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In some cases I found that $$\csc(x)= \lim\limits_{k\rightarrow \infty}\sum_{n=-k}^{k}(-1)^{n}\frac{1}{x-n\pi}$$

Is anything to prove or disprove that?

Michael Hardy
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Jow Chieh
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    Aside: the motivation for the formula (as I understand it) is that that both sides have the same poles and also have the same residues at those poles. So the difference is a (periodic) entire function, and what's left to show is that difference is actually zero. Like everything else in complex analysis, I think there's a spiffy way to prove this with a contour integral. –  Sep 03 '15 at 05:30
  • also note that $$\sec x =\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{x-\frac{1}{2} \pi (2 n-1)}$$ – Math-fun Sep 03 '15 at 07:38

1 Answers1

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Recall that the sine function can be represented as the infinite product

$$\sin x=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right) \tag 1$$

Taking the logarithmic derivative of $(1)$, we obtian

$$\begin{align} \cot x&=\frac1x+2x\sum_{n=1}^{\infty}\frac{1}{x^2-n^2\pi^2}\\\\ &=\sum_{n=-\infty}^{\infty}\frac{x}{x^2-n^2\pi^2}\\\\ &=\frac12\sum_{n=-\infty}^{\infty}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \end{align}$$

Then, using the trigonometric identity $\csc x=\cot(x/2)-\cot x$ reveals that

$$\begin{align} \csc x&=\frac12\sum_{n=-\infty}^{\infty}\left(\frac{2}{x-2n\pi}-\frac{1}{x-n\pi}+\frac{2}{x+2n\pi}-\frac{1}{x+n\pi}\right)\\\\ &=\sum_{n=-\infty}^{\infty}\left(\frac{2}{x-2n\pi}-\frac{1}{x-n\pi}\right)\\\\ &=\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{x-n\pi} \end{align}$$

as was to be shown!

Mark Viola
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  • It's pretty much clear, but one question is that where can I find the equation $ \sin x=x\prod_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right) $? – Jow Chieh Sep 03 '15 at 06:37
  • @JowChieh take a look here: https://en.wikipedia.org/wiki/Sinc_function#Properties – Math-fun Sep 03 '15 at 07:25