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I saw on wikipedia that he consider $$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$$

The roots are given by $x=\pm n\pi$ and thus (to me) $$\frac{\sin x}{x}=(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)(x-3\pi)(x+3\pi)...$$

How can he get $$\frac{\sin x}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right)\left(1+\frac{x}{3\pi}\right)...\ \ \ ??$$

idm
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    The leading coefficient is missing in your expansion. Evaluate the LHS at e.g. $\frac{\pi}{2}$ to find it. – Clement C. Aug 26 '15 at 15:16
  • See Titcmarch's book: Theory of functions and use of residues. They are there and much more. – Adelafif Aug 26 '15 at 16:01
  • One is tempted to surmise that $\prod_{n \in \mathbb{Z}\backslash{0}} x-n\pi = \prod_{n \in \mathbb{Z}\backslash{0}} 1-\frac{x}{n\pi}$ simply because the two polynomials share all the same roots... is this sufficient? – 727 Aug 26 '15 at 16:08
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    @LJL: It's not clear how $\prod (x-n\pi)$ could converge unless $x=n\pi$ for some integer $n$. Otherwise, for fixed $x$, the factors grow without bound (and the signs eventually oscillate). On the other hand, the factors of $\prod (1 - x/(n\pi))$, converge to $1$ for fixed $x$ as $n\to \infty$. –  Aug 26 '15 at 17:41

1 Answers1

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Euler's proof is a bit adhoc. He uses the fact that for finite polynomials if:

$$(x-r_1)\dots(x-r_n)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$

then $$\dfrac{1}{r_1}+\dots+\dfrac{1}{r_n}=\dfrac{-a_1}{a_0}$$

Then he assumes the result is true for power series, and uses:

$$\dfrac{\sin(\sqrt x)}{\sqrt x}=1-\dfrac{x}{3!}+\dfrac{x^2}{5!}+\dots$$

so $r_k=(k\pi)^2$ and $a_0=1$ and $a_1=\dfrac{-1}{6}$.

Reference

JMP
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