Series solutions of algebraic equations

We first illustrate the idea with the quadratic and then go on to the quintic.

Quadratic

Consider the generic quadratic polynomial q[x] = a x^2 + b x + c. Let r[y] be a branch of the inverse function of q in a small neighborhood about y = 0, meaning that q[r[y]] = 0 for any sufficient small y. Then r[0] is clearly a root of q. In order to find r[0] as a function of a, b and c, we will compute the formal power series of r[y] in y at zero; for this it suffices to invert q[x] as a formal power series:

s = InverseSeries[Series[a x^2 + b x + c, {x, 0, 9}], y]

This gives nine terms of the series, which we evaluate at zero to get r[0]:

Normal[s] /. y -> 0

The general term appears to be:

tau = -c^(k + 1) / b^(2k + 1) a^k Binomial[2 k, k]/(k + 1);

We can sum this series to get a closed-form solution for one of the roots of the original quadratic equation:

<<Algebra`SymbolicSum`

SymbolicSum[tau, {k, 0, Infinity}]

Quintic

Now we show how this method works for a quintic of the form:

t^5 - t - rho

Proceeding as above, we compute:

s = InverseSeries[Series[t^5 - t - rho, {t, 0, 29}], y];
Normal[s] /. y -> 0

In version 2.3 this gives the first eight terms in the series for one of the roots. The general term of this series can be written as:

tau = - rho^(4k + 1) Binomial[5k, k] / (4k + 1);

This series can be summed in hypergeometric functions:

SymbolicSum[tau, {k, 0, Infinity}]

It is possible to get the other four roots by building a Riemann surface for this function. In principle this method will work for a polynomial of any degree.