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Math Help - Asymptotic relation to a function

  1. #1
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    Asymptotic relation to a function

    Given the function,
    \zeta (M,s)=\zeta (x)\prod_{p\leq M}(1-\frac{1}{p^{s}})
    where the product is taken over primes p.
    How do you find the asymptotic relation
    \ln \zeta (7,s)\sim \frac{1}{11^{s}}

    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Asymptotic relation to a function

    Quote Originally Posted by mathematicaphoenix View Post
    Given the function,
    \zeta (M,s)=\zeta (x)\prod_{p\leq M}(1-\frac{1}{p^{s}})
    where the product is taken over primes p.
    How do you find the asymptotic relation
    \ln \zeta (7,s)\sim \frac{1}{11^{s}}

    Thanks
    By definition is...

    \zeta(M,s)= \zeta(s)\ \prod_{p \le M} (1-\frac{1}{p^{s}}) = \frac{\prod_{p \le M} (1-\frac{1}{p^{s}})}{\prod_{p} (1-\frac{1}{p^{s}})}= \frac{1}{\prod_{p >M} (1-\frac{1}{p^{s}})} (1)

    Proceeding as in...

    http://www.mathhelpforum.com/math-he...on-194803.html

    ... You find...

    \ln \zeta(M,s)= - \sum_ {p>M} \ln (1-\frac{1}{p^{s}}) =

    = \sum_{p>M} (\frac{1}{p^{s}} + \frac{1}{2\ p^{2 s}} + \frac{1}{3\ p^{3 s}} +...) (2)

    For M=7 You obtain as in the post I indicated ...

    \lim_{s \rightarrow + \infty} 11^{s}\ \ln \zeta(7,s) = 1 (3)

    ... that can be written as...

    \ln \zeta(7,s) \sim \frac{1}{11^{s}} (4)

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Asymptotic relation to a function

    Thanks. Actually I tried first as in the other post. But applied laws of logarithm first an couldn't get (1).
    Anyway I got it now. Thanks a lot once again.

    Regards
    Nabigh
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