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

Suppose that

x^{m+1}+ax−b= 0. Show thatUse this formula to find a solution to

x^{5}+ 4x+ 2 = 0 to four decimal places of accuracy. Whenm= 0 this series reduces to the geometric series. Write this sum as a hypergeometric series.

There are several things I find interesting about this exercise.

- 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 _{4}*F*_{3}.

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

*x*^{5} + 4*x* + 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

*x*^{5} + 3*x* − 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 *x*^{5} + 3*x* − 4.