Prove that roots of trig polynomials are denumerable

[Mr Davis 97]

Homework Statement

Prove that the roots of trigonometric polynomials with integer coefficients are denumerable.

Homework Equations

The Attempt at a Solution

The book does not define what a trig polynomial is, but I am assuming it is something of the form ##\displaystyle a_0 + \sum^N_{n=1}a_n \cos (nx) + \sum^N_{n=1}b_n \sin(nx)##. With normal polynomials, we have access to the fundamental theorem of algebra. I am not sure what we have access to; I guess we have access to the fact that the function must be periodic.

My argument might come down to showing that there are a finite number of roots in a period, and showing that there are denumerable periods, which would show that the number of roots is denumerable. Is this on the right track?

 

[fresh_42]

I would do it the other way around. First show that ##\cos nx## and ##\sin nx## have countably many roots, perhaps with Euler's formula, and then show that all possible combinations doesn't change this cardinality, because ##N\cdot \aleph_0 = \aleph_0 + \aleph_0 = \aleph_0 \cdot \aleph_0 = \aleph_0##.

 

[Ray Vickson]

Homework Statement

Prove that the roots of trigonometric polynomials with integer coefficients are denumerable.

Homework Equations

The Attempt at a Solution

The book does not define what a trig polynomial is, but I am assuming it is something of the form ##\displaystyle a_0 + \sum^N_{n=1}a_n \cos (nx) + \sum^N_{n=1}b_n \sin(nx)##. With normal polynomials, we have access to the fundamental theorem of algebra. I am not sure what we have access to; I guess we have access to the fact that the function must be periodic.

My argument might come down to showing that there are a finite number of roots in a period, and showing that there are denumerable periods, which would show that the number of roots is denumerable. Is this on the right track?

OK, but if you have proved the result for ordinary polynomials, you have it as well for trig polynomials, because
$$\sin (nx) = \frac{1}{2i} \left( (e^{ix})^n - (e^{ix})^{-n} \right)\; \text{and} \; \cos(nx) = \frac{1}{2} \left( (e^{ix})^n + (e^{ix})^{-n} \right)$$
Thus, if ##z = e^{ix}## we have
$$a_0 + \sum_{n=1}^N (a_n \cos(nx) + b_n \sin(nx) ) = \frac{1}{z^N} \left[ a_0 z^N + \sum_{n=1}^N z^{N-n} \left(\frac{1}{2} a_n (z^{2n}+1) + \frac{1}{2i}b_n (z^{2n} - 1) \right) \right] $$