<s>[INST] When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:

Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?
 [/INST] This problem is equivalent to proving that there exists some real number $r$ such that 
\begin{equation}
\lim_{n \to \infty} f_n(r) = 0 
\end{equation}
where $f_n = g_n^0(x)$ and $g^k_n$ is defined by the recurrence 
\begin{equation}
g_n^k(x) = 
\begin{cases} \sqrt {x+k} & \text{if }n=1 \\ 
\sqrt{x + k +  g^{k+1}_{n-1}(x)}  & \text{o.w.} \\ 
\end{cases}. 
\end{equation}
Therefore, by definition, we have 
\begin{equation}
f_n(x) = 
\begin{cases} \sqrt {x} & \text{if }n=1 \\ 
\sqrt{x +  g^{1}_{n-1}(x)}  & \text{o.w.} \\ 
\end{cases},
\end{equation}
which after proving the following:
\begin{equation}
 g_n^1(x)  = g_n^0(x+1) ,
\end{equation} 
 by a simple induction, gives us that
\begin{equation}
f_n(x) = 
\begin{cases} \sqrt {x} & \text{if }n=1 \\ 
\sqrt{x +  f_{n-1}(x+1)}  & \text{o.w.} \\ 
\end{cases}.
\end{equation}
Let us suppose that $r_n - f_{n-1}(r_n+1) =  \epsilon_n$ and that $\lim_{n \to \infty} \epsilon_n  = 0 $ then we have that 
\begin{equation}
\lim_{n \to \infty} f_{n}(r_n) = 0.
\end{equation} 
One possible route to find this is to consider the growth of the sequence
\begin{equation}
\delta_n(x)  =  f_n(x+1) - f_n(x). 
\end{equation}
Therefore given a good approximation to $r_n$, an educated guess for $r_{n+1} $ should be some
\begin{equation}
r_{n+1} \in [r_{n} - 1, r_{n}],
\end{equation}
this is because
\begin{equation}
f_{n+1}^2(r_{n} - \epsilon) = r_{n} - \epsilon + f_n(r_{n} - \epsilon) +\delta_n(r_{n} - \epsilon)  
\end{equation}
and 
\begin{equation}
\delta_n(r_{n} - \epsilon) =   f_n(r_{n} +1- \epsilon) - f_n(r_{n} - \epsilon) =  f_n(r_{n} +(1-\epsilon)) - f_n(r_{n} - \epsilon)
\end{equation}
gives us that we should be looking for some 
\begin{equation}
-\delta_n(r_{n} - \epsilon)  \approx r_{n} - \epsilon + f_n(r_{n} - \epsilon)
\end{equation}
which is equivalent to saying 
\begin{equation}
-\delta_n(r_{n} - \epsilon)  - f_n(r_{n} - \epsilon) \approx r_{n} - \epsilon 
\end{equation}
or 
\begin{equation}
-f_n(r_{n} + (1- \epsilon))  \approx r_{n} - \epsilon, 
\end{equation}
or 
\begin{equation}
f_n(r_{n} + (1- \epsilon))  \approx \epsilon - r_n .
\end{equation}
which is a "sort-of" fixed point equation. Now we are getting somewhere! 
We now have a method to recursively approximate solutions from previous solutions; i.e.
\begin{equation}
f_n(r_{n} + (1- \epsilon)) =   \epsilon -r_{n}  \implies f_{n+1}(r_{n} - \epsilon) = 0  . 
\end{equation}
Let us compute $r_n$ for small values to verify the solution. 


*

*$f_1(x) = \sqrt x $ is the base case and it has the obvious solution $r_1 = 0$.

*Since $f_1(0) = 0$ and $\sqrt{1- \frac{\sqrt5  -1}{2}} = \frac{\sqrt5  -1}{2} $ we have an exact solution in the form of $r_1 = 0 $, $\epsilon = \frac{\sqrt5  -1}{2} $, and $r_2 = r_1 - \epsilon $; i.e. $f_2 (r_1 - \epsilon) = f_2\left(\frac{ 1-\sqrt5 }{2}\right) = 0$.

*Let $\epsilon = \frac{1}{2} (3 - \sqrt{5})$ then one can easily check that both $f_2(r_2 +1 - \epsilon) = \epsilon - r_2 $ and $f_3(r_2  - \epsilon) =f_3(-1) = 0 . $

*Let $\epsilon = 0.153761 $ then one can easily check that both $f_2(r_3 + 1 - \epsilon) \approx \epsilon - r_3 $ and $f_4(r_3  - \epsilon) = f_3(-1.153761) \approx 0 . $
In conclusion, we have the following recursion relation for the roots

\begin{equation}
r_{n+1} = 
\begin{cases} 0 & \text{if }n=1 \\ 
 r_n - \epsilon  & \text{if } f_n(r_n +1 - \epsilon) =  r_n - \epsilon  \\ 
\end{cases};
\end{equation}

however, it is possible that a recursive analysis of the following two functions 


*

*$f'_n(x)$

*$\delta_n(x)$
could give a better recursive bound $b_n  < b_{n-1}<1 $ on the intervals
\begin{equation}
r_{n+1} \in [r_n-b, r_n]
\end{equation}
</s>