partial fractions for polynomials


This entry precisely states and proves the existence and uniqueness of partial fraction decompositions of ratios of polynomialsMathworldPlanetmathPlanetmathPlanetmath of a single variableMathworldPlanetmath, with coefficients over a field.

The theory is used for, for example, the method of partial fraction decomposition for integrating rational functions over the reals (http://planetmath.org/ALectureOnThePartialFractionDecompositionMethod).

The proofs involve fairly elementary algebra only. Although we refer to Euclidean domains in our proofs, the reader who is not familiar with abstract algebra may simply read that as “set of polynomials” (which is one particular Euclidean domain).

Also note that the proofs themselves furnish a method for actually computing the partial fraction decomposition, as a finite-time algorithm, provided the irreduciblePlanetmathPlanetmath factorization of the denominator is known. It is not an efficient way to find the partial fraction decomposition; usually one uses instead the method of making substitutions into the polynomials, to derive linear constraints on the coefficients. But what is important is that the existence proofs here justify the substitution method. The uniqueness property proved here might also simplify some calculations: it shows that we never have to consider multipleMathworldPlanetmathPlanetmath solutions for the coefficients in the decomposition.

Theorem 1.

Let p and q0 be polynomials over a field, and n be any positive integer. Then there exist unique polynomials α1,,αn,β such that

pqn=β+α1q+α2q2++αnqn,degαj<degq. (1)
Proof.

Existence has already been proven as a special case of partial fractions in Euclidean domains; we now prove uniqueness. Suppose equation (1) has been given. Multiplying by qn and rearranging,

p=βqn+r1,r1=α1qn-1++αn,degr1<degqn.

But according to the division algorithmPlanetmathPlanetmath for polynomials (also known as long division), the quotientPlanetmathPlanetmath and remainder polynomial after a division (by qn in this case) are unique. So β must be uniquely determined. Then we can rearrange:

p-βqn=α1qn-1+r2,r2=α2qn-2++αn,degr2<degqn-1.

By uniqueness of division again (by qn-1), α1 is determined. Repeating this process, we see that all the polynomials αj and β are uniquely determined. ∎

Theorem 2.

Let p and q0 be polynomials over a field. Let q=ϕ1n1ϕ2n2ϕknk be the factorization of q to irreducible factors ϕi (which is unique except for the ordering and constant factors). Then there exist unique polynomials αij,β such that

pq=β+i=1kj=1niαijϕij,degαij<degϕi. (2)
Proof.

Existence has already been proven as a special case of partial fractions in Euclidean domains; we now prove uniqueness. Suppose equation (2) has been given. First, multiply the equation by q:

p=βq+i,jαijqϕij.

The polynomial sum on the far right of this equation has degree <q, because each summand has degree deg(αijq/ϕij)<degϕi+degq-jdegϕidegq. So the polynomial sum is the remainder of a division of p by q. Then the quotient polynomial β is uniquely determined.

Now suppose si and si are polynomials of degree <ϕini, such that

i=1ksiϕini=i=1ksiϕini. (3)

We claim that si=si. Let q1=ϕ1n1 and q2=q/q1, and write

s1q1+uq2=i=1ksiϕini=i=1ksiϕini=s1q1+uq2,

for some polynomials u and u. Rearranging, we get:

(s1-s1)q2=(u-u)q1.

In particular, q1 divides the left side. Since q1=ϕ1n1 is relatively prime from q2, it must divide the factor (s1-s1). But deg(s1-s1)<degq1, hence s1-s1 must be the zero polynomialMathworldPlanetmath. That is, s1=s1.

So we can cancel the term s1/ϕ1n1=s1/ϕ1n1 on both sides of equation (3). And we could repeat the argument, and show that s2 and s2 are the same, s3 and s3 are the same, and so on. Therefore, we have shown that the polynomials si in the following expression

pq-β=i=1ksiϕini,degsi<degϕini

are unique. In particular, si is the following numerator that results when the fractions αij/ϕij are put under a common denominator ϕnii:

si=j=1niαijϕini-j.

But by the uniqueness part of Theorem 1, the decomposition

siϕini=βi+j=1niαijϕij,degαij<degϕi

uniquely determines αij. (Note that the proof of Theorem 1 shows that βi=0, as degsi<degϕini.) ∎

Title partial fractions for polynomials
Canonical name PartialFractionsForPolynomials
Date of creation 2013-03-22 15:40:22
Last modified on 2013-03-22 15:40:22
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 6
Author stevecheng (10074)
Entry type Result
Classification msc 12E05
Synonym partial fraction decomposition of rational functions
Synonym partial fractions for rational functions
Related topic PartialFractionsOfExpressions
Related topic ALectureOnThePartialFractionDecompositionMethod