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 2018-10-16, 14:03 #1 miket   May 2013 10012 Posts Latest progress on Riemann hype Apoptosis Tranlated by machine: The Riemann conjecture, one of the most difficult conjectures to prove in mathematics, was proclaimed by the 82-year-old famous Chinese mathematician Professor Li Zhong. The report was made in Room N913, South Building, Academy of Mathematics and Systems Sciences, Chinese Academy of Sciences, from 14:30 to 16:30 on October 13, 2018. Below is the material I wrote after hearing the report in October 13th. I went to listen to Professor Li Zhong's lecture this afternoon. Read the proof of Riemann's conjecture. Li Zhongli used Riech metric to prove the Riemann hypothesis strictly. The relationship between his proof and the mathematician Atiyah's proof can be summarized as follows: the two men have the same idea, but Atiyah's proof is incomplete in the strict sense that it is not appropriate to use a quantity. Li Zhongli used Riech metric to prove that he was strict. The two person is completely independent in the process of writing. He had previously given a lecture at Tsinghua University. He called his work his own conjecture and gave proof. He was asked if he was Riemann's hypothesis. He did not answer. Li Zhong's report at Tsinghua University was before Atiyah announced. This report is based on the proof of Riemann hypothesis. Li Zhong believes that the first thing to prove is not important. The key is that the Riemann hypothesis has been proved strictly. I personally think it is also important to prove first. Mr. Li's achievements will greatly encourage young scholars in China. After hearing this report, I think there should be no problem in the correctness of the results. The above is my experience as a layman. A famous mathematician, former vice-president, professor and doctoral supervisor of the School of Mathematical Sciences of Peking University, and former Secretary-General of the Chinese Mathematical Association, Mr. Peng Lizhong, wrote a comment on this point (which was later forwarded to the retired teachers of Peking University by Professor Li Zhong): Riech metric (Li-metric) was defined and Atiyah constant was proved to be finite under this measure. The mathematical constants of RH are proved. Li proved in Atiyah that Riemann function is bounded under li metric and RH is established. PS; It seems on Li's paper,the Riech metric is Riech points. Last fiddled with by miket on 2018-10-16 at 14:09 Reason: small fix
 2018-10-16, 16:50 #2 CRGreathouse     Aug 2006 3·1,993 Posts Matthew R. Watkins collects purported proofs and disproofs of the Riemann hypothesis here: http://empslocal.ex.ac.uk/people/sta...a/RHproofs.htm You might consider sending him this one; it would be his ninth for 2018, if I count correctly.
 2018-10-16, 17:19 #3 Till     "Tilman Neumann" Jan 2016 Germany 13×37 Posts All attempts I looked at so far try to prove that RH is true. Can we take that as some kind of evidence that it is true? ;-)
2018-10-16, 17:31   #4
CRGreathouse

Aug 2006

3·1,993 Posts

Quote:
 Originally Posted by Till All attempts I looked at so far try to prove that RH is true. Can we take that as some kind of evidence that it is true? ;-)
Most of the entries on Watkins' list are claimed as proofs rather than disproofs, but see Bredakis (disproof), Pozdnyakov (unprovable), Aizenberg (disproof), Chun-Xuan (disproof), and Hwang (disproof).

2018-10-16, 23:43   #5
Dr Sardonicus

Feb 2017
Nowhere

3·5·73 Posts

Quote:
 Originally Posted by Till All attempts I looked at so far try to prove that RH is true. Can we take that as some kind of evidence that it is true? ;-)
Perhaps not, but there have been a number of results increasingly "pointing toward" RH being true. One of the earliest is the Bohr-Landau Theorem (Niels Bohr's brother Harald). It says that, for any

$\delta > 0$

all but an "infinitesimal proportion" of the zeroes

$\sigma + i*t$

in the critical strip satisfy

$|\sigma - \frac{1}{2}| < \delta$

(That is, as T increases without bound, the proportion of zeroes with |t| < T not satisfying the inequality tends to 0.)

One of the stronger results known is that AT LEAST TWO FIFTHS OF THE ZEROS OF THE RIEMANN ZETA FUNCTION ARE ON THE CRITICAL LINE. In fact, this result (improving a result of Levinson) shows that at least 2/5 of the zeroes in the critical strip are simple and lie on the critical line.

Also, as noted in a recent thread about RH, analogs of RH in other contexts have in fact been proven.

Last fiddled with by Dr Sardonicus on 2018-10-16 at 23:48

 2018-10-17, 14:17 #6 Till     "Tilman Neumann" Jan 2016 Germany 13·37 Posts Thanks for the replies, very interesting to me! Another thought: If it could be shown that RH is unprovable, would that have implications on the PvsNP problem? (maybe unprovable, too ?) Last fiddled with by Till on 2018-10-17 at 14:37 Reason: PvsNP
2018-10-17, 15:22   #7
Dr Sardonicus

Feb 2017
Nowhere

3·5·73 Posts

Quote:
 Originally Posted by Till If it could be shown that RH is unprovable, would that have implications on the PvsNP problem? (maybe unprovable, too ?)
I don't know much at all about this sort of thing, but it seems to me that if RH were not true, its falsity could be proven by a single counterexample (i.e. a zero in the critical strip that is not on the critical line).

Therefore, if it were unprovable ("formally undecidable"), it would have to be true.

 2018-10-17, 17:09 #8 Till     "Tilman Neumann" Jan 2016 Germany 13×37 Posts Your first sentence is a verbial inversion of a simpler logical statement, I'ld say. The simple statement would be: Counterexample exists => RH is false Inversion gives: RH is true => There is no counterexample But this does not imply any statement like RH is undecidable => ? But what I am really asking is if there are implications about PvsNP. My thoughts going around the line that * prime factorization is intimately related to RH, and * prime factorization is an important case study for PvsNP Last fiddled with by Till on 2018-10-17 at 17:13 Reason: simpler statement named so
2018-10-18, 07:08   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2×23×137 Posts

Quote:
 Originally Posted by Till But this does not imply any statement like RH is undecidable => ?
I basically agree, but we can say this:

If we prove that RH is undecidable => We can never find a counterexample, and we can't show that no counterexamples exist.

So I'm not sure how that could ever be shown to be so. How could we prove that it is impossible to find a counterexample, and at the same time not be able to prove no counterexamples exist? That doesn't feel right to me. So I'd say that it isn't possible to prove that RH is undecidable.

2018-10-18, 07:52   #10
axn

Jun 2003

112×43 Posts

Quote:
 Originally Posted by retina I basically agree, but we can say this: If we prove that RH is undecidable => We can never find a counterexample, and we can't show that no counterexamples exist. So I'm not sure how that could ever be shown to be so. How could we prove that it is impossible to find a counterexample, and at the same time not be able to prove no counterexamples exist? That doesn't feel right to me. So I'd say that it isn't possible to prove that RH is undecidable.
Till disagrees with Dr.S.
Retina agrees with Till.
Retina's explanation basically spells out Dr.S's position.

2018-10-18, 08:48   #11
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2·23·137 Posts

Quote:
 Originally Posted by axn Till disagrees with Dr.S. Retina agrees with Till. Retina's explanation basically spells out Dr.S's position. My head explodes.
I guess I misunderstood something there.

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