Find power series

**angel.white** · April 24th 2008, 02:17 PM

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

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

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

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

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

So radius = 2

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

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

**Moo** · April 24th 2008, 02:20 PM

Hello,

I've learnt to do this way :

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

And then determine $a_n$ by induction.

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

**Mathstud28** · April 24th 2008, 02:31 PM

Originally Posted by angel.white

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

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

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

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

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

So radius = 2

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

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

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

**PaulRS** · April 24th 2008, 04:05 PM

Originally Posted by angel.white

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

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

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

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

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

So radius = 2

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

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

Given $A(x)=\sum_{n=0}^{\infty}{a_n\cdot{x^n}}$ and $B(x)=\sum_{n=0}^{\infty}{b_n\cdot{x^n}}$

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

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

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

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