The question asks:
Find a Power Series representation and Radius of Convergence for: $f(x)=\dfrac{x}{9+x^2}$
I see that it is pretty straight forward that if I re-write this by factoring out a 9, and use the Geometric Series:
$f(x)=\dfrac{x}{9\cdot\left(1+\dfrac{x^2}{9}\right)}=\dfrac{x}{9}\cdot\dfrac{1}{1-\left(-\dfrac{x^2}{9}\right)}=\dfrac{x}{9}\cdot\displaystyle\sum_{n=0}^\infty\left(-\dfrac{x^2}{9}\right)^n=\sum_{n=0}^\infty(-1)^n\dfrac{1}{9^{n+1}} x^{2n+1}$
Radius of Convergence is $|x^2/9|<1$, which is $|x|<3$.
So my question is, why this isn't also a valid answer:
$f(x) = \dfrac{x}{x^2\left(\dfrac{9}{x^2}+1\right)} = \dfrac{1}{x} \cdot \dfrac{1}{1-\left(-\dfrac{9}{x^2}\right)} = \dfrac{1}{x}\cdot\displaystyle\sum_{n=0}^\infty \left(-\dfrac{9}{x^2}\right)^n=\sum_{n=0}^\infty(-1)^n\dfrac{9^n}{x^{2n+1}}$
This second one seems to have a Radius of Convergence $|9/x^2|<1$, which is when $|x|>3$.
Does it have to do with $0$ being in the domain of $f$, so that factoring out $x^2$ wouldn't work? (The series I found is not defined at $x=0$). I think, but I'm not really sure how to show that the second series is equal to $f$, but only for values of $|x|>3$, would that not mean that the second series covers more of $f$ since the first is only limited to $|x|<1$?
Does the second series still count as a Power Series (even though $x$ is being raised to negative powers since it's in the denominator)?
They almost seem related to each other some how, they are pretty close to the reciprocal of each other.
Both series are valid representations of the function, just in different regions of the complex plane. The first is an expansion around the origin, while the second can be seen as an expansion around infinity. It is called a Laurent series.