What is rational fraction?

Example 9.2 (Reading off poles, orders and degree)

Let $F = \dfrac{X^3 - X}{X^4 - 2X^3 + X^2}$. Factor both layers: numerator $X(X-1)(X+1)$, denominator $X^2(X-1)^2$; cancel the common factor $X(X-1)$:

Poles: $0$ and $1$, both simple — the orders are read on the reduced denominator, so the apparent double roots of the original one are irrelevant. Degree: $\deg F = 1 - 2 = -1$, visible asymptotically ($xF(x) \to 1$ as $x \to \infty$). The degree behaves like the polynomial degree ($\deg FG = \deg F + \deg G$, $\deg(F + G) \leq \max$), a bookkeeping rule used constantly in the coefficient hunts below: each “limit of $xF(x)$” argument is a degree count in disguise.

Example 9.4 (Reduce first, then divide)

Find the integer part of $F = \dfrac{X^3 + 1}{X^2 - 1}$. Dividing blindly: $X^3 + 1 = (X^2 - 1)X + (X + 1)$, so $F = X + \frac{X + 1}{X^2 - 1}$. But the fraction was not reduced: $X^3 + 1 = (X + 1)(X^2 - X + 1)$ and $X^2 - 1 = (X+1)(X-1)$ share the factor $X + 1$, and

the same integer part $X$, but the fractional part collapses to a single brick, and the “pole” at $-1$ was never a pole at all. Always reduce to lowest terms before hunting poles: the poles of $F$ are the roots of the reduced denominator. (The integer part is insensitive to the simplification, as the uniqueness in Proposition 9.3 guarantees.)

Example 9.8

Decompose $F = \dfrac{1}{X(X-1)(X-2)}$. Three simple poles; cover-up at each:

so $F = \dfrac{1/2}{X} - \dfrac{1}{X-1} + \dfrac{1/2}{X-2}$. Check at $X = 3$: directly, $F(3) = \frac{1}{3\cdot2\cdot1} = \frac16$; from the decomposition, $\frac{1/2}{3} - \frac{1}{2} + \frac{1/2}{1} = \frac16 - \frac12 + \frac12 = \frac16$.