Question 112
Let $f^{(n)}$ denote the $n$-fold iterate of a function, i.e.
$$
f^{(1)}(x) = f(x), \quad f^{(2)}(x) = f(f(x)), \quad f^{(3)}(x) = f(f(f(x))), \dots
$$
and let $f : [0, \infty) \to [0, \infty)$ be such that
$$
f^{(2)}(x) = s f(x) + t x, \quad x \ge 0.
$$
Assuming only that $s, t \in \mathbb{R}$ are non-zero, show that if
$$
f^{(n+m)}(p) = f^{(m)}(p), \quad \text{for some } n, m \in \mathbb{N}, p \in [0, \infty),
$$
then $f(p) = p$.
$$
f^{(1)}(x) = f(x), \quad f^{(2)}(x) = f(f(x)), \quad f^{(3)}(x) = f(f(f(x))), \dots
$$
and let $f : [0, \infty) \to [0, \infty)$ be such that
$$
f^{(2)}(x) = s f(x) + t x, \quad x \ge 0.
$$
Assuming only that $s, t \in \mathbb{R}$ are non-zero, show that if
$$
f^{(n+m)}(p) = f^{(m)}(p), \quad \text{for some } n, m \in \mathbb{N}, p \in [0, \infty),
$$
then $f(p) = p$.