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I was on Mathworld some time ago when I read this from http://mathworld.wolfram.com/RiemannHypothesis.html:

The Riemann hypothesis was computationally tested and found to be true for the first 200000001 zeros by Brent et al. (1982), covering zeros sigma+it in the region 0 < t < 81702130.19.

My question is: How can you be sure that you haven't missed any zeros? It seems to me that it is impossible because for any fixed t one would have to check all real sigma values between 0 and 1. And even if there was some way to do that one would still need to test all real values of t between 0 and 81702130.19. Do they have a list of "candidate zeros" that they would just try out?

Thanks in advance.

mtheorylord
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  • What are "the first" so and so zeros?? From what do you, or they, begin to count? – DonAntonio Aug 25 '16 at 22:44
  • You'll have to ask them. I don't know the answer, just the link. – mtheorylord Aug 25 '16 at 22:47
  • Does it have anything to do with gram points? – mtheorylord Aug 25 '16 at 22:48
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    @mtheorylord The short answer is that there is are analytic formulas that allow one to estimate the number of zeros within a given rectangular region. This needs to be computed using high-precision arithmetic, but once you can prove that the error is $<0.5$ you can safely round that estimate to the nearest integer. This is how you can be sure you haven't missed any. – Erick Wong Aug 25 '16 at 22:52
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    Read the paper by Brent et al. Then ask here if you don't understand why their method is complete. You get a close vote from me for not attempting to answer your question for yourself. – Rob Arthan Aug 25 '16 at 22:54
  • Why the hell are there four votes to close as "unclear what you're asking"? It's perfectly clear what the OP is asking: "help me understand this line from mathworld; here are my (probably wrong) thoughts." –  Aug 26 '16 at 19:47
  • And, @RobArthan, "read the paper, then come back"?? This isn't mathoverflow; we shouldn't expect everyone here to be able to just go read and understand a 1980's journal article. This is a place where the math community can hopefully help interested learners interpret those results. And the OP did try to answer the question for themselves; they described their thoughts in the post. –  Aug 26 '16 at 19:52
  • @MikeHaskel: I wrote: "read the paper ... and then ask here if you don't understand ...". What's so difficult about reading a journal article to find out whether you need help to understand it? – Rob Arthan Aug 26 '16 at 20:13
  • What paper? Could you add the link to the paper you're referring to? I couldn't find the answer after researching online. – mtheorylord Aug 26 '16 at 22:36

1 Answers1

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Let $N(T)$ be the number of non-trivial zeros up to height $T$ : $N(T) = \#\{ \rho \ \mid \ 0 < Im(\rho) < T \ \}$ and $N_0(T)$ those lying on the critical line. The Riemann hypothesis is that $N(T) = N_0(T)$ for every $T$.

  • you need to understand the functional equation $\xi(s) = \xi(1-s)$ where $\xi(s) = A(s) \zeta(s)$ and $A(s) = \frac{1}{2}s (s-1) \pi^{-s/2} \Gamma(s/2) $. Together with $\xi(s) = \overline{\xi(\overline{s})}$ it shows that $\xi(1/2+it)$ is real. Hence it has one zero at every sign change.

  • and the argument principle showing that $$2 N(T) = \frac{1}{2i\pi} \oint_{\begin{array}{l}2- i T\to 2+ i T \to\\ -1+iT \to -1-iT \to 2-iT\end{array}} \frac{\xi'(s)}{\xi(s)}ds = \frac{2}{\pi}\text{arg } A(1/2+iT) + \frac{2}{\pi} \text{arg } \zeta(1/2+iT)$$ where $\text{arg } f(s) = \text{Im}(\log f(s))$ is defined by starting with $\text{arg } f(2) = 0$, and following $\log f(s)$ analytically on $2+it, t \in [0,T]$, and then on $\sigma+iT,\sigma \in [2,1/2]$ (assuming $f(s)$ has no zero on $Re(s) > 1$ and $Im(s) = T$ and that $f(2) > 0$)

Everything is explained for example in Titchmarsh's book "the theory of Riemann zeta function", and how to estimate all these in practice for the first few zeros, using the function $Z(t)$. At the end, you can bound $N(T)$ and $N_0(T)$ within an accuracy $< 1/2$, and show they are equal.

reuns
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  • Thanks. This answer is surprising to me. I thought it would some kind of brute-force. – mtheorylord Aug 26 '16 at 00:34
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    @mtheorylord the important part is that you can't show in general an analytic function $f(s)$ has a zero whose real part is exactly $1/2$ simply by approximating it. but $\zeta(s)$ is special because of its functional equation making $A(s) \zeta(s)$ real on $Re(s) = 1/2$, so proving there is a zero with real part $1/2$ reduces to computing sign changes. – reuns Aug 26 '16 at 01:41
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    It might be worth remarking that if, hypothetically, $\zeta$ had a double zero at some $1/2+it_0$, it is not clear that we could ever know this. We would be able to determine that there were two zeroes inside boxes of the form $[1/2-\delta,1/2+\delta] \times [t_0-\delta, t_0+\delta]$ for smaller and smaller delta, but it is not clear that we would ever know that they were exactly on top of each other, or that they were exactly on the critical line. – David E Speyer Jan 30 '23 at 20:47