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$\dfrac1{x^2+3}=x^{-2}\left(1+\dfrac{3}{x^2}\right)^{-1} = x^{-2}\left(1-\dfrac{3}{x^2}+\dfrac{9}{x^4}+\cdots\right)$

But this can also be factored into:

$\dfrac{1}{3}\left(1+\dfrac{x^2}{3}\right)^{-1} = \dfrac{1}{3}\left(1-\dfrac{x^2}{3}+\dfrac{x^4}{9}+\cdots\right)$

What's wrong here? The two answers are the opposite of each other.

MPW
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user61871
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    The second is useful for "smallish" $|x|$. The first is useful for large $|x|$. Neither is a partial fraction decomposition in the usual sense of the term. – André Nicolas Jan 27 '15 at 02:07
  • I've worked on a few more problems, and am I right to assume that if you want descending powers of x, or "large |x|" then you put the x term as the denominator, and vice-versa?

    When you say useful for large |x|, that means it's a better approximation?

     – user61871 Jan 27 '15 at 02:49
  • The second series converges only for $|x|\lt \sqrt{3}$. It gives no useful information for larger $|x|$. The first is called an asymptotic expansion. I am mentioning the term so you can read about such things on Wikipedia or elsewhere. Your first expansion shows that for large $|x|$, $\frac{1}{x^2}$ is a decent approximation, and $\frac{1}{x^2}-\frac{3}{x^4}$ is better, and so on. – André Nicolas Jan 27 '15 at 02:58

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