Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

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.

share|improve this answer
add comment

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.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.