$\displaystyle f (x) = 4/ x^2 - 2x $ = 4 ( 1/ (x^2 - 2x)) = 4 ( 1/ 1- (1-x)^2)
Follow Math Help Forum on Facebook and Google+
Originally Posted by harveyo $\displaystyle f (x) = 4/ x^2 - 2x $ = 4 ( 1/ (x^2 - 2x)) = 4 ( 1/ 1- (1-x)^2) More context needed. Also: $\displaystyle f (x) = \frac{4}{x^2 - 2x}=-\frac{2}{x(1-x/2)} $ should give a series expansion for $\displaystyle |x|<2$ and $\displaystyle x\ne 0$ CB
Context: write the power series expansion, indicate an interval of convergence when center c = 1
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 $\displaystyle x=1.$
View Tag Cloud