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Math Help - Growth rate of Prime zeta function

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    Growth rate of Prime zeta function

    What would be the asymptotic growth rate of \zeta _{p}{(s)} or for the special cases \zeta _{p}{(2)} or \zeta _{p}{(3)}
    I tried using the relation {p}_{n}\sim n\ln (n) but couldn't get it to work. How many terms would you require to compute \zeta _{p}{(2)} and \zeta _{p}{(3)} to 20 digits using direct computation??

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    Re: Growth rate of Prime zeta function

    Quote Originally Posted by mathematicaphoenix View Post
    What would be the asymptotic growth rate of \zeta _{p}{(s)} or for the special cases \zeta _{p}{(2)} or \zeta _{p}{(3)}
    I tried using the relation {p}_{n}\sim n\ln (n) but couldn't get it to work. How many terms would you require to compute \zeta _{p}{(2)} and \zeta _{p}{(3)} to 20 digits using direct computation??

    By definition is...

    P(s)=\sum_{p} \frac{1}{p^{s}} (1)

    ... and it is immediate to verify that...

    \lim_{s \rightarrow \infty} 2^{s}\ P(s)=1 (2)

    ... so that is...

    P(s) \sim \frac{1}{2^{s}} (3)

    Kind regards

    \chi \sigma
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    Re: Growth rate of Prime zeta function

    Quote Originally Posted by chisigma View Post
    By definition is...

    P(s)=\sum_{p} \frac{1}{p^{s}} (1)

    ... and it is immediate to verify that...

    \lim_{s \rightarrow \infty} 2^{s}\ P(s)=1 (2)

    ... so that is...

    P(s) \sim \frac{1}{2^{s}} (3)

    Kind regards

    \chi \sigma
    I got that result. But how would you use (3) to calculate the the number of terms required to get P(s) to a given accuracy using direct calculation.
    For example, for \zeta (s) I have already found that, to get D digits of accuracy you would require
    n\approx 10^{\frac{D}{s}}
    terms. I want to do a similar thing with P(s). Can you help?

    Thanks
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    MHF Contributor chisigma's Avatar
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    Re: Growth rate of Prime zeta function

    Quote Originally Posted by mathematicaphoenix View Post
    I got that result. But how would you use (3) to calculate the the number of terms required to get P(s) to a given accuracy using direct calculation.
    For example, for \zeta (s) I have already found that, to get D digits of accuracy you would require
    n\approx 10^{\frac{D}{s}}
    terms. I want to do a similar thing with P(s). Can you help?

    Thanks
    If You intend to use that formula, then what You can do is to extimate the minimum value of s that gives an error < 10^{-D}. If You write...

    P(s)= \frac{1}{2^{s}} + \varepsilon (s) (1)

    ... then the condition is approximately...

    \frac{2}{3^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 3} (2)

    If You uses two terms of the series writing P(s) \sim \frac{1}{2^{s}} + \frac{1}{3^{s}} then You write...

    P(s)= \frac{1}{2^{s}} + \frac{1}{3^{s}}+ \varepsilon (s) (3)

    ... and the condition is approximately...

    \frac{2}{5^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 5} (4)

    Now You can proceed increasing the terms till to arrive to the required accuracy...

    Kind regards

    \chi \sigma
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    Re: Growth rate of Prime zeta function

    Quote Originally Posted by chisigma View Post
    If You intend to use that formula, then what You can do is to extimate the minimum value of s that gives an error < 10^{-D}. If You write...

    P(s)= \frac{1}{2^{s}} + \varepsilon (s) (1)

    ... then the condition is approximately...

    \frac{2}{3^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 3} (2)

    If You uses two terms of the series writing P(s) \sim \frac{1}{2^{s}} + \frac{1}{3^{s}} then You write...

    P(s)= \frac{1}{2^{s}} + \frac{1}{3^{s}}+ \varepsilon (s) (3)

    ... and the condition is approximately...

    \frac{2}{5^{s}}< 10^{-D} \implies s> D\ \frac{\ln 10 + \ln 2}{\ln 5} (4)

    Now You can proceed increasing the terms till to arrive to the required accuracy...

    Kind regards

    \chi \sigma
    Isn't there a way to derive a simple formula such as that for the \zeta (s)
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Growth rate of Prime zeta function

    Quote Originally Posted by mathematicaphoenix View Post
    Isn't there a way to derive a simple formula such as that for the \zeta (s)
    Of course the approximate relation n \approx 10^{\frac{D}{s}} valid for \zeta(s) is valid for P(s) using p instead of n...

    p \approx 10^{\frac{D}{s}} (1)

    Kind regards

    \chi \sigma
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    Re: Growth rate of Prime zeta function

    Quote Originally Posted by chisigma View Post
    Of course the approximate relation n \approx 10^{\frac{D}{s}} valid for \zeta(s) is valid for P(s) using p instead of n...

    p \approx 10^{\frac{D}{s}} (1)

    Kind regards

    \chi \sigma
    If I apply (1) to get an approximation for 20 digits of P(2) I get a value that's 10 orders of magnitude less. I should get 10^{20} (P. Sebah et.al.) What should I do?

    Thanks
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