# Verifying a relation involving Bernoulli polynomials

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.

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

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.

• 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, – user12802 Oct 9 '13 at 15:55
• @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. – Slade Oct 9 '13 at 17:33
• 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 – user12802 Oct 13 '13 at 11:54
• @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. – Slade Oct 13 '13 at 12:05