TITLE: Show that $f^{(n)}(0)=0$ for $n=0,1,2, \dots$
QUESTION [15 upvotes]: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ an infinitely many times differentiable function and $f(\frac{1}{n})=0$ for each $n \in \mathbb{N}$. Show that $f^{(n)}(0)=0$ for $n=0,1,2, \dots$
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Could you give me some hint what I could do?? I got stuck right now..

REPLY [3 votes]: I would stick with gniourf_gniourf's solution, but a neat alternative is to use the fact that $$\lim_{h\to0} \sum_{i=0}^n {n \choose i} (-1)^{n-i} f(ih) = f^{(n)}(0)$$ now take $h = 1/m!$ to get $i/m! = 1/( 1 \cdot 2 \cdots (i-1) \cdot (i+1) \cdots m)$ in the arguments of $f$. In particular $f$ vanishes at $i/m!$.
Thus $f^{(n)}(0)$ is the limit of the zero sequence, and is zero itself.