Let $f \in C^{\infty}(\mathbb{R}, \mathbb{R})$ such that $f(0) = \underset{+\infty}{\lim} f$.
Show that there is a sequence $(a_{n})$ such that $\forall n \geq 1$, $f^{(n)}(a_{n}) = 0$.
I get the idea for n = 1, but after that, i can’t really find the answer ...
Thank you
Assume for the sake of contradiction that the statement is false. Because of OP’s post, there exists some $n \geq 2$ minimal such that $f^{(n)}$ has no root. Then we can assume it is non-negative. As $f^{(n-1)}$ vanishes and is increasing, there is a real number $y$ such that $f^{(n-1)}(y) > 0$.
It follows, by Taylor with integral remainder, as $f^{(n)} \geq 0$, that for $x \geq y$, $f(x) \geq \sum_{k=0}^{n-1}{\frac{f^{(k)}(y)}{k!}(x-y)^k}$. But the RHS goes to infinity as $y$ goes to infinity, which contradicts the assumption.