power series expansion

**harveyo** · January 11th 2010, 08:34 PM

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

**Jhevon** · January 11th 2010, 08:41 PM

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.

**harveyo** · January 11th 2010, 09:11 PM

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.

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

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

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

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

?????????

**HallsofIvy** · January 12th 2010, 05:15 AM

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 $\frac{a}{1- r}$ .)

**harveyo** · January 15th 2010, 04:44 PM

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 .....

?

**awkward** · January 16th 2010, 06:58 AM

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).

**Jhevon** · January 16th 2010, 01:56 PM

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

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