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

I would like some reference about this infinitely nested radical expansion for all real numbers between $0$ and $2$.

I’ll use a shorthand for this expansion, as a string of signs, $+$ or $-$, with infinite periods denoted by brackets.

$$2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$$

$$1=\sqrt{2 – \sqrt{2 – \sqrt{2 – \dots}}}=(-)$$

$$0=\sqrt{2 – \sqrt{2 + \sqrt{2 + \dots}}}=-(+)$$

$$\phi=\sqrt{2 + \sqrt{2 – \sqrt{2 + \sqrt{2 – \dots}}}}=(+-)$$

$$\frac{1}{\phi}=\sqrt{2 – \sqrt{2 + \sqrt{2 – \sqrt{2 + \dots}}}}=(-+)$$

In general, the expansion can be found by a very easy algorithm:

  • take any number in $(0,2)$, square it
  • if the result $>2$ write $+$, if the result $<2$ write $-$
  • subtract $2$ from the result, square, repeat

If on some step we get $2$ exactly, we just write $(+)$ and the expansion is finished.

Examples:

$$\pi-2=–+-++-+-+++++++-+-+———+-+–+–+–+++—++++ \dots=1.141592653589793 \dots$$

Basically, $50$ terms of our expansion gave only $15$ correct decimal digits for $\pi$. But considering the expansion can be coded as binary, it’s not so bad.

The convergence plot, and two binary plots for this $50$ terms can be seen below:

enter image description here


$$e-1=+—–+++-++-+—++-++++-+—++-+++-++++-++++—++ \dots=1.71828182845905 \dots$$

enter image description here


Do you know any reference about this expansion? Can every real number between $0$ and $2$ be expanded this way?

Is number $2$ special in this case, or can we make a similar expansion using some other number (and other power for the root)?


Edit

Now that I think about it, we can use the general expansion for $x \in [0,a]$:

$$x=\left(a \pm \left(a \pm \left(a \pm \dots \right)^p \right)^p \right)^p$$

$$a=2^{\frac{p}{1-p}}$$

For example:

$$\frac{1}{4}=\left(\frac{1}{4} + \left(\frac{1}{4} + \left(\frac{1}{4} + \dots \right)^2 \right)^2 \right)^2$$

$$\frac{3}{4}-\frac{\sqrt{2}}{2}=\left(\frac{1}{4} – \left(\frac{1}{4} – \left(\frac{1}{4} – \dots \right)^2 \right)^2 \right)^2$$

etc.

However, this case $p=2,~~~a=\frac{1}{4}$ is not just a random example, it’s the only rational expansion of this kind. So I would say it’s more important than the titular root expansion.


Edit

An interesting article that connects the nested roots of this kind to Chebyshev polynomials: http://www.sciencedirect.com/science/article/pii/S0022247X12003344

Answer

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.

Attribution
Source : Link , Question Author : Yuriy S , Answer Author : Sangchul Lee

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