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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.

Big Papa
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2 Answers2

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$f(x)=\frac{-3}{x^2+1}+\frac{3}{x^2-9}$

$\frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$

$\frac{3}{x^2-9}=\frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$

Combine to get $f(x)=-4/3+(80/27)x^2-........$

I'll let you finish.

herb steinberg
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  • I am very sorry, this is a completely new topic to me. How would I go about solving this then? would the first term be the -4/3+(80/27)(x)^2 and substituting something in for x? Again this is a new topic for me so I don't know the proper techniques. – Big Papa Dec 07 '18 at 22:46
  • @Elijah What I gave you are the first two terms of what you are looking for. To get the next two just add the coefficients of the $x^4$ and $x^6$ terms from the two given expressions. $-3-1/241$ and $3-1/2187$. – herb steinberg Dec 07 '18 at 23:10
  • I like this so much that I discarded my purely formal approach. – Lubin Dec 07 '18 at 23:17
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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.

zwim
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