partial fractions for polynomials

partial fractions for polynomials

This entry precisely states and proves
the existence and uniqueness of

partial fraction decompositions

of ratios of

polynomials

of a single

variable

, 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

irreducible

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

multiple

solutions

for the coefficients in the decomposition.

Theorem 1

.

Let

p

and

q

≠

0

be polynomials over a field,
and

n

be any

positive integer

.
Then there exist unique polynomials

α

1

,

…

,

α

n

,

β

such that

p

q

n

=

β

+

α

1

q

+

α

2

q

2

+

⋯

+

α

n

q

n

,

deg

⁡

α

j

<

deg

⁡

q

.

(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

q

n

and rearranging,

p

=

β

⁢

q

n

+

r

1

,

r

1

=

α

1

⁢

q

n

-

1

+

⋯

+

α

n

,

deg

⁡

r

1

<

deg

⁡

q

n

.

But according to the

division algorithm

for polynomials (also known as long division), the

quotient

and remainder polynomial
after a division (by

q

n

in this case) are unique.
So

β

must be uniquely determined.
Then we can rearrange:

p

-

β

⁢

q

n

=

α

1

⁢

q

n

-

1

+

r

2

,

r

2

=

α

2

⁢

q

n

-

2

+

⋯

+

α

n

,

deg

⁡

r

2

<

deg

⁡

q

n

-

1

.

By uniqueness of division again (by

q

n

-

1

),

α

1

is determined.
Repeating this process, we see that all the polynomials

α

j

and

β

are uniquely determined.
∎

Theorem 2

.

Let

p

and

q

≠

0

be polynomials over a field.
Let

q

=

ϕ

1

n

1

⁢

ϕ

2

n

2

⁢

⋯

⁢

ϕ

k

n

k

be the factorization of

q

to irreducible factors

ϕ

i

(which is unique except for the ordering and constant factors).
Then there exist unique polynomials

α

i

⁢

j

,

β

such
that

p

q

=

β

+

∑

i

=

1

k

∑

j

=

1

n

i

α

i

⁢

j

ϕ

i

j

,

deg

⁡

α

i

⁢

j

<

deg

⁡

ϕ

i

.

(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

=

β

⁢

q

+

∑

i

,

j

α

i

⁢

j

⁢

q

ϕ

i

j

.

The polynomial sum on the far right of this equation
has degree

<

q

, because
each summand has degree

deg

⁡

(

α

i

⁢

j

⁢

q

/

ϕ

i

j

)

<

deg

⁡

ϕ

i

+

deg

⁡

q

-

j

⋅

deg

⁡

ϕ

i

≤

deg

⁡

q

.
So the polynomial sum is the remainder of a division of

p

by

q

.
Then the quotient polynomial

β

is uniquely determined.

Now suppose

s

i

and

s

i

′

are polynomials of degree

<

ϕ

i

n

i

,
such that

∑

i

=

1

k

s

i

ϕ

i

n

i

=

∑

i

=

1

k

s

i

′

ϕ

i

n

i

.

(3)

We claim that

s

i

=

s

i

′

. Let

q

1

=

ϕ

1

n

1

and

q

2

=

q

/

q

1

,
and write

s

1

q

1

+

u

q

2

=

∑

i

=

1

k

s

i

ϕ

i

n

i

=

∑

i

=

1

k

s

i

′

ϕ

i

n

i

=

s

1

′

q

1

+

u

′

q

2

,

for some polynomials

u

and

u

′

.
Rearranging, we get:

(

s

1

-

s

1

′

)

⁢

q

2

=

(

u

′

-

u

)

⁢

q

1

.

In particular,

q

1

divides the left side.
Since

q

1

=

ϕ

1

n

1

is relatively prime from

q

2

, it must divide
the factor

(

s

1

-

s

1

′

)

. But

deg

⁡

(

s

1

-

s

1

′

)

<

deg

⁡

q

1

,
hence

s

1

-

s

1

′

must be the

zero polynomial

. That is,

s

1

=

s

1

′

.

So we can cancel the term

s

1

/

ϕ

1

n

1

=

s

1

′

/

ϕ

1

n

1

on both sides of
equation (

3

).
And we could repeat the

argument

,
and show that

s

2

and

s

2

′

are the same,

s

3

and

s

3

′

are the same, and so on.
Therefore, we have shown that
the polynomials

s

i

in the following

expression

p

q

-

β

=

∑

i

=

1

k

s

i

ϕ

i

n

i

,

deg

⁡

s

i

<

deg

⁡

ϕ

i

n

i

are unique. In particular,

s

i

is
the following numerator that results when the fractions

α

i

⁢

j

/

ϕ

i

j

are put under a common denominator

ϕ

n

i

i

:

s

i

=

∑

j

=

1

n

i

α

i

⁢

j

⁢

ϕ

i

n

i

-

j

.

But by the uniqueness part of

Theorem

1

,
the decomposition

s

i

ϕ

i

n

i

=

β

i

+

∑

j

=

1

n

i

α

i

⁢

j

ϕ

i

j

,

deg

⁡

α

i

⁢

j

<

deg

⁡

ϕ

i

uniquely determines

α

i

⁢

j

.
(Note that the proof of Theorem

1

shows that

β

i

=

0

, as

deg

⁡

s

i

<

deg

⁡

ϕ

i

n

i

.)
∎

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

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