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.


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.

  • 2
    $\begingroup$ All right, I was on the right track. Thanks' $\endgroup$ – Rodrigo A. Pérez Jul 26 '16 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.