Timeline for linear independence of $\sin(k \pi / m)$
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jul 19, 2016 at 15:03 | comment | added | Todd Trimble | @JarekKuben You are quite right; thank you very much. I have fixed the error. | |
| Jul 19, 2016 at 15:02 | history | edited | Todd Trimble | CC BY-SA 3.0 |
fixed an error pointed out in the last comment
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| Jul 18, 2016 at 19:56 | comment | added | Jarek Kuben | $\exp(\pm k\pi i/m)$ is primitive $2m$-th root of unity only if $k$ is odd. That would be OK if $m$ was even, but then $N$ wouldn't be squarefree. | |
| Jul 18, 2016 at 15:10 | vote | accept | user95204 | ||
| Jul 18, 2016 at 14:43 | history | undeleted | Todd Trimble | ||
| Jul 18, 2016 at 14:42 | history | edited | Todd Trimble | CC BY-SA 3.0 |
resurrected an answer in response to comments
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| Jul 18, 2016 at 13:03 | history | deleted | Todd Trimble | via Vote | |
| Jul 18, 2016 at 13:02 | comment | added | Vladimir Dotsenko | The number of primitive roots is equal to the degree of the extension so they are linearly independent if and only if they form a basis. For example, for $N=4$ the two primitive roots are $i$ and $-i$. | |
| Jul 18, 2016 at 12:59 | comment | added | Todd Trimble | @VladimirDotsenko Basis or not, don't we still have linear independence? I'm not seeing anything in Lang speaking against this, although maybe I'm overlooking something (?). | |
| Jul 18, 2016 at 12:47 | comment | added | Vladimir Dotsenko | Dear Todd, this is confusing. Primitive roots of unity only form a basis for a square-free $N$. (See math.stackexchange.com/questions/87290/… for instance.) | |
| Jul 18, 2016 at 12:24 | history | answered | Todd Trimble | CC BY-SA 3.0 |