Revision as of 17:28, 25 December 2014

Problem

Show, by using the Fourier series, that $$\sum_{k=1}^\infty \frac{1}{k^n} = - \frac{(2 \pi i)^n}{2(n!)} B_n,\ n \in 2 \mathbb{Z}_{>0}$$ where \(B_n\) are the Bernoulli numbers defined as $$\frac{z}{e^z - 1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!}$$

Hint: consider \( B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R} \)

Solution

As suggested in the problem, we will have a look at the convergent fourier series $$B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R}$$ with Fourier coefficients $$c_k = \begin{cases} 0 & \text{for } k = 0\\ \frac{1}{k^n} & \text{for } k \neq 0 \end{cases}$$

It is obvious that \( B_n(x) \) is \(2\pi\)-periodic and, by differentiating every term separately, that $$B'_n = iB_{n-1}$$

We will first show that \( B_2(x) = f(x) := \pi^2/3 - \pi x + x^2/2 \) for \( x \in [0, 2\pi]\).

Recall the definition of the Fourier coefficients for \(2\pi\)-periodic functions in \(C^\infty(\mathbb{R})\): $$c_k = \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx$$

We can easily verify the following identities: $$\begin{align} \frac{1}{2\pi} \int_0^{2\pi} e^{-ikx} \: dx &\ =\ \begin{cases} 1 & \text{for } k = 0\\ 0 & \text{for } k \neq 0 \end{cases} \\ \frac{1}{2\pi} \int_0^{2\pi} x \: e^{-ikx} \: dx &\ =\ \begin{cases} \pi & \text{for } k = 0\\ \frac{i}{k} & \text{for } k \neq 0 \end{cases} \\ \frac{1}{2\pi} \int_0^{2\pi} x^2 \: e^{-ikx} \: dx &\ =\ \begin{cases} \frac{4}{3} \pi^2 & \text{for } k = 0\\ \frac{2(1+ik\pi)}{k^2} & \text{for } k \neq 0 \end{cases} \end{align}$$

Calculating the Fourier coefficients for \(f(x)\) then yields $$\begin{align} \hat f(k) &= \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx\\ &= \frac{1}{2\pi} \left( \frac{\pi^2}{3} \int_0^{2\pi} e^{-ikx} \: dx - \pi \int_0^{2\pi} x \: e^{-ikx} \: dx + \frac{1}{2} \int_0^{2\pi} x^2 \: e^{-ikx} \right) \\ &= \begin{cases} 0 & \text{for } k = 0\\ \frac{1}{k^2} & \text{for } k \neq 0 \end{cases} \end{align}$$ which proves that \(B_2(x) = f(x)\) for \(x \in [0, 2\pi]\).

Thus, we have shown that the \(B_n|_{[0, 2\pi]}\) are polynomials, which are recursively defined by \(B'_n = iB_{n-1}\) and \(\int_0^{2\pi} B_n(x) dx = -i(B_{n+1}(2\pi) - B_{n+1}(0)) = 0\). We also add $$B_1(x) = -iB'_2(x) = i\pi - ix \qquad B_0(x) = -iB'_1(x) = -1$$

By induction over \(n\), we will now prove that $$B_n(x) = \sum_{k=0}^n B_{n-k}(0) \frac{(ix)^k}{k!}$$ Base case for \(n=0\): $$B_0(0) \frac{(ix)^0}{0!} = B_0(0) = -1 = B_0(x)$$ Inductive step (\(n\!-\!1 \rightarrow n\)): $$\begin{align} B_n(x) &= \int_0^x iB_{n-1}(\tilde x)\ d\tilde x + B_n(0) \\ &= \int_0^x i\left(\sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{(i\tilde x)^k}{k!} \right) d\tilde x + B_n(0) \\ &= \left( \sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{i^{k+1}}{k!} \int_0^x \tilde x^k\ d\tilde x \right) + B_n(0) \\ &= \left( \sum_{k=0}^{n-1} B_{(n-1)-k}(0) \frac{i^{k+1}}{k!} \frac{x^{k+1}}{k+1} \right) + B_n(0) \\ &= \left( \sum_{k=0}^{n-1} B_{n-(k+1)}(0) \frac{(ix)^{k+1}}{(k+1)!} \right) + B_n(0) \\ &= \left( \sum_{k=1}^n B_{n-k}(0) \frac{(ix)^k}{k!} \right) + B_n(0) \\ &= \sum_{k=0}^n B_{n-k}(0) \frac{(ix)^k}{k!} \end{align}$$

What's this all leading up to, you may ask. Find the answer here in next weeks sequel. Nik (talk) 18:28, 25 December 2014 (CET)