I have been looking into analytic continuation of regular recurrence relations into the negative, real, and complex domains ($\mathbb{N} \mapsto \mathbb{Z}, \mathbb{R},\mathbb{C}$). In doing so, we’ve come upon a well-known $\sqrt{2}$ nested radical along with solution of the recurrence. The present solutions of the nested radical agree with other known solutions, but go much further because of the aforementioned analytic continuation. Here is what we did.
We begin with the recurrence relation
$$f_n=f_{n-1}^2-2$$
With two known solutions
$$f_n=\cos(\cos^{-1}(f_0/2)\cdot2^n)$$ $$f_n=b^{2^n}+b^{-2^n}$$
where $b=\bigg(f_0+\sqrt{f_0^2-4}\bigg)/2$. You can think of this as a Binet-type solution. (Actually, there is a third solution where you would substitute $\cosh$ for $\cos$.) All of these solutions are, in fact, identical. And just as an aside, the are two cases, for $f_0=3,7$, for which $b=\varphi^2,\varphi^4$, respectively, where $\varphi$ is the golden ratio.
Now we seek to extend this to negative indices to determine if the above solutions are valid when substituting $-n$ for $n$. Thus, we write
$$ \begin{align} &f_{n-1}=\sqrt{2+f_n}\\ &f_{-1}=\sqrt{2+f_0}\\ &f_{-2}=\sqrt{2+f_1}=\sqrt{2+\sqrt{2+f+0}}\\ &\vdots\\ &f_{-n}=\sqrt{2+\sqrt{2+\sqrt{2\ldots+\sqrt{2+f_0}}}}\\ \end{align} $$
This is obviously the well-known $\sqrt{2}$ nested radical.
We can get a closed form solution for $f_{-n}$ as follows. Let $f_n=2\cos g_n$ and utilize the trigonometric identity
$$\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}$$
to find $$f_{-n}=\cos(\cos^{-1}(f_0/2)\cdot2^{-n})$$
This is functionally identical to the solution for $f_n$. We could also show that
$$f_{-n}=b^{2^{-n}}+b^{-2^{-n}}$$
Now, in the literature here and/or here, it is stated that the nested radical solution for $f_0\in[-2,2]$ is the cosine solution and for $|f_0|\ge 2$ it is the Binet solution. It is obvious that they are avoiding the arccosine of a number greater than unity or a imaginary root. We don’t think that is necessary. In fact, we posit that for any $z\in\mathbb{C}$
$$R_n(z)= \sqrt{2+\sqrt{2+\sqrt{2\ldots+\sqrt{2+z}}}}=\left\{ \begin{array} \\\cos(\cos^{-1}(z/2)\cdot2^{-n})\\ b^{2^{-n}}+b^{-2^{-n}} \end{array}\right. $$
where $b=\bigg(z+\sqrt{z^2-4}\bigg)/2$.
This solution has been tested vigorously with randomly chosen positive and negative integers, real and complex values of $z$ So, the question is, how can we prove this rigorously?
Of course, this also validates the analytic continuation of the recurrence. We have demonstrated analytic continuation of many recurrences (i.e., $\mathbb{N}\mapsto\mathbb{C} $) with complete success. But this is still a work in progress. All suggestions and comments are welcome.