I'm trying to locate the first 1000 zeros of $\zeta(s)$ and not sure about the best way to go about it. I was considering the Newton-Raphson method but can'd find a good way to code $\zeta'(s)$ in python as I can't find a functional equation anywhere.

I'm using a simple 'while' loop to locate sign changes of $Z(t)$ to count zeros (would also appreciate if anyone knows a quicker way to count the zeros of $\zeta(s)$ btw) and have found 649 zeros below 1000 and 10,142 zeros below 10,000.

I now need to locate these zeros so I can sum the arguments of each complex zero. I'm led to believe this will give me $$\int_{\gamma} \zeta(s) ~ ds$$ where $\gamma$ is the contour of length 1000 or 10,000 respectively. ie: $$\int_{\gamma} \zeta(s) ~ ds = 2\pi \cdot \sum_{s' \in S} \arg{(s')}$$ where $S$ is the set of all zeros below 1000 or 10,000 around the critical strip. I will first calculate this then back it up rigorously.

Thus if I am able to locate all zeros and sum their arguments, then divide this number by $2 \pi$ I will be able to directly compare this result with the number of zeros found earlier, the goal being to verify the Riemann Hypothesis up to a certain height.

Any help very much appreciated!


Check out Andrew Odlyzko's work.His main page is here:


Here is a table of zeros:


Check out this paper for algorithms:


And for large zeros, this one:


  • 3
    $\begingroup$ Mathematica gives the results instantly: N@Zeta[8] gives the 8th zero, for instance. $\endgroup$ – David G. Stork Feb 7 '18 at 23:08
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    $\begingroup$ "The eighth zero"? Under what order? The absolute value of its imaginary part? $\endgroup$ – DonAntonio Feb 7 '18 at 23:51
  • $\begingroup$ thanks, I am familiar with Odlyzko's work but I don't see how that paper helps me locate zeros of the zeta function using newton raphson or otherwise... $\endgroup$ – kingee Feb 10 '18 at 22:41
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    $\begingroup$ @DonAntonio That's correct. $\endgroup$ – Klangen Oct 15 '18 at 10:37
  • $\begingroup$ There is also the Mathematica function ZetaZero[] $\endgroup$ – Klangen Oct 15 '18 at 10:38

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