Timeline for A sum involving roots of unity
Current License: CC BY-SA 4.0
18 events
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| Feb 8, 2019 at 11:24 | history | edited | Chitsai Liu | CC BY-SA 4.0 |
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| Feb 6, 2019 at 3:13 | comment | added | GH from MO | I added a proof of Nemo's identity. The whole thing is mysterious and beautiful! | |
| Feb 6, 2019 at 3:11 | answer | added | GH from MO | timeline score: 5 | |
| Feb 6, 2019 at 2:11 | history | edited | Chitsai Liu | CC BY-SA 4.0 |
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| Feb 6, 2019 at 1:29 | vote | accept | Chitsai Liu | ||
| Feb 5, 2019 at 23:19 | answer | added | Fedor Petrov | timeline score: 10 | |
| Feb 5, 2019 at 13:25 | comment | added | Chitsai Liu | Based on the comments from @Nemo, it suffices to show that $$ \sum_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1, $$ where $y=e^{\frac{2\pi i}{6n+4}}$ is the primitive $(6n+4)$th root of unity. | |
| Feb 5, 2019 at 11:23 | review | Suggested edits | |||
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| Feb 5, 2019 at 8:37 | comment | added | Nemo | Now if one specifies w as the primitive (3n+2)th root of unity then the first sum on the rhs due to the symmetry $k\to 2n+1-k$ equals the sum under consideration $$\sum _{k=1}^{2 n+1} \frac{(-1)^k w^{k (3 k+1)/2}}{1-w^{3 k}}=\sum _{k=0}^{2 n} \frac{(-1)^k w^{\frac{1}{2} (k+2) (3 k+1)}}{1-w^{3 k+1}}$$ thus allows to solve for this sum in terms of 4 simpler sums. I believe this 4 simple sums can be calculated using the symmetry $k\to 2n+1-k$. | |
| Feb 5, 2019 at 8:22 | comment | added | Nemo | I proved $$\sum _{k=1}^{2 n+1} \frac{(-1)^k w^{k (3 k+1)/2}}{1-w^{3 k}}=-\sum _{k=0}^{2 n} \frac{(-1)^k w^{\frac{1}{2} (k+2) (3 k+1)}}{1-w^{3 k+1}}+\\\frac{1}{2} \sum _{k=0}^{2 n} \frac{(-1)^k w^{(3 k+1) (n+1)}}{1-w^{(3k+1)/2}}+\frac{1}{2} \sum _{k=0}^{2 n} \frac{w^{(3 k+1) (n+1)}}{w^{(3k+1)/2}+1}+\frac{1}{2} \sum _{k=1}^{2 n+1} \left(\frac{(-1)^k w^{k/2}}{1-w^{3k/2}}+\frac{w^{k/2}}{w^{3k/2}+1}\right)$$ for arbitrary $w$. | |
| Feb 5, 2019 at 5:38 | comment | added | Fedor Petrov | It looks like a "q-version" (whatever this means) of the known congruence $1-1/2+1/3-1/4+\dots+1/(2n+1)\equiv 0 \pmod {3n+2}$ provided that $3n+2$ is prime. (For $n=659$ this was proposed on IMO in year $3n+2=1979$.) The standard solution is $S=(1+1/2+\dots+1/(2n+1))-2(1/2+\dots+1/(2n))=1/(n+1)+\dots+1/(2n+1)$ and now the symmetry $1/x+1/(3n+2-x)\equiv 0 \pmod {3n+2}$. I do not see how to modify it. | |
| Feb 5, 2019 at 2:47 | history | edited | GH from MO |
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| Feb 5, 2019 at 1:40 | comment | added | Chitsai Liu | @Seva, It seems to be difficult to prove this sum is real. | |
| Feb 4, 2019 at 18:19 | comment | added | Seva | Can you prove at least that your sum is rational? If so, you can average it over all primitive roots of unity of degree 3n+2. | |
| Feb 4, 2019 at 12:39 | history | edited | Chitsai Liu |
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| Feb 4, 2019 at 6:21 | comment | added | darij grinberg | This reminds me of Daniel Shanks, A short proof of an identity of Euler (if not for the denominators on the left hand side...). Wondering if there is anything behind this? | |
| Feb 4, 2019 at 5:13 | history | edited | Chitsai Liu | CC BY-SA 4.0 |
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| Feb 4, 2019 at 4:52 | history | asked | Chitsai Liu | CC BY-SA 4.0 |