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This is essentially a definition question.

Given a rational function $\frac{p(x)}{q(x)}$, what would the $x^k$ coefficient of this rational function mean (in particular for the negative $k$'s). Is there some sort of expansion $$\frac{p(x)}{q(x)} = \sum_{-\infty}^\infty a_kx^k$$ where one would say $a_k$ is the coefficient of $x^k$?

How would you find the coefficient? A very simple rational function would be $\frac{(x - 1)}{(x - 2)(x - 3)}$. How would you find the $x^{-1}$ coefficient of this?

Thanks for any help and clarifications.

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up vote 1 down vote accepted

There is no such expansion that converges everywhere. To see this, consider what happens to both sides of your equation as $x\to 0$ and $x\to \infty$.

You can, however, expand $p(x)/q(x)$ as different series that converge around different values of $x$. For instance, near $0$ you have $$\frac{x-1}{(x-2)(x-3)} = \sum_{k=0}^{\infty}\left[ \frac{1}{2}\left(\frac{1}{2}\right)^k - \frac{2}{3} \left(\frac{1}{3}\right)^k\right] x^k,$$ which you can derive by using partial fractions and the formula for geometric series. Similarly expanding $p(y^{-1})/q(y^{-1})$ with respect to $y$ will give you a power series in $x^{-1}$ valid as $x\to \infty$:

$$\begin{align*}\frac{p(y^{-1})}{q(y^{-1})} &= -\frac{1}{6} + \frac{2}{3} \frac{1}{1-3y} - \frac{1}{2} \frac{1}{1-2y}\\ &= -\frac{1}{6} + \sum_{k=0}^\infty \left[\frac{2}{3} 3^k - \frac{1}{2} 2^k\right] y^k \\ &= -\frac{1}{6} + \sum_{k=0}^\infty \left[\frac{2}{3} 3^k - \frac{1}{2} 2^k\right] x^{-k}\end{align*}.$$

So in some sense you could say that the coefficient of $x^{-1}$ is 1, near infinity.

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Every rational function has an expansion of the form $$\sum_{k=r}^{\infty}a_kx^k$$ where $r$ may be positive, zero, or negative, but is in any case finite. If $r\ge0$ then the coefficient of $x^{-1}$ is zero, as shown by user7530's series for your example. If $r\lt0$ then $a_{-1}$ can be found by the following procedure: let $s=-r$, multiply the rational function by $x^s$, differentiate $s-1$ times, evaluate at $x=0$, and divide by $s-1$ factorial.

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For deriving the series expansion of a general rational function, one way of proceeding would be to take the polynomial and rational parts of the original rational function (similar to taking integer and fractional parts of a number) through, say, long division, and then performing the series expansion on the rational part... – J. M. Oct 17 '11 at 11:55

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