Timeline for Vanishing of a sum of roots of unity
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 17, 2022 at 6:25 | vote | accept | Denis Serre | ||
| Apr 16, 2022 at 17:41 | answer | added | François Brunault | timeline score: 15 | |
| Apr 16, 2022 at 17:40 | comment | added | LSpice | @GHfromMO's response referenced above. | |
| Apr 16, 2022 at 17:29 | history | became hot network question | |||
| Apr 16, 2022 at 17:24 | comment | added | GH from MO | @FrançoisBrunault Raising to a higher power of $2$ makes the calculation easier. See my resposne. | |
| Apr 16, 2022 at 16:33 | history | edited | GH from MO |
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| Apr 16, 2022 at 16:11 | answer | added | GH from MO | timeline score: 12 | |
| Apr 16, 2022 at 11:00 | comment | added | François Brunault | This is close to a Gauss sum, but not quite (summation up to $N-1$ instead of $2N-1$). One can try to work in the ring $R=\mathbb{Z}[z]/(2)$ (reduction mod 2). If $S$ denotes the sum in question, then in $R$ we have $S^2 = \sum_{k=0}^{N-1} \zeta_N^{2k^2+k}$ with $\zeta_N=z^2$. This last sum is a generalized quadratic Gauss sum. If $N$ is odd then it can be computed and it seems that some power of it is an odd integer, hence $S \neq 0$. This probably doesn't work for $N$ even. | |
| Apr 16, 2022 at 8:32 | comment | added | YiFan | Numerically, it seems like the sum tends towards $\sqrt{N/2}+2^{-3/2}(-1)^{N}$ to pretty good accuracy. I suspect it should be possible to prove something like this using some clever number theoretic argument, maybe. At least a lower bound of the form $a\sqrt{N}$ with $0<a<2^{-1/2}$ should be quite doable, though I'm not seeing how at the moment. | |
| Apr 16, 2022 at 7:52 | history | asked | Denis Serre | CC BY-SA 4.0 |