Expansion of $ \frac{1}{|\vec r -\vec r'|} $ I would like to show that:
$$ \frac{1}{|\vec r -\vec r'|} =\frac{1}{r} + \frac{\vec r'\cdot r'}{r^3}+\frac{3 ((\vec r \cdot \vec r)^2 -\vec r^2 \vec r'^2 )}{2r^5} +\dots$$
What I derived so far is:
$$\frac{1}{|\vec r -\vec r'|}= \sum_{n=0}^{\infty} \frac{1}{n!} (-\vec r' \cdot \nabla_r )^n \frac{1}{r} $$
The last part is however unclear to me, since I don't know how to rewrite, e.g. $(-\vec r' \cdot \nabla_r )$
 A: The notation $(-\vec r'\cdot \nabla_r)$ means "first take the gradient with respect to $r$, then do the dot product with $-\vec r'$. For example
$$ (-\vec r'\cdot \nabla_r)^2 f(r) = -\vec r' \cdot \left[ \nabla_r(-\vec r'\cdot \nabla_r f(r)) \right].$$
So let's have a look at the first few summands. Starting with $n=0$, we get just $\frac 1 r$, since $0!$ is $1$ and the operator is applied $0$ times. Now $n=1$:
$$
(-\vec r'\cdot\nabla_r)\frac 1 r = -\vec r' \cdot \left(\nabla_r \frac{1}{r}\right) = -\vec r' \cdot \left(\frac{-\vec r}{r^3}\right) = \frac{\vec r'\cdot\vec r}{r^3}.
$$
Here we used the fact that the gradient of $\frac 1 r=\frac{1}{|\vec r|}$ is $\frac{-\vec r}{r^3}$. For $n=2$ we can use what we already calculated:
$$
\frac 1 {2!} (-\vec r'\cdot \nabla_r)^2 \frac 1 r
= \frac 1 {2!} (-\vec r'\cdot \nabla_r) \frac{\vec r'\cdot\vec r}{r^3}
= \frac {-\vec r'} {2} \cdot \nabla_r\left[
\frac{\vec r'\cdot\vec r}{r^3} \right] = \dots
$$
Here you need to involve the product rule for gradients of dot products, it involves the vector gradient which is in this case the same as the Jacobian matrix. Have you done calculations like that before?
A: $\vec r' \cdot \nabla_{\vec r}$ is just a directional derivative. A good set of identities for using directional derivatives can be helpful.  In particular, for any vectors $\vec a, \vec b$,
$$\begin{align*}\vec a \cdot \nabla_{\vec r} r &= \vec a \cdot \hat r \\ \vec a \cdot \nabla_{\vec r}\vec r &= \vec a \end{align*}$$
You should also be familiar with using the chain rule:
$$\vec a \cdot \nabla_{\vec r} f(r) = [\vec a \cdot \nabla_{\vec r} r] f'(r) = \vec a \cdot \hat r f'(r)$$
These simple results should be all you need to evaluate any derivative you come across.  For example, if you need to evaluate $\vec a\cdot \nabla_\vec{r} \hat r$, it can be written as
$$\begin{align*} \vec a \cdot \nabla_{\vec r} \hat r &= \vec a \cdot \nabla_\vec{r} \frac{\vec r}{r}\end{align*}$$
You can use the product rule to separate this into two pieces:
$$\vec a \cdot \nabla_\vec{r} \frac{\vec r}{r} =\frac{1}{r}  [\vec a \cdot \nabla_\vec{r} \vec r] + \vec r [\vec a \cdot \nabla_\vec{r} \frac{1}{r} ]$$
The first term can be attacked with the second identity I've given you; the second term can be attacked using the chain rule and the first identity.
