$\displaystyle

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

$

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

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

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- Jan 11th 2010, 09:24 PMharveyopower series expansion II
$\displaystyle

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

$

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

= 4 ( 1/ 1- (1-x)^2) - Jan 11th 2010, 10:46 PMCaptainBlack
- Jan 12th 2010, 03:06 AMharveyo
Context: write the power series expansion, indicate an interval of convergence when center c = 1

- Jan 12th 2010, 04:42 AMKrizalid
Okay well, in order to do that, I'll give you an example:

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