# Why is a factorial always present in the denominator of the hypergeometric function?

The sum definition of the generalized hypergeometric function is:

$${_p}F_q (a_1, \dots, a_p;b_1, \dots,b_q;z)=\sum\limits_{n=0}^\infty \frac{(a_1)_k (a_2)_k\cdots (a_p)_k z^k}{(b_1)_k (b_2)_k\cdots (b_q)_k k!}$$

where $(a)_k=a(a+1)\cdots (a+k-1)=\frac{\Gamma(a+k)}{\Gamma(a)}$ is the Pochhammer symbol.

Using the current definition, if one does not want a factorial in the denominator, one must cancel it out by adding a factorial to the numerator.

If the factorial was not in the denominator by definition, it be easily added if needed by adding a $1$ as a parameter, ${_p}F_q (a_1, \dots, a_p;b_1, \dots,b_{q-1}, 1;z)$.

and simply left as it is if it is not needed.

My question is, why is $k!$ present in the denominator of the hypergeometric function?

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Well, one wants ${}_0 F_0({-}\mid z)$ to be $\exp\,z$, seeing that the hypergeometric functions are intended to be generalizations... –  Ｊ. Ｍ. Aug 10 '12 at 14:52
@J.M. But why? What's wrong with having it as ${_0}F_1 (;1;z)$? –  Argon Aug 10 '12 at 14:54
a conjecture : differentiation properties ! Euler provided following O.D.E. : $$x(1-x)\frac{d^2y}{dx^2}+(\gamma-(\alpha+\beta+1)x)\frac {dy}{dx}-\alpha\beta y=0$$ and remarked that term by term differentiation gave : $$\frac d{dx}{}_2F_1\left(\begin{array}{c} \alpha&& \beta\ \\ & \gamma\end{array};\ x\right)=\frac {\alpha\beta}{\gamma}{}_2F_1\left(\begin{array}{c} \alpha+1&& \beta+1\ \\ & \gamma+1\end{array};\ x\right)$$ (The nth derivative of the hypergeometric series gives too the nth coefficients). Earlier papers could teach more but Euler's work is huge and Gauss' hidden... –  Raymond Manzoni Aug 10 '12 at 16:02
You have just answered it yourself: Let ${_p}\tilde F_q (a_1, \dots, a_p; b_1, \dots, b_q; z)$ be the version without the factorial in the denominator, then ${_p}F_q (a_1, \dots, a_p; b_1, \dots, b_q; z) = {_p}\tilde F_{q+1} (a_1, \dots, a_p; b_1, \dots, b_q,1; z)$. So the used definition is merely convention. –  Alexander Thumm Aug 10 '12 at 16:05
"What's wrong with having..." - doesn't having the simplest case be unsymmetric not look right to you? –  Ｊ. Ｍ. Aug 11 '12 at 1:00