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$\newcommand{\a}{\alpha} \newcommand{\bb}{\mathbb} \newcommand{\b}{\beta}$ Let \begin{align*} p(x, t) = x^n + \a_{n-1}(t) x^{n-1} + \dots + \a_1(t) x + \a_0 \end{align*} be a monic polynomial where coefficients $\{\a_0(t), \dots, \a_{n-1}(t)\}$ are real-valued continuous functions over $t \in \bb R$. In particular, each $\a_j(t)$ is polynomial in $t$ with real coefficients.

My question is: could we be able to find $n$ continuous complex-valued functions $\{\b_0(t), \dots, \b_{n-1}(t)\}$ over $t \in \mathbb R$ such that for each $t$, $\{\b_j(t)\}$ constitute the roots of the monic polynomial $x^n + \a_{n-1}(t) x^{n-1} + \dots + \a_1(t) x + \a_0$? I think the answer is positive since we are working over domain $\mathbb R$. If this is true, are these functions polynomials in $t$ (probably with complex coefficients)?

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    $\begingroup$ The functions $\beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$. $\endgroup$ – Federico Nov 27 '18 at 18:03
  • $\begingroup$ Moreover, see math.stackexchange.com/questions/940653/… $\endgroup$ – Federico Nov 27 '18 at 18:05
  • $\begingroup$ @Federico: Thanks. I see the functions could not be polynomial. But do they exist? The link was considering the domain $\mathbb C$ whereas here the domain of interest is $\mathbb R$. $\endgroup$ – user1101010 Nov 27 '18 at 18:26
  • $\begingroup$ The roots of $p(\,\cdot\,,t)$ are a continuous function $\mathbb R\to \mathbb C^n/\mathrm{perm}$, where the quotient is w.r.t. permutations. The fact that the domain is $\mathbb R$ implies that you can lift this to a continuous function $\beta:\mathbb R\to\mathbb R^n$ representing the roots $\endgroup$ – Federico Nov 27 '18 at 18:27
  • $\begingroup$ Basically, you have troubles only when multiple roots coincide. When they separate again (if they do), you just decide arbitrarily which $\beta_i$ tracks which root $\endgroup$ – Federico Nov 27 '18 at 18:30
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If all the coefficients of a polynomial are continuous functions over a real interval, then there is a selection of continuous functions that constitute the roots of the polynomial.

This is a result of Kato (as far as I know; it may be traced back to earlier publications.)

However, if the coefficients of a polynomial are continuous functions over a domain that contains the origin in the complex plane, then roots each as a continuous function do not exist in general. Note that roots are always continuous in the sense of topology. (F. Zhang)

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  • $\begingroup$ Unless you provide a reference, I don't think that this shoud be considered an answer. And what's the meaning of thae last sentence? $\endgroup$ – José Carlos Santos Jan 29 at 17:57

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