# closed-form expression for roots of a polynomial

It is often said colloquially that the roots of a general polynomial of degree $5$ or higher have "no closed-form formula," but the Abel-Ruffini theorem only proves nonexistence of algebraic closed-form formulas. And I remember reading somewhere that the roots of quintic equations can be expressed in terms of the hypergeometric function.

What is known, beyond Abel-Ruffini, about closed-form formulas for roots of polynomials? Does there exist a formula if we allow the use of additional special functions?

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–  Robert Israel Feb 1 '13 at 4:46
Transcendent functions do the job sometimes. For example, the roots of $X^n - 1$ are given by $e^{2 \pi k / n}$. –  Hans Giebenrath Feb 1 '13 at 6:50
@HansGiebenrath, roots of $X^n -1$ count as obtained by extraction of radicals. –  Andreas Caranti Feb 1 '13 at 7:22
Indeed the Galois group of $X^n-1$ is not only solvable, but also abelian. –  JSchlather Feb 1 '13 at 7:26
I'm sorry. You are right. I was aiming at CM-fields and extensions thereof generated by values of $j$. I think it should be possible to construct a non-solvable extension generated by the value of $j$. –  Hans Giebenrath Feb 1 '13 at 9:32

As you might already know, solutions to the quintic can be expressed in terms of either ${}_4 F_3$ hypergeometric functions or Jacobi theta functions. See King or Prasolov/Solovyev for details.
For polynomials of higher degree, there is also a general formula for the roots, due to Umemura. The formulae involve the multidimensional generalization of the Jacobi theta functions (the Riemann theta function), and are a bit unwieldy; see Umemura's paper if you want more details. See also this preprint for a solution of the reduced polynomial equation $x^n-x-\alpha=0$ in terms of hypergeometric functions.
Doesn't the lagrange inversion formula give you a 'formula' for the roots? (as an expansion from some point $a$ nearby)