0
$\begingroup$

The famous plot of $\zeta(1/2+it)$ for real $t$ seems to show that this function gets a non zero real value exactly once between two consecutive Riemann zeros. Moreover, letting $\rho_{i}$ the $i$-th critical zero of positive imaginary part, it seems that such a non zero real value is attained at around $1/2(\rho_{i}+\rho_{i+1})$.

My question is: are there rigorous proofs for this phenomenon?

Improve this question
$\endgroup$
  • $\begingroup$ Not research level. See french's wikipedia. Assuming $S(t) = \lfloor S(t) \rfloor+o(\frac{1}{\ln t})$ would make the Gram law true for $t$ large enough. $\endgroup$ – reuns Jul 10 '17 at 21:03
  • $\begingroup$ French wikipedia is often harder to read. That's why I didn't check it. $\endgroup$ – Sylvain JULIEN Jul 10 '17 at 21:08
  • $\begingroup$ The $\zeta(s)$'s french article is much better than every others. It explains almost everything in Titchmarsh's book. $\endgroup$ – reuns Jul 10 '17 at 21:11
8
$\begingroup$

$\zeta(1/2+i g_n)$ is nonzero real iff $g_n$ if Gram point, which is efficienly computable and you can use binary search to count up to given $t$.

Your claim "his function gets a non zero real value exactly once between two consecutive Riemann zeros" is Gram's law, which is known to be false.

Improve this answer
$\endgroup$
  • $\begingroup$ Thank you for your answer. Is it known that $g_{n+l}\sim\dfrac{\rho_{n}+\rho_{n+1}}{2}$ for some integer $l$? I didn't find this in the link you provided. Moreover has the curvature of $\zeta(1/2+it)$ at Gram points been studied? $\endgroup$ – Sylvain JULIEN Jul 10 '17 at 9:50
  • $\begingroup$ @SylvainJULIEN I don't know. $\endgroup$ – joro Jul 10 '17 at 10:26
4
$\begingroup$

This is an answer to the first question in your comment above. (For notational convenience only I'll assume the Riemann Hypothesis.) You need to be careful regarding $\sim$. Since $\gamma_n\sim \frac{2\pi n}{\log n}$, $\gamma_n\sim\gamma_{n+1}$ is true. I claim $g_n\sim \frac{2\pi n}{\log n}$ as well: With $N_g(T)$ the number of Gram points to height $T$, one has $$ N_g(T)=\frac{T}{2\pi}\log T+O(T) $$ see Lemma (2.1) in the recent paper by Trudgian. So for some $c>0$, $$ \frac{T}{2\pi}\log T-cT<N_g(T)<\frac{T}{2\pi}\log T+cT. $$ Solving $\frac{T}{2\pi}\log T-cT=n$ for $T$ gives that $$ T=\frac{2\pi n}{W(2\pi n\exp(-2\pi c))}\sim \frac{2\pi n}{\log(n)}, $$ where $W(x)$ is the Lambert $W$-function, ProductLog in Mathematica (and Mathematica helps with the limits.) Similarly for the $\frac{T}{2\pi}\log T+cT$ term. So solving $N_g(T)=n$ for $T$ with $T=g_n$ gives the same asymptotic.

So $\gamma_n\sim g_n\sim \gamma_{n+1}\sim \frac{\gamma_n+\gamma_{n+1}}{2}$ trivially.

Improve this answer
$\endgroup$
  • $\begingroup$ Ok, I will try to be more accurate then. For a set of values of $ n $ such that Gram's law holds, let $ \delta_{n} : =\gamma_{n+1}-\gamma_{n} $, $\delta_{n}^{+} : =\gamma_{n+1}-\vert g_{n}\vert $ , $\delta_{n}^{-} : =\vert g_{n}\vert-\gamma_{n} $. Is there a positive constant $ A $ such that $ \vert\delta^{+}_{n}-\delta^{-}_{n}\vert =o(\delta_{n}^{-A}) $? $\endgroup$ – Sylvain JULIEN Jul 10 '17 at 19:22
  • $\begingroup$ Time limit exceeded. $ \vert g_{n}\vert $ denotes the ordinate of the Gram point between $ 1/2+i\gamma_{n} $ and $ 1/2+i\gamma_{n+1} $. $\endgroup$ – Sylvain JULIEN Jul 10 '17 at 19:32
  • $\begingroup$ @SylvainJULIEN The fact that 'Gram's Law' fails to be true means that $g_n$ does not necessarily lie between $\gamma_n$ and $\gamma_{n+1}$. $\endgroup$ – Stopple Jul 10 '17 at 19:39
  • $\begingroup$ Ok, but what I'm interested in is the case when there is only one Gram point between two consecutive critical zeroes and in this case whether this Gram point is 'close' in some precise sense I tried to define above to the middle of the segment defined by those two consecutive zeroes. $\endgroup$ – Sylvain JULIEN Jul 10 '17 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.