Timeline for Positivity of partial Dirichlet series for a quadratic character?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2023 at 0:49 | comment | added | Kimball | Ah, okay, thanks for the clarification. | |
| Jan 16, 2023 at 17:29 | history | became hot network question | |||
| Jan 16, 2023 at 13:59 | history | edited | Zhang | CC BY-SA 4.0 |
added 193 characters in body
|
| Jan 16, 2023 at 13:54 | comment | added | Zhang | @Kimball:in this question, I am interested in the sum of the first $N$ terms of the Dirichlet series, where $N$ is the conductor of the character, NOT an arbitrary integer independent of $\chi$. | |
| Jan 16, 2023 at 13:38 | comment | added | Ofir Gorodetsky | I believe here the parameter $m$ is taken to be the conductor of $\chi$ which is not addressed by the answer at MSE. The work of Granville-Soundararajan mentioned there says: for given $m$, one can find a quadratic character $\chi$ with $\sum_{n \le m} \chi(n)/n$ negative. In OP's problem, $m$ is the conductor of $\chi$ so the variables $m$ and $\chi$ are 'coupled'. OP's problem turns out to be much easier than the results of Granville-Soundararajan (at least conditionally). | |
| Jan 16, 2023 at 13:28 | comment | added | Kimball | What do you mean the answer there was a partial answer? It tells you the answer is no. What more do you want? | |
| Jan 16, 2023 at 13:27 | comment | added | reuns | Greg Martin's answer on MSE is the right one. | |
| Jan 16, 2023 at 13:09 | vote | accept | Zhang | ||
| Jan 16, 2023 at 11:56 | answer | added | Ofir Gorodetsky | timeline score: 5 | |
| Jan 16, 2023 at 9:25 | history | asked | Zhang | CC BY-SA 4.0 |