In my previous post, I defined
$$f(\color{blue}{m},\color{red}{n})= \sqrt{2\color{blue}{-}\sqrt{2\color{blue}{-}\cdots\color{blue}{-} \sqrt{2\color{red}{+}\sqrt{2\color{red}{+}\cdots\color{red}{+} \sqrt{2\color{blue}{-}\sqrt{2\color{blue}{-}\cdots\color{blue}{-} \sqrt{2\color{red}{+}\sqrt{2\color{red}{+}\cdots\color{red}{+} \sqrt{2\color{blue}{-}\cdots}}}}}}}}}$$ where there are $\color{blue}{m}$ consecutive "minus signs", followed by $\color{red}{n}$ consecutive "plus signs", and so on.
I will list some interesting values of $f(m,n)$ where $m = (n+1)$
$$ \begin{align} f(1,0)&=2\cos\left(\frac{\pi}{3}\right)\\ f(2,1)&=2\cos\left(\frac{2\pi}{7}\right)\\ f(3,2)&=2\cos\left(\frac{4\pi}{11}\right)\\ f(4,3)&=2\cos\left(\frac{40\pi}{127}\right)\\ f(5,4)&=2\cos\left(\frac{176\pi}{513}\right)\\ \end{align} $$
$f(n+1,n) = 2\cos(\frac{2^n\times2^{n+1}+(-1)^n}{3\times2^{2n+1}+(-1)^n})\pi$
EDITOR's note: the OP will be responsible for fixing the rest of his post.
Here is the beautiful part
When $n = 0$
$$2\cos(\frac{2^n\times2^{n+1}+(-1)^n}{3\times2^{2n+1}+(-1)^n})\pi = 2\cos(\frac{\pi}{3})$$
When $n = \infty$
$$\lim_{n\to\infty} 2\cos(\frac{2^n\times2^{n+1}+(-1)^n}{3\times2^{2n+1}+(-1)^n})\pi = 2\cos(\frac{\pi}{3})$$
Therefore cosine angle for $n = 0$ and $n = \infty$ in solving above specific pattern of cyclic infinite nested square roots of 2, the result is same i.e. $2\cos\frac{\pi}{3}$
It is amazing to note that between $0$ and $\infty$ we get so many finite values of cosine angles solving cyclic infinite nested square roots of 2 of above specific pattern.
Question: Significance of such cyclic infinite nested square roots of 2 ?
Partially I got the answer for Numerator which forms specific number sequence as here
In the denominator of radian value, it is 3,7,33,127,513,... is the integer sequence, I'm not able to identify. Please help me in this regard. Thanks in advance.
