q-analog confusion

Question

In this pdf https://cpb-us-w2.wpmucdn.com/blog.nus.edu.sg/dist/2/3912/files/2014/10/Chapter9-20ycqjr.pdf page $5$ equation $(9.11)$, I understand why:

$$\prod(1+xq^{k-1})=\sum_{k=0}^{\infty}\frac{q^{k(k-1)/2}}{(q)_k}x^k$$ But I have trouble seeing why (9.12) holds, for instance it says substitute $x=-1$, and and multiply $\frac{(-1)^n}{(q)_n}$ to the equation:

$$0=\sum _{k=0}^n\frac{(q)_n}{(q)_k(q)_{n-k}}q^{k(k-1)/2}x^k$$

Which makes sense, then when it goes to comparing coefficients to get $A(x)B(x) = 1$ which is equation $(9.12)$ in the pdf, I just dont see why this is the case and why this holds.

I would appreciate the help in trying to help me understand why this equation $(9.12)$ holds.

(Note I have attached the pdf for your reference)

Answer 1

The pdf file has the equation you are interested in: $$ A(x)B(x) = 1. \tag{9.12} $$ These two products were defined earlier as: $$ A(x) = \sum_{m=0}^\infty \frac{(-1)^mx^m}{(q)_m}\quad \text{ and } \quad B(x) = \sum_{m=0}^\infty \frac{q^{m(m-1)/2}}{(q)_m}x^m. $$ For brevity, define $\;a_k:=(-1)^k/(q)_k, \;$ $\;b_k:=q^{k(k-1)/2}/(q)_k,\;$ and $\;c_n := \sum_{k=0}^n a_kb_{n-k}.\;$ Then $$ A(x) = \sum_{k=0}^\infty a_kx^k,\; B(x) = \sum_{k=0}^\infty b_kx^k,\; C(x) := A(x)B(x) = \sum_{n=0}^\infty c_nx^n.$$ Notice that $\;C(x) = 1\;$ means that $\;c_0=1\;$ and $\;c_n=0\;$ for all $n\ge 1.\;$ But $\;a_0=b_0=1\;$ implies $\;c_0=1\;$ and the equation (with typo corrected) $$ 0 = \sum_{k=0}^n \frac{(-1)^{n-k}}{(q)_{n-k}} \cdot \frac{q^{k(k-1)/2}}{(q)_k},\quad n\ge 1. \tag{9.11} $$ is just exactly $\;c_n=0\;$ for all $n\ge 1.\;$

Answer 2

The deduction follows eq. (9.8) which is $$\prod_{k=1}^{n}(1+xq^{n-1})=\sum_{k=0}^n\left[n\atop k\right]_q q^{k(k-1)/2}x^k.$$

The product includes $1+x$ as a factor. Setting $x=-1$ gives zero. Therefore $$\sum_{k=0}^n(-1)^k\left[n\atop k\right]_q q^{k(k-1)/2}=0.$$ That's the same as $$\sum_{k=0}^n(-1)^k\frac{(q)_n}{(q)_k(q)_{n-k}} q^{k(k-1)/2}=0.$$ Multiplying by $(-1)^n/(q)_n$ gives $$\sum_{k=0}^n\frac{(-1)^{n-k}q^{k(k-1)/2}}{(q)_k(q)_{n-k}} =0\tag{*}.$$ This is not what the anonymous pdf suggests, as it has another typo there.

$(*)$ can be rewritten as $\sum_{k=0}^n a_{n-k}b_k=0$ where $a_k=(-1)^k/(q)_k$ and $b_k=q^{k(k-1)/2}/(q)_k$. If we define $A(x)=\sum_{k=0}^\infty a_kx^k$ and $B(x)=\sum_{k=0}^\infty b_kx^k$ then the coefficient of $x^n$ in $A(x)B(x)$ is $\sum_{k=0}^n a_{n-k}b_k$. This is zero, for $n>0$, and so $A(x)B(x)$ reduces to its constant term, which is $a_0b_0=1$.