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I've been trying to prove the following: A trigonometric polynomial $P$ with degree $N>0$ on $\mathbb{T}$ has at most $2N$ zeroes.

So if $P(x)=\sum_{n=-N}^{N}c_ne^{inx}$, my idea was to somehow use that $\cos(nx)$ and $\sin(nx)$ have $2n$ roots in order to do this, but I'm kinda stuck on the way to write it down. Am I actually even allowed to use it like that, or should I prove that first? The more I think about it the less it wants to make sense though, since I can't write it down properly. I've read on Wikipedia that this was in a Numerical Analysis book form Powell (1981), haven't been able to access it though.

PLY
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You can think of this as a complex function

$$P(z) = \sum_{n = -N}^N c_n z^n$$

on the circle, where $z = e^{ix}$. On the other hand, we can think of this as the restriction of a function on $\mathbb{C}$ with the same definition. Now $z^NP(z)$ is a complex polynomial of degree $\le 2N$, and therefore has at most $2N$ zeros on all of $\mathbb{C}$, some of which may be in $\mathbb{T}$. Dividing by $z^N$ gives the claim.

  • I see, so what you're doing is multiplying by $z^N$ in order for $z^N P(z)$ to be a complex polynomial and then use the fundamental theorem of algebra. What I still question then is how we can just divide both sides by $z^N$. If I understood correctly we now find $\deg(z^NP(z))\leq 2N$, but how can we justify this division by $z^N$ to remain at the same number of zeroes for $P(z)$? I thought that division like this to find amount of zeroes only works when you divide by a constant. – PLY Apr 25 '18 at 09:46
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    @Kirigiri If $P(z) = 0$ then $f(z) = 0$, where $f(z) = z^N P(z)$. There are at most $2N$ numbers satisfying the first condition, and we don't actually have to divide (although it's not a problem dividing - $z^N$ is non-zero on $\mathbb{T}$). –  Apr 25 '18 at 18:53
  • Thank you for clarifying! I indeed forgot that we are dealing with $e^{ix}$ and it will never be equal to zero for set domain. – PLY Apr 25 '18 at 22:59