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$$\sin x = x\prod_{n=1}^\infty \left[1-\frac{x^2}{n^2\pi^2}\right]$$

If you apply $\log$ to both sides and derivate:

$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \frac{1}{1-\frac{x^2}{n^2\pi^2}}\right]$$

I have to make the expansion:

$$\frac{1}{1-\frac{x^2}{n^2\pi^2}} = 1 + {\left(\frac{x^2}{n^2\pi^2}\right)} + \left(\frac{x^2}{n^2\pi^2}\right)^2 + \left(\frac{x^2}{n^2\pi^2}\right)^3 + \cdots \tag{1}$$

But for this, I'm gonna expand the infinite sum:

$$\cot x = \frac{1}{x} - \left(\frac{2x}{1^2\pi^2} \frac{1}{1-\frac{x^2}{1^2\pi^2}} + \frac{2x}{2^2\pi^2} \frac{1}{1-\frac{x^2}{2^2\pi^2}} + \frac{2x}{3^2\pi^2} \frac{1}{1-\frac{x^2}{3^2\pi^2}}+\cdots\right)$$

So I can understand the boundaries of this expansion (cause the series $\frac{1}{1-x}=1 + x + x^2 + x^3 + \cdots$ as long as $|x|<1$.

So, am I right in saying that I can do the expansion $(1)$ as long as $|x|<\pi$? Because for $x=\pi$ we have: $$\frac{1}{1-\frac{\pi^2}{1^2\pi^2}} = \frac{1}{1-1}$$ And if $|x|<\pi$ the other terms like:

$$\frac{1}{1-\frac{\pi^2}{2^2\pi^2}}$$ Can be expanded by $(1)$ too.

So:

$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \left(1 + {\left(\frac{x^2}{n^2\pi^2}\right)} + \left(\frac{x^2}{n^2\pi^2}\right)^2 + \left(\frac{x^2}{n^2\pi^2}\right)^3 + \cdots \right)\right]\tag{$|x|<\pi$}$$ $$x \cot x = 1 - 2\sum_{n=1}^\infty \left[\frac{x^2}{n^2\pi^2} + \frac{x^4}{n^4\pi^4} + \frac{x^6}{n^6\pi^6} + \frac{x^8}{n^8\pi^8} + \cdots\right]$$

And finally the famous: $$x \cot x = 1 - 2\sum_{n=1}^{\infty} \left[\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right]\tag{$|x|<\pi$}$$

Am I right with the boundaries for $x$?

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    $\begingroup$ Yes, you're right. For a meromorphic function, the radius of convergence is always given by the distance from the closest singularity, $\pm \pi$ in your case. $\endgroup$ – Jack D'Aurizio Aug 8 '14 at 17:04
  • $\begingroup$ (+1) This is the same as $(1)$ in this answer. $\endgroup$ – robjohn Aug 8 '14 at 22:50
  • $\begingroup$ Is this basically the way to compute every even zeta function? Very nice $\endgroup$ – Yuriy S Mar 13 '16 at 2:45
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Actualy (see [1] pg. 75) $$ \cot(z)=\frac{1}{z}-\sum^{\infty}_{n=1}\frac{(-1)^{n-1}2^{2n}B_{2n}}{(2n)!}z^{2n-1}\textrm{, }|z|<\pi $$ and ([1] pg.807) $$ \zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n-1}}{2(2n)!}B_{2n}\textrm{, }n\in\textbf{N} $$ where $B_{2n}$ are the Bernoulli numbers i.e. $$ \frac{x}{e^x-1}=\sum^{\infty}_{n=0}\frac{B_n}{n!}x^n\textrm{, }|x|<2\pi $$ Your expansion is very right. But unfortunately this is quite known. In [1] you can find more of these results. For example $$ \log\left(\frac{\sin(z)}{z}\right)=-\sum^{\infty}_{n=1}\frac{\zeta(2n)}{n}\frac{z^{2n}}{\pi^{2n}} $$

Actualy for an analytic function in $(-1,1)$ which is also absolutly convergent at 1 we have $$ \sum^{\infty}_{k=1}\left(f\left(\frac{x}{2\pi i k}\right)+f\left(\frac{-x}{2\pi i k}\right)-2f(0)\right)=2\sum^{\infty}_{k=1}\frac{f^{(2k)}(0)}{(2k)!}\frac{B_{2k}}{(2k)!}x^{2k}\textrm{, }|x|<2\pi $$

[1]: Milton Abramowitz and Irene A. Stegun. "Handbook of Mathematical functions". Dover publications, inc., New York. (1972)

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  • 1
    $\begingroup$ With $\frac{B_k}{k!}$ defined as the coefficients of $\frac{x}{e^x-1}$ note that $\zeta(2k)=\frac{(2\pi)^{2k}(-1)^{k-1}}{2(2k)!}B_{2k}$ implies $x \cot x = 1 - 2\sum_{k=1}^{\infty} \left[\zeta(2k)\frac{x^{2k}}{\pi^{2k}}\right] = 1- x\sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \frac{1}{1-\frac{x^2}{n^2\pi^2}}\right]$ and $\sin x = x\prod_{n=1}^\infty \left[1-\frac{x^2}{n^2\pi^2}\right]$ $\endgroup$ – reuns Jul 21 at 7:10
  • $\begingroup$ @ renus. Yes indeed. $\endgroup$ – Nikos Bagis Jul 22 at 8:41

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