All real numbers in $[0,2]$ can be represented as $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$

Solution 1:

Here is a possible explanation. Let $\alpha \in [0, \pi/2]$ and define $\epsilon_1, \epsilon_2, \cdots$ by $ \epsilon_i = \operatorname{sgn}( \cos ( 2^i \alpha )) \in \{-1, 1\}$. Here, we take the convention that $\operatorname{sgn}(0) =1 $. Then applying the identity $2\cos\theta = \operatorname{sgn}(\cos\theta) \sqrt{2 + 2\cos(2\theta)}$ repeatedly, we have

$$ 2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ \cdots + \epsilon_n \sqrt{2 + \smash[b]{2\cos(2^{n+1} \alpha)} }}}}. $$

This can be used to show that, with an appropriate definition of infinite nested radical, the following identity

$$ 2\cos \alpha = \sqrt{2 + \epsilon_1 \sqrt{2 + \epsilon_2 \sqrt{ 2 + \cdots }}} $$

is true. This shows that any real number between $[0, 2]$ can be written as an infinite nested radical of the desired form. Moreover, if we denote $x = 2\cos\alpha$, then

• $\epsilon_1 = \operatorname{sgn}(2\cos (2\alpha)) = \operatorname{sgn}(x^2 - 2)$,
• $\epsilon_2 = \operatorname{sgn}(2\cos (4\alpha)) = \operatorname{sgn}((x^2 - 2)^2 - 2)$,

and likewise. This explains why signs are determined by OP's algorithm.

Solution 2:

Peculiar observation

If we define a binary number $b = b_1b_2\cdots b_n$ with digits mapped to the symbols like this: $$b_k = \begin{cases}0 \text{ if } (-) \text{ at position } k\\1 \text{ if } (+) \text{ at position } k\end{cases}$$ Then if we run the algorithm proposed in the question, looping

x(k) = x(k-1)^2-2;
b(k) = (x(k)>0);

the vector b will get logical values corresponding to bits 1 and 0 of the binary number above and we can calculate it for the linear space of $x\in[0,1]$. If we do this we can then calculate each number as the scalar product $$[1/2,1/4,\cdots,1/2^k]b$$ and if we then plot it, it will look like

Which is kind of a peculiar plot having a bit of a discontinuous and fractal structure. I think the largest discontinuity is around $x = \sqrt{1/2}$ but I have no theoretical explanation why..

edit as pointed out by Sangchul Lee this seems similar to Tent Map

Solution 3:

Your algorithm pretty much shows that there exists an expansion for every number in $[0,2]$.

If we replace $2$ by $a>2$ (and keep the square root), we will fail because we need that squaring a number from the interval $[u,v]$ produces a number that is either in $a+[u,v]$ or in $a-[u,v]$. So we must have $u=0$ and $v\ge a$ and $v^2\le a+v$. The last two imply $a\le v\le 2$.

If we additionally switch to $k$th roots, the condition becomes that $v\ge a$ and $v^k\le a+v$, hence $a\le v\le\sqrt[k-1]2$.