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Math Help - [SOLVED] Logarithmic derivation

  1. #1
    Newbie Klaus's Avatar
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    [SOLVED] Logarithmic derivation

    Good Morning everybody,
    I'd like to show the equality
    \frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^{ms}}
    By logarithmic differenciation(using Euler product for zeta), I get
    \frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}
    But I can't figure how to reach the last step, to the double sum.
    Could someone help me ?
    Thank you
    Last edited by Klaus; May 3rd 2008 at 05:20 AM. Reason: LaTex error, thank you Isomorphism.
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by Klaus View Post
    Good Morning everybody,
    I'd like to show the equality
    \frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^ms}
    By logarithmic differenciation(using Euler product for zeta), I get
    \frac{\zeta'(s)}{\zeta(s)}=-\sum_p (\ln(1-p^{-s}))'=-\sum_p \frac{p^{-s}\ln p}{1-p^{-s}}
    But I can't figure how to reach the last step, to the double sum.
    Could someone help me ?
    Thank you

    Shouldnt it read \frac{\zeta'(s)}{\zeta(s)}=-\sum_p\sum_{m=1} \frac{\ln p}{p^{ms}} ??

    You are nearly there

    Can you see the geometric progression below?
     -\sum_p \frac{p^{-s}}{1-p^{-s}}=-\sum_p \sum_m (p^{-s})^m
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  3. #3
    Newbie Klaus's Avatar
    Joined
    May 2008
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    Oh yeah, that's it !
    I hadn't thought of geometric progessions
    Thank you very much Isomorphism(what a cool name!)
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