1. ## Find power series

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

$\displaystyle f(x) = \frac {x^3}{(x-2)^2}$

This is what I did:
$\displaystyle = x^3 \left( \frac {1}{x-2} \right)^2$

$\displaystyle = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2$

$\displaystyle = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2$

But I don't know where to go from here, the answer says:

$\displaystyle \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n$

2. Hello,

I've learnt to do this way :

$\displaystyle \frac{x^3}{(x-2)^2}=\sum_{n=0}^{+\infty} a_n x^n$

And then determine $\displaystyle a_n$ by induction.

I don't have time to see if it works here though...

3. Originally Posted by angel.white
Find a power series representation for the function and determine the radius of convergence:

$\displaystyle f(x) = \frac {x^3}{(x-2)^2}$

This is what I did:
$\displaystyle = x^3 \left( \frac {1}{x-2} \right)^2$

$\displaystyle = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2$

$\displaystyle = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2$

But I don't know where to go from here, the answer says:

$\displaystyle \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n$
Consider this $\displaystyle \frac{1}{(x-2)^2}=\frac{-D[\frac{1}{x-2}]}{dx}$ find the power series for $\displaystyle \frac{1}{x-2}$ differentiate it adn take the opposite

4. Originally Posted by angel.white
Find a power series representation for the function and determine the radius of convergence:

$\displaystyle f(x) = \frac {x^3}{(x-2)^2}$

This is what I did:
$\displaystyle = x^3 \left( \frac {1}{x-2} \right)^2$

$\displaystyle = x^3 \left( -\frac12 * \frac {1}{1-\frac x2}\right)^2$

$\displaystyle = x^3 \left( -\frac12 * \sum_{n=0}^\infty \left( \frac x2 \right)^n\right)^2$

But I don't know where to go from here, the answer says:

$\displaystyle \sum_{n=3}^\infty \frac {n-2}{2^{n-1}} x^n$
Given $\displaystyle A(x)=\sum_{n=0}^{\infty}{a_n\cdot{x^n}}$ and $\displaystyle B(x)=\sum_{n=0}^{\infty}{b_n\cdot{x^n}}$

We have: $\displaystyle A(x)\cdot{B(x)}=\sum_{n=0}^{\infty}{\left(\sum_{k= 0}^n{a_k\cdot{b_{n-k}}}\right)\cdot{x^n}}$

Applying this: $\displaystyle \left(\sum_{k=0}^{\infty}{\frac{x^n}{2^n}}\right)^ 2=$$\displaystyle \sum_{n=0}^{\infty}{\left(\frac{n+1}{2^n}\right)\c dot{x^n}}$

To do this exercise you can consider that: $\displaystyle \frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}{n\cdot{x^n}}$ when |x|<1