# Basel problem derivation

The basis for this derivation is the assumption that the properties of finite polynomials hold true for infinite series. First, consider the Taylor series expansion of the sine function:

$\displaystyle\sin x=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n+1}=x-% \frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots$

Dividing through by x, we get:

$\displaystyle\frac{\sin x}{x}=1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}% {7!}+\cdots$

The roots of the function $\sin x/x$ occur at $x=n\pi$, where $n=\pm 1,\pm 2,\pm 3,\ldots$. Let us assume that we can express this infinite series as a product  of linear factors given by its roots, just as we do for finite polynomials:

 $\displaystyle\frac{\sin x}{x}$ $\displaystyle=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-% \frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\left(1-\frac{x}{3\pi}\right% )\left(1+\frac{x}{3\pi}\right)\cdots$ $\displaystyle=\left(1-\frac{x^{2}}{\pi^{2}}\right)\left(1-\frac{x^{2}}{4\pi^{2% }}\right)\left(1-\frac{x^{2}}{9\pi^{2}}\right)\cdots.$

If we multiply out this product and collect all the $x^{2}$ terms, we get:

 $\displaystyle-\left(\frac{1}{\pi^{2}}+\frac{1}{4\pi^{2}}+\frac{1}{9\pi^{2}}+% \cdots\right)=-\frac{1}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}$

From the original infinite series of $\sin x/x$ the coefficient of $x^{2}$ is $-1/3!$. Equating both expressions gives:

 $\displaystyle-\frac{1}{6}=-\frac{1}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}$
 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$

This was the method Euler used to solve the Basel problem  . There are more rigorous and modern derivations, but this one is a simple, straight forward, and easily understood one.

Title Basel problem derivation BaselProblemDerivation 2013-03-22 18:40:50 2013-03-22 18:40:50 curious (18562) curious (18562) 5 curious (18562) Derivation msc 11A25