How to prove ${}_2\phi_1(q^{-n},b;c;q,q)=\frac{(c/b;q)_n}{(c;q)_n}b^n.$

Question

Firstly, we have already known the one of $q$-analogues of Vandermonde's formula, which is $${}_2\phi_1(q^{-n},b;c;q,cq^n/b)=\frac{(c/b;q)_n}{(c;q)_n}.$$ And there is a hint, when we change the order of summation it deduces that $${}_2\phi_1(q^{-n},b;c;q,q)=\frac{(c/b;q)_n}{(c;q)_n}b^n.$$ What confused me is that the some detail steps were omitted.

Answer 1

It not so hard if you know what to do. However, it involves some calculations. (which is almost always the case with $q$-analogues special functions) Unfortunately I'm missing a $q$-power somewhere, so you have some homework finding the mistake. Still I'm sure this is the way that they want you to solve it.

By the $q$-Chu-Vandermonde formula we have $$ {}_2 \phi_1(q^{-n}, b; c; q, cq^n/b) = \frac{(c/b;q)_n}{(c;q)_n}. $$ Reversing the order of summation of the left hand side of the $q$-Chu-Vandermonde formula we obtain $$ \sum_{k=0}^{n} \frac{(q^{-n}, b; q)_{n-k}}{(q,c;q)_{n-k}} (cq^n/b)^{n-k} = \frac{(c/b;q)_n}{(c;q)_n}. $$ Note that we have $$ \frac{(a;q)_{n-k}}{(b;q)_{n-k}} = \frac{(a;q)_{n}}{(b;q)_{n}} \frac{(q^{1-n}/b;q)_k}{(q^{1-n}/a;q)_k} (b/a)^k. $$ If we apply this equality twice on the reversed summation of the $q$-Chu-Vandermonde formula the left hand side is equal to $$ \sum_{k=0}^{n} \frac{(q^{-n};q)_n}{(q;q)_n} \frac{(q^{-n};q)_{k}}{(q;q)_k} \frac{(b;q)_n}{(c;q)_n} \frac{(q^{1-n}/c;q)_k}{(q^{1-n}/b;q)_k} q^{k(n+1)} (c/b)^k (c/b)^{n-k} q^{n^2 - kn}. $$ Simplifying the left hand side using $(q^{-n};q)_n = (-1)^n q^{\binom{n}{2} - n^2} (q;q)_n$ and putting everything independent of $k$ to the right hand side yields $$ \sum_{k=0}^n \frac{(q^{-n}, q^{1-n}/c; q)_k}{(q, q^{1-n}/b; q)_k} q^k = (-1)^n \frac{(c/b;q)_n}{(b;q)_n} q^{-\binom{n}{2}} (c/b)^n q^{-n^2}. $$ Substitute $b, c \mapsto q^{1-n}/b, q^{1-n}/c$ and use $(q^{1-n}/c;q)_n = (c;q)_n (-q^{1-n}/c)^n q^{\binom{n}{2}}$ so that $$ {}_2 \phi_1(q^{-n}, b; c; q, q) = \frac{(c/b;q)_n}{(c;q)_n} b^n q^{-n}. $$ This last $q^n$ power should cancel with something. I'm sure I made some calculation mistake somewhere. But basically this is the idea. Notice that the book of Gasper and Rahman, named Basic Hypergeometric Series, contains an appendix with a list of useful identities and manipulations for $q$-Pochhammer symbols.