Q:

To prove f'(0)=f''(0)=0.  Given F is twice differential

Let $f:\mathbb{R}\to\mathbb{R}$ is twice differential function such that  $f(\frac{1}{2^n})=0 \; \forall n \in \mathbb {N} $.show $f'(0)=f''(0)=0$
since $f$ is continous ,this gives me $f(0)=0$,also $f'(0)$ exist,and $\frac{1}{2^n}\to 0$.using this i find $f'(0)=0$.but having trouble in proving $f''(0)=0$.any help ? 

A:

Rolle's Theorem gives us a sequence $(x_n)$ with $x_n \to 0$ and $f'(x_n)=0$.
This shows (knowing $f''(0)$ exists) $$f''(0) = \lim_{n \to \infty} \frac{f'(x_n)-f'(0)}{x_n-0} = \lim_{n \to \infty} \frac{0}{x_n} = 0.$$
Additional note: If $f$ is known to be $n$ times differentiable, we can iterate this argument and deduce $f^{(d)}(0)=0$ for all $d \leq n$. In particular, if $f$ is analytic, we deduce $f \equiv 0$ (which is of course the Identity theorem).