# How to show that two trigonometric polynomials of degree $n$ combined have at most $2n$ zeros?

I am already aware of this question: Prove the following trigonometric polynomial has $2n$ zeros

But it's not the same.

Let be $$P(x) = \sum_{k=0}^{n} a_k \cos (kx)$$ and $$\tilde{P}(x) = \sum_{k=0}^n a_k \cos((n - k)x)$$.

If we denote $$Z(Q)$$ the number of zeros of $$Q$$ (we count without multiplicities, so: $$Z(X^2) = 1$$), then, how can we show that, that over $$[0, 2\pi[$$ :

$$\begin{equation*} Z(P) + Z(\tilde{P}) \leq 2n \end{equation*}$$

I know that I can show that: $$\sum_{k=0}^n a_k \cos(kx)$$ have at most $$2n$$ zeros, using an imaginary exponential form and the fact that $$x \mapsto e^{ix}$$ is a bijection from $$[0, 2\pi[$$ to $$\mathbb{U}$$ the set of complex of module $$1$$.

But I don't know how to relate those two polynomials to an exponential imaginary form, I tried to expand $$\cos((n - k)x)$$ to create $$\sin, \cos$$ expressions, but without luck.

• @JackD'Aurizio My bad, I meant without – Raito Apr 23 at 18:15

The claim is false. If we take $$P(x)=1-\cos(x)+\cos(2x)$$ we have $$n=2$$ and $$\tilde{P}(x)=P(x)$$.
$$Z(P)=4$$ then leads to $$Z(P)+Z(\tilde{P})=8\gg 4$$.
• What if we ask about distinct roots of $Z(P), Z(\tilde P)$? I think that if $a_k$ are real there is a good chance the result may be true in the sense that the number of distinct roots of $P\tilde P=0$ is still at most $2n$, but of course I could be wrong... – Conrad Apr 24 at 12:38
• @Conrad: that was my suspect, in a deleted comment I wondered about the original exercise being about the distinct roots of $P\tilde{P}$. – Jack D'Aurizio Apr 24 at 15:26