If $f \in C^{\infty}$ and $f^{(k)}(0)=0$ for all integers $k \ge 0$, then $f \equiv 0$. I thought this was true since,
$f(x)=f(0)+f'(0)x+f''(0) \frac {x^2}{2!} + \dots$
But I am wrong.
Where did I make mistake?
 A: Taylor's theorem : Let $f : [a,b] \rightarrow \mathbb{R}$ be a function of $\mathcal{C}^{n+1}[a,b]$, let $c \in [a,b]$ and $h \in \mathbb{R}$ such that $c+h \in [a,b]$. Then
$$ f(c+h) = \sum\limits_{k=0}^n \frac{f^{(n)}(c)}{k!}h^k + R_{n+1}(h),$$
where
$$ \big| R_{n+1}(h)\big| \leq \sup\limits_{x \in [c,c+h]} \big| f^{(n+1)}(x)\big| \cdot \frac{|h|^{n+1}}{(n+1)!}.$$
Interesting case : Consider the Taylor series of 
$$f(x) := \begin{cases}e^{-\frac{1}{x^2}}~~x \neq 0; \\ 0 ~~~~~~~~ x = 0; \end{cases}$$
For all $n > 0$ we have $f^{(n)}(0) = 0$. So
$$\sum\limits_{k=0}^\infty \frac{f^{(k)}(0)}{k!}h^k =0.$$
Also
$$ f(h) = 0 + R_{n+1}(h).$$
Hence the function is somehow "included in the error term" which does NOT converge to $0$.
Moral : A function is not necessarily equal to its Taylor series.
A: Your relation is true for x=0, where it just says 0=0, but not necessarily true for any other x
functions for which that relation is true in a neighborhood of some x are called (real) analytic at x and that is a much stronger property
