Taylor expansion of 1/distance

[mertcan]

Hi, I would like to express that r and r' are vectors in the attachment and let's say that r is observer distance vector r' is source distance vector. By the way I know this is taylor expansion (for instance if there was only x component (scalar form) I would not any ask question ). But I do not understand when we include vectors, why there is a dot product of r and r'?? Also I know the taylor expansion of multiple scalar variables but I do not know how to derive taylor expansion of multiple vector variables ( I think 1/distance is a scalar function comprised of 2 different vectors here )?

 

[Khashishi]

You can write the vectors in component form, although this isn't a very elegant solution. It's very straightforward though.
$\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{1}{\sqrt{\sum_i(r_i - r'_i)^2}}$
If you do a taylor expansion around $r_x'=r_y'=r_z'=0$
you get $\frac{1}{r} + \sum_i{\left[\partial_i \left(\frac{1}{\sqrt{\sum_j (\mathbf{r}-\mathbf{r}')^2}} \right) r_i' \right] }$ (partial operates only on the next term in paren)
$= \frac{1}{r} + \sum_i{\left[\frac{-1}{2} \left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right)^{-3/2} \partial_i \left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right) r_i' \right] }$ (chain rule)
$= \frac{1}{r} + \sum_i{\left[\frac{-1}{2} \left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right)^{-3/2} (-2) \left(r_i-r_i' \right) r_i' \right] }$ (chain rule)
The derivative was supposed to be evaluated at r' = 0, so it becomes
$= \frac{1}{r} + \sum_i{\left[\left(\sum_j (\mathbf{r}-\mathbf{r}')^2 \right)^{-3/2} \left(r_i \right) r_i' \right] }$ (chain rule)
Let $r=\sqrt{\sum_i(r_i - r'_i)^2}$
$= \frac{1}{r} + \frac{1}{r^3} r_i r_i'$
That's the same as $\frac{1}{r} + \frac{1}{r^3} \mathbf{r} \cdot \mathbf{r}_i'$