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






share|cite|improve this question















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.






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


















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












$begingroup$

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
    2 hours ago










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










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325963%2fwhy-is-so-much-work-done-on-numerical-verification-of-the-riemann-hypothesis%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

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.






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












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.






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.






$endgroup$



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.









share|cite|improve this answer



share|cite|improve this answer








edited yesterday





















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.









answered yesterday









user1728user1728

27614




27614




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.





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.






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











  • $begingroup$
    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.






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
    2 hours ago










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










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    45 mins 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





















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.









answered yesterday









NellNell

3488




3488




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.





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.






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











  • $begingroup$
    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
    2 hours ago










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










  • $begingroup$
    I didn't say $n$ was all that high...
    $endgroup$
    – Nell
    45 mins 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
2 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
2 hours ago












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




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

















draft saved

draft discarded
















































Thanks for contributing an answer to MathOverflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f325963%2fwhy-is-so-much-work-done-on-numerical-verification-of-the-riemann-hypothesis%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

getting Checkpoint VPN SSL Network Extender working in the command lineHow to connect to CheckPoint VPN on Ubuntu 18.04LTS?Will the Linux ( red-hat ) Open VPNC Client connect to checkpoint or nortel VPN gateways?VPN client for linux machine + support checkpoint gatewayVPN SSL Network Extender in FirefoxLinux Checkpoint SNX tool configuration issuesCheck Point - Connect under Linux - snx + OTPSNX VPN Ububuntu 18.XXUsing Checkpoint VPN SSL Network Extender CLI with certificateVPN with network manager (nm-applet) is not workingWill the Linux ( red-hat ) Open VPNC Client connect to checkpoint or nortel VPN gateways?VPN client for linux machine + support checkpoint gatewayImport VPN config files to NetworkManager from command lineTrouble connecting to VPN using network-manager, while command line worksStart a VPN connection with PPTP protocol on command linestarting a docker service daemon breaks the vpn networkCan't connect to vpn with Network-managerVPN SSL Network Extender in FirefoxUsing Checkpoint VPN SSL Network Extender CLI with certificate

NetworkManager fails with “Could not find source connection”Trouble connecting to VPN using network-manager, while command line worksHow can I be notified about state changes to a VPN adapterBacktrack 5 R3 - Refuses to connect to VPNFeed all traffic through OpenVPN for a specific network namespace onlyRun daemon on startup in Debian once openvpn connection establishedpfsense tcp connection between openvpn and lan is brokenInternet connection problem with web browsers onlyWhy does NetworkManager explicitly support tun/tap devices?Browser issues with VPNTwo IP addresses assigned to the same network card - OpenVPN issues?Cannot connect to WiFi with nmcli, although secrets are provided

대한민국 목차 국명 지리 역사 정치 국방 경제 사회 문화 국제 순위 관련 항목 각주 외부 링크 둘러보기 메뉴북위 37° 34′ 08″ 동경 126° 58′ 36″ / 북위 37.568889° 동경 126.976667°  / 37.568889; 126.976667ehThe Korean Repository문단을 편집문단을 편집추가해Clarkson PLC 사Report for Selected Countries and Subjects-Korea“Human Development Index and its components: P.198”“http://www.law.go.kr/%EB%B2%95%EB%A0%B9/%EB%8C%80%ED%95%9C%EB%AF%BC%EA%B5%AD%EA%B5%AD%EA%B8%B0%EB%B2%95”"한국은 국제법상 한반도 유일 합법정부 아니다" - 오마이뉴스 모바일Report for Selected Countries and Subjects: South Korea격동의 역사와 함께한 조선일보 90년 : 조선일보 인수해 혁신시킨 신석우, 임시정부 때는 '대한민국' 국호(國號) 정해《우리가 몰랐던 우리 역사: 나라 이름의 비밀을 찾아가는 역사 여행》“남북 공식호칭 ‘남한’‘북한’으로 쓴다”“Corea 대 Korea, 누가 이긴 거야?”국내기후자료 - 한국[김대중 前 대통령 서거] 과감한 구조개혁 'DJ노믹스'로 최단기간 환란극복 :: 네이버 뉴스“이라크 "韓-쿠르드 유전개발 MOU 승인 안해"(종합)”“해외 우리국민 추방사례 43%가 일본”차기전차 K2'흑표'의 세계 최고 전력 분석, 쿠키뉴스 엄기영, 2007-03-02두산인프라, 헬기잡는 장갑차 'K21'...내년부터 공급, 고뉴스 이대준, 2008-10-30과거 내용 찾기mk 뉴스 - 구매력 기준으로 보면 한국 1인당 소득 3만弗과거 내용 찾기"The N-11: More Than an Acronym"Archived조선일보 최우석, 2008-11-01Global 500 2008: Countries - South Korea“몇년째 '시한폭탄'... 가계부채, 올해는 터질까”가구당 부채 5000만원 처음 넘어서“‘빚’으로 내몰리는 사회.. 위기의 가계대출”“[경제365] 공공부문 부채 급증…800조 육박”“"소득 양극화 다소 완화...불평등은 여전"”“공정사회·공생발전 한참 멀었네”iSuppli,08年2QのDRAMシェア・ランキングを発表(08/8/11)South Korea dominates shipbuilding industry | Stock Market News & Stocks to Watch from StraightStocks한국 자동차 생산, 3년 연속 세계 5위자동차수출 '현대-삼성 웃고 기아-대우-쌍용은 울고' 과거 내용 찾기동반성장위 창립 1주년 맞아Archived"중기적합 3개업종 합의 무시한 채 선정"李대통령, 사업 무분별 확장 소상공인 생계 위협 질타삼성-LG, 서민업종인 빵·분식사업 잇따라 철수상생은 뒷전…SSM ‘몸집 불리기’ 혈안Archived“경부고속도에 '아시안하이웨이' 표지판”'철의 실크로드' 앞서 '말(言)의 실크로드'부터, 프레시안 정창현, 2008-10-01“'서울 지하철은 안전한가?'”“서울시 “올해 안에 모든 지하철역 스크린도어 설치””“부산지하철 1,2호선 승강장 안전펜스 설치 완료”“전교조, 정부 노조 통계서 처음 빠져”“[Weekly BIZ] 도요타 '제로 이사회'가 리콜 사태 불러들였다”“S Korea slams high tuition costs”““정치가 여론 양극화 부채질… 합리주의 절실””“〈"`촛불집회'는 민주주의의 질적 변화 상징"〉”““촛불집회가 민주주의 왜곡 초래””“국민 65%, "한국 노사관계 대립적"”“한국 국가경쟁력 27위‥노사관계 '꼴찌'”“제대로 형성되지 않은 대한민국 이념지형”“[신년기획-갈등의 시대] 갈등지수 OECD 4위…사회적 손실 GDP 27% 무려 300조”“2012 총선-대선의 키워드는 '국민과 소통'”“한국 삶의 질 27위, 2000년과 2008년 연속 하위권 머물러”“[해피 코리아] 행복점수 68점…해외 평가선 '낙제점'”“한국 어린이·청소년 행복지수 3년 연속 OECD ‘꼴찌’”“한국 이혼율 OECD중 8위”“[통계청] 한국 이혼율 OECD 4위”“오피니언 [이렇게 생각한다] `부부의 날` 에 돌아본 이혼율 1위 한국”“Suicide Rates by Country, Global Health Observatory Data Repository.”“1. 또 다른 차별”“오피니언 [편집자에게] '왕따'와 '패거리 정치' 심리는 닮은꼴”“[미래한국리포트] 무한경쟁에 빠진 대한민국”“대학생 98% "외모가 경쟁력이라는 말 동의"”“특급호텔 웨딩·200만원대 유모차… "남보다 더…" 호화病, 고질병 됐다”“[스트레스 공화국] ① 경쟁사회, 스트레스 쌓인다”““매일 30여명 자살 한국, 의사보다 무속인에…””“"자살 부르는 '우울증', 환자 중 85% 치료 안 받아"”“정신병원을 가다”“대한민국도 ‘묻지마 범죄’,안전지대 아니다”“유엔 "학생 '성적 지향'에 따른 차별 금지하라"”“유엔아동권리위원회 보고서 및 번역본 원문”“고졸 성공스토리 담은 '제빵왕 김탁구' 드라마 나온다”“‘빛 좋은 개살구’ 고졸 취업…실습 대신 착취”원본 문서“정신건강, 사회적 편견부터 고쳐드립니다”‘소통’과 ‘행복’에 목 마른 사회가 잠들어 있던 ‘심리학’ 깨웠다“[포토] 사유리-곽금주 교수의 유쾌한 심리상담”“"올해 한국인 평균 영화관람횟수 세계 1위"(종합)”“[게임연중기획] 게임은 문화다-여가활동 1순위 게임”“영화속 ‘영어 지상주의’ …“왠지 씁쓸한데””“2월 `신문 부수 인증기관` 지정..방송법 후속작업”“무료신문 성장동력 ‘차별성’과 ‘갈등해소’”대한민국 국회 법률지식정보시스템"Pew Research Center's Religion & Public Life Project: South Korea"“amp;vwcd=MT_ZTITLE&path=인구·가구%20>%20인구총조사%20>%20인구부문%20>%20 총조사인구(2005)%20>%20전수부문&oper_YN=Y&item=&keyword=종교별%20인구& amp;lang_mode=kor&list_id= 2005년 통계청 인구 총조사”원본 문서“한국인이 좋아하는 취미와 운동 (2004-2009)”“한국인이 좋아하는 취미와 운동 (2004-2014)”Archived“한국, `부분적 언론자유국' 강등〈프리덤하우스〉”“국경없는기자회 "한국, 인터넷감시 대상국"”“한국, 조선산업 1위 유지(S. Korea Stays Top Shipbuilding Nation) RZD-Partner Portal”원본 문서“한국, 4년 만에 ‘선박건조 1위’”“옛 마산시,인터넷속도 세계 1위”“"한국 초고속 인터넷망 세계1위"”“인터넷·휴대폰 요금, 외국보다 훨씬 비싸”“한국 관세행정 6년 연속 세계 '1위'”“한국 교통사고 사망자 수 OECD 회원국 중 2위”“결핵 후진국' 한국, 환자가 급증한 이유는”“수술은 신중해야… 자칫하면 생명 위협”대한민국분류대한민국의 지도대한민국 정부대표 다국어포털대한민국 전자정부대한민국 국회한국방송공사about korea and information korea브리태니커 백과사전(한국편)론리플래닛의 정보(한국편)CIA의 세계 정보(한국편)마리암 부디아 (Mariam Budia),『한국: 하늘이 내린 한 폭의 그림』, 서울: 트랜스라틴 19호 (2012년 3월)대한민국ehehehehehehehehehehehehehehWorldCat132441370n791268020000 0001 2308 81034078029-6026373548cb11863345f(데이터)00573706ge128495