g(x) = 2 / (1-x)^3 center c=0

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- Jan 11th 2010, 08:34 PMharveyopower series expansion
g(x) = 2 / (1-x)^3 center c=0

- Jan 11th 2010, 08:41 PMJhevon
- Jan 11th 2010, 09:11 PMharveyo
- Jan 12th 2010, 05:15 AMHallsofIvy
You haven't done

**anything**that jhevon suggested!

He suggested that you write out the power series for $\displaystyle \frac{1}{1- x}$ and differentiate**that**twice!

What is the power series for $\displaystyle \frac{1}{1- x}$?

(Hint: the sum of the geometric series $\displaystyle \sum_{n=0}^\infty ar^n$ is $\displaystyle \frac{a}{1- r}$.) - Jan 15th 2010, 04:44 PMharveyo
power series

1/ 1-x = 1 + x +x^2 + x^3 ......

therefore

1/ (1-x)^2 = 1+ 2x +3x^2 +4x^3 ......

now

1/(1-x)^3 = 2 + 6x + 12x^2 .....

? - Jan 16th 2010, 06:58 AMawkward
- Jan 16th 2010, 01:56 PMJhevon
the left side of your third equation is incorrect. and when i said differentiate the power series for 1/(1 - x) i was talking about differentiating the expression behind the summation sign in $\displaystyle \sum_{n = 0}^\infty x^n$, for $\displaystyle |x|<1$, twice

or you can follow awkwards suggestion if you learned the negative binomial theorem