record-ID\":\"<urn:uuid:a003aeb2-a838-417c-9ed1-efc42da3a12b>\",\"Content-Length\":\"92947\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f0605d52-bde2-4509-934a-223c8f6ab3b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a0a76ce-0944-498e-8599-9b5d05ce3737>\",\"WARC-IP-Address\":\"23.62.230.138\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/class-8-rd-sharma-solutions-chapter-11-time-and-work-exercise-11-1-set-2/?ref=rp\",\"WARC-Payload-Digest\":\"sha1:IKYSCMUTLT6EXNSPFFMCFWFGOPNZUZDU\",\"WARC-Block-Digest\":\"sha1:W3K6ULFDL3X6HB7RFJFI77TUPYZ3QMOJ\",\"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-20210414004149-20210414034149-00312.warc.gz\"}"} |
https://airspeedvelocity.net/2014/07/14/a-better-way-to-hit-bottom/ | [
"# A better way to hit bottom\n\nAfter my previous post on finding minimums, a couple of people pointed out that `reduce` would be much better suited to the task. They’re absolutely right, but I ignored that for the last post since I had another agenda. Plus, the combination of `min` and `reduce` is an interesting one with its own wrinkles, that deserves a post of its own, so here it is.\n\n`reduce` takes a sequence, a starting value, and a combining function, and uses that function to boil the sequence down to a single value. For example, to total up the values in an array:\n\n```let a = [2, 4, 6, 8]\n// starting with 0, add each value\n// of 'a' to a running total\nlet sum = reduce(a, 0, +)\n// sum = 20\n```\n\nThat certainly sounds like what we want for finding the minimum value in a sequence, with `min` as the combining function (as recently made possible in beta 3):\n\n```let a = [5, 4, 3, 8]\n// but what to put in the starting value?\nlet the_min = reduce(a, ???, min)\n```\n\nWe need to pick a starting value to use on the first call to `min`. Unlike with addition, there’s no natural constant to start a search for a minmum value. You could start with the maximum value for an `Int`, but that smells like a hack.\n\nUsing the first element of the sequence would do it:\n\n```var gen = a.generate()\nif let first = gen.next() {\nreduce(a, first, min)\n}\n```\n\nBut this means the first thing this code does every time is redundantly compare the first element to itself. Strictly speaking, that’s fine from a functional behaviour point of view, and even the performance consequences amortize to negligible, but it’s not very satisfying.\n\nThis version avoids that problem:\n\n```var gen = a.generate()\nif let first = gen.next() {\nreduce(gen, first, min)\n}\n```\n\nWhy does that work? Because `Array.GeneratorType` happens to be both a `Generator` and a `Sequence`. In fact, in a quick scan through the Swift standard library, I couldn’t spot any generator that wasn’t also a sequence. But just because they always are doesn’t mean they have to be, which is why implementing a generic function `tail` that returns all but the first element of a sequence is surprisingly tricky – but let’s leave that for a later post.\n\nSo, armed with our ability to use reduce to find a minimum, let’s implement our solution from the previous post that finds the minimum numeric value in an array of strings:\n\n```let string_ints = [\"dog\", \"12\", \"cat\", \"4\", \"200\", \"elephant\"]\n\nvar gen = string_ints.generate()\nif let first = gen.next() {\nreduce(gen, first) { (lhs, rhs) in\nswitch (lhs.toInt(), rhs.toInt()) {\ncase (_, nil): return lhs\ncase (nil, _): return rhs\ncase (let i, let j): return i < j ? lhs : rhs\n}\n}\n}\n```\n\nOK that works, but this is pretty ugly with the generator on top of that convoluted closure. And remember last time I mentioned there were still a couple of issues – if `string_ints` contains no integer strings, it’ll just return the first non-integer string. And it does every integer conversion twice.\n\nDo we still need the generator? Our closure that chooses between sequence elements now discards `nil` values as soon as it finds a non-`nil` one. That means that we could supply `nil` as our initial argument. This also means a returned value of `nil` could stand in for “not found”, either because the sequence was empty, or because it contained no strings convertible to numbers.\n\nBut before we make that change, let’s neaten things up by pulling out the code that does the `nil` discrimination into its own function.\n\nHere is `combineMaybe`, a function that takes a function that combines two values (like `min`), and returns a new function that chooses between two optionals, always preferring the non-`nil` one before resorting to the supplied function if neither value is `nil`:1 2\n\n```func combineMaybe<T>\n( combine: (T,T)->T )\n-> (T?,T?)->T? {\nreturn { lhs, rhs in\nswitch(lhs, rhs) {\ncase (_, nil): return lhs\ncase (nil, _): return rhs\ndefault: return combine(lhs!,rhs!)\n}\n}\n}\n```\n\n`combineMaybe` is a function decorator – a function that takes another function, and enhances it in some useful way, returning the enhanced function.\n\nWe can now use the standard library version of `min`, enhanced with our decorator, as the combiner function to reduce, supplying `nil` as the starting value. An initial call to `map` to first convert the strings to integers generates the input sequence:\n\n```let string_ints = [\"dog\", \"12\", \"cat\", \"4\", \"200\", \"elephant\"]\nlet maybe_ints = map(string_ints) { \\$0.toInt() }\n\nif let the_min = reduce( maybe_ints, nil, combineMaybe(min)) {\n// the_min = 4\n}\nelse {\n// handle both empty input or no numerics in input\n}\n```\n\nThis fixes the bug of returning a non-numeric string when that’s all the sequence contains. And it only converts each number once, in the call to `map`.3 4\n\nIf instead we wanted to pull out the maximum values, all that is needed is to replace the passing of `min` into `combineMaybe` with `max`, with everything else remaining the same.\n\nIn fact, it’ll even work with other binary functions as well. If you replaced `min` with `+`, it’ll add all the numeric values in the array, while still giving you a chance to catch if no numbers were found in the sequence (sometimes zero is still not the right answer for an empty sequence, even one containing integers rather than strings).\n\n1. The “maybe”-based name comes from Haskell, where `maybe` is a function that takes a default value, a function, and an optional type (which are called Maybe types).\n2. As ever, really wishing switch was an expression that returned the value of its cases when they were single expressions.\n3. Ok, so this doesn’t fulfil the original requirement of returning a string rather than an integer (from the previous article). I think we can keep that as a subsequent step, I was mostly being antagonistic about that part.\n4. Haskell also has a function, `mapMaybe`, that is like a `map` to an optional value, followed by a `filter` that removes all the resulting nil values. That would be a good building block for solving this problem too, though it still wouldn’t help with the starting value for reduce.\n\n## One thought on “A better way to hit bottom”\n\n1.",
null,
"Portable Innovations (@portableinno) says:\n\n> though it still wouldn’t help with the starting value for reduce.\n\nFor the problem with reduce requiring an initial value you could package your generator based approach (liked that!) into a new function reduce1 (similar to foldl1 or foldr1 in Haskell).\nI had to wrap the generator into GeneratorOf(), though, because the generator of a sequence isn’t a Sequence itself:\n\n/// reduce `sequence` using the function `combine`\nfunc reduce1(sequence: S, combine: (E, E) -> E) -> E? {\nvar gen = sequence.generate()\nif let first = gen.next() {\nreturn reduce(GeneratorOf(gen), first, combine)\n}\nelse {\nreturn nil\n}\n}\n\nLike “switch” “if” should really be an expression…\nBut we can use map instead for a more cons ice expression:\n\nfunc reduce1(sequence: S, combine: (E, E) -> E) -> E? {\nvar gen = sequence.generate()\nreturn gen.next().map { first in reduce(GeneratorOf(gen), first, combine) }\n}\n\nI chose handling the empty sequence case with an optional return type instead of requiring a non empty sequence argument like Haskell’s foldl1 does (don’t know why they didn’t choose to use Maybe).\n\n-Thorsten"
] | [
null,
"https://i1.wp.com/pbs.twimg.com/profile_images/1077093916/Twitter-Profilbild_normal.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.8703091,"math_prob":0.9284063,"size":7097,"snap":"2020-45-2020-50","text_gpt3_token_len":1681,"char_repetition_ratio":0.1380234,"word_repetition_ratio":0.063897766,"special_character_ratio":0.24827392,"punctuation_ratio":0.13154173,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9565597,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-25T04:47:22Z\",\"WARC-Record-ID\":\"<urn:uuid:04979041-95f4-4858-92d6-409b5a2d4d53>\",\"Content-Length\":\"89970\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5cc70c70-5ba8-4832-bcc6-4e06fbf3cff5>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c4ee673-4ab5-4012-9530-edda96952eaf>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://airspeedvelocity.net/2014/07/14/a-better-way-to-hit-bottom/\",\"WARC-Payload-Digest\":\"sha1:36PVZDF2YCSQO56CCSOWNHKVFA755NON\",\"WARC-Block-Digest\":\"sha1:C3P7H5XL5ZEMTZFOWKRQFA2SRJ2ZEEZD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141181179.12_warc_CC-MAIN-20201125041943-20201125071943-00443.warc.gz\"}"} |
https://sources.debian.org/src/brian/1.4.3-1/docs_sphinx/electrophysiology_models.txt/ | [
"File: electrophysiology_models.txt\n\npackage info (click to toggle)\nbrian 1.4.3-1\n• links: PTS, VCS\n• area: main\n• in suites: sid, stretch\n• size: 23,436 kB\n• sloc: python: 68,707; cpp: 29,040; ansic: 5,182; sh: 111; makefile: 61\n file content (197 lines) | stat: -rw-r--r-- 9,015 bytes parent folder | download | duplicates (3)\n 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197 .. currentmodule:: brian Electrophysiology: models ========================= The electrophysiology library contains a number of models of electrodes, amplifiers and recording protocols to simulate intracellular electrophysiological recordings. To import the electrophysiology library:: from brian.library.electrophysiology import * There is a series of example scripts in the examples/electrophysiology folder. Electrodes ---------- Electrodes are defined as resistor/capacitor (RC) circuits, or multiple RC circuits in series. Define a simple RC electrode with resistance Re and capacitance Ce (possibly 0 pF) as follows:: el=electrode(Re,Ce) The ``electrode`` function returns an :class:`Equations` object containing the electrode model, where the electrode potential is ``v_el`` (the recording), the membrane potential is ``vm``, the electrode current entering the membrane is ``i_inj`` and command current is ``i_cmd``. These names can be overriden using the corresponding keywords. For example, a membrane equation with a .5 nA current injected through an electrode is defined as follows:: eqs=Equations('dv/dt=(-gl*v+i_inj)/Cm : volt')+electrode(50*Mohm,10*pF,vm='v',i_cmd=.5*nA) Specify ``i_cmd=None`` if the electrode is only used to record (no current injection). More complex electrodes can be defined by passing lists of resistances and capacitances, e.g.:: el=electrode([50*Mohm,20*Mohm],[5*pF,3*pF]) Amplifiers ---------- Current-clamp amplifier ^^^^^^^^^^^^^^^^^^^^^^^ A current-clamp amplifier injects a current through an intracellular electrode and records the membrane potential. Two standard circuits are included to compensate for the electrode voltage: bridge compensation and capacitance neutralization (see e.g. the `Axon guide `__). The following command:: amp=current_clamp(Re=80*Mohm,Ce=10*pF) defines a current-clamp amplifier with an electrode modelled as a RC circuit. The function returns an :class:`Equations` object, where the recording potential is ``v_rec``, the membrane potential is ``vm``, the electrode current entering the membrane is ``i_inj`` and command current is ``i_cmd``. These names can be overriden using the corresponding keywords. For implementation reasons, the amplifier always includes an electrode. Optionally, bridge compensation, can be used with the ``bridge`` keyword and capacitance neutralization with the ``capa_comp`` keyword. For example, the following instruction defines a partially compensated recording:: amp=current_clamp(Re=80*Mohm,Ce=10*pF,bridge=78*Mohm,capa_comp=8*pF) The capacitance neutralization is a feedback circuit, so that it becomes unstable if the feedback capacitance is larger than the actual capacitance of the electrode. The bridge compensation is an input-dependent voltage offset (``bridge*i_cmd``), and thus is always stable (unless an additional feedback, such as dynamic clamp, is provided). Note that the bridge and capacitance neutralization parameters can be variable names, e.g.:: amp=current_clamp(Re=80*Mohm,Ce=10*pF,bridge='Rbridge',capa_comp=8*pF) and then the bridge compensation can be changed dynamically during the simulation. Voltage-clamp amplifier ^^^^^^^^^^^^^^^^^^^^^^^ The library includes a single-electrode voltage-clamp amplifier, which clamps the potential at a given value and records the current going through the electrode. The following command:: amp=voltage_clamp(Re=20*Mohm) defines a voltage-clamp amplifier with an electrode modelled as a pure resistance. The function returns an :class:`Equations` object, where the recording current is ``i_rec``, the membrane potential is ``vm``, the electrode current entering the membrane is ``i_inj`` and command voltage is ``v_cmd`` (note that ``i_rec`` = - ``i_inj``). These names can be overriden using the corresponding keywords. For implementation reasons, the amplifier always includes an electrode. Electrode capacitance is not included, meaning that the capacitance neutralization circuit is always set at the maximum value. The quality of the clamp is limited by the electrode or ''series'' resistance, which can be compensated in a similar way as bridge compensation in current-clamp recordings. Series resistance compensation consists in adding a current-dependent voltage offset to the voltage command. Because of the feedback, that compensation needs to be slightly delayed (with a low-pass circuit). The following example defines a voltage-clamp amplifier with half-compensated series resistance and compensation delay 1 ms:: amp=voltage_clamp(Re=20*Mohm,Rs=10*Mohm,tau_u=1*ms) The ``tau_u`` keyword is optional and defaults to 1 ms. Acquisition board ^^^^^^^^^^^^^^^^^ An acquisition board samples a recording and sends a command (e.g. injected current) at regular times. It is defined as a :class:`NeuronGroup`. Use:: board=AcquisitionBoard(P=neuron,V='V',I='I',clock) where ``P`` = neuron group (possibly containing amplifier and electrode), ``V`` = potential variable name, ``I`` = current variable name, ``clock`` = acquisition clock. The recording variable is then stored in ``board.record`` and a command is sent with the instruction ``board.command=I``. Discontinuous current clamp ^^^^^^^^^^^^^^^^^^^^^^^^^^^ The discontinuous current clamp (DCC) consists in alternatively injecting current and measuring the potential, in order to measure the potential when the voltage across the electrode has vanished. The sampling clock is mainly determined by the electrode time constant (the sampling period should be two orders of magnitude larger than the electrode time constant). It is defined and used in the same way as an acquisition board (above):: board=DCC(P=neuron,V='V',I='I',frequency=2*kHz) where ``frequency`` is the sampling frequency. The duty cycle is 1/3 (meaning current is injected during 1/3 of each sampling step). Discontinuous voltage clamp ^^^^^^^^^^^^^^^^^^^^^^^^^^^ The discontinuous voltage clamp or single-electrode voltage clamp (SEVC) is an implementation of the voltage clamp using a feedback current with a DCC amplifier. It is defined as the DCC:: board=SEVC(P=neuron,V='V',I='I',frequency=2*kHz,gain=10*nS) except that a gain parameter is included. The SEVC injects a negative feedback current I=gain*(Vcommand-V). The quality of the clamp improves with higher gains, but there is a maximum value above which the system is unstable, because of the finite temporal resolution. The recorded current is stored in ``board.record`` and the command voltage is sent with the instruction ``board.command=-20*mV``. With this implementation of the SEVC, the membrane is never perfectly clamped. A better clamp is obtained by adding an integral controller with the keyword ``gain2=10*nS/ms``. The additional current J(t) is governed by the differential equation dJ/dt=gain2*(Vcommand-V), so that it ensures perfect clamping in the stationary state. However, this controller does not improve the settling time of the clamp, but only the final voltage value. Active Electrode Compensation ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ The electrophysiology library includes the Active Electrode Compensation (AEC) technique described in Brette et al (2008), `High-resolution intracellular recordings using a real-time computational model of the electrode `__, Neuron 59(3):379-91. It can be applied offline, or online, using the models of experimental setup described above. (for dynamic-clamp or voltage-clamp recordings, the electrode compensation must be done online). An AEC board is initialized in the same way as an acquisition board:: board=AEC(neuron,'V','I',clock) where ``clock`` is the acquisition clock. The estimation phase typically looks like:: board.start_injection() run(2*second) board.start_injection() run(100*ms) board.estimate() where white noise is injected for 2 seconds (default amplitude .5 nA). You can change the default amplitude and DC current as follows: ``board.start_injection(amp=.5*nA,DC=1*nA)``. After estimation, the kernel is stored in ``board.Ke``. The following options can be passed to the function ``estimate``: ``ksize`` (default 150 sampling steps), ``ktail`` (default 50 sampling steps) and ``dendritic`` (default ``False``, use ``True`` is the recording is a thin process, i.e., axon or dendrite). Online compensation is then switched on with ``board.switch_on()`` and off with ``board.switch_off()``. For example, to inject a .5 nA current pulse for 200 ms, use the following instructions:: board.switch_on() board.command=.5*nA run(200*ms) board.command=0*nA run(150*ms) board.switch_off() During the simulation, the variable ``board.record`` stores the compensated potential."
] | [
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http://www.mkpv.top/how-to-calculate-flow-rate-given-pressure-and-diameter-2/ | [
"",
null,
"flow rate and pressure relationship First of all,flow = flow rate pipe inner diameter pipe inner diameter 蟺4. Therefore, the flow rate and the flow rate basically know one to calculate the other parameter.\n\n• ### How do you find the flow rate of a pipe?\n\n• First of all, flow = flow rate pipe inner diameter pipe inner diameter 蟺4. Therefore, the flow rate and the flow rate basically know one to calculate the other parameter. But if the pipe diameter D and the pressure P in the pipe are known, can the flow rate be calculated?\n\n• ### Is the flow rate related to pressure pressure and pipe diameter?\n\n• Is the flow rate related to pressure, flow rate, and pipe diameter? From the perspective of qualitative analysis, the relationship between pressure and flow in the pipeline is proportional. That is, the greater the pressure, the greater the flow rate. The flow rate is equal to the velocity multiplied by the section.\n\n• ### What is the use of flow rate calculator?\n\n• The flow rate calculator can also calculate the mass flow rate of a fluid given its density is known. Input and output support metric and imperial measurement units. This pipe flow rate calculator calculates the volumetric flow rate ( discharge rate) a gas or fluid (liquid) going through a round or rectangular pipe of known dimensions.\n\n• ### How to calculate pipe diameter?\n\n• To calculate pipe diameter, you have to know flow velocity and flow rate. If you know mass flow rate, then fluid density has to be known. If the flowing fluid is gas, then instead of density, you have to know gas constant, absolute pressure, and temperature. Density is calculated using the equation for the perfect gas."
] | [
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https://de.maplesoft.com/support/help/maple/view.aspx?path=RegularChains/Separant&L=G | [
"",
null,
"Separant - Maple Help\n\nRegularChains\n\n Separant\n separant of a nonconstant polynomial",
null,
"Calling Sequence Separant(p, R)",
null,
"Parameters\n\n R - polynomial ring p - polynomial of R",
null,
"Description\n\n • The function call Separant(p,R) returns the separant of p with respect to the variable ordering of R, that is, the derivative of p with respect to its main variable.\n • It is assumed that p is nonconstant polynomial, and that the characteristic of R is zero.\n • This command is part of the RegularChains package, so it can be used in the form Separant(..) only after executing the command with(RegularChains). However, it can always be accessed through the long form of the command by using RegularChains[Separant](..).",
null,
"Examples\n\n > $\\mathrm{with}\\left(\\mathrm{RegularChains}\\right):$\n > $R≔\\mathrm{PolynomialRing}\\left(\\left[x,y,z\\right]\\right)$\n ${R}{≔}{\\mathrm{polynomial_ring}}$ (1)\n > $p≔\\left(y+1\\right){x}^{3}+\\left(z+4\\right)x+3$\n ${p}{≔}\\left({y}{+}{1}\\right){}{{x}}^{{3}}{+}\\left({z}{+}{4}\\right){}{x}{+}{3}$ (2)\n > $\\mathrm{MainVariable}\\left(p,R\\right)$\n ${x}$ (3)\n > $\\mathrm{Separant}\\left(p,R\\right)$\n ${3}{}{{x}}^{{2}}{}{y}{+}{3}{}{{x}}^{{2}}{+}{z}{+}{4}$ (4)\n\nChange the ordering of the variable.\n\n > $R≔\\mathrm{PolynomialRing}\\left(\\left[z,y,x\\right]\\right)$\n ${R}{≔}{\\mathrm{polynomial_ring}}$ (5)\n > $p≔\\mathrm{expand}\\left(\\left(y+1\\right){x}^{3}+\\left(z+4\\right)x+3\\right)$\n ${p}{≔}{{x}}^{{3}}{}{y}{+}{{x}}^{{3}}{+}{x}{}{z}{+}{4}{}{x}{+}{3}$ (6)\n > $\\mathrm{MainVariable}\\left(p,R\\right)$\n ${z}$ (7)\n > $\\mathrm{Separant}\\left(p,R\\right)$\n ${x}$ (8)"
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"https://bat.bing.com/action/0",
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"https://de.maplesoft.com/support/help/maple/arrow_down.gif",
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https://www.edupristine.com/blog/debt-capital-market | [
"## Introduction to Debt Capital Market i.e. DCM.\n\nDebt Capital Market (DCM) is a desk that trades debt securities. Debt securities typically means government bonds, corporate bonds, CDs, municipal bonds, preferred stock, collateralized securities (such as CDOs, CMOs, CMBSs) and zero-coupon securities.\n\nAsk the veterans of Wall Street traders who deals in the most complex of derivative products or ask a neighbourhood vegetable vendor, about the most important thing in the trading business. The plain and simple answer is- Price. So let us begin the journey to calculate your bonuses at DCM desk with ‘ Price’ of bond.\n\nA bond mainly has 3 features.\n\n1. The promise: The promise to payback the borrowed amount (let us call it face value)\n\n2. The Timeline: The tenor of redemption and\n\n3. The Coupon: The synonym of Interest.\n\nBy definition:\n\nPrice of Bond = Sum of the Present Value (PV) of Coupon Payments+ PV of Principal Payments\n\nThe following excel helps you calculate the price. It just says that if a company issues a 10 years bond with a coupon of 10% (which translate to 100 of annuity/PMT/annual Coupon).\n\nPV function of excel just want 4 things from you:\n\n1. Rate (Coupon)\n\n2. Nper (No of coupon payments which is 10 for an annual coupon paying bond like this. For a semi-annual redemption bond the Nper should be 2*10 i.e. 20)\n\n3. PMT (Annual coupon payment in this case, it could be semi-annual/month coupon payment also, for a zero coupon bond it is zero)\n\n4. FV (Promised amount in the end of tenor)\n\nThis cell B5 just tells you the Price of the Bond at the time of issuance using PV function. No surprises the Price of the Bond at the time of issuance is the issuance amount only (i.e.1000).",
null,
"Let us move to the ‘Trading’ part. One thing to remember about DCM Traders is that they think in terms of yield only. If a Trader has hung up the phone without saying this word then he must have been on a personal call. So what is Yield BTW?\n\nLet us first try a mathematician’s answer.\n\nYield=Coupon/Price of Bond\n\nWith this formula only, I bet nobody can make out why it is so important. So let me elaborate further. Typically when a trader says yield he means: Yield to Maturity. Investopedia defines it as: ”The rate of return anticipated on a bond if held until the end of its lifetime.”\n\nThe Keyword here is –‘Anticipation’. Every Trader has some different anticipation of yield. So let us cook a perfect deal. Let us assume that you have found a buyer for this bond with a yield expectation of 9.5%. As his yield expectation is lower so the Price has to be higher than 1000.One key point is that coupon will remain at 10%.Applying the same excel formula, for him the Price would work out to be 1031 as shown below:",
null,
"Now we need to understand there are 2 types of Prices of bond: ‘Dirty Price’ and ‘Clean Price’. Dirty Price includes the accrued interest/coupon from the date of last coupon payment till the Trade date.\n\nClean Price =Dirty Price- Accrued Interest\n\nLet us begin with a simple example. The following excel has a 10 years Bond with 1000 as Face Value, 10% coupon payment which falls due on 31 March of every year and the yield expectation is 9.5%.The transaction date is 30-June-2015.Trick here is to include the transaction date in the repayment schedule in chronological manner and put 0 (zero) against it as Pay-out. Since we have the exact date wise schedule so this time we should use XNPV function to calculate the Dirty Price of Bond. Rate =Yield to maturity of buyer (9.5%). Values and Dates should include the transaction date and final redemption date.",
null,
"Now it is time to calculate the Accrued Interest which is the interest between 31 March 2015 and 30 June 2015 @ of 10% (Coupon Rate).The Accrued Interest is calculated as 27. So the Clean Price is 1052-25=1027.\n\nDCM desk has 3 typical revenue streams:\n\n1. Capital Gain\n\n2. Transaction Fee\n\n3. Net Interest Income (Coupon-Cost of fund).\n\nCapital gain is the most important number (read ‘bonus defining number’) to showcase the work desk has done for the year. Actually, it is the holy grail of DCM desk. It is what differentiates Bond from a tradition loan product. For first secondary trade of a bond, Capital Gain is the difference between the Face Value and the Clean Price. For follow on trades it is the difference between two Clean Prices.\n\nAnother important number to showcase the performance is known as RoE (Return on Equity) or RoNW (Return on Net worth).\n\nRoE= (Capital Gain +Transaction Fee +Net Interest Income)/Total Equity employed\n\nTotal equity employed can be derived from the total issuance size and the debt equity ratio (d:e) of the institution/bank. RoE in north of 15% is considered as a great number on the appraisal books.\n\nTotal Equity employed=Bond issuance/ (1+d:e)\n\nThe below chart summarizes the trade at one place:",
null,
""
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https://www.bartleby.com/solution-answer/chapter-28-problem-57e-single-variable-calculus-early-transcendentals-volume-i-8th-edition/9781305270343/let-fxx3-a-if-a-0-use-equation-2-7-5-to-find-fa-b-show-that-f0-does-not-exist/185e4359-e4d4-11e8-9bb5-0ece094302b6 | [
"",
null,
"",
null,
"",
null,
"Chapter 2.8, Problem 57E",
null,
"### Single Variable Calculus: Early Tr...\n\n8th Edition\nJames Stewart\nISBN: 9781305270343\n\n#### Solutions\n\nChapter\nSection",
null,
"### Single Variable Calculus: Early Tr...\n\n8th Edition\nJames Stewart\nISBN: 9781305270343\nTextbook Problem\n\n# Let f ( x ) = x 3 . (a) If a ≠ 0, use Equation 2 .7 .5 to find f'(a).(b) Show that f'(0) does not exist.(c) Show that f ( x ) = x 3 . has a vertical tangent line at (0, 0).(Recall the shape of the graph of f. See Figure 1.2.13.)\n\n(a)\n\nTo determine\n\nTo find: The derivative of the function f(x) at x=a if a0.\n\nExplanation\n\nGiven:\n\nThe given function is, f(x)=x3.\n\nResult Used:\n\nThe derivative of a function f at x=a, denoted by f(x), is\n\nf(a)=limxaf(x)f(a)xa (1)\n\nDifference of cube formula: x3y3=(xa)(x2+y2+xy).\n\nCalculation:\n\nObtain the derivative of the function f(x) at x=a.\n\nCompute f(a) by using the equation (1),\n\nf(a)=limxaf(x)f(a)xa=limxax3a3xa=limxa(x13a13)xa=limxa(x13a13)xa×(x23+a13x13+a23)(x23+a13x13+a23)\n\nApply the difference of cube formula in the numerator as follows,\n\nf(a)=limxa(x13)3(a13\n\n(b)\n\nTo determine\n\nTo Show: The function f(0) does not exist.\n\n(c)\n\nTo determine\n\nTo show: The y=x3 has a vertical tangent line at (0,0)\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#### Are A={1,2,3,4} and B={x:x1} disjoint?\n\nMathematical Applications for the Management, Life, and Social Sciences\n\n#### In Exercises 110, evaluate the expression. 4(18!)\n\nFinite Mathematics for the Managerial, Life, and Social Sciences\n\n#### True or False: is a geometric series.\n\nStudy Guide for Stewart's Multivariable Calculus, 8th\n\n#### Sometime, Always, or Never:\n\nStudy Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th",
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https://www.cquestions.com/2010/06/write-c-program-to-print-pascal.html | [
"## INDEX\n\n### Write a c program to print Pascal triangle.\n\n1. Pascal triangle in c without using array\n\n2. C code to print Pascal triangle\n\n3. Simple c program for Pascal triangle\n\n4. C program to generate Pascal triangle\n\n5. Pascal triangle program in c language\n\n6. C program to print Pascal triangle using for loop\n\n#include<stdio.h>\n\nlong fact(int);\nint main(){\nint line,i,j;\n\nprintf(\"Enter the no. of lines: \");\nscanf(\"%d\",&line);\n\nfor(i=0;i<line;i++){\nfor(j=0;j<line-i-1;j++)\nprintf(\" \");\n\nfor(j=0;j<=i;j++)\nprintf(\"%ld \",fact(i)/(fact(j)*fact(i-j)));\nprintf(\"\\n\");\n}\nreturn 0;\n}\n\nlong fact(int num){\nlong f=1;\nint i=1;\nwhile(i<=num){\nf=f*i;\ni++;\n}\nreturn f;\n}\n\nSample output:\n\nEnter the no. of lines: 8\n1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1\n1 5 10 10 5 1\n1 6 15 20 15 6 1\n1 7 21 35 35 21 7 1\n\n10. Write a c program to add two numbers without using addition operator.\n11. Write a c program to subtract two numbers without using subtraction operator.\n15. Write a c program to solve quadratic equation.\n18. Write a c program which passes structure to function.\n28. Write a c program which takes password from user.\n29. Write a scanf function in c which accept sentence from user.\n30. Write a scanf function in c which accept paragraph from user.",
null,
"Anonymous said...\n\ngood site\n\nUnknown said...\n\n#include\nint main()\n{\nint i,j,k,n,m;\nprintf(\"Enter the range \");\nscanf(\"%d\",&n);\nm = n-1;\nfor(i=1;i<=n;i++)\n{\nk = i;\nj = m;\nwhile(j)\n{\nprintf(\" \");\nj--;\n}\nwhile(k)\n{\nprintf(\"*\");\nprintf(\" \");\nk--;\n}\nprintf(\"\\n\");\nm--;\n}\n}\n\ndude your code is right but this a code for pyramid\n\nEinstein's Descendents said...\n\nplease write a program to display all the perfect and prime numbers within a range.\n(I tried them, but after entering the range, the screen holds showing nothing.. on further clicking the program execution exits)\n\nGUNAVATHIPALANISAMY said...\n\nhow to say a number is perfect or not?",
null,
"rose_preethi said...\n\nThe 6 is a perfect number , because we can divide a 6 by 1,2,3 and if we add 1+2+3 we will get an answer as 6... When the sum of divisior of a particular number is equal to the number then we say that is a perfect number..\nAnother Example is 28 which is divided by 1,2,4,7,14 and when we add the number we will get as 28, 1+ 2 + 4 + 7 + 14 = 28\n\nPriyanka kumari said...\n\nHi Pavani,\nWhich compiler are you using?",
null,
"preema& reema said...\n\ngood site,thank u.......\n\nwat is pascal triangle\n\nsandhya said...\n\nconfusin\n\nsandhya said...\n\nincomplete,,,,,,,,,,,,, jst luk 1st for loop of i\n\nRakesh Rajwar said...\n\ngud clean example\n\nUnknown said...\n\nsimple way with out using function....\n\n#include\n#include\n\nvoid main()\n{\nint i,j,n,c,k,space;\nclrscr();\nprintf(\"Enter the limit \");\nscanf(\"%d\",&n);\nprintf(\"\\n\\n\");\nspace=n;\nfor(i=0;i<=n;i++)\n{\nc=1;\nfor(k=space;k>=0;k--)\nprintf(\" \");\n\nspace--;\nfor(j=0;j<=i;j++)\n{\nprintf(\"%d \",c);\nc=(c*(i-j)/(j+1));\n}\nprintf(\"\\n\");\n}\ngetch();\n}\n\nParminder Singh said...\n\nwhy hav u used long datatype for fact() function... why d things don't work up with int as a return type???",
null,
"farzana said...\n\nfunction get more confusion....",
null,
"farzana said...\n\nfunction get more confusion....",
null,
"Anonymous said...\n\nlots of confusion.....\n\nUnknown said...\n\ncan u please give me solution for the following pattern\n1\n2 2\n3 3 3\n4 4 4 4\n\nUnknown said...\n\nsend code for fallowing o/p::\n1\n2 3\n4 5 6\n7 8 9 10",
null,
"Anonymous said...\n\n#include\nvoid main()\n{\nint x,y;\nfor(x=1;x<=4;x++)\n{\nfor(y=1;y<=x;y++)\nprintf(\"%2d\",x);\nprintf(\"\\n\");\n}\n\n}",
null,
"Anonymous said...\n\ncan u please give me solution for the following pattern...\n1 2 3 4 5\n1 2 3 4\n1 2 3\n1 2\n1\n\nPriyanka kumari said...\n\nC program to print different shapes using printf function",
null,
"Anonymous said...\n\ncan u pls give me solution for fibonacci series in use for loop...",
null,
"Anonymous said...\n\nthank u",
null,
"Anonymous said...\n\ncan u pls provide ans for dis\n5987\n987\n87\n7\n\nUnknown said...\n\n1\n1 1\n1 2 1\n1 3 3 1\n1 4 6 4 1\n1 5 10 10 5 1\n1 6 15 20 15 6 1",
null,
"Anonymous said...\n\n#include\nint main()\n{\nint i,j,n,c;\nprintf(\"Enter the limit \");\nscanf(\"%d\",&n);\nprintf(\"The pascal triangle is print in floyd's triangle\\n\\n\");\nfor(i=0;i<n;i++)\n{\nc=1;\nfor(j=0;j<=i;j++)\n{\nprintf(\"%d \",c);\nc=(c*(i-j)/(j+1));\n}\nprintf(\"\\n\");\n}\nreturn;\n}\n\nby Muthusamy.S GCT",
null,
"Anonymous said...\n\n#include\nint main()\n{\nint r,n,c;\nprintf(\"Enter the limit \");\nscanf(\"%d\",&n);\nint s=n;\nfor(r=0;r<n;r++)\n{\nfor(c=1;c<=s;c++)\n{\nprintf(\"%d \",c);\n}\ns--;\nprintf(\"\\n\");\n}\nreturn;\n}\nby Muthusamy.s from GCT",
null,
"Anonymous said...\n\n#include\nint main()\n{\nint r,n,c;\nprintf(\"Enter the limit \");\nscanf(\"%d\",&n);\nint s=n;\nfor(r=0;r<n;r++)\n{\nfor(c=1;c<=s;c++)\n{\nprintf(\"%d \",c);\n}\ns--;\nprintf(\"\\n\");\n}\nreturn;\n}\nby Muthusamy.s from GCT\n\nsrinath reddy said...\n\nstack operations are used for storing elements in a stack. The input method for stack is last in first out. Thank you for sharing this program.\nregards:\nsrinath reddy.\nadmin of Programming Tutorials for Beginners\n\nUnknown said...\n\n#include\n#include\n#include\nvoid main()\n{\nint a,n,n1,num,sum=0,r,i,j;\nclrscr();\nprintf(\"enter the power of 11\");\nscanf(\"%d\",&n);\nfor(i=0;i0)\n{\nr=num%10;\n//sum=sum+r;\n\nprintf(\" %d \",r);\nnum=num/10;\n}\nprintf(\"\\n\");\n\n}\n\ngetch();\n}",
null,
"Anonymous said...\n\ncan someone tell me whats wrong with this code for generating pascal triangle .iwish to print the pascal triangle line by line without having to induct spaces among elements\n\nint fact1(int j)\n{\nint k=1,i;\nfor(i=j;i>=0;i-- ){\nk=i*k;\n}\nreturn k;\n}\n\nint main()\n{\nint res,line,k;\nfor(line=0;line<7;line+=1){\nfor(k=0;k<line;k++)\n{\nres=fact1(line)/fact1(line-k)*fact1(k);\nprintf(\"%d\\t\",res);\n}\nprintf(\"\\n\");\n}\nreturn 0;\n}",
null,
"Anonymous said...\n\n#include\n\nint main()\n{\nint n, i, c, a = 1;\n\nprintf(\"Enter the number of rows of Floyd's triangle to print\\n\");\nscanf(\"%d\", &n);\n\nfor (i = 1; i <= n; i++)\n{\nfor (c = 1; c <= i; c++)\n{\nprintf(\"%d \",a);\na++;\n}\nprintf(\"\\n\");\n}\n\nreturn 0;\n}",
null,
"Anonymous said...\n\n#include\nint main ()\n{\nint i,j,a=1;\nfor(i=1;i<=4;i++){\nfor(j=1;j<=i;j++){\nprintf (\"%d \",a);\n}\na++;\nprintf (\"\\n\");\n}\ngetchar();\nreturn 0;\n}",
null,
"Anonymous said...\n\n****** *******\n***** *****\n*** ***\n** **\n* *",
null,
"Anonymous said...\n\nThanx a lot darling...!!!!!!!!\n\nUnknown said...\n\nnice but plz give the many pascal question that we butter understand about the pascal.....\n\nblogger said...\n\nI have this one works 100% i coded this my self although it's little foolish\n#include\nmain()\n{\nint i, j, b,a,k,n,count;\nprintf(\"enter the limit \\n\") ;\nscanf(\"%d\", & n) ;\n\nfor(i=0;i0;k--)\nprintf(\" \");\nprintf(\"1 \\n\\n\");\n}\nif(j==1)\n{\nfor(k=2*(n-i) ;k>0;k--)\nprintf(\" \");\nprintf(\"1 \");\n}\n}\ncount=0;\nk=0;\nfor(j=1;j<i;j++)\n{\na=1;\na[i]=1;\nb[k]=a[j]+a[j+1];\nprintf(\" %d \", b[k]) ;\nk++;\ncount++;\n}\nfor(j=2,k=0;k<=i-1;j++, k++)\na[j]=b[k] ;\nfor(j=1;j<=i;j++)\n{\nif(j==i)\n{\nprintf(\" 1\");\nprintf(\"\\n\\n\") ;\n}\n}\n}\nreturn 0 ;\n}\n\nblogger said...\n\nUnknown said...\n\n#include\n#include\n\nint main()\n{\nint a;\nint no;\nprintf(\"Enter the number of rows for pascal triangle : \");\nscanf(\"%d\",&no); //taking input\nfor(int i=1; i<=no; i++)\n{\nfor(int j=1; j<=i; j++)\n{\nif(j==1||j==i)\n{\na[i][j]=1;\n}\nelse // taking arrays value\n{\na[i][j]=a[i-1][i-j]+a[i-1][i-j+1];\n}\n}\n}\nfor(int i=1; i<=no; i++)\n{\nfor(int k=1; k<=no-i; k++)\n{\nprintf(\" \");\n}\nfor(int j=1; j<=i; j++)\n{\nprintf(\"%d \",a[i][j]); //printing pascal triangle\n}\nprintf(\"\\n\");\n}\ngetch();\n}\n\nUnknown said...\n\nsimple easy approach\n\n#include\nvoid main()\n{\nint i,j,x,n,s;\nprintf(\"Enter The Number : \");\nscanf(\"%d\",&n);\nfor(i=0;i<=n;i++)\n{\nx=1;\nfor(s=1;s<=n-i;s++)\nprintf(\"\\t\");\nfor(j=1;j<=i+1;j++)\n{\nprintf(\"%d\\t\\t\",x);\nx=x*(i-j+1)/j;\n}\nprintf(\"\\n\");\n}\n}\n\nHarendra said...\nThis comment has been removed by the author.\nHarendra said...\n\n#include\nmain()\n{\nint i,j,n,k,temp,c;\n\nprintf(\"Enter the number of lines\\n\");\nscanf(\"%d\",&n);\nprintf(\"\\n\\n\");\ntemp=n;\nfor(i=0;i<=n;i++,printf(\"\\n\"))\n{\nc=1;\nfor(k=temp;k>=0;k--)\nprintf(\" \");\ntemp--;\nfor(j=0;j<=i;j++)\n{\nprintf(\"%d \",c);\nc=(c*(i-j)/(j+1));\n}\n}\n\n}\n\nUnknown said...\nThis comment has been removed by the author.\nUnknown said...\n\nplz tell me code of this patten:\n\n1*2*3*4*17*18*19*20\n-5*6*7*14*15*16\n--8*9*12*13\n---10*11\n\nUnknown said...\n\nUnknown said...\n\nhow to do this???\n\nA1234\nAB123\nABC12\nABCD1\nABCDE\n\nhemanth said...\n\nSmall correction : if we give n rows it will print n+1 rows ,so write I<n in first for loop",
null,
"Anonymous said...\n\n1\n1\n2\n12\n3\n123\n4\n1234\n\nUnknown said...\n\nif num=3 then\n1\n2*3\n4*5*6\n4*5*6\n2*3\n1\n\nUnknown said...\n\ncan u please give me solution of this question?\nQuestion: Write program using loop to draw 50 stars in five rows.Through pascal programming language.\n\nUnknown said...\n\ncan u please give me solution of this question?\nQuestion: Write program using loop to draw 50 stars in five rows.Through pascal programming language.\n\nUnknown said...\n\nusing System;\nusing System.Collections.Generic;\nusing System.Linq;\nusing System.Text;\n\nnamespace ConsoleApplication39\n{\nclass Program\n{\nstatic void Main(string[] args)\n{\nint i, j;\nfor(i=0 ;i<5 ;i++)\n{\nfor(j=5 ;j>i;j--)\n{\nConsole.Write(\"*\");\n\n}\nConsole.Write(\" \");\nfor (j = 5; j > i; j--)\n{\nConsole.Write(\"*\");\n\n}\n}\n\n}\n}\n\n}\n\nUnknown said...\n\ncan you assist me to create a pascal program to output the digital pyramid\n\nUnknown said...\n\nany 1 can solve tis plzzzzz\n1\n12\n123\n12345 tis is a o/p u can gv a sol. plzzzz gv me soon\n\nUnknown said...\n\nHow can we execute with out using factorial.\n\nUnknown said...\n\nruchi is it a star pattern program\n\nUnknown said...\n\ncan u help me to write a program in c\n\nrite a program to print the first n terms in the series\n2, 6, 11, 10, 15, 21, 18, 24, 31, 26, 33, 41 ....\n\nInput and Output Format:\nInput consists of a single integer 'n' that corresponds to the number of terms to be printed.\n\nOutput consists of n terms in the series separated by a single space. There is a trailing space at the end of the series.\n\nSample Input:\n12\nSample Output:\n2 6 11 10 15 21 18 24 31 26 33 41"
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http://matcha-tee.me/cnc-programming-basics/cnc-programming-basics-tutorials-pdf/ | [
"Cnc Programming Basics Tutorials Pdf",
null,
"cnc programming basics tutorials pdf.\n\ncnc machine programming tutorial in hindi basics tamil pdf examples tutorials video,cnc coding tutorials programming basics pdf download pattern repeating canned cycle basic sample program,cnc programming basics pdf download with g code definitive free tutorial coding tutorials ppt,cnc programming lessons tutorial in hindi basics codes with g code definitive free,cnc coding tutorials machine programming tutorial in hindi lessons 4 basic milling,cnc programming basics pdf download tamil guide to cad for what is g code tutorial,cnc machine programming tutorial in hindi basics tamil pdf with g code definitive free,cnc coding tutorials programming code example basics ppt,cnc programming basics tutorial textbook pdf examples incremental tutorials machine in hindi,well set out code programming training centre cnc basics tutorial tutorials pdf tamil."
] | [
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"http://matcha-tee.me/wp-content/uploads/2019/07/cnc-programming-basics-tutorials-pdf.jpg",
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https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/ | [
"The Triskele Curve\n\nloopspace\n\n2nd August 2018\n\n# 1 Introduction\n\nI recently came across Triskele Globes. It was sufficiently near to Christmas that I thought I'd have a go at making one as they look quite festive. I was right: they do look quite festive. However, looking very closely at them then I wasn't quite sure that the fold was along the right curve. There seemed to be a couple of points of deformation, particularly at the corners. So I thought I'd have a look at figuring out what the curve ought to be.\n\n# 2 Defining the Curve in 3D\n\nLet's generalise. Suppose we wrinkle a piece of paper along its length so that its cross section is the same all through.",
null,
"Let's assume that we start with a piece of paper lying in the positive quadrant of the $x-y$–plane up against the axes. We then wrinkle it in the $y$–direction, meaning that as you travel in the $y$–direction then the paper changes height but along the $x$–direction then it stays constant. We keep it above the positive quadrant and so that its edges stay above the axes (though we give ourselves leave to move it vertically).\n\nThis means that the paper is described by a parametrisation of the form $\\left(s,f\\left(t\\right),g\\left(t\\right)\\right)$.",
null,
"We want to fold it straight down so that the folded part has constant cross section in the $z$–direction. Once it has been folded, the bottom edge of the paper will be contained in a horizontal plane since we can only flex the paper in one direction at a time. We can therefore arrange it so that the bottom edge (once folded) lies in the $x-y$–plane.",
null,
"Pre-folding, this edge was in the $y-z$–plane and post-folding it is in the $x-y$–plane. Therefore the distance along the paper from the fold to the $y-z$–plane prior to folding is the same as the distance along the paper from the fold to the $x-y$–plane after folding. That is to say, the fold lies on the plane $x=z$.\n\nOn the paper, that translates to $s=g\\left(t\\right)$. Thus the fold curve is $\\left(g\\left(t\\right),f\\left(t\\right),g\\left(t\\right)\\right)$.\n\nAfter folding, this then continues down to the $x-y$–plane and the parametrisation of the folded part is therefore $\\left(g\\left(t\\right),f\\left(t\\right),s\\right)$. A full parametrisation of the paper is therefore:\n\n $\\left(\\mathrm{max}\\left(s,g\\left(t\\right)\\right),f\\left(t\\right),\\mathrm{min}\\left(s,g\\left(t\\right)\\right)\\right)$\n\n# 3 Defining the Curve in 2D\n\nThe previous section defines the curve in space, but what we want to do is to draw it on the paper at the outset. To do that, we need to draw it on the paper and then \"unwrinkle\" the paper. If we let $\\left(u,v\\right)$ be the paper's natural coordinates then in the pre-folded parametrisation, we have $u=s$ and $v$ is the arc length along $\\left(f\\left(t\\right),g\\left(t\\right)\\right)$. That is,\n\n $v\\left(t\\right)={\\int }_{0}^{t}\\sqrt{f\\text{'}\\left(r{\\right)}^{2}+g\\text{'}\\left(r{\\right)}^{2}}dr$\n\nHence the curve $\\left(g\\left(t\\right),f\\left(t\\right),g\\left(t\\right)\\right)$ unwrinkles to the curve:\n\n$\\left(g\\left(t\\right),{\\int }_{0}^{t}\\sqrt{f\\text{'}\\left(r{\\right)}^{2}+g\\text{'}\\left(r{\\right)}^{2}}dr\\right)$\n 1\n\nWhen the cross-section is given by a height function – that is, $f\\left(t\\right)=t$ – the fold curve is $\\left(g\\left(t\\right),t,g\\left(t\\right)\\right)$ in 3D space and on the paper is given by:\n\n$\\left(g\\left(t\\right),{\\int }_{0}^{t}\\sqrt{1+g\\text{'}\\left(r{\\right)}^{2}}dr\\right)$\n 2\n\n# 4 Special Case: Ellipse\n\nWhen making the Triskele Globes, each piece of paper has the following cross section prior to the folding.",
null,
"This consists of two straight parts (which form the inside of the globe) and two elliptical ends (which will get folded down). The part we are most interested in is the elliptical end.\n\nIf we scale our measurements so that the inner cube has length $2$, then the parametrisation of the ellipse is:\n\n $\\left(\\mathrm{sin}\\left(\\varphi \\right),b\\mathrm{cos}\\left(\\varphi \\right)\\right),\\phantom{\\rule{2em}{0ex}}$\n\nwith $0. The parametrisation of the paper is therefore:\n\n $\\left(s,\\mathrm{sin}\\left(\\varphi \\right),b\\mathrm{cos}\\left(\\varphi \\right)\\right)$\n\nTo get the fold curve, we need to figure out the curve:\n\n $\\left(b\\mathrm{cos}\\left(\\varphi \\right),{\\int }_{-\\pi /2}^{\\varphi }\\sqrt{{\\mathrm{cos}}^{2}\\left(\\psi \\right)+b{\\mathrm{sin}}^{2}\\left(\\psi \\right)}d\\psi \\right)$\n\nThe integral simplifies (slightly) to:\n\n $\\begin{array}{rl}{\\int }_{-\\pi /2}^{\\varphi }\\sqrt{{\\mathrm{cos}}^{2}\\left(\\psi \\right)+{b}^{2}{\\mathrm{sin}}^{2}\\left(\\psi \\right)}d\\psi & ={\\int }_{0}^{\\varphi }\\sqrt{1-\\left(1-{b}^{2}\\right){\\mathrm{sin}}^{2}\\left(\\psi \\right)}d\\psi \\\\ & ={\\int }_{-\\pi /2}^{\\varphi }\\sqrt{1-m{\\mathrm{sin}}^{2}\\left(\\psi \\right)}d\\psi \\end{array}$\n\nwhere ${m}^{2}=1-{b}^{2}$.\n\nThis is an incomplete elliptic integral of the 2nd kind. Unless $b=1$, it has no closed form solution. Fortunately, there are numerical solutions and the scipy module for python can compute them. The program curve.py uses this to produce a table of coordinates for the paper curve which can be read in by the LaTeX document triskele.tex to produce the template; the version for $b=.8$ is triskele.pdf.\n\nBelow this is plotted with the approximation of an arc for comparison. It should be noted that they are very close.",
null,
"If $b=1$ (whence the cross section is circular) then the integral simplifies to\n\n ${\\int }_{-\\pi /2}^{\\varphi }\\sqrt{{\\mathrm{cos}}^{2}\\left(\\psi \\right)+{\\mathrm{sin}}^{2}\\left(\\psi \\right)}d\\psi ={\\int }_{-\\pi /2}^{\\varphi }d\\psi =\\varphi +\\pi /2$\n\nand so the fold curve has equation\n\n $\\left(\\mathrm{cos}\\left(\\varphi \\right),\\varphi +\\pi /2\\right)$\n\nwhich we can plot directly to obtain the following curve, overlaid again with an arc so show how the separation has increased.",
null,
"Note that the case where $b=1$ means that the ends are circular and that the two folds meet at the centre.\n\n# 5 Finding Functions\n\nIn Equation 1 there are three functions: $f\\left(t\\right)$, $g\\left(t\\right)$, and the arc length function, say $h\\left(t\\right)$. These are related via a Pythagorean relationship on the derivatives that states that:\n\n$h\\text{'}\\left(t{\\right)}^{2}=f\\text{'}\\left(t{\\right)}^{2}+g\\text{'}\\left(t{\\right)}^{2}$\n 3\n\nThus the problem of finding such triples $\\left(f,g,h\\right)$ which have a closed form expression (for some suitable definition thereof) is closely related to a functional version of finding Pythagorean triples.\n\nThe fact that we are working with derivatives is what makes this a non-trivial question. As we have already seen, this problem is complicated even for an ellipse.\n\nTrigonometric and hyperbolic functions form a natural source of these triples:\n\n $f\\left(t\\right)$ $g\\left(t\\right)$ $h\\left(t\\right)$ $\\mathrm{sin}\\left(t\\right)$ $\\mathrm{cos}\\left(t\\right)$ $t$ $\\mathrm{ln}|\\mathrm{cos}\\left(t\\right)|$ $t$ $\\mathrm{ln}|\\mathrm{sec}\\left(t\\right)+\\mathrm{tan}\\left(t\\right)|$ $\\mathrm{cosh}\\left(t\\right)$ $t$ $\\mathrm{sinh}\\left(t\\right)$ ${\\mathrm{tan}}^{-1}\\left(\\mathrm{sinh}\\left(t\\right)\\right)$ $\\mathrm{ln}\\left(\\mathrm{cosh}\\left(t\\right)\\right)$ $t$ $\\frac{1}{\\sqrt{1+x}}$ $\\frac{1}{\\sqrt{1-x}}$ ${\\mathrm{sin}}^{-1}\\left(t\\right)$"
] | [
null,
"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/paper.png",
null,
"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/WrinkledPaper.png",
null,
"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/FoldedPaper.png",
null,
"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/triskele-figure0.png",
null,
"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/triskele-figure1.png",
null,
"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/triskele-figure2.png",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.95761,"math_prob":0.99992025,"size":4531,"snap":"2020-45-2020-50","text_gpt3_token_len":1001,"char_repetition_ratio":0.15286061,"word_repetition_ratio":0.029339854,"special_character_ratio":0.20988744,"punctuation_ratio":0.08938547,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999521,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-30T01:55:35Z\",\"WARC-Record-ID\":\"<urn:uuid:dd02b5d4-da84-4f09-a4a6-4e37493785db>\",\"Content-Length\":\"24971\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c9d318ae-9016-4ce7-b16d-883678df20f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:4ba4460d-ea14-48c7-b81d-dc8322e2f27a>\",\"WARC-IP-Address\":\"109.123.120.183\",\"WARC-Target-URI\":\"https://loopspace.mathforge.org/CountingOnMyFingers/Triskele/\",\"WARC-Payload-Digest\":\"sha1:J6SSNT36BRQRNS2BAIMI2DY4P43IAHQM\",\"WARC-Block-Digest\":\"sha1:WKJ4N3ODVHH26BEFWSM3GWCNPV3P4KRK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107906872.85_warc_CC-MAIN-20201030003928-20201030033928-00507.warc.gz\"}"} |
https://aitopics.org/mlt?cdid=news%3AA539E8D5&dimension=concept-tags | [
"to\n\n### ONLamp.com: Building Decision Trees in Python\n\nThe decision tree in the figure is just one of many decision tree structures you could create to solve the marketing problem. The task of finding the optimal decision tree is an intractable problem. For those of you who have taken an analysis of algorithms course, you no doubt recognize this term. For those of you who haven't had this pleasure (he says, gritting his teeth), essentially what this means is that as the amount of test data used to train the decision tree grows, the amount of time it takes to do so grows as well--exponentially. While it may be nearly impossible to find the smallest (or more fittingly, the shallowest) decision tree in a respectable amount of time, it is possible to find a decision tree that is \"small enough\" using special heuristics.\n\n### Why do Decision Trees Work?\n\nDecision trees are a type of recursive partitioning algorithm. Decision trees are built up of two types of nodes: decision nodes, and leaves. The decision tree starts with a node called the root. If the root is a leaf then the decision tree is trivial or degenerate and the same classification is made for all data. For decision nodes we examine a single variable and move to another node based on the outcome of a comparison.\n\n### Interactive Decision Tree Software for Customer Service\n\nBrands today have detailed processes with dynamic changes to serve millennials across the globe. As a CX Head, it is important to use a decision tree software to ensure that the right information is given to customers across assisted and digital channels. Knowmax is a decision tree software that helps you create workflows and publish them across all customer touchpoints using an interactive UI backed with robust analytics.\n\n### Provably robust boosted decision stumps and trees against adversarial attacks\n\nThe problem of adversarial robustness has been studied extensively for neural networks. However, for boosted decision trees and decision stumps there are almost no results, even though they are widely used in practice (e.g. We show in this paper that for boosted decision stumps the \\textit{exact} min-max robust loss and test error for an $l_\\infty$-attack can be computed in $O(T\\log T)$ time per input, where $T$ is the number of decision stumps and the optimal update step of the ensemble can be done in $O(n 2\\,T\\log T)$, where $n$ is the number of data points. Moreover, the robust test error rates we achieve are competitive to the ones of provably robust convolutional networks. Papers published at the Neural Information Processing Systems Conference.\n\n### Optimal Sparse Decision Trees\n\nDecision tree algorithms have been among the most popular algorithms for interpretable (transparent) machine learning since the early 1980's. The problem that has plagued decision tree algorithms since their inception is their lack of optimality, or lack of guarantees of closeness to optimality: decision tree algorithms are often greedy or myopic, and sometimes produce unquestionably suboptimal models. Hardness of decision tree optimization is both a theoretical and practical obstacle, and even careful mathematical programming approaches have not been able to solve these problems efficiently. This work introduces the first practical algorithm for optimal decision trees for binary variables. The algorithm is a co-design of analytical bounds that reduce the search space and modern systems techniques, including data structures and a custom bit-vector library."
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.9405278,"math_prob":0.9348356,"size":3337,"snap":"2020-34-2020-40","text_gpt3_token_len":676,"char_repetition_ratio":0.14071406,"word_repetition_ratio":0.003773585,"special_character_ratio":0.18849266,"punctuation_ratio":0.08681135,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9636318,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T05:22:35Z\",\"WARC-Record-ID\":\"<urn:uuid:6346656a-efb5-4d35-9d4a-c653fe7ae9e2>\",\"Content-Length\":\"86542\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:315a0432-2740-464f-9abf-674f0bc589b7>\",\"WARC-Concurrent-To\":\"<urn:uuid:85602720-0381-482f-ab20-457c01a8e84f>\",\"WARC-IP-Address\":\"35.188.181.171\",\"WARC-Target-URI\":\"https://aitopics.org/mlt?cdid=news%3AA539E8D5&dimension=concept-tags\",\"WARC-Payload-Digest\":\"sha1:TRSQ64MGSLNXT3RWL2IP6BXSUTO67PL5\",\"WARC-Block-Digest\":\"sha1:UNY42AXBH6GXPALFKVTZUMVJDLCXSOCV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735860.28_warc_CC-MAIN-20200804043709-20200804073709-00503.warc.gz\"}"} |
https://bcallaway11.github.io/Courses/ECON_4750_Fall_2021/code_challenge4.html | [
"As we talked about in class, one common way to measure the distance between two numbers $$a$$ and $$b$$ is to compute\n\n$distance = (a-b)^2$\n\nFor this code challenge, you need to write a function that\n\n• Takes in two vectors of numbers (the vectors will have the same number of elements in them)\n\n• Computes the distance between corresponding elements in each vector (i.e., the distance between the first element in the the first vector and the first element in the second vector, etc.)\n\n• Returns the average distance between corresponding elements in each vector.\n\nRules:\n\n• You cannot load any external packages, but any base R functions are allowed\n\nTo win\n\n• You must email me your function [email protected]\n\n• I’ll call the function on two vectors of numbers. If it computes the correct average distance between two vectors, then you win.\n\nSolution below…\n\nmean_distance <- function(x,y) {\ndistance <- (x-y)^2\nmean(distance)\n}\n\nmean_distance(c(10,3,5), c(0.1,3,0))\n## 41.00333"
] | [
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https://help.scilab.org/docs/5.4.1/pt_BR/fec_properties.html | [
"Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nChange language to: English - Français - 日本語 - Русский\n\nPlease note that the recommended version of Scilab is 6.0.2. This page might be outdated.\nSee the recommended documentation of this function\n\nAjuda Scilab >> Biblioteca de Gráficos > 2d_plot > fec properties\n\n# fec properties\n\ndescription of the fec entities properties\n\n### Description\n\nThe Fec entity is a leaf of the graphics entities hierarchy. It represents 2D finite elements plots (see `fec`, `Sgrayplot`).\n\nparent:\n\nThis property contains the handle of the parent. The parent of the fec entity should be of the type `\"Axes\"` or `\"Compound\"`.\n\nchildren:\n\nThis property contains a vector with the `children` of the handle. However, `Fec` handles currently do not have any `children`.\n\nvisible:\n\nThis field contains the `visible` property value for the entity . It should be `\"on\"` or `\"off\"` . By default, the plot is visible, the value's property is `\"on\"`. If `\"off\"` the plot is not drawn on the screen.\n\ndata:\n\nThis is a three column matrix `[x,y,f]`, where `x(i)` and `y(i)` are the coordinates of the `i`'th node. `f(i)` is the value associated to the node `i`.\n\ntriangles:\n\nThis is a five column matrix `[tn,n1,n2,n3,flag]`. `tn(j)` is the triangle number. `n1(j)`, `n2(j)` and `n3(j)` are the index of the nodes which constitute the triangle. (`flag(j)` is not used).\n\nz_bounds:\n\nThis vector of size 2, `[zmin,zmax]`, gives the z values associated with the first and the last color (of the current colormap). More exactly if the colormap have `nc` colors and if we note ```dz = (zmax-zmin)/nc``` ,then the part of the triangulation where `zmin + (i-1)dz <= z < zmin + i dz` is filled with the color `i`). By default the `z_bounds` property value is `[0,0]`. In this case, the `zmin` and `zmax` are autommatically set to the minumum and maximum of the `func` argument.\n\noutside_color:\n\nThis vector of size 2, `[cmin, cmax]`, defines the color used when nodes values are outside the `z_bounds = [zmin,zmax]` interval. When node values are lower than `zmin` the color with index `cmin` is used. When node values are greater than `zmax` the color with index `cmax` is used. By default, the `outside_color` property value is `[0,0]`. In this case, `cmin` and `cmax` are automatically set to the two bounds of the colormap. If `cmin` or `cmax` are negative, then values outside z_bounds interval are not displayed, they appear transparent.\n\ncolor_range:\n\nThis vector of size 2, `[rmin, rmax]`, allows to use only a part of the colormap for display. `rmin` and `rmax` stand for colormap indices. If they are both greater than 1, then the actual colormap used to display the fec entity is `colormap(rmin:rmax)` where `colormap` is the colormap of the parent figure. By default, the `color_range` property value is `[0,0]`. In this case, the whole colormap is used.\n\nline_mode:\n\nIf `\"on\"`, the wireframe enclosing triangles is drawn. If `\"off\"`, only the inside of triangles are drawn.\n\nforeground:\n\nThis color index specifies the color of the mesh. If `line_mode` property is `\"on\"`, the wireframe is drawn using this color.\n\nclip_state:\n\nThis field contains the `clip_state` property value for the fec. It should be :\n\n• `\"off\"` this means that the fec is not clipped.\n\n• `\"clipgrf\"` this means that the fec is clipped outside the Axes box.\n\n• `\"on\"` this means that the fec is clipped outside the rectangle given by property clip_box.\n\nclip_box:\n\nThis field is to determinate the `clip_box` property. By Default its value should be an empty matrix if clip_state is \"off\". Other cases the vector `[x,y,w,h]` (upper-left point width height) defines the portions of the fec to display, however `clip_state` property value will be changed.\n\nuser_data:\n\nThis field can be use to store any scilab variable in the fec data structure, and to retreive it.\n\n### Examples\n\n```x=-10:10; y=-10:10;m =rand(21,21);\nSgrayplot(x,y,m);\na=get(\"current_axes\");\nf=a.children.children(1)\nf.data(:,3)=(1:size(f.data,1))';\na.parent.color_map=hotcolormap(64);```\n\n• set — Ajusta um valor de propriedade de uma objeto entidade gráfica ou de um objeto Interface do Usuário (User Interface)\n• get — Recupera um valor de propriedade de uma entidade de gráficos ou um objeto Interface do Usuário.\n• delete — Deleta uma entidade gráfica e seus galhos\n• fec — Esboço pseudo-colorido triangular de uma função definida por uma malha triangular\n• Sgrayplot — esboço 2d suave de uma superfície utilizando cores\n• graphics_entities — Descrição das estruturas de dados de entidades gráficas"
] | [
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.5427465,"math_prob":0.9183136,"size":4683,"snap":"2019-35-2019-39","text_gpt3_token_len":1308,"char_repetition_ratio":0.13293439,"word_repetition_ratio":0.042666666,"special_character_ratio":0.24450138,"punctuation_ratio":0.14527027,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9557393,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-16T00:56:42Z\",\"WARC-Record-ID\":\"<urn:uuid:4554c0d0-dffa-4fb5-830b-fc055e826996>\",\"Content-Length\":\"33814\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:924ed2e6-baa6-462f-97d5-d494973071af>\",\"WARC-Concurrent-To\":\"<urn:uuid:fd600ff0-6efc-4d2b-bec4-445158c9968c>\",\"WARC-IP-Address\":\"176.9.3.186\",\"WARC-Target-URI\":\"https://help.scilab.org/docs/5.4.1/pt_BR/fec_properties.html\",\"WARC-Payload-Digest\":\"sha1:4K24N4FM63P3B7BMNQIAUWTQMG4VHZHQ\",\"WARC-Block-Digest\":\"sha1:WINAHORCVF7RB3EK7K25YUDMM75AYNS2\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514572439.21_warc_CC-MAIN-20190915235555-20190916021555-00250.warc.gz\"}"} |
http://nokm.ecbu.pw/crank-nicolson-method-python.html | [
"# Crank Nicolson Method Python\n\nSee for more details. the method is implicit, i. gz and run pip install sinope-. Le vy models Christoph Schwab Computational Methods for Quantitative Finance. Crank-Nicolson time discretization for the Heat equation with Dirichlet boundary conditions. In 1D, an N element numpy array containing the intial values of $$\\psi$$ at the spatial grid points. The Crank-Nicolson method is a method of numerically integrating ordinary. In numerical analysis, the Crank–Nicolson method is a finite difference method used for numerically solving the heat equation and similar partial differential equations. Several additional tools written in C++, Python or bash scripting language are also included for convenience. Rothwell 1, *,JonathanL. shown that the application of the finite element method in conjunction. Now it's time for a confession: In the real-world of using computers to derive numerical solutions to differential equations, no-one actually uses Euler's Method. The equation is stable if Real(λ) ≤ 0. evolve half time step on x direction with y direction variance attached where Step 2. The trapezoid method can be derived from the trapezoid rule for integration. This Demonstration shows the application of the Crank–Nicolson (CN) method in options pricing. (2009) Efficient uncertainty quantification with the polynomial chaos method for stiff systems. Crank-Nicolson splits the difference between Forward and Backward difference schemes. Anaconda 5. Box 94079, 1090 GB Amsterdam, Netherlands Abstract A widely-used approach in the time integration of initial-value problems for time-dependent partial differential equations (PDEs) is the method of lines. In this I have extended the same problem to 2 dimensional with the help of Alternate direction implicit method. The default method is first order implicit but may be changed to Crank-Nicholson by changing h. To clarify nomenclature, there is a physically important difference between convection and advection. Table 1 reports pricing absolute errors (analytic price - model price), relative er-. Solution of Partial differential equations: classification, Laplace equation, Finite difference method. I have researched on the implicit methods on solving the equation, it uses something called Crank-Nicolson Method, but it too looked like its too complex. Hi, I am trying to make again my scholar projet. The Crank-Nicolson scheme will lead to the following approximations for the derivatives in Equation 1 :. See for more details. We will illustrate the property and advantages of these methods by comparing the performance of them. I have managed to code up the method but my solution blows up. The Crank-Nicolson finite difference method represents an average of the implicit method and the explicit method. The convective and diffusive terms are temporally integrated with a second-order Adams-Bashforth scheme and a second-order Crank-Nicolson scheme, respectively. evolve another half time step on y. En mathématiques, en analyse numérique, la méthode de Crank-Nicolson est un algorithme simple permettant de résoudre des systèmes d'équations aux dérivées partielles. Operator splitting techniques Divide-and-conquer strategy: decompose unwieldy (systems of) PDEs into simpler subproblems and treat them individually using specialized numerical algorithms Differential splitting ∂u ∂t +Lu = 0, L = XS s=1 Ls Discretization order: time, space (operator splitting is applied to L before the discretization in space). 4 Stiffness and Stability In addition to having a stable problem, i. Definitions. To illustrate that Euler's Method isn't always this terribly bad, look at the following picture, made for exactly the same problem, only using a step size of h = 0. Codes Lecture 19 (April 23) - Lecture Notes. ##2D-Heat-Equation. A class for pricing European options using the implicit method of finite differences; The Crank-Nicolson method. These choices are by no means the only approximations that. (5) Relax and hold steady: elliptic problems. Demonstrates the method of Steepest Decent to find the minimum of a function z = f(x,y). Compare the solution with the exact solution:. using backward Euler or Crank-Nicolson schemes; see . Note: Citations are based on reference standards. Python/Matplotlib Code # Crank-Nicolson method to solve the heat equation. Derivation of PDE for Asian options. Here I used the Crank-Nicolson scheme to model a fault scarp with a vertical offset of 10 m. λ 0 ··· 0 λ (1 − 2λ) Try to do implicit method and Crank-Nicolson method, where Crank-Nicolson takes uxx with the average of the centered-difference formula at j and j + 1. Introduction to Numerical Methods Lecture notes for MATH 3311 Jeffrey R. In particular, the Black-Scholes option pricing model can be transformed into a partial differential equation and numerical solution for option pricing can be approximated using the Crank-Nicolson difference scheme. Helmholtz decomposition; Least-squares ; Galerkin ; Leap-frog method (second order, explicit) Crank-Nicolson method (second order, implicit) Lax-Wendroff method (second order, explicit) Third-order explicit Taylor-Galerkin method; Fourth-order implicit Taylor-Galerkin method. fmin_bfgs (f, x0 [, fprime, args, gtol, norm, …]) Minimize a function using the BFGS algorithm. The leapfrog method Next: The Crank-Nicolson method Up: FINITE DIFFERENCING IN (omega,x)-SPACE Previous: Explicit heat-flow equation A difficulty with the given program is that it doesn't work for all possible numerical values of. Its shortcomings, discussed in detail in the last lab, nameley its inaccuracy and its slowness, are just too great. Several additional tools written in C++, Python or bash scripting language are also included for convenience. Explicit method, implicit method, Crank-Nicolson method. The implicit part involves solving a tridiagonal system. An Introduction to the Finite Element Method (FEM) for Differential Equations tions and numerical methods are the only way to solve the differential equa-. As time progresses, the two “heat sources” (or sinks) at the sides are kept at constant low temperature. Week 3: Parabolic equation in 2D, Explicit & Crank-Nicolson method, Alternating direction Implicit method (ADI), Elliptic equations, Solution of Poisson equation with Example, Successive over Relaxation (SOR) method, Solution of Elliptic equation by using ADI method, Example. We will derive and use Numerov's method, which is a very elegant. I'm using Neumann conditions at the ends and it was advised that I take a reduced matrix and use that to find the interior points and then afterwards. Dari problem di atas, maka dapat di buat programnya. The convective and diffusive terms are temporally integrated with a second-order Adams-Bashforth scheme and a second-order Crank-Nicolson scheme, respectively. 2 are given. HELP!!!!!*****I've looked everywhere on website to solve my coursework problem, however our matlab teacher is a piece of crap, do nothing in class just reading meaningless handouts----- here is the question----- Write a Matlab script program (or function) to implement the Crank-Nicolson finite difference method based on the equations described in appendix. Convertible Bond Pricer. That said, variations of this method are certainly used (for example Crank-Nicolson and Runge-Kutta, so the time spent reading this chapter is not a total waste! Video Explanation Here is a video describing the forward Euler method:. The free boundary values we use in the algorithm of the compact finite difference method three can only obtain accuracy of 1 5000 when h = 0. 2 A numerical solution to the ODE in eq. coming boundary conditions may be analyzed, and all are treated as nat. development of a computer code capable of solving the neutron transport. The implicit method: Defining Vn j V(S j;t n) let us work systematically through equation (1) to obtain the. I have an extremely simple solver written for the Schroedinger equation but with imaginary time, which transforms it basically into the diffusion. Temporal Discretization (Backward Euler, Forward Euler, Crank-Nicholson, Runge-Kutta) 5. The methods of choice are upwind, downwind, centered, Lax-Friedrichs, Lax-Wendroff, and Crank-Nicolson. This method is also called Crank-Nicolson,'' especially when it is used in the context of partial differential equations. Math6911 S08, HM Zhu 6. Ali Bahce adlı kişinin profilinde 5 iş ilanı bulunuyor. pyplot as pl import numpy as np. SymPy is a Python library for symbolic mathematics, with the ambition to offer a full-featured computer algebra system (CAS). If you want to see some examples of Juypter Notebooks before continuing further, here are some examples (1) numerical intergration with the Trapezoid Rule, (2) numerical integration with the Crank Nicolson method, (3) symbolic calculations. Das Verfahren wurde Mitte des 20. COMMENT on programming languages: One free alternative to MATLAB is the mathematical and scientific support built around the Python scripting language, especially SciPy. Crank Nicolson method. Crank-Nicolson Finite Difference Method - A MATLAB Implementation. For the space discretization, either finite element methods or finite difference methods will work. Introduction to Numerical Methods Lecture notes for MATH 3311 Jeffrey R. Crank-Nicolson Implicit Scheme Tridiagonal Matrix Solver via Thomas Algorithm In the previous tutorial on Finite Difference Methods it was shown that the explicit method of numerically solving the heat equation lead to an extremely restrictive time step. Iterative Methods for Solving Linear Systems Iterative methods formally yield the solution x of a linear system after an infinite number of steps. Hi, just curious, what is the best model to price Convertibles? Say, if I use Crank Nicolson, how can I deal with the constraints at each. 1) can be written as. The Crank-Nicolson scheme cannot give growing amplitudes, but it may give oscillating amplitudes in time. An Introduction to the Finite Element Method (FEM) for Differential Equations tions and numerical methods are the only way to solve the differential equa-. BLSPRICEFDAM Black-Scholes put and call pricing for American Options using the Crank-Nicholson finite difference solution of Black-Scholes Partial differential equation. The numerical resolution is based on the Galerkin finite element discretization of the collective space and the Crank-Nicolson scheme for time. Screening system quantitative developer. via Finite Difference Methods (MatLab) Use the following IBM Put option figures to do so; Current IBM spot price (As of November 28th 2015): S0=£138. Crank-Nicolson time discretization for the Heat equation with Dirichlet boundary conditions. The Crank Nicholson method takes the left side of equation one and turns it into (2) Where i is the position of the temperature u, and n is the time index of the temperature u. Es ist ein implizites Verfahren 2. Daniel Duffy and the course is delivered entirely online by QuantNet. I'm writing a code in python to evolve the time-dependent Schrodinger equation using the Crank-Nicolson scheme. NEURON's computational engine employs special algorithms that achieve high efficiency by taking advantage of the structure of the equations that describe neuronal properties (for examples see (Hines, 1984), discussion of staggered Crank-Nicholson method in chapter 4 of The NEURON Book (Carnevale and Hines, 2006)). We focus on the case of a pde in one state variable plus time. The explicit method 100 Writing the FiniteDifferences class 101 Writing the FDExplicitEu class 103 The implicit method 105 Writing the FDImplicitEu class 106 The Crank-Nicolson method 108 Writing the FDCnEu class 110 Pricing exotic barrier options 111 A down-and-out option 111 Writing the FDCnDo class 112. • Explicit, implicit, Crank-Nicolson! • Accuracy, stability! • Various schemes! Multi-Dimensional Problems! • Alternating Direction Implicit (ADI)! • Approximate Factorization of Crank-Nicolson! Splitting! Outline! Solution Methods for Parabolic Equations! Computational Fluid Dynamics! Numerical Methods for! One-Dimensional Heat. Let Overstock. On the other hand the method is only first order (slow convergence). This is a signi cant increase above the Crank Nicolson method. This example demonstrates the solution of a particular nonlinear time-dependent fourth-order equation, known as the Cahn-Hilliard equation. Programming the finite difference method using Python Submitted by benk on Sun, 08/21/2011 - 14:41 Lately I found myself needing to solve the 1D spherical diffusion equation using the Python programming language. These choices are by no means the only approximations that. I am currently trying to solve a basic 2D heat equation with zero Neumann boundary conditions on a circle. Geophysical flow simulations have evolved sophisticated Implicit-Explicit time stepping methods (based on fast slow wave. Daniel Duffy and QuantNet. It does not look at the interval variable, nor is this plot taken into account in the counting done by the update() method if an interval variable was specified. Solution of Partial differential equations: classification, Laplace equation, Finite difference method. Peter Leitner & Stefan Hofmeister Crank-Nicolson using MPI Wednesday, May 10, 2017 13 / 13 Title Solution of the Time-dependent Schrödinger Equation using the Crank-Nicolson algorithm with MPI on a 2-D regular Cartesian grid - Seminar on High Performance Computing 2 Summer term 2017. The approach is based on the generalized Crank-Nicolson method supplemented with an Euler-MacLaurin expansion for the time-integrated nonhomogeneous term. Small stochastic perturbations and round off errors. Definitions. Python Tutorial: In class code development tutorial demonstrating tools needed for the course. Helmholtz decomposition; Least-squares ; Galerkin ; Leap-frog method (second order, explicit) Crank-Nicolson method (second order, implicit) Lax-Wendroff method (second order, explicit) Third-order explicit Taylor-Galerkin method; Fourth-order implicit Taylor-Galerkin method. Browse other questions tagged numerical-analysis finite-difference python boundary-conditions crank-nicolson or ask your own question. I am quite experienced in MATLAB and, therefore, the code implementation looks very close to possible implementation in MATLAB. Time stepping for nite horizon European vanillas 8. ) Usual methods of numerical integration (Runge-Kutta, Adams-Bashforth, etc. Initial Conditions may be specified as either constant or subdomain expressions for the flow variables, a potential flow solution, or use an existing solution. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 1 Getting started. Finite di erence method for heat equation Praveen. The equation is evaluated halfway between the old (m) and new (m+1) time levels. As others have pointed out, choosing between implicit and explicit methods depends solely on the type of problem you are solving. This tutorial presents MATLAB code that implements the Crank-Nicolson finite difference method for option pricing as discussed in the The Crank-Nicolson Finite Difference Method tutorial. Blog Announcing Stack Overflow's New CEO, Prashanth Chandrasekar!. Cahn-Hilliard equation¶. (Is the Crank-Nicolson method stable when r > 1 ?) Solution 4. 2014) CFD Python has a new home on GitHub. using backward Euler or Crank-Nicolson schemes; see . secondorder = 2. Its helpful to students of Computer Science, Electrical and Mechanical Engineering. In terms of stability and accuracy, Crank Nicolson is a very stable time evolution scheme as it is implicit. Linear multistep methods are used for the numerical solution of ordinary differential equations, in particular the initial value problem The Adams-Bashforth methods and Adams-Moulton methods are described on the Linear multistep method page. Using Python To Solve Comtional Physics Problems. Ordnung und numerisch stabil. Some background. implicit/explicit methods, Crank-Nicolson,. A class for pricing European options using the Crank-Nicolson method of finite differences; Pricing exotic barrier options. I was born in Drogheda, Ireland in 1952. Solving Fisher's nonlinear reaction-diffusion equation in python. En el campo del análisis numérico, el método de Crank-Nicolson es un método de diferencias finitas usado para la resolución numérica de ecuaciones en derivadas parciales, tales como la ecuación del calor. Here are the two scripts I used in class on February 10 demonstrating numerical convergence for Forward Euler and Crank-Nicolson time-stepping: and test_heat_cn_dirichlet. fadvance integrates the equation over the dt step by calling all the BREAKPOINT blocks of models at t+dt/2 twice with v+. Qualifications * Ph. Crank–Nicolson method In numerical analysis, the Crank–Nicolson method is a finite difference method used for numerically solving the heat equation and similar partial differential equations. Numerical Methods For Engineers. In this case the solution is exponentially decaying. LopesCardozo,E. via Finite Difference Methods (MatLab) Use the following IBM Put option figures to do so; Current IBM spot price (As of November 28th 2015): S0=£138. See for yourself why shoppers love our selection and award-winning customer service. Given (tn, yn), the forward Euler method (FE) computes yn+1 as. The time-stepping scheme relies on a semi-implicit Crank-Nicolson for the linear terms of the MHD equations and a Adams-Bashforth scheme for the non-linear terms and the Coriolis force. 1 Practical example of Pricing European Swaptions under. Thesis: Linear and Nonlinear Reaction-Diffusion Model Analysis of Brain Glioma Growth using Crank-Nicolson Scheme Based on Python. MS Visual Studio Community 2017. m Crank–Nicolson method for the heat equation. (5) Relax and hold steady: elliptic problems. fmin_bfgs (f, x0 [, fprime, args, gtol, norm, …]) Minimize a function using the BFGS algorithm. From our previous work we expect the scheme to be implicit. As with any finite-difference method, the Crank-Nicolson scheme involves discretizing the PDE in the time and underlying directions and approximating the derivative terms. Implicit vs. The implicit Crank-Nicholson method is significantly better in terms of stability than the Euler method for ordinary differential equations. The approach is based on the generalized Crank-Nicolson method supplemented with an Euler-MacLaurin expansion for the time-integrated nonhomogeneous term. Methodology in numerical. Here are the two scripts I used in class on February 10 demonstrating numerical convergence for Forward Euler and Crank-Nicolson time-stepping: and test_heat_cn_dirichlet. The particularity of MAC scheme is the. Let's denote the time at the nth time-step by tn and the computed solution at the nth time-step by yn, i. Research Experience for Undergraduates. Cette méthode utilise les différences finies pour approcher une solution du problème : elle est numériquement stable , et quadratique pour le temps. Trace Driven Simulation of Network traffic of a Non-Stationary two Server System März 2015 – September 2015; ARIMA, ARCH, GARCH for Financial time series Prediction. BLSPRICEFDAM Black-Scholes put and call pricing for American Options using the Crank-Nicholson finite difference solution of Black-Scholes Partial differential equation. Solves the one-dimensional wave equation. We requires to implement different solvers including forward Euler, backward Euler with Cholesky, backward Euler with SOR, Crank-Nicolson with SOR and Crank-Nicolson with projected SOR, then apply them to different finite schemes to compute option. Also, the u_glob function works in an element by element fashion such that coordinates at the boundaries between elements appear twice. A python script that displays an animation of an electron propagation and its interaction with arbitrary potential. It is similar to the (standard) Euler method, but differs in that it is an implicit method. Vetzal z, and G. Minimize a function using a nonlinear conjugate gradient algorithm. This is matlab code. It does not look at the interval variable, nor is this plot taken into account in the counting done by the update() method if an interval variable was specified. Cambridge, second edition, 204 References 19. The free boundary values we use in the algorithm of the compact finite difference method three can only obtain accuracy of 1 5000 when h = 0. And for that i have used the thomas algorithm in the subroutine. 5 Python s Visualization Tools 13. Week 3: Parabolic equation in 2D, Explicit & Crank-Nicolson method, Alternating direction Implicit method (ADI), Elliptic equations, Solution of Poisson equation with Example, Successive over Relaxation (SOR) method, Solution of Elliptic equation by using ADI method, Example. Since the WCS-FDTD method can be considered as an approximate factorization of the Crank—Nicolson scheme, the derived equations of CPML involve the calculation of auxiliary terms at two time steps which are different from that of conventional FDTD CPML method. The Crank–Nicolson method is often applied to diffusion problems. To clarify nomenclature, there is a physically important difference between convection and advection. \"\"\" This program solves the heat equation u_t = u_xx with dirichlet boundary condition u(0,t) = u(1,t) = 0 with the Initial Conditions u(x,0) = 10*sin( pi*x ) over the domain x = [0, 1] The program solves the heat equation using a finite difference method where we use a center difference method in space and Crank-Nicolson in time. Huang and Fehler's method is equivalent to applying the Crank–Nicholson scheme to the FD part in the FFD method of Ristow and Ruhl. Steepest Decent Animation. Heat Equation In 2d Pygimli Geophysical Inversion And. 02: As you can see, the accuracy of this numerical solution is much higher than before, but so is the amount of work needed!. pyplot as pl import numpy as. (2009) Smoothing of Crank–Nicolson scheme for the two-dimensional diffusion with an integral condition. Methodology in numerical. Finite horizon American contracts IIIPricing in incomplete markets 9. The explicit method 100 Writing the FiniteDifferences class 101 Writing the FDExplicitEu class 103 The implicit method 105 Writing the FDImplicitEu class 106 The Crank-Nicolson method 108 Writing the FDCnEu class 110 Pricing exotic barrier options 111 A down-and-out option 111 Writing the FDCnDo class 112. I'm writing a code in python to evolve the time-dependent Schrodinger equation using the Crank-Nicolson scheme. For example, in the integration of an homogeneous Dirichlet problem in a rectangle for the heat equation, the scheme is still unconditionally stable and second-order accurate. This function performs the split-step Fourier method to solve the 1D time-dependent Schrödinger equation for a given potential. Resolution of 1D heat equation with implicit Euler method. I tried to make the question as detailed as possible. Example code implementing the Crank-Nicolson method in MATLAB and used to price a simple option is provided. Crank-Nicolson for a European put was introduced before, to better master this technique, i share another sample code using Crank-Nicholson finite difference for American option. Here I used the Crank-Nicolson scheme to model a fault scarp with a vertical offset of 10 m. 3 Pricing European Swaptions under the Hull-White model. Siauw and A. Apply the Crank-Nicolson method with and obtain temperature distributions for. This example demonstrates the solution of a particular nonlinear time-dependent fourth-order equation, known as the Cahn-Hilliard equation. Finite Difierence Method 3 2 Finite difierence approximations A flnite difierence method typically involves the following steps: 1. 1) at the point The approximation formula for time derivative is given by and for spatial derivative (15. The Baruch MFE Program provides a teaching assistant to each student,. Applications in Computational Finance: Applications and test cases (Black Scholes pricing and Greeks), Monte Carlo methods, finite difference methods (Euler, Crank-Nicolson), lattice methods, exact methods (Barone-Adesi-Whaley, bonds, swaps, swaptions). BTCS), and Crank-Nicolson (CN) methods. There is an online module available for a course on numerical methods covering finite difference methods . Numerical Methods in Quantum Mechanics Corso di Laurea Magistrale in Fisica Interateneo Trieste { Udine Anno accademico 2018/2019 Paolo Giannozzi University of Udine Contains software and material written by Furio Ercolessi1 and Stefano de Gironcoli2 1Formerly at University of Udine 2SISSA - Trieste Last modi ed March 28, 2019. Python/Matplotlib Code # Crank-Nicolson method to solve the heat equation. 01 , for the option prices obtained by. I have managed to code up the method but my solution blows up. 1 Modi ed Euler Method Numerical solution of Initial Value Problem: dY dt = f(t;Y) ,Y(t n+1) = Y(t n) + Z t n+1 tn f(t;Y(t))dt: Approximate integral using the trapezium rule:. Examples in Matlab and Python. Math6911 S08, HM Zhu 6. The leapfrog method Next: The Crank-Nicolson method Up: FINITE DIFFERENCING IN (omega,x)-SPACE Previous: Explicit heat-flow equation A difficulty with the given program is that it doesn't work for all possible numerical values of. uni-dortmund. • American option pricing: Utilized MATLAB to implement finite difference methods (explicit, implicit, Crank-Nicolson schemes) for pricing American options under regime switching framework, and compared the results with the Least Squares Monte Carlo simulation approach. I did this in 1D and in 2D by writing four classes in Python using scipy. The Crank Nicholson method takes the left side of equation one and turns it into (2) Where i is the position of the temperature u, and n is the time index of the temperature u. Option numerically using Crank-Nicholson and Secant methods • Calculated. This program solves dUdT - k * d2UdX2 = F(X,T) over the interval [A,B] with boundary conditions. Helmholtz decomposition; Least-squares ; Galerkin ; Leap-frog method (second order, explicit) Crank-Nicolson method (second order, implicit) Lax-Wendroff method (second order, explicit) Third-order explicit Taylor-Galerkin method; Fourth-order implicit Taylor-Galerkin method. Minimize a function using a nonlinear conjugate gradient algorithm. However, a successful theory of. View June Sun’s profile on LinkedIn, the world's largest professional community. Simulation methods in nance. value print ‘c ‘,. Read detailed description of Quantitative Methods for Finance Professional Certificate by NYIF with reviews, dates, location and price with the help of Coursalytics. Solving the Diffusion Equation Explicitly This post is part of a series of Finite Difference Method Articles. 5, r max = 0. m to solve the semi-discretized heat equation with ode15s and compare it with the Crank-Nicolson method for different time step-sizes. The conservation of the norm by the Crank-Nicholson method is remarkable (variation always less than10−11%). Its shortcomings, discussed in detail in the last lab, nameley its inaccuracy and its slowness, are just too great. I would also like to add that this is the first time that I have done numerical computing like this and I don't have a lot of experience with PDE's and finite. In its most simple formulation, the spectral methods require periodic boundary conditions in both dimensions. (6) gives N simultaneous equations for the N unknowns in terms of the given boundary values. This project mainly focuses on -Method for the initial boundary heat equation. Many finance textbooks have a few chapters on the finite-difference method to various levels of sophistication [Wil06, AP10, FR08, Cla11, Hir13] and some [Sey09] extend coverage to finite-element methods. In the model problem, the scheme is second order accurate. Numerical Methods for Partial Differential Equations: Finite Difference and Finite Volume Methods focuses on two popular deterministic methods for solving partial differential equations (PDEs), namely finite difference and finite volume methods. ##2D-Heat-Equation. Stochastic volatility models 10. The natura. Programming the finite difference method using Python Submitted by benk on Sun, 08/21/2011 - 14:41 Lately I found myself needing to solve the 1D spherical diffusion equation using the Python programming language. Here, nre f is the reference index. In 1D case crank nicolson is used for better convergence and results. Das Verfahren wurde Mitte des 20. The two processes are coupled together. Computational Methods for (Quantitative) Finance This University course focused on numerical solutions for some Quantitative Finance problems. STABILITY ANALYSIS OF THE CRANK-NICOLSON-LEAP-FROG METHOD WITH THE ROBERT-ASSELIN-WILLIAMS TIME FILTER NICHOLAS HURL , WILLIAM LAYTON†, YONG LI‡, AND CATALIN TRENCHEA§ Abstract. Crank Nicolson Algorithm ( Implicit Method ) BTCS ( Backward time, centered space ) method for heat equation ( This is stable for any choice of time steps, however it is first-order accurate in time. Notice for instance the abscence of ';'. Several additional tools written in C++, Python or bash scripting language are also included for convenience. • Used explicit method, implied method and Crank-Nicolson method respectively to solve partial differential equations such as Black-Scholes equation to calculate option price Risk Management Python Project. Consider the heat equation ut=κuxx with boundary conditions of u(x,0)=0u(0,t)=100u(l,t)=0 Numerical analysis by pyton can be done with import numpy as np import matplotlib. The explicit method for the heat-equation involved a forward difference term for the time derivative and a centred second derivative for the second space derivative:. In 2D, a NxM array is needed where N is the number of x grid points, M the number of y grid. The field is the domain of interest and most often represents a physical structure. sparse import diags def Crank_Nicolson(dy,ny,dt,nt,k,T,ntout): Tout =. In sections 3, 4, and 5, we. evolve another half time step on y direction with x direction variance attached. The CN method is a central-time, central-space (CTCS) finite-difference method (FDM) for numerically solving partial differential equations (PDE). Computation of American Option, Manjula R - 103010510, May 2015 6. Crank-Nicolson method, 132, 147 D. Crank-Nicolson for a European put was introduced before, to better master this technique, i share another sample code using Crank-Nicholson finite difference for American option. Numerical Methods For Engineers. and advanced programming in Python using Numpy . This method separates the value of American options into two parts. Gandhi School Ancol Elementary, Junior, and Senior High School Science 1 - 12. All Courses; First Year. 336 Spring 2006 Numerical Methods for Partial Differential Equations Prof. clc clear MYU=1; A=1; N=100; M=100; LX=1; LY=1; DX=LX/M; DY=LY/N; %-----INITILIZATION--MATRIX-----t=1; for i=1:M;. As you will see, this method is appropriate only for mildly stiff systems. They both result in Tridiagonal Symmetric Toeplitz matrices. In this report, I give some details for implement-ing the Finite Element Method (FEM) via Matlab and Python with FEniCs. A down-and-out option; A class for pricing down-and-out-options using the Crank-Nicolson method of finite. My degrees are B. In order to derive the Crank-Nicolson (CN) scheme for the Black Scholes equation, it is incumbent that we elucidate the implicit and explicit finite difference methods of which CN is an equally weighted average. Your code isn't an implementation of Crank-Nicolson method, but a implementation of method of lines. The approach is based on the generalized Crank-Nicolson method supplemented with an Euler-MacLaurin expansion for the time-integrated nonhomogeneous term. We will derive and use Numerov's method, which is a very elegant. Crank-Nicolson Method. The three main numerical ODE solution methods (LMM, Runge-Kutta methods, and Taylor methods) all have FE as their simplest case, but then extend in different directions in order to achieve higher orders of accuracy and/or better stability properties. The above formula in eqn. 1 Statistics, Time Series, omputation Finance, erivative Pricing, Algorithmic Trading Review in R, Python Ron Wu Last update: 4/25/16 Table of Contents. To solve this problem implicitly, I transform the problem into the spectral domain, solve the 2D diffusion equation with a Crank-Nicholson method, and compare the results to explicit finite difference and explicit spectral methods. Hence, this method is not stable for inviscid fluid dynamics. By applying methods based solely on the PDE, we gain an increase in accuracy on the order of 10 7. I didn't know how to deal with the potential so I looked around and found a way from this question, which I have verified from a couple other sources. This function performs the Crank-Nicolson scheme for 1D and 2D problems to solve the inital value problem for the heat equation. The finite volume method is a method for representing and evaluating partial differential equations in the form of algebraic equations [LeVeque, 2002; Toro, 1999]. AMath 586 / ATM 581 Homework #4 using the Crank-Nicolson method. Crank-Nicolson Implicit Scheme Tridiagonal Matrix Solver via Thomas Algorithm In the previous tutorial on Finite Difference Methods it was shown that the explicit method of numerically solving the heat equation lead to an extremely restrictive time step. with finite difference methods exclusively. This scheme is called the Crank-Nicolson. 4 Time-dependent Schrödinger equation 13. value # print outputs print ‘b ‘, R. An Introduction to the Finite Element Method (FEM) for Differential Equations tions and numerical methods are the only way to solve the differential equa-. It is similar to the (standard) Euler method, but differs in that it is an implicit method. Jørgensen,D. Finite Difference Heat Equation using NumPy. Finite Difference Approach to Option Pricing 20 February 1998 CS522 Lab Note 1. The equation is evaluated halfway between the old (m) and new (m+1) time levels. その大まかな位置づけを. A package for solving parabolic differential equations, in particular the Schrodinger equation and the Gross-Pitaevskii equation using the Crank Nicolson method. I'm writing a code in python to evolve the time-dependent Schrodinger equation using the Crank-Nicolson scheme. from mpl_toolkits. For example, we can discretize the time derivative term ∂u using backward Euler or Crank-Nicolson ∂t schemes; see . Crank Nicolson Algorithm-Numerical Analysis-MATLAB Code, Exercises for Mathematical Methods for Numerical Analysis and Optimization. At each step they require the computation of the residualofthesystem. worksheet Attach:newton. Crank-Nicolson time discretization for the Heat equation with Dirichlet boundary conditions. The equation is stable if Real(λ) ≤ 0. In numerical analysis, the Crank-Nicolson method is a finite difference method used for numerically solving the heat equation and similar partial differential equations. Various techniques are in use to deal with the time dependency (e. The Crank–Nicolson method is often applied to diffusion problems. An Introduction to the Finite Element Method (FEM) for Differential Equations tions and numerical methods are the only way to solve the differential equa-. Finite Difference Heat Equation using NumPy. Barone-Adesi & Whaley. Dari problem di atas, maka dapat di buat programnya. FD1D_HEAT_EXPLICIT is a FORTRAN90 library which solves the time-dependent 1D heat equation, using the finite difference method in space, and an explicit version of the method of lines to handle integration in time. coming boundary conditions may be analyzed, and all are treated as nat. Divergence of orbits and decay of correlations. This demo is implemented in a single Python file, demo_cahn-hilliard."
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https://mirror.las.iastate.edu/CRAN/web/packages/univariateML/vignettes/overview.html | [
"When dealing with univariate data you want to do one or more of\n\n• Find a good model for the data.\n• Estimate parameters for your candidate models.\n\nThe unvariateML package has a fast and reliable functions to help you with these tasks. The core of the package are more than 20 functions for fast and thoroughly tested calculation of maximum likelihood estimates for univariate models.\n\n• Compare the fit of your candidate models with AIC or BIC.\n• Look at QQ plots or PP plots of your data.\n• Plot the data together with density estimates.\n• Compute confidence intervals using parametric bootstrap.\n\nThis vignette shows you how to use the tools of univariateML to do exploratory data analysis.\n\n# Mortality in Ancient Egypt\n\nThe dataset egypt contains contains the age at death of 141 Roman era Egyptian mummies. Our first task is to find a univariate model that fits this data.\n\nlibrary(\"univariateML\")\n\n## # A tibble: 6 x 2\n## age sex\n## <dbl> <chr>\n## 1 1.5 male\n## 2 1.83 male\n## 3 2 male\n## 4 2 male\n## 5 3 male\n## 6 3 male\n\nhist(egypt$age, main = \"Mortality in Ancient Egypt\", freq = FALSE)",
null,
"## Comparing Many Models with AIC The AIC is a handy and easy to use model selection tool, as it only depends on the log-likelihood and number of parameters of the models. The \\code{AIC} generic in R can take multiple models, and the lower the \\code{AIC} the better. Since all the data is positive we will only try densities support on the positive half-line. AIC(mlbetapr(egypt$age),\nmlexp(egypt$age), mlinvgamma(egypt$age),\nmlgamma(egypt$age), mllnorm(egypt$age),\nmlrayleigh(egypt$age), mlinvgauss(egypt$age),\nmlweibull(egypt$age), mlinvweibull(egypt$age),\nmllgamma(egypt$age)) ## df AIC ## mlbetapr(egypt$age) 2 1312.464\n## mlexp(egypt$age) 1 1249.553 ## mlinvgamma(egypt$age) 2 1322.949\n## mlgamma(egypt$age) 2 1234.772 ## mllnorm(egypt$age) 2 1263.874\n## mlrayleigh(egypt$age) 1 1260.217 ## mlinvgauss(egypt$age) 2 1287.124\n## mlweibull(egypt$age) 2 1230.229 ## mlinvweibull(egypt$age) 2 1319.120\n## mllgamma(egypt$age) 2 1314.187 The Weibull and Gamma models stand out with an AIC far below the other candidate models. To see the parameter estimates of mlweibull(egypt$age) just print it:\n\nmlweibull(egypt$age) ## Maximum likelihood estimates for the Weibull model ## shape scale ## 1.404 33.564 mlweibull(egypt$age) is a univariateML object. For more details about it call summary:\n\nsummary(mlweibull(egypt$age)) ## ## Maximum likelihood for the Weibull model ## ## Call: mlweibull(x = egypt$age)\n##\n## Estimates:\n## shape scale\n## 1.404158 33.563564\n##\n\nmodel_select(egypt$age, models = c(\"gamma\", \"weibull\")) ## Maximum likelihood estimates for the Weibull model ## shape scale ## 1.404 33.564 ## Quantile-quantile Plots Now we will investigate how the two models differ with quantile-quantile plots, or Q-Q plots for short. qqmlplot(egypt$age, mlweibull, datax = TRUE, main = \"QQ Plot for Ancient Egypt\")\n# Can also use qqmlplot(mlweibull(egypt$age), datax = TRUE) directly. qqmlpoints(egypt$age, mlgamma, datax = TRUE, col = \"red\")\nqqmlline(egypt$age, mlweibull, datax = TRUE) qqmlline(egypt$age, mlgamma, datax = TRUE, col = \"red\")",
null,
"The Q-Q plot shows that neither Weibull nor Gamma fits the data very well.\n\nIf you prefer P-P plots to Q-Q plots take a look at ?ppplotml instead.\n\n## Plot Densities\n\nUse the plot, lines and points generics to plot the densities.\n\nhist(egypt$age, main = \"Mortality in Ancient Egypt\", freq = FALSE) lines(mlweibull(egypt$age), lwd = 2, lty = 2, ylim = c(0, 0.025))\nlines(mlgamma(egypt$age), lwd = 2, col = \"red\") rug(egypt$age)",
null,
"## Confidence Intervals with Parametric Bootstrap\n\nNow we want to get an idea about the uncertainties of our model parameters. Do to this we can do a parametric bootstrap to calculate confidence intervals using either bootstrapml or confint. While bootstrapml allows you to calculate any functional of the parameters and manipulate them afterwards, confint is restricted to the main parameters of the model.\n\n# Calculate two-sided 95% confidence intervals for the two Gumbel parameters.\nbootstrapml(mlweibull(egypt$age)) # same as confint(mlweibull(egypt$age))\n\n## 2.5% 97.5%\n## shape 1.248255 1.606138\n## scale 29.463496 38.198904\n\nbootstrapml(mlgamma(egypt$age)) ## 2.5% 97.5% ## shape 1.3311650 2.04828111 ## rate 0.0421817 0.06914691 These confidence intervals are not directly comparable. That is, the scale parameter in the Weibull model is not directly comparable to the rate parameter in the gamma model. So let us take a look at a a parameter with a familiar interpretation, namely the mean. The mean of the Weibull distribution with parameters shape and scale is scale*gamma(1 + 1/shape). On the other hand, the mean of the Gamma distribution with parameters shape and rate is shape/rate. The probs argument can be used to modify the limits of confidence interval. Now we will calculate two 90% confidence intervals for the mean. # Calculate two-sided 90% confidence intervals for the mean of a Weibull. bootstrapml(mlweibull(egypt$age),\nmap = function(x) x*gamma(1 + 1/x),\nprobs = c(0.05, 0.95))\n\n## 5% 95%\n## 27.69290 33.55712\n\n# Calculate two-sided 90% confidence intervals for the mean of a Gamma.\nbootstrapml(mlgamma(egypt$age), map = function(x) x/x, probs = c(0.05, 0.95)) ## 5% 95% ## 27.39642 34.17312 We are be interested in the quantiles of the underlying distribution, for instance the median: # Calculate two-sided 90% confidence intervals for the two Gumbel parameters. bootstrapml(mlweibull(egypt$age),\nmap = function(x) qweibull(0.5, x, x),\nprobs = c(0.05, 0.95))\n\n## 5% 95%\n## 23.07959 28.96785\n\nbootstrapml(mlgamma(egypt$age), map = function(x) qgamma(0.5, x, x), probs = c(0.05, 0.95)) ## 5% 95% ## 22.03612 27.49912 We can also plot the bootstrap samples. hist(bootstrapml(mlweibull(egypt$age),\nmap = function(x) x*gamma(1 + 1/x),\nreducer = identity),\nmain = \"Bootstrap Samples of the Mean\",\nxlab = \"x\",\nfreq = FALSE)",
null,
"## Density, CDF, quantiles and random variate generation\n\nThe functions dml, pml, qml and rml can be used to calculate densities, cumulative probabilities, quantiles, and generate random variables. Here are $$10$$ random observations from the most likely distribution of Egyptian mortalities given the Weibull model.\n\nset.seed(313)\nrml(10, mlweibull(egypt$age)) ## 25.90552 59.64456 13.36882 44.29378 12.22563 17.66144 54.57633 22.86824 ## 11.48328 19.94814 Compare the empirical distribution of the random variates to the true cumulative probability. set.seed(313) obj = mlweibull(egypt$age)\nq = seq(0, max(egypt\\$age), length.out = 100)\nplot(q, pml(q, obj), type = \"l\", ylab = \"Cumulative Probability\")\nr = rml(100, obj)\nlines(ecdf(r))",
null,
""
] | [
null,
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null,
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",
null,
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null,
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QDgQdoFAB6kXQDgQdoFAB6kXQDgQdoFAB6kXQDgQdoFAB6kXQDgQdoFAB6kXQDgQdoFAB6kXQDgQdoFAB6kXUAGViHfSRwLpF1ABnYvP5zOPdxJu4AcrMO+ST0KSLuAHASszhsNpF1ADlavT+YlbBKGbZ99eXsyU4y0C5DHLlU4nbvuk3YBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4kHYBgAdpFwB4/A4Iryj6SSBnFAAAAABJRU5ErkJggg==",
null,
"data:image/png;base64,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",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.6343085,"math_prob":0.98076683,"size":6363,"snap":"2022-27-2022-33","text_gpt3_token_len":2139,"char_repetition_ratio":0.16574933,"word_repetition_ratio":0.09808102,"special_character_ratio":0.32406098,"punctuation_ratio":0.14976229,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9976665,"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\":\"2022-07-04T08:37:11Z\",\"WARC-Record-ID\":\"<urn:uuid:18ced7be-4adf-44e8-93d6-4ddb62c52354>\",\"Content-Length\":\"50708\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3107075-f78a-4837-baec-c8c83fcf1fdd>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9ce3b23-1a50-407e-be44-44b4fba97aad>\",\"WARC-IP-Address\":\"129.186.90.88\",\"WARC-Target-URI\":\"https://mirror.las.iastate.edu/CRAN/web/packages/univariateML/vignettes/overview.html\",\"WARC-Payload-Digest\":\"sha1:DL24YDBBPBYHYG4PNO5IJ3E7K7IDVMEI\",\"WARC-Block-Digest\":\"sha1:QJ5LMHDR5OEYGU53H3JKCKKKQLTJLWDD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104364750.74_warc_CC-MAIN-20220704080332-20220704110332-00779.warc.gz\"}"} |
https://dsp.stackexchange.com/questions/46311/is-system-ht-ut-ut-1-lti-or-not | [
"# Is system $h(t) = u(t)-u(t-1)$ LTI or not?\n\n$$y(t) = \\int_{0}^{\\infty} x(\\tau)h(t-\\tau) \\ \\mathrm{d}\\tau$$ where $h(t) = u(t)-u(t-1)$.\n\nI have some problem with understanding whether system LTI or LTV just looking the impulse response of the system. Is there any short-cut to identify the system time-variance looking the impulse function?\n\n• LTV systems do not have one impulse response, they have infinitely many. – Rodrigo de Azevedo Jan 12 '18 at 10:38\n\n$h(t)$ can be rewritten as $\\mathrm{rect}(t-0.5)$. So basically your output signal can be calculated as:\n\n$$y(t) = \\int_{0}^{\\infty} x(\\tau)\\mathrm{rect}(t-0.5-\\tau) \\ \\mathrm{d}\\tau$$\n\nNote that if $t<0$, then $t-0.5-\\tau<-0.5$, as $\\tau\\in(0,\\infty$). Ergo,\n\n$$|t-0.5-\\tau|>|0.5|$$\n\nThis means that the $\\mathrm{rect}()$ function will be zero if $t$ takes negative values. Therefore, we get that $y(t)=0 \\ \\forall t<0$.\n\nFor $t>0$, all we have to do is convolve the rectangular window with $x(t)u(t)$, as the integral begins at $0$ and not $-\\infty$.\n\nIf we call the window $w(t)$, we can express $y(t)$ as:\n\n$$y(t)= \\left\\{ \\begin{array}{ll} [x(t)u(t)]*w(t) & \\mbox{if } t > 0 \\\\ 0 & \\mbox{if } t < 0 \\end{array} \\right.$$\n\nNow, could you determine whether the system is time-invariant or not?\n\nHint: try appling a delay to the input signal, and see if its output is equal to a delayed version of the output itself.\n\n• My comment was only valid for a discrete-time system; either I misread or the notation was changed, so now it's continuous-time and the claim $u(t)-u(t-1)=\\delta(t)$ is not correct anymore (if $t$ is a continuous variable). – Matt L. Jan 11 '18 at 15:24\n• @MattL. Thanks for that, I messed up big time. I've already corrected it, I believe now the answer makes sense. – Tendero Jan 11 '18 at 15:51\n\nIf a system is described by an impulse response $h(t)$, then by definition it is an LTI system. Because impulse response of the form $h(t) = \\mathcal{T} \\{ \\delta(t) \\}$ only exists for LTI systems.\n\n• text book says it is time varying i am also thinking like you – Nail Tosun Jan 11 '18 at 12:04\n• @NailTosun What textbook? – Tendero Jan 11 '18 at 12:23\n• I edited the question – Nail Tosun Jan 11 '18 at 13:46"
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.75219536,"math_prob":0.9993661,"size":909,"snap":"2019-43-2019-47","text_gpt3_token_len":328,"char_repetition_ratio":0.100552484,"word_repetition_ratio":0.0,"special_character_ratio":0.37733772,"punctuation_ratio":0.1273585,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99986875,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T16:52:55Z\",\"WARC-Record-ID\":\"<urn:uuid:95706b64-628e-463e-b7e6-e90d23532f4e>\",\"Content-Length\":\"147274\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d1093f28-bb1c-4de8-97be-c2d416d4020e>\",\"WARC-Concurrent-To\":\"<urn:uuid:03b66e0c-aa5a-4bcc-8765-54db779524cc>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://dsp.stackexchange.com/questions/46311/is-system-ht-ut-ut-1-lti-or-not\",\"WARC-Payload-Digest\":\"sha1:E2562UMQWXE5GNO3OCWF6YNUTAG2EXMC\",\"WARC-Block-Digest\":\"sha1:4LWUH25DALJYBDLALI6AQNLUAUZQ27LI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670156.86_warc_CC-MAIN-20191119144618-20191119172618-00457.warc.gz\"}"} |
https://ai.stackexchange.com/questions/9226/steps-for-final-logistic-regression-modal | [
"# Steps for final Logistic Regression Modal\n\nI am new for machine learning and I am tried to understand basic steps to get final modal of Logistic Regression.\n\nI know Logistic Regression is supervisory learning technique. Therefore we want to give training data to training the modal. According to my understanding, I can take steps to get the final modal as follows.\n\nStep 1 - Make an algorithm. For Logistic Regression is y = sigmoid(W x + B) function. Get zero or some value for W and B.\n\nStep 2 - Give sample known training data and find W and B values. x(1), x(2)...x(m) inputs and get y(1), y(2)...y(m) outputs. Gradient Descent use for find W and B that minimizes the cost function.\n\nStep 3 - Then apply W and B values which I found.y^ = sigmoid(W x + B). And then again apply sample known training data to get the predicting of y^.\n\nStep 4 - Get the final modal to test unknown data.\n\nare these steps right? Please give me basic fundamental steps to understand supervisory learning technique. I would like to know un-supervisory learning technique steps also.\n\n• can you clarify your step 3? I do not understand what it is trying to say. – skim Dec 7 '18 at 14:46\n\nIn general these steps are correct, but are few clarifications would be good. For supervised learning, you can:\n\n1. pick model to use\n2. split data into training and test set\n3. train model, or determine weights by optimization of our error function (inputting in training data)\n4. use trained model to predict on test data\n5. utilize metrics of interest to evaluate our model (e.g. MSE for continuous outcomes, AUC for binary)\n\nThis however is a very elementary pipeline to train a supervised machine learning model. There are many other steps taken to ensure our model is as robust as possible (like cross validation, feature selection, hyperparameter optimization, model selection...). I would strongly recommend learning about these other steps and seeing how they integrate with the above pipeline.\n\nFor unsupervised learning, the process depends a lot on the task, but in general it involves optimizing some objective function using the data provided. This is usually done using all the data in one go (not split into training and test) unlike supervised learning. This can be seen more clearly with examples, take methods like PCA and k clustering; the steps taken to apply these methods very much depend on the method used.\n\nGoing further, I would recommend Andrew Ng's coursera course on machine learning. It is very accessible and a great introduction to this field, covering both unsupervised/supervised learning."
] | [
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] | {"ft_lang_label":"__label__en","ft_lang_prob":0.89289314,"math_prob":0.87932813,"size":1016,"snap":"2019-51-2020-05","text_gpt3_token_len":235,"char_repetition_ratio":0.11462451,"word_repetition_ratio":0.0,"special_character_ratio":0.23917323,"punctuation_ratio":0.12206573,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9786394,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-29T09:42:08Z\",\"WARC-Record-ID\":\"<urn:uuid:f2b10639-1fc0-495e-92bc-383645c08c68>\",\"Content-Length\":\"132262\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d2864a96-8d53-475a-8486-caa725affc3a>\",\"WARC-Concurrent-To\":\"<urn:uuid:b55c6cbd-72e9-4b0d-8987-17d1757659aa>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://ai.stackexchange.com/questions/9226/steps-for-final-logistic-regression-modal\",\"WARC-Payload-Digest\":\"sha1:2JQ4KA3YMTM5Z7524LB2RDEBSTQAK2VF\",\"WARC-Block-Digest\":\"sha1:2Q47XEBAE34JSG5NTJOI5YJD3UKMJMWB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251789055.93_warc_CC-MAIN-20200129071944-20200129101944-00148.warc.gz\"}"} |
http://www.157110.com/tdxgs/2138.html | [
"# 熊市抄底指标\n\n• ## 软件介绍\n\n• 网友评论\n• 下载地址",
null,
"",
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"### 公式源码\n\nX:=MA(AMOUNT,5)/MA(V,5)/100;\n\nVAR3:=(X-MA(AMOUNT,125)/MA(V,125)/100)/X;\n\nVAR4:=MA((LLV(L,45)-C)/(HHV(H,45)-LLV(L,45))*100,3);\n\nVAR5:=C/MA(C,60)-1;\n\nDRAWTEXT_FIX(1,0.00,0.00,0,'小白公式提供,更多公式登陆 157110.COM 下载'),COLORRED;\n\n• 热门标签"
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"http://d.157110.com/d/file/20190410/image/1554858437105522.png_oss.png",
null,
"http://d.157110.com/d/file/20190410/image/1554858446117635.png_oss.png",
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] | {"ft_lang_label":"__label__zh","ft_lang_prob":0.7267415,"math_prob":0.99820656,"size":928,"snap":"2020-34-2020-40","text_gpt3_token_len":762,"char_repetition_ratio":0.1417749,"word_repetition_ratio":0.0,"special_character_ratio":0.4234914,"punctuation_ratio":0.2581818,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9796111,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-29T13:36:13Z\",\"WARC-Record-ID\":\"<urn:uuid:2b3a4d85-252e-4542-8285-2f55f08d108c>\",\"Content-Length\":\"24533\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:01415ea6-06cb-478c-a39e-323e7f9d584c>\",\"WARC-Concurrent-To\":\"<urn:uuid:32ccc32e-b83c-429f-ae84-8d86693a6d5d>\",\"WARC-IP-Address\":\"47.97.44.108\",\"WARC-Target-URI\":\"http://www.157110.com/tdxgs/2138.html\",\"WARC-Payload-Digest\":\"sha1:PZDVQ3O3KM3FDF7W3KKA4CWE4SOEYBN3\",\"WARC-Block-Digest\":\"sha1:EISBMMA7E6BKZN4XBVVT2Q4QQIBHBJF3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401643509.96_warc_CC-MAIN-20200929123413-20200929153413-00707.warc.gz\"}"} |
http://www.everyscience.com/Chemistry/Physical/Molecular_Rotation/c.1036.php | [
"# Introduction to Rigid Rotors\n\nIn a discussion of rotational energy levels, a very important property is the moment of inertia, I, of the molecule about any particular rotational axis. The moment of inertia of a molecule is generated by taking the mass of each atom in the molecule, multiplying it by the square of its perpendicular distance from the rotational axis, and summing these values together. i.e.\n\nNote that the rotational axis must be one that passes through the centre of mass of the molecule.\n\nThe moment of inertia of a molecule is a measure of how difficult it is to rotationally accelerate the molecule – the larger the moment of inertia, the smaller the increase in angular momentum for a given applied torque.\n\nThe moment of inertia depends upon the masses of the atoms present and the molecular geometry, indicating that rotational spectroscopy will be able to give information about bond lengths and angles.\n\nIn general, the rotational properties of any molecule can be expressed using the moments of inertia about three mutually perpendicular axes. Conventionally, these axes are labelled Ia, Ib and Ic, choosing the axes in such a way that I³ I³ I. Note that for linear molecules, the moment of inertia around the molecular axis is zero, as all the atoms lie on the axis of rotation so are at zero distance from it.\n\nFor our purposes we shall make the supposition that molecules are rigid rotors, bodies that do not distort under the stress of rotation. Rigid rotors can be classified into four types:\n\nSpherical rotors have three equal moments of inertia.\nSymmetric rotors have two equal moments of inertia.\nLinear rotors have one moment of inertia (that around the molecular axis) equal to zero.\nAsymmetric rotors have three different moments of inertia.\n\nThe rotational energy levels of a rigid rotor may be obtained by construction and solution of the appropriate Schrodinger equation, but there is a much simpler approach that may be used. Classical mechanics gives expressions for the energy of a rotating body in terms of the angular momentum, and we may obtain the analogous quantum mechanical expressions by substitution of the quantum expressions for angular momentum.\n\nThe classical expression for a body rotating about a given axis with angular velocity ω is:\n\n(Note the similarity to the classical expression for linear kinetic energy, E = mv2/2 – the moment of inertia is the rotational equivalent of the mass, and the angular velocity replaces the linear velocity.) A body free to rotate about three mutually perpendicular axes has an energy given by:\n\nwhere the letters a, b and c distinguish the three rotational axes. Since the classical angular momentum is given by J = Iω, it follows that:\n\nNow, in the discussion of the quantum mechanics of rotation in three dimensions, we stated that the magnitude of the angular momentum was given by a quantum number l, which was restricted to positive integral values (and zero). We may use this result generally in our discussion of rigid rotors."
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https://www.intechopen.com/chapters/57521 | [
"Open access peer-reviewed chapter\n\n# State-Space Models for Binomial Time Series with Excess Zeros\n\nBy Fan Tang and Joseph E. Cavanaugh\n\nSubmitted: April 30th 2017Reviewed: September 28th 2017Published: December 20th 2017\n\nDOI: 10.5772/intechopen.71336\n\n## Abstract\n\nCount time series with excess zeros are frequently encountered in practice. In characterizing a time series of counts with excess zeros, two types of models are commonplace: models that assume a Poisson mixture distribution, and models that assume a binomial mixture distribution. Extensive work has been published dealing with modeling frameworks based on Poisson-type approaches, yet little has concentrated on binomial-type methods. To handle such data, we propose two general classes of time series models: a class of observation-driven ZIB (ODZIB) models, and a class of parameter-driven ZIB (PDZIB) models. The ODZIB model is formulated in the partial likelihood framework, which facilitates model fitting using standard statistical software for ZIB regression models. The PDZIB model is conveniently formulated in the state-space framework. For parameter estimation, we devise a Monte Carlo Expectation Maximization (MCEM) algorithm, with particle filtering and particle smoothing methods employed to approximate the intractable conditional expectations in the E-step of the algorithm. We investigate the efficacy of the proposed methodology in a simulation study, which compares the performance of the proposed ZIB models to their counterpart zero-inflated Poisson (ZIP) models in characterizing zero-inflated count time series. We also present a practical application pertaining to disease coding.\n\n### Keywords\n\n• autocorrelation\n• count time series\n• observation-driven models\n• parameter-driven-models\n• particle methods\n• zero-inflation\n\n## 1. Introduction\n\nCount time series with excess zeros are commonly encountered in a variety of research fields. In principle, both zero-inflation and autocorrelation may be present in such series. Failing to adequately accommodate temporal dynamics and a high frequency of zeros can lead to incorrect inferential conclusions. Developing a general modeling framework that accounts for these characteristics poses a daunting challenge.\n\nIn characterizing data comprised of counts with excess zeroes, two types of models are commonplace: a model that assumes a Poisson mixture distribution, and a model that assumes a binomial mixture distribution. A considerable literature exists for regression models based on the zero-inflated Poisson (ZIP) distribution to deal with count data that are independently distributed . Many researchers have extended the classical ZIP model to analyze repeated measures data by incorporating independent random effects, as these can account for within-subject correlation and between-subject heterogeneity [2, 3]. To deal with count time series with excess zeros, some researchers have proposed parameter-driven ZIP models that accommodate the temporal dynamics by incorporating correlated random effects, which can be represented by a latent autoregressive process [4, 5]. However, for data arising from a binomial mixture distribution, a survey of the literature for analogous frameworks reflects an absence of work dealing with binomial time series with excess zeros. To handle such data, we propose two general classes of models: a class of observation-driven ZIB (ODZIB) models, and a class of parameter-driven ZIB (PDZIB) models. The inspiration for the two proposed modeling frameworks arises from the work of Hall , Yau et al. , and Yang et al. [5, 7].\n\nDepending on how the temporal correlation is conceptualized, count time series models can be classified as either observation-driven or parameter-driven . For the former, serial correlation is characterized by specifying that the conditional mean of the current response depends explicitly on its past values . For the latter, such correlation is characterized through an unobservable underlying process . In this chapter, we employ the partial likelihood framework to formulate the ODZIB model, as this largely simplifies parameter estimation with negligible loss of information. The ODZIB model can be viewed as an extension of the observation-driven binomial model . Such a model is often fit using standard statistical software available for classical ZIB regression models. For the PDZIB model, we employ a state-space approach, as this framework allows for the investigation of the underlying latent processes that govern the temporal correlation and zero inflation. Due to the non-Gaussian distribution of the count response, and the non-linear nature of modeling its conditional mean, traditional state-space methods using the Kalman filter and the Kalman smoother are not available for parameter estimation. We thereby adopt a Monte Carlo Expectation Maximization (MCEM) algorithm based on the particle filter and the particle smoother .\n\nThe remainder of the chapter is organized as follows. In Section 2, we briefly introduce a class of observation-driven models for a zero-inflated count time series that arises from a binomial mixture. Section 3 proposes a class of parameter-driven models in the state-space framework, and presents the MCEM algorithm devised to fit such models. A comprehensive simulation study is provided in Section 4. In Section 5, we illustrate the proposed methodology through a practical application. Section 6 concludes with a brief discussion.\n\n## 2. Observation-driven ZIB models\n\n### 2.1. ZIB models\n\nA popular approach for modeling independent zero-inflated binomial data is the ZIB model proposed by Hall . This model assumes that data are generated from a mixture distribution, comprised of a binomial distribution and a degenerate distribution at zero. For response variable Y, let yidenote the observation for subject i, i = 1, 2, …, n. The probability mass function for the ZIB model is defined as follows:\n\nfyiπiωi=ωi+1ωi1πini,ifyi=0,1ωiniyiπiyi1πiniyi,ifyi>0.E1\n\nHere, ωiis the zero-inflation parameter, and πiis the intensity parameter representing the probability of success, both modeled via logit link functions:\n\nlogitωi=xi1Τγ,E2\nlogitπi=xi2Τβ.E3\n\nIn the preceding, xi1 and xi2 are sets of explanatory variables for the corresponding vectors of regression coefficients γand β. The Expectation Maximization (EM) algorithm or the Newton-Raphson method can be used to obtain the parameter estimates.\n\n### 2.2. Observation-driven ZIB models\n\nIn this section, we introduce an autoregressive model for binomial time series with excess zeros based on an observation-driven approach. We retain the same model structure as that introduced in Section 2.1 to account for the binomial mixture, yet we employ lagged responses as covariates to resolve the temporal correlation. The proposed model can be viewed as an extension of the binomial time series model presented by Kedem and Fokianos .\n\nLet ytdenote the binomial count response. Define the information set\n\nFt1=σyt1yt2xtE4\n\nso as to represent all that is known to the observer at time tabout the response and any relevant covariate processes. Thus, the vector xtrepresents a collection of past and possibly present time-dependent covariates that are observed at time t − 1. In the present setting, xtmay be viewed as either fixed or random. Conditioning on the information Ft1, the response is assumed to follow a ZIB distribution with probability mass function defined as follows:\n\nftytFt1πtωt=ωt+1ωt1πtnt,ifyt=0,1ωtntytπtyt1πtntyt,ifyt>0.E5\n\nSimilarly, ωtand πtrepresent the zero-inflation parameter and the intensity parameter, respectively. Both parameters are modeled via logit link functions. Specifically, we assume that\n\nlogitωt=x1,tΤγ,E6\nlogitπt=x2,tΤβ+j=1pφjytj,E7\n\nwhere x1, tand x2, tare sets of time-dependent explanatory variables for the corresponding vectors of regression coefficients γ and β, and φ = [φ1, …, φp]Τis a vector of autoregressive coefficients corresponding to the past responses [yt − 1, …, yt − p]Τ. For simplicity, we treat the zero-inflation parameter ωtas a constant that does not vary over time. In the observation-driven ZIB model, serial correlation is accommodated by introducing lagged values of the response to the linear predictor.\n\nThe partial data likelihood of the observed series is\n\nPLθ=t=1nftytFt1,E8\n\nwhere θ = [β, φ, γ]Τis the vector of unknown parameters. The partial likelihood does not require the derivation of the joint distribution of the response and the covariates, and is largely simplified relative to the full likelihood. This approach facilitates conditional inference for a fairly large class of transitional processes where the response depends on its past values.\n\nThe log-likelihood for the observation-driven ZIB model is\n\nlogPLθ=t=1nlogωtIyt=0+1ωtntytπtyt1πtntyt.E9\n\nThe vector θ^obtained by maximizing the partial likelihood is called the maximum partial likelihood estimator (MPLE).\n\nSimilar to Section 2.1, we can apply the EM algorithm or the Newton–Raphson method to obtain the MPLE. This estimation process can be conveniently conducted in practice using standard software tools available for fitting classical ZIB models. In SAS, we can use the finite mixture models (FMM) procedure to fit the observation-driven ZIB model, while we can use function gamlss in the package generalized additive models for location scale and shape (GAMLSS) for model fitting in R. Hypothesis testing for θis carried out through the partial likelihood method. The common tests are based on Wald statistics, score statistics, and partial likelihood ratio statistics. All of these tests are conducted based on the framework for classical maximum likelihood inference.\n\n## 3. Parameter-driven ZIB models\n\n### 3.1. Model formulation\n\nAn alternative approach to describe binomial time series with excess zeros is based on parameter-driven ZIB models. This class of models can be viewed as an analogue of the parameter-driven ZIP models presented by Yang et al. .\n\nTo account for temporal dynamics in the series, we introduce a latent stationary autoregressive process {zt} of order p(AR(p)):\n\nzt=i=1pφizti+εt.E10\n\nHere, εtis a Gaussian white noise process with a mean of 0 and a variance of σ2. Additionally, φiexplains how the past state zt − irelates to the current state zt.\n\nLet ytbe the observed count at time t. Given the current state zt, the positive count response ytis assumed to follow a ZIB distribution with a probability mass function defined as\n\nftytztπtωt=ωt+1ωt1πtnt,ifyt=0,1ωtntytπtyt1πtntyt,ifyt>0.E11\n\nSimilar to the previous model parameterizations, ωtand πtrepresent the zero-inflation parameter and the intensity parameter, respectively. Both parameters are modeled via logit link functions and could be time-varying. To relate the intensity parameter πtto the latent component zt, we use the model\n\nlogitπt=xtΤβ+zt,E12\n\nwhere xtis a set of explanatory variables observed at time t, and βis the corresponding vector of regression coefficients. In the present setting, xtis assumed fixed. For simplicity, we treat the zero-inflation parameter ωtas a constant that does not vary over time.\n\nFor the parameter-driven ZIB model, the conditional mean and variance of the response variable ytare given by\n\nEYtzt=1ωtntπt,E13\nVarYtzt=1ωtntπt1πt1ωtnt.E14\n\nObviously, the presence of zero-inflation (ωt > 0) not only explains the excess zeros in the series, but also introduces overdispersion. Additionally, the correlated random effects ztcontribute to the extra variance.\n\nWe can write the parameter-driven ZIB model in the following hierarchical form:\n\nstst1NpΦst1Σ,E15\nutBernoulliω,E16\nytst,utBinomialnt1utπt,E17\n\nwhere st = [zt, …, zt − p + 1]Τis a p-dimensional state vector with ztbeing its first element, utis an unobservable membership indicator that determines whether the response comes from a degenerate distribution or an ordinary binomial distribution, Φis an unknown transition matrix, and Σis the covariance matrix of the state noise process st. The process stis initiated with a normal vector s0 that has mean μ0 and covariance matrix Σ0. Diffuse priors are often assigned to s0 in practice. Given the two unobserved latent processes stand ut, we can conceptualize a sequential update of the response variable yt.\n\nIn Eq. (15), Φand Σare p × pmatrices defined as follows:\n\nΦ=φ1φ2φp1φp100001000010,Σ=σ2000000000000000.E18\n\nThe transition matrix Φgoverns the generation of the state vector stfrom the past state st − 1 for time points t = 1, …, n. Note that the covariance matrix Σin Eq. (18) is not positive definite. This is both legitimate and common in the state-space modeling approach.\n\n### 3.2. Parameter estimation via MCEM algorithm\n\n#### 3.2.1. Model fitting\n\nTo fit the parameter-driven ZIB model, in principle, one would first obtain the marginal likelihood of the observed data y1, …, ynby integrating out unobserved components. However, because of the presence of correlated random effects and the non-Gaussian nature of the response, these integrals are not analytically tractable. Therefore, approximations or numerical solutions for the maximum likelihood estimates (MLEs) are necessary. Instead of obtaining the MLEs based on the marginal likelihood, we propose an EM algorithm , which relies on the complete-data likelihood to estimate the parameters.\n\nLet y1 : t = [y1, y2, …, yt]Τdenote the vector of observed data from time point 1 through t. In a similar fashion, let s0 : t = [s0, s1, …, st]Τand u1 : t = [u1, u2, …, ut]Τdenote the vectors of two latent processes, respectively, over the same time frame. Let θ = [ω, βΤ, φΤ, σ]Τdenote the vector of unknown parameters.\n\nTo develop an EM algorithm for parameter estimation of the mixture model, Eqs. (15)(17), we begin by formulating the complete-data likelihood; i.e., the joint density of s0 : n, u1 : n, and y1 : n. The two latent processes s0 : nand u1 : nare considered missing. Based on the state-space representation, the complete-data likelihood may be orthogonally decomposed as follows:\n\nLcθ=fs0:nu1:ny1:n=fs0:nu1:nfy1:ns0:nu1:n=fs0:nfu1:nfy1:ns0:nu1:n=fs0t=1nfstst1t=1nfutt=1nfytstut.E19\n\nHere, the initial state vector s0 is assumed to be normally distributed with mean vector μ0 and covariance matrix Σ0. In implementing the algorithm, we set μ0 = 0 and Σ0 = Ip, as the effect of the starting values of μ0 and Σ0on the estimated parameters θis negligible.\n\nUp to an additive constant, the complete-data log-likelihood is given by\n\nlcθ=n2logσ212σ2t=1nztφΤst12+t=1nutlogω+1utlog1ω+t=1n1utytxtΤβntlog1+expxtΤβ+zt.E20\n\nThe complete-data log-likelihood can be described as the sum of three functionally independent parameter forms, such that lc(θ) = l(φ, σ| st) + l(ω| ut) + l(β| st, ut), resulting in ease of the maximization in the M-step for each set of parameters.\n\nWith the implementation of the EM algorithm, we need to compute the conditional expectation of lc(θ) given the observed data y1 : n. Deriving an analytical form for the conditional expectation is not feasible due to the nonlinear forms in the latent variables and the response, as well as the non-Gaussian distributions of the response and the latent indicators. There are many numerical methods available to approximate the conditional expectation, such as the Markov chain Monte Carlo (MCMC) algorithm , the MCEM algorithm [7, 25], the penalized quasi-likelihood (PQL) method , and integrated nested Laplace approximations (INLA) . Following Yang et al. , we develop an MCEM algorithm to approximate the conditional expectation.\n\nTo simplify the notation, we let Atj, btj, ctj, dtj, etj, and ftjdenote the conditional expectations of st1st1Τ, ztst − 1, zt2, ut, 1utlog1+expxtΤβ+zt, and 1utexpxtΤβ+zt/1+expxtΤβ+ztevaluated at θ(j), respectively. In the Monte Carlo E-step of the algorithm, we first compute the conditional expectation of lc(θ):\n\nQθθj=Elcθy1:nθj=n2logσ212σ2t=1nctj2φΤbtj+φΤAtjφ+t=1ndtjlogω+1dtjlog1ω+t=1n1dtjytxtΤβntetj,E21\n\nwhere particle filtering and smoothing techniques are used to approximate the conditional expectations. The details of the particle methods for the parameter-driven ZIB model are presented in Section 3.3.\n\nThe following partial derivatives are applied to maximize Q(θ| θ(j)) in the M-step:\n\nQω=1ωt=1ndtj11ωt=1n1dtj,E22\nQφ=1σ2t=1nbtjAtjφ,E23\nQσ=nσ+1σ3t=1nctj2φΤbtj+φΤAtjφ,E24\nQβ=Elcθy1:nθjβ=Elcθβy1:nθj=Et=1n1utytnt1utexpxtΤβ+zt1+expxtΤβ+ztxty1:nθj=t=1n1dtjytntftjxt.E25\n\nAt the jthiteration, we obtain the following closed-form solutions for ω(j + 1), φ(j + 1), and σ(j + 1):\n\nωj+1=1nt=1ndtj,E26\nφj+1=t=1nAtj1t=1nbtj,E27\n\nσj+1=1nt=1natjt=1nbtjΤt=1nAtj1t=1nbtj.E28\n\nIn addition, we can easily compute β(j + 1) through iterative algorithms such as Broyden-Fletcher-Goldfarb-Shanno (BFGS). Once we acquire the particle smoothers from the smoothing step, we can obtain the MCEM estimates by plugging in the sample means of the functions of particle smoothers for the conditional expectations.\n\nTo offset the slow convergence and to reduce the computational cost of the EM algorithm, starting with good initial parameters is essential. For the proposed parameter-driven ZIB model, we suggest using the estimates of the parameters from a classical ZIB model or from the observation-driven ZIB model discussed in Section 2.2.\n\n#### 3.2.2. Standard errors\n\nStandard errors of the parameter estimators can be obtained either by using the inverse of the observed information to approximate the variance/covariance matrix, or by employing a collection of replicated bootstrapped parameter estimates. Given the computational cost of the MCEM algorithm, we pursue the first approach by applying Louis's formula to compute the observed information matrix Io(θ). Based on the missing information principle, we have\n\nIoθ=IcθImθ,E29\n\nwhere Ic(θ) and Im(θ) are defined as follows:\n\nIcθ=E2lcθθΤy1:n,E30\nImθ=ElcθlcθΤy1:nElcθy1:nElcθΤy1:n.E31\n\nThe first-order derivatives of lc(θ) are given by\n\nlcω=1ωt=1nut11ωt=1n1ut,E32\nlcβ=t=1n1utytntexpxtΤβ+zt1+expxtΤβ+ztxt,E33\nlcφ=1σ2t=1nztφΤst1st1,E34\nlcσ=nσ+1σ3t=1nztφΤst12.E35\n\nThe second-order derivatives of lc(θ) are given by\n\n2lcωω=1ω2t=1nut11ω2t=1n1ut,E36\n2lcββΤ=t=1n1utntexpxtΤβ+zt1+expxtΤβ+zt2xtxtΤ,E37\n2lcφφΤ=1σ2t=1nst1st1Τ,E38\n2lcσσ=nσ23σ4t=1nztφΤst12,E39\n2lcφσ=2σ3t=1nztφΤst1st1.E40\n\nAgain, particle filtering and smoothing techniques are used to approximate the conditional expectations in Ic(θ) and Im(θ).\n\nIn principle, the variance/covariance matrix can be approximated by taking the inverse of the observed information matrix. However, the computation of the inverse is often problematic. As indicated by Kim and Stoffer , the observed information matrix is not guaranteed to be numerically positive definite. To address this problem, we slightly modify Louis’s formula by introducing a slack variable ξ, such that\n\nIoθ=Icθ1ξImθ,E41\n\nwhere ξis a non-negative variable ranging from 0 to 1. In practice, we can iteratively increase this value until the observed information matrix can be inverted.\n\n### 3.3. Particle methods\n\nParticle filtering and particle smoothing belong to the class of sequential Monte Carlo (SMC) methods . These particle methods can be viewed as the non-linear and non-Gaussian extensions of the popular Kalman filtering and smoothing algorithms for traditional state-space models. Rather than yielding a single estimate for the filter or the smoother, as computed through conventional Kalman filtering and smoothing, particle methods provide a set of particles with associated weights to approximate the conditional densities governing the filters and smoothers. Implemented via sequential importance sampling (SIS), in the E-step of the EM algorithm, particle methods provide approximate solutions to the intractable integrals corresponding to the conditional expectations of functions of the latent components given the observed data. However, sample degeneracy is a typical problem for SIS methods. In particular, degeneracy occurs when particles have small weights or even negative weights, rendering their contributions to the conditional density negligible. Resampling (e.g., bootstrapping) offers a recourse for eliminating particles with negligible effects. Kim provides an elegant treatment of particle filtering and smoothing for state-space models.\n\n#### Particle filtering\n\nFor the parameter-driven ZIB model, we implement particle filtering by first generating s00iNpμ0Σ0. Then for t = 1, …, n:\n\n(F.1) Generate stt1iNpΦst1t1iΣand utt1iBernoulliω.\n\n(F.2) Compute the filtering weights\n\nqtt1intyt1utt1iπtt1iyt11utt1iπtt1intyt,E42\n\nwhere logitπtt1i=xtΤβ+ztt1iand ztt1iis the first element of stt1i.\n\n(F.3) Generate sttiuttiby resampling stt1iutt1iwith replacement based on the preceding filtering weights.\n\nAs a byproduct of the particle filtering, the observed-data log-likelihood can be approximated by\n\nt=1nlog1Ni=1Nqtt1i,E43\n\nwhere Nis the number of particles in the filtering step.\n\n#### Particle smoothing\n\nNext, we employ the particle smoothing algorithm proposed by Godsill et al. to obtain the conditional expectations of the functions of the latent variables given the complete set of observed data. In this step, we first choose snnrunnr=snniunniwith probability qnn1i. Then for t = n − 1, …, 1:\n\n(S.1) Calculate the smoothing weights\n\nqtniqtt1iexp12σ2zt+1niφΤstti2ωut+1ni1ω1ut+1ni.E44\n\n(S.2) Choose stnrutnr=sttiuttiwith probability qtni.\n\nWe obtain independent realizations by repeating the preceding process for r = 1, …, R.\n\n## 4. Simulation studies\n\nIn this section, we investigate through simulation two salient issues pertaining to the proposed modeling frameworks. In the first part, we explore the convergence of the MCEM algorithm through simulated examples, and investigate the finite sample distributional properties of the parameter estimators through a comprehensive simulation study. In the second part, we present a simulation study to compare the performance of the proposed ZIB models to their counterpart ZIP models in characterizing zero-inflated count time series.\n\n### 4.1. Evaluation of the MCEM algorithm\n\nWe consider time series data simulated from four different parameter-driven models: ZIB + AR(2), binomial + AR(2), ZIB + AR(1), and binomial + AR(1). The sample size is set to 300 and the number of cases ntfor each time point is set to 30. All of the models feature the following linear predictor:\n\nlogitπt=β0+β1x1,t+zt,E45\n\nwhere x1, tis a covariate series generated from a standard uniform distribution. The true parameters for the most complicated model ZIB + AR(2) are as follows:\n\nω=0.3,β0=2,β1=3,φ1=0.8,φ2=0.6,andσ=0.5.E46\n\nFor the rest of the models considered, the corresponding parameters are set to 0 if no such a form is included. Autoregressive (AR) coefficients are chosen to assure stationarity of the series. In fitting the models, the number of particle filters (N) is set to 500 and the number of particle smoothers (R) is set to 300. We stop the MCEM algorithm after 300 iterations. Table 1 presents the parameter estimates for the simulated data corresponding to the four parameter-driven models.\n\nωβ0β1φ1φ2σ\nTrue0.3002.000−3.0000.800−0.6000.500\nBinomial + AR(1)1.984−2.9680.8000.540\nZIB + AR(1)0.2832.124−2.9300.7810.563\nBinomial + AR(2)1.989−3.0120.852−0.6200.499\nZIB + AR(2)0.2931.992−2.8720.831−0.5760.506\n\n### Table 1.\n\nTrue and estimated parameters for the simulated examples.\n\nFigure 1 shows the trace plots of the log-likelihood for the four fitted parameter-driven models. Note that the log-likelihood of the MCEM algorithm is not strictly increasing at each iteration due to the introduction of Monte Carlo errors. However, the log-likelihood stabilizes after a few dozen iterations with slight fluctuations around the maximal value. Figure 2 shows the trace plots for the parameter estimates from the most complex fitted model, ZIB + AR(2). The plots indicate that the parameter estimates converge to the MLEs quickly with negligible fluctuations. The trace plots of the parameter estimates for the other three models exhibit similar patterns (results not shown). In practice, we recommend always checking the trace plots of the estimates to assess convergence of the MCEM algorithm.",
null,
"Figure 1.Trace plots of the log-likelihood for fitted parameter-driven models based on simulated data.",
null,
"Figure 2.Trace plots of the estimated parameters for the fitted ZIB + AR(2) model.\n\nWe next investigate the finite sample distributional properties of the parameter estimators from the MCEM algorithm. We consider the same parameter-driven models presented in the preceding simulated example. For each model structure, 500 replications are generated based on sample sizes of 200 and 500. We employ the proposed MCEM algorithm to fit models based on these replications, and record the subsequent parameter estimates and their standard errors. As the MECM algorithm is computationally expensive, we set the number of particles for both filters and smoothers to 200, and the stopping iteration for the MCEM algorithm at 100. In Tables 23, we provide the simulation results based on the most complex model, ZIB + AR(2).\n\nTrueMeanMedianESDASE\nω0.3000.2990.2950.0320.032\nβ02.0001.9991.9920.1390.166\nβ1−3.000−2.992−2.9800.2350.224\nφ10.8000.7430.7570.1200.165\nφ2−0.600−0.563−0.5720.1040.145\nσ0.5000.5040.5080.0630.098\n\n### Table 2.\n\nSummary statistics for replicated parameter estimates from fitted ZIB + AR(2) models with sample size 200.\n\nTrueMeanMedianESDASE\nω0.3000.3000.2990.0200.020\nβ02.0002.0022.0020.0810.104\nβ1−3.000−3.006−3.0070.1380.139\nφ10.8000.7540.7540.0760.095\nφ2−0.600−0.566−0.5730.0670.086\nσ0.5000.5090.5100.0390.057\n\n### Table 3.\n\nSummary statistics for replicated parameter estimates from fitted ZIB + AR(2) models with sample size 500.\n\nIn general, the mean and median of the estimates converge to the true parameters, with a minor degree of negative bias associated with the estimation of the AR coefficients. The empirical standard deviations (ESDs) are reasonably close to the average asymptotic standard errors (ASEs). Therefore, the standard errors calculated by Louis’s method prove to be sufficient. As the sample size increases from 200 to 500, the bias for the estimation of the AR coefficients attenuates, and the standard errors tend to diminish. The two behaviors indicate that weak convergence holds. The results for the other three parameter-driven models are analogous to those presented in Tables 23. Tables 49 show the simulation results for the binomial + AR(2) model, ZIB + AR(1) model, and binomial + AR(1) model, respectively.\n\nTrueMeanMedianESDASE\nβ02.0001.9982.0030.1080.167\nβ1−3.000−3.002−3.0100.1790.174\nφ10.8000.7830.7830.0870.101\nφ2−0.600−0.593−0.5960.0800.094\nσ0.5000.4960.4940.0520.062\n\n### Table 4.\n\nSummary statistics for replicated parameter estimates from fitted binomial + AR(2) models with sample size 200.\n\nTrueMeanMedianESDASE\nβ02.0001.9951.9890.0700.101\nβ1−3.000−2.994−2.9950.1130.108\nφ10.8000.7910.7910.0570.063\nφ2−0.600−0.593−0.5950.0530.059\nσ0.5000.4980.4960.0320.038\n\n### Table 5.\n\nSummary statistics for replicated parameter estimates from fitted binomial + AR(2) models with sample size 500.\n\nTrueMeanMedianESDASE\nω0.3000.2990.2990.0310.032\nβ02.0001.9711.9710.2080.251\nβ1−3.000−2.982−2.9690.1990.210\nφ10.8000.7630.7700.0730.067\nσ0.5000.5000.5020.0560.063\n\n### Table 6.\n\nSummary statistics for replicated parameter estimates from fitted ZIB + AR(1) models with sample size 200.\n\nTrueMeanMedianESDASE\nω0.3000.2990.2990.0200.021\nβ02.0001.9841.9890.1350.168\nβ1−3.000−2.992−2.9910.1330.132\nφ10.8000.7810.7850.0410.040\nσ0.5000.5000.4990.0350.040\n\n### Table 7.\n\nSummary statistics for replicated parameter estimates from fitted ZIB + AR(1) models with sample size 500.\n\nTrueMeanMedianESDASE\nβ02.0002.0062.0240.1920.233\nβ1−3.000−2.987−2.9880.1650.167\nφ10.8000.7820.7880.0540.056\nσ0.5000.4970.4960.0510.052\n\n### Table 8.\n\nSummary statistics for replicated parameter estimates from fitted binomial + AR(1) models with sample size 200.\n\nTrueMeanMedianESDASE\nβ02.0001.9971.9960.1250.168\nβ1−3.000−2.997−2.9940.1060.106\nφ10.8000.7870.7890.0350.035\nσ0.5000.4990.4990.0300.033\n\n### Table 9.\n\nSummary statistics for replicated parameter estimates from fitted binomial + AR(1) models with sample size 500.\n\nDifference in AICLevel of empirical support for model with larger AIC\n0−2Substantial\n4−7Considerably less\n>10Essentially none\n\n### Table 10.\n\nGuidelines for assessing AIC differences.\n\nModelAICωβ0β1φ1φ2σ\nPDZIB(1)922.980.248−3.349−0.249−0.2230.430\n(0.034)(0.051)(0.120)(0.160)(0.044)\nPDZIP(1)923.310.248−3.389−0.242−0.2410.410\n(0.034)(0.051)(0.116)(0.166)(0.043)\nODZIB(1)1039.800.341−3.184−0.319−0.007\n(0.061)(0.024)(0.086)(0.002)\nODZIP(1)1030.040.341−3.224−0.309−0.007\n(0.061)(0.046)(0.084)(0.004)\nPDZIB(2)922.980.248−3.359−0.237−0.1200.2640.426\n(0.034)(0.054)(0.126)(0.166)(0.153)(0.046)\nPDZIP(2)924.090.248−3.395−0.230−0.1190.2630.402\n(0.034)(0.052)(0.118)(0.178)(0.158)(0.045)\nODZIB(2)1038.110.341−3.250−0.275−0.0080.007\n(0.061)(0.033)(0.088)(0.002)(0.002)\nODZIP(2)1028.490.341−3.288−0.266−0.0070.007\n(0.061)(0.058)(0.087)(0.004)(0.004)\n\n### Table 11.\n\nModel fitting results for eight different zero-inflated models.\n\nThe normality of the parameter estimators is assessed by Q-Q plots based on the sets of replicated estimates (figures not shown). For the most complex ZIB + AR(2) model, approximate normality holds for the finite sample distribution of the parameter estimators, with slightly non-normal tail behavior (thick or thin) evident for the estimated AR coefficients. As the sample size is increased from 200 to 500, this non-normal behavior is attenuated. Similar patterns are observed for the other three parameter-driven models.\n\n### 4.2. Model comparison\n\nAs previously mentioned, based on a Poisson mixture distribution, extensive methodology has been published to deal with count time series with excess zeros. In addition, the Poisson distribution provides an accurate approximation to the binomial distribution when the sample size is large and the success probability is small. Therefore, one may question whether Poisson-type models are sufficient for approximating binomial-type models when data are generated from a binomial mixture distribution. In this section, we try to address this question through a simulation study.\n\nTwo different types of ZIB models are proposed in this work: the parameter-driven ZIB model, and the observation-driven ZIB model. To evaluate the propriety of the binomial-type models, we consider two corresponding Poisson-type counterparts: the parameter-driven ZIP model, and the observation-driven ZIP model. We assess the performance of the four models under two scenarios: first, where data are generated from the parameter-driven ZIB model, and second, where data are generated from the observation-driven ZIB model.\n\nTo denote the parameter-driven ZIB/ZIP model with an AR(p) latent process, we use PDZIB(p)/PDZIP(p). Similarly, we use ODZIB(p)/ODZIP(p) to denote the observation-driven ZIB/ZIP model with plagged responses employed as covariates.\n\nIn the first scenario, data are generated from a PDZIB(2) model having the same form as that provided in Section 4.1. To reduce the computational burden associated with fitting the models, 100 replicated series of length 200 are generated. We fit four different zero-inflated models to each of the series. For the two parameter-driven models, we specify a latent autoregressive process of order two, and employ the MECM algorithm to fit the models. For the two observation-driven models, we incorporate the lagged responses yt − 1 and yt − 2 to account for the temporal correlation, and employ the Newton–Raphson algorithm to fit the models.\n\nIn the second scenario, data are generated from an ODZIB(2) model featuring the following structures:\n\nlogitπt=β0+β1x1,t+φ1yt1+φ2yt2,andlogitω=γ0.E47\n\nHere, x1, tis a covariate series generated from a standard uniform distribution, and φ1 and φ2 are the autoregressive coefficients for the lagged responses yt − 1 and yt − 2, respectively. The values of the true parameters are the same as those for the parameter-driven model.\n\nAgain, we generate 100 replications of length 200 based on the preceding model. The same four zero-inflated models are fit to each of the replications. The Akaike information criterion (AIC) is used to guide the selection of an optimal model in both scenarios. To evaluate the magnitude of the absolute difference in AIC values, Burnham and Anderson provide the following guidelines (Table 10).\n\nThus, a difference in AIC values of two or more is considered meaningful, and a difference of 10 or more is considered pronounced.\n\nFigure 3 illustrates the performance of the four zero-inflated models, in terms of AIC differences, when data are generated from a PDZIB(2) model. The PDZIB(2) model serves as the reference for model comparison. Each point represents the difference in the AIC value between the target model and the reference model. As evident from the figure, the PDZIB(2) model markedly outperforms the other three models for all 100 replications, with AIC differences over 50. Although vastly inferior to the PDZIB(2) model, the PDZIP(2) model performs better than the two observation-driven models. The ODZIB(2) performs the worst among the four models considered. Parameter-driven models clearly exhibit a substantial advantage over observation-driven models when the underlying data arise via a parameter-driven approach.",
null,
"Figure 3.AIC differences of zero-inflated fitted models relative to parameter-driven ZIB fitted models.\n\nFigure 4 shows the performance of the four zero-inflated models, in terms of AIC differences, when data are generated from an ODZIB(2) model. Similarly, the ODZIB(2) model serves as the reference. The ODZIB(2) model easily performs the best among all four models for all 100 replications, reflecting a substantial improvement in model fit over the other three models based on AIC differences (>20). Compared to the two parameter-driven models, the ODZIP(2) model accommodates the data much more appropriately. Between the two parameter-driven models, the PDZIB(2) model is substantially favored over the PDZIP(2) model. Thus, observation-driven models markedly outperform parameter-driven models when the underlying data arise via an observation-driven approach.",
null,
"Figure 4.AIC differences of zero-inflated fitted models relative to observation-driven ZIB fitted models.\n\nWe close this section with a brief discussion of issues germane to model selection. These issues are relevant not only in evaluating the results of the preceding simulations, but also in facilitating the choice of a model in practice.\n\nFirst, one may question which class of models should be considered when coping with binomial time series data with excess zeros. In the simulation sets, the fitted parameter-driven models markedly outperform the fitted observation-driven models when data are generated via a parameter-driven approach. Although parameter-driven models are computationally expensive to fit, observation-driven models do not appear to provide an adequate characterization of the data in such settings. Additionally, unlike observation-driven models, parameter-driven models provide a description of the underlying latent processes that govern the temporal correlation and zero inflation. Observation-driven models, in contrast, outperform parameter-driven models when the underlying data are generated via an observation-driven approach. In general, the selection of the class of models depends on the conceptualization of the model structure and the perceived value of recovering and investigating the underlying latent processes. However, in the context of zero-inflated count time series, since an understanding of the phenomenon that gives rise to the data will rarely inform the practitioner as to whether the parameter-driven or observation-driven conceptualization is more appropriate, we recommend the use of AIC or an alternate likelihood-based selection criterion in choosing between these two model classes.\n\nSecond, one may question which distribution should be used when dealing with count time series with excess zeros. The Poisson-type model with an offset is often considered an appropriate approximating model for a binomial-type model when the sample size is large and the success probability is low. However, in the presence of zero inflation, our simulation results indicate the necessity of using binomial-type models over their Poisson counterparts when the underlying distribution is actually a binomial mixture. In practice, if the dynamics of the phenomenon that gives rise to the data do not inform the underlying data generating distribution, we again recommend the use of AIC or another likelihood-based criterion in choosing an appropriate distribution.\n\n## 5. Application\n\nIn this section, to illustrate our proposed methodology, we consider an application pertaining to the diagnosis coding of a severe disease, Kaposi’s sarcoma (KS). The application concerns the assessment of a particular level change for a primary KS diagnosis. The data used are extracted from the Healthcare Cost and Utilization Project (HCUP) database. We identify all hospitalizations during the period from January 1998 through December 2011 during which a primary or secondary diagnosis of KS is received. For case ascertainment, we use the International Classification of Diseases, 9thRevision, Clinical Modification (ICD-9-CM), code 176. We then aggregate all cases of KS by month to produce a national sample of the monthly KS hospitalizations. The data consist of monthly counts of both primary and overall KS hospitalizations from January 1998 to December 2011. The sample size for both KS series is 168. Figure 5 shows both the primary KS count time series and the overall KS count time series. In the latter, the overall KS count serves as the denominator for the binomial-type model and the offset for the Poisson-type model.",
null,
"Figure 5.Monthly time series plots of primary KS hospitalizations (top panel) and overall KS hospitalizations (bottom panel) from January 1998 to December 2011.\n\nA coding change was implemented in early 2008, during which many hospitals may have modified the coding convention by switching the primary code to secondary, as this modification may lead to an increase in hospital reimbursements. During the study period, a large number of zero counts is observed and data among adjacent points seem to be highly correlated. Since the primary KS count series exhibits a relatively large degree of zero-inflation (appropriately 25% of the values are zero), we apply our proposed ZIB models to characterize the data.\n\nOur analysis focuses on two objectives. First, we aim to model the dynamic pattern of the primary KS series; in particular, we are interested in determining the appropriate order of the autoregressive process embedded in the series, and evaluate whether there is a significant level change at January 2008. Second, we aim to compare the performance of our proposed ODZIB(p) and PDZIB(p) models to their counterpart ODZIP(p) and PDZIP(p) models.\n\nFor potential autocorrelation structures, we let pbe either 1 or 2. As a result, we consider eight candidate models in total. Each of the models features an indicator to represent an intervention in January 2008, which allows us to test whether there is significant level change at this time period.\n\nSpecifically, for the two PDZIB(p) models, we employ the following linear predictor:\n\nlogitπt=β0+β1xt+zt,E48\nzt=i=1pφizti+εt,E49\n\nwhere tis a discrete time index, and xt = I(t > 2008) is a dummy variable indicating whether the index tis greater than the predefined change point (January 2008). Thus, β1 reflects the level change in KS counts due to the coding practice, and the φidenote the coefficients for the autoregressive process.\n\nFor the two ODZIB(p) models, we employ the following linear predictor:\n\nlogitπt=β0+β1xt+i=1pφiyti,E50\n\nwhere β1 and φireflect parameters analogous to those defined for the parameter-driven setting.\n\nIn addition, we consider four comparable Poisson-type models based on the work by Yang et al. [5, 7]. For the two PDZIP(p) models, we employ the linear predictor\n\nlogμt=lognt+β0+β1xt+zt,E51\nzt=i=1pφizti+εt.E52\n\nFor the two ODZIP(p) models, we employ the linear predictor\n\nlogμt=lognt+β0+β1xt+i=1pφiyti.E53\n\nHere, ntserves as an offset variable representing the overall number of KS diagnoses. AIC is used to guide the selection of the optimal model.\n\nTable 11 features results for the eight fitted candidate models. The parameter estimates along with their standard errors are presented. All eight models indicate a significant level change for the primary KS series after the introduction of the potential coding change practice (β1 < 0). Among the first four models, which feature an autocorrelation structure of order one, parameter-driven models are deemed superior to observation-driven models, with AIC differences over 100. The PDZIB(1) model is slightly favored over the PDZIP(1) in terms of the AIC value. We observe similar patterns in the last four models, which feature an autocorrelation structure of order two. Among the parameter-driven models, adding a second order to the autocorrelation offers little improvement in model fit, since the increase in goodness-of-fit is offset by a decrease in parsimony. Therefore, the best model appears to be PDZIB(1).\n\n## 6. Conclusion\n\nCount time series featuring a preponderance of zeros are commonly encountered in a variety of scientific applications. In characterizing such series, modeling frameworks that assume a Poisson mixture distribution have been extensively studied. However, minimal work has been focused on modeling frameworks that assume a binomial mixture distribution. When data are more naturally assumed to arise from the latter, a Poisson-type model with an offset is often employed; however, the propriety of such an approximation is unclear.\n\nWe propose two general classes of models to effectively characterize a count time series that arises from a zero-inflated binomial mixture distribution. The observation-driven ZIB model, formulated in the partial likelihood framework, is fit using the Newton–Raphson algorithm. The parameter-driven ZIB model, formulated in the state-space framework, is fit using the MCEM algorithm. When data are generated from a binomial mixture, our proposed ZIB models outperform their Poisson-type counterparts. We illustrate our methodology with an application that assesses a particular level change for a diagnosis code.\n\nFuture work involves extending the current frameworks to the zero-inflated beta-binomial (ZIBB) model. Both observation-driven and parameter-driven ZIBB models can be formulated and fit based on methodological developments similar to those presented in this work. However, weak identifiability could arise as a potentially problematic issue in fitting the parameter-driven ZIBB model, as not only the overdispersion explicitly induced by the beta distribution but also the correlated random effects account for any excess variability in the data . In addition, we could consider more complicated correlation structures by incorporating moving average components in the linear predictors for parameter-driven models. Such an extension necessitates non-trivial revisions to the state-space model formulation and the complete-data likelihood, which warrant further investigation.\n\nchapter PDF\nCitations in RIS format\nCitations in bibtex format\n\n## More\n\n© 2017 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nFan Tang and Joseph E. Cavanaugh (December 20th 2017). State-Space Models for Binomial Time Series with Excess Zeros, Time Series Analysis and Applications, Nawaz Mohamudally, IntechOpen, DOI: 10.5772/intechopen.71336. Available from:\n\n### Related Content\n\n#### Time Series Analysis and Applications\n\nEdited by Nawaz Mohamudally\n\nNext chapter\n\n#### Ensemble Prediction of Stream Flows Enhanced by Harmony Search Optimization\n\nBy Milan Cisty and Veronika Soldanova\n\n#### State of the Art Virtual Reality and Augmented Reality Knowhow\n\nEdited by Nawaz Mohamudally\n\nFirst chapter\n\n#### Introductory Chapter: Enhancing Augmented Reality User Experience (AR-UX) with Design Thinking\n\nBy Nawaz Mohamudally\n\nWe are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities."
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https://medcalc.org/manual/responsiveness.php | [
"MedCalc",
null,
"",
null,
"# Responsiveness\n\n Command: Statistics",
null,
"Agreement & responsiveness",
null,
"Responsiveness\n\n## Description\n\nAllows to calculate various indices for responsiveness, which is the ability to detect any change.\n\n## Required input",
null,
"• 1st and 2nd measurement: the variables for a 1st and 2nd measurement.\n• Filter: an optional filter to include only a selected subgroup of subjects (rows).\n• Options\n• Paired data: select this option when the variables for 1st and 2nd measurements contain paired data (measurements are repeated on the same subjects). If the 2 measurements are independent, unselect this option.\n• Calculate differences as: option to calculate differences as 1st minus 2nd measurement (default is 2nd minus 1st).\n• Advanced: the confidence intervals for the different indices of responsiveness are estimated using the bias-corrected and accelerated (BCa) bootstrap (Efron, 1987; Efron & Tibshirani, 1993). Click for bootstrapping options such as number of replications and random-number seed.\n\n## Results",
null,
"### Summary statistics\n\n• The sample size, mean, variance and standard deviation of the 1st and 2nd measurement.\n\n### Difference\n\n• The average difference between the two measurements with the pooled standard deviation and (in case of paired observations) the standard deviation of the paired differences.\n\n### Indices of responsiveness\n\n• Effect size (ES) using baseline SD: this is the average difference divided by the standard deviation of the 1st measurement (this is Glass' Δ).\n• Effect size (ES) using pooled SD: this is the average difference divided by the pooled standard deviation of both measurements (this is Cohen's d).\n• Standardized response mean (SRM): this is the average difference divided by the standard deviation of the differences between the paired measurements.\n\n## Literature\n\n• Efron B (1987) Better Bootstrap Confidence Intervals. Journal of the American Statistical Association 82:171-185.\n• Efron B, Tibshirani RJ (1993) An introduction to the Bootstrap. Chapman & Hall/CRC.",
null,
"• Husted JA, Cook RJ, Farewell VT, Gladman DD (2000) Methods for assessing responsiveness: a critical review and recommendations. Journal of Clinical Epidemiology 53:459-168.",
null,
"• Norman GR, Wyrwich KW, Patrick DL (2007) The mathematical relationship among different forms of responsiveness coefficients. Quality of Life Research 16:815-822."
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https://chemistry.stackexchange.com/questions/64445/when-are-two-orbitals-orthogonal/64454 | [
"# When are two orbitals orthogonal?\n\nWhen are two orbitals considered to be orthogonal?\n\nIt will be helpful if you can provide an example molecule (or molecular orbitals in a molecule) since I can't really think of a scenario where molecular orbitals can be exactly 90° to each other.\n\nI'm not well-versed in quantum mechanics, so would prefer a simple answer - but all explanations are welcome.\n\n• Orbital means one-electron function. Typically their orthogonality is made by construction. – Rodriguez Dec 14 '16 at 4:18\n\nOne atomic (or molecular) orbital is said to be orthogonal to another atomic (or molecular) orbital if there is no interaction between the electrons in one orbital with the electrons (wavefunction) in the other orthogonal orbital.\n\nAcetylene is an example of a molecule in which the two pi bonds are orthogonal to each other (see the light-blue pi bond and the purple pi bond in the figure below). The two pi bonds are perpendicular to each other and there is no interaction of the electrons in one of acetylene's pi bonds with the electrons in the other pi bond.",
null,
"By the same token the two p orbitals on one of the carbons are also orthogonal to each other. The electron in one p orbital does not interact with the electron in the other p orbital on the same carbon.\n\nIn ethylene the pi bond is orthogonal (perpendicular) to the sigma skeleton as shown in the figure below. Since the orbitals are orthogonal there is no interaction between the sigma and pi electrons (wavefunctions).",
null,
"On the other hand, orbitals that interact and form a bond are not orthogonal. The two ${sp^2}$ orbitals on the two carbons in ethylene that overlap to form the $\\ce{C-C}$ sigma bond are not orthogonal; the electrons interact and form a bond.\n\n• This answer leads to impression that orthogonal orbitals must not overlap, but it is wrong: e.g. $1\\mathrm s$ and $2\\mathrm s$ orbitals overlap considerably, but are orthogonal. – Ruslan Dec 14 '16 at 18:42\n• @Ruslan Thanks for clarifying. I've edited the answer accordingly, hopefully it reads better now. – ron Dec 14 '16 at 20:20\n• @Ruslan Related question and answer here. – hBy2Py Dec 14 '16 at 20:35\n• I think it is wrong to say that the electrons in the pi bonds in acetylene do not interact. – Martin - マーチン Sep 1 '17 at 9:27\n• @ron you seem to be confusing spatial orthogonality with mathematical orthogonality. And those drawings grossly misrepresent the shapes of pi orbitals such that it appears there is no spatial overlap when in fact there is a great deal (but no net overlap) – Andrew Jul 15 '20 at 12:23\n\nUnfortunately, the sense in which orbitals are orthogonal is more or less impossible to define rigorously without recourse to functions of some kind. So, I'll give an explanation a shot using some simple, 1-D functions to illustrate the concept, followed by the pictorial orbital example you've asked for.\n\nAt a basic level, in order to have any two functions be orthogonal, you need to define a \"mixing rule\" for them. The most useful mixing rule for orbitals turns out to be integrating the product of the two functions$^\\dagger$ over some defined domain.\n\nSo, consider two real 1-D functions, $f(x)$ and $g(x)$, defined over an interval $x=[a,b]$. The mixing rule $\\Pi\\,_a^b\\!\\left(f,g\\right)$ I'll define as follows:\n\n$$\\Pi\\,_a^b\\!\\left(f,g\\right) \\rightarrow \\int_a^b{f(x)g(x)\\, dx}$$\n\nSimply put, two functions $f(x)$ and $g(x)$ are orthogonal under the mixing rule $\\Pi\\,_a^b$ (or any other mixing rule!) if and only if $\\Pi\\,_a^b\\!\\left(f,g\\right) = 0$.\n\nA pretty standard example of two orthogonal functions in this framework would be the following:\n\n\\begin{align*} f(x) &= \\cos(2\\pi x) \\\\ g(x) &= \\sin(2\\pi x) \\\\ a &= 0 \\\\ b &= 1 \\end{align*}\n\nThe integral to check from the above is:\n\n$$\\Pi\\,_0^1\\!\\left(f, g\\right) \\rightarrow \\int_0^1{\\cos(2\\pi x)\\, \\sin(2\\pi x)\\, dx}$$\n\nSince this is Stack Exchange and not a math exam, I will now cheat and just link to Wolfram Alpha to demonstrate that the above integral is, indeed, zero. Thus, on the domain $x=[0,1]$, $\\sin(2\\pi x)$ and $\\cos(2\\pi x)$ are orthogonal.\n\nOn the other hand, to pick some random examples, $f(x)=-x$ and $g(x)=\\ln(x)$ are not orthogonal on the interval $x=[0,1]$.\n\n$$\\Pi\\,_0^1\\!\\left(f,g\\right) \\rightarrow \\int_0^1{\\left(-x\\right)\\ln(x)\\, dx} = 0.25$$\n\nSimilarly, the original two functions are not orthogonal on the interval $x=[0.2,0.9]$:\n\n$$\\Pi\\,_{0.2}^{0.9}\\!\\left(f, g\\right) \\rightarrow \\int_{0.2}^{0.9}{\\cos(2\\pi x)\\, \\sin(2\\pi x)\\, dx} = -0.0444852$$\n\nOk, enough with the math lesson. How does this help define orthogonal orbitals?\n\nWell, orbitals, both atomic and molecular, are (usually) defined just as as 3-D functions over all space. If you have two orbitals $\\phi\\!\\left(\\vec r\\right)$ and $\\psi\\!\\left(\\vec r\\right)$ and want to know if they're orthogonal, you use a mixing rule somewhat similar to $\\Pi$ above and calculate the appropriate integral. If the resulting value is zero, then the orbitals are orthogonal.\n\nFor example, consider a $1\\mathrm s$ and a $2\\mathrm p$ orbital (yes, these are atomic orbitals, but the principle holds for molecular orbitals, too):",
null,
"Here, the colors of the two lobes of the $2\\mathrm p$ orbital are important: one color, say orange, represents positive function values; the other, say blue, represents negative values. For the $1\\mathrm s$ orbital, though, it's all orange, or all positive).\n\nWhat happens if, per $\\Pi\\!\\left(\\phi, \\psi\\right)$, you center these two orbitals on top of one another, multiply their values together, and integrate over all space? Well, the half of the $\\mathrm p$ orbital that is positive will result in positive contributions to the integral of $\\Pi$, and the half of the $\\mathrm p$ orbital that is negative will provide negative contributions to the $\\Pi$ integral. And, further, due to the spherical symmetry of the $\\mathrm s$ orbital and the reflection symmetry of the $\\mathrm p$ orbital, those two positive and negative contributions will exactly cancel. Thus, the $1\\mathrm s$ and $2\\mathrm p$ atomic orbitals are orthogonal.\n\nEpilogue: So, given all of this, why do we also say that vectors that are $90^\\circ$ from one another are orthogonal?\n\nWell, it just involves a different mixing rule (the domain is implicit, deriving from the dimensionality of the two vectors, which must be the same):\n\n$$\\Gamma\\!\\left(\\vec v_1, \\vec v_2\\right) \\rightarrow \\vec v_1\\cdot\\vec v_2$$\n\nUnder the (extraordinarily common) mixing rule $\\Gamma\\!\\left(\\vec v_1, \\vec v_2\\right)$, the two vectors $\\vec v_1$ and $\\vec v_2$ are said to be orthogonal when $\\vec v_1\\cdot\\vec v_2=0$.\n\nNote: In theory, I could have just linked to this question on Math.SE. However, none of the answers there were really couched in suitably readable terms.\n\n$^\\dagger$ Technically, integrating the product of one function with the complex conjugate of the other: $\\int_a^b{f^*(x)g(x)\\,dx}$\n\n• Nice answer. I'm with you saying that explaining orthogonality without at least some formulas really don't make sense. – Fl.pf. Dec 14 '16 at 5:30\n• See also my answer on What does orthogonality mean in function space? question at Math.SE for a bit of illustration of the concept. – Ruslan Dec 14 '16 at 20:09\n• Regarding the epilogue: I gave up on maths some years ago, but from what I understand, if one generalises to \"infinite-dimensional vectors\" the sum in the dot product becomes an integral, and the generalisation to a complex vector space results in the complex conjugation. So the dot product can actually be thought of as a special case of the inner product. – orthocresol Sep 1 '17 at 8:02\n• You're exactly right, @orthocresol. I'm pretty sure these can all be considered Hilbert spaces, where continuous functions are of an infinite-dimensional character. – hBy2Py Sep 1 '17 at 10:49\n\nI wonder if you truly do mean molecular orbitals. Let's start with atomic orbitals. We model an isolated atom as having three P orbitals, each a \"dumb bell\" oriented along one of three perpendicular axes (x, Y and z if you like). There are other orbitals of course, such as s and possibly d, and other shells which have the same types of orbitals but with different sizes and numbers of nodes. But focusing on the three perpendicular p-orbitals (which I'm sure you've seen an image of), I hope you can see why we call them orthogonal. They don't overlap, and electrons in different p-orbitals essentially don't interact for our purposes.\n\nNow, the spherical s-orbital sitting in the center of those p's doesn't seem \"perpendicular\" to them; but nonetheless, the s does not interact with the p's. So we say the s-orbital is orthogonal to the p-orbitals, just as the p's are to each other. Orthogonal here means more than just perpendicular in space, it also refers to the mathematical relationship of these shapes.\n\nMolecular orbitals are formed when atomic orbitals overlap. Often we also include orbital hybridization in this model. Molecular orbitals tend to look a bit more complex than our familiar atomic orbitals, and you can't always perceive them as appearing physically perpendicular in space. However, as I hinted at, orthogonality is a more abstract property. In the proper mathematical context, we can still call these orbitals orthogonal, and it still has physical significance.\n\nHere's a more detailed source that is probably on your level: https://www.decodedscience.org/atomic-interactions-wavefunctions-orthogonality-orbitals-hybridization-bonding/55459\n\n(Though I must point out they might be using \"molecular orbital\" a bit too broadly, personally I think M.O.s like sigma and pi bonding and antibonding should be distinguished from the atomic orbitals of s, p, d, and f.)\n\nNow, if you want to know why s-orbitals are spherical and p-orbitals are dumbbell-shaped, we're going to have to say the W word. ;)"
] | [
null,
"https://i.stack.imgur.com/Ejy5z.jpg",
null,
"https://i.stack.imgur.com/c88wy.jpg",
null,
"https://i.stack.imgur.com/coh5s.jpg",
null
] | {"ft_lang_label":"__label__en","ft_lang_prob":0.81927735,"math_prob":0.99131966,"size":4313,"snap":"2021-04-2021-17","text_gpt3_token_len":1244,"char_repetition_ratio":0.14458111,"word_repetition_ratio":0.021374045,"special_character_ratio":0.29492232,"punctuation_ratio":0.15115005,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9981633,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,10,null,10,null,10,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-24T16:52:20Z\",\"WARC-Record-ID\":\"<urn:uuid:5ba52e44-52d9-4627-a8b2-4887d9dac3cd>\",\"Content-Length\":\"183511\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a8f92b0-b552-486b-ac7a-3f33438b9af7>\",\"WARC-Concurrent-To\":\"<urn:uuid:94e17e88-15a3-4524-90fc-ee1af75f900f>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://chemistry.stackexchange.com/questions/64445/when-are-two-orbitals-orthogonal/64454\",\"WARC-Payload-Digest\":\"sha1:MRYRLS57SJMEAGRLYGKZXSNOTWCQ46C3\",\"WARC-Block-Digest\":\"sha1:QWIIN44ZEUHS4Q756AGSPKL2RJ5IMSRB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703549416.62_warc_CC-MAIN-20210124141945-20210124171945-00582.warc.gz\"}"} |
https://community.khronos.org/t/terrain-triangle-question/25429 | [
"# Terrain triangle question\n\nThis is an easy one, I’m sure…maybe you guys can help me.\n\nI’ve got my terrain represented as a linked list of triangles. I want to find out, using the coordinates of my moving object(car), which triangle it is inside of. So I want to traverse my list until I find the triangle that the car is in.\n\nDo any of you guys know an equation or expression to represent this relationship? How can I tell if the car’s (x,z) coordinates ly in the polygons area (triangle)?\n\nThis way, I can figure out what the Y coordinate of my car should be, which I don’t have a problem doing. It’s just figuring out which polygon (triangle) is in the region of the car.\n\nfor each of the three edges of the triangle, calculate the edge plane, that is going along an edge and along the triangle normal. If the point lies on the positive side of the plane, then it’s not in the triangle. If the point lies inside the negative side of all the three planes, the point is inside the triangle.\n\nbool PointInTriangle(float P, float V)\n{\nfor(int i = 0; i < 3; i ++)\n{\n// edge direction\nfloat E = { V[(i+1)%3] - V[i], V[(i+1)%3] - V[i] };\n\n`````` // relative position of point to edge start\nfloat D = { P - V[i], P - V[i] };\n``````\n`````` //edge normal, pointing to the outside of the triangle\nfloat En = { -E.y, E.x };\n\n// distance of point from edge plane\nfloat dp = En * D + En * D;\n``````\n`````` // point on the positive side of plane (with a threhold)\nif (dp > 0.0001f) return false;\n}\n\nreturn true;\n``````\n\n}\n\nif your triangles are on a terrain and therefore roughly pointing up, as you said, use the (x, z) coords of the car and the (x, z) coords of the triangles, that’s why the algorithm is in 2D.\n\nIf it is in 3D, you need to find which of the x, y, z axis the normal roughly points towards to.\n\nexample Normal of tri is (0.1, -0.1, -0.81), use (x, y) coordinates of the vertices to avoid having your triangle fold into a line in the intersection test, and give erroneous results (due to the theshold value, which you need to use to avoid gaps between triangles).\n\n[This message has been edited by oliii (edited 03-31-2003).]"
] | [
null
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https://doc.sagemath.org/html/en/reference/categories/sage/categories/magmas_and_additive_magmas.html | [
"The category of sets $$(S,+,*)$$ with an additive operation ‘+’ and a multiplicative operation $$*$$\n\nEXAMPLES:\n\nsage: from sage.categories.magmas_and_additive_magmas import MagmasAndAdditiveMagmas\nCategory of magmas and additive magmas\n\n\nThis is the base category for the categories of rings and their variants:\n\nsage: C.Distributive()\nCategory of distributive magmas and additive magmas\nCategory of rngs\nCategory of semirings\nCategory of rings\n\n\nThis category is really meant to represent the intersection of the categories of Magmas and AdditiveMagmas; however Sage’s infrastructure does not allow yet to model this:\n\nsage: Magmas() & AdditiveMagmas()\nJoin of Category of magmas and Category of additive magmas\n\nsage: Magmas() & AdditiveMagmas() # todo: not implemented\nCategory of magmas and additive magmas\n\nclass CartesianProducts(category, *args)#\nextra_super_categories()#\n\nImplement the fact that this structure is stable under Cartesian products.\n\nDistributive#\nclass SubcategoryMethods#\n\nBases: object\n\nDistributive()#\n\nReturn the full subcategory of the objects of self where $$*$$ is distributive on $$+$$.\n\nA magma and additive magma $$M$$ is distributive if, for all $$x,y,z \\in M$$,\n\n$x * (y+z) = x*y + x*z \\text{ and } (x+y) * z = x*z + y*z$\n\nEXAMPLES:\n\nsage: from sage.categories.magmas_and_additive_magmas import MagmasAndAdditiveMagmas\nCategory of distributive magmas and additive magmas\n\n\nNote\n\nGiven that Sage does not know that MagmasAndAdditiveMagmas is the intersection of Magmas and AdditiveMagmas, this method is not available for:\n\nsage: Magmas() & AdditiveMagmas()\nJoin of Category of magmas and Category of additive magmas\n\n\nStill, the natural syntax works:\n\nsage: (Magmas() & AdditiveMagmas()).Distributive()\nCategory of distributive magmas and additive magmas\n\n\nthanks to a workaround implemented in Magmas.SubcategoryMethods.Distributive():\n\nsage: (Magmas() & AdditiveMagmas()).Distributive.__module__\n'sage.categories.magmas'\n\n\nReturn None.\n\nIndeed, this category is meant to represent the join of AdditiveMagmas and Magmas. As such, it defines no additional structure.\n\nEXAMPLES:\n\nsage: from sage.categories.magmas_and_additive_magmas import MagmasAndAdditiveMagmas\n\nsage: from sage.categories.magmas_and_additive_magmas import MagmasAndAdditiveMagmas"
] | [
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