## Generalized hypergeometric function

### Function

 definition $??$ definition ${}_pF_q[a_1,…,a_p; b_1,…,b_q](z):=\sum_{n=0}^\infty c_n z^n$ with $c_n = \dfrac{1}{n!}\dfrac{\prod_{k=1}^p a_k^{\overline{n}}}{\prod_{j=1}^q b_j^{\overline{n}}}$

#### Discussion

##### Definition

The coefficient can more explicitly written as $c_n = \prod_{m=0}^{n-1}\dfrac{1}{(1+m)}\dfrac{\prod_{k=1}^p(a_k+m)}{\prod_{j=1}^q(b_j+m)}$

or written down in Terms of Gamma functions. The version I chose above seems most elementary to me.

Above we used the rising factorial $x^{\overline{n}} := x^{\overline{n},1}$, where

$x^{\overline{n},k} := \prod_{m=0}^{n-1}(x + m\,{k})=x\cdot(x+k)\cdot(x+2k)\cdots(x+(n-1)\,k)$.

Note that, really, $\dfrac{z^n}{n!}=\dfrac{z^{\overline{n},0}}{1^{\overline{n},1}}$, so ${}_pF_q$ is the infinite sum of fractions of products of rising factorials of fixed rising distance ($k=1$ for the factors of the coefficients and $k=0$ for $z$).

##### Generalization

The function is a special case of Normalized Fox-Wright function. There, the distored factorial $\prod_{m=0}^{n-1}(a_k+m)$, a product up to $n-1$, is replaced by a “product” up to some more general number. This Expression is given in terms of a fraction of Gamma functions.

##### Example

For a given index $I$, setting $a_I=1$ results in

$\prod_{m=0}^{n-1} (a_I+m) = \prod_{m=0}^{n-1} (m+1) = n!$

and thus we can switch from an exponential generating function form $\dfrac{z^n}{n!}$ to $z^n$. So let's consider $a_2=1$.

Now further, for $a_1=1$ and $b_1=2$ we get a factor

$\dfrac{\prod_{m=0}^{n-1} (a_1+m)}{\prod_{m=0}^{n-1} (b_1+m)} = \dfrac{n!}{(n+1)!} = \dfrac{1}{n+1}$

so that

${}_2F_1[1,1;2](z) = \sum_{n=0}^\infty \dfrac{1}{n+1} z^n = \dfrac{1}{(-z)}\sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}(-z)^k = -\dfrac{1}{z}\log(1-z)$

##### Motivation

Say you have to solve the differential equation $f’(x)=f(x)$ with $f(0)=1$. You naturally make the ansatz

$f(x) = 1 + c_1\, x + c_2\, x^2 + c_3\, x^3 + \dots$,

$f’(x) = c_1 + 2\,c_2\, x + 3\,c_3\, x^2 + \dots$.

Comparing coefficients, this implies that the solution to $f’(x)=f(x)$ must have, for example $3\,c_3=c_2$. In fact all coefficients are determined this way, by the recursive relation

$\dfrac{c_{n+1}}{c_n} = \dfrac{1}{n+1}$

With the polynomial $q(n) := n+1$, this means $c_n = \frac{1}{\prod_{k=0}^{n-1} q(k)} c_0 = \dfrac{1}{n!}$ and hence $f(x) = \sum_{n=0}^\infty \dfrac{1}{n!} x^n$.

Such an approach to solve a differential equation will often look like this. A whole lot of function have series coefficients $c_n$, such that

$\dfrac{c_{n+1}}{c_n} = \dfrac{p(n)}{q(n)}$

where p and q are some polynomials. Any (arbitrary product of) polynomials of an integer n can be written as a product of terms $(a_i-n)$. So define the generalized hypergeometric function

${}_pF_q[a_1,…,a_p; b_1,…,b_q](z) :=1 + \dfrac{a_1\dots a_p}{b_1\dots b_q}\dfrac{z}{1!} + \dfrac{a_1(a_1+1)\dots a_p(a_p+1)}{b_1(b_1+1)\dots b_q(b_q+1)}\dfrac{z^2}{2!}+\dots$

to catch them all!

They are the solutions to differential equations with recursive character.

##### Differential equation

The function solves the following quite general differential equation of oder which is of order $\mathrm{(p,q+1)}$:

$\dfrac{{\rm{d}}}{{\rm{d}}z}D_bf(z) = D_af(z)$

with

$D_b := \prod_{n=1}^{q}\left(z\dfrac{{\rm{d}}}{{\rm{d}}z} + b_n-1\right)$,

$D_a := \prod_{n=1}^{p}\left(z\dfrac{{\rm{d}}}{{\rm{d}}z} + a_n\right)$.