I am having trouble getting an analytical form for this recurrence (say call it problem R1):
$$ a_n = 2a_{n-1}\sqrt{1-a^2_{n-1}} \quad \text{for $n > 0$ with $a_0=1/2$} $$
But a related recurrence problem (R2) that could help solve the above is as follows (from register allocation algorithms):
$$ a_n= a^2_{n-1} -2 \quad \text{for n > 0} $$
which from the textbook am reading can be solved by substituting $a_n$ with: $ a_n = b_n + \frac{1}{b_n} $ so that we'll arrive at the formula:
$$ b_n + \frac{1}{b_n} = b^2_{n-1} + \frac{1}{b^2_{n-1}} $$
from which we can solve (using quadratic equations) that $b_n=b^2_{n-1}$.. Upon telescoping, we'll get $b_n=b_0^{2^n}$, where $b_0$ is the initial value. I somehow think that problem R2 is very much related to problem R1, but I could not get transform R1 into a form similar to R2... I've been stuck in this problem for days.. Any help?
$$a_n = 2a_{n-1}\sqrt{1-a^2_{n-1}}$$ Let $a_n=\sin(b_n)$ to make $$\sin(b_n)=\sin(2b_{n-1})\implies b_n=2b_{n-1}\implies b_n=2^{n-1} c\implies a_n=\sin(2^{n-1} c)$$
Use $a_0=\frac 12$