Finding the first non-zero terms of a power series I have the function:
$f(x) = \frac{30}{(x^2 + 1)(x^2-9)}$
I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.
 A: $f(x)=\frac{-3}{x^2+1}+\frac{3}{x^2-9}$
$\frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$
$\frac{3}{x^2-9}=\frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$
Combine to get $f(x)=-4/3+(80/27)x^2-........$
I'll let you finish.
A: You can start by converting to rational parts.
$\displaystyle f(x)=\frac a{x-3}+\frac b{x+3}+\frac c{x^2+1}$ in order to have a sum of usual power series :
http://hardycalculus.com/calcindex/IE_powerseriestables.htm
Then substitute in $\displaystyle \sum (\pm 1)^nu^n=\frac 1{1\mp u}$ by factoring $3$ and get $u=\frac x3$.
For the other series substitute $u=x^2$ instead.
Gather all the coefficients with the same power of $x$ to get the final power series $\sum a_nx^n$ and determine its radius of convergence.
Edit: I just saw your problem is just determining first coefficients, not all of them. But even though the methodology presented here allow to find the first coefficients quite quickly, just limit yourself to small powers of $x$.
An alternative method is to calculate $f(0),f'(0),f''(0), ...$ and so on.
