# Taylor series

1. Oct 12, 2009

### industrygiant

I can't work out how to calculate the Taylor series for

$$\frac{1}{|R-r|}$$

when R>>r, but they are both vectors. We were told to expand in r/R but I did the step below and I'm not sure where to go from there

I got to

$$\frac{1}{R \sqrt{1 - (2R.r)/R^2 + (r^2)/(R^2)}}$$

I also know the result in first order is

$$\frac{1}{R} + \frac{R.r}{R^3}$$

2. Oct 12, 2009

### kuruman

Simplify and rewrite your expression as
$$\frac{1}{R \sqrt{1 - 2r/R + (r/R)^2}}$$

Now define a (small) quantity ε =r/R in terms of which you have

$$\frac{1}{R \sqrt{1 - 2r/R + (r/R)^2}} =\frac{1}{R}f(\epsilon)$$

Where
$$f(\epsilon)=(1-2\epsilon+\epsilon^2)^{-1/2}$$

Expand f(ε) in Taylor series the usual way, then replace ε with r/R.