I am reading Lascota and Mackey's "Chaos, Fractals, and Noise" (which is great). In an early chapter, they define the following operator (really this is a special case of the Frobenius-Perron operator, but that's not relevant to my question.) For a function $f:[0,1]\to \mathbb R$, define $Pf := \frac{1}{4 \sqrt{1-x}}(f(\frac{1}{2}-\frac{1}{2}\sqrt{1-x})+f(\frac{1}{2}+\frac{1}{2}\sqrt{1-x})$. Now, begin with the initial distribution $f=1$. Then one can calculate that: $$Pf = \frac{1}{2 \sqrt{1-x}}$$ $$P^2 f =\frac{1}{8 \sqrt{x(1-x)}}(\sqrt{2-2\sqrt{1-x}}+\sqrt{2+2\sqrt{1-x}})$$ $$P^3 f=\frac{1}{16 \sqrt{x(1-x)}}\left(\sqrt{2-\sqrt{2+2 \sqrt{1-x}}}+\sqrt{2+\sqrt{2+2 \sqrt{1-x}}}+\sqrt{2-\sqrt{2-2 \sqrt{1-x}}}+\sqrt{2+\sqrt{2-2 \sqrt{1-x}}}\right)$$
etc.
The part that is baffling me is that they then claim that $\lim_{n\to \infty}P^n f=f_* (x)=\frac{1}{\pi \sqrt{x(1-x)}}$. It is not so difficult to check that $Pf_* (x)=f_* (x)$, which makes it plausible that $f_*$ is indeed the limit, but I have no idea how one would arrive at such a nice result.
I have tried directly solving the functional equation $Pf=f$ to derive that $f$ must be $\frac{1}{\pi \sqrt{x(1-x)}}$, but this doesn't seem to be the right approach. I'm thinking there must be some nice way to find some kind of nested radical expression for $1/\pi$ that is relevant here, but I don't know how to do this. It seems possibly related to this expression for $\pi$ but again, I don't see the connection. I appreciate any insight you have!