Timeline for are these polynomials or rationals functions?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2016 at 16:12 | vote | accept | T. Amdeberhan | ||
| Nov 21, 2016 at 3:11 | vote | accept | T. Amdeberhan | ||
| Nov 21, 2016 at 16:11 | |||||
| Nov 21, 2016 at 3:08 | history | edited | T. Amdeberhan | CC BY-SA 3.0 |
There are two questions, so this is to help readers.
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| Nov 21, 2016 at 3:03 | comment | added | Christian Remling | @AlexeyUstinov: Yes, we also need to observe that we've found all the zeros, but Fedor pointed that out already. | |
| Nov 21, 2016 at 3:01 | comment | added | Alexey Ustinov | It will be better to add that for general $t$ equation (1) has no multiple zeros. | |
| Nov 21, 2016 at 0:43 | comment | added | Christian Remling | @darijgrinberg: Yes, I guess if I use it sufficiently many times, I'll end up with a formula for the quotient. | |
| Nov 20, 2016 at 22:27 | comment | added | darij grinberg | You should be able to use that "evaluated" formula to simplify the denominator $P_1\left(x\right) + \cdots + P_n\left(x\right)$ without recourse to trig, right? | |
| Nov 20, 2016 at 22:22 | comment | added | Christian Remling | @FedorPetrov: Noticed that too :). I'm using that the exponent is odd in the final step, $\sin^{2k} nt$ would not be a linear combination of sines. | |
| Nov 20, 2016 at 22:17 | comment | added | Fedor Petrov | I was not right, by the way:) | |
| Nov 20, 2016 at 22:17 | comment | added | Fedor Petrov | I would rewrite (2) as $2\sin(Nt/2)\sin((N+1)t/2)=0$, this immediately yields that $t=2\pi k/N$ or $t=2\pi k/(N+1)$ with $k\in \mathbb{Z}$. We actually need to check also that all the roots of polynomial in the denominator are simple, but this is also clear as we found $N$ different roots $2\cos (2\pi k/N),2\cos (2\pi k/(N+1))$. | |
| Nov 20, 2016 at 21:59 | history | answered | Christian Remling | CC BY-SA 3.0 |