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Can anyone explain to solve the identity posted by my friend $$2\cos12°= \sqrt{2+{\sqrt{2+\sqrt{2-\sqrt{2-...}}} }}$$ which is an infinite nested square roots of 2. (Pattern $++--$ repeating infinitely)

Converging to finite nested radical of $2\cos12° = \frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$

The finite nested radical, I was able to derive $\cos12° = \cos(30-18)°$ as follows

$$\cos30°\cdot\cos18° + \sin30°\cdot\sin18°$$ $$= \frac{√3}{2}\cdot\frac{\sqrt{2+2\cos36°}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2-2\cos36°}}{2}$$ Where $\cos18° = \frac{\sqrt{2+2\cos36°}}{2}$ (by Half angle cosine formula) and $\sin18° = \frac{\sqrt{2-2\cos36°}}{2}$ (solving again by half angle cosine formula) $2\cos36° =\frac{ \sqrt5 +1}{2}$ which is golden ratio

$\frac{\sqrt3}{2}\cdot\frac{\sqrt{10+2\sqrt5}}{4}+ \frac{1}{2}\cdot\frac{\sqrt{5}-1}{4} = \frac{\sqrt{30+6\sqrt5}}{8}+ \frac{\sqrt5-1}{8}$

Further steps finally lead to the finite nested radical

Method actually I tried to solve infinite nested square roots of 2 is as follows.

$2\cos\theta = \sqrt{2+2\cos2\theta}$ and $2\sin\theta = \sqrt{2-2\cos2\theta}$

Now simplifying infinite nested square roots of 2, we will get the following as simplified nested radical $$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos12°}}}}$$

Simplifying step by step as follows

$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\sin6°}}}$ then

$2\cos12° = \sqrt{2+\sqrt{2+\sqrt{2-2\cos84°}}}$ (by $\sin\theta = \cos(90-\theta)$

$2\cos12° = \sqrt{2+\sqrt{2+2\sin42°}}$

$2\cos12° = \sqrt{2+\sqrt{2+2\cos48°}}$

$2\cos12° = \sqrt{2+2\cos24°}$

$2\cos12° = 2\cos12°$

We are back to $\sqrt1$

Actually this is how I got stuck!

But for infinite nested square roots of 2(as depicted), if I run program in python I am able to get good approximation ( Perhaps if we run large number of nested square roots in python we get more number of digits matching the finite nested radical), because I'm not able get anywhere solving such a kind of infinite cyclic nested square roots of 2.

Dear friends, is there anyway to find the solution by any other means like solving infinite nested square roots

Thanks in advance.

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  • $\begingroup$ I am not sure how you derived to the equation. The basic equation I derive is - $s^2=2+s$ $\endgroup$ – Moti Sep 8 at 22:57
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If the value of the radical is $x$, then we have $$x=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}}\tag1$$ Repeated squaring gives $$\left(\left(\left(x^2-2\right)^2-2\right)^2-2\right)^2=2-x\tag2$$

Now, $(2)$ has $8$ solutions, and notice for all choices of the first three signs in $(1)$, repeated squaring gives $(2)$. Thus, the solutions of $(2)$ are the eight solutions to $$x=\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm\sqrt{2-x}}}}$$

So, we must first show that $2\cos12^\circ$ satisfies $(2)$, and then to show that it is the root given by the choice of signs in the question.

To verify that $2\cos12^\circ$, we use the formula $$(2\cos\theta)^2-2 = 2(2\cos^2\theta-1)=2\cos2\theta\tag3$$ Then setting $x=2\cos12^\circ$, $(3)$ gives $$\begin{align} x^2-2&=2\cos24^\circ\\ (x^2-2)^2-2&=2\cos48^\circ\\ ((x^2-2)^2-2)^2-2&=2\cos96^\circ\\ (((x^2-2)^2-2)^2-2)-2&=2\cos192^\circ=-2\cos12^\circ=-x\\ \end{align}$$ as required.

ADDENDUM

Since $0\leq x\leq 2$, there is a value $0\leq\theta\leq\frac\pi2$ such that $x=2\cos\theta$. The argument above gives $2\cos16\theta=-2\cos\theta$ so either $$16\theta=(2n+1)\pi+\theta$$or$$16\theta=(2n+1)\pi-\theta$$ The condition $0\leq\theta\leq\frac\pi2$ gives $8$ possibilities for $\theta$: either$$\theta=\frac{(2n+1)\pi}{15},\ n=0,1,2,3$$ or $$\theta=\frac{(2n+1)\pi}{17},\ n=0,1,2,3$$ so if you can sort the values of the nested radicals for the $8$ choices of sign in decreasing order, you not only evaluate the given infinite nested radical, but $7$ more. For example, it seem clear that choosing the choosing the $3$ plus signs would give the larges value, which would mean that the value of that radical would be $2\cos\frac\pi{17}$. If the second largest value comes from choosing the first $2$ signs as $+$ and the third as $-$, that would complete the proof for the original question.

Actually, there's still something missing. For this argument to work, we also have to show that the infinite nested radical converges for all $8$ sign choices. I'm having trouble coming up with an economical way to do that.

I carried out the numerical work, and found that $$\begin{align} 2\cos\frac{7\pi}{15}&= \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}\\ 2\cos\frac{7\pi}{17}&= \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2-\cdots}}}}\\ 2\cos\frac{5\pi}{15}&= \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}\\ 2\cos\frac{5\pi}{17}&= \sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\\ 2\cos\frac{3\pi}{15}&= \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\cdots}}}}\\ 2\cos\frac{3\pi}{17}&= \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}\\ 2\cos\frac{\pi}{15}&= \sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\cdots}}}}\\ 2\cos\frac{\pi}{17}&= \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}\\ \end{align}$$

Still haven't found a nice argument for convergence, though it's clear numerically that all sequences converge rapidly.

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Somehow I got the answer from my subsequent post (after a long homework for cyclic infinite nested square roots of 2)

Sivakumar Krishnamoorthi (https://math.stackexchange.com/users/686991/sivakumar-krishnamoorthi), Solving cyclic infinite nested square roots of 2 as cosine functions, URL (version: 2020-09-26): https://math.stackexchange.com/q/3841605

$2\cos48°$ or $2\cos\frac{4}{15}π$ is cyclic infinite nested square roots of 2 of form $cin\sqrt2[2-2+]$ i.e. $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+...}}}}$

According to half angle cosine formula within next 2 steps we get $2\cos12°$ or $2\cos\frac{π}{9}$ as $cin\sqrt2[2+2-]$ i.e. cyclic infinite nested square roots of 2 as $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-...}}}}$

Therefore $2\cos12°$ as a multiple of 3 it can be solved as nested radicals as in LHS of my question and on RHS it is cyclic infinite nested square roots of 2 which is also equivalent. As both the result belong to same $\cos \frac{π}{15}$ cyclic infinite nested square roots of 2 on RHS and finite nested radical are the same in terms of results.

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