# 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?

-
Sure, this is a Taylor series expansion, just in terms of an operator. In $\mathbb R^n$, you would use $a \cdot \nabla$ for some direction vector $a$. – Muphrid Jan 14 '13 at 4:53
Hmm, the log of the Pascal-matrix can be seen as derivative operator. Then in turn its exponential (the pascal-matrix itself) might be th answer to your question (I'm not sure as I'm still not really familiar with the notation/concepts as it occurs in your question) – Gottfried Helms Jan 14 '13 at 7:20

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 $$\begin{eqnarray*} f({\bf x}) &=& \left. \sum_{n=0}^\infty \frac{1}{n!} [({\bf x}-{\bf a})\cdot \nabla_{{\bf x'}}]^n f({\bf x'}) \right|_{{\bf x'}={\bf a}} \\ &=& \left. \exp[({\bf x}-{\bf a}) \cdot \nabla_{{\bf x'}}] f({\bf x'}) \right|_{{\bf x'}={\bf a}}. \end{eqnarray*}$$
how do you get around the fact that $(x-a)$ does not commute with $\nabla$? – orlandpm Jan 14 '13 at 6:13
@orlandpm: $(x-a)$ does commute with $\nabla_{x'}$! (don't forget to look at the subscript) – Fabian Jan 14 '13 at 10:16
@orlandpm: As Fabian has helpfully indicated, the del operator is with respect to the primed coordinates so $({\bf x}-{\bf a})$ and $\nabla_{{\bf x'}}$ commute. – user26872 Jan 14 '13 at 23:10
Yeah, got it now. Didn't see what was going on with the $x'$ at first glance – orlandpm Jan 14 '13 at 23:28