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Bernoulli polynomials evaluated at 1/4

  1. Apr 11, 2010 #1
    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.
     
  2. jcsd
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