Timeline for Geometry of numbers argument: counting integers with some linear condition
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 31, 2016 at 22:35 | comment | added | Johnny T. | I was wondering would a similar argument work if I change the definition of $U(Z)$ to be the number of $v$ with $|v|< ZA$ and $\|\lambda (v - t_0)\| <ZA^{-1}$, where $t_0$ is some integer? Thank you very much! | |
| Jun 22, 2016 at 11:03 | comment | added | Fedor Petrov | I think, inequality $U(Z_1)\gg (Z_1/Z_2)^2 U(Z_2)$ is less or more clear: partition the arc $\|z\|<Z_2A^{-1}$ onto arcs $\delta_1,\dots,\delta_k$ of length $2Z_1A^{-1}$, and also partition integers between 1 and $Z_2A$ onto segments $\Delta_1,\dots,\Delta_k$ of length $Z_1A$, $k$ is about $Z_2/Z_1$. By pigeonhole principle there exist indices $u,v$ such that at least $U(Z_2)/k^2$ elements of $\Delta_u$ belong to $\delta_v$, this implies $U(Z_1)\geqslant U(Z_2)/k^2$. | |
| Jun 22, 2016 at 9:47 | history | answered | js21 | CC BY-SA 3.0 |