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

Apr 23, 2018

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {2}^{n - 1} n {x}^{n + 1}$

#### Explanation:

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

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Temporarily disregard the ${x}^{2}$ and consider

$f \left(x\right) = {x}^{2} \frac{1}{1 - 2 x} ^ 2$.

Take the integral of $\frac{1}{1 - 2 x} ^ 2$:

$\int \frac{\mathrm{dx}}{1 - 2 x} ^ 2$

Quick substitution:

$u = 1 - 2 x$

$\mathrm{du} = - 2 \mathrm{dx} , - \frac{1}{2} \mathrm{du} = \mathrm{dx}$

$- \frac{1}{2} \int {u}^{-} 2 \mathrm{du} = \frac{1}{2 u} = \frac{1}{2 \left(1 - 2 x\right)}$

Thus, knowing that differentiating this integrated expression returns the original $\frac{1}{1 - 2 x} ^ 2 ,$ we can say

$f \left(x\right) = {x}^{2} \frac{d}{\mathrm{dx}} \frac{1}{2 \left(1 - 2 x\right)}$

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

$f \left(x\right) = {x}^{2} \frac{d}{\mathrm{dx}} \frac{1}{2} \cdot \frac{1}{1 - 2 x}$

We can easily relate $\frac{1}{1 - 2 x}$ to $\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$:

$\frac{1}{1 - 2 x} = {\sum}_{n = 0}^{\infty} {\left(2 x\right)}^{n} = {\sum}_{n = 0}^{\infty} {2}^{n} {x}^{n}$

So,

$f \left(x\right) = {x}^{2} \frac{d}{\mathrm{dx}} \frac{1}{2} {\sum}_{n = 0}^{\infty} {2}^{n} {x}^{n}$

We can absorb the $\frac{1}{2}$ in:

$f \left(x\right) = {x}^{2} \frac{d}{\mathrm{dx}} {\sum}_{n = 0}^{\infty} {2}^{n - 1} {x}^{n}$

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

$f \left(x\right) = {x}^{2} {\sum}_{n = 0}^{\infty} \frac{d}{\mathrm{dx}} {2}^{n - 1} {x}^{n}$

$f \left(x\right) = {x}^{2} {\sum}_{n = 1}^{\infty} {2}^{n - 1} n {x}^{n - 1}$

Multiply in the ${x}^{2} :$

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {2}^{n - 1} n {x}^{n - 1 + 2}$

$f \left(x\right) = {\sum}_{n = 1}^{\infty} {2}^{n - 1} n {x}^{n + 1}$