Question

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

Answer 1

`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)`