Results 1 to 3 of 3

Thread: hurwitz zeta function

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    16

    hurwitz zeta function

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


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

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

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

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

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Quote Originally Posted by klw289 View Post
    I'm trying to calculate $\displaystyle \zeta$(4,$\displaystyle \frac{1}{2}$)


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

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

    So $\displaystyle \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...

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

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

    ... so that is...

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Renji Rodrigo's Avatar
    Joined
    Sep 2009
    From
    Rio de janeiro
    Posts
    38
    A little generalization xD

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


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

    we have


    $\displaystyle \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 .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. zeta function
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Dec 7th 2010, 06:09 PM
  2. More on the zeta function....
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Mar 1st 2010, 12:15 PM
  3. zeta function
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Feb 27th 2010, 02:11 PM
  4. Zeta Function Proof
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: Jan 28th 2010, 05:17 PM
  5. Riemann Zeta function - Zeta(0) != Infinity ??
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 8th 2009, 12:50 AM

Search Tags


/mathhelpforum @mathhelpforum