Prove that a trigonometric polynomial has 2n roots

Asked Jun 03 '24 at 11:35

Active Jun 03 '24 at 12:18

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$0<a_0<a_1<a_2<...<a_n$ Prove:$$a_0+a_1\cos\theta+...+a_n\cos n\theta $$ has 2n distinct roots in $(0,2\pi)$

This is a question in my textbook,the author leaves a hint.That is we should firstly prove that

$$P_n(z)= \sum_{k=0}^{n}a_k z^k \ has \ n \ roots \ in \ B(0,1),$$ B(0,1)={$z\in \mathbb{C}:|z|<1$}. I don't know how to prove this,may I use Rouche theorem or argument principle? I have almost no ideas at all.

edited Jun 03 '24 at 11:50

Anne Bauval

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asked Jun 03 '24 at 11:35

MathNoob

1

It is not clear whether your question is how to prove the hint, or how do derive the claim from it. – Anne Bauval Jun 03 '24 at 11:48
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Does this answer your question? Showing that the roots of a polynomial with descending positive coefficients lie in the unit disc. (found using the MSE search engine). – Anne Bauval Jun 03 '24 at 11:49

Prove that a trigonometric polynomial has 2n roots

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