Xrhrft

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?

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 $nge 5$.

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 $kge 5$, 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:

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

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!