The Basel problem via Fourier series

The Basel problem asks to compute the summation of the reciprocal of the natural numbers squared, and it is known that

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

While there are numerous known proofs (our presentation resembles Proof 9 most closely), we give a self-contained one that motivates the usage of Fourier analysis for periodic functions.

Recall that a function $f:\mathbb{R}\rightarrow\mathbb{C}$ is $2\pi$ -periodic if for every $x \in \mathbb{R}$ , $f(x+2\pi)=f(x)$ . From Fourier analysis, every such function can be expressed as a series of cosines and sines. More precisely,

If $f$ is continuous and $2\pi$ -periodic, then

$\left\Vert f - \sum_{n=-\infty}^{\infty} \hat{f}(n) e_n\right\Vert_2 = 0,$

where $e_n(x) = e^{inx}$ , and $\hat{f}(n)$ is the $n$ -th Fourier coefficient equal to $\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx} dx$ .

We say that $\sum_{n=-\infty}^{\infty} \hat{f}(n) e_n$ is the Fourier series of $f$ . When we define the inner product of two functions to be $\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) \overline{g(x)} dx$ , then it is easily seen that the “characters” $\{e_n\}$ form an orthonormal system, and the $n$ -th Fourier coefficient can be viewed as the projection of $f$ along the $n$ -th character. The Fourier theorem asserts that $f$ can be approximated in $L_2$ -norm by its Fourier series. In fact, by applying the Weierstrass M-test, we can make the following statement with a stronger convergence guarantee:

If $f$ is continuous and $2\pi$ -periodic and $\sum_{n=-\infty}^{\infty} \left|\hat{f}(n)\right|$ is convergent, then its Fourier series $\sum_{n=-\infty}^{\infty} \hat{f}(n) e_n$ converges uniformly to $f$ .

Furthermore, the Fourier transform $\hat{f}$ itself encodes many of the properties of the original function $f$ . In particular, we note that the symmetry of $f$ is preserved in its Fourier transform.

If $f$ is even, then for every $n \in \mathbb{Z}$ , $\hat{f}(-n) = \hat{f}(n)$ .
If $f$ is odd, then for every $n \in \mathbb{Z}$ , $\hat{f}(-n) = -\hat{f}(n)$ .

Now we have the machinery to compute $\sum_{n=1}^{\infty} 1/n^2$ .

(Of Basel) Define $f:[-\pi, \pi]\rightarrow \mathbb{R}$ to be $f(x)=|x|$ . We compute the Fourier coefficient of $f$ . First note that

$\hat{f}(0) = \frac{1}{2\pi}\int_{-\pi}^{\pi} |x| dx = \frac{\pi}{2}.$

For $n \neq 0 \in \mathbb{Z}$ , we have

$2\pi\hat{f}(n) = \int_{0}^{-\pi}xe^{-inx} dx + \int_{0}^{\pi}xe^{-inx} dx.$

Using integration by parts, we have for every $a, b \in \mathbb{R}$ ,

$\begin{eqnarray*} \int_a^b xe^{-inx}dx & = & \frac{x e^{-inx}}{-in}\Biggr|_a^b - \int_a^b \frac{e^{-inx}}{-in}dx \\ & = & \frac{x e^{-inx}}{-in}\Biggr|_a^b + \frac{e^{-inx}}{n^2}\Biggr|_a^b. \end{eqnarray*}$

Then

$\begin{eqnarray*} 2\pi\hat{f}(n) & = & \frac{-\pi(-1)^n}{-in} + \frac{(-1)^n -1 }{n^2} + \frac{\pi(-1)^n}{-in} + \frac{(-1)^n -1 }{n^2} \\ & = & \frac{2((-1)^n - 1)}{n^2}. \end{eqnarray*}$

Thus,

$\hat{f}(n) = \begin{cases} \frac{-2}{\pi n^2} & \text{if $n$ is odd,} \\ 0 & \text{if $n$ is even and nonzero,} \\ \frac{\pi}{2} & \text{if $n=0$.}$

From Euler’s formula, the Fourier series of $f$ can be written as

$\hat{f}(0) + \sum_{n=1}^{\infty} \left[ (\hat{f}(n) + \hat{f}(-n)) \cos(nx) + i(\hat{f}(n) - \hat{f}(-n)) \sin(nx) \right].$

Since $f$ is even, by Proposition, $\hat{f}(-n)=\hat{f}(n)$ . From our calculation, the Fourier series of $f$ is explicitly

$\frac{\pi}{2} + \sum_{n\geq1, n \text{ odd}} \frac{-4}{\pi n^2} \cos(nx).$

Note that $\sum_{n=1}^{\infty} 1/n^2$ is convergent from the Cauchy condensation test. Thus, by the Convergence Theorem, we conclude that

$\frac{\pi}{2} + \sum_{n\geq1, n \text{ odd}} \frac{-4}{\pi n^2} \cos(nx) \longrightarrow |x|,$

where the convergence is uniform (and in particular, pointwise). Setting $x=0$ , we see that

$\sum_{n\geq1, n \text{ odd}} \frac{1}{n^2} = \frac{\pi^2}{8}.$

Lastly, observe that

$\begin{eqnarray*} \sum_{n=1}^{\infty} \frac{1}{n^2} & = & \sum_{n\geq1, n \text{ even}} \frac{1}{n^2} + \sum_{n\geq1, n \text{ odd}} \frac{1}{n^2} \\ & = & \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{\pi^2}{8}, \end{eqnarray*}$

finishing the proof.

Victor Chen

mathematics blog

The Basel problem via Fourier series

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