How do you find a power series representation for x^2 / ( 1 - 2x )^2?

1 Answer
Apr 23, 2018

f(x)=sum_(n=1)^oo2^(n-1)nx^(n+1)

Explanation:

The idea is to relate this expression to the known power series expansion

1/(1-x)=sum_(n=0)^oox^n

Temporarily disregard the x^2 and consider

f(x)=x^2 1/(1-2x)^2.

Take the integral of 1/(1-2x)^2:

intdx/(1-2x)^2

Quick substitution:

u=1-2x

du=-2dx, -1/2du=dx

-1/2intu^-2du=1/(2u)=1/(2(1-2x))

Thus, knowing that differentiating this integrated expression returns the original 1/(1-2x)^2, we can say

f(x)=x^2d/dx1/(2(1-2x))

Let's find the power series representation for the differentiated expression:

f(x)=x^2d/dx1/2*1/(1-2x)

We can easily relate 1/(1-2x) to 1/(1-x)=sum_(n=0)^oox^n:

1/(1-2x)=sum_(n=0)^oo(2x)^n=sum_(n=0)^oo2^nx^n

So,

f(x)=x^2d/dx1/2sum_(n=0)^oo2^nx^n

We can absorb the 1/2 in:

f(x)=x^2d/dxsum_(n=0)^oo2^(n-1)x^n

Differentiate the summation with respect to x, recalling that differentiating the summation causes the index to shift up by 1:

f(x)=x^2sum_(n=0)^ood/dx2^(n-1)x^n

f(x)=x^2sum_(n=1)^oo2^(n-1)nx^(n-1)

Multiply in the x^2:

f(x)=sum_(n=1)^oo2^(n-1)nx^(n-1+2)

f(x)=sum_(n=1)^oo2^(n-1)nx^(n+1)