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

I tried it with the Taylor series of $\sin(x)$ but I failed.

Is there any help?


marked as duplicate by Start wearing purple, mrf, Davide Giraudo, Norbert, azimut Jul 19 '13 at 11:10

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    $\begingroup$ This is a prime example of the Weierstrass Factorization Theorem. en.wikipedia.org/wiki/Weierstrass_factorization_theorem, although you may be looking for a more elementary proof than this. $\endgroup$ – Eric Auld Jul 19 '13 at 0:47
  • $\begingroup$ @EricAuld thanks but it is not clear for me and it is better if there is another one :) $\endgroup$ – mhd.math Jul 19 '13 at 0:55
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    $\begingroup$ You should check out Journey Through Genius by William Dunham. He goes through the intuition behind this factorization when Euler (?) did it but when it was first realized, there was little to no rigor to the arguments. The "proof" he presents that Euler used is good enough to get the idea, though. $\endgroup$ – Cameron Williams Jul 19 '13 at 0:55
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    $\begingroup$ You should probably read this excellent article :-) cornellmath.wordpress.com/2007/07/13/… $\endgroup$ – TenaliRaman Jul 19 '13 at 1:38
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    $\begingroup$ Euler argued that because $\frac{\sin \pi x}{\pi x}$ has roots at $x = \pm1, \pm2, \pm3, ...$, the Factor theorem allows you to write $\frac{\sin \pi x}{\pi x} = (1 - x)(1 + x)(2 - x)(2 + x)...(n - x)(n + x)... = (1 - x^2)(2^2 - x^2)...(n^2 - x^2)... = \prod_{n=1}^{\infty}(1-\frac{x^2}{n^2})$. $\endgroup$ – Matthew Hampsey Jul 19 '13 at 2:47

You can start from the identity

$$ \frac{\pi \cot(\pi x)}{x} = \sum_{n=-\infty}^{\infty}\frac{1}{x^2-n^2} \quad x\neq \mathbb{Z}$$

$$ \implies \frac{\pi \cot(\pi x)}{x}= \frac{1}{x^2}+2\sum_{n=1}^{\infty}\frac{1}{x^2-n^2}$$

$$ \implies {\pi \cot(\pi x)}= \frac{1}{x}+2\sum_{n=1}^{\infty}\frac{x}{x^2-n^2}. $$

Now, you can advance by integrating both sides w.r.t. $x$.


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