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January 4

Factoring trigonometric polynomials

Let

$T(x)=a_0+\sum _{n=1}^N{a}_n\cos (nx)+\sum _{n=1}^N{b}_n\sin (nx)(x\in ℝ)$

be a real trigonometric polynomial of degree N and let T have 2N roots {z₁, ... ,z_2N} in the interval [0, 2π). Does it follow that

$T(x)=k\prod _{n=1}^{2N}\sin \frac{x-z_n}{2}$

for some k?

Also, if a₀ = 0, what can you say about the roots {z₁, ... ,z_2N}? (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 $e^{i{z}_n}$. So the corresponding polynomials coincide up to a multiplicative factor $k$, as you stated. pma 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 {e^iz₁, ... ,e^iz_2N} is 0. For N=1 this is e^iz₁+e^iz₂ = 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)

$T(x):=\frac{a_0}{2}+\sum _{n=1}^N{a}_n\cos (nx)+\sum _{n=1}^N{b}_n\sin (nx)=\frac{1}{2}\sum _{n=-N}^N{c}_n{e}^{inx},$

where $c_n:=a_n-i{b}_n$ and $c_{-n}=\overline{c_n}$ for $0\le n\le N$, and $P(z):=\sum _{n=0}^{2N}{c}_{n-N}{z}^n\in ℂ[z]$ with roots $\{e^{i{z}_1},\dots ,e^{i{z}_{2N}}\}$. Then the $N$-degree coefficient $c_0=a_0$ vanishes iff

$\sum _J\exp \left(i\sum _{n\in J}{z}_n\right)=0$, 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 $\{e^{i{z}_1},\dots ,e^{i{z}_{2N}}\}$, even for $N=2$... pma 00:02, 6 January 2019 (UTC)

I think that you meant $c_n:=(a_n-i{b}_n)/2$? I tried to obtain the expression for $k$ and in the end obtained the following expression:

$T(x)=(i{a}_N+b_N)\exp (i\sum _{n=1}^{2N}{z}_n/2)\prod _{n=1}^{2N}\sin \frac{x-z_n}{2}$.

Therefore

$k=(i{a}_N+b_N)\exp \left(i\sum _{n=1}^{2N}{z}_n/2\right)$

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).pma 22:48, 6 January 2019 (UTC)