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Math Help - asymptotic relation to a function

  1. #1
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    asymptotic relation to a function

    How do you find an asymptotic relation to a given type of function. I was reading a paper by P. Sebah and X. Gourdon , and in that paper they gave the asymptotic relation

    \ln (\zeta (sn))\sim \frac{1}{2^{sn}}

    also it is mentioned that to get 20 digits, only

    \frac{20}{\log_{10}(4)}\approx 33

    are required. I didn't know how get these results. Can someone help me?
    Last edited by mathematicaphoenix; December 31st 2011 at 05:56 PM.
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  2. #2
    Grand Panjandrum
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    Re: asymptotic relation to a function

    Quote Originally Posted by mathematicaphoenix View Post
    How do you find an asymptotic relation to a given type of function. I was reading a paper by P. Sebah and X. Gourdon , and in that paper they gave the asymptotic relation

    \ln (\zeta (s_n))\sim \frac{1}{2^{s_n}}

    also it is mentioned that to get 20 digits, only

    \frac{20}{\log_{10}(4)}\approx 33

    are required. I didn't know how get these results. Can someone help me?
    What is sn or s_n ?

    CB
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: asymptotic relation to a function

    Quote Originally Posted by mathematicaphoenix View Post
    How do you find an asymptotic relation to a given type of function. I was reading a paper by P. Sebah and X. Gourdon , and in that paper they gave the asymptotic relation

    \ln (\zeta (s_n))\sim \frac{1}{2^{s_n}}

    also it is mentioned that to get 20 digits, only

    \frac{20}{\log_{10}(4)}\approx 33

    are required. I didn't know how get these results. Can someone help me?
    Remembering the 'Euler product' ...

    \frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}}) (1)
    ... You derive immediately ...

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

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

    Now observing (2) it is not too difficult to see that…

    \lim_{s \rightarrow + \infty} 2^{s}\ \ln \zeta(s) = 1 (3)

    … and (3) is equivalent to write…

    \ln \zeta(s) \sim \frac{1}{2^{s}} (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; December 31st 2011 at 07:28 AM.
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  4. #4
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    Re: asymptotic relation to a function

    Quote Originally Posted by CaptainBlack View Post
    What is sn or s_n ?

    CB
    It's sn
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  5. #5
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    Re: asymptotic relation to a function

    Quote Originally Posted by chisigma View Post
    Remembering the 'Euler product' ...

    \frac{1}{\zeta(s)} = \prod_{p} (1-\frac{1}{p^{s}}) (1)
    ... You derive immediately ...

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

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

    Now observing (2) it is not too difficult to see that…

    \lim_{s \rightarrow + \infty} 2^{s}\ \ln \zeta(s) = 1 (3)

    … and (3) is equivalent to write…

    \ln \zeta(s) \sim \frac{1}{2^{s}} (4)

    Kind regards

    \chi \sigma
    Thanks chisigma. But I am still fuzzy about how you got
    \lim_{s \rightarrow + \infty} 2^{s}\ \ln \zeta(s) = 1
    NOTE: There is a typing mistake in the original question. Correct form is
    \ln \zeta (sn)\sim \frac{1}{2^{sn}}

    Also can you help me on how to get
    \frac{20}{\log_{10}4}
    this is the approximation to ge 20 digits in base 10!!

    Thanks
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: asymptotic relation to a function

    Writing 'term by term' we have...

    2^{s}\ \ln \zeta(s)= - 2^{s}\ \sum_{p} \ln (1-\frac{1}{p^{s}}) =

    = \frac{2^{s}}{2^{s}} + \frac{2^{s}}{2\ 2^{2 s}} + \frac{2^{s}}{3\ 2^{3 s}} + ...

    + \frac{2^{s}}{2^{s}} + \frac{2^{s}}{2\ 3^{2 s}} + \frac{2^{s}}{3\ 3^{3 s}} + ...

    +...

    + \frac{2^{s}}{p^{s}} + \frac{2^{s}}{2\ p^{2 s}} + \frac{2^{s}}{3\ p^{3 s}} + ... (1)

    ... and observing (1) You can see that if n tends to infinity the first term remains 1 and all other terms tend to 0...

    How good is the approximation for s 'large enough' given by the expression  \ln \zeta(s) \sim \frac{1}{2^{s}} must be verified, if possible, by direct computation...

    Kind regards

    \chi \sigma
    Last edited by chisigma; January 1st 2012 at 01:02 AM.
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  7. #7
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    Re: asymptotic relation to a function

    Thank you very much. I got it now.
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