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$0 < a_0 < a_1 < \cdots < a_n$.
Prove that $a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta$ has $2n$ different zeros, $\theta \in (0,2\pi)$.
[Hint: First prove that $P_n(z)=a_o+a_1z+a_2z^2+\cdots+a_nz^n$ has $n$ zeros in unit ball $B(0,1)$.]

This is an assignment I copied from my textbook. It's in the section "The Argument Principle & Rouche Theorem".
Though I followed this hint, I still can't see how this would imply the desired conclusion. Help needed.

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    $\begingroup$ The complex function part is treated here :(math.stackexchange.com/q/188039). I discovered there that it is named "Eneström–Kakeya theorem" $\endgroup$
    – Jean Marie
    May 5, 2017 at 20:08
  • $\begingroup$ @JeanMarie Thanks for your help! But I still can't figure out how this result relates to my original problem concerning cos. $\endgroup$
    – xixumei
    May 6, 2017 at 1:08
  • $\begingroup$ About the previous comment, yes, this hint refers to a special case of Eneström–Kakeya theorem, and you can prove it easily, but the expression $a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta$ is vague for me. Is there something missing here? $\endgroup$
    – Nosrati
    May 6, 2017 at 5:26
  • $\begingroup$ @MyGlasses I directly copied this problem from my textbook. $\endgroup$
    – xixumei
    May 6, 2017 at 5:28
  • $\begingroup$ Obviously this is the real part of $a_0+a_1z+\dots+a_nz^n$ when $|z=1|$. So can we relate the zeros of an holomorphic function on a bounded domain and the zeros of its real part on the boundary of the domain? $\endgroup$
    – Régis
    May 6, 2017 at 9:58

2 Answers 2

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Direct solution via intermediate value theorem

Multiply with the positive (on $(0,2\pi)$) factor $2\sin\fracθ2$ to get the equivalent equation \begin{multline} 0=g(θ)=(a_0-a_1)\sin\tfrac12θ+(a_1-a_2)\sin\tfrac32θ+…\\…+(a_{n-1}-a_n)\sin((n-\tfrac12)θ)+a_n\sin((n+\tfrac12)θ) \end{multline} At the extremal points of $\sin((n+\tfrac12)θ)$, which are $$θ_k=\frac{2k+1}{2n+1}\pi, ~~~ k=0,1,...,2n,$$ the value $g(θ_k)$ has the same sign $(-1)^k$ as its last term. This is because it dominates the sum of the other terms, as $$a_n>a_n-a_0=\sum_{k=0}^{n-1}|a_k-a_{k+1}|.$$ This is evidence for $2n$ sign alternations in the function value and thus at least $2n$ real roots inside the given interval $(0,2\pi)$.

Solution according to the hint

Per the hint, try to locate all the roots inside the unit circle. As the coefficient sequence is separated by inequalities, one can modify the coefficients slightly without destroying this defining property. Then multiplying with a linear factor with a root at $1$ gives $$ q(z)=(z-1)p(rz)=r^{n}a_nz^{n+1}+(r^{n-1}a_{n-1}-r^{n}a_n)z^n+...+(a_0-ra_1)z-a_0. $$ The roots of $q(z)$ are contained in a circle of radius $$ R=\max(1,r^{-n}|a_n|^{-1}(|r^{n-1}a_{n-1}-r^{n}a_n|+...+|a_0-ra_1|+|a_0|))=1. $$ This bound is valid as long as $ra_{k+1}\ge a_k$, $k=0,..,n-1$. There is some $r<1$ that satisfies this finite number of inequalities.

So if $z$ is a root of $p$, then $z/r$ is a root of $q$, thus $|z/r|\le 1$, $|z|<r$. All roots of $p(z)$ are well inside the unit circle.

The path $p(e^{iθ})$, $θ\in[0,2\pi)$ of the image of one rotation along the unit circle has winding number $n$ around zero. Which means it crosses the positive real half axis at least $n$ times and also the negative real half-axis at least $n$ times. These crossing points are also roots for $f(θ)=Re(p(e^{iθ}))$, the function under consideration. Thus $$ f(θ)=a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta $$ has at least $2n$ roots in $(0,2\pi)$. Note that $f(0)=a_n+...+a_1+a_0>0$.

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  • $\begingroup$ So you didn't use the given hint? $\endgroup$
    – xixumei
    May 7, 2017 at 23:34
  • $\begingroup$ No, but you can prove that hint via multiplication $q(z)=(z-1)p(z)=a_nz^{n+1}-(a_n-a_{n-1})z^n-…-(a_1-a_0)z-a_0$ which has an outer root radius of $1$ by the Lagrange bound. $\endgroup$ May 7, 2017 at 23:43
  • $\begingroup$ Yeah I now know how to prove the hint. Still having difficulty relating the hint to the original problem. $\endgroup$
    – xixumei
    May 7, 2017 at 23:50
  • $\begingroup$ The idea of the winding number might help, the image of the unit circle crosses the real axis twice per winding. $\endgroup$ May 8, 2017 at 6:28
  • $\begingroup$ How does your second argument demonstrate simple zeros? $\endgroup$ Mar 15, 2021 at 0:33
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The problem you have provided does not appear to be a simple application of Rouche's theorem or equivalent. I would be curious what the author's thoughts were for a proof. In particular, I believe the problem is equivalent to the following non-trivial theorem.

Theorem. If $0<a_0<a_1<\cdots<a_n$ and $p(z)=a_0+\cdots+a_nz^n$, then $$q(z):=z^{-n}p(z^{-1})+z^np(z)$$ has only simple zeros located on the unit circle.

The equivalency can be easily verified. Indeed, if we define $$c(z):=a_0+a_1\cos(z)+a_2\cos(2z)+⋯+a_n\cos(nz)\ \ \ ,$$ then the statement "$c(z)$ has $2n$ simple zeros in the interval $(0,2\pi)$" is equivalent to the statement that "$q(z)$ has $2n$ simple zeros on the unit circle". This is because $\cos(z)=(e^{iz}+e^{-iz})/2$, hence $$c(z)=e^{-inz}/2*q(e^{iz})\ \ \ .$$ Thus $c(z)$ has a zero in $(0,2\pi)$ if and only if $q(z)$ has a zero on the unit circle.

I would gladly welcome a simple proof of the above theorem with an application of Rouche's theorem or similar, but this appears to not be the case. In fact, if you generalize $p(z)$ to polynomials with zeros in the unit disc and throw in a rotation coefficient of $|a|=1$, the above theorem is precisely the complete characterization of all polynomials with zeros on the unit circle, something that was only established in 1995. See the first answer here To prove this complex polynomial has all zeros on unit circle, Chen (J. of Math. Anal. and Appl. vol 190, 714-724 (1995)).

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