# Tag Info

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A positive definite matrix is such that $X^TAX\geq0$ for all column vectors $X$, and $X^TAX=0$ only for $X=0$. Therefore, if $A$ has not full rank, there is $Y\neq 0$ such that $AY=0$, but then $Y^TAY=0$ with $Y\neq 0$ contradicting the above requirements. As for roots (eigenvalues) if there is a negative root $\lambda$, then there is a corresponding ...

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I use python with matplotlib library, it's an incredible tool to plotting 2D and 3D functions

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For daily Maths jobs, I recommend Maple $18$, however; Matlab rules parts of Maths powerfully. Using Maple at the class is easy (at least for me). For example, when I want to describe some implicit functions at the class, I may do as follows: [> with(plots): [> implicitplot3d((x^2+y^2)*(x^2+z^2)*(y^2+z^2) = 1, x = -4 .. 4, y = -4 .. 4, z = -4 .. 4, ...

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Case $1$: $xe^x\leq1$ Then $\int\dfrac{1}{e^{-x}-x}dx$ $=\int\dfrac{e^x}{1-xe^x}dx$ $=\int\sum\limits_{n=0}^\infty x^ne^{(n+1)x}~dx$ $=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}n!x^ke^{(n+1)x}}{(n+1)^{n-k+1}k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

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In Mathematica, it seems the syntax would be, HypergeometricPFQ[{a/b,1},{c,d},h] Checking how Maple translates that, mmainput := HypergeometricPFQ[{a/b,1},{c,d},h]: MmaTranslator:-FromMma(mmainput); hypergeom([a/b, 1], [c, d], h) Passing HypergeometricPFQ[{3/4,1},{2,3},1.1] to wolframalpha.com agrees with Maple evaluating ...

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According to Wolfram: $\displaystyle{\frac{1}{e^{-x}-x}=1+2x+\frac{7x^2}{2}+\frac{37x^3}{6}+\cdots}$ therefore the integral can be written like that : $$\displaystyle{\int \frac{1}{e^{-x}-x}\, dx=\int \left(1+2x+\frac{7x^2}{2}+\frac{37x^3}{6}+\cdots \right )\, dx=x+x^2+\frac{7x^3}{6}+\frac{37x^4}{24}+\cdots}$$

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It may not be possible to get a nice clean answer. To me, it looks like your best bet is to use a Taylor series. To refresh your memory, the Taylor series of a function centered at $x=a$ is: $$\sum _{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n,$$ where $f^{(n)}(a)$ is the $n^{\mathrm{th}}$ derivative of $f$ at $a$. Since Taylor series are polynomials, it ...

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According to http://calculus-geometry.hubpages.com/hub/List-of-Functions-You-Cannot-Integrate-No-Antiderivatives, the function $$f: x \mapsto \frac{1}{e^x + x}$$ does not have an antiderivative expressible in terms of elementary functions, i.e. as algebraic combinations and compositions of polynomials, $\ln$, and $\exp$. Your integrand is simply $f(-x)$. ...

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This would be more appropriate on mathematica.stackexchange.com but, for the time being: residueSystem[n_Integer?Positive] := Select[Range[n], GCD[#, n] == 1 &]; residueSystem[10] (* Out: {1, 3, 7, 9} *)

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Use the sum formula $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$ and compute the Fourier coefficients on just the part with the variables.

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