(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The problem is to prove the identity

[tex]B_k(1/4) = 2^{-k} B_k(1/2)[/tex]

for even k.

2. Relevant equations

The Bernoulli polynomials [tex]B_k(y)[/tex] are defined by the generating function relation

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

3. The attempt at a solution

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

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

The left side is

[tex]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}[/tex]

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

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

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

[tex]2^{1-k} B_k - B_k = B_k(1/2)[/tex]

or

[tex]B_k(1/2) = (2^{1-k} - 1)B_k.[/tex]

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

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# Homework Help: Bernoulli polynomials evaluated at 1/4

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