For each $a\in(0,\frac{1}{2})$, we define a sequence $\{x_{k}\}_{k\ge1}$ where $x_1=a$ and $2x_{k+1}=\left(\frac{x_k}{1-x_k}\right)^2$. Evaluate $$\prod_{k=1}^\infty(1-x_k)$$
I first tried to use trigonometric substitution but couldn’t find a suitable one so I found $x_2=\frac{a^2}{2(1-a)^2},x_3=\frac{a^4}{2(a^2-4a+2)^2}$. No pattern is visible to me to write $x_k$ explicitly in terms of k due to which I’m stuck here. What can be the way to proceed?
Given $x_1 =a \in\left(0, \frac{1}{2} \right)$, the sequence is decreasing and bounded in $\left(0, \frac{1}{2} \right)$ because $$2(x_{n+1}-x_n)=\frac{x_n(2-x_n)(2x_n-1)}{(x_n-1)^2}<0 $$ The limit $l$ of the sequence is the solution of $$2l = \left(\frac{l}{1-l}\right)^2 \implies l =0 \hspace{1cm} \text{(we note that } l<\frac{1}{2})$$
We have $$ x_{n+1}-\frac{1}{2} = \frac{1}{2}\left(\frac{x_n}{1-_n}\right)^2-\frac{1}{2}=\left( x_n-\frac{1}{2} \right) \cdot \frac{1}{(1-x_n)^2} \implies \color{red}{(1-x_n)=\sqrt{\frac{x_{n}-\frac{1}{2}}{x_{n+1}-\frac{1}{2}}}} $$ then $$\color{red}{\prod_{n=1}^{+\infty}(1-x_n)}=\lim_{n\infty}\sqrt{\frac{x_{1}-\frac{1}{2}}{x_{n+1}-\frac{1}{2}}} =\sqrt{\frac{a-\frac{1}{2}}{l-\frac{1}{2}}} \color{red}{= \sqrt{1-2a}} $$