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I got a little question regarding the Bernoulli polynomials/numbers. Basically, I want to show that $$B_n(0) = -\frac{B_n(2\pi i)^n}{n!}$$ Where $B_n(x)$ the $n$-th Bernoulli polynomial and $B_n$ the $n$-th Bernoulli number). I have gotten quite far, I think, but now I seem to be stuck:

I got that $\sum_{k=1}^n B_{n-k}(0)\frac{(2\pi i)^k}{k!}=0=\sum_{k=1}^n\frac{B_{n-k}}{(n-k)!k!}$. Now, I read up on it a bit, and usually what I find is something along the lines of "we compare the coefficients and find the above identity". However, I don't see how we can arrive at that conclusion (I see that we can reindex the sums, such that it looks about right though). My main problem is: how can we know that if we look at the k-th component in the sum, they have to match up. Silly example, but $1+2+3=3+2+1$, there, the sum is the same, but the elements don't match up (if you don't reorganize them).

Can somebody help me out, I'd appreciate it!

graydad
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Hannelore
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  • What is your definition of Bernoulli polynomials? – Olivier Bégassat Jan 06 '15 at 23:05
  • Usually, they are defined over a generating function, however, in our problem we did it by taking $\sum \frac{1}{k^n}e^{i k x}$ and then assuming that this is a fourier series of the bernoulli polynomials, then finding them over an ansatz.The definition is: $B_n(x) = \sum_{k=0}^n B_{n-k}(0) \frac{(i x)^k}{k!}$. – Hannelore Jan 06 '15 at 23:27

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