How to prove or disprove that the set of roots of the trigonometric polynomials with integer coefficients is denumerable

cardinalselementary-set-theorypolynomialsreal-analysis

How can I prove that the set of roots of the trigonometric polynomials WITH integer coefficients is denumerable ? The concept of trigonometric polynomials is also not clear to me.

I know how to show the roots of the polynomials in one variable with rational coefficients are denumerable.

Can anyone please help me to show this?

Best Answer

Trigonometric polynomial is a polynomial of the form $$W(x)=a_0 +\sum_{k=1}^n (a_k \cos kx +b_k \sin kx)$$ If we substitute $$\cos kx =\frac{e^{ikx} +e^{-ikx}}{2} \mbox{ } \sin kx =\frac{e^{ikx} -e^{-ikx}}{2i}$$ where $i^2 =-1.$

And then $z=e^{ix}$ then the equation $$W(x) =0$$ is equivalent to some equation $$P(z) =\sum_{l=0}^{2n} c_l z^l =0$$ But from the Fundamental Theorem of Algebra the equation $$P(z)=0$$ has at most $2n$ solutions say $\{z_1, z_2, z_3,....z_{2n}\}$ and each equation $$\cos x =\frac{z_s +\overline{z_s }}{2} \mbox{ } \sin x =\frac{z_s -\overline{z_s}}{2i}$$ has atm most countable solutions for $s=1,2,...,2n.$

Therefore the set $$\{ x:W(x) =0\}$$ is countable as finite union of countable sets.

Related Solutions

[Math] Show that the set of polynomials with rational coefficients is countable.

Let us begin by stating that the set of rationals is countable, as Z² is countable and every rational number can be expressed as an ordered pair of integers.

Then we can enumerate the polynomials as follows:

1.Start with n=1
2.For every m on [1,n] list the next order-m polynomial
3.Increase n by 1 and return to step 2.

For a proper enumeration of each order of polynomials (which must exist as, Q being countable, Qⁿ is countable for all finite n and every order-n rational-coefficient polynomial can be expressed as an ordered n-tuple), every polynomial of rational coefficients will eventually appear exactly once.

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[Math] Show that the set of polynomials with rational coefficients is countable.

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