# Product of trigonometric polynomials is a trigonometric polynomial

A trigonometric polynomial was defined as $$f(x) = \frac{a_0}{2} + \sum_{k=1}^{n}(a_k \cos(kx) + b_k \sin(kx))$$

I heard somewhere that trigonometric polynomials have a ring structure, i.e. a product of two polynomials is again a polynomial.

I was wondering how can one see it. A direct attempt to multiply two polynomials seem to result in a mess.

Edit:

Using a suggestion from nicomezi from below I go to the complex representation:

$$f(x) = \frac{a_0}{2} + \frac{1}{2} \sum_{k=1}^{n}(e^{ikx}z_k + e^{-ikx} \bar z_k )$$

We can also "hide" the first term:

$$f(x)=\frac{1}{2} \sum_{k=0}^{n}(e^{ikx}z_k + e^{-ikx} \bar z_k ), z_0 = a_0$$

Now when we multiply two polynomials of degrees $n$ and $m$ we get a sum of terms of the form:

$$\frac{1}{4}(e^{ikx}z_k + e^{-ikx} \bar z_k )(e^{ilx}z_l + e^{-ilx} \bar z_l )= \frac{1}{4}(e^{i(k+l)x}z_kz_l + e^{-i(k+l)x}\overline{z_k z_l}) + \frac{1}{4}(e^{i(k-l)x}z_k\bar z_l + e^{-i(k-l)x} \overline{z_k \bar z_l}), 1 \le k \le n, 1 \le l \le m$$

So each term is a polynomial of degree $(k+l)$, thus the whole product is a polynomial of degree $(n+m)$.

There is no need to multiply two polynomials, that does, indeed, result in a mess. Instead, all you need to show is that:

• $\cos(kx)\cos(lx)$ is a trigonometric polynomial,

• $\cos(kx)\sin(lx)$ is a trigonometric polynomial, and

• $\sin(kx)\sin(lx)$ is a trigonometric polynomial.

The rest comes from linearity.

Let's check $\cos(kx)\sin(lx)$ is a trigonometric polynomial; the rest of the cases are very similar.

Since $e^{ikx}=\cos(kx)+i\sin(kx)$, $\cos(kx)=\frac{1}{2}\left(e^{ikx}+e^{-ikx}\right)$. Similarly, $\sin(lx)=\frac{1}{2}\left(ie^{-ilx}-ie^{ilx}\right)$. Therefore, $$\cos(kx)\sin(lk)=\frac{1}{4}\left(ie^{i(k-l)x}+ie^{-i(k+l)x}-ie^{i(k+l)x}-ie^{i(l-k)x}\right).$$ Since our answer should be real, we are only interested in the real part of this, which is: $$\cos(kx)\sin(lx)=-\frac{1}{2}\sin((k-l)x)+\frac{1}{2}\sin((k+l)x)$$

I believe the neatest way to see this is through the Fourier transform.

You can think of the series that you have denoted as a collection of frequencies that you are in possession of. Multiplying the two series in the time domain is equivalent to convolution in the frequency domain.

If you are not aware of what convolution is, it is in some sense a rule of combining two functions $f$ and $g$ such that $f$ is "smeared" according to $g$'s pattern.

• thanks for your effort, I must admit that I am at the very beginning of my studies of Fourier series and Fourier transform is something above my level yet. But if I believe my class instructor I should be able to find an answer to this question "at my level of knowledge".
– zesy
May 3, 2016 at 8:38
• Another way but less elegant is to consider the reals trigonometric polynomials as a particular case of complex trigonometric polynomials, which are easier to manipulate. Then you may be able to prove it quite easily with mathematical induction. May 3, 2016 at 8:46