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.
Upper bound on differences of consecutive zeta zeros
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0Mind including a link to the MO question? – 2012-04-30
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0http://mathoverflow.net/questions/84989/upper-bounds-on-the-difference-of-consecutive-zeta-zeros – 2012-04-30
1 Answers
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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0Come on give me points people, I worked on that answer and I answered his question! – 2012-10-20
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0Let's get the points flowing – 2012-10-20
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1Voting is a capricious thing. – 2012-10-20
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0Very 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. – 2012-10-22
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0@ 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 ! – 2012-10-22