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The average gap $\delta_n=|\gamma_{n+1}-\gamma_n|$ between consecutive zeros $(\beta_n+\gamma_n i,\beta_{n+1}+\gamma_{n+1}i)$ of Riemann's zeta function is $\frac{2\pi}{\log\gamma_n}.$ There are many papers giving lower bounds to $$ \limsup_n\ \delta_n\frac{\log\gamma_n}{2\pi} $$ unconditionally or on RH or GRH. (The true value is believed to be $+\infty.$) I'm interested in an upper bound on the smaller quantity $\delta_n$. I asked the question on MathOverflow but have not yet found an effective bound. Both unconditional results and those relying on the RH are interesting.

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There are two things:

1) On RH you have the clean bound $$ \delta_n \leq \pi ( 1 + o(1)) / \log\log \gamma_n $$ as $n \rightarrow \infty$, due to Goldston and Gonek. If the $o(1)$ bothers you, you can remove it by re-working the details in their (short) paper (see http://www.math.sjsu.edu/~goldston/article38.pdf in particular see Corollary 1).

Unconditionally, you have the point-wise bound due to Littlewood, $$ \delta_n \leq C / \log\log\log \gamma_n $$ I am not aware of anybody working out the explicit value of the constant $C$ in this case.

2) You can get better bounds if you are interested in bounds valid for ``most'' zeros. For example it is known that $$ \sum_{T \leq n \leq 2T} \delta_n^{2k} \asymp T (\log T)^{-2k} $$ This allows you to get good bounds for most $\delta_n$'s which are as good as $\Psi(\gamma_n) / \log \gamma_n$ with a $\Psi(x)$ going to infinity arbitrarily slowly.

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  • $\begingroup$ Come on give me points people, I worked on that answer and I answered his question! $\endgroup$ – blabler Oct 20 '12 at 4:47
  • $\begingroup$ Let's get the points flowing $\endgroup$ – blabler Oct 20 '12 at 4:47
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    $\begingroup$ Voting is a capricious thing. $\endgroup$ – robjohn Oct 20 '12 at 8:35
  • $\begingroup$ Very nice! Absolutely deserving of the points. I am looking at extremal behavior of the gaps, so the first part is what interests me. I had not seen the paper. $\endgroup$ – Charles Oct 22 '12 at 13:18
  • $\begingroup$ @ blabler and robjohn : How is it known : the equation from 2) with the sigma ? Voting capricious ? I disagree. The entire MSE is based upon voting ! And I feel sorry for blabler. He deserves more votes and I am also sorry for me. Why ? Because I lost a bounty to this question and he missed my bounty , so we both lost ! Now that bounty thing did not work and it has failed for me before , now that is capricious ! Are you in a bad mood perhaps ? Welcome to MSE blabler ! $\endgroup$ – mick Oct 22 '12 at 16:15

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