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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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, ... ,z_2N} 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, ... ,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 $e^{iz_{n}}$. 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)

$T(x):={a_{0} \over 2}+\sum _{n=1}^{N}a_{n}\cos(nx)+\sum _{n=1}^{N}b_{n}\sin(nx)={\mathbf {1 \over 2} }\sum _{n=-N}^{N}c_{n}e^{inx},$

where $c_{n}:=a_{n}-ib_{n}$ and $c_{-n}={\overline {c_{n}}}$ for $0\leq n\leq N$, and $P(z):=\sum _{n=0}^{2N}c_{n-N}z^{n}\in \mathbb {C} [z]$ 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 $N=2$... pma 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).pma 22:48, 6 January 2019 (UTC)