# Basic Linear Approximation

I have come across the need to quickly perform linear approximations, for example I ran across this simplification provided r << d (I think maybe it should be r >> d).

$2(r + d)^{-2} - r^{-2} - (r + 2d)^{-2} = 2 - \frac{2d}{r} - 1 - 1 + \frac{2d}{r}$. Could someone explain how to do this kind of computation? I don't see where that simplification came from.

What I do know is that $(r + d)^{2} = r^{2}(1 + \frac{d}{r})^{2} \approx r^{2}(1 + \frac{2d}{r})$ provided r >> d. When I use that to simplify the expression above, I do not get the result above.

we have $$(big + small)^n = (big)^n + n \, (big)^{n-1}\, small + \frac{n(n-1)}{2}(big)^{n-2}(small)^2+\cdots$$
if we look at $$(r+d)^{-2} = r^{-2} -2r^{-3}d + 3r^{-4}d^2+\cdots \text{ for } r >> d.$$
if you want to compute \begin{align}2(r+d)^{-2} - r^{-2} - (r+2d)^{-2} &= 2\left(r^{-2} - 2r^{-3} d + 3r^{-4}d^2+\cdots\right) - r^{-2}-\left(r^{-2}-2r^{-3}(2d)+3r^{-4}(2d)^2+\cdots\right)\\ &=-\frac{6d^2}{r^4}+\cdots \end{align}