Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

up vote 4 down vote accepted

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 $$\begin{eqnarray*} f({\bf x}) &=& \left. \sum_{n=0}^\infty \frac{1}{n!} [({\bf x}-{\bf a})\cdot \nabla_{{\bf x'}}]^n f({\bf x'}) \right|_{{\bf x'}={\bf a}} \\ &=& \left. \exp[({\bf x}-{\bf a}) \cdot \nabla_{{\bf x'}}] f({\bf x'}) \right|_{{\bf x'}={\bf a}}. \end{eqnarray*}$$

share|improve this answer
    
how do you get around the fact that $(x-a)$ does not commute with $\nabla$? –  orlandpm Jan 14 '13 at 6:13
2  
@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
1  
Yeah, got it now. Didn't see what was going on with the $x'$ at first glance –  orlandpm Jan 14 '13 at 23:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.