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I have a univariate polynomial of degree $n$ (where $n$ is larger than $4$). The real-valued coefficients of the polynomial depend on a parameter $\psi$, i.e. $$p_\psi(x)=a_n(\psi) x^n+a_{n-1}(\psi) x^{n-1}+\ldots+a_1(\psi)x+a_0(\psi).$$ Ideally, I would like to compute the roots of this polynomial, $r_i(\psi)$, $i\in\{1,\ldots,n\}$, and take the limit of $\psi$ to zero $${\hat r}_i=\lim_{\psi\rightarrow 0} r_i(\psi)$$ However, since the degree of the polynomial is too high, I cannot get a solution in closed form.

A feasible alternative would be the following: Compute the roots $r^\ast_j$ of the limit of the polynomial $$p^\ast(x)=\lim_{\psi\rightarrow 0} p_\psi(x).$$ In my case, the coefficients are such that a) the leading coefficient $a_n(\psi)$ will converge to zero and b) the limiting polynomial $p^\ast(x)$ will factor into low order polynomials.

My question is: under which conditions do the two calculations lead to the same answer, i.e. ${\hat r}_i=r^\ast_j$, for the roots that exist in both cases? Any answers or references would be much appreciated.

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On the assumption that the limits you state exist, each of the coefficients $a_k(\psi)$ is a continuous function of $\psi$ near $0$, and then the two ways to get the roots will necessarily yield the same result.

Edit: As the comments point out, the further condition that the leading coefficient $a_n(ψ)$ will converge to zero implies that at least one of the "limits" $\lim_{\psi\rightarrow 0} r_i(\psi)$ does not exist; so the above answer does not deal with the case in hand. However, by simply ignoring the instances where $|r_i(\psi)|\to\infty$, and assuming that the remaining limits exist, the remaining roots will be the same under each approach.

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    $\begingroup$ However, the fact that $a_n(\psi)\to 0$ implies that one of the roots of $a_n(\psi)$ tends to $\infty$ ("in order to" account for the fewer roots $p^*$ has) $\endgroup$
    – Hagen von Eitzen
    May 21, 2016 at 16:53
  • $\begingroup$ This answer presupposes that you include complex roots of $p_\psi$, even if you're only interested in real roots in the end. For example, let $p_\psi(x)=-\psi^2 x^4+x^2+\psi^2$. This has (for real $\psi$) two real roots, which go to $\pm\infty$ as $\psi\to0$, and it has two imaginary roots which tend to $0$ as $\psi\to0$. The limit of $p_\psi(x)$ as $\psi\to0$ is $x^2$, so its root, a double root at $0$, comes only from the complex roots of $p_\psi$. (The example also shows that in the comment by @HagenvonEitzen "one of the roots" means "at least one of the roots".) $\endgroup$
    – Andreas Blass
    May 21, 2016 at 17:09
  • $\begingroup$ Thank you very much for these excellent comments. They are very much appreciated. Does anybody have a proof or reference for these results? $\endgroup$
    – Tom M.
    May 23, 2016 at 19:46
  • $\begingroup$ @TomM.: A proof can be found in the answer to this question $\endgroup$
    – John Bentin
    May 23, 2016 at 21:17
  • $\begingroup$ @JohnBentin: Thanks a lot for the pointer! $\endgroup$
    – Tom M.
    May 24, 2016 at 20:35

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