Timeline for cosine of rational multiples of Pi take values of equal difference
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2017 at 15:14 | comment | added | Andrew | @FelipeVoloch Felipe has suggested a reference and it looks promising. I will read it but as a physicist I will need additional time to learn and digest the mathematics there. | |
| Apr 16, 2017 at 15:05 | comment | added | Andrew | continued: Hence we need to have $\zeta^{m_3}$ cancel with $\zeta^{N-m_1-1}$ and $\zeta^{m_1}$ cancel with $\zeta^{N-m_3-1}$ while $2\zeta^{m_2}$ cancels with $2\zeta^{N-m_2-1}$. Indeed, it is pretty hard to satisfy all these conditions, and when $m_1-m_2=m_2-m_3$ and $\frac{2m_2+1}{N}=\frac{1}{2}$ as in my original idea, these conditions are satisfied and we can achieve zero for the polynomial. This is kind of a loose argument but maybe there is a way to make the loose statement strict ... | |
| Apr 16, 2017 at 15:05 | comment | added | Andrew | Observing $\zeta^{m_3}-2\zeta^{m_2}+\zeta^{m_1}+\zeta^{N-m_3-1}-2\zeta^{N-m_2-1}+\zeta^{N-m_1-1}=0$, the last three terms are no longer conjugates of the first three, but shifted by -1 in the power, so it is "unlikely" that each term such as $\zeta^{m_3}$ can still cancel with $\zeta^{N-m_3-1}$. On the other hand the terms associated with the $m_2$'s have a different coefficient, i.e. 2, than the rest, so they have to cancel each other. To be continued | |
| Apr 16, 2017 at 13:17 | comment | added | Andrew | @AnthonyQuas Also the coefficients of the polynomial are not random, so the limit on the degree may be refined further. | |
| Apr 16, 2017 at 13:07 | comment | added | Andrew | @AnthonyQuas I woke up this morning and suddenly understood your statement! I also noticed that these 6 powers are not randomly distributed on the circle, so maybe the 5N/6 limit can be further refined. I will also check Felipe's reference to see if it helps. Your answer is a significant progress and thank you very much! | |
| Apr 16, 2017 at 3:11 | comment | added | Anthony Quas | I'll illustrate my answer by an example. Let $N=67$ and imagine that $\zeta^6 - 2\zeta^{13}+\zeta^{18}+\zeta^{60}-2\zeta^{53}+\zeta^{48}=0$. There are non-zero coefficients in the 6th, 13th, 18th, 48th, 53rd and 60th powers of $\zeta$. The gap between 60th and 6th powers is 13 (going around the circle), which is smaller than the gap between the 18th and 48th. , so I'll multiply the entire equation by $\zeta^{-48}=\zeta^{19}$ to obtain $1-2\zeta^5+\zeta^{12}+\zeta^{25}-2\zeta^{32}+\zeta^{37}=0$. The degree of this polynomial is less than $5N/6$. | |
| Apr 15, 2017 at 23:19 | comment | added | Felipe Voloch | Now that's been reduced to a sum of eight roots of unity, it suffices to apply theorem 6 of Conway and Jones, Acta Arith. XXX (1976) 229-240 to find all solutions. | |
| Apr 15, 2017 at 22:15 | comment | added | Andrew | I am stuck at the last step: " if you set the power that immediately follows the longest gap to be the constant term, the polynomial you arrive at has degree at most 5N/6." Otherwise this is a very nice proof for the numbers that satisfy $\phi(N)>5N/6$. I will think more along this line to see if there can be a proof or counter proof for non-primes. | |
| Apr 15, 2017 at 20:45 | comment | added | Anthony Quas | I suppose so! Since the OP already dealt with $N\le 8$, I wasn't too worried about that case. | |
| Apr 15, 2017 at 20:27 | comment | added | Mikhail Borovoi | Do you mean "which is true if $N$ is prime and $N>5\,$"? I think that $\phi(5)=4<\frac{5}{6}\cdot 5$. | |
| Apr 15, 2017 at 19:43 | history | answered | Anthony Quas | CC BY-SA 3.0 |