It is always like a wind of fresh air to see some serious science here.
- Economists reading this thread in awe.
A proof of the Riemann Hypothesis on the horizon.
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Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
https://math.stackexchange.com/questions/3783816/integral-of-an-analytic-function-also-analytic
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The linked post is a different setup and not relevant to what is considered in that preprint.
Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
https://math.stackexchange.com/questions/3783816/integral-of-an-analytic-function-also-analytic
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Heard that this paper was discussed on the sidelines of the analytic number theory conference at the Universite de Montreal, held in celebration of Andrew Granville's birthday. Can someone in Montreal confirm this ?
Would love to know more about it.
I think the paper is a sincere effort. If it is incorrect, I would like to know where the errors are.
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Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
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Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
Non-analyst here: According to https://math.stackexchange.com/questions/3495194/condition-for-integral-of-analytic-function-to-be-analytic, the footnote on page 4 at least holds for integrals of complex-analytic functions. I don't know if the argument can be adapted for real-analytic integrands, which is the case that the preprint requires.
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Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
Non-analyst here: According to https://math.stackexchange.com/questions/3495194/condition-for-integral-of-analytic-function-to-be-analytic, the footnote on page 4 at least holds for integrals of complex-analytic functions. I don't know if the argument can be adapted for real-analytic integrands, which is the case that the preprint requires.
Any real analytic function on some open subset of the real line can be extended to a complex analytic function on some open subset of the complex plane, so if the domain of the complex analytic extension is sufficiently big enough for what the author is trying to do that should be good enough. It looks like the author would basically get from integration by parts a sum of multiples of arctangents and arctangent anti-derivatives as the anti-derivative of the integrand.
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It’s a very different setup. In the complex setup, being analytic is the same as being differentiable, which is an infinitesimal condition. Not so for real analytic functions, where you have convergence issues. Differentiability obviously plays well with integration.
Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
Non-analyst here: According to https://math.stackexchange.com/questions/3495194/condition-for-integral-of-analytic-function-to-be-analytic, the footnote on page 4 at least holds for integrals of complex-analytic functions. I don't know if the argument can be adapted for real-analytic integrands, which is the case that the preprint requires.
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Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
I understand the argument that the integral of a *real* analytic function is not necessarily analytic. Nevertheless, this only points to a possible place in the paper where things could go wrong. It remains to be checked whether, for the specific function considered in the paper, analyticity indeed breaks down after integration.
Of course, I believe chance for this paper to be right is extremely slim. The shaky statement in the footnote does not boost confidence, either. But if someone has made a sincere effort and we even bother to bring a discussion on it to EJMR, perhaps we owe a little bit to the author to explain, in more concrete terms, where the errors are.
The author of the paper is actually very young, was still an undergraduate back in 2016. (He once posted a comment to Terence Tao's blog.) This is not a 62-year-old PhD student kind of troll.
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Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
I understand the argument that the integral of a *real* analytic function is not necessarily analytic. Nevertheless, this only points to a possible place in the paper where things could go wrong. It remains to be checked whether, for the specific function considered in the paper, analyticity indeed breaks down after integration.
Of course, I believe chance for this paper to be right is extremely slim. The shaky statement in the footnote does not boost confidence, either. But if someone has made a sincere effort and we even bother to bring a discussion on it to EJMR, perhaps we owe a little bit to the author to explain, in more concrete terms, where the errors are.
The author of the paper is actually very young, was still an undergraduate back in 2016. (He once posted a comment to Terence Tao's blog.) This is not a 62-year-old PhD student kind of troll.Or…maybe the onus is on someone claiming to prove a massive result actually to provide the key steps of the proof?
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Haha, I didn’t notice the footnote, which seems to have been posted after I posted my question here.
If the issue is what I said, it’s about something abstract. There’s a logical leap in applying unique continuation. It would be nice to see that on a numerical plot, but I don’t know how reasonable that is.
And part of what makes one uneasy about this paper is exactly its abstraction. A zeta zero violating the Riemann hypothesis is incontrovertible. But here it’s said that its existence is forced by a mix of theory and classical PNT-level estimates.
That said, the guy seems talented, and these comments should be taken as constructive feedback. Also, don’t work on such problems (without very good reason), and do get a Ph.D.
Algebraist here, so take this skeptically. Say we have a real analytic function on R^2 (or maybe {(x,y) with x>c} for some c). Say f decays reasonably quickly along vertical lines (like 1/|y|^(1+\epsilon), say). Then we can integrate along vertical strips to get a function on R. Is it again real analytic? I’m not seeing why it should be true.
Maybe this isn't explicitly stated enough, but this question suggests an error in the paper.
I understand the argument that the integral of a *real* analytic function is not necessarily analytic. Nevertheless, this only points to a possible place in the paper where things could go wrong. It remains to be checked whether, for the specific function considered in the paper, analyticity indeed breaks down after integration.
Of course, I believe chance for this paper to be right is extremely slim. The shaky statement in the footnote does not boost confidence, either. But if someone has made a sincere effort and we even bother to bring a discussion on it to EJMR, perhaps we owe a little bit to the author to explain, in more concrete terms, where the errors are.
The author of the paper is actually very young, was still an undergraduate back in 2016. (He once posted a comment to Terence Tao's blog.) This is not a 62-year-old PhD student kind of troll. -
being a professional number theorist
Suppose f is a univariate irreducible polynomial with integer coefficients and has degree at least two. Are there infinitely many primes p such that f mod p has no root in Z/pZ? Explain why.
x^2 + 1 does not have a root if p mod 4 = 3
I guess there are infinitely many such primes.
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Actually, it’s obvious there are infinitely many such primes. If not, let them be p1,…,pn be all primes satisfying pi mod 4=3 and pi>3.
let q=(2p1*…pn)^2+3
Clearly q mod 4 = 3
So, at least one prime factor z of q has to satisfy z mod 4 =3, and z cannot be 3 nor any of the pi.