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I have the following infinite function$$F(x)=1+\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2}}{(q;q)_n}=\sum\limits_{n=0}^{\infty}\dfrac {x^nq^{n^2}}{(q;q)_n}\tag1$$$$G(x)=(xq;q)_{\infty}F(x)=(xq;q)_{\infty}\left(\sum\limits_{n=0}^{\infty}\dfrac {x^nq^{n^2}}{(q;q)_{n}}\right)\tag2$$$$F(x)=F(xq)+xq\cdot F(xq^2)\tag3$$ Apparently, $G(x)$ can be represented in terms of $G(x)$ as$$G(x)=(1-xq)G(xq)+xq(1-xq)(1-xq^2)G(xq^2)$$ However, I don't know how to represent $G(x)$ in terms of $G(x)$.

Question: Using $(1),(2),(3)$, how would you express $G(x)$ in terms of $G(x)$?

I'm not sure where to start. I started with $G(x)=(xq;q)_{\infty}F(x)$ and substituted $(3)$ to get $G(x)=(xq;q)_{\infty}(F(xq)+xq\cdot F(xq^2))$$$\begin{align*}G(x) & =(xq;q)_{\infty}F(xq)+(xq;q)_{\infty}xq\cdot F(xq^2)\tag4\\ & =(xq;q)_{\infty}\sum\limits_{n=0}^{\infty}\dfrac {x^nq^{n^2+n}}{(q;q)_n}+(xq;q)_{\infty}\sum\limits_{n=1}^{\infty}\dfrac {x^nq^{n^2}}{(q;q)_n}(1-q^n)\end{align*}$$ However I don't think that is getting me anywhere.

I need help after that. I'm not the best with infinite functions.

1 Answers 1

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By the functional equation

$$F(x) = F(xq) + xqF(xq^2)$$

we obtain

\begin{align} G(x) &= (xq;q)_\infty F(xq) + xq(xq;q)_\infty F(xq^2) \\ &= (1-xq)(xq^2;q)_\infty F(xq) + xq(1-xq)(1-xq^2)(xq^3;q)_\infty F(xq^2)\\ &=(1-xq)G(xq) + xq(1-xq)(1-xq^2)G(xq^2) \end{align}

since $G(xq) = (xq^2;q)_\infty F(xq)$ and $G(xq^2) = (xq^3;q)_\infty F(xq^2)$.