This seems like it should be simple but I can't understand it. I'm reading this paper about spiral structure in galaxies. It says "this local shrinkage has increased the density, μ, per unit surface area in that neighbourhood by a fraction ε ... so it should feel an additional gravitational force of the order" and there's an approximation as follows
$$G \mu L^2 \left[(L - \epsilon L)^{-2} - L^{-2}\right] \approx \epsilon G \mu$$ (Equation 1 in the paper)
$G$ is the gravitational constant and $\epsilon$ should be a fraction; when I try and work through to get $\epsilon G \mu$ (purely for my own understanding) I get
$$G \mu \left(\frac{1}{\epsilon^2 - 2 \epsilon} - 1\right)$$
but I can't approximate the part in the brackets to $\epsilon$? Am I being stupid?
Let $$ f(x) = \frac{1}{(1-x)^2} $$ For $x < 1$ we can approximate $f$ using its Taylor expansion $$ f(x) = 1 + x f'(0) + \frac{x^2}{2} f''(0) + \dots $$ We derivate $f$ to arrive at $$ f'(x) = \frac{2}{(1-x)^3} \quad f''(x) = \frac{6}{(1-x)^4} $$ which inserted into our approximation $$ f(\epsilon) = 1 + 2 \epsilon + 3\epsilon^2 + O(\epsilon^3) $$ This in turn can be inserted into the gravitational equation to get $$ G \mu \left[ \frac{1}{(1-x)^2} - 1\right] = G \mu \left[2 \epsilon + 3\epsilon^2 + O(\epsilon^3) \right] $$ When $\epsilon$ is small the two latter terms can safely be dropped. Now I'm not sure where the $2$ disappears and also your expression, unlike the one in the book, has a singularity at $0$ so I'm not sure it's the same expression.