Timeline for Heuristic argument for the Riemann Hypothesis
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| S Sep 1, 2019 at 0:30 | history | suggested | CommunityBot | CC BY-SA 4.0 |
Typo in formula (+ other spelling, to reach character limit)
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| Sep 1, 2019 at 0:20 | review | Suggested edits | |||
| S Sep 1, 2019 at 0:30 | |||||
| Aug 31, 2019 at 21:51 | comment | added | Lagrida Yassine | $\dfrac{1}{\zeta(s)} = s \displaystyle \int_{1}^{+ \infty} \dfrac{M(x)}{x^{s+1}} dx , \quad \Re(s) > 1$ .. the equality is true also for $s=1$ (Von Mongoldt 1897). | |
| Aug 31, 2019 at 20:58 | comment | added | Jesse Elliott | Gonek and Ng have conjectured that $\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x}(\log \log \log x)^{5/4}}$ (resp., $\liminf_{x \to \infty} \frac{M(x)}{\sqrt{x}(\log \log \log x)^{5/4}}$) is finite and positive (resp., finite and negative), which would imply that $M(x) = \Omega_{\pm} (\sqrt{x})$. Good and Churchhouse and L\'evy have made (I think less plausible) conjectures that imply that $\limsup_{x \to \infty} \frac{|M(x)|}{(x \log \log x)^{1/2}}$ is finite and positive, which also implies $M(x) = \Omega_{\pm} (\sqrt{x})$. The law of iterated logarithm is at play here I think. | |
| Aug 31, 2019 at 20:27 | comment | added | burtonpeterj | What does $|M(n)| = O(\sqrt{n})$ say about structure in the zeroes? | |
| Aug 31, 2019 at 20:26 | comment | added | Will Sawin | @burtonpeterj Probably based on heuristics on the linear independence / random distribution of the zeroes. | |
| Aug 31, 2019 at 18:45 | comment | added | burtonpeterj | Let's write $M$ for the Mertens function. Apparently the best lower bounds for $\limsup_{n \to \infty} |M(n)|n^{-\frac{1}{2}}$ are less than $2$. Moreover, recent progress on this seems incremental. Given that, why is it considered so implausible that $|M(n)| = O(\sqrt{n})$? | |
| Aug 31, 2019 at 18:22 | comment | added | Marc | @burtonpeterj This sum is sometimes called the Mertens function. According to Wikipedia, there is a conjecture by Steve Gonek that this sum should by $O(\sqrt{n} (\log \log \log n)^{5/4})$, so it would grow even a little bit slower than the coin flips heuristic would suggest. | |
| Aug 31, 2019 at 17:42 | comment | added | burtonpeterj | For iid coin flips the law of the iterated logarithm gives that the sum is $O(\sqrt{n \log \log(n)})$ with probability $1$. Is it known that the partial sums of the Mobius function are not $O(\sqrt{n \log\log(n)})$? | |
| Aug 31, 2019 at 16:21 | history | answered | JoshuaZ | CC BY-SA 4.0 |