<s>[INST] Recursion - series examples 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.
 [/INST] \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*}
</s>