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So I was interested in the following function's domain:

$$f_n(x)=\sqrt{1-\sqrt{2-\sqrt{3-{\sqrt{...-\sqrt{n-x}}}}}}$$

So I started just looking at small values of $n$ and seeing the lengths of the domains and the actual domains themselves.

$$\operatorname{Dom{[f_1(x)]}}=(-\infty,1]\rightarrow |\operatorname{Dom{[f_1(x)]}}|=\infty$$

$$\operatorname{Dom{[f_2(x)]}}=[1,2]\rightarrow |\operatorname{Dom{[f_2(x)]}}|=1$$

$$\operatorname{Dom{[f_3(x)]}}=[-1,2]\rightarrow |\operatorname{Dom{[f_3(x)]}}|=3$$

$$\operatorname{Dom{[f_4(x)]}}=[0,4]\rightarrow |\operatorname{Dom{[f_4(x)]}}|=4$$

$$\operatorname{Dom{[f_5(x)]}}=[-11,5]\rightarrow |\operatorname{Dom{[f_5(x)]}}|=16$$

$$\operatorname{Dom{[f_6(x)]}}=[-19,6]\rightarrow |\operatorname{Dom{[f_6(x)]}}|=25$$

$$\operatorname{Dom{[f_7(x)]}}=[-29,7]\rightarrow |\operatorname{Dom{[f_7(x)]}}|=36$$

$$\operatorname{Dom{[f_8(x)]}}=[-41,8]\rightarrow |\operatorname{Dom{[f_8(x)]}}|=49$$

$$\operatorname{Dom{[f_9(x)]}}=[-55,9]\rightarrow |\operatorname{Dom{[f_9(x)]}}|=64$$

So outside the first couple of upper bound values, it seems then that the upper bounds of the domain of $f_n(x)$ is $n$ itself while the length of the domain seems to be $(n-1)^2$.

I looked up the sequence of upper bounds in the OEIS and there was no pattern associated with it and so I wondered if there was a closed form for the upper bound of $f_n(x)$ and same for the length. Looking up the sequence $\{1,3,4,16,25,36,49,64,...\}$ in OEIS gives no resultant pattern, and again was wondering if there is a closed form expression for the length of the domain of $f_n(x)$.

Is there a better approach to determining a sequence of upper and lower bounds for this function and is there a more rigorous way to determine exactly the domain of the $n$th function?

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EDIT: There was an error in my original solution, I incorrectly said the Domain was (for $n \ge 5$)

$$\text{Dom}[f_n(x)] = [n-n^2,n]$$

when the right answer is

$$\text{Dom}[f_n(x)] = [n-(n-1)^2,n]$$

It doesn't change the nature of the proof, but I've corrected it below.


Once can prove with mathematical induction that

$$\text{Dom}[f_n(x)] = [n-(n-1)^2,n]$$

holds for $n\ge 5$.

They key idea is the recursion

$$f_{n+1}(x) = f_n(\sqrt{n+1-x}).$$

That means in order to determine $\text{Dom}[f_{n+1}(x)]$, we must determine when $\sqrt{n+1-x}$ is defined and when $\sqrt{n+1-x} \in \text{Dom}[f_{n}(x)]$.

That means

$$\text{Dom}[f_{n+1}(x)] = (-\infty,n+1] \cap \{x: \sqrt{n+1-x} \in \text{Dom}[f_{n}(x)]\}$$

This formula, starting with the known $\text{Dom}[f_1]=(-\infty,1]$, generates the sequence as given in the question. It is easy to check by hand that the given sequence is correct for $n=1,2,3,4$ and $5$. This proves the start of the inductive process ($n=5$).

Now to the inductive step: Assume the forumula is correct for $n=k$. We now know that

$$\text{Dom}[f_{k+1}(x)] = (-\infty,k+1] \cap \{x: \sqrt{k+1-x} \in \text{Dom}[f_k(x)]\}$$

By the induction hypothesis $\text{Dom}[f_k(x)]=[k-(k-1)^2,k]$. Since $k\ge 5$, we have $k-(k-1)^2 \le 0$. That means $\sqrt{k+1-x} \in [k-(k-1)^2,k]$ is equivalent to

$$\sqrt{k+1-x} \le k$$

which is equivalent to

$$ 0 \le k+1-x \le k^2$$

which is equivalent to

$$(k+1)-(k+1-1)^2= -k^2+k+1 \le x \le k+1$$

Since $[-k^2+k+1,k+1] \subset (-\infty,k+1]$, we proved what we wanted to prove in the induction step:

$$\text{Dom}[f_{k+1}(x)]=[(k+1)-k^2,k+1]$$

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The straightforward approach is to let $x_0 = x$ and $x_{k+1} = \sqrt{n-k - x_k}$ for all $0 \le k < n$. Then $x_n = f_n(x)$ is the desired quantity. It is defined if and only if $x_k \le n-k$ for all $k \ge 0$. By the recursion, all but the first case are equivalent to $\sqrt{n-k - x_k} \le n-k-1$ i.e. $$x_k \ge n-k - (n-k-1)^2 \tag{*}$$ for all $k \ge 0$. In particular, $n - (n-1)^2 \le x \le n$ for $k = 0$.

It remains to show that if $x$ is in this range, then every $x_k$ satisfies $(*)$. Can you go from here? I've got to run, but I'll be back later to update my answer if necessary.

Update: This actually doesn't seem to be very straightforward. Ingix's solution is much better!

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