Timeline for Does $ M(x)=O(\sqrt{x}) $ if and only if the De Bruijn-Newman constant is negative?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 22, 2018 at 20:24 | comment | added | GH from MO | I did not say that these two statements are equivalent. I said that if $M(x)=\sqrt{x}$ is false (as expected) then it is equivalent to $\Lambda<0$ (which is false by the preprint quoted). My claim is trivial, because any two false statements are equivalent. | |
| Jan 22, 2018 at 19:47 | comment | added | Sungjin Kim | @GHfromMO Hello GH, could you please elaborate on equivalence of $M(x)=O(\sqrt x)$ and $\Lambda<0$, or include a reference? | |
| Jan 20, 2018 at 2:26 | comment | added | GH from MO | @SylvainJULIEN: You are welcome. Actually, I was (or rather: we are) lucky as Rogders and Tao just proved the conjecture. There is a lot of activity and amazing progress in analytic number theory these days, thanks to great mathematicians captivated by the subject. | |
| Jan 19, 2018 at 10:00 | comment | added | Sylvain JULIEN | Wonderful ! I would not have thought that Newman's conjecture would be proven in my lifetime. Maybe a proof of RH will follow...Thank you very much for your answer. | |
| Jan 19, 2018 at 9:57 | vote | accept | Sylvain JULIEN | ||
| Jan 19, 2018 at 3:36 | history | edited | GH from MO | CC BY-SA 3.0 |
edited body
|
| Jan 19, 2018 at 3:30 | history | answered | GH from MO | CC BY-SA 3.0 |