How to put a series in hypergeometric form

I skipped a step in the previous post, not showing how a series was put into the form of a hypergeometric function.

Actually I skipped two steps. I first said that a series was not obviously hypergeometric, and yet at first glance it sorta is.

I’d like to make up for both of these omissions, but with a simpler example. It looks like a hypergeometric series, and it is, but it’s not the one you might think. The example will be the function below.

f(x)  = \sum_{n=0}^\infty \binom{5n}{n} \frac{x^n}{n!}

Rising powers

Hypergeometric functions are defined by series whose coefficients are rising powers.

The nth rising power of a is

(a)n = a (a+1) (a + 2) … (a + n – 1).

More on rising powers here. The coefficient of xn in a series defining a hypergeometric function is a ratio of nth rising powers of constants.

Why the example isn’t obvious

You can write factorials as rising powers: n! = (1)n. The binomial coefficient in our example is a ratio of rising powers:

(5n)! = (1)5n

in the numerator and

n! (4n)! = (1)n (1)4n

in the denominator. So why aren’t we done? Because although these are all rising powers, they’re not all nth rising powers. The subscripts on the (a)n are all different: 5n, 4n, and n.

So is our function not hypergeometric? It is, but we’ll have to introduce more parameters.

Rational polynomials

Another way to define hypergeometric series is to say that the ratio of consecutive coefficients is a rational function of the index n. This is very succinct, but not as explicit as the definition in terms of rising powers, though they’re equivalent.

In addition to brevity, this definition has another advantage: the hypergeometric parameters are the negatives of the zeros and poles of said rational function. The zeros are the upper parameters and the poles are the lower parameters.

This is how you put a series in hypergeometric form, if it has such a form. It’s also how you test whether a series is hypergeometric: a series is hypergeometric if and only if the ratio of consecutive terms is a single rational function of the index.

Back to our example

The ratio of the (n+1)st term to the nth term in our series is

\begin{align*} \frac{t_{n+1}}{t_n} &= \left.\binom{5(n+1)}{n+1} \frac{z^{n+1}}{(n+1)!} \middle/{\binom{5n}{n}} \frac{z^n}{n!} \right. \\ &= \frac{(5n+5)!}{(n+1)! \,(4n+4)!} \,\frac{n!\,(4n)!}{(5n)!} \, \frac{n!}{(n+1)!}\frac{5^5}{4^4} z\\ &= \frac{(5n+5)(5n+4)(5n+3)(5n+2)(5n+1)}{(4n+4)(4n+3)(4n+2)(4n+1)(n+1)^2} \, \frac{5^5}{4^4} z \end{align*}

and the final expression is a rational function of n. We can read the hypergeometric function parameters directly from this rational function. The upper parameters are 1, 4/5, 3/5, 2/5, and 1/5. The lower parameters are 1, 3/4, 2/4, 1/4, and 1. Identical factors in the upper and lower parameters cancel each other out, so we can remove the 1 from the list of upper parameters and remove one of the 1’s from the list of lower parameters [1].

So our example function is

{}_4 F_4\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5};\,\, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1; \,\,\frac{3125}{256}x \right)

The order of the upper parameters doesn’t matter, and neither does the order of the lower parameters, though it matters very much which ones are upstairs and which are downstairs. The subscript to the left of the F gives the number of upper parameters and the subscript to the right gives the number of lower parameters. These subscripts are redundant since you could just count the number of parameters between the semicolons, but they make it easier to tell at a glance what class of hypergeometric function you’re working with.

Spot check

Let’s evaluate our original series and our hypergeometric series at a point to make sure they agree. I chose x = 0.01 to stay within the radius of convergence of the hypergeometric series.

Both

    Sum[Binomial[5 n, n] (0.01^n)/n!, {n, 0, Infinity}]

and

    HypergeometricPFQ[
        {1/5, 2/5, 3/5, 4/5}, 
        {1/4, 2/4, 3/4, 1},
        0.01*3125/256]

return the same value, namely 1.05233.

Related posts

[1] The extra factor of n! in the denominator of hypergeometric functions complicates things. It seems like an unfortunately historical artifact to me, but maybe it’s a net benefit for reasons I’m not thinking of. When we look at zeros of the numerator and denominator, we have a single 1 on top and three ones on bottom. The 1 on top cancels out one of the 1s on bottom, but why don’t we have two 1s in the lower parameters? Because one of them corresponds to the extra n! in the denominator.

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