Timeline for Why is so much work done on numerical verification of the Riemann Hypothesis?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2020 at 21:54 | comment | added | mike | Got it! Thanks again! | |
| Jul 22, 2020 at 21:41 | comment | added | Conrad | @mike check this freely available paper for more details about $S$ researchgate.net/publication/… | |
| Jul 22, 2020 at 21:35 | comment | added | mike | Nice to know. Thanks a lot! | |
| Jul 22, 2020 at 21:32 | comment | added | Conrad | @mike if I remember correctly there is no known $T$ for which $|S(T)|$ is bigger than $4$ or $5$ although it is known to be unbounded - the speculation is that the true behavior of the zeroes is found where $S$ is fairly large | |
| Jul 22, 2020 at 21:21 | comment | added | Conrad | @mike - there is a $O(1/T)$ exact formula involving $S(T)=\arg zeta(1/2+iT)$ with good estimates for that one too | |
| Jul 22, 2020 at 20:08 | comment | added | mike | 2:The error term in Riemann–von Mangoldt formula (in wikipedia.org) is larger than 1. $\left|N(T)-\left(\frac{T}{2\pi}\log\frac{T}{2\pi e}-\frac{7}{8}\right)\right|<0.137\log T+0.443\log\log T+4.350,\text{ for } T>2.$. How do we use $N(T)$ from this formula to compare the number of numerical zeros from 1:? | |
| Mar 21, 2019 at 17:44 | comment | added | Greg Martin | Re 3: In particular, the higher that RH has been verified, the better explicit numerical bounds one can get on error terms for prime-counting functions. | |
| Mar 21, 2019 at 16:30 | comment | added | Nell | I'd say the point is not just that $Z(t)$ can be expressed somewhat more simply, but that it is in general simple to prove that a real function has a zero in a small interval - if the sign changes, there has to be a zero in there. | |
| Mar 21, 2019 at 15:39 | history | answered | Conrad | CC BY-SA 4.0 |