We know that according Riemann hypothesis all non trivial zeros of dzeta function lie on (0.5, x) line on complex surface. I wonder how Reieman found that idea. Does he just found first few zeros by brute force method and since they all lie on 1/2 line, he says that probably all of them has this property? Or there are deeper reasons to say that? Or other third possibility: Riemann was just a genius and nobody has idea how he found his hypothesis?
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2Riemann was a genius yes. However, “finding” a hypothesis is a strange turn of phrase. You are speaking as if he proved it; he of course did not prove it, but felt as if it would be only natural for it to be true. This is due to the functional equation; note that $1/2=1-1/2$ – FShrike Dec 25 '21 at 13:13
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Even Ramanujan had his knowledge not "out from nothing". He just didn't want to tell other people how he arrived at his results. But how Riemann arrived at this remarkable conjecture acutally is a good question (+1). Hope he did not want to hide it as Ramanujan did. – Peter Dec 25 '21 at 13:18
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2Sorry for pointing out your typo "Reiermann". This sounds really bad in German. Riemann had deeper reasons for his conjecture (see his own article about it). – Dietrich Burde Dec 25 '21 at 13:22
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@DietrichBurde typo removed – Andrew123 Dec 25 '21 at 13:30
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2Did you look up Riemann's papers to see where this hypothesis first arose? Did you check what Riemann himself might have written about this question? – Lee Mosher Dec 25 '21 at 13:34
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Best is to look in Edwards book as he unpacks the Riemann paper, the claims there, the aftermath ( eg Riemann unpublished computations found decades later etc); the gist of it is that Riemann thought it "likely" but said he tried to prove it, got nowhere and moved to other things – Conrad Dec 25 '21 at 13:42
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1@LeeMosher, no because as just hobbyst, I absolutely has no chance to understand Riemann, I'm looking for explanation expressed in popular science language. Is it improper forum to talk in such language? – Andrew123 Dec 25 '21 at 13:44
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1It's not improper, but, a better question would include an explanation of what efforts you have made in your own investigations of the question. – Lee Mosher Dec 25 '21 at 13:56
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@LeeMosher I’ve studied almost nothing about RH, but I always thought it was a natural observation due to inspection of the functional equation, excluding the trivial integer roots: $\zeta(s)=0\iff\zeta(1-s)=0,,s\notin\Bbb Z$, from which the line of reflective symmetry at $\Re s=\frac{1}{2}$ would be a natural, or aesthetic, suggestion for such zeroes. – FShrike Dec 25 '21 at 13:58
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It seems the main consequence of $RH$ once proved it, will be on the distribution of prime numbers which could be something opposite to the "secret" with some big primes in Criptography. Fortunately, there is a Generalized $RH$ about the $L-$functions on rings of integers of algebraic extensions which could be useful for Criptography in another way. – Piquito Dec 25 '21 at 15:34
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@Piquito A proof of the Riemann hypothesis would not be a danger for the RSA-method. Nor would it help to solve problems concerning prime gaps. We still would not be able to construct new huge primes efficiently (that means without actually applying a primality test). – Peter Dec 25 '21 at 16:41
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@Peter: I have read that in some serious source some time ago, I have not specifically referred to the RSA-method but to the keys (not so public service) that use very large prime numbers. The latter is not indecipherable but requires a certain time to decipher it and if after this time you change the problem, it will no longer serve those who do not want to use it. If $RH$ is demonstrated then the certain time mentioned will be shortened considerably, therefore it should be abandoned. This is said by specialist cryptographers. – Piquito Dec 25 '21 at 17:19
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@Piquito This is strange , since the evidence for the truth of RH is high anyway. Which difference does it make whether it is actually proven ? – Peter Dec 26 '21 at 08:41
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@Peter.- I agree with what you say. Regards. (I let myself to add, however, that some theorical considerations cannot be established without proof of what is used). – Piquito Dec 26 '21 at 23:01
1 Answers
Well, I happened to read Riemann's 1859 paper just yesterday out of curiosity, so here's my brief answer (warning: I know very little about the history, and even less about the progress. All I can offer you here is a surface-level answer). In the paper, he defines his famous zeta function $\zeta(s)$ and talks about prime numbers and so on. As a brief description of the first few (3-4) pages, here's what happens:
He defines $\zeta(s)$ for $\text{Re}(s)>1$ using the familiar series definition $\sum_{n=1}^{\infty}\frac{1}{n^s}$. The reason he's interested in this function is that it is very nicely related to prime numbers, namely $\zeta(s)=\prod(1-p^{-s})^{-1}$, the product taken over all positive primes (Euler's formula).
He talks about relationship to the Gamma function $\Gamma(s)$ through the identity $\Gamma(s)\zeta(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\,dx$ (though in the paper, the notation $\Pi(s-1)$ is used instead of $\Gamma(s)$).
Next, he immediately goes on to talk about analytic continuation of $\zeta(s)$ be deforming the contour of integration; using Jacobi's theta identity blablabla.
While doing the analytic continuation, he comes across an expression and gives it a name. In the paper (page 3), he defines $s=\frac{1}{2}+it$ and he defines the function \begin{align} \xi(t)&:=\Pi\left(\frac{s}{2}\right)(s-1)\pi^{-s/2}\zeta(s)\\ &:=\Gamma\left(\frac{s}{2}+1\right)(s-1)\pi^{-s/2}\zeta(s)\\ &=\frac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s) \end{align} This was Riemann's original $\xi(t)$ function. Nowadays this would be denoted at $\Xi(t)$, while expression on the right is what we define as $\xi(s)$, so Riemann's original function is actually $\xi_{\text{new}}\left(\frac{1}{2}+is\right)$. But for the rest of this, I shall stick to Riemann's notation. This is an entire function of $t$ (holomorphic on $\Bbb{C}$), and he now talks about zeros of this function, which I quote:
The number of roots of $\xi(t)=0$ whose real parts lie between $0$ and $T$ is approximately \begin{align} =\frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi}; \end{align} because the integral $\int d\log\xi(t)$, taken in a positive sense around the region consisting of the values of $t$ whose imaginary parts lie between $\frac{1}{2}i$ and $-\frac{1}{2}i$ and whose real parts lie between $0$ and $T$, is (up to a fraction of the order of magnitude of the quantity $\frac{1}{T}$) equal to $\left(T\log\frac{T}{2\pi}-T\right)i$; this integral however is equal to the number of roots of $\xi(t)=0$ lying within in this region, multiplied by $2\pi i$. One now finds indeed approximately this number of real roots within these limits, and it is very probable that all roots are real. Certainly one would wish for a stricter proof here; I have meanwhile temporarily put aside the search for this after some fleeting futile attempts, as it appears unnecessary for the next objective of my investigation.
The last bold is mine. The assertion that $\xi(t)=0$ having real roots is the same as the roots of $\zeta(s)$ lying on the critical line $\text{Re}(s)=\frac{1}{2}$ (recall Riemann set $s=\frac{1}{2}+it$), which is precisely what we today call Riemann's hypothesis. So, Riemann seemed to be considering zeros of his $\xi(t)$ function, and got some estimates on how many zeros lie in a certain region, and hypothesized that all the roots must in fact be real.

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3This survey makes clear that the conjecture did not come out "from nothing". It even emerged quite naturally. (+1) – Peter Dec 25 '21 at 14:20
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Like one of the most stunning conjecture, Birch and Swinnerton-Dyer about L-functions of elliptic curves defined by a cubic equation, which becomes of hard computations. But Fermat's Last Theorem, this ex-conjecture could not have come but from intuition (in the language used here, "from nothing"). – Piquito Dec 25 '21 at 15:19