Prove the following lemma: The function $$r:x \mapsto \begin{cases} e^{-{1\over x}}, & \text{if $x>0$} \\ 0, & \text{if $x \le 0$} \end{cases}$$ is $C^{\infty}$ (and x=0),that has necessarily zero Taylor series in $x=0$, i.e. $$(r^{j}(0)=0))$$ for all $j$.
Any ideas how to prove this lemma?
Your $r(x)$ is the Wikipedia example of an Non-analytic smooth function.
For $x>0$ we know, that $r$ is $C^\infty$. The interesting part is $x=0$. Let us therefore consider the derivatives of $\exp(-1/x)$. We know, that \begin{align} \frac{d}{dx}e^{-{1}/{x}} = x^{-2}e^{-1/x}= \frac{1}{M_1(x)} e^{-1/x} \end{align} where $M_1$ is some monomial. Repeating this process, you should be able to show, that \begin{align} \frac{d^n}{dx^n}e^{-{1}/{x}} = \sum_k \frac{1}{M_k(x)} e^{-1/x} \end{align} Next, you have to show that ${M_k(x)} e^{-1/x}\rightarrow 0$ for $x \rightarrow 0$ independent of the order of the monomial, this is equivalent to showing, that $\exp(-x)x^k\rightarrow 0$ for $x \rightarrow \infty$.
Therefore, $r$ is $C^\infty$ with $\frac{d^j}{dx^j}r(0) = 0$.