Thread: Bernoulli Polynomial

By considering $\displaystyle \overset{1}{\underset{0}{\int}}xB_{n}(x)\, dx$.

I am trying to prove that

$\displaystyle B_{n+1}=(n+1)\overset{n}{\underset{r=0}{\sum}}\lef t(\begin{array}{c}
n\\
r\end{array}\right)\frac{B_{r}}{n-r+2}$


but failing miserably, despite trying to integrate by parts.

Any takers?

Originally Posted by Cairo

By considering $\displaystyle \overset{1}{\underset{0}{\int}}xB_{n}(x)\, dx$.

I am trying to prove that

$\displaystyle B_{n+1}=(n+1)\overset{n}{\underset{r=0}{\sum}}\lef t(\begin{array}{c}
n\\
r\end{array}\right)\frac{B_{r}}{n-r+2}$


but failing miserably, despite trying to integrate by parts.

Any takers?

What definition are you taking as the Bernoulli polynomials?

Sorry. I should have said that

$\displaystyle \frac{ze^{xz}}{e^{z}-1}=\overset{\infty}{\underset{n=0}{\sum}}\frac{B_{ n}(x)}{n!}z^{n}$

where

$\displaystyle B_{n}(x)=\overset{\infty}{\underset{k=0}{\sum}}\le ft(\begin{array}{c}
n\\
k\end{array}\right)B_{k}x^{n-k}$.