User:GFauxPas/Sandbox/Zeta2/Informal Proof

Rigorous justifications of the algebraic manipulations following require more advanced mathematics. As presented, this is an informal proof. However, it has historical significance.

We have that if $a$ is the root of a polynomial, $\displaystyle \left({1-\frac x a}\right)$ is a factor of that polynomial.

Following that line of reasoning,


 * $\displaystyle \sin x = Ax \prod \left({1 - \frac x { n\pi }}\right)$

where the product is taken over all $n \in \Z \setminus \left\{{0}\right\}$, and $A$ is some constant.

That is,

That $\dfrac {\sin x}x \to 1$ as $x \to 0$ is a well known limit. Letting $x$ tend to $0$ in the above equation shows that $A = 1$.

That is,

$\displaystyle \sin x = x \left({1 - \frac{x^2}{\pi^2}}\right) \left({1 - \frac{x^2}{4 \pi^2}}\right) \left({1 - \frac{x^2}{9 \pi^2}}\right) \cdots$

Recall the series expansion of $\sin x$:


 * $\displaystyle \sin x = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + \cdots$

We equate the cube terms of the series and the product:

Equating coefficients: