Linearly independent derivatives

Question

Let $f \in C^{\infty}(\mathbb{R};\mathbb{R})$ be bounded and strictly monotonically increasing, whose $n^{th}$-derivative does not vanish (except possibly, on a finite number of points). Then, for every $n \in \mathbb{N}$, is the set of derivatives $$ \left\{f^n\right\}_{n \in \mathbb{N}} $$ linearly independent?

Answer 1

Proof by contradiction: Let us suppose that there is a $n\in\mathbb{N}_0$ such that

$f^{(n+1)}=\sum_0^n a_k f^{(k)}$

This is a differential equation that is linear homogeneus with constant coefficients. Its solution it's a linear combination of terms like $x^ke^{ax}(\sin(bx)+\cos(bx))$, with $k\in \mathbb{N}, a,b\in\mathbb{R}$.

This contradicts either $f≠P(x)$ or $f$ being monotonous and/or bounded

Note: a nice example to the theorem is that $\arctan^{(n)}(x)$ is linearly independent. Knowing this one might ask: what is the span of this subset of $C^{\infty}$?