# find a power series representation of f(x)

• Dec 5th 2012, 01:09 PM
bkbowser
find a power series representation of f(x)
i am to use $\frac{1}{1-x}=\sum{x^n}=1+x+x^2+...$ provided $|x|<1$ to find a power series representation for $f(x)=\frac{x^3}{(x-2)^2}$

so i want to foil the binomial, move the cube, and convert the statement into the form of $\frac{1}{1-x}$

considering only $\frac{1}{(x-2)^2}$

i get $\frac{1}{x^2-4x+4}=\frac{1}{4-(4x-x^2)}=\frac{1}{4}*\frac{1}{1-(x-\frac{x^2}{4})}$

so $\frac{1}{4}*\frac{1}{1-(x-\frac{x^2}{4})}=\frac{1}{4}*\sum{(x-\frac{x^2}{4})^n}$

now i'm at a loss has to how to get it into the form $\sum{c_n*x^n}$

integrating/differentiating term-by-term doesn't seem to be helpful as each makes the situation far more complicated.
• Dec 5th 2012, 01:20 PM
bkbowser
Re: find a power series representation of f(x)
actually i should be able to take the derivative of $\frac{1}{(x-2)^2}$

so $d/dx\frac{1}{(x-2)^2}=\frac{1}{(x-2)}$

then $d/dx\sum{x^n}=\frac{-1}{2}\frac{1}{(1-x)}$

and $d/dx\frac{-1}{2}\sum{x^n}=\frac{-1}{2}\sum{nx^{n-1}}$

and then just multiply in the cube, no?
• Dec 5th 2012, 01:53 PM
bkbowser
Re: find a power series representation of f(x)
Quote:

Originally Posted by bkbowser
actually i should be able to take the derivative of $\frac{1}{(x-2)^2}$

so $d/dx\frac{1}{(x-2)^2}=\frac{1}{(x-2)}$

then $d/dx\sum{x^n}=\frac{-1}{2}\frac{1}{(1-x)}$

and $d/dx\frac{-1}{2}\sum{x^n}=\frac{-1}{2}\sum{nx^{n-1}}$

and then just multiply in the cube, no?

sorry there's a few simple things wrong here.

$d/dx\frac{1}{(x-2)^2}=\frac{1}{(x-2)}$ should be;

$\frac{1}{(x-2)^2}=d/dx\frac{1}{(x-2)}$

$=d/dx\frac{1}{(x-2)}=d/dx\frac{-1}{2}*\frac{1}{(1-\frac{x}{2})}=d/dx\frac{-1}{2}*\sum\frac{x^n}{2^n}$

then after reintroducing the cube;

$x^3*\frac{-1}{2}*\sum\frac{x^n}{2^n}=-\sum{\frac{x^{n+3}}{2^{n+1}}}$