Timeline for Incomplete Kloosterman sum
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 13, 2014 at 14:04 | comment | added | Will Sawin | @GHfromMO It is now! | |
| Jul 13, 2014 at 11:15 | comment | added | GH from MO | I noticed your $e(x)$, but I thought this was your short-hand for $e_p(x)$. | |
| Jul 13, 2014 at 1:27 | vote | accept | Farzad Aryan | ||
| Jul 13, 2014 at 1:27 | vote | accept | Farzad Aryan | ||
| Jul 13, 2014 at 1:27 | |||||
| Jul 13, 2014 at 1:26 | vote | accept | Farzad Aryan | ||
| Jul 13, 2014 at 1:26 | |||||
| Jul 12, 2014 at 21:18 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jul 12, 2014 at 21:18 | comment | added | Will Sawin | Thanks GH and Noam! No one has yet pointed out that I used $e(x)$ instead of $e\left(\frac{x}{p}\right)$, which would be more consistent. I typed this answer quickly! | |
| Jul 12, 2014 at 21:13 | comment | added | Noam D. Elkies | …and there's a stray ")" in each instance of the denominator $e(b)-1$. | |
| Jul 12, 2014 at 21:09 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 2 characters in body
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| Jul 12, 2014 at 7:23 | comment | added | GH from MO | Also, in your last sentence, $f(0,0)$ is not $p$ but $p-1$. | |
| Jul 12, 2014 at 7:16 | comment | added | GH from MO | In your third display, $f(a-1,b-1)$ should be $f(a+1,b+1)$. This is because the Fourier transform of $\mathbf 1_{\{xy=1\}}e(x+y)$ equals $f(a+1,b+1)$. | |
| Jul 12, 2014 at 5:31 | comment | added | Alexey Ustinov | Sorry, this full sum saves a $\log$ indeed. I was accustomed to think about arbitrary lines. | |
| Jul 12, 2014 at 5:19 | comment | added | Will Sawin | I don't think so - my second set of sums is over all $x$ from $1$ to $p$. I'm not using the geometric progression formula. | |
| Jul 12, 2014 at 5:15 | comment | added | Alexey Ustinov | You forget about $e(a)-1$ in denominator. The error term is $O(\sqrt p\log^2p)$. | |
| Jul 12, 2014 at 0:54 | history | answered | Will Sawin | CC BY-SA 3.0 |