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Math Help - [SOLVED] Von-Mangoldt Formula

  1. #1
    Junior Member raheel88's Avatar
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    [SOLVED] Von-Mangoldt Formula

    Consider the Von Mangoldt explicit formula, namely,

    \Psi (x) = x - \sum \frac{x^\rho}{\rho} + O (1)

    How can I prove that, assuming the Riemann Hypothesis to be true, then,

    \frac{\Psi (x) - x}{\sqrt x} = -2\sum \frac{sin(\gamma_n t)}{\gamma}

    where \gamma_i is the imaginary part of the i-th non-trivial zeta zero.

    I've had a decent stab at it but I get stuck in trying to manipulate the sum once I've divided by \sqrt x

    Thanks in advance!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by raheel88 View Post
    Consider the Von Mangoldt explicit formula, namely,

    \Psi (x) = x - \sum \frac{x^\rho}{\rho} + O (1)

    How can I prove that, assuming the Riemann Hypothesis to be true, then,

    \frac{\Psi (x) - x}{\sqrt x} = -2\sum \frac{sin(\gamma_n t)}{\gamma}

    where \gamma_i is the imaginary part of the i-th non-trivial zeta zero.

    I've had a decent stab at it but I get stuck in trying to manipulate the sum once I've divided by \sqrt x

    Thanks in advance!
    Try this.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Try this.
    Are you familiar with the formula  \frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov  erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b  eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))<br />
 ?
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  4. #4
    Junior Member raheel88's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Are you familiar with the formula  \frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov  erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b  eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))<br />
?
    yes, that's easy to prove

    the step i'm trying to get my head around is when you absorb the cosine term into the O (1) term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by raheel88 View Post
    yes, that's easy to prove

    the step i'm trying to get my head around is when you absorb the cosine term into the O (1) term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?
    You cant assume its  O(1) though for sine since the term in the series only has  \gamma_n in the denominator, not  \gamma_n^2 like the other series.
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  6. #6
    Junior Member raheel88's Avatar
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    that's plausible...thanks for your help chiph588.

    you saved the day once again!
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by raheel88 View Post
    that's plausible...thanks for your help chiph588.

    you saved the day once again!
    I should also mention this works because  \gamma_n\sim\frac{2\pi n}{\log(n)} .

    This is my signature below!
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  8. #8
    Junior Member raheel88's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I should also mention this works because  \gamma_n\sim\frac{2\pi n}{\log(n)} .

    This is my signature below!
    yeh i spotted that...very intersting result! I've not come across it before in my course, so a proof would be nice whenever you're not too busy.

    regards
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by raheel88 View Post
    yeh i spotted that...very intersting result! I've not come across it before in my course, so a proof would be nice whenever you're not too busy.

    regards
    Are you familiar with the estimation theorem for  N(t) where  N(t) is the number of zeros of  \zeta(s) with ordinate less than  t ?

    The proof follows from that. I'll be glad to show you if you have seen this. I can't think of any other way.
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  10. #10
    Junior Member raheel88's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Are you familiar with the estimation theorem for  N(t) where  N(t) is the number of zeros of  \zeta(s) with ordinate less than  t ?

    The proof follows from that. I'll be glad to show you if you have seen this. I can't think of any other way.
    You mean N(t) \sim \frac{t\ln t}{2\pi} ?
    Sure!
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  11. #11
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by raheel88 View Post
    You mean N(t) \sim \frac{t\ln t}{2\pi} ?
    Sure!
    Ok cool!

    First note that  N(\gamma_n-1)\leq n \leq N(\gamma_n+1) .

    From the theorem you just mentioned, we get  2\pi N(\gamma_n\pm1)\sim(\gamma_n\pm1)\log(\gamma_n\pm1  )\sim\gamma_n\log(\gamma_n) .

    Thus by the squeeze theorem we have  2\pi n\sim\gamma_n\log(\gamma_n) .

    So  \log(2\pi n)\sim\log(\gamma_n\log(\gamma_n))\implies \log(2\pi)+\log(n)\sim\log(\gamma_n)+\log(\log(\ga  mma_n))  \implies \log(n)\sim\log(\gamma_n) .

    Finally we see  2\pi n\sim\gamma_n\log(\gamma_n)\implies 2\pi n\sim\gamma_n\log(n) \implies \gamma_n\sim\frac{2\pi n}{\log(n)} .
    Last edited by chiph588@; June 9th 2010 at 08:20 AM.
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  12. #12
    Junior Member raheel88's Avatar
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    Nice!
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