# Basel Problem using Fourier Series

A very famous mathematical problem known as the “Basel Problem” is solved by Euler in 1734. Basically, it asks for the exact value of $\sum_{n=1}^\infty\frac{1}{n^2}$.

Three hundred years ago, this was considered a very hard problem and even famous mathematicians of the time like Leibniz, De Moivre, and the Bernoullis could not solve it.

Euler showed (using another method different from ours) that $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6},$ bringing him great fame among the mathematical community. It is a beautiful equation; it is surprising that the constant $\pi$, usually related to circles, appears here.

## Squaring the Fourier sine series

Assume that $\displaystyle f(x)=\sum_{n=1}^\infty b_n\sin nx.$

Then squaring this series formally,
\begin{aligned} (f(x))^2&=(\sum_{n=1}^\infty b_n\sin nx)^2\\ &=\sum_{n=1}^\infty b_n^2\sin^2 nx+\sum_{n\neq m}b_nb_m\sin nx\sin mx. \end{aligned}

To see why the above hold, see the following concrete example:
\begin{aligned} (a_1+a_2+a_3)^2&=(a_1^2+a_2^2+a_3^2)+(a_1a_2+a_1a_3+a_2a_1+a_2a_3+a_3a_1+a_3a_2)\\ &=\sum_{n=1}^3 a_n^2+\sum_{n\neq m}a_na_m. \end{aligned}

## Integrate term by term

We assume that term by term integration is valid.
$\displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx+\frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx.$

Recall that $\displaystyle \int_{-\pi}^\pi \sin nx\sin mx\,dx=\begin{cases}0 &\text{if }n\neq m\\ \pi &\text{if }n=m \end{cases}.$

So
\begin{aligned} \frac 1\pi\int_{-\pi}^{\pi}\sum_{n=1}^\infty b_n^2\sin^2{nx}\,dx&=\frac 1\pi\sum_{n=1}^\infty b_n^2(\int_{-\pi}^\pi\sin^2 nx\,dx)\\ &=\frac 1\pi\sum_{n=1}^\infty b_n^2 (\pi)\\ &=\sum_{n=1}^\infty (b_n)^2. \end{aligned}

Similarly
\begin{aligned} \frac{1}{\pi}\int_{-\pi}^\pi\sum_{n\neq m}b_nb_m\sin nx\sin mx\,dx&=\frac 1\pi\sum_{n\neq m}b_nb_m(\int_{-\pi}^{\pi}\sin nx\sin mx\,dx)\\ &=\frac 1\pi\sum_{n\neq m}b_nb_m(0)\\ &=0. \end{aligned}

So $\displaystyle \frac 1\pi\int_{-\pi}^\pi (f(x))^2\,dx=\sum_{n=1}^\infty (b_n)^2.$ (Parseval’s Identity)

## Apply Parseval’s Identity to $f(x)=x$

By Parseval’s identity,
$\displaystyle \frac{1}{\pi}\int_{-\pi}^\pi x^2\,dx=\sum_{n=1}^\infty(\frac{2(-1)^{n+1}}{n})^2.$

Simplifying, we get $\displaystyle \frac 1\pi\cdot\left[\frac{x^3}{3}\right]_{-\pi}^\pi=\sum_{n=1}^\infty\frac{4}{n^2}.$
\begin{aligned} \frac 1\pi(\frac{2\pi^3}{3})&=\sum_{n=1}^\infty \frac{4}{n^2}\\ \frac{\pi^2}{6}&=\sum_{n=1}^\infty\frac{1}{n^2}. \end{aligned}

## Author: mathtuition88

https://mathtuition88.com/

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