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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.

Especially Lime
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xixumei
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    The complex function part is treated here :(https://math.stackexchange.com/q/188039). I discovered there that it is named "Eneström–Kakeya theorem" – Jean Marie May 05 '17 at 20:08
  • @JeanMarie Thanks for your help! But I still can't figure out how this result relates to my original problem concerning cos. – xixumei May 06 '17 at 01:08
  • 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? – Nosrati May 06 '17 at 05:26
  • @MyGlasses I directly copied this problem from my textbook. – xixumei May 06 '17 at 05:28
  • 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? – Régis May 06 '17 at 09:58
  • @Régis Yeah. But I can't see how this observation leads to the conclusion that the trigonometric polynomial has 2n different zeros. – xixumei May 06 '17 at 10:54
  • Well if for instance it holds that $f$ has $n$ zeros on $\Omega$ implies $Re(f)$ has $2n$ zeros on $\partial \Omega$ we would be done. I don't know if this is true, but doesn't look quite unreasonable. – Régis May 06 '17 at 11:26

2 Answers2

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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, show that all the roots of $$p(z)=a_0+a_1z+...+a_nz^n$$ are inside the unit circle. As the elements of the coefficient sequence are 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 $f(0)=a_n+...+a_1+a_0>0$.

Lutz Lehmann
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  • So you didn't use the given hint? – xixumei May 07 '17 at 23:34
  • 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. – Lutz Lehmann May 07 '17 at 23:43
  • Yeah I now know how to prove the hint. Still having difficulty relating the hint to the original problem. – xixumei May 07 '17 at 23:50
  • The idea of the winding number might help, the image of the unit circle crosses the real axis twice per winding. – Lutz Lehmann May 08 '17 at 06:28
  • How does your second argument demonstrate simple zeros? – Bobby Ocean Mar 15 '21 at 00:33
  • @BobbyOcean : t does not. That the winding number is $n$ means that there are at least $n$ crossings in positive orientation on each the positive and negative half-axis. The resulting $2n$ roots can have multiplicities greater than $1$, I'm not sure that one can control that with a simple argument. – Lutz Lehmann Mar 15 '21 at 00:45
  • Oh, maybe I misread, I thought the statement was to prove that the zeros were different from each other and I didn't see how your argument achieved that. – Bobby Ocean Mar 15 '21 at 00:46
  • There are $n$ rotations around the origin. Each of them has a different segment of $[0,2\pi)$ as pre-image. Each segment contains 2 roots of the real part, one with positive and one with negative imaginary part. Thus $2n$ distinct roots. – Lutz Lehmann Mar 15 '21 at 00:48
  • I am sorry, I am not following, how you conclude "Thus 2n distinct roots". You don't know their distinct. They could have multiplicity higher than 1. I don't see how the winding number or argument theorem alone give you simple zeros. The OP question seems to indicate to show that there are "2n different zeros". Am I misunderstanding? – Bobby Ocean Mar 15 '21 at 00:52
  • $p$ maps the unit circle to a path that winds $n$ times around the origin, without crossing the origin. Despite all possible back-and-forth, in each circumnavigation of the origin the path has to cross each coordinate half-axis at least once in the order of a counter-clockwise rotations. Each such intersection is a root of the real or imaginary part. – Lutz Lehmann Mar 15 '21 at 00:57
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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)).

Bobby Ocean
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