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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)?

user1101010
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    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 https://math.stackexchange.com/questions/940653/writing-the-roots-of-a-polynomial-with-varying-coefficients-as-continuous-functi?rq=1 – 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
  • @Federico: Thanks. By $\pi_1(\mathbb R)$, you mean the fundamental group? I need to check my topology book for the property you are using. Do you have a name for this property? – user1101010 Nov 27 '18 at 18:32
  • @Federico: Can we conclude that these $\beta_j(t)$'s are analytic? – user1101010 Nov 27 '18 at 18:36
  • No: the example $x^2+t$ has roots $\pm\sqrt{-t}$ for $t\leq0$ and $\pm i\sqrt{t}$ for $t\geq0$. They are not even differentiable at $t=0$ – Federico Nov 27 '18 at 18:37
  • maybe they have some regularity when they are distinct. you can try to find some information maybe in https://www.jstor.org/stable/2689304?seq=1#page_scan_tab_contents. i havent read it – Federico Nov 27 '18 at 18:39
  • Theorem 3.5 of http://users.ices.utexas.edu/~alen/articles/polyroots.pdf just gives the continuity of simple root, without any regularity – Federico Nov 27 '18 at 18:41
  • Ok, i've found this that seems to treat your case: https://arxiv.org/pdf/math/0611633.pdf – Federico Nov 27 '18 at 18:43
  • see also https://arxiv.org/pdf/1309.2151.pdf – Federico Nov 27 '18 at 18:44
  • @Federico: Thanks so much. – user1101010 Nov 27 '18 at 18:44
  • they show absolute continuity of the roots, which is not really much regularity... – Federico Nov 27 '18 at 18:45
  • Ok, last link: here are slides summarizing some known results: https://www.uv.es/wfav2013/wfav2013_archivos/index_archivos/charlas/Rainer_WFAV2013.pdf – Federico Nov 27 '18 at 18:49
  • @Federico: Thanks again for your comments and the files linked are very helpful to me. This question was to clarify something I was attempting to work over anther question here https://math.stackexchange.com/questions/3015124/zeros-of-largest-root-of-a-parametrized-family-of-polynomials. Would you take a look at? Maybe it is much simpler than I thought. – user1101010 Nov 27 '18 at 18:49
  • The roots $f_j(z)$ of a polynomial with coefficients in $\mathbb{C}(z)$ are locally analytic except at finitely many points, those being poles and branch points. To show it you can search for the $z$ where $f_j'(z)$ is well-defined, implying that $f_j$ is holomorphic thus analytic (see in $f(z)^2 = z$ how "locally analytic" cannot be strengthened) – reuns Nov 27 '18 at 19:18
  • @reuns: Thanks. But how do we know there are locally analytic "except at only finitely many points"? – user1101010 Nov 27 '18 at 19:40
  • They gave many hints above. The map from the roots to the coefficients is $C^1$. To invert locally it suffices to look at the linearization (partial derivatives). Now for coefficients that are analytic functions, it suffices to show the inverse is locally holomorphic, which can be done by formally differentiating the equation that $f$ satisfies to obtain an equation for $f'$. – reuns Nov 27 '18 at 19:46
  • Let $\mathbb{C}(z)$ the field of rational functions and $P_z(y ) =\sum_{n=0}^N a_n(z) y^n \in \mathbb{C}(z)[y]$ irreducible. Pick a root $ P_z(f(z)) = 0$ so $f(z)$ is in an algebraic extension of $\mathbb{C}(z)$. Differentiating $f'(z)=\frac{-\sum_{n=0}^N a_n'(z) f(z)^n}{P_z'(f(z))}$. Then $\gcd(P_z,P_z')=1$ so $P_z(y)u_z(y)+P_z'(y)v_z(y)=1$ and $P_c'(f(c)) = 0$ implies $P_c(f(c))\frac{u_c(f(c))}{v_c(f(c))} \ne 0 $ and since $P_c(f(c)) = 0$ it means $c$ is a pole of $\frac{u_c(f(c))}{v_c(f(c))}$. But they have only finitely many poles. – reuns Nov 27 '18 at 21:46
  • @Federico Why not an official answer? – Paul Frost Nov 27 '18 at 23:31

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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)

Fuzhen Zhang
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