Could you explain to me why

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

I've read in this article http://twoplusonet.wordpress.com/2011/06/24/an-elegant-result/

that firstly we write

$$\frac{\sin x}{x}\text{ as a polynomial }k\left(1-\frac{x}{a_1}\right)\left(1-\frac{x}{a_2}\right)\left(1-\frac{x}{a_3}\right)\cdots\tag{*}$$

Then, due to the fact that sin$x$ has roots at $_-^+ \pi, _-^+ 2\pi,\ldots$, we can write

$$k\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2 \pi}\right)\cdots\tag{**}$$

Since $\lim _{x\rightarrow 0} \frac {\sin x}{x}=1$, we have $k=1$ . Using the difference of two squares we get the formula above.

My question is: (why) is that enough? In the article mentioned before the author says that between writing $\frac{\sin x}{x}$ as a polynomial * and (**) we make a "little jump of faith (and let the analysts deal with the consequences)". Does that mean something has been omitted here?

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    $\begingroup$ Polynomials, by definition, have a finite number of roots. You are trying to get an infinite product expansion for $\sin$, and there are many details that have been left out in the blog post you linked to. So yes, much has been omitted. $\endgroup$ – Potato Feb 21 '13 at 9:07
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    $\begingroup$ What Potato said is very true. But, also, shouldn't the "difference of two squares" mean that every $x$ in the first expression should really be an $x^2$? $\endgroup$ – Patch Feb 21 '13 at 9:35
  • $\begingroup$ Yes, you are right. I'll correct it right now. $\endgroup$ – Hagrid Feb 21 '13 at 10:49
  • $\begingroup$ Thank you for your comments. Do you know where I could find a more detailed version of this? $\endgroup$ – Hagrid Feb 21 '13 at 11:42
  • $\begingroup$ @Potato : Is that really "by definition"? $\endgroup$ – Michael Hardy Feb 21 '13 at 15:14

The result is actually quite deep.

The function $\sin(x)$ extends to the complex plane as the function $z\mapsto \sin z = \frac{\exp(iz) - \exp(-iz)}{2}$ where $\exp$ is the complex exponential. This function is in particular holomorphic on the entire complex plane.

Since it has a simple zero at $z = 0$, we can divide $\sin(z) / z$ to recover another holomorphic function on the complex plane. Then we can apply the Weierstrass factorization theorem (in fact, the Hadamard form), using that by definition $\sin z$ has order 1 as an entire function, which implies that

$$ \frac{\sin(z)}{z} = G(z) \cdot \prod_{k = 1}^\infty \left( 1 - \frac{z^2}{(k\pi)^2}\right)$$

where $G(z) = \exp (az + b)$ is a nowhere vanishing entire function. Observing that $\sin z / z$ is an even function of $z$ (that is: $\sin(-z)/(-z) = \sin z / z$) we see that $G(z)$ must be even, requiring that $a = 0$. The constant $b$ can be fixed by evaluating both sides at $z = 0$.


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