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Math Help - hurwitz zeta function

  1. #1
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    hurwitz zeta function

    I'm trying to calculate \zeta(4, \frac{1}{2})


    I used \zeta (s,a) = \Gamma (1-s) I(s,a)

    Am I right by saying \Gamma (1-s) is undefined for s=4 and
    I(4, 1/2) = 0 as s is a positive whole number

    So \zeta (4, 1/2) =0

    Am I doing this right or am I just making a mess of this

    Thanks in advance
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by klw289 View Post
    I'm trying to calculate \zeta(4, \frac{1}{2})


    I used \zeta (s,a) = \Gamma (1-s) I(s,a)

    Am I right by saying \Gamma (1-s) is undefined for s=4 and
    I(4, 1/2) = 0 as s is a positive whole number

    So \zeta (4, 1/2) =0

    Am I doing this right or am I just making a mess of this

    Thanks in advance
    Better chance You have using the relation...

    \zeta(s,\frac{1}{2}) = \sum_{n=0}^{\infty} (n+\frac{1}{2})^{-s} = 2^{s}\ \sum_{n=0}^{\infty} (2n+1)^{-s} =

    = 2^{s}\ (1-2^{-s})\ \zeta (s)= (2^{s}-1)\ \zeta(s) (1)

    ... so that is...

    \zeta(4,\frac{1}{2}) = 15\ \zeta(4) = \frac{\pi^{4}}{6}

    Kind regards

    \chi \sigma
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  3. #3
    Junior Member Renji Rodrigo's Avatar
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    A little generalization xD

    Calculate
     \zeta(2n,\frac{1}{2})
    where n \in N.


    \zeta(2n,\frac{1}{2})=\sum_{k=0}^{\infty}\frac{1}{  (k+\frac{1}{2})^{2n}} =2^{2n}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{2n}}


    if you know
    \sum^{\infty}_{k=0}\frac{1}{(2k+1)^{2n}}=\frac{B_{  2n}(-1)^{n+1}.[4^{n}-1]\pi^{2n}}{2(2n)!}

    we have


    \zeta(2n,\frac{1}{2}) =\frac{4^nB_{2n}(-1)^{n+1}.[4^{n}-1]\pi^{2n}}{2(2n)!}

    where B_n is the n-th Bernoulli number .
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