Taylor Expansion of e^{i \vec{k} \cdot \vec{r}}

In summary, the conversation discusses the general formula for Taylor expansion and whether it can be applied to e^{i \vec{k} \cdot \vec{r}}. It is determined that the expression is not in the correct form for Taylor expansion and the possibility of expanding it for specific values of the variables is considered. It is suggested that expanding ex with x=ik·r would be a possible approach.
  • #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?
 
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  • #2
latentcorpse said:
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
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
latentcorpse said:
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
latentcorpse said:
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
 

1. What is the Taylor Expansion of e^{i \vec{k} \cdot \vec{r}}?

The Taylor Expansion of e^{i \vec{k} \cdot \vec{r}} is a mathematical series that represents a complex exponential function with a vector argument. It is used to approximate the value of this function at a given point by using a sum of terms with increasing powers of the vector argument.

2. How is the Taylor Expansion of e^{i \vec{k} \cdot \vec{r}} calculated?

The Taylor Expansion is calculated by taking derivatives of the function at a specific point and plugging them into the formula for the Taylor series approximation. The more terms you include in the series, the more accurate the approximation will be.

3. What is the significance of e^{i \vec{k} \cdot \vec{r}} in physics?

e^{i \vec{k} \cdot \vec{r}} is commonly used in physics to describe wave phenomena, such as in the study of electromagnetic fields and quantum mechanics. It is also used in Fourier analysis to break down complex signals into simpler components.

4. Can the Taylor Expansion of e^{i \vec{k} \cdot \vec{r}} be used for any value of k and r?

No, the Taylor Expansion is only valid for small values of the vector arguments k and r. If the values are too large, the approximation will become less accurate and may even diverge.

5. How does the Taylor Expansion of e^{i \vec{k} \cdot \vec{r}} relate to other mathematical concepts?

The Taylor Expansion is closely related to the Maclaurin series, which is a special case of the Taylor series where the point of expansion is 0. It is also related to the Euler's formula, which states that e^{ix} = cos(x) + i*sin(x) and is used to represent complex numbers in polar form.

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