Proving of the multiplication theorem for Bernoulli polynomial

Question

How the expression below can be proven:

$$B_n(mx) = m^{n−1} \sum\limits_{k=0}^{m-1}B_n\left(x+\frac{k}{m}\right)$$

Where $B_n(x)$ is Bernoulli polynomial

I know it is already proved by Joseph Ludwig Raabe, but I don`t know how exactly.

Answer 1

I use the generating function $\frac{te^{xt}}{e^t-1}=\sum_{n=0}^{\infty} B_n(x)\frac{t^n}{n!}$ and $\frac{1}{e^{mt}-1}=\frac{1+e^t+e^{2t} +... +e^{(m-1)t}}{e^m-1}$ $$ \sum_{n=0}^{\infty} B_n(mx)\frac{t^n}{n!} =\frac{te^{mxt}}{e^t-1} =\frac{te^{mxt}(1+e^t+e^{2t}+....+e^{(m-1)t})}{e^{mt}-1} =\sum_{k=0}^{m-1}\frac{te^{(mx+k)t}}{e^{mt}-1} $$ $$ =\frac{1}{m}\sum_{k=0}^{m-1}\frac{(mt)e^{(x+\frac{k}{m})(mt)}}{e^{mt}-1} =\frac{1}{m}\sum_{k=0}^{m-1}\sum_{n=0}^{\infty} B_n(x+\frac{k}{m})\frac{(mt)^n}{n!} =\sum_{n=0}^{\infty}\left [ m^{n-1}\sum_{k=0}^{m-1} B_n\left(x+\frac{k}{m}\right) \right ]\frac{t^n}{n!} $$ comparing coefficients you get your proof.

Thanks for the problem ,it was fun.

Edit1: typo pointed out by @Herman

Answer 2

By induction over $n$.

If $n=1$ we have from definition of $B_1(x)=x-1/2$,

\begin{eqnarray*} B_1(mx) = mx - \frac{1}{2}, \end{eqnarray*} On the other hand, the right hand side with $n=1$ is

\begin{eqnarray*} \sum_{k=0}^{m-1} B_1 \left ( x +\frac{k}{m} \right ) &=& \sum_{k=0}^{m-1} \left ( x+ \frac{k}{m} - \frac{1}{2} \right ) \\ &=& mx + \frac{1}{m} \frac{m(m-1)}{2} - \frac{m}{2} \\ &=& mx - \frac{1}{2} \end{eqnarray*}

Assume that for $n-1$ the equality holds.

Define $f_n(x) = B_n(mx) - m^{n-1} \sum_{k=0}^{m-1} B_n \left ( x + \frac{k}{m} \right )$. Since $dB_n(x)/dx= n B_{n-1}(x)$, The derivative of $f_n$ is given by

\begin{eqnarray*} f'_n(x) &=& m n B_{n-1}(mx) - m^{n-1} \sum_{k=0}^{m-1} \frac{n}{m} B_{n-1} \left ( x + \frac{k}{m} \right ) \\ &=& m n f_{n-1}(x) \\ &=& 0 \quad \text { by the induction hypothesis on $n-1$ }. \end{eqnarray*}

Hence $f_n(x)=c_n$ for $c_n$ constant. To find the constant note that since $\int_0^1 B_n(x) dx = 0$, then

\begin{eqnarray*} \int_0^{1/m} f_n(x) &=& \int_0^{1/m} \left [ B_n(mx) - m^{n-1} \sum_{k=0}^{m-1} B_n \left ( x + \frac{k}{m} \right ) \right ] dx \\ &=& \frac{1}{m} \int_0^1 B_n(y) dy - m^{n-1} \sum_{k=0}^{m-1} \int_0^{1/m} B_n \left ( x + \frac{k}{m} \right ) dx \\ &=& 0 + m^{n-1} \sum_{k=0}^{m-1} \int_{k/m}^{(k+1)/m} B_n \left ( y \right ) dy \\ &=& 0 + m^{n-1} \sum_{k=0}^{m-1} \int_0^1 B_n \left ( y \right ) dy \\ &=& 0. \end{eqnarray*}

So, since the integral of a constant in the interval $[0,1/m]$, $m > 0$, is 0, that constant has to be zero. Hence $f_n(x)=0$ as desired.

Answer 3

I will give a non-inductive proof. We begin by considering the case $k > 1$ (the case $k = 1$ is trivial). Recall that the Fourier series for $B_k(x)$ on the interval $[0, 1]$ is given by $$B_k(\{x\}) = k! \sum_{m \neq 0} (2 \pi i m)^{-k} e(mx).$$ Assume first that $0 < x < 1/q$. Then, $$B_k(qx) q^{1-k} = k! \sum_{m \neq 0} (2 \pi i m q)^{-k} q e(mqx) = k! \sum_{\substack{s \equiv 0 \pmod{q} \\ s \neq 0}} (2 \pi i s)^{-k} q e(sx).$$ Note that the last sum only includes multiples of $q$. However, we can have a sum over the integers using the fact that $$\sum_{0 \leq a \leq q-1} e(s(x + a/q)) = \begin{cases} 0, & \text{if } s \not\equiv 0 \pmod{q} \\ qe(sx), & \text{if } s \equiv 0 \pmod{q} \end{cases}.$$ Hence, $$B_k(qx) q^{1-k} = \sum_{0 \leq a \leq q-1} k! \sum_{m \neq 0} (2 \pi i m)^{-k} e(m(x + a/q)).$$ We can see that each term in the inner sum corresponds to the Fourier series representation of $B_k(x + a/q)$ since $0 < x + a/q < 1$ for $0 \leq a \leq q-1$. The result for arbitrary $x$ follows from this and the observation that $$ f_{k,q}(x) := B_k(qx) q^{1-k} - \sum_{0 \leq a \leq q-1} B_k(x + a/q)$$ is a polynomial in $x$ that vanishes over an entire interval, namely $(0,1/q)$. Because a polynomial can have an uncountable number of roots only if it is identically zero, we conclude that $f_{k,q}(x) \equiv 0$.