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I would appreciate help, please, as to how to verify this relation from Kato's "Fermat's Dream" p.96.

He say: By the definition of $B_n(x)$, the Bernoulli polynomial, we have

$$\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n = \frac{u e^{xu}}{e^u - 1}$$

The definition for Bernoulli polynomials is, for $n \in \mathbb{N}$ is

$$B_n(x) = \sum_{k=0}^{n} \binom{n}{k}B_k x^{n - k}$$

I am trying

$$\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n = \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}B_k x^{n - k} u^n$$

Also for Bernoulli numbers there is,

$$\sum_{k=0}^{\infty}\frac{B_k}{n!} x^k = \frac{x}{e^x -1}$$

I don't know if what I am trying is on the right track. Either way, I would appreciate help pulling it all together. Thanks very much.

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Remember that the method of generating functions works with formal power series. Using the generating series for the Bernoulli numbers and the exponential series we have: $$\frac{u e^{xu}}{e^u - 1} = \frac{u}{e^u - 1} \cdot e^{xu} = \left(\sum_{k=0}^{\infty}\frac{B_k}{k!} u^k\right) \left(\sum_{k=0}^{\infty}\frac{(xu)^k}{k!}\right) $$ Now the rule for multiplication of power series gives: \begin{align}\left(\sum_{k=0}^{\infty}\frac{B_k}{k!} u^k\right) \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}u^k\right) &= \sum_{n=0}^{\infty}\left( \sum_{k=0}^{n} \frac{B_k}{k!} \frac{x^{n-k}}{(n-k)!} \right) u^n \\ &= \sum_{n=0}^{\infty}\left(\frac{1}{n!} \sum_{k=0}^{n}n! \frac{B_k}{k!} \frac{x^{n-k}}{(n-k)!} \right) u^n \\ &= \sum_{n=0}^{\infty}\frac{ \sum_{k=0}^{n} {n \choose k}B_k x^{n-k}}{n!} u^n\\ &=:\sum_{n=0}^{\infty}\frac{B_n(x)}{n!}u^n \end{align} And from this you can read-off the formula for the Bernoulli polynomials $$B_n(x) = \sum_{k=0}^{n} {n \choose k}B_k x^{n - k}$$

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I am going to basically steal gammatester's answer and say it backwards, which I think is a much more intuitive way to look at things.

You are trying to compute $G(u) = \sum_{n=0}^\infty \frac{B_n(x)}{n!}u^n$, which is exactly the exponential generating function of the sequence $\{B_0 (x), B_1 (x), B_2(x), \ldots\}$ (we can, and should, treat $x$ as a formal symbol rather than a variable—it could just as well be $\pi$ in fact).

The sequence is defined by $B_n (x) = \sum_{k=0}^n {n\choose k} B_k x^{n-k}$. But $c_n =\sum_{k=0}^n {n\choose k} a_k b_{n-k}$ is a type of convolution product of sequences which behaves behaves remarkably well with exponential generating functions. Specifically, if $F$ and $G$ are the e.g.f.'s of $\{a_n\}$ and $\{b_n\}$, respectively, then $FG$ is the exponential generating function of $\{c_n\}$.

This means that we can reduce our problem to two much simpler problems:

  1. Find the exponential generating function of $\{x^n\}$.
  2. Find the exponential generating function of $\{B_n\}$.

But #1 is straightforward: $\sum_{n=0}^\infty \frac{x^n}{n!} u^n = e^{xu}$. And #2 has already been given to you!—$\frac{u}{e^u-1}$ (make sure, in both cases, to remember that $u$, not $x$, is the variable for our functions).

So, an experienced generatingfunctionologist can look at this problem and instantly read off that the answer is the product of those two functions.

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  • $\begingroup$ Dear user33433 - I have been quite torn as to which of the two excellent answers to accept. I find it a difficult situation since, e.g., in this specific case, your answer is a particularly instructive way to look at the problem. In this regard, I posted a question on meta: meta.math.stackexchange.com/questions/11217/…. And I would be delighted to make you the first recipient if such a mechanism becomes available. With regards, $\endgroup$ – user12802 Oct 9 '13 at 15:55
  • $\begingroup$ @Andrew I thank you for your kind words, and I graciously yield to my competitor. Don't worry; there's more where that came from. $\endgroup$ – Slade Oct 9 '13 at 17:33
  • $\begingroup$ Hi again. The question I posted on meta instantaneously got more downvotes than fleas on a dog - or some more fitting analogy. I am taking matters into my own hands. After studying up on egf's I can much more appreciate the merits of your answer. In fact it introduced me to a great concept that I otherwise would probably never have heard of. As an uneducated self-studier, I derive a great deal of benefit from the generosity and patience of others. So +50 for a great lesson and the wit - generatingfunctionologist - from the Wilf title. It takes 24 hrs to make official even counted to seconds $\endgroup$ – user12802 Oct 13 '13 at 11:54
  • $\begingroup$ @Andrew Much obliged. I'll throw in a bit of trivia: generatingfunctionology is one of the very few words with every vowel in it, including Y. $\endgroup$ – Slade Oct 13 '13 at 12:05

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