Let $(a_n)$ be a sequence of real numbers, then exists a function $f \in C^\infty(\mathbb{R})$ with compact support such that $f^{(n)}(0)=a_n$ for all $n\in\mathbb{N}$
Since this problem belongs to the Taylor polynomial chapter, my first attempt consisted in considering the Taylor polynomial centered at $0$. We know that $f$ has compact support, then $f$ is bounded and $f(x)=0$ for all $x \in \mathbb{R}$. Using the infinitesimal formula we have that $f(0)=\sum_{n=0}^{\infty}f^{(n)}(0)\frac{(0-x_0)^n}{n!}$ or every $x_0$. So, for every iteration, we have constants $a_0, a_1, ..., a_n$ such that $a_n = \frac{f^{(n)}(0)}{n!}$. Can I just say: let $a_n = n!a_n$? Here I don't know how to keep on (if I'm in the right path at all). I would really appreciate if someone could help me or give a hint.