Generalization of the q-Binomial Theorem

Question

The following is the well-known q-Binomial Theorem:

For all $n \geq 1$, we have: $$ \prod_{j=1}^{n}(1+xq^j) = \sum_{k = 0}^{n} q^{k(k+1)/2} {n \choose k}_q x^k $$

I am not too familiar with the proof of the theorem and I was wondering if the following generalization also holds or if the equality does not hold term by term but only by the sums:

Generalization: For all $n \geq 1$ and $1 \leq m \leq n$, we have: $$ \prod_{j = m}^{n}(1+xq^j) = \sum_{k = m}^{n} q^{k(k+1)/2} {n \choose k}_q x^k $$

(EDIT) Thanks to the commenter, I now realize that the generalization is false. Is there a similar formula which expresses $\prod_{j = m}^{n}(1+xq^j)$ as a power series of $x$?

I would appreciate any suggestions to relevant literature.

Answer 1

The $q$-binomial theorem as was taught to me is $$ {}_1 \phi_0 (a; -; q, z) := \sum_{n = 0}^{\infty} \frac{(a;q)_n}{(q;q)_n} z^n = \frac{(az;q)_{\infty}}{(z;q)_{\infty}}, $$ see chapter 1.3 of Basic hypergeometric series of Gasper and Rahman. In particular $$ \prod_{j = 1}^{n} (1 + xq^j) = \frac{(-xq;q)_{\infty}}{(-xq^{n+1};q)_{\infty}} = {}_1 \phi_0(q^{-n};-;q,-xq^{n+1}) = \sum_{k = 0}^{n} \frac{(q^{-n};q)_k}{(q;q)_k} (-xq^{n+1})^k. $$ Simplifying the coefficients of $x^k$ on the left hand side of the equation yields $$ \frac{(q^{-n};q)_k}{(q;q)_k}(-1)^k q^{nk+k} = (-1)^k q^{\binom{n}{k} - nk} \frac{(q;q)_n}{(q;q)_{n-k}(q;q)_k} (-1)^k q^{nk+k} = q^{\binom{n+1}{k}} \binom{n}{k}_q, $$ hence your formula of the $q$-binomial theorem is a special case.

To find a solutions for your generalization we write $$ \prod_{j = m}^{n} (1 + xq^j) = \frac{(-xq^m;q)_{\infty}}{(-xq^{n+1};q)_{\infty}} = {}_1 \phi_0 (q^{-(n-m+1)};-;q,-xq^{n+1}) = \sum_{k=0}^{n-m+1} \frac{(q^{-(n-m+1)};q)_k}{(q;q)_{k}} (-xq^{n+1})^k. $$ Simplifying the coefficient of $x^k$ on the left hand side gives $$ \frac{(q^{-(n-m+1)};q)_k}{(q;q)_{k}} (-q^{n+1})^k = q^{\binom{k}{2} + mk} \binom{n-m+1}{k}_q. $$ This yields $$ \prod_{j=m}^n (1 + xq^j) = \sum_{k=0}^{n-m+1} q^{\binom{k}{2} + mk} \binom{n-m+1}{k}_q x^k, $$ modulo maybe some calculation mistakes. ;)