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Math Help - Sum over Complex Zeros

  1. #1
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    Sum over Complex Zeros

    I am given \psi(x)=x-\sum{\frac{x^\rho}{\rho}}+O(1) where \rho denotes complex zeros.

    And also given \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))) where \rho=\beta+i\gamma

    Assume the Riemann Hypothesis is true, and write the complex zeros as \frac{1}{2}+i\gamma_{1},\frac{1}{2}+i\gamma_{2} . . . . . . . . .

    Then show that \frac{\psi(x)-x}{\sqrt{x}}= -2\sum\limits_{n = 1}^\infty \frac{sin(\gamma_{n}logx)}{\gamma_{n}}+O(1)

    Any help is greatly appreciated.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If, as You have written,...

    \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))) (1)

    ... and for all complex zeroes is \beta=\frac{1}{2}, then all the terms of the sum \sum (\frac{x^{\rho}}{\rho}+ \frac{x^{\overline{\rho}}}{\overline\rho}) contain the term x^{\beta}= \sqrt{x}...

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by skamoni View Post
    I am given \psi(x)=x-\sum{\frac{x^\rho}{\rho}}+O(1) where \rho denotes complex zeros.

    And also given \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))) where \rho=\beta+i\gamma

    Assume the Riemann Hypothesis is true, and write the complex zeros as \frac{1}{2}+i\gamma_{1},\frac{1}{2}+i\gamma_{2} . . . . . . . . .

    Then show that \frac{\psi(x)-x}{\sqrt{x}}= -2\sum\limits_{n = 1}^\infty \frac{sin(\gamma_{n}logx)}{\gamma_{n}}+O(1)

    Any help is greatly appreciated.
    Lemma: If  \rho is a root of  \zeta(s) ,  \overline{\rho} is too. (This is not too hard to verify.)

    So we have  \psi(x) = x-\sum{\frac{x^\rho}{\rho}}+O(1) = x-\sum_{\Im{\rho}>0}{\left(\frac{x^\rho}{\rho}+\frac  {x^{\overline{\rho}}}{\overline{\rho}}\right)}+O(1  )
    (I'm a little scared about being able to change order of summation up above... maybe someone else should take a look at this line to confirm its validity.)

     \psi(x)-x = -\sum_{\gamma>0}\frac{2\sqrt{x}}{\frac{1}{4}+\gamma  ^2}(\frac{1}{2}\cos(\gamma\log(x))+\gamma\sin(\gam  ma\log(x)))+O(1)
     = -2\sqrt{x} \sum_{n=1}^{\infty} \frac{1}{\frac{1}{4}+\gamma_n^2}(\frac{1}{2}\cos(\  gamma_n\log(x))+\gamma_n\sin(\gamma_n\log(x)))+O(1  )

    Since we have  O(1) in our equation, we can drop the  \frac{1}{4} in the denominator.
    Now split up the sum.

     \frac{\psi(x)-x}{\sqrt{x}} = -2 \sum_{n=1}^{\infty} \frac{\cos(\gamma_n\log(x))}{2\gamma_n^2}-2\sum_{n=1}^{\infty}\frac{\gamma_n\sin(\gamma_n\lo  g(x))}{\gamma_n^2}+O(1)

    Since  |\cos(x)|\leq 1 \; \forall \; x\in\mathbb{R} and we have  \gamma_n^2 in our denominator, the first sum gets absorbed into the  O(1) term.

    Thus we get  \frac{\psi(x)-x}{\sqrt{x}}= -2\sum\limits_{n = 1}^\infty \frac{sin(\gamma_{n}\log(x))}{\gamma_{n}}+O(1) .
    Last edited by chiph588@; March 19th 2010 at 09:02 AM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post

    So we have  \psi(x) = x-\sum{\frac{x^\rho}{\rho}}+O(1) = x-\sum_{\Im{\rho}>0}{\left(\frac{x^\rho}{\rho}+\frac  {x^{\overline{\rho}}}{\overline{\rho}}\right)}+O(1  )
    (I'm a little scared about being able to change order of summation up above... maybe someone else should take a look at this line to confirm its validity.)
    There's a chance I may be able to help...but more for my own curiosity, what did you do in the above? I don't actually see what happened
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    There's a chance I may be able to help...but more for my own curiosity, what did you do in the above? I don't actually see what happened
    The first sum is summing over all the roots of  \zeta(s) . So since we if  \rho is a root then  \overline{\rho} is a root too, we can just look in the upper half plane (i.e. when  \Im{\rho}>0 ) for roots and sum the conjugate pairs together.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    The first sum is summing over all the roots of  \zeta(s) . So since we if  \rho is a root then  \overline{\rho} is a root too, we can just look in the upper half plane (i.e. when  \Im{\rho}>0 ) for roots and sum the conjugate pairs together.
    Ok, I think I get it. So you're saying that since \zeta(\rho)=0\implies \zeta(\overline{\rho})=0 it is obvious that every zero comes in pairs \{\rho,\overline{\rho}\} where one is such that \text{Im}\text{ }<0 and the other positive. So \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho<0}f(\rho)+\sum_{\rho,\text{ }\Im \rho>0}f(\rho) =\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}+\sum_{\rho,\text{ }\rho>0}f(\overline{\rho})=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right  \}?
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Ok, I think I get it. So you're saying that since \zeta(\rho)=0\implies \zeta(\overline{\rho})=0 it is obvious that every zero comes in pairs \{\rho,\overline{\rho}\} where one is such that \text{Im}\text{ }<0 and the other positive. So \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho<0}f(\rho)+\sum_{\rho,\text{ }\Im \rho>0}f(\rho) =\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}+\sum_{\rho,\text{ }\rho>0}f(\overline{\rho})=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right  \}?
    I don't understand some of your intermediate steps, but \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right  \} is correct.

     \rho is complex, so  \rho>0 doesn't make sense.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    I don't understand some of your intermediate steps, but \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right  \} is correct.
    \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)- The sum over all equals the sum over the ones in the upper plane plus those in the lower plane.

    All the ones in the lower plane have a unique one in the upper plane (it's conjugate). So, to hit all the ones in the lower plane we sum over all the conjugates of the ones in the upper.

    So \sum_{\rho,\text{ }\Im \rho}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)

    But clearly this is equal to

    \sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im\rho>0}f(\overline{\rho})- We just changed in the second sum where the conjugation occurs.

    But, we can add these two together to get \sum_{\rho,\text{ }\Im \rho>0}\left\{f(\rho)+f(\overline{\rho})\right\}.


    This is assuming everything converged.
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    \sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)- The sum over all equals the sum over the ones in the upper plane plus those in the lower plane.

    All the ones in the lower plane have a unique one in the upper plane (it's conjugate). So, to hit all the ones in the lower plane we sum over all the conjugates of the ones in the upper.

    So \sum_{\rho,\text{ }\Im \rho}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)

    But clearly this is equal to

    \sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im\rho>0}f(\overline{\rho})- We just changed in the second sum where the conjugation occurs.

    But, we can add these two together to get \sum_{\rho,\text{ }\Im \rho>0}\left\{f(\rho)+f(\overline{\rho})\right\}.


    This is assuming everything converged.
    Yes that's correct.
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Yes that's correct.
    Well, I don't see anything wrong with that. That said, I am NOT very well-versed in advanced CA.
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  11. #11
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    Quote Originally Posted by skamoni View Post
    Assume the Riemann Hypothesis is true, and write the complex zeros as \frac{1}{2}+i\gamma_{1},\frac{1}{2}+i\gamma_{2} . . . . . . . . .
    I think it would be interesting to work it through assuming the Riemann Hypothesis is false. Recall if it's false, then instead of a conjugate pair of zeros, we have a pair of conjugate pair: \sigma\pm \gamma i and 1/2+\sigma\pm \gamma i with 0<\sigma<1/2.

    . . . guess I need to check to make sure Von Mangoldt's expression for \psi(x) is valid independent of the placement of zeros in the critical strip. I think it is since it can be derived using the Residue Theorem (all the terms are sums of residues).
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