Solving finite and infinite nested square roots of 2 - yet another interesting approach

Question

Consider the following consecutive equalities:

$\sqrt2=2\cos(\frac{1}{4})\pi$

$\sqrt{2-\sqrt2}=2\cos(\frac{3}{8})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2}}}=2\cos(\frac{5}{16})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}=2\cos(\frac{11}{32})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}=2\cos(\frac{21}{64})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}=2\cos(\frac{43}{128})\pi$

Here is the beautiful part

$\frac{3}{8} = \frac{1}{4}+\frac{1}{8}$

$\frac{5}{16} = \frac{3}{8}-\frac{1}{16}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}$

$\frac{11}{32} = \frac{5}{16}+\frac{1}{32}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}$

$\frac{21}{64} = \frac{11}{32}-\frac{1}{64}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}$

$\frac{43}{128} = \frac{21}{64}+\frac{1}{128}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}$

If extended to infinity $\sqrt{2-\sqrt{2-\sqrt{2...}}} = 2cos(\frac{1}{3})\pi$ (we can calculate the sum of infinite Geometric progression and we get there result as $1\over3$)

We can easily guess that if a nested square root has $n$ $2$’s then a denominator of the corresponding cosine argument is $2^{n+1}$.

Most beautiful part is getting the numerator which is $2^{n+1} \over 3$ approximated to nearest integer.[or approximation to floor and making it as very next odd number(if the floor is even number)--> and this is better ]

Subsequent numbers having numerator are as follows

$\frac{2^3}{3}=3;\frac{2^4}{3}=5;\frac{2^5}{3}=11;\frac{2^6}{3}=21;\frac{2^7}{3}=43;\frac{2^8}{3}=85;\frac{2^9}{3}=171;\frac{2^{10}}{3}=341...$

Therefore for the finite number of nested square roots of 2 like this, it is the easy way to evaluate the angle inside the cosine function!

One more exciting part is association with integer sequence of type Jacobsthal sequence $a(n-1) + 2a(n-2)$, with $a(0) = 0, a(1) = 1$.

Still more interesting part is Fibonacci like pattern from above integer sequence

My question is can we link geometric progression to simplify different patterns of the finite or infinite nested square roots of 2, like this?

To expand my question to understand

Let us consider finite nested square roots of 2 in simplified way

$\sqrt{2-\sqrt2}$ as $n\sqrt2(1-)$ and $\sqrt{2-\sqrt{2+\sqrt2}}$ as $n\sqrt2(1-1+)$

If we have finite nested square roots of 2 such as $n\sqrt2(1-2+)$ repeated signs inside nested radical 'n' times and infinitely, how can we get the angle?. What I have shown is simplest example. Let us consider $n\sqrt2(3-1+7-)$ or $n\sqrt2(2-3+4-1+)$ and so on. Is there any easier way to find the cosine angle?

Answer 1

I don't quite understand the phrasing of your question, and in particular I don't quite understand if you're asking for a proof of these identities or not. Here is a proof. We can reverse-engineer the result we need by asking what relation between two angles $\alpha_n, \alpha_{n+1}$ is implied by

$$\sqrt{2 - 2 \cos \alpha_n} = 2 \cos \alpha_{n+1};$$

here $\alpha_n$ and $\alpha_{n+1}$ are two consecutive angles among the sequence of angles you've listed. Squaring and using the cosine double-angle formula $\cos 2 \theta = 2 \cos^2 \theta - 1$ gives

$$2 - 2 \cos \alpha_n = 4 \cos^2 \alpha_{n+1} = 2 + 2 \cos 2 \alpha_{n+1}$$

which gives $\cos \alpha_n = - \cos 2 \alpha_{n+1}$. If $\alpha_i \in \left[ 0, \frac{\pi}{2} \right]$, which holds for all of your angles, then this gives

$$\alpha_{n+1} = \frac{\pi - \alpha_n}{2}.$$

It's not hard to see that this generates all of the angles you've listed, starting from $\alpha_0 = \frac{\pi}{4}$, and that this sequence converges to the unique fixed point $\alpha_{\infty} = \frac{\pi}{3}$ of the map $\alpha \mapsto \frac{\pi - \alpha}{2}$. We have the explicit closed form

$$\alpha_n = \frac{\pi}{3} - \frac{\pi}{12} \left( - \frac{1}{2} \right)^n$$

which can be proven by rewriting the recurrence relation as $\alpha_{n+1} - \frac{\pi}{3} = \left( \alpha_n - \frac{\pi}{3} \right) \left( - \frac{1}{2} \right)$. Adding these two fractions gives

$$\alpha_n = \left( \frac{2^{n+2} - (-1)^n}{3 \cdot 2^{n+2}} \right) \pi$$

and then we note that $2^{n+2} \equiv (-1)^{n+2} \equiv (-1)^n \bmod 3$ so the numerator is always divisible by $3$, and dividing it by $3$ gives $\frac{2^{n+2} - (-1)^n}{3}$ which is the closest integer to $\frac{2^{n+2}}{3}$ as you observed.

Answer 2

Let $\{a_n\}$ be a sequence of consecutive nested square roots of $2$, that is $a_n=\sqrt{2\pm a_{n-1}}$ for each $n\ge n_0$. Assuming $a_{n-1}=2\cos x_{n-1}$ for $x_{n-1}\in [0,\pi]$, we obtain $a_n=2\cos x_n$, where $x_n=x_{n-1}/2$, if $a_n=\sqrt{2+a_{n-1}}$ and $x_n=(\pi-x_{n-1})/2$, if $a_n=\sqrt{2-a_{n-1}}$. This way the sequence of signs transforms to a sequence of angles. The latter looks close to a geometric series only in special cases. When all signs are “$+$” then $\{x_n\}$ is a geometric series. When all signs are “$-$ then $(\pi/3-x_n)=-(\pi/3-x_{n-1})/2$. That is, $\{\pi/3-x_n\}$ is a geometric series.