Timeline for Speed of convergence for Weyl's Equidistribution theorem
Current License: CC BY-SA 3.0
2 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 12, 2014 at 23:00 | comment | added | user58955 | The exponent is $-(2+\epsilon)$ in Roth's theorem, so I suppose one would pick $H = N^{1/3}$ to minimise the right-hand size. | |
| Jun 12, 2013 at 6:47 | history | answered | blober | CC BY-SA 3.0 |