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The q-shifted factorial is defined as $(a;q)_n := (1-a)(1-aq)\ldots(1-aq^{n-1})$. It is supposed to be an analog of the Pochhammer symbol, or falling factorial: $x(x-1)\ldots(x-n+1)$. But the formulas do not seem to agree in format! (Moreover, whoever writes about this topic usually assumes unconsciously that the reader already is familiar with the notation, thereby turning a simple topic into a mess of formulas)

I think I have a correct interpretation for how these two are analogous, replacing $a$ with $q$, but am not clear as to the details. Since $(1-q^r) = [r]_q (1-q)$, aren't a bunch of factors $1/(1-q)$ hidden? And what about the $x$?

Please clarify the correct meaning of this analogy.

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I can imagine this is somewhat confusing in the beginning. Take $a = q^x$, then $$ \lim_{q \to 1} \frac{(1 - q^x)}{(1 - q)} = x, $$ using l'hopital. Therefore we can connect the $q$-shifted factorial to the raising (!) factorial $$ \lim_{q \to 1} \frac{(q^x; q)_n}{(1 - q)^n} = x(x + 1)(x + 2) \ldots (x + n - 1). $$ One of the main usages for this connection is for the basic hypergeometric series $$ {}_2\phi_1(a,b,c;q,z) := \sum_{n=0}^{\infty} \frac{(a;q)_n(b;q)_n}{(c;q)_n(q;q)_n}z^n $$ If we substitute $a = q^{\alpha}, b = q^{\beta}, c = q^{\gamma}$ and assume that $|z| < 1$ such that the basic hypergeometric series converge the limit $q \to 1$ gives the ordinary hypergeometric series, i.e. $$ \lim_{q \to 1} \sum_{n=0}^{\infty} \frac{(q^{\alpha};q)_n(q^{\beta};q)_n}{(q^{\gamma};q)_n(q;q)_n}z^n = \lim_{q \to 1} \sum_{n=0}^{\infty} \frac{(q^{\alpha};q)_n}{(1 - q)^n}\frac{(q^{\beta};q)_n}{(1 - q)^n}\frac{(1 - q)^n}{(q^{\gamma};q)_n}\frac{(1 - q)^n}{(q;q)_n}z^n = {}_2F_1(\alpha, \beta, \gamma; z), $$ where ${}_2F_1$ is the defined here https://en.wikipedia.org/wiki/Hypergeometric_function. There are many examples identities involving basic hypergeometric series that are $q$-analogues of well known hypergeometric identities. Which means that with the substitution defined above, taking the limit $q \to 1$, we obtain the original hypergeometric identity. For example a couple of them can be found here: http://homepage.tudelft.nl/11r49/documents/wi4006/qseries.pdf.

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    $\begingroup$ All right, I was on the right track. Thanks' $\endgroup$ – Rodrigo A. Pérez Jul 26 '16 at 21:26

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