Timeline for Perfectly balanced sets of complex numbers
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2020 at 6:29 | vote | accept | Anton Klyachko | ||
| Jan 2, 2020 at 6:14 | comment | added | Sungjin Kim | Yes, that is correct. | |
| Jan 2, 2020 at 5:37 | comment | added | Anton Klyachko | Yes, I see. So, you prove that a set $X$ is perfectly balanced (i.e. satisfies the condition from my question) if and only if $\{x^k\;|\;x\in X\}=\root^p\of c$ for some integer $k$, prime $p$, and complex $c$, right? It would be the answer to a more sensible version of Question 2. | |
| Jan 1, 2020 at 8:48 | comment | added | Sungjin Kim | Happy new year! The asymmetric ones has a proper subset whose sum of powers vanish infinitely often. So, they are not inclusion-minimal. | |
| Jan 1, 2020 at 8:42 | comment | added | Sungjin Kim | In your problem, the requirement that any proper subset gives finitely many $n$'s with $n$th power sum vanish, allows us to conclude that your set $X$ must be of the form. | |
| Jan 1, 2020 at 7:28 | comment | added | Anton Klyachko | Happy New Year, Sungjin! In Step 3, do you mean that any inclusion-minimal vanishing sum of roots of unity is of the form $g+g\zeta_p+\cdots + g\zeta_p^{p-1}$? It seems strange because in the paper you cited some asymmetric inclusion-minimal vanishing sums are discussed... | |
| Dec 27, 2019 at 1:21 | history | edited | Sungjin Kim | CC BY-SA 4.0 |
added 52 characters in body
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| Dec 26, 2019 at 23:28 | history | edited | Sungjin Kim | CC BY-SA 4.0 |
added 28 characters in body
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| Dec 26, 2019 at 23:16 | history | edited | Sungjin Kim | CC BY-SA 4.0 |
completed proof.
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| Dec 25, 2019 at 15:55 | comment | added | Sungjin Kim | The next one in the iteration, we consider $x_1^b+x_2^b+x_3^b=0$ or $x_3^b+x_4^b=0$. Rewriting the summation expression for $s_{an+b}$ again, gives shorter sum. | |
| Dec 25, 2019 at 15:50 | comment | added | Anton Klyachko | Thanks! But how $x_1^b+x_2^b+x_3^b+x_4^b=0$ "yields a shorter vanishing sum"? | |
| Dec 25, 2019 at 15:49 | comment | added | Sungjin Kim | If $x_1^b+x_2^b=0$ and $x_1^a=x_2^a$ then $s_{an+b}=0$ gives $\sum_{x\in X} x^{an+b} = \sum_{j=1}^k x_j^{an+b}= (x_1^b+x_2^b)x_1^{an}+\sum_{j=3}^k x_j^{an+b}$ | |
| Dec 25, 2019 at 15:11 | history | edited | Sungjin Kim | CC BY-SA 4.0 |
added 158 characters in body
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| Dec 24, 2019 at 21:29 | history | answered | Sungjin Kim | CC BY-SA 4.0 |