Timeline for Sharpening of Lindelöf hypothesis
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2012 at 3:27 | answer | added | anonnn | timeline score: 4 | |
| Sep 7, 2012 at 9:41 | history | edited | Stopple |
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| Sep 7, 2012 at 2:17 | comment | added | GH from MO | Just a historical remark: there were many exponents (at least 10) between Weyl's 1/6 and Bombieri-Iwaniec's 9/56, see p.118 in Titchmarsh: The theory of the Riemann zeta function (2nd edition, Oxford, 1986). | |
| Sep 6, 2012 at 18:17 | comment | added | user9072 | Regarding your 2nd com.: For RH one could prove ever larger zero free regions, or ever larger proportions of zeros on the line. Regarding 1st: I am a quite a bit out of my comfort-zone commenting on this, so I only comment and do/did not include it in answer. I think (but this could be wrong) that there might not be such a link; the reason being that while zeros of zeta are of course important for the behavior on the critical line they are more so in a commulative sense. Thus it might be that if RH would only fail by very little this might not have enough influence to detect it this way. | |
| Sep 6, 2012 at 17:23 | comment | added | Bazin | Something that I like with $(LH)$: you can try to improve the $\epsilon$, whereas $(RH)$ seems a different sort of game, all or nothing, no gradual approach. | |
| Sep 6, 2012 at 17:21 | comment | added | Bazin | Thanks for the answers. Let me reformulate my question: the estimate $$ \ln\bigl(\vert\zeta(\frac{1}{2}+it)\vert\bigr)=O\bigl((\ln t\ln\ln t)^{1/2}\bigr) $$ seems compatible with the remarks, but I guess that its logical relationship with $(RH)$ is not clear. | |
| Sep 6, 2012 at 14:42 | answer | added | user9072 | timeline score: 6 | |
| Sep 6, 2012 at 14:24 | answer | added | Micah Milinovich | timeline score: 12 | |
| Sep 6, 2012 at 12:06 | history | asked | Bazin | CC BY-SA 3.0 |