<s>[INST] The function $(1+x)(1+x^4)(1+x^{16})(1+x^{64})\cdots =\prod_{n\geq 0} (1+x^{4^n})$ Let $$f(x)=(1+x)(1+x^4)(1+x^{16})(1+x^{64})\cdots=\prod_{n\geq 0} (1+x^{4^n})$$
Then what is $f^{-1}(\frac{8}{5f(3/8)})?$
The answer should be a rational number.
My attempt: I tried to take a log of the expression to turn it into a sum, but that did not simplify. So I am clueless on how to proceed. Any suggestions?
 [/INST] It's hard to say things about $f(x)$ for general $x$. But we have 
\begin{align}
(1-x) f(x) f(x^2) &= (1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)\dotsb \\
 &= (1-x^2)(1+x^2)(1+x^4)(1+x^8)\dotsb \\
 &= (1-x^4)(1+x^4)(1+x^8) \dotsb \\
 &= (1-x^8)(1+x^8) \dotsb \\
 &= \dots = 1.
\end{align}
(Assuming convergence.)
If $x = f^{-1}\left(\frac{8}{5f(3/8)}\right)$, then $\frac58 \cdot f(\frac38) \cdot f(x)=1$, and the above identity tells us that $\frac58 \cdot f(\frac38) \cdot f(\frac{9}{64}) = 1$. So one possible answer is $\frac{9}{64}$.
To verify that this solution is valid (i.e., to verify convergence) and that it is the only solution, note that $$\log f(x) = \sum_{n\ge 1} \log\left(1 + x^{4^n}\right) \le \sum_{n \ge 1} x^{4^n}$$ which converges for $|x| < 1$, and $f$ is increasing (because each factor is increasing) and therefore injective.
</s>