How to arrive at this approximation?

Question

I encountered an equation:

$$\frac{1}{(ja + \delta{z_{n}} - \delta{z_{n-j}})^2} + \frac{1}{(-ja + \delta{z_{n}} - \delta{z_{n+j}})^2}$$

can someone tell me how it approximates to:

$$-2\left[\frac{\delta{z_{n}} -\delta{z_{n-j}}}{(ja)^3} - \frac{\delta{z_{n}} -\delta{z_{n+j}}}{(ja)^3}\right]$$

I'd really appreciate the help!

Answer 1

This is the reason, that $(1+\epsilon)^n\approx 1+n\epsilon$ if $\epsilon$ is small. $$(A+\delta x)^{-2}=A^{-2}(1+\delta\frac xA)^{-2}\\ \approx A^{-2}(1-2\delta\frac xA)\\ =A^{-2}-2\delta\frac {x}{A^3}$$

Answer 2

Take out the factor $(\frac{1}{ja})^2$ as a common factor, giving you

$(\frac{1}{ja})^2 \times [(1+ [\frac{\delta{z_n} - \delta{z_{n-j}}}{ja}])^{-2} + (1- [\frac{\delta{z_n} - \delta{z_{n-j}}}{ja}])^{-2} ]$.

I'm assuming that you have $ ja >> 1 $, in which case you can expand the terms $(1 \pm (\frac{\delta{z_n} -\delta{z_{n-j}}}{ja}))^{-2}$ (since $(\frac{\delta{z_n} -\delta{z_{n-j}}}{ja})^2 << 1)$ by the binomial expansion to get the desired result.