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The series for some day to day functions such as $\tan z$ and $\cot z$ involve them. So does the series for $\dfrac{z}{e^z-1}$ and the Euler Maclaurin summation formula. How can it be analitically shown that their generating functions are:

$$\frac{z}{e^z-1}=\sum\limits_{n=0}^\infty B_n \frac{z^n}{n!}$$

$$\operatorname{sech} z =\sum\limits_{n=0}^\infty E_n \frac{z^n}{n!}$$

as an introductory example?

I've read a little about $A_n$ numbers and their relation to the tangent and secant, which might lead to a new question, but I guess I'm OK with a proof of the first two series.

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What in the world are "$A_n$ numbers"?!? Anyway, why not just look up Bernoulli and Euler numbers on Wikipedia? – KCd Apr 1 '12 at 5:02
@KCd $A_n$ numbers are the alternating permutation numbers, as called in Wikipedia. There is no proof there, that's why I'm asking. – Pedro Tamaroff Apr 1 '12 at 5:13
What definitions are you using for $B_n$ and $E_n$? I ask because the series you give are often used as the definitions and therefore are (by definition!) correct. – Gerry Myerson Apr 1 '12 at 6:12
@GerryMyerson Oh! I see. What I mean is, how do you show that they generate the given funciton? – Pedro Tamaroff Apr 1 '12 at 6:15
If you define $B_n$ to be the coefficients in the Taylor series for $z/(e^z-1)$, then showing they generate that function is just a matter a pointing to the definition. And many sources define $B_n$ exactly that way. That's why I asked you, how are you defining these numbers? – Gerry Myerson Apr 1 '12 at 6:21
up vote 4 down vote accepted

For the Bernoulli numbers, take a look at Graham, Knuth, & Patashnik, Concrete Mathematics, Sections 6.5 and 7.6. In 6.5 they define the Bernoulli numbers $B_k$ by the implicit recurrence $$\sum_{j=0}^m\binom{m+1}jB_j=[m=0]\tag{1}$$ for all $m\ge 0$. (The righthand side is an Iverson bracket.) They then prove the identity $$\sum_{k=0}^{n-1}k^m=\frac1{m+1}\sum_{k=0}^m\binom{m+1}kB_kn^{m+1-k}\;.$$

In 7.6, on exponential generating functions, they rewrite $(1)$ by substituting $n$ for $m+1$ and adding $B_n$ to both sides to get $$\sum_k\binom{n}kB_k=B_n+[n=1]\tag{2}$$ for all $n\ge 0$. The lefthand side of $(2)$ is the binomial convolution of $\langle B_n:n\in\omega\rangle$ and the constant $1$ sequence. The egf of the constant $1$ sequence is just $e^z$; let $$\widehat B(z)=\sum_{n\ge 0}B_n\frac{z^n}{n!}\;,$$ the egf of $\langle B_n:n\in\omega\rangle$. Then $$\sum_k\binom{n}kB_k=\widehat B(z)e^z\;.$$ On the other hand, the egf of the righthand side of $(2)$ is $$\sum_{n\ge 0}\Big(B_n+[n=1]\Big)\frac{z^n}{n!}=\widehat B(z)+z\;,$$ so $$\widehat B(z)e^z=\widehat B(z)+z\;,$$ and $$\widehat B(z)=\frac{z}{e^z-1}\;.$$ The relationship with the Bernoulli polynomials is explored further in the next few pages.

I can’t help with the Euler (secant) numbers: I’ve only ever seen them defined as the coefficients in the Maclaurin expansion of $\operatorname{sech} x$. You might look at Exercise 7.41 and its solution, however, since it shows the connection between the up-down numbers and the tangent and secant functions.

Added: Suppose that $\widehat A(x)$ and $\widehat B(x)$ are exponential generating functions for $\langle a_n:n\in\omega\rangle$ and $\langle b_n:n\in\omega\rangle$, respectively, so that $$\widehat A(x)=\sum_{n\ge 0}\frac{a_n}{n!}x^n\quad\text{ and }\quad\widehat B(x)=\sum_{n\ge 0}\frac{b_n}{n!}x^n\;.$$

Now let $$c_n=\sum_k\binom{n}ka_kb_{n-k}\;;$$ the sequence $\langle c_n:n\in\omega\rangle$ is the binomial convolution of $\langle a_n:n\in\omega\rangle$ and $\langle b_n:n\in\omega\rangle$. Let $\widehat C(x)$ be the egf of this binomial convolution. Then

$$\begin{align*} \widehat C(x)&=\sum_{n\ge 0}\frac{c_n}{n!}x^n\\ &=\sum_{n\ge 0}\sum_k\left(\frac{a_n}{k!}\cdot\frac{b_{n-k}}{(n-k)!}\right)x^n\\ &=\left(\sum_{n\ge 0}\frac{a_n}{n!}x^n\right)\left(\sum_{n\ge 0}\frac{b_n}{n!}x^n\right)\\ &=\widehat A(x)\widehat B(x)\;. \end{align*}$$

Just as ordinary convolution of sequences is reflected in the product of their ordinary generating functions, binomial convolution of sequences is reflected in the product of their exponential generating functions.

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I'm surprised you wrote an answer like this without up-voting the question. – Michael Hardy Apr 1 '12 at 21:25
@Michael: Sheer oversight on my part, now remedied. – Brian M. Scott Apr 1 '12 at 21:28
@Brian Great answer. But, shouldn't the second equation be $k^m$ instead of $k_m$? And, could you add a little on the binomial convolution? It's new to me. Also, I can't understand what "egf" and "rhe" is. Thank you! – Pedro Tamaroff Apr 2 '12 at 0:45
@Peter: You’re right: $k_m$ was a typo for $k^m$, and Rhe was a typo for The, both now fixed. I thought that egf would be clear in context: it’s a semi-standard abbreviation for exponential generating function. – Brian M. Scott Apr 2 '12 at 0:57
@Brian OH! Right. I'm just lost on the binomial convolution of $⟨B_n:n \in \omega⟩$. Should I check what it is on the book? – Pedro Tamaroff Apr 2 '12 at 1:03

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