Timeline for Statements going against the grain of Riemann Hypothesis (R.H.) [closed]
Current License: CC BY-SA 3.0
13 events
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Feb 25, 2017 at 22:45 | comment | added | reuns | If $M(n) = O(n^{a})$ then there is $N,C$ such that for every $n > N$ : $|M(n)| < C n^{a}$. Of course $M(n) > n^{1-\epsilon}$ for infinitely many $n$ contradicts this statement | |
Jun 29, 2016 at 21:14 | comment | added | Jérôme JEAN-CHARLES | OK I agree that my question are a bit simplistic, I thought a mere specific epsilon was enough. Considering how far we are ( N/logN) I feel justified in my mistake. Sorry for the disturbance though. | |
Jun 29, 2016 at 4:43 | history | closed |
Franz Lemmermeyer Peter Humphries Jan-Christoph Schlage-Puchta Wolfgang Ryan Budney |
Not suitable for this site | |
S Jun 29, 2016 at 0:13 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
spurious ) brackets plus characters to meet 6 requirement, bold font fixed, grammar improved, removed spurious closing parentheses
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S Jun 29, 2016 at 0:13 | history | suggested | Henry | CC BY-SA 3.0 |
spurious ) brackets plus characters to meet 6 requirement
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Jun 29, 2016 at 0:07 | review | Suggested edits | |||
S Jun 29, 2016 at 0:13 | |||||
Jun 28, 2016 at 19:50 | answer | added | aosjckajsd | timeline score: 4 | |
S Jun 28, 2016 at 18:30 | history | suggested | Martin Sleziak | CC BY-SA 3.0 |
Riemmann -> Riemann
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Jun 28, 2016 at 18:18 | review | Suggested edits | |||
S Jun 28, 2016 at 18:30 | |||||
Jun 28, 2016 at 18:17 | answer | added | GH from MO | timeline score: 5 | |
Jun 28, 2016 at 17:53 | comment | added | Peter Humphries | Yes to Q1, no to Q2, of course, because then the statement that $M(x) \ll_{\varepsilon} x^{1/2 + \varepsilon}$ would be false, and this is equivalent to RH. Basically, if $\limsup_{x \to \infty} |M(x)|x^{-\Theta} > 0$ for some $\Theta > 1/2$, then RH is false. | |
Jun 28, 2016 at 17:38 | review | Close votes | |||
Jun 29, 2016 at 4:43 | |||||
Jun 28, 2016 at 16:47 | history | asked | Jérôme JEAN-CHARLES | CC BY-SA 3.0 |