What's the limit of $\sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $?

Question

Let's look at the continued radical

$ R = \sqrt{2 + \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2 + ...}}}}}} $

whose signs are defined as $ (+, -, +, -, -, + ,-, -, -,...)$, similar to the sequence $101001000100001...$, where

$1 = +$

$0 = - $

This radical seems to converge to a constant approximately equal to $1.567883...$.

The question is: Is it possible to find this limit $R$ in closed form?

Remark: In the article "On the periodic continued radicals of 2 and generalization for Vieta’s product", it is proved that a periodic sequence of signs composed of nested square roots of two converges to $2\sin(q\pi)$ for some rational number $q$. I have tried with non periodic sequences of plus and minus, and they also converge to numbers between $0$ and $2$. if this radical has a closed form, It can be the sine of an irrational multiple of $\pi$, since both are transcendental numbers.

Answer 1

Let $R$ be the iterated square root in question

One easily check that any finite expression of the form $$\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\cdots}}}$$ is between $0$ and $2$. So $0\le R\le 2$.

Suppose that $a=2\cos t$. Then $$\sqrt{2+a}=\sqrt{2(1+\cos t)}=\sqrt{4\cos^2(a/2)}=2\cos\frac a2$$ and $$\sqrt{2-a}=\sqrt{2(1-\cos t)}=\sqrt{4\sin^2(a/2)}=2\sin\frac a2 =2\cos\left(\frac\pi2-\frac a2\right).$$ Therefore $$\begin{align} \cos^{-1}\frac R2&=\frac12\cos^{-1}\frac12\sqrt{2-\sqrt {2+\sqrt{2 -\sqrt{2-\sqrt{2-\cdots}}}}}\\ &=\frac{\pi}4-\frac14\cos^{-1} \frac12\sqrt {2+\sqrt{2 -\sqrt{2-\sqrt{2-\cdots}}}}\\ &=\frac{\pi}4-\frac18\cos^{-1} \frac12\sqrt{2 -\sqrt{2-\sqrt{2-\cdots}}}\\ &=\frac{\pi}4-\frac{\pi}{16}+\frac1{16}\cos^{-1} \frac12 \sqrt{2-\sqrt{2-\cdots}}\\ \end{align} $$ etc. So we can get a series for $\cos^{-1}(R/2)$. I think it might be something like $$\frac\pi2\sum_{n=0}^\infty\frac{(-1)^n}{2^{n(n+1)/2}}$$ which is some sort of theta function.

Answer 2

For any function $f$ and sequence of functions $f_1, f_2, \cdots, f_n$, let $$\mathop{\bigcirc}_{k=1}^n f_k \stackrel{def}{=} f_1 \circ f_2 \circ \cdots \circ f_n \quad\text{ and }\quad f^{\circ n} \stackrel{def}{=} \mathop{\bigcirc}_{k=1}^n f = \underbrace{f\circ f \circ \cdots \circ f}_{n \text{ times}} $$ be a short hand of composing the functions in given order.

Consider following maps: $$\begin{align} \psi :&\quad [0,1] \ni \theta \quad\mapsto\quad 2\cos\left(\frac{\pi}{2}\theta\right) \in [0,2]\\ \phi_{\pm} :&\quad [0,2] \ni x\quad \mapsto\quad \sqrt{ 2 \pm x } \in [0,2] \end{align} $$

The infinite radical at hand can be interpreted as picking an arbitrary $x \in [0,2]$ and study following limit:

$$\lim_{n\to\infty} \left( \mathop{\bigcirc}_{k=1}^n \phi_{+}\circ \phi_{-}^{\circ k}\right)(x)$$ As functions, it is not hard to verify following equalities $$\begin{align} \psi^{-1}\circ\phi_{+}\circ\psi \;=&\quad [0,1] \ni \theta &\mapsto&\quad \frac{\theta}{2} \in [0,1]\tag{*1a}\\ \psi^{-1}\circ\phi_{-}\circ\psi \;=&\quad [0,1] \ni \theta &\mapsto&\quad 1 - \frac{\theta}{2} \in [0,1]\tag{*1b} \end{align} $$ Let $\alpha = -\frac12.\,$ Apply $(*1b)$ $k$ times followed by $(*1a)$, we get $$\begin{align} \left(\psi^{-1}\circ\phi_{+}\circ\phi_{-}^{\circ k}\circ\psi\right)(\theta) &= \frac12\left(1 + \alpha + \alpha^2 + \cdots + \alpha^{k-1} + \alpha^k\theta\right) = \frac12\left(\frac{1 - \alpha^k}{1 - \alpha} + \alpha^k \theta\right)\\ &= \frac12\left(\frac23(1-\alpha^k) + \alpha^k\theta\right) = \frac13 + \alpha^{k+1}\left(\frac23 - \theta\right) \end{align} $$ From this, we find

$$\left(\psi^{-1}\circ \left(\mathop{\bigcirc}_{k=1}^n \phi_{+}\circ\phi_{-}^{\circ k}\right)\circ\psi\right)(\theta) = \frac13 + \alpha^2\left[ \frac13 - \alpha^3\left[ \cdots \left[ \frac13 - \alpha^{n+1}\left[\frac23-\theta\right] \right] \cdots \right] \right]\\ = \frac13 \left [ 1 + \sum_{k=2}^n(-1)^k \alpha^{\frac{k(k+1)}{2}-1} \right] + (-1)^{n+1} \alpha^{\frac{(n+1)(n+2)}{2}-1}\left(\frac23-\theta\right) $$ Since $|\alpha| < 1$, the $\theta$ dependent term in above expression drops off exponentially.
Independent of choice of $\theta \in [0,1]$, we have $$\lim_{n\to\infty} \left(\psi^{-1}\circ \left(\mathop{\bigcirc}_{k=1}^n \phi_{+}\circ\phi_{-}^{\circ k}\right)\circ\psi\right)(\theta) = \frac13 \left[ 1 - 2\sum_{k=2}^\infty(-1)^k \alpha^{\frac{k(k+1)}{2}}\right]$$ This means the infinite radical is well defined and has value

$$2\cos\left[\frac{\pi}{6}\left(1 - 2\sum_{k=2}^\infty(-1)^k \alpha^{\frac{k(k+1)}{2}}\right)\right]$$

Numerically, this expression evaluates to $\approx 1.567883223337111$, matching the number mentioned in question.

Answer 3

First aid.

$ \sqrt {2_1 - \sqrt {2_2 \pm \sqrt {2_3 \pm \ldots \pm \sqrt {2_n }}}} = 2 \sin \left ( \dfrac {90 ^ \circ (2a + 1)} {2 ^ n} \right ) $

$ \sqrt {2_1 + \sqrt {2_2 \pm \sqrt {2_3 \pm \ldots \pm \sqrt {2_n }}}} = 2 \cos \left ( \dfrac {90 ^ \circ (2a + 1)} {2 ^ n} \right ) $

If $ n \le 2 $ then $ a = 0 $

If $ n \ge 2 $ then $ 0 \le a \le 2 ^ {n - 2} - 1 $

If $ n, a $ are known, then the signs $ s_t = \pm 1 $ lying between the terms $ 2_2 $ and $ 2_n $ are calculated from the relationships

$ r = round(a / 2 ^ {n - t}) $

$ s_t = (- 1) ^ r , t = 2, 3, \ldots , n - 1 $

if we match $ + = 0, - = 1 $ , then the binary digits which are among the terms $ 2_2 $ and $ 2_n $ form a number $ b $ , which is closely related to the $ a $ parameter:

● Binary representations of $ a, b $ always have the same number of digits.

● Each value of $ a $ is assigned to a single value of $ b $, regardless of the $ n $ value.

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