This blog originates from a question in the final exam of Applied Stochastic Process this semester. The question focused on the Karhunen-Loève expansion of Brownian Bridge. Soon I was surprised to find that it provides a derivation for the Basel problem:

$$\begin{align} \sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6} \end{align}$$

Brownian Bridge and KL Expansion

A standard bridge is defined as

$$\begin{align} B_t=W_t-tW_1,\quad t\in [0,1] \end{align}$$

where $W_t$ is a standard Wiener Process. Autocorrelation function of $B_t$ is

$$\begin{align} R(s,t)=\min\{s,t\}-st \end{align}$$

The KL expansion is given by the Mercer's decomposition of $R(s,t)$

$$\begin{align} R(s,t)=\sum_{i}\lambda _i\phi _i(s)\phi _i(t) \end{align}$$

in which eigen functions $\phi _i$s are given by

$$\begin{align} \int _0^1 R(s,t)\phi _i(s) \,\mathrm{d}s = \lambda _i\phi _i(t) \end{align}$$

My Solution in the Exam

Here I post my solution in the exam: substitute in the expression of $R(s,t)$

$$\begin{align} \lambda _i\phi _i(t)=&\int _0^1 (\min\{s,t\}-st)\phi _i(s) \,\mathrm{d}s\\ =&\int _0^t s\phi _i(s) \,\mathrm{d}s + t\int _t^1 \phi _i(s) \,\mathrm{d}s - t\int _0^1 s\phi _i(s) \,\mathrm{d}s \end{align}$$

take differentiation $\dfrac{\mathrm{d}^{} }{\mathrm{d}t^{}}$ to get

$$\begin{align} \lambda _i\dfrac{\mathrm{d}^{} }{\mathrm{d}t^{}}\phi_i(t)=& t\phi _i(t) + \left( \int _t^1\phi _i(s) \,\mathrm{d}s + t\phi _i(t) \right) - \int _0^1 s\phi _i(s) \,\mathrm{d}s\\ =&\int _t^1\phi _i(s) \,\mathrm{d}s - \int _0^1 s\phi _i(s) \,\mathrm{d}s \end{align}$$

one more differentiation $\dfrac{\mathrm{d}^{} }{\mathrm{d}t^{}}$

$$\begin{align} \lambda _i\dfrac{\mathrm{d}^{2} }{\mathrm{d}t^{2}}\phi _i(t) = -\phi _i(t) \end{align}$$

with boundary condition $\phi _i(0)=\phi _i(1)=0$, which has solution

$$\begin{align} \phi _n(t)=\sqrt{2}\sin(n\pi t),\quad \lambda _n=\left(\dfrac{1}{n\pi}\right)^2 \end{align}$$

i.e. the Mercer expansion is

$$\begin{align} R(s,t) = \min\{s,t\}-st = \sum_{n=1}^\infty \dfrac{2}{n^2\pi^2}\sin(n\pi s)\sin(n\pi t) \end{align}$$

Basel Problem

What makes it interesting is that we can consider the function value $R(1/2,1/2)$

$$\begin{align} R(\dfrac{1}{2},\dfrac{1}{2})=\dfrac{1}{4}=&\sum_{n=1}^\infty \dfrac{2}{n^2\pi^2}\sin(\dfrac{n\pi}{2})\sin(\dfrac{n\pi}{2})\\ =&\sum_{n\in\text{odd}}^\infty \dfrac{2}{n^2\pi^2}\\ =&\dfrac{2}{\pi^2}\left(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\ldots \right) \end{align}$$

On the other hand notice that

$$\begin{align} \dfrac{1}{4}\sum_{n=1}^\infty \dfrac{1}{n^2} = \sum_{n=1}^\infty \dfrac{1}{(2n)^2}=\left(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\ldots \right) \end{align}$$

we have

$$\begin{align} \sum_{n=1}^\infty \dfrac{1}{n^2}-\dfrac{1}{4}\sum_{n=1}^\infty \dfrac{1}{n^2} = & \left(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\ldots \right) - \left(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\ldots \right)\\ =&\left(\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\ldots \right)=\dfrac{\pi^2}{8}\\ \Rightarrow \sum_{n=1}^\infty \dfrac{1}{n^2}=& \dfrac{4}{3}\dfrac{\pi^2}{8} = \dfrac{\pi^2}{6} \end{align}$$

Appendix: Euler's Method

Euler studied the function $\mathrm{sinc} \pi t=\dfrac{\sin \pi t}{\pi t}$, which has roots at $\pm 1,\pm 2,\pm 3,\ldots$, which means the polynomial form of $\mathrm{sinc}\pi t$ should looks like

$$\begin{align} \dfrac{\sin \pi t}{\pi t} =& (1+t)(1-t)(1+\dfrac{t}{2})(1-\dfrac{t}{2})\ldots\\ =&(1-t^2)(1-\dfrac{t^2}{4})\ldots \\ =&1- t^2\sum_{n=1}^\infty \dfrac{1}{n^2} + o(t^2) \end{align}$$

according to taylor series, it should also have expansion of the form

$$\begin{align} \dfrac{\sin \pi t}{\pi t} =&\dfrac{\pi t - \dfrac{1}{6}(\pi t)^3 + o(t^3)}{\pi t} = 1 - \dfrac{\pi^2}{6}t^2+o(t^2) \end{align}$$

by comparing the $t^2$ term we obtain that

$$\begin{align} \sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6} \end{align}$$