Solution
= Solution
\b[The <iterated function> is $f^{(n)}(x)=\sqrt{x^2+n}$ for every positive integer $n$.] The exponent here denotes repeated <function composition>, not an ordinary power or a <derivative>.
For the base case $n=1$, the formula is the definition. If it holds at $n$, then
$$
f^{(n+1)}(x)=f(f^{(n)}(x))
=\sqrt{\left(\sqrt{x^2+n}\right)^2+1}
=\sqrt{x^2+n+1}.
$$
This proves the claim by <mathematical induction>, for every real input $x$.