26 CHAPTER 1. SOME PREREQUISITE TOPICS

With this, the general partial fractions theorem is next. From Theorem 1.13.9, everypolynomial q(λ ) has a factorization of the form ∏

Mi=1 pi (λ )

mi where the pi (λ ) are irre-ducible, meaning they cannot be factored further. Thus the polynomials pi (λ ) are distinctand relatively prime as is every subset having at least two of these pi (λ ).

Proposition 1.14.2 Let a(λ )b(λ ) be any rational function. Then it is of the form

a(λ )

∏Mi=1 pi (λ )

mi

where the pi (λ ) are distinct irreducible polynomials, meaning they can’t be factored anyfurther as described in the chapter and each mi is a nonnegative integer.

Then there are polynomials q(λ ) and nki (λ ) with the degree of nki (λ ) less than thedegree of pi (λ ) or nki (λ ) = 0, such that

a(λ )b(λ )

= q(λ )+M

∑i=1

mi

∑k=1

nki (λ )

pi (λ )k (1.2)

Proof: Suppose first that ∑Mi=1 mi = 1. Then the rational function is of the form a(λ )

p(λ ) andthis can be placed in the desired form by an application of the division algorithm as above.Suppose now that this proposition is true if ∑

Mi=1 mi ≤ n for some n ≥ 1 and suppose you

havea(λ )b(λ )

=a(λ )

∏Mj=1 p j (λ )

m j,

M

∑j=1

m j = n+1, each m j ≥ 0

If some m j = n+ 1, then one obtains the situation of Proposition 1.14.1. Therefore, itsuffices to assume that no m j = n+1 so there are at least two m j which are nonzero.

Every subset of the {p1 (λ ) , p2 (λ ) , ..., pM (λ )} having at least two pi (λ ) is relativelyprime because these polynomials are all irreducible. Therefore, there are polynomialsbi (λ ) such that bi (λ ) = 0 if mi = 0 and ∑

Mi=1 bi (λ ) pi (λ ) = 1. Then multiply by this

to obtain

a(λ )b(λ )

=a(λ )

∏Mj=1 p j (λ )

m j=

a(λ )∑Mi=1 bi (λ ) pi (λ )

∏Mj=1 p j (λ )

m j=

M

∑i=1

a(λ )bi (λ ) pi (λ )

∏Mj=1 p j (λ )

m j

Now in the ith term of the sum, the pi (λ ) in the top cancels with exactly one of the fac-tors in the bottom or else the term is 0. It follows that the original a(λ )

b(λ ) is of the form

∑Ni=1

âi(λ )

∏Mj=1 p j(λ )

mi j where ∑Mj=1 mi j ≤ n. By induction applied to each of the terms in this

sum, one obtains a(λ )b(λ ) equal to an expression of the form in 1.2. ■

Proposition 1.14.3 The partial fractions expansion is unique.

Proof: Subtracting, you get

q(λ )− q̂(λ ) =M

∑i=1

mi

∑k=1

n̂ki (λ )−nki (λ )

pi (λ )k