My problem is the following, I have ${1}/{R}={1}/{|\vec{r}-\vec{r}'(t)|}$ which can be expanded as
$1/R=1/r+\vec{r}\cdot\vec{r}'(t)/r^3+...$
How do I do this expansion? This was a part of a derivation from my course in electromagnetism, and I spent too much time on this. I have written $1/|\vec{r}-\vec{r}'(t)|=1/\sqrt{\vec{r}^2-\vec{r}'(t)^2}$ and have Taylor expanded wrt $t$, but get the wrong result.
The correct way to expand this is
$$\frac{1}{R} = \frac{1}{\sqrt{r^2-2\vec{r}\cdot\vec{r}'+r'^2}} = \frac{1}{r\sqrt{1-2\frac{\vec{r}\cdot\vec{r}'}{r^2}+\frac{r'^2}{r^2}}}$$
and using the Taylor expansion
$$(1+x)^{-1/2}= 1-\frac{1}{2}x+O(x^2)$$
we obtain
$$\frac{1}{R} = \frac{1}{r}\left( 1 + \frac{\vec{r}\cdot\vec{r}'}{r^2}-\frac{r'^2}{2r^2} + O((r'/r)^2)\right)$$
But the last term before the big O is also of order $(r'/r)^2$ so
$$\frac{1}{R} = \frac{1}{r}\left( 1 + \frac{\vec{r}\cdot\vec{r}'}{r^2} + O((r'/r)^2)\right)$$
or
$$\frac{1}{R} \approx \frac{1}{r} + \frac{\vec{r}\cdot\vec{r}'}{r^3} \; .$$