taylor's formula in $\mathbb{R}^n$ as exponential of derivative operator I hope this question is not too vague.  There is some relationship between the formal operator $$e^\frac{d}{dx} = 1 + \frac{d}{dx} + \frac{1}{2!}\frac{d^2}{dx^2} + \ldots  $$
and taylor's formula.  Is there a precise relationship that can be stated here?
Specifically, I was taught a mnemonic once for taylors formula in multiple dimensions in terms of a similar exponential.  Does anybody know what I am referring to?
 A: In one dimension we have the Maclaurin series 
$$\begin{eqnarray*}
f(x) 
    &=& \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n \\
    &=& \left.\exp\left(x \frac{d}{dx'}\right)f(x')\right|_{x'=0} 
\end{eqnarray*}$$
and, more generally, the Taylor series 
$$\begin{eqnarray*}
f(x) 
    &=& \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n \\
    &=& \left.\exp\left[(x-a) \frac{d}{dx'}\right]f(x')\right|_{x'=a}.
\end{eqnarray*}$$
Likewise, in $\mathbb{R}^n$ 
$$\begin{eqnarray*}
f({\bf x}) 
    &=& \left.
\sum_{n=0}^\infty \frac{1}{n!} ({\bf x}\cdot \nabla_{{\bf x'}})^n f({\bf x'})
\right|_{{\bf x'}={\bf 0}} \\
    &=& \left.
\exp({\bf x}\cdot \nabla_{{\bf x'}}) f({\bf x'})
\right|_{{\bf x'}={\bf 0}} 
\end{eqnarray*}$$
and 
$$\begin{eqnarray*}
f({\bf x}) 
    &=& \left.
\sum_{n=0}^\infty \frac{1}{n!} [({\bf x}-{\bf a})\cdot \nabla_{{\bf x'}}]^n f({\bf x'})
\right|_{{\bf x'}={\bf a}} \\
    &=& \left.
\exp[({\bf x}-{\bf a}) \cdot \nabla_{{\bf x'}}] f({\bf x'})
\right|_{{\bf x'}={\bf a}}. 
\end{eqnarray*}$$
