部分分式的理論
The Theory of Partial Fractions

In our disscussions on the integration of rational functions, we stated a few theorems on partial fractions. The aim of this appendix is to prove them.

Our proofs will be base on the Euclidean algorithm (輾轉相除法) to find the HCF (Highest Common Factor, 最高公因式) of two non-zero polynomials r₀(x) and r₁(x). Without loss of generality we may assume that $\deg r_1\leq \deg r_0$, where $\deg r$ means the degree (次數) of the polynomial r. The Euclidean algorithm in turn is based upon the division algorithm (長除法) which provides us with a unique quotient polynomial q₁(x)and a remainder polynomial r₂(x) such that

$$r_0(x)=q_1(x)r_1(x)+r_2(x) , \ \ \ \ \ \ \ \ deg r_2<deg r_1, \eqno(1)$$

If r₂(x)=0, then it is quite clear that r₁(x) is the desired HCF. Otherwise from (1) it is also clear that the HCF of r₁ and r₂is the same as the HCF of r₀ and r₁, and we may repeat the process to obtain a sequence of polynomials $r_0, r_1, r_2, \ldots, r_n, r_{n+1}=0$ with decreasing degrees such that

$$r_{k-1}(x)=q_{k}(x)r_{k}(x)+r_{k+1}(x), \ \ \ \ \ \ k=1, 2, \ldots, n \eqno(2)$$

for some polynomials q_k. The last non-zero polynomial r_n is the HCF of r₀ and r₁.
Lemma 1. Let r_n(x) be the HCF of the two non-zero polynomials r₀(x) and r₁(x). Then there exist two polynomails a(x) and b(x) such that

$$r_n(x)=a(x)r_0(x)+b(x)r_1(x). \eqno(3)$$

證明. We use the notations of the discussions above. Then from (2), r(x) can be written in the form

r_n(x)=A(x)r_n-1(x)+B(x)r_n-2(x)

with A(x)=-q(x) and B(x)=1. Assume that r_n(x) can be expressed in the form

$$r_n(x)=A(x)r_{k}(x)+B(x)r_{k-1}(x), \ k\geq 2.$$

Then from (2) we get

r_n(x)=[B(x)-A(x)q_k-1(x)]r_k-1(x)+A(x)r_k-2(x),

which is again of the form

r_n(x)=A(x)r_n-1(x)+B(x)r_n-2(x)

with a different A(x) and a different B(x). Lemma 1 then follows from mathematical induction. Q. E. D.

Two non-zero polynomials f(x) and g(x) are called relatively prime (互質) if their HCF is a non-zero constant. It follows immediately from Lemma 1 that if f(x) and g(x)are relatively prime then there exist polyvomials a(x) and b(x) such that

a(x)f(x)+b(x)g(x)=1.

Lemma 2. Let f(x), g(x) and r(x) be polynomials such that
(a) f(x) and g(x) are relatively prime, and
(b) f(x) is a factor of r(x)g(x), then f(x) is also a factor of r(x).

證明. From the remarks after Lemma 1 we see the existence of polynomials a(x) and b(x) such that

a(x)f(x)+b(x)g(x)=1.

Hence

r(x)=a(x)r(x)f(x)+b(x)r(x)g(x).

Since f(x) is a factor of both terms on the right, it is also a factor of r(x). Q. E. D.
Lemma 3. If f(x) and g(x) are relatively prime polynomials, deg f > 0, deg g > 0, and if h(x) is another polynomial satisfying

$$\textup{deg}\ h < \textup{deg}\ f + \textup{deg}\ g,$$

then there exist polynomials r(x) and s(x) such that deg r < deg f, deg s < deg g, and

$$\frac{h(x)}{f(x)g(x)}=\frac{r(x)}{f(x)}+\frac{s(x)}{g(x)}.$$

Further the polynomials r(x) and s(x) are uique.

證明. By Lemma 1, there are polynomials a(x) and b(x) such that

a(x)f(x)+b(x)g(x)=1.

Hence $\hbox to 2 truecm{\hfill}$ $$\frac{h(x)}{f(x)g(x)} =\frac{b(x)h(x)}{f(x)}+\frac{a(x)h(x)}{g(x)}$$. to 4truecm Applying long divisions to the two fractions on the right-hand side, we get polynomials q(x), r(x) and s(x) with deg r < deg f and deg s < deg g, such that

$$\frac{h(x)}{f(x)g(x)}=g(x)+\frac{r(x)}{f(x)}+\frac{s(x)}{g(x)}.$$

Clearing up the denominators, we have

h(x)=q(x)f(x)g(x)+r(x)g(x)+s(x)f(x).

If $q(x) \neq 0$, then the degree of the first term on the right-hand side is at least deg f + deg g, while the degree of every other term in this equation is less than deg f + deg g. This is impossible. Consequently q(x)=0 and

$$\frac{h(x)}{f(x)g(x)}=\frac{r(x)}{f(x)}+\frac{s(x)}{g(x)}.$$

Next assume that we also have

$$\frac{h(x)}{f(x)g(x)}=\frac{R(x)}{f(x)}+\frac{S(x)}{g(x)}.$$

with deg R < deg f and deg S < deg g. Then

[R(x)-r(x)]g(x)=-[S(x)-s(x)]f(x).

By Lemma 2, f(x) divides R(x) -r(x). But deg (R-r) < deg f. Hence R(x)-r(x)=0. Similarly S(x)-s(x)=0. This proves the uniqueness. Lemma 4. Let f(x) and g(x) be polynomials, $0 \leq$ deg f < n deg g, where n is a positive integer. Then there are polynomials c₁(x), c₂(x) , $\cdots, c_n(x)$ , all of degrees < deg g, such that

$$\frac{f(x)}{[g(x)]^n}=\frac{c_1(x)}{g(x)}+\frac{c_2(x)}{[g(x)]^2} + \cdots +\frac{c_n(x)}{[g(x)]^n}.$$

證明. The Lemma is trivial if n=1. Assume that it is true for n-1. Long division will yield two polynomials q(x) and c(x) such that f(x) =q(x)g(x)+c_n(x), where deg q= deg f- deg g<(n-1) deg g and deg c< deg g. Application of the induction assumption to q(x) and division of both sides by g(x) will produce Lemma 4 for n.

Now we return to the four basic theorems concerning partial fractions mentioned in the text:

證明. This is but a restatement of the fundamental theorem of algebra, which can be proved in a course on functions of a complex variables (複變數函數論). According to this theorem, Q(x) can be factored into a product of of the form

$$c (x-\alpha_1) (x-\alpha_2) \cdots (x-\alpha_n),$$

where c is a real constant and $\alpha_1, \alpha_2, \cdots , \alpha_n$ are roots of Q(x). These numbers $\alpha_k$are in general complex numbers and they are not necessarily distinct. The non-real roots occur in conjugate pairs. Each real root $\alpha_k$ yields a real linear factor, and each conjuate pair of complex roots yield an irreducible ( 不可約 ) real quadratic factor. Hence we have this theorem.

Observe that no two of the linear and quadratic factors have a common root. Hence they are pairwise relatively prime. 定理5 is thus reduced to a consequence of Lemma 3.

Both theorems follow from Lemma 4 as simple corollaries.