Timeline for Residues of $\frac{1}{\prod_{i=1}^n (x-P_i)^{e_i}}$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 16, 2018 at 16:22 | comment | added | Fedor Petrov | @Huy this is up to you, but I do not consider the question being answered. | |
| Sep 16, 2018 at 16:19 | vote | accept | Huy Dang | ||
| Sep 16, 2018 at 17:41 | |||||
| Sep 16, 2018 at 7:11 | comment | added | Huy Dang | @FelipeVoloch You are right. One can extend the conjecture to the case $e_i$ large and change the condition to $\sum_{i=1}^n \overline{e}_i < p+n$ where $ \overline{e}_i \equiv e_i \pmod p$ and $1 <\overline{e}_i \le p$. | |
| Sep 16, 2018 at 7:01 | comment | added | Fedor Petrov | @FelipeVoloch exactly, and I even tried to sketch the proof:) | |
| Sep 16, 2018 at 6:48 | comment | added | Felipe Voloch | Residues equal to zero is not the same as having a rational antiderivative in positive characteristic, e.g. $x^{p-1}$. But the conditions $\sum e_i < p+n, e_i>1$ imply that each $e_i < p-1$ so it's OK. | |
| Sep 16, 2018 at 5:30 | vote | accept | Huy Dang | ||
| Sep 16, 2018 at 12:36 | |||||
| Sep 15, 2018 at 23:40 | comment | added | Huy Dang | That is a very smart argument. I haven’t thought about $p$ does not divide $a-b$. Thank you so much! | |
| Sep 15, 2018 at 23:38 | vote | accept | Huy Dang | ||
| Sep 15, 2018 at 23:38 | |||||
| Sep 15, 2018 at 23:26 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
added 15 characters in body
|
| Sep 15, 2018 at 23:19 | history | answered | Fedor Petrov | CC BY-SA 4.0 |