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I have been attempting to produce a series expression for the roots of high degree polynomial using the Lagrange Inversion theorem. I am curious about the statement from the Wikipedia page on Bring Radicals.

The roots of the polynomial $x^5 + px +q$ can be expressed in terms of the Bring radical as $$\sqrt[4]{p}\,\operatorname{BR}\left(p^{-\frac{5}{4}}q\right)$$ where $$\operatorname{BR}(a) = -f^{-1}(a) = \sum_{k=0}^\infty \binom{5k}{k} \frac{(-1)^{k+1} a^{4k+1}}{4k+1}$$

I feel comfortable with the derivation of the Lagrange inversion theorem using residues; however, I do not see the justification for this formula of the roots. $$\sqrt[4]{p}\,\operatorname{BR}\left(p^{-\frac{5}{4}}q\right)$$ I also read that the roots must be numerically checked for issues regarding the indeterminacy of the sign due to complex conjugation. I am aware that the solution of a sextic equation in terms of Kampe de Feriet series requires a second free parameter.

Question: What would stop someone from directly factoring an $n^{\text{th}}$ order polynomial using the Lagrange inversion theorem? $$f(x) = \sum_{m=0}^{n} a_m x^m$$ $$g(x) = \sum_{m=1}^{n} a_m x^m$$ Why could one not simply find the roots of $f(x)$ by computing $g^{-1}(-a_0)$? Would the number of free parameters cause the number of possible sign combinations to blow-up exponentially? If so, I could write a script that would brute force them. Given that there is little to no literature on the subject, I guess that the point of failure is more subtle. Would the convergence of the series be so poorly conditioned that it would become useless for most intensive purposes?

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    $\begingroup$ The LIF provides the inverse about zero of $f(z) = z + tz^{n+1}$, and this generates essentially the Catalan sequence of numbers OEIS A000108 for $n=1$ and the (aerated) Fuss-Catalan sequences for $n>1$ (cf. oeis.org/A001764). $\endgroup$ Commented Aug 14, 2022 at 8:09

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