Let $f_{1}(x)=\sqrt{1-x}$, and for integers $n \geq 2$, let $$f_{n}(x)=f_{n-1}\left(\sqrt{n^2 - x}\right).$$ Let $N$ be the largest value of $n$ for which the domain of $f_n$ is nonempty. For this value of $N,$ the domain of $f_N$ consists of a single point $\{c\}.$ Compute $c.$
When I first saw this I thought there would be a trivial recursion, solvable by listing first few terms, finding a pattern and/or manipulating the equation into a nicer form. I then realized this would be impossible. Nothing is really popping out of me as a possible, nice method to use. Please supply a solution. Thanks.
$$f_1(x) = \sqrt{1-x}$$
so the domain of $f_1$ is $(-\infty, 1]$. Then $$f_2(x) = \sqrt{1-\sqrt{4-x}}$$
so the domain of $f_2$ is the set of all the $x$ satisfying $\sqrt{4-x} \leq 1$, i.e. $0 \leq 4-x \leq 1$, i.e. $3 \leq x \leq 4$. Then $$f_3(x) = \sqrt{1-\sqrt{4-\sqrt{9-x}}}$$
so the domain of $f_3$ is the set of all the $x$ satisfying $3 \leq \sqrt{9-x} \leq 4$, i.e. $9 \leq 9-x \leq 16$, i.e. $-7 \leq x \leq 0$. Then $$f_4(x) = \sqrt{1-\sqrt{4-\sqrt{9-\sqrt{16-x}}}}$$
so the domain of $f_4$ is the set of all the $x$ satisfying $-7 \leq \sqrt{16-x} \leq 0$, i.e. $\sqrt{16-x} =0$, i.e. $x=16$. Then $$f_5(x) = \sqrt{1-\sqrt{4-\sqrt{9-\sqrt{16-\sqrt{25-x}}}}}$$ so the domain of $f_5$ is the set of all the $x$ satisfying $\sqrt{25-x}=16$, i.e. $x=-231$. Then $$f_6(x) = \sqrt{1-\sqrt{4-\sqrt{9-\sqrt{16-\sqrt{25-\sqrt{36-x}}}}}}$$ the domain of $f_6$ is the set of all the $x$ satisfying $\sqrt{36-x}=-231$. So it is empty.
So you have that $N=5$, and $c=-231$.