# Expansion

• December 30th 2010, 11:46 PM
Bart
1 Attachment(s)
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?

Attachment 20298http://www.mathhelpforum.com/math-he...isc/pencil.png
• December 31st 2010, 01:12 AM
CaptainBlack
Quote:

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?

Attachment 20298http://www.mathhelpforum.com/math-he...isc/pencil.png

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

CB
• December 31st 2010, 10:03 AM
Bart
Quote:

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

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.
• December 31st 2010, 05:12 PM
mr fantastic
Quote:

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.

$\displaystyle (D - md)^{-2} = \left( D \left[ 1 - \frac{md}{D}\right] \right)^{-2} = ....$
• January 1st 2011, 04:08 PM
Bart
Now I see it, you just use the rule A^n * B^n = (A*B)^n.
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