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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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    At least for $p(x)=x^5+x+a$ this seems to be possible using [Bring's radical](https://en.wikipedia.org/wiki/Bring_radical#Series_representation).2016-03-05

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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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    Hi Beni. The theorem is very interesting, in the sense that it is what motivates me to ask about infinite series...2011-06-13
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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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    wh$i$ch begs the quest$i$on: 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)2011-06-13
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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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    Sorry, I only meant convergence near the origin. I see that is not quite what the OP wants though.2011-06-13