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As I found in wikipedia Riemann Hypothesys has been verified numerically by X. Gourdon (2004) up to 10000000000000 ($10^{13}$) zeroes.

I have a few question about how they did it. I tried to read on their official website (numbers.computation.free.fr), but could not find all answers. if I use notation $\zeta(s)$.

  1. Do I correctly understand, that they did not just find zeroes, but they verified Riemann Hypothesis? In other words, if they claim that they do it up to some number (for example $10^{13}$), from scientific point of view it does not make sense to verify it below this value and it does make sense only to go above this number?

  2. Do I correctly understand that they verified up to $10^{13}$ zeros (or up to zero with index $10^{13}$), but not up to Im(s)=$10^{13}$? If yes, how to understand up to which Im(s) they did it?

  3. Is there an explicit formula for zero with index n? What does it mean when they mention "The first column contains a zero index n"?

  4. Do I correctly understand, that to verify it, they first calculated number of zeros like described here and then tried to find all these zeroes and then they checked if Zeta function change sign in Re(s)=1/2. If Zeta function change sign in Re(s)=1/2 with specific Im(s), that means zero can иe only in the point where Re(s)=1/2.

Zlelik
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  • I think they proved that the first $10^{13}$ non-trivial zeros of $\zeta(s)$ have real part $1/2$. Here the zeros of $\zeta(s)$ are ordered by the absolute value of $s$. – principal-ideal-domain Aug 20 '17 at 21:08
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    recommend Edwards http://store.doverpublications.com/0486417409.html inexpensive – Will Jagy Aug 20 '17 at 21:09
  • @principal-ideal-domain Maybe you forgot to put link for "Here"? – Zlelik Aug 20 '17 at 21:18
  • @Zlelik With "here" I meant my first sentence where I talked about "first" which implies that they are ordered in some way. – principal-ideal-domain Aug 20 '17 at 21:20
  • Yes for 1. 2. 4. Do you have any question then ? There is no explicit formula for the $n$th zero except the obvious one in term of the inverse of $t \mapsto \vartheta(t)+ \text{arg}(\zeta(1/2+it)$ which can be approximated with $\vartheta^{-1}(t)$ (see this) – reuns Aug 20 '17 at 22:12
  • @reuns Thanks a lot. Almost all my questions answered. The only one I have, if they tell that they tested $10^{13}$ zeroes, how to find until which value of Im(s) they did it? Let say if I want to continue this work, It does not make sense to start from Im(s)=0, It make sense to start where they finished. But In this case I have to find all $10^{13}$ zeroes again or try to inverse formula for number of zeroes. – Zlelik Aug 21 '17 at 06:15
  • See the link if they checked the first $N$ zeros it means they checked for $\Im(s) \in (0, 2\pi\frac{N}{\log N})$. You need the argument principle to prove there are no zeros off the critical line for $\Im(s) \in [a,b]$. – reuns Aug 21 '17 at 16:14
  • @Zlelik: You may use directly Gourdon's paper or revert the asymptotic formula from reuns' first link as done by alpha to obtain the approximate value for the $n$-th zero as $;t_n\approx 2\pi,\exp(W((n-7/8-1/2)/e)+1)$. – Raymond Manzoni Aug 21 '17 at 19:37
  • @Zlelik: The error doesn't exceed $1$ for the $10^5$ first zeros (from my $1/2$ subtraction). For $n=10^{13}$ we get nearly $2445999556030$ which you may also find in other links.

    Excellent continuation,

     – Raymond Manzoni Aug 21 '17 at 19:38
  • Please don't use generic terms like "question" in the title. All posts on this site are questions – imagine how the main page would look and how inefficient it would be if everyone did that. – joriki Jun 01 '20 at 11:06

1 Answers1

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1,4. You can't talk about sign changes of $\zeta$, as it is complex-valued. There's a normalized $\zeta$, sometimes denoted $\Xi$, that is real-valued for real inputs, and $\Xi(\gamma)=0$ if and only if $\zeta(1/2+i\gamma)=0$. One finds zeros of $\zeta$ on the critical line by finding sign changes of $\Xi$ on the real line. A sign change indicates a zero, but conceivably a triple zero (or even higher multiplicity). And a double-zero would not have a sign change at all.

There is an integral that gives the number of zeros (with multiplicity) $\rho=\beta+i \gamma$ with $|\gamma|\leq H$. Since this must be a whole number, one can numerically evaluate the integral to within $1/2$ and get the exact count of zeros. Combining, the two types of information, Gourdon knows that he found all of the zeros and they are all on the critical line and are all simple zeros (single).

2,3. Let $N(T)$ be the number of zeros with height at most $H$. We know (from work of Trudgian) that $$\left| N(T) - \left(\frac{T}{\pi}\log \frac{T}{2\pi e} +\frac 74 \right) \right| < 0.34 \log(T) + 4$$ for $T>100$. Gourdon worked to $N(T)=2\cdot 10^{13}$ (they come in pairs), and solving that for $T$ gives $2445999556028 \leq T \leq 2445999556032$.

Kevin O'Bryant
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