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

For each $r \in \bb R$, let $L(r)$ denote the largest root. My question is whether $L(r)$ is a polynomial in $r$. By Vieta's formula, for each $r$, we can represent all the roots as a polynomial in $r$. I am not sure if some crossing happened in the roots, this is still true.

user1101010
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1 Answers1

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Consider $n = 2,$ and the case $f(r, z) = z^2 - rz = z(z-r)$, i.e., $a_1 = -1, a_0 = 0$.

The largest root is $r$ (for $r \ge 0$), but $0$ for $r < 0$. The function

$$ h(r) = \begin{cases} r & r \ge 0 \\ 0 & r < 0 \end{cases} $$ is not differentiable at zero, hence not a polynomial.

John Hughes
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  • Thanks for your nice answer. May I ask another question: I am looking for the zeros of $L(r) - 1$ and your example shows it is not a polynomial in $t$. But I am still hoping the number of the zeros is bounded. Could you comment on this part? Thanks. – user1101010 Nov 26 '18 at 18:36