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Taylor expansion

  • #1
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How do you Taylor expand [itex]e^{i \vec{k} \cdot \vec{r}}[/itex]

the general formula is [itex]\phi(\vec{r}+\vec{a})=\sum_{n=0}^{\infty} \frac{1}{n!} (\vec{a} \cdot \nabla)^n \phi(\vec{a})[/itex]

but [itex]\vec{k} \cdot \vec{r}[/itex] isn't of the form [itex]\vec{r}+\vec{a}[/itex] is it?
 

Answers and Replies

  • #2
144
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How do you Taylor expand [itex]e^{i \vec{k} \cdot \vec{r}}[/itex]

the general formula is [itex]\phi(\vec{r}+\vec{a})=\sum_{n=0}^{\infty} \frac{1}{n!} (\vec{a} \cdot \nabla)^n \phi(\vec{a})[/itex]

but [itex]\vec{k} \cdot \vec{r}[/itex] isn't of the form [itex]\vec{r}+\vec{a}[/itex] is it?
I can only answer your second question. No, as far as I know, it isn't. In the former case (the scalar product) you multiply the corresponding vector components and in the latter you add them.
 
  • #3
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in fact k.r is just going to be a scalar so there is in fact no variable in that expression - how can we taylor expand it at all?
 
  • #4
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in fact k.r is just going to be a scalar so there is in fact no variable in that expression - how can we taylor expand it at all?
Is this for a specific k and a specific r? If it really is a constant, then there is no reason to Taylor expand in the first place. However, the same could be said for exp(x), if we are only talking about some specific value of x. k and or r could still be variables here, even though their dot product has a constant value, for specific k and specific r. You need to know what variable(s) you are expanding wrt first.
 
  • #5
Redbelly98
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How do you Taylor expand [itex]e^{i \vec{k} \cdot \vec{r}}[/itex]
When I read that, I assume you just expand ex in the usual fashion, where x=ik·r
 

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