# Series representation of x^2/(a^3-x^3)

• Feb 9th 2010, 08:33 PM
buckeye1973
Series representation of x^2/(a^3-x^3)
Hi all,

I'm having trouble with the series representation of a function.

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

First I tried breaking this up into partial fractions, then I realized I'm not very good at doing partial fractions... But I think I found another path that maybe I'm supposed to use:

Consider that:
$\displaystyle \int \frac{x^2}{a^3-x^3} dx = -\frac{1}{3} ln(a^3 - x^3) + C$

I think this somehow relates to the integral of the alternating geometric series (switching variable for clarity):
$\displaystyle \int \frac{1}{1+u} du = ln(1 + u) + C$

I know I can integrate the terms of the geometric series to get the natural log, but I also know I can't just plug $\displaystyle u = a^3 - x^3 - 1$, and now I'm not sure what to do. Am I headed down the right tree or barking up the wrong path? Do I need to return to the partial fractions approach instead?

Thanks again for any help,

Brian
• Feb 9th 2010, 09:48 PM
Pulock2009
log(1+x) can be easily expanded. it was somewht like this log(1+x)=1+x^2/2-x^3/3+.....(i am not sure please correct me if i am wrong). since u have got an expression in log u should be able to expand it......
• Feb 9th 2010, 09:50 PM
Pulock2009
then differentiate it once again and u should get the series representation as my common sense says!.
• Feb 9th 2010, 09:52 PM
Drexel28
Quote:

Originally Posted by buckeye1973
Hi all,

I'm having trouble with the series representation of a function.

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

First I tried breaking this up into partial fractions, then I realized I'm not very good at doing partial fractions... But I think I found another path that maybe I'm supposed to use:

Consider that:
$\displaystyle \int \frac{x^2}{a^3-x^3} dx = -\frac{1}{3} ln(a^3 - x^3) + C$

I think this somehow relates to the integral of the alternating geometric series (switching variable for clarity):
$\displaystyle \int \frac{1}{1+u} du = ln(1 + u) + C$

I know I can integrate the terms of the geometric series to get the natural log, but I also know I can't just plug $\displaystyle u = a^3 - x^3 - 1$, and now I'm not sure what to do. Am I headed down the right tree or barking up the wrong path? Do I need to return to the partial fractions approach instead?

Thanks again for any help,

Brian

The integration method works ok (don't forget to differentiate).

How about this though? $\displaystyle f(x)=\frac{x^2}{a^3-x^3}=\frac{x^2}{a^3}\frac{1}{1-\left(\frac{x}{a}\right)^3}$. Now, just for the sake of clarity try calling $\displaystyle z=\left(\frac{x}{a}\right)^3$. Look familiar?
• Feb 10th 2010, 05:28 AM
buckeye1973
Quote:

Originally Posted by Pulock2009
log(1+x) can be easily expanded. it was somewht like this log(1+x)=1+x^2/2-x^3/3+.....(i am not sure please correct me if i am wrong). since u have got an expression in log u should be able to expand it......

Actually, it's $\displaystyle 1-x^2/2+x^3/3-x^4/4+...$ I already knew this part, you've basically restated my question.

Quote:

Originally Posted by buckeye1973
I know I can integrate the terms of the [alternating] geometric series to get the natural log

• Feb 10th 2010, 05:39 AM
buckeye1973
Quote:

Originally Posted by Drexel28
How about this though? $\displaystyle f(x)=\frac{x^2}{a^3-x^3}=\frac{x^2}{a^3}\frac{1}{1-\left(\frac{x}{a}\right)^3}$. Now, just for the sake of clarity try calling $\displaystyle z=\left(\frac{x}{a}\right)^3$. Look familiar?

Ah yes, that does it. I wasn't seeing the basic algebra to get my original expression into the known geometric series form. Thank you very much!

Brian
• Feb 10th 2010, 06:28 AM
Pulock2009
ok! i get u: u were talking about expanding it as a geometric series with a common ratio and all that......