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Math Help - Expansion

  1. #1
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    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?

    Expansion-test.jpg
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  2. #2
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    Quote Originally Posted by Bart View Post
    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?

    Click image for larger version. 

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

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    (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.
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  4. #4
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    Quote Originally Posted by Bart View Post
    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} = ....
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  5. #5
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    Now I see it, you just use the rule A^n * B^n = (A*B)^n.
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