# Infinite series representation for root of polynomials?

Given a polynomial $p(x)=a_nx^n+\dots+a_1x+a_0$, can every root of the polynomial be represented as $\sum_{k=0}^\infty b_k$ with the $b_k$'s being a function of $a_0,\dots,a_n$ using only elementary operations of arithmetic and taking roots?

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What field are the $a_i$ over, and what coefficients are the $b_i$ allowed to have? – Qiaochu Yuan Jun 13 '11 at 14:58
I don't want to restrict that (if different fields yield different answers that's very interesting) but one can assume we talk about the field of rational numbers, for a concrete case. – Gadi A Jun 13 '11 at 15:51
Sorry if this is a dumb question, but: Would this not be equivalent of asking about finding a sequence $c_k$ (instead of a series) that converges to the root, each term being an 'elemental' function of $a_i$? IF so, couldn't we think $c_k$ as the values provided by some (complex) root-finding iterative algorithm (say, Durand–Kerner)? – leonbloy Jun 13 '11 at 16:28
At least for $p(x)=x^5+x+a$ this seems to be possible using Bring's radical. – Watson Mar 5 at 10:26

I think this is true at least formally if you allow the $b_i$ to have coefficients in $\overline{\mathbb{Q}}$. This is because, if $K$ is an algebraically closed field of characteristic $0$, then the field of Puiseux series with coefficients in $K$ is also algebraically closed, and by iterating this construction for each coefficient $a_i$ I think we get the desired result abstractly, although I am not sure what one can say about actual (as opposed to formal) convergence.

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According to en.wikipedia.org/wiki/Puiseux_series there is actual convergence in $\mathbb{C}$. – Akhil Mathew Jun 13 '11 at 16:14
@Akhil: this is true in sufficiently small neighborhoods, but I don't know if one can get global convergence even for quintic polynomials. See, for example, the explicit series roots given at qchu.wordpress.com/2010/10/08/… . – Qiaochu Yuan Jun 13 '11 at 16:16
Sorry, I only meant convergence near the origin. I see that is not quite what the OP wants though. – Akhil Mathew Jun 13 '11 at 16:27

The difficult part is to get a good a priori estimate $\Omega\subset{\mathbb C}$ of the set $S$ of roots. Starting with any $z_0\in\Omega$, e.g., with rational coordinates, Newton's rule $$z_{n+1}:=z_n-{p(z_n)\over p'(z_n)}\qquad(n\geq 0)\ ,$$ i.e., $$b_0=z_0, \qquad b_{n+1}:=-{p(z_n)\over p'(z_n)}\qquad(n\geq 0),$$ provides a series $\sum_{k\geq 0} b_k$ converging to a point $\zeta\in S$ where the $b_k$ depend rationally on the coefficients of $p$ (and the chosen point $z_0$).

There is a famous paper by Smale on this: "The fundamental theorem of algebra and complexity theory", Bulletin of the American Mathematical Society ${\bf 4}$ (1981), 1–36.

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In this approach it seems the problem is to find a radical function of the coefficients that makes for a good choice of $z_0$. No obvious candidates come to mind... – Qiaochu Yuan Jun 13 '11 at 18:54
– t.b. Jun 13 '11 at 18:55
which begs the question: is there some iterative algorithm for finding all the complex roots of a polynomial (including the choosing of the starting point/s) that is guaranteed to find all roots? (disregarding numerical errors) – leonbloy Jun 13 '11 at 21:19

Isn't this theorem related? As I understand the question, the OP asks wether it can be solvable by radicals, and Abel's theorem states that for polynomial equations of degree $n \geq 5$ there is no general solution.

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The OP asks about a series, this implies infinite terms - Abel's theorem has nothing to say about that. – leonbloy Jun 13 '11 at 16:33
Ok. Sorry then.. – Beni Bogosel Jun 13 '11 at 16:36
Hi Beni. The theorem is very interesting, in the sense that it is what motivates me to ask about infinite series... – Gadi A Jun 13 '11 at 16:59