Maximise $y$ with respect to $x$ for $y=\prod_{k=1}^{\infty}(1-x^{-k})$

Question

$$y=\prod_{k=1}^{\infty}(1-x^{-k})$$ I want to maximise this function. So far I have: $$\ln(y)=\sum_{k=1}^{\infty}\ln(1-x^{-k})$$ $$\frac{dy}{dx}\cdot\frac{1}{y}=\sum_{k=1}^{\infty}\frac{kx^{-k-1}}{1-x^{-k}}$$ $$\frac{dy}{dx}\cdot\frac{x}{y}=\sum_{k=1}^{\infty}\frac{kx^{-k}}{1-x^{-k}}$$ $$\frac{dy}{dx}\cdot\frac{x}{y}=\sum_{k=1}^{\infty}\frac{k}{x^{k}-1}$$ For a turning point, $\frac{dy}{dx}=0$, therefore $$0=\sum_{k=1}^{\infty}\frac{k}{x^{k}-1}$$ Which I acknowledge is probably difficult to solve in any way other than numerically. That $x^k-1$ term has an $x-1$ factor, though I doubt that will help.

Answer 1

Let $f(x)=\prod_{k\geq 1}(1-x^k)$, defined over $I=(-1,1)$. We have: $$\frac{d}{dx}\log f(x) = \frac{f'(x)}{f(x)}=\sum_{k\geq 1}\frac{d}{dx}\log(1-x^k)=-\sum_{k\geq 1}\frac{k x^{k-1}}{1-x^k}=-\frac{1}{x}\sum_{n\geq 1}\sigma_1(n)\, x^n$$ where $\sigma_1(n)=\sum_{d\mid n}d$ is the sum-of-divisors function. To locate the maximum of $f(x)$, that occurs for some $x\in(-1,0)$, it is sufficient to solve with some numerical algorithm (for instance, Newton's method) the equation: $$ 0 = \sum_{n\geq 1}\sigma_1(n)\,x^n. $$ We have that the real zero of the last equation is around $x\approx -0.41118$, so your function attains its maximum when $\color{red}{x\approx -2.432}$.