Extracting Bernoulli polynomials from their generating function

Question

The generating function for Bernoulli polynomials is

$$ \frac{te^{tx}}{e^t-1} = \sum_{n=0}^\infty B_n(x) \frac{t^n}{n!}$$

The only way that I know of to get the coefficients out of this is to use Taylor's theorem. But to use Taylor's theorem, it needs to have derivatives at zero, and this doesn't, because there is always a $e^t-1$ in some denominator, which at zero is undefined.

However, if instead of evaluating things at zero, I take the limit as $t \to 0$, then it looks like it works. As an example, for the second "term",

$$ \lim_{t\to 0} \left(\frac{te^{tx}}{e^t-1} \right)^{(1)}\frac{t^0}{n!} = \frac{x-1/2}{n!} $$

Which works, since $B_1(x) = x - 1/2$.

My questions are,

1) Does this actually work, and if so, why? Are there other cases where we have formal power series which sum to something that doesn't have a Taylor series at zero?

2) Is there a better way to extract Bernoulli polynomials from that generating function (not using some other formula)?

Answer 1

As we are not permitted to use anything else we start from the definition $$\frac{te^{tx}}{e^t-1} = \sum_{n\ge 0} B_n(x) \frac{t^n}{n!}$$ and obtain $$B_n(x) = \frac{n!}{2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \frac{te^{tx}}{e^t-1} \; dt$$ and hence $$[x^q] B_n(x) = \frac{n!}{q!\times 2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n+1}} \frac{t^{q+1}}{e^t-1} \; dt \\ = \frac{n!}{q!\times 2\pi i} \int_{|t|=\epsilon} \frac{1}{t^{n-q+1}} \frac{t}{e^t-1} \; dt.$$ We see that this is zero for $q>n$ and moreover, $$[x^q] B_n(x) = \frac{n!}{q!} \frac{B_{n-q}}{(n-q)!} = {n\choose q} B_{n-q}.$$

This means we have reduced the problem of computing Bernoulli polynomials to the problem of computing Bernoulli numbers, since as we now know, $$B_n(x) = \sum_{q=0}^n {n\choose q} B_{n-q} x^q.$$

This can be done by differentiating the generating function $$\frac{t}{e^t-1} = \sum_{n\ge 0} B_n \frac{t^n}{n!}$$ to get $$\frac{1}{e^t-1} - \frac{t}{(e^t-1)^2} e^t = \sum_{n\ge 1} B_n \frac{t^{n-1}}{(n-1)!}.$$

The left is $$\frac{1}{e^t-1} - \frac{1}{e^t-1} \frac{te^t}{e^t-1} \\ = \frac{1}{t} + \left(-\frac{1}{t} + \frac{1}{e^t-1}\right) - \frac{1}{t} \frac{te^t}{e^t-1} - \left(-\frac{1}{t} + \frac{1}{e^t-1}\right) \frac{te^t}{e^t-1}.$$

Extracting coefficients from these exponential generating functions we obtain for $n\ge 0$

$$B_{n+1} = \frac{B_{n+1}}{n+1} - \frac{B_{n+1}(1)}{n+1} - \sum_{q=0}^n {n\choose q} \frac{B_{q+1}}{q+1} B_{n-q}(1).$$

Observe that $$\frac{te^t}{e^t-1} = t + \frac{t}{e^t-1}$$

and hence $$B_{n+1}(1) = B_{n+1} \quad\text{except for}\quad B_1(1) = 1 + B_1.$$

This gives for $n\ge 1$ $$B_{n+1} = - {n\choose n-1} \frac{B_n}{n} - \sum_{q=0}^n {n\choose q} \frac{B_{q+1}}{q+1} B_{n-q} \\ = - B_n - \frac{B_{n+1}}{n+1} - \sum_{q=0}^{n-1} {n\choose q} \frac{B_{q+1}}{q+1} B_{n-q}$$ or $$B_{n+1}\frac{n+2}{n+1} = - B_n - \sum_{q=0}^{n-1} {n\choose q} \frac{B_{q+1}}{q+1} B_{n-q}$$

and finally $$B_{n+1} = - \frac{n+1}{n+2} B_n - \frac{n+1}{n+2} \sum_{q=0}^{n-1} {n\choose q} \frac{B_{q+1}}{q+1} B_{n-q}.$$

This recurrence lets us calculate the $B_n$ starting from $B_0=1$ and $B_1=-\frac{1}{2}.$

Addendum Wed Apr 22 22:47:07 CEST 2015. Since I did not have a reference at hand when I wrote the above I failed to see that there is a much simpler identity which results from writing $$t = (e^t-1) \sum_{n\ge 0} B_n \frac{t^n}{n!}.$$

This is $$t = e^t \sum_{n\ge 0} B_n \frac{t^n}{n!} - \sum_{n\ge 0} B_n \frac{t^n}{n!}.$$

Extracting coefficients for $n\ge 2$ and observing the convolution of two exponential generating functions in the first term we have $$0 = \sum_{q=0}^n {n\choose q} B_q - B_n = \sum_{q=0}^{n-1} {n\choose q} B_q.$$ This yields $$B_{n-1} {n\choose n-1} = -\sum_{q=0}^{n-2} {n\choose q} B_q$$ or $$B_{n-1} = -\frac{1}{n} \sum_{q=0}^{n-2} {n\choose q} B_q \quad\text{or}\quad B_m = -\frac{1}{m+1} \sum_{q=0}^{m-1} {m+1\choose q} B_q.$$