Wikipedia:Reference desk/Archives/Mathematics/2019 January 4

= January 4 =

Factoring trigonometric polynomials
Let
 * $$T(x) = a_0 + \sum_{n=1}^N a_n \cos (nx) + \sum_{n=1}^N b_n \sin(nx) \qquad (x \in \mathbb{R})$$

be a real trigonometric polynomial of degree N and let T have 2N roots {z1, ... ,z2N} in the interval [0, 2π). Does it follow that
 * $$T(x) = k\prod_{n=1}^{2N} \sin{{x-z_n} \over 2}$$

for some k?

Also, if a0 = 0, what can you say about the roots {z1, ... ,z2N}? (For example if N=1, a0 = 0 implies z1=z2±π.) --RDBury (talk) 11:15, 4 January 2019 (UTC)


 * Yes. If we write $$z:=e^{ix}$$, we see that both expressions on the RH-sides (the trigonometric polynomial and the sine product) have the form $$P(z)/z^N$$ for some $$2N$$-degree complex polynomial $$P(z)$$,  and in both cases, the polynomial  has $$2N$$ distinct complex roots $$e^{iz_n}$$. So the corresponding polynomials coincide up to a multiplicative factor $$k$$, as you stated. pm a  14:45, 5 January 2019 (UTC)
 * Thanks, I had a feeling converting to complex variables was the key but I was still missing some steps. By similar reasoning, I think for part 2 the answer is that the Nth symmetric polynomial in {eiz1, ... ,eiz2N} is 0. For N=1 this is eiz1+eiz2 = 0 which reduces easily to z1=z2±π, butfor N≥2 the relationship isn't that simple. --RDBury (talk) 21:50, 5 January 2019 (UTC)
 * I agree... We may write (using a more standard $$a_0/2$$ instead of $$a_0$$ for the constant term of the trigonometric polynomial)
 * $$T(x) := {a_0\over 2} + \sum_{n=1}^N a_n \cos (nx) + \sum_{n=1}^N b_n \sin(nx)= {\mathbf {1\over2}} \sum_{n=-N}^N c_n e^{inx},$$
 * where $$c_n:=a_n-ib_n$$ and $$c_{-n} =\overline{c_{n}}$$ for $$0\le n\le N$$, and  $$P(z):=\sum_{n=0}^{2N} c_{n-N} z^n\in\Complex[z]$$ with roots $$\{e^{iz_1},\dots,e^{iz_{2N}}\}$$. Then the $$N$$-degree coefficient $$c_0=a_0$$ vanishes iff
 * $$\sum_J\exp\big(i\sum_{n\in J}z_n\big)=0$$, the sum being extended over all subsets $$J$$ of $$\{1,2,\dots,2N\}$$ of cardinality $$N,$$ but I can't see a more geometric equivalent condition on the inscribed $$2N$$-gon with vertices $$\{e^{iz_1},\dots,e^{iz_{2N}}\}$$, even for $$N=2$$... pm a 00:02, 6 January 2019 (UTC)
 * I think that you meant $$c_n:=(a_n-ib_n)/2$$? I tried to obtain the expression for $$k$$ and in the end obtained the following expression:
 * $$T(x) = (ia_N+b_N)\exp({i\sum_{n=1}^{2N}z_n/2})\prod_{n=1}^{2N} \sin{{x-z_n} \over 2}$$.
 * Therefore
 * $$k = (ia_N+b_N)\exp\left({i\sum_{n=1}^{2N}z_n/2}\right)$$
 * As for $$a_0$$ I can only observe that it can not be arbitrary large because otherwise there will be no roots. Ruslik_ Zero 17:43, 6 January 2019 (UTC)
 * Yes, thank you, in fact I forgot a factor 1/2 in front of the sum (fixed now).pm a 22:48, 6 January 2019 (UTC)