Fourier basis and Parseval's identity

We calculate

$$\langle e_m,e_n\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-imx}e^{inx}\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n-m)x}\,dx .$$

If $n=m$, the integrand is $1$ and the result is $1$. If $n\neq m$, an antiderivative of $e^{i(n-m)x}$ is $\frac{e^{i(n-m)x}}{i(n-m)}$, which takes the same value at $\pm\pi$ because $e^{i(n-m)\pi}=e^{-i(n-m)\pi}=(-1)^{n-m}$; the integral vanishes. The family is therefore orthonormal, and since it is total, it is a Hilbert basis: every $f\in L^2(-\pi,\pi)$ can be written $f=\sum_{n\in\Z}c_ne_n$, with $c_n=\langle e_n,f\rangle$, where the series converges in the $L^2$ norm.

We have $c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}x\,e^{-inx}\,dx$. For $n=0$, the integrand is odd and $c_0=0$. For $n\neq0$, integrate by parts, differentiating $x$ and integrating $e^{-inx}$:

$$\int_{-\pi}^{\pi}x\,e^{-inx}\,dx=\Bigl[\frac{x\,e^{-inx}}{-in}\Bigr]_{-\pi}^{\pi}+\frac{1}{in}\int_{-\pi}^{\pi}e^{-inx}\,dx=\frac{\pi(-1)^n+\pi(-1)^n}{-in}+0=\frac{2\pi i(-1)^n}{n},$$

where we used $e^{\mp in\pi}=(-1)^n$ and $\frac{1}{-i}=i$. Thus

$$c_n=\sqrt{2\pi}\,\frac{i(-1)^n}{n},\qquad |c_n|^2=\frac{2\pi}{n^2}\quad(n\neq0).$$

Parseval's identity states that $\norm f^2=\sum_{n\in\Z}|c_n|^2$. On the one hand, $\norm f^2=\int_{-\pi}^{\pi}x^2\,dx=\frac{2\pi^3}{3}$. On the other hand, each integer $n\geq1$ appears twice, for $n$ and $-n$, and

$$\sum_{n\in\Z}|c_n|^2=2\sum_{n\geq1}\frac{2\pi}{n^2}=4\pi\sum_{n\geq1}\frac{1}{n^2}.$$

Equating the two expressions gives

$$\sum_{n\geq1}\frac{1}{n^2}=\frac{2\pi^3}{3}\cdot\frac{1}{4\pi}=\frac{\pi^2}{6}.$$

This is the celebrated Basel problem, solved by Euler in 1735, obtained here as a consequence of the Pythagorean theorem in infinite dimensions.

The function $f_N$ is the orthogonal projection of $f$ onto the finite-dimensional subspace $F_N$ spanned by $e_{-N},...,e_N$, since the projector onto this subspace is $\sum_{|n|\leq N}\ket{e_n}\bra{e_n}$. By the best-approximation result established by the Pythagorean theorem, $f_N$ is the element of $F_N$ closest to $f$. The error is the norm of the orthogonal component, which is again calculated from the remaining coefficients using Parseval's identity:

$$\norm{f-f_N}^2=\sum_{|n|>N}|c_n|^2=4\pi\sum_{n>N}\frac{1}{n^2}.$$

Comparison with the integral $\int_N^\infty\frac{dx}{x^2}=\frac1N$ gives $\norm{f-f_N}^2\simeq\frac{4\pi}{N}$ for large $N$: the error tends to zero, but slowly. The slow $1/n$ decay of the coefficients arises because the periodic extension of the function $f(x)=x$ is discontinuous at $\pm\pi$. As with the Fourier transform on the real line, the smoother a function is, the faster its Fourier coefficients decay.