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.


this is an application of the binomial theorem/identity. you will need to go up to the third term.

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 $$\begin{align}2(r+d)^{-2} - r^{-2} - (r+2d)^{-2} &= 2\left(r^{-2} - 2r^{-3} d + 3r^{-4}d^2+\cdots\right) - r^{-2}-\left(r^{-2}-2r^{-3}(2d)+3r^{-4}(2d)^2+\cdots\right)\\ &=-\frac{6d^2}{r^4}+\cdots \end{align}$$

  • $\begingroup$ I like your answer. I think the document I was reading factored out an r^-2 leaving us with two -1 terms as I wrote above. I couldn't understand where they came from. The document was poorly written, it's not actually -1, -1, but r^-2, r^-2 which still cancels. Thanks. $\endgroup$ – user7348 May 25 '15 at 3:53
  • $\begingroup$ @user7348, you are welcome. $\endgroup$ – abel May 25 '15 at 3:54

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