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Math Help - power series expansion II

  1. #1
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    power series expansion II

     <br />
 f (x) = 4/ x^2 - 2x <br />

    = 4 ( 1/ (x^2 - 2x))

    = 4 ( 1/ 1- (1-x)^2)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by harveyo View Post
     <br />
f (x) = 4/ x^2 - 2x <br />

    = 4 ( 1/ (x^2 - 2x))

    = 4 ( 1/ 1- (1-x)^2)
    More context needed.

    Also:

     <br />
f (x) = \frac{4}{x^2 - 2x}=-\frac{2}{x(1-x/2)} <br />

    should give a series expansion for |x|<2 and x\ne 0

    CB
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  3. #3
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    Context: write the power series expansion, indicate an interval of convergence when center c = 1
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  4. #4
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    Okay well, in order to do that, I'll give you an example:

    We have \frac1{1-x}=\sum_{j=0}^{\infty}x^j, then put x\mapsto x-1 and you'll get the series centered at x=1.
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