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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.

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  • $\begingroup$ @JackD'Aurizio My bad, I meant without $\endgroup$ – Raito Apr 23 at 18:15
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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 $Z(P)+Z(\tilde{P})=8\gg 4$.

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  • $\begingroup$ 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... $\endgroup$ – Conrad Apr 24 at 12:38
  • $\begingroup$ @Conrad: that was my suspect, in a deleted comment I wondered about the original exercise being about the distinct roots of $P\tilde{P}$. $\endgroup$ – Jack D'Aurizio Apr 24 at 15:26

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