$f,g$ analytic on $I\subset \mathbb{R}$. If exist $a\in I$ such that $f=g$ and $f^{(n)}=g^{(n)}$then we have $f(x)=g(x)$ for every $x \in I$. 
Problem
Let $f,g$ analytic on an open interval $I\subset \mathbb{R}$. If exist
some $a\in I$ such that $f(a)=g(a)$ and $f^{(n)}(a)=g^{(n)}(a)$ for
all $n \in \mathbb{N}$ then we have $f(x)=g(x)$ for every $x \in I$.
Show this is no longer true if you just suppose $f,g$ class $C^\infty
> $

Well, my idea is just write these series and show that if all terms are the same, these functions well be the same as well. But my main problem is understanding this fundamental difference between analytic or $ C^\infty $.
The core of this problem is being able to write the series of these functions?
 A: For $C^{\infty}$ but non analytic functions, there is a standard example, that is
$$f(x)=\begin{cases}e^{-\frac{1}{x^2}}\ \ &x\neq 0\\
0\ \ &x=0\end{cases}\\
g(x)=0$$
It is a good exercise to prove that $f(x)$ is $C^\infty$ and that $\forall_ n\ f^{(n)}(0)=0$. Thus we have two different infinitely differentiable  function, equal and with n-th derivative equal in $0$.
Note that, obviously, $f(x)\neq \sum \frac{f^{(n)}(0)}{n!}x^n=0$
This is the fundamental difference between being smooth and being analytic: smooth function can be approximated with an error $O(x^n)$ for any $n$, but this does not mean that the error in the approximation tends to $0$. That condition is only satisfied by analytic functions. 
In our example, the Taylor expantion has an error $\frac{f^{(n+1)}(\zeta)}{(n+1)!}\zeta^{(n+1)}$, and this term does not go to zero, since $f^{(n)}$  grows bigger for every $n$
As a note, it might be useful to observe that actually we have our example is "quite nice", that is our function is non analytic only in $0$. There are some functions (like the Fabius function) which are smooth but nowhere analytic. The math behind those example is quite hard though.
