Let, $f$ : $\mathbb{R}$$\rightarrow$$\mathbb{R}$ be a twice differentiable function such that $f$($\frac{1}{2^n}$)= $0$ $\forall$ $n$ $\in$$\mathbb{N}$. Show that, $f'(0)=0$=$f''(0)$.
Help me how to proceed. from definition of $f'$ I am unable to do anything!
More generally:
Proposition. Let $f\colon\mathbb R\to\mathbb R$ be differentiable and $x_n\to0$ a sequence in $\mathbb R\setminus \{0\}$ with $x_n\to 0$, such that $f(x_n)=0$ for all $n\in\mathbb N$.
Proof: Let $a=f'(0)$. Assume $a\ne 0$. Then with $\epsilon:=\frac{|a|}2$ there exists $\delta>0$ such that for all $x$ with $0<|x|<\delta$, we have $\left|\frac{f(x)-f(0)}{x}-a\right|<\epsilon$. By continuity of $f$, $f(0)=\lim_{n\to\infty}f(x_n)=0$, hence $f(x)\ne0$. For $n$ big enoough, we have $0<|x_n|<\delta$, contradiction. $_\square$.
The proprosition gives us immediately $f'(0)=0$. By Rolle, there are $x_n\in(\frac1{2^{n+1}},\frac1{2^n})$ where $f'(x_n)=0$, so the same argument applied to $f''(0)$.
In fact, we see that if $f$ is $k$ times differentiable, then $f^{(k)}(0)=0$.
We have that $$\lim_{n \to +\infty} \frac{1}{2^n}=0$$
Since $f$ is differentiable, it is continuous, so $$f(0)=\lim_{x \to 0} f(x)=\lim_{n \to +\infty} f(\frac{1}{2^n})=0$$.
On the other hand, $$f'(0)= \lim_{h \to 0}\frac{f(h)-f(0)}{h}$$ exists, so $$f'(0)=\lim_{n \to +\infty} \frac{f(\frac{1}{2^n})-0}{\frac{1}{2^n}}=0$$
