# power series expansion

• Jan 11th 2010, 09:34 PM
harveyo
power series expansion
g(x) = 2 / (1-x)^3 center c=0
• Jan 11th 2010, 09:41 PM
Jhevon
Quote:

Originally Posted by harveyo
g(x) = 2 / (1-x)^3 center c=0

notice that $\frac 2{(1 - x)^3}$ is the second derivative of $\frac 1{1 - x}$, and we know the power series for that. simply derive the power series for $\frac 1{1 - x}$ twice and that's your answer.
• Jan 11th 2010, 10:11 PM
harveyo
Quote:

Originally Posted by Jhevon
notice that $\frac 2{(1 - x)^3}$ is the second derivative of $\frac 1{1 - x}$, and we know the power series for that. simply derive the power series for $\frac 1{1 - x}$ twice and that's your answer.

$2/ (1-x)^3 = 2 * 1/(1-x)^3$

$= 2 * -1/2 d/dx [1/(1-x)^2]$

$= -1 d/dx [1/(1-x)^2]$

$= -1 d/dx [-1/(1-x)]$

?????????
• Jan 12th 2010, 06:15 AM
HallsofIvy
You haven't done anything that jhevon suggested!

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

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

(Hint: the sum of the geometric series $\sum_{n=0}^\infty ar^n$ is $\frac{a}{1- r}$.)
• Jan 15th 2010, 05:44 PM
harveyo
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, 07:58 AM
awkward
Quote:

Originally Posted by harveyo
g(x) = 2 / (1-x)^3 center c=0

Another approach:

Write $g(x) = 2 (1-x)^{-3}$

and apply the Binomial Theorem (for negative exponents).
• Jan 16th 2010, 02:56 PM
Jhevon
Quote:

Originally Posted by harveyo
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 .....

?

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 $\sum_{n = 0}^\infty x^n$, for $|x|<1$, twice

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