Why is so much work done on numerical verification of the Riemann Hypothesis?Consequences of the Riemann hypothesisExceptional zeros and Liouville's $lambda$ functionThe Riemann Hypothesis and the Langlands programHow many proofs of the Weil conjectures are there?Quasicrystals and the Riemann HypothesisQuestions on de Branges' work on the Riemann hypothesisLargest known zero of the Riemann Zeta functionNumerical Evidence for Grand Riemann Hypothesis?Riemann Hypothesis and Euler product“Long-standing conjectures in analysis … often turn out to be false”

Why is so much work done on numerical verification of the Riemann Hypothesis?


Consequences of the Riemann hypothesisExceptional zeros and Liouville's $lambda$ functionThe Riemann Hypothesis and the Langlands programHow many proofs of the Weil conjectures are there?Quasicrystals and the Riemann HypothesisQuestions on de Branges' work on the Riemann hypothesisLargest known zero of the Riemann Zeta functionNumerical Evidence for Grand Riemann Hypothesis?Riemann Hypothesis and Euler product“Long-standing conjectures in analysis … often turn out to be false”













20












$begingroup$


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).










share|cite|improve this question











$endgroup$







  • 5




    $begingroup$
    Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers.
    $endgroup$
    – Nell
    yesterday






  • 14




    $begingroup$
    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
    yesterday







  • 34




    $begingroup$
    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.
    $endgroup$
    – Richard Stanley
    yesterday






  • 1




    $begingroup$
    Note there is some computable $C$ such that $psi(n) =1_x > 1 log lfloor exp(n-sum_ < Cn frac(n+1/2)^rhorho)+frac12(1- log 2pi-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
    yesterday







  • 5




    $begingroup$
    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
    $endgroup$
    – literature-searcher
    yesterday















20












$begingroup$


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).










share|cite|improve this question











$endgroup$







  • 5




    $begingroup$
    Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers.
    $endgroup$
    – Nell
    yesterday






  • 14




    $begingroup$
    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
    yesterday







  • 34




    $begingroup$
    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.
    $endgroup$
    – Richard Stanley
    yesterday






  • 1




    $begingroup$
    Note there is some computable $C$ such that $psi(n) =1_x > 1 log lfloor exp(n-sum_ < Cn frac(n+1/2)^rhorho)+frac12(1- log 2pi-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
    yesterday







  • 5




    $begingroup$
    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
    $endgroup$
    – literature-searcher
    yesterday













20












20








20


2



$begingroup$


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).










share|cite|improve this question











$endgroup$




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).







nt.number-theory analytic-number-theory riemann-zeta-function riemann-hypothesis






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









YCor

28.4k484138




28.4k484138










asked yesterday









TomTom

288210




288210







  • 5




    $begingroup$
    Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers.
    $endgroup$
    – Nell
    yesterday






  • 14




    $begingroup$
    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
    yesterday







  • 34




    $begingroup$
    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.
    $endgroup$
    – Richard Stanley
    yesterday






  • 1




    $begingroup$
    Note there is some computable $C$ such that $psi(n) =1_x > 1 log lfloor exp(n-sum_ < Cn frac(n+1/2)^rhorho)+frac12(1- log 2pi-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
    yesterday







  • 5




    $begingroup$
    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
    $endgroup$
    – literature-searcher
    yesterday












  • 5




    $begingroup$
    Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers.
    $endgroup$
    – Nell
    yesterday






  • 14




    $begingroup$
    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
    yesterday







  • 34




    $begingroup$
    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.
    $endgroup$
    – Richard Stanley
    yesterday






  • 1




    $begingroup$
    Note there is some computable $C$ such that $psi(n) =1_x > 1 log lfloor exp(n-sum_ < Cn frac(n+1/2)^rhorho)+frac12(1- log 2pi-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
    yesterday







  • 5




    $begingroup$
    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
    $endgroup$
    – literature-searcher
    yesterday







5




5




$begingroup$
Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers.
$endgroup$
– Nell
yesterday




$begingroup$
Is there really such a huge amount of (human) work on numerical verifications? I am not aware of that many papers.
$endgroup$
– Nell
yesterday




14




14




$begingroup$
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
yesterday





$begingroup$
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
yesterday





34




34




$begingroup$
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.
$endgroup$
– Richard Stanley
yesterday




$begingroup$
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.
$endgroup$
– Richard Stanley
yesterday




1




1




$begingroup$
Note there is some computable $C$ such that $psi(n) =1_x > 1 log lfloor exp(n-sum_ < Cn frac(n+1/2)^rhorho)+frac12(1- log 2pi-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
yesterday





$begingroup$
Note there is some computable $C$ such that $psi(n) =1_x > 1 log lfloor exp(n-sum_ < Cn frac(n+1/2)^rhorho)+frac12(1- log 2pi-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
yesterday





5




5




$begingroup$
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
$endgroup$
– literature-searcher
yesterday




$begingroup$
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
$endgroup$
– literature-searcher
yesterday










3 Answers
3






active

oldest

votes


















26












$begingroup$

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 $8cdot 10^14$. 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.






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user1728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    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
    yesterday







  • 1




    $begingroup$
    @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
    yesterday


















13












$begingroup$

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






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$endgroup$








  • 2




    $begingroup$
    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
    yesterday






  • 5




    $begingroup$
    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.
    $endgroup$
    – Greg Martin
    yesterday


















13












$begingroup$

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.






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Nell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    4 hours ago










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3 Answers
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votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









26












$begingroup$

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 $8cdot 10^14$. 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.






share|cite|improve this answer










New contributor




user1728 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$












  • $begingroup$
    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
    yesterday







  • 1




    $begingroup$
    @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
    yesterday















26












$begingroup$

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 $8cdot 10^14$. 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.






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  • $begingroup$
    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
    yesterday







  • 1




    $begingroup$
    @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
    yesterday













26












26








26





$begingroup$

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 $8cdot 10^14$. 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.






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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 $8cdot 10^14$. 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.







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New contributor




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edited yesterday





















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answered yesterday









user1728user1728

27614




27614




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  • $begingroup$
    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
    yesterday







  • 1




    $begingroup$
    @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
    yesterday
















  • $begingroup$
    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
    yesterday







  • 1




    $begingroup$
    @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
    yesterday















$begingroup$
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
yesterday





$begingroup$
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
yesterday





1




1




$begingroup$
@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
yesterday




$begingroup$
@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
yesterday











13












$begingroup$

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






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    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
    yesterday






  • 5




    $begingroup$
    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.
    $endgroup$
    – Greg Martin
    yesterday















13












$begingroup$

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






share|cite|improve this answer









$endgroup$








  • 2




    $begingroup$
    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
    yesterday






  • 5




    $begingroup$
    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.
    $endgroup$
    – Greg Martin
    yesterday













13












13








13





$begingroup$

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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered yesterday









ConradConrad

37616




37616







  • 2




    $begingroup$
    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
    yesterday






  • 5




    $begingroup$
    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.
    $endgroup$
    – Greg Martin
    yesterday












  • 2




    $begingroup$
    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
    yesterday






  • 5




    $begingroup$
    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.
    $endgroup$
    – Greg Martin
    yesterday







2




2




$begingroup$
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
yesterday




$begingroup$
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
yesterday




5




5




$begingroup$
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.
$endgroup$
– Greg Martin
yesterday




$begingroup$
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.
$endgroup$
– Greg Martin
yesterday











13












$begingroup$

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.






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New contributor




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$endgroup$












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    4 hours ago















13












$begingroup$

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.






share|cite|improve this answer










New contributor




Nell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    4 hours ago













13












13








13





$begingroup$

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.






share|cite|improve this answer










New contributor




Nell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



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.







share|cite|improve this answer










New contributor




Nell is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer








edited yesterday





















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answered yesterday









NellNell

3489




3489




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  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    4 hours ago
















  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    4 hours ago















$begingroup$
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
5 hours ago




$begingroup$
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
5 hours ago












$begingroup$
I didn't say $n$ was all that high...
$endgroup$
– Nell
4 hours ago




$begingroup$
I didn't say $n$ was all that high...
$endgroup$
– Nell
4 hours ago

















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