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 is \mathbb{Z}\text{-periodic} if for every n \in \mathbb{Z}, f(x+n)=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 \mathbb{Z}-periodic, then

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

where e_n(x) = e^{2\pi inx}, and \hat{f}(n) is the n-th Fourier coefficient equal to \int_0^1 f(x)e^{-2\pi inx} dx.

When we define the inner product of two functions to be \int_0^1 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 then asserts that f can be approximated in L_2-norm by any sufficiently large orthonormal system. In fact, by applying the Weierstrass M-test, we can make the following statement with a stronger convergence guarantee:

If f is continuous and \mathbb{Z}-periodic and \sum_{n=-\infty}^{\infty} \left|\hat{f}(n)\right| is convergent, then \sum_{n=-\infty}^{\infty} \hat{f}(n) e_n converges uniformly to f.

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

(Of Basel) Let f be a periodic function which is equal to (1-2x)^2 when x \in [0,1], and extended to the real line by f(x) = f(x \text{ mod } 1). We compute the Fourier coefficient of f.

    \[\hat{f}(0) = \int_{0}^{1} f(x) dx = \frac{(1-2x)^3}{-6}\Biggr|_0^1 = \frac{1}{3}\]

For n \neq \mathbb{Z}, we have

    \[\hat{f}(n) = \int_{0}^{1}(1-4x+4x^2)e^{-2\pi inx} dx. \]

Note that \int_0^1 e^{-2\pi inx} dx = 0. Using integration by parts, we have

    \begin{eqnarray*} \int_0^1 xe^{-2\pi inx}dx & = & \frac{-1}{2\pi in} \left(x e^{-2\pi inx}\Biggr|_0^1 - \int_0^1 e^{-2\pi inx}dx \right) \\ & = & \frac{-1}{2\pi in} \end{eqnarray*}

and similarly

    \begin{eqnarray*} \int_0^1 x^2e^{-2\pi inx}dx & = & \frac{-1}{2\pi in} \left(x^2 e^{-2\pi inx}\Biggr|_0^1 - \int_0^1 2x e^{-2\pi inx}dx \right) \\ & = & \frac{-1}{2\pi in} \left( 1 + \frac{1}{\pi i n} \right) \\ & = & \frac{-1}{2\pi in} + \frac{1}{2\pi^2 n^2} \end{eqnarray*}

Thus, \hat{f}(n) = \frac{2}{\pi^2 n^2}. Note that \sum_{n=1}^{\infty} 1/n^2 is convergent from the Cauchy condensed test. Thus, we can conclude that

    \[{ \frac{1}{3} + 2\cdot \sum_{n=1}^{\infty} \frac{2}{\pi^2 n^2} e^{2\pi inx} \longrightarrow (1-2x)^2, \]

where the convergence is uniform (and in particular, pointwise). Setting x=0 finishes the proof.

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