A proof of a criterion of vanishing of a function. Let $f\in C^{\infty}[-1,1]$, and there exists a constant $M>0$ s.t 
$$|f^{(j)}(x)|\le M \forall j \in \mathbb{Z}\forall x\in [-1,1]$$
Prove that if $f(1/k)=0$ for each $k\in \mathbb{N}$ then $f=0$.
I am thinking of using a Taylor expansion, I need somehow to show that all the coefficients vanish.
So if I take $f(1/k)=0$ I take $k\to \infty$ then $f(0)=0$ but how to do the rest of the coefficients get to vanish?
I don't see it.
 A: Your idea is correct. In fact $f$ has a Taylor expansion
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n .$$
Note we have $f(x) -  \sum_{n=0}^N\frac{f^{(n)}(0)}{n!}x^n = \frac{f^{(N+1)}(\xi)}{(N+1)!}x^{N+1}$ with some $\xi$ between $0$ and $x$, that is $\lvert f(x) -  \sum_{n=0}^N\frac{f^{(n)}(0)}{n!}x^n \rvert \le \frac{M}{(N+1)!}$ which proves convergence.
Now let us consider the complex power series $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$. It has radius of convergence $R = \infty$ because $1/R = \limsup \sqrt[n]{\lvert\frac{f^{(n)}(0)}{n!}\rvert} \le \limsup \sqrt[n]{\lvert\frac{M}{n!}\rvert} = 0$. Therefore the power series converges to a holomorphic function $F : \mathbb C \to  \mathbb C$. We have $F(x) = f(x)$ for $x \in [-1,1]$. Moreover we have $F(1/k) = 0$ for all $k \in \mathbb N$ and this implies $F = 0$ (identity theorem for holomorphic functions), hence $f = 0$.
Edited on request:
Here is a proof without using complex analysis. We shall show that all $f^{(n)} (0)= 0$ which proves $f = 0$.
We construct inductively sequences $s^n = (\xi_k^n)$, $n =0,1,2,3,\dots$, of points in $(0, 1]$ such that


*

*For each $n$, the sequence $(\xi_k^n)$ is strictly decreasing and $\lim_{k\to\infty}\xi_k^n = 0$.

*$f^{(n)}(\xi_k^n) = 0$ for all $n, k$.
This will show that $f^{(n)}(0) = \lim_{k\to\infty}f^{(n)}(\xi_k^n) = 0$.
For $n = 0$ we take $\xi_k^0 = 1/k$. Now assume that we have constructed the sequences $s^0,\dots,s^n$. To construct $s^{n+1}$, we proceed as follows.
We know that $f^{(n)}(\xi_k^n) = f^{(n)}(\xi_{k+1}^n) = 0$. By Rolle's theorem we find $\xi_k^{n+1} \in (\xi_{k+1}^n,\xi_k^n)$ such that $f^{(n+1)}(\xi_k^{n+1}) = 0$. Then $(\xi_k^{n+1})$ is the desired sequence.
A: A function satisfies your assumption on the derivatives is real analytic, and so the set of its zeroes cannot have an accumulation point in its domain.
I leave it to you to find the proofs of these classical theorems.
