Timeline for How small can a sum of a few roots of unity be?
Current License: CC BY-SA 3.0
7 events
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| S Jul 14, 2013 at 4:09 | history | suggested | C.S. | CC BY-SA 3.0 |
$J=\{1,5,9,17,18\}\cup\left(\frac{n}{2}+\{2,3,11,15,19\}$ changed to $J=\{1,5,9,17,18\}\cup\left(\frac{n}{2}+\{2,3,11,15,19\}\right)$
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| Jul 14, 2013 at 4:03 | review | Suggested edits | |||
| S Jul 14, 2013 at 4:09 | |||||
| Nov 15, 2010 at 20:51 | comment | added | Gerry Myerson | @Denis, OP didn't ask for distinct roots of unity so I assume it's no issue. Tarry-Escott just means $\sum^ma_i^r=\sum^mb_i^r$ for $r=1,\dots,k$ with no $a_i=b_j$, preferably with $m$ as small as possible. The paper shows how to go from such an equation to a small sum of $2m$ roots of unity. It's known $m$ can be taken to be something like $k^2$, and the paper indicates how small a sum that leads to. | |
| Nov 15, 2010 at 11:53 | comment | added | Denis Serre | @Gerry. My purpose was to sum distinct roots of unity. Isn't it an issue ? Bwt, I did not-find the term Tarry-Escott in your paper. | |
| Nov 15, 2010 at 11:51 | history | edited | Denis Serre | CC BY-SA 2.5 |
added 388 characters in body; added 3 characters in body
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| Nov 15, 2010 at 11:17 | comment | added | Gerry Myerson | I'm not sure I understand the construction, but I think it's unnecessarily complicated. Just take a sum of 10 $n$th roots of modulus $O(n^{-5})$ and square it to get a sum of 100 (not necessarily distinct) $n$th roots of modulus $O(n^{-10})$. Or raise it to the power $r$ to get $10^r$ $n$th roots whose sum has modulus $O(n^{-5r})$. But don't the known Tarry-Escott solutions mentioned in my paper do better than that? | |
| Nov 15, 2010 at 10:19 | history | answered | Denis Serre | CC BY-SA 2.5 |