# If coefficients are polynomial functions over $\mathbb R$ of a monic polynomial, can we find $n$ continuous functions that constitute the roots?

$$\newcommand{\a}{\alpha} \newcommand{\bb}{\mathbb} \newcommand{\b}{\beta}$$ 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)?

• The functions $\beta_i$ will not be polynomials in general. Consider for instance $p(x,t)=x^2+t$. – Federico Nov 27 '18 at 18:03
• Moreover, see math.stackexchange.com/questions/940653/… – Federico Nov 27 '18 at 18:05
• @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 $\mathbb R$. – user1101010 Nov 27 '18 at 18:26
• 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 – Federico Nov 27 '18 at 18:27
• 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 – Federico Nov 27 '18 at 18:30