Bernoulli Polynomial

**Cairo** · March 31st 2011, 09:53 AM

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

I am trying to prove that

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

but failing miserably, despite trying to integrate by parts.

Any takers?

**Drexel28** · March 31st 2011, 01:39 PM

Originally Posted by Cairo

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

I am trying to prove that

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

but failing miserably, despite trying to integrate by parts.

Any takers?

What definition are you taking as the Bernoulli polynomials?

**Cairo** · March 31st 2011, 09:40 PM

Sorry. I should have said that

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

where

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

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