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Math Help - A series for Zeta^2

  1. #1
    jlb
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    A series for Zeta^2

    I have "derived" (I am probably not the 1st one to do so) the following equation for the norm of Zeta^2:


    \| \zeta \left( a, ib \right) \| ^2=\zeta \left( 2a, 0 \right) \left( 1 + \sum^{\infty}_{j=2} \frac{2^{\omega \left(j\right) } \cos(n_1 b \ln(p_1)) \ldots \cos(n_k b \ln(p_k))}{j^a} \right)

    Where each term j has prime factorization j={p_1}^{n_1} {p_2}^{n_2}..., and \omega \left(j\right) is the number of different prime factors. For a=Re\left( s \right) > 1, the right-hand sum of converges absolutely.

    I am trying to figure out how to show if the series diverges/converges or semi-converges for 0.5< a < 1. (Obviously, if the series did converge in that range that would prove the RH so it is a bit unlikely to do so...but I do not know how to approach the proof.)

    The only reference to that series I could find is a much simpler form, in the special case of b=0 where it reduces to the known series (e.g. displayed in Wolfram, and in Hardy and Wright 5th edition):

    \frac {  \| \zeta \left( a \right) \| ^2   }   { \zeta \left( 2a \right)   }  = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}

    Has anyone seen the above series and where? Thank you!


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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by jlb View Post
    I have "derived" (I am probably not the 1st one to do so) the following equation for the norm of Zeta^2:


    \| \zeta \left( a, ib \right) \| ^2=\zeta \left( 2a, 0 \right) \left( 1 + \sum^{\infty}_{j=2} \frac{2^{\omega \left(j\right) } \cos(n_1 b \ln(p_1)) \ldots \cos(n_k b \ln(p_k))}{j^a} \right)

    Where each term j has prime factorization j={p_1}^{n_1} {p_2}^{n_2}..., and \omega \left(j\right) is the number of different prime factors. For a=Re\left( s \right) > 1, the right-hand sum of converges absolutely.

    I am trying to figure out how to show if the series diverges/converges or semi-converges for 0.5< a < 1. (Obviously, if the series did converge in that range that would prove the RH so it is a bit unlikely to do so...but I do not know how to approach the proof.)

    The only reference to that series I could find is a much simpler form, in the special case of b=0 where it reduces to the known series (e.g. displayed in Wolfram, and in Hardy and Wright 5th edition):

    \frac {  \| \zeta \left( a \right) \| ^2   }   { \zeta \left( 2a \right)   }  = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}

    Has anyone seen the above series and where? Thank you!


    In an analytic number theory course I've taken, there was a homework problem that asked us to show \frac {  \zeta^2 \left( a \right)  }   { \zeta \left( 2a \right)   }  = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}  , for  \Re(a)>1 .

    Proof:

     \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a} = \prod_{p} \sum_{k=0}^\infty \frac{2^{\omega(p^k)}}{p^{ak}} since  \frac{2^{\omega \left(j\right) }} {j^a} is multiplicative.

     = \prod_{p} \left(1+2\sum_{k=1}^\infty \frac{1}{p^{ak}}\right) = \prod_{p} \left(1+\frac{2p^{-a}}{1-p^{-a}}\right)

     = \prod_{p} \frac{1+p^{-a}}{1-p^{-a}} = \prod_{p} \frac{1-p^{-2a}}{(1-p^{-a})^2}
     = \frac{\zeta^2(a)}{\zeta(2a)} ,  \Re(a)>1



    With that said I'm wondering how you got  \| \zeta \left( a \right) \| ^2 instead of  \zeta(a)^2 like I did...
    Last edited by chiph588@; June 9th 2010 at 11:59 AM.
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  3. #3
    jlb
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    Quote Originally Posted by chiph588@ View Post
    In an analytic number theory course I've taken, there was a homework problem that asked us to show \frac {  \zeta^2 \left( a \right)  }   { \zeta \left( 2a \right)   }  = \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a}  , for  \Re{a}>1 .

    Proof:

     \sum^{\infty}_{j=1} \frac{2^{\omega \left(j\right) }} {j^a} = \prod_{p} \sum_{k=0}^\infty \frac{2^{\omega(p^k)}}{p^{ak}} since  \frac{2^{\omega \left(j\right) }} {j^a} is multiplicative.

     = \prod_{p} \left(1+2\sum_{k=1}^\infty \frac{1}{p^{ak}}\right) = \prod_{p} \left(1+\frac{2p^{-a}}{1-p^{-a}}\right)

     = \prod_{p} \frac{1+p^{-a}}{1-p^{-a}} = \prod_{p} \frac{1-p^{-2a}}{(1-p^{-a})^2}
     = \frac{\zeta^2(a)}{\zeta(2a)} ,  \Re{a}>1



    With that said I'm wondering how you got  \| \zeta \left( a \right) \| ^2 instead of  \zeta(a)^2 like I did...
    To get the norm of Zeta(s), s being complex and s= a+ ib,

    I derived the following identity:

     \| \frac {1}{1-1/p^{s}} \|^2  =  \left( \frac {1}{1-1/p^{2a}} \right)<br />
 \left(1 + 2 \sum^{\infty}_{j=1} \frac {\cos \left( jb \ln {p} \right)}{p^{ja}} \right) \tag{1}


    where the right-hand sum is absolutely convergent for  Re\left( s \right) > 0. Note that equation (1) is not the Fourier expansion of the left-hand side, but a real function of 2 independent variables.


    I then take equation (1) and use it in an Euler product to get  \| \zeta \left( s \right) \| ^2 .
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