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Let $a=(a_0,a_1,...,a_n)\in \mathbb R^{n+1}$ and let $P_a(x)=\displaystyle\sum_{k=0}^{n}a_k \cos (kx) $ and let $b=(a_n,a_{n-1},...,a_0)$.

If $Z_a$ is the number of roots of $P_a$ on $[0,2\pi]$

then $$Z_a+Z_b \geq 2n$$

In this post Roots of trigonometric polynomial, an answer was given to me.I have tried to look for an elementary solution and at the same time study complex analysis and use the argument principle.

If $\gamma$ is the circle centered around $0$ with radius $1$, $$\int_{\gamma} \frac {f'(z)}{f(z)}dz =2\pi i(\textit{Winding Number of f around 0})$$

According to what I have read, it is necessary that $f$ does not vanish at the curve, which could be the case for my specific problem. I have no idea how to take this into account.

May I ask if someone could explain the steps for proving the lower bound on $Z_a+Z_b$ or propose an alternative solution that is more elementary? Thank you in advance!

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2 Answers 2

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Mohammed,

I thought about this a bit more together with a student of mine, and we were able to work it out. There is a way to generalize the argument principle in the case that the function has zeroes on the curve, and it is enough to handle this case. We thought this was interesting enough to make into a small paper, and if you are interested you can find it here:

https://arxiv.org/abs/2006.12545

As you can see, we thanked you in the paper for bringing the problem to our attention. By any chance, would you mind telling me now where the problem comes from?

Thanks, Greg

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Hmm, you're right, there is a bit of an issue with my previous proof. What I had in mind was the following. Using the notation from my previous answer, if $Q_a$ has zeroes on the unit circle, say $w_1, \ldots , w_r$, then we can write

$$ Q_a(z) = \Big(\prod_{j=1}^r (z-w_j) \Big)R(z), $$

where $R(z)$ is a polynomial of degree $n-r$ which does not vanish on the unit circle. Then

$$ Q_b(z) = z^n Q_a(1/z) = z^n \Big(\prod_{j=1}^r (1/z-w_j) \Big)R(1/z)= K \Big(\prod_{j=1}^r (1/w_j-z) \Big)z^{n-r} R(1/z), $$

where $K= \prod_{j=1}^r w_j$. $Q_b$ then has zeroes at $1/w_1, \ldots , 1/w_r$, which are all points on the unit circle as well. So you have the $r$ zeroes $w_1, \ldots , w_r$, and $r$ more $1/w_1, \ldots , 1/w_r$, and then you can apply the argument principle argument to $R(z)$, which now you know doesn't vanish on the circle, to get another $2(n-r)$, and when you add them up you get at least $2n$ zeroes.

But I had overlooked the following problem. What if $Q_a$ has repeated zeroes on the unit circle? A good example is $a_0=1, a_1 = -2, a_2 =1$, so that $Q_a(z) = (z-1)^2$. Since you wouldn't count the zero at $\theta = 0$ twice, my method only gives one zero there, and one more for $Q_b$, and then there is no winding once you factor it out, so you don't get the $4$ zeroes you require. However, your statement is still correct, since the image of the unit circle under $(z-1)^2$ does intersect the imaginary axis in other places, it just isn't covered by my winding argument.

So, I think my method works when there are no repeated zeroes on the unit circle, but doesn't seem to when there are. I suspect it can be adapted somehow, but don't immediately see it.

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  • $\begingroup$ Hmm, actually I think even the simple roots give a problem. I think my method only works when there are no roots on the unit circle. $\endgroup$
    – user387394
    Apr 16, 2020 at 23:33

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