# power series expansion II

• January 11th 2010, 09:24 PM
harveyo
power series expansion II
$
f (x) = 4/ x^2 - 2x
$

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

= 4 ( 1/ 1- (1-x)^2)
• January 11th 2010, 10:46 PM
CaptainBlack
Quote:

Originally Posted by harveyo
$
f (x) = 4/ x^2 - 2x
$

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

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

More context needed.

Also:

$
f (x) = \frac{4}{x^2 - 2x}=-\frac{2}{x(1-x/2)}
$

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

CB
• January 12th 2010, 03:06 AM
harveyo
Context: write the power series expansion, indicate an interval of convergence when center c = 1
• January 12th 2010, 04:42 AM
Krizalid
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 $x=1.$