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$\newcommand{\a}{\alpha} \newcommand{\bb}{\mathbb} \newcommand{\b}{\beta}$ The question originates from a comment by reuns of this question Geometry of the set of coefficients such that monic polynomials have roots within unit disk.

Let $\a = (\a_0, \dots, \a_{n-1}) \in \bb R^n$ be fixed and $\a \neq 0$. We consider the parametrized family of monic polynomials with degree $n$ $$ f(r, z) = z^n + r\a_{n-1} z^{n-1} + \dots + r \a_1 z + r \a_0,$$ where $r \in \bb R$.

For each $r \in \bb R$, let $L(r)$ denote the largest modulus of roots of $f(r, z)$. $L(r)$ is not a polynomial in $r$ (Is the largest root of a parametrized family of polynomials again polynomial?). But $L(r)$ should be a continuous function over $r$. I am wondering whether the number of zeros of $L(r) - 1$ is finite.

Following was some wrong attempt. See the question and comments here If coefficients are polynomial functions over $\mathbb R$ of a monic polynomial, can we find $n$ continuous functions that constitute the roots?. I think the zeros should be finite. Informally, we can consider $$ f(r, z) = z^n + \a_{n-1}(r) z^{n-1} + \dots + \a_1(r) + \a_0(r).$$ This just rewrites $r \a_i$ as a function over $r$. Since we only consider $r \in \bb R$, there exists $n$ continuous functions $\b_0(r), \dots, \b_{n-1}(r)$, that for each $r \in \bb R$, constitutes the roots of $f(r, z)$. By Vieta's formula, all $\b_j(r)$ should be polynomial functions in $r$ (This part I feel is problematic). Then for each $j$, $\b_j(r) - 1$ would have at most $n$ zeros. The worst scenario would be these zeros happening in separate manner in the sense: the zeros of $\b_j(r) -1 $ are those $r$'s that $\b_j(r)$ also is the largest root of modulus. This gives us $n^2$ zeros of $L(r) -1 $. Is this argument correct? Could a tighter bound be provided?

Edit: Before I started the bounty, there is one answer by Akari Gale which I am not sure of . This is one of the reasons I started a bounty. As of now, there is one downvote of that answer. Could someone leave a comment on why the answer is not OK? Thanks.

user1101010
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If $L(r) = 1$ for some $r$, then $z=cos(x) + isin(x)$ is a root of $f(r, z)$ for some $x \in \mathbb{R}$. We can write $$0 = z^n + r\alpha_{n-1}z^{n-1} + ... + r\alpha_1z + r\alpha_0 \Rightarrow r =-\frac{-1}{\alpha_{n-1}z^{-1} + ... + \alpha_1z^{-n+1} + \alpha_0z^{-n}}$$ so $$\alpha_{n-1}z^{-1} + ... + \alpha_1z^{-n+1} + \alpha_0z^{-n} \in \mathbb{R} \Rightarrow $$ $$\alpha_{n-1}\sin(-x) + ... + \alpha_1 \sin((-n+1)x) + \alpha_0 \sin(-nx) = 0 \Rightarrow$$ $$\alpha_{n-1} \sin(x) + ... + \alpha_1 \sin((n-1)x) + \alpha_0 \sin(nx) = 0$$ It can be written in the form $P(\sin(x), \cos(x)) = 0$ for $P \in \mathbb{R}[X, Y]$ using multiple-angle formulas. Or $P(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}) = 0$

If this polynomial after multiplication by $(1 + t^2)^n$ is not zero, then for r there is only a finite number of possible values.

Akari Gale
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  • First, thanks for attempting the question. But the roots can be complex. – user1101010 Nov 27 '18 at 19:10
  • I edited the answer – Akari Gale Nov 27 '18 at 20:37
  • Could you elaborate how your write the equation in $y$ as a polynomial equation for $P \in \mathbb R[X, Y]$? Also I do not understand your last paragraph. Thanks. – user1101010 Nov 27 '18 at 20:40
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    Using this http://mathworld.wolfram.com/Multiple-AngleFormulas.html. Since r is expressed in terms of t, a finite number of roots means a finite number of possible values for r – Akari Gale Nov 27 '18 at 20:43
  • So the number of zeros is at most $2n$ since the degree is $2n$? – user1101010 Nov 27 '18 at 20:49
  • theoretically, the polynomial can be equal to zero, then the roots will be infinitely many. I don't know if this is possible – Akari Gale Nov 27 '18 at 20:57
  • +1 Looks good to me. Minor sign mistake in the first line. And you can actually remove the minus sign in the second, no need to rename $y=-x$. Lastly, you could use Taylor expansion near $0$ to prove $P=0$ can't happen. And thanks to have found a solution to something I was not able to prove in another question :) (as I think this question comes from a comment in this other thread) – Jean-Claude Arbaut Nov 29 '18 at 19:35
  • Yes. I didn't link the question because I already linked too many. Will edit it to mention it. – user1101010 Nov 29 '18 at 19:58
  • oh that's true. I admit sometimes such inaccuracies in the evenings. Thank you – Akari Gale Nov 29 '18 at 20:14
  • What you computed is a condition on $r$ for the polynomial to have one root of absolute value $1$, it need not be the largest. However, one of these $r$ will have $L(r)=1$ as there are are large roots for large $r$ and only small roots for small $r$, so $L(r)$ must pass through $1$ somewhere, but it does not answer if a non-monotonous $L(r)$ is possible. – Lutz Lehmann Dec 02 '18 at 09:59
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    To have a root of absolute value 1 is a necessary condition. This is the answer to the question about the finiteness of solutions of the equation. – Akari Gale Dec 02 '18 at 11:21