Closed form for the recurrence $c_0=-1,\ c_{n+1}=\sqrt{\frac{c_n+1}{2}}$ that computes $\cos(\pi/2^n)$

Question

Let $c_n:=\cos(\pi/2^n)$ for $n\geq 0$. These values can be computed using the following recurrence (I can explain why if needed): $$c_0=-1,\ c_{n+1}=\sqrt{\frac{c_n+1}{2}}.$$

Is there a closed form for this recurrence? It there is, how can one find it? If there isn't, how can one prove it?

Answer 1

In fact you have, $$ \cos x= \sqrt{\frac{\cos 2x+1}{2}}~~~~~\text{for $0\le x\le \pi/2$}$$ Hence, $$ c_{n+1}=\cos \frac{\pi}{2^{n+1}} = \sqrt{\frac{\cos \frac{\pi}{2^{n}}+1}{2}}= \sqrt{\frac{c_n+1}{2}}$$

Below is how you can find it.

First see that for all $n>1$ we have $0\le c_n\le 1$ therefore there exists $a_n \in(0,\frac\pi2)$ such that

$$\color{blue}{c_n = \cos a_n!!!}$$

Now what is the relation bwt $a_n$ and $a_{n+1}?$

we have, $$\cos a_{n+1} =c_{n+1}=\sqrt{\frac{c_n+1}{2}}= \sqrt{\frac{\cos a_n+1}{2}} = \cos\frac{a_n}{2}$$

Basically we obtain $$a_{n+1} = \frac{a_n}{2}$$

which is a geometric sequence and therefore, $$a_n = \frac{a_1}{2^{n-1}}$$

Check that $$ \cos a_1 = 0\implies a_1 =\frac\pi2$$

This gives $$\color{red}{c_n = \cos\frac{\pi}{2^n}~~n\ge 1}$$