Timeline for Prove an inequality related to sums of Legendre symbols
Current License: CC BY-SA 4.0
12 events
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| Apr 17, 2022 at 2:55 | vote | accept | math110 | ||
| Apr 12, 2022 at 16:32 | history | edited | Denis Serre | CC BY-SA 4.0 |
added 24 characters in body; edited title
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| Apr 12, 2022 at 8:39 | history | became hot network question | |||
| Apr 12, 2022 at 2:06 | answer | added | Will Sawin | timeline score: 18 | |
| Apr 12, 2022 at 1:44 | history | edited | math110 | CC BY-SA 4.0 |
deleted 4 characters in body
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| Apr 12, 2022 at 1:37 | history | edited | math110 | CC BY-SA 4.0 |
deleted 6 characters in body
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| Apr 12, 2022 at 1:35 | comment | added | David E Speyer | To me, this bound seem's likely to be true but not close to tight. The maximum of $\sum_{1 \leq i<j \leq p} x_i x_j$ subject to the same inequalities is $\tfrac{p-1}{2p}$, and I would think you could get more mileage out of the alternation of the Legenrdre symbol than just an $1+O(1/p)$ relative improvement. But I don't have a proof. | |
| Apr 12, 2022 at 0:49 | history | edited | math110 | CC BY-SA 4.0 |
edited body
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| Apr 12, 2022 at 0:46 | comment | added | math110 | Preliminary writing, is said to be the study of this quadratic inequality, so the Internet is not yet found, thank you | |
| Apr 12, 2022 at 0:43 | comment | added | LSpice |
What is "this paper"? \\ Also, MO generally better receives questions that are not worded in the imperative (such as "How can one prove this inequality involving Legendre sums?" rather than "Prove this inequality"). \\ Finally, there is a TeX command \genfrac designed for Legendre-symbol-type commands. In this case, \genfrac(){}{}{i - j}p will do it. I have edited accordingly (but not for the title, where I just slightly cleaned up the grammar).
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| Apr 12, 2022 at 0:43 | history | edited | LSpice | CC BY-SA 4.0 |
`\genfrac`
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| Apr 12, 2022 at 0:34 | history | asked | math110 | CC BY-SA 4.0 |