Timeline for Character sums over prime
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 1 at 10:38 | history | edited | gmvh |
edited tags
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| Oct 1 at 3:42 | history | edited | LSpice | CC BY-SA 4.0 |
Capitalise title; remaining typo; `{align*}`
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| Oct 1 at 3:39 | comment | added | LSpice | Re, then does $\sum_{p = 1}^N$ mean the sum over the primes in $[1, N]$, or over the first $N$ primes? | |
| Oct 1 at 1:11 | comment | added | Farzad Aryan | I was hoping to see if there are better results than, $N \gg q^{1/\epsilon}$. The sum is over primes. | |
| Oct 1 at 1:05 | comment | added | Farzad Aryan | Regarding the Siegel zero, assume the strong form of it, that $L(s, \chi)$ has no zero on the real line. | |
| Sep 30 at 20:12 | comment | added | LSpice | Is $\sum_{p = 1}^N$ meant to be a sum only over primes? | |
| Sep 30 at 15:31 | comment | added | Ofir Gorodetsky | Both Theorem 11.16 and Exercise 11.3.1.1, in the case of no exceptional zero, boil down to equation (11.26). On GRH you can of course take $N$ to be much smaller. | |
| Sep 30 at 15:01 | comment | added | Ofir Gorodetsky | If 'no Siegel zero' means 'no exceptional zero', you have cancellation when $q=N^{o(1)}$. Indeed, this is addressed by Theorem 11.16 of Montgomery-Vaughan and Exercise 11.3.1.1 in page 382. | |
| Sep 30 at 14:56 | comment | added | Will Sawin | I think the precise form of the "no Siegel zeroes" assumption could matter a lot here. | |
| Sep 30 at 14:51 | history | edited | kodlu | CC BY-SA 4.0 |
fixed typos
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| Sep 30 at 14:00 | history | asked | Farzad Aryan | CC BY-SA 4.0 |