How to put a series in hypergeometric form

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I skipped a step in the previous post, I did not show how a series was inserted into the form of a hypergeometric function.

I actually skipped two steps. First of all I said that a series was not clearly hypergeometric, and yet at first glance it is.

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

Rising forces

Hypergeometric functions are defined by series whose coefficients are ascending intensities.

God NThe rising power of A he

(A)N = A (A+1) (A + 2)… (A + N – 1).

More on forces rising here. The coefficient of XN In a series defining a hypergeometric function is a ratio of NThe rising forces of the regulars.

Why the example is not clear

You can write factorial as ascending bones: N! = (1)N. The binomial coefficient in our example is the ratio of immigrant forces:

(5N)! = (1)5N

In the counter and

N! (4N)! = (1)N (1)4N

In the denominator. So why did we not finish? Because although all of these are ascending forces, they are not all NThe rising forces. Subscribers to (A)N Everyone is different: 5N, 4N, And N.

So is our function not hypergeometric? It is yes, but we will have to introduce additional parameters.

Rational polynomials

Another way to define hypergeometric series is to say that the ratio of successive coefficients is a rational function of the index N. It is very concise, but not as explicit as the definition in terms of rising powers, although they are equivalent.

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

This is how you put a series in a hypergeometric way, if it has such a shape. This is also how you test whether a series is hypergeometric: a series is hypergeometric if and only if the ratio of successive members is one rational function of the index.

Back to our example

The ratio of (N+1) The first term for NThe 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 parameters of the hypergeometric function directly from this rational function. The top 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 that we can remove the 1 from the list of upper parameters and remove one of the upper parameters. The 1st from the list of lower parameters [1].

So our sample 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, and also the order of the lower parameters, do not matter, although it is very important which ones up and which ones down. The subscription to the left of and Gives the number of upper parameters and the subtitle on the right gives the number of lower parameters. These subscriptions are redundant because you can simply count the number of parameters between semicolons, but they make it easier to know at a glance what kind of hypergeometric function you are working with.

Sample test

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


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


        1/5, 2/5, 3/5, 4/5, 
        1/4, 2/4, 3/4, 1,

Return the same value, i.e. 1.05233.

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[1] The additional cause of N! In the denominator of hypergeometric functions complicates things. It seems to me like a historical object unfortunately, but maybe it’s a net benefit for reasons I can not think of. When we look at zeros of the numerator and denominator, we have 1 single up and three Those at the bottom. The 1 above eliminates one of the 1 at the bottom, but why do we not have two 1s in the lower parameters? Because one of them is suitable for addition N! In the denominator.




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