Power Series of x^2/(a^3-x^3)

[runningeagle]

Hi,

PART A
I'm looking at f(x)=[attachment=1:2vh2mtca]1.gif[/attachment:2vh2mtca]

A is a non zero constant

I want to find a power sereis representation for f(x) and find its interval of convergence.

I think I should relate f(x) somehow with 1/1-x but I am unsure how to factor, integrate, or derive to get there.

I integrated with u-subsitution to get -1/3(1/a^3-x^3), but I don't know where to go from here.

PART B

The next function g(x)=[attachment=0:2vh2mtca]log.gif[/attachment:2vh2mtca] where log is the natural logarithm.

I took the derivative to get x^2[1/(a^3-x^3)] Now I am unsure what to do to get this to relate to 1/(1-X)

 

[galactus]

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

Rewrite as \(\displaystyle \frac{x^{2}}{a^{3}\left(1-(\frac{x}{a})^{3}\right)}\)

Note, this looks something like the closed form for the geometric series we need.

\(\displaystyle \frac{1}{a^{3}}x^{2}+\frac{1}{a^{6}}x^{5}+\frac{1}{a^{9}}x^{8}+.......=\frac{1}{a^{3}}\sum_{k=0}^{\infty}x^{2}(\frac{x}{a})^{3k}=\sum_{k=0}^{\infty}\frac{x^{3k+2}}{a^{3k+3}}\)


\(\displaystyle \frac{-1}{3}log(a^{3}-x^{3})\)

\(\displaystyle =-\frac{log(a^{3})}{3}+\frac{x^{3}log(e)}{3a^{3}}+\frac{x^{6}log(e)}{6a^{6}}+...\)

Can you find this one?. I gave a nudge on the first. Think about the series for ln(1-x).

 

[BigGlenntheHeavy]

Another Way:

\(\displaystyle Using \ taylor \ F-3,9 \ on \ my \ trusty \ TI-89, \ I \ get:\)

\(\displaystyle taylor\bigg(\frac{x^{2}}{a^{3}-x^{3}},x,20,0 \bigg) \ gives,\)

\(\displaystyle \frac{x^{2}}{a^{3}} \ + \ \frac{x^{5}}{a^{6}} \ + \ \frac{x^{8}}{a^{9}} \ + \ \frac{x^{11}}{a^{12}} \ + \ \ \frac{x^{14}}{a^{15}} \ + \ \frac{x^{17}}{a^{18}} \ + \ \frac{x^{20}}{a^{21}}\)

\(\displaystyle Which \ readily \ gives \ the \ pattern \ \sum_{n=0}^{\infty}\frac{x^{3n+2}}{a^{3n+3}}.\)

\(\displaystyle Hence, \ f(x) \ = \ \frac{x^{2}}{a^{3}-x^{3}} \ = \ \sum_{n=0}^{\infty}\frac{x^{3n+2}}{a^{3n+3}}, \ |x| \ < \ a, \ a\ne \ 0\)

 

[BigGlenntheHeavy]

\(\displaystyle For \ the \ second \ one, \ I \ get, \ using \ my \ trusty \ TI-89;\)

\(\displaystyle f(x) \ = \ \frac{-1}{3}ln|a^{3}-x^{3}| \ = \ -ln|a| \ + \ \sum_{n=0}^{\infty}\frac{x^{3n+3}}{(3n+3)a^{3n+3}}, \ |x| \ < \ a, \ a \ \ne \ 0.\)