Closed form of $\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}}$

Question

How to prove the following formula,

$$\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}} = \begin{cases} 2 \cos\Big[\frac{1}{2^{n}}\Big(\pi + \arctan{\frac{\sqrt{x(4-x)}}{x-2}}\Big)\Big] & 0 < x < 2 \\ 2 \cos\Big[\frac{1}{2^{n}}\arctan{\frac{\sqrt{x(4-x)}}{x-2}}\Big] & 2 < x \leq 4 \\ 2 \cosh\Big[\frac{1}{2^{n}}\text{arctanh}{\frac{\sqrt{x(x-4)}}{x-2}}\Big] & x \geq 4 \end{cases} \ $$ where $x \in \mathcal{R}$ and $n$ is the number of time the square root symbol $\sqrt{\text{ }\text{ }}$ appears.

This formula is a generalization of the following formula, which can be found here and here, and which can be obtained by taking the above general formula in the limit $x = 2$.

$$\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{2}}}} = 2\cos\Big(\frac{\pi}{2^{n+1}}\Big)$$

score 1 · Answer 1 · answered Jul 17 '20 at 15:56

1

Hint If $a=2\cos(y)$ then $$2+a= 2 + 2\cos(y)= 4 \cos^2(\frac{y}{2}) \,.$$

So write $\sqrt{x}=2 \cos(\alpha)$ and use the above formula:

$$2 +\sqrt{x}= (2 \cos(\frac{\alpha}{2}))^2 \\ 2+\sqrt{2+\sqrt{x}}=2+2\cos(\alpha/2)= (2\cos(\alpha/4))^2 \\ ...$$

answered Jul 17 '20 at 15:56

N. S.

134,609

How could one find the formula without knowing it first ? – The Esoteric Mathematician Jul 17 '20 at 16:16
@Zakhurf The basis for this is the half-angle formula $\cos^2\left(\frac x2\right)=1+\cos x$ which follows from the double-angle formula. This is a more-or-less well-known identity. – mrtaurho Jul 17 '20 at 16:26
@Zakhurf That's the well known half angle formula. – N. S. Jul 17 '20 at 16:35

Closed form of $\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}}$

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