# Why do nth roots (radicals) have closed forms whilst other polynomial roots do not?

The nth root function - $\sqrt[n]{a_{0}}$ - may be seen as an arithmetic operation (the inverse of the $pow(x, n)$ function) but it can also be interpreted as computing the roots of a specific class of polynomial - polynomials of the form $x^n - a_{0} = 0$ (which are cyclotomic if n is prime). Galois theory is a direct result of the fact that we only allow such roots (or "radicals") to have a closed form notation, whereas general polynomial roots are given no such notation and are thus, in general, "unsolvable by radicals" if the polynomial is of quintic degree or higher. Why is this?

As far as I am aware, such nth roots are not any easier or faster to calculate numerically than the roots of a general polynomial. The same methods are applied to both problems which involve the computation of irrational numbers. So what is special about nth roots as compared with the roots of other polynomials? Is there a logical reason that only such "pure" nth-root polynomial roots have a closed form or is just an accident of history?

In particular, I am considering a simple generalization of the nth root function from a binary function to a k+1-ary function that takes the last k coefficients of a polynomial as parameters, not just the very last coefficient. It would be an interesting question to see the resulting Galois theory for different values of k other than k = 1.

Note: The Bring radical or ultraradical is similar to this idea and can be used to solve otherwise unsolvable quintic equations, but is specific to quintic polynomials.

• $n$th powers have one big numerical advantage over a random polynomial root: They can be calculated using the binomial series. – Semiclassical Aug 13 '14 at 4:29
• Wow, really? I thought Newton's method and its refinements were the most efficient way to compute nth roots. Thanks for the info! – calculemur Aug 13 '14 at 4:48
• Another advantage of $n$th powers is that they're bijective over the non-negative reals, which is what makes their inverses well-defined. In order to define inverse functions for more general polynomials, you'd at least have to first solve the non-trivial problem of where the inverse is allowed to go, – David H Aug 13 '14 at 4:49
• I'm sorry, I should have been explicit that I was talking about polynomial roots over the complex numbers and not the real numbers. The proposed generalization of the nth root function has a bijective inverse defined as the computation of k symmetric polynomials in the roots of the equation. Every unique set of coefficients defines a new set of roots and vice versa and thus the inverse is well defined. – calculemur Aug 13 '14 at 5:03
• Re your last comment: Not quite. Every ordered set of coefficients gives rise to an unordered set of roots. Or in other words, any permutation of the roots leaves the sequence of coefficients of the polynomial untouched. The relation is $n!$ to one. – Jyrki Lahtonen Aug 13 '14 at 5:24