# Bernoulli polynomials evaluated at 1/4

1. Apr 11, 2010

### zpconn

1. The problem statement, all variables and given/known data

The problem is to prove the identity

$$B_k(1/4) = 2^{-k} B_k(1/2)$$

for even k.

2. Relevant equations

The Bernoulli polynomials $$B_k(y)$$ are defined by the generating function relation

$$\frac{xe^{xy}}{e^x-1} = \sum_{k=0}^{\infty} \frac{B_k(y) x^k}{k!}.$$

3. The attempt at a solution

I have obtained an identity for $$B_k(1/2)$$. From the defining relation we find

$$\frac{xe^{x/2}}{e^x-1} = \sum_{k=0}^{\infty} \frac{B_k(1/2)x^k}{k!}.$$

The left side is

$$x\frac{e^{x/2} + 1}{e^x - 1} - \frac{x}{e^x-1} = \frac{x}{e^{x/2} - 1} - \frac{x}{e^x-1}$$

and this is, using the defining relation for the Bernoulli polynomials and for the ordinary Bernoulli numbers,

$$\sum_{k=0}^{\infty} \frac{2 B_k (x/2)^k}{k!} - \sum_{k=0}^{\infty} \frac{B_k x^k}{k!}$$.

For absolute clarity, in this expression $$B_k$$ is the kth Bernoulli number, not polynomial. Now equating coefficients of $$x^k / k!$$ yields

$$2^{1-k} B_k - B_k = B_k(1/2)$$

or

$$B_k(1/2) = (2^{1-k} - 1)B_k.$$

I'm convinced the same approach will work for $$B_k(1/4)$$ but the algebra gets a bit messy, so I'm having difficulty seeing the solution.