**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 
$$
(\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[)
(\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))\qquad (1)
$$
(ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of eq. (1) 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).  

(iii) Even if the LHS of (1) converges for all $x,t\in \mathbb{R}$, $f$ need not be analytic. Consider the following function (classic in theory of distributions)  
$$
f(x)=0\mbox{ if } x\notin ]-1,1[\mbox{ and } f(x)=e^{\frac{1}{1-x^2}} \mbox{ if } x\in ]-1,1[
$$