Represent Power Series of a function

Question

Hi could anyone help me answer this question

Find the power series representation for the function and determine the radius of convergence

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

After getting 1/(1-2x)^2 I do not know how to convert it to a power series.

Answer 1

Some starting material: Using that $\sum_{n=0} ^\infty x^n = \frac{1}{1-x}$, we can write $$ f(x) = \frac{x^2}{1 - 4x + 4x^2} = x^2 \cdot \frac{1}{1 - (4x - 4x^2)} = x^2 \sum_{n=0} ^\infty 4^n x^n(1-x)^n = \sum_{n=0} ^\infty 4^n x^{n+2}(1-x)^n. $$ I imagine you can take it from here.

Answer 2

You have $$ |z|<1\implies \frac {1}{1-z} = \sum_{n=0}^\infty z^n;\\ \frac d{dx}\frac {1}{1-x} = \frac {1}{(1-x)^2}; |x|<1\implies \frac {1}{(1-x)^2} = \sum_{n=0}^\infty (n+1)x^n.\\ $$ Now make the substitution $x\to 2x$: $$|x|<\frac 12\implies \frac {1}{(1-2x)^2} = \sum_{n=0}^\infty (n+1)2^nx^n\\ \implies \frac {x^2}{(1-2x)^2} = \sum_{n=0}^\infty (n+1)2^nx^{n+2}. $$

Then you see that for $x=1/2$, $(n+1)2^nx^{n+2}\nrightarrow 0$: the radius is $1/2$.