# Taylor expansion of scalar fields [closed]

Starting of with electrodynamics I have to compute the taylor expansion around $$\vec{r} = 0$$ of

$$\psi (\vec{r}) = |\vec{r} - \vec{r_0}|^{\frac{3}{2}}$$ where $$\vec{r_0}$$ is a constant vector up to second order and

$$\psi(\vec{r}) = e^{i\vec{k}\vec{r}}$$ where $$\vec{k}$$ is a constant vector up to arbitrary order.

I don't have problems with multidimensional taylor expansions as long as there are no vectors involved and functions look like $$f(x,y,z) = y \cdot \sin(xz) + xz^2$$ for example. Therefore, in the cases above I feel lost.

Can someone explain how to solve the exercise?

## closed as off-topic by AccidentalFourierTransform, John Rennie, user191954, Aaron Stevens, stafusaOct 22 '18 at 22:36

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• By $e^{i\vec{k}\vec{r}}$ do you mean $e^{i\vec{k}\cdot\vec{r}}$? – probably_someone Oct 21 '18 at 12:51
• That's what I mean – offline Oct 21 '18 at 13:05

Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$$, $$\vec{k}=k_x\hat{x}+k_y\hat{y}+k_z\hat{z}$$, and $$\vec{r_0}=x_0\hat{x}+y_0\hat{y}+z_0\hat{z}$$) and write the expressions out that way, it might be clearer:

$$\psi(\vec{r})\to\psi(x,y,z)=\left(\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}\right)^{3/2}$$

$$\psi(\vec{r})\to\psi(x,y,z)=e^{i(xk_x+yk_y+zk_z)}$$

Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $$r$$, $$r_0$$, $$k$$ of each of the vectors, in which case you would get:

$$\psi(\vec{r})\to\psi(r,\theta)=\left(\sqrt{r^2+r_0^2-2rr_0\cos\theta}\right)^{3/2}$$

$$\psi(\vec{r})\to\psi(r,\theta)=e^{irk\cos\theta}$$

where $$\theta$$ is, in the first example, the angle between $$\vec{r}$$ and $$\vec{r_0}$$ (in other words, the axis of rotational symmetry of the function is located along $$\vec{r_0}$$), and is, in the second example, the angle between $$\vec{r}$$ and $$\vec{k}$$ (so that the axis of rotational symmetry is along $$\vec{k}$$).