Demonstration for the equal number of odd and unequal partitions of an integer

Question

I'm having some problems trying to resolve one exercise from The art and craft of problem solving by Paul Zeitz. What this problem asks you is to prove that $F(x)$ is equal to 1 for all $x$, where $$F(x) = (1-x)(1+x)(1-x^3)(1+x^2)(1-x^5)(1+x^3)(1-x^7)(1+x^4)\cdots$$ This would demonstrate that the number of unequal partitions and odd partitions of an integer is the same since the generating function $U(x)$ for the number of unequal partitions of an integer $n$ is $$U(x) = \sum_{n=0}^\infty u_nx^n = (1+x)(1+x^2)(1+x^3)\cdots$$ and the generating function $V(n)$ for the number of odd partitions of an integer $n$ is $$V(x) = \sum_{n=0}^\infty v_nx^n = (1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+x^{15}+\cdots)\cdots = \frac{1}{(1-x)(1-x^3)(1-x^5)\cdots}$$ and so $F(x)=\frac{U(x)}{V(x)}$

The way the exercise suggests to carry out the demonstration is to prove that $F(x)$ is invariant under the substitution $x\mapsto x^2$. This is easy to chech $$F(x^2) = \prod_{k=1}^\infty (1-(x^2)^{2k-1})(1+(x^2)^k)=\prod_{k=1}^\infty (1-x^{2k-1})(1+x^{2k-1})(1+x^{2k}) = F(x)$$

Now the idea is to keep iterating the substitution and I guess that the solution is that for values of $x$ such that $|x| < 1$ all the $x$ terms go to $0$, but I see two problems with this:

1. This wouldn't work for values of $x$ greater than $1$ or less than $-1$. We don't care about these values because the function does not converge when evaluated with them? Why?
2. After iterating the substitution $n$ times we get $$F(x) = F(x^{2^n}) = \prod_{k=1}^\infty (1-x^{2^n(2k-1)})(1+x^{2^nk})$$ and although the $x$ terms go to $0$ for the previously mentioned values of $x$ as $n$ goes to infinity, it is still an infinite product and thus an indeterminate form $1^\infty$ (I think) which I haven't been able to resolve.

Answer 1

We already know from OPs derivation \begin{align*} F(x)=F\left(x^2\right)=\cdots=F\left(x^{2^q}\right)\qquad\qquad q\in\mathbb{N}_0\tag{1} \end{align*}

We set $F(x)=\sum_{n=0}^\infty a_nx^n$ and obtain from (1) by coefficient comparison \begin{align*} F(x)&=F\left(x^2\right)\\ \sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\infty}a_nx^{2n}\qquad\Longrightarrow\qquad a_{2k+1}=0\qquad\forall k\geq 0\tag{2} \end{align*}

We see the coefficients of odd powers of $x$ in $F(x)$ are zero. Let's have a look at \begin{align*} &a_0+a_1x+a_2x^2+\color{blue}{a_3x^3}+\cdots\\ &\qquad=a_0+a_1x^2+a_2x^4+\color{blue}{a_3x^6}+\cdots\\ &\qquad=a_0+a_1x^4+a_2x^8+\color{blue}{a_3x^{12}}+\cdots\\ &\qquad=a_0+a_1x^8+a_2x^{16}+\color{blue}{a_3x^{24}}+\cdots\\ &\qquad\ \ \vdots \end{align*}

We observe by iteratively using $F(x)=F\left(x^2\right)=F\left(x^4\right)=\cdots$ that since for instance $a_3=0$, also all coefficients \begin{align*} a_6&=[x^6]F(x)=[x^6]F\left(x^2\right)=[x^3]F(x)=0\\ a_{12}&=[x^{12}]F(x)=[x^{12}]F\left(x^4\right)=[x^3]F(x)=0\\ a_{24}&=[x^{24}]F(x)=[x^{24}]F\left(x^8\right)=[x^3]F(x)=0\\ &\ \ \vdots \end{align*} and in general all coefficients of the form \begin{align*} a_{3\cdot2^q}=[x^{3\cdot 2^q}]F(x)=[x^{3\cdot 2^q}]F(x^{2^q})=[x^3]F(x)=0\qquad\qquad q\geq 0 \end{align*} are zero. Here we use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ of a series.

We conclude: Each natural number $m\geq 1$ has a unique representation \begin{align*} \color{blule}{m=2^q\cdot r}\qquad\qquad q\geq 0, r\ \text{odd} \end{align*} as product of a power of $2$ (if $q\geq 1$) with an odd natural number $r$. It follows from (1) and (2) \begin{align*} \color{blue}{[x^{m}]F\left(x^{2^q}\right)}=[x^{2^{q}r}]F\left(x^{2^q}\right)=[x^r]F(x)\color{blue}{=0}\qquad\qquad q\geq 0 \end{align*}

Since from OPs representation of \begin{align*} F(x)=\prod_{k=1}^{\infty}\left(\left(1-x^{2k-1}\right)\left(1+x^k\right)\right) \end{align*} we also know that $[x^0]F(x)=1$, the claim \begin{align*} \color{blue}{F(x)=1} \end{align*} follows.