I have noticed that there is a huge amount of work which has been done on numerically verifying the Riemann hypothesis for larger and larger non-trivial zeroes.

I don't mean to ask a stupid question, but is there some particular reason that numerical verifications give credence to the truth of the Riemann hypothesis or some way that the computations assist in proving the hypothesis (as we know, historically hypotheses and conjectures have had numerical verification to the point where it seemed that they must be true but the conjectures then turned out to be false, especially hypotheses related to prime numbers and things like that).

Is there something special about this hypothesis which makes this kind of argument more powerful than normal? Would one be able to use these arguments somewhere in the case for a proof of the hypothesis or would they never be used in the proof at all (and yes, until it is proven we cannot know that, sure).

  • 7
    $\begingroup$ Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers. $\endgroup$
    – Nell
    Mar 21, 2019 at 14:24
  • 27
    $\begingroup$ Regarding your last paragraph: The proof of the Ternary Goldbach Conjecture proceeded in two steps: 1) prove the result for all numbers bigger than a certain, known number $n$, and 2) use a computer to verify the result for all numbers less than $n$. It is conceivable that a similar method of proof would work for the Riemann hypothesis. $\endgroup$
    – user1729
    Mar 21, 2019 at 14:26
  • 62
    $\begingroup$ The Riemann hypothesis may be false. Even if one feels that the chance that RH is false is very small, the expected payoff for finding a counterexample is still large. $\endgroup$ Mar 21, 2019 at 15:20
  • 2
    $\begingroup$ Note there is some computable $C$ such that $\psi(n) =1_{x > 1} \log \lfloor \exp(n-\sum_{|\Im(\rho)| < Cn} \frac{(n+1/2)^\rho}{\rho})+\frac12(1- \log 2\pi-\log(1-(n+1/2)^{-2}))\rfloor $. The RH aims at replacing $Cn$ by $C' n^{1/2} \log^3 n$. From the small non-trivial zeros you know $\psi(n)$ for $n$ small, and from $\psi(n)$ for $n$ small and the knowledge that the RH is true for $\Im(s)$ small, you know the approximate imaginary part of the non-trivial zeros. $\endgroup$
    – reuns
    Mar 21, 2019 at 17:04
  • 8
    $\begingroup$ Zagier famously gave 300 million zeros as the level of experiment needed to convince him, and when computations hit 200 million, the computationalists were cajoled to extend this so that a bet with Bombieri could be won. maths-people.anu.edu.au/~brent/pub/pub081.html $\endgroup$ Mar 21, 2019 at 18:14

4 Answers 4


People are interested in computing the zeros of $\zeta(s)$ and related functions not only as numerical support for RH. Going beyond RH, there are conjectures about the vertical distribution of the nontrivial zeros (after "unfolding" them to have average spacing 1, assuming they are on a vertical line to begin with).

Odlyzko found striking numerical support for such conjectures by making calculations with zeros very high up the critical line: hundreds of millions of zeros around the $10^{20}$-th zero. See the Katz--Sarnak article here and look at the picture on the second and fourth pages. These vertical distribution conjectures do not look convincing by working with low-lying zeros.

If you're not interested in considering large-scale statistics of the zero locations, there is a small refinement of RH worth keeping in mind since the calculations supporting RH are based on it: the (nontrivial) zeros of $\zeta(s)$ are expected to be simple zeros. This has always turned out to be the case in numerical work, and the methods used to confirm all zeros in a region lie exactly on -- not just nearby -- the critical line would not work in their current form if a multiple zero were found. The existence of a multiple zero on the critical line would of course not violate RH, but if anyone did detect one because a zero-counting process doesn't work out (say, suggesting there's a double zero somewhere high up the critical line), I don't know if there is an algorithm waiting in the wings that could be used to prove a double zero exists if a computer suggests a possible location. I think it is more realistic to expect a computer to detect a multiple zero than to detect a counterexample to RH. Of course I really don't expect a computer to detect either such phenomena, but if I had to choose between them...

From Wikipedia's table on its RH page, the latest exhaustive numerical checks on RH (all zeros up to some height) go up to around the $10^{13}$-th zero. There are other conjectures that have been tested numerically far beyond $10^{13}$ data points, e.g., the $3x+1$ problem has been checked for all positive integers up to $80 \cdot 2^{60} \approx 10^{19}$, Goldbach's conjecture has been checked for the first $2 \cdot 10^{18}$ even numbers greater than $2$, and the number of twin prime pairs found so far is over $8\cdot 10^{14}$. With such examples in mind, I would not agree that the numerical testing of RH is out of line with how far people are willing to let their computers run to test other open problems.

  • $\begingroup$ One caveat: it is my impression that the check that goes the farthest ($10^{13}$ zeros) is more in the nature of an empirical check, rather than a rigorous proof that all of the first $10^{13}$ zeros do lie on the critical line. It's still valid experimental evidence towards RH, but I would not use this particular verification in a proof. $\endgroup$
    – Nell
    Mar 21, 2019 at 17:49
  • 3
    $\begingroup$ @Nell I looked at the paper at numbers.computation.free.fr/Constants/Miscellaneous/… and you're correct: while they did a confirmation of the main calculation by adjusting some free parameters and redoing it again to find the same number of zeros in various locations as they found the first time, they admit at the end of Section 1.1 that the only way to verify the calculation is to do it all a second time (unlike integer factorization, for which success can be checked by direct multiplication). They say their work is not a strict mathematical proof. $\endgroup$
    – user1728
    Mar 21, 2019 at 20:57
  • 1
    $\begingroup$ @user1728, are you an alias of user1729 (who also commented above)? $\endgroup$
    – Alex M.
    Mar 23, 2019 at 20:07

I would add a few more comments to the very pertinent ones above:

1: We are lucky to have two things that work in our favor - an excellent representation of $\zeta$ on the critical line by a simple real function (simple up to a good approximation, approximation usually called the Riemann Siegel formula) - the Hardy function, $Z(t)$ - multiplied by a function of absolute value $1$, so critical zeros of a very complicated transcendental complex function ($\zeta(s)$) are also zeros of a much simpler real function, $Z(t)$, zeros that can be determined to high accuracy.

2: We are also lucky to have a very accurate formula (Riemann-von Mangoldt) that determines with perfect accuracy the number of zeros in the critical strip up to a fixed bound on the imaginary part, so putting 1 and 2 together we conclude that RH is true up to high imaginary part bounds by computing the zeros on the critical line with 1 and showing that there are these many zeros in the full strip up to that level by 2:

3: There is a duality between $\zeta$ non-trivial zeroes and primes that allows to at least try investigate some problems about primes using $\zeta$ zeroes instead, so having a huge database of such could be quite useful at least potentially

  • 4
    $\begingroup$ I'd say the point is not just that $Z(t)$ can be expressed somewhat more simply, but that it is in general simple to prove that a real function has a zero in a small interval - if the sign changes, there has to be a zero in there. $\endgroup$
    – Nell
    Mar 21, 2019 at 16:30
  • 7
    $\begingroup$ Re 3: In particular, the higher that RH has been verified, the better explicit numerical bounds one can get on error terms for prime-counting functions. $\endgroup$ Mar 21, 2019 at 17:44
  • 1
    $\begingroup$ @mike - there is a $O(1/T)$ exact formula involving $S(T)=\arg zeta(1/2+iT)$ with good estimates for that one too $\endgroup$
    – Conrad
    Jul 22, 2020 at 21:21
  • 1
    $\begingroup$ @mike if I remember correctly there is no known $T$ for which $|S(T)|$ is bigger than $4$ or $5$ although it is known to be unbounded - the speculation is that the true behavior of the zeroes is found where $S$ is fairly large $\endgroup$
    – Conrad
    Jul 22, 2020 at 21:32
  • 1
    $\begingroup$ @mike check this freely available paper for more details about $S$ researchgate.net/publication/… $\endgroup$
    – Conrad
    Jul 22, 2020 at 21:41

Part of the point is that such numerical checks can be demonstrations of the efficiency of this or that new algorithm. However, it is also the case that a finite check (that all the zeroes of $\zeta(s)$ with $\Im(s)\leq T$, say, lie on the critical line) can be used in actual proofs of other statements, provided that it is rigorous.

For that matter, computing the first $n$ zeroes of the Riemann zeta function can be used to disprove another conjecture. Take, for instance,

Odlyzko, A. M., & te Riele, H. J. J. (1985). Disproof of the Mertens conjecture. Journal für die reine und angewandte Mathematik, 357, 138-160.

  • 2
    $\begingroup$ The reference to the work of Odlyzko and the Riele is a bit misleading - they only used the first 2000 zeros, but they needed to compute them to very high precision, 100 digits. $\endgroup$
    – Stopple
    Mar 22, 2019 at 23:44
  • $\begingroup$ I didn't say $n$ was all that high... $\endgroup$
    – Nell
    Mar 23, 2019 at 1:03

Although merely calculating the nontrivial zeros cannot lead to a direct proof of the Riemann hypothesis, these calculations are still helpful at creating numerical bounds for arithmetical functions.

For instance, by calculating the first 25000 nontrivial zeros of $\zeta(s)$, Rosser and Schoenfeld in 1962 were able to deduce the bound (see Theorem 1 of this paper):

$$ {x\over\log x}\left(1+{1\over2\log x}\right)<\pi(x)<{x\over\log x}\left(1+{3\over2\log x}\right) $$

for $x\ge59$. This might not be the latest result, but it is possible that we can refine this estimate if more calculations of zeros are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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