Question

how do you get the power series of 2/((x-1)^3)?

Answer 1

i know it involves calc.

Answer 2

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.

Answer 3

\[2/(x-1)^3=-2/(1-x)^3\]
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)=
\[-2(1-x)/(1-x)^4=-2/(1-x)^3=\sum_{2}^{\infty}n(n-1)x^{n-2}\]

Answer 4

\[1/1-x = \sum_{n=0}^{\infty} x ^{n}\]

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.