**Taylor Formula and displacement operator:** I (too often) see in papers (mathematical physics but a recent paper by mathematicians also) the statement 

> Let $D=\frac{d}{dx}$ be the derivation operator. Then, for all $f\in C^\infty(\mathbb{R})$, 
$$
e^{tD}[f](x)=f(x+t)
$$

which is false (take any $\phi\in C^\infty(\mathbb{R})$ with compact support, for instance).
 
**Remarks** (i) The function $f\in C^\infty(\mathbb{R})$ is analytic over $\mathbb{R}$ iff 
$$
(\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[)
(\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))
$$
(ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of the formula do not converge otherwise $f$ would be the restriction of an entire function (which e.g. $\frac{1}{1+x^2}$ is not, for example).