1. ## Expansion

For a physics-exercise (56b) I have to use the expansion (1+x)^n ~= 1 + n*x when x^2 << 1

Given is that D >> md , how do I make the expansion of (D-md)^-2?

2. Originally Posted by Bart
For a physics-exercise (56b) I have to use the expansion (1+x)^n ~= 1 + n*x when x^2 << 1

Given is that D >> md , how do I make the expansion of (D-md)^-2?

$\displaystyle (D-md)^{-2}=D^{-2}(1-\frac{md}{D})^{-2}\approx D^{-2} (1+(-2)(-\frac{md}{D}))$

CB

3. Originally Posted by CaptainBlack
$\displaystyle (D-md)^{-2}=D^{-2}(1-\frac{md}{D})^{-2}\approx D^{-2} (1+(-2)(-\frac{md}{D}))$

CB
Thanks. I understood that I have to rewrite this formula so I get the form of (1 + x)^-2
The problem is that I don't know how to perform the first step, I have difficulty with the algebra because of that power.

4. Originally Posted by Bart
Thanks. I understood that I have to rewrite this formula so I get the form of (1 + x)^-2
The problem is that I don't know how to perform the first step, I have difficulty with the algebra because of that power.
The person who set this question has assumed basic skill in algebra.

$\displaystyle \displaystyle (D - md)^{-2} = \left( D \left[ 1 - \frac{md}{D}\right] \right)^{-2} = ....$

5. Now I see it, you just use the rule A^n * B^n = (A*B)^n.