Show that $f^{(n)}(0)=0$ for $n=0,1,2, \dots$ 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..
 A: Assume $f$ is  differentiable and  there are $a_n \rightarrow 0$ such that $f(a_n)=0$ then appying the mean value to $[a_n, a_{n+1}]$ we get 
$b_n \in [a_n, a_{n+1}]$  
$$f(a_{n+1})-f(a_n)=f^{\prime}(b_n)(a_{n+1}-a_n)$$
 so $f^{\prime}(b_n)=0$. 
This should solve the problem.
A: 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.
A: A straightforward approach using Rolle's Theorem:
Let $n\in\mathbb{N}^*$. Since $f$ is differentiable on $\left(\dfrac1{n+1},\dfrac1n\right)$ and continuous on $\left[\dfrac1{n+1},\dfrac1n\right]$, there exists $c_n\in\left(\dfrac1{n+1},\dfrac1n\right)$ such that $f'(c_n)=0$.
This shows that there exists a sequence $(c_n)_{n\in\mathbb{N}^*}$ that is decreasing and $\lim\limits_{n\to+\infty}c_n=0$ such that for all $n\in\mathbb{N}^*$, $f'(c_n)=0$. Since $f'$ is continuous at $0$, we must have:
$$f'(0)=\lim_{n\to+\infty}f'(c_n)=0.$$
Now repeat the process.
