Recursion - series examples

Question

Given: $$p(n+1) = 2^n \cdot \sqrt{2\cdot {\left( 1 - \sqrt{1 - \left(\frac{p(n)}{2^n}\right)^2}\right)}}$$ And: $$p(2) = 2\cdot \sqrt{2}$$

Find $p(n)$. I an unable to come up with a generalization for $p(n)$. Please help.

Answer 1

\begin{align*} P_{2} &= 2\sqrt{2} \\ P_{3} &= 4\sqrt{2-\sqrt{2}} \\ P_{4} &= 8\sqrt{2-\sqrt{2+\sqrt{2}}} \\ P_{5} &= 16\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2}}}} \end{align*}

$P_{n}$ looks like the perimeter of regular $2^{n+1}$-sided polygon of "diameter" $1$:

$$P_{n}=2^{n} \sin \frac{\pi}{2^{n}}$$

It's not difficult to verify by considering:

\begin{align*} 2^{n}\sqrt{2 \left( 1-\sqrt{1-\sin^{2} \frac{\pi}{2^{n}}} \right)} &= 2^{n}\sqrt{2 \left( 1-\cos \frac{\pi}{2^{n}} \right)} \\ &= 2^{n}\sqrt{4\sin^{2} \frac{\pi}{2^{n+1}}} \\ &= 2^{n+1}\sin \frac{\pi}{2^{n+1}} \end{align*}