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Zeta function over primes

  1. May 15, 2009 #1
    where could i get some info about the function

    [tex] \sum_{p} p^{-s}=P(s) [/tex]

    * the functional equation relating P(s) and P(1-s)

    * the relation with Riemann zeta
     
  2. jcsd
  3. May 15, 2009 #2
    You can express it as a summation over Riemann zeta's multiplied by a Möbius function. We have:

    [tex]\zeta(s) = \sum_{r_{1},r_{2}\ldots}\prod_{j}p_{j}^{-sr_{j}}[/tex]

    where [itex]p_{j}[/itex] is the jth prime and the [itex]r_{j}[/itex] in the summation range from zero to infinity. Summing over the [itex]r_{j}[/itex] gives:

    [tex]\zeta(s)= \prod_{p}\frac{1}{1-p^{-s}}[/tex]

    Take the log of both sides:

    [tex]\log\left[\zeta(s)\right]= -\sum_{p}\log\left(1-p^{-s}\right)[/tex]

    Expand the logarithm and sum over the primes p:

    [tex]\log\left[\zeta(s)\right]=\sum_{k=1}^{\infty}\frac{P(ks)}{k}[/tex]

    You can then invert this relation to find the [itex]P(s)[/itex] using Möbius inversion.
     
  4. May 15, 2009 #3
    So, you find:

    [tex]P(s) = \log\left[\zeta(s)\right] - \sum_{p}\frac{\log\left[\zeta(ps)\right]}{p} + \sum_{p_{1}<p_{2}}\frac{\log\left[\zeta(p_{1}p_{2}s)\right]}{p_{1}p_{2}}- \sum_{p_{1}<p_{2}<p_{3}}\frac{\log\left[\zeta(p_{1}p_{2}p_{3}s)\right]}{p_{1}p_{2}p_{3}}+\cdots[/tex]
     
  5. May 16, 2009 #4
    I think [tex]P(s)[/tex] as defined above by Count Iblis, can be written as

    [tex]P(s)= \log \zeta(s)+\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}(-1)^{m+1}\frac{\log\zeta\Big(s\prod_{k=0}^{m}p_{n+k}\Big)}{\prod_{k=0}^{m} p_{n+k}}.
    [/tex]

    I assume that's for [tex]\textnormal{Re}(s)>1[/tex].
     
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