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It seems that $\cos(\pi x)$ is given by $$\cos(\pi x)=\frac{\sinh(\pi)}{\pi}\sum_{n=0}^{\infty}\left(\frac{\left(-1\right)^{n}}{1+\left(x-n\right)^{2}}-\frac{\left(-1\right)^{n\ }}{1+\left(x+n+1\right)^{2}}\right).$$ I found this by playing around in desmos. Do you know how we could prove this? Has a similar result to this been published elsewhere on the internet? I can't find anything.

Peanutlex
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1 Answers1

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The correct value of the sum reads: $$ \sum_{n=-\infty}^\infty\frac{(-1)^n}{1+(x-n)^2}= \frac{\pi\sinh(\pi)\cos(\pi x)}{\sinh^2(\pi)+\sin^2(\pi x)},\tag1 $$ which (slightly) deviates from your claim. It would be identical if $\sin^2(\pi x)$ in the denominator were removed. Here the difference of the claimed and actual expressions is shown:

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The simplest way to prove the expression (1) is using the Mittag-Leffler's theorem. Indeed both sides of the equality have the same sets of simple poles at $z^\pm_n=n\pm i$ with the same corresponding residues $\operatorname{res}_{z^\pm_n}(f)=\pm\frac{(-1)^n}{2i}$.

A more constructive approach: $$\begin{align} \sum_{n=-\infty}^\infty\frac{(-1)^n}{1+(x-n)^2} &=\frac1{2i}\sum_{n=-\infty}^\infty(-1)^n\left[\frac{1}{x-n-i}-\frac{1}{x-n+i}\right]\tag2\\ &=\frac\pi{2i}\left[\frac{1}{\sin(\pi(x-i))}-\frac{1}{\sin(\pi(x+i))}\right]\tag3\\ &=\frac\pi{2i}\frac{2\cos(\pi x)\sin(\pi i)} {[\sin(\pi x)\cos(\pi i)]^2-[\cos(\pi x)\sin(\pi i)]^2}\tag4\\ &=\frac{\pi\sinh(\pi)\cos(\pi x)} {\cosh^2(\pi)\sin^2(\pi x)+\sinh^2(\pi)\cos^2(\pi x)}\tag5\\ &=\frac{\pi\sinh(\pi)\cos(\pi x)}{\sinh^2(\pi)+\sin^2(\pi x)}.\tag6 \end{align} $$

Explanation:

$(2)\to(3)$: $\displaystyle\sum_{n=-\infty}^\infty\frac{(-1)^n}{z-n}=\frac\pi{\sin(\pi z)}$;

$(3)\to(4)$: $\displaystyle\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$;

$(4)\to(5)$: $\displaystyle\sin(ix)=i\sinh(x),\; \cos(ix)=\cosh(x)$;

$(5)\to(6)$: $\displaystyle \cosh^2(x)=\sinh^2(x)+1$.

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  • Thank you very much this is exactly the sort of thing I was looking for. How did you calculate the correct expression for the sum? – Peanutlex Mar 20 '20 at 12:47
  • Also, do you mind taking a look at this other question I asked? Do you think we could use what you've shown here to answer this other question? https://math.stackexchange.com/questions/3578963/expressing-a-function-in-terms-of-other-functions – Peanutlex Mar 20 '20 at 13:11
  • @Peanutlex I have added a construction of the resulting expression. The simplest way to prove the implication $(2)\to(3)$ is probably still ML theorem (with evident poles and residues of the rhs). – user Mar 21 '20 at 20:16