# Is there a gamma-like function for the q-factorial?

I'm looking at quantum calculus and just trying to understand what is going with this subject. Looking at the q-factorial made me wonder if this function could take all real or even complex numbers in the same way that $\Gamma (z)$ works as an extension of $f(n) =n!$. Since, I need practice with both $\Gamma$ and q-analogs, would it be a good project to try to recreate $\Gamma (z)$ in this new setting or is the whole project lacking in sanity, mathematical soundness?

Also, a minor question, why is there sometimes a coefficient in q-analog expansions as in this expression:

$(a;q)_n = \prod_{k=0}^{n-1} (1-aq^k)=(1-a)(1-aq)(1-aq^2)\cdots(1-aq^{n-1}).$

I'm a bit embarrassed not to know, but non of the lit. I have explains it, I'll just randomly see it tossed in there from time to time... and it really throws me off.

Thank you.

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A Google search for "q-Gamma function" reveals mathworld.wolfram.com/q-GammaFunction.html . –  Qiaochu Yuan Mar 30 '11 at 17:41
I have no idea why I could not find that. Thanks. –  a little don Mar 30 '11 at 18:08
I think I'll try to develop it on my own now that I know it exists. ;) –  a little don Mar 30 '11 at 18:08
That coefficient is the q-Pochhammer symbol: mathworld.wolfram.com/q-PochhammerSymbol.html –  deoxygerbe Mar 30 '11 at 18:15
Also google Bhargava's generalization of the gamma function: this generalization includes both the usual gamma function and the q-gamma function as special cases. Google "Bhargava gamma" and the first couple of links will be to his very readable and beautiful paper. –  Barry Smith Mar 30 '11 at 19:25

q-Gamma function $\Gamma_q$ generalized Euler $\Gamma$ function by replacing the recurrence identity with it's q-deformation: $$\Gamma(x+1) = x \Gamma(x) \quad \Longrightarrow \quad \Gamma_q(x+1) = [x]_q \Gamma_q(x)$$ where $[x]_q = \frac{1-q^x}{1-q}$ is a q-number.
When $x$ is a positive integer, and assuming $\Gamma_q(1) = 1$, we get $$\Gamma_q(n+1) = \prod_{k=1}^n \frac{1-q^k}{1-q} = \frac{(q,q)_n}{(1-q)^n} = \left\{ \begin{array}{cc} \frac{(q,q)_\infty}{(1-q)^n (q,q^n)_\infty} & |q|<1 \\ \frac{ q^{n(n-1)/2} (q^{-1},q^{-1})_\infty}{(1-q^{-1})^n (q^{-1},q^{-n})_\infty} & |q|>1 \end{array} \right.$$ The last expression is now defined for $n \in \mathbb{C}$, as long as $n$ is not a non-positive integer, as $(q,q^{-k})_\infty = 0$ for $k \in \mathbb{Z}^+$.