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.
Best Answer
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 $Z(P)+Z(\tilde{P})=8\gg 4$.