# Quintic trinomial root

This post looks at an exercise from Special Functions, exercise 6 in Appendix E.

Suppose that xm+1 + axb = 0. Show that Use this formula to find a solution to x5 + 4x + 2 = 0 to four decimal places of accuracy. When m = 0 this series reduces to the geometric series. Write this sum as a hypergeometric series.

• It’s a possibly useful result.
• It’s an application of Lagrange’s inversion formula; that’s where the series comes from.
• The equation has m+1 roots. What’s special about the root given by the series?
• The sum is not obviously hypergeometric.
• What happens when the sum diverges? Is it a useful asymptotic series? Does the analytic continuation work?

I want to address the last two points in this post. Maybe I’ll go back and address the others in the future. To simplify things a little, I will assume m = 4.

The parameters of the hypergeometric series are not apparent from the expression above, nor is it even apparent how many parameters you need. It turns out you need m upper parameters and m-1 lower parameters. Since we’re interested in m = 4, that means we have a hypergeometric function of the form 4F3. We can evaluate this expression in Mathematica as

    f[a_, b_] := (b/a) HypergeometricPFQ[
{1/5, 2/5, 3/5, 4/5},
{1/2, 3/4, 5/4},
-3125 b^4 / (256 a^5)
]


When we evaluate f[4, -2] we get -0.492739, which is the only real root of

x5 + 4x + 2 = 0.

Recall that the sign on b is negative, so we call our function with b = -2.

Now lets, try another example, solving for a root of

x5 + 3x – 4 = 0.

If we plug a = 3 and b = 4 into the series at the top of the post, the series diverges immediately, not giving a useful asymptotic series before it diverges.

The series defining our hypergeometric function diverges for |z| > 1, and we’re evaluating it at z = -3125/243. However, the function can be extended beyond the unit disk by analytic continuation, and when we ask Mathematica to numerically evaluate the root with by running

    N{ f[3, 4] ]

we get 1, which is clearly a root of x5 + 3x – 4.