## anonymous 5 years ago how do you get the power series of 2/((x-1)^3)?

1. anonymous

i know it involves calc.

2. anonymous

You can use the formula from geometric series, knowing that [1/(1-x)] is equal to the sum from 0 to infinity of x^n, provided that the absolute value of your x is less than 1. Just reformulate your original function into the form of (1/1-x) and substitute your values in to the summation.

3. anonymous

$2/(x-1)^3=-2/(1-x)^3$ we know thatf(x)= $1/(1-x)=\sum_{0}^{\infty}x^n$ f'(x)=$1/(1-x)^2=\sum_{1}^{\infty}n(x)^{n-1}$ f''(x)= $-2(1-x)/(1-x)^4=-2/(1-x)^3=\sum_{2}^{\infty}n(n-1)x^{n-2}$

4. anonymous

$1/1-x = \sum_{n=0}^{\infty} x ^{n}$ For example you can just take your 2/((x-1)^3) and rearrange it into 2* 1/(x-1)^3. Then you can pull out the 2 and work your power series expansion as 2 times the sum of the series for the function 1/(x-1)^3. You can throw the 2 back in there when you're done just to make it pretty.