A property of Bernoulli numbers

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A property of Bernoulli numbers

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Gauler

#1 h Mar 31, 2025, 6:15 PM

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Define the Bernoulli numbers $B_n$ as
$\frac{x}{e^x-1}=\sum_{n\ge 0}B_n \frac{x^n}{n!}$ and the Bernoulli polynomials $B_n(t)$ as
$\frac{xe^{tx}}{e^x-1}=\sum_{n\ge 0} B_n(t)\frac{x^n}{n!}$ for $t\in \mathbb R$ .

From these definitions, prove that
$B_n\left(t\right)=\sum _{m=0}^n\binom{n}{m}B_{n-m}t^m$ and
$\int_a^x B_n(t) dt=\frac{B_{n+1}(x)-B_{n+1}(a)}{n+1}$ for all $n\ge 1$ .

This post has been edited 1 time. Last edited by Gauler, Mar 31, 2025, 6:29 PM

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Etkan

#2 h Mar 31, 2025, 7:03 PM • 2 Y

Y by MihaiT, Gauler

Gauler wrote:

Define the Bernoulli numbers $B_n$ as
$\frac{x}{e^x-1}=\sum_{n\ge 0}B_n \frac{x^n}{n!}$ and the Bernoulli polynomials $B_n(t)$ as
$\frac{xe^{tx}}{e^x-1}=\sum_{n\ge 0} B_n(t)\frac{x^n}{n!}$ for $t\in \mathbb R$ .

From these definitions, prove that
$B_n\left(t\right)=\sum _{m=0}^n\binom{n}{m}B_{n-m}t^m$ and
$\int_a^x B_n(t) dt=\frac{B_{n+1}(x)-B_{n+1}(a)}{n+1}$ for all $n\ge 1$ .

We have $\begin{align*}\sum \limits _{n=0}^\infty \left (\sum \limits _{m=0}^n\binom{n}{m}B_{n-m}t^m\right )\frac{x^n}{n!} & =\sum \limits _{n=0}^\infty \sum \limits _{m=0}^n\binom{n}{m}B_{n-m}t^m\frac{x^n}{n!} \\ & =\sum \limits _{m=0}^\infty \sum \limits _{n=m}^\infty \binom{n}{m}B_{n-m}t^m\frac{x^n}{n!} \\ & =\sum \limits _{m=0}^\infty t^m\sum \limits _{n=m}^\infty \binom{n}{m}B_{n-m}\frac{x^n}{n!} \\ & =\sum \limits _{m=0}^\infty t^m\sum \limits _{n=0}^\infty \binom{n+m}{m}B_n\frac{x^{n+m}}{(n+m)!} \\ & =\sum \limits _{m=0}^\infty t^m\sum \limits _{n=0}^\infty \frac{(n+m)!}{n!m!}B_n\frac{x^nx^m}{(n+m)!} \\ & =\sum \limits _{m=0}^\infty t^m\sum \limits _{n=0}^\infty B_n\frac{x^n}{n!}\frac{x^m}{m!} \\ & =\sum \limits _{m=0}^\infty t^m\frac{x^m}{m!}\sum \limits _{n=0}^\infty B_n\frac{x^n}{n!} \\ & =\sum \limits _{m=0}^\infty \frac{(tx)^m}{m!}\frac{x}{e^x-1} \\ & =\frac{x}{e^x-1}\sum \limits _{m=0}^\infty \frac{(tx)^m}{m!} \\ & =\frac{x}{e^x-1}e^{tx} \\ & =\frac{xe^{tx}}{e^x-1} \\ & =\sum \limits _{n=0}^\infty B_n(t)\frac{x^n}{n!}, \end{align*}$ which implies that $B_n(t)=\sum \limits _{m=0}^n\binom{n}{m}B_{n-m}t^m$ for all $n\in \mathbb{N}_0$ . Hence $\begin{align*}\int \limits _a^xB_n(t)\,dt & =\int \limits _a^x\sum \limits _{m=0}^n\binom{n}{m}B_{n-m}t^m\,dt \\ & =\sum \limits _{m=0}^n\binom{n}{m}B_{n-m}\int \limits _a^xt^m\,dt \\ & =\sum \limits _{m=0}^n\binom{n}{m}B_{n-m}\frac{x^{m+1}-a^{m+1}}{m+1} \\ & =\sum \limits _{m=0}^n\frac{n!}{m!(n-m)!}B_{n-m}\frac{x^{m+1}-a^{m+1}}{m+1} \\ & =\sum \limits _{m=0}^n\frac{(n+1)!}{(m+1)!((n+1)-(m+1))!}B_{(n+1)-(m+1)}\frac{x^{m+1}-a^{m+1}}{n+1} \\ & =\sum \limits _{m=0}^n\binom{n+1}{m+1}B_{(n+1)-(m+1)}\frac{x^{m+1}-a^{m+1}}{n+1} \\ & =\sum \limits _{m=1}^{n+1}\binom{n+1}{m}B_{(n+1)-m}\frac{x^m-a^m}{n+1} \\ & =\sum \limits _{m=0}^{n+1}\binom{n+1}{m}B_{(n+1)-m}\frac{x^m-a^m}{n+1} \\ & =\frac{B_{n+1}(x)-B_{n+1}(a)}{n+1} \end{align*}$ for all $n\in \mathbb{N}_0$ .

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